1LIFE ASSURANCE MATHEMATICS W.F.Scott c (1999 W.F.Scott Department of Mathematical Sciences King’s College University of Aberdeen Aberdeen AB24 3UE U.K. 2 Preface This book consists largely of material written for Parts A2 and D1 of the U.K. actuarial exami- nations (old system). It is hoped that the material given here will prove useful for much of Subjects 104 and 105 of the new examinations and certain similar examinations at universities, and that it may also be useful as a general reference work on life assurance mathematics. William F. Scott. November 10, 1999 Contents 1 NON-SELECT LIFE TABLES 9 1.1 Survivorship functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.2 Probabilities of death and survival . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.3 The force of mortality, µ x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.4 The expectation of life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.5 The assumption of a uniform distribution of deaths . . . . . . . . . . . . . . . . . . . 17 1.6 Central death rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 1.7 Laws of mortality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 1.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2 SELECT LIFE TABLES 29 2.1 What is selection? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.2 Construction of select tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 2.3 The construction of A1967-70. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.4 Some formulae for the force of mortality. . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.5 Select tables used in examinations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3 ASSURANCES 37 3.1 A general introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.2 Whole life assurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.3 Commutation functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.4 The variance of the present value of benefits . . . . . . . . . . . . . . . . . . . . . . . 39 3.5 Assurances payable at the end of the year of death. . . . . . . . . . . . . . . . . . . . 41 3.6 Assurances payable at the end of the 1/m of a year of death. . . . . . . . . . . . . . 44 3.7 Temporary and deferred assurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3.8 Pure endowments and endowment assurances . . . . . . . . . . . . . . . . . . . . . . 46 3.9 Varying assurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 3.10 Valuing the benefits under with profits policies . . . . . . . . . . . . . . . . . . . . . 50 3.11 Guaranteed bonus policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 3.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 3.13 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 4 ANNUITIES 61 4.1 Annuities payable continuously . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 4.2 Annuities payable annually . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 4.3 Temporary annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 4.4 Deferred annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 3 4 CONTENTS 4.5 Annuities payable m times per annum . . . . . . . . . . . . . . . . . . . . . . . . . . 68 4.6 Complete annuities (or “annuities with final proportion”) . . . . . . . . . . . . . . . 70 4.7 Varying annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 4.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 4.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 5 PREMIUMS 81 5.1 Principles of premium calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 5.2 Notation for premiums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 5.3 The variance of the present value of the profit on a policy. . . . . . . . . . . . . . . . 83 5.4 Premiums allowing for expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 5.5 Premiums for with profits policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 5.6 Return of premium problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 5.7 Annuities with guarantees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 5.8 Family income benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 5.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 5.10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 6 RESERVES 95 6.1 What are reserves? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 6.2 Prospective reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 6.3 Net premium reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 6.4 Retrospective reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 6.5 Gross premium valuations and asset shares. . . . . . . . . . . . . . . . . . . . . . . . 102 6.6 The variance of L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 6.7 Zillmerised reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 6.8 Full preliminary term reserves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 6.9 Reserves for with-profits policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 6.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 6.11 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 7 APPLICATIONS OF RESERVES 119 7.1 Surrender values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 7.2 Paid-up policy values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 7.3 Alterations and conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 7.4 The actual and expected death strains . . . . . . . . . . . . . . . . . . . . . . . . . . 124 7.5 Mortality profit and loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 7.6 Other sources of profit and loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 7.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 7.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 8 EXTRA RISKS 137 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 8.2 A constant addition to the force of mortality . . . . . . . . . . . . . . . . . . . . . . 137 8.3 A variable addition to the force of mortality . . . . . . . . . . . . . . . . . . . . . . . 139 8.4 Rating up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 8.5 Debts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 8.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 8.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 CONTENTS 5 9 PROFIT-TESTING 151 9.1 Principles of profit-testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 9.2 Cash flow calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 9.3 The profit vector and the profit signature . . . . . . . . . . . . . . . . . . . . . . . . 155 9.4 The assessment of profits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 9.5 Some theoretical results about ¦σ t ¦ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 9.6 Withdrawals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 9.7 The actual emergence of profits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 9.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 9.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 10 STATIONARY POPULATIONS 171 10.1 Some Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 10.2 The Central Death Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 10.3 Relationships Between m x and q x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 10.4 Stationary Funds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 10.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 10.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 11 JOINT-LIFE FUNCTIONS 185 11.1 Joint-Life Mortality Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 11.2 Select Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 11.3 Extensions to More than 2 Lives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 11.4 The Joint Expectation of Life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 11.5 Monetary Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 11.6 Last Survivor Probabilities (two lives only) . . . . . . . . . . . . . . . . . . . . . . . 195 11.7 Last Survivor Monetary Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 11.8 Reserves for Last Survivor Assurances . . . . . . . . . . . . . . . . . . . . . . . . . . 198 11.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 11.10Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 12 CONTINGENT ASSURANCES 205 12.1 Contingent Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 12.2 Contingent Assurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 12.3 Premiums and Reserves for Contingent Assurances . . . . . . . . . . . . . . . . . . . 210 12.4 A Practical Application – The Purchase of Reversions . . . . . . . . . . . . . . . . . 211 12.5 Extension to Three Lives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 12.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 12.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 13 REVERSIONARY ANNUITIES 219 13.1 Reversionary Annuities Payable Continuously . . . . . . . . . . . . . . . . . . . . . . 219 13.2 Reversionary Annuities Payable Annually or m thly . . . . . . . . . . . . . . . . . . . 220 13.3 Widow’s (or Spouse’s) Pension on Death after Retirement . . . . . . . . . . . . . . . 221 13.4 Actuarial Reduction Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 13.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 13.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 6 CONTENTS 14 PROFIT TESTING FOR UNIT-LINKED POLICIES 231 14.1 Unit-Linked Policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 14.2 Mechanics of the Unit Fund . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 14.3 The Sterling Fund (or Sterling Reserves) . . . . . . . . . . . . . . . . . . . . . . . . . 233 14.4 The Assessment of Profits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 14.5 Zeroisation of the Profit Signature . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 14.6 Withdrawals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 14.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 14.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 15 MULTIPLE-DECREMENT TABLES 247 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 15.2 The Associated Single-Decrement Tables . . . . . . . . . . . . . . . . . . . . . . . . . 249 15.3 The Relationships between the Multiple-Decrement Table and its Associated Single- Decrement Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 15.4 Dependent Rates of Exit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 15.5 Practical Construction of Multiple-Decrement Tables . . . . . . . . . . . . . . . . . . 253 15.6 Further Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 15.7 Generalization to 3 Modes of Decrement . . . . . . . . . . . . . . . . . . . . . . . . . 257 15.8 “Abnormal” Incidence of Decrement . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 15.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 15.10Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264 16 FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES 267 16.1 Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 16.2 The Use of “Defective” Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 16.3 Evaluation of Mean Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 16.4 Benefits on Death by a Particular Cause . . . . . . . . . . . . . . . . . . . . . . . . . 272 16.5 Extra Risks Treated as an Additional Mode of Decrement . . . . . . . . . . . . . . . 272 16.6 Calculations Involving a Change of State . . . . . . . . . . . . . . . . . . . . . . . . . 274 16.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 16.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 17 MULTIPLE-STATE MODELS 281 17.1 Two Points of View . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 17.2 Kolmogorov’s Forward Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 17.3 Life Tables as Stochastic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 17.4 Sickness Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 18 SICKNESS FUNCTIONS 287 18.1 Rates of Sickness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 18.2 Valuing Sickness Benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 18.3 Various Other Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 18.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 18.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 Appendix The Manchester Unity Experience 1893-97 . . . . . . . . . . . . . . . . . . . . . 301 CONTENTS 7 19 PENSION FUNDS 303 19.1 General Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 19.2 Valuation Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 19.3 Service Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 19.4 Salary Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 19.5 The Value of Future Contributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 19.6 The Value of Pension Benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 19.7 Fixed Pension Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 19.8 Average Salary Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 19.9 Final Salary Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 19.10Lump Sums on Retirement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 19.11Death and Withdrawal Benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 19.12Return of Contributions on Death or Withdrawal . . . . . . . . . . . . . . . . . . . . 318 19.13Spouse’s Benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 19.14Preserved Pensions on Leaving Service . . . . . . . . . . . . . . . . . . . . . . . . . . 322 19.15Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 19.16Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 Appendix Formulae for valuing a return of contributions . . . . . . . . . . . . . . . . . . . 331 APPENDICES 333 A Some notes on examination technique . . . . . . . . . . . . . . . . . . . . . . . . . . 333 B Some technical points about the tables used in examinations . . . . . . . . . . . . . 334 C Some common mistakes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 D Some formulae for numerical integration . . . . . . . . . . . . . . . . . . . . . . . . . 336 SUPPLEMENT 337 8 CONTENTS Chapter 1 NON-SELECT LIFE TABLES 1.1 Survivorship functions We consider a certain population from age α to a “limiting age” w , i.e. the youngest age to which no-one survives. (In theoretical work we may have w = ∞.) Let (x) be a shorthand notation for “a life aged (exactly) x”; we define the survivorship function s(x) = Pr¦ (α) survives to age x ¦ (1.1.1) In practice one usually uses the function l x = l α s(x) (1.1.2) where l α is called the “radix” of the table, and is usually a large number such as 1, 000, 000. If lives are considered from birth, we naturally have α = 0, and we have l x > 0 for x < w, l w = 0, unless w is infinite, in which case l x → 0 as x → ∞. The function l x is assumed to be continuous. By formulae (1.1.1) and (1.1.2) , Pr¦ (α) survives to age x ¦ = l x l α (1.1.3) Example 1.1.1. English Life Table No.12 - Males refers to the male population of England and Wales in 1970-72. In this table α = 0 , w may be taken to be about 106, and the radix of the table is 100, 000. We observe (for example) that, since l 50 = 90, 085 , the probability that a new-born child will survive to age 50 is s(50) = l 50 /l 0 = 0.90085, or 90.1%. We now consider survival from age α in terms of a random variable. Let T (or T(x) if the age x is not clear; T x is sometimes used, but this may be confused with the symbol T x = l x ◦ e x used in stationary population problems) be the variable representing the future lifetime (in years, including fractions) of (x). According to our definitions, Pr¦ T(α) ≤ t ¦ = the probability that (α) will die within t years = 1 − the probability that (α) will survive for at least t years = 1 −s(α +t) (1.1.4) 9 10 CHAPTER 1. NON-SELECT LIFE TABLES 1.2 Probabilities of death and survival We now consider how to deal with lives aged between α and w. To simplify matters, we shall suppose here that w = ∞. The variables ¦T(x) : x ≥ α¦ are assumed to be related as follows: for each x 1 , x 2 , x 3 such that α ≤ x 1 ≤ x 2 ≤ x 3 , Pr¦T(x 1 ) > x 3 −x 1 ¦ = Pr¦T(x 1 ) > x 2 −x 1 ¦Pr¦T(x 2 ) > x 3 −x 2 ¦ (1.2.1) In particular, when x 1 = α, x 2 = x and x 3 = x +t, we have Pr¦T(x) > t¦ = Pr¦T(α) > x −α +t¦ Pr¦T(α) > x −α¦ = s(x +t) s(x) = l x+t l x (1.2.2) We introduce the important notation: t p x = Pr¦T(x) > t¦ = Pr¦ (x) survives to age x +t¦ (1.2.3) t q x = Pr¦T(x) ≤ t¦ = Pr¦ (x) dies between ages x and x +t¦ (1.2.4) When t = 1 , it may be omitted, so that p x = Pr¦ (x) survives for at least a year ¦ (1.2.5) q x = Pr¦ (x) dies within a year ¦ (1.2.6) q x is called the (q-type) rate of mortality at age x. (The m-type, or central, rate of mortality, m x , is considered in section 1.6.) It is clear that, for all t ≥ 0 , t p x + t q x = 1 (1.2.7) and, by the formula (1.2.2), t p x = s(x +t) s(x) = l x+t l x (1.2.8) and t q x = 1 − l x+t l x (1.2.9) That is, the (cumulative) distribution function of the variable T is F(t) = Pr¦T ≤ t¦ = t q x = 1 − l x+t l x , if t ≥ 0 0 , if t < 0 (1.2.10) We may say that “the expected number of survivors at age x + t among l x lives aged x is l x+t ” because the chance that a given life aged x will survive to age x + t is l x+t l x . In practice, the word “expected” is sometimes omitted, but it is always to be understood; assuming that the l x lives are independent, the number of survivors at age x +t will follow a binomial distribution with mean l x t p x = l x+t and variance l x t p x (1 − t p x ) = l x+t 1 − l x+t l x 1.2. PROBABILITIES OF DEATH AND SURVIVAL 11 (Even if the lives are not independent, the formula for the expected number of deaths is correct.) Note that p x = l x+1 l x (1.2.11) q x = 1 − l x+1 l x = d x l x (1.2.12) where d x = l x −l x+1 (1.2.13) = the (expected) number of deaths at age x last birthday among l x lives aged x Example 1.2.1. The following extract from English Life Table No. 12 - Males illustrates the functions l x , d x , p x and q x . age x l x d x p x q x 0 100,000 2,499 0.97551 0.02449 1 97,551 153 0.99843 0.00157 2 97,398 96 0.99901 0.00099 3 97,302 67 0.99931 0.00069 4 97,235 60 0.99938 0.00062 (The high death rate at age 0, relative to the rates at ages 1 to 4, is quite noticeable.) We also observe that, for all ages x 3 ≥ x 2 ≥ x 1 ≥ α , we have the “rule of multiplication of probabilities of survival”: x 3 −x 1 p x 1 = x 2 −x 1 p x 1 x 3 −x 2 p x 2 (1.2.14) (which is merely a re-statement of equation (1.2.1). ) Note. The definition of t p x is sometimes taken to be Pr¦(α) survives to age x +t [ (α) survives to age x ¦, which equals Pr¦T(α) > x −α +t and T(α) > x −α¦ Pr¦T(α) > x −α¦ = Pr¦T(α) > x −α +t¦ Pr¦T(α) > x −α¦ = Pr¦T(x) > t¦ in agreement with our earlier definition. Further, t p x is sometimes written as S x (t). When α = 0 we sometimes write S 0 (t) = S(t), so S(t) and s(t) are equal. The notation S(t) is much used in survival analysis, in which t denotes the time lived by a patient since the start of a given treatment. In reliability engineering the notation R(t) may be used instead of S(t). 12 CHAPTER 1. NON-SELECT LIFE TABLES 1.3 The force of mortality, µ x This may be defined as the “instantaneous rate of mortality”, i.e. µ x = lim h→0+ h q x h (1.3.1) Theorem 1.3.1. Suppose that µ x is continuous on [α, w). We have µ x = − l x l x (1.3.2) Proof. we have µ x = lim h→0+ −[s(x +h) −s(x)] hs(x) = −s + (x) s(x) where s + (x) denotes the R.H. derivative of s(x). But s(x) and s + (x) = −s(x)µ x are continuous on [α, w), so s (x) exists and equals −s(x)µ x (see McCutcheon and Scott, An Introduction to the Mathematics of Finance, Appendix 1. ) Hence µ x = − s (x) s(x) = − l x l x We remark that µ x = − d dx log l x (1.3.3) Hence − x α µ y dy = x α d dy (log l y ) dy = [ log l y ] x α = log l x −log l α = log l x l α from which we obtain the important formulae: l x = l α exp − x α µ y dy (1.3.4) and s(x) = exp − x α µ y dy (1.3.5) It follows that t p x = l x+t l x = exp − x+t x µ y dy = exp − t 0 µ x+r dr (1.3.6) If µ x is piecewise continuous, we may apply the above formulae over each age-range and combine the results; the above formulae remain true, except that l x does not exist at the points at which 1.3. THE FORCE OF MORTALITY, µ x 13 µ x is not continuous (the R.H. and L.H. derivatives of l x differ at these points.) Alternatively, we may regard the actual position as closely approximated by a model in which the force of mortality is continuous. We may now express the probability density function of the variable T (the future lifetime of (x)) in terms of the force of mortality. The p.d.f. f(t) of T is the derivative of the distribution function F(t), i.e. f(t) = F (t) = d dt 1 − l x+t l x , if t > 0 0 , if t < 0 By the chain rule d dt l x+t = l x+t = −l x+t µ x+t (by formula (1.3.2)) so we obtain f(t) = l x+t µ x+t l x = t p x µ x+t , if t > 0 0 , if t < 0 (1.3.7) In view of the following general formula (connecting the d.f. and p.d.f.): F(t) = t −∞ f(r) dr we have t q x = t 0 r p x µ x+r dr (1.3.8) An alternative notation Some statisticians refer to the force of mortality as the “hazard rate” or “transition intensity.” Let f(t) and F(t) denote the p.d.f. and d.f. respectively of T = T(x) . The hazard rate at age x +t (often considered to be “at time t years after entry to assurance (say) at age x”) is defined in terms of f(t) and F(t) by the formula h(t) = f(t) 1 −F(t) (1.3.9) which equals µ x+t (by formulae (1.2.7) and (1.3.7).) If α = 0 and x = 0, we naturally have T = the future lifetime of a new-born child, and µ t = f(t) 1 −F(t) = hazard rate at age t years (1.3.10) Example 1.3.1. Suppose that, for x ≥ α, µ x = cδx δ−1 (δ > 1, c > 0) (1.3.11) (This is “Weibull’s law of mortality”.) Find a simple expression for S x (t) = t p x . 14 CHAPTER 1. NON-SELECT LIFE TABLES Solution. s(x) = exp − x α µ y dy = exp − x α cδy δ−1 dy = exp −c y δ x α = exp −cx δ . exp cα δ hence t p x = S x (t) = exp −c (x +t) δ −x δ Numerical estimation of µ x from l x If l x is known only at integer values of x, we may use numerical differentiation to estimate µ x . The formula f (x) f(x + 1) −f(x −1) 2 (1.3.12) which is exact if f is a cubic between x −1 and x + 1 , may be used to show that µ x − l x l x = l x−1 −l x+1 2l x (1.3.13) Deferred probabilities The symbol m [ indicates deferment for m years; for example, m [ n q x = P¦(x) will die between ages x +m and x +m+n ¦ (1.3.14) By elementary probability, this equals Pr¦(x) will die before age x +m+n¦ −Pr¦(x) will die before age x +m¦ = m+n q x − m q x = m p x − m+n p x = l x+m −l x+m+n l x That is, m [ n q x = l x+m −l x+m+n l x (1.3.15) This may be remembered by the following rule: of l x lives aged x, l x+m −l x+m+n is the (expected) number of deaths aged between ages x +m and x +m+n. If n = 1 it may be omitted, so we have m [q x = Pr¦(x) will die between the ages x +m and x +m+ 1 ¦ = d x+m l x (1.3.16) We remark that there is no such thing as m [ n p x . We may also use the result that m [ n q x = Pr¦(x) survives to age x +m¦.Pr¦(x +m) dies within n years¦ = m p x . n q x+m (1.3.17) 1.4. THE EXPECTATION OF LIFE 15 Example 1.3.2. Consider the force of mortality to be as in Example 1.3.1. Find an expression for the chance that (x) will die between ages x +m and x +m+n. Solution. We require: m [ n q x = l x+m −l x+m+n l x = s(x +m) −s(x +m+n) s(x) = exp[−c(x +m) δ ] −exp[−c(x +m+n) δ ] exp[−cx δ ] 1.4 The expectation of life Let T be the future lifetime (in years, including fractions) of (x). The mean of T is written ◦ e x , and is called the complete expectation of life at age x. That is, ◦ e x = E(T) (1.4.1) = ∞ 0 t t p x µ x+t dt (1.4.2) by formula 1.3.7. If w < ∞ , the upper limit of the integral should be replaced by w −x. Theorem 1.4.1. ◦ e x = ∞ 0 t p x dt (1.4.3) (with ∞ replaced by w −x if w is finite). Proof. By integration by parts, with u = t , v = − t p x , ◦ e x = ∞ 0 t t p x µ x+t dt = [−t t p x ] ∞ 0 − ∞ 0 − t p x dt = ∞ 0 t p x dt since it may be shown that t t p x →0 as t →∞ (using the fact that E(T) < ∞. ). We may also use the formula Var(T) = E(T 2 ) −[E(T)] 2 (1.4.4) where E(T 2 ) = ∞ 0 t 2 t p x µ x+t dt = −t 2 t p x ∞ 0 + ∞ 0 2t t p x dt (on setting u = t 2 , v = − t p x ) = 2 ∞ 0 t t p x dt 16 CHAPTER 1. NON-SELECT LIFE TABLES (since t 2 t p x →∞ , because E(T 2 ) < ∞ ). Hence Var(T) = 2 ∞ 0 t t p x dt −( ◦ e x ) 2 (1.4.5) The median future lifetime of (x) is the solution, t, of the equation F(t) = 0.5 i.e., t q x = 0.5 i.e., l x+t = 0.5 l x (1.4.6) x +t may be estimated by linear interpolation, as in the next example. Example 1.4.1. Find the median future lifetime of (10), according to English Life Table No.12 - Males. Solution. We must find the value of t such that l 10+t = 0.5 l 10 = 48, 469.5 On inspecting the tables, 10 + t lies between 72 and 73. By linear interpolation in y = 10 + t , we have l y −l 72 l 73 −l 72 = y −72 i.e. y = 72 + 48, 469.5 −48, 625 45, 430 −48, 625 = 72.05, so t = 62.05 years. The modal future lifetime of (x) is the time, t 0 years, at which the p.d.f. t p x µ x+t (or, equivalently, the function l x+t µ x+t ) reaches its maximum. The modal age at death of (x) is y = x + t 0 , which is the value of y for which l y µ y attains its maximum in the range y ≥ x. (The function l y µ y is sometimes called the “curve of deaths”.) If l y µ y has a unique maximum at age 80 (say), this is the modal age at death for all lives aged under 80, and hence the modal future lifetime of a life aged x < 80 is 80 −x. (See Exercise 1.7 for an example in which there is a unique modal age at death.) We now consider the discrete random variable K (or K(x) if the age x is not clear) defined by K = the integer part of T = the number of complete years to be lived in the future by (x) (1.4.7) Now it follows by formula 1.3.16 that Pr¦K = k¦ = k [q x ( k = 0, 1, 2, ...) (1.4.8) This variable is used in many actuarial calculations. In particular, the curtate expectation of life at age x, written e x , is the mean of K ; that is e x = ∞ ¸ k=0 k k [q x = ∞ ¸ k=1 k. k [q x (1.4.9) 1.5. THE ASSUMPTION OF A UNIFORM DISTRIBUTION OF DEATHS 17 Theorem 1.4.2. e x = ∞ ¸ k=1 k p x = l x+1 +l x+2 +. . . l x (1.4.10) Proof. e x = d x+1 l x + 2 d x+2 l x + 3 d x+3 l x +. . . = 1 l x [(d x+1 +d x+2 +d x+3 +. . .) + (d x+2 +d x+3 +. . .) + (d x+3 +. . .) +. . .] = l x+1 +l x+2 +l x+3 +. . . l x [since l y = (l y −l y+1 ) + (l y+1 −l y+2 ) +. . . = d y +d y+1 +. . . ] We may also evaluate Var(K) by the formula Var(K) = E(K 2 ) −[E(K)] 2 = ∞ ¸ k=1 k 2 d x+k l x −(e x ) 2 (1.4.11) As is clear by general reasoning, ◦ e x e x + 1 2 (1.4.12) A more precise approximation may be obtained from the Euler-Maclaurin formula, which we shall discuss later. 1.5 The assumption of a uniform distribution of deaths Let x be fixed. We may say there is a uniform distribution of deaths (U.D.D.) between ages x and x + 1 if, for 0 ≤ t ≤ 1, l x+t = (1 −t)l x +t l x+1 (1.5.1) (i.e. l y is linear for x ≤ y ≤ x + 1) This equation may be written in the form l x+t = l x −td x (0 ≤ t ≤ 1) (1.5.2) Theorem 1.5.1. The following conditions are each equivalent to the assumption of U.D.D. between ages x and x + 1: t q x = t q x (0 ≤ t ≤ 1) (1.5.3) t p x µ x+t = q x (0 ≤ t < 1) (1.5.4) 18 CHAPTER 1. NON-SELECT LIFE TABLES Proof. Assume 1.5.2 holds. Then t p x = 1 −t q x (1.5.5) But t p x = 1 − t q x , so t q x = t q x ; therefore 1.5.3 holds. This argument may be reversed, so 1.5.2 and 1.5.3 are equivalent. Now suppose that (1.5.3) holds. By differentiation with respect to t, we obtain (1.5.4), and we may obtain (1.5.3) from (1.5.4) by integration. Example 1.5.1. In a certain non-select mortality table, there is a uniform distribution of deaths between any two consecutive integer ages. Find formulae in terms of l 30 , l 31 and l 32 for (i) 1.5 p 30.5 (ii) µ 30.5 Solution. (i) l 32 l 30.5 = l 32 1 2 (l 30 +l 31 ) since l 30+t is linear for 0 < t < 1 (ii) t p 30 µ 30+t = q 30 for 0 ≤ t ≤ 1, so µ 30.5 = q 30 1 2 p 30 = l 30 −l 31 l 30 1 2 = l 30 −l 31 1 2 (l 30 +l 31 ) 1.6 Central death rates For simplicity of notation, we again suppose that the limiting age of our mortality table is infinity. Define T x = ∞ 0 l x+t dt (1.6.1) = ∞ x l y dy (on setting y = x +t) from which we obtain ◦ e x = ∞ x l y dy l x = T x l x and hence T x = l x ◦ e x (1.6.2) We also define L x = 1 0 l x+t dt (1.6.3) 1.7. LAWS OF MORTALITY 19 Note that L x = x+1 x l y dy = ∞ x l y dy − ∞ x+1 l y dy = T x −T x+1 (1.6.4) Assume that there is a Uniform Distribution of Deaths (U.D.D) between the ages x and x + 1, i.e. l x+t is linear for 0 ≤ t ≤ 1. We have L x = 1 0 l x+t dt = 1 2 (l x +l x+1 ) (1.6.5) (as the trapezoidal rule is exactly correct, not just an approximation). Since l x+1 = l x −d x , we have L x = l x − 1 2 d x (1.6.6) Formulae (1.6.5) and (1.6.6) are sometimes used as approximations when U.D.D. does not hold. The central death rate at age x is defined as m x = d x L x (1.6.7) Note the following important approximation: m x = 1 0 l x+t µ x+t dt 1 0 l x+t dt µ x+ 1 2 (1.6.8) Relationships between m x and q x If there is U.D. of D. between ages x and x + 1, formula (1.6.6) shows that m x = d x L x = d x l x − 1 2 d x = q x 1 − 1 2 q x (1.6.9) We may rearrange this equation to get q x = m x 1 + 1 2 m x (1.6.10) If U.D.D. does not hold, these results may be used as approximations. 1.7 Laws of mortality The term ‘law of mortality’ is used to describe a mathematical expression for µ x (or possibly q x or m x ) which may be explained from biological or other arguments (rather than being just a best- fitting curve.) The most famous law of mortality is that of Gompertz (1825), who postulated that 20 CHAPTER 1. NON-SELECT LIFE TABLES µ x satisfies the following simple differential equation: dµ x dx = kµ x (x ≥ α) This may be solved to give µ x = Bc x (x ≥ α) (1.7.1) Gompertz’ Law is often found to be quite accurate (at least as a first approximation) for ages over about 25 or 30, the value of c usually being between 1.07 and 1.12. Example 1.7.1. Show that, if Gompertz’ Law holds for all ages greater than or equal to α, there is a positive constant g such that t p x = g c x (c t −1) for x ≥ α, t ≥ 0 (1.7.2) Solution. t p x = exp ¸ − t 0 µ x+r dr = exp ¸ −Bc x t 0 c r dr = exp ¸ − Bc x (c t −1) log c = g c x (c t −1) with g = exp ¸ − B log c In 1860 Makeham suggested the addition of a constant term to Gompertz’ formula for µ x , giv- ing Makeham’s law: µ x = A+Bc x (x ≥ α) (1.7.3) Example 1.7.2. Show that, if Makeham’s law holds for all ages greater than or equal to α, there are positive constants s and g such that t p x = s t g c x (c t −1) for x ≥ α, t ≥ 0 (1.7.4) Solution. t p x = exp ¸ − t 0 µ x+r dr = exp ¸ − t 0 (A+Bc x c r )dr = exp[−At]g c x (c t −1) (using Example 1.7.1) = s t g c x (c t −1) with g = exp ¸ − B log c , s = e −A 1.7. LAWS OF MORTALITY 21 Weibull’s law of mortality has already been mentioned in Example 1.3.1. In practice it is usu- ally less successful than those of Gompertz and Makeham in representing human mortality. The fitting of laws of mortality is complicated by the fact that mortality rates may be varying with time. If a law of mortality holds, certain mortality and monetary functions may be evaluated analyt- ically (rather than numerically), but this point is of little practical importance in the computer age. There are also certain simplifications in the evaluation of joint-life functions (that is, functions depending on the survival of more than one life.) 22 CHAPTER 1. NON-SELECT LIFE TABLES Exercises 1.1 Calculate the following probabilities on the basis of English Life Table No. 12-Males. (i) 40 p 25 , (ii) 60 [ 10 q 25 , (iii) the probability that (30) survives for at least 10 years, (iv) the probability that (40) survives to age 65, (v) the probability that (50) dies within 10 years, (vi) the probability that (50) fails to reach age 70, (vii) the probability that (60) dies between ages 80 and 85, (viii) the probability that (60) dies within the first five years after retiring at age 65. In addition, express (i) and (ii) in words. 1.2 On the basis of E.L.T. No. 12 - Males, find the probability that a life aged 30 will (i) survive to age 40, (ii) die before reaching age 50, (iii) die in his 50th year of age, i.e. between ages 49 and 50, (iv) die between his 40th birthday and his 50th birthday, (v) die either between exact ages 35 and 45 or between exact ages 70 and 80. 1.3 A man aged 50 has just retired because of ill health. Up to exact age 58 he will be subject to a constant force of mortality of 0.019803 p.a., after which his mortality will be that of E.L.T. No. 12 - Males. Find the probability that he will (i) die before age 55, (ii) live to age 65, (iii) die between ages 55 and 60. 1.4 For the first 5 years after arrival in a certain country, lives are subject to a constant force of mortality of 0.005. Thereafter lives are subject to mortality according to English Life Table No. 12- Males with an addition of 0.039221 to the force of mortality. (i) A life aged exactly 30 has just arrived in the country. (a) Show that the probability that the life will survive to age 35 is 0.97531. (b) Find the probability that the life will survive to age 60. (ii) What is the probability that a life aged exactly 33 who has been in the country for 3 years will die between ages 50 and 51? (Assume that these lives will remain in the given country.) 1.5 For a certain animal population, l x = l 0 (1 +x) 2 (x ≥ 0) Calculate (i) the complete expectation of life at birth, (ii) the force of mortality at age 1 year, (iii) the chance that a newly-born animal will die between ages 1 and 2 years. 1.8. EXERCISES 23 1.6 Suppose that t p x µ x+t is decreasing for 0 ≤ t ≤ 1. Show that q x < µ x . 1.7 Suppose that Gompertz’ law, µ x = Bc x , holds for all x ≥ α, c being greater than 1. Assume that µ α < log c. (i) Give a formula for s(x). (ii) Show that l x µ x attains a maximum when µ x = log c, and has no other stationary points. 1.8 (Difficult) Suppose that there is a “uniform distribution of deaths” from age x to age x + 1, i.e. l x+t is linear in t for 0 ≤ t ≤ 1. Show that, for all 0 ≤ s ≤ t ≤ 1, t−s q x+s = t −s 1 −sq x q x 1.9 (Difficult) Suppose that the “Balducci hypothesis” holds from age x to age x + 1, i.e. 1−t q x+t = (1 −t)q x for 0 ≤ t ≤ 1. Show that, for all 0 ≤ s ≤ t ≤ 1, t−s q x+s = t −s 1 −(1 −t)q x q x 1.10 Suppose that, for some a > 0, s(x) = (1 −x/w) a (0 ≤ x ≤ w) Give simple formulae for (i) µ x , (ii) ◦ e x (Note finite limiting age), (iii) 10 p 70 , (iv) 40 [ 5 q 35 1.11 A certain group of lives now aged 60 experience mortality according to a(55) males ultimate with addition to the force of mortality. The addition is 0.0005 at age 60, increasing linearly to 0.0025 at age 80, at which level the addition remains constant. Find the probability that a life aged exactly 60 dies within 20 years. 24 CHAPTER 1. NON-SELECT LIFE TABLES Solutions 1.1 (i) 40 p 25 = l 65 l 25 = 0.71528 (ii) 60 [ 10 q 25 = l 85 −l 95 l 25 = 0.10021 (iii) 10 p 30 = l 40 l 30 = 0.98452 (iv) 25 p 40 = l 65 l 40 = 0.73025 (v) 10 q 50 = l 50 −l 60 l 50 = 0.12389 (vi) 20 q 50 = l 50 −l 70 l 50 = 0.39162 (vii) 20 [ 5 q 60 = l 80 −l 85 l 60 = 0.16173 (viii) 5 [ 5 q 60 = l 65 −l 70 l 60 = 0.17338 In words, (i) is “the probability that a life aged 25 will survive to age 65.” (ii) is “the probability that a life aged 25 will survive to age 85 and die before age 95.” 1.2 (i) 10 p 30 = l 40 l 30 = 0.98452 (ii) 20 q 30 = l 30 −l 50 l 30 = 0.05437 (iii) 19 [ q 30 = d 49 l 30 = 0.00613 (iv) 10 [ 10 q 30 = l 40 −l 50 l 30 = 0.03889 (v) 5 [ 10 q 30 + 40 [ 10 q 30 = (l 35 −l 45 ) + (l 70 −l 80 ) l 30 = 0.35789 1.3 Let us use an asterisk to denote the mortality table of the life concerned, while l x etc. refers to E.L.T. 12-Males. Let k = 0.019803. (i) 5 q ∗ 50 = 1 − 5 p ∗ 50 = 1 −exp(−5k) = 0.09427 1.9. SOLUTIONS 25 (ii) 15 p ∗ 50 = 8 p ∗ 507 p ∗ 58 = exp(−8k) l 65 l 58 = 0.71188 (iii) 5 [ 5 q ∗ 50 = 5 p ∗ 50 − 10 p ∗ 50 = 5 p ∗ 50 − 8 p ∗ 502 p ∗ 58 = exp(−5k) −exp(−8k) l 60 l 58 = 0.08540 1.4 (i) (a) Prob. of survival for 5 years = exp − 5 0 0.005dt = 0.97531 (b) 0.97531 exp (−0.039221 25) l ELT 60 l ELT 35 = 0.3051 (ii) Prob. = prob. of survival for 17 years - prob. of survival for 18 years = exp(−2 0.005) exp (−0.039221 15) l ELT 50 l ELT 35 −exp(−2 0.005) exp (−0.039221 16) l ELT 51 l ELT 35 = 0.02377 1.5 (i) ◦ e 0 = ∞ 0 l x l 0 dx = ∞ 0 (1 +x) −2 dx = ¸ −1 1 +x ∞ 0 = 1 (ii) µ x = − l x l x = 2 1 +x , so µ 1 = 1 (iii) 1 | q 0 = l 1 −l 2 l 0 = 1 4 − 1 9 = 0.13889 1.6 Let f(t) = t q x −µ x .t (0 ≤ t ≤ 1) f (t) = d dt t 0 r p x µ x+r dr −µ x = t p x µ x+t −µ x < 0 for 0 < t < 1 Now f(0) = 0, and f(t) is decreasing for 0 ≤ t ≤ 1, so f(1) < 0; therefore q x < µ x . 1.7 (i) For x ≥ α, t p x = exp ¸ − t 0 Bc x+s ds = exp −Bc x (c t −1)/ log c Therefore s(x) = x−α p α = exp [−B(c x −c α )/ log c]. 26 CHAPTER 1. NON-SELECT LIFE TABLES (ii) It follows that l x = k. exp [−Bc x / log c] for some k. Therefore d dx (l x µ x ) =l x Bc x log c +Bc x k exp [−Bc x / log c] [−Bc x log c] log c =l x µ x (log c −Bc x ) (∗) =0 when µ x = Bc x = log c Let x 0 be the unique point at which this occurs. Notice that (by equation (∗) above) d dx (l x µ x ) > 0 for x < x 0 , and is negative for x > x 0 . Therefore x 0 is a maximum point of l x µ x . 1.8 Under U.D.D., l x+t = tl x+1 + (1 −t)l x (0 ≤ t ≤ 1) and similarly for l x+s . Therefore l x+s −l x+t l x = (t −s)l x −(t −s)l x+1 l x = (t −s)q x Therefore t−s q x+s = (t −s)q x 1 −sq x , since l x l x+s = 1 1 −sq x . 1.9 t−s q x+s = 1 − l x+t l x+s = 1 − l x+1 l x+s − l x+t l x+s 1 − l x+1 l x+t = 1−s q x+s −(1 − t−s q x+s ) 1−t q x+t Therefore t−s q x+s = 1−s q x+s − 1−t q x+t 1 − 1−t q x+t (on rearranging). Now apply “Balducci” to the R.H.S. of the equation. This gives the desired result. 1.10 (i) µ x = − d dx [log s(x)] = a w −x (ii) ◦ e x = w−x 0 t p x dt = w x l y dy/l x = w x (1 −y/w) a dy (1 −x/w) a Substitute z = 1 −y/w to obtain (calculus exercise!) the result that ◦ e x = w −x a + 1 (iii) 10 p 70 = s(80) s(70) = w −80 w −70 a (iv) 40 [ 5 q 35 = s(75) −s(80) s(35) = 1 − 75 w a − 1 − 80 w a 1 − 35 w a 1.9. SOLUTIONS 27 1.11 Addition to force of mortality = 0.0005 + 0.0001t (0 ≤ t ≤ 20) Prob. of survival for 20 years = exp − 20 0 (µ 60+t + 0.0005 + 0.0001t)dt = 20 p 60 exp(−0.003) on integration = 0.4108 ∴ Ans. = 0.5892 28 CHAPTER 1. NON-SELECT LIFE TABLES Chapter 2 SELECT LIFE TABLES 2.1 What is selection? In the previous chapter we considered mortality rates to depend only on age: life tables of this form are sometimes called aggregate life tables. We now consider the situation when mortality rates (or the force of mortality ) depend on two factors: (i) age, and (ii) the time (duration) since a certain event, known as “selection”. One important example of “selection” is the acceptance of a proposal for life assurance at normal rates of premium: the mortality rates of lives who have been recently accepted for life assurance at normal rates may be expected to be lower than those of the general population. After a certain period the difference in mortality between those who have been accepted for life assurance and the general population of the same age decreases, but it is not correct to say that the effect of selection “wears off” entirely (since the general population contains some lives who would never have been accepted for life assurance at normal premium rates). This point may be confirmed by a comparison between the mortality rates of English Life Table No.12 - Males and A1967-70 ultimate, which we will discuss below. (The process by which life offices decide whether to accept lives for assurance, and on what terms, is called “underwriting”.) Another form of selection is the “self-selection” exercised by those who buy annuities: it may nor- mally be assumed that such lives are in good health (for their age), for the purchase of an annuity is otherwise likely to be a poor investment. In these examples, the mortality of those selected is lower than that of the general population, particularly in the period soon after selection. In “reverse selection”, mortality rates are higher than those of the general population (or some other reference group). An example of reverse selection is early retirement due to ill-health. After a certain period from the date of ill-health retirement, the mortality of these lives may be expected to become closer to that of lives who retired in normal health, or who are not yet retired. 29 30 CHAPTER 2. SELECT LIFE TABLES 2.2 Construction of select tables Let us, for definiteness, consider “selection” to be the acceptance of a proposer for life assurance at normal rates by a life office. Let x = age at entry to assurance, i.e. at the date of selection t = duration (in years) since the date of selection The current (or attained) age of such a life is y = x +t We write “[x] +t” as a shorthand notation for a “life aged x at selection and duration t years since selection”. Thus for example, h q [x]+t = Pr¦“[x] +t” will die within h years¦ = Pr¦a life aged x +t, who has selected t years ago, will die within h years¦ (2.2.1) µ [x]+t = the force of mortality of a life aged x +t who has selected t years ago = lim h→0+ h q [x]+t h (2.2.2) The select period. We assume that there is a period, s years, such that the mortality of those selected at least s years ago depends only on the attained age, x +t. That is, q [x]+t = q x+t (t ≥ s) (2.2.3) µ [x]+t = µ x+t (t ≥ s) (2.2.4) where q x+t and µ x+t refer to those who were selected at least s years ago: such people are called ultimate lives. For each fixed entry age x, we may define the family of random variables. T([x] +t) = the future lifetime of “[x] +t” (t ≥ 0) Regarding x as fixed, we may construct a “life table” for those selected at age x by methods similar to those of chapter 1 (with t in place of x, and α = 0). We have, for example, h q [x]+t = Pr¦T([x] +t) ≤ h¦ = 1 − l [x]+t+h l [x]+t where l [x]+t is the expected number number of survivors at age x + t of l [x] lives selected at age x. But instead of fixing each radix l [x] arbitrarily, we choose them to be such that l [x]+t depends only on x + t when t ≥ s. More precisely, we first construct a life table for the ultimate lives by the methods given in chapter 1; the function l [x]+t is then constructed to be such that l [x]+t = l x+t (t ≥ s) (2.2.5) This ensures that relationships such as h p [x]+t = l [x]+t+h l [x]+t (2.2.6) 2.3. THE CONSTRUCTION OF A1967-70. 31 (which are found by replacing “x” by “[x] +t” in the formulae of chapter 1) are true. We may omit the square brackets enclosing x in expressions such as h p [x]+t and l [x]+t+h if t ≥ s or t + h ≥ s respectively. We now give formulae for l [x]+t for 0 ≤ t ≤ s. If q [x]+t is given for t = 0, 1, 2, . . . , s−1 , we proceed recursively, using the formulae l [x]+t+1 l [x]+t = 1 −q [x]+t (t = s −1, , s −2, . . . , 0) (2.2.7) That is l [x]+s−1 = l [x]+s /(1 −q [x]+s−1 ) l [x]+s−2 = l [x]+s−1 /(1 −q [x]+s−2 ) l [x] = l [x]+1 /(1 −q [x] ) (2.2.8) If µ [x]+t is given for 0 ≤ t ≤ s , we use the formula l [x]+t = l x+s exp − s t µ [x]+r dr (t < s) (2.2.9) 2.3 The construction of A1967-70. We illustrate the construction of select tables by reference to A1967-70, which has a select period of 2 years. This table refers to the mortality of male assured lives in the U.K. during the period 1967 to 1970, and is based on data collected by the Continuous Mortality Investigation Bureau (C.M.I.B.). Mortality rates were extended to young ages (including “ultimate” lives aged under 2) for which no data existed. The “building blocks” of the table were the mortality rates: q y (y = 0, 1, 2, . . . ) q [x]+t (x = 0, 1, 2, . . . ; t = 0, 1) The stages of construction of l y and l [x]+t were as follows: 1. l 0 was fixed arbitrarily at 34,489. 2. l 1 , l 2 , l 3 , . . . , were computed as shown (working downwards in the third column of Table 2.3.1). 3. By working to the left from l 2 , l [0]+1 and l [0] were found; similarly for l [1]+1 and l [1] , etc. select select duration 0 duration 1 ultimate 34, 489 = l 0 34, 489(1 −q 0 ) = 34, 463.8 = l 1 l [0] = l [0]+1 1−q [0] ← l [0]+1 = l 2 1−q [0]+1 ← 34, 463.8(1 −q 1 ) = 34, 440.4 = l 2 ← 34, 440.4(1 −q 2 ) = 34, 418.7 = l 3 . . . 32 CHAPTER 2. SELECT LIFE TABLES Table 2.3.1 2.4 Some formulae for the force of mortality. We assume that l [x]+t and µ [x]+t are continuous in t (t ≥ 0). µ [x]+t = lim h→0+ h q [x]+t h = lim h→0+ l [x]+t −l [x]+t+h hl [x]+t = − d dt l [x]+t l [x]+t (t ≥ 0) (2.4.1) (the derivative is, strictly speaking, a L.H. derivative, but arguments similar to those given in chapter 1 show that the l [x]+t is differentiable in t ). We may thus write µ [x]+t = f (t) (t ≥ 0) where f(t) = −log(l [x]+t ). We now consider how best to estimate µ [x]+t for t = 0, 1, 2, . . . , s −1 from a table of l [x]+t (t = 0, , 1, 2, . . . , s). To estimate µ [x] = µ [x]+0 , we may use the differentiated form of Newton’s forward difference formula, which gives f (0) ∆f(0) − 1 2 ∆ 2 f(0) (2.4.2) where ∆f(0) = f(1) −f(0) ∆ 2 f(0) = f(2) −2f(1) +f(0) To estimate µ [x]+t , where t = 1, 2, . . . , s −1 , we may use the formula f (t) f(t + 1) −f(t −1) 2 to obtain µ [x]+t − 1 2 log(l [x]+t+1 /l [x]+t−1 ) (2.4.3) 2.5 Select tables used in examinations. Formulae and Tables for Actuarial Examinations give 1. A1967-70 (select period 2 years) and 2. a(55) males and females. Note that a(55) consists of 2 tables, a(55) males and a(55) females. In each case the select period is 1 year, and functions are given only for ages 60 and over. These tables were constructed on the basis of C.M.I.B. data and were intended to be suitable for those purchasing annuities in about 1955, an allowance being made for improvements in mortality . One should be careful not to look up “select” values (such as q [x] )when ultimate values (such as q x ) are required, or vice versa. In the case of a(55), one should be careful not to look up the “males” table when the “females” table is required, and vice versa. 2.6. EXERCISES 33 Exercises 2.1 A mortality table has a select period of three years. (i) Find expressions in terms of the life table functions l [x]+t and l y for (a) q [50] (b) 2 p [50] (c) 2 [ q [50] (d) 2 [ 3 q [50]+1 (ii) Calculate 3 p 53 given that: q [50] = 0.01601, 2 q [50] = 0.96411, 2 [ q [50] = 0.02410, 2|3 q [50]+1 = 0.09272 2.2 In its premium rate basis, an office assumes a 3-year select period. Functions on this table are indicated by an asterisk. The table is such that: q ∗ [x+7] = q ∗ [x+3]+1 = q ∗ [x]+2 = q ∗ x+1 and ultimate mortality follows A1967-70 ultimate. Assuming further that l ∗ y = l y on A1967-70 ultimate, calculate l ∗ [45]+t for t = 0, 1 and 2. 2.3 A certain life table has a select period of 1 year. At each integer age x , the select rate mortality is 50% of the ultimate rate. Calculate e [60] , given that e 60 = 17.5424 and q 60 = 0.0142. 2.4 Explain briefly the concept of selection in relation to mortality tables. Define the term “select period” and, using the A1967-70 table as an example, explain how a select table differs in construction from an aggregate table such as English Life Table No. 12 - Males. 34 CHAPTER 2. SELECT LIFE TABLES Solutions 2.1 (i) (a) q [50] = l [50] −l [50]+1 l [50] = .01601 (A) (b) 2 p [50] = l [50]+2 l [50] = .96411 (B) (c) 2| q [50] = l [50]+2 −l 53 l [50] = .02410 (C) (d) 2|3 q [50]+1 = l 53 −l 56 l [50]+1 = .09272 (D) (ii) Put l 50 = 100, 000 (say); then l [50] = 98, 399 from (A) l [50]+2 = 96, 411 from (B) l 53 = 94, 001 from (C) l 56 = 84, 877 from (D) Then 3 p 53 = l 56 l 53 = .90294. 2.2 l ∗ [45]+2 = l 48 1 −q ∗ [45]+2 = l 48 1 −q 46 = 33, 033 l ∗ [45]+1 = l [45]+2 1 −q ∗ [45]+1 = 33, 033 1 −q 43 = 33, 101 l ∗ [45] = l ∗ [45]+1 1 −q ∗ [45] = 33, 101 1 −q 39 = 33, 144 2.3 e [60] = l 61 +l 62 + . . . l [60] = l 60 l [60] l 61 +l 62 + . . . l 60 = l 60 l [60] e 60 But q [60] = 0.5q 60 , which implies that l [60] = l 61 1 − 1 2 q 60 Therefore l 60 l [60] = l 60 (1− 1 2 q 60 ) l 61 = (1− 1 2 q 60 ) 1−q 60 = 1.007202 so e [60] = 1.007202 17.542 = 17.668 2.4 Mortality rates depend on the age x at “selection” [e.g. entry to assurance, purchase of an annuity (self selection)] and duration t years since selection. Thus q [x]+t = Pr¦ a life aged x+t 2.7. SOLUTIONS 35 who was selected t years ago will die within a year ¦, etc. The select period, s years, is such that q [x]+t = q x+t for t ≥ s i.e. for lives who were selected at least s years ago (these being called “ultimate” lives) mortality depends only on the attained aged. The l [x]+t function is constructed to be such that l [x]+t = l x+t (t ≥ s) In A1967-70, s = 2. In a non-select or aggregate life table, such as E.L.T. No. 12 - Males, mortality depends on one variable only (age) and there is no “selection”. Note. It is not correct to say that the effects of selection “wear off” entirely (and that the mortality of ultimate lives is the same as that of the general population). If this were true, A1967-70 ultimate would resemble ELT 12 - Males, and it does not. The general population includes some lives who would never have been acceptable for assurance cover (at normal premium rates). 36 CHAPTER 2. SELECT LIFE TABLES Chapter 3 ASSURANCES 3.1 A general introduction A life assurance (or life insurance) policy is a contract which promises to pay a specified sum, S say, on the death of a given life (the life assured) at any future time, or between certain specified dates. The death benefit S is called the sum assured; it may be level (constant) or it may increase or decrease in a manner specified in the contract. Policies under which payments depend on the death or survival of more than one life may also be issued. In the case of with profits policies, the sum assured may be increased by additions, called bonuses, which depend on the experience of the office and its bonus distribution policy. With profits policies may be found in traditional or unitised form (see later discussion). One also encounters unit- linked policies, in which the benefits are directly linked to the performance of certain assets, e.g. a certain portfolio of equities. If the benefits (and premiums) are completely specified in money terms in the contract, the policy is said to be without profits (or non-profit). Policies may also provide endowment benefits, i.e. sums payable on the survival of the life (or lives) assured until a certain date (or dates). In this chapter we shall consider contracts issued on the life of one person. The benefits and expenses are paid for by premiums, which we shall discuss later. 3.2 Whole life assurances Suppose that, when calculating the value of benefits, the life office assumes that its funds will earn interest at a constant rate, i per annum. The corresponding force of interest per annum is δ = log(1 +i) and the present value of 1 due at time t years is v t = (1 +i) −t = e −δt (3.2.1) Now consider a life aged x, who is subject to a certain non-select mortality table, and let T be the future lifetime of (x). A whole life assurance is a policy providing a certain sum assured, S, on the death of (x) at any future date. The present value of this benefit (assumed for the moment to be payable immediately on death) is Z = g(T) = Sv T (3.2.2) 37 38 CHAPTER 3. ASSURANCES The mean of this variable, E(Z), is called the mean (or expected) present value (M.P.V.) of the whole life assurance benefit, and is E(Z) = ∞ 0 g(t)f(t) dt (3.2.3) where f(t) is the p.d.f. of T (for t > 0); hence E(Z) = S ∞ 0 v t t p x µ x+t dt (3.2.4) When S = 1 , we write E(Z) = ∞ 0 v t t p x µ x+t dt = ¯ A x (3.2.5) where A stands for “assurance” and the bar indicates that the sum assured is payable immediately on the death of (x). The M.P.V. of the sum of £S payable immediately on the death of (x) is therefore S ¯ A x Example 3.2.1. Find the M.P.V. of a whole life assurance of £10, 000 , payable immediately on the death of (40), according to E.L.T. No. 12-Males with interest at 4% p.a. Solution. M.P.V. = 10, 000 ¯ A 40 on E.L.T. No. 12-Males, 4% interest = 10, 000 0.31981 from Tables = £3, 198.10 Select tables For a select life (i.e. a person who has just been selected) aged x, we replace x by [x], and for an “[x] +t” life (i.e. a life now aged x +t who was selected t years ago) we replace x by [x] +t. Thus, letting T = T([x]) denote the future lifetime of “[x]”, we write ¯ A [x] = M.P.V. of v T = ∞ 0 v t t p [x] µ [x]+t dt (3.2.6) If T = T([x] +t) = future lifetime of “[x] +t” , we have ¯ A [x]+t = M.P.V. of v T = ∞ 0 v r r p [x]+t µ [x]+t+r dr (3.2.7) 3.3. COMMUTATION FUNCTIONS 39 3.3 Commutation functions Functions such as ¯ A x may easily be evaluated by numerical integration, and are often tabulated directly at various rates of interest. (In fact, a modern computer can easily compute ¯ A x directly at any rate of interest required, e.g. 4.032% ). Commutation functions are numerical devices (developed by Griffith Davies and others) which allow the calculation of certain common assurance (and annuity) values at a specified rate of interest from a small number of columns, labelled D x , ¯ M x etc. In view of their importance in the Tables for Actuarial Examinations we shall look at them in detail. At this point we consider only whole life assurances payable immediately on death. Define D x = v x l x (3.3.1) ¯ C x = 1 0 v x+t l x+t µ x+t dt (3.3.2) ¯ M x = ∞ ¸ t=0 ¯ C x+t (3.3.3) Now ¯ A x = ∞ 0 v x+t l x+t µ x+t dt v x l x = ¯ C x + ¯ C x+1 + ¯ C x+2 + . . . D x (on writing r+1 r v x+t l x+t µ x+t dt = 1 0 v x+r+u l x+r+u µ x+r+u du by the change of variable u = t −r) so that ¯ A x = ¸ ∞ t=0 ¯ C x+t D x = ¯ M x D x (3.3.4) We shall consider commutation functions again in relation to temporary and deferred assurances, etc. 3.4 The variance of the present value of benefits We recall that Z = g(T) = Sv T is the present value of S due immediately on the death of (x), and that the mean of Z is S ¯ A x . What is the variance of Z? Answer. We have var(Z) = E(Z 2 ) −[E(Z)] 2 = ∞ 0 S 2 v 2t t p x µ x+t dt − S ¯ A x 2 = S 2 ¸ ∞ 0 (v ∗ ) t t p x µ x+t dt − ¯ A x 2 (where v ∗ = v 2 ) = S 2 ¯ A ∗ x − ¯ A x 2 (3.4.1) 40 CHAPTER 3. ASSURANCES where the rate of interest i ∗ is such that v ∗ = 1 1 +i ∗ = v 2 = 1 (1 +i) 2 i.e. i ∗ = 2i +i 2 (3.4.2) Note that the force of interest, δ ∗ , corresponding to i ∗ is such that v ∗ = e −δ ∗ = v 2 = e −2δ i.e. δ ∗ = 2δ (3.4.3) The standard deviation of Z is, of course, found by taking the square root of var(Z). If we consider a block of n identical whole life policies on independent lives aged x, the total present value of the assurance benefits is Z = Z 1 +Z 2 + . . . +Z n where, for i = 1, 2, . . . , n, Z i = present value of benefit under ith policy = Sv T (i) with T (i) = future lifetime of ith life. Since the variables ¦T (i) ¦ are assumed to be independent, the variance of Z is Var(Z) = n ¸ i=1 Var(Z i ) = n ¸ i=1 S 2 ¯ A ∗ x − ¯ A x 2 = nS 2 ¯ A ∗ x − ¯ A x 2 (3.4.4) and so the standard deviation of the total present value is S √ n ¯ A ∗ x − ¯ A x 2 (3.4.5) We remark that the mean present value of a group of policies is the sum of their separate M.P.V’s even if the lives are not independent. Formulae 3.4.4 and 3.4.5 may be generalised to cover the case when the sums assured and/or ages of the lives are different (see exercise 3.3). Jensen’s inequality Let g(t) be convex, which is the case of g (t) ≥ 0 for all t > 0. Jensen’s inequality shows that E[g(T)] ≥ g[E(T)] (3.4.6) An application. Let g(t) = e −δt , where δ > 0. We have g (t) = δ 2 e −δt ≥ 0 for all t > 0 3.5. ASSURANCES PAYABLE AT THE END OF THE YEAR OF DEATH. 41 By Jensen’s inequality, E(v T ) ≥ v E(T) but E(T) = ◦ e x and E(v T ) = ¯ A x so we have shown that ¯ A x ≥ v ◦ e x (3.4.7) That is, at any positive rate of interest, v ◦ e x (which is the present value of £1 on the assumption that (x) dies when he reaches his expectation of life) is less than or equal to the mean present value, ¯ A x . Example 3.4.1. Verify formula 3.4.7 for x = 50 on the basis of E.L.T. No.12-Males, 4% p.a. interest. Solution. ◦ e 50 = 22.68 so v ◦ e 50 = 0.41085 which is less than ¯ A 50 = 0.44429 3.5 Assurances payable at the end of the year of death. Suppose that the death benefit S is payable at the end of the year of death (years being measured from the date of issue of the policy). It is now convenient to use the random variable K = [T] = the integer part of T which has discrete probabilities Pr¦K = k¦ = k [q x (k = 0, 1, 2, . . . ) The present value of the assurance is now Z = g(K) = Sv K+1 which has mean E[g(K)] = ∞ ¸ k=0 Sv k+1 k [q x (3.5.1) = SA x 42 CHAPTER 3. ASSURANCES where A x = ∞ ¸ k=0 v k+1 k [q x (3.5.2) We define the commutation functions C y = v y+1 d y (3.5.3) M y = ∞ ¸ k=0 C y+k (3.5.4) and find that A x = ¸ ∞ k=0 v x+k+1 d x+k D x = M x D x (3.5.5) Example 3.5.1. Show that the variance of Z (= Sv K+1 ) is S 2 A ∗ x −(A x ) 2 (3.5.6) where ∗ indicates at the rate of interest 2i +i 2 . Solution. Var(Z) =E(Z 2 ) −[E(Z)] 2 =S 2 ¸ ∞ ¸ k=0 (v ∗ ) k+1 k [q x −(A x ) 2 ¸ =S 2 A ∗ x −(A x ) 2 The relationship between A x and ¯ A x . Assuming that deaths occur on average mid-way through each policy year (i.e. the year between two consecutive policy anniversaries), benefits payable at the end of the year of death will be received, on average, 6 months later than those immediately payable on death. We thus have the approximate relationship ¯ A x (1 +i) 1 2 A x (3.5.7) This may be confirmed by establishing the following approximate relationship ¯ C x (1 +i) 1 2 C x (3.5.8) which follows from the mean value theorem for integrals. Formulae (3.3.3) and (3.5.4) now show that ¯ M x (1 +i) 1 2 M x (3.5.9) 3.5. ASSURANCES PAYABLE AT THE END OF THE YEAR OF DEATH. 43 and hence ¯ A x = ¯ M x D x (1 +i) 1 2 M x D x = (1 +i) 1 2 A x (3.5.10) Theorem 3.5.1. If there is a uniform distribution of deaths between the ages x + k and x + k + 1 (for k = 0, 1, 2, . . . ), we have ¯ A x = i δ A x (3.5.11) Proof. It is sufficient (since ¯ A x = ¸ ∞ k=0 ¯ C x+k D x and A x = [ ¸ ∞ k=0 C x+k ]/ D x ) to show that ¯ C y = i δ C y for y = x +k (k = 0, 1, 2, . . . ) Now, since we have U.D. of D. between ages y and y + 1, l y+t µ y+t = d y (0 ≤ t < 1) Therefore ¯ C y = 1 0 v y+t l y+t µ y+t dt = d y v y 1 0 v t dt = d y v y 1 −v δ = d y v y+1 i δ = i δ C y Note. By the mathematics of finance, if i is small we have i δ 1 + 1 2 i and (1 +i) 1 2 1 + 1 2 i so we sometimes find the following approximations (which are suitable only when i is small): ¯ A x = (1 + 1 2 i)A x ¯ C x = (1 + 1 2 i)C x ¯ M x = (1 + 1 2 i)M x (3.5.12) 44 CHAPTER 3. ASSURANCES 3.6 Assurances payable at the end of the 1/m of a year of death. Suppose that £1 is payable at the end of the 1/m year (measured from the issue date) following the death of (x); for example, if m = 12 , the sum assured is payable at the end of the month of death. The position is illustrated in Figure 3.6.1 below, in which ∗ indicates the time when (x) dies. - 0 [ 1 m [ 2 m [ . . . . . . . . . r m [ (r+1) m [ death occurs here J J^ payment is made here time (years) Figure 3.6.1: The mean present value of this benefit is A (m) x = ∞ ¸ r=0 v r+1 m r m | 1 m q x (3.6.1) As m →∞ , we find that (as one would expect from general reasoning) A (m) x −→ ¯ A x (3.6.2) Corresponding to formula 3.5.7 , we have the useful approximation A x v 1/(2m) ¯ A x (3.6.3) where the factor of v 1/(2m) allows for interest for 1/(2m) year, which is the average “delay” in receiving death benefit. The function A (m) x is not often used in practice. 3.7 Temporary and deferred assurances A term (or temporary) assurance contract pays a sum assured of £S, say, on the death of (x) within the term of the policy, which we usually write as n years. If the benefit is payable immediately on death, the present value (as a random variable) is Z = g(T) = Sv T if T ≤ n 0 if T > n The mean of this variable is written as S ¯ A 1 x:n , the “1” indicating that the status (x) must fail before the status n (a fixed period of n years) for a payment to be due (and that the payment is due when this event occurs). Thus ¯ A 1 x:n = n 0 v t t p x µ x+t dt (3.7.1) This may be evaluated by numerical integration or by commutation functions, i.e. ¯ A 1 x:n = n−1 ¸ t=0 ¯ C x+t D x = ¯ M x − ¯ M x+n D x (3.7.2) 3.7. TEMPORARY AND DEFERRED ASSURANCES 45 (provided that n is an integer) If the sum assured is payable at the end of year of death (if this occurs within the n year term), we have the present value Z = g(K) = Sv K+1 if K < n 0 if K ≥ n The M.P.V. of this benefit is written as S A 1 x:n , and we have A 1 x:n = n−1 ¸ t=0 v t+1 t [ q x (3.7.3) (It is assumed that n is an integer.) In terms of commutation functions, A 1 x:n = n−1 ¸ t=0 C x+t D x = M x −M x+n D x (3.7.4) In view of the relationship between C x and ¯ C x , we have the approximations ¯ A 1 x:n (1 +i) 1 2 A 1 x:n i δ A 1 x:n (3.7.5) A deferred assurance provides a sum of £S (say) on the death of (x) if this occurs after a certain period, called the deferred period (or, more correctly, the period of deferment), which we usually write as n years. If the benefit is payable immediately on death of (x) after the deferred period has elapsed, the present value is Z = g(T) = Sv T if T > n 0 if T < n if the sum assured is payable immediately on death, or Z = g(K) = Sv K+1 if K ≥ n 0 if K < n if the sum assured is payable at the end of year of death. The corresponding M.P.V.’s are written as n [ ¯ A x = v n n p x ¯ A x+n (3.7.6) and n [A x = v n n p x A x+n (3.7.7) We have the formulae n [ ¯ A x = ∞ n v t t p x µ x+t dt (3.7.8) = ∞ ¸ t=n ¯ C x+t D x = ¯ M x+n D x (3.7.9) 46 CHAPTER 3. ASSURANCES and n [A x = ∞ ¸ t=n v t+1 t| q x (3.7.10) = ∞ ¸ t=n C x+t D x = M x+n D x (3.7.11) In view of the relationship between ¯ C x and C x , we have the approximations: n [ ¯ A x (1 +i) 1 2 n [A x i δ n [A x (3.7.12) The evaluation of term and deferred assurance functions is often simplified by the observation that a whole life policy may be thought of as a combination of a term assurance and a deferred assurance, so that A x = A 1 x:n + n [A x (3.7.13) and ¯ A x = ¯ A 1 x:n + n [ ¯ A x (3.7.14) 3.8 Pure endowments and endowment assurances We now consider a pure endowment of £1 of term n years, i.e. a policy providing the sum of £1 at time n years if (x) is then alive. The present value of this benefit is Z = v n if T ≥ n 0 if T < n and hence the M.P.V. is v n n p x (since Pr¦T ≥ n¦ = n p x ). This M.P.V. is written as n E x or A 1 x:n or ¯ A 1 x:n i.e. n E x = A 1 x:n = ¯ A 1 x:n = v n n p x (3.8.1) In terms of commutation functions we have the useful result that n E x = D x+n D x (3.8.2) Returning to the formulae (3.7.6) and (3.7.7) , we see that the M.P.V. of a deferred assurance may be written as the product of n E x and ¯ A x+n or A x+n . That is, the value is obtained by multiplying the M.P.V. of the assurance at the “vesting date” (when the life attains age x+n) by a pure endowment factor. We finally consider an endowment assurance of term n years with sum assured £S. In this policy, the sum assured is payable if (x) dies within n years or on maturity of the contract at time 3.8. PURE ENDOWMENTS AND ENDOWMENT ASSURANCES 47 n years, whichever occurs first. It follows that an endowment assurance is a combination of a term assurance and a pure endowment (of the same term). If we assume that the death benefit is payable immediately on death, if death occurs within n years, the present value of the benefits is Z = g(T) = Sv T if T < n Sv n if T ≥ n The mean of Z is written S ¯ Ax:n .Thus ¯ Ax:n = ¯ A 1 x:n + ¯ A 1 x:n (3.8.3) = ¯ A 1 x:n + n E x = ¯ M x − ¯ M x+n +D x+n D x (3.8.4) Similarly, if the death benefit is payable at the end of the year of death, the present value is Z = g(K) = Sv K+1 if K < n Sv n if K ≥ n and the M.P.V. is written as S Ax:n . Thus Ax:n = A 1 x:n + A 1 x:n (3.8.5) = A 1 x:n + n E x = M x −M x+n +D x+n D x (3.8.6) In view of the fact that the term D x+n occurs in formulae (3.8.4) and (3.8.6), it is not correct to write ¯ Ax:n (1 +i) 1 2 Ax:n A COMMON MISTAKE Correct relationships are (for example): ¯ A x:n (1 +i) 1 2 A 1 x:n +A 1 x:n (3.8.7) and ¯ A x:n (1 +i) 1 2 (M x −M x+n ) +D x+n D x (3.8.8) Select tables The adjustments to the formulae when a select table is used are straightforward: one need only replace ‘x +t’ by ‘[x] +t’ (or ‘[x]’ if the life assured has just been selected) and dispense with [ ] if the duration since selection is at least equal to the select period, s years, and in compound interest functions. For example, C [x]+t = v x+t+1 d [x]+t Example 3.8.1. On the basis of A1967-70 select mortality and 4% p.a. interest, calculate the mean present value of each of the following assurance benefits for a life aged 30: 48 CHAPTER 3. ASSURANCES (i) A whole life assurance for £10, 000, payable immediately on death; (ii) A 20-year term assurance for £50, 000, payable at the end of the year of death; (iii) A 20-year endowment assurance for £50, 000, with the death benefit payable immediately on death; (iv) A deferred temporary assurance for £100, 000, payable at the end of the year of death, if death occurs between ages 40 and 50 exactly. [Use the factor (1+i) 1 2 for accelerating payments from end of year of death to the moment of death.] Solution. (i) 10, 000 ¯ A [30] = 10, 000(1.04) 1 2 A [30] = £1, 934.87 (ii) 50, 000 (M [30] −M 50 ) D [30] = £1, 012.90 (iii) (1.04) 1 2 1, 012.90 + 50, 000 D 50 D [30] = £23, 070.56 (iv) 100, 000 (M 40 −M 50 ) D [30] = £1, 360.91 3.9 Varying assurances Suppose that a contract provides the sum of β(t) immediately on the death of (x) at time t years. The present value of this benefit is Z = g(T) = β(T) v T (T > 0) and hence the M.P.V. is ∞ 0 v t β(t) t p x µ x+t dt (3.9.1) If β(t) = t for all t > 0 , the M.P.V. is written as ( ¯ I ¯ A) x , i.e. ( ¯ I ¯ A) x = ∞ 0 tv t t p x µ x+t dt (3.9.2) 3.9. VARYING ASSURANCES 49 If β(t) = [t] +1 , where [t] indicates the integer part of t, the M.P.V. is written as (I ¯ A) x , and it may easily be shown that (I ¯ A) x = ∞ ¸ t=0 (t + 1) ¯ C x+t D x = ¯ C x + ¯ C x+1 + + ¯ C x+1 + ¯ C x+2 + + D x = ¯ M x + ¯ M x+1 + D x = ¯ R x D x (3.9.3) where ¯ R x = ∞ ¸ k=0 ¯ M x+k (3.9.4) In view of the fact that the average benefit is 1 2 less when β(t) = t than in the present case, we have the approximation ( ¯ I ¯ A) x (I ¯ A) x − 1 2 ¯ A x (3.9.5) We now consider the case when a death benefit of β(k) in year k is payable at the end of the year of death. The present value is Z = g(K) = β(K + 1) v K+1 (K = 0, 1, 2, . . .) and hence the M.P.V. is ∞ ¸ k=0 β(k + 1)v k+1 k [ q x (3.9.6) When β(k) = k for k = 1, 2, . . . , the M.P.V. is written (IA) x , so we have (IA) x = ∞ ¸ k=0 (k + 1)v k+1 k [ q x (3.9.7) In terms of commutation functions we have (IA) x = ∞ ¸ k=0 (k + 1)C x+k D x = M x +M x+1 + D x = R x D x (3.9.8) where R x = ∞ ¸ k=0 M x+k (3.9.9) 50 CHAPTER 3. ASSURANCES In view of the approximations ¯ C x (1 +i) 1 2 C x , etc., we find that ¯ R x i δ R x (1 +i) 1 2 R x (3.9.10) and (I ¯ A) x i δ (IA) x (1 +i) 1 2 (IA) x (3.9.11) We may also define similar symbols for increasing temporary and endowment policies, e.g. (IA) 1 x:n = n−1 ¸ k=0 (k + 1)v k+1 k [ q x = n−1 ¸ k=0 (k + 1)C x+k D x = [C x + 2C x+1 + ] −[C x+n + 2C x+n+1 + ] −n[C x+n +C x+n+1 + ] D x = R x −R x+n −nM x+n D x (3.9.12) and (IA) x:n = (IA) 1 x:n +n n E x (3.9.13) 3.10 Valuing the benefits under with profits policies So far we have studied without profits contracts, for which the premiums and benefits are com- pletely fixed (in money terms). In with profits contracts, the premiums are higher but the policy- holder has the right to a share of the profits of the office. In the U.K., the usual system for sharing profits is by means of additions to the basic benefit (sum assured, pension, etc.) in the form of bonuses. These may be either reversionary bonuses, which are usually added annually and are not subsequently reduced, or terminal bonuses, which are paid only on death or maturity (and possibly on surrender) and are not guaranteed to continue. Nowadays one may also encounter unitised with profit policies. These are not the same as unit-linked policies: the maturity proceeds of the latter are linked to the performance of a unit trust or internal fund of the life office, while for U.W.P. policies the office credits each policy with internal units which are subject to certain guarantees. In this book we shall study reversionary bonuses only. There are 3 systems of adding bonuses in common use: (1) the simple bonus system; (2) the compound bonus system, and (3) the two-tier (or supercompound) bonus system. 3.10. VALUING THE BENEFITS UNDER WITH PROFITS POLICIES 51 (1) Simple bonuses. In this system, bonuses are calculated only on the basic sum assured (B.S.A.), which we shall denote by S. Consider a whole life with profits policy issued t years ago to a life then aged x. Suppose that bonuses vest annually in advance, let the bonuses already vested be B, and suppose that a bonus is about to be added. the formula used to calculate new bonus addition is new bonus = S rate of bonus p.a. (3.10.1) To value the benefits we must estimate the rate of simple bonus p.a. (applying now and for the forseeable future) as b (say). The total sum assured is thus S +B + bS in year 1 from the present time, S +B + 2bS in year 2 from the present time, and so on, giving a total benefit in year k of S +B +kbS (3.10.2) If the benefits are payable at the end of the year of death, the M.P.V. of the benefits is ∞ ¸ k=0 [S +B + (k + 1)bS] v k+1 k [q x+t =(S +B)A x+t +bS(IA) x+t (3.10.3) Notes. (1) At the inception of the policy, we have t = 0 and B = 0. (2) If the benefits are payable immediately on death, the M.P.V. is (S +B) ¯ A x+t +bS(I ¯ A) x+t (3.10.4) (3) In the case of an n-year endowment assurance with profits, the benefits on maturity are normally equal to those on death in the final year. This leads to the following formulae for the value of benefits: at end of year of death : (S +B)A x+t:n−t +bS(IA) x+t:n−t (3.10.5) immediately on death : (S +B) ¯ A x+t:n−t +bS(I ¯ A) x+t:n−t (3.10.6) (2) Compound bonuses In this system, the formula for bonus additions is new bonus = ¸ ¸ ¸ ¸ basic sum assured plus bonuses already vesting ¸ rate of bonus p.a. (3.10.7) Assuming a bonus rate of b p.a. applies now and for the forseeable future, the benefits will be (S +B)(1 +b) in year 1 (S +B)(1 +b) 2 in year 2 and so on, giving (S +B)(1 +b) k in year k (3.10.8) 52 CHAPTER 3. ASSURANCES If the benefits are payable at the end of the year of death, their M.P.V. is ∞ ¸ k=0 (S +B)(1 +b) k+1 v k+1 k [q x+t =(S +B) ∞ ¸ k=0 1 +b 1 +i k+1 k [q x+t =(S +B)A ∗ x+t (3.10.9) where A ∗ is at rate of interest i −b 1 +b (3.10.10) If the benefits are payable immediately on death, their M.P.V. is approximately (S +B)(1 +i) 1 2 A ∗ x+t (3.10.11) NOT (S +B)A ∗ x+t For endowment assurances, the corresponding formulae are (S +B)A ∗ x+t:n−t (3.10.12) and (S +B) (1 +i) 1 2 A ∗ 1 x+t:n−t +A ∗ 1 x+t:n−t (3.10.13) Notes. (1) In formula (3.10.13), we observe that the pure endowment part is as in formula (3.10.12): only the term assurance part is “accelerated” (2) At inception, we have t = 0 and B = 0. (3) Supercompound (two-tier) bonuses The formula is new bonus =(B.S.A.) bonus rate p.a. in respect of B.S.A. + bonuses already vesting ¸ ¸ bonus rate p.a. in respect of bonuses already vesting ¸ (3.10.14) Let us assume that the bonus rate p.a. in respect of the B.S.A. and previous bonus additions are a and b respectively (now and for the forseeable future). Let B(k) = total bonus in year k (from the present time), and let us define B(0) = B. We have B(1) = B(1 +b) +aS B(2) = B(1 +b) 2 +aS[1 + (1 +b)] and so on, giving the conjecture B(k + 1) = B(1 +b) k+1 +aS s k+1 b (3.10.15) 3.10. VALUING THE BENEFITS UNDER WITH PROFITS POLICIES 53 Proof by induction. Since B(1) = (aS +bB) +B = B(1 +b) +aS the result is true for k = 0. Now assume formula (3.10.15) is true for B(k +1), k being some value. We have B(k + 2) = [aS +bB(k + 1)] +B(k + 1) = (1 +b) (1 +b) k+1 B +aS (1 +b)s k+1 b + 1 = (1 +b) k+2 B +aS s k+2 b so the result is also true for B(k + 2), as required. It follows that, in this bonus system, the M.P.V. of a whole life assurance is ∞ ¸ k=0 [S +B(k + 1)]v k+1 k [q x+t (assuming death benefits are payable at end of the year of death) = ∞ ¸ k=0 S +aS ¸ (1 +b) k+1 −1 b +B(1 +b) k+1 k [q x+t = ∞ ¸ k=0 S 1 − a b + a b S +B (1 +b) k+1 ¸ k [q x+t =S 1 − a b A x+t + a b S +B A ∗ x+t (3.10.16) where A ∗ x+t is at rate i −b 1 +b Notes. (1) If the sum assured is payable immediately on death, the M.P.V. is approximately S 1 − a b ¯ A x+t + a b S +B (1 +i) 1 2 A ∗ x+t (3.10.17) (2) At inception of the policy, we have t = 0 and B = 0. (3) The “ordinary” compound bonus system is a special case of the two-tier system with a = b. (4) For an endowment assurance the formulae are: at end of year of death : S 1 − a b A x+t:n−t + a b S +B A ∗ x+t:n−t (3.10.18) immediately on death : S 1 − a b ¯ A x+t:n−t + a b S +B (1 +i) 1 2 A ∗ 1 x+t:n−t +A ∗ 1 x+t:n−t (3.10.19) (Note that in formula (3.10.19) the term assurance benefits are “accelerated” but the pure endowment parts are the same as in formula (3.10.18).) 54 CHAPTER 3. ASSURANCES 3.11 Guaranteed bonus policies As their name suggests, these are not with profits contracts, as they provide additions to the basic benefit at rates which are fixed at the outset and do not depend on future the experience of the life office. But many of the formulae used to value these benefits are the same as those we have derived in the previous section. Example 3.11.1. A company issues 20-year endowment assurances each with a basic sum assured of £1, 000 to male lives aged 45. Guaranteed simple reversionary bonuses at the rate of 2.25% of the current sum assured vest on pay- ment of each annual premium. Alternatively, at the outset of a policy, a life assured may elect that compound reversionary bonuses should be added to the policy instead of simple reversionary bonuses. The sum assured and added bonuses will be payable at maturity, or at the end of the year of death, if earlier. Derive an expression from which can be calculated the guaranteed rate of compound bonus the company should offer so that the value of the benefits at the outset is the same. Solution. Let b be the annual rate of guaranteed compound bonus. The equation of value for b is: 1000 [A 45:20 + 0.0225(IA) 45:20 ] = 1000A ∗ 45:20 at rate i ∗ = i−b 1+b This may be solved for i ∗ , and hence b may be found. 3.12. EXERCISES 55 Exercises 3.1 Evaluate (a) ¯ A 60 and (b) A 60 on the bases: (i) A1967-70 ultimate, 4% p.a. interest; (ii) E.L.T. No.12 - Males, 4% p.a. interest. 3.2 (i) Show that A x = vq x +vp x A x+1 (ii) Given that p 60 = 0.985, p 61 = 0.98, i = 0.05 and A 62 = 0.6 , evaluate A 61 and A 60 3.3 Consider n lives now aged x 1 , x 2 , . . . , x n respectively. Let Z be the total present value of a contract providing the sum of £S i immediately on death of (x i ), i = 1, 2, . . . , n. The n lives are subject to the same non-select mortality table, Table B, and the interest is taken to be fixed at rate i p.a. (i) Show that E(Z) = S 1 ¯ A x 1 + +S n ¯ A x n on Table B with interest at rate i p.a. (ii) Assuming further that the future lifetimes T(x i ) of the lives are independent variables, show that Var(Z) = n ¸ j=1 S 2 j [ ¯ A ∗ x j −( ¯ A x j ) 2 ] where ∗ refers to an interest rate of 2i +i 2 p.a. 3.4 Using commutation functions or otherwise calculate the values of the following: (i) A [40]:10 on A1967-70, 4% p.a. interest; (ii) A 1 30:20 on A1967-70 Ultimate, 4%; (iii) ¯ A 1 30:20 on A1967-70 Ultimate, 4%; (iv) ¯ A 30:20 on A1967-70 Ultimate, 4%; (v) ¯ A 30:20 on English Life Table No.12 Males, 4%. 3.5 A life aged 50 who is subject to the mortality of the A1967-70 Select table, effects a pure endowment policy with a term of 20 years for a sum assured of £10,000. (i) Write down the present value of the benefits under this contract, regarded as a random variable. (ii) Assuming an effective rate of interest of 5% per annum, calculate the mean and the variance of the present value of the benefits available under this contract. 3.6 What are the random variables (in terms of K = curtate future lifetime of (x)) whose means are represented by the following symbols? (i) n E x (ii) A 1 x:n (iii) A 1 x:n 3.7 A life aged exactly 60 wishes to arrange for a payment to be made to a charity in 10 years’ time. If he is still alive at that date the payment will be £1000. If he dies before the payment date, the amount given will be £500. Assuming an effective interest rate of 6% per annum and 56 CHAPTER 3. ASSURANCES mortality according to ELT No.12-Males, calculate the standard deviation of the present value of the liability. 3.8 (Difficult) You are given that (i) 1000 (IA) 50 =4,996.75 , (ii) 1000 A 1 50:1 =5.58, (iii) 1000 A 51 = 249.05 and (iv) i =0.06. Calculate 1000(IA) 51 . 3.13. SOLUTIONS 57 Solutions 3.1 (i) (a) ¯ A 60 = (1 + 0.04) 1 2 A 60 = (1.04) 1 2 0.51726 = 0.52750 (b) A 60 = 0.51726 from tables (ii) (a) ¯ A 60 = 0.58317 from tables (b) A 60 = (1.04) − 1 2 ¯ A 60 = 0.57185 3.2 (i) A x = ∞ ¸ k=0 k [ q x v k+1 = q x v + ∞ ¸ k=1 p x k−1 [q x+1 v k+1 = q x v + p x v ∞ ¸ k=0 k [q x+1 v k+1 = q x v + p x vA x+1 (ii) A 61 = 1 1.05 (0.02 + 0.98 0.6) = 0.57905 A 62 = 1 1.05 (0.015 + 0.985 0.57905) = 0.55749 3.3 (i) We have Z = n ¸ j=1 S j v T j so E(Z) = n ¸ j=1 S j E(v T j ) = n ¸ j=1 S j ¯ A x j (ii) By independence of T x j , we have Var(Z) = n ¸ j=1 S 2 j var(v T j ) = n ¸ j=1 S 2 j [ ¯ A ∗ x j −( ¯ A x j ) 2 ] 3.4 (i) M [40] −M 50 +D 50 D [40] = 1904.8595 −1767.5555 + 4597.0607 6981.5977 = 0.67812 (ii) M 30 −M 50 D 30 = 1981.9552 −1767.5555 10433.310 = 0.020550 58 CHAPTER 3. ASSURANCES (iii) ¯ A 1 30:20 (1 +i) 1 2 A 1 30:20 = 0.020956 (iv) ¯ A 30:20 = ¯ A 1 30:20 + D 50 D 30 = 0.46157 (v) A 1 30:20 ¯ A 1 30:20 (1 +i) 1 2 = (1 +i) − 1 2 ¯ M 30 − ¯ M 50 D 30 = 0.03286 3.5 (i) P.V. = 10, 000v 20 if T ≥ 20 0 if T < 20 where T = future lifetime of (50). (ii) E(P.V.) = ( 20 p [50] v 20 + (1 − 20 p [50] ) 0)10, 000 = 10, 000 0.725539 1.05 −20 = 2734.48 E[(P.V.)] 2 = 20 p [50] (10, 000v 20 ) 2 = 0.725539 10, 000 2 1.05 −40 = 10, 305, 968 ⇒V ar(P.V.) = E[(P.V.) 2 ] −[E(P.V.)] 2 = 2, 828, 587 = 1, 681.84 2 3.6 (i) Z = v n if K ≥ n 0 if K < n (ii) Z = 0 if K ≥ n v K+1 if K < n (iii) Same as 3.6(i) 3.7 Let X payment to be made at time 10 years. X = 1000 with prob. 10 p 60 500 with prob. (1 − 10 p 60 ) ∴ E(X) = 1000 10 p 60 + 500(1 − 10 p 60 ) = 500(1 + 10 p 60 ) = £847.207 Var(X) = 1000 2 10 p 60 + 500 2 (1 − 10 p 60 ) −(mean) 2 = 1, 000, 000[ 10 p 60 + 1 4 (1 − 10 p 60 )] −(mean) 2 = 1, 000, 000[0.25 + 0.75 10 p 60 ] −(mean) 2 = 770, 811.1 −(847.207) 2 = 53, 051.39 ∴ s.d. of X = 230.33 Present value is v 10 (X) and so has s.d. 230.33v 10 at 6% = £128.615 3.13. SOLUTIONS 59 3.8 (IA) 50 = C 50 + 2C 51 + D 50 = A 1 50:1 + (C 51 +C 52 + ) + (C 51 + 2C 52 + ) D 50 = A 1 50:1 + D 51 D 50 (A 51 + (IA) 51 ) Now vq 50 = A 1 50:1 = 0.00558 ∴ D 51 D 50 = v −vq 50 = 0.93782 ∴ 4.99675 = 0.00558 + 0.93782[0.24905 + (IA) 51 ] ∴ (IA) 51 = 5.07305 ∴ 1, 000(IA) 51 = 5, 073 60 CHAPTER 3. ASSURANCES Chapter 4 ANNUITIES 4.1 Annuities payable continuously Let (x) be entitled to £1 p.a. for life, payable “continuously”, i.e. the rate of payment at time t years is £1 p.a. if (x) is then alive. The present value of this benefit, expressed as a random variable, is g(T) = ¯ a T for T > 0 (4.1.1) This variable has mean ¯ a x , i.e. ¯ a x = E(¯ a T ) = ∞ 0 ¯ a t t p x µ x+t dt (4.1.2) An alternative definition is: ¯ a x = ∞ 0 v t . .. . interest (or discounting) factor t p x . .. . probability of survival of (x) for t years dt (4.1.3) To show that these definitions are equivalent, we may use integration by parts, as follows: ∞ 0 ¯ a t . .. . ¯ a t = t 0 v r dr So d dt (¯ a t ) = v t t p x µ x+t . .. . d dt (− t p x ) = d dt ( t q x ) = t p x µ x+t dt 61 62 CHAPTER 4. ANNUITIES ∞ 0 ¯ a t t p x µ x+t dt = ¯ a t (− t p x ) ∞ 0 + ∞ 0 v t t p x dt = ∞ 0 v t t p x dt Note the following conversion relationship (i.e. a formula connecting assurance and annuity values): ¯ A x = 1 −δ¯ a x (4.1.4) which follows from: ¯ a x = ∞ 0 ¯ a t t p x µ x+t dt = ∞ 0 1 −v t δ t p x µ x+t dt = 1 δ (1 − ¯ A x ). We may also argue as follows. By the Mathematics of Finance, 1 . .. . Capital invested = δ¯ a T . .. . p.v. of interest + v T . .. . p.v. of return of capital (for all T > 0) Take expected values of each side, which gives 1 = δ¯ a x + ¯ A x The variance of the present value of an annuity. Since ¯ a T = 1 −v T δ we have Var(¯ a T ) = 1 δ 2 Var(v T ) = ¯ A ∗ x − ¯ A 2 x δ 2 (4.1.5) where ∗ indicates “at the rate of interest 2i +i 2 p.a.”; the corresponding force of interest is δ ∗ = 2δ, so some writers use 2 ¯ A x rather than ¯ A ∗ x . Example 4.1.1. Express Var(¯ a T ) in terms of ¯ a x and ¯ a ∗ x . Solution. ¯ A ∗ x = 1 −(2δ)¯ a ∗ x since force of interest is 2δ so, by formula (4.1.5), Var(¯ a T ) = 1 −2δ(¯ a ∗ x ) −(1 −δ¯ a x ) 2 δ 2 = 1 −2δ(¯ a ∗ x ) −1 −δ 2 (¯ a x ) 2 + 2δ¯ a x δ 2 = 1 δ 2(¯ a x − ¯ a ∗ x ) −δ(¯ a x ) 2 4.2. ANNUITIES PAYABLE ANNUALLY 63 Commutation Functions. Define D x = v x l x (as stated in Chapter 3) (4.1.6) ¯ D x = 1 0 v x+t l x+t dt (4.1.7) ¯ N x = ∞ ¸ t=0 ¯ D x+t (4.1.8) Example 4.1.2. Show that ¯ a x = ¯ N x D x (4.1.9) Solution. ¯ N x = ∞ ¸ t=0 ¯ D x+t but (on change of variable) ¯ D x+t = t+1 t v x+r l x+r dr for t = 0, 1, 2, . . ., so ¯ N x = ∞ 0 v x+r l x+r dr from which we obtain ¯ N x D x = ∞ 0 v t t p x dt = ¯ a x . 4.2 Annuities payable annually Now consider the cases when an annuity of £1 p.a. is payable (i) annually in advance, (ii) annually in arrear. (The former is called an annuity-due.) In case (i), the p.v. of the benefit is g(K) = ¨ a K+1 (4.2.1) In case (ii), it is g(K) = a K = ¨ a K+1 −1 (4.2.2) where we take a 0 = 0. The means are denoted by ¨ a x and a x respectively. Note that ¨ a x = 1 +a x VITAL! (4.2.3) (since the annuitant gets extra immediate payment of 1 in case (i)). Now ¨ a x =E ¨ a K+1 = ∞ ¸ k=0 ¨ a k+1 k [q x (4.2.4) But ¨ a K+1 = 1 −v K+1 d (K = 0, 1, 2 . . . ) 64 CHAPTER 4. ANNUITIES so ¨ a x =E ¸ 1 −v K+1 d = 1 −E(v K+1 ) d = 1 −A x d (4.2.5) We thus have the conversion relationship A x = 1 −d¨ a x (4.2.6) Important formulae for a x and ¨ a x . We note the following results: a x = ∞ ¸ t=1 v t t p x (4.2.7) ¨ a x = ∞ ¸ t=0 v t t p x (4.2.8) It is sufficient to prove (4.2.8) since a x = ¨ a x −1 then gives (4.2.7). Two Proofs (1) Regard the annuity as sum of pure endowments due at times 0,1,2,. . .. Thus ¨ a x = 0 E x + 1 E x + 2 E x . . . = ∞ ¸ t=0 v t t p x (2) By the definition, ¨ a x = ∞ ¸ k=0 ¨ a k+1 k [q x = ¨ a 1 d x l x + ¨ a 2 d x+1 l x + ¨ a 3 d x+2 l x +. . . = 1 l x ¸ d x + (1 +v)d x+1 + (1 +v +v 2 )d x+2 +. . . ¸ = ¸ (d x +d x+1 +d x+2 +. . . ) +v(d x+1 +d x+2 +. . . ) +v 2 (d x+2 +. . . ) +. . . ¸ l x = l x +vl x+1 +v 2 l x+2 +. . . l x = ∞ ¸ t=0 v t t p x , as required. Evaluation of ¨ a x by Commutation Functions Define N x = ∞ ¸ t=0 D x+t 4.3. TEMPORARY ANNUITIES 65 We obtain ¨ a x = v x l x +v x+1 l x+1 +. . . v x l x = D x +D x+1 +. . . D x = N x D x Remark If we require a x , use a x = ¨ a x −1 or a x = N x+1 D x (since N x+1 = D x+1 +D x+2 +. . . ) The variance of ¨ a K+1 . We use the formula Var ¨ a K+1 = Var 1 −v K+1 d = 1 d 2 Var v K+1 = A ∗ x −(A x ) 2 d 2 (4.2.9) where ∗ indicates the rate of interest i ∗ = 2i +i 2 p.a. Note For an annuity payable in arrear, we use the result that Var (a K ) = Var ¨ a K+1 −1 = Var ¨ a K+1 (4.2.10) 4.3 Temporary annuities Let us first consider the “continuous payments” case, and suppose that (x) is entitled to £1 p.a. payable continuously for at most n years. (that is, payments cease when x dies or after n years, whichever is earlier.) The p.v. is g(T) = ¯ a T if T < n ¯ a n if T ≥ n = ¯ a min{T,n} (4.3.1) The mean of g(T) is ¯ a x:n = n 0 ¯ a t t p x µ x+t dt + ¯ a n ∞ n t p x µ x+t dt = n 0 v t t p x dt (4.3.2) by integration by parts. (Check this!) Note. THERE ARE NO SUCH THINGS AS ¯ a1 x:n and ¯ a x: 1 n !! Commutation Functions. If n is an integer, ¯ a x:n = n 0 v x+t l x+t dt v x l x = ¯ N x − ¯ N x+n D x (4.3.3) 66 CHAPTER 4. ANNUITIES If payments are made annually in advance, and are limited to at most n payments, we obtain ¨ a x:n = mean of ¨ a K+1 if K < n ¨ a n if K ≥ n = E ¨ a min{K+1,n} (4.3.4) = n−1 ¸ k=0 ¨ a k+1 k [q x + ¨ a n n p x which may be shown to be equal to 1 +vp x +v 2 2 p x + +v n−1 n−1 p x (i.e., the sum of the values of n pure endowments each for £1) = D x +D x+1 + +D x+n−1 D x Hence ¨ a x:n = N x −N x+n D x (4.3.5) If the payments are made annually in arrear, we obtain a x:n = mean of a K if K < n a n if K ≥ n = E a min{K,n} (4.3.6) leading to a x:n = D x+1 +D x+2 + +D x+n D x = N x+1 −N x+n+1 D x (4.3.7) Note that (a) ¨ a x:n = 1 +a x:n−1 (4.3.8) (b) ¨ a x:n and ¨ a [x]:n are tabulated for certain values of x+n in the A1967-70 section of “Formulae and Tables.” Example 4.3.1. By expressing A x:n and ¨ a x:n as expectations of appropriate random variables, or otherwise, prove the conversion relationships (i) A x:n = 1 −d¨ a x:n , and (ii) ¯ A x:n = 1 −δ¯ a x:n . 4.4. DEFERRED ANNUITIES 67 Proof. (i) By Maths. in Finance, ¨ a min{K+1,n} = 1 −v min{K+1,n} d . Take expected values to obtain ¨ a x:n = 1 −A x:n d which gives the required result. (ii) Take expected values in the equation ¯ a min{T,n} = 1 −v min{T,n} δ . 4.4 Deferred annuities These are annuities which commence in m (say) years’ time, provided that the annuitant is then active. Thus m [¯ a x = M.P.V. of annuity of 1 p.a. to (x), payable con- tinuously, beginning in m years’ time (4.4.1) It is often best to evaluate m [¯ a x by the formula m [¯ a x = D x+m D x . .. . pure endowment factor ¯ a x+m . .. . annuity factor at age x +m (4.4.2) Similarly, m [¨ a x = D x+m D x ¨ a x+m (4.4.3) and so on. Note that ¨ a x = ¨ a x:n + n [¨ a x (4.4.4) and m [¨ a x = D x+m D x N x+m D x+m = N x+m D x (4.4.5) N.B. Pensions are (essentially) deferred annuities. Select Tables The modifications needed for select tables are straightforward, and are illustrated in Table 4.4.1 below. 68 CHAPTER 4. ANNUITIES Type of Annuity Symbol In terms of t p [x] In terms of Commutation Functions Immediate a [x] ∞ ¸ t=1 v t t p [x] N [x]+1 D [x] Annuity-due ¨ a [x] ∞ ¸ t=0 v t t p [x] N [x] D [x] Temporary (n years) a [x]:n n ¸ t=1 v t t p [x] N [x]+1 −N x+n+1 D [x] (n ≥ 1) Temporary annuity-due (n years’ payments) ¨ a [x]:n n−1 ¸ t=0 v t t p [x] N [x] −N x+n D [x] (n ≥ 2) Deferred (m years) m [a [x] ∞ ¸ t=m+1 v t t p [x] N x+m+1 D [x] (m ≥ 1) Deferred annuity-due (m years) m [¨ a [x] ∞ ¸ t=m v t t p [x] N x+m D [x] (m ≥ 2) Table 4.4.1: Expression for Values of Single-Life Curtate Annuities. Life aged x. Select Mortality Table (select period 2 years) 4.5 Annuities payable m times per annum We require the following formula from numerical analysis: The Euler-Maclaurin formula: ∞ 0 f(t) dt ∞ ¸ t=0 f(t) − 1 2 f(0) + 1 12 f (0) (4.5.1) Woolhouse’s formula may be then deduced: 1 m ∞ ¸ t=0 f t m ∞ ¸ t=0 f(t) − m−1 2m f(0) + m 2 −1 12m 2 f (0) (4.5.2) (We have assumed that f(t) →0 and f (t) →0 as t →∞.) Example 4.5.1. Use the Euler-Maclaurin formula to deduce an approximate formula for ¯ a x in terms of ¨ a x 4.5. ANNUITIES PAYABLE M TIMES PER ANNUM 69 Solution. Let f(t) = v t t p x = exp − t 0 (δ +µ x+r ) dr . Hence f (t) = −(δ +µ x+t ) exp − t 0 (δ +µ x+r ) dr . ∴ f(0) = 1 and f (0) = −(δ +µ x ). By E.-M. ¯ a x ¨ a x − 1 2 − 1 12 (µ x +δ) or a x + 1 2 − 1 12 (µ x +δ) (4.5.3) In practice the final term is usually ignored, giving ¯ a x ¨ a x − 1 2 = a x + 1 2 (4.5.4) Example 4.5.2. Find an approximate formula for ◦ e x in terms of e x . Solution. Set i = 0 in example 4.5.1. This gives ◦ e x e x + 1 2 − 1 12 µ x (4.5.5) The final term is usually omitted, giving ◦ e x e x + 1 2 (4.5.6) The symbols ¨ a (m) x and a (m) x refer to the expected present values of an annuity of 1 per annum payable monthly in advance and monthly in arrear respectively. Thus ¨ a (m) x = ∞ ¸ t=0 1 m v t m t m p x (4.5.7) and a (m) x = ∞ ¸ t=0 1 m v t m t m p x (4.5.8) Example 4.5.3. Find an approximate formula for ¨ a (m) x in terms of ¨ a x . Solution. Apply Woolhouse’s formula to f(t) = v t t p x . This gives ¨ a (m) x ¨ a x − m−1 2m − m 2 −1 12m 2 (µ x +δ) (4.5.9) In practice one usually uses ¨ a (m) x ¨ a x − m−1 2m (4.5.10) Note. a (m) x = ¨ a (m) x − 1 m = ¨ a x − m−1 2m − 1 m (ignoring the final term) = a x + m−1 2m (4.5.11) 70 CHAPTER 4. ANNUITIES Temporary m thly annuities Define ¨ a (m) x:n = M.P.V. of £1 p.a. payable m thly in advance to (x) for at most n years = ¨ a (m) x − n [¨ a (m) x (4.5.12) = ¨ a (m) x − D x+n D x ¨ a (m) x+n ¸ ¨ a x − m−1 2m − D x+n D x ¸ ¨ a x+n − m−1 2m = ¨ a x:n − m−1 2m 1 − D x+n D x . .. . CARE! DO NOT OMIT THIS TERM (4.5.13) We also have a (m) x: n = a x:n + m−1 2m 1 − D x+n D x (4.5.14) 4.6 Complete annuities (or “annuities with final proportion”) Suppose that (x) is to receive 1 p.a. monthly in arrear with a final payment immediately on death at time r since the last full payment, i.e. on death at time t m +r, where t = 0, 1, 2, . . . and 0 ≤ r < 1 m This final payment varies from 0 to 1 m and hence is on average about 1 2m . This annuity is called “complete” or “with final proportion”, and we define ◦ a (m) x = M.P.V. of complete annuity a (m) x + 1 2m ¯ A x (4.6.1) When m = 1 we may write ◦ a (m) x = ◦ a x , giving ◦ a x a x + 1 2 ¯ A x (4.6.2) More accurate formulae for ◦ a x . If we have U.D. of D in each of the age-ranges x to x+1, x+1 to x + 2, etc., we have the exact formulae ◦ a x = a x + i −δ δ 2 A x = a x + i −δ iδ ¯ A x (4.6.3) Proof. The final payment is shown below: Hence (using the formula for a varying assurance) 4.7. VARYING ANNUITIES 71 final payment (on death of (x) ) 6 - . . . . . . 1 2 3 4 0 1 time (years) ◦ a x = a x + ∞ ¸ t=0 v t t p x . .. . pure endowment factor to age x +t 1 0 rv r r p x+t µ x+t+r . .. . value at time t of death benefit in year t + 1 dr By U.D. of D., ◦ a x = a x + ∞ ¸ t=0 v t t p x q x+t 1 0 rv r dr . .. . = ( ¯ I¯ a) 1 = ¯ a 1 −v δ = a x + ∞ ¸ t=1 v t+1 t [q x (1 +i) ¸ ¯ a 1 −v δ = a x +A x ¯ s 1 −1 δ = a x +A x (1 +i) −1 δ −1 /δ = a x +A x i −δ δ 2 The second part of (4.6.3) follows from ¯ A x = i δ A x (which holds by U.D.D.) 4.7 Varying annuities Suppose, firstly, that payments are made continuously so long as (x) survives at the rate b(t) p.a. at time t. The present value of these payments (as a random variable depending on the future lifetime T of (x)) is (by the Mathematics of Finance) g(T) = T 0 v t b(t) dt (4.7.1) The mean present value is thus E[g(T)] = ∞ 0 g(t) t p x µ x+t dt (4.7.2) 72 CHAPTER 4. ANNUITIES Using the following important formula (obtained from integration by parts): ∞ 0 u(t) t 0 v(r) dr dt = ∞ 0 v(t) ∞ t u(r) dr dt (∗) we obtain the alternative expression E[g(T)] = ∞ 0 v t . .. . interest factor b(t) . .. . rate of payment at time t t p x . .. . probability of survival dt (4.7.3) To prove this, let v(t) = v t b(t) and u(t) = t p x µ x+t so that ∞ t u(r) dr = t p x then use formula (∗) Example 4.7.1. Define ( ¯ I¯ a) x = M.P.V. of an increasing annuity to (x) in which the rate of payment at time t is t. Find an expression for ( ¯ I¯ a) x in terms of an integral. Solution. Let b(t) = t, t ≥ 0. This gives g(T) = ( ¯ I¯ a) T , and we have ( ¯ I¯ a) x = M.P.V. of annuity to (x) with payment at rate t p.a. at time t = ∞ 0 tv t t p x dt (4.7.4) An approximation. Let f(t) = tv t t p x , (t ≥ 0). By Euler-Maclaurin, ( ¯ I¯ a) x ∞ ¸ t=0 tv t t p x − 1 2 f(0) + 1 12 f (0) But f(0) = 0 and f (0) = [v t t p x +t d dt (v t t p x )] t=0 = 1, so ( ¯ I¯ a) x ∞ ¸ t=1 tv t t p x + 1 12 = (Ia) x + 1 12 (see definition of (Ia) x below) (4.7.5) Now suppose that b(t) = [t] + 1 = 1 in year 1 2 in year 2 . . . 4.7. VARYING ANNUITIES 73 The M.P.V. of the corresponding annuity is written (I¯ a) x . Thus (I¯ a) x = ∞ 0 ([t] + 1)v t t p x dt = ∞ 0 + ∞ 1 + ∞ 2 v t t p x dt = ¯ a x + 1 [¯ a x + 2 [¯ a x +. . . = ¯ N x + ¯ N x+1 +. . . D x = ¯ S x D x (4.7.6) where ¯ S x = ∞ ¸ t=0 ¯ N x+t (4.7.7) Now suppose that benefits are paid annually so long as (x) survives, the benefit at time t years being b(t) (t = 0, 1, 2, . . . ). The present value of the benefits (regarded as a random variable) is g(K) = K ¸ t=0 v t b(t) (4.7.8) since the last payment is made at time K. (If b(0) > 0, this is a variable annuity-due). The M.P.V. of the varying annuity is thus E[g(K)] = ∞ ¸ k=0 g(k) k [q x = 1 l x ¦g(0)d x +g(1)d x+1 +. . . ¦ = 1 l x ¦g(0)[d x +d x+1 +. . . ] + [g(1) −g(0)][d x+1 +d x+2 +. . . ] +[g(2) −g(1)][d x+2 +d x+3 +. . . ] +. . . ¦ = 1 l x ¸ b(0)l x +vb(1)l x+1 +v 2 b(2)l x+2 +. . . ¸ E[g(K)] = ∞ ¸ t=0 b(t) . .. . benefit at time t v t . .. . interest factor t p x . .. . survival factor (4.7.9) This formula forms the basis of “spreadsheet” calculations for pension schemes, etc. [This may also be found by summing the pure endowments] 74 CHAPTER 4. ANNUITIES Applications Let b(t) = t + 1 (t = 0, 1, 2, . . . ). The M.P.V. is defined as (I¨ a) x , so we have (I¨ a) x = ∞ ¸ t=0 (t + 1)v t t p x = ¸ ∞ ¸ t=0 (t + 1)D x+t ¸ /D x = D x + 2D x+1 +. . . D x = (D x +D x+1 +. . . ) + (D x+1 +D x+2 +. . . ) +. . . D x = N x +N x+1 +. . . D x = S x D x (4.7.10) where S x = ∞ ¸ t=0 N x+t (4.7.11) Similarly, when b(t) = t (t = 1, 2, . . . ) we get the M.P.V. (Ia) x = ∞ ¸ t=1 tv t t p x = ∞ ¸ t=1 t D x+t D x = S x+1 D x (4.7.12) Example 4.7.2. Show that (Ia) x = (I¨ a) x − ¨ a x Solution. (Ia) x = ∞ ¸ t=0 tv t t p x = ∞ ¸ t=0 (t + 1 −1)v t t p x = ∞ ¸ t=0 (t + 1)v t t p x − ∞ ¸ t=0 v t t p x =(I¨ a) x − ¨ a x Temporary increasing annuities 4.7. VARYING ANNUITIES 75 Define (I¨ a) x:n = M.P.V. of payments of t + 1 at time t (t = 0, 1, 2, . . . , n − 1), provided that (x) is then alive = n−1 ¸ t=0 (t + 1)v t t p x = ¸ n−1 ¸ t=0 (t + 1)D x+t ¸ /D x = D x + 2D x+1 + + (n −1)D x+n−2 +nD x+n−1 D x = (D x + 2D x+1 +. . . ) D x − (D x+n + 2D x+n+1 +. . . ) D x − (nD x+n +nD x+n+1 +. . . ) D x = S x −S x+n −nN x+n D x (4.7.13) Similarly, (Ia) x:n = D x+1 + 2D x+2 + +nD x+n D x = (D x+1 + 2D x+2 +. . . ) D x − (D x+n+1 + 2D x+n+2 +. . . ) D x − (nD x+n+1 +nD x+n+2 +. . . ) D x = S x+1 −S x+n+1 −nN x+n+1 D x Also, ¯ I¯ a x:n = n 0 tv t t p x dt (I¯ a) x:n = ¯ S x − ¯ S x+n −n ¯ N x+n D x (4.7.14) Note that (for example) m [ (I¨ a) x:n = D x+m D x (I¨ a) x+m:n which enables us to express m [ (I¨ a) x:n in terms of commutation functions, viz m [ (I¨ a) x:n = S x+m −S x+m+n −nN x+m+n D x Conversion relationships for increasing annuities and assurances ¯ a x = δ ¯ I¯ a x + ¯ I ¯ A x (1) ¨ a x = d (I¨ a) x + (IA) x (2) ¨ a x = δ (I¯ a) x + I ¯ A x (3) Proof To show (1), take expected values on each side of the maths of finance formula ¯ a T = δ ¯ I¯ a T +Tv T 76 CHAPTER 4. ANNUITIES To show (2), take expected values on each side of ¨ a K+1 = d (I¨ a) K+1 + (K + 1)v K+1 To show (3), we need to use the formula (I¯ a) t = ¨ a [t]+1 −([t] + 1)v t δ for all t ≥ 0 (∗∗) where (I¯ a) t = p.v. of an annuity-certain payable continuously at rate 1 p.a. in year 1, 2 p.a. in year 2, . . . , ceasing at time t exactly. = t 0 v r ([r] + 1) dr (∗∗) holds for t = 0, 1, 2, 3, . . . by McCutcheon and Scott, formula 3.6.6. It may be proved for general t ≥ 0 by letting t = n +r ( n integer, 0 ≤ r < 1) and observing that (I¯ a) t = (I¯ a) n + n+r n v s (n + 1) ds = ¨ a n −nv n δ + (n + 1)v n 1 −v r δ = ¨ a n+1 −(n + 1)v t δ Note. There are also conversion relationships for temporary increasing annuities, e.g. ¯ a x:n = δ( ¯ I¯ a) x:n + ( ¯ I ¯ A) x:n 4.8. EXERCISES 77 Exercises 4.1 The following is an extract from a life table with a select period of 1 year. Age x l [x] l x+1 age x + 1 55 90,636 90,032 56 56 89,739 89,132 57 88,151 58 87,094 59 85,874 60 84,586 61 Evaluate ¨ a 56:5 and ¨ a [56]:5 at 5 % per annum interest. 4.2 Given that a x = 20, a x:n = 18 and a x+n = 8, find the values of n E x and ¨ a x:n . 4.3 (i) Write down an expression for a x in terms of v, p x , and a x+1 . (ii) On a certain select mortality table the select period is one year. Express a [x] in terms of v, p [x] and a x+1 . Given that q [x] = .6q x and that at 4 1 4 % p.a. interest a 45 = 15.719 and a 46 = 15.509, find the value of a [45] (at the same rate of interest). 4.4 Using the A1967-70 table with 4% p.a. interest find the values of ¨ a [40]:30 , ¨ a [39]+1:30 , ¨ a 40:30 , (I¨ a) 40 , 5 [¨ a [40]:25 , 5 [(Ia) 40:25 4.5 (i) Find the present value of a deferred annuity of £1000 p.a. to a man aged 40. Payments commence on his 60th birthday, if he is then alive, and continue thereafter annually for life. Basis: A1967-70 ultimate mortality, 4 % p.a. interest. (ii) As above, but select mortality (at entry.) (iii) As above, but A1967-70 ultimate mortality to exact age 60 and a(55) males ultimate mortality above exact age 60. 4.6 According to a certain mortality table, which has a select period of 1 year and is such that q [x] = 0.6q x for each x, a 70 at 10% interest = 5.641, and, a 71 at 10% interest = 5.449. Find a [70] at 10% interest. 78 CHAPTER 4. ANNUITIES Solutions 4.1 ¨ a 56:5 = l 56 +vl 57 + +v 4 l 60 l 56 = 4.451 ¨ a [56]:5 = l [56] +vl 57 + +v 4 l 60 l [56] = 4.463 4.2 a x = a x:n + n E x a x+n =⇒ n E x = 1 4 ¨ a x:n = a x:n − n E x + 1 = 18.75 4.3 (i) a x = vp x (1 +a x+1 ) (ii) Note that with a one-year select period a [45] = vp [45] ¨ a 46 = (1.0425) −1 16.509p [45] . (a) We must determine p [45] from the given data. Now a 45 = vp 45 ¨ a 46 i.e. p 45 = (1 +i)a 45 ¨ a 46 = 1.0425 15.719 16.509 = .9926136 Hence q 45 = 0.0073864 Then q [45] = 0.6q 45 = 0.0044318 Hence p [45] = 1 −q [45] = 0.9955682 From (a) we then get a [45] = 15.7658 4.4 ¨ a [40]:30 = 16.960 directly tabulated ¨ a [39]+1:30 = (N [39]+1 −N 70 )/D [39]+1 = 16.953 ¨ a 40:30 = 16.949 directly tabulated (I¨ a) 40 = S 40 /D 40 = 276.468 5 [¨ a [40]:25 = (N 45 −N 70 )/D [40] = 12.342 5 [(Ia) 40:25 = (S 46 −S 71 −25N 71 )/D 40 = 122.337 4.9. SOLUTIONS 79 4.5 (i) 1000.N 60 /D 40 = 5130.08, say £5130 (ii) 1000.N 60 /D [40] = 5133.68, say £5134 (iii) 1000. D 60 D 40 .¨ a ∗ 60 , where the factor D 60 D 40 is evaluated on A1967-70 mortality and the factor ¨ a ∗ 60 is evaluated on a(55) males mortality. The value is thus 1000 .40873 12.625 = 5160.22, say £5160. 4.6 a 70 = v(1 − q 70 )¨ a 71 , so 1 − q 70 = 0.96218. Hence q [70] = 0.6q 70 = 0.022918, and q [70] = v(1 −q [70] )¨ a 71 = 5.730 80 CHAPTER 4. ANNUITIES Chapter 5 PREMIUMS 5.1 Principles of premium calculations Premiums calculated without an allowance for expenses are called net premiums, whilst premiums which are actually charged are called office or gross premiums. Office premiums are usually cal- culated with an explicit allowance for expenses, and in some cases for a profit to the office. A life assurance policy may be issued with (a) a single premium payable at the date of issue, or (b) reg- ular premiums payable in advance and usually (but not always) of level amount. The frequency of regular payments may be, for example, yearly or monthly, and the maximum number of premiums payable may be limited to (for example) 20. Premiums should, of course, cease on the death of the assured life, when the policy matures, or when there is no longer any possibility of future benefits: for example, on the expiry of the term of a temporary (term) assurance policy. Premiums are usually calculated by the equivalence principle, which may be stated as follows: E(Z) = 0 (5.1.1) where Z = present value of profit to the life office on the contract In some cases an explicit loading for profit is included in the calculation; that is, we replace equation (5.1.1) by E(Z) = B (5.1.2) where B = M.P.V. of profit on the contract. In the absence of information to the contrary, however, we shall assume that the equivalence principle (5.1.1) is to be used, although in some cases the use of “conservative” assumptions regarding mortality, interest and expenses means that the office has an implicit margin of expected profit. The rate of interest is usually taken to be fixed (not random). Example 5.1.1. Consider a whole life assurance of 1 payable immediately on the death of (x), and suppose that premiums are payable continuously at rate P p.a. until the death of (x). Give a formula for P in terms of annuity functions, assuming that the equivalence principle applies. Solution. Let Z = g(T) = P.V. of profit to office on this contract = P¯ a T −v T 81 82 CHAPTER 5. PREMIUMS Since we require that E¦g(T)¦ = 0, we have E¦P¯ a T −v T ¦ = 0 ∴ P.E (¯ a T ) = E v T ∴ P¯ a x = ¯ A x ∴ P = ¯ A x ¯ a x = 1 ¯ a x −δ using the conversion relationship ¯ A x = 1 −δ¯ a x The equations (5.1.1) and (5.1.2) may be expressed as equations of payments, or equations of value, i.e. M.P.V. of premiums =M.P.V. of benefits + M.P.V. of expenses (if any) + M.P.V. of profit to the office (if any) (5.1.3) In example 5.1.1, the equation of value is P¯ a x = ¯ A x 5.2 Notation for premiums The International Actuarial Notation (see Formulae and Tables for Actuarial Examinations) should be used, at least for straightforward policies. If the policy is complicated it is best to just use the symbol P (or P , P ) for the premium (single or regular). The general symbols P, P , P may be used for any sum assured, but the standard symbols P x , ¯ P( ¯ A x ) , etc. , refer to a sum assured of £1. Some of the more important rules of the International Notation are the following: 1. The symbols P( ), or ¯ P( ), or P (m) ( ) indicate the level net annual premium for the benefit indicated in the brackets. Premiums are assumed to continue for as long as the contract can provide benefits, i.e. n years for a n-year contract, provided of course that the life assured is still alive. P, ¯ P, P (m) indicate that payments are made annually in advance, continuously and m thly in advance respectively. 2. If premiums are limited to, at most t years’ payments (where t < the term of contract, i.e. the term of the benefits), we write t P( ), t ¯ P( ), t P (m) ( ). e.g. 10 P(A x:20 ) = net annual premium for 20-year EA with S.A. =1 (payable at end of year of death or on survival for 20 years.), limited to at most 10 payments (i.e. payments cease on death or after 10 payments are made, whichever occurs first) = A x:20 ¨ a x:10 3. When benefits are payable at end of year of death we may (optionally) shorten the notation as follows: (e.g.) P(A x ) = P x 5.3. THE VARIANCE OF THE PRESENT VALUE OF THE PROFIT ON A POLICY. 83 Table 5.2.1 relates to a mortality table with select period 2 years, for example A1967-70, and refers to policies in which premiums are payable annually in advance. 5.3 The variance of the present value of the profit on a policy. Consider again the policy of Example 5.1.1. We have g(T) = p.v. of profit to office on contract = P¯ a T −v T (We need not assume the equivalence principle holds: if it does, E[g(T)] = 0.) Note that g(T) = P 1 −v T δ −v T = P δ − P +δ δ v T ∴ Var[g(T)] = P +δ δ 2 Var(v T ) = P +δ δ 2 ¯ A ∗ x . .. . at rate 2i +i 2 − ¯ A x 2 (5.3.1) Note the idea of “collecting” all the terms involving v T before taking the variance: a similar technique may be used for endowment assurance policies (but not generally). 5.4 Premiums allowing for expenses Most expenses may be classified as either: (a) initial (incurred at the outset only) , or (b) renewal (incurred on the payment of later premiums). There may also be expenses of payment of benefits (especially for pensions and annuities) and the expenses of maintaining records of policies with continuing benefits after premiums have ceased. Expenses may also be divided into: (i) commission payments, and (ii) other costs. Some offices employ a policy fee system, whereby a fixed addition of (say) £15 is added to the annual premium; for example the office annual premium for a policy with a sum assured for £20,000 may be quoted as “£10 per £1,000 plus a policy fee of £15”, giving £20 10 + £15 = £215. This system reflects the fact that certain administrative costs do not depend on the size of the benefit. Example 5.4.1. Consider n-year endowment assurance without profits with sum assured £1 issued to a select life aged x. Expenses are e per premium payment (including the first) plus I at issue date (so the total initial expense is I +e ). Find formulae for the level office annual premium, P . 84 CHAPTER 5. PREMIUMS N e t A n n u a l P r e m i u m p a y a b l e t h r o u g h o u t d u r a t i o n o f c o n t r a c t N e t A n n u a l P r e m i u m L i m i t e d t o t p a y m e n t s T y p e o f A s s u r a n c e S y m b o l I n t e r m s o f A a n d ¨a I n t e r m s o f c o m m u t a t i o n F u n c t i o n s ( n ≥ 2 ) S y m b o l I n t e r m s o f A a n d ¨a I n t e r m s o f c o m m u t a t i o n F u n c t i o n s ( n ≥ 2 a n d t ≥ 2 ) T e m p o r a r y P 1 [ x ] : n A 1 [ x ] : n ¨a [ x ] : n M [ x ] − M x + n N [ x ] − N x + n t P 1 [ x ] : n A 1 [ x ] : n ¨a [ x ] : t M [ x ] − M x + n N [ x ] − N x + t D e f e r r e d T e m p o r a r y P m [ A 1 [ x ] : n m [ A 1 [ x ] : n ¨a [ x ] : m + n M x + m − M x + m + n N [ x ] − N x + m + n t P m [ A 1 [ x ] : n m [ A 1 [ x ] : n ¨a [ x ] : t M x + m − M x + m + n N [ x ] − N x + t W h o l e - l i f e P [ x ] A [ x ] ¨a [ x ] M [ x ] N [ x ] t P [ x ] A [ x ] ¨a [ x ] : t M [ x ] N [ x ] − N x + t D e f e r r e d W h o l e - l i f e P ( m [ A [ x ] ) m [ A [ x ] ¨a [ x ] M x + m N [ x ] t P ( m [ A [ x ] ) m [ A [ x ] ¨a [ x ] : t M x + m N [ x ] − N x + t E n d o w m e n t A s s u r a n c e P [ x ] : n A [ x ] : n ¨a [ x ] : n M [ x ] − M x + n + D x + n N [ x ] − N x + n t P [ x ] : n A [ x ] : n ¨a [ x ] : t M [ x ] − M x + n + D x + n N [ x ] − N x + t T a b l e 5 . 2 . 1 : E x p r e s s i o n s f o r A n n u a l P r e m i u m s . S e l e c t M o r t a l i t y T a b l e ( s e l e c t p e r i o d 2 y e a r s ) N o t e . W h e n n < 2 o r t < 2 , t h e s e f o r m u l a e s h o u l d b e s u i t a b l y m o d i fi e d . 5.5. PREMIUMS FOR WITH PROFITS POLICIES 85 Solution. The equation of value is P ¨ a [x]:n = A [x]:n +e¨ a [x]:n +I (5.4.1) where P = level annual premium. Suppose further that e = kP (i.e. renewal expenses are a proportion k of the office premium): we have (1 −k)P ¨ a [x]:n = A [x]:n +I ∴ P = A [x]:n +I (1 −k)¨ a [x]:n = 1 (1 −k) ¸ P [x]:n + I ¨ a [x]:n (5.4.2) Sometimes I is a proportion of the first premium, cP , say. This leads to (1 −k)P ¨ a [x]:n = A [x]:n +cP ∴ P = A [x]:n (1 −k)¨ a [x]:n −c (For example, k = 2 1 2 % and c = 47 1 2 % if expenses are 50% of the first premium and 2 1 2 % of all subsequent premiums .) 5.5 Premiums for with profits policies In early days of life assurance, premiums for policies with the right to participate in the profits of the life office were calculated on conservative assumptions but without an explicit allowance for bonus declarations. In modern conditions there is often an explicit allowance for possible future bonus rates for with profits (or “participating”) policies, as illustrated in the next example. Example 5.5.1. A with-profits whole-life assurance is about to be issued to a man aged 45. The basic sum assured is £12, 000. The office assumes that at the start of each policy year there will be bonus additions at the rate of 1% per annum compound. The basic sum assured and bonuses are payable immediately on the man’s death. The policy has half-yearly premiums, payable for at most 20 years. Calculate the half-yearly premium on the following basis: (Note: at 3.75% per annum interest, the value of A [45] is 0.34587). Solution. Let annual premium be P. Value of benefits 12, 000(1 +i) 1 2 [1.01vq [45] + (1.01) 2 v 2 1 [q [45] + ] = 12, 000(1 +i) 1 2 A ∗ [45] (see formula (3.10.11)) where ∗ is at rate i −0.01 1 + 0.01 3.96% By linear interp., A ∗ [45] 0.32804. ∴ Value of benefits = £4033.69 86 CHAPTER 5. PREMIUMS Equation of value is 0.96P ¨ a (2) [45]:20 . .. . 12.362 = 0.16P ¨ a (2) [45]:1 . .. . 0.98753 + 200 + 4033.69 ∴ P = £361.58 Hence the half-yearly premium is £180.79 5.6 Return of premium problems Consider, as an example, a policy issued to (x) with level annual premiums, P, providing: (i) £1, 000 on survival for n years, and (ii) a return of all premiums paid, at the end of the year of death, if (x) dies within n years. Assume firstly that the premiums are returned without interest (sometimes described as “Return No Interest”, abbreviated to R.N.I.). If there are no expenses, the equation of value for P is: P¨ a x:n = 1000 D x+n D x . .. . survival benefit + P(IA) 1 x:n . .. . return of premiums on death (5.6.1) This may be solved for P. Note. If the premiums are returned immediately on death, replace (IA) 1 x:n by (I ¯ A) 1 x:n . Now suppose that the premiums are returned with compound interest at rate j p.a. In practice, such policies are often described as “ Return With Interest” (R.W.I.). If the death benefit is paid at the end of the year of death, the term P(IA) 1 x:n in equation (5.6.1) must be replaced by n−1 ¸ k=0 v k+1 k [q x P¨ s k+1 j ¸ . .. . death benefit in year k + 1 = P n−1 ¸ k=0 ¸ (1 +j) k+1 −1 d j v k+1 k [q x = P [j/(1 +j)] . ¦A ∗1 x:n −A 1 x:n ¦ (5.6.2) where ∗ indicates the rate of interest i−j 1+j Notes. 1. If the death benefit is payable immediately on death, formula (5.6.2) should be multi- plied by (1 +i) 1 2 /(1 +j) 1 2 = (1 +i ∗ ) 1 2 . 2. If i (the rate of interest used to discount premiums and benefits in the equation of value) equals j, some of these problems may be simplified: if there are no expenses, mortality (within n years) may be ignored giving the compound interest equation P¨ a n = 1000v n 5.7. ANNUITIES WITH GUARANTEES 87 This result may best be proved by reference to reserves, which we shall discuss later. Example 5.6.1. A life office sells policies each providing a cash sum at age 65. Premiums of £1, 000 are payable annually in advance during the deferred period. On the death of the policyholder during the period of deferment, the premiums paid are returned immediately without interest. In respect of a life now aged 45, find the cash sum at age 65, given that the office uses the following basis: A1967-70 select 4% p.a. interest expenses are ignored Solution. Let C be available at age 65. The equation of value is 1000¨ a [45]:20 = C D 65 D [45] + 1000(I ¯ A) 1 [45]:20 ∴ C = 1000 (N [45] −N 65 ) −(1.02)(R [45] −R 65 −20M 65 ) D 65 = 1000[76, 722.7 −1.02 7, 407.24] 2144.1713 = £32, 258 5.7 Annuities with guarantees Annuities are sometimes sold with the provision that payments will certainly continue until their total equals the purchase price, B say (or possibly some proportion of this purchase price.) Let us consider an annuity of £1 p.a., payable continuously with this guarantee to a life aged x at the issue date. Ignoring expenses, the equation of value for B is B = ¯ a B + B [¯ a x (5.7.1) Theorem 5.7.1. Equation (5.7.1) has a unique solution Proof. Let f(B) = ¯ a B + B [¯ a x −B (B ≥ 0) = B 0 v t dt + ∞ B v t t p x dt −B Hence f (B) = v B −v B t p x −1 = v B B q x −1 < 0 for all B > 0 Now f(0) = ¯ a x > 0 and, as B → ∞, f(B) → −∞ (as ¯ a x → 1 δ and B [¯ a x → 0). It follows that f(B) = 0 has a unique solution 88 CHAPTER 5. PREMIUMS Note. The purchase price of an annuity with this guarantee may be considerably larger than for an ordinary annuity. If the annuity instalments are paid annually in arrear we must solve the equation B = a n + n [a x (5.7.2) subject to the condition n−1 < B ≤ n. (Here n is the guarantee period, which must be an integer.) This is solved by trial and error. Another type of guaranteed annuity is that in which the balance (if any) of the purchase price (or a proportion of it) over the total annuity payments received is paid immediately on the death of the annuitant. For example, if the guarantee consists of a payment of the balance of 85% of the purchase price over the total annuity instalments received, equation (5.7.1) would be replaced by B = ¯ a B + n 0 [0.85B −t]v t t p x µ x+t dt (5.7.3) where n = 0.85B (the time when the annuity payments equal 85% of the purchase price). Example 5.7.1. A man aged 65 buys a guaranteed annuity payable continuously for a purchase price of £32, 258. The annuity payments are guaranteed to continue until the total payments reach £20, 000. The office issuing the contract uses the following basis: A1967-70 ultimate; 4% p.a. interest; expenses are ignored. Let n be the guarantee period. Give an equation for n, and prove that this equation has a unique solution, which lies between 6 and 7 years. Solution. Let Z be the annual rate of payment of the annuity. We have the equation 32, 258 = Z(¯ a n + n [¯ a 65 ) (1) subject to nZ = 20, 000 (2) Replace Z by 20, 000/n in (1) to obtain ¯ a n + n [¯ a 65 = 32, 258 20, 000 n = 1.6129n i.e. solve f(n) = ¯ a n + n [¯ a 65 −1.6129n = 0 Now f(0) = ¯ a 65 > 0, and, as n →∞, f(n) →−∞. f (n) = d dn n 0 v t dt + ∞ n v t t p 65 dt −1.6129n = v n −v n n p 65 −1.6129 = v n n p 65 −1.6129 < 0, since, for n > 0, v n n p 65 < 1 ∴ f(n) decreases, so f(n) = 0 has a unique solution. 5.8. FAMILY INCOME BENEFITS 89 Try n = 7 f(n) = 6.12136 + 1, 289.7567 2, 144.171 7.772 −11.2903 < 0 Try n = 6 f(n) = 5.34626 + 1, 401.6093 2, 144.171 8.112 −9.6774 > 0 Hence 6 < n < 7 5.8 Family income benefits A family income benefit of term n years is a series of instalments payable from the date of death of the assured life, if he dies within n years, for the balance of the term. It may be considered as a decreasing term assurance in which the benefit on death is an annuity-certain for the balance of the term. Case 1.The benefits are payable continuously Suppose that F.I.B. payments are at rate £B per annum, and the life assured is aged x at the issue date. The present value of the benefits is B(¯ a n − ¯ a T ) if T < n 0 if T ≥ n (5.8.1) The P.V. of the benefits may thus be written in the form B ¯ a n − ¯ a min{T,n} (5.8.2) and hence their M.P.V. is B(¯ a n − ¯ a x:n ) (5.8.3) Case 2.The benefits made m thly in arrear, beginning at the end of the 1/m year of death (mea- sured from the issue date.) Consider the combination of a F.I.B. and an m thly temporary annuity of term n years, payable in arrear. It is clear that their total present value is Ba (m) n no matter when (x) dies. Hence the M.P.V. of the family income benefit is B a (m) n −a (m) x: n (5.8.4) Case 3. As in case 2, but with payments beginning immediately on death. As payments are received on average 1 2m year earlier than in case 2, the M.P.V. is approximately B(1 +i) 1 2m a (m) n −a (m) x: n (5.8.5) Premiums. These are found by an equation of value in the usual way. There is, however, a danger that, if premiums are payable for the full n-year term, the policy may have a negative reserve at the later durations (see later discussion of reserves.) This means that the policyholder may be able to lapse the contract leaving the office with a loss, and possibly obtain the same benefits more cheaply by effecting a new policy. In practice, the effect of expenses and other factors may make this possibility hardly profitable, especially if the F.I.B. is part of a more general assurance contract, as in Example 5.8.1 below. The possibility of lapse option may be avoided by making the premium - paying term shorter than the benefit term. 90 CHAPTER 5. PREMIUMS Example 5.8.1. Ten years ago a woman aged exactly 35 effected an assurance policy by level annual premiums payable for a maximum of 25 years. The policy provided the following benefits: (i) a whole life assurance benefit of £2, 000 payable at the end of the year of death, (ii) a family income benefit of term 25 years, with payments of £300 per month, beginning im- mediately on death, if this occurs within 25 years. The final payment is made in the month ending 25 years after the issue date. Calculate the annual premium on the basis given below: A1967-70 6% interest expenses are 3% of all premiums. Solution. Let P be the annual premium. 0.97P¨ a 35:25 = 2000A 25 + 3600(1 +i) 1 24 a (12) 25 −a (12) 35: 25 P = 2000 0.12187 + 3600(1.06) 1 24 (13.1312 −12.8527) 0.97 13.282 = £96.93 Mortgage protection policies are similar to those for family income benefits, although the death benefits are payable in one sum rather than in instalments over balance of the term. The similarity arises from the fact that the loan outstanding after the t th payment has been made is equal to the value of the future loan instalments (see M c Cutcheon and Scott, “An Introduction to the Mathematics of Finance”, Table 3.8.1). 5.9. EXERCISES 91 Exercises 5.1 A life aged 40 effects a 25-year without profits endowment assurance policy with a sum assured of £50,000 (payable at the end of the year of death or on survival to the end of the term). Level premiums are payable annually in advance throughout the term of the policy or until earlier death of the life assured. Calculate the level premium, P, using the following premium basis Mortality: A1967-70 Ultimate; Interest: 6% p.a. Expenses: none 5.2 An office issues a large number of 25-year without-profit endowment assurances on lives aged exactly 40. Level annual premiums are payable throughout the term, and the sum assured of each policy is £10,000, payable at the end of the year of death or on survival to end of the term. The office’s premium basis is: A1967-1970 ultimate; 4% p.a. interest; expenses are 5% of each annual premium including the first, with additional initial ex- penses of 1% of the sum assured. Calculate the annual premium for each policy. 5.3 A 5-year temporary assurance, issued to a woman aged 55, has a sum assured of £50,000 in the first year, reducing by £10,000 each year. The sum assured is payable at the end of the year of death. Level premiums, limited to at most 3 years’ payments, are payable annually in advance. Calculate the annual premium. Basis: A1967-1970 select mortality 4% p.a. interest expenses are 10% of all office premiums 5.4 A life office sells immediate annuities, using English Life Table No. 12 - Males, 4% p.a. interest with no expenses as the premium basis. Assuming that the mortality of annuitants does follow this table, that investments will earn 4% per annum, and that expenses are negligible, find the probability that the office will make a profit on the sale of an annuity payable continuously to a life aged 55. 5.5 (i) Let g(T) be the present value of the profit to the life office, at the issue date, in respect of an n-year without profits endowment assurance to (x) with sum assured (payable immediately on death if this occurs within n years) and premium P per annum, payable continuously for the term of the policy. Expenses are ignored in all calculations. (a) Write down an expression for g(T). (b) Derive expressions for (1) the mean, and (2) the variance of g(T). (c) For what value of P is the mean of g(T) equal to zero? (ii) An office issues a block of 400 without profits endowment assurances, each for a term of 25 years, to lives aged exactly 35. The sum assured under each policy is £10, 000 and the premium is £260 per annum, payable continuously during the term. The sum assured is payable immediately on death, if death occurs within the term of the policy. 92 CHAPTER 5. PREMIUMS Assuming that the office will earn 4% interest per annum, that the future lifetime of the lives may be described statistically in terms of the A1967-70 ultimate table, and that expenses may be ignored, find (a) the mean present value of the profit to the office on the block of policies, and (b) the standard deviation of the present value of this profit. ¯ A 35:25 = 0.15646 on A1967-70 ultimate 8.16% 5.6 A life office issued a certain policy to a life aged 40. The benefits under this contract are as follows: On death before age 60: an immediate lump sum of £1, 000 On survival to age 60: an annuity of £500 p.a., payable continuously for the remaining lifetime of the policyholder. Level annual premiums are payable continuously until age 60 or earlier death. Premiums are calculated according to the following basis: Mortality: English Life Table No.12-Males Interest: 4% p.a. Expenses: Nil Calculate the annual premium. 5.10. SOLUTIONS 93 Solutions 5.1 Equation of value is: M.P.V. of premiums = M.P.V. of benefits P¨ a 40:25 = 50000A 40:25 at A67-70 ultimate, 6% p.a.interest =⇒ P 13.081 = 50000 0.25955 (see page 67 of tables) =⇒ P = £992.09 5.2 Let the annual premium be P . We have 100 + 10000A 40:25 = 0.95P ¨ a 40:25 ∴ P = 10000A 40:25 + 100 0.95¨ a 40:25 = £276.70 5.3 Let the annual premium be P. M.P.V. of assurance benefits is (from first principles) 10000 D [55] ¸ 5C [55] + 4C [55]+1 + 3C 57 + 2C 58 +C 59 ¸ [= 60000A 1 [55]:5 −10000(IA) 1 [55]:5 ] M.P.V. of premiums less expenses is 0.9P¨ a [55]:3 = 0.9P [N [55] −N 58 ] D [55] ∴ P = 10000¦5C [55] + 4C [55]+1 + 3C 57 + 2C 58 +C 59 ¦ 0.9[N [55] −N 58 ] = 10000 371.511 0.9 10449 = £395.06 5.4 Consider £1 p.a. of annuity. The purchase price is ¯ a 55 and the office will make a profit if death occurs before time t, where ¯ a 55 = ¯ a t at 4% That is, 1 −v t δ = ¯ a 55 , so t = log [1 −δ¯ a 55 ] log v i.e. t = log ¯ A 55 log v = log ¯ A 55 −δ = 16.99 The probability of making a profit is thus t q 55 = 1 − l 55+t l 55 = 1 − l 71.99 l 55 = 0.434 (or 43.4%) (by interpolation) 5.5 (i) (a) g(T) = P¯ a T −Sv T if T < n P¯ a n −Sv n if T ≥ n 94 CHAPTER 5. PREMIUMS (b) (1) E[g(T)] = P¯ a x:n −S ¯ A x:n (2) Write g(T) = P ¸ 1 −h(T) δ −Sh(T) where h(T) = v T if T < n v n if T ≥ n = P δ − P δ +S h(T) ∴ Var[g(T)] = P δ +S 2 Var[h(T)] = P δ +S 2 ¯ A ∗ x:n − ¯ A x:n 2 where ∗ indicates the rate of interest 2i +i 2 p.a. (c) By (b)(1), E[g(T)] = 0 when P = S ¯ A x:n ¯ a x:n (ii) (a) First consider 1 policy. E[g(T)] = 260¯ a 35:25 −10, 000 ¯ A 35:25 = 260 15.542 . .. . using ¯ a 35:25 ¨ a 35 − 1 2 − D 60 D 35 (¨ a 60 − 1 2 ) − 10, 000[ 1 −δ¯ a 35:25 . .. . 0.39043 ] = 136.60 ∴ For 400 policies, M.P.V. of profit = £54, 640 (b) For 1 policy, variance of p.v. of profit is 260 δ + 10, 000 2 ¯ A 8.16% 35:25 . .. . 0.15646 −( ¯ A 35:25 . .. . 0.39043 ) 2 = (1.662915 10 4 ) 2 0.004024415 = 1, 112, 866 ∴ s.d. for 400 policies = 20 1.662915 10 4 √ 0.004024415 = £21, 098 5.6 Let P = annual premium. P¯ a 40:20 = 1, 000 ¯ A 1 40:20 + 500 ¯ N 60 D 40 ∴ P = 95.84 + 2, 040.81 13.261 = £161.12 Chapter 6 RESERVES 6.1 What are reserves? Reserves, or policy values, are sums of money held by financial institutions such as life offices and pension funds to cover the difference between the present value of future liabilities (including expenses) and the present value of future premiums or contributions. Alternatively, the reserve of a contract may be considered as an accumulation of past premiums less expenses and the cost of death claims (and other benefits). Reserves are required for various purposes, e.g. (1) to pay surrender values (or transfer values in a pension fund); (2) to work out the revised premium or sum assured if a policy is altered or converted to another type; (3) for inclusion in statutory returns to the Department of Trade and Industry (or other supervisory bodies) for the purpose of demonstrating the office’s solvency; (4) for internal office calculations to decide the bonus rates of with-profits policies, the distribution of profits to shareholders, etc. The bases used to calculate “reserves” for each of these purposes may be different, and some practical considerations are beyond our present scope. 6.2 Prospective reserves Consider a life assurance policy issued t years ago to a life then aged x. Define the NET LIABILITY or PROSPECTIVE LOSS of the office in respect of this policy to be the random variable L = present value of future benefits + present value of future expenses − present value of future premiums (6.2.1) The reserve or policy value of the contract (calculated prospectively, i.e. by reference to future cashflows) is defined as t V = E(L) (6.2.2) = M.P.V. of future benefits and expenses −M.P.V. of future premiums (6.2.3) 95 96 CHAPTER 6. RESERVES If expenses may be ignored, we have t V = M.P.V of future benefits −M.P.V of future premiums (6.2.4) The mortality, interest and expense assumptions used to evaluate t V are known as the reserving basis. This may or may not agree with the premium basis. If these bases agree (or are assumed to agree) and there are no expenses, we obtain net premium reserves, which we shall consider in the next section. By convention, the reserve t V is calculated just before receipt of any premium then due: the reserve just after payment of this premium is t V +P −e (6.2.5) where P is the premium paid at duration t years and e is the expense then incurred. Are negative reserves allowable? Certain formulae may give negative values of t V, at least for some policies and at early durations t. A negative reserve should not normally be held because the life office is thereby treating the contract as an asset: if the policy is discontinued there is no way in which the policyholder can be made to pay money to the office! Similarly, the reserves stated in Statutory Returns should not be negative, although negative reserves may be permissible in certain internal office calculations. The general rule is therefore that, if a formula gives a negative value of t V, this should be replaced by zero. Policies should in general be designed so that negative reserves do not arise (cf. the discussion of family income benefit policies in Section 5.8). 6.3 Net premium reserves These are reserves calculated without allowance for expenses, and on the assumption that the pre- mium and the reserve bases agree. (In some cases, the actual premium basis is different, and the premiums valued are calculated on the reserve basis). A net premium reserve basis is unambiguously specified by (i) a mortality table, and (ii) a rate of interest. By formula (6.2.4), t V = net premium reserve = M.P.V. of future benefits - M.P.V. of future premiums (6.3.1) where the valuation assurance and annuity factors are calculated on the specified mortality and interest basis. The premiums valued are also calculated on this basis. Example 6.3.1. Consider a whole life policy with sum assured £1 without profits, payable imme- diately on the death of (x). The policy was issued t years ago by level annual premiums payable continuously throughout life. Find a formula for the net premium reserve t V (on a given mortality and interest basis). 6.3. NET PREMIUM RESERVES 97 Solution. Let L be the net liability (a random variable). We have L = v U −P¯ a U where U = future lifetime of (x +t), and P = ¯ P( ¯ A x ). ∴ t V = E(L) = ¯ A x+t − ¯ P( ¯ A x )¯ a x+t . Notation. The International Notation for net premium reserves is similar to that for premiums, with the addition of “t” to indicate the duration. If premiums are limited to (say) h years of payment, the symbol h is placed above the duration t, as shown in Table 6.3.1. The general symbol t V may be used for any sum assured and any reserve, whether net premium or not, but t V x , t ¯ V ( ¯ A x ), etc., refer to net premium reserves for a policy with a sum assured of £1. If select mortality tables are used, we write t V [x] , etc. CONVENTIONS (1) The symbols t V x , etc., refer to the reserves just before payment of the premium due at time t (if a premium is then payable.) The net premium reserve just after receipt of this premium is of course t V x + P x , etc. Similar considerations apply if premiums are payable half-yearly, monthly, etc. (2) By formula (6.3.1) with t = 0, we have 0 V = 0 for all net premium reserves. In the case of n-year endowment assurances or pure endowments, it is not always clear whether to take n V as zero or the sum assured (S, say), i.e. whether to assume that the sum assured has or has not already been paid. It appears that the convention n V = 0 is used in profit testing but not elsewhere. Type of Policy Notation Prospective Formula for Reserve Whole Life Assurance t ¯ V( ¯ A x ) ¯ A x+t − ¯ P( ¯ A x )¯ a x+t n-year Term Assurance t ¯ V( ¯ A 1 x:n ) ¯ A 1 x+t:n−t − ¯ P( ¯ A 1 x:n )¯ a x+t:n−t n-year Endowment Assurance t ¯ V( ¯ Ax:n ) ¯ A x+t:n−t − ¯ P( ¯ Ax:n )¯ a x+t:n−t h-Payment Years Whole Life Assurance h t ¯ V ( ¯ A x ) ¯ A x+t − h ¯ P( ¯ A x )¯ a x+t:h−t t < h ¯ A x+t t ≥ h h-Payment Years n-year Endowment Assurance h t ¯ V ( ¯ A x:n ) ¯ A x+t:n−t − h ¯ P( ¯ A x:n )¯ a x+t:h−t t < h ¯ A x+t:n−t t ≥ h n-year Pure Endowment t ¯ V(A 1 x:n ) A 1 x+t:n−t − ¯ P(A 1 x:n )¯ a x+t:n−t Table 6.3.1: Net Premium Reserves (premiums payable continuously) 98 CHAPTER 6. RESERVES Type of Policy Symbol Prospective Policy Value Retrospective Policy Value Whole Life Assurance t V x A x+t −P x ¨ a x+t D x D x+t P x ¨ a x:t −A 1 x:t Endowment Assurance t V x:n A x+t:n−t −P x:n ¨ a x+t:n−t D x D x+t P x:n ¨ a x:t −A 1 x:t Temporary Assurance t V 1 x:n A 1 x+t:n−t −P 1 x:n ¨ a x+t:n−t D x D x+t P 1 x:n ¨ a x:t −A 1 x:t Pure Endowment t V 1 x:n A 1 x+t:n−t −P 1 x:n ¨ a x+t:n−t D x D x+t [P 1 x:n ¨ a x:t ] Table 6.3.2: Net Premium Reserves (premiums payable annually) Example 6.3.2. (i) What is the International Notation for the reserve of Example 6.3.1 ? (ii) Express this reserve in terms of annuity functions. Solution. (i) t ¯ V ( ¯ A x ) (ii) t ¯ V ( ¯ A x ) = ¯ A x+t − ¯ P( ¯ A x )¯ a x+t = 1 −δ¯ a x+t −( 1 ¯ a x −δ)¯ a x+t using conversion relationships = 1 − ¯ a x+t ¯ a x (6.3.2) Reserves at non-integer durations Let us consider (for example) a whole life non-profit policy with sum assured £1 payable at the end of the year of death, effected by (x) by level annual premiums, P x . The net premium reserve at integer duration t is t V x just before premium is paid t V x +P x just after premium is paid Now consider duration r + k, integer r, 0 < k < 1. The reserve r+k V x is estimated by linear interpolation between r V x +P x and r+1 V x , i.e. r+k V x (1 −k)( r V x +P x ) +k r+1 V x (0 < k < 1) (6.3.3) [The actual value is v 1−k [ 1−k q x+r+k + (1 − 1−k q x+r+k )A x+r+1 ] −P x v 1−k (1 − 1−k q x+r+k )¨ a x+r+1 6.4. RETROSPECTIVE RESERVES 99 which is complicated to evaluate.] Similar formulae apply for other classes of business with premi- ums payable annually in advance. Reserves when premiums are payable mthly Suppose that premiums are payable mthly in advance (e.g. m=12, which corresponds to monthly premiums, which are very common in practice.) To simplify matters let us suppose that the sum assured is payable immediately on death: this avoids the calculation of awkward assurance factors at non-integer ages. Consider a policy providing £1,000 immediately on the death of (x), effected by mthly premiums during the lifetime of (x). The annual premium, payable mthly in advance, is P (m) ( ¯ A x ) = P , say, and the reserve at duration t (where t is an integer multiple of 1 m ) is t V = 1, 000 ¯ A x+t −P¨ a (m) x+t just before payment of the premium then due, or t V + P m just after payment of this premium. The reserve at duration t = r + k m (where r is an integer multiple of 1 m and 0 < k < 1) is usually estimated by linear interpolation between r V + P m and r+ 1 m V . NOTE One may use standard symbols, e.g. t V (12) ( ¯ A x ) = ¯ A x+t −P (12) ( ¯ A x )¨ a (12) x+t t V (4) ( ¯ A 1 x:10 ) = ¯ A 1 x+t:10−t −P (4) ( ¯ A 1 x:10 )¨ a (4) x+t:10−t if t is an integer multiple of 1 12 and 1 4 respectively, but these are seldom employed. 6.4 Retrospective reserves So far we have calculated the reserve of a policy by referring to future cashflows, assuming that the contract will not be surrendered. In practice, the policyholder may wish to surrender the contract, and in that event will, at least in the early years of the policy, probably expect a surrender value related to the accumulation of his premiums less expenses and the cost of life assurance cover. Such a surrender value is related to the retrospective reserve of the contract, which is obtained by accumulating the premiums less expenses and the cost of benefits, of a hypothetical large group of identical policies whose mortality follows the office’s tables exactly (i.e. without random fluctuations), and then dividing the hypothetical funds among the survivors. Let us again assume that there are no expenses and that the premium and reserve bases agree. Consider, for example, a whole-life policy, with sum assured £1 payable at the end of the year of death, issued t years ago by level annual premiums to a life then aged x. The (prospective) net premium reserve (on a specified mortality and interest basis, e.g. A1967-70 ultimate, 6% p.a. interest) just before payment of the premium now due is t V x = A x+t −P x ¨ a x+t The corresponding retrospective reserve is found by accumulating the funds of (say) l x identical policies until time t, and sharing the money out among the survivors. Since mortality is assumed to follow the table exactly, there are d x deaths in the first policy year, d x+1 in the second policy year, and so on, with l x+t survivors at time t. Interest is assumed to be earned at the rate i p.a. used in 100 CHAPTER 6. RESERVES the premium and the reserving bases, so the accumulated funds at time t are [l x P x (1 +i) t +l x+1 P x (1 +i) t−1 + +l x+t−1 P x (1 +i)] −[d x (1 +i) t−1 +d x+1 (1 +i) t−2 + +d x+t−1 ] =l x (1 +i) t [P x ¨ a x:t −A 1 x:t ] On division by l x+t (the assumed number of survivors at policy duration t), we find that retrospective reserve at duration t years = D x D x+t [P x ¨ a x:t −A 1 x:t ] More generally, we have retrospective reserve at duration t years = D x D x+t M.P.V. (at issue date) of premiums, minus M.P.V. of benefits and expenses, in the first t years ¸ ¸ ¸ (6.4.1) If there are no expenses, we have retrospective reserve at duration t years = D x D x+t M.P.V. (at issue date) of premiums, less benefits in first t years ¸ ¸ (6.4.2) Some examples of formulae for retrospective reserves are given in Table 6.3.2 above. Theorem 6.4.1. If the premium and the reserving bases agree, the prospective and the retrospective reserves of a given policy are equal. Proof. We illustrate the argument by means of the whole life policy discussed earlier in this section. The retrospective and prospective reserves at policy duration t are V R = D x D x+t [P x ¨ a x:t −A 1 x:t ] and V P = A x+t −P x ¨ a x+t respectively. Hence D x+t D x (V R −V P ) = P x ¨ a x:t −A 1 x:t − D x+t D x A x+t + D x+t D x P x ¨ a x+t = P x ¨ a x −A x (using the facts that ¨ a x = ¨ a x:t + t [¨ a x and A x = A 1 x:t + t [A x .) But P x ¨ a x − A x = 0, by the equation of value for the level annual premium, so we have proved the desired result. This argument may easily be extended to cover other policies, including situations in which there are expenses. 6.4. RETROSPECTIVE RESERVES 101 It follows that, if the premium and reserving bases agree (or are assumed to agree in reserve calcu- lations), one need not specify whether a prospective or retrospective reserve is required. In practice, however, the prospective method is more often used. Example 6.4.1. Give formulae for the following net premium reserves in terms of other monetary functions: (i) t V x:n , (ii) t ¯ V ( ¯ A x:n ), (a) by the prospective method, and (b) the retrospective method. Solution. (i) (a) A x+t:n−t −P x:n ¨ a x+t:n−t (b) D x D x+t ¦P x:n ¨ a x:t −A 1 x:t ¦ (ii) (a) ¯ A x+t:n−t − ¯ P( ¯ A x:n )¯ a x+t:n−t (b) D x D x+t ¦ ¯ P( ¯ A x:n )¯ a x:t − ¯ A 1 x:t ¦ Example 6.4.2. Show that (i) t V x:n = 1 − ¨ a x+t:n−t ¨ a x:n (6.4.3) (ii) t ¯ V ( ¯ A x:n ) = 1 − ¯ a x+t:n−t ¯ a x:n (6.4.4) Solution. (i) t V x:n = A x+t:n−t −P x:n ¨ a x+t:n−t = 1 −d¨ a x+t:n−t −( 1 ¨ a x:n −d)¨ a x+t:n−t = 1 − ¨ a x+t:n−t ¨ a x:n 102 CHAPTER 6. RESERVES (ii) t ¯ V ( ¯ A x:n ) = ¯ A x+t:n−t − ¯ P( ¯ A x:n )¯ a x+t:n−t = 1 −δ¯ a x+t:n−t −( 1 ¯ a x:n −δ)¯ a x+t:n−t = 1 − ¯ a x+t:n−t ¯ a x:n 6.5 Gross premium valuations and asset shares. We now allow for the possibility of expenses in the premium and reserving bases, which are no longer assumed to be the same. The reserves obtained by the formula (6.2.3) are called prospective gross premium reserves, whereas those obtained by a retrospective accumulation of premiums less expenses and the cost of life assurance benefits are called retrospective gross premium reserves. (In practice, however, the phrase “gross premium reserve” usually refers to prospective reserves only.) If the premium and reserve bases do not agree, the two methods will normally give different results. An important feature of gross premium reserves are the facts that the prospective reserve at inception is not normally equal to zero, and the retrospective gross premium reserve of an n-year policy just after it matures is not generally equal to zero. We begin by considering prospective gross premium reserves. The premium basis (or the premi- ums themselves) must be specified, together with the reserving basis, which gives (i) a mortality table, (ii) an interest rate, and (iii) the allowance for future expenses. If the allowance for future expenses is a proportion k of future premiums (including any due now), the formula for the prospective reserve is t V = M.P.V. of future benefits −M.P.V. of future premiums multiplied by the factor (1 −k) (6.5.1) Example 6.5.1. Consider the whole life policy discussed in section 6.4 above, and suppose that (i) the premium basis is A1967-70 ultimate, 6% p.a. interest, with no allowance for expenses; and (ii) reserves are to be calculated prospectively by the gross premium method on the following basis: mortality: A1967-70 ultimate; interest: 4% p.a.; expenses: 5% of all future premiums. Find a formula for the reserve at duration t years (just before receipt of the premium then due.) 6.5. GROSS PREMIUM VALUATIONS AND ASSET SHARES. 103 Solution The annual premium is P = P x on A1967-70 ultimate, 6% p.a. so the formula for the reserve is t V = A x+t −0.95P ¨ a x+t where the factors A x+t and ¨ a x+t are on A1967-70 ultimate, 4% interest. If a retrospective gross premium reserve is required, one must specify the allowance for past expenses. If, in the policy of example 6.5.1, one allowed for expenses of 10% of all past premiums, the formula for the retrospective gross premium reserve would be t V = D x D x+t 0.9P ¨ a x:t −A 1 x:t where the annuity and assurance factors, and D x /D x+t , are on A1967-70 ultimate at 4% p.a. inter- est (or whatever mortality and interest basis is to be used.) Asset shares The asset share of a life assurance policy is a retrospective gross premium reserve calculated on the basis of the mortality, interest and expenses actually experienced by the office. In order to illustrate the calculations, let us consider an n-year policy issued t years ago to (x), and define S t = sum assured in year t (payable at end of year of death); P t = premium payable at the start of policy year t; e t = actual expenses incurred at the start of policy year t; i t = rate of interest earned by the office in policy year t; q [x]+t = observed mortality rate among lives aged x + t who entered assurance at age x. Let t V denote the asset share at duration t of the policy, before payment of the premium then due. We clearly have 0 V = 0, and the following relation holds: t+1 V = ( t V +P t+1 −e t+1 )(1 +i t+1 ) −death cost, D t+1 0 ≤ t ≤ n −1 (6.5.2) To calculate the “death cost”, we must consider the given policy’s share of the cost of making the asset share (at the end of the policy year) up to the sum assured, S t+1 , for those policies which became claims in the year. Since the proportion of policies in force at the start of the year which became claims was q [x]+t , the death cost is D t+1 =q [x]+t (S t+1 − t+1 V) (6.5.3) (which is negative if S t+1 < t+1 V). Notes 1. As will be stated in the next chapter, the quantity S t+1 − t+1 V is called the death strain at risk of the policy in year t + 1. 2. By substituting formula (6.5.3) in formula (6.5.2), we may obtain an expression for t+1 V in terms of t V and other known quantities. In practice, however, one may perhaps replace D t+1 by a suitable approximate value and use formula (6.5.2) directly. 104 CHAPTER 6. RESERVES 3. The asset share at policy duration n years may of course differ from the maturity value, leading to a profit or loss to the office at that time. 4. In the unlikely event that mortality , interest and expenses follow a certain valuation basis exactly, the asset share of a policy will equal the retrospective gross premium reserve on that basis. 5. Asset shares are particularly important for with profit policies, as their surrender and maturity values are not fixed in advance and the asset share gives one method of deciding “fair” surrender and maturity values. A number of other considerations are, however, also involved, and we shall not pursue the discussion of asset shares. Net premium versus gross premium reserves A full description of the advantages and disadvantages of the various approaches is outside the scope of this course, but we make some remarks. It would at first seem obvious that a prospective gross premium valuation should be used, as this represents the mean of the present value of the net liability. But this assumes that the policy will not be surrendered: the policyholder may do this at any time, and (at least in the early years of the contract) will expect a surrender value related to the accumulation of his past premiums less expenses, i.e. a retrospective gross premium reserve. This suggests that the office should hold reserves equal to the greater of (i) a prospective premium reserve, and (ii) a retrospective gross premium reserve (or asset share), the bases being perhaps different (reflecting expected future conditions and actual past conditions respectively.) The position is further complicated by the possible existence of guaranteed surrender values and the difficulty of estimating future conditions. In practice, a net premium valuation at a low rate of interest may be used, either by choice or because this is required by the supervisory authorities: such a basis (though apparently artificial, since the premium value is unrelated to the actual premium being charged) may have the property that the corresponding reserves are comparable with, or generally greater than, the more “scientific” reserves given above. Similar arguments lead to the “net premium method” for with-profit policies (provided that terminal bonuses are allowed for separately): cf. Section 6.9 below. The reason for the use of a low rate of interest in net premium reserves is that, by Lidstone’s theorem, net premium reserves normally increase as the rate of interest falls. The supervisory authorities may be prepared to allow Zillmerisation (see Section 6.7 below) of the reserves (subject to limits on I), but many offices prefer to publish unmodified net premium reserves, in case Zillmerisation is seen as a sign of financial weakness. The above discussion refers mainly to the calculation of reserves for solvency testing, but reserves are also used for internal purposes: deciding upon bonus levels, alterations to policies, etc. For these purposes the office may use other bases. The office must also allow for random fluctuations from the mean numbers of deaths (especially when there is a small number of policies with very large sums assured or death strains at risk) and consider the exchange or reinsurance of large risks. 6.6 The variance of L We return to the definition of the random variable L given by formula (6.2.1). Its mean is defined as the reserve t V, but in certain problems we also require its variance, Var(L). How may this best be calculated? 6.6. THE VARIANCE OF L 105 Let us consider a policy providing £S immediately on the death of a life aged x at issue, with premiums of P p.a. payable continuously throughout life. The random variable L is defined as L = prospective loss on contract (at duration t) = Sv U −P¯ a U (6.6.1) where U is the future lifetime of the policyholder, who is now aged x +t, and expenses are ignored. Now we cannot use the formula Var(L) = Var(Sv U ) + Var(P¯ a U ) wrong! (because Sv U and P¯ a U are not independent), so we express L in terms of v U . That is, L = Sv U −P 1 −v U δ = S + P δ v U − P δ from which we obtain Var(L) = S + P δ 2 Var(v U ) = S + P δ 2 ¯ A ∗ x+t − ¯ A x+t 2 (6.6.2) where ∗ indicates a rate of interest of 2i +i 2 p.a. This technique may also be used for certain other types of contracts. We now consider a number of life assurance policies, labelled from j = 1 to j = N. The total reserves for the group of policies is N ¸ j=1 E[L j ] = V 1 +V 2 + +V N (6.6.3) where L j and V j are, respectively, the net liability and the reserve of the jth policy. If we further assume that the lives assured are independent, Var ¸ N ¸ j=1 L j ¸ = N ¸ j=1 Var(L j ) (6.6.4) where Var(L j ) may be calculated by (say) formula (6.6.2). If we now suppose that all the policies are identical (i.e. they are of the same sum assured, type and duration, and were issued at the same date to N different lives all of the same age), the average of the net liabilities is ¯ L = ¸ N ¸ j=1 L j ¸ /N (6.6.5) which has mean E(L) and variance Var(L)/N, where L is the net liability of a given policy. These results are sometimes used in solvency calculations, etc. It is sometimes also assumed that N is so large that the distribution of N ¸ j=1 L j or ¯ L is approximately normal (by the central limit theorem and related results), although this may only be accurate for very large values of N. 106 CHAPTER 6. RESERVES 6.7 Zillmerised reserves Let us now suppose that the premium basis is the same as the reserving basis (or is assumed to be the same for the purpose of calculating reserves.) As stated earlier, the retrospective and prospective methods give the same results, but reserves may not be “net premium reserves” because of the effect of expenses. In certain important classes of policy, however, the reserve is obtainable by means of a simple adjustment to the net premium reserve: the corresponding reserve is called a “Zillmerised reserve”, “Zillmerised net premium reserve” or sometimes “modified net premium reserve”. Let us consider an n-year endowment assurance for a sum assured of £1 without profits, payable at the end of the year of death (if this occurs within the term of the policy.) There are level premiums of P , payable annually in advance for n years or until earlier death of the policyholder, who was aged x at the issue date. The reserve is required at duration t, just before payment of the premium then due. The premium and reserving basis includes the following allowances for expenses: expenses of e on the payment of each premium, with additional initial expenses of I (so the total initial expense is I +e). Theorem 6.7.1. The reserve for the above policy is t V Z = (1 +I) t V −I (6.7.1) where t V denotes the corresponding net premium reserve, i.e. t V = t V x:n =A x+t:n−t −P x:n ¨ a x+t:n−t =1 − ¨ a x+t:n−t ¨ a x:n Proof. We first note that the annual premium P is such that P ¨ a x:n = A x:n +e¨ a x:n +I ∴ P −e = I +A x:n ¨ a x:n Now consider the reserve at duration t by the prospective method. t V Z = M.P.V. of future benefits and expenses less premiums = A x+t:n−t −(P −e)¨ a x+t:n−t = A x+t:n−t − ¸ I +A x:n ¨ a x:n ˙ a x+t:n−t = A x+t:n−t −P x:n ¨ a x+t:n−t − I ¨ a x:n ˙ a x+t:n−t (where P x:n is the net annual premium) = t V x:n −I ¸ ¨ a x+t:n−t ¨ a x:n Now we use the fact that (by formula 6.4.3) t V x:n = 1 − ¨ a x+t:n−t ¨ a x:n 6.7. ZILLMERISED RESERVES 107 to give the formula t V Z = t V x:n −I [1 − t V x:n ] = (1 +I) t V x:n −I as required. Notes 1. The case of a whole life policy is similarly dealt with (put n = ∞ in the above formulae.) 2. A similar argument may be used to establish formula (6.7.1) if the sum assured is payable immediately on death and premiums are payable continuously. In this case, t V = t ¯ V( ¯ A x:n ). 3. Zillmer’s formula does not in general hold for other classes of policy, nor when the premium due at time t has been paid. (It is, however, sometimes used in practice for all policies and all durations.) 4. When the duration t is short (e.g. t = 1) Zillmer’s formula may give a negative reserve, which should, as a rule, be replaced by zero, i.e. a policy should not be treated as an asset to the office. Note that formula (6.7.1) gives 0 V Z = −I which is correct if the additional initial expense I is thought of as being disbursed before the first premium is received: if it is not, one should write 0 V Z = 0 (as is always assumed in profit-testing: see later.) 5. If the sum assured is not £1, we naturally multiply formula (6.7.1) by the sum assured: for example, the reserve per £1000 sum assured of a whole life without profits policy might be quoted as 1020 t V x −20 (corresponding to I= 2% ). Example 6.7.1. Ten years ago life office issued a 20-year endowment assurance without profits to (35). The sum assured is £10,000, payable at the end of the year of death (or on survival for 20 years), and premiums are payable annually in advance. The basis for premiums and reserves is: A1967-70 ultimate; 6% p.a. interest; expenses are 3% of all office premiums (including the first) with additional initial expenses of 1.5% of the sum assured. Calculate (i) the annual premium, and (ii) the reserve, (a) just before receipt of the premium now due, and 108 CHAPTER 6. RESERVES (b) just after receipt of the premium now due. Solution. (i) The office premium is P = 10000 0.97 ¸ A 35:20 + 0.015 ¨ a 35:20 = £288.67 Note that tables at 6% interest are limited, so we use these formulae: ¨ a 35:20 = ¨ a 35 −v 20 l 55 l 35 ¨ a 55 = 11.9969 and A 35:20 = 1 −d¨ a 35:20 (ii) One may use Zillmer’s formula in this case (or other methods). The reserve is (a) 10000[1.015 10 V 35:20 −0.015] = £3497 (b) £3497 + 0.97 £288.76 = £3777 (see formula (6.2.5)) Example 6.7.2. Suppose that the policyholder of example 6.7.1 were to surrender his policy (before paying the premium now due) and the office grants a surrender value equal to the reserve, as defined in that question. What yield per annum would the policyholder have obtained? Solution. The equation of value is 288.67¨ s 10 = 3497 at rate i=yield p.a. ∴ ¨ s 10 = 12.11 We use the compound interest tables, i.e. i ¨ s 10 0.03 11.81 0.04 12.49 Interpolate: i −0.03 0.04 −0.03 12.11 −11.81 12.49 −11.81 = 0.44 ∴ i 3.44% 6.8. FULL PRELIMINARY TERM RESERVES. 109 6.8 Full preliminary term reserves. We again assume that the premium and reserving bases agree, and suppose that the reserve for a whole life or endowment assurance policy is such that 1 V = 0 (6.8.1) That is, the reserve just before payment of the second annual premium is zero. Let us again suppose that the sum assured is £1, payable at the end of the year of death, and that expenses are as in the previous section. We must have 1 V = A x+1:n−1 −(P −e)¨ a x+1:n−1 = 0. Hence P −e = A x+1:n−1 ¨ a x+1:n−1 = P x+1:n−1 Hence t V = A x+t:n−t −P x+1:n−1 ¨ a x+t:n−t = A (x+1)+(t−1):(n−1)−(t−1) −P x+1:n−1 ¨ a (x+1)+(t−1):(n−1)−(t−1) = t−1 V x+1:n−1 = the net premium reserve for the corresponding policy on a life aged x + 1 at entry with term n −1 years at duration t −1 This reserve is called the Full Preliminary Term reserve, which we shall write as t V FPT . We have shown that t V FPT = t−1 V x+1:n−1 (6.8.2) In the case of a whole life policy, we may set n = ∞ to obtain t V FPT = t−1 V x+1 (6.8.3) If the sum assured is not £1, we naturally multiply these expressions by the sum assured. Similar formulae hold if the sum assured is payable immediately on death, when premiums are payable continuously or monthly, etc., and for certain other types of policy. What level of initial expenses leads to a F.P.T. reserve? We observe that, by the retrospective method, 1 V FPT = D x D x+1 (P −e)¨ a x:1 −I −A 1 x:1 = 0 from which we obtain P = e +I +A 1 x:1 (6.8.4) This states that the first premium is exactly sufficient to pay the initial expenses, I + e, and the cost of the first year’s life assurance cover, A 1 x:1 . The use of full preliminary term reserves was suggested by T. B. Sprague in 1870. Other actu- aries (particularly in North America) have devised formulae for “modified full preliminary term reserves”, in which the allowance for total initial expenses may be smaller than that of formula (6.8.4), but we do not pursue this topic. 110 CHAPTER 6. RESERVES 6.9 Reserves for with-profits policies We shall deal with “traditional” with profits policies only, and ignore terminal bonuses (although these may in practice be very important.) Let us consider an n-year endowment assurance issued t years ago to a life then aged x with basic sum assured S (payable, with attaching bonuses, at the end of the year of death or on maturity.) We suppose that the policy was issued with level annual premiums, and that the premium now due has not yet been paid. Reversionary bonuses are added on the payment of each premium, and we denote the total bonus added to date by B (hence the total death benefit in the year just ending is S +B.) The principal methods used for calculating reserves for with profits policies are: (1) the net premium method; (2) the bonus reserve (or gross premium) method; and (3) the asset share method. (1) The net premium method. The reserve is taken to be the net premium reserve for corre- sponding non profit policy, plus the mean present value of the bonuses already declared. In the case of the endowment assurance policy discussed above, the reserve is given by the formulae: t V WP = S t V x:n +BA x+t:n−t (6.9.1) = (S +B)A x+t:n−t −SP x:n ¨ a x+t:n−t (6.9.2) where all the actuarial functions are calculated on the given mortality and interest basis. The rationale of this method is that the additional premiums for a with profits policy (relative to the corresponding without profits policy) are considered to have “earned” the bonus B so far declared, so an additional reserve is required to cover the value of these bonuses. If the sum assured is payable immediately on death, we have t V WP = (S +B) ¯ A x+t:n−t −P¨ a x+t:n−t (6.9.3) where P is the annual premium for the corresponding non-profit contract, i.e. P = S ¯ A x:n ¨ a x:n = SP( ¯ A x:n ) (6.9.4) Similar formulae may be given for whole life policies and for policies with monthly premiums, etc. Example 6.9.1. On 1 st January 1993 a life office issued a with-profit assurance endowment policy with a term of 10 years to each of 100 male lives then aged exactly 50. The basic sum as- sured under each contract was £20, 000, and the basic sum assured, plus any bonuses, is payable on maturity or immediately on earlier death. Level annual premiums are payable in advance throughout the term. Among the group of policies there were 2 deaths during 1993 and 4 deaths during 1994. There were no lapses or surrenders during 1993 or 1994. For this group of policies, the office has declared a compound reversionary bonus of 3 3 4 % per annum vesting in advance on each 1 st January from outset. The office values the policies by a net premium method, using A1967-70 ultimate mortality and interest at 3%p.a. Calculate the total reserves which the office held for the group of policies on 31 st December 1994. 6.9. RESERVES FOR WITH-PROFITS POLICIES 111 Solution. We require 94reserve of one policy. Using the net premium method for with profits policies, reserve per policy = S[ 2 V( ¯ A 50:10 )] +B ¯ A 52:8 where S = 20, 000, B = 20, 000[(1.0375) 2 −1] (i.e. bonuses already vesting) i.e. reserve per policy = 20, 000(1.0375) 2 ¯ A 52:8 −P¨ a 52:8 where P = net premium for corresponding non-profit policy = 20, 000 ¯ A 50:10 ¨ a 50:10 = 1, 760.94 [where we have used ¯ A 50:10 (A 1 50:10 )(1.03) 1 2 +A 1 50:10 = (1.03) 1 2 (A 50:10 −v 10 10 p 50 ) +v 10 10 p 50 = 0.068 + 0.68419 = 0.75219 ] ∴ Reserves needed = 94 [21, 528.125 ¯ A 52:8 . .. . = (1.03) 1 2 (A 52:8 −v 8 8 p 52 ) +v 8 8 p 52 = (1.03) 1 2 (0.06142) + 0.73329 = 0.79562 −1, 760.94¨ a 52:8 ] = £443, 417 (2) The bonus reserve method. As the name suggests, reserves are calculated on the assumption that reversionary bonuses will be declared at certain annual rates. The value of the benefits is therefore calculated by the formulae given in Section 3.10 above, according to the system of bonus declarations used by the office, and the reserve is then found by subtracting the value of the office premiums less projected future expenses. If, in the case of the endowment assurance policy discussed above, it is assumed that future bonuses will be at the rate b per annum on the compound system and a proportion k of future premiums will be absorbed in expenses, a formula for the reserve is t V WP = (S +B)A ∗ x+t:n−t −(1 −k)P ¨ a x+t:n−t (6.9.5) where ∗ indicates the rate of interest (i − b)/(1 + b). Similar formulae may be derived for other policies. Since the premium valued is the actual (or office) premium for the policy, this method is some- times referred to as the gross premium method for with profits policies. As might be expected, the reserves calculated by the bonus reserve method depend greatly on the assured future levels of bonus, and in the early years of a contract the method may give reserves which are quite unrelated to the accumulation of premiums. 112 CHAPTER 6. RESERVES (3) The asset share method. This has been described (for non-profit policies) in Section 6.5 above; the formulae are almost unchanged for with profit policies, the only difference being the fact that the sum assured in each policy year includes the vested bonuses. Since these only affect the death cost, the reserves given by the asset share method do not depend greatly on past bonus declarations. A complete description of the advantages and disadvantages of these methods of calculating reserves is beyond the scope of this book. 6.10. EXERCISES 113 Exercises 6.1 (i) Express t V x in terms of A x and A x+t . Hence, or otherwise, find the values of n V x and n V x+n , given that 1 −A x+2n = A x+2n −A x+n = A x+n −A x (ii) Calculate 1 V 40 given that P 40 = .01536, p 40 = .99647 and i = .05. 6.2 Consider an n-year pure endowment policy, issued to (x), with sum assured 1 and with annual premiums payable throughout the duration of the policy. In the event of death within the n years, all premiums paid will be returned without interest at the end of the year of death. Obtain expressions for the reserve at duration t (i) prospectively and (ii) retrospectively. Using commutation functions, show that your expressions are equal. (Assume that the premium due at time t has not yet been paid.) 6.3 (Difficult) Given that P x = 0.02, n V x = 0.06, and P 1 x:n = 0.25, find P 1 x:n . 6.4 Ten years ago a life office issued a 20-year endowment assurance without profits to (35). The sum assured is £10000, payable at the end of the year of death (or on survival for 20 years), and premiums are payable annually in advance. Using A1967-70 ultimate 6%, and ignoring expenses, calculate (i) the annual premium; (ii) the reserve, assuming that the premium now due has been paid. 6.5 (Difficult) You are given: (i) P x = 0.01212 (ii) 20 P x = 0.01508 (iii) P 1 x:10 = 0.06942 (iv) 10 V x = 0.11430 Calculate 20 10 V x . 6.6 A whole life assurance with sum assured £100,000 payable at the end of the year of death was purchased by a life aged 30. The policy has annual premiums payable throughout life. The basis for calculating reserves for this policy is as follows: net premium method: A1967-70 ultimate, 5% p.a. interest. Estimate the policy value at duration 28 1 4 by interpolation. 6.7 Describe the following terms briefly: (i) net premium reserves; (ii) Zillmerised reserves; 114 CHAPTER 6. RESERVES (iii) Full Preliminary Term reserves; and (iv) gross premium reserves, in each case giving suitable formulae in respect of a whole life assurance issued t years ago to a life then aged x, with level premiums payable annually in advance throughout the term and with a sum assured of £1 payable at the end of the year of death. Assume that the premium now due has not yet been paid. 6.11. SOLUTIONS 115 Solutions 6.1 (i) t V x = 1 − ¨ a x+t ¨ a x = d¨ a x −d¨ a x+t d¨ a x = A x+t −A x 1 −A x The given information implies that A x+n = 2A x+2n −1 and A x = 2A x+n −A x+2n = 3A x+2n −2 Hence n V x = A x+n −A x 1 −A x = 1 −A x+2n 3 −3A x+2n = 1 3 and n V x+n = A x+2n −A x+n 1 −A x+n = 1 −A x+2n 2 −2A x+2n = 1 2 (ii) Retrospective method gives 1 V 40 = [P 40 −vq 40 ](1 +i) l 40 l 41 = P 40 (1 +i) −q 40 p 40 = 0.01264 6.2 The equation of value for the annual premium P is: P¨ a x:n = n E x +P(R x −R x+n −nM x+n )/D x or P(N x −N x+n ) = D x+n +P(R x −R x+n −nM x+n ) (∗) The expressions for the reserves are as follows: Retrospective: t V = [P(N x −N x+t ) −P(R x −R x+t −tM x+t )] /D x+t (∗∗) Prospective: t V = n−t E x+t +tPA 1 x+t:n−t +P(IA) 1 x+t:n−t −P¨ a x+t:n−t = [D x+n +tP(M x+t −M x+n ) + P(R x+t −R x+n −[n −t]M x+n ) − P(N x+t −N x+n )] /D x+t (∗ ∗ ∗) The expressions (∗∗) and (∗ ∗ ∗) are equal if and only if PN x −PR x = D x+n −tPM x+n −PR x+n −P[n −t]M x+n +PN x+n i.e. if and only if D x+n +P(R x −R x+n −nM x+n ) = P(N x −N x+n ) which is true by (∗). 6.3 We know that n V x = 1 − ¨ a x+n ¨ a x = 0.06, so ¨ a x+n ¨ a x = 0.94 116 CHAPTER 6. RESERVES 1 ¨ a x:n − 1 ¨ a x = ¨ a x − ¨ a x:n ¨ a x ¨ a x:n = n E x ¨ a x+n ¨ a x ¨ a x:n = 0.94P 1 x:n = 0.235 P x:n −P x = 1 ¨ a x:n −d − 1 ¨ a x −d = 1 ¨ a x:n − 1 ¨ a x = 0.235 so P x:n = 0.255 P 1 x:n = P x:n −P 1 x:n = 0.005 6.4 (i) The annual premium is 10000P 35:20 = 10000 1 ¨ a 35:20 −d = £267.51 since ¨ a 35:20 = ¨ a 35 −v 20 l 55 l 35 ¨ a 55 = 11.997 (ii) The reserve is 10000 10 V 35:20 = 10000 1 − ¨ a 45:10 ¨ a 35:20 = £3593 since ¨ a 45:10 = ¨ a 45 −v 10 l 55 l 45 ¨ a 55 = 7.687 ∴ Reserve after premium is paid = £3593 + £267.51 = £3861 (to nearest £1) 6.5 The prospective method does not work. By the retrospective method, (reminder: 20 10 V x refers to limited payments policy) 20 10 V x − 10 V x = D x D x+10 [ 20 P x ¨ a x:10 −P x ¨ a x:10 ] (since benefits are the same under both policies) = ( 20 P x −P x ) D x ¨ a x:10 D x+10 = 20 P x −P x P 1 x:10 = 0.04264 ∴ 20 10 V x = 0.15694 6.6 28 V = 100, 000 28 V 30 = 33, 101 29 V = 100, 000 29 V 30 = 34, 754 Interpolate between 28 V + P and 29 V, where P is the annual premium, i.e. P = 100, 000 P 30 = 724.00. This gives 28 1 4 V = 3 4 (33, 101 + 724) + 1 4 (34, 754) = £34, 057 6.11. SOLUTIONS 117 6.7 (i) Net premium reserve = reserve calculated ignoring expenses and assuming premium basis agrees with re- serving basis = t V x = A x+t −P x ¨ a x+t in example given (ii) Zillmerised reserve = (1 +I) t V −I per unit sum assured (where t V = net prem. reserve, I = additional initial expenses per £1 sum assured) = (1 +I) t V x −I in example given. (iii) F.P.T. reserve = net premium reserve for corresponding policy of term n −1, entry age x + 1, duration t −1 = t−1 V x+1:n−1 per unit sum assured for E.A. = t−1 V x+1 in example given (iv) Gross prem. reserve = value of future benefits − value of future gross premiums less expenses. = A x+t −(1 −k)P ¨ a x+t in example, where P = office A.P. and k = propn. of premiums allowed for future expenses. 118 CHAPTER 6. RESERVES Chapter 7 APPLICATIONS OF RESERVES 7.1 Surrender values Most life assurance policies (except term assurances) provide for a sum of money known as a sur- render value to be paid to the holder if the payment of premiums is discontinued and the contract terminated. If the scale of surrender values is fixed in advance, the policy is said to have guaranteed surrender values. For example, a whole life contract (with level annual premiums payable through- out life) may be issued with guaranteed surrender values equal to the full preliminary term reserve, i.e. (SV ) t = S. t−1 V x+1 (7.1.1) where x is the age at inception of the policy, t is the duration, (SV) t is the surrender value, S is the sum assured, and t V x is calculated on a specified mortality and interest basis. (It is assumed that the premium due at time t is unpaid.) More usually, however, surrender values are not guaranteed, although there may be guaranteed minimum surrender values, as is the case for Industrial Assurance policies in the U.K. (i.e. those in which the premiums are collected regularly from the home of the policyholder.) Surrender values are normally related in a simple way to the reserve of the contract, but the method of calculating the reserve for this purpose may be different from those used for other purposes, e.g. statutory returns to the Department of Trade and Industry. Surrender value scales must take several factors into account: (i) the experience of the office since the policy was issued, (ii) the expected future experience, (iii) consistency between different classes of business, (iv) competition between life offices (and possibly with a market for “second hand” life policies), and (v) guaranteed minimum surrender values (if applicable.) A full discussion of all points involved is beyond our present scope. In some cases surrender values are calculated from paid up policy values, which we discuss in the next section. Transfer values in pension schemes correspond to surrender values of life assurance policies: a transfer value is, however, payable only to another pension scheme and not to the scheme member directly. 119 120 CHAPTER 7. APPLICATIONS OF RESERVES 7.2 Paid-up policy values Instead of surrendering his policy, a policyholder who wishes to stop paying the premiums may ask for the contract to continue in force with reduced benefits (payable when the original benefits were to be paid.) The reduced benefit is called the paid-up sum assured (P.U.S.A.). If the original sum assured on death differs from that on maturity, the paid-up benefits will normally be in the same ratio. The general symbol t W denotes the paid-up sum assured at duration t years, assuming that the premium then due is unpaid. If expenses are ignored and all calculations are conducted on the same mortality and interest basis, net premium paid-up policy values may be defined, using the notation t W x , t W x:n , etc. For example, t W x = the paid-up sum assured at duration t in respect of a whole-life assurance issued to a life then aged x by level annual premiums, with sum assured £1 payable at the end of the year of death (7.2.1) Suppose that, at the date of discontinuance of premiums of this whole life policy, the life office wishes to make the mean net liability for the altered contract the same as that of the original contract. This leads to the equation t V x = t W x A x+t (7.2.2) from which we obtain t W x = t V x A x+t (7.2.3) Similar calculations hold for other policies, e.g. t W x:n = t V x:n A x+t:n−t (7.2.4) More generally, we obtain the following general formula for the P.U.S.A.: t W = t V A (7.2.5) where t V is the reserve or surrender value and A is an assurance factor, such as A x+t or ¯ A x+t:n−t . (We note that no expenses have been allowed for at or after the date of alteration: if such an allowance is desired, the formulae should be modified.) In the important case of endowment assurances, the proportionate rule is sometimes used in practice, i.e. t W = t n S (7.2.6) where S is the original sum assured, t is the total number of premiums actually paid and n is the number originally payable. Surrender values of endowment assurances (or other policies) are sometimes found from the P.U.S.A. by the formula (SV ) t = (P.U.S.A.)A (7.2.7) where A is the appropriate assurance factor at age x +t. In particular, if the proportionate rule for endowment assurances is used, the surrender value at duration t years is sometimes taken as t n SA x+t:n−t (7.2.8) 7.3. ALTERATIONS AND CONVERSIONS 121 if the sum assured is payable at the end of the year of death, if this occurs within the balance of the term, n −t years, or t n S ¯ A x+t:n−t (7.2.9) if the sum assured is payable immediately on death within this period. Example 7.2.1. (i) Four years ago a life then aged 35 effected a 25-year without profits endowment assurance by annual premiums for a sum assured of £50,000 payable at the end of the year of death or on survival for 25 years. He surrendered the policy just before paying the premium now due. The life office calculated the premiums for this policy on the basis of A1967-70 select at 4% interest with the following expense loadings: 2.5% of each premium, including the first, plus a further initial expense of 1% of the sum assured. The office also calculates reserves on this basis, and surrender values are equal to 95% of reserves. (a) Calculate the office annual premium. (b) Calculate the surrender value of the policy. (ii) Suppose that the policyholder of (i) had made his policy paid-up instead of surrendering it, and that the office calculates paid-up values by the proportionate rule. Calculate the yield per annum (to the nearest 0.1%) that he would have obtained on his entire transaction, assuming that he survives to maturity date. Solution. (i) (a) Let A.P. be P . We have 0.975P ¨ a [35]:25 = 50000(A [35]:25 + 0.01) ∴ P = £1289.10 (b) Reserve = 50000A 39:21 −0.975P ¨ a 39:21 = £4790.00 ∴ S.V. = 0.95 reserve = £4550.50 (ii) PUSA=50000 4/25 = 8000. Solve P ¨ s 4 = 8000v 21 i.e. (1 +i) 21 ¨ s 4 = 6.20588 By trials and interpolation, i 1.9%. 7.3 Alterations and conversions Life offices are frequently asked to alter a policy; for example, a policyholder with an endowment assurance maturing at age 65 may request that the policy should be changed to an endowment 122 CHAPTER 7. APPLICATIONS OF RESERVES assurance maturing at age 60. The policyholder might also request his policy to be converted to another class of business: for example, from a whole life assurance to an endowment assurance. The usual rule when carrying out these calculations is to equate the reserves before and after conversion. If V 1 and V 2 denote the reserves before and after the conversion respectively, the rule may be stated as follows: V 1 = V 2 (7.3.1) If there are expenses of C for the conversion itself, the formula becomes V 1 −C = V 2 (7.3.2) The bases used to calculate V 1 and V 2 may in practice be different, and V 2 must be calculated prospectively by the gross premium method. To show that formulae (7.3.1) and (7.3.2) hold, we argue as follows. At the date of conversion, the policyholder’s original policy is worth V 1 on immediate surrender. (In practice, however, the reserve V 1 used in conversion calculations may exceed the surrender value, as the office is not losing business on a conversion or alteration.) Imagine that V 1 is applied as a special single premium towards the new contract. The equation of value of the new contract is thus V 1 + M.P.V. of future premiums (including any due now) = M.P.V. of future benefits + future expenses + C Rearranging this equation gives: V 1 −C = M.P.V. of future benefits + expenses − premiums = V 2 The office then solves this equation for the unknown quantity (usually the new sum assured or premium.) An example of the use of formula (7.3.1) has already been encountered in equation (7.2.2) above. Example 7.3.1. Ten years ago, a man now aged 40 effected a whole of life assurance for a sum assured of £10000, payable at the end of the year of death, by level annual premiums. The premiums were calculated using A1967-70 ultimate 4% and an allowance for expenses of 50% of the first year’s premium and 5% of each subsequent premium. Immediately before payment of the eleventh annual premium the man requests that the policy be converted into an endowment assurance maturing on his sixtieth birthday, with the annual premium remaining unaltered. Calculate his revised sum assured using a full preliminary term reserve (on A1967-70 ultimate, 4% interest) for finding the reserve of the original policy, and A1967-70 ultimate, 4% interest, expenses of 5% of premiums for calculating the reserve of the new policy. Solution. Annual premium is P , where 0.95P ¨ a 30 = 10000A 30 + 0.45P. ∴ P = £97.13. V 1 = Reserve of original policy = 10000 9 V 31 . .. . 1 − ¨ a 40 ¨ a 31 = £949.85 7.3. ALTERATIONS AND CONVERSIONS 123 Equation of value at date of conversion is reserve before conversion = (prosp.) reserve after conversion ∴ 949.85 = SA 40:20 − 0.95 97.13 . .. . allowing 5% of premiums for expenses ¨ a 40:20 ∴ S = £4717 Example 7.3.2. An office issued a ten-year endowment assurance to a man aged exactly 45. The monthly premium was £25 during the first five years increasing to £50 thereafter. The sum assured, payable immediately on death, was calculated using A1967-70 select 4%, allowing for expenses of £1 per month with additional initial expenses of 2 1 2 % the sum assured. After five years the man requested that the premium remain at £25 per month for the next five years, with the death benefit staying unaltered. Calculate the reduced sum payable on survival using the premium basis and allowing for a £30 alteration charge. Solution. Let S =original sum assured. 300¨ a (12) [45] :10 + 300 D 50 D [45] ¨ a (12) 50 :5 = S 0.025 + ¯ A [45]:10 + 12¨ a (12) [45] :10 ∴ S = £4871 (using ¨ a (12) [45] :10 = 8.154, ¨ a (12) 50 :5 = 4.489, D 50 D [45] = 0.8093, ¯ A [45]:10 = 0.6808 ) Let S.A. on survival be reduced to S . Equate reserves before and after conversion [allowing for cost of conversion]: 4871 ¯ A 50:5 −(600 −12)¨ a (12) 50 :5 −30 = 4871 ¯ A 1 50:5 +S D 55 D 50 −(300 −12)¨ a (12) 50 :5 ∴ (4871 −S ) D 55 D 50 = 300¨ a (12) 50 :5 + 30 ∴ S = £3144 Example 7.3.3. In a certain country, the Universities operate the following superannuation scheme for academic staff, of whom there are two grades, Lecturer and Professor. Lecturers retire at age 65 with a lump sum of £30, 000, while Professors retire at age 70 with a pension of £12, 000 per annum payable monthly in advance. Staff pay for their benefits by means of level annual premiums, payable annually in advance until retirement or earlier death. Premiums are returned without interest on death before retirement, and withdrawals may be ignored. The employers do not contribute to the scheme. Death benefits are paid at end of year of death. The benefits are provided by policies issued by a life office which uses the following basis for all calculations: A1967-70 ultimate 124 CHAPTER 7. APPLICATIONS OF RESERVES 4% per annum interest expenses are ignored. A Professor cannot be demoted to Lecturer. (i) Calculate the annual premium payable by a Lecturer on entry to the scheme at 25. (ii) Calculate the reserve, just before payment of the premium now due, for a Lecturer aged 45 with 20 years’ service. (iii) The Lecturer of (ii) is now promoted to Professor. Calculate his or her new annual premium, the first due at once. Solution. (i) Let the annual premium be P. The equation of value is P¨ a 25:40 = 30000A 1 25:40 +P(IA) 1 25:40 =⇒ P = £274.44 (ii) V 1 = 30000 D 65 D 45 + 20PA 1 45:20 +P(IA) 1 45:20 −P¨ a 45:20 = £8536 (iii) Let the new premium be P . We have, V 1 = V 2 That is, 8536 = value of future benefits −future premiums for new contract = 12000 D 70 D 45 ¨ a (12) 70 + 20PA 1 45:25 +P (IA) 1 45:25 −P ¨ a 45:25 =⇒ P = £1525.87 7.4 The actual and expected death strains Consider N identical policies on lives aged x at entry and now aged x + t, and assume that the premiums now due are unpaid. The death strain at risk for a given contract in the coming policy year is S − t+1 V (7.4.1) where S denotes the sum payable on death (assumed payable at the end of the policy year) and t+1 V denotes the reserve at the end of the policy year. Since the life office may be considered to be holding the reserve t+1 V for each policy in force at time t + 1 in an imaginary “bank account” for the policyholder, the death strain at risk measures the extra cash required if he dies in policy year t +1. Let θ x+t denote the number of deaths in the coming year. Note that (in the notation used for mortality investigations) N = E x+t = the q-type exposed to risk at age x +t (since each life is exposed to the risk of death for the whole policy year) 7.5. MORTALITY PROFIT AND LOSS 125 Assuming independence of the lives we have θ x+t ∼ binomial(N, q x+t ) which has mean Nq x+t . Now consider the random variable A.D.S. = actual death strain = θ x+t (S − t+1 V) (7.4.2) The mean of this variable (even without the assumption of independence of the lives) is EDS = expected death strain = Nq x+t (S − t+1 V) (7.4.3) Note. Once the number of deaths θ x+t , is known (at the end of the policy year), ADS ceases to be a random variable: it is then a realisation of the variable. The position may be generalised to cover cases in which policies considered are not identical, provided that their premiums are due on the same date. The actual death strain is ADS = ¸ death claims in policy year t + 1 (S − t+1 V) (7.4.4) and the expected death strain is EDS = ¸ all policies in force at the start of policy year t + 1 q x+t (S − t+1 V) (7.4.5) Note. The sum assured S may depend on the policy year, t + 1. 7.5 Mortality profit and loss Consider a block of business in force at the start of policy year t + 1, the premiums now due being unpaid, and suppose that the life office holds funds equal to the total reserves for this block of busi- ness. Let us (i) ignore expenses, (ii) suppose that the premium and reserving bases agree, and (iii) assume that interest will be earned in policy year t + 1 at the valuation rate, i p.a. The mortality profit in policy year t +1 is defined as the funds at the end of the year less the money needed to pay death claims and to set up reserves for the survivors. Hence mortality profit = ¸ all in force at the start of year t + 1 ( t V +P)(1 +i) − ¸ death claims S − ¸ survivors t+1 V (7.5.1) Hence 126 CHAPTER 7. APPLICATIONS OF RESERVES mortality profit = ¸ all in force at the start of year ( t V +P)(1 +i) − ¸ deaths (S − t+1 V) − ¸ all in force at the start of year t+1 V (7.5.2) Theorem 7.5.1. Under the present assumptions, the mortality profit to the life office in respect of this group of policies, i.e. the difference between the actual funds at the end of the policy year and the funds required to pay claims and set up reserves, is equal to E.D.S. minus A.D.S.; that is, mortality profit in policy year t + 1 = E.D.S. −A.D.S. (7.5.3) Proof. Consider a particular policy in the group. We have equation ( t V +P)(1 +i) = Sq x+t + (1 −q x+t ) t+1 V (7.5.4) which may be thought of as saying that the accumulated funds at the end of the year, ( t V +P)(1 +i) , must provide: (i) the expected cost of death claims, Sq x+t , plus (ii) the expected cost of setting up reserves for the survivors, (1 −q x+t ) t+1 V. Equation (7.5.4) may be rearranged to give ( t V +P)(1 +i) = t+1 V +q x+t (S − t+1 V) (7.5.5) On summing over all policies in force at the start of the year, we find that ¸ all in force at the start of year ( t V +P)(1 +i) = ¸ all in force at the start of year t+1 V + EDS (7.5.6) Hence, by equation (7.5.2), mortality profit = EDS −ADS as required. Notes. 1. The sum assured S and the premium P may depend on the policy year, t + 1. 2. Equation (7.5.4) may be verified directly for particular policy types: e.g., for a whole life policy with sum assured £1, ( t V x +P x )(1 +i) = q x+t + (1 −q x+t ) t+1 V x (7.5.7) Example 7.5.1. On 1st January 1988, a certain life office sold 100 10-year without profits endow- ment assurance policies to lives then aged 50. Premiums are payable annually in advance, and the sum assured under each contract is £30000, payable on maturity or at the end of the year of previous death. The office calculates premiums and reserves on the following basis: 7.5. MORTALITY PROFIT AND LOSS 127 mortality: A1967-70 ultimate interest: 5% per annum expenses: nil All 100 policies were still in force on 1st January 1994, and 2 policyholders died during 1994. Calculate the mortality profit or loss to the office during 1994 in respect of this business, at 31 December 1994. Solution. Death strain at risk for each policy at 31.12.94 = 30000 −30000 7 V 50:10 = 30000 ¨ a 57:3 ¨ a 50:10 = 10753 Hence mortality profit = E.D.S. −A.D.S. = (100q 56 −2) 10753 = −11378 (Loss of £11378) Example 7.5.2. On the 1st January 1979 a life office issued a number of 20-year non-profit endow- ment assurance policies, with annual premiums payable in advance throughout the term and sums assured payable on maturity or at the end of the year of death, to lives then aged exactly 45. At 31st December 1993, total sums assured of £4,000,000 remained in force. During 1993 sums assured of £50,000 became death claims, and there were no other exits in 1993. The office calculates premiums, and maintains net premium reserves, on the basis given below. Calculate the profit or loss from mortality for this group of policies for the year ending 31st December 1993. Basis: mortality: A1967-70 ultimate interest: 4% per annum expenses: nil. Solution. Sums assured in force on 1.1.1993 = £4050000 ∴ Expected death strain = q 59 (4050000)[1 − 15 V 45:20 ] = q 59 (4050000) ¨ a 60:5 ¨ a 45:20 Actual death strain = 50000(1 − 15 V 45:20 ) = 50000 ¨ a 60:5 ¨ a 45:20 128 CHAPTER 7. APPLICATIONS OF RESERVES ∴ mortality profit = E.D.S. −A.D.S. = (4050000 0.01299373 −50000) 4.489 13.488 = £874 Some generalisations 1. If expenses are included in the calculations, we need only alter t V + P to t V + P − e, where e is the expense incurred at the beginning of policy year t + 1. The equation corresponding to (7.5.4) is ( t V +P −e)(1 +i) = Sq x+t + (1 −q x+t ) t+1 V (7.5.8) and formula (7.5.3) remains correct. (It is assumed that the actual expenses are as allowed for in the basis used for premiums and reserves. If not, one obtains expense profits or losses, which we do not discuss here.) 2. If the premium basis differs from the reserving basis, formula (7.5.3) remains true provided that reserves are calculated by a gross premium method (either retrospective or prospective.) This follows from the fact that equation (7.5.8) remains true (with mortality, interest and expenses according to the reserving basis and P on the premium basis.) Example 7.5.3. For several years an insurance company has issued a large number of special en- dowment assurances. Each policy matures at exact age 65 and is effected by annual premiums payable on each January 1st throughout the term. The sum assured, payable at the end of the year of death during the term, is one half of the sum assured that will be paid if the policyholder survives to age 65. Details of the policies in force on 31 December 1988 are as follows: Exact age Total sums assured payable on death (£) Total annual premiums (£) 55 3500000 250000 The claims in 1989 were on policies with the following total sums assured and premiums: Total sums assured payable on death (£) Total annual premiums (£) 50000 4000 Assuming that at 31.12.88 the office’s funds were equal to its reserves, calculate the mortality profit or loss in 1989, given that the following reserving basis is used: prospective method, gross premium reserve using A1967-70 ult., 4% p.a. interest, no expenses. Solution. Let EDS and ADS denote the expected death strain and the actual death strain in 1989. Then EDS = ¸ all j q 55 [S j −¦S j (A 56:9 + 9 E 56 ) −P j ¨ a 56:9 ¦] 7.6. OTHER SOURCES OF PROFIT AND LOSS 129 where the summation is over all policies in force at the start of the year, i.e. EDS = q 55 ( ¸ S j ) −( ¸ S j )(A 56:9 + 9 E 56 ) + ( ¸ P j )¨ a 56:9 = q 55 (694048) (since ¸ all j S j = 3500000 and ¸ all j P j = 250000) = 5859 The actual death strain is obtained by a summation of the death strains at risk over those policies which become claims. Thus ADS = ¸ Claims [S j −¦S j (A 56:9 + 9 E 56 ) −P j ¨ a 56:9 ¦] = ( ¸ Claims S j ) −( ¸ Claims S j )(A 56:9 + 9 E 56 ) + ( ¸ Claims P j )¨ a 56:9 = 13082 (since ¸ Claims S j = 50000 and ¸ Claims P j = 40000) The mortality profit is thus = 5859−13082=−7223, i.e. a loss of £7,223. 7.6 Other sources of profit and loss Although some of these topics may also be considered under the heading of “profit-testing”, we shall give certain straightforward results on interest and surrender profits here. Interest profits. Suppose that the office earns interest at a rate i p.a. in policy year t + 1, the interest rate in the reserving basis being i p.a. Assuming as before that the funds at the end of policy year t equal the reserves for the business under consideration, the interest profit is [ ¸ all in force at the start of policy year t + 1 ( t V +P −e)](i −i) (7.6.1) Example 7.6.1. On 1st January 1978 a special twenty-year endowment assurance policy for an initial sum assured of £20,000 was issued to a life then aged 40 exactly. The sum assured, which is payable at the end of the year of death or on survival to maturity date, increased to £25,000 after 10 years and to £30,000 on maturity. The annual premium was P for the first ten years, and thereafter increased to 1.25P. The office’s basis for premiums and reserves is A1967-70 ultimate 4% per annum interest expenses are ignored (i) Calculate the initial annual premium P, and the reserve for the contract at (a) 31st December 1989, and (b) 31st December 1990. 130 CHAPTER 7. APPLICATIONS OF RESERVES (ii) Find the profit or loss from mortality for the year ending 31st December 1990 for a group of such policies, all effected on 1st January 1978 by lives then aged 40 exactly, given that one policyholder dies in 1990 and 120 policies remained in force at 31st December 1990. There were no withdrawals in 1990. (iii) The office found that it made neither profit nor loss in 1990 in respect of the policies of (ii) above. What rate of interest did it earn in 1990? (Expenses are negligible.) Solution. (i) Let initial premium be P. P[¨ a 40:20 + 0.25 10 [¨ a 40:10 ] = 20000A 40:20 + 5000 10 [A 40:10 + 5000 D 60 D 40 ∴ P = 20000M 40 + 5000M 50 −25000M 60 + 30000D 60 N 40 + 0.25N 50 −1.25N 60 = £907.00 (a) 12 V = 25000A 52:8 + 5000 D 60 D 52 −1.25P¨ a 52:8 = £14085 (b) 13 V = [( 12 V + 1.25P)(1.04) −25000q 52 ] 1 −q 52 = £15772 (this may also be calculated directly) (ii) E.D.S. = 121q 52 (25000 − 13 V) = £6734 A.D.S. = 25000 − 13 V = £9228 ∴ mortality profit = 6374 −9228 = −£2494 (iii) Let interest rate earned be i . The interest profit is (i −0.04)[121]( 12 V + 1.25P) = 2494 ∴ i = 0.0414, or 4.14% Surrender profits. Assume that surrenders may take place only at the end of a policy year (just before payment of any premium then due). The surrender profit is ¸ surrenders ( t+1 V −S.V.) (7.6.2) 7.6. OTHER SOURCES OF PROFIT AND LOSS 131 where S.V. is the surrender value paid. Summary of profit and loss Profit or loss for year = accumulation of funds at end of year −cost of death claims −cost of surrenders −cost of setting up reserves for remaining policies = Mortality profit + interest profit + surrender profit (7.6.3) Example 7.6.2. On 1st January 1977 an office issued a block of without profits endowment assur- ance policies and a block of without profits pure endowment policies to men aged exactly 40. Each policy was for a term of 20 years and a sum assured of £1000. Premiums are payable annually in advance during the term of the policies and death claims are paid on 31st December each year. On 1st January 1987 there were 1000 endowment assurance policies and 5000 pure endowment policies in force. During the year to 31st December 1987 sixty endowment assurance policies became claims by death and twenty pure endowment policies were terminated, without payment of any sum assured, by the deaths of the assured lives. On 31st December 1987 ten endowment assurance policies and five pure endowment policies were surrendered. Calculate the profit or loss arising during the year 1987, showing separately the contributions from mortality and from surrender if the office uses the following basis: Premiums and reserves: A1967-70 ultimate at 4% p.a. interest Surrender values: Net premium reserve on A1967-70 ultimate at 6% p.a. interest Ignore expenses. Solution. Let us consider each type of policy separately. Endowment Assurances. Reserve per policy at end of 1987 = 1000 11 V 40:20 = 1000 1 − ¨ a 51:9 ¨ a 40:20 = 1000 1 − 7531 13.764 = 452.85 ∴ Death strain at risk per policy in force at start of 1987 is 1000 −452.85 = 547.15 ∴ E.D.S. = 1000q 50 547.15 = 26202 A.D.S. = 60 547.15 = 32829 132 CHAPTER 7. APPLICATIONS OF RESERVES ∴ mortality profit = E.D.S. −A.D.S. = −6627 Surrender profit=10(Reserve-S.V.), where S.V. = 1000 11 V 40:20 at 6 % = 1000 1 − 7.029 11.871 = 407.88 ∴ Surrender profit = 449.70 Pure Endowments. Annual premium is P = 1000 D 60 D 40 ¨ a 40:20 = 29.70 Reserve per policy at end of 1987 is P N 40 −N 51 D 51 = 29.70 132002 −68970 4399.08 = 425.55 Death strain at risk is −425.55 per policy. E.D.S. = 5000q 50 (−425.55) = −10189 A.D.S. = −20 425.55 = −8511 ∴ mortality profit = E.D.S. −A.D.S. = −1678 S.V.= net premium reserve at 6%. It is easier to find the reserve prospectively. Define P = net premium at 6% p.a. interest = 1000v 20 l 60 l 40 ¨ a 40:20 = 279.25 11.871 = 23.52 ∴ Reserve = 1000v 9 l 60 l 51 −P ¨ a 51:9 at 6% p.a. interest = 546.87 −165.32 = 381.55 ∴ Surrender profit is 5(425.55-381.55)=220.00 Summary Mortality profit = −£8305 Surrender profit = £670 Total profit = −£7635 (a loss of £7,635) 7.7. EXERCISES 133 Exercises 7.1 Explain briefly how a life office calculates the new sum assured or annual premium on the alteration or conversion of an existing policy. 7.2 A life office issued a whole life without profits policy to a woman aged 30, with sum assured of £110,000. Premiums were payable annually in advance throughout life, and the sum assured was payable at the end of the year of death. Immediately before payment of the 15th premium, the policyholder requests that the policy be converted to a without profits endowment assurance for the same sum assured, payable on maturity at age 60 or at the end of the year of death, if before age 60. The office calculates premiums and maintains reserves on the basis of A1967-70 ultimate mortality and 4% per annum interest, with expenses of 4% of each premium. Expenses of alteration may be ignored. Find the revised annual premium. 7.3 On 1 January 1972 an office issued a large number of 40-year endowment assurance policies, each with sum assured £1000 (payable at the end of the year of death, if this occurs within the term) to a group of lives all then aged 25. On 31 December 1988 there were 8567 policies still in force. During 1989 there were 13 deaths among the policyholders. Find the actual death strain, the expected death strain and the mortality profit or loss for the business in 1989, using the following basis for all calculations: mortality: A1967-70 select interest: 4% p.a. expenses: none 7.4 On 1 January 1991, a life office issued a number of identical special endowment assurances to lives then aged 45. Each policy had a term of 15 years and level annual premiums were payable throughout the term. On survival to the end of the term, a sum assured of £5000 is payable. On death before the end of the term, a sum assured of £1000 is payable at the end of the year of death and, in addition, all premiums paid are returned without interest. The premium basis for these policies was as follows: mortality: A1967-70 select interest: 4% p.a. expenses: Nil (i) Show that the annual premium for each of these policies is £242.27. (ii) Calculate the reserve, on the premium basis, on 1 January 1996 for one of these policies, assuming that the premium then due is unpaid. (iii) On 1 January 1995, a total of 10,000 of these policies were still in force. In 1995, 47 of these lives died, ten policies were surrendered, and the life office earned 4% interest on its investments. Expenses were negligible in 1995. The surrenders all took place at the end of 1995, and the office gave surrender values equal to 95% of the reserve on the premium basis. Calculate the profit or loss to the office in 1995 from this block of business in respect of (a) mortality , and 134 CHAPTER 7. APPLICATIONS OF RESERVES (b) surrenders. 7.5 A life office has issued a 5-year decreasing temporary assurance to a life aged 60. The sum assured, payable at the end of the year of death, is £100,000 if the life dies in the first year, £90,000 if the life dies in the second year, and so on, decreasing by £10,000 each year. Level annual premiums are payable throughout the term of the policy. The premium and reserve basis for this policy is as follows: mortality: A1967-70 Ultimate interest: 4% p.a. expenses: Nil (i) Calculate the annual premium for the policy. (ii) Write down a relationship between the reserve at integer duration t and the reserve at duration t + 1. 7.8. SOLUTIONS 135 Solutions 7.1 The office calculates V 1 , the reserve for the original policy at the date of conversion. The office then sets up the equation V 1 −C = V 2 where C = cost of alteration (if any) and V 2 = (prospective) reserve of new policy. Solve for unknown quantity (usually a sum assured or premium.) 7.2 Let P = original A.P. = 110000 1 −0.04 P 30 Reserve before conversion = V 1 = 110, 000A 44 −0.96P¨ a 44 = 110, 000 14 V 30 = 110, 000 ¸ 1 − ¨ a 44 ¨ a 30 = 16, 923 Let new A.P. be P . We solve for P in the equation V 1 = V 2 , i.e. V 1 = 110, 000A 44:16 −0.96P ¨ a 44:16 ∴ P = 110, 000 .54703 −16, 923 0.96 11.777 = £3, 825 7.3 On 31/12/88 the duration is 17 years. The reserve at the end of 1989 (per unit sum assured) therefore equals 18 V [25]:40 = 1 − ¨ a 43:22 /¨ a [25]:40 = .28397 Hence in 1989 we have, per policy in force at the start of the year, a death strain at risk of (1000-283.97)=£716.03. The expected death strain per policy is 716.03 q 42 = 1.31105. For the block of business, therefore, we have: total expected death strain = 8567 1.31105 = 11, 234.52 total actual death strain = 13 716.03 = 9, 308.4 The 1989 mortality profit is therefore (11,234.52−9,308.4)=£1,926.14. 136 CHAPTER 7. APPLICATIONS OF RESERVES 7.4 (i) Let A.P. be P. P¨ a [45]:15 (= 11.250) = 1, 000A [45]:15 (= 0.56732) + 4000 D 60 D [45] (= 0.50271) +P (IA) 1 [45]:15 (= 0.60855) ∴ P = £242.27 (ii) 5 V = 1000A 50:10 + 4000 D 60 D 50 + 5PA 1 50:10 +P(IA) 1 50:10 −P¨ a 50:10 = £1, 349 (iii) (a) D.S.A.R. in 1995 in respect of 1 policy = (1000 + 5P) − 5 V = £861.97 EDS = 10, 000q 49 DSAR = £36, 714 ADS = 47 DSAR = £40, 513 ∴ mortality profit = EDS −ADS = −£3, 798 (loss of £3,798) (b) Surrender profit per policy = 0.05 reserve ∴ Total surrender profit = 10 0.05 5 V = £675. 7.5 (i) Let the Annual Premium be P. P¨ a 60:5 = 110, 000A 1 60:5 −10, 000(IA) 1 60:5 which gives P = £1, 347.03 (ii) ( t V +P)(1.04) = q 60+t (100, 000 −10, 000t) + (1 −q 60+t ) t+1 V (since the sum assured on death in year t + 1 is 100, 000 −10, 000t). Chapter 8 EXTRA RISKS 8.1 Introduction It is usual in practice for a life office to accept the majority of proposers for life assurance policies on “normal” (or “standard”) terms, the remainder being either declined or accepted on special terms, i.e. they are accepted but with higher premiums or reduced benefits. Those accepted on special terms are considered to be subject to an extra mortality risk due to a health impairment, a dangerous occupation or a dangerous spare-time activity (such as motor racing.) The assessment of extra risks is usually carried out by experienced underwriters using a “numerical rating system”; we shall not go into this, but shall assume that the addition to the normal rates or force of mortality is known. Some offices make extensive use of “rating up”: that is impaired lives are treated as normal lives a number of years older than actual age. (Females are sometimes treated as males 5, say, years younger, although this rule is very inaccurate at certain young ages.) Smokers are sometimes treated as non-smokers 6 years older. 8.2 A constant addition to the force of mortality Let µ [x]+t refer to the force of mortality at duration t years for a life who was selected for assurance to age x on the office’s normal terms. (If a non-select table is used, omit [ ].) Let n years be the duration of the policy, and let µ ∗ [x]+t be the corresponding force of mortality for a life accepted on special terms. We now assume that there is k > 0 such that: µ ∗ [x]+t = µ [x]+t +k (0 ≤ t ≤ n) (8.2.1) Hence t p ∗ [x] = exp ¸ − t 0 µ ∗ [x]+r dr = exp ¸ − t 0 (µ [x]+r +k) dr = e −kt t p [x] (8.2.2) 137 138 CHAPTER 8. EXTRA RISKS Also, t E ∗ [x] = v t t p ∗ [x] = e −(δ+k)t t p [x] = e −δ t t p [x] (8.2.3) where δ = δ +k (8.2.4) and hence i = e δ −1 = e δ+k −1. Since an annuity factor is a sum or integral of pure endowment factors, we have (for example) ¨ a ∗ [x] = ¨ a [x] on normal mortality but with force of interest δ = δ +k (8.2.5) This rule does not apply to assurances and premiums, but one may sometimes use conversion relationships to express these in terms of annuity factors, as in the following example. Example 8.2.1. A certain life office uses the following basis for calculating premiums for assurance policies on lives accepted at normal rates: A1967-70 ultimate 4% interest expenses are ignored A certain proposer, aged 50, for a 15-year endowment assurance without profits by annual premiums is considered by the office to be subject to the mortality of the office’s normal table with an addition of 0.019048 to the force of mortality for the next 15 years. The sum assured is £10,000, payable at the end of the year of death, or on survival for 15 years. Calculate the addition to the normal annual premium. Solution. Let ∗ indicate special mortality basis. Extra annual premium is 10, 000[P ∗ 50:15 −P 50:15 ] Note that P ∗ 50:15 = 1 ¨ a ∗ 50:15 −d by a conversion relationship = 1 ¨ a 50:15 0.06 −d since δ .04 + 0.019048 = δ .06 But d is still at 4% interest. Hence the extra premium is 10, 000 ¸ 1 ¨ a 50:15 .06 − 1 ¨ a 50:15 .04 = £108.51 8.3. A VARIABLE ADDITION TO THE FORCE OF MORTALITY 139 In some problems we must evaluate assurance factors directly, without the use of conversion rela- tionships; e.g. ( ¯ I ¯ A) ∗1 x:n = M.P.V. of an increasing temporary assurance for an impaired life aged x at entry = n 0 tv t t p ∗ x µ ∗ x+t dt = n 0 tv t ( t p x e −kt )(µ x+t +k) dt = n 0 t (v ) t . .. . v is at rate i (with δ = δ +k) t p x (µ x+t +k) dt = ( ¯ I ¯ A) 1 x:n at rate of interest i +k( ¯ I¯ a) x:n at rate of interest i (8.2.6) A formula for the expectation of life of an impaired life We have o e ∗ x = E(T ∗ ) = ∞ 0 t p ∗ x dt (by formula (1.4.3)) = ∞ 0 t p x e −kt dt = ¯ a x at force of interest k p.a. (8.2.7) Note also that E[(T ∗ ) 2 ] = 2 ∞ 0 t. t p ∗ x dt (by integration by parts) = 2 ∞ 0 t. t p x e −kt dt = 2( ¯ I¯ a) x at force of interest k p.a. (8.2.8) Hence Var(T ∗ ) = 2( ¯ I¯ a) x −(¯ a x ) 2 at force of interest k p.a. (8.2.9) Remark. To find the median future lifetime of the impaired life, we solve the equation t q ∗ x = 0.5 i.e. t p ∗ x = l x+t e −kt l x = 0.5 8.3 A variable addition to the force of mortality Let us now suppose that there is ξ(t) ≥ 0 such that µ ∗ [x]+t = µ [x]+t +ξ(t) (0 ≤ t ≤ n) (8.3.1) 140 CHAPTER 8. EXTRA RISKS We obtain the formula t p ∗ [x] = t p [x] exp − t 0 ξ(r) dr (8.3.2) In some cases t 0 ξ(r) dr may be calculated analytically, as in the next example. Example 8.3.1. A certain impaired life aged 75 experiences mortality according to a(55) males ultimate with an addition to the force of mortality. The addition is 0.005 at age 75, increasing linearly to 0.020 at age 90. (i) Find a formula for the probability that this life will be alive at age 75 +t years (0 ≤ t ≤ 15). (ii) Estimate, by approximate integration using the trapezoidal rule (not repeated), the single premium for a temporary assurance of £50,000 payable immediately on death of (75), if this occurs within 15 years, on the following basis: mortality: as described above interest: 10% p.a. expenses: 2% of the single premium. Solution. (i) µ ∗ 75+t = µ 75+t + 0.005 + 0.001t (0 ≤ t ≤ 15) ∴ t p ∗ 75 = t p 75 exp ¸ − t 0 (0.005 + 0.001r) dr = t p 75 exp −(0.005t + 0.0005t 2 ) (ii) Let single premium be P. 0.98P = 50, 000 15 0 v t t p ∗ 75 µ ∗ 75+t dt 50, 000 15 2 µ ∗ 75 +v 15 15 p ∗ 75 µ ∗ 90 = 50, 000 7.5 [0.06550 + 0.1104 0.23939 0.2636] ∴ P = £27, 730 8.4 Rating up In practice, impaired lives are often considered to experience the mortality of a “normal” life r years older (where r depends on the severity of the impairment.) This allows the life office to base all its premiums on one set of tables. 8.4. RATING UP 141 A practical method of finding “r” is as follows. Suppose that the normal mortality table follows Gompertz’ law, so there are constants B, c such that µ x+t = Bc x+t (0 ≤ t ≤ n) (This is usually a fairly accurate representation of mortality for all ages over about 30.) Suppose that the impairment is such that µ ∗ x+t = (1 +θ)µ x+t (0 ≤ t ≤ n) (8.4.1) for some constant, θ. (This formula is usually suitable for “medical” impairments; for accident risks it is generally better to add a constant to the normal force of mortality.) Hence µ ∗ x+t = (1 +θ)Bc x+t = Bc (x+r)+t (0 ≤ t ≤ n) where c r = 1 +θ, i.e. r = log(1 +θ) log c (8.4.2) In practice c is usually about 1.08, so we have the approximate rule r = log(1 +θ) log 1.08 (8.4.3) For example, if θ = 0.4 (corresponding to a 40% increase in the force of mortality), r log 1.4 log 1.08 = 4.37 For practical purposes this value of r would be rounded to 4 or 5 years. This rule is not precise, but it may be sufficiently accurate in practice since the value of θ may in general be estimated only rather approximately. Example 8.4.1. A certain life office’s premium basis for lives accepted at normal rates is as follows: Mortality: A1967-70 ultimate Interest: 4% per annum Expenses: none A certain impaired life, aged 50, is considered to be subject to the office’s normal mortality rates for a life 10 years older. This life requires a contract providing the following benefits: (i) £30, 000 at the end of the year of death, if within 10 years; (ii) on survival for 10 years, the sum of £(30, 000 +S). The impaired life finds that the level annual premium for this policy is the same for a policy providing identical benefits for a life his age accepted at normal rates. Find S. 142 CHAPTER 8. EXTRA RISKS Solution. normal premium = 30, 000A 50:10 +S D 60 D 50 ¨ a 50:10 = 30, 000P 50:10 +S D 60 D 50 ¨ a 50:10 “impaired life” = 30, 000P 60:10 +S D 70 D 60 ¨ a 60:10 Equate these to give S = 30, 000(P 60:10 −P 50:10 ) D 60 D 50 ¨ a 50:10 − D 70 D 60 ¨ a 60:10 = 30, 000(0.08986 −0.08339) 0.075689 −0.068168 = £25, 809 8.5 Debts Instead of paying higher premiums, a proposer accepted on special terms may elect to pay the normal premiums with reduced death benefits. Such reductions in the death benefits are called “debts” or “liens”, and may be either (a) a level amount; or (b) diminishing in arithmetic progression to zero in the final policy year. We first give an example involving a level debt. (Note that a debt is a reduction in the death benefit only: the maturity benefit is not reduced.) Example 8.5.1. A life office calculates annual premiums for without profits endowment assurances using the following basis: Mortality: A1967-70 select Interest: 4% Initial expenses: 50% of the first premium Renewal expenses: 5% of each premium after the first (i) Calculate the annual premium payable for a 25 year endowment assurance taken out by a male life aged exactly 45 for a sum assured of £50,000. The death benefit is payable at the end of the year of death. (ii) A man aged exactly 45 effects a policy identical to that in (i) above but rated up 7 years. (a) Calculate the level extra premium, payable throughout the term of the policy. (b) Alternatively, the life office offers to charge the standard premium but to impose a level debt. Calculate the amount of the debt. 8.5. DEBTS 143 Solution. (i) Let A.P. be P. 0.95P¨ a [45]:25 = 50, 000A [45]:25 + 0.45P ∴ P = £1, 492.41 (ii) (a) Consider a “normal” (select) life aged 52. Let P be premium. 0.95P ¨ a [52]:25 = 50, 000A [52]:25 + 0.45P ∴ P = 1, 752.32 ∴ Extra premium = £259.91 (b) Let death benefit be reduced by D. The equation of value is 0.95P¨ a [52]:25 = 50, 000A [52]:25 −DA 1 [52]:25 + 0.45P ∴ D = £12, 649 Now suppose that we are dealing with an n-year endowment assurance, and that the debt diminishes in arithmetical progression to zero in the final policy year; that is, the debts run as follows: (n −1)D in year 1 (n −2)D in year 2 .................. D in year n −1 0 in year n Consequently the M.P.V. of the debts, at the issue date, is D¦nA ∗ 1 [x]:n −(IA) ∗ 1 [x]:n ¦ (8.5.1) where ∗ indicates impaired lives’ mortality; the death benefits are taken to be payable at the end of the year of death (if these benefits are payable immediately on death, a bar should be placed over the A symbols.) The calculations are illustrated in the next example. Example 8.5.2. A life office issues 20-year without-profit endowment assurance policies providing a sum assured of £8,000 on maturity or at the end of the year of earlier death. Level monthly premiums are payable in advance throughout the term or until earlier death. A certain impaired life aged 45 is considered by the office to have the mortality rates of a “nor- mal” (select) life 5 years older. Assuming that the ordinary premium appropriate to this life’s actual age is payable, calculate the initial amount of the debt (i.e. the deduction from the sum assured) such that the debt reduces uniformly each year to nil in the last policy year. Basis: Mortality of “normal” lives: A1967-70 select Interest: 4% 144 CHAPTER 8. EXTRA RISKS expenses: 30% of premiums in first year 5% of later premiums Solution. Let P 12 be “normal” monthly premium. The equation of value for P is 0.95P¨ a (12) [45]:20 −0.25P¨ a (12) [45]:1 = 8000A [45]:20 P = 8000 [0.48051] .95 ¨ a [45]:20 − 11 24 1 − D 65 D [45] −0.25 1 − 11 24 1 − D [45]+1 D [45] ¸ = 3844.08 ¦0.95[13.507 −0.45833 0.62253] −0.25[1 −0.45833 0.04014]¦ = 3, 844.08 12.5606 −0.24541 = £312.14 (£26.01 per month) Let the initial debt be 19D. The equation of value for D is 0.95 312.14¨ a (12) [50]:20 −0.25 312.14¨ a (12) [50]:1 = 8000A [50]:20 − 20DA 1 [50]:20 −D(IA) 1 [50]:20 D = 8000 0.49664 −312.14 0.95 ¸ ¨ a [50]:20 −0.45833 1 − D 70 D [50] +0.25 312.14 ¸ 1 −0.45833 1 − D [50]+1 D [50] 20(M [50] −M 70 ) −(R [50] −R 70 −20M 70 ) D [50] = 3973.12 −296.53[13.087 −0.45833 0.66887] + 78.035[0.98111] 1 4581.322 [20 758.289 −9360.22] = 259.898 1.2672 = 205.09 ∴ Initial Debt is 19D = £3, 896.71 8.6. EXERCISES 145 Exercises 8.1 An explorer aged exactly 57 has just made a proposal to a life office for a whole life assurance with a sum assured of £10,000 payable at the end of the year of death. For lives accepted at normal rate, level annual premiums are payable until death under this policy. The explorer is about to undertake a hazardous expedition which will last three years. The life office estimates that during these three years the explorer will experience a constant addition of 0.02871 to the normal force of mortality, but after three years will experience normal mortality. The life office quotes a level extra premium payable for the first three years. Calculate this level extra premium on the following basis: normal mortality: A1967-70 ultimate interest: 3% per annum expenses: none 8.2 An impaired life aged exactly 55 wishes to effect a without profit endowment assurance for a sum assured of £1,000 payable at the end of 10 years or at the end of the year of earlier death. Level annual premiums are payable throughout the term of the policy. Special terms are offered on the assumption that the life will experience mortality which can be represented by: (a) for the first five years, a constant addition of 0.009569 to the normal force of mortality, and (b) for the remaining five years, the mortality of a life 8 years older. The life office quotes a level extra premium payable throughout the term. Calculate this level extra premium. Basis: normal mortality: A1967-70 ultimate interest: 3% per annum expenses: none 8.3 A group of impaired lives now aged 40 experience mortality according to A1967-70 ultimate with an addition to the force of mortality. The addition is 0.0005 at age 40, increasing linearly to 0.0025 at age 60, at which level the addition remains constant. Find the probability that an impaired life aged exactly 40 (i) will die within 20 years; (ii) will die within 30 years; (iii) will die between 20 and 30 years from the present time. 8.4 (i) Can you envisage circumstances under which an office could offer an impaired life who wishes to pay the “normal” premiums a level debt but not a diminishing debt? Hint Consider a life who is very severely impaired, and think of an approximate rela- tionship between (a) the level debt and (b) the initial debt when debts decrease linearly to zero. 146 CHAPTER 8. EXTRA RISKS (ii) A certain proposer, aged 40, for a 20-year endowment assurance by annual premiums with sum assured £40,000 without profits (payable at the end of the year of death) is considered by your office to be subject to the mortality of the normal table with an addition of 10 years to the age. Your office’s basis for calculating premiums for “normal” lives is A1967-70 ultimate 4% p.a. interest expenses of 6% of all premiums. (a) The proposer asks to pay the annual premium for a “normal” life aged 40, and suggests that the office should make the normal death benefits subject to a debt which decreases in arithmetical progression to zero in the final year. Calculate the initial debt. (b) Suppose that the proposer also asks your office to quote a level debt. Calculate the level debt. 8.5 The mortality of a certain impaired life aged exactly 60 may be represented as follows: (1) at ages up to exact age 65, there is a constant addition of 0.009569 to the “normal” force of mortality, which is that of A1967-70 select; and (2) at ages between exact age 65 and exact age 70, mortality follows that of a “normal” (ultimate) life 4 years older. Suppose that this life requires a 10-year endowment assurance without profits with sum assured £10, 000 payable on maturity or at the end of the year of earlier death. The life wishes, however, to pay the “normal” level annual premium for his age. Calculate the level debt which would be deducted from the normal death benefit, using the basis given below: mortality: as given above interest: 4% p.a. expenses: 2 1 2 % of all premiums, with an additional initial expense of 1% of the sum assured (before deducting any debt.) 8.7. SOLUTIONS 147 Solutions 8.1 Let P be the normal annual premium. Then ¨ a 57 P = 10, 000A 57 at 3% P 377.41 Let E be the extra premium charged for 3 years. Then, the equation of value will be (* denotes impaired mortality): (P +E)¨ a ∗ 57:3 +P 3 [¨ a ∗ 57 = 10, 000A ∗ 57 where ¨ a ∗ 57:3 3% = ¨ a 57:3 6% = 2.804 3 [¨ a ∗ 57 = 1.06 −3 l 60 l 57 ¨ a 60 at 3% = 11.076 A ∗ 57 = A ∗ 1 57:3 + 3 E ∗ 57 A 60 at 3% = A ∗ 57:3 − 3 E ∗ 57 + 3 E ∗ 57 A 60 at 3% = 1 −d 3% ¨ a 57:3 6% −1.06 −3 l 60 l 57 (1 −A 60 at 3%) = 0.59572 Hence E = 10, 000A ∗ 57 −P(¨ a ∗ 57:3 + 3 [¨ a ∗ 57 ) ¨ a 57:3 = 256.3207 8.2 P NORMAL = 1000P 55:10 = £85.84 P ∗ = 1000 ¸ 1 ¨ a ∗ 55:10 . .. . ¨ a ∗ 55:10 = ¨ a ∗ 55:5 + [ 5 E ∗ 55 ]¨ a ∗ 60:5 = ¨ a 55:5 at 5% +v 5 l 60 l 55 at 5% [¨ a 68:5 at 4%] = 7.677 − d . .. . at 4% ¸ ∴ P ∗ = £91.79 so extra annual premium = £5.95 8.3 Let µ x refer to A67-70 ult., µ ∗ x to impaired lives. µ ∗ 40+t = µ 40+t + 0.0005 + 0.0001t (0 ≤ t ≤ 20) µ 40+t + 0.0025 (t > 20) 148 CHAPTER 8. EXTRA RISKS (i) 20 p ∗ 40 = exp − 20 0 (µ 40+t + 0.0005 + 0.0001t) dt = 20 p 40 exp − ¸ 0.0005t + 1 2 (0.0001t 2 ) 20 0 ¸ = l 60 l 40 exp ¦−(0.01 + 0.02)¦ = 0.895579 0.9704455 = 0.86911 ∴ ans = 0.1309 (ii) 30 p ∗ 40 = 0.86911 10 p ∗ 60 = 0.86911 l 70 l 60 exp[−10 0.0025] = 0.66656 ∴ Ans = 0.3334 (iii) 0.3334−0.1309=0.2025 8.4 (i) The debt cannot, of course, exceed the “normal” sum assured on death. The level debt must be (very roughly) about one-half of the initial debt (since the average dim. debt is one-half of the initial debt), so we may have cases in which: initial debt (when debts diminish) > S.A. > level debt (in cases of serious impairment) (ii) (a) Normal A.P. = 40, 000P 40:20 0.94 = 1454.89 If initial debt is 19D, we have the equation of value 0.94(1454.89)¨ a 50:20 = 40, 000A 50:20 −D¦20A 1 50:20 −(IA) 1 50:20 ¦ i.e. D = 40, 000 0.49818 −0.94 1454.89 13.047 20 0.16819 −2.04074 = 1, 575.26 ∴ Initial debt = £29, 929.94 (£29, 930) (b) Replace D¦20A 1 50:20 −(IA) 1 50:20 ¦ in above by the term DA 1 50:20 , to give D = 2084.17 0.16819 = £12, 392 8.5 Let P be the normal A.P. 0.975P ¨ a [60]:10 . .. . 7.884 = 10, 000 A [60]:10 . .. . 0.69677 + 100 ∴ P = £919.45 8.7. SOLUTIONS 149 Let ∗ indicate special mortality. ¨ a ∗ [60]:10 = ¨ a [60]:5 0.05 . .. . 4.452 + v 5 0.05 l 65 l [60] . .. . 0.72603 ¨ a 69:5 0.04 . .. . 4.29496 = 7.570 Let level debt be D. 0.975P ¨ a ∗ [60]:10 . .. . 7.570 = 10, 000 A ∗ [60]:10 . .. . 1 −d¨ a ∗ [60]:10 = 0.708835 − D A ∗ 1 [60]:10 . .. . 0.708835 −[v 5 .05 v 5 .04 ] l 65 l [60] l 74 l 69 = 0.708835 −0.478114 = 0.230719 + 100 ∴ D = 402.119 0.230719 = £1, 743 150 CHAPTER 8. EXTRA RISKS Chapter 9 PROFIT-TESTING 9.1 Principles of profit-testing In this book we shall consider “conventional” life assurance business only, not unit-linked policies (which are considered in part D1.) The main new ideas in profit-testing are as follows: (1) The sale of a without-profits policy is considered by the life office as an “investment” on the part of the shareholders and/or with-profits policyholders. (In a mutual life office there are no shareholders.) Similarly, the sale of a with-profits policy may be considered to be an “investment” on the part of the office’s shareholders, though the position is complicated by the fact that future profits are shared between the with-profits policyholders and the shareholders. (2) The office calculates the expected net cash flow, and the expected profit, in each policy year in two stages. Stage A. The net cash flow in each policy year, and the profit after allowing for appropriate reserves, of a given contract is calculated on the assumption that the policy is still in force at the start of that year. The resulting net cash flow is called the in force net cash flow, and the vector of profits (indexed by the policy year, t) is called the profit vector, which is written as ¦(PRO t )¦. Stage B. The estimated net cash flow per policy sold in policy year t is called the initial net cash flow for that year, and is found by multiplying the “in force” net cash flow by the probability that the policy is still in force at the start of the tth policy year, t−1 p x . (We assume here that the age at entry to assurance is x, and that a non-select mortality table is used.) The corresponding vector of profits (after allowing for reserves) is called the profit signature, which is written as ¦σ t ¦. (3) All cash flows are considered at the end of the policy year, so premiums less expenses are accumulated to the end of the appropriate policy year at the rate of interest which the office assumes it will earn on the life funds. (4) Random fluctuations in mortality are ignored; that is, each policy is considered as if it were one of a large number of identical contracts whose experience exactly follows the mortality table used by the office in its projections. 151 152 CHAPTER 9. PROFIT-TESTING 9.2 Cash flow calculations We consider an n-year “generalised endowment assurance”, issued to (x), and use the following notation: P t = premium payable in policy year t (at the start of that year, i.e. payable at time t −1 years from the issue date); e t = expenses assumed to be incurred in policy year t (at the start of that year); D t = death benefit in policy year t (paid at the end of the policy year); S t = survival benefit at the end of policy year t (S t is usually zero for t < n); i = rate of interest p.a. which the life office assumes it will earn on its life funds; The symbols q x+t , t−1 p x , etc, refer to the mortality table which the life office assumes will apply to the policyholders under consideration. Note. In general, three distinct bases are used in profit-testing: Basis 1: the premium basis, which is used only to calculate the premiums; Basis 2: the reserving basis, which is used only to calculate the reserves; Basis 3: the experience basis, which is that which, the office assumes, will be experienced. In some cases all three of these bases agree, or two of them agree. Note that the symbols e t , i and q x+t given above all refer to Basis 3, while P t is calculated using Basis 1. We now define, for 1 ≤ t ≤ n, (CF) t = the “in force” expected net cash flow in policy year t (9.2.1) We clearly have (CF) t = (P t −e t )(1 +i) . .. . accumulation of premium less expenses to end of year − q x+t−1 . .. . expected cost of death benefits D t − p x+t−1 . .. . expected cost of survival benefits S t (9.2.2) The “initial” expected net cash flow in year t, i.e. the net cash flow in year t per policy sold, is found from the important equation: “initial” net cash flow in year t = t−1 p x . .. . prob. that policy will still be in force at time t −1, i.e. at start of policy year t (CF) t (9.2.3) Example 9.2.1. An endowment assurance policy, with sum assured £5000, term five years and level annual premiums, is issued to a life aged 55. The annual premium is calculated on the following basis: 9.2. CASH FLOW CALCULATIONS 153 Mortality: A1967-70 ultimate Interest: 6% per annum Initial expenses: £250 Renewal expenses (associated with the payment of the second and each subsequent premium): £42 at the time of payment of the second premium, increasing thereafter by 5% per annum (compound). The death benefit is payable at the end of the year of death. (i) Show that the annual premium is £948.74 (ii) Assume that in calculating cash flows and profit signature for the policy the office uses the premium basis. On this basis determine for each year of the policy’s duration (a) the ‘in force’ expected cash flow, and (b) the ‘initial’ expected cash flow. Solution. (i) Let the annual premium be P. Then P¨ a 55:5 = 5000A 55:5 + 250 + 4 ¸ t=1 42(1.05) t−1 t p 55 v t (where all functions are on A1967-70 ultimate, 6%) , from which it follows that P = 5000 0.75171 + 250 + 152.62 4.386 = £948.74 154 CHAPTER 9. PROFIT-TESTING (ii) We draw up the following table: (1) (2) (3) (4) (5) (6) Year Premium Expenses Interest Death Cost Survival Cost t P t e t .06(P t −e t ) 5000q 54+t S t p 54+t 1 948.74 250.00 41.92 42.20 0.00 2 948.74 42.00 54.40 47.10 0.00 3 948.74 44.10 54.28 52.50 0.00 4 948.74 46.30 54.14 58.45 0.00 5 948.74 48.62 54.01 64.95 4935.05 (7) (8) (9) In force expected cash flow Prob. in force Initial expected cash flow (CF) t t−1 p 55 698.46 1.00000 698.46 914.04 .99156 906.33 906.42 .98222 890.30 898.13 .97191 872.90 −4045.87 .96054 −3886.22 Column(7) = (2) −(3) + (4) −(5) −(6) Column(9) = (7) (8), where column (8) gives the probability that the policy will be in force at the start of year t. 9.3. THE PROFIT VECTOR AND THE PROFIT SIGNATURE 155 9.3 The profit vector and the profit signature We now suppose that the office maintains suitable reserves t V at the end of each policy year. We assume that 0 V = 0 and n V = 0 (i.e., any survival benefit payable when the contract matures at time n years has been paid.) Corresponding to the “in force” and “initial” net cash flows, we have the profit vector and profit signature; that is, (PRO) t = cash which will, according to office’s projections, be transferred from the life fund to the with-profits policyhold- ers/shareholders at the end of policy year t, per policy in force at the start of that year (9.3.1) σ t = as for (PRO) t , but per policy sold (9.3.2) The relationship between these quantities is therefore σ t = t−1 p x (PRO) t (9.3.3) Calculation of (PRO) t We observe that (PRO) t = t−1 V(1 +i) . .. . accum. of reserves until end of year + (CF) t . .. . “in force” net cash flow in year t − p x+t−1 . t V . .. . money needed for reserves at end of the year (9.3.4) (since the chance that a given policy in force at the start of the policy year remains in force at the end is p x+t−1 .) This formula is sometimes written in the form (PRO) t = (CF) t + i t−1 V .... interest on reserve − (IR) t . .. . increase in reserves required (9.3.5) where (IR) t = p x+t−1 . t V − t−1 V (9.3.6) = money needed at end of the year for reserves, less money held at the start We may then calculate ¦σ t ¦ by formula (9.3.3). Example 9.3.1. Consider the policy described in example 9.2.1. Assuming that the premium and experience basis are as described in that question and that the reserves held are as follows: 0 V=0 156 CHAPTER 9. PROFIT-TESTING 1 V=919 2 V=1,876 3 V=2,873 4 V=3,914 5 V=0 Calculate the profit vector and profit signature. Solution. We draw up the following table: (1) (2) (3) (4) (5) (6) Year Reserve at start of year Survival prob. Increase in reserve Interest on reserve In force expected cash flow t t−1 V p 54+t (IR) t .06 t−1 V (CF) t 1 0.00 .99156 911.24 0.00 698.46 2 919.00 .99058 939.33 55.14 914.04 3 1876.00 .98950 966.83 112.56 906.42 4 2873.00 .98831 995.25 172.38 898.13 5 3914.00 .98701 −3914.00 234.84 −4045.87 (7) (8) (9) In force profit for year Prob. in force Profit signature (PRO) t t−1 p 55 σ t −212.78 1.00000 −212.78 29.85 .99156 29.60 52.15 .98222 51.22 75.26 .97191 73.15 102.97 .96054 98.91 Column(4) = p 54+t . t V − t−1 V Column(7) = (6) + (5) −(4) Column(9) = (7) (8), where column (8) is the probability that the policy will be in force at the start of year t. 9.4. THE ASSESSMENT OF PROFITS 157 Example 9.3.2. A life office issues a 3-year without profits endowment assurance policy to a life aged 62. The sum assured of £1, 500 is payable on maturity or at end of the year of death, if within 3 years, and there are level annual premiums of £472.50 payable in advance. The office uses the following “experience” basis: mortality: A1967-70 ultimate interest: 6% per annum initial expenses: £20 renewal expenses: £5 at the beginning of the second and third policy years. The office’s reserve basis is as follows: net premium method, using A1967-70 ultimate mortality and 3% p.a. interest. Determine the profit signature of this policy. Solution. We first work out the “in force” net cash flows, (CF) t . (1) (2) (3) (4) (5) (6) t Prem. Exp. Interest Death cost Maturity (CF) t = (1) + (2) = 0.06((1) −(2)) = 15, 000q 61+t Cost +(3) −(4) −(5) 1 472.50 20 27.15 26.62 0 453.03 2 472.50 5 28.05 29.48 0 466.07 3 472.50 5 28.05 ←−−−−− 1, 500 −−−−−→ −1,004.45 Now we find the profit vector, (PRO) t , and the profit signature, σ t . (1) (2) (3) (4) (5) (6) (7) Reserve at start of year Interest on Reserve p 61+t (IR) t (PRO) t = t−1 p 62 σ t t = 0.06(1) (CF) t + (2) −(4) = (5) (6) 1 0 0 0.98225 468.12 −15.09 1 −15.09 2 476.58 28.60 0.98035 466.07 14.71 0.98225 14.45 3 975.71 . .. . by net premium method do not use office premium in valuation 58.54 0.97826 −975.71 29.80 0.96295 28.70 9.4 The assessment of profits The profit signature ¦σ t ¦ can be assessed in one or more of the following ways. 158 CHAPTER 9. PROFIT-TESTING (1) One could work out the Internal Rate of Return (or yield) by solving the equation n ¸ t=1 v t σ t = 0 (9.4.1) (the internal rate of return, i 0 , being the solution of this equation.) (2) The shareholders may value the net profits at a certain rate of interest, j per annum. This rate is called the Risk Discount Rate, and may reflect uncertainties in ¦σ t ¦, with j normally higher than i. The net present value of the profits is thus NPV(j) = n ¸ t=1 v t σ t at rate j (9.4.2) (3) The Profit Margin is defined as n.p.v. of profits n.p.v. of premiums , both at some rate of interest, i m , say = ¸ n t=1 v t σ t ¸ n−1 t=0 P t+1t p x v t , at rate i m (9.4.3) (i m is often equal to the risk discount rate, j.) Example 9.4.1. A life office issues a three-year non-profit endowment assurance policy to a man aged 30. The sum assured is £60,000 on maturity or at the end of the year of earlier death. Level premiums of £19,000 are payable annually in advance. The office maintains reserves as follows: policy year reserve at end of policy year 1 £19,000 2 £38,000 The office expects that its life funds will earn interest at 7% p.a. over the next 3 years. The office expects expenses to be as follows: initial expenses: 10% of the first year’s premium, renewal expenses: 2% of later premiums. Mortality is expected to follow A1967-70 ultimate. Calculate (i) the profit signature; (ii) the net present value at the issue date of the profit to the office, using a risk discount rate of 10% p.a. 9.5. SOME THEORETICAL RESULTS ABOUT ¦σ T ¦ 159 Solution. (i) We work out the net cash flows, (CF) t , per policy in force at the start of the policy year. (1) (2) (CF) t t (P t −e t ) (P t −e t )(1 +i) death + mat. costs = (1) −(2) 1 17,100 18,297 39.22 18,258 2 18,620 19,923 40.26 19,833 3 18,620 19,923 60,000 −40,077 [ death cost = 60, 000q 30+t−1 mat.cost (in final year only) = 60, 000p 30+t−1 ] We now work out the profit vector. (1) (2) (3) (4) t t V (CF) t i. t−1 V p 30+t−1 . t V − t−1 V (PRO) t = (1) + (2) −(3) 1 19,000 18,258 0 18,987 −729 2 38,000 19,883 1330 18,975 2,238 3 0 −40,077 2660 −38,000 583 Hence profit signature is: −729 2, 238 0.999346 = 2, 237 583 0.998676 = 582 (ii) −729v + 2, 237v 2 + 582v 3 at 10% = £1, 623 9.5 Some theoretical results about ¦σ t ¦ Theorem 9.5.1. If all three bases defined in section 9.2 agree, σ t = (PRO) t = 0 for all t Proof. This follows from the formula (PRO) t = t−1 V(1 +i) + (CF) t −p x+t−1 . t V = 0 which is merely a restatement of the equation ( t−1 V +P t −e t )(1 +i) = q x+t−1 D t +p x+t−1 S t +p x+t−1 . t V which holds (cf chapter 7) since all three bases agree. Theorem 9.5.2. If the premium basis and experience basis agree, the internal rate of return earned by the office on a sale of a policy is equal to the rate i p.a. earned by the office’s life funds. 160 CHAPTER 9. PROFIT-TESTING Proof. We must show that n ¸ t=1 v t σ t = 0 at rate i p.a. Now we have n ¸ t=1 v t σ t = n ¸ t=1 v t [ t−1 p x (CF) t + t−1 p x . t−1 V(1 +i) − t p x . t V] But n ¸ t=1 v t t−1 p x (CF) t = 0 as this is merely a restatement of the equation of value for calculating premiums. Hence n ¸ t=1 v t σ t = n ¸ t=1 v t t−1 p x (1 +i) t−1 V − n ¸ t=1 v t t p x . t V = n−1 ¸ r=1 v r r p x . r V − n ¸ t=1 v t t p x . t V = 0, using the fact that 0 V = n V = 0 Corollary. If the life office wishes to obtain an I.R.R. on the sale of a policy in excess of the rate of interest it believes it will earn on the life funds, premiums must be higher than those calculated on the “experience” basis. Example 9.5.1. What is the I.R.R. associated with the profit signature of example 9.3.1? Solution. Since the premium and reserving bases agree, the I.R.R. must be i = 6% p.a. 9.6 Withdrawals So far we have ignored the possibility of surrender. We shall now assume that surrenders may occur, but only at the end of a policy year (just before payment of the premium then due). Let w t denote the chance that a policy will be surrendered at the end of year t (t = 1, 2, . . . , n − 1). We may assume that w n = 0, since the policyholder will receive the maturity benefits (if any) at that time. Let (SV) t denote surrender value at time t. We observe that (PRO) t is as before, except that there are additional profits due to the surrender of (p x+t−1 )w t . .. . prob. that a policy in force at time t −1 will be surrendered at time t [ t V −(SV) t ] . .. . profit at time t per surrender at that time (9.6.1) 9.6. WITHDRAWALS 161 Hence (PRO) t = profit vector at time t, allowing for surrenders = (PRO) t +p x+t−1 w t [ t V −(SV) t ] (9.6.2) Note that (PRO) t = (PRO) if t V = (SV) t for t = 1, 2, . . . , n −1. We must also adjust σ t , to allow for (i) the change from (PRO) to (PRO) , and (ii) the changed probability of the policy being in force at time t −1. We now have t−1 p x = probability that policy is in force at time t −1 = t−1 p x (1 −w 1 )(1 −w 2 ) . . . (1 −w t−1 ) (9.6.3) Hence σ t = revised profit signature = t−1 p x (PRO) t (9.6.4) Example 9.6.1. Consider the policy of Example 9.3.1, but now assume that at each of the durations 1,2,3,4 years, 5% of the surviving policyholders will surrender, and that the S.V. is 98% of the office’s reserve. Find the new profit signature. Solution. We work out the revised profit vector, (PRO) t : t (PRO) t +p 55+t−1 (0.05)(0.02 t V)∗ 1 −212.78 + 0.99156 .05 .02 919 = −211.87 2 29.85 + 0.99058 .05 .02 1, 876 = 31.71 3 52.15 + 0.98950 .05 .02 2, 873 = 54.99 4 72.26 + 0.98831 .05 .02 3, 914 = 79.13 5 102.97 + 0 = 102.97 * when t = 5, (PRO) t = (PRO) t . Hence we have: t (PRO) t t−1 p x σ t 1 −211.87 1 −211.87 2 31.71 .99156 .95 29.87 3 54.99 .98222 .95 2 48.75 4 79.13 .97191 .95 3 65.93 5 102.97 .96054 .95 4 80.56 Notes 1. The I.R.R. allowing for withdrawals may be found by solving the equation n ¸ t=1 v t σ t = 0 (9.6.5) 2. Even if (PRO) t = (PRO) t for t = 1, 2, . . . , n, ¦σ t ¦ is not the same as ¦σ t ¦ because the factors ¦ t−1 p x ¦ are not the same as ¦ t−1 p x ¦. 3. Unlike mortality rates, which are fairly predictable by actuaries, withdrawal rates depend to a great extent on economic and commercial factors which are less easy to forecast accurately. 162 CHAPTER 9. PROFIT-TESTING 9.7 The actual emergence of profits The actual profits emerging from the sale of a given policy depend, of course, on whether the policyholder dies (or surrenders the policy) during the term, and the year in which such an event occurs. Let us suppose here that there are no withdrawals, and use the symbols i t and e t to denote the actual interest rate earned by the life funds in year t, and the actual expenses incurred at the beginning of that policy year. The actual year-end profits emerging in the n years (maximum) of the policy are as follows: (1) If the policy is no longer in force at the beginning of policy year t, the profit emerging in that year is zero; (2) If the policy is still in force at the beginning of the year, the profit is ( t−1 V +P t −e t )(1 +i t ) −D t (9.7.1) if the life dies during the year, and ( t−1 V +P t −e t )(1 +i t ) −S t − t V (9.7.2) if the life survives the year. These formulae may be easily modified if the policy is altered, as in the following example. Example 9.7.1. A life office issues a 5-year with-profits endowment assurance policy to a life aged exactly 60. The policy has a basic sum assured of £10,000 payable at the end of the year of death or at the maturity date. Level premiums are payable annually is advance throughout the term of the policy. Simple reversionary bonuses vest at the start of each year, including the first. The premium is calculated according to the following basis: mortality: A1967-70 select interest: 4% per annum simple reversionary bonuses at the rate of 4% per annum are assumed initial expenses: 60% of the first premium renewal expenses: 5% of each premium after the first (i) Show that the premium is equal to £2,627. (ii) The office holds net premium reserves using a rate of interest of 3% per annum and A1967-70 ultimate mortality . Calculate the profit signature for this policy, assuming that the office will earn interest at 7% per annum on its assets, mortality follows the A1967-70 ultimate table, and expenses and bonuses will follow the premium basis. 9.7. THE ACTUAL EMERGENCE OF PROFITS 163 (iii) Immediately before the fourth premium was due (before the fourth bonus declaration) the policy was made paid-up, with no entitlement to further bonuses. The paid up sum assured was 60% of the benefits immediately before the alteration, including declared bonuses. The policyholder survived to the maturity date. Interest was earned on the life funds was at 6% per annum over the period of the contract, and bonuses in the first three years followed the premium assumptions. Expenses followed the premium assumptions up to the alteration date, and no expenses were incurred after the policy was made paid-up. For each of the five years of the policy term, calculate the actual year-end profit earned on the policy. Solution. (i) Let P be the annual premium. M.P.V. of premiums less expenses is 0.95P¨ a [60]:5 −0.55P = 3.7563P M.P.V. of benefits is 10, 000A [60]:5 + 400(IA) [60]:5 = 10, 000 0.82565 + 400 ¸ R [60] −R 65 −5M 65 + 5D 65 D [60] = 9867.88 Hence P = £2, 627.02 (ii) The reserves held are: 1 V = 10, 400A 61:4 −10, 000P 60:5 ¨ a 61:4 at 3% interest = 10, 000 1 − ¨ a 61:4 ¨ a 60:5 + 400A 61:4 = 2191.60 2 V = 10, 000 1 − ¨ a 62:3 ¨ a 60:5 + 800A 62:3 = 4, 475.67 3 V = 10, 000 1 − ¨ a 62:3 ¨ a 60:5 + 1200A 63:2 = 6, 862.31 4 V = 10, 000 1 − ¨ a 64:1 ¨ a 60:5 + 1600A 64:1 = 9, 366.17 164 CHAPTER 9. PROFIT-TESTING This gives the following table: Year Premiums Expenses Reserve t−1 V at start Interest Cost of Claims Cost of Reserves at End Profit per policy in force at start of year t (1) (2) (3) (4) (5) (6) (7) 1 2627.02 1576.21 0 73.56 150.10 2159.97 −1185.71 2 2627.02 131.35 2191.60 328.11 172.95 4404.00 438.43 3 2627.02 131.35 4475.67 487.99 198.80 6740.51 520.02 4 2627.02 131.35 6862.31 655.06 227.99 9182.08 602.97 5 2627.02 131.35 9366.17 830.33 12,000 0 692.17 Col. (4) = 0.07[(1) −(2) + (3)] Col. (5) = cost of death and maturity claims (including bonuses) Col. (6) = p 60+t−1 . t V Col. (7) = (1) −(2) + (3) + (4) −(5) −(6) Hence the profit signature for the contract is (−1185.71, 483.43p 60 . . . 692.17 4 p 60 ) = (−1185.71, 432.10, 504.31, 574.37, 646.38) (iii) We draw up the following table: Year P t −e t Reserve at start t−1 V Interest Reserve at end t V Survival benefits Profit t (1) (2) (3) (4) (5) (6) 1050.81 0 63.05 2191.60 0 −1077.74 2 2495.67 2191.60 281.24 4475.67 0 492.84 3 2495.67 4475.67 418.28 6862.31 0 527.31 4 0 6862.31 411.74 6524.25∗ 0 749.80 5 0 6524.25 391.46 0 6720 195.71 ∗6720A 64:1 at 3% = 6524.25; the sum assured after alteration is 0.6[10, 000 +3 400] = 6720, and the office calculates reserves by the net premium method. Col. (3) = 0.06[(1) + (2)] Col. (6) = (1) + (2) + (3) −(4) −(5) 9.8. EXERCISES 165 Exercises 9.1 (i) In the context of profit-testing, explain the difference between the “profit vector” and the “profit signature”. (ii) A certain life office sells assurance policies with term 3 years to lives aged 70. For each policy, the profit vector is estimated to be (−50, 30, 30). Given that the mortality of the policyholder is expected to follow A1967-70 ultimate, calculate (a) the profit signature per policy sold; (b) the net present value of the profit to the office on the basis of a risk discount rate of 8% per annum. 9.2 A life office issues a 5-year guaranteed bonus endowment assurance policy to a life aged 60, with basic sum assured £30,000. The sum assured, with attaching bonuses, is payable at the end of the year of death or at maturity. Level premiums are payable annually in advance. The office holds net premium reserves, using the basis A1967-70 ult at 3% p.a. Interest on premiums and reserves is expected to be earned at an effective rate of 8% p.a. Bonuses will be declared annually at a rate of 3% of the basic sum assured. Bonuses vest at the start of each policy year. Expenses of 40% of the first year’s premiums and 5% of subsequent years’ premiums will be incurred. Mortality is expected to follow A1967-70 ultimate. Withdrawals may be ignored. (i) For each policy year, calculate the total sum assured. (ii) For each duration t = 1, 2, 3, 4 years, calculate the reserve ( t V) immediately before pay- ment of the premium then due, given the following values on A1967-70 ultimate, 3% p.a. interest: (IA) 60:5 = 4.1853 (IA) 61:4 = 3.4681 (IA) 62:3 = 2.6973 (IA) 63:2 = 1.8672 (IA) 64:1 = v = 0.97087 (iii) (Difficult.) Calculate the annual premium (P) required for the shareholders/with profits policyholders to achieve an internal rate of return of 12% p.a. on the sale of this contract. 9.3 Ten years ago a life office issued a large block of 10-year without profits endowment assurances to lives then aged 30. Each policy was effected by annual premiums, and had a sum assured of £40,000, payable on survival or at the end of the year of death. The office’s premium basis was A1967-70 ultimate 5% interest expenses of 2% of all premiums with additional initial expenses of 0.5% of the sum assured. It was found that mortality , interest and expenses followed these assumptions, but there were surrenders just before payment of the premiums due at durations 1,2 and 3 years. At each of these times, 3% of the surviving policyholders surrendered their contracts, and were given a surrender value equal to the office’s reserve, which was calculated on the premium basis, minus a surrender penalty of £40, which the office transferred to the surplus account. By using a profit-testing approach, or otherwise, calculate the surplus accruing to the office at the end of each of the first three policy years, per policy sold. Hint. Note that (PRO) t = 0, so profits arise only from surrenders. 166 CHAPTER 9. PROFIT-TESTING 9.4 Your office is considering the issue of 3-year annual-premium endowment assurance policies without profits to lives aged 62. In respect of a policy with sum assured £10, 000, payable at the end of the year of death (if within 3 years) or on maturity, calculate the net present value of the profit signature on the following assumptions: premium basis: mortality: A1967-70 ultimate interest: 6% p.a. expenses: 3% of all premiums. reserve basis: net premium method using A1967-70 ultimate, 4% p.a. interest rate of interest to be earned in life fund: 8% p.a. expenses: 3% of office premiums mortality: A1967-70 ultimate risk discount rate: 10% p.a. 9.5 A life office issues 3-year term assurance policy to a man aged exactly 59. The sum assured is £15, 000, payable at the end of the year of death. Level premiums are payable annually in advance. Expenses are expected to be as follows: initial expenses: £10 renewal expenses: £2 incurred at the beginning of the 2 nd and each subsequent policy year. It is assumed that interest of 7% per annum will be earned on the life funds, and that mortality follows the A1967-70 ultimate table. The risk discount rate used by the office is 15% per annum. The office calculates the annual premium by requiring that the net present value of the expected profit on each policy is equal to 20% of one office premium. Calculate the office premium on each of the following reserving bases: (i) The office holds zero reserves at each year-end. (ii) The office sets up a reserve at each year-end (except the last) equal to 80% of one office premium. 9.9. SOLUTIONS 167 Solutions 9.1 (i) The profit vector refers to expected profits per policy in force at start of year, whilst the profit signature refers to profits per policy sold. Relationship is σ t = (PRO) t prob. that policy is in force at time t −1 (ii) (a) σ t = t−1 p 70 (PRO) t , so σ 1 = −50 σ 2 = 30 l 71 l 70 = 28.83 σ 3 = 30 l 72 l 70 = 27.59 (b) N.P.V. = σ 1 v +σ 2 v 2 +σ 3 v 3 at 8% = −50v + 28.83v 2 + 27.59v 3 = −46.30 + 24.72 + 21.90 = £0.32 9.2 Note: In (ii), we need the increasing assurance factors (A 65+t:5−t and ¨ a 65+t:5−t are directly tabulated) (i) policy year t total S.A. 1 30,900 2 31,800 3 32,700 4 33,600 5 34,500 (ii) The formula is t V = (30, 000 + 900t)A 60+t:5−t + 900(IA) 60+t:5−t −P 1 ¨ a 60+t:5−t on A67-70 ult. 3% where P 1 = net premium on A67-70 ult. 3% = 30, 000A 60:5 + 900(IA) 60:5 ¨ a 60:5 = 30, 000 0.86682 + 900 4.1853 4.572 = £6511.67 This leads to the table (note that 0 V = 5 V = 0): t t V 1 30, 900A 61:4 + 900(IA) 61:4 −P 1 ¨ a 61:4 = 6, 353 2 31, 800 62:3 + 900(IA) 62:3 −P 1 ¨ a 62:3 = 12, 947 3 32, 700 63:2 + 900(IA) 63:2 −P 1 ¨ a 63:2 = 19, 811 4 33, 600 64:1 + 900(IA) 64:1 −P 1 ¨ a 64:1 = 26, 983 (iii) We first work out (CF) t (in terms of P): 168 CHAPTER 9. PROFIT-TESTING (1)prs. less exp., with interest (2)death costs (3)survival costs (1)−(2)−(3) t (P −e t )(1.08) q 59+t D t p 59+t S t (CF) t 1 0.648P 445.96 0 0.648P −445.96 2 1.026P 509.23 0 1.026P −509.23 3 1.026P 580.42 0 1.026P −580.42 4 1.026P 660.39 0 1.026P −660.39 5 1.026P 34500 . .. . since sum paid on death or survival 1.026P −34, 500 We now work out (PRO) t : (1)interest on reserve (2)increase in reserve (3)cash flow (1)-(2)+(3) t 0.08 t−1 V (IR) t = p 59+t . t V − t−1 V (CF) t (PRO) t 1 0 6261.31 0.648P −445.96 0.648P −6707 2 508.24 6386.67 1.026P −509.23 1.026P −6388 3 1035.76 6512.36 1.026P −580.42 1.026P −6057 4 1584.88 6641.66 1.026P −660.39 1.026P −5717 5 2158.64 −26983.00 1.026P −34, 500 1.026P −5358 We have the equation 5 ¸ t=1 v t t−1 p 60 (PRO) t = 0 at 12% interest i.e. (0.648P −6707)v +(1.026P −6388)(0.985568)v 2 +(1.026P −6057)(0.969785)v 3 +(1.026P −5717)(0.952572)v 4 +(1.026P −5358)(0.93385)v 5 = 0 ∴ 3.2577P = 21, 488 ∴ P = £6, 596 (Rough check: right order of magnitude:- P =· P 1 , and 5P · 34, 500 (maturity benefit)). 9.3 Profits arise only in respect of surrenders at times 1,2,3. The profit at time t years (t = 1, 2, 3) per policy in force at start of year is (PRO) t = (1 −q 30+t−1 )(0.03) (40) . .. . surrender penalty = 1.2p 30+t−1 The probability of being in force at the start of year t is t−1 p 30 = t−1 p 30 (1 −0.03) t−1 t=1,2,3 ∴ Expected surplus arising at time t, per policy sold, is σ t = (1.2)0.97 t−1 t p 30 9.9. SOLUTIONS 169 t 1.2(0.97) t−1 t p 30 = σ t 1 1.2 0.99935 = £1.20 2 1.2 0.97 0.99868 = £1.16 3 1.2 0.97 2 0.99798 = £1.13 9.4 Let P = annual premium P = 10, 000P 62:3 .06 0.97 = 10, 000 0.30263 0.97 = £3, 119.90 Reserves: 0 V = 0, 1 V = 10, 000 1 V 62:3 .04 = 10, 000 1 − ¨ a 63:2 ¨ a 62:3 at 4% = 3, 146.38 2 V = 10, 000 1 − 5.138 6.519 = 6, 472.66, 3 V = 0 Net cash flows (ignoring reserves) t (CF) t 1 0.97P(1.08) −10, 000q 62 = 3, 090.91 2 0.97P(1.08) −10, 000q 63 = 3, 071.86 3 0.97P(1.08) −10, 000 = −6, 731.59 Profit vector and signature t t V (PRO) t = (CF) t +i. t−1 V −[p 62+t−1 . t V − t−1 V] t−1 p 62 σ t 1 3,146.38 3, 090.91 + 0 −3, 090.53 = 0.38 1 0.38 2 6,472.66 3, 071.86 + 251.71 −3, 199.06 = 124.51 0.98225 122.30 3 0 −6, 731.59 + 517.81 + 6, 472.66 = 258.88 0.96294 249.29 ∴ N.P.V. = 0.38v + 122.30v 2 + 249.29v 3 at 10% = £288.71 170 CHAPTER 9. PROFIT-TESTING 9.5 (i) Let A.P. be P (1) (2) (3) (4) (5) (6) Year Premiums Interest Expected (CF) t = (PRO) t t−1 p 59 σ t = (4) (5) less expenses = 0.07 ×(1) death costs = (1) + (2) −(3) 1 P−LO 0.07P −0.7 194.91 1.07P−205.61 1 1.07 P−205.61 2 P−2 0.07P −0.14 216.49 1.07P−218.63 0.987006 1.0561P−215.79 3 P−2 0.07P −0.14 240.20 1.07P−242.34 0.972761 1.0409P−235.74 We solve the equation: 0.2P = vσ 1 +v 2 σ 2 +v 3 σ 3 at 15% ∴ P = 205.61 1.15 −1 + 215.79 1.15 −2 + 235.74 1.15 −3 −0.2 + 1.07 1.15 −1 + 1.0561 1.15 −2 + 1.0409 1.15 −3 = 496.96 2.2133 = £224.53 (ii) (1) (2) (3) (4) (5) (6) Year Reserve Interest (CF) t Reserve Increase in (PRO) t = at start on reserve at end reserve of year = 0.07 ×(1) of year = p 59+t−1 . t V − t−1 V = (2) + (3) −(5) 1 0 0 1.07P−205.61 0.8P 0.7896 0.28039P−205.61 2 0.8P 0.056P 1.07P−218.63 0.8P −0.01155 1.13755P−218.63 3 0.8P 0.056P 1.07P−242.34 0 −0.8 1.926 P−242.34 (7) σ t = t−1 p 59 .(PRO) t 0.28039P−205.61 1.12277P−215.76 1.87354P−235.74 We solve the equation: 0.2P = vσ 1 +v 2 σ 2 +v 3 σ 3 at 15% ∴ P = £233.90 Chapter 10 STATIONARY POPULATIONS 10.1 Some Definitions We recall that ◦ e x = complete expectation of life at age x = E(T) where T refers to the future lifetime of (x) in years, including fractions. Note that ◦ e x = ∞ 0 t. t p x µ x+t dt (10.1.1) Integrating by parts gives ◦ e x = [t(− t p x )] ∞ 0 + ∞ 0 t p x dt = ∞ 0 t p x dt = ∞ 0 l x+t l x dt (10.1.2) (as it may be shown that lim t→∞ (t. t p x ) = 0 when ◦ e x is finite.) We recall also that e x = curtate expectation of life at age x = E(K) where K is the integer part of T. We find that e x = ∞ ¸ t=1 t p x (10.1.3) and, using the Euler–Maclaurin formula, e x · ◦ e x − 1 2 171 172 CHAPTER 10. STATIONARY POPULATIONS Define T x = ∞ 0 l x+t dt = ∞ x l y dy (on setting y = x +t) from which we obtain ◦ e x = ∞ x l y dy l x = T x l x and hence T x = l x ◦ e x (10.1.4) A stationary population is an idealised large population such that (in an old-fashioned mathe- matical notation) the number of lives in the population between ages x and x + dx is always equal to kl x dx where k is called the scaling factor and l x is according to some life table. By integrating over all ages greater than or equal to x, we can see that Total population aged ≥ x = ∞ x kl y dy = k.T x (10.1.5) The position is illustrated in Figure (10.1.1). Figure 10.1.1: a stationary population (It is nearly always advisable to draw a diagram in stationary population questions.) In any time- interval of length t years, the number of lives attaining a specified exact age x is kl x .t, so the rate at which lives attain age x is kl x per annum, i.e. kl x lives ‘flow’ continuously over exact age x each year. Note that (if lives are considered at all ages) there will be kl 0 births each year. Also, the num- ber of deaths each year between ages x 1 and x 2 (x 1 < x 2 ) is k(l x 1 −l x 2 ) (10.1.6) The number of lives in the population between any two ages x 1 and x 2 (x 1 < x 2 ) is k(T x 1 −T x 2 ) (10.1.7) 10.1. SOME DEFINITIONS 173 Example 10.1.1. The membership of a certain learned society is stationary at 11,500. Members enter only at exact age 50. They are subject to the mortality of English Life Table No.12 - Males, and there are no withdrawals. What is the annual number of entrants ? Solution One knows that k.T 50 = 11, 500 So kl 50 ◦ e 50 = 11, 500 The annual number of entrants is kl 50 = 11, 500 ◦ e 50 = 507.05 or 507. Example 10.1.2. A large company has for many years maintained a staff in a stationary condition by recruiting 500 annual entrants at exact age 20, uniformly over the year. If the staff retire at age 60, there are no withdrawals, and English Life No. 12 – Males mortality is experienced, find: (a) the size of the staff, (b) the number of staff who retire each year, (c) the number of pensioners. Solution (a) We have kl 20 = 500. The total number of staff is k(T 20 −T 60 ) = 500 l 20 (l 20 ◦ e 20 −l 60 ◦ e 60 ) = 19, 115. (b) Number who retire each year = kl 60 = 500 l 20 l 60 = 410 (c) Number of pensioners = kT 60 = 500 l 20 l 60 ◦ e 60 = 6, 172. 174 CHAPTER 10. STATIONARY POPULATIONS Some more symbols We let L x = the number of lives between ages x and x + 1 in a stationary population with scaling factor k = 1. = T x −T x+1 = 1 0 l x+t dt (10.1.8) We may use the approximations L x · l x+ 1 2 · 1 2 (l x +l x+1 ) Another definition is ◦ e x:n = average lifetime lived by (x) between ages x and x +n = E(T ∗ ) where the random variable T ∗ is defined as follows: T ∗ = T if (x) dies within n years n if (x) lives for n years. The variable T ∗ has p.d.f. t p x µ x+t for 0 < t < n, and there is discrete probability n p x that T ∗ = n. Hence E(T ∗ ) = n 0 t. t p x µ x+t dt +n. n p x = [−t. t p x ] n 0 + n 0 t p x dt +n. n p x (by integration by parts) = n 0 t p x dt Hence ◦ e x:n = T x −T x+n l x = ◦ e x − l x+n l x ◦ e x+n (10.1.9) (Notice that ¯ a x:n = ◦ e x:n if i = 0 .) Similarly, e x:n = average number of complete years lived by (x) between ages x and x +n = n ¸ t=1 t p x = l x+1 +l x+2 +... +l x+n l x =e x − l x+n l x e x+n (10.1.10) 10.2. THE CENTRAL DEATH RATE 175 Example 10.1.3. You are given that l x = l 1 ¸ 1 − log x 9 log 10 (1 ≤ x ≤ 11) Find ◦ e 1:10 . Solution ◦ e 1:10 = 10 0 l t+1 l 1 dt = 11 1 l x l 1 dx = 11 1 ¸ 1 − log x 9 log 10 dx = 10 − 1 9 log 10 [xlog x −x] x=11 x=1 = 10 − 11 log 11 −10 9 log 10 = 9.2097. The average age at death of those who die between ages x and x + n may be worked out as follows. We have E(T ∗ ) = E(T ∗ [T ∗ < n)Pr¦T ∗ < n¦ +E(T ∗ [T ∗ ≥ n)Pr¦T ∗ ≥ n¦ so that ◦ e x:n = E(T ∗ [T ∗ < n) n q x +n. n p x Hence the average age at death of those who die between ages x and x +n is x +E(T ∗ [T ∗ < n) =x + ◦ e x:n −n. n p x n q x =x + T x −T x+n −nl x+n l x −l x+n (10.1.11) Note that this result refers to a randomly-chosen life aged x: we need not assume that there is a stationary population. 10.2 The Central Death Rate Define h m x = the central death rate between ages x and x +h = l x −l x+h x+h x l y dy = number of deaths each year between ages x and x +h population aged between x and x +h in a stationary population If h = 1, it is omitted, giving m x = d x L x (10.2.1) 176 CHAPTER 10. STATIONARY POPULATIONS Note that m x = 1 0 l x+t µ x+t dt 1 0 l x+t dt · µ x+ 1 2 (10.2.2) 10.3 Relationships Between m x and q x Assume that there is a Uniform Distribution of Deaths (U.D.D.) between ages x and x+1, i.e. l x+t is linear for 0 ≤ t ≤ 1. We have L x = 1 0 l x+t dt = 1 2 (l x +l x+1 ) (10.3.1) (as the trapezoidal rule is exactly correct, not just an approximation). Since l x+1 = l x −d x , we have L x = l x − 1 2 d x (10.3.2) Hence m x = d x L x = d x l x − 1 2 d x = q x 1 − 1 2 q x (10.3.3) We may rearrange this equation to get q x = m x 1 + 1 2 m x (10.3.4) If U.D.D. does not hold, these results may be used as approximations. 10.4 Stationary Funds Consider a pension scheme, etc., such that B = annual benefit outgo for scheme (plus expenses, if any) and C = annual contribution income are constant. If the funds, F, are such that the interest on them pays B −C, then the funds will also remain constant. There are two cases: (a) If income and outgo are received and paid continuously, then δ.F = B −C (10.4.1) (B, C are rates of payment per annum). (b) If income and outgo are received and paid only at the end of each year, then iF = B −C (10.4.2) Note: In case (b), F denotes the fund at the start of the year. 10.4. STATIONARY FUNDS 177 Example 10.4.1. Each year for many years a life office has issued 10,000 temporary assurance policies each with a term of ten years and a sum assured of £5,000 to lives aged 25 exactly. One third of those who survive to age 35 then effect a without profits whole life policy for the same sum assured as the term policy, and one quarter effect a 25-year without profits endowment assurance for twice that sum assured. All premiums are payable annually in advance and death claims are paid at the end of the year of death. Policies are issued uniformly throughout the year. The office calculates premiums on A1967-70 ultimate 4%, ignoring expenses. If the office’s ex- perience follows this basis, calculate the size of the fund held for these contracts. Solution The term assurance premium = 5, 000 M 25 −M 35 N 25 −N 35 = £3.29. Whole life premium = 5, 000P 35 = £56.85. Endowment assurance premium = 10, 000P 35:25 = £245.30. Annual premium income is 10, 000 ¸ 1 +e 25 − l 35 l 25 (1 +e 35 ) 3.29 + 10, 000 l 35 l 25 1 3 (1 +e 35 ) 56.85 + 10, 000 l 35 l 25 1 4 ¸ 1 +e 35 − l 60 l 35 (1 +e 60 ) 245.30 = 328, 003 + 7, 634, 065 + 14, 769, 057 = £22, 731, 125. Annual rate of payment of claims is 10000 ¸ 5000 l 25 −l 35 l 25 + l 35 l 25 1 3 5000 + 1 4 10, 000 = £41, 723, 716. Hence δ.F = 41, 723, 716 −22, 731, 125 = 18, 992, 591 where i = 0.04. Therefore Fund = 18992591 δ = £484, 249, 000. 178 CHAPTER 10. STATIONARY POPULATIONS Exercises 10.1 The staff of a large company is maintained as a stationary population by 500 new entrants each year at exact age 20. One third of those reaching age 30 leave immediately. Of the remainder, 1 4 of those attaining age 60 retire immediately and the survivors retire at age 65. The only other decrement is death. Calculate (i) the number of staff, (ii) the number of deaths in service each year. Basis: English Life Table No. 12 - Males 10.2 A certain country’s school system provides education for all children between the ages of 5 and 16 exactly. The country’s population is stationary, there being 100,000 births per year, uniformly distributed over the year. The population of the country is subject to the mortality of English Life Table No. 12 - Males. (i) Find the number of pupils at any given time. (ii) The country’s teacher training colleges are such that a constant flow of new entrants to the profession is maintained. Teachers are recruited uniformly over the year, and the ratio of pupils to teachers is 20 to 1. All teachers enter the profession at age 21 and retire at age 60, there being no withdrawals. Find the annual number of new teachers recruited. 10.3 For many years a company has recruited, uniformly over each year, 200 employees on their 20th birthdays and a fixed number of additional employees on their 25th birthdays. Mor- tality has followed English Life Table No. 12 - Males. Employees may retire on their 60th or 65th birthdays, and one third of employees reaching their 60th birthdays retire on that date. Employees leave the company only through death or retirement, and the total number of employees is 10,000. Find the total number of new recruits each year. 10.4 The male population of a certain country has been stationary for many years, there being 100,000 male births per annum, spread uniformly over the year; the mortality of males follows English Life Table No.12 - Males, and migration may be ignored. (a) Calculate the size of the country’s male population at any time. (b) The government of the country has decided to introduce a social security plan, under which all employed men between ages 15 and 65 must contribute a fixed sum every week, the same sum being also payable by their employers. Men over age 65 will receive a pension of 100 units of currency per week, and those sick or unemployed between ages 15 and 65 also receive this amount. If it may be assumed that at any time 95% of men between ages 15 and 65 are employed, while the remaining 5% are sick or unemployed, calculate the weekly contribution payable by each employed worker. The scheme is to be financed on a pay-as-you-go basis (no fund is built up) and administrative costs are to be ignored. 10.5 A large manufacturing company has for many years staffed one of its divisions by the recruit- ment, uniformly over each year, of 1,000 staff at exact age 20. At the end of one year in the job, new staff are examined for suitability, and 20% are dismissed. All employees are assessed at age 35, and 50% are immediately moved out of the division. At age 40, all remaining em- ployees are moved out of the division. Death is the only other reason for leaving the division. Staff experience mortality according to English Life Table No. 12 - Males. Calculate the number of staff in the division. 10.5. EXERCISES 179 10.6 In a certain country, all civil servants are recruited on their 25th birthdays. (Birthdays are assumed to occur uniformly over the calendar year). Of those who reach exact age 40, 10% obtain employment in private companies and leave the civil service immediately. All remaining civil servants retire when they reach age 60 or die in service before this age. Civil servants have for many years experienced the mortality of English Life Table No. 12 - Males, and will do so for the indefinite future. The population of civil servants has been stationary for many years, and civil servants are recruited at the rate of 3,000 each year. (i) How many civil servants are in service at any given date? (ii) The government of the country has just decided to reduce the size of the civil service by 10%. This is to be done by immediately lowering the retirement age of civil servants. Find the new retirement age (to the nearest month), assuming that there are no other changes. 10.7 In a mortality table with a one-year select period, the following relationships apply at age x: l [x] = 1 2 (l x +l x+1 ) t q [x] = t.q [x] (0 ≤ t ≤ 1) (i) Express p [x] in terms of p x . (ii) Express e [x] in terms of e x and p x . (iii) Express ◦ e [x] in terms of ◦ e x+1 and p x . (iv) Express m [x] in terms of p x . 10.8 For many years a life office has issued a steady flow of 10-year endowment assurances without profits to lives aged 55. Premiums are payable continuously and the sum assured is payable immediately on death, if death occurs before age 65. Premiums are calculated on the following basis: mortality: English Life Table No. 12 – Males interest: 4% expenses: 5% of all office premiums, plus 25% of office premiums in the first year. Mortality, interest and expenses have always been in accordance with this basis. The sums assured issued each year have always been £5,000,000, spread uniformly over the year. No policies terminate except by death or maturity. In respect of this business, calculate (i) the annual rate of gross premium income, (ii) the annual rate of payment of maturity claims, (iii) the annual rate of payment of death claims, and (iv) the size of the fund which has been built up. 180 CHAPTER 10. STATIONARY POPULATIONS Solutions 10.1 (i) kl 20 = 500. Number of Staff = k T 20 − 1 3 T 30 − 1 6 T 60 − 1 2 T 65 = 500 l 20 l 20 ◦ e 20 − 1 3 l 30 ◦ e 30 − 1 6 l 60 ◦ e 60 − 1 2 l 65 ◦ e 65 = 15, 361 (ii) Number of deaths is 500 − 500 l 20 1 3 l 30 + 1 6 l 60 + 1 2 l 65 = 89. 10.2 (i) kl 0 = number of births each year = 100,000 As l 0 = 100, 000, k = 1. Number of pupils = k(T 5 −T 16 ) = l 5 ◦ e 5 −l 16 ◦ e 16 = 1, 066, 409 (ii) Number of teachers = 1 20 1, 066, 409 = 53, 320. Let annual number of recruits be c. Then the equation is c l 21 (T 21 −T 60 ) = 53, 320 Hence c = 1, 431 teachers per annum. 10.3 kl 20 = 200 employees enter at age 20 per annum. Let cl 25 be the number who enter at age 25 per annum. Then the number of employees is k T 20 − 1 3 T 60 − 2 3 T 65 +c T 25 − 1 3 T 60 − 2 3 T 65 = 10, 000 where k = 200 l 20 . Solving this for c gives c = 10, 000 −8157.81 3, 447, 483 = 0.00053436 Hence number recruited each year = 200 +cl 25 = 251 10.4 (a) kl 0 = 100, 000 Hence k = 1. 10.6. SOLUTIONS 181 Total male population = kT 0 = 100, 000 ◦ e 0 = 6, 809, 000 (b) Number of pensioners = kT 65 Number aged between 15 and 65 who receive benefit = 0.05k(T 15 −T 65 ). Number of contributors = 0.95k(T 15 −T 65 ). Let the employee s contribution be c per week. Hence 2c (number of contributors) = 100 (no. of beneficiaries) Therefore c = 100[T 65 + 0.05(T 15 −T 65 )] 2 0.95(T 15 −T 65 ) T 15 = 5, 352, 735 and T 65 = 818, 455 giving c = 12.13 units. 10.5 kl 20 = number of staff recruited per annum = 1, 000 Number of staff = k(T 20 −T 21 ) + 0.8k(T 21 −T 35 ) + 0.4k(T 35 −T 40 ) = 1000 l 20 (T 20 −0.2T 21 −0.4T 35 −0.4T 40 ) = 14, 060 10.6 (i) kl 25 = 3, 000 Number of civil servants is k(T 25 −0.1T 40 −0.9T 60 ) = 3000 l 25 (l 25 ◦ e 25 −0.1l 40 ◦ e 40 −0.9l 60 ◦ e 60 ) = 94713 (ii) Number of civil servants required is 0.9 94713 = 85, 242 Let x be the new age of retirement. Then 3000 l 25 (T 25 −0.1T 40 −0.9T x ) = 85, 242 Hence T x = 1, 524, 475 T 55 = 1, 602, 333 T 56 = 1, 516, 571 So using linear interpolation gives x · 55.91 · 55 years and 11 months 182 CHAPTER 10. STATIONARY POPULATIONS 10.7 (i) p [x] = l x+1 l [x] = l x+1 1 2 (l x +l x+1 ) = 2p x 1 +p x (ii) e [x] = l x+1 +l x+2 +... l [x] = l x l [x] e x = p [x] p x e x = 2 1 +p x e x (iii) ◦ e [x] = 1 l [x] [ ∞ 0 l [x]+t dt] = 1 l [x] [ 1 0 l [x]+t dt + ∞ 1 l x+t dt] = 1 l [x] [ 1 2 (l [x] +l x+1 ) +l x+1 . ◦ e x+1 ] (as t q [x] = t.q [x] , l [x]+t is linear for 0 ≤ t ≤ 1) = 1 2 (1 +p [x] ) +p [x] . ◦ e x+1 = 1 2 1 + 3p x 1 +p x + 2p x 1 +p x ◦ e x+1 (iv) m [x] = 1 0 l [x]+t µ [x]+t dt 1 0 l [x]+t dt = l [x] −l x+1 1 2 (l [x] +l x+1 ) = 1 −p [x] 1 2 (1 +p [x] ) = 2(1 −p x ) 1 + 3p x (using part (i)) 10.8 (i) Let P be the office annual premium per £1 sum assured. We have 0.95P ¯ a 55:10 = A 55:10 + 0.25P ¯ a 55:1 Hence P = 0.100663. The total sums assured in force at any given time are 5, 000, 000 l 55 (T 55 −T 65 ) (£5,000,000 “flows across” age 55, so the scaling factor is 5, 000, 000 l 55 .) Total sums assured = 5, 000, 000 ◦ e 55 − l 65 l 55 ◦ e 65 = £45, 618, 854. 10.6. SOLUTIONS 183 Hence rate of gross premium income = £4, 592, 130 p.a. (ii) 5, 000, 000 l 65 l 55 = £3, 985, 870 (iii) 5, 000, 000 −3, 985, 870 = £1, 014, 130 (iv) Annual rate of payment of benefits = £5, 000, 000 Annual rate of payment of expenses = 5% of gross premium income + 25% of gross premium income in first policy year = 0.05 4, 592, 130 + 0.25 ¸ 5, 000, 000 ◦ e 55 − l 56 l 55 ◦ e 56 0.100663 = £355, 209 Let funds be F. We have δF = 5, 000, 000 + 355, 209 −4, 592, 130 Hence F = £19, 456, 000 (final figures are unreliable.) 184 CHAPTER 10. STATIONARY POPULATIONS Chapter 11 JOINT-LIFE FUNCTIONS 11.1 Joint-Life Mortality Tables Consider 2 independent lives aged x and y respectively, subject to the mortality of Tables I and II respectively. Let T = min¦T 1 , T 2 ¦, where T 1 = future lifetime of (x) and T 2 = future lifetime of (y). That is, T is the “joint future lifetime” of (x), (y), i.e. the time until the first of them dies. Remark If (x) and (y) are a married couple, or business partners, the assumption of independence of T 1 , T 2 is questionable: consider, for example, a car accident in which both lives are killed. But in practice independence is normally assumed. Define t p I II xy = Pr¦both (x), (y) will survive for t years at least¦ = t p I x t p II y (by independence of T 1 , T 2 ) (11.1.1) and t q I II xy = Pr¦T ≤ t¦ = the distribution function of T (11.1.2) = 1 − t p I II xy (11.1.3) Note A common error is to write t q I II xy = t q I x t q II y If Table I = Table II, we may omit the superscripts, so (for example) t p xy = t p x t p y on the table under consideration. Note In the a(55) tables, the notation t p xy , etc. refers to a male life aged x and a female life aged y (the male life comes first.) That is, Table I is a(55) males ultimate, Table II is a(55) females ultimate. 185 186 CHAPTER 11. JOINT-LIFE FUNCTIONS From now on we will omit “I, II” (unless needed.) Define l xy = l x l y (x ≥ α 1 , y ≥ α 2 , where α 1 , α 2 are the youngest ages in tables I, II respectively). By equation 11.1.1, t p xy = l x+t:y+t l xy Note that t q xy = 1 − t p xy = 1 − t p x . t p y = 1 −(1 − t q x )(1 − t q y ) = t q x + t q y − t q x . t q y (11.1.4) Using equation 11.1.3, t q xy = l xy −l x+t:y+t l xy and d xy = l xy −l x+1:y+1 . The force of mortality is defined as µ xy = lim h→0 + h q xy h = lim h→0 + h q x h + h q y h − h q x h h q y h h = µ x +µ y (11.1.5) (since h q x h →µ x , h q y h →µ y as h →0 + ). Example 11.1.1. Given l xy = 1, 000 l x+10:y = 960 l x:y+10 = 920 calculate the probability that, of two lives aged x and y, only one will survive for 10 years. Solution 10 p x = l x+10:y l x:y = 0.96 10 p y = l x:y+10 l xy = 0.92 Required probability = 10 p x (1 − 10 p y ) + 10 p y (1 − 10 p x ) = 0.1136. 11.1. JOINT-LIFE MORTALITY TABLES 187 Theorem t p xy = exp(− t 0 µ x+r:y+r dr) (11.1.6) Proof. From (11.1.1), t p xt p y = exp[− t 0 µ x+r dr] exp[− t 0 µ y+r dr] = exp[− t 0 (µ x+r +µ y+r ) dr] = exp[− t 0 µ x+r:y+r dr] by (11.1.5) The probability density function (pdf ) of T. Note that F(t) = Pr¦T ≤ t¦ = t q xy is the distribution function of T. (When t < 0, the distribution function of T is zero). Now f(t) = p.d.f. of T = F (t) = d dt [1 − t p xy ] (for t > 0) = − d dt ¸ exp − t 0 µ x+r:y+r dr (from (11.1.6)). = exp ¸ − t 0 µ x+r:y+r dr µ x+t:y+t (using Chain Rule) Thus f(t) = t p xy µ x+t:y+t (t > 0) 0 (t < 0) (11.1.7) In particular, we must have ∞ 0 t p xy µ x+r:y+r dr = ∞ 0 t p xy (µ x+t +µ y+t ) dt = 1 since the integral of a p.d.f. is always equal to 1. The discrete variable K = integer part of T This is the number of complete years to be lived in the future by the “joint life status” of (x, y). Note that K = min¦K 1 , K 2 ¦. 188 CHAPTER 11. JOINT-LIFE FUNCTIONS where K 1 = integer part of T 1 , K 2 = integer part of T 2 . This variable is discrete, with probabilities Pr¦K = k¦ = Pr¦1st death occurs between times k and k + 1¦ = k [q xy = k p xy − k+1 p xy (by an argument similar to that for a single life) = l x+k:y+k −l x+k+1:y+k+1 l xy = d x+k:y+k l xy (k = 0, 1, 2 . . .) (11.1.8) 11.2 Select Tables We replace (x), (y) by [x] + r, [y] + u (referring to lives aged x + r, y + u respectively who were selected r, u years ago respectively). Define l [x]+r:[y]+u = l [x]+r l [y]+u which leads to t p [x]+r:[y]+u = l [x]+r+t:[y]+u+t l [x]+r:[y]+u Example 11.2.1. On A67 −70 calculate 3 p [59]+1:[69]+1 . Solution 3 p [59]+1:[69]+1 = l [59]+4:[69]+4 l [59]+1:[69]+1 = l 63 .l 73 l [59]+1:[69]+1 = 0.856. We also define µ [x]+r:[y]+u = lim h→0 + h q [x]+r:[y]+u h = µ [x]+r +µ [y]+u The p.d.f. of T = future lifetime of ([x] +r, [y] +u) is f(t) = t p [x]+r:[y]+u .µ [x]+r+t:[y]+u+t (t ≥ 0) 0 (t < 0). 11.3 Extensions to More than 2 Lives We extend the definitions of T, t p xy , etc., as follows: consider 3 lives aged x, y, z respectively, whose future lifetimes are T 1 , T 2 , T 3 respectively (assumed to be independent). t p xyz = Pr¦(x), (y) and (z) all survive for at least t years¦ = t p x . t p y . t p z = l x+t:y+t:z+t l xyz 11.3. EXTENSIONS TO MORE THAN 2 LIVES 189 where l xyz = l x l y l z for all x, y, z ≥ α 1 , α 2 , α 3 (α 1 , α 2 , α 3 being the youngest ages in tables I, II, III). Other functions follow similarly; in particular µ xyz = lim h→0 + h q xyz h = µ x +µ y +µ z and T = min¦T 1 , T 2 , T 3 ¦ has p.d.f. f(t) = t p xyz µ x+t:y+t:z+t = t p xyz (µ x+t +µ y+t +µ z+t ) (t ≥ 0) 0 (t < 0) Example 11.3.1. Evaluate ∞ 0 t p xxxx µ x+t:x+t dt Solution ∞ 0 t p xxxx µ x+t:x+t dt = 0.5 ∞ 0 t p xxxx (4µ x+t ) dt. = 0.5. To find the probability that exactly one life out of three survives for t years, use the formula Pr¦(x) survives but (y), (z) die¦ +Pr¦(y) survives but (x), (z) die¦ +Pr¦(z) survives but (x), (y) die¦ (note that these events are mutually exclusive) = t p x (1 − t p y )(1 − t p z ) + t p y (1 − t p x )(1 − t p z ) + t p z (1 − t p x )(1 − t p y ) Similar calculations apply to Pr¦exactly one life dies¦ and other possibilities. Also Pr¦at least 1 dies within t years¦ = t q xyz = 1 −Pr¦all 3 survive¦ = 1 − t p xyz = 1 − t p xt p yt p z = 1 −(1 − t q x )(1 − t q y )(1 − t q z ). Example 11.3.2. The probability that exactly one of three lives aged x will survive for n years is 27 times the probability that all three will die within n years. Find the probabilities that (a) at least two will survive n years (b) at least one will die within n years Solution (a) 3 n p x (1 − n p x ) 2 = 27(1 − n p x ) 3 thus n p x = 9(1 − n p x ) hence n p x = 0.9, so n q x = 0.1. 190 CHAPTER 11. JOINT-LIFE FUNCTIONS By the binomial distribution, the probability of at least two survivors = 3 2 ( n p x ) 2 (1 − n p x ) + ( n p x ) 3 = 3( n p x ) 2 −2( n p x ) 3 = 0.972 (b) Pr¦at least one will die¦ = 1 −( n p x ) 3 = 1 −(0.9) 3 = 0.271 11.4 The Joint Expectation of Life Define ◦ e xy = the complete joint expectation of life for the pair aged (x, y) = E(T), where T = min¦T 1 , T 2 ¦ as before. Hence ◦ e xy = ∞ 0 t. t p xy .µ x+t:y+t dt Proceeding as in single-life case, using integration by parts, we get ◦ e xy = ∞ 0 t p xy dt (11.4.1) We also have e xy = curtate joint expectation of life of (x, y) = E(K) where K = integer part of T = ∞ ¸ t=1 t p xy (11.4.2) The Euler–Maclaurin formula gives the approximation e xy · ◦ e xy − 1 2 The variance of T is calculated as follows: Var(T) = E(T 2 ) −[E(T)] 2 = ∞ 0 t 2 . t p xy µ x+t:y+t dt −( ◦ e xy ) 2 = ∞ 0 2t. t p xy dt − ¸ ∞ 0 t p xy dt 2 (Use integration by parts; similar to the single-life case). 11.5. MONETARY FUNCTIONS 191 11.5 Monetary Functions Notation is very similar to single-life case (changing ‘x’ to ‘xy’). See “International Actuarial Nota- tion” in “Formulae and Tables for Actuarial Examinations”. (a) Annuities We define ¯ a xy = mean present value (m.p.v.) of an annuity of £1 per annum (p.a.) payable continuously so long as both (x), (y) are alive = E[¯ a T ] where T = future lifetime of joint-life status = ∞ 0 ¯ a¯ t| . t p xy µ x+t:y+t dt = ∞ 0 v t t p xy dt (11.5.1) (the proof is similar to single-life case; note that v = 1 1+i ). Note also that ¨ a xy = m.p.v. of an annuity of £1 p.a. payable at the start of each year so long as both (x), (y) are alive = E[¨ a K+1 ] where K = integer part of T = ∞ ¸ t=0 v t t p xy . (11.5.2) Other joint-life annuity functions are a xy = ∞ ¸ t=1 v t . t p xy = ¨ a xy −1 ¨ a xy:n = ¨ a xy − n p xy v n ¨ a x+n:y+n Woolhouse’s approximation for annuities payable m thly hold for joint lives, i.e. ¨ a (m) xy · ¨ a xy − m−1 2m ¨ a (m) xy:n · ¨ a xy:n − m−1 2m (1 − n p xy v n ) The Euler–MacLaurin formula gives ¯ a xy · ¨ a xy − 1 2 · a xy + 1 2 . Evaluation of annuities using commutation functions. Commutation functions are not often used, and the only functions available in the examination 192 CHAPTER 11. JOINT-LIFE FUNCTIONS tables are D xy = v 1 2 (x+y) l xy (11.5.3) N xy = ∞ ¸ t=0 D x+t:y+t (11.5.4) From these we get n E xy = m.p.v. of pure endowment of £1 payable at time n if (x), (y) are both alive = v n . n p xy = D x+n:y+n D xy and ¨ a xy = ∞ ¸ t=0 v t l x+t:y+t l xy = N xy D xy . Note: In A67 −70 we only find D xx and N xx (i.e. equal ages only.) (b) Assurances ¯ A xy = m.p.v. of £1 payable immediately on the failure of the joint-life status = E(v T ) where T = future lifetime of joint-life status = ∞ 0 v t . t p xy µ x+t:y+t dt (11.5.5) This can be evaluated numerically by approximate integration. Note that ¯ a xy = E 1 −v T δ = 1 − ¯ A xy δ so we have the “conversion” relationship ¯ A xy = 1 −δ¯ a xy . We also have A xy = E[v K+1 ] = ∞ ¸ t=0 v t+1 t [q xy (11.5.6) and it will now be shown that A xy = 1 −d¨ a xy Proof ¨ a xy = E[¨ a K+1 ] = E 1 −v K+1 d = 1 d (1 −A xy ) 11.5. MONETARY FUNCTIONS 193 So A xy = 1 −d.¨ a xy An approximation which can often be used is A xy · (1 +i) − 1 2 ¯ A xy Write ¯ A 1 .... xy :n to indicate that a sum assumed is payable on the first death of (x), (y) if within n years. Example 11.5.1. Show how one can evaluate ¯ A 1 .... xy :n on the a(55) tables (male/female.) Solution. Method 1: ¯ A 1 .... xy :n · (1 +i) 1 2 A 1 .... xy :n = (1 +i) 1 2 [A xy:n −v n n p xy ] (as in single-life case). = (1 +i) 1 2 1 −d¨ a xy:n −v n l x+n l x l y+n l y Now use ¨ a xy:n = ¨ a xy −v n l x+n l x l y+n l y ¨ a x+n:y+n and finally use ¨ a xy = a xy + 1. Method 2: ¯ A 1 .... xy :n = ¯ A xy:n −v n n p xy = (1 −δ¯ a xy:n ) −v n l x+n l x l y+n l y · 1 −δ(a xy + 1 2 ) −v n l x+n l x l y+n l y ¸ 1 −δ(a x+n:y+n + 1 2 ) . (c) Premiums We employ equations of value in the same way as for single-life policies. If there are no expenses we obtain net premiums, e.g., in the International Actuarial Notation, ¯ P( ¯ A xy ) = ¯ A xy ¯ a xy = 1 ¯ a xy −δ If there are expenses, calculate office premiums as for single-life policies by setting up an equation of value. The general symbols P and P may be used for any net or gross premium. 194 CHAPTER 11. JOINT-LIFE FUNCTIONS (d) Reserves Calculated as for single-life policies, e.g., t ¯ V ( ¯ A xy ) = net premium reserve at duration t years for the joint-life assurance of £1, payable immediately on the first death of two lives aged x and y at the issue date, affected by annual premiums payable continuously during the joint-lifetime. = ¯ A x+t:y+t − ¯ P( ¯ A xy )¯ a x+t:y+t (using prospective method). Note: The formula for retrospective reserve in this example is D xy D x+t:y+t [P¯ a xy:t −A 1 .... xy :t ]. Also, one may show that t V (A xy ) = 1 − ¯ a x+t:y+t ¯ a xy using conversion relationships. We use the general symbol t V for the reserve at duration t, just before payment of any premium then due. Exam tables (i) A67-70 gives: ¨ a xx , N xx , D xx , ¨ a [xx] (= ¨ a [x]:[x] ), ¨ a xxx , ¨ a xxxx all at 4% p.a. interest. (ii) a(55) gives a xy at 2-year intervals for male aged x and female aged y (x, y ≥ 60 only), at 4%, 6% and 8% interest. Interpolation is needed to get (for example) a 61:69 , as follows: a 61:69 · 1 4 [a 60:68 +a 62:68 +a 60:70 +a 62:70 ]. Example 11.5.2. Two lives, aged 50 and 60, effect a 10-year temporary joint life assurance with sum assured £60,000. The sum assured is payable immediately on the first death, provided that this occurs within 10 years of the issue date. The policy has annual premiums, payable during the joint lifetime of both lives for at most 10 years. (i) Express ¯ A 1 50 : 60:10 as an integral. Estimate the value of the integral by the following form of Simpson’s rule: 10 0 f(t) dt = 10 6 [f(0) + 4f(5) +f(10)] (ii) Hence, or otherwise, estimate the values of A 50:60:10 and ¨ a 50:60:10 , and calculate the annual premium for the policy. Basis: A1967 −70 ultimate mortality 4% p.a. interest No expenses. 11.6. LAST SURVIVOR PROBABILITIES (TWO LIVES ONLY) 195 Solution (i) ¯ A 1 50 : 60:10 = 10 0 v t . t p 50:60 (µ 50+t +µ 60+t ) dt. Simpson’s rule gives an approximate answer of ¯ A 1 50 : 60:10 = 0.2238. (ii) A 50:60:10 = A 1 50 : 60:10 +v 10 l 60 l 50 l 70 l 60 · (1.04) − 1 2 . ¯ A 1 50 : 60:10 +v 10 l 70 l 50 = 0.70792. ¨ a 50:60:10 = 1 −A 50:60:10 d = 7.594. The premium P p.a. is P = 60000 ¯ A 1 50 : 60:10 ¨ a 50:60:10 = £1, 768.24 11.6 Last Survivor Probabilities (two lives only) Define, t p xy = Pr¦at least 1 of (x), (y) will be alive in t years’ time¦ = 1 − t q xy . where t q xy = Pr¦last survivor of (x), (y) dies within t years¦. By elementary probability theory, t p xy = t p x + t p y − t p xy and hence t q xy = t q x + t q y − t q xy These formulae are true even if (x), (y) are not independent. If they are independent, we may write t p xy = t p x t p y giving t p xy = t p x + t p y − t p x t p y (11.6.1) and t q xy = t q x t q y . (11.6.2) Note that t q xy is the distribution function of the variable T max = max¦T 1 , T 2 ¦. and the p.d.f. of T max = time to death of last survivor is f(t) = d dt ( t q xy ) = d dt ( t q x ) + d dt ( t q y ) − d dt ( t q xy ) 196 CHAPTER 11. JOINT-LIFE FUNCTIONS Assume independence of T 1 , T 2 . We get: f(t) = t p x µ x+t + t p y µ y+t − t p xt p y (µ x+t +µ y+t ) (t > 0) Now the “force of mortality” has to be defined as follows: µ xy (t) = hazard rate at time t = f(t) 1 −F(t) where f(t) = the p.d.f. of T max , F(t) = the distribution function of T max . This gives µ xy (t) = t p x µ x+t + t p y µ y+t − t p x . t p y (µ x+t +µ y+t ) t p xy Note This is not often used in practice. 11.7 Last Survivor Monetary Functions (a) Annuities Define, for example, ¨ a xy = m.p.v. of an annuity of £1 p.a. payable annually in advance so long as at least one life is alive. On multiplying the equation t p xy = t p x + t p y − t p xy by v t , and summing or integrating over t, one obtains the following relationships for annuity functions ¨ a xy = ¨ a x + ¨ a y − ¨ a xy ¯ a xy = ¯ a x + ¯ a y − ¯ a xy (11.7.1) ¨ a (m) xy = ¨ a (m) x + ¨ a (m) y − ¨ a (m) xy We may also define temporary last survivor annuity functions, e.g. ¨ a xy:n = n−1 ¸ t=0 v t . t p xy = n−1 ¸ t=0 v t [ t p x + t p y − t p xy ] = ¨ a x:n + ¨ a y:n − ¨ a xy:n Note: these formulae hold even without the assumption of independence of the lives. (b) Assurances Taking expected values on each side of the relationship v T 1 +v T 2 = v min{T 1 ,T 2 } +v max{T 1 ,T 2 } 11.7. LAST SURVIVOR MONETARY FUNCTIONS 197 one obtains ¯ A xy = m.p.v. of a last survivor assurance of £1 payable immediately on the death of (x), (y), = ¯ A x + ¯ A y − ¯ A xy (11.7.2) Similarly, when the sum assured is payable at the end of the year of death, we have v K 1 +1 +v K 2 +1 = v min{K 1 +1,K 2 +1} +v max{K 1 +1,K 2 +1} and hence A xy = A x +A y −A xy (11.7.3) (which is still true even when the lives are not independent.) Note also the conversion relationships A xy = A x +A y −A xy = (1 −d¨ a x ) + (1 −d¨ a y ) −(1 −d¨ a xy ) = 1 −d¨ a xy and, similarly, ¯ A xy = 1 −δ¯ a xy , etc. Example 11.7.1. Evaluate A 40:40 on A1967 −70 ultimate, 4% interest. Solution A 40:40 = 1 −d¨ a 40:40 = 1 −d[2¨ a 40 − ¨ a 40:40 ] = 1 −d[2 18.894 −17.052] = 0.20246 (c) Premiums. We may use an equation of value to calculate net or gross premiums. For example, using the International Actuarial Notation, we have P xy = net annual premium payable annually in advance for a policy providing £1 at the end of the year of the death of the second to die of (x), (y) = A xy ¨ a xy Using the conversion relationships given above, the following equations can be derived: ¯ P( ¯ A xy ) = ¯ A xy ¯ a xy = 1 ¯ a xy −δ and P xy = 1 ¨ a xy −d. 198 CHAPTER 11. JOINT-LIFE FUNCTIONS 11.8 Reserves for Last Survivor Assurances Consider a whole-life assurance providing £S at the end of the year of death of the last survivor of (x), (y). We ignore expenses and assume that the premium and reserve bases are the same. Then Annual premium, P = SP xy = S ¸ 1 ¨ a xy −d The value of the prospective reserve at duration t, just before payment of the premium then due, depends on whether or not both lives are still alive at time t. Let t V 1 = reserve assuming both lives are still alive t V 2 = reserve assuming (x) now alive but (y) has died t V 3 = reserve assuming (y) now alive but (x) has died We find that t V 1 = S A x+t:y+t −P xy ¨ a x+t:y+t t V 2 = S [A x+t −P xy ¨ a x+t ] t V 3 = S [A y+t −P xy ¨ a y+t ] Theorem When the first life dies, the prospective reserve will increase (from t V 1 to t V 2 or t V 3 ). Proof t V 1 = S (1 −d¨ a x+t:y+t ) −P xy ¨ a x+t:y+t = S 1 −(P xy +d)¨ a x+t:y+t < S [1 −(P xy +d)¨ a x+t ] (as ¨ a x+t < ¨ a x+t:y+t ) = S [(1 −d¨ a x+t ) −P xy ¨ a x+t ] = S [A x+t −P xy ¨ a x+t ] = t V 2 ( and similarly for t V 3 ) If it is not known whether one or both of (x), (y) survive at time t (and in practice this may be the case since one death does not result in a claim or a change in the premium), one may value the policy as a ‘weighted average’ of t V 1 , t V 2 and t V 3 , with weights proportional to the probabilities t p xy , t p x (1 − t p y ) and t p y (1 − t p x ) respectively. Hence the weights are t p xy t p xy , t p x (1 − t p y ) t p xy , t p y (1 − t p x ) t p xy respectively in order for them to add up to 1. Therefore the weighted average reserve is V = t p xy t p xy t V 1 + t p x (1 − t p y ) t p xy t V 2 + t p y (1 − t p x ) t p xy t V 3 Finally, it can be shown that the retrospective reserve is equal to the weighted average reserve, i.e. V = 1 v t . t p xy S P xy ¨ a xy:t −A 1 .... xy :t ¸ ¸ ¸. 11.8. RESERVES FOR LAST SURVIVOR ASSURANCES 199 Example 11.8.1. A last-survivor policy provided £10, 000 immediately on the death of the sec- ond to die of a man aged 65 and his wife aged 60. Premiums were payable monthly in advance so long as at least one of the couple survived. The office which issued the policy used the following basis: mortality : a(55)ultimate, males/females as appropriate interest : 8% per annum , expenses : 2.5% of all office premiums . (i) Find the monthly premium. (ii) Just after the payment of the first premium, the man died. Find the office’s prospective reserve just after this event, on the premium basis. Solution (i) Let annual premium, payable monthly, be P . 0.975P ¨ a (12) m 65: f 60 = 10, 000(1 +i) 1 2 A m 65: f 60 ¨ a (12) m 65: f 60 · ¨ a m 65: f 60 − 11 24 = 7.228 ¨ a (12) m 65 · ¨ am 65 − 11 24 = 7.942 ¨ a (12) f 60 · ¨ a f 60 − 11 24 = 9.836 Hence ¨ a (12) m 65: f 60 · ¨ a m 65: f 60 − 11 24 = 11.0075 − 11 24 = 10.549 A m 65: f 60 = 1 −d¨ a m 65: f 60 = 0.18463 Hence P = 10, 000 (1.08) 1 2 0.18463 0.975 10.549 = £186.55 Therefore monthly premium = £15.55 . (ii) Reserve = 10000(1 +i) 1 2 A f 60 −0.975P a (12) f 60 = 2, 476.96 −1, 773.82 (using A f 60 = 1 −d.¨ a f 60 , and a (12) f 60 = a f 60 + 11 24 ) Hence Reserve = £694.15. 200 CHAPTER 11. JOINT-LIFE FUNCTIONS Exercises 11.1 Given that n p x = 0.3, n p y = 0.4, n p z = 0.6, find the probability that, of the lives (x), (y) and (z), (a) none will survive n years (b) exactly one will survive n years (c) at least one will survive n years. 11.2 Prove that (IA) xy = ¨ a xy −d.(I¨ a) xy 11.3 Express in terms of p x , p y , and p z the probabilities that, of three lives (x), (y) and (z), (a) all three will survive one year (b) at least one will survive one year (c) exactly two will survive one year (d) at least two will survive one year 11.4 Derive the formula e xy = E(K) = ∞ ¸ t=1 t p xy where K = integer part of T (T = min¦T 1 , T 2 ¦) using Pr¦K = k¦ = k [q xy . 11.5 Evaluate A 75:75 on the basis of A1967 −70 ultimate at 4% interest. 11.6 The probability that at least one of three lives aged 60 will survive to age 65 is eight times the probability that exactly one will survive to age 65. Assuming that the 3 lives are independent and subject to the same table of mortality, find the probability that exactly one life will survive to age 65. 11.7 (i) Define t p xy and show that t p xy = t p x + t p y − t p xy (ii) Hence, or otherwise, show that ¨ a xy = ¨ a x + ¨ a y − ¨ a xy 11.8 12 years ago a man then aged 48 effected a without profits whole life assurance for £10,000 (payable at the end of the year of death) by annual premiums. The premium now due is unpaid. He now wishes to alter the policy so that the same sum assured will be payable at the end of the year of the first death of himself and his wife, who is 4 years older than himself. Calculate the revised office annual premium, ceasing on the first death, if the office uses the following basis for premiums and reserves. mortality: A1967-70 ultimate, rated down 4 years for female lives, interest: 4% per annum, expenses: 3% of all office premiums including the first, with additional initial expenses of 1 1 2 % of the sum assured. (This additional initial expense is not charged again on the conversion of an existing policy, providing that the sum assured does not increase.) 11.9 Consider the random variable L equal to the present value of £1 payable immediately on (i) the first death of (x) and (y), and (ii) the second death, in each case at a given rate of interest i p.a. Show that, in case (i), var(L) = ¯ A ∗ xy −( ¯ A xy ) 2 where ∗ indicates a rate of interest of 2i +i 2 p.a., and give a corresponding result for case (ii). 11.10. SOLUTIONS 201 Solutions 11.1 (a) (1 − n p x )(1 − n p y )(1 − n p z ) = 0.7 0.6 0.4 = 0.168 (b) n p x (1 − n p y )(1 − n p z ) + n p y (1 − n p x )(1 − n p z ) + n p z (1 − n p x )(1 − n p y ) = 0.3(0.6)(0.4) + 0.4(0.7)(0.4) + 0.6(0.7)(0.6) = 0.436 (c) 1 −(1 − n p x )(1 − n p y )(1 − n p z ) = 1 −0.168 = 0.832. 11.2 Let K = curtate future lifetime of (x, y) (I¨ a) xy = E (I¨ a) K+1 = E ¸ ¨ a K+1 −(K + 1)v K+1 d ¸ = 1 d [¨ a xy −(IA) xy ] (as (IA) xy = E[(K + 1)v K+1 ]) Hence (IA) xy = ¨ a xy −d(I¨ a) xy . 11.3 (a) p x . p y . p z (b) 1 −(1 −p x )(1 −p y )(1 −p z ) (c) p x p y (1 −p z ) +p x p z (1 −p y ) +p y p z (1 −p x ) (d) p x p y (1 −p z ) +p x p z (1 −p y ) +p y p z (1 −p x ) +p x p y p z 11.4 e xy = curtate joint expectation of life = ∞ ¸ k=1 k. k [q xy = 1 l xy ∞ ¸ k=1 k.d x+k:y+k = 1 l xy [d x+1:y+1 + 2d x+2:y+2 + 3d x+3:y+3 + ] = 1 l xy [(d x+1:y+1 +d x+2:y+2 + ) + (d x+2:y+2 +d x+3:y+3 + ) + ] = 1 l xy [l x+1:y+1 +l x+2:y+2 + ] = 1 l xy ∞ ¸ t=1 l x+t:y+t = ∞ ¸ t=1 t p xy 11.5 ¯ A 75:75 = 2 ¯ A 75 − ¯ A 75:75 · 2(1.04) 1 2 A 75 −(1.04) 1 2 [1 −d.¨ a 75:75 ] = 0.64676. 202 CHAPTER 11. JOINT-LIFE FUNCTIONS 11.6 Let 5 p 60 = p. We have (using binomial distribution) 1 −(1 −p) 3 = 8 3p(1 −p) 2 Hence 3p −3p 2 +p 3 = 24p(1 −p) 2 3 −3p +p 2 = 24 −48p + 24p 2 So 23p 2 −45p + 21 = 0 p = 45 ± (−45) 2 −4 23 21 46 = 45 ± √ 93 46 = 0.76862 (ignoring root > 1) Hence probability of exactly one survivor = 3p(1 −p) 2 = 0.12345 11.7 (i) t p xy = Pr¦ the last survivor of (x) and (y) survives for t years¦ = Pr¦(x) survives for t years but (y) dies¦ +Pr¦(y) survives for t years but (x) dies¦ +Pr¦(x) and (y) both survive for t years¦ = t p x (1 − t p y ) + t p y (1 − t p x ) + t p xt p y = t p x + t p y − t p xt p y (ii) ¨ a xy = ∞ ¸ t=0 t p xy v t = ∞ ¸ t=0 v t ( t p x + t p y − t p xt p y ) = ∞ ¸ t=0 v t t p x + ∞ ¸ t=0 v t t p y − ∞ ¸ t=0 v t t p xy = ¨ a x + ¨ a y − ¨ a xy 11.8 V 1 = reserve of original policy = 10, 000[1.015 12 V 48 −0.015] (using Zillmer’s formula). = 10, 000[1.015(1 − ¨ a 60 ¨ a 48 ) −0.015] = £2, 341.89 (there is no need to find the original premium) Let P be the revised annual premium. Equation: V 1 = 10, 000A 60:60 −0.97P¨ a 60:60 = 10, 000(1 −d.¨ a 60:60 ) −0.97P¨ a 60:60 = 6175.77 −9.6447P Hence P = £397.51 11.9 In case (i), L = v T where T = min¦T 1 , T 2 ¦. We have var (L) = E(L 2 ) −[E(L)] 2 = E[(v ∗ ) T ] −( ¯ A xy ) 2 = ¯ A ∗ xy −( ¯ A xy ) 2 . 11.10. SOLUTIONS 203 The corresponding result for case (ii) is var (L) = ¯ A ∗ xy −( ¯ A xy ) 2 . 204 CHAPTER 11. JOINT-LIFE FUNCTIONS Chapter 12 CONTINGENT ASSURANCES 12.1 Contingent Probabilities Suppose we have 2 independent lives (x), (y), subject to the same mortality table. (If these tables are different, it is easy to make the necessary adjustments in the formulae.) Define t q 1 xy = Pr¦(x) dies within t years, and before (y)¦ i.e. (x) dies within t years and (y) dies after this event (not necessarily within t years). Let T 1 = future lifetime of (x) with pdf f 1 (t 1 ) = t 1 p x µ x+t 1 (t 1 > 0) T 2 = future lifetime of (y) with pdf f 2 (t 2 ) = t 2 p y µ y+t 2 (t 2 > 0) Then t q 1 xy = Pr¦T 1 ≤ t and T 1 ≤ T 2 ¦ = 0<t 1 ≤t t 1 ≤t 2 f 1 (t 1 )f 2 (t 2 ) dt 1 dt 2 = t 0 ¸ ∞ t 1 t 1 p x µ x+t 1 t 2 p y µ y+t 2 dt 2 dt 1 = t 0 t 1 p x µ x+t 1 t 1 p y dt 1 = t 0 r p xy µ x+r dr (12.1.1) If t = 1 we may omit it, giving q 1 xy = 1 0 t p xy µ x+t dt Note also that ∞ q 1 xy = ∞ 0 t p xy µ x+t dt = Pr¦(x) dies before (y)¦ If x = y, we clearly have t q 1 x:x = t 0 r p xx µ x+r dr = 1 2 t 0 r p xx µ x+r:x+r dr = 1 2 t q xx (12.1.2) 205 206 CHAPTER 12. CONTINGENT ASSURANCES By general reasoning, the following result holds: t q 1 xy + t q x 1 y = t q xy (12.1.3) Note also that ∞ q 1 xy + ∞ q x 1 y = ∞ q xy = 1 One may also define deferred contingent probabilities, and we find that t [q1 x:y = t p xy .q 1 x+t:y+t and t [ ∞ q1 x:y = t p xy . ∞ q 1 x+t:y+t Consideration of the second death We also have the definition, t q 2 xy = Pr¦(x) dies after (y) and within t years¦ By calculations similar to those for t q 1 xy , we have t q 2 xy = t 0 (1 − r p y ) r p x µ x+r dr = t q x − t q 1 xy Thus, as expected by general reasoning, t q x = t q 1 xy + t q 2 xy (12.1.4) which follows from Pr¦T 1 ≤ t¦ = Pr¦T 1 ≤ t and T 1 ≤ T 2 ¦ +Pr¦T 1 ≤ t and T 1 > T 2 ¦ (the two terms on the R.H.S. are the probabilities of mutually exclusive events.) 12.2 Contingent Assurances Consider a benefit of £1 payable immediately on the death of (x) if this occurs before the death of (y). The M.P.V. (at a specified rate of interest) is: ¯ A1 xy = E ¸ v T 1 if T 1 ≤ T 2 0 if T 1 > T 2 = t 1 ≤t 2 v t 1 ( t 1 p x µ x+t 1 )( t 2 p y µ y+t 2 ) dt 2 dt 1 = ∞ 0 v t 1 t 1 p x µ x+t 1 ∞ t 1 t 2 p y µ y+t 2 dt 2 dt 1 = ∞ 0 v t 1 t 1 p x µ x+t 1 ( t 1 p y ) dt 1 = ∞ 0 v t t p xy µ x+t dt (12.2.1) This is often evaluated by approximate integration. 12.2. CONTINGENT ASSURANCES 207 Example 12.2.1. Using A1967-70 mortality and 4% interest estimate, by approximate integration, the value of ¯ A 1 50:60 [Assume l 108 = 0, so that the integral is over a range of 48 years. Break this into 4 sub-intervals and use Simpson’s rule over each.] Solution ¯ A 1 50:60 = ∞ 0 v t t p 50 µ 50+t t p 60 dt (at 4% interest) = 48 0 f(t) dt where f(t) = v t t p 50:60 µ 50+t . Using Simpson’s rule ¯ A 1 50:60 = 2 [f(0) + 4f(6) + 2f(12) + 4f(18) + 2f(24) 4f(30) + 2f(36) + 4f(42) +f(48)] = 0.1546. Temporary contingent assurance functions A1 x:y:n = m.p.v. of £1 payable immediately on death of (x) if this occurs before the death of (y) and within n years = n 0 v t t p xy µ x+t dt (12.2.2) Note also that ¯ A1 xx = 1 2 ¯ A xx (12.2.3) (since there is a 50% chance that the first “x” of (x, x) will be the first to die). We may therefore evaluate contingent assurances on 2 equal ages from tables of ¯ A xx . As expected by general reasoning, we also have A xy = A1 xy +A x 1 y (12.2.4) Example 12.2.2. (a) Estimate, by approximate integration, the value of a contingent assurance of £100,000 payable immediately on the death of a female aged 60, provided a male aged 50 is still alive and provided her death occurs within 15 years. The mortality of the female follows a(55) ultimate (females), and that of the male follows A1967-70 ultimate. The interest rate is 7 1 2 % per annum and expenses are ignored. Note. A very accurate approximation to the value is not expected. (b) It has been suggested that a policy providing the above benefit should be issued by annual premiums ceasing on the death of the female life or after 15 years, whichever is earlier. State with reasons whether you agree with this suggestion. If you do not, suggest a more suitable premium- paying term, other than issuing the contract by a single premium. 208 CHAPTER 12. CONTINGENT ASSURANCES Solution (a) Value = 100, 000 ¯ A m 50: 1 f 60:15 = 100, 000 15 0 v t t p 50:60 µ 60+t dt Evaluate integral by (for example) the three-eighths rule: 15 0 f(t) dt · 15 8 [f(0) + 3f(5) + 3f(10) +f(15)] = 0.12657. Hence value = £12, 657 (b) No: premiums must also cease on death of (50), otherwise they may still be payable with no prospect of claim: negative prospective reserve. Restrict premium-paying period so that premiums cease on first death or after 15 years, if earlier. We may also encounter contingent assurances payable on the second death. Define ¯ A2 xy =m.p.v. of £1 payable immediately on death of (x) if this occurs after the death of (y). By a similar argument to that for ¯ A 1 xy , ¯ A 2 xy = ∞ 0 v t t p x µ x+t (1 − t p y ) dt Hence ¯ A x = ¯ A 1 xy + ¯ A 2 xy (12.2.5) as expected. We may also encounter temporary contingent assurances payable on the second death, e.g. the m.p.v. of £1 payable immediately on death of (x) within n years, provided that (x) dies after (y), equals n 0 v t t p x µ x+t (1 − t p y ) dt = A 1 x:n −A 1 xy:n Note also that ¯ A 2 xx = 1 2 ¯ A xx (12.2.6) since there is a fifty per cent chance that the first “x” will be the last survivor. Example 12.2.3. Using the fact that ¯ A 1 xy + ¯ A x 1 y = ¯ A xy , find on A1967 −70 ultimate at 4% interest the values of (i) ¯ A 60: 1 60 , (ii) ¯ A 60: 2 60 and (iii) ¯ A 60: 1 60:10 . 12.2. CONTINGENT ASSURANCES 209 Solution. (i) ¯ A 60: 1 60 = 0.5 ¯ A 60:60 · 0.5(1.04) 1 2 [1 −d¨ a 60:60 ] · 0.31491 (ii) Use A 60: 2 60 = A 60 −A 60: 1 60 to obtain ¯ A 60: 2 60 = 0.21259 (iii) ¯ A 60:6 1 0:10 = 1 2 ¯ A 1 60 : 60:10 · 1 2 (1.04) 1 2 ¸ A 60:60 − D 70:70 D 60:60 A 70:70 = 0.15513 The symbols A 1 xy , etc. These are the same as ¯ A 1 xy , etc., but with the benefit payable at the end of the year of death. An exact expression is A 1 xy = ∞ ¸ t=0 v t+1 t p xy q 1 x+t:y+t (12.2.7) In practice, we may use the approximations A 1 xy · (1 +i) − 1 2 ¯ A 1 xy , etc. (12.2.8) Note the following results, similar to those given above for A1 xy , etc.: A xy = A1 xy +A x 1 y A x = A 1 xy +A 2 xy A 1 xx = 1 2 A xx A 2 xx = 1 2 A xx The variance of the present value of a contingent assurance Let Z = present value of £1 payable on death of (x), if this occurs before the death of (y) = v T 1 if T 1 ≤ T 2 where v = 1 1+i 0 if T 1 > T 2 As we have shown above, this has mean E(Z) = ¯ A 1 xy 210 CHAPTER 12. CONTINGENT ASSURANCES The variance of Z is E(Z 2 ) −[E(Z)] 2 where E(Z 2 ) = E ¸ (v 2 ) T 1 if T 1 ≤ T 2 0 if T 1 > T 2 = E ¸ (v ∗ ) T 1 if T 1 ≤ T 2 0 if T 1 > T 2 where v ∗ = 1 1 +i ∗ with i ∗ = i 2 + 2i = ¯ A ∗ 1 xy at rate of interest i 2 + 2i (or force of interest 2δ) Hence Var(Z) = ¯ A ∗ 1 xy −( ¯ A 1 xy ) 2 (12.2.9) 12.3 Premiums and Reserves for Contingent Assurances Premiums are calculated by the usual equation of value (including expenses if necessary). Reserves are usually calculated prospectively, making allowance for any deaths which have already occurred. Example 12.3.1. Two lives (A and B) are both aged 30. Calculate, on the basis of A1967 − 70 ultimate mortality and 4% p.a. interest, the annual premium, payable during the lifetime of A, to provide an insurance of £1000 payable at the end of the year of death of A, provided A dies after B. Find the policy value (on the premium basis) after 10 years (before the premium then due is paid) if (i) only A is then alive; (ii) both lives are then alive. Solution A 2 30:30 = A 30 −A 1 30:30 = A 30 − 1 2 A 30:30 = (1 −d¨ a 30 ) − 1 2 (1 −d¨ a 30:30 ) = 1 2 −d(¨ a 30 − 1 2 ¨ a 30:30 ) Hence premium = A 2 30:30 ¨ a 30 1000 = 1000 ¸ 1 2¨ a 30 −d 1 − 1 2 ¨ a 30:30 ¨ a 30 = 1000.(0.00327) = 3.27 = P, say. Hence the reserves are as follows: (i) 10 V = 1000A 40 −P¨ a 40 if only A is alive = 211.56 (ii) 10 V = 1000A 2 40:40 −P¨ a 40 if both are alive = 39.49 12.4. A PRACTICAL APPLICATION – THE PURCHASE OF REVERSIONS 211 Lapse options One should avoid having a negative prospective reserve at any duration, as this may lead to a lapse option against the office. In particular, one should ensure that premiums cease as soon as there is no possibility of future benefits. (See Example 12.2.2 above.) 12.4 A Practical Application – The Purchase of Reversions Definitions A reversion is a contract providing a sum of money payable on the death of a certain life, (y). An absolute reversion is a reversion in which the sum is paid under all circumstances, whilst in a contingent reversion the sum is paid only if certain other lives are (or are not) then alive. Example 12.4.1. J. Brown(50) will receive £1,000,000 on the death of his mother (aged 80), provided that he is then alive. He wishes to sell his “interest”. What is it worth? Solution Suppose that a purchaser uses a certain rate of interest, i p.a., and assumes that a certain mortality table applies to both J. Brown and to his mother. (Note that the purchaser might assume different tables for the lives). The interest is contingent since it requires (50) to be alive when (80) dies. Hence M.P.V. = 1, 000, 000 ¯ A 50: 1 80 In practice, the purchaser will want to be sure of getting the money when (80) dies, and will “plug the gap” with an insurance policy (which pays out on the death of (80) if she dies second.) Suppose that the life office issuing the insurance policy uses the same mortality and interest basis as the purchaser, and also ignores expenses. Single premium to buy insurance = 1, 000, 000 ¯ A 50:8 2 0 Hence purchase price = 1, 000, 000 ¯ A 80 − cost of premium for insurance policy = 1, 000, 000 ¯ A 80 − ¯ A 50: 2 80 = 1, 000, 000 ¯ A 50: 1 80 (as found before) Note:- The life office issuing the policy would examine the health (and any other risk factors) of (50) very carefully. 212 CHAPTER 12. CONTINGENT ASSURANCES 12.5 Extension to Three Lives We briefly mention the following symbols and formulae: ∞ q1 xyz = Pr¦(x) dies first of (x), (y), (z)¦ = ∞ 0 t p x µ x+t t p y t p z dt (12.5.1) ∞ q2 xy 1 z = Pr¦(x) dies second, (y) having died first¦ = ∞ 0 t p x µ x+t (1 − t p y ) t p z dt = ∞ q1 xz − ∞ q1 xyz (12.5.2) ∞ q2 xyz = Pr¦(x) dies second of (x), (y), (z)¦ = ∞ q2 xy 1 z + ∞ q2 xyz 1 (since either (y) or (z) dies first) = ∞ q1 xy + ∞ q1 xz −2 ∞ q1 xyz using formula (12.5.2) Similar definitions and relationships apply for contingent assurances, e.g. A2 xyz = m.p.v. of £1 payable immediately on the death of (x) if he dies second =A1 xy +A1 xz −2A1 xyz More complicated functions may be defined and evaluated, but we do not pursue this topic. 12.6. EXERCISES 213 Exercises 12.1 Which (if any) of the following statements are correct? A. ¯ A xx = 2 ¯ A 1 xx B. ¯ A xx = 2 ¯ A x C. ¯ A xx = 2 ¯ A x − ¯ A 1 xx D. ¯ A xx = ¯ A 1 xx + ¯ A 2 xx 12.2 Adams (aged 40) and Brown (aged 50) are two business partners. Adams wishes to provide for the sum of £80,000 to be paid immediately on Brown’s death if Brown predeceases him within ten years, and effects a policy providing this benefit by single premium. The life office issuing the contract employs the following basis: Mortality (both lives) : A1967 −70 ultimate Interest : 6% Expenses: 2% of the single premium. Using Simpson’s rule, or otherwise, estimate the single premium payable by Adams. 12.3 Estimate the value of ∞ q 1 74:84 on the basis of A1967 −70 ultimate mortality. (Assume l 108 = 0, so that the integral is over a range of 24 years. Break this into 3 sub-intervals and use Simpson’s Rule over each.) 12.4 (i) Express A 2 xx in terms of ¨ a x , ¨ a xx and the rate of interest. (ii) Smith and Jones are both aged 60. A life office has been asked to issue a special joint-life assurance policy providing £10,000 at the end of the year of death of the first to die of these two lives. In addition, if Smith is the second to die, a further £5,000 will be payable at the end of the year of his death. The policy is to have annual premiums payable during the joint lifetime of Smith and Jones. Calculate the annual premium on the following basis: A1967 −70 ultimate mortality 4% interest expenses are 5% of all premiums, with an additional initial expense of £100. (iii) Write down (but do NOT evaluate) formulae for the reserve at duration 10 years (imme- diately before payment of the premium then due) on the premium basis, if (a) both Smith and Jones are alive; and (b) Jones has died but Smith is alive. 12.5 A policy providing the sum of £100,000 immediately on the death of (x) if she dies before (y) is to be issued by a life office to a group of trustees. (i) Ignoring expenses, write down an expression for the single premium in terms of an integral. (ii) The trustees suggest that level annual premiums should be payable in advance until the death of the last survivor of (x) and (y). 214 CHAPTER 12. CONTINGENT ASSURANCES (a) Ignoring expenses, give a formula for the annual premium. (b) Would you advise the life office to issue the policy with premiums payable as suggested? Give reasons for your answer. 12.6 Estimate, by the trapezoidal rule or another suitable rule for approximate integration, the single premium for a temporary contingent assurance of £50,000 payable immediately on the death of Mrs Smith (aged 60), provided that this event occurs within 5 years and that her husband (aged 50) is alive at the date of her death. Mrs Smith is subject to the mortality of a(55) ultimate (females) and Mr Smith is subject to the mortality of A1967 − 70 ultimate. An interest rate of 7.5% p.a. is to be used, and allowance is to be made for expenses of 6% of the single premium. (Note A very accurate answer is not expected.) 12.7 Define the following functions in words, and give an expression for each of them in terms of an integral. (i) ∞ q 1 xy (ii) ¯ A 2 xy (iii) ¯ A 1 x:y:n 12.8 The chief of a certain tribe holds that office until age 50 or earlier death, and may be succeeded only by a person aged from 36 to 45. (A person aged exactly 36 is eligible, but a person aged exactly 45 is not.) The customs of the tribe require that a chief’s successor be his oldest eligible brother; if there is no eligible brother, the position of chief is given to someone from outside the previous chief’s family, who are then permanently debarred from becoming chief. The present chief is aged exactly 47 and has two brothers, aged exactly 37 and 33 respectively. The chief and his brothers may be regarded as independent lives subject to the mortality of a given table. Obtain an expression, in terms of quantities of the form n p x , n q 1 xy only, for the probability that (33) will become chief. 12.9 Your life office has been asked to quote a single premium for a contingent assurance policy providing £300,000 immediately on the death of a woman now aged 80 within 15 years, provided that at the date of her death a man now aged 60 has died. Your office uses the following basis: mortality : males - a(55)ultimate (males) females - a(55) ultimate (females) interest : 8% per annum expenses : 10% of the single premium. (i) Assuming that the two lives are independent, write down a formula for the single premium in terms of an integral. (ii) State a suitable non-repeated rule of approximate integration for evaluating this integral. (You are NOT required to carry out the evaluation.) (iii) Would you subject the male life to stringent underwriting procedures? Give brief reasons for your answer. 12.7. SOLUTIONS 215 Solutions 12.1 Only A is correct. 12.2 Let single premium be P. P = 80000 10 0 v t t p 50 µ 50+t t p 40 dt + 0.02P at 6% interest = 80000 0.98 10 0 f(t) dt where f(t) = v t t p 50 µ 50+t . t p 40 · 80000 0.98 10 6 [f(0) + 4 f(5) +f(10)] using Simpson’s rule = £4, 686. 12.3 ∞ q 1 74:84 = ∞ 0 t p 74 µ 74+t . t p 84 dt = 24 0 f(t) dt where f(t) = t p 74 µ 74+t . t p 84 · 4 3 [f(0) + 4f(4) + 2f(8) + 4f(12) + 2f(16) + 4f(20) +f(24)] = 0.2907. 12.4 (i) A 2 xx = 1 2 A xx = A x − 1 2 A xx (or use A 2 xx = A x −A 1 xx = A x − 1 2 A xx ); = (1 −d¨ a x ) − 1 2 (1 −d¨ a xx ) = 1 2 − i 1 +i (¨ a x − 1 2 ¨ a xx ) (ii) Let the annual premium be P, and set x = 60. 0.95P¨ a xx = 10, 000A xx + 5, 000A 2 xx + 100 = 10, 000[1 −d¨ a xx ] + 5, 000 ¸ 1 2 −d(¨ a x − 1 2 ¨ a xx ) + 100 Putting x = 60 gives 0.95P (9.943) −100 =10, 000 ¸ 1 − 0.04 1.04 9.943 + 5, 000[0.5 − 0.04 1.04 (12.551 −0.5 9.943)] =6175.77 + 1042.40 = 7218.17 Hence P = 7318.17 9.44585 = £774.75 216 CHAPTER 12. CONTINGENT ASSURANCES (iii) (a) Reserve = 10, 000A 70:70 + 5, 000A 70: 2 70 −0.95P¨ a 70:70 (b) Reserve = 5, 000A 70 12.5 (i) 100, 000 ∞ 0 v t t p xy µ x+t dt (ii) (a) Annual premium, P = 100000 ∞ 0 v t t p xy µ x+t dt ¨ a xy . (b) No, because premiums should not continue after first death. If (x) dies first, the benefit is paid and the policy may be lapsed, and if (y) dies first there is no possibility of benefit and the policy be lapsed. There is thus a lapse option on the first death. 12.6 Let single premium = P. Then 0.94P = 50, 000 ¯ A 1 f 60: m 50:5 at 7 1 2 % interest = 50, 000 5 0 v t t p f 60 µ f 60+t . t pm 50 dt By the trapezoidal rule. 5 0 v t t p f 60 µ f 60+t . t pm 50 dt · 0.041726 Hence P · 50000 0.041726 0.94 = £2, 219 12.7 (i) ∞ q 1 xy = Pr¦(x) will die before (y)¦ = ∞ 0 t p xy µ x+t dt (ii) ¯ A 2 x:y = m.p.v. of a contingent assurance of £1 payable immediately on death of (x), provided this occurs after the death of (y). = ∞ 0 v t t p x µ x+t (1 − t p y ) dt (iii) ¯ A1 x:y:n = m.p.v. of a temporary contingent assurance of £1 payable immediately on death of (x), provided that this occurs within n years, and before (y) dies. = n 0 v t . t p xy µ x+t dt 12.7. SOLUTIONS 217 12.8 Let chief = (y) = (47), brothers (x 1 ) = (37), (x 2 ) = (33). (x 2 ) becomes chief if and only if (1) he survives for 3 years and (y) “retires” then, (x 1 ) having died. (2) (x 1 ) becomes chief (by succeeding (y) on his death or “retirement”) and dies after time 3 and before time 12, leaving (x 2 ) alive. Probability of event (1) is 3 p y (1 − 3 p x 1 ) 3 p x 2 Probability of event (2) is Pr¦(x 1 ) dies between times 3 and 12, leaving (x 2 ) alive¦ = 12 3 t p x 1 µ x 1 +t t p x 2 dt = 9 0 r+3 p x 1 µ x 1 +r+3 r+3 p x 2 dr = 3 p x 1 x 2 9 0 r p x 1 +3:x 2 +3 µ x 1 +r+3 dr = 3 p x 1 3 p x 2 . 9 q 1 x 1 +3:x 2 +3 12.9 (i) annual premium = 300, 000 15 0 v t t p f 80 µ f 80+t (1 − t pm 60 ) dt 0.90 (ii) The three-eighths rule would be suitable, since it avoids evaluation of the integrand when t = 7 1 2 . (iii) Yes. If (60) dies soon, there is a high chance that (80) will die before age 95 and so give rise to a claim. The office must check: (1) the health of (60); (2) whether he has any occupational or other risks, e.g. participation in a dangerous sport. Note. The sum assured is large enough to justify the costs of a medical examination. 218 CHAPTER 12. CONTINGENT ASSURANCES Chapter 13 REVERSIONARY ANNUITIES 13.1 Reversionary Annuities Payable Continuously Consider an annuity of £1 p.a. payable continuously to (y) after the death of (x). The present value of this reversionary annuity is Z = ¯ a U − ¯ a T if U > T 0 if U ≤ T where T = future lifetime of (x), with p.d.f. = t p x µ x+t (t > 0); U = future lifetime of (y), with p.d.f. = u p y µ y+u (u > 0). Note. If (x) and (y)’s mortality rates follow different tables, this should be indicated. Define ¯ a x|y = m.p.v. of Z = u>t (a u − ¯ a t )( u p y µ y+u )( t p x µ x+t ) dudt = ∞ 0 ¸ ∞ t (¯ a u − ¯ a t ) u p y µ y+u du t p x µ x+t dt (13.1.1) Theorem ¯ a x|y = ¯ a y − ¯ a xy (13.1.2) Proof It is easiest to proceed indirectly, as follows. Observe that Z + ¯ a min(T,U) = (¯ a U − ¯ a T ) + ¯ a T if U > T ¯ a U if U ≤ T = a U Taking expected values on both sides gives E(Z) + ¯ a xy = ¯ a y Another important formula ¯ a x|y = ∞ 0 v t t p x µ x+tt p y ¯ a y+t dt 219 220 CHAPTER 13. REVERSIONARY ANNUITIES Proof ¯ a x|y = ∞ 0 v t t p y (1 − t p x ) dt = ∞ 0 f(t)g (t) dt where f(t) = 1 − t p x = t q x (which is such that f (t) = t p x µ x+t ) and g(t) = − t [¯ a y = − ∞ t v r r p y dr (which is such that g (t) = v t t p y ) Using integration by parts ¯ a x|y = [f(t)g(t)] ∞ 0 − ∞ 0 f (t)g(t) dt = [(− t [¯ a y ) t q x ] ∞ 0 + ∞ 0 t p x µ x+t ( t [¯ a y ) dt = ∞ 0 t p x µ x+t v t t p y ¯ a y+t dt. Evaluation of reversionary annuities Using the Euler–Maclaurin formula we have ¯ a x|y = ¯ a y − ¯ a xy · (a y + 1 2 ) −(a xy + 1 2 ) = a y −a xy = a x|y Example 13.1.1. Calculate an appoximate value of ¯ a m 65| f 69 on the a(55) tables at 8% interest. Solution ¯ a m 65| f 69 · a f 69 −a m 65: f 69 = 7.533 −5.877 (using interpolation) = 1.656 13.2 Reversionary Annuities Payable Annually or m thly We first assume that payments are made m thly , the first payment being at the end of the 1 m year (measured from the issue date) following the death of (x), and use the symbol a (m) x|y to refer to this case. Using the equation a (m) x|y +a (m) xy = a (m) y it can be seen that a (m) x|y = m.p.v. of reversionary annuity of £1 p.a. payable monthly to (y) after the death of (x) = a (m) y −a (m) xy 13.3. WIDOW’S (OR SPOUSE’S) PENSION ON DEATH AFTER RETIREMENT 221 When m = 1, it may be omitted, giving a x|y = a y −a xy By Woolhouse’s formula a (m) x|y · (a y + m−1 2m ) −(a xy + m−1 2m ) = a y −a xy = a x|y . An alternative approach is to regard this reversionary annuity as a collection of “pure endow- ments” (payable at times 1 m , 2 m , ) which are paid if (x) has died and (y) is alive, i.e. a (m) x|y = ∞ ¸ t=1 1 m v t m Pr¦(x) has died but (y) is alive at time t m ¦ = 1 m ∞ ¸ t=1 v t m (1 − t m p x ). t m p y = a (m) y −a (m) xy (as before). Suppose now that a m thly reversionary annuity (of £1 p.a.) begins immediately on the death of (x). Since the payments begin on average 1 2m year earlier than in the case discussed previously, the m.p.v. is approximately (1 +i) 1 2m a (m) x|y (13.2.1) An exact formula is ∞ 0 v t t p x µ x+t·t p y ¨ a (m) y+t dt · ∞ 0 v t t p xy µ x+t ¯ a y+t + 1 2m dt =¯ a x|y + 1 2m ¯ A1 xy ·a x|y + 1 2m ¯ A1 xy (13.2.2) In practice this gives results similar to (13.2.1). 13.3 Widow’s (or Spouse’s) Pension on Death after Retire- ment Many pension schemes provide a spouse’s pension on the death of the member in service (D.I.S. = Death In Service) and/or on death after retirement (D.A.R. = Death After Retirement) We consider D.A.R. only at this stage and suppose that a male employee retires at age 65. Now consider 2 different cases:- Case 1 A widow’s pension of £1 p.a. is payable on the death of (65) only if he was married to the same woman at retirement. In this case, the m.p.v. at age 65 per married man is ¯ am 65| f y (assuming pension is payable continuously) 222 CHAPTER 13. REVERSIONARY ANNUITIES where y = average age of wife of member aged 65 = 65 −d, where d = age difference between husband and wife (approximately 3 years in practice) Hence the value of widow’s pension for each member retiring at age 65 (marital status unknown) is h 65 ¯ a m 65| f 65−d (13.3.1) where h x = the probability that a man aged x is married. Note. We ignore the possibility of divorce (or assume that the ex-wife still gets pension). Case 2 (now more common) The widow’s pension is payable to any widow: we use the “collective” approach. Suppose, for example, that the widow’s pension of £1 p.a. is payable monthly in advance, beginning immediately on death of her husband. The m.p.v. for a man retiring at age 65 (marital status then unknown) is ∞ 0 v t t p m 65 µ m 65+t h 65+t ¨ a (12) f 65−d+t dt (13.3.2) where ¨ a (12)f 65−d+t = m.p.v. of widow’s annuity to the widow of a man dying at age 65 +t, if widow is assumed to be d years younger than husband Example 13.3.1. A life office sells “personal pensions” policies under which the benefits for men on retirement at age 65 consist of: (a) a member’s pension (payable monthly in advance for 5 years certain and for life thereafter), and (for men married at age 65 only) (b) a spouse’s pension (payable monthly in advance, beginning immediately on the death of the member) of half the member’s pension. The possibility of divorce of men aged over 65 may be ignored, and post-retirement marriages do not give rise to spouse’s pension. The member’s contributions are invested in certain unitised with-profits funds, and it is assumed that, in respect of a certain Mr Brown’s policy, the fund available to purchase pension at age 65 will be £100,000. Suppose that the life office uses the following basis to calculate the amount of pension which may be purchased at retirement in respect of a given fund: mortality of males: a(55) males ultimate mortality of females: a(55) females ultimate interest: 8% per annum expenses: 1% of the fund (at age 65) 13.3. WIDOW’S (OR SPOUSE’S) PENSION ON DEATH AFTER RETIREMENT 223 Men who are not married at age 65 need not buy spouse’s pension, but married men must buy this. Calculate Mr Brown’s expected monthly pension if (i) he is assumed to be single at age 65; and (ii) he is assumed to be married at age 65, and his wife is 3 years younger. Solution. (i) Let P be annual pension, payable monthly in advance. 0.99 100, 000 = P(¨ a (12) 5 + 5 [¨ a (12) 65 ) on males’ table = P( i d (12) a 5 +v 5 l 70 l 65 ¸ ¨ a 70 − 11 24 ) = P(4.1637 + 0.68058 0.86621 6.8097) = 8.1782P Hence P = £12, 105, so monthly pension = £1, 008.75 (ii) The equation of value is now 99, 000 · P(8.1782) + 1 2 P(1.08) 1 24 a f 62 −a m 65: f 62 · P(8.1782 + 1.2026) Hence P = £10, 553, so monthly pension = £879.42 Notes (1) There may be rules to try to exclude benefits for widows of “deathbed marriages”. (2) There may be problems if the man was married more than once (and his ex-wives are still alive). In the U.K. the last wife receives all the pension. (3) If a wife is very much younger than her husband, there may be an “actuarial reduction” (see later). (4) Widow’s pensions might cease on remarriage, but this rule is no longer common. (5) h 65+t is sometimes assumed to be “piecewise continuous”, e.g. h 65+t = 0.85 for 0 ≤ t < 1 0.87 for 1 ≤ t < 2 0.89 for 2 ≤ t < 3 0.90 for t ≥ 3. The integral in (13.3.2 ) may be evaluated approximately by a sum, i.e. ∞ 0 v t t p m 65 µ m 65+t (h 65+t )¨ a (12) f 65−d+t dt · 0.85v 1 2 q m 65 ¨ a (12) f 65−d+ 1 2 + 0.87v 1 1 2 1 [qm 65 ¨ a (12) f 65−d+1 1 2 + 0.89v 2 1 2 2 [qm 65 ¨ a (12) f 65−d+2 1 2 + 0.9 ¸ v 3 1 2 3 [qm 65 ¨ a (12) f 65−d+3 1 2 +v 4 1 2 4 [qm 66 ¨ a (12) f 65−d+4 1 2 + ¸ (assuming that the man dies on average half-way through each year of age). 224 CHAPTER 13. REVERSIONARY ANNUITIES 13.4 Actuarial Reduction Factors Suppose that, in a pension policy or scheme, the rules state that a reduction applies to the normal widow’s pension if wife is more than 10 years younger than her husband. (This rule may also apply to widow’s D.I.S. pensions). Suppose a member dies aged x, leaving widow aged y (y < x −10). The value of this widow’s pension is calculated to be the same (actuarially speaking) as for a widow aged x − 10 (whose pension is not, of course, reduced.) Let the widow aged y get £R p.a. for each £1 p.a. of ‘normal’ widow’s pension. Then (Full widow’s pension).R¨ a (12) f y = (Full widow’s pension).¨ a (12) f x−10 This gives R = actuarial reduction factor = ¨ a (12) f x−10 ¨ a (12) f y 13.5. EXERCISES 225 Exercises 13.1 (i) Define the following symbols in words, and give a formula in terms of an integral for each of them: (a) A x 1 y (b) A2 xy (c) a y|x (ii) Consider the following sets of payments: (1) £1 immediately on the death of (y) if (y) dies before (x), and (2) an income of £δ p.a. payable continuously to (x) after the death of (y), plus £1 immedi- ately on the death of (x) if this occurs after that of (y). Prove that the present values (at force of interest δ p.a.) of (1) and (2) are equal. Hence write down a relationship involving A x 1 y , A2 xy and a y|x . 13.2 A special life policy on 2 lives aged x and y respectively provides cash sums of £10,000 and £20,000 immediately on the first and second deaths respectively. In addition, an annuity at the rate of £1,000 per annum will be paid continuously, commencing immediately on the first death and ceasing immediately on the second death. Obtain an expression for the mean present value of the benefits in terms of joint-life and single- life annuity functions and the force of interest. Ignore expenses. 13.3 An office issues a policy on the lives of a woman aged 60 and her husband aged 64. Under this policy, level premiums are payable annually in advance for 20 years or until the first death of the couple, if earlier. On the first death of the couple, the survivor will receive an annuity of £10,000 per annum, payable weekly, beginning immediately on the first death. Calculate the annual premium if the office uses the basis given below: Mortality males: a(55) males ultimate females: a(55) females ultimate Expenses: 20% of the first premium 5% of each premium after the first Interest: 6% per annum. 13.4 A special annuity, payable yearly in arrear, is effected on the lives of a man aged x and his wife aged y. The conditions of payment are: (a) so long as both survive the rate of payment will be £3,000 per annum; (b) if the wife dies first, the rate of payment will be £2,000 per annum until the man’s death; (c) payments at the rate of £3,000 per annum will continue for six years certain after the death of the husband, the first payment being at the end of the year of his death, and will be reduced thereafter to £1,500 per annum during the lifetime of the wife. Obtain an expression for the present value of this annuity in terms of single and joint-life annuity factors, life table and compound interest functions. Assume that the same (non- select) table of mortality is appropriate for the two lives. 226 CHAPTER 13. REVERSIONARY ANNUITIES 13.5 A single-premium policy provides the following benefits to a husband and wife each aged 40. (1) An annuity of £5,000 per annum, payable continuously, commencing on the husband’s death within 25 years, or on his survival for 25 years, and continuing so long as either husband or wife is alive. (2) A return of half the single premium without interest immediately on the death of the husband within 25 years, provided that his wife has already died. The office issuing the contract uses the following basis: mortality : A1967-70 ultimate interest : 4% per annum expenses are ignored. Calculate the single premium. 13.6 A husband and wife, aged 70 and 64 respectively, effect a policy under which the benefits are (1) a lump sum of £10,000 payable immediately on the first death, and (2) a reversionary annuity of £5,000 p.a. payable continuously throughout the lifetime of the surviving spouse after the death of the first. Level premiums are payable annually in advance until the first death. Calculate the annual premium on the undernoted basis: Males’ Mortality: a(55) males ultimate Females’ Mortality: a(55) females ultimate Interest: 8% p.a. Expenses: 10% of all premiums Ignore the possibility of divorce. 13.6. SOLUTIONS 227 Solutions 13.1 (i) (a) The m.p.v. of £1 payable immediately on the death of (y), if this occurs before that of (x). A x 1 y = ∞ 0 v t t p xy µ y+t dt (b) The m.p.v. pf £1 payable immediately on the death of (x) if this occurs after that of (y). A2 xy = ∞ 0 v t t p x µ x+t (1 − t p y ) dt (c) The m.p.v. of a reversionary annuity of £1 p.a. payable continuously to (x) after the death of (y). a y|x = ∞ 0 v t t p y µ y+t t p x a x+t dt (other expressions also possible) (ii) Consider payment (1). Suppose it is invested (at force of interest δ p.a.) to give income so long as (x) lives (assuming (y) dies first, otherwise there is no payment (1)). On the death of (x), if after that of (y), the capital (£1) is paid immediately. Payments (2) thus have the same present value as payment (1) (both being random variables depending on the future lifetimes of (x) and (y)). Take means of these present values to get A x 1 y = δ.¯ a y|x +A2 xy 13.2 Benefit = 10, 000A xy + 20, 000A xy + 1000(¯ a x|y + ¯ a y|x ) = 10, 000( ¯ A xy + 2A x + 2A y −2A xy ) + 1000(¯ a x + ¯ a y −2¯ a xy ) = 10, 000(2A x + 2A y −A xy ) + 1000(¯ a x + ¯ a y −2¯ a xy ) = 10, 000(2 −2δ¯ a x + 2 −2δ¯ a y −1 +δ¯ a xy ) + 1000(¯ a x + ¯ a y −2¯ a xy ) = (1000 −20, 000δ)(¯ a x + ¯ a y ) + (10, 000δ −2000)¯ a xy + 30, 000 13.3 Let annual premium be P. Then 0.95P¨ a m 64: f 60:20 −0.15P = 10000[¯ a m 64| f 60 + ¯ a f 60| m 64 ] = 10000[¯ a f 60 − ¯ a m 64: f 60 + ¯ am 64 − ¯ a m 64: f 60 ] · 10000[a f 60 +am 64 −2a m 64: f 60 ] = 42230 ¨ a m 64: f 60:20 = ¨ a m 64: f 60 −v 20 l m 84 l m 64 l f 80 l f 60 ¨ a m 84: f 80 = 8.5709 Hence P = 42230 0.95 8.5709 −0.15 = £5, 284 . 13.4 (a) value of benefit = 3000a xy . (b) value of benefit = 2000(a x −a xy ) 228 CHAPTER 13. REVERSIONARY ANNUITIES (c) value of benefit is value of (1) an annuity-certain (of 3000 p.a., for 6 years) beginning at end of year of death of (x), plus (2) 1500 p.a., payable at times t (t ≥ 7) if (y) alive and (x) dead 6 years previously = 3000¨ a 6 A x + 1500 ∞ ¸ t=7 v t t p y (1 − t−6 p x ) = 3000¨ a 6 (1 −d¨ a x ) + 1500v 6 6 p y (a y+6 −a x:y+6 ) Add (a), (b), (c) to find total value of the annuity. 13.5 Let single premium be P. Consider benefit (1) with payment at rate £1 p.a. There are two cases: (a) husband dies within 25 years; (b) husband survives for 25 years. (Indicate “m,f” to clarify which life is which.) In case (a), M.P.V. is 25 0 v t t p m 40 µ m 40+t·t p f 40 ¯ a f 40+t dt In case (b), M.P.V. is v 25 25 p m 40 ¸ 25 p f 40 a m 65: f 65 + (1 − 25 p f 40 )am 65 =v 25 25 pm 40 25 p f 40 (¯ a f 65 − ¯ a m 65: f 65 ) +v 25 25 pm 40 ¯ am 65 =v 25 25 pm 40 25 p f 40 ∞ 0 v t t p m 65: f 65 µ m 65+t ¯ a f 65+t dt +v 25 25 pm 40 ¯ am 65 Therefore total M.P.V. is ∞ 0 v t t p m 40: f 40 µ m 40+t ¯ a f 40+t dt +v 25 25 pm 40 ¯ am 65 =¯ a m 40| f 40 +v 25 25 pm 40 ¯ am 65 Benefit (1) = 5000(¯ a 40|40 ) + 5000v 25 25 p 40 ¯ a 65 = 5000(¯ a 40 − ¯ a 40:40 ) + 5000v 25 25 p 40 ¯ a 65 · 5000[¨ a 40 − ¨ a 40:40 +v 25 25 p 40 (¨ a 65 − 1 2 )] = 5000(1.842 + 3.1418) = 24, 919. Benefit (2) = P 2 25 0 v t t p 40 µ 40+t (1 − t p 40 ) dt = P 2 [A 1 40:25 −A 1 40:40:25 ] = P 2 [A 1 40:25 − 1 2 A 1 40 : 40:25 ] = P 2 (1.04) 1 2 ¸ A 40:25 − D 65 D 40 − 1 2 1 −d.¨ a 40:40:25 − D 65:65 D 40:40 = P 2 (0.094993 −0.5 0.17511) = 0.003719P 13.6. SOLUTIONS 229 Hence equation of value is P = 24919 + 0.003719P Therefore P = £25, 012. 13.6 Let annual premium be P . Value of benefits = 10, 000A m 70: f 64 + 5000(¯ a m 70| f 64 + ¯ a f 64| m 70 ) = 10, 000(1 −δ¯ a m 70: f 64 ) + 5000(¯ am 70 + ¯ a f 64 −2¯ a m 70: f 64 ) · 10, 000[1 −δ(a m 70: f 64 + 1 2 )] + 5000[am 70 +a f 64 −2a m 70: f 64 ] = 5, 323.85 + 18, 450 = 23, 773.85 Value of premiums less expenses = 0.9P¨ a m 70: f 64 = 5.9184P . Hence P = 23773.85 5.9184 = £4, 017 . 230 CHAPTER 13. REVERSIONARY ANNUITIES Chapter 14 PROFIT TESTING FOR UNIT-LINKED POLICIES 14.1 Unit-Linked Policies Most of the money paid in premiums by the policyholders is used to purchase “units”: that is, money is placed in a unitised investment of some kind (U.K. equities, property, etc.). The assets underlying each policyholder’s units form the Unit Fund of the policy, the value of which may be calculated from the unit price and the number of units held. Unit prices are quoted at two levels, the “Bid Price” (at which the units can be sold) and the “Offer Price” (at which the units must be bought). The proceeds on maturity (at policy duration n) or earlier surrender are usually equal to the bid value of the units. On death, the benefits are usually equal to the value of the units (at bid price), subject to a minimum death benefit. In addition, the office holds a balancing account, called the Sterling Fund (or Sterling Reserves), into which are paid deductions from the premiums for expenses and fund management charges, and from which it pays the actual office expenses and death guarantee costs. The office may also transfer profits/losses from the sterling reserves to the shareholders or with- profits policyholders (who may be considered to be “investing” in the sale of the unit-linked policies). They may not take money from the unit fund as this belongs entirely to the policyholders. 14.2 Mechanics of the Unit Fund Unit prices are quoted at 2 prices, the bid price and the offer price, which is an artificial higher price. Define 1 −λ = Bid Price Offer Price where λ is called the Bid/Offer Spread (λ being perhaps 0.05). The bid price is the “real” price and all valuation calculations use the bid price. We use the following notation: 231 232 CHAPTER 14. PROFIT TESTING FOR UNIT-LINKED POLICIES P t = the office premium in year t (t = 1, 2, ..., n) (which is actually paid at time t −1) a t = the allocation proportion in year t = the proportion of the premium P t which is allocated to buying units (at the offer price) The cost of allocation in year t is the money actually used by the office to buy units, that is (1 −λ)a t P t . Hence P t −(1 −λ)a t P t may be transferred to the sterling reserves as a deduction for expenses. Define c t = the fund management charge in year t = a charge made at the end of year t, usually a percentage of the value of the unit fund, which is transferred to the sterling fund. The accumulation of the unit fund Let i u be the assumed rate of growth per annum of the unit fund. Define F t =the value of the unit fund at time t (after deduction of the fund management charge, but before payment of any premium then due) asssuming that the policy is still in force. Define F 0 = 0 Then F t = [F t−1 + (1 −λ)a t P t ] (1 +i u ) −c t (14.2.1) Suppose that the fund management charge is a proportion m of the unit fund; we have c t = m[F t−1 + (1 −λ)a t P t ] (1 +i u ) (14.2.2) and thus F t = (1 −m) [F t−1 + (1 −λ)a t P t ] (1 +i u ) (14.2.3) Note. In some cases c t may be a fixed sum rather than a proportion of the fund. Example 14.2.1. A life office issues a large block of 3-year unit-linked endowment assurances under which 80% of the first year’s premium and 101% of subsequent premiums are invested in units at the offer price. The bid price of the units is 95% of the offer price. The units are subject to an annual management charge of 0.75% of the bid value of the fund at the end of each policy year. The annual premium is £1,000 and unit prices are assumed to grow at 9% per annum. Calculate the bid value of the units at the end of each year, according to the office’s projections. 14.3. THE STERLING FUND (OR STERLING RESERVES) 233 Solution (1) (2) (3) (4) (5) Cost of Fund brought Fund at end Allocation forward from start of year t P t (1 −λ)a t F t−1 +P t (1 −λ)a t before F.M.C. F.M.C. F t 1 760.00 760.00 828.40 6.21 822.19 2 959.50 1,781.69 1,942.04 14.57 1,927.47 3 959.50 2,886.97 3,146.80 23.60 3,123.20 (3) = (1.09) (2) (4) = 0.0075 (3) (5) = (3) −(4). 14.3 The Sterling Fund (or Sterling Reserves) Money is assumed to earn interest at rate i s p.a. in the sterling reserves. Define e t = projected expenses for office in year t (payable at the start of year t) c t = fund management charge (as before) (DG) t = death guarantee cost in year t = q x+t−1 (S t −F t ) if S t > F t 0 if S t ≤ F t where x = the age of the policyholder at the start of the policy, and S t = the guaranteed minimum death benefit in year t. We suppose that the policy is one of a large number of similar unit-linked contracts on lives whose mortality follows a specified table, random variations being ignored. Define (SCF) t = the expected net cash flow in the sterling fund in year t per policy in force at the start of the year = the “in force” net cash flow The “initial” expected net cash flow in the sterling fund in year t; that is the expected net cash flow per policy sold, is given by t−1 p x .(SCF) t (t = 1, 2, ..., n) (14.3.1) The formula for (SCF) t is (SCF) t = [P t −(1 −λ)a t P t −e t ](1 +i s ) +c t −(DG) t (14.3.2) Maturity Bonuses In some policies, there may be a maturity bonus in the form of a fixed sum, or a proportion of the bid value of the fund at maturity. This money must come from the sterling fund, so we must adjust (SCF) n as follows: (SCF) n = “normal” (SCF) n (as in (14.3.2)) −p x+n−1 (Maturity Bonus) (14.3.3) 234 CHAPTER 14. PROFIT TESTING FOR UNIT-LINKED POLICIES The Profit Vector and Profit Signature Define (PRO) t = the profit vector = the expected net profit to the office in year t per policy in force at the start of the year σ t = the profit signature = the expected net profit to the office in year t per policy sold = t−1 p x .(PRO) t (14.3.4) There are 2 cases to consider. Case 1 If there is no need (or desire) to maintain sterling reserves at the end of each policy year, we have (PRO) t = (SCF) t for t = 1, 2, ..., n Case 2 Suppose now that the office wishes to maintain sterling reserves of t V at the end of year t (t = 1, 2, ..., n −1). It is assumed that 0 V = n V = 0. The calculations are very similar to those for conventional profit-testing. The profit vector is (PRO) t = (SCF) t +i s t−1 V −(IR) t (14.3.5) where (IR) t = increase in reserves = p x+t−1 t V − t−1 V Therefore (PRO) t = (SCF) t + (1 +i s ) t−1 V −p x+t−1 t V (14.3.6) In both cases the profit signature is found by equation 4.3.4, i.e. σ t = t−1 p x (PRO) t Example 14.3.1. (continued from Example 14.2.1) Suppose that, in addition to the information in example 14.2.1, the death benefit, payable at the end of the year of death, is the greater of twice the annual premium and the bid value of the units. The office expects to incur initial expenses on these policies of 20% of the first premium. Renewal expenses are expected to be £20, payable at the beginning of each policy year after the first. The mortality rate at each age is assumed to be 0.003. Sterling reserves are assumed to earn interest at 4% per annum. Ignore the possibility of withdrawal. Assuming that the office holds zero sterling reserves at the end of each policy year, calculate the profit signature. 14.4. THE ASSESSMENT OF PROFITS 235 Solution (1) (2) (3) (4) (5) (6) Premium less cost Accumulation Death Guarantee In Force of allocation Expenses of Sterling Costs FMC cash flow t P t −P t (1 −λ)a t e t Fund in year (DG) t c t (SCF) t 1 240.00 200 41.60 3.53 6.21 44.28 2 40.50 20 21.32 0.22 14.57 35.67 3 40.50 20 21.32 0 23.60 44.92 (3) = [(1) −(2)] (1.04) (4) = 0.003(2000 −F t ) (6) = (3) −(4) + (5) (7) (8) (9) Profit Vector Profit signature t (PRO) t t−1 p x σ t 1 44.28 1 44.28 2 35.67 0.997 35.56 3 44.92 0.994 44.65 (7) = (6) (as no reserves held) (8) = (0.997) t−1 (9) = (7) (8) 14.4 The Assessment of Profits The profit signature ¦σ t ¦ can be assessed in one or more of the following ways. (1) One could work out the Internal Rate of Return (or yield) by solving n ¸ t=1 v t σ t = 0 where v = 1 1 +j (the internal rate of return, j, being the solution of this equation.) (2) The shareholders may value the net profits at a certain rate of interest, j per annum. This rate is called the Risk Discount Rate, and may reflect uncertainties in ¦σ t ¦, with j normally higher than i s . The net present value of the profits is thus NPV (j) = n ¸ t=1 v t σ t at rate j. (3) The Profit Margin is defined as n.p.v. of profits n.p.v. of premiums , both at some rate of interest, i m say = ¸ n t=1 v t σ t ¸ n−1 t=0 P t+1 t p x v t , at rate i m . 236 CHAPTER 14. PROFIT TESTING FOR UNIT-LINKED POLICIES Example 14.4.1. Find the net present value of the profit signature in example 4.3.1 at a risk discount rate of 10%. Solution n.p.v. of profit = 3 ¸ t=1 v t σ t at 10% = 44.58v + 35.56v 2 + 44.65v 3 = 103.46. 14.5 Zeroisation of the Profit Signature Suppose that the profit signature, σ t , with no provision for sterling reserves at the end of each year (i.e. t V = 0 for all t), is of the following form: (σ t ) = ¸ ¸ ¸ ¸ ¸ 26.22 −8.05 −5.18 −2.29 0.62 ¸ (14.5.1) It may be considered undesirable for the shareholders to take a profit of £26.22 at time 1, as there may not be enough money to cover the “negative profits” at times 2, 3 and 4. Thus the profit taken by the shareholders in year 1 should be reduced in order to avoid negative sterling fund profits in years 2, 3 and 4. This process is called Zeroisation. Let X be the sum needed at time 1 to pay the negative cash flows at times 2, 3 and 4 when the Sterling Reserves earn interest at rate i s = 4 1 2 % p.a. (for example). Then X = 8.05v + 5.18v 2 + 2.29v 3 at rate 4 1 2 % = £14.46 Note X is the amount required at time 1, and hence the negative cash flows are discounted to time 1 (not the start of the policy). Hence the shareholders may take a profit at time 1 of 26.22 −14.46 = £11.76. There will now be no need for capital injections (from the shareholders to the Sterling Reserves) at times 2, 3 and 4. The shareholders may still take a profit of £0.62 at time 5. The zeroised profit signature ¦σ t ¦ is thus (σ t ) = ¸ ¸ ¸ ¸ ¸ 11.76 0 0 0 0.62 ¸ Notes 14.6. WITHDRAWALS 237 (1) This process may be carried out even if σ 1 < 0. (2) We may calculate the revised profit vector using the equation (PRO) t = σ t t−1 p x (3) We may also calculate the sterling reserves, t V , implied by zeroisation. Example 14.5.1. Using the profit signature (14.5.1), and supposing that q x+t−1 = 0.01 for t = 1, 2, 3, 4, 5, calculate the sterling reserves in each year that are implied by zeroisation of the profit signature. Solution (1) (2) (3) (4) (5) (6) Original profit Zeroised profit Remainder of Accum. Prob. that signature signature net-cash flow of (3) policy is in force t σ t σ t σ t −σ t at i s t p x t V 1 26.22 11.76 14.46 14.46 0.99 14.61 2 -8.05 0 -8.05 7.06 0.980 7.20 3 -5.18 0 -5.18 2.19 0.970 2.26 4 -2.29 0 -2.29 0 0.961 0 5 0.62 0.62 0 0 0.951 0 (6) = (4) (5) Observe that column (4) gives the sterling fund per policy sold, and the probability of being in force at time t is t p x . Thus Funds needed at time t per policy sold = t V Pr¦ policy is still in force at time t¦. 14.6 Withdrawals So far we have ignored the possibility of surrender. Now assume that the surrender of a policy may occur only at the end of a policy year (just before payment of the premium then due). Define w t = the probability that a policy will be surrendered at the end of year t (t = 1, 2, ..., n −1) Assume that w n = 0 as the policyholder will receive the maturity benefit at that time. The chance that a policy in force at the start of year t is surrendered at the end of year t is therefore, p x+t−1 w t Define (SV ) t = the surrender value at time t which is usually equal to F t , the bid value of the policy’s units, in some cases minus a surrender penalty of (say) £10 or 1% of F t . 238 CHAPTER 14. PROFIT TESTING FOR UNIT-LINKED POLICIES The revised profit vector, (PRO) t , allowing for withdrawals is (PRO) t = (PRO) t +p x+t−1 w t [F t −(SV ) t + t V ] (14.6.1) where (PRO) t is as before. If t V = 0 for all t, then obviously (PRO) t = (PRO) t +p x+t−1 w t [F t −(SV ) t ]. (14.6.2) Note that if there is no surrender penalty (i.e. (SV ) t = F t ), (PRO) t = (PRO) t . The profit signature, σ t , must also be adjusted to allow for surrenders; we have σ t = t−1 p x (PRO) t (14.6.3) where t−1 p x = Pr¦ policy is in force at time t −1, allowing for withdrawals¦ = t−1 p x (1 −w 1 )(1 −w 2 )...(1 −w t−1 ) if t ≥ 2 1 if t = 1 Note Even if (SV ) t = F t , and hence (PRO) t = (PRO) t , σ t will still differ from σ t as the probability of the policy still being in force in each year will be different, when t > 1. The profit signature allowing for withdrawals, ¦σ t ¦, may be zeroised in the same way as before. When calculating the reserves, t V , implied by zeroisation, the probability of the policy being in force at time t should be changed from t p x to t p x (1 −w 1 )(1 −w 2 )...(1 −w t ) Example 14.6.1. A life office issues a large number of 3-year unit-linked endowment policies to men aged 65, under each of which level annual premiums of £1,000 are paid. 80% of the first premium and 105% of each subsequent premium is invested in units at the offer price. There is a bid/offer spread in unit values, the bid price being 95% of the offer price. A fund management charge of 0.5% of the bid value of the policyholder’s fund is deducted at the end of each policy year, before payment of any benefits then due. The death benefit, which is payable at the end of the year of death, is £3,000 or the bid value of the units if greater. The maturity value is equal to the bid value of the units. The office incurs expenses of £100 at the start of the first year and £20 at the start of each of the second and third years. Mortality is assumed to follow A1967-70 ultimate. It is assumed that, at the end of each of the first two policy years, 2% of the surviving policyholders withdraw. The withdrawal benefit is 98% of the bid value of the units, after deducting the management charge. 14.6. WITHDRAWALS 239 (a) Assuming that the growth in the unit value is 7% p.a. and that the office holds unit reserves equal to the bid value of units and zero Sterling Reserves at the end of each year, calculate the profit emerging at the end of each policy year per policy sold. Sterling Reserves are assumed to earn interest at 6% p.a. (b) Calculate the revised profit emerging at the end of each year if the office takes a smaller profit in year 1 in order to ensure that the profit emerging in the second and third policy years is zero. Solution (a) We first work out the unit fund, F t , per policy sold, assuming it remains in force. Policy Cost of Funds brought Funds on end Unit fund Year allocation forward from start before deduction F.M.C. at end of year t P t (1 −λ)a t P t (1 −λ)a t +F t−1 of charge c t F t 1 760.00 760.00 813.20 4.07 809.13 2 997.50 1806.63 1933.09 9.67 1923.42 3 997.50 2920.92 3125.38 15.63 3109.75 We now calculate (PRO) t = (SCF) t and then find (PRO) t +p x+t−1 w t (0.02F t ), where x = 65. Policy Prem. less cost Accumulation D. G. In force Year of allocation Expenses in Sterling Fund costs F.M.C. cash flow t P t −P t (1 −λ)a t e t at 6% interest (DG) t c t (SCF) t 1 240 100 148.40 -52.65 4.07 99.82 2 2.50 20 -18.55 -28.57 9.67 -37.45 3 2.50 20 -18.55 0 15.63 -2.92 (PRO) t p x+t−1 w t (0.02F t ) (PRO) t t−1 p x σ t 99.82 0.32 100.14 1 100.14 -37.45 0.75 -36.70 0.95645 -35.10 -2.92 0 -2.92 0.91245 -2.66 where t−1 p x = t−1 p x (1 −w 1 )...(1 −w t−1 ) with w 3 = 0. (b) Let reduction in profit at time 1 be X. Then X = 35.10v + 2.66v 2 at i s = 6% = 35.48. Hence profit taken at time 1 is 100.14 −35.48 = £64.66 240 CHAPTER 14. PROFIT TESTING FOR UNIT-LINKED POLICIES Exercises 14.1 A life office issues a three-year unit-linked endowment policy to a life aged exactly 60. The annual premium is £2,000, payable at the start of each year. The allocation proportion is 90% in year 1 and 97% thereafter. At the end of year of death during the term, the policy pays the higher of £10,000 and the bid value of units allocated to the policy, after deduction of the fund management charge. A bonus of 2% of the (bid) value of the unit fund is payable at maturity. The life office makes the following assumptions in projecting future cash flows: Mortality A1967-70 ultimate Initial expenses: £300 Renewal expenses: £50, incurred at the start of the second and the third years Fund management charge: 2% per annum, taken at the end of each year before payment of any benefits Sterling fund interest rate: 4% per annum Bid/offer spread: 6% Unit fund growth rate: 10% per annum. Construct tables to show the following: (i) the growth of the unit fund; (ii) the profit signature, assuming that no sterling reserves are held; (iii) the profit signature after taking into account sterling reserves, given that the sterling reserves per policy are to be £36.48 before receipt of the premium due at time 1 year and £78.64 before receipt of the premium due at time 2 years. In each case, indicate clearly how you calculate your table entries. Ignore the possibility of surrenders. 14.2 (a) In the context of profit-testing of unit-linked business, define the following terms briefly: (i) the Unit Fund, (ii) the Sterling Reserves, (iii) the profit vector of a policy, (iv) the profit signature of a policy, (v) the risk discount rate, and (vi) zeroisation of Sterling Reserves. (b) A life office issues a large number of identical 4-year annual premium unit-linked endow- ment assurances to lives aged 65. According to the office’s calculations, the profit vector per policy sold, ignoring withdrawals and assuming that no Sterling reserves are maintained at the end of each year, is as follows (£): 191.12 −111.45 −3.28 10.95 The office’s mortality basis is A1967-70 ultimate, and Sterling Reserves earn interest at 5% per annum. Calculate (i) the profit signature per policy sold, ignoring any need to maintain Sterling Reserves at the end of each year, and 14.7. EXERCISES 241 (ii) the profit signature per policy sold if Sterling reserves are zeroised. (c) The office now wishes to make an allowance for surrenders. It assumes that, at the end of the first and the second policy years, 3% of the surviving policyholders will surrender (just before payment of the second and third annual premiums respectively.) Surrender values are equal to the value of the policyholder’s units (after deduction of fund management charges), with a surrender penalty of £10. Calculate (i) the revised profit signature per policy sold, ignoring any need to maintain Sterling Reserves at the end of each year, (ii) the revised profit signature per policy sold if the Sterling Reserves are zeroised, and (iii) the net present value, at a risk discount rate of 15% per annum, of the revised profit signature per policy sold, assuming that the Sterling Reserves are zeroised. 14.3 (a) If a profit test for a unit-linked policy reveals negative cash flows in the second or any subsequent policy year, it is customary to eliminate these negative values by setting up sterling reserves at the end of each year. Describe briefly the technique (“zeroisation”) by which these reserves are calculated. (b) An office issues a 3-year unit-linked policy with a yearly premium of £500. The death benefit, payable at the end of the year of death, is £1,000 or the bid value of units if greater. The maturity value is the bid value of the units at maturity. 95% of each premium is invested in units at the offer price. The bid price of units is 95% of the offer price. Management charges of 1 4 % of the bid value of the units are deducted at the end of each year (before payment of death and maturity claims). The office expects to incur expenses of £75 at the start of the first year and £25 at the start of each subsequent year. Using a profit testing analysis, calculate for a life aged 60 at entry (i) the expected profit in each of the 3 years per policy in force at the beginning of the year, (ii) the net present value at the issue date of the expected profit from one policy assuming a risk discount rate of 10% per annum. Assume that the unit fund grows at 8% per annum (before deduction of management charges), that sterling reserves need not be maintained at the end of each year, and that the possibility of surrender may be ignored. The mortality of policyholders follows A1967-70 ultimate and sterling reserves earn interest at 6% per annum during each policy year. 14.4 If a profit test for a unit-linked policy reveals negative cash flows in the second or any sub- sequent policy year, it is customary to eliminate these negative values by setting up sterling reserves at the end of each year. Calculate the sterling reserves required at the end of each policy year, per policy then in force, for a 3- year policy for which the profit signature (with no allowance for sterling reserves at the end of each year) is (250, - 100, - 50), given that the rate of mortality is 0.01 per annum at each age and sterling reserves earn interest at 8% p.a. 14.5 An office issues a unit-linked endowment assurance with annual premium £400 and term five years to a life aged 60 who is subject to A1967-70 ultimate mortality. The sum assured, payable at the end of year of death or at the maturity date, is the bid value of the units held, subject to a guaranteed minimum death benefit of £2,000. The allocation proportion is 70% for the first annual premium and 98% for all subsequent annual premiums. For units the bid/offer spread is 5% and the annual rate of management charge is 0.75%. In determining the sterling reserves necessary for the policy the office makes the following assumptions: 242 CHAPTER 14. PROFIT TESTING FOR UNIT-LINKED POLICIES Initial expenses: £125 Renewal expenses (associated with the payment of the second and each subsequent premium) £20 increased by 7% p.a. compound from the outset of the policy. Growth rate for units: 7% p.a. Interest rate for sterling fund: 4% p.a. (a) On this basis (i) construct a table showing the growth of the unit fund over the duration of the policy, and (ii) construct a table showing the growth of the sterling fund in the absence of reserves. (b) Hence determine the sterling reserves which should be held by the office to eliminate the sterling fund negative cash flows in the second and subsequent years of the policy’s duration. (c) Consider the unit-linked policy described above. Suppose, however, that the growth rate for units will be 10% p.a., that the sterling fund interest rate will be 6% p.a., and that the inflationary growth rate for renewal expenses will be 4% p.a. (from the outset of the policy). (i) Construct a table showing the growth of the unit fund over the duration of the policy and (ii) Construct a table showing the growth of the sterling fund in the absence of reserves. (iii)Assuming that the office sets up the sterling fund reserves found above, determine the resulting sterling fund profit vector and signature. Find also the internal rate of return corresponding to the profit signature. 14.8. SOLUTIONS 243 Solutions 14.1 (i) (1) t P t (1 −λ)a t P t (1 −λ)a t +F t−1 (1) 1.1 F.M.C. F t 1 1692 1692 1861.20 37.22 1823.98 2 1823.60 3647.58 4012.34 80.25 3932.09 3 1823.60 5755.69 6331.26 126.63 6204.63 (ii) (2) (3) Maturity t P t −P t (1 −λ)a t e t [(2) −(3)] 1.04 (DG) t FMC Cost (SCF) t 1 308 300 8.32 118.00 37.22 0 -72.46 2 176.40 50 131.46 97.17 80.25 0 114.54 3 176.40 50 131.46 67.37 126.63 121.89 68.83 (Maturity cost in year 3 = 0.02(6204.63) p 62 .) t (PRO) t t−1 p 60 σ t 1 -72.46 1 -72.46 2 114.54 0.98557 112.89 3 68.83 0.96979 66.75 (iii) (PRO) t = (SCF) t + (1 +i s ) t−1 V − t V.p x+t−1 . t (SCF) t (1.04) t−1 V p 59+t . t V (PRO) t t−1 p 60 σ t 1 -72.46 0 35.95 -108.41 1 -108.41 2 114.54 37.94 77.38 75.10 0.98557 74.02 3 68.83 81.79 0 150.62 0.96979 146.07 14.2 (a) Simple definition of each term required. Check definitions with text. (b) (i) t (PRO) t t−1 p 65 σ t 1 191.12 1 191.12 2 -111.45 0.97597 -108.77 3 -3.28 0.95007 -3.12 4 10.95 0.92226 10.10 (ii) Let X be the amount withheld at time 1 to cover negative cash flows at times 2 and 3. X = 108.77v + 3.12v 2 at 5% interest = 106.42 Hence σ t = ¸ ¸ ¸ 84.70 0 0 10.10 ¸ (c) (i) Notice that F t −(SV ) t = 10 for all t. Hence (PRO) t = (PRO) t + 10w t p 64+t 244 CHAPTER 14. PROFIT TESTING FOR UNIT-LINKED POLICIES t (PRO) t 10w t p 64+t (PRO) t t−1 p 65 σ t 1 191.12 0.29 191.41 1 191.41 2 -111.45 0.29 -111.16 0.94669 -105.23 3 -3.28 0 -3.28 0.89392 -2.93 4 10.95 0 10.95 0.86775 9.50 Notice that w t = 0.03 for t = 1, 2 0 for t = 3, 4 (ii) Retain X at time 1. X = 105.23v + 2.93v 2 at 5% interest = 102.88 Hence zeroised profit signature is ¸ ¸ ¸ 88.53 0 0 9.50 ¸ (iii) NPV = 88.53v + 9.50v 4 at 15% interest = £82.41. 14.3 (a) Zeroisation is the process whereby the profit in the first year is reduced to pay for any future negative cash flows (as explained in text). (b) (i) (1) t P t (1 −λ)a t P t (1 −λ)a t +F t−1 (1) 1.08 FMC F t 1 451.25 451.25 487.35 1.22 486.13 2 451.25 937.38 1012.37 2.53 1009.84 3 451.25 1461.09 1577.98 3.94 1574.04 (2) (3) t P t −P t (1 −λ)a t e t [(2) −(3)] 1.06 (DG) t FMC (SCF) t = (PRO) t 1 48.75 75 -27.82 7.42 1.22 -34.02 2 48.75 25 25.17 0 2.53 27.70 3 48.75 25 25.17 0 3.94 29.11 (ii) t (PRO) t t−1 p 60 σ t 1 -34.02 1 -34.02 2 27.70 0.98557 27.30 3 29.11 0.96979 28.23 Hence net present value = 3 ¸ t=1 σ t v t at 10% interest = −34.02v + 27.30v 2 + 28.23v 3 = £12.84. 14.4 σ t = ¸ 250 −100 −50 ¸ 14.8. SOLUTIONS 245 Let X be the sum retained in year 1 to cover the later negative cash flows. Then X = 100v + 50v 2 at 8% interest = £135.46 Hence the zeroised profit signature is σ t = ¸ 114.54 0 0 ¸ (1) Accumulation t σ t σ t σ t −σ t of (1) t p x t V 1 250 114.54 135.46 135.46 0.99 136.83 2 -100 0 -100 46.30 0.980 47.24 3 -50 0 -50 0 0.970 0 14.5 (a) (i) (1) t P t (1 −λ)a t P t (1 −λ)a t +F t−1 (1) 1.07 FMC F t 1 266.00 266.00 284.62 2.13 282.49 2 372.40 654.89 700.73 5.26 695.47 3 372.40 1067.87 1142.62 8.57 1134.05 4 372.40 1506.45 1611.90 12.09 1599.81 5 372.40 1972.21 2110.26 15.83 2094.43 (ii) (2) (3) t P t −P t (1 −λ)a t e t [(2) −(3)] 1.04 (DG) t FMC (SCF) t 1 134 125 9.36 24.79 2.13 -13.30 2 27.60 21.40 6.45 20.89 5.26 -9.18 3 27.60 22.90 4.89 15.37 8.57 -1.91 4 27.60 24.50 3.22 7.87 12.09 7.44 5 27.60 26.22 1.44 0 15.83 17.27 t (PRO) t t−1 p 60 σ t 1 -13.30 1 -13.30 2 -9.18 0.98557 -9.05 3 -1.91 0.96979 -1.85 4 7.44 0.95257 7.09 5 17.27 0.93385 16.13 (b) Let X be the amount that has to be withheld at time 1 to pay for the negative cash flows at times 2 and 3. Hence X = 9.05v + 1.85v 2 at 4% interest = £10.41 So the zeroised profit signature is σ t = ¸ ¸ ¸ ¸ ¸ −23.71 0 0 7.09 16.13 ¸ 246 CHAPTER 14. PROFIT TESTING FOR UNIT-LINKED POLICIES (4) Accumulation of t σ t −σ t (4) at 4% interest t p x t V 1 10.41 10.41 0.98557 10.56 2 -9.05 1.78 0.96979 1.84 3 -1.85 0 0.95257 0 4 0 - 0 5 0 - 0 (c) (i) t P t (1 −λ)a t P t (1 −λ)a t +F t−1 (4)1 1 FMC F t 1 266.00 266.00 292.60 2.19 290.41 2 372.40 662.81 729.09 5.47 723.62 3 372.40 1096.02 1205.62 9.04 1196.58 4 372.40 1568.98 1725.88 12.94 1712.94 5 372.40 2085.34 2293.87 17.20 2276.67 (ii) (5) (6) t P t −P t (1 −λ)a t e t [(5) −(6)] 1.06 (DG) t FMC (SCF) t 1 134.00 125.00 9.54 24.67 2.19 -12.94 2 27.60 20.80 7.21 20.44 5.47 -7.76 3 27.60 21.63 6.33 14.26 9.04 1.11 4 27.60 22.50 5.41 5.64 12.94 12.71 5 27.60 23.40 4.45 0 17.20 21.65 (iii) t (SCF) t t V (1.06) t−1 V p 59+t t V (PRO) t t−1 p 60 σ t 1 -12.94 10.56 0 10.41 -23.35 1 -23.35 2 -7.76 1.84 11.19 1.81 1.62 0.98557 1.60 3 1.11 0 1.95 0 3.06 0.96979 2.97 4 12.71 0 0 0 12.71 0.95257 12.11 5 21.65 0 0 0 21.65 0.93385 20.22 This uses (PRO) t = (SCF) t + (1 +i s ) t−1 V −p x+t−1 . t V Let the internal rate of return be j p.a. Then j solves 5 ¸ t=1 σ t v t = 0 where v = 1 1 +j . i.e. −23.35v + 1.60v 2 + 2.97v 3 + 12.11v 4 + 20.22v 5 = 0. Solving this equation by trials and interpolation gives v = 0.87173. Hence j = 0.147 = 14.7% p.a. Chapter 15 MULTIPLE-DECREMENT TABLES 15.1 Introduction Consider a body of lives subject to two “modes of decrement”, α and β. For example, consider a group of bachelor employees of a large company, from which men can leave by either mode α, marriage, or by mode β, leaving the company (mortality being ignored). Take a bachelor employee aged x, and let T 1 =time to marriage of (x), whether or not he is still an employee of the company and T 2 =time until (x) leaves the service of the company, whether he is then married or not (At this point it is assumed that all bachelors eventually marry.) Let T = min¦T 1 , T 2 ¦ =time until exit from the group of bachelor employees, by either mode α (marriage) or mode β (withdrawal from service), whichever comes first, of a bachelor employee aged x Note This is similar to the joint-life situation in Chapter 11. Define t (ap) x =the probability that (x) will “survive” for at least t years with respect to both modes of decrement. (In the above example, he will not marry or leave the service within t years.) =Pr¦T ≥ t¦ 247 248 CHAPTER 15. MULTIPLE-DECREMENT TABLES and t (aq) x =Pr¦T < t¦ =Pr¦(x) leaves before time t, either by mode α or mode β¦ In the example above t (aq) x = the probability that (x) gets married or leaves service (or does both these things) within t years We may proceed through the development of “life tables”, by merely putting “a” in front of the various functions. For example, the function, (al) x , x ≥ x 0 , is constructed to be such that t (ap) x = (al) x+t (al) x , (x ≥ x 0 , t ≥ 0) When t = 1 it may be omitted, giving (aq) x = (ad) x (al) x = (al) x −(al) x+1 (al) x and t [(aq) x = (ad) x+t (al) x Also, define the “force of exit” from the double-decrement table (by whichever mode occurs first) by (aµ) x = lim h→0 + h (aq) x h It follows as for ordinary life tables that t (ap) x = exp ¸ − t 0 (aµ) x+r dr The probability density function of T is t (ap) x (aµ) x+t , t ≥ 0 0 , t < 0 and its distribution function is t (aq) x = t 0 r (ap) x (aµ) x+r dr (t ≥ 0) We also define (am) x =central rate of “total decrement” (modes α and β together) = (ad) x (aL) x = 1 0 (al) x+t (aµ) x+t dt 1 0 (al) x+t dt 15.2. THE ASSOCIATED SINGLE-DECREMENT TABLES 249 15.2 The Associated Single-Decrement Tables Each mode of decrement may be thought of as possessing its own “life table”, denoted by, for example, l α x or l β x . Thus t p α x = Pr¦(x) does not exit by mode α before time t, whether exit by mode β happens or not¦ ( = Pr¦(x) does not marry within t years ¦ in our example.) t p β x = Pr¦(x) does not exit by mode β before time t, whether exit by mode α happens or not¦ ( = Pr¦(x) does not leave service within t years¦ in our example.) Note that t p α x = Pr¦T 1 ≥ t¦ and t p β x = p r ¦T 2 ≥ t¦. We also have t q α x = Pr¦(x) exits by mode α before time t, whether exit by mode β happens or not¦ = Pr¦T 1 ≤ t¦ = 1 − t p α x . Similarly t q β x = Pr¦T 2 ≤ t¦ = 1 − t p β x . The functions, t q α x , t q β x are called the independent rates (or probabilities) of exit within t years at age x in their respective single-decrement tables. 15.3 The Relationships between the Multiple-Decrement Ta- ble and its Associated Single-Decrement Tables It is usually assumed that the variables T 1 and T 2 are independent. This gives t (ap) x = Pr¦T 1 ≥ t and T 2 ≥ t¦ = Pr¦T 1 ≥ t¦Pr¦T 2 ≥ t¦ = t p α xt p β x (15.3.1) Hence we may construct the function (al) x from l α x , l β x . In particular t (ap) x = (al) x+t (al) x = l α x+t l α x l β x+t l β x for all x ≥ x 0 , t ≥ 0 250 CHAPTER 15. MULTIPLE-DECREMENT TABLES Put x = x 0 , y = x 0 +t to obtain (al) y = k.l α y .l β y (y ≥ x 0 ) where k is a constant. If we choose the radix (al) x 0 to be equal to l α x 0 .l β x 0 , then (al) y = l α y .l β y for all y ≥ x 0 ( as k = 1) Some Important Formulae t (aq) x = 1 − t (ap) x = 1 −(1 − t q α x )(1 − t q β x ), assuming independence of T 1 and T 2 , = t q α x + t q β x − t q α x . t q β x If t = 1 , (aq) x = q α x +q β x −q α x q β x (15.3.2) Also (aµ) x = lim h→0 + h (aq) x h = lim h→0 + ¸ h q α x + h q β x − h q α x . h a β x h = lim h→0 + h q α x h + lim h→0 + h q β x h + lim h→0 + h q α x h h q β x h h = µ α x +µ β x (similar to joint-life argument). 15.4 Dependent Rates of Exit Define t (aq) α x = Pr¦(x) exits by mode α within time t, exit by mode β not having previously occurred¦ In terms of our example, this is the probability that (x) gets married within t years while still an employee of the company. We have t (aq) α x = Pr¦T 1 < t and T 1 < T 2 ¦ where T 1 , T 2 have p.d.f.’s f 1 (t 1 ) = t 1 p α x µ α x+t 1 , t 1 > 0, f 2 (t 2 ) = t 2 p β x µ β x+t 2 , t 2 > 0, respectively . So 15.4. DEPENDENT RATES OF EXIT 251 t (aq) α x = t 1 <t 2 t 1 <t f 1 (t 1 ).f 2 (t 2 ) dt 1 dt 2 = t 0 ∞ t 1 t 1 p α x µ α x+t 1 . t 2 p β x µ β x+t 2 dt 2 dt 1 = t 0 t 1 p α x µ α x+t 1 ∞ t 1 t 2 p β x µ β x+t 2 dt 2 dt 1 = t 0 t 1 p α x µ α x+t 1 . t 1 p β x dt 1 Thus t (aq) α x = t 0 r p α x .µ α x+r . r p β x dr (15.4.1) = t 0 r (ap) x µ α x+r dr When t = 1, (aq) α x = 1 0 t p α x µ α x+t . t p β x dt. (aq) α x , (aq) β x are called the dependent rates (or probabilities) of exit at age x by mode α, β respec- tively. (The word ‘dependent’ indicates that (aq) α x depends not only on mode α but also on mode β). Example 15.4.1. A population is subject to 2 modes of decrement, α and β. For 30 ≤ x ≤ 32, it is known that µ α x = 0.05 and µ β x = (35 −x) −1 For an individual in this population at exact age 30, calculate (a) the probability that he will still be in the population at age 32. (b) the probability that he will leave by mode α before age 32. Solution (a) t p α 30 = exp − t 0 µ α 30+r dr = e −0.05t t p β 30 = exp − t 0 [35 −(30 +r)] −1 dr = exp − t 0 1 5 −r dr = exp[log(5 −r)] t 0 = 5 −t 5 . Thus 2 (ap) 30 = 2 p α 30 . 2 p β 30 = e −0.1 3 5 = 0.5429 252 CHAPTER 15. MULTIPLE-DECREMENT TABLES (b) 2 (aq) α 30 = 2 0 t p α 30 µ α 30+t . t p β 30 dt = 2 0 0.05e −0.05t 5 −t 5 dt = 0.05 2 0 e −0.05t dt −0.01 2 0 t.e −0.05t dt = (1 −e −0.1 ) + 2 5 e −0.1 −4(1 −e −0.1 ) (using integration by parts) = 0.07645. Theorem If there is a Uniform Distribution of Decrements between ages x and x + 1 in each single-decrement table, then (aq) α x = q α x 1 − 1 2 q β x and (aq) β x = q β x 1 − 1 2 q α x Proof Under a Uniform Distribution of Decrements (U.D. of D.) for a life table, (a) l x+t is linear (0 ≤ t ≤ 1) (b) t q x = t.q x (0 ≤ t ≤ 1) (c) t p x µ x+t = q x (0 ≤ t ≤ 1) Hence (aq) α x = 1 0 t p α x .µ α x+t . t p β x dt = 1 0 q α x (1 − t q β x ) dt (using U.D. of D. for mode α) = q α x 1 0 (1 −t.q β x ) dt (using U.D. of D. for mode β) = q α x ¸ t − t 2 2 q β x t=1 t=0 = q α x (1 − 1 2 q β x ). Note It is often assumed that U.D. of D. holds approximately in each single-decrement table, so the above results may be used as approximations. Example 15.4.2. In a double-decrement table with 2 causes of decrement, death (d) and withdrawal (w), the central rates of decrement at a certain age x are m d x = 0.01 m w x = 0.2 Show that if the central rate of withdrawal was doubled, the dependent rate of mortality would be reduced by approximately 0.00076. 15.5. PRACTICAL CONSTRUCTION OF MULTIPLE-DECREMENT TABLES 253 Solution Assuming U.D. of D., q d x · m d x 1 + 1 2 m d x = 0.009950 q w x · m w x 1 + 1 2 m w x = 0.18182 Hence (aq) d x · q d x (1 − 1 2 .q w x ) = 0.00905 Now suppose m w x is increased to 0.4. Then q w x · 0.4 1 + 0.2 = 0.3333. Hence (aq) d x · q d x (1 − 1 2 0.3333) = 0.00829 i.e. a reduction of 0.00076. Getting q α x , q β x from (aq) α x , (aq) β x Assuming U.D. of D., we have shown that (aq) α x = q α x (1 − 1 2 q β x ), (i) (aq) β x = q β x (1 − 1 2 q α x ) (ii) Substituting q β x = (aq) β x 1− 1 2 q α x from (ii) into (i) gives a quadratic equation for q α x , i.e. (q α x ) 2 +q α x [(aq) β x −(aq) α x −2] + 2(aq) α x = 0. This can be solved (rejecting any solution outside the range 0 and 1) to find q α x , and hence q β x may be found from (i) or (ii). 15.5 Practical Construction of Multiple-Decrement Tables We define (ad) α x =the expected number of exits by mode α between ages x and x + 1 in a multiple-decrement table with (al) x lives at age x =(al) x (aq) α x Note also that (aq) x = (aq) α x + (aq) β x 254 CHAPTER 15. MULTIPLE-DECREMENT TABLES To prove these results, we note that t (aq) α x + t (aq) β x = Pr¦T 1 ≤ t and T 1 ≤ T 2 ¦ +Pr¦T 2 ≤ t and T 2 ≤ T 1 ¦ = Pr¦min¦T 1 , T 2 ¦ ≤ t¦ = t (aq) x . and hence (putting t = 1, which can be omitted) (ad) x = (al) x (aq) x = (ad) α x + (ad) β x Also, t [(aq) α x = Pr¦(x) leaves by mode α in the multiple-decrement table between ages x +t and x +t + 1¦ = t (ap) x (aq) α x+t = (ad) α x+t (al) x Procedure There are 3 basic steps in the construction of a multiple-decrement table. Step 1 Choose a radix (al) x 0 where x 0 = the youngest age considered. (For example, (al) x 0 may be 10,000 or 100,000). Step 2 Calculate q α x , q β x (x = x 0 , x 0 + 1, ...) and hence evaluate (aq) α x , (aq) β x (x = x 0 , x 0 + 1, ...), assuming U.D. of D. Step 3 Calculate (ad) α x 0 = (al) x 0 (aq) α x 0 and (ad) β x 0 = (al) x 0 (aq) β x 0 Then find (al) x 0 +1 = (al) x 0 −[(ad) α x 0 + (ad) β x 0 ] = (al) x 0 −(ad) x 0 Repeat this procedure for x 0 + 1, x 0 + 2 and so on. Example 15.5.1. A certain population is subject to 2 modes of decrement, d = death i = permanent disability 15.5. PRACTICAL CONSTRUCTION OF MULTIPLE-DECREMENT TABLES 255 The independent rates of mortality are in accordance with A67-70 ult., and there is an indepen- dent rate of disablement of 0.01 at each age x (60 ≤ x ≤ 62). Construct a multiple-decrement table for ages x = 60, 61, 62 (and include the number of survivors at age 63.) Solution Use U.D. of D. to find (aq) d x = q d x (1 − 1 2 q i x ) and (aq) i x = q i x (1 − 1 2 q d x ) Choose a radix of (al) x 0 = 100, 000 (for example) x q d x q i x (aq) d x (aq) i x (al) x (ad) d x (ad) i x 60 0.01443 0.01 0.01436 0.00993 100,000 1,436 993 61 0.01601 0.01 0.01593 0.00992 97,571 1,554 968 62 0.01775 0.01 0.01776 0.00991 95,049 1,679 942 63 92,428 Applications of Multiple-Decrement Tables 1. The probabilities of various events can be calculated. Example 15.5.2. Using the multiple-decrement table constructed in example 15.5.1, calculate (a) The probability that a person aged exactly 60 is alive and not disabled at age 63. (b) The probability that a person aged 60 becomes disabled within 3 years. Solution (a) 3 (ap) 60 = (al) 63 (al) 60 = 92428 100000 = 0.92428 (b) 3 (aq) i 60 = (ad) i 60 + (ad) i 61 + (ad) i 62 (al) 60 = 993 + 968 + 942 100, 000 = 0.02903 2. Financial calculations can be carried out using multiple-decrement tables, as will be shown in the next chapter. 256 CHAPTER 15. MULTIPLE-DECREMENT TABLES 15.6 Further Formulae The Identity of the forces Define (aµ) α x = lim h→0 + h (aq) α x h Theorem (aµ) α x = µ α x for all x (15.6.1) Proof Assume that µ α x and µ β x are continuous. Then (aµ) α x = lim h→0 + h (aq) α x h = lim h→0 + h 0 t (ap) x µ α x+t dt h Now, by L’Hˆopital’s rule, (aµ) α x = lim h→0 + d dh [ h 0 t (ap) x µ α x+t dt] d dh (h) = lim h→0 + h (ap) x µ α x+h 1 = µ α x (as lim h→0 + h (ap) x = 1) This result is called the identity of the forces. Note that (aµ) x = lim h→0 + h (aq) x h = lim h→0 + ¸ h (aq) α x h + h (aq) β x h = (aµ) α x + (aµ) β x = µ α x +µ β x Central Rates of Decrement The dependent central rate of decrement by mode α at age x is given by (am) α x = (ad) α x (aL) x = l 0 (al) x+t (aµ) α x+t dt 1 0 (al) x+t dt = 1 0 (al) x+t µ α x+t dt 1 0 (al) x+t dt (from (15.6.1)) · µ α x+ 1 2 Compare this to the corresponding independent central rate of decrement: 15.7. GENERALIZATION TO 3 MODES OF DECREMENT 257 m α x = 1 0 l α x+t µ α x+t dt 1 0 l α x+t dt · µ α x+ 1 2 Therefore it is normally assumed that (am) α x · m α x Note We may also construct tables using select rates of decrement, using q d x = q [x] , q d x+1 = q [x]+1 , ... for example. 15.7 Generalization to 3 Modes of Decrement Suppose there are now 3 modes of decrement, α, β and γ. By generalising the results for 2 modes, it can be shown that t (ap) x = t p α x . t p β x . t p γ x and (aq) α x = 1 0 t p α x µ α x+t . t p β x . t p γ x dt Theorem Under U.D. of D. of each mode of decrement in its single-decrement table (aq) α x = q α x [1 − 1 2 (q β x +q γ x ) + 1 3 q β x q γ x ] Proof (aq) α x = 1 0 t p α x µ α x+t . t p β x . t p γ x dt = 1 0 q α x (1 − t q β x )(1 − t q γ x ) dt (using U.D. of D. on α) = q α x 1 0 (1 −t.q β x )(1 −t.q γ x ) dt (using U.D. of D. on β and γ) = q α x 1 0 1 −t(q β x +q γ x ) +t 2 q β x .q γ x dt = q α x ¸ t − t 2 2 (q β x +q γ x ) + t 3 3 q β x .q γ x t=1 t=0 = q α x ¸ 1 − 1 2 (q β x +q γ x ) + 1 3 q β x .q γ x . Note also that the identity of the forces remains true, and (aµ) x = µ α x +µ β x +µ γ x . 258 CHAPTER 15. MULTIPLE-DECREMENT TABLES In the practical construction of a multiple-decrement table, we find the values of (al) x , (ad) α x , (ad) β x , (ad) γ x and use (al) x+1 = (al) x −[(ad) α x + (ad) β x + (ad) γ x ]. Example 15.7.1. The members of a large company’s manual workforce are subject to three modes of decrement, death, withdrawal and promotion to supervisor. It is known that these workers’ independent rates of mortality are those of English Life Table No. 12 - Males, the independent withdrawal rate is 0.04 at each age, and their independent promotion rate is 0.02 at age 50 and 0.03 at age 51. (a) Draw up a service table for manual workers from age 50 to age 51 with a radix of 100,000 at age 50, including the value of (al) 52 . (b) Calculate the probability that a life aged exactly 50 will gain promotion within 2 years. Solution (a) Let d = death, p = promotion, w = withdrawal. x q d x q w x q p x (aq) d x (aq) w x (aq) p x (al) x (ad) d x (ad) w x (ad) p x 50 0.00728 0.04 0.02 0.00706 0.03946 0.01953 100,000 706 3946 1953 51 0.00823 0.04 0.03 0.00795 0.03924 0.02928 93,395 742 3665 2735 52 86,253 This table assumes U.D. of D., i.e. (aq) d x = q d x [1 − 1 2 (q w x +q p x ) + 1 3 q w x .q p x ]. with similar formulae for withdrawal and promotion rates. (b) 2 (aq) p 50 = (ad) p 50 + (ad) p 51 (al) 50 = 1953 + 2735 100, 000 = 0.04688. 15.8 “Abnormal” Incidence of Decrement Suppose that there are 2 modes of decrement, α and β operating between ages x and x+1. Assume mode β operates “smoothly” (i.e. µ β x exists and is continuous) between ages x and x + 1. However, mode α only operates at a particular age, x +k (0 ≤ k ≤ 1). That is t p α x = 1 for t < k 1 −q α x for t > k. It follows that µ α x+t is not defined for t = k and hence formulae such as (aq) α x = 1 0 t p α x . t p β x µ α x+t dt 15.8. “ABNORMAL” INCIDENCE OF DECREMENT 259 cannot be used. To deal with this “abnormal” mode of decrement, we argue that from age x to x + k, mode β operates by itself. Then, at age x +k, there is a chance q α x of exit by mode α, so (aq) α x = Pr¦(x) survives to age x +k under mode β only¦ Pr¦(x) leaves at age x +k by mode α, given survival until then¦ = k p β x .q α x = q α x (1 − k q β x ) If (as usually the case) U.D. of D. for mode β is assumed (aq) α x = q α x (1 −k.q β x ) (15.8.1) To obtain (aq) β x , we may use (aq) β x = (aq) x −(aq) α x = [q α x +q β x −q α x .q β x ] −q α x (1 −k.q β x ) So (aq) β x = q β x [1 −(1 −k)q α x ] (15.8.2) Special Cases 1. If k = 1 2 , we obtain the familiar formulae (aq) α x = q α x (1 − 1 2 q β x ) and (aq) β x = q β x (1 − 1 2 q α x ) 2. If k = 0 (aq) α x = q α x and (aq) β x = q β x (1 −q α x ) 3. If k = 1, (i.e. we consider exits by mode α to occur just before reaching age x + 1) (aq) α x = q α x (1 −q β x ) and (aq) β x = q β x . Example 15.8.1. Military personnel attend a training camp for an intensive course which involves 3 weeks of continual exercises. Each soldier remains in the camp for exactly 3 weeks, provided he is not hospitalised because 260 CHAPTER 15. MULTIPLE-DECREMENT TABLES of injury or “failed” by one of the instructors and sent back to his base. The independent weekly rates of decrement are as follows: Hospitalised Week through injury Being failed 1 0.078 0.132 2 0.102 0.092 3 0.058 0.043 (i) Of a group of 1,000 soldiers who start the course, calculate the number who will successfully complete the course, the number who will be hospitalised and the number who will be failed. Assume a uniform distribution over each week of hospitalisation and failure. Anyone sent to hospital or failed leaves the camp immediately and does not return. There are no other modes of decrement. (ii) Some time later the course is altered so that in the first week the instructors test the sol- diers only at the start of the sixth day, and the independent weekly rate of being failed in week 1 alters to 0.160. During weeks 2 and 3, testing takes the form of continuous assessment as previously, with no change to the independent weekly rates of being failed. Calculate the revised numbers of soldiers successfully completing the course, being hospitalised and being failed, assuming nothing else changes. Solution (i) Consider “age” as time (in weeks) since entry to camp. Assume U.D. of D., which gives (aq) h x = q h x (1 − 1 2 q f x ) and (aq) f x = q f x (1 − 1 2 q h x ) where h = hospitalised, f = failed. x q h x q f x (aq) h x (aq) f x (al) x (ad) h x (ad) f x 0 0.078 0.132 0.07285 0.12685 1000 73 127 1 0.102 0.092 0.09731 0.08731 800 78 70 2 0.058 0.043 0.05675 0.04175 652 37 27 3 588 Hence number who complete the course = 588. (ii) Here, k = 5 7 (test is at end of 5 th day), so (aq) h x = q h x (1 − 2 7 q f x ) and (aq) f x = q f x (1 − 5 7 q h x ) for x = 0 only. q f 0 is now 0.16, but all the other values of q f x and q h x are unchanged. x q h x q f x (aq) h x (aq) f x (al) x (ad) h x (ad) f x 0 0.078 0.16 0.07443 0.15109 1000 74 151 1 0.102 0.092 0.09731 0.08731 775 75 68 2 0.058 0.043 0.05675 0.04175 632 37 26 3 570 Hence now the number who complete the course = 570. 15.8. “ABNORMAL” INCIDENCE OF DECREMENT 261 Application to Profit-Testing Let α = withdrawal, and β = mortality. It is assumed that withdrawals occur only at the end of a policy year. That is, mode α only operates at time k = 1. In chapter 14, the following results were used: Pr¦ policy in force at time t −1 will be surrendered at time t¦ = (1 −q mortality x+t−1 ).q withdrawal x+t−1 where q withdrawal x+t−1 was denoted by w t and Pr¦ death occurs in policy year given it is in force at start of year¦ = q mortality x+t−1 These formulae correspond exactly to those given in this section when k = 1. Extension to 3 modes of decrement With 3 modes of decrement, formulae similar to (15.8.1) and (15.8.2) may be awkward to derive. These problems are best handled from first principles. Often one mode (α, say) operates at exact age x and the others (β and γ) operate uniformly over the year of age from x to x + 1. This can be dealt with by defining (al) x = number of lives at age x before mode α operates, and (al) + x = number of lives at age x after mode α operates. Then the numbers of exits by modes β and γ between ages x and x + 1 can be worked out as if there were no other modes of decrement. Note however that (ad) β x = (al) + x (aq) β x where (aq) β x = q β x (1 − 1 2 q γ x ) Similar calculations apply to “abnormal” exits at the end of the year. 262 CHAPTER 15. MULTIPLE-DECREMENT TABLES Example 15.8.2. Among the employees of a certain firm retirement may take place at or after age 57, but is compulsory at age 60. The independent rates of mortality of the employees are those of A1967-70 ultimate. 20% of those attaining age 57 retire at once. Leaving aside these retirements, the central rates of retirement are as follows: Central Rate Age of Retirement 57 0.09 58 0.08 59 0.05 There are no withdrawals or ill-health retirements after age 50. Construct a service table for employees from age 57 to age 60, with a radix of 100,000 employees attaining age 57. Solution Create a special mode of decrement, r , to refer to retirement at exact age x. Then, (ad) r 57 = 20, 000, leaving (al) + 57 lives who are subject to ‘normal’ retirements and death, both of which are approxi- mately “U.D. of D.” So construct a table of (aq) d x and (aq) r x (ignoring mode r ). x q d x q r x · m r x 1+ 1 2 m r x (aq) d x (aq) r x 57 0.01050 0.08612 0.01004 0.08567 58 0.01169 0.07692 0.01124 0.07647 59 0.01299 0.04878 0.01267 0.04846 Now construct the multiple-decrement table, noting that (ad) r x = (al) + x (aq) r x . x (al) x (ad) r x (al) + x (ad) r x (ad) d x 57 100,000 20,000 80,000 6,854 803 58 72,343 0 72,343 5,532 813 59 65,998 0 65,998 3,198 836 60 61,964 61,964 0 Note We may take (al) 60 = (ad) r 60 since all survivors at age 60 retire at once. 15.9. EXERCISES 263 Exercises 15.1 Let α 1 , α 2 be the modes of decrement in a double-decrement table. Suppose that α 1 is uniformly distributed over the year of age from x to x + 1 in its associated single- decrement table, and µ α 2 x+t = c for 0 ≤ t ≤ 1. Find formulae for (aq) α 1 x and (aq) α 2 x in terms of q α 1 x and q α 2 x . 15.2 For a certain group of married women, for whom remarriage is not permitted, the dependent q- type rate of widowhood at each integer age x from 70 to 72 inclusive is twice the corresponding dependent q-type rate of mortality. The independent rates of mortality of wives follow a(55) ultimate (females). (i) Using a radix of 100,000 and assuming a uniform distribution of each mode of decrement in its associated single-decrement table, construct a double-decrement table for married women from age 70 to age 72 inclusive, giving also the value of (al) 73 , the number of wives at age 73. (ii) Find the probabilities that a wife now aged 70 will (a) be alive and married at age 73; and (b) be widowed within 3 years. 15.3 In a certain country, widowed and divorced men are subject to the following independent q-type rates of decrement: mortality: English Life Table No. 12 - Males remarriage: rates depend on the age at, and the duration since, the end of former marriage; the following table is an extract from these rates: exact age at end duration duration of former 0 1 marriage year year 50 0.050 0.025 51 0.045 0.023 52 0.042 0.020 Calculate the probability that a man aged exactly 50 whose marriage has just ended will remarry within 2 years. 15.4 A large industrial company recruits a constant number of school leavers aged exactly 18 years on 1 July each year. Upon joining, workers undergo training for one year. Of those who complete this period of training, ten per cent fail a final test of competence and are dismissed. Employees may also leave service voluntarily at any time. The central rate of voluntary withdrawal from service is 0.15 for trainees and 0.10 at each age for fully trained employees. The occupation is hazardous and all workers, including trainees, are exposed to the risk of injury. The independent q-type rate of injury is 0.051219 at age 18 and 0.050030 at ages 19 and above. An employee who is injured is transferred to alternative work with a subsidiary company, at a relocation cost of £1,000. The mortality of all employees follows English Life Tables No.12 - Males. The number of employees attaining age 21 each year is 500. (i) Construct a service table covering the first 3 years of employment with the original com- pany, distinguishing between those about to take the final test of competence and those who pass it. [Regard failing the test as a special mode of decrement ] (ii) How many people are recruited on each 1 st July? 264 CHAPTER 15. MULTIPLE-DECREMENT TABLES Solutions 15.1 α 1 is uniformly distributed, so t p α 1 x µ α 1 x+t = q α 1 x for 0 ≤ t ≤ 1 µ α 2 x+t = c, so t p α 2 x = exp[− t 0 µ α 2 x+t dt] = e −ct for 0 ≤ t ≤ 1 (aq) α 1 x = 1 0 t p α 1 x µ α 1 x+t . t p α 2 x dt = t 0 q α 1 x e −ct dt = q α 1 x ¸ − 1 c e −ct t=1 t=0 = q α 1 x (1 −e −c ) c = q α 1 x ¸ −q α 2 x log(1 −q α 2 x ) as c = −log(1 −q α 2 x ) (aq) α 2 x = (aq) x −(aq) α 1 x = [q α 1 x +q α 2 x −q α 1 x q α 2 x ] −q α 1 x ¸ −q α 2 x log(1 −q α 2 x ) (from above) 15.2 (i) (aq) w x = 2(aq) d x for x = 70, 71, 72 where w = widowhood d = death. Using U.D. of D. (aq) w x = q w x (1 − 1 2 q d x ) = 2q d x (1 − 1 2 q w x ) = 2(aq) d x Hence q w x = 2q d x 1+ 1 2 q d x Therefore (aq) d x = q d x (1− 1 2 q d x ) 1+ 1 2 q d x , and (aq) w x = 2(aq) d x . x q d x (aq) d x (aq) w x (al) x (ad) d x (ad) w x 70 0.02307 0.02254 0.04509 100,000 2,254 4,509 71 0.02559 0.02494 0.04989 93,237 2,326 4,651 72 0.02839 0.02760 0.05519 86,260 2,380 4,761 73 79,119 (ii)(a) 3 (ap) 70 = 79119 100,000 = 0.79119 (b) 3 (aq) w 70 = (ad) w 70 +(ad) w 71 +(ad) w 72 (al) 70 = 0.1392. 15.3 Denote d = death 15.10. SOLUTIONS 265 r = remarriage Notice that q r 51 = q [50]+1 , i.e. duration 1 year from age 50 at end of former marriage. x q d x q r x (aq) d x (aq) r x (al) x (ad) d x (ad) r x 50 0.00728 0.050 0.00710 0.04982 100,000 710 4,982 51 0.00823 0.025 0.00813 0.02490 94,308 767 2,348 91,193 2 (aq) r 50 = (ad) r 50 + (ad) r 51 (al) 50 = 0.0733. 15.4 (i) Denote d = death w = withdrawal (excluding those who fail test) f = those who fail test i = injury Use q w x · m w x 1 + 1 2 m w x and in view of U.D. of D. use formulae such as (aq) d x = q d x [1 − 1 2 (q w x +q i x ) + 1 3 q w x q i x ] for deaths, injuries and withdrawals, and treat the test as an “abnormal” mode of exit. x q d x q i x q w x (aq) d x (aq) i x (aq) w x 18 0.00112 0.05122 0.13953 0.001016 0.04762 0.13588 19 0.00117 0.05003 0.09524 0.001087 0.04762 0.09280 20 0.00119 0.05003 0.09524 0.001105 0.04762 0.09280 Define (al) 19 = number of people before test (at age 19 exactly) (al) + 19 = number of people who pass test. with (al) + 19 = 0.9(al) 19 Let (al) 18 = 10, 000 (for example) x (al) x (ad) f x (al) + x (ad) d x (ad) i x (ad) w x 18 10,000 0 10,000 10 476 1359 19 8,155 816 7,339 8 349 681 20 6,301 0 6,301 7 300 585 21 5,409 Notice that (ad) f x = (al) x −(al) + x (ii) 10,000 entrants give 5,409 employees after 3 years. Hence 924 entrants are needed to get, on average, 500 employees aged exactly 21. 266 CHAPTER 15. MULTIPLE-DECREMENT TABLES Chapter 16 FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES 16.1 Principles One may value cash flows, calculate premiums and find reserves using the same ideas as are used when there is just one mode of decrement (death). Notation is best considered from first principles, and there are no commutation functions (except for pensions, which are covered later). The procedure is generally as follows: Stage 1 Construct a multiple-decrement table. Stage 2 Write down an equation of value of the form m.p.v. of premiums = m.p.v. of benefits + m.p.v. of expenses (This assumes that the expected present value of the profits to the office is zero. If it is not, add a suitable profit term to the equation.) Then solve the equation of value for the unknown quantity (for example, the annual premium or the sum assured payable on death) Note It should normally be assumed that exits occur, on average, in the middle of each policy year, and that premiums cease if a life exits by any mode of decrement. 16.2 The Use of “Defective” Variables A defective random variable, T, is such that lim t→∞ F(t) = k < 1 where F(t) is the distribution function of T. For example, let T = time of exit of (x) by mode α in a double-decrement table, the other mode of exit being β. Then 267 268CHAPTER 16. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES F(t) = Pr¦T ≤ t¦ = t (aq) α x = t 0 r p α x µ α x+r . r p β x dr Since ∞ (aq) α x + ∞ (aq) β x = 1, we have lim t→∞ F(t) = lim t→∞ t (aq) α x < 1 A defective variable T may have a probability density function, f(t) = F (t). In the above example, f(t) = t p α x µ α x+t . t p β x , t > 0 0, t < 0 With defective variables, we still have results such as E[g(T)] = ∞ −∞ g(t)f(t) dt For example, consider a benefit of £1 payable immediately on exit by mode α to a life now aged x. The m.p.v. of this benefit is E[v T ] = ∞ 0 v t . t p α x µ α x+t . t p β x dt Notice the similarities to the joint-life and contingent assurance functions in chapters 11 and 12. 16.3 Evaluation of Mean Present Values This can either be done by integrals or sums. (A) Integrals The theory of ‘defective’ variables is used to value the m.p.v. of benefits on exit by a given mode of decrement. We also consider the m.p.v. of premium payments. (i) Consider a double-decrement table with exits by mode α and mode β, and suppose that there is a benefit of £S payable immediately on the exit of (x) by mode α within n years. The present value of this benefit, as a random variable, is Z = Sv T if T < n 0 if T > n. where T = time to exit by mode α in the double-decrement table. T is a defective variable with p.d.f. t p α x .µ α x+t . t p β x (t > 0). Hence the m.p.v. of this benefit is S n 0 v t . t p α x µ α x+t . t p β x dt (16.3.1) 16.3. EVALUATION OF MEAN PRESENT VALUES 269 (ii) Consider premiums of £P per annum, payable continuously for at most n years while (x) remains a member of the group under consideration. The mean present value is P n 0 v t (al) x+t (al) x dt = P n 0 v t . t (ap) x dt = P n 0 v t . t p α x . t p β x dt (16.3.2) (assuming 2 modes of decrement, α and β) Example 16.3.1. Suppose there are 2 modes of decrement, death (d) and withdrawal (w), and there is a constant force of withdrawal of k per annum. Calculate the value of premiums of £P per annum payable continuously for at most n years while (x) remains a member of the group. Solution µ w x+t = k for all t. Hence t p w x = e −kt for all t. m.p.v. of premiums = P n 0 v t . t p d x . t p w x dt = P n 0 e −δt . t p d x .e −kt dt (as v t = e −δt where δ is the force of interest) = P n 0 e −(δ+k)t . t p d x dt = P¯ a x:n at force of interest δ = δ +k (B) Sums Sums can be used to value death and other benefits and premium payments either exactly (if financial transactions occur at the end of policy years) or by approximating integrals (16.3.1) and (16.3.2) if benefits are payable immediately or premiums are paid continuously. (i) Consider a benefit of £S payable at the end of the year of exit of (x) by mode α within n years. The m.p.v. of this benefit is S ¸ v (ad) α x (al) x +v 2 (ad) α x+1 (al) x + +v n (ad) α x+n−1 (al) x . If the benefit is payable immediately, then exits can be assumed to occur, on average, mid-way through year. The m.p.v. of the benefit is S ¸ v 1 2 (ad) α x (al) x +v 3 2 (ad) α x+1 (al) x + +v n− 1 2 (ad) α x+n−1 (al) x (ii) Consider an annual premium of £P per annum, payable in advance while (x) is still a member of the group, for at most n years. The m.p.v. is P ¸ 1 +v (al) x+1 (al) x +v 2 (al) x+2 (al) x + +v n−1 (al) x+n−1 (al) x 270CHAPTER 16. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES If the premiums are payable continuously, expression (16.3.2) can be approximated to a sum. The m.p.v. is now P ¸ v 1 2 (al) x+ 1 2 (al) x +v 3 2 (al) x+ 3 2 (al) x + +v n− 1 2 (al) x+n− 1 2 (al) x ¸ Note (al) x+ 1 2 , (al) x+ 3 2 , etc. must often be found by interpolation. Example 16.3.2. A multiple-decrement table referring to mortality (d) and withdrawal (w) from a life assurance contract is x (al) x (ad) w x (ad) d x 60 10,000 64 126 61 9,810 61 131 62 9,618 48 134 63 9,436 Suppose a 3-year term assurance is issued to a life aged 60, providing £20,000 at the end of the year of death. If expenses consist of 5% of each premium, calculate the annual premium, P, payable in advance while a policyholder, for at most 3 years. Interest is at 4% per annum. Solution m.p.v. of benefits = 20, 000 (al) 60 [v.(ad) 60 +v 2 (ad) 61 +v 3 (ad) 62 ] = 722.79 m.p.v. of premiums = P (al) 60 [(al) 60 +v(al) 61 +v 2 (al) 62 ] = 2.8325P. Hence P solves 0.95 (2.8325P) = 722.79 Therefore P = £268.61 per annum. Benefits other than cash sums In some cases, the benefits on exit by a certain mode of decrement consists not of a cash sum but an annuity or some other benefit. In such cases, treat the m.p.v. of the annuity (or other benefit) at the date of exit as if it were paid out at that time. (The annuity will often have to be evaluated by interpolation if it begins mid-way through a year of age.) Example 16.3.3. A life office issues policies to lives aged under 60 providing the following benefits: (i) on becoming permanently disabled before age 60, an annuity of £2,000 per annum payable weekly for life and £20,000 immediately on death, and (ii) immediately on death before age 60 while not permanently disabled, £20,000. Calculate the office annual premium, payable weekly and ceasing on death, on permanent disability or on reaching age 60, for a life aged 58 if the office uses the following basis: 16.3. EVALUATION OF MEAN PRESENT VALUES 271 Mortality: the independent rates of mortality of those not permanently disabled are those of A1967-70 ultimate; the permanently disabled are subject to the mortality of English Life Table No.12 - Males with the age rated up by 6 1 2 years; Permanent disability: a constant independent rate of 0.006; Interest: 4% per annum; Expenses: 2 1 2 % of all office premiums, plus £50 at the issue date. Solution Construct service table with d = death and i = disability, assuming U.D. of D. x q d x q i x (aq) d x (aq) i x (al) x (ad) d x (ad) i x 58 0.01169 0.006 0.01165 0.00596 100,000 1,165 596 59 0.01299 0.006 0.01295 0.00596 98,238 1,272 585 60 96,381 Value of benefits: (i) v 1 2 (ad) i 58 (al) 58 2, 000¯ a ih 58 1 2 + 20, 000 ¯ A ih 58 1 2 +v 3 2 (ad) i 59 (al) 58 2, 000¯ a ih 59 1 2 + 20, 000 ¯ A ih 59 1 2 (ii) 20, 000 ¸ v 1 2 (ad) d 58 (ad) 58 +v 3 2 (ad) d 59 (al) 58 = 468.34 For benefit (i) ¯ a ih 58 1 2 = ¯ a 65 on E.L.T. 12 - Males = 8.918 ¯ A ih 58 1 2 = ¯ A 65 on E.L.T. 12 - Males = 0.65023 Similarly ¯ a ih 59 1 2 = ¯ a 66 = 8.587 ¯ A ih 59 1 2 = ¯ A 66 = 0.66323. So value of (i) = 180.24 + 167.89 = 348.13. Let P be the annual premium. The value of premiums less expenses is 0.975P ¸ v 1 2 (al) 58 1 2 (al) 58 +v 3 2 (al) 59 1 2 (al) 58 ¸ −50 (al) 58 1 2 · 1 2 [(al) 58 + (al) 59 ] = 99119 (al) 59 1 2 · 97310 Hence value of premiums less expenses is 1.8422P −50. Therefore 1.8422P −50 = 468.34 + 348.13 . Consequently P = £470.35. 272CHAPTER 16. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES 16.4 Benefits on Death by a Particular Cause Sometimes a policy will provide death benefits if (x) dies from a particular cause. One may write µ α x+t = force of mortality from cause α, and µ β x+t = force of mortality from all other causes. When these are added together to give (aµ) x+t , this will often be the force of mortality for a given life table. Example 16.4.1. A national newspaper recently advertised “free insurance for subscribers”, whereby a benefit of £10,000 would be paid immediately on accidental death (mode α) within n years. Assume µ α x+t = 0.0005 for t ≥ 0 (for all x) and mortality due to all causes follows A1967-70 ult. Assuming a given rate of interest and ignoring expenses, calculate the annual premium payable continuously by the newspaper to provide this “free” policy. Solution Let the annual premium be P. m.p.v. of benefits = 10000 n 0 v t . t p α x µ α x+t . t p β x dt = 10000 0.0005 n 0 v t . t p A67−70ult x dt = 5¯ a x:n on A67-70 ult. m.p.v. of premiums = Pa x:n Thus P = £5 per annum (or £0.42 per month.) Note: Observe that P does not depend on the mortality table (for all causes), nor the rate of interest, nor the term of n years. Also notice that £0.42 is a net premium, whereas an office premium (allowing for expenses) would be larger. 16.5 Extra Risks Treated as an Additional Mode of Decre- ment Suppose that in addition to “normal” mortality (mode α), a group of lives are subject to certain additional hazards (mode β). Mode β may refer to: (a) Certain occupational hazards; (b) Risks associated with a leisure activity, such as motor racing; (c) Some medical conditions. One may wish to calculate premiums for policies which pay out only on death from the “addi- tional” cause, in which case the formulae in Section 16.3 may be used. If the death benefit is paid on any cause of death, the problem can be treated as an “extra risk” question, as covered in the Actuarial Subject A2. Example 16.5.1. A certain life office’s premium basis for policies accepted at normal rates is: 16.5. EXTRA RISKS TREATED AS AN ADDITIONAL MODE OF DECREMENT 273 A1967-70 select, 4% interest, expenses are ignored. A proposer, aged 45, for temporary assurance ceasing at age 65 is subject to an extra occupational hazard which is considered to be equivalent to an addition of 0.009569 to the force of mortality at all ages. The sum assured, which is payable immediately on death, is £10,000. (a) Calculate the level annual premium, payable throughout the term of the policy. (b) The proposer requests that, in the event of death occurring as a result of the special occupa- tional hazard, the sum assured should be doubled, and offers to pay an additional single premium at the outset for this extra cover. Calculate this single premium. Solution (a) Consider α = normal mortality, β = extra occupational mortality Value of benefits is 10000 20 0 v t t p α [45] . t p β [45] (µ α [45]+t +µ β [45]+t ) dt = 10, 000 20 0 v t e −kt . t p [45] (µ [45]+t +k) dt (where k = 0.009569) = 10, 000 ¯ A ∗ 1 [45]:20 where * indicates normal plus extra mortality = 10, 000(1.04) 1 2 A ∗ 1 [45]:20 = 10, 000(1.04) 1 2 [(1 −d.¨ a ∗ [45]:20 ) −A ∗ [45]: 1 20 ] Now use the rule (as in extra risks) that the rate of interest (in annuity and pure endowment functions) may be altered to allow for the addition to the force of mortality. ¨ a ∗ [45]:20 is at force of interest δ = δ +k where δ = log(1.04) = 0.03922 (rate of interest = 4%) So δ = 0.048790 and hence, i = e δ −1 = 0.05 = 5%. Hence m.p.v. of benefits is 10, 000(1.04) 1 2 [1 −d.¨ a [45]:20 0.05 −A [45]: 1 20 0.05 ] = £2, 102.87 (using A [45]: 1 20 0.05 = l 65 l [45] v 20 0.05 ) So annual premium = 2102.87 ¨ a ∗ [45]:20 = 2102.87 ¨ a [45]:20 0.05 = £167.78 (b) The m.p.v. of this benefit is found by considering exit by mode β only. This gives a single 274CHAPTER 16. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES premium of 10, 000 20 0 v t . t p α [45] . t p β [45] .µ β [45]+t dt = 10, 000 20 0 e −δt .e −kt . t p [45] .k dt (k = 0.009569) = 10, 000k¯ a ∗ [45]:20 = 10, 000k¯ a [45]:20 0.05 · 95.69 ¸ ¨ a [45]:20 0.05 − 1 2 (1 − l 65 l [45] v 20 0.05 ) · 95.69 12.1899 = £1, 166.45 16.6 Calculations Involving a Change of State Often there is interest in the probabilities of survival for a life who leaves a multiple-decrement table and continues in another state. These problems may often be treated from first principles, rather than by the construction of a multi-state model. Example 16.6.1. Manual workers between the ages of 50 and 60 are subject to 2 modes of decre- ment, d = death, and p = promotion to foreman. Foremen are subject to the mortality of another table, T , with functions l x , etc... Find the probabilities that (a) A manual worker aged 50 will be alive and a foreman at age 52. (b) A manual worker aged 50 will be promoted to foreman but then die before age 52. Solution (a) The exact formula is 2 0 t p d 50 . t p p 50 µ p 50+t l 52 l 50+t dt (the expression in brackets giving survival as a foreman to age 52), which approximates to (ad) p 50 (al) 50 l 52 l 50 1 2 + (ad) p 51 (al) 50 l 52 l 51 1 2 (b) The probability is approximately (ad) p 50 (al) 50 ¸ 1 − l 52 l 50 1 2 ¸ + (ad) p 51 (ad) 50 ¸ 1 − l 52 l 51 1 2 ¸ 16.7. EXERCISES 275 Exercises 16.1 The following is an extract from a multiple-decrement table referring to mortality and with- drawal from certain life assurance contracts, these modes being referred to as ‘d’ and ‘w’ respectively. age, x (al) x (ad) w x (ad) d x 50 15,490 24 51 51 15,415 21 58 52 15,336 15 60 (i) What is the probability that a policyholder aged 51 will withdraw before attaining age 53? (ii) Suppose that a single-premium three-year term assurance contract is to be issued to a life aged 50, subject to the mortality and withdrawal rates shown in the above table. The sum assured is £50,000, payable immediately on death, and the benefit on withdrawal is zero. On the basis of rate of interest of 6% p.a., and allowing for expenses of 10% of the single premium, calculate the single premium for the above policy. (iii) Suppose now that the office issuing the policy of (ii) above wishes to introduce surrender values for these three-year contracts. The surrender value is to be equal to 0.5 per cent of the single premium for each week between the date of withdrawal and the end of the term. Assuming that surrenders in each policy year take place on average half-way through the year, write down an equation of value from which you could calculate the revised single premium. (Do NOT proceed to the evaluation of this premium.) 16.2 Employees of a certain company are given the opportunity of early retirement immediately after they complete a 3-year overseas assignment. Those who undertake this assignment effect a 3-year policy providing the following benefits payable at the end of the year of claim: (a) on death or ill-health retirement during the term, the sum of £6,000; (b) on survival as an employee of the company to the end of the term, a lump sum; (c) on withdrawal from the company during the second year an amount equal to 1 1 4 times the annual premium, and on withdrawal from the company during the third year an amount equal to 2 1 2 times the annual premium. No benefit is payable on withdrawal from the company in the first year. The following multiple-decrement table is applicable to employees going overseas at exact age 47. (d = death, w = withdrawal, i = ill-health retirement) x (al) x (ad) d x (ad) i x (ad) w x 47 100,000 853 13,059 23,007 48 63,081 616 11,604 8,468 49 42,393 478 9,035 1,875 50 31,005 Using an interest rate of 4% p.a., calculate the lump sum for an employee going overseas at exact age 47 who pays level annual premiums of £2,000 in advance. Ignore expenses. 16.3 An organization recruits new employees at exact age 20. The employees serve an apprentice- ship during the first two years of employment. The central rate of withdrawal of apprentices 276CHAPTER 16. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES at ages 20 and 21 is 0.05 per annum, and withdrawals are uniformly spread over the year of age (in the single-decrement table for withdrawals). At exact age 22, apprentices join the permanent staff, who retire at exact age 60. The force of withdrawal of permanent staff is 0.0094787 per annum. Both apprentices and permanent staff experience mortality according to A1967-70 ultimate. The employer pays an immediate benefit of £2,000 on withdrawal and £8,000 on death in service to permanent staff. The death benefit (but not the withdrawal benefit) is paid to apprentices. (a) Construct a double-decrement table for apprentices between ages 20 and 22 with a radix of 100,000 at age 20. (b) Calculate, using an interest rate of 5% per annum, the value of the benefits in respect of a new employee aged 20. 16.4 A life office sells policies to lives aged 63 which provide a benefit of £50 per week, ceasing at age 65 or earlier death, to those becoming permanently disabled before age 65. The following basis is used to calculate the single premium for this contract: Independent rates of mortality of policyholders (not disabled): A1967-70 select (at entry) Force of disablement: 0.01 p.a. Mortality of disabled lives: A1967-70 ultimate, rated up by 7 years Rate of interest: 4% p.a. Expenses are ignored. Weekly benefits may be taken as payable continuously. Evaluate the single premium for this contract. 16.5 In a certain country, married men are subject to the following independent q-type rates of decrement: Mortality: A1967-70 ultimate Widowhood: wives of married men are subject to A1967-70 ultimate Divorce: 0.02 per annum at all ages Widowed and divorced men are subject to the following independent q-type rates of decrement: Mortality: English Life Table No. 12 - Males; Remarriage: rates depend on the age at, and the duration since, the end of the former marriage; the following table is an extract from these rates: age at end duration duration of former 0 1 Marriage year year 50 1 2 0.050 0.025 51 1 2 0.045 0.023 52 1 2 0.042 0.020 53 1 2 0.040 0.018 Calculate the probability that a married man aged 50, whose wife is also aged 50, will attain age 52 as a widower who has not remarried between ages 50 and 52. 16.8. SOLUTIONS 277 Solutions 16.1 (i) 2 (aq) w 51 = (ad) w 51 + (ad) w 52 (al) 51 = 0.002335 (ii) Let single premium be P. Then P solves 0.9P = 50, 000 ¸ (ad) d 50 v 1 2 + (ad) d 51 v 1 1 2 + (ad) d 52 v 2 1 2 (al) 50 ¸ at 6% interest = 50, 000 0.009977 = 498.86 Hence P = £554.29. (iii) Let P be the premium allowing for surrender values. There are on average 52.18 weeks in 1 year. So, the surrender value in year 1 is on average 52.18 0.005 2.5P . The average surrender value in year 2 is 52.18 0.005 1.5P , and so on. The equation of value is 0.9P = value of death benefit (as above) + 52.18 0.005 (al) 50 P [v 1 2 (ad) w 50 (2.5) +v 1 1 2 (ad) w 51 (1.5) +v 2 1 2 (ad) w 52 (0.5)] 16.2 Let the lump sum be X. m.p.v. premiums = 2000 ¸ 1 + (al) 48 (al) 47 v + (al) 49 (al) 47 v 2 = 3996.99. m.p.v. benefits: (a) 6, 000 (al) 47 [v((ad) d 47 + (ad) i 47 ) +v 2 ((ad) d 48 + (ad) i 48 ) +v 3 ((ad) d 49 + (ad) i 49 )] = 1987.92 (b) X.v 3 (al) 50 (al) 47 = 0.27563(X) (c) 2000 (al) 47 [1.25(ad) w 48 .v 2 + 2.5(ad) w 49 .v 3 ] = 279.07 So X solves 3996.99 = 2266.99 + 0.27563(X) Hence X = £6276.53 16.3 (a) q w 20 · m w 20 1 + 1 2 m w 20 = 0.05 1.025 = 0.04878 = q w 21 278CHAPTER 16. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES Using (aq) d x = q w x (1 − 1 2 q d x ), the double-decrement table is, x q d x q w x (aq) d x (aq) w x (al) x (ad) d x (ad) w x 20 0.000889 0.04878 0.000868 0.04876 100,000 87 4876 21 0.000841 0.04878 0.000821 0.04876 95,037 78 4634 22 90,325 (b) Consider value of benefits for employee attaining age 22. Let k = 0.0094787. Value of death benefit is 8000 38 0 v t e −kt . t p d 22 µ d 22+t dt = 8000 ¯ A ∗ 1 22:38 at force of interest δ = δ +k = 0.058269 = 8000 ¯ A 1 22:38 0.06 = 8000(1.06) 1 2 ¸ A 22:38 0.06 − l 60 l 22 v 38 0.06 = 206.80 Value of withdrawal benefit is 2000 38 0 v t e −kt . t p d 22 .k dt = 2000.k¯ a 22:38 ,0.06 = 2000 0.0094787 ¸ ¨ a 22:38 − 1 2 (1 − l 60 l 22 v 38 0.06 ) = 285.64 Value of benefit in first 2 years, valued at age 20, is 8000 (al) 20 [v 1 2 (ad) d 20 +v 1 1 2 (ad) d 21 ] = 12.59 Hence value of benefits for new employee is 12.59 +v 2 (al) 22 (al) 20 [206.80 + 285.64] = £416.03 16.4 Let d = death and i = disablement. Note that q d 63 = q [63] and q d 64 = q [63]+1 on A67-70 tables. Also, q i 63 = q i 64 = 1 −e −0.01 = 0.00995 x q d x q i x (aq) d x (aq) i x (al) x (ad) d x (ad) i x 63 0.00839 0.00995 0.00835 0.00991 100,000 835 991 64 0.01245 0.00995 0.01239 0.00989 98,174 1216 971 65 95,987 m.p.v. benefits · 50 52.18 (al) 63 v 1 2 (ad) i 63 .¯ a ∗ 63 1 2 :1 1 2 +v 1 1 2 (ad) i 64 ¯ a ∗ 64 1 2 : 1 2 where 16.8. SOLUTIONS 279 ¯ a ∗ 63 1 2 :1 1 2 = ¯ a 70 1 2 :1 1 2 on A67-70 ultimate · 1 2 [¯ a 70:2 + ¯ a 71:1 ] = 1 2 [1.84904 + 0.96010] = 1.4046 ¯ a ∗ 64 1 2 :1 1 2 = ¯ a 71 1 2 : 1 2 on A67-70 ultimate · 1 2 (0.96010) = 0.48005 Hence single premium = £47.08. 16.5 Denote α = mortality of married men β = mortality of wives γ = divorce. Use formula of the form (aq) α x = q α x [1 − 1 2 (q β x +q γ x ) + 1 3 q β x q γ x ] x q α x q β x q γ x (aq) β x 50 0.004789 0.004789 0.02 0.004730 51 0.005377 0.005377 0.02 0.005309 A multi-decrement table can be used, but short cuts can be taken. (ap) 50 = (1 −q α 50 )(1 −q β 50 )(1 −q γ 50 ) = 0.97064. So, Pr¦ married man aged 50 becomes widower between ages 51 and 52¦ = 1 [(aq) β 50 = (ap) 50 (aq) β 51 = 0.005154 Pr¦ widower aged 50 1 2 attains age 52 without remarrying¦ = l 52 l 50 1 2 (1 −0.050)(1 − 1 2 0.025) (using E.L.T. 12-Males) = 0.92674. (i) Pr¦ widower aged 51 1 2 attains age 52 without remarrying¦ = l 52 l 51 1 2 (1 − 1 2 0.045) = 0.97322. (ii) Hence required probability is (aq) β 50 (i) + (ap) 50 (aq) β 51 (ii) = 0.004730 0.92674 + 0.005154 0.97322 = 0.009399 280CHAPTER 16. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES Chapter 17 MULTIPLE-STATE MODELS 17.1 Two Points of View There are essentially two approaches to the theory of multiple-decrement tables: (a) Traditionalist - uses the underlying single-decrement tables; (b) Modernist - uses continuous-time stochastic processes with a finite number of states. Both view-points have advantages and disadvantages, and both produce the same practical re- sults. In chapters 15 and 16, the traditional approach and notation was used. This chapter shall briefly take a more modernist approach, and show the differences between the two methods. A development of the modernist approach for multiple-state models is covered in Actuarial Subject D2. 17.2 Kolmogorov’s Forward Equations Suppose there are n possible states. For example, n = 3 in the following system, which refers to a double-decrement table: Define p ij (x, x +t) =Pr¦(x) will be in state j at age x +t given that he is in state i at age x¦ Figure 17.2.1: a double-decrement model 281 282 CHAPTER 17. MULTIPLE-STATE MODELS Note that the initial conditions are p ij (x, x) = 1 if i = j 0 if i = j. The forces of transition, µ ij (y), are defined as µ ij (y) = lim h→0 + p ij (y, y +h) h (i = j) Also, µ i (y) = ¸ i=j µ ij (y) = the force of transition from state i to any other state at age y Note If all these forces are constant, this is a homogeneous chain. If the forces of transition vary with age, it is called inhomogeneous. Kolmogorov’s Forward Equations (for the general case in which there are n states) give the following system for each fixed x: d dt p ij (x, x +t) = −µ j (x +t)p ij (x, x +t) + ¸ ν=j µ νj (x +t)p iν (x, x +t) (1 ≤ i, j ≤ n, t ≥ 0) (17.2.1) The three-state model of Figure 17.2.1 is such that µ 1 (y) = µ 12 (y) +µ 13 (y) µ 2 (y) = 0 µ 3 (y) = 0 (as no one leaves states 2 or 3) so Kolmogorov’s Forward Equations give d dt p 12 (x, x +t) = p 11 (x, x +t)µ 12 (x +t) (17.2.2) and d dt p 13 (x, x +t) = p 11 (x, x +t)µ 13 (x +t) (17.2.3) Note also that p 11 (x, x +t) = 1 −p 12 (x, x +t) −p 13 (x, x +t) (17.2.4) and hence d dt p 11 (x, x +t) = − d dt p 12 (x, x +t) − d dt p 13 (x, x +t) 17.3. LIFE TABLES AS STOCHASTIC PROCESSES 283 Adding equations (17.2.2) and (17.2.3) gives d dt [p 12 (x, x +t) +p 13 (x, x +t)] = p 11 (x, x +t)[µ 12 (x +t) +µ 13 (x +t)] = p 11 (x, x +t)µ 1 (x +t) And hence, using equation (17.2.4), d dt p 11 (x, x +t) = −p 11 (x, x +t)µ 1 (x +t) (17.2.5) (Equation (17.2.5) also follows directly from equation (17.2.1) with i = j = 1.) This is a straightforward first-order differential equation. Using the initial condition p 11 (x, x) = 1, (17.2.5) can be solved to give the unique solution p 11 (x, x +t) = exp[− t 0 µ 1 (x +r) dr] (17.2.6) This corresponds exactly to a result derived in chapter 15, i.e. t (ap) x = exp[− t 0 (aµ) x+r dr] = exp[− t 0 (µ α x+r +µ β x+r ) dr] where µ α x+r corresponds to µ 12 (x +r), and µ β x+r corresponds to µ 13 (x +r) . Returning to the differential equation (17.2.2) and solving with the initial condition p 12 (x, x) = 0 gives p 12 (x, x +t) = t 0 p 11 (x, x +r)µ 12 (x +r) dr Again this corresponds to a “traditional” formula in chapter 15, i.e. t (aq) α x = t 0 r (ap) x µ α x+r dr 17.3 Life Tables as Stochastic Processes Life tables just contain one mode of decrement (death) and hence the continuous-time model is Note that µ 12 (y) = the force of mortality at age y = µ y in “traditional” notation Proceeding as in section 17.2, but with only one mode of decrement, we find that Kolmogorov’s Forward Equations may be solved to give p 11 (x, x +t) = exp ¸ − t 0 µ 12 (x +r) dr In “traditional” notation, this is just the familiar result that t p x = exp ¸ − t 0 µ x+r dr 284 CHAPTER 17. MULTIPLE-STATE MODELS Figure 17.3.1: the life table model Remark In the stochastic processes approach, it is assumed that the Chapman–Kolmogorov equations hold. That is, p ij (x, z) = n ¸ k=1 p ik (x, y)p kj (y, z) (17.3.1) for all x ≤ y ≤ z, and for the n states in the continuous-time model. In Figure 17.3.1 there are only 2 states (alive and dead), and clearly p 21 (x 1 , x 2 ) = 0 for all x 1 ≤ x 2 and p 22 (x 1 , x 2 ) = 1 for all x 1 ≤ x 2 Equation (17.3.1) with i = 1 and j = 1 is just a re-statement of the following axiom of the traditional life table: z−x p x = y−x p x . z−y p y 17.4 Sickness Models The Continuous Mortality Investigation Bureau (C.M.I.B.) has used the following multiple-state model for sickness rates: Define σ y = the force of sickness at age y, p y,z = the force of recovery at age y and duration of sickness z, µ y = the force of mortality of healthy lives at age y, ν y,z = the force of mortality of sick lives at age y and duration of sickness z The various forces of transition have been graduated using Permanent Health Insurance data. Suppose one wishes to calculate probabilities such as p 12 (x, x +t) = Pr¦ a healthy life aged x will be sick at time t¦. The existence of the parameter z complicates the multiple-state model: Kolmogorov’s Forward Equations have to be modified, as the forces of recovery and mortality of sick lives depend on 17.4. SICKNESS MODELS 285 Figure 17.4.1: a sickness model duration of sickness as well as age. This leads to “semi-Markov” processes which are more difficult to handle than the cases so far discussed. A further complication is caused by the need (in practical applications) to consider “deferred periods”. For example, one may wish to calculate Pr¦ a healthy life aged x will be sick at time t and has been sick for at least d weeks (d is the deferred period)¦ A possible simplification One possible way to simplify this model is to initially ignore mortality and to consider only forward transitions between “healthy” and “sick” states. Define T 1 , T 2 , T 3 , ... to be the times (in years) spent as 1. a healthy life (who has not been sick since age x), 2. a sick life (who has been sick exactly once), 3. a healthy life (who has been sick exactly once), and so on. A pictorial representation of this is as follows: In this set-up, one can work out Pr¦ a healthy life aged x will be sick at age x +t¦ = ¸ j=2,4,6,... Pr¦ a life in state 1 at age x will be in state j at time t¦ Each term may be evaluated, using the joint density function f(t 1 , t 2 , ..., t j ) of the variable (T 1 , T 2 , ..., T j ). That is, Pr¦ a life aged x in state 1 will be in state j at time t¦ =Pr¦T 1 +... +T j−1 < t, but T 1 +... +T j > t¦ = a multiple integral (we omit the details) Advantages of this method: 1. The mathematics is not so difficult to follow. 2. The deferred period may be easily handled by modifying the range of the multiple integrals. Disadvantages: 1. There can be difficulties in evaluating the multiple integrals for large values of j. 286 CHAPTER 17. MULTIPLE-STATE MODELS Figure 17.4.2: sick and healthy states In practice, the Manchester Unity System is often used to calculate sickness functions; this will be covered in Chapter 18. Chapter 18 SICKNESS FUNCTIONS 18.1 Rates of Sickness The force of sickness, ¯ z x , at age x is the probability that a life aged exactly x is “sick” (according to the rules of the scheme). In practice, sickness benefit must be claimed for short periods (e.g. days or weeks) during which time the life is either “sick” or “not sick”. Consider these periods to be of length h years. Define z (h) x = Pr¦(x) is entitled to sickness benefit for the time period until age x +h¦ It is supposed that, uniformly on any bounded age interval, lim h→0 + z (h) x = ¯ z x If benefit is paid at the rate of £1 per week, this amounts to £52.18 per year on average. Suppose that benefit is payable in advance over intervals of length h years, where h = 1 n , i.e. each year is divided into n intervals. The expected cash to be paid in sickness benefit between ages x and x + 1 for a life now aged x is n−1 ¸ j=0 jh p x (52.18h)z (h) x+jh Letting n →∞ (or h →0 + ) gives s x = expected cash paid out in sickness benefit = 52.18 1 0 t p x ¯ z x+t dt (18.1.1) s x is called the annual rate of sickness at age x: it is the expected number of weeks of sickness between ages x and x + 1 for a life now aged x. We also define z x = the central rate of sickness at age x = 52.18 1 0 t p x ¯ z x+t dt 1 0 t p x dt (18.1.2) 287 288 CHAPTER 18. SICKNESS FUNCTIONS Equations (18.1.1) and (18.1.2) give the approximations: s x · 52.18. 1 2 p x ¯ z x+ 1 2 (18.1.3) z x · 52.18¯ z x+ 1 2 (18.1.4) and hence s x · 1 2 p x z x (18.1.5) “Formulae and Tables for Actuarial Examinations” makes use of the Manchester Unity Expe- rience 1893-97, Occupational Groups AHJ. Fuller details of this experience can be found in the Appendix to this chapter. Sickness by Duration Benefits may depend on the duration of sickness. Define ¯ z m/n x = Pr¦(x) is “sick” and has been sick for more than m weeks but not more than m+n weeks.¦ Similar modifications are used to define z m/n x and s m/n x . For example, ¯ z 13/13 x = Pr¦(x) is “sick” with duration of sickness more than 13 weeks but less than 26 weeks¦. Note that ¯ z 13 x = ¯ z 0/13 x = Pr¦(x) is “sick” and has been sick for less than 13 weeks¦ Also, ¯ z 104/all x = Pr¦(x) is sick with duration of sickness greater than 104 weeks, or 2 years¦. The Tables give the following values: z 13 x , z 13/13 x , z 26/26 x , z 52/52 x , z 104/all x Furthermore, z x = z all x = z 13 x +z 13/13 x +z 26/26 x +z 52/52 x +z 104/all x Some More Definitions (a) The Deferred Period This is the time between falling sick and being able to claim sickness benefits. In Permanent Health Insurance, one encounters the notation D1, D4, etc. which refers to the length of the deferred period in weeks. For example, in a D1 policy, a member must be sick for 1 week before being allowed to receive sickness benefit. (b) The Waiting Period This is the time between joining a friendly society or sickness benefit scheme and being able to claim sickness benefits. This time-period is often 6 months or a year. (c) The Off-Period This is the minimum time that must elapse between 2 bouts of sickness in order for them not to be considered as the same bout of sickness for benefit calculations. 18.2. VALUING SICKNESS BENEFITS 289 If the sickness benefit falls with duration of sickness, there may be a temptation for people to temporarily “recover” and then to start claiming benefit again at the higher initial rate. To prevent this, an off-period is specified so that if 2 spells of sickness are separated by less than the off-period, the later spell is treated, for benefit purposes, as a continuation of the first spell. In the Manchester Unity experience, the off-period is 1 year. If benefits do not fall as duration of sickness increases there is no need for an off-period rule. 18.2 Valuing Sickness Benefits Consider a life aged x, subject to a certain mortality table, and suppose that sickness benefit will be payable at the rate of £1 (per week) during all sickness within the next T years. Consider the age-range from x to x + T to be split into nT short intervals, each of length h years, and suppose that sickness benefit is payable in advance on sickness during any of these short-age intervals. The mean present value of the sickness benefit is thus nT−1 ¸ j=0 v jh jh p x (52.18h)z (h) x+jh Letting n →∞ gives m.p.v. of sickness benefit = 52.18 T 0 v t . t p x ¯ z x+t dt (18.2.1) This can be approximated by a sum, i.e. m.p.v. of sickness benefit · T−1 ¸ t=0 v t+ 1 2 . t+ 1 2 p x (52.18¯ z x+t+ 1 2 ) = T−1 ¸ t=0 v t+ 1 2 . t+ 1 2 p x z x+t (18.2.2) Notes 1. If sickness benefits continue throughout life (although this is unusual as it is not easy to define “sickness” among the very old), we may let T →∞, giving m.p.v. of sickness benefit = 52.18 ∞ 0 v t . t p x ¯ z x+t dt (18.2.3) · ∞ ¸ t=0 v t+ 1 2 . t+ 1 2 p x z x+t (18.2.4) 2. If the sickness benefit is payable for sickness of duration greater than m weeks but less than m+n weeks, replace ¯ z x and z x by ¯ z m/n x and z m/n x , respectively. The M.P.V. of a sickness benefit may be evaluated by direct evaluation of a sum, by approximate integration or by commutation functions. Commutation functions Define H x = 52.18 1 0 v x+t l x+t ¯ z x+t dt (18.2.5) 290 CHAPTER 18. SICKNESS FUNCTIONS This can be approximated to give H x · 52.18v x+ 1 2 l x+ 1 2 ¯ z x+ 1 2 (18.2.6) and so H x · D x+ 1 2 z x (18.2.7) where D x = l x v x , as usual. We also define K x = ∞ ¸ t=0 H x+t (18.2.8) so that the m.p.v. of a sickness benefit of £1 per week for life to (x) is approximately ∞ ¸ t=0 D x+t+ 1 2 D x z x+t (from (18.2.4)) = ¸ ∞ t=0 H x+t D x = K x D x (18.2.9) Note K x = K all x = K 13 x + K 13/13 x + K 26/26 x + K 52/52 x + K 104/all x , where H 13 x and K 13 x , etc., are defined by replacing z x by z 13 x , etc., in the formulae for H x and K x . Commutation functions are available in “Formulae and Tables” on the following basis only: • English Life Tables, No.12 - Males • 4% interest • Manchester Unity 1893-97 (A, H, J) Temporary and Deferred benefits If sickness benefit ceases at a certain age, say 65, then the m.p.v. of sickness benefit of £1 per week to (x) is K x −K 65 D x (18.2.10) If benefits are deferred for n years, we obtain K x+n D x rather than K x D x This is used in connection with waiting periods. If there is a waiting period of 6 months for sickness benefits, replace K x by K x+ 1 2 (interpolate in the Tables.) 18.2. VALUING SICKNESS BENEFITS 291 Example 18.2.1. A life office is proposing to issue 3-year sickness benefit policies to lives aged 30. The benefits are £50 per week during sickness within the next three years. There is no waiting period and the off-period is as in the Tables provided. Find the single premium on each of the following bases: mortality: English Life Table No.12 - Males interest: (i) 4% p.a., (ii) 5% p.a. sickness: Manchester Unity 1893-97 (AHJ) expenses: none. Solution (i) m.p.v. of benefits = 50 K 30 −K 33 D 30 = 50 ¸ 1784760 −1706624 29372 = £133.01 (ii) Calculate value of benefits from first principles. The value is 50 l 30 v 1 2 0.05 l 30 1 2 z 30 +v 1 1 2 0.05 l 33 1 2 z 31 +v 2 1 2 0.05 l 32 1 2 z 32 The value of l 30 1 2 , l 31 1 2 , l 32 1 2 can be found by interpolation using English Life Tables, No.12 - Males, to give m.p.v. of benefits = 50 95265 249682 = £131.05 If sickness benefit is payable for sickness lasting more than m weeks but less than m+n weeks, replace K x and H x by K m/n x and H m/n x respectively. For example, K 13/all x −K 13/all 65 D x = m.p.v. of a sickness benefit of £1 per week payable up to age 65 on sickness lasting more than 13 weeks Since K 13/all x is not given directly in the Tables, one must use K 13/all x = K 13/13 x +K 26/26 x +K 52/52 x +K 104/all x Example 18.2.2. A friendly society issued a policy providing the following benefits to a man aged exactly 25 at entry: (a) on death at any time before age 60, the sum of £4,000 payable immediately; (b) on survival to age 60, an annuity of £8 per week payable weekly in advance for as long as he survives; (c) on sickness, an income benefit to be payable during sickness of £32 per week for the first 6 months reducing to £16 per week for the next 18 months and to £8 per week thereafter. Sickness 292 CHAPTER 18. SICKNESS FUNCTIONS benefit is not payable after age 60. There is no waiting period. Premiums are payable monthly in advance for at most 35 years, and are not waived during periods of sickness. The society uses the following basis to calculate premiums. Find the monthly premium. mortality: English Life Table No.12 - Males sickness: Manchester Unity Sickness Experience 1893-97, Occupation Group AHJ interest: 4% p.a. expenses: none Solution Value of benefits: (a) 4000 ¯ A 1 25:35 = 4000 M 25 −M 60 D 25 = 268.05 (b) 52.18 8 N 60 D 25 = 926.65 (c) 32 K 26 25 −K 26 60 D 25 + 16 K 26/78 25 −K 26/78 60 D 25 + 8 K 104/all 25 −K 104/all 60 D 25 = 660.92 Let P = annual premium payable monthly. Then the equation of value is P¨ a (12) 25:35 = 268.05 + 926.65 + 660.92 = 1855.62 ¨ a (12) 25:35 = ¨ a (12) 25 − D 60 D 25 ¨ a (12) 60 = (¯ a 25 + 1 24 ) − D 60 D 25 ¯ a 60 + 1 24 = 18.495. Hence P = £100.33 and monthly premium = £8.36. 18.3 Various Other Points (a) A possible adjustment Consider a policy providing a sickness benefit of £10 per week on sickness lasting more than 2 years for an entrant aged x, with all benefits ceasing at age 65. One might want to adjust the formula 10 ¸ K 104/all x −K 104/all 65 D x ¸ (18.3.1) to 10 ¸ K 104/all x+2 −K 104/all 65 D x ¸ (18.3.2) since the benefit cannot be received in the first 2 years. This point is discussed in the Appendix to this Chapter, and it is concluded that neither formula is exactly right, so the use of adjusted version (18.3.2) is usually optional. The only exception (in which (18.3.2) is accurate and (18.3.1) is not) occurs when the term of the policy is very short. For example when x = 63, formula (18.3.2) gives zero, which is correct. 18.3. VARIOUS OTHER POINTS 293 (b) Waiver of Premium Benefits Suppose that a policy has a weekly premium of £P but this is waived during sickness (or during sickness of a certain duration). This is handled by assuming that all premiums are paid by the policyholder, but there is an additional sickness benefit of £P per week. Also, as a general rule, expenses will apply even when premiums are waived. If this is not the case, one should adjust the equation of value appropriately. Example 18.3.1. A friendly society issues sickness insurance policies which provide income during periods of sickness as follows: (a) £100 per week while a sickness has duration in excess of 13 weeks but less than 1 year; (b) £75 per week while a sickness has duration in excess of 1 year but less than 2 years; (c) £50 per week while a sickness has duration in excess of 2 years. All benefits cease at age 65. Premiums are payable weekly until age 65 and are waived when sickness benefit is being paid. There is no waiting period. Calculate the weekly premium for a life aged 38 at entry. Basis: mortality: E.L.T. No.12 (Males) sickness: Manchester Unity Sickness Experience 1893-97, Occupation Group AHJ. interest: 4% p.a. expenses: 50% of all premiums payable in the first year, plus 10% of all premiums payable after the first year. Solution Let the weekly premium be P. m.p.v. of (premiums-expenses) is 0.9 52.18P¯ a 38:27 −0.4 52.18Pa 38:1 = 713.98P (i) m.p.v. of benefits and premiums waived is 100[ K 13/39 38 −K 13/39 65 D 38 ] + 75[ K 52/52 38 −K 52/52 65 D 38 ] + 50[ K 104/all 38 −K 104/all 65 D 38 ] +P[ K 13/all 38 −K 13/all 65 D 38 ] = 784.80 + 267.63 + 543.00 +P 22.276 = 1595.43 + 22.276P (ii) Setting (i) = (ii) gives P(713.98 −22.276) = 1595.43 Hence P = £2.31 (c) Reserves These are calculated prospectively or retrospectively (usually the former) in a way similar to that used for life policies. If the premium and reserve bases agree, the prospective and retrospective reserves are equal. If the bases differ, one must specify how to find the reserve. 294 CHAPTER 18. SICKNESS FUNCTIONS Example 18.3.2. Consider a policy issued to (x) providing a sickness benefit of £10 per week ceasing at age 65. (a) Give a formula for the weekly premium, P, ceasing at age 65 (ignoring expenses). (b) Calculate the reserve at duration t years by (i) the prospective method, and (ii) the retrospective method, on the premium basis. Solution (a) 52.18P¯ a x:65−x = 10 K x −K 65 D x (b) (i) t V = 10 K x+t −K 65 D x+t −52.18P¯ a x+t:65−x−t (ii) t V = D x D x+t 52.18P¯ a x:t −10 K x −K x+t D x Note that (i) and (ii) are equal, by an argument similar to that used for life policies. (d) Understanding the Tables for Actuarial Exams 1. Pages 82-83 give z 13 x , z 13/13 x , z 26/26 x , z 52/52 x , z 104/all x , z x (= z all x ) 2. Pages 84-85 give K 13 x D x , K 13/13 x D x , K 26/26 x D x , K 52/52 x D x , K 104/all x D x , K x D x 3. Pages 86-87 give D x , K 13 x , K 13/13 x , K 26/26 x , K 52/52 x , K 104/all x 18.4. EXERCISES 295 Exercises 18.1 (i) In a combined sickness and mortality table K x+1 = 554, 405 z x = 1.129 D x = 24, 510 D x+1 = 23, 425 Estimate K x . (ii) An office offers an optional waiver of premium benefit on sickness of any duration in respect of a 25-year with or without profits endowment assurance policy with weekly premiums payable for 25 years or until earlier death. There is a waiting period of 12 months for the waiver of premium benefit, and, during the second year of the policy, only half the premium (including the extra premium for the waiver benefit) is waived. The sum assured under the endowment policy is payable immediately on death, or on survival until the end of the term. Using the basis given below, calculate the percentage by which the normal weekly premium (i.e. the premium for a policy without the waiver benefit) for a life aged exactly 30 at entry should be increased in order to provide the waiver benefit. mortality: English Life Table No.12-Males; sickness: Manchester Unity 1893-97, Occupation Group AHJ; interest: 4% per annum; expenses: 15% of each extra premium for the waiver benefit. 18.2 A policy issued by a life office to a male life aged exactly 35 is subject to level weekly premiums ceasing at exact age 65. If the man has been sick for 6 months or more when a premium falls due, the premium is waived. The policy provides the following benefits: (a) on survival to exact age 65, an annuity of £5,000 per annum payable monthly in advance, (b) on death before age 65, a return of all premiums paid (including those waived during sickness) together with compound interest at 4% per annum to the date of death. There is no waiting period and the off periods are the same as those underlying the tables in Formulae and Tables for Actuarial Examinations. Calculate the weekly premium. Basis: English Life Table No.12-Males, Manchester Unity Sickness Experience 1893-97, Oc- cupation Group AHJ, interest 4% per annum, no expenses. 18.3 (a) Ten years ago, a man then aged exactly 30 effected an insurance policy providing sickness benefits of £100 per week for the first six months of sickness, £50 per week for the remainder of the first year and £30 per week thereafter, with benefit ceasing at age 60. Calculate the weekly premium payable to age 50 on the following basis: Mortality: English Life Table No. 12 - Males; Sickness: Manchester Unity 1893-97, Occupation Group AHJ; Interest: 4% per annum; Expenses: 10% of each premium. (b) The man now wishes to alter his policy so that premiums will in future be waived during all periods of sickness. Calculate the revised premium payable assuming that the alteration basis follows the premium basis above. Note. Expenses are incurred even when premiums are waived. 296 CHAPTER 18. SICKNESS FUNCTIONS 18.4 A certain friendly society recruits only married men aged under 55. The society provides the following benefits: (i) immediately on the death of the member at any age, £2,000, (ii) during the first 10 years of membership, immediately on the death of the member’s wife before her husband, £500, (iii) on survival of the member to age 65, an annuity of £10 per week for life, and (iv) during any spell of sickness of the member before age 65, £10 per week reducing to £6 per week after 3 months’ sickness. The basis for all calculations is: mortality of members: E.L.T. No.12 - Males mortality of wives: E.L.T. No. 12 - Males with an age-deduction of 5 years interest: 4% per annum sickness: Manchester Unity 1893-97 AHJ expenses: 5% of all contributions, including those waived during sickness Wives of members are taken as being of the same age as their husbands; there is no benefit on the death of wives of marriages taking place after entry to the society, and the possibility of divorce is to be ignored. There is no waiting period and the off-period is as in the Tables provided. The possibility of withdrawal is ignored. (a) Calculate the weekly contribution rate, waived during sickness and ceasing at age 65 or the previous death of the member, for an entrant aged 25. (b) Calculate the reserve for a member aged 35 who joined at age 25. Note Use Simpson’s rule for approximate integration, i.e. b a f(t) dt · b −a 6 [f(a) + 4f( a +b 2 ) +f(b)] 18.5 A Friendly Society issues policies providing the following benefits: (i) A sickness benefit of £25 per week for the first 13 weeks of sickness and £12.50 per week thereafter, benefit ceasing at age 60. Contributions are waived during sickness. (ii) On death before age 60, a lump sum of £1,000 plus a return of contributions (including any waived) without interest. (iii) On survival to age 60, a lump sum of £2,000. Contributions are payable by level weekly amounts until age 60. There is a six-month waiting period for the sickness benefit (including the premium waiver) and the off-period may be assumed to be the same as that underlying the tables in “Formulae and Tables for Actuarial Examinations”. (a) Calculate the weekly contribution payable by a new member aged 35 on the basis given below. (b) Calculate the reserve (on the basis given below) to be held for this member five years after he joins the Society. Basis: English Life Table No. 12 - Males Manchester Unity Sickness Experience 1893-97, Occupation Group AHJ interest 4% per annum. expenses are 5% of all premiums (including those waived). 18.4. EXERCISES 297 18.6 A friendly society provides the following benefits: (i) On sickness, £40 per week for the first 26 weeks and £50 per week for the next 26 weeks. No sickness benefit is payable after age 65. (ii) On attaining age 65 or immediately on earlier death, the sum of £3,000. Members contribute £1 per week, ceasing at age 65 or earlier death, but this is waived during periods of sickness (whether benefit is payable or not.) Calculate the reserve which should be held for a member aged 50, using the following basis: English Life Table No.12 - Males, 4% per annum interest, Manchester Unity Sickness Experience 1893-97 (AHJ). There is no waiting period, and off-periods are as in “Formulae and Tables for Actuarial Examinations”. 18.7 On 1 January 1995 a friendly society had in force a large group of policies providing certain sickness and life assurance benefits. These policies had all been issued 10 years ago to lives now aged 45. The benefits cease when the policyholders reach age 65, and the policies have level weekly premiums payable throughout the term of the policy. Each policy provides an income of £20 per week during any period of sickness which has lasted more than 6 months, and a lump sum of £20,000 immediately on death before age 65. Premiums are waived during sickness (even when no benefit is payable), and the off-period is as in the Manchester Unity sickness tables. The basis for the calculation of premiums and reserves is: Mortality: E.L.T. 12 - Males Interest: 4% p.a. Sickness: Manchester Unity Experience 1893-97 (Occupation Group AHJ) Expenses: 20% of all premiums, including those waived (a) Calculate the weekly premium for each of these policies. (b) Calculate the reserve per policy in force on 1 January 1995. 298 CHAPTER 18. SICKNESS FUNCTIONS Solutions 18.1 (i) K x = H x +K x+1 · D x+ 1 2 .z x +K x+1 · 1 2 (D x +D x+1 )z x +K x+1 = 581, 464 (ii) Let weekly premium for the basic policy be P, and let k.P be the extra premium for the waiver benefit. The equation of value is 52.18 0.85kP¯ a 30:25 = P(1 +k) ¸ 1 2 (K 31 −K 32 ) + (K 32 −K 55 ) D 30 (note the waiting period of 1 year). Hence 690.07k = (1 +k)20.81 Therefore k = 0.0311 = 3.11%. Note One does not need to calculate the actual premium P in this example. 18.2 Let weekly premium be P. The equation of value is 52.18P.¯ a 35:30 = 5000 D 65 D 35 ¨ a (12) 65 + 30 0 v t . t p 35 µ 35+t (52.18P¯ s¯ t| ) dt +P K 26/all 35 −K 26/all 65 /D x (i) Note that 30 0 v t . t p 35 µ 35+t (52.18P¯ s¯ t| ) dt = 52.18P 30 0 1 −v t δ t p 35 µ 35+t dt = 52.18P δ 30 q 35 − ¯ A 1 35:30 Hence (i) gives 869.13P = 9, 994.55 + 203.26P + 16.8815P Hence P = £15.40. 18.3 (a) Let weekly premium be P. The equation of value is 52.18 0.9P¯ a 30:20 = 100 K 26 30 −K 26 60 D 30 + 50 K 26/26 30 −K 26/26 60 + 30 K 52/all 30 −K 52/all 60 D 30 Hence 640.51P = 1711.35 + 125.21 + 242.74 and so P = £3.25 per week. 18.5. SOLUTIONS 299 (b) Let extra premium be E per week. The equation of value for E is (E + 3.25) K 40 −K 50 D 40 = 52.18 0.9E¯ a 40:10 Therefore (E + 3.25) 13.077 = 382.67E and so E = 0.115, say £0.12. Hence revised premium is £3.37 per week. 18.4 (a) Let weekly contribution be P. m.p.v. benefits: (i) 2000 ¯ A 25 = 377.64 (ii) 500 10 0 v t . t p 20 µ 20+t . t p 25 dt · 500 0.008638 = 4.319 (using Simpson’s Rule) (iii) 10 52.18 D 65 D 25 ¯ a 65 = 693.28 (iv) (6 +P) K 25 −K 65 D 25 + 4 K 13 25 −K 13 65 D 25 = 254.23 + 31.853P Hence the equation of value is 52.18 0.95P¯ a 25:40 = 377.64 + 4.319 + 693.28 + 254.23 + 31.853P Hence 927.54P = 1329.46 and so, P = £1.43. (b) 10 V = 2000 ¯ A 35 + 52.18 10 ¯ N 65 D 35 + (6 + 1.43) K 35 −K 65 D 35 + 4 K 13 35 −K 13 65 D 35 −0.95 1.43 52.18¯ a 35:30 = 537.38 + 1038.18 + 270.88 + 62.55 −1180.71 = £728.28 The reserve may also be calculated by the retrospective method. 18.5 (a) Let the weekly contribution be P. The equation of value is 0.95 52.18P¯ a 35:25 = 1, 000 ¯ A 1 35:25 + 52.18P( ¯ I ¯ A) 1 35:25 + 2000 D 60 D 35 + (12.5 +P) K 35 1 2 −K 60 D + 12.5 K 13 35 1 2 −K 13 60 D 35 (i) Use, ( ¯ I ¯ A) 1 35:25 · (I ¯ A) 1 35:25 − 1 2 ¯ A 1 35:25 and find K 35 1 2 and K 13 35 1 2 by linear interpolation. 300 CHAPTER 18. SICKNESS FUNCTIONS Equation (i) gives 759.52P = 86.28 + 71.38P + 625.57 + (12.5 +P)28.066 + 165.92 and so P = £1.86 per week (b) 5 V = (1000 + 5 52.18P) ¯ A 1 40:20 + (52.18 1.86)( ¯ I ¯ A) 1 40:20 + 2000 D 60 D 40 + (12.5 +P) K 40 −K 60 D 40 + 12.5 K 13 40 −K 13 60 D 40 −0.95 52.18P¯ a 40:20 (Prospectively) Using P = 1.86, this gives 5 V = 142.35 + 113.99 + 768.11 + 413.06 + 59.53 −1222.68 = £374.36 18.6 Using the prospective method, Reserve = 40 K 26 50 −K 26 65 D 50 + 50 K 26/26 50 −K 26/26 65 D 50 + K 50 −K 65 D 50 + 3000 ¸ ¯ A 1 50:15 + D 65 D 50 −52.18¯ a 50:15 = 681.85 + 209.36 + 39.08 + 1775.84 −542.88 = £2163.25 18.7 (a) Let the weekly premium be P. The equation of value is 0.8 52.18P¯ a 35:30 = 20000 ¯ A 1 35:30 + 20 K 26/all 35 −K 26/all 65 D 35 +P K 35 −K 65 D 35 This gives 695.30P = 2472.44 + 337.63 + 36.46P Hence P = £4.27 per week. (b) Using the prospective method 10 V = 20000 ¯ A 1 45:20 + 20 K 26/all 45 −K 26/all 65 D 45 +P K 45 −K 65 D 45 −0.8 52.18P¯ a 45:20 With P = 4.27, this gives 10 V = 3176.82 + 416.27 + 168.42 −2285.91 = £1475.60 18.5. SOLUTIONS 301 Appendix: The Manchester Unity Experience 1893-97 1. Until fairly recently, when the C.M.I. published statistics concerning P.H.I. (Permanent Health Insurance), the principal published set of sickness rates were those derived from the experience in 1893-97 of the Independent Order of Oddfellows, Manchester Unity (a large “affiliated order” friendly society, i.e. an association of lodges with some measure of central control.) The members were all male. 2. Friendly societies existed largely to protect wage-earners in times of adversity, there being no State benefits before 1908 (unless one counts the Workmen’s Compensation Acts of 1897, which covered injuries at work.) There was (and to some extent still is) a wide variety of societies ranging in size and financial stability, and not all societies provided sickness benefits. From 1911 to 1946, sickness benefits for wage-earners were provided by Approved Societies (friendly societies or life offices approved under the National Insurance Act, 1911). A part of their income was provided by a Government grant, and they had to use a valuation basis specified by the Government Actuary. Societies with favourable experience were able to pro- vide additional (in practice, usually dental and ophthalmic) benefits. Since 1946 the State has provided welfare benefits (including sickness benefits) directly, either in return for national insurance contributions or under a means test. The friendly societies have generally declined in importance; many have disappeared, and others concentrate on social activities. A certain number (particularly of centralized societies) continue, and indeed prosper, in modern con- ditions: under the Friendly Societies Act 1992, they are able to conduct a greater variety of business. 3. The Manchester Unity 1893-97 investigation was carried out by Alfred W. Watson, and pub- lished in 1903. A calendar year system was used, lives being classified according to age nearest birthday at the start of the calendar year. The unadjusted sickness rates are of the form ˆ z x = no. of weeks of sickness at age x last birthday E c x These rates were graduated by an adjusted-average formula to produce the published rates, z x . (The statistical basis of the graduation of sickness rates is more complex than for mortality rates, and we do not attempt a discussion.) The rates were also subdivided by periods of sickness, giving z 13 x , z 13/13 x , z 26/26 x , z 52/52 x , z 104/all x . The off-period assumed in the investigation was 1 year, this being the actual off-period for most of the lodges. 4. Members were also subdivided according to the following occupational groups: A: Agricultural Workers B: Outdoor Tradesmen and Labourers C: Railwaymen D: Seamen and Fishermen E: Quarry Workers F: Iron and Steel Workers G: Miners H: Rural workers not included in A-G J: Urban workers not included in A-G Sickness rates were calculated for the four groups AHJ, BCD, EF, G, it being found that sickness rates were lowest in the first group and increased as one moved from one group to the 302 CHAPTER 18. SICKNESS FUNCTIONS next. The experience of occupation group AHJ is given in Tables for Actuarial Examinations. Since the middle classes and well-to-do did not (in general) join friendly societies providing sickness benefits, this represents the experience of wage-earners in 1893-97, excluding those in the more hazardous occupations. (In some cases, however, sickness rates were high because the jobs demanded a high level of physical fitness.) 5. Mortality investigations were also conducted, with separate tables for 3 geographical areas and for urban and rural areas, but these have long been out of date. It is usual to combine the M.U. sickness rates with a more modern mortality table (e.g. English Life Table No. 12- Males, as used in Tables for Actuarial Examinations.) 6. The sickness rates of M.U. (whole society) were found to be remarkably similar to those for employed men in the national insurance scheme in 1953-58. 7. Strictly speaking, sickness rates should be “select”, i.e. they should depend on duration of membership as well as on attained age, since new entrants are not normally accepted unless they are reasonably healthy. Another point is illustrated by the example of a new entrant aged x who is to receive £20 per week for the first 26 weeks of sickness, £10 for the second 26 weeks and £5 per week for the remainder of sickness, subject to a waiting period of 6 months. The value of the benefits might be adjusted from 20K 26 x+ 1 2 + 10K 26/26 x+ 1 2 + 5K 52/all x+ 1 2 D x (i) to 20K 26 x+ 1 2 + 10K 26/26 x+1 + 5K 52/all x+1.5 D x (ii) to allow for the fact that no “26/26” benefit can be received before age x + 1 and no “52/all” benefit can be received before age x + 1.5. But formula (ii) is not quite right; consider, for example, the “26/26” benefits, which are now valued as 10[D x+1.5 z 26/26 x+1 +D x+2.5 z 26/26 x+2 +...] D x Now z 26/26 x+1 is based on an experience (M.U.) in which the exposed to risk included some recent entrants who could not claim “26/26” benefit because they had not been eligible for sickness benefit for more than 6 months. That is, z 26/26 x+1 should be slightly adjusted upwards relative to that of M.U. Similarly, z 26/26 x+2 should be slightly increased, so formula (ii) understates the expected value of the benefits. It is therefore better to use the formula 10[0.5D x+0.5 z 26/26 x +D x+1.5 z 26/26 x+1 +...] D x since the “extra” first term allows approximately for the fact that the later terms are too low. This argument leads us to use sickness rates without deferment (except for the waiting period), as in formula (i), except perhaps when benefits cease soon after the entry age. In practice, the uncertain definition of “sickness” and other factors are likely to be of much greater importance in the estimation of sickness rates. But when comparing the actual weeks of sickness claim at the later durations in a given experience with that expected on the basis of Manchester Unity (say), one must bear in mind any differences in the proportions of members whose durations of membership are so short that they cannot claim. Thus, for example, an experience consisting mainly of recent entrants may be expected to have low rates of sickness at the later durations. Chapter 19 PENSION FUNDS 19.1 General Introduction Pension schemes may be described (in broad terms) as either: (a) defined - benefit schemes, or (b) defined - contribution schemes. Defined - benefit schemes These are pension schemes whereby the pension and other benefits are set out in the rules of the scheme. Most schemes of this type provide benefits which depend on the ‘final salary’. For example, in a typical U.K. public sector final salary scheme, the annual pension is 1 80 the final salary per year of service. There may be various other benefits such as death benefits and a lump sum on retirement. The benefits are paid for by a combination of (i) the employee’s contributions (e.g. a fixed percentage of salary) and (ii) the employer’s contributions, which may vary from time to time according to actuarial advice. Remarks 1. A “non-contributory” pension scheme is one in which the employees do not contribute (e.g. U.K. civil service). 2. An “insured” pension scheme is one in which benefits are secured by contracts with a life office. Defined - contribution schemes These are also called “money purchase” schemes; the contributions of both employees and em- ployer are fixed (often as percentages of the salary). The pension benefit is what these combined contributions can buy, usually after investment (either directly by the scheme or by life office con- tracts) until retirement age. Obviously, the annual pension depends on the rate of return obtained on the employer’s and member’s contributions and the terms on which pension may be purchased at retirement (unless pension is purchased directly when the contributions are received). 19.2 Valuation Principles Only defined-benefit schemes shall be considered, and the mean present value of the future benefits and future contributions of an “active” member aged x will be calculated. (“Active” refers to a 303 304 CHAPTER 19. PENSION FUNDS member who has not yet retired). The reserve for each member is calculated prospectively. That is, reserve = mean present value of future benefits - mean present value of future contributions (of both employee and employer) Notes 1. The rate of interest, i per annum, used in valuing the benefits and contributions is normally gross (free of tax). i is called the valuation interest rate. 2. Expenses are usually ignored, as they are paid for separately by the employer. 3. A service table is required. This is a multiple-decrement table with various modes of decrement: death, ill-health retirement, etc. 4. If benefits or contributions depend on the employee’s salary, a salary scale is required to estimate future salaries from current salaries. 5. Sometimes one may require to find the employer’s contribution rate by setting the reserve equal to zero at entry to the scheme of a new entrant, or for a group of new entrants. 19.3 Service Tables The following notation is used, as in “Formulae and Tables in Actuarial Examinations”: l x = number of members at exact age x d x = number of deaths at age x last birthday i x = number of “ill-health” retirements at age x last birthday r x = number of “age” retirements at age x last birthday w x = number of withdrawals at age x last birthday Notes 1. “Age” retirements are the retirements at or above the minimum normal retirement age (NRA) of the scheme. “Age” retirements may be concentrated at an exact age (for example 60 or 65) or may be spread uniformly between ages 60 and 65, or both. 2. Members often have to retire by a certain age (often 65). In the “Formulae and Tables”, “age” retirements occur only from age 60 onwards, and all members must retire at age 65 at the latest. 3. To calculate the value of benefits one may require separate mortality tables for “age” retirement pensioners and ill-health retirement pensioners. We may also have to deal with benefits for people with “deferred” pensions in the scheme who no longer work for the company, so a mortality table for them is sometimes required (see later.) 19.4 Salary Scales Assume that salaries are revised continuously. Define ¦¯ s x ¦, the salary scale function, to be such that the salary rate per annum at age x + t of a life now aged x with current salary rate £(SAL) per annum is (SAL) ¯ s x+t ¯ s x (19.4.1) Also define s x = x+1 x ¯ s y dy = 1 0 ¯ s x+t dt (19.4.2) 19.4. SALARY SCALES 305 Using approximate integration in (19.4.2), we have s x · ¯ s x+ 1 2 · 1 2 [¯ s x + ¯ s x+1 ] (19.4.3) “Formulae and Tables” gives values of s x (not ¯ s x ), so we find ¯ s x by linear interpolation: ¯ s x · 1 2 [s x−1 +s x ] · s x− 1 2 (19.4.4) Note To estimate ¯ s 65 , one must use ¯ s 64 1 2 · s 64 and ¯ s 63 1 2 · s 63 and then employ linear extrapolation: see Example 19.4.1 below. The salary to be earned by (x) between ages x +t and x +t + 1, given that he is an active member during this age-interval, is estimated as: (SAL) s x+t ¯ s x (19.4.5) where SAL is the current salary rate (in £) earned by (x). We also find that: (SAL) ¯ s x+t ¯ s x = assumed salary rate per annum at exact age x +t (19.4.6) Adjustments 1. If SAL refers to the earnings received in the past year (i.e. between ages x −1 and x), adjust the denominator from ¯ s x to s x−1 . 2. If SAL refers to the expected earnings in the coming year (i.e. between ages x and x +1), adjust the denominator from ¯ s x to s x . Example 19.4.1. (a) Consider a person now aged exactly 25 whose annual salary rate is currently £9,192. Estimate (i) his annual salary rate at exact age 53, (ii) his earnings between exact ages 64 and 65, (iii) the average amount earned by him each year between exact ages 60 and 65, (iv) his annual salary rate at exact age 65. If he dies at age 57 last birthday, determine the average values of (v) his annual salary rate at the moment of death, (vi) the total amount earned over his last year of life. (b) Calculate revised answers for (i) to (vi) assuming that the person aged exactly 25 is expected to earn £9,192 in the coming year. Assume that salaries are revised continuously, and use the pension table in “Formulae and Tables for Actuarial Exams” with 4% p.a. interest. Solution (a) (i) 9192 ¯ s 25 ¯ s 53 · 9192 s 24 1 2 s 52 1 2 = 9192 4.77 1.80 = £24, 359 (ii) 9192 ¯ s 25 s 64 · 9192 s 24 1 2 s 64 = £27, 576 (iii) 9192 ¯ s 25 s 60 +s 61 +s 62 +s 63 +s 64 5 · £27, 198. 306 CHAPTER 19. PENSION FUNDS (iv) Annual salary rate at 64 is 9192 ¯ s 64 ¯ s 25 · 27, 499 Annual salary rate at 64 1 2 is 9192 ¯ s 64 1 2 ¯ s 25 · 27, 576 Hence by linear extrapolation the annual salary rate at age 65 is approximately 27, 576 + (27, 576 −27, 499) = £27, 653 (v) On average he will die at exact age 57 1 2 . Hence answer is 9192 ¯ s 25 ¯ s 57 1 2 · 9192 ¯ s 25 s 57 = £25, 891 (vi) 9192 ¯ s 25 s 56 1 2 · £25, 738. (b) Adjusting the denominator from ¯ s 25 to s 25 results in multiplying each answer by a factor of ¯ s 25 s 25 · s 24 1 2 s 25 · 0.962567. This gives answers of: (i) £23, 447 (ii) £26, 544 (iii) £26, 180 (iv) £26, 618 (v) £24, 922 (vi) £24, 775 Estimation of ¯ s x and s x ¯ s x is usually estimated by a product of 2 factors, one to allow for future inflationary increases and one to take account of “career progression”. The inflation factor is usually of the form (1 + e) x , where e is the assumed annual rate of future salary escalation. It is often assumed that i −e · the “real” rate of interest (assuming wage inflation and price inflation are about the same) ·2% (or some similar figure) where i is the valuation interest rate and future salary inflation is assumed to be e per annum. 19.5 The Value of Future Contributions There are 2 main ways of calculating contributions: (a) contributions are at the rate of k% of salary; (b) contributions are at a fixed rate of £F per annum (payable continuously). (a) The mean present value of the future contributions (of employee, employer or both) at rate k% of salary for a member age x with current salary rate of £SAL per annum is k 100 SAL ¯ s x 65−x 0 v t l x+t l x ¯ s x+t dt This is usually approximated to k 100 SAL ¯ s x 64−x ¸ t=0 v t+ 1 2 l x+t+ 1 2 l x s x+t (using ¯ s x+ 1 2 · s x ) (19.5.1) (Note the summation is up to “64 −x”). Now define D x = v x l x , (as previously defined) 19.5. THE VALUE OF FUTURE CONTRIBUTIONS 307 ¯ D x = 1 0 D x+t dt · D x+ 1 2 · 1 2 [D x +D x+1 ] s ¯ D x = s x ¯ D x and s ¯ N x = ¸ 64−x t=0 s ¯ D x+t One can now evaluate (19.5.1) by means of commutation functions, giving a mean present value of k 100 SAL ¯ s x 64−x ¸ t=0 v x+t+ 1 2 .l x+t+ 1 2 v x l x s x+t = k 100 SAL ¯ s x 64−x ¸ t=0 D x+t+ 1 2 D x s x+t · k 100 SAL ¯ s x D x 64−x ¸ t=0 s ¯ D x+t = k 100 SAL ¯ s x D x s ¯ N x . So, the m.p.v. of future contributions is k 100 SAL ¯ s x D x s ¯ N x (19.5.2) Notes 1. If SAL refers to the past year, change ¯ s x to s x−1 in the denominator. 2. If SAL refers to the coming year, change ¯ s x to s x in the denominator. Example 19.5.1. Consider a life aged 35 with current salary rate £10,000 who contributes 5% of his salary to a pension scheme. Using the Examination Tables, calculate the mean present value of the employee’s future contributions. Solution m.p.v. of contributions is 5 100 10000 ¯ s 35 D 35 s ¯ N 35 · 500 417, 224 1 2 (2.98 + 3.08) 7232 = £9520. (b) Suppose contributions are independent of salary and that these are a fixed annual sum of £F, 308 CHAPTER 19. PENSION FUNDS payable continuously. Then the mean present value of future contributions is F 65−x 0 v t l x+t l x dt · F 64−x ¸ t=0 v t+ 1 2 l x+t+ 1 2 l x = F 64−x ¸ t=0 D x+t+ 1 2 D x · F 1 D x 64−x ¸ t=0 ¯ D x+t = F ¯ N x D x (19.5.3) where ¯ N x = ¸ 64−x t=0 ¯ D x+t . Example 19.5.2. If a life aged 35 contributes £500 each year to his pension scheme, calculate the value of his future contributions. Solution m.p.v. = 500 ¯ N 35 D 35 = £7, 096. Suppose now that contributions are k% of the “pensionable salary”, where pensionable salary is defined as “salary - A”, A being a fixed amount. Assuming that the current salary rate, SAL, already exceeds A, the m.p.v. of these contributions is k 100 SAL ¯ s x s ¯ N x D x − k 100 A ¯ N x D x (19.5.4) Example 19.5.3. Employees contribute to a pension scheme at a rate of 6% of “pensionable salary”, where “pensionable salary” equals actual annual salary rate less £2000 (to allow for other pension arrangements). Find the present value of the future contributions payable by a member aged exactly 30 whose current annual salary rate is £15,500. Solution m.p.v. · 15500 0.06 s 29 1 2 D 30 s ¯ N 30 −2000 0.06 ¯ N 30 D 30 = 18, 377.7 −1, 606.5 = £16, 771 19.6 The Value of Pension Benefits Pension benefits are usually of one of the following forms: (a) A fixed pension of £P per year of service; (b) A fraction of the average salary per year of service; 19.7. FIXED PENSION SCHEMES 309 (c) A fraction of the final salary per year of service. Notes 1. Years of service normally include fractions counting pro rata. 2. The benefits on the date of age retirement or ill-health retirement are valued by multiplying the annual pension by an appropriate annuity function. For example, the benefit on age retirement at age 65 is annual pension ¯ a r 65 where r indicates age retirement mortality rates. Similarly, ¯ a i x refers to a life retiring at age x due to ill-health. Ill-health retirement mortality is usually heavier than that for age retirements. 3. Although the symbol ¯ a r 65 is usually used, the benefit may be payable in various ways, e.g. monthly in advance, and is possibly subject to a guarantee that at least 5 years’ payments will be made. In this example, ¯ a r 65 = ¨ a (12) 5 + 5 [¨ a (12) 65 on a suitable mortality table. 19.7 Fixed Pension Schemes Consider a fixed pension of £P per year of service, including fractions pro rata, for a life now aged x with n years’ past service. Ill-health and age retirements will be considered separately. Ill-health retirements The m.p.v. of the benefit is 64−x ¸ t=0 v t+ 1 2 i x+t l x (n +t + 1 2 )P¯ a i x+t+ 1 2 This can be divided into 2 terms: 1. The value of the Past Service Pension (P.S.P.), which is the pension earned in respect of past service with the employer; in this case the P.S.P. is nP; 2. The value of the Future Service Pension (F.S.P.), which is the pension which will be earned in future; in this case the F.S.P. is (t + 1 2 )P if retirement occurs (due to ill-health) at age x +t + 1 2 . So the m.p.v. is nP 64−x ¸ t=0 v t+ 1 2 i x+t l x ¯ a i x+t+ 1 2 +P 64−x ¸ t=0 v t+ 1 2 i x+t l x (t + 1 2 )¯ a i x+t+ 1 2 (19.7.1) Define the following commutation functions: C ia x = v x+ 1 2 i x ¯ a i x+ 1 2 , M ia x = 64−x ¸ t=0 C ia x+t ¯ M ia x = M ia x − 1 2 C ia x , and ¯ R ia x = 64−x ¸ t=0 ¯ M ia x+t 310 CHAPTER 19. PENSION FUNDS It follows that ¯ R ia x = 64−x ¸ t=0 (t + 1 2 )C ia x+t Proof. ¯ R ia x = ¯ M ia x + ¯ M ia x+1 +... + ¯ M ia 64 = ( 1 2 C ia x +C ia x+1 +... +C ia 64 ) + ( 1 2 C ia x+1 +C ia x+2 +... +C ia 64 ) +... + 1 2 C ia 64 = 1 2 C ia x + 1 1 2 C ia x+1 +... + (64 −x + 1 2 )C 64 = 64−x ¸ t=0 (t + 1 2 )C ia x+t . These commutation functions can now be used to calculate (19.7.1). The value of the P.S.P. is nP 64−x ¸ t=0 v x+t+ 1 2 v x l x i x+i ¯ a i x+t+ 1 2 = nP 1 D x 64−x ¸ t=0 C ia x+t = nP M ia x D x The value of the F.S.P. is P 64−x ¸ t=0 v x+t+ 1 2 i x+t v x l x (t + 1 2 )¯ a i x+t+ 1 2 = P 1 D x 64−x ¸ t=0 (t + 1 2 )C ia x+t = P ¯ R ia x D x Hence the value of all ill-health pension benefits is P ¸ n.M ia x + ¯ R ia x D x (19.7.2) Age retirements The valuation of benefits caused by age retirements is very similar to that for ill-health retirements, but with a final term corresponding to age retirement at exact age 65. The m.p.v. of age retirement benefits is 64−x ¸ t=0 v t+ 1 2 r x+t l x (n +t + 1 2 )P¯ a r x+t+ 1 2 +v 65−x r 65 l x (n + 65 −x)P¯ a r 65 (19.7.3) 19.8. AVERAGE SALARY SCHEMES 311 Again, this may be separated into the P.S.P. and the F.S.P. terms. Define the commutations C ra x = v x+ 1 2 r x ¯ a r x+ 1 2 , x < 65 v 65 r 65 ¯ a r 65 , x = 65 M ra x = ¸ 65−x t=0 C ra x+t (note the summation is up to “65 −x”) ¯ M ra x = M ra x − 1 2 C ra x and ¯ R ra x = 64−x ¸ t=0 ¯ M ra x+t = 64−x ¸ t=0 (t + 1 2 )C ra x+t + (65 −x)C ra 65 (using a very similar proof to that for ill-health retirements). Then the value of the P.S.P. is nP 1 D x ¸ 64−x ¸ t=0 v x+t+ 1 2 r x+t ¯ a r x+t+ 1 2 +v 65 r 65 ¯ a r 65 ¸ = nP M ra x D x The value of the F.S.P. is P 1 D x ¸ 64−x ¸ t=0 (t + 1 2 )v x+t+ 1 2 r x+t ¯ a r x+t+ 1 2 + (65 −x)v 65 r 65 ¯ a r 65 ¸ = P 1 D x ¸ 64−x ¸ t=0 (t + 1 2 )C ra x+t + (65 −x)C ra 65 ¸ = P ¯ R ra x D x Hence the value of all age retirement pension benefits is P ¸ n.M ra x + ¯ R ra x D x . So if one defines M (i+r)a x = M ia x + M ra x , and similarly for other commutation functions, then the value of a fixed pension of £P per annum payable for any retirement is P ¸ n.M (i+r)a x + ¯ R (i+r)a x D x ¸ (19.7.4) for a member aged x with n years’ past service. 19.8 Average Salary Schemes Suppose that the annual pension on retirement is 1 60 (total salary in service of company) = 1 60 (total past salary, T.P.S.) + 1 60 (total future salary). 312 CHAPTER 19. PENSION FUNDS Thus the benefits can be separated into the Past Service Pension and the Future Service Pension. Value of P.S.P. This is just the value of a fixed pension of 1 60 (T.P.S.) so, using the functions defined in the previous section, m.p.v. of past service benefits = T.P.S. 60 M (i+r)a x D x (19.8.1) Value of F.S.P. Consider ill-health retirement first. The salary to be earned between ages x +t and x +t + 1 2 is estimated as 1 2 (SAL) s x+t ¯ s x The m.p.v. of benefits is therefore 1 60 64−x ¸ t=0 v t+ 1 2 i x+t l x SAL ¯ s x (s x +s x+1 +... +s x+t−1 + 1 2 s x+t )¯ a i x+t+ 1 2 = SAL 60¯ s x D x 64−x ¸ t=0 (s x +s x+1 +... +s x+t−1 + 1 2 s x+t )C ia x+t (19.8.2) Define the commutation functions s ¯ M ia x = s x ¯ M ia x and s ¯ R ia x = 64−x ¸ t=0 s ¯ M ia x+t The ‘summation’ in (19.8.2) can be re-written as s x ( 1 2 C ia x +C ia x+1 +... +C ia 64 ) +s x+1 ( 1 2 C ia x+1 +C ia x+2 +... +C ia 64 ) +...s 64 1 2 C ia 64 = s x ¯ M ia x +s x+1 ¯ M ia x+1 +... +s 64 ¯ M ia 64 So the value of the F.S.P. is SAL 60¯ s x D x 64−x ¸ t=0 s x+t ¯ M ia x+t = SAL 60¯ s x D x 64−x ¸ t=0 s ¯ M ia x+t = SAL 60¯ s x D x s ¯ R ia x Consider age retirements now. The m.p.v. of the future service pension is very similar to the summation in (19.8.2), with all ‘i’ terms being replaced by ‘r’ terms and with an extra term relating to retirements at exactly age 65; this means that a final term of SAL 60¯ s x D x (s x +s x+1 +... +s 64 )C ra 65 19.9. FINAL SALARY SCHEMES 313 has to be added. On defining the commutation functions s ¯ M ra x = s x ¯ M ra x and s ¯ R ra x = ¸ 64−x t=0 s ¯ M ra x+t and using a similar argument to that used for ill-health retirements, the value of the F.S.P. is found to be SAL 60¯ s x D x s ¯ R ra x Thus the value of all benefits for an average salary scheme is TPS 60 M (i+r)a x D x + SAL 60¯ s x s ¯ R (i+r)a x D x (19.8.3) Example 19.8.1. A pension scheme provides each member who retires (whether for “age” or “ill- health” reasons) with an annual pension of 1 60 th of his average annual income over a member’s entire service, for each year of service. Fractions of years of service are included when calculating the amount of pension payable. If contributions are paid entirely by the employer, calculate the appropriate contribution rate (as a percentage of salary) for a new entrant aged 20. Solution Let k% be the contribution rate. Then k must solve k 100 SAL ¯ s 20 s ¯ N 20 D 20 = SAL 60¯ s 20 s ¯ R (i+r)a 20 D 20 Hence, k = s ¯ R (i+r)a 20 0.6 s ¯ N 20 = 5.25%. 19.9 Final Salary Schemes Suppose that the annual pension is given by the formula 1 80 “final salary” per year of service. “Final salary” is usually defined as the average annual salary in the last m years of service. Define z x = 1 m (s x−m +s x−m+1 +... +s x−1 ). In “Formulae and Tables”, m = 3, and hence z x = 1 3 (s x−3 +s x−2 +s x−1 ) The method of valuing benefits is similar to that in section 19.7, allowing for salary factors. Ill-health retirements The m.p.v. of benefits for a member aged x with n years past service is 64−x ¸ t=0 SAL 80¯ s x z x+t+ 1 2 (n +t + 1 2 )v t+ 1 2 i x+t l x ¯ a i x+t+ 1 2 (19.9.1) 314 CHAPTER 19. PENSION FUNDS Define the commutation functions z C ia x = z x+ 1 2 C ia x , z M ia x = 64−x ¸ t=0 z C ia x+t , z ¯ M ia x = z M ia x − 1 2 z C ia x , and z ¯ R ia x = 64−x ¸ t=0 z ¯ M ia x+t = 64−x ¸ t=0 (t + 1 2 ) z C ia x+t (note similarity to definitions in section 19.7) Expression (19.9.1) becomes SAL 80¯ s x D x 64−x ¸ t=0 (n +t + 1 2 ) z C ia x+t and, following the usual steps, we find that the m.p.v. of the benefits is equal to SAL 80¯ s x n. z M ia x D x + SAL 80¯ s x z ¯ R ia x D x (19.9.2) This shows the terms for past and future service pensions separately. Age retirements Again this is similar to the fixed pension case, but with salary factors inserted. The new commutation functions are z C ra x = z x+ 1 2 C ra x if x < 65, z 65 C ra 65 if x = 65, z M ra x = 65−x ¸ t=0 z C ra x+t (note the upper limit of 65 −x) z ¯ M ra x = z M ra x − 1 2 z C ra x z ¯ R ra x = 64−x ¸ t=0 z ¯ M ra x+t We find that the m.p.v. of benefits due to age retirements is SAL 80¯ s x D x ¸ 64−x ¸ t=0 (n +t + 1 2 ) z C ra x+t + (n + 65 −x) z C ra 65 ¸ = SAL 80¯ s x n. z M ra x D x + SAL 80¯ s x z ¯ R ra x D x (19.9.3) Thus the value of all pension benefits for a final salary scheme is SAL 80¯ s x D x n. z M (i+r)a x + z ¯ R (i+r)a x (19.9.4) 19.9. FINAL SALARY SCHEMES 315 Example 19.9.1. Three members of a pension scheme whose age nearest birthday is 45 have the following annual rates of salary and exact numbers of years of past service: A : £15,000 20 years B : £12,000 10 years C : £14,000 5 years For these members find the present value of a pension, payable on age-retirement or on ill-health retirement, of 1 100 th of the average salary in the final 3 years before retirement for each year of service, including fractions. Give separately the values of the past-service and future-service benefits. Solution Value of P.S.P. = 1 100 ¸ 3 i=1 n i (SAL) i ¯ s 45 D 45 z M (i+r)a 45 · 1 100 490000( z M ia 45 + z M ra 45 ) 1 2 (s 44 +s 45 )D 45 = 29, 071 Value of F.S.P. = 1 100 ¸ 3 i=1 (SAL) i ¯ s 45 D 45 z ¯ R (i+r)a 45 · 1 100 41000 1 2 (s 44 +s 45 )D 45 ( z ¯ R ia 45 + z ¯ R ra 45 ) = 42, 929. Hence total benefits have m.p.v. £72,000. Remarks 1. If retirement is possible within m years, the factor z x should be adjusted to allow for the actual past salary progression. Also, if someone retires with less than m years’ service, the final salary must be redefined as the average over a shorter period. (In calculations these points are sometimes ignored.) 2. If “final salary” means the salary rate at the date of retirement, then we change z x to ¯ s x . Also, if m = 1, z x = s x−1 . But one must not change z ¯ R (i+r)a x to s ¯ R (i+r)a x , as the latter refers to average salary schemes. 3. If fractions of a year are not included when calculating the pension value, it is usually sufficiently accurate to adjust the n years’ past service to n− 1 2 . If there is no past service, consider the member as having “− 1 2 ” year past service. So one must substitute (n − 1 2 ). z M (i+r)a x for n. z M (i+r)a x in formula (19.9.4) (even when n = 0). 4. Benefits may depend on a “pensionable salary” equal to salary minus some fixed sum, say £A. If so, then the m.p.v. of the pension benefit in a final salary scheme is now SAL 80¯ s x D x n. z M (i+r)a x + z ¯ R (i+r)a x − A 80 n.M (i+r)a x − ¯ R (i+r)a x D x 5. It may be the case that the number of years of service are restricted to, say, 40 for pension purposes. This will affect lives joining at ages under 25, assuming a latest retirement age of 65. There are 2 cases to consider: 316 CHAPTER 19. PENSION FUNDS (a) If n < 40, then leave the P.S.P. unchanged but restrict the F.S.P. . In the final salary scheme mentioned above, the value of the F.S.P. is altered to SAL 80¯ s x D x z ¯ R (i+r)a x − z ¯ R (i+r)a x+40−n (19.9.5) (b) If n ≥ 40, then restrict the past service to 40 years and make the F.S.P. zero. Hence the value of pension benefits is 40 80 SAL ¯ s x D x z M (i+r)a x (19.9.6) Example 19.9.2. An executive pension scheme provides a pension of 1 45 of final salary for each year of scheme service, with a maximum of 2 3 of final salary, upon retirement due to age between the ages of 60 and 65. Final salary is defined as salary in the 3 years prior to retirement. A director, now aged 47 exactly has 14 years of past service and expects to earn £80,000 over the coming year. Using the symbols defined in the Formulae and Tables for Actuarial Examinations, what is the expected present value of the future service pension on age retirement for this member? Solution As the maximum pension is 2 3 = 30 45 of the final salary, service is restricted to a maximum of 30 years for pension purposes. Hence, as n is 14, m.p.v. of F.S.P. = 80000 45 z ¯ R ra 47 − z ¯ R ra 63 s 47 D 47 (notice that £80,000 is the salary next year and not the salary rate, and hence we put s 47 in the denominator). 19.10 Lump Sums on Retirement Suppose there is a cash payment on retirement of 3 times the annual pension, where the pension is 1 80 final salary per year of service (as in section 19.9). Define the commutation functions C i x = i x v x+ 1 2 z C i x = z x+ 1 2 C i x , and so on, changing the annuity factor in all commutation functions (both “age” or “ill-health”) to 1. These functions are tabulated (on the same basis as in “Formulae and Tables”) in the Supplement to this book. Using nearly the same arguments as before, the m.p.v. of the lump sum on retirement is 3 SAL 80¯ s x D x (n. z M i+r x + z ¯ R i+r x ) (19.10.1) If the annual pension is a fixed sum of £P per year of service, then a lump sum of 3P on retirement has m.p.v. 3P n.M i+r x + ¯ R i+r x D x (19.10.2) 19.11. DEATH AND WITHDRAWAL BENEFITS 317 19.11 Death and Withdrawal Benefits The benefits on death in service usually consist of one or more of: 1. a fixed sum, or a certain multiple of salary; 2. a return of the employee’s contributions, accumulated at a rate of interest j (where j may be zero); 3. a spouse’s pension. We begin by considering the first type of benefit. Suppose the death benefit is 2 the salary rate at date of death. The benefit for a person aged x, with current salary rate SAL, has m.p.v. 2(SAL) 64−x ¸ t=0 v t+ 1 2 d x+t l x s x+t ¯ s x (19.11.1) Define the commutation functions C d x = v x+ 1 2 d x , s C d x = s x C d x M d x = 64−x ¸ t=0 C d x+t s M d x = 64−x ¸ t=0 s C d x+t Expression (19.11.1) can be written in the form 2 SAL ¯ s x D x 64−x ¸ t=0 s x+t C d x+t and hence the value of the benefit is 2 SAL ¯ s x s M d x D x (19.11.2) If the benefit on death is just a fixed amount, £B, then the m.p.v. is B 64−x ¸ t=0 v t+ 1 2 d x+t l x B M d x D x (19.11.3) Withdrawals Similar formulae may be developed (replacing the ‘d’ with ‘w’) but withdrawal benefits are usually in the form of a return of contributions or a deferred pension. (We shall discuss the formulae for valuing these benefits later.) Note The commutation functions M d x , M w x , etc. are tabulated in the Supplement to this book. Note that M d x corresponds to j M d x when j = 0 (page v of the Supplement) and M w x corresponds to j M w x when j = 0 (page vi of Supplement). 318 CHAPTER 19. PENSION FUNDS Example 19.11.1. A company has a pension scheme which provides a pension upon retirement of 1 80 th of the final salary per year of service. In addition the sum of £10,000 is paid on death in service of a member. If all contributions are paid by the employer, find the contribution rate as a percentage of salary required for a new entrant aged 40 with a salary rate of £10,000. Use the basis of “Formulae and Tables” (with the supplement); final salary is the average annual salary in the 3 years prior to retirement. Solution Let k be the contribution rate per cent. Then k solves 0.01k 10, 000 ¯ s 40 D 40 s ¯ N 40 = 10, 000 80¯ s 40 D 40 z ¯ R (i+r)a 40 + 10, 000 M d 40 D 40 Thus, 0.01k 172628.7 = 14827.60 + 895.6 Hence k = 9.11%. 19.12 Return of Contributions on Death or Withdrawal Suppose that the employee’s contributions are returned with compound interest at rate j per annum if the member leaves service. Let TPC be the member’s total past contributions to the scheme. The following additional commutation functions are defined: j C w x = (1 +j) x+ 1 2 v x+ 1 2 w x , j M w x = 64−x ¸ t=0 j C w x+t j ¯ M w x = j M w x − 1 2 j C w x sj ¯ M w x = s x j ¯ M w x sj ¯ R w x = 64−x ¸ t=0 (1 +j) −(x+t+ 1 2 )sj ¯ M w x+t Similar functions for returns of contributions on death are defined by substituting ‘d’ for ‘w’. Suppose that the employees contribute at rate k% of salary. Then the m.p.v. of the return of contributions on withdrawal for a member aged x is (TPC)(1 +j) −x j M w x D x + k 100 SAL ¯ s x D x sj ¯ R w x (19.12.1) Proof. See the Appendix to this Chapter. Note The values of j M w x , sj ¯ R w x , etc. are given for j = 0.03 and j = 0 in the Supplement to this book. If the employee’s contributions are fixed at £A per annum, payable continuously, the m.p.v. of the return of contributions on withdrawal is (TPC)(1 +j) −x j M w x D x +A j ¯ R w x D x (19.12.2) 19.12. RETURN OF CONTRIBUTIONS ON DEATH OR WITHDRAWAL 319 (see the Appendix to this Chapter.) Note When j = 0 one may omit the ‘j’ from the commutation functions, so if contributions are returned without interest their mean present value is (TPC) M w x D x + k 100 SAL ¯ s x D x s ¯ R w x (19.12.3) if the employee contributes k% of salary, and (TPC) M w x D x +A ¯ R w x D x (19.12.4) if he contributes at the fixed rate of £A p.a. If employee’s contributions are returned on death (with or without interest), formulae (19.12.1) to (19.12.4) are changed by just substituting ‘d’ for ‘w’. Furthermore, if contributions are returned on either death or withdrawal, one may adjust formula (19.12.1) to (TPC)(1 +j) −x j M d+w x D x + k 100 SAL ¯ s x D x sj ¯ R d+w x (19.12.5) where j M d+w x = j M d x + j M w x and sj ¯ R d+w x = sj ¯ R d x + sj ¯ R w x , as expected. (Similar remarks apply to formulae (19.12.2) to (19.12.4).) Example 19.12.1. You are actuary to a pension scheme which provides the following benefits: (a) a pension of 1 80 final salary for each year of service (including fractions pro rata) on retire- ment for “age” or “ill-health” reasons. (b) a lump sum on retirement of 3 times the annual pension. Your actuarial basis is that given in the “Formulae and Tables”, and you calculate reserves prospec- tively, ignoring expenses. Final salary is the average annual salary in the 3 years before retirement. Members pay contributions at the rate of 2% of salary. The total contribution rate is assessed for each member separately, and is the proportion of salary which will pay for the benefits, when the member joins, i.e. the prospective reserve at the entry date is zero. 1. Find the employer’s contribution rate for a member joining at age 20. 2. Find the reserve held for a member aged 45, with salary at the rate of £20,000 per annum, who joined the employer at age 20. 3. Your pension scheme accepts transfer values from other pension schemes. Suppose that a life aged 45 joins the scheme, bringing a transfer value of £10,000. The member’s current salary rate is £15,000 p.a., and he asks for his transfer value to be used to give him “added years” of service, i.e. to credit him with n years of past service. Find n, given that, in the event of death in service, the transfer value will be returned with compound interest at 3% p.a. Note There are no withdrawals over age 45 in the Tables, so we need not specify what happens to the transfer value in that event. 320 CHAPTER 19. PENSION FUNDS Solution 1. Let k be total contribution rate per cent for a new member aged 20. Then k 100 SAL ¯ s 20 D 20 s ¯ N 20 = SAL 80¯ s 20 D 20 [ z ¯ R (i+r)a 20 + 3. z ¯ R i+r 20 ] Thus k = 100 80 ¸ z ¯ R (i+r)a 20 + 3. z ¯ R i+r 20 s ¯ N 20 ¸ = 7.443. So employer contributes at rate 7.443%−2% = 5.443% of salary. 2. Reserve is 20, 000 80¯ s 45 D 45 [25. z M (i+r)a 45 + z ¯ R (i+r)a 45 + 3(25. z M i+r 45 + z ¯ R i+r 45 )] − 20, 000 ¯ s 45 D 45 (0.0744) s ¯ N 45 = £59, 585. 3. Let n years of past service be credited. The equation of value is 10000 = 15000n 80¯ s 45 D 45 ( z M (i+r)a 45 + 3. z M i+r 45 ) + 10000(1.03) −45 j M d 45 D 45 where j = 0.03 Hence 10, 000 = 1420.47.n + 1394.42 Therefore n = 6.06 years (The transfer value is treated as if it were the T.P.C.) 19.13 Spouse’s Benefits A pension scheme may provide the following benefits (among others): (a) a spouse’s death in service (D.I.S.) pension, and/or (b) a spouse’s death after retirement (D.A.R.) pension. If spouse’s pension ceases on remarriage, a double-decrement table (for death and remarriage) should be constructed and used in place of the mortality table for spouses. (a) Spouse’s D.I.S. Consider (for example) a male member aged x of a pension scheme providing an annuity payable (say) monthly in advance to his widow if he dies in service. As in chapter 13, we define h x = probability that a man is married at age x and d = the average age difference (in years) between husband and wife, i.e. average value of (husband’s age - wife’s age). The m.p.v. of this benefit is 64−x ¸ t=0 v t+ 1 2 d x+t l x h x+t+ 1 2 (spouse’s pension p.a.)¨ a (12) f x+t+ 1 2 −d (19.13.1) 19.13. SPOUSE’S BENEFITS 321 The size of spouse’s pension may depend on the member’s salary at or near the date of death and the number of years of service. Note. Commutation functions are complicated and are usually ignored. (b) Spouse’s D.A.R. One must consider the following 2 cases: (i) Widow’s pension is only payable if the widow was married to the scheme member when he retired. (Post-retirement marriages do not count for benefit purposes.) (ii) Any widow may receive the pension. The spouse’s D.A.R. pension is valued by first considering the appropriate value at the retirement date of the member, and then allowing for survivorship and interest before retirement. For example, consider a man retiring at age 65 in normal health. Assuming that spouse’s pensions are payable continuously, the value at that age of the spouse’s D.A.R. benefit is as follows: Case (i) : (annual widow’s pension).h 65 ¯ a m 65| f 65−d (19.13.2) Case (ii) : (annual widow’s pension). ∞ 0 v t . t p m 65 µ m 65+t h 65+t ¯ a f 65−d+t dt (19.13.3) Example 19.13.1. A pension fund provides the following benefits for widows of male members: (a) on death of the member in service, a widow’s pension of annual amount equal to one-third of the member’s salary rate at the time of his death; and (b) on death of the member after age or ill-health retirement, a widow’s pension of 1 120 th of the member’s average salary in the 3 years immediately preceding retirement for each year of service, fractions of a year counting pro rata. Develop formulae for valuing the widow’s benefits for a male member aged x with n years’ past service (including fractions). You may assume that all age-retirements take place at exact age 60, between ages 60 and 65, or at exact age 65, and that ill-health retirements may take place at any age between 35 and 60. Widows’ pensions on death after retirement are payable to any widow (not just to a widow who was married to the member when he retired.) Widows’ pensions are payable monthly in advance and do not cease on remarriage. You may assume that a service table has already been constructed, and that the member’s current salary rate per annum is SAL. Commutation functions need not be developed. Solution (a) Define ¯ s x , s x , h x as previously. Let z x = 1 3 (s x−3 +s x−2 +s x−1 ) as before. Let y = average age of wife of (x) = x −d Then value of benefit is SAL 3 64−x ¸ t=0 v t+ 1 2 d x+t l x ¯ s x+t+ 1 2 ¯ s x h x+t+ 1 2 ¨ a (12) y+ f t+ 1 2 322 CHAPTER 19. PENSION FUNDS (b) Consider ill-health retirement first. m.p.v. = SAL 120 59−x ¸ t=0 v t+ 1 2 i x+t l x z x+t+ 1 2 ¯ s x (n +t + 1 2 )a IH x+t+ 1 2 where a IH x+t+ 1 2 = ∞ 0 v r . r p ih x+t+ 1 2 µ ih x+t+ 1 2 +r h x+t+ 1 2 +r ¨ a (12) y+t+ 1 2 +r dr (“ih” indicates that a mortality table for men retiring due to ill-health is employed). For age retirements, there are terms for retirement (i) between ages 60 and 65, (ii) at exact ages 60 and 65. (i) value = SAL 120 64−x ¸ t=60−x v t+ 1 2 r x+t l x z x+t+ 1 2 ¯ s x (n +t + 1 2 )a NH x+t+ 1 2 where a NH x+t+ 1 2 is as for a IH x+t+ 1 2 , but with a mortality table for age retirements. (ii) value = SAL 120 v 60−x r 60 l x z 60 ¯ s x (n + 60 −x)a NH 60 + SAL 120 v 65−x r 65 l x z 65 ¯ s x (n + 65 −x)a NH 65 (r x refers to retirements at exact ages 60, 65; note that l 65 = r 65 ). 19.14 Preserved Pensions on Leaving Service Suppose that (for example) “early leavers” get a deferred annual pension of 1 60 final salary per year of service, vesting at age 65. ‘Final’ salary refers to the average salary in the 3 years before leaving service. In practice, this pension may escalate between the date of leaving and age 65; suppose that escalation is at rate j per annum compound. Also, one must specify a mortality table for the early leavers: let us use the notation ˆ l x for their life table. The m.p.v. of this withdrawal benefit for a life age x is SAL 60 64−x ¸ t=0 v 65−x w x+t l x (n +t + 1 2 ) z x+t+ 1 2 ¯ s x (1 +j) 65−(x+t+ 1 2 ) ˆ l 65 ˆ l x+t+ 1 2 ¯ a 65 (19.14.1) where ¯ a 65 is the annuity factor, with the pension possibly payable monthly, etc. To evaluate (19.14.1) by commutation functions, one needs to define suitable ‘new’ commutation functions. That is, the expression (19.14.1) is equal to SAL 60¯ s x D x 64−x ¸ t=0 (n +t + 1 2 ) z C wa x+t where z C wa x = ˆ l 65 a 65 v 65 ¸ w x z x+ 1 2 (1 +j) 65−x− 1 2 ˆ l x+ 1 2 ¸ 19.14. PRESERVED PENSIONS ON LEAVING SERVICE 323 We now define z M wa x = 64−x ¸ t=0 z C wa x+t , z ¯ M wa x = z M wa x − 1 2 z C wa x and z ¯ R wa x = 64−x ¸ t=0 z ¯ M wa x+t (Note the similarities with the definitions used in previous sections.) The present value of the deferred pension is SAL 60¯ s x D x n. z M wa x + z ¯ R wa x (19.14.2) Note The notation used here is not standard, but is an example of “do-it-yourself” commutation functions. 324 CHAPTER 19. PENSION FUNDS Exercises 19.1 Contributions to a pension scheme by employees are made at a rate of 5% of salary when aged under 35, 6% between ages 35 and 45, and 7 1 2 % when aged 45 or over. Calculate the present value of the future contributions payable by a member aged exactly 30 who in the past year has received a total salary of £12,718. 19.2 A company pension scheme provides the following benefits for all members: (1) a pension on retirement (on grounds of ill-health or of age) of one-eightieth of final pen- sionable salary for each year of service (including fractions), (2) a lump sum on retirement of 3 times the annual pension, (3) on death in service, a lump sum of £30,000, (4) on withdrawal from service, a return of the employee’s contributions, accumulated at 3% per annum compound. Final pensionable salary is defined as the average annual salary in the three years immediately before retirement. All members who reach age 65 retire immediately. Employees contribute to the scheme at the rate of 3% of salary, payable continuously. Salaries are revised continuously. The employer’s contribution rate is assessed for each member sep- arately, and is such that the prospective reserve for each new entrant is zero. Expenses are ignored. (i) (a) Derive a formula, in terms of suitable commutation functions, for valuing benefit (1) above in respect of a new entrant aged x with annual salary rate SAL. (You need not define the service table functions.) (b) In respect of a new entrant aged x with annual salary rate SAL, give formulae for valuing benefits (2), (3) and (4) above, using suitable commutation functions. (You need not derive the formulae.) (c) Hence find a formula for the employer’s contribution rate for a new member aged x and a starting salary rate of £10,000 p.a. (ii)(a) Using the basis given in the pension fund section of the Formulae and Tables (and the supplement), find the value of each of the benefits (1), (2), (3) and (4) for a new entrant aged 45 with salary rate £10,000 per annum. (b) Hence or otherwise find the employer’s contribution rate for this new member. 19.3 The pension scheme of a certain company provides an annual pension on retirement (for ‘age’ or ‘ill- health’ reasons) of amount equal to one per cent of the member’s total earnings throughout his service. The pension is payable weekly. In addition, in the event of a member dying in service there is payable at the time of death a lump sum of £30,000. There is no benefit on withdrawal. The company pays a constant percentage of all the members’ salaries into the pension fund. The percentage is that which will exactly cover the cost of benefits for a new entrant to the fund at age 30 with an initial salary rate of £10,000 per annum. Contributions are payable continuously, and the employees do not contribute to the scheme. Expenses are negligible. (a) Calculate the contribution rate paid by the company, assuming the last retirement age is 65. (b) A valuation of the fund is to be conducted. For each active member of the scheme there is recorded 19.15. EXERCISES 325 (i) the age nearest birthday (which is regarded as the member’s exact age) at the valuation date, (ii) the annual salary rate at the valuation date, and (iii) the total past earnings in service (prior to the valuation date.) For each age, the totals of (ii) and (iii) are recorded and the following is an extract from the data. Age x No. of members Total past earnings Total of annual salary aged x for members aged x rates for members aged x £ £ 25 11 302,100 70,100 Assuming that the basis of the Tables provided is appropriate, find the liability at the valuation date for the benefits payable to the members aged 25, and determine whether the future contributions payable in respect of these members are more or less than sufficient to cover the benefits. 19.4 A pension scheme provides each member who retires (for any reason) with annual pension equal to 1 60 final salary per year of service. Final salary is the average income over the last 3 years of service, and fractions of a year of service are not included when calculating the pension. Assuming that equal contributions are payable by the member and his employer, that in the event of death in service a benefit is payable equal to the return without interest of both the member’s and the employer’s contributions, and that in the event of withdrawal from service a return without interest is made of the member’s contributions, calculate the appropriate contribution rate payable by both the member and his employer in respect of a new entrant aged 40. 19.5 You are consulting actuary to a small pension scheme, which has just been established. You have decided to use the pension fund tables in Formulae and Tables for Actuarial Examinations as the basis for all calculations. The scheme provides the following benefits to employees: (i) on retirement (for ill-health reasons or otherwise), a pension of 1 80 th of annual pension- able salary, averaged over the previous three years, for each year of future service including fractions; (ii) on withdrawal or death in service, a return of the member’s contributions, accumulated at 3% per annum compound interest. Employees contribute 2% of salary to the scheme. Pensionable salary is defined as salary less £4000. Salaries are revised continuously, and contributions are made continuously. Details of the current membership are as follows. member age current salary rate (£) 1 45 30,000 2 45 20,000 3 35 10,000 4 35 10,000 (a) The employer has decided to contribute the proportion of total salaries which, together with the employees’ contributions, will exactly pay for the benefits. Calculate the employer’s contribution rate. (b) A new employee, aged 35 and with current salary rate £8,000, is about to be hired. Calculate the surplus or deficiency in the pension fund after this new member joins. 326 CHAPTER 19. PENSION FUNDS 19.6 It is desired to set up a pension scheme for the group of employees described below. For each member there is recorded his exact age, his exact length of service with the company, and his annual rate of salary. Past Rate of Age Service Salary (Years) p.a. £ 25 4 8,800 25 6 8,500 25 5 8,500 25 3 8,600 25 1 8,600 Past Rate of Age Service Salary (Years) p.a. £ 35 12 12,400 35 18 12,500 35 15 12,800 35 5 12,500 35 10 12,400 Past Rate of Age Service Salary (Years) p.a. £ 45 25 15,000 45 15 14,000 45 10 13,000 45 20 13,200 45 5 13,000 The scheme will provide pensions of 1 60 th of “pensionable salary” for each year of service (fractions of a year being included) and on death or withdrawal from service a return will be made of the member’s contributions with 3% compound interest. All members will contribute at the same rate and the employer will contribute the same amount as each member. The contribution rate will be such as to provide exactly the benefits for future service. The basis of “Formulae and Tables” is to be used. (a) Assuming that pensionable salary is the average annual earnings over the three-year period ending on the retirement date, calculate the contribution rate payable by each member. (b) Calculate also the total liability for past service benefits. The employer wishes to meet this liability by paying additional contributions proportional to future salary payments. At what rate should these additional contributions be made? (c) Immediately after the scheme is set up as described above the 45 year old member with 5 years of service withdraws. Is the position of the fund improved or worsened by this with- drawal? 19.7 The pension scheme of a large company provides the following benefits, among others: (1) on death in service of married members: a spouse’s pension of one-third of the member’s annual salary at the date of death; (2) on death after normal retirement of the member at age 65 or after ill-health retirement at an earlier age: a spouse’s pension of 1% of annual salary at the date of retirement for each year of scheme membership (including fractions of a year); no pension is payable if the marriage has taken place after the member’s retirement date. Assume that a service table has been constructed, and that the proportion of members mar- ried at exact y is h y . Assume further that spouses are the same age as members, and that a unisex mortality table is used for all calculations, with the age rated up by 5 years on ill-health retirement. You are given the age nearest birthday, current salary rate and years of past service (including fractions) of each member. Spouse’s pension is payable monthly in advance, beginning immediately on the death of the member. (i) Using the rate of interest i per annum, find formulae for the mean present value of each of the benefits (1) and (2) above for a member aged x (x < 65). You are NOT required to construct commutation functions. (ii) Suppose that the scheme’s rules are to be changed so that benefit (2) is to be payable to any surviving spouse. Show how to modify the formulae of (i) to accommodate this change. 19.16. SOLUTIONS 327 Solutions 19.1 12, 718 s 29 D 30 ¦0.05 s ¯ N 30 + 0.01 s ¯ N 35 + 0.015 s ¯ N 45 ¦ = 12, 718 1.30162 = £16, 554 19.2 (i)(a) m.p.v. = SAL 80¯ s x D x ¸ 64−x ¸ t=0 (t + 1 2 ) z C ia x+t + (t + 1 2 ) z C ra x+t + (65 −x) z C ra 65 ¸ = SAL 80¯ s x z ¯ R (i+r)a x D x (b)(2): 3.SAL 80¯ s x z ¯ R i+r x D x (3) 30000 M d x D x (4) 0.03 SAL ¯ s x D x sj ¯ R w x where j = 0.03 (c) Let employer’s contribution rate be k for this member. Then (k + 0.03) SAL ¯ s x D x s ¯ N x = above benefits. So k + 0.03 = 1 s ¯ N x ¸ 1 80 z ¯ R (i+r)a x + 3 80 z ¯ R i+r x + 3¯ s x M d x + 0.03. sj ¯ R w x (ii)(a) Values of benefits are: (1) 13,088 (2) 3,642 (using Supplement) (3) 3,016 (4) 0 Hence total value of benefits = £19,746. (b) value of contributions = 142580(k + 0.03) Hence k + 0.03 = 19746 142580 and hence k = 10.85%. 19.3 (a) Let k be the contribution rate. Then k SAL ¯ s 30 D 30 s ¯ N 30 = SAL 100¯ s 30 s ¯ R (i+r)a 30 D 30 + 30000 M d 30 D 30 where SAL = 10, 000. Hence k = ¯ s 30 s ¯ N 30 ¸ 0.01 s ¯ R (i+r)a 30 ¯ s 30 + 3.M d 30 ¸ = 0.0535 = 5.35% 328 CHAPTER 19. PENSION FUNDS (b) Value of benefits for the group is 302100 100 M (i+r)a 25 D 25 + 70100 100¯ s 25 s ¯ R (i+r)a 25 D 25 + 11 30000 M d 25 D 25 = 2977.37 + 54730.83 + 11811.67 = £69, 528 Value of future contributions is 0.0535 70100 ¯ s 25 D 25 s ¯ N 25 = £76, 718. Hence contributions are more than sufficient. 19.4 Let k% of salary be the contribution rate for both the member and the employer. Then if the member has a salary rate of £SAL, 2k 100 SAL ¯ s 40 D 40 s ¯ N 40 = SAL 60¯ s 40 D 40 ¸ z ¯ R (i+r)a 40 − 1 2 z M (i+r)a 40 + k 100 SAL ¯ s 40 D 40 s ¯ R w 40 + 2k 100 SAL ¯ s 40 D 40 s ¯ R d 40 Hence 6342.42k = 35513.07 + 13.74k + 562.35k. Therefore k = 6.16% 19.5 (a) Construct the following table, using j = 0.03: Member age salary SAL 80¯ s x z ¯ R (i+r)a x D x 4000 80 ¯ R (i+r)a x D x 0.02 SAL ¯ s x sj ¯ R d+w x D x 1 45 30,000 39,264 3,999 899 2 45 20,000 26,176 3,999 599 3 35 10,000 15,189 3,498 549 4 35 10,000 15,189 3,498 549 95,818 14,994 2,596 Let the employer contribute k% of salary, so m.p.v. of contributions is (k + 0.02) ¸ 50000 ¯ s 45 D 45 s ¯ N 45 + 20000 ¯ s 35 D 35 s ¯ N 35 = 1093716(k + 0.02) Therefore k + 0.02 = 95818−14994+2596 1093716 and hence k = 5.627%. (b) Reserve for new member is 8000 80¯ s 35 z ¯ R (i+r)a 35 D 35 − 4000 80 ¯ R (i+r)a 35 D 35 + 0.02 8000 ¯ s 35 D 35 . sj ¯ R d+w 35 − 8000 ¯ s 35 D 35 0.07627 s ¯ N 35 = 12, 151 −3, 498 + 440 −11, 617 = −2, 524. Hence there is now a surplus of £2,524. 19.6 (a) For each age x, Value of Past Service Pension = (n×SAL) 60¯ s x z M (i+r)a x D x 19.16. SOLUTIONS 329 Value of Future Service Pension = SAL 60¯ s x z ¯ R (i+r)a x D x Value of return of contributions = SAL 100¯ s x sj ¯ R d+w x D x (where j = 0.03) per 1% of salary return of x ¸ SAL ¸ (n SAL) P.S.P. F.S.P. contributions (per 1% of salary) 25 43,000 163,100 7,776 76,783 2,223.7 35 62,600 752,300 55,421 126,780 1,720.5 45 68,200 1,044,000 103,230 119,012 1,022.7 166,427 322,575 4,966.9 The value of the contributions is, per 1% of contributions, ¸ SAL 100¯ s x s ¯ N x D x (1) x Value of (1) 25 8,791.2 35 11,919.1 45 9,724.0 30,434.3 Let members’ contribution be k%. The equation of value is thus 2k 30, 434.3 = 322, 575 + 4, 966.9k Hence k = 5.77%. (b) The past service liability is £166,427. Let the contribution rate per cent needed to pay for this be p. Then p solves, p 30, 434.3 = 166, 427 Hence p = 5.468%. (c) The reserve for the withdrawn member is 13, 000 60¯ s 45 D 45 z ¯ R (i+r)a 45 + 5 13000 60¯ s 45 D 45 z M (i+r)a 45 + 0.0577 13000 ¯ s 45 D 45 sj ¯ R d+w 45 −(2 0.0577 + 0.05468) 13000 ¯ s 45 D 45 s ¯ N 45 = 30, 238 −31, 526 = −1, 288 So the position of the pension fund has worsened by £1,288. 19.7 (i) Benefit (1) has value SAL 3 64−x ¸ t=0 v t+ 1 2 d x+t l x h x+t+ 1 2 ¨ a (12) x+t+ 1 2 s x+t ¯ s x Benefit (2): On normal retirement the value is SAL 100¯ s x v 65−x .¯ s 65 (n + 65 −x) r 65 l x h 65 a (12) 65|65 (1 +i) 1 24 On ill-health retirement, the value is SAL 100 64−x ¸ t=0 v t+ 1 2 ( s x+t ¯ s x ) i x+t l x (n +t + 1 2 )[h x+t+ 1 2 a (12) x+t+ 1 2 +5|x+t+ 1 2 (1 +i) 1 24 ] 330 CHAPTER 19. PENSION FUNDS (ii) Benefit (1): no change Benefit (2): For normal retirement change h 65 a (12) 65|65 (1 +i) 1 24 to ∞ 0 v t . t p 65 µ 65+t h 65+t ¨ a (12) 65+t dt For ill-health retirement, change h x+t+ 1 2 a (12) x+t+ 1 2 +5|x+t+ 1 2 (1 +i) 1 24 to ∞ 0 v r . r p x+t+ 1 2 +5 µ x+t+ 1 2 +5+r h x+t+ 1 2 +r ¨ a (12) x+t+ 1 2 +r dr 19.16. SOLUTIONS 331 Appendix: Formulae for valuing a return of contributions Suppose employee’s contributions are to be returned on withdrawal with interest at rate j per annum compound. Consider a member aged x whose current salary rate is SAL and whose total past contributions, accumulated at rate j p.a., are TPC. Assume that the employee will in future contribute k% of salary. The value of the return of future contributions is k 100 SAL ¯ s x 64−x ¸ t=0 v t+ 1 2 w x+t l x [s x (1 +j) t +s x+1 (1 +j) t−1 +... +s x+t−1 (1 +j) + 1 2 s x+t ] = k 100 SAL ¯ s x l x ¦v 1 2 w x 1 2 s x +v 1 1 2 w x+1 [s x (1 +j) + 1 2 s x+1 ] +v 2 1 2 w x+2 [s x (1 +j) 2 +s x+1 (1 +j) + 1 2 s x+2 ] +... +v 64 1 2 −x w 64 [s x (1 +j) 64−x +s x+1 (1 +j) 64−x−1 +... +s 63 (1 +j) + 1 2 s 64 ]¦ Collecting the coefficients of s x , s x+1 , ... gives k 100 SAL ¯ s x l x ¦s x [ 1 2 v 1 2 w x +v 1 1 2 w x+1 (1 +j) +... +v 64 1 2 −x w 64 (1 +j) 64−x ] +s x+1 [ 1 2 v 1 1 2 w x+1 +v 2 1 2 w x+2 (1 +j) +... +v 64 1 2 −x w 64 (1 +j) 64−x−1 ] +... +s 64 [ 1 2 v 64 1 2 −x w 64 ]¦ Define j C w x = (1 +j) x+ 1 2 v x+ 1 2 w x j M w x = 64−x ¸ t=0 j C w x+t j ¯ M w x = j M w x − 1 2 j C w x sj ¯ M w x = s x . j ¯ M w x sj ¯ R w x = 64−x ¸ t=0 (1 +j) −(x+t+ 1 2 ) . sj ¯ M w x+t The value of the return of future contributions can be written as k 100 SAL ¯ s x v x l x ¦s x (1 +j) −(x+ 1 2 ) ¸ 1 2 (v(1 +j)) x+ 1 2 w x + (v(1 +j)) x+1 1 2 w x+1 +... + (v(1 +j)) 64 1 2 w 64 +s x+1 (1 +j) −(x+1 1 2 ) ¸ 1 2 (v(1 +j)) x+1 1 2 w x+1 + (v(1 +j)) x+2 1 2 w x+2 +... + (v(1 +j)) 64 frac12 w 64 +... +s 64 (1 +j) −64 1 2 ¸ 1 2 (v(1 +j)) 64 1 2 w 64 ¦ = k 100 SAL ¯ s x D x . sj ¯ R w x 332 CHAPTER 19. PENSION FUNDS The value of the return of past contributions with interest at rate j is (TPC) 64−x ¸ t=0 v t+ 1 2 (1 +j) t+ 1 2 w x+t l x = (TPC) D x 64−x ¸ t=0 (1 +j) −x (v(1 +j)) x+t+ 1 2 w x+t = (TPC)(1 +j) −x j M w x D x Hence the total value is (TPC)(1 +j) −x j M w x D x + k 100 SAL ¯ s x D x sj ¯ R w x Suppose now that the employee’s contributions do not depend on salary, but are instead £A each year, payable continuously. The value of the return of future contributions is obtained in a similar way to the previous derivation, with the salary scale function taken as 1. This gives a present value of A j ¯ R w x D x where j ¯ R w x = 64−x ¸ t=0 (1 +j) −(x+t+ 1 2 ) . j ¯ M w x+t . Appendix A Some notes on examination technique Each candidate should spend some time going over his or her notes, the textbook and (especially) past examination papers, making a note of subjects and formulae of particular importance. For this purpose a set of postcards/computer cards is useful: each topic may be summarised on one postcard, enabling the candidate to learn the most important facts and then have a “revision aid”. (He or she should ask a friend to choose a card at random, to give the title and to ask for a description of the topic.) In actuarial examinations (particularly in life contingencies) it is important to separate the basic ideas or “rationale” of the solution from (i) technical aspects, such as the correct use of actuarial tables and commutation functions, and (ii) arithmetic. It is sometimes necessary to omit (ii) or even (i) under extreme time-pressure, but one should leave space to return to the question later if time permits. On the other hand, it is satisfying to give a complete answer, and this should be attempted if time is not short. One should keep the left-hand pages of the examination book free for rough working and arithmetic. On first seeing the examination paper, one should tick the questions about which one is most confident - especially bookwork. Leave the less familiar questions until later. One should ensure that an approximate time-scheme (say, 15 minutes per 10-mark question) is adhered to, within reasonable limits. Routine arithmetic (or even looking up the tables) may sometimes be left till near the end, when one is too tired to think out new ideas but can still do (or check) arithmetic. As in musical performances, it is essential to get off to a good start, by being in the right frame of mind and picking the most suitable questions to attempt first. 333 Appendix B Some technical points about the tables used in examinations 1. Avoid excessive reliance on commutation functions. In theoretical/statistical questions, it is nearly always best to avoid the use of commutation func- tions. 2. In the A1967-70 section of “Formulae and Tables” commutation functions are given only at 4% interest. For other interest rates one most use the more limited tables provided, e.g. ¨ a x:n at 6% = ¨ a x −v n l x+n l x ¨ a x+n (There is no point writing v n l x+n l x as D x+n D x since D x and D x+n are not given at 6% in the tables.) 3. Remember to make use of the functions maturing at ages 60, 65, etc. This saves time. 4. If only limited tables are available, one must proceed directly, using suitable approximations. For example, suppose that one is asked to evaluate ¯ A 1 50.5:2.5 at 10% interest on A1967-70 ultimate. This may be done by the trapezoidal rule as follows: 2.5 0 v t t p 50.5 µ 50.5+t dt 2.5 2 ¸ µ 50.5 +v 2.5 l 53 µ 53 l 50.5 where µ 50.5 and l 50.5 are estimated by linear interpolation. 5. Remember that for a(55) there are two tables - male and female (each with a select period of 1 year.) Assume that males and females are subject to the table of the appropriate sex (unless the question states otherwise.) 6. Be careful (especially in A1967-70, but also in a(55)) that you distinguish between “select” and “ultimate” tables and use the correct rate of interest. 334 Appendix C Some common mistakes WRONG CORRECT 1. ¯ A x:n (1 +i) 1 2 A x:n ¯ A x:n (1 +i) 1 2 A 1 x:n +A 1 x:n = (1 +i) 1 2 (M x −M x+n ) +D x+n D x 2. ¯ a x i δ a x ¯ a x a x + 1 2 or ¨ a x − 1 2 3. ¨ a (m) x: n i d (m) a x:n ¨ a (m) x: n ¨ a x:n − m−1 2m 1 − D x+n D x 4. a 1 x:n , a 1 x:n do not exist 5. t 0 c r dr = c r t 0 or ¸ c r+1 r + 1 t 0 t 0 c r dr = ¸ c r log c t 0 (Note that c r = e r log c ) 6. Mean present value = v T Present value (as random variable) = ¯ A x M.P.V. = ¯ A x P.V. = v T 7. var(P¯ a T −v T ) = var(P¯ a T ) −var(v T ) or var(P¯ a T ) + var(v T ) var P (1 −v T ) δ −v T = var ¸ − P +δ δ v T + P δ = P +δ δ 2 var(v T ) 8. ¨ a x:n = 1 +a x:n ¨ a x:n = 1 +a x:n−1 9. µ x = l x l x µ x = − l x l x 10. t p x = exp t 0 µ x+r dr t p x = exp − t 0 µ x+r dr 11. a x:n = N x −N x+n D x a x:n = N x+1 −N x+n+1 D x 335 Appendix D Some formulae for numerical integration 1. Mid-point rule b a f(x)dx (b −a)f a +b 2 2. Trapezoidal rule b a f(x)dx b −a 2 [f(a) +f(b)] 3. Simpson’s rule b a f(x)dx b −a 6 ¸ f(a) + 4f a +b 2 +f(b) 4. Three-eighths rule b a f(x)dx b −a 8 ¸ f(a) + 3f 2a +b 3 + 3f a + 2b 3 +f(b) 336 336 SUPPLEMENT age x C i x M i x R i x z C i x z M i x z R i x 20 0.0 188.8 6835.5 0.0 904.4 33753.7 TABLES FOR 21 0.0 188.8 6646.7 0.0 904.4 32849.3 LUMP SUM 22 0.0 188.8 6457.9 0.0 904.4 31944.9 BENEFITS ON 23 0.0 188.8 6269.1 0.0 904.4 31040.5 RETIREMENT 24 0.0 188.8 6080.3 0.0 904.4 30136.1 25 0.0 188.8 5891.5 0.0 904.4 29231.7 Ill-health 26 0.0 188.8 5702.8 0.0 904.4 28327.3 retirement functions 27 0.0 188.8 5514.0 0.0 904.4 27422.9 28 0.0 188.8 5325.2 0.0 904.4 26518.5 29 0.0 188.8 5136.4 0.0 904.4 25614.1 30 0.0 188.8 4947.6 0.0 904.4 24709.7 31 0.9 188.8 4758.8 2.2 904.4 23805.3 32 0.8 187.9 4570.5 2.2 902.2 22902.0 33 0.8 187.1 4383.0 2.2 900.0 22000.9 34 1.6 186.3 4196.3 4.4 897.8 21101.9 35 1.5 184.7 4010.8 4.4 893.5 20206.3 36 1.4 183.2 3826.8 4.3 889.1 19315.0 37 1.8 181.8 3644.3 5.8 884.8 18428.1 38 1.8 180.0 3463.5 5.7 879.0 17546.2 39 1.7 178.2 3284.4 5.7 873.3 16670.1 40 2.0 176.5 3107.1 7.0 867.6 15799.6 41 2.0 174.4 2931.6 6.9 860.6 14935.5 42 2.3 172.5 2758.1 8.2 853.7 14078.3 43 2.2 170.2 2586.8 8.1 845.5 13228.7 44 2.4 168.0 2417.6 9.4 837.3 12387.3 45 2.4 165.6 2250.8 9.2 828.0 11554.7 46 2.6 163.2 2086.4 10.4 818.7 10731.3 47 2.9 160.7 1924.5 12.2 808.3 9917.8 48 3.1 157.7 1765.3 13.3 796.2 9115.5 49 3.3 154.6 1609.1 14.3 782.9 8326.0 50 3.9 151.3 1456.2 17.1 768.6 7550.3 51 4.2 147.4 1306.9 19.2 751.5 6790.2 52 5.0 143.2 1161.6 22.9 732.4 6048.3 53 5.4 138.2 1020.9 25.3 709.5 5327.4 54 6.3 132.8 885.4 29.8 684.1 4630.6 55 6.9 126.5 755.7 33.5 654.3 3961.3 56 7.7 119.6 632.6 38.0 620.8 3323.7 57 8.7 111.9 516.9 43.3 582.8 2721.9 58 10.0 103.2 409.4 50.3 539.5 2160.8 59 11.4 93.2 311.2 58.3 489.2 1646.5 60 13.6 81.8 223.7 70.2 430.8 1186.5 61 13.7 68.2 148.7 71.5 360.6 790.8 62 15.3 54.4 87.4 80.7 289.1 465.9 63 18.0 39.1 40.7 95.5 208.4 217.1 64 21.1 21.1 10.6 112.9 112.9 56.5 337 SUPPLEMENT age x C r x M r x R r x z C r x z M r x z R r x 20 0.0 1524.0 65671.0 0.0 8048.7 347094.1 TABLES FOR 21 0.0 1524.0 64147.0 0.0 8048.7 339045.4 LUMP SUM 22 0.0 1524.0 62623.1 0.0 8048.7 330996.8 BENEFITS ON 23 0.0 1524.0 61099.1 0.0 8048.7 322948.1 RETIREMENT 24 0.0 1524.0 59575.2 0.0 8048.7 314899.5 25 0.0 1524.0 58051.2 0.0 8048.7 306850.8 Age retirement 26 0.0 1524.0 56527.2 0.0 8048.7 298802.2 functions 27 0.0 1524.0 55003.3 0.0 8048.7 290753.5 28 0.0 1524.0 53479.3 0.0 8048.7 282704.9 29 0.0 1524.0 51955.4 0.0 8048.7 274656.2 30 0.0 1524.0 50431.4 0.0 8048.7 266607.6 31 0.0 1524.0 48907.4 0.0 8048.7 258558.9 32 0.0 1524.0 47383.5 0.0 8048.7 250510.3 33 0.0 1524.0 45859.5 0.0 8048.7 242461.6 34 0.0 1524.0 44335.6 0.0 8048.7 234413.0 35 0.0 1524.0 42811.6 0.0 8048.7 226364.3 36 0.0 1524.0 41287.6 0.0 8048.7 218315.7 37 0.0 1524.0 39763.7 0.0 8048.7 210267.0 38 0.0 1524.0 38239.7 0.0 8048.7 202218.4 39 0.0 1524.0 36715.8 0.0 8048.7 194169.7 40 0.0 1524.0 35191.8 0.0 8048.7 186121.1 41 0.0 1524.0 33667.8 0.0 8048.7 178072.4 42 0.0 1524.0 32143.9 0.0 8048.7 170023.8 43 0.0 1524.0 30619.9 0.0 8048.7 161975.1 44 0.0 1524.0 29096.0 0.0 8048.7 153926.5 45 0.0 1524.0 27572.0 0.0 8048.7 145877.8 46 0.0 1524.0 26048.0 0.0 8048.7 137829.2 47 0.0 1524.0 24524.1 0.0 8048.7 129780.5 48 0.0 1524.0 23000.1 0.0 8048.7 121731.8 49 0.0 1524.0 21476.2 0.0 8048.7 113683.2 50 0.0 1524.0 19952.2 0.0 8048.7 105634.5 51 0.0 1524.0 18428.2 0.0 8048.7 97585.9 52 0.0 1524.0 16904.3 0.0 8048.7 89537.2 53 0.0 1524.0 15380.3 0.0 8048.7 81488.6 54 0.0 1524.0 13856.3 0.0 8048.7 73439.9 55 0.0 1524.0 12332.4 0.0 8048.7 65391.3 56 0.0 1524.0 10808.4 0.0 8048.7 57342.6 57 0.0 1524.0 9284.5 0.0 8048.7 49294.0 58 0.0 1524.0 7760.5 0.0 8048.7 41245.3 59 0.0 1524.0 6236.5 0.0 8048.7 33196.7 60 378.5 1524.0 4712.6 1952.7 8048.7 25148.0 61 212.4 1145.4 3377.9 1107.4 6096.0 18075.7 62 112.3 933.0 2338.7 591.0 4988.6 12533.4 63 93.8 820.7 1461.8 498.0 4397.5 7840.4 64 77.7 726.9 688.1 415.5 3899.6 3691.8 65 649.2 649.2 3484.1 3484.1 338 SUPPLEMENT Functions for valuing (at interest rate i per annum) refunds on death of contributions with compound interest at interest rate j per annum i = .04 j = 0.03 x j M d x j R d x sj R d x = ¸ 64 y=x j C d y = ¸ 64 y=x (1 +j) −(y+.5) . j M d y = ¸ 64 y=x (1 +j) −(y+.5) . sj M d y 20 3108.18 31393.46 87740.98 25 2878.57 23700.42 76568.15 30 2700.59 17508.94 63315.76 35 2524.59 12500.67 49478.97 40 2338.12 8479.14 36324.46 45 2120.85 5291.34 24305.71 50 1804.15 2861.89 13941.80 55 1333.41 1177.45 6005.02 60 655.64 243.17 1286.33 61 490.11 147.36 784.29 62 350.45 79.12 423.29 63 225.80 33.70 181.20 64 108.85 8.09 43.67 Notes. (1) j C d x = (1 +j) x+.5 .(1 +i) −(x+.5) .d x (2) j M d x = ¸ 64−x t=0 j C d x+t (3) j M d x = j M d x − 1 2 j C d x (4) sj M d x = s x j M d x 339 SUPPLEMENT Functions for valuing (at interest rate i per annum) refunds on withdrawal of contributions with compound interest at interest rate j per annum i = .04 j = 0.03 x j M w x j R w x sj R w x = ¸ 64 y=x j C w y = ¸ 64 y=x (1 +j) −(y+.5) . j M w y = ¸ 64 y=x (1 +j) −(y+.5) . sj M w y 20 43547.50 119414.16 207941.08 25 19045.16 42216.97 98759.32 30 8041.95 13393.17 38299.39 35 3006.98 3281.50 10746.75 36 2399.53 2334.90 7831.22 37 1875.23 1608.25 5520.49 38 1426.97 1063.28 3732.97 39 1049.20 666.53 2391.95 40 735.82 388.85 1425.63 41 482.26 204.89 767.03 42 284.70 92.43 353.18 43 140.12 31.95 124.58 44 46.19 6.20 24.67 Notes. (1) j C w x = (1 +j) x+.5 .(1 +i) −(x+.5) .w x (2) j M w x = ¸ 64−x t=0 j C w x+t (3) j M w x = j M w x − 1 2 j C w x (4) sj M w x = s x j M w x 340 SUPPLEMENT Functions for valuing (at interest rate i per annum) refunds on death of contributions with compound interest at interest rate j per annum i = .04 j = 0 x j M d x j R d x sj R d x = ¸ 64 y=x j C d y = ¸ 64 y=x (1 +j) −(y+.5) . j M d y = ¸ 64 y=x (1 +j) −(y+.5) . sj M d y 20 792.85 18575.75 56093.05 25 674.16 14932.38 50789.94 30 594.98 11769.88 44005.80 35 527.62 8965.83 36247.43 40 466.07 6483.42 28117.45 45 404.38 4303.49 19889.59 50 326.91 2467.10 12048.69 55 227.54 1071.57 5469.33 60 104.08 232.42 1229.62 61 76.40 142.17 756.75 62 53.72 77.11 412.57 63 34.07 33.21 178.59 64 16.18 8.09 43.67 Notes. (1) j C d x = (1 +j) x+.5 .(1 +i) −(x+.5) .d x (2) j M d x = ¸ 64−x t=0 j C d x+t (3) j M d x = j M d x − 1 2 j C d x (4) sj M d x = s x j M d x 341 SUPPLEMENT Functions for valuing (at interest rate i per annum) refunds on withdrawal of contributions with compound interest at interest rate j per annum i = .04 j = 0 x j M w x j R w x sj R w x = ¸ 64 y=x j C w y = ¸ 64 y=x (1 +j) −(y+.5) . j M w y = ¸ 64 y=x (1 +j) −(y+.5) . sj M w y 20 20579.08 103387.17 180950.89 25 7847.29 37164.60 87256.75 30 2916.96 12076.18 34605.99 35 971.53 3051.10 10002.54 36 758.81 2185.93 7337.82 37 580.57 1516.24 5208.20 38 432.61 1009.65 3546.59 39 311.55 637.57 2288.96 40 214.05 374.77 1374.41 41 137.46 199.01 745.20 42 79.53 90.52 345.94 43 38.36 31.58 123.14 44 12.40 6.20 24.67 Notes. (1) j C w x = (1 +j) x+.5 .(1 +i) −(x+.5) .w x (2) j M w x = ¸ 64−x t=0 j C w x+t (3) j M w x = j M w x − 1 2 j C w x (4) sj M w x = s x j M w x 342 2 Preface This book consists largely of material written for Parts A2 and D1 of the U.K. actuarial examinations (old system). It is hoped that the material given here will prove useful for much of Subjects 104 and 105 of the new examinations and certain similar examinations at universities, and that it may also be useful as a general reference work on life assurance mathematics. William F. Scott. November 10, 1999 Contents 1 NON-SELECT LIFE TABLES 1.1 Survivorship functions . . . . . . . . . . . . 1.2 Probabilities of death and survival . . . . . 1.3 The force of mortality, µx . . . . . . . . . . 1.4 The expectation of life . . . . . . . . . . . . 1.5 The assumption of a uniform distribution of 1.6 Central death rates . . . . . . . . . . . . . . 1.7 Laws of mortality . . . . . . . . . . . . . . . 1.8 Exercises . . . . . . . . . . . . . . . . . . . 1.9 Solutions . . . . . . . . . . . . . . . . . . . 2 SELECT LIFE TABLES 2.1 What is selection? . . . . . . . . . . . . 2.2 Construction of select tables . . . . . . . 2.3 The construction of A1967-70. . . . . . . 2.4 Some formulae for the force of mortality. 2.5 Select tables used in examinations. . . . 2.6 Exercises . . . . . . . . . . . . . . . . . 2.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . deaths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 9 9 12 15 17 18 19 22 24 29 29 30 31 32 32 33 34 37 37 37 39 39 41 44 44 46 48 50 54 55 57 61 61 63 65 67 3 ASSURANCES 3.1 A general introduction . . . . . . . . . . . . . . . . . . 3.2 Whole life assurances . . . . . . . . . . . . . . . . . . . 3.3 Commutation functions . . . . . . . . . . . . . . . . . 3.4 The variance of the present value of benefits . . . . . . 3.5 Assurances payable at the end of the year of death. . . 3.6 Assurances payable at the end of the 1/m of a year of 3.7 Temporary and deferred assurances . . . . . . . . . . . 3.8 Pure endowments and endowment assurances . . . . . 3.9 Varying assurances . . . . . . . . . . . . . . . . . . . . 3.10 Valuing the benefits under with profits policies . . . . 3.11 Guaranteed bonus policies . . . . . . . . . . . . . . . . 3.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 3.13 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 4 ANNUITIES 4.1 Annuities payable continuously 4.2 Annuities payable annually . . 4.3 Temporary annuities . . . . . . 4.4 Deferred annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . death. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 4.5 4.6 4.7 4.8 4.9 Annuities payable m times per annum Complete annuities (or “annuities with Varying annuities . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . final . . . . . . . . . . . . . . . . . proportion”) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . 68 70 71 77 78 81 81 82 83 83 85 86 87 89 91 93 95 95 95 96 99 102 104 106 109 110 113 115 119 119 120 121 124 125 129 133 135 137 137 137 139 140 142 145 147 5 PREMIUMS 5.1 Principles of premium calculations . . . . . . 5.2 Notation for premiums . . . . . . . . . . . . . 5.3 The variance of the present value of the profit 5.4 Premiums allowing for expenses . . . . . . . . 5.5 Premiums for with profits policies . . . . . . 5.6 Return of premium problems . . . . . . . . . 5.7 Annuities with guarantees . . . . . . . . . . . 5.8 Family income benefits . . . . . . . . . . . . . 5.9 Exercises . . . . . . . . . . . . . . . . . . . . 5.10 Solutions . . . . . . . . . . . . . . . . . . . . 6 RESERVES 6.1 What are reserves? . . . . . . . . . . . . . . 6.2 Prospective reserves . . . . . . . . . . . . . 6.3 Net premium reserves . . . . . . . . . . . . 6.4 Retrospective reserves . . . . . . . . . . . . 6.5 Gross premium valuations and asset shares. 6.6 The variance of L . . . . . . . . . . . . . . . 6.7 Zillmerised reserves . . . . . . . . . . . . . . 6.8 Full preliminary term reserves. . . . . . . . 6.9 Reserves for with-profits policies . . . . . . 6.10 Exercises . . . . . . . . . . . . . . . . . . . 6.11 Solutions . . . . . . . . . . . . . . . . . . . 7 APPLICATIONS OF RESERVES 7.1 Surrender values . . . . . . . . . . . . 7.2 Paid-up policy values . . . . . . . . . . 7.3 Alterations and conversions . . . . . . 7.4 The actual and expected death strains 7.5 Mortality profit and loss . . . . . . . . 7.6 Other sources of profit and loss . . . . 7.7 Exercises . . . . . . . . . . . . . . . . 7.8 Solutions . . . . . . . . . . . . . . . . . . . . . . on a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . policy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 EXTRA RISKS 8.1 Introduction . . . . . . . . . . . . . . . . . . . 8.2 A constant addition to the force of mortality 8.3 A variable addition to the force of mortality . 8.4 Rating up . . . . . . . . . . . . . . . . . . . . 8.5 Debts . . . . . . . . . . . . . . . . . . . . . . 8.6 Exercises . . . . . . . . . . . . . . . . . . . . 8.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS 9 PROFIT-TESTING 9.1 Principles of profit-testing . . . . . . . . . 9.2 Cash flow calculations . . . . . . . . . . . 9.3 The profit vector and the profit signature 9.4 The assessment of profits . . . . . . . . . 9.5 Some theoretical results about {σt } . . . . 9.6 Withdrawals . . . . . . . . . . . . . . . . 9.7 The actual emergence of profits . . . . . . 9.8 Exercises . . . . . . . . . . . . . . . . . . 9.9 Solutions . . . . . . . . . . . . . . . . . . 10 STATIONARY POPULATIONS 10.1 Some Definitions . . . . . . . . . 10.2 The Central Death Rate . . . . . 10.3 Relationships Between mx and qx 10.4 Stationary Funds . . . . . . . . . 10.5 Exercises . . . . . . . . . . . . . 10.6 Solutions . . . . . . . . . . . . . 5 151 151 152 155 157 159 160 162 165 167 171 171 175 176 176 178 180 185 185 188 188 190 191 195 196 198 200 201 205 205 206 210 211 212 213 215 219 219 220 221 224 225 227 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 JOINT-LIFE FUNCTIONS 11.1 Joint-Life Mortality Tables . . . . . . . . . 11.2 Select Tables . . . . . . . . . . . . . . . . . 11.3 Extensions to More than 2 Lives . . . . . . 11.4 The Joint Expectation of Life . . . . . . . . 11.5 Monetary Functions . . . . . . . . . . . . . 11.6 Last Survivor Probabilities (two lives only) 11.7 Last Survivor Monetary Functions . . . . . 11.8 Reserves for Last Survivor Assurances . . . 11.9 Exercises . . . . . . . . . . . . . . . . . . . 11.10Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 CONTINGENT ASSURANCES 12.1 Contingent Probabilities . . . . . . . . . . . . . . . . . 12.2 Contingent Assurances . . . . . . . . . . . . . . . . . . 12.3 Premiums and Reserves for Contingent Assurances . . 12.4 A Practical Application – The Purchase of Reversions 12.5 Extension to Three Lives . . . . . . . . . . . . . . . . 12.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 12.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 REVERSIONARY ANNUITIES 13.1 Reversionary Annuities Payable Continuously . . . . . . . 13.2 Reversionary Annuities Payable Annually or mthly . . . . 13.3 Widow’s (or Spouse’s) Pension on Death after Retirement 13.4 Actuarial Reduction Factors . . . . . . . . . . . . . . . . . 13.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 14 PROFIT TESTING FOR UNIT-LINKED POLICIES 14.1 Unit-Linked Policies . . . . . . . . . . . . . . . . . . . . 14.2 Mechanics of the Unit Fund . . . . . . . . . . . . . . . . 14.3 The Sterling Fund (or Sterling Reserves) . . . . . . . . . 14.4 The Assessment of Profits . . . . . . . . . . . . . . . . . 14.5 Zeroisation of the Profit Signature . . . . . . . . . . . . 14.6 Withdrawals . . . . . . . . . . . . . . . . . . . . . . . . 14.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 14.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS 231 231 231 233 235 236 237 240 243 247 247 249 249 250 253 256 257 258 263 264 267 267 267 268 272 272 274 275 277 281 281 281 283 284 287 287 289 292 295 298 301 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 MULTIPLE-DECREMENT TABLES 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 The Associated Single-Decrement Tables . . . . . . . . . . . . . . 15.3 The Relationships between the Multiple-Decrement Table and its Decrement Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.4 Dependent Rates of Exit . . . . . . . . . . . . . . . . . . . . . . . 15.5 Practical Construction of Multiple-Decrement Tables . . . . . . . 15.6 Further Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.7 Generalization to 3 Modes of Decrement . . . . . . . . . . . . . . 15.8 “Abnormal” Incidence of Decrement . . . . . . . . . . . . . . . . 15.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.10Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Associated Single. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TABLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT 16.1 Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 The Use of “Defective” Variables . . . . . . . . . . . . . . . . . . . . . . . 16.3 Evaluation of Mean Present Values . . . . . . . . . . . . . . . . . . . . . . 16.4 Benefits on Death by a Particular Cause . . . . . . . . . . . . . . . . . . . 16.5 Extra Risks Treated as an Additional Mode of Decrement . . . . . . . . . 16.6 Calculations Involving a Change of State . . . . . . . . . . . . . . . . . . . 16.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 MULTIPLE-STATE MODELS 17.1 Two Points of View . . . . . . . . 17.2 Kolmogorov’s Forward Equations . 17.3 Life Tables as Stochastic Processes 17.4 Sickness Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 SICKNESS FUNCTIONS 18.1 Rates of Sickness . . . . . . . . . . . . . . . . . 18.2 Valuing Sickness Benefits . . . . . . . . . . . . 18.3 Various Other Points . . . . . . . . . . . . . . . 18.4 Exercises . . . . . . . . . . . . . . . . . . . . . 18.5 Solutions . . . . . . . . . . . . . . . . . . . . . Appendix The Manchester Unity Experience 1893-97 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11Death and Withdrawal Benefits . . . . . . 19. . . . . . 19. . . . . . . . . . . . . . .12Return of Contributions on Death or Withdrawal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2 Valuation Principles . . D Some formulae for numerical integration . . . . . . . . . . . . . 19. . . . . . . . . . . .1 General Introduction . . . . . . . . 19. . . . . . 19. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19. . . . . . . . . . . . . . . . . . 19. . . . in examinations . . . . . . . . . . . . . . . . . 19.8 Average Salary Schemes . . . . . . . . .3 Service Tables . . . . . . . . . . . . . . . . . . . . . . . . . .5 The Value of Future Contributions . . . . . . . . . . . . . . Appendix Formulae for valuing a return of contributions APPENDICES A Some notes on examination technique . . . . . . . . . . . . . . . . . . . . . . . .6 The Value of Pension Benefits . . . . .13Spouse’s Benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . SUPPLEMENT 7 303 303 303 304 304 306 308 309 311 313 316 317 318 320 322 324 327 331 333 333 334 335 336 337 . . . . . . 19. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19. . . . . . . . . .16Solutions . . . . . . . . . . . . . . . . . . . . . . . .9 Final Salary Schemes . . . 19. . . . . . . . . . . . . . . . . . . . . . B Some technical points about the tables used C Some common mistakes . . . . . . . . . .7 Fixed Pension Schemes . . . . . . . .4 Salary Scales . . . . . . . . . . . . . . . . . . . . .14Preserved Pensions on Leaving Service . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19. . . . . . .CONTENTS 19 PENSION FUNDS 19. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .10Lump Sums on Retirement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19. . . . . . . . . . . . . . . . . . . . . 19. . . . . . .15Exercises . . . . . . . . . . . . . . . . 8 CONTENTS . 000.Chapter 1 NON-SELECT LIFE TABLES 1.1. unless w is infinite.1%. we define the survivorship function s(x) = P r{ (α) survives to age x } In practice one usually uses the function lx = lα s(x) (1. According to our definitions. the probability that a new-born child will survive to age 50 is s(50) = l50 /l0 = 0.1.1.Males refers to the male population of England and Wales in 1970-72. i.1 Survivorship functions We consider a certain population from age α to a “limiting age” w .1. By formulae (1.12 . P r{ (α) survives to age x } = lx lα (1.1.1) and (1.90085. If lives are considered from birth. we naturally have α = 0. the youngest age to which no-one survives. and the radix of the table is 100.1. 000. English Life Table No.e. (In theoretical work we may have w = ∞. w may be taken to be about 106.2) .2) (1. P r{ T (α) ≤ t } = the probability that (α) will die within t years = 1 − the probability that (α) will survive for at least t years = 1 − s(α + t) 9 (1. in which case lx → 0 as x → ∞. The function lx is assumed to be continuous. lw = 0. Let T (or T (x) if the age x ◦ is not clear.3) Example 1. In this table α = 0 .1. We now consider survival from age α in terms of a random variable. Tx is sometimes used. but this may be confused with the symbol Tx = lx ex used in stationary population problems) be the variable representing the future lifetime (in years. or 90.) Let (x) be a shorthand notation for “a life aged (exactly) x”. since l50 = 90. 000. and is usually a large number such as 1.4) .1. 085 . and we have lx > 0 for x < w. We observe (for example) that. including fractions) of (x).1) where lα is called the “radix” of the table. 6) qx is called the (q-type) rate of mortality at age x.2. The variables {T (x) : x ≥ α} are assumed to be related as follows: for each x1 . it may be omitted. for all t ≥ 0 .6. the word lx “expected” is sometimes omitted.2. by the formula (1.9) That is.8) and t qx =1− lx+t lx lx+t lx (1. the number of survivors at age x + t will follow a binomial distribution with mean lx · t px = lx+t and variance lx · t px (1 − t px ) = lx+t 1 − lx+t lx .2.5) (1.3) (1.2.2) (1. we shall suppose here that w = ∞. mx . so that px = P r{ (x) survives for at least a year } qx = P r{ (x) dies within a year } (1.2. assuming that the lx lives are independent.2.2.7) and. NON-SELECT LIFE TABLES 1.10 CHAPTER 1.2. if . is considered in section 1. In practice. or central. To simplify matters. t px = s(x + t) lx+t = s(x) lx (1.2. x2 = x and x3 = x + t.2. x3 such that α ≤ x1 ≤ x2 ≤ x3 . but it is always to be understood. we have P r{T (x) > t} = P r{T (α) > x − α + t} s(x + t) lx+t = = P r{T (α) > x − α} s(x) lx (1. the (cumulative) distribution function of the variable T is F (t) = P r{T ≤ t} = t qx = 1− 0 . when x1 = α.2).) It is clear that.4) When t = 1 . rate of mortality.1) We introduce the important notation: t px t qx = P r{T (x) > t} = P r{ (x) survives to age x + t} = P r{T (x) ≤ t} = P r{ (x) dies between ages x and x + t} (1.10) We may say that “the expected number of survivors at age x + t among lx lives aged x is lx+t ” because the chance that a given life aged x will survive to age x + t is lx+t .2 Probabilities of death and survival We now consider how to deal with lives aged between α and w. P r{T (x1 ) > x3 − x1 } = P r{T (x1 ) > x2 − x1 }P r{T (x2 ) > x3 − x2 } In particular. (The m-type. x2 .2. t px + t qx = 1 (1. if t≥0 t<0 (1. 02449 0. t px is sometimes written as Sx (t).12) where dx = lx − lx+1 = the (expected) number of deaths at age x last birthday among lx lives aged x (1. px and qx .2.2. The notation S(t) is much used in survival analysis.2.00069 0.302 97. When α = 0 we sometimes write S0 (t) = S(t). The definition of t px is sometimes taken to be P r{(α) survives to age x + t | (α) survives to age x }. age x 0 1 2 3 4 lx 100.97551 0.1).551 97.2.00157 0.13) Example 1. Further.499 153 96 67 60 px 0. so S(t) and s(t) are equal.) We also observe that.398 97. in which t denotes the time lived by a patient since the start of a given treatment.1.11) (1.00099 0.235 dx 2.2.14) x3 −x1 px1 = x2 −x1 px1 · x3 −x2 px2 (which is merely a re-statement of equation (1. PROBABILITIES OF DEATH AND SURVIVAL 11 (Even if the lives are not independent.99931 0. the formula for the expected number of deaths is correct.2. which equals P r{T (α) > x − α + t} P r{T (α) > x − α + t and T (α) > x − α} = P r{T (α) > x − α} P r{T (α) > x − α} = P r{T (x) > t} in agreement with our earlier definition. relative to the rates at ages 1 to 4.00062 (The high death rate at age 0.99843 0. ) Note. .99938 qx 0. we have the “rule of multiplication of probabilities of survival”: (1.Males illustrates the functions lx .99901 0. The following extract from English Life Table No. is quite noticeable.1. for all ages x3 ≥ x2 ≥ x1 ≥ α .) Note that px = lx+1 lx lx+1 dx qx = 1 − = lx lx (1.2.000 97. 12 . dx . In reliability engineering the notation R(t) may be used instead of S(t). w). An Introduction to the Mathematics of Finance.3.3.12 CHAPTER 1. except that lx does not exist at the points at which . But s(x) and s+ (x) = −s(x)µx are continuous on [α.3.2) µx = lim −s+ (x) −[s(x + h) − s(x)] = h→0+ hs(x) s(x) where s+ (x) denotes the R.3. Appendix 1.1) Theorem 1.1. We have µx = − Proof. we have lx lx (1.3. ) Hence s (x) l µx = − =− x s(x) lx We remark that µx = − Hence x d log lx dx x (1. h→0+ h (1.4) and x s(x) = exp − α µy dy (1. i. derivative of s(x). NON-SELECT LIFE TABLES 1. Suppose that µx is continuous on [α.3) − α µy dy = d (log ly ) dy dy α x = [ log ly ]α = log lx − log lα lx lα = log from which we obtain the important formulae: x lx = lα exp − α µy dy (1.e. w).3 The force of mortality.6) If µx is piecewise continuous. the above formulae remain true. so s (x) exists and equals −s(x)µx (see McCutcheon and Scott. we may apply the above formulae over each age-range and combine the results.5) It follows that t px = lx+t = exp − lx x+t t µy dy x = exp − 0 µx+r dr (1.3.3.H. µx µx = lim h qx This may be defined as the “instantaneous rate of mortality”. 9) 1 − F (t) which equals µx+t (by formulae (1. and µt = f (t) = hazard rate at age t years 1 − F (t) (1.3.3.2)) .8) An alternative notation Some statisticians refer to the force of mortality as the “hazard rate” or “transition intensity. f (t) of T is the derivative of the distribution function F (t). c > 0) (1.11) (This is “Weibull’s law of mortality”.7) In view of the following general formula (connecting the d.3. The hazard rate at age x + t (often considered to be “at time t years after entry to assurance (say) at age x”) is defined in terms of f (t) and F (t) by the formula f (t) h(t) = (1. if t < 0 By the chain rule d lx+t = lx+t dt = −lx+t µx+t so we obtain f (t) = lx+t µx+t lx (by formula (1.1.d.): t F (t) = −∞ f (r) dr we have t qx t = 0 r px µx+r dr (1. and L.f. derivatives of lx differ at these points.3. µx = cδxδ−1 (δ > 1.f.3. µx 13 µx is not continuous (the R.f.2.7). . We may now express the probability density function of the variable T (the future lifetime of (x)) in terms of the force of mortality.3.3. for x ≥ α.d. if t > 0 .H.1.f. we naturally have T = the future lifetime of a new-born child. i. if t < 0 = t px µx+t 0 (1.3. The p.H.) Alternatively. d 1 − lx+t . if t > 0 lx f (t) = F (t) = dt 0 .10) Example 1.f.” Let f (t) and F (t) denote the p. THE FORCE OF MORTALITY.) Find a simple expression for Sx (t) = t px . respectively of T = T (x) .e. and p. Suppose that. we may regard the actual position as closely approximated by a model in which the force of mortality is continuous.d. and d.7) and (1.3.) If α = 0 and x = 0. If n = 1 it may be omitted. NON-SELECT LIFE TABLES s(x) = exp − α x µy dy cδy δ−1 dy x α = exp − α = exp −c y δ = exp −cx hence t px δ .14) By elementary probability.n qx+m (1.3. x CHAPTER 1. m |n qx = P {(x) will die between ages x + m and x + m + n } (1.3. The formula f (x + 1) − f (x − 1) f (x) (1.16) We remark that there is no such thing as We may also use the result that m |n qx = P r{(x) survives to age x + m}. lx+m − lx+m+n is the (expected) number of deaths aged between ages x + m and x + m + n.P r{(x + m) dies within n years} = m px .3. for example.12) 2 which is exact if f is a cubic between x − 1 and x + 1 .15) lx This may be remembered by the following rule: of lx lives aged x. may be used to show that µx − lx lx−1 − lx+1 = lx 2lx (1.14 Solution.13) Deferred probabilities The symbol m | indicates deferment for m years.17) . lx+m − lx+m+n (1.3. so we have m |n qx = m |qx = P r{(x) will die between the ages x + m and x + m + 1 } dx+m = lx m |n px .3.3. exp cαδ = Sx (t) = exp −c (x + t)δ − xδ Numerical estimation of µx from lx If lx is known only at integer values of x. (1. this equals P r{(x) will die before age x + m + n} − P r{(x) will die before age x + m} = m+n qx − m qx = m px − m+n px lx+m − lx+m+n = lx That is. we may use numerical differentiation to estimate µx . ex = 0 ◦ ∞ t px dt (1.4. the upper limit of the integral should be replaced by w − x. and is called the complete expectation of life at age x.4.3. Find an expression for the chance that (x) will die between ages x + m and x + m + n. ex = E(T ) ∞ ◦ (1.1. If w < ∞ .3. v = −t px ) ∞ + 0 2t · t px dt =2 0 t · t px dt . By integration by parts. THE EXPECTATION OF LIFE 15 Example 1. The mean of T is written ex .4. ex = 0 ◦ ∞ t · t px µx+t dt = [−t · t px ]0 − ∞ ∞ ∞ −t px dt 0 = 0 t px dt since it may be shown that t · t px → 0 as t → ∞ (using the fact that E(T ) < ∞. including fractions) of (x).4 The expectation of life ◦ Let T be the future lifetime (in years.2) = 0 t · t px µx+t dt by formula 1. ). Proof. That is.1. with u = t .4. We may also use the formula where E(T 2 ) = 0 ∞ Var(T ) = E(T 2 ) − [E(T )]2 (1. We require: m |n qx lx+m − lx+m+n lx s(x + m) − s(x + m + n) = s(x) exp[−c(x + m)δ ] − exp[−c(x + m + n)δ ] = exp[−cxδ ] = 1. Theorem 1.4) t2 · t px µx+t dt ∞ 0 ∞ = −t2 · t px (on setting u = t2 .3.2. v = −t px .7.1. Solution.1) (1. Consider the force of mortality to be as in Example 1.3) (with ∞ replaced by w − x if w is finite).4.4. 4. 1.5 − 48.) If ly µy has a unique maximum at age 80 (say). .k |qx (1.) We now consider the discrete random variable K (or K(x) if the age x is not clear) defined by K = the integer part of T = the number of complete years to be lived in the future by (x) Now it follows by formula 1.e.1. NON-SELECT LIFE TABLES ∞ t · t px dt − (ex )2 ◦ (1.8) (1. this is the modal age at death for all lives aged under 80. t.5) The median future lifetime of (x) is the solution.7) This variable is used in many actuarial calculations.4.4. is the mean of K . We must find the value of t such that l10+t = 0.4.3.7 for an example in which there is a unique modal age at death. lx+t = 0. according to English Life Table No.e. 625 i.05 years. In particular. and hence the modal future lifetime of a life aged x < 80 is 80 − x. as in the next example. Hence Var(T ) = 2 0 CHAPTER 1. at which the p.5 i. 10 + t lies between 72 and 73.f. Find the median future lifetime of (10).5 i. 469.) (1. the curtate expectation of life at age x. The modal future lifetime of (x) is the time. so t = 62.16 (since t2 · t px → ∞ . By linear interpolation in y = 10 + t .16 that P r{K = k} = k |qx ( k = 0.4.d. (1.05. Solution. t0 years. t px µx+t (or.. 469. we have ly − l72 = y − 72 l73 − l72 48.e.5 lx x + t may be estimated by linear interpolation.4.5 l10 = 48. that is ∞ ∞ ex = k=0 k · k |qx = k=1 k. 2.6) Example 1. 625 = 72. 430 − 48.. (See Exercise 1.12 Males. t qx = 0. because E(T 2 ) < ∞ ). equivalently. the function lx+t µx+t ) reaches its maximum.. written ex . (The function ly µy is sometimes called the “curve of deaths”.5 On inspecting the tables. y = 72 + 45. of the equation F (t) = 0.9) .. The modal age at death of (x) is y = x + t0 . which is the value of y for which ly µy attains its maximum in the range y ≥ x. ) + (dx+3 + . . = dy + dy+1 + .4.e.5.) lx + (dx+2 + dx+3 + . lx lx lx 1 [(dx+1 + dx+2 + dx+3 + . THE ASSUMPTION OF A UNIFORM DISTRIBUTION OF DEATHS Theorem 1. .5.D.5.12) 2 A more precise approximation may be obtained from the Euler-Maclaurin formula.2. . ex = = dx+1 dx+2 dx+3 +2 +3 + .5. = lx [since ly = (ly − ly+1 ) + (ly+1 − ly+2 ) + .) between ages x and x + 1 if.D. ly is linear for x ≤ y ≤ x + 1) This equation may be written in the form lx+t = lx − tdx (0 ≤ t ≤ 1) (1.5.4) .1) (i. ex ◦ ex + 1. for 0 ≤ t ≤ 1.2) Theorem 1. lx (1.1. We may say there is a uniform distribution of deaths (U.4.D..11) As is clear by general reasoning. We may also evaluate Var(K) by the formula Var(K) = E(K 2 ) − [E(K)]2 ∞ ] = k=1 k2 dx+k − (ex )2 lx (1.4. . 1 (1.10) Proof. which we shall discuss later. .D. . . .. ex = k=1 17 ∞ k px = lx+1 + lx+2 + . . .4. The following conditions are each equivalent to the assumption of U. lx+t = (1 − t)lx + t lx+1 (1. . .5 The assumption of a uniform distribution of deaths Let x be fixed.5.3) (1. . between ages x and x + 1: t qx t px µx+t = t · qx = qx (0 ≤ t ≤ 1) (0 ≤ t < 1) (1. .1.) + . . .] lx+1 + lx+2 + lx+3 + . 5 (ii) µ30. Find formulae in terms of l30 . so t qx = t · qx .5 Solution.5 l32 since l30+t is linear for 0 < t < 1 + l31 ) l30 − l31 q30 = = p30 l30 1 2 l30 − l31 1 2 (l30 + l31 ) 1 2 (l30 (ii) t p30 µ30+t = q30 for 0 ≤ t ≤ 1. l31 and l32 for (i) 1.3) from (1. Define ∞ Tx = 0 ∞ lx+t dt ly dy (on setting y = x + t) x (1.5. Assume 1. This argument may be reversed. we again suppose that the limiting age of our mortality table is infinity. Then t px CHAPTER 1.5.5.5) But t px = 1 − t qx .1) = from which we obtain ex = ◦ ∞ ly dy x lx Tx = lx and hence Tx = lx ex We also define 1 ◦ (1.4). NON-SELECT LIFE TABLES = 1 − t · qx (1.6. and we may obtain (1.3 holds.5.3) .5.5.5.1.5 p30. there is a uniform distribution of deaths between any two consecutive integer ages.5.5.18 Proof.2 and 1. so µ30.6. By differentiation with respect to t. we obtain (1.3 are equivalent.6. so 1.4) by integration. therefore 1. In a certain non-select mortality table.2 holds.2) Lx = 0 lx+t dt (1.3) holds. Example 1.5 = 1 2 1.6 Central death rates For simplicity of notation. (i) l32 = l30. Now suppose that (1.5. D.6.9) (1.6.e.D. we have 1 Lx = l x − 2 d x (1. 1.6.6) Formulae (1.10) If U.D. not just an approximation).7 Laws of mortality The term ‘law of mortality’ is used to describe a mathematical expression for µx (or possibly qx or mx ) which may be explained from biological or other arguments (rather than being just a bestfitting curve.D.D) between the ages x and x + 1.6.6) are sometimes used as approximations when U. does not hold.6.6.6. does not hold.6.6) shows that mx = = dx Lx dx lx − 1 dx 2 qx = 1 1 − 2 qx We may rearrange this equation to get qx = mx 1 1 + 2 mx (1.4) = Tx − Tx+1 Assume that there is a Uniform Distribution of Deaths (U.5) (as the trapezoidal rule is exactly correct.6.) The most famous law of mortality is that of Gompertz (1825). of D. who postulated that . lx+t is linear for 0 ≤ t ≤ 1. We have 1 Lx = 0 lx+t dt = 1 (lx + lx+1 ) 2 (1. i. The central death rate at age x is defined as mx = dx Lx (1.D.8) Relationships between mx and qx If there is U. LAWS OF MORTALITY Note that x+1 19 Lx = x ∞ ly dy ∞ = x ly dy − x+1 ly dy (1.5) and (1. between ages x and x + 1.7) Note the following important approximation: mx = 1 l µ dt 0 x+t x+t 1 l dt 0 x+t µx+ 1 2 (1. these results may be used as approximations. Since lx+1 = lx − dx . formula (1.6.1.7.D. there are positive constants s and g such that t px = st g c x (ct −1) for x ≥ α.07 and 1. t t px = exp − 0 t µx+r dr (A + Bcx cr )dr x = exp − 0 = exp[−At]g c = st g c x (ct −1) (using Example 1.2) Solution. there is a positive constant g such that t px = gc x (ct −1) for x ≥ α. Show that. the value of c usually being between 1.7. s = e−A log c (ct −1) with g = exp − .7.1. t ≥ 0 (1. Show that.7.3) Example 1.1) Gompertz’ Law is often found to be quite accurate (at least as a first approximation) for ages over about 25 or 30. if Gompertz’ Law holds for all ages greater than or equal to α.12. t t px = exp − 0 µx+r dr t = exp −Bcx = exp − = gc x cr dr Bc (c − 1) log c with g = exp − B log c 0 x t (ct −1) In 1860 Makeham suggested the addition of a constant term to Gompertz’ formula for µx .7.4) Solution.1) B .7.2. giving Makeham’s law: µx = A + Bcx (x ≥ α) (1.7. NON-SELECT LIFE TABLES µx satisfies the following simple differential equation: dµx = kµx dx This may be solved to give µx = Bcx (x ≥ α) (x ≥ α) (1.7.20 CHAPTER 1. Example 1. t ≥ 0 (1. if Makeham’s law holds for all ages greater than or equal to α. 3. If a law of mortality holds.) .1.7. but this point is of little practical importance in the computer age. LAWS OF MORTALITY 21 Weibull’s law of mortality has already been mentioned in Example 1. In practice it is usually less successful than those of Gompertz and Makeham in representing human mortality.1. There are also certain simplifications in the evaluation of joint-life functions (that is. certain mortality and monetary functions may be evaluated analytically (rather than numerically). functions depending on the survival of more than one life. The fitting of laws of mortality is complicated by the fact that mortality rates may be varying with time. 1. express (i) and (ii) in words. ..019803 p. die between ages 55 and 60. live to age 65. (iv) the probability that (40) survives to age 65. No. die either between exact ages 35 and 45 or between exact ages 70 and 80. lives are subject to a constant force of mortality of 0.a. between ages 49 and 50. 1. find the probability that a life aged 30 will (i) (ii) (iii) (iv) (v) 1.97531. 12 . the force of mortality at age 1 year. A man aged 50 has just retired because of ill health.039221 to the force of mortality.2 On the basis of E. the chance that a newly-born animal will die between ages 1 and 2 years. 12 . (b) Find the probability that the life will survive to age 60. 12.22 CHAPTER 1. In addition. die in his 50th year of age. (viii) the probability that (60) dies within the first five years after retiring at age 65. (v) the probability that (50) dies within 10 years. (ii) 60 |10 q25 . Find the probability that he will (i) (ii) (iii) die before age 55. i.T. (ii) What is the probability that a life aged exactly 33 who has been in the country for 3 years will die between ages 50 and 51? (Assume that these lives will remain in the given country. Thereafter lives are subject to mortality according to English Life Table No. (i) 40 p25 . (a) Show that the probability that the life will survive to age 35 is 0. (iii) the probability that (30) survives for at least 10 years.5 For a certain animal population. die between his 40th birthday and his 50th birthday. (vii) the probability that (60) dies between ages 80 and 85.T.Males with an addition of 0. 12-Males. (i) A life aged exactly 30 has just arrived in the country.e. Up to exact age 58 he will be subject to a constant force of mortality of 0.) 1.L. lx = Calculate (i) (ii) (iii) l0 (1 + x)2 (x ≥ 0) the complete expectation of life at birth. NON-SELECT LIFE TABLES Exercises 1.Males.005. No. die before reaching age 50. (vi) the probability that (50) fails to reach age 70.1 Calculate the following probabilities on the basis of English Life Table No.L.Males.3 survive to age 40. after which his mortality will be that of E.4 For the first 5 years after arrival in a certain country. 1−t qx+t = (1 − t)qx for 0 ≤ t ≤ 1. t−s qx+s = t−s qx 1 − sqx 1.e. Show that qx < µx . Show that. at which level the addition remains constant. (i) (ii) Give a formula for s(x). Find the probability that a life aged exactly 60 dies within 20 years. for all 0 ≤ s ≤ t ≤ 1.6 Suppose that t px µx+t is decreasing for 0 ≤ t ≤ 1. 1. increasing linearly to 0. i. 40 |5 q35 1.9 (Difficult) Suppose that the “Balducci hypothesis” holds from age x to age x + 1.0025 at age 80. ◦ ex (Note finite limiting age). µx = Bcx . Assume that µα < log c.8 (Difficult) Suppose that there is a “uniform distribution of deaths” from age x to age x + 1.10 Suppose that. holds for all x ≥ α. lx+t is linear in t for 0 ≤ t ≤ 1. Show that lx µx attains a maximum when µx = log c.e. 1. Show that. and has no other stationary points. Give simple formulae for (i) (ii) (iii) (iv) s(x) = (1 − x/w)a (0 ≤ x ≤ w) µx .8. c being greater than 1.7 Suppose that Gompertz’ law. t−s qx+s = t−s qx 1 − (1 − t)qx 1.11 A certain group of lives now aged 60 experience mortality according to a(55) males ultimate with addition to the force of mortality. . EXERCISES 23 1.0005 at age 60. i.1. for some a > 0. 10 p70 . The addition is 0. for all 0 ≤ s ≤ t ≤ 1. 39162 = 0. (i) ∗ 5 q50 = 1 − 5 p∗ = 1 − exp(−5k) 50 = 0.98452 = 0.T.16173 = 0.” is “the probability that a life aged 25 will survive to age 85 and die before age 95. refers to E.00613 = 0. 12-Males.71528 l85 − l95 l25 = 0.24 CHAPTER 1.17338 60 |10 q25 10 p30 = = = = 25 p40 10 q50 l50 − l60 l50 l50 − l70 l50 l80 − l85 l60 20 q50 20 |5 q60 = 5 |5 q60 = l65 − l70 l60 In words.10021 = 0.3 Let us use an asterisk to denote the mortality table of the life concerned.” 10 p30 20 q30 l40 l30 l30 − l50 = l30 = = d49 l30 = l40 − l50 l30 = 0.05437 = 0. while lx etc.73025 = 0.019803.1 (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) 40 p25 = l65 l25 = l40 l30 l65 l40 = 0.03889 19 |q30 10 |10 q30 5 |10 q30 + 40 |10 q30 = = 0. (i) (ii) 1.35789 (l35 − l45 ) + (l70 − l80 ) l30 1.12389 = 0.L.09427 .98452 = 0. NON-SELECT LIFE TABLES Solutions 1.2 (i) (ii) (iii) (iv) (v) is “the probability that a life aged 25 will survive to age 65. Let k = 0. of survival for 18 years = exp(−2 × 0. of survival for 5 years = exp − ELT 0.4 (i) (a) Prob.1.97531 (b) 0.039221 × 16) = 0.005) exp (−0.prob.08540 1.6 Let f (t) = t qx − µx .005) exp (−0.5 (i) (ii) (iii) e0 = ∞ lx −1 dx = (1 + x)−2 dx = l0 1+x 0 0 l 2 µx = − x = . of survival for 17 years . 1.039221 × 15) l50 ELT l35 l51 ELT l35 ELT ELT − exp(−2 × 0.02377 1.13889 l0 4 9 1. and f (t) is decreasing for 0 ≤ t ≤ 1. so µ1 = 1 lx 1+x ◦ ∞ ∞ =1 0 1| q0 = 1 1 l1 − l2 = − = 0.7 (i) For x ≥ α.t (0 ≤ t ≤ 1) d t f (t) = r px µx+r dr − µx = t px µx+t − µx < 0 for 0 < t < 1 dt 0 Now f (0) = 0.3051 l35 (ii) Prob. therefore qx < µx . = prob. so f (1) < 0.9.005dt 0 = 0. t px t = exp − 0 Bcx+s ds = exp −Bcx (ct − 1)/ log c Therefore s(x) = x−α pα = exp [−B(cx − cα )/ log c].039221 × 25) l60 ELT = 0. SOLUTIONS l65 l58 25 (ii) (iii) ∗ 15 p50 ∗ 5 |5 q50 = 8 p∗ 7 p∗ = exp(−8k) 50 58 = 0.97531 exp (−0. .71188 = 5 p∗ − 10 p∗ = 5 p∗ − 8 p∗ 2 p∗ 50 50 50 50 58 = exp(−5k) − exp(−8k) l60 l58 5 = 0. d a 1. since = . and is negative for x > x0 . lx+t = tlx+1 + (1 − t)lx (0 ≤ t ≤ 1) and similarly for lx+s . of the equation. 1. Therefore d (lx µx ) =lx Bcx log c dx [−Bcx log c] + Bcx k exp [−Bcx / log c] log c =lx µx (log c − Bcx ) =0 when µx = Bcx = log c (∗) Let x0 be the unique point at which this occurs. 1 − sqx lx+s 1 − sqx = 1− = lx+t lx+1 lx+t lx+1 =1− − 1− lx+s lx+s lx+s lx+t qx+s − (1 − t−s qx+s )1−t qx+t 1−s t−s qx+s Therefore t−s qx+s = 1−s qx+s − 1−t qx+t (on rearranging)..26 (ii) CHAPTER 1. NON-SELECT LIFE TABLES It follows that lx = k. exp [−Bcx / log c] for some k. This gives the desired result. Notice that (by equation (∗) above) d (lx µx ) > 0 for x < x0 . Therefore lx+s − lx+t (t − s)lx − (t − s)lx+1 = = (t − s)qx lx lx t−s qx+s Therefore 1. 1 − 1−t qx+t Now apply “Balducci” to the R.S. dx Therefore x0 is a maximum point of lx µx .H.8 Under U.10 (i) µx = − [log s(x)] = dx w−x (ii) ex ◦ w−x w = = 0 w (1 x t px dt = x ly dy/lx − y/w)a dy (1 − x/w)a Substitute z = 1 − y/w to obtain (calculus exercise!) the result that w−x ◦ ex = a+1 (iii) 10 p70 = s(80) = s(70) w − 80 w − 70 a (iv) 40 |5 q35 1− s(75) − s(80) = = s(35) 75 a w − 1− 35 a w 80 a w 1− .D.9 = (t − s)qx lx 1 .D. 5892 .4108 ∴ Ans.9.0001t 20 (0 ≤ t ≤ 20) Prob.1.0005 + 0.11 Addition to force of mortality = 0.0001t)dt on integration = 20 p60 exp(−0.003) = 0. SOLUTIONS 27 1. = 0. of survival for 20 years = exp − 0 (µ60+t + 0.0005 + 0. NON-SELECT LIFE TABLES .28 CHAPTER 1. Males and A1967-70 ultimate. or who are not yet retired. We now consider the situation when mortality rates (or the force of mortality ) depend on two factors: (i) age. After a certain period the difference in mortality between those who have been accepted for life assurance and the general population of the same age decreases.1 What is selection? In the previous chapter we considered mortality rates to depend only on age: life tables of this form are sometimes called aggregate life tables.) Another form of selection is the “self-selection” exercised by those who buy annuities: it may normally be assumed that such lives are in good health (for their age). and on what terms. (The process by which life offices decide whether to accept lives for assurance. which we will discuss below.12 . mortality rates are higher than those of the general population (or some other reference group). for the purchase of an annuity is otherwise likely to be a poor investment. the mortality of these lives may be expected to become closer to that of lives who retired in normal health. One important example of “selection” is the acceptance of a proposal for life assurance at normal rates of premium: the mortality rates of lives who have been recently accepted for life assurance at normal rates may be expected to be lower than those of the general population. In “reverse selection”. particularly in the period soon after selection. the mortality of those selected is lower than that of the general population. known as “selection”. but it is not correct to say that the effect of selection “wears off” entirely (since the general population contains some lives who would never have been accepted for life assurance at normal premium rates). After a certain period from the date of ill-health retirement.Chapter 2 SELECT LIFE TABLES 2. and (ii) the time (duration) since a certain event. In these examples. This point may be confirmed by a comparison between the mortality rates of English Life Table No. An example of reverse selection is early retirement due to ill-health. 29 . is called “underwriting”. 3) (2. Thus for example.30 CHAPTER 2. h q[x]+t = P r{T ([x] + t) ≤ h} l[x]+t+h =1− l[x]+t where l[x]+t is the expected number number of survivors at age x + t of l[x] lives selected at age x. SELECT LIFE TABLES 2. T ([x] + t) = the future lifetime of “[x] + t” (t ≥ 0) Regarding x as fixed. will die within h years} = the force of mortality of a life aged x + t who has selected t years ago h q[x]+t = lim h→0+ h (2. we first construct a life table for the ultimate lives by the methods given in chapter 1. We assume that there is a period. More precisely.1) (2. such that the mortality of those selected at least s years ago depends only on the attained age.2.4) where qx+t and µx+t refer to those who were selected at least s years ago: such people are called ultimate lives. we choose them to be such that l[x]+t depends only on x + t when t ≥ s.2. the function l[x]+t is then constructed to be such that l[x]+t = lx+t This ensures that relationships such as h p[x]+t (t ≥ s) (2. We have. we may define the family of random variables.6) . s years. at the date of selection t = duration (in years) since the date of selection The current (or attained) age of such a life is y =x+t We write “[x] + t” as a shorthand notation for a “life aged x at selection and duration t years since selection”.2 Construction of select tables Let us. x + t.2. and α = 0). That is.5) = l[x]+t+h l[x]+t (2.2.2. i. who has selected t years ago.e. consider “selection” to be the acceptance of a proposer for life assurance at normal rates by a life office.2) The select period. h q[x]+t µ[x]+t = P r{“[x] + t” will die within h years} = P r{a life aged x + t. Let x = age at entry to assurance. But instead of fixing each radix l[x] arbitrarily. q[x]+t = qx+t µ[x]+t = µx+t (t ≥ s) (t ≥ s) (2.2. for definiteness. we may construct a “life table” for those selected at age x by methods similar to those of chapter 1 (with t in place of x. for example. For each fixed entry age x. we use the formula l[x]+t = exp − lx+s s µ[x]+r dr t (t < s) (2. select duration 0 l select duration 1 [0]+1 l[0] = 1−q[0] ← ··············· 2 l[0]+1 = 1−ql[0]+1 ← ··············· ← ultimate 34. l1 . and is based on data collected by the Continuous Mortality Investigation Bureau (C. 440.4 34. . t = 0. we proceed recursively. l3 .7 . We now give formulae for l[x]+t for 0 ≤ t ≤ s. If q[x]+t is given for t = 0. 489(1 − q0 ) = 34. l[0]+1 and l[0] were found. 1. 463.). . .2. 3. . 1. etc.9) 2.3 The construction of A1967-70. THE CONSTRUCTION OF A1967-70. 2. 0) (2. 2. . .I.B.2. 2. ) q[x]+t (x = 0. 440. 1.1). . . By working to the left from l2 . . . 2.3. .3. .7) l[x]+t That is l[x]+s−1 = l[x]+s /(1 − q[x]+s−1 ) l[x]+s−2 = l[x]+s−1 /(1 − q[x]+s−2 ) ······························ l[x] = l[x]+1 /(1 − q[x] ) (2. Mortality rates were extended to young ages (including “ultimate” lives aged under 2) for which no data existed. We may omit the square brackets enclosing x in expressions such as h p[x]+t and l[x]+t+h if t ≥ s or t + h ≥ s respectively. using the formulae l[x]+t+1 = 1 − q[x]+t (t = s − 1. similarly for l[1]+1 and l[1] . l0 was fixed arbitrarily at 34. .8 34. We illustrate the construction of select tables by reference to A1967-70. during the period 1967 to 1970. . which has a select period of 2 years. 489 34. . were computed as shown (working downwards in the third column of Table 2. . . 418. s − 2. 1) The stages of construction of ly and l[x]+t were as follows: 1. 463.K.M.8(1 − q1 ) = 34. . = l0 = l1 = l2 = l3 .489. .2. . l2 . .2. The “building blocks” of the table were the mortality rates: qy (y = 0.8) If µ[x]+t is given for 0 ≤ t ≤ s .4(1 − q2 ) = 34. . 31 (which are found by replacing “x” by “[x] + t” in the formulae of chapter 1) are true. This table refers to the mortality of male assured lives in the U. s − 1 . . A1967-70 (select period 2 years) and 2. Formulae and Tables for Actuarial Examinations give 1.3) f (t + 1) − f (t − 1) 2 2.1 CHAPTER 2. . 2. a L. derivative. strictly speaking. a(55) males and females.1) (the derivative is. which gives 1 f (0) ∆f (0) − ∆2 f (0) (2. . We now consider how best to estimate µ[x]+t for t = 0. . . In each case the select period is 1 year.M. we may use the differentiated form of Newton’s forward difference formula.3.2) 2 where ∆f (0) = f (1) − f (0) ∆2 f (0) = f (2) − 2f (1) + f (0) To estimate µ[x]+t . We assume that l[x]+t and µ[x]+t are continuous in t (t ≥ 0). 1. s). .4 Some formulae for the force of mortality.32 Table 2. data and were intended to be suitable for those purchasing annuities in about 1955. One should be careful not to look up “select” values (such as q[x] )when ultimate values (such as qx ) are required. We may thus write µ[x]+t = f (t) (t ≥ 0) where f (t) = − log(l[x]+t ). To estimate µ[x] = µ[x]+0 . a(55) males and a(55) females. h q[x]+t µ[x]+t = lim h→0+ h l[x]+t − l[x]+t+h = lim h→0+ hl[x]+t d − l[x]+t = dt (t ≥ 0) l[x]+t (2. where t = 1. but arguments similar to those given in chapter 1 show that the l[x]+t is differentiable in t ). . and vice versa.H.4.5 Select tables used in examinations. These tables were constructed on the basis of C. SELECT LIFE TABLES 2. . . . Note that a(55) consists of 2 tables. or vice versa. In the case of a(55).B. . s − 1 . . we may use the formula f (t) to obtain µ[x]+t 1 − log(l[x]+t+1 /l[x]+t−1 ) 2 (2. an allowance being made for improvements in mortality . 2. 2. one should be careful not to look up the “males” table when the “females” table is required.4. 1. . s − 1 from a table of l[x]+t (t = 0.I. .4. and functions are given only for ages 60 and over. 01601. At each integer age x . 2.Males. = 0.0142.5424 and q60 = 0. an office assumes a 3-year select period. . 2 q[50] 2|3 q[50]+1 = 0. (i) Find expressions in terms of the life table functions l[x]+t and ly for (a) (b) (c) (d) (ii) q[50] 2 p[50] 2 | q[50] 2 |3 q[50]+1 Calculate 3 p53 given that: q[50] 2 |q[50] = 0. Calculate e[60] . given that e60 = 17.1 A mortality table has a select period of three years. The table is such that: ∗ ∗ ∗ ∗ q[x+7] = q[x+3]+1 = q[x]+2 = qx+1 ∗ and ultimate mortality follows A1967-70 ultimate.02410. Define the term “select period” and. = 0. the select rate mortality is 50% of the ultimate rate.3 A certain life table has a select period of 1 year. explain how a select table differs in construction from an aggregate table such as English Life Table No. EXERCISES 33 Exercises 2.2 In its premium rate basis. calculate l[45]+t for t = 0. 2. Assuming further that ly = ly on A1967-70 ∗ ultimate. using the A1967-70 table as an example. 12 .4 Explain briefly the concept of selection in relation to mortality tables.96411.2.6. 1 and 2. Functions on this table are indicated by an asterisk.09272 2. = · l[60] l60 l60 = e60 l[60] But q[60] = 0. 399 from (A) = 96.4 Mortality rates depend on the age x at “selection” [e.90294. SELECT LIFE TABLES Solutions 2.g. 877 from (D) l56 l53 Then 3 p53 = 2. 144 1 − q39 2. purchase of an annuity (self selection)] and duration t years since selection. l[60] l60 l61 + l62 + .34 CHAPTER 2. .1 (i) (a) (b) q[50] = 2 p[50] l[50] − l[50]+1 l[50] l[50]+2 l[50] l[50]+2 − l53 l[50] = l53 − l56 l[50]+1 = .5q60 .3 e[60] = l61 + l62 + . 033 = = 33. .01601 = . . 411 from (B) = 94.2 ∗ l[45]+2 = ∗ l[45]+1 = = . 101 ∗ 1 − q[45]+1 1 − q43 ∗ l[45]+1 ∗ 1 − q[45] ∗ l[45] = = 33. then = 98.96411 (A) (B) = (c) 2| q[50] = = .007202 so e[60] = 1. . l48 l48 = = 33.09272 (D) Put l50 l[50] l[50]+2 l53 l56 = 100. entry to assurance.02410 (C) (d) (ii) 2|3 q[50]+1 = .668 (1− 1 q ) 2. which implies that l61 l[60] = 1 1 − 2 q60 Therefore l60 l[60] = l60 (1− 1 q60 ) 2 l61 2 60 = 1−q60 = 1. Thus q[x]+t = Pr{ a life aged x + t . 001 from (C) = 84.007202 × 17. 000 (say). 033 ∗ 1 − q[45]+2 1 − q46 l[45]+2 33.542 = 17. 101 = 33. etc.Males. If this were true.e. The select period. and it does not.L. . The general population includes some lives who would never have been acceptable for assurance cover (at normal premium rates).7. such as E. The l[x]+t function is constructed to be such that l[x]+t = lx+t (t ≥ s) In A1967-70.2. mortality depends on one variable only (age) and there is no “selection”. Note. In a non-select or aggregate life table. 12 . No.Males. SOLUTIONS 35 who was selected t years ago will die within a year }.T. s years. A1967-70 ultimate would resemble ELT 12 . It is not correct to say that the effects of selection “wear off” entirely (and that the mortality of ultimate lives is the same as that of the general population). for lives who were selected at least s years ago (these being called “ultimate” lives) mortality depends only on the attained aged. s = 2. is such that q[x]+t = qx+t for t ≥ s i. SELECT LIFE TABLES .36 CHAPTER 2. or between certain specified dates. it may be level (constant) or it may increase or decrease in a manner specified in the contract. a certain portfolio of equities. sums payable on the survival of the life (or lives) assured until a certain date (or dates). Policies may also provide endowment benefits. when calculating the value of benefits. The death benefit S is called the sum assured. With profits policies may be found in traditional or unitised form (see later discussion). and let T be the future lifetime of (x).e. In this chapter we shall consider contracts issued on the life of one person.Chapter 3 ASSURANCES 3.1) Now consider a life aged x. 3. e.g. which depend on the experience of the office and its bonus distribution policy. in which the benefits are directly linked to the performance of certain assets. the policy is said to be without profits (or non-profit). A whole life assurance is a policy providing a certain sum assured. called bonuses. S. The benefits and expenses are paid for by premiums. the life office assumes that its funds will earn interest at a constant rate.2 Whole life assurances Suppose that.1 A general introduction A life assurance (or life insurance) policy is a contract which promises to pay a specified sum.2. who is subject to a certain non-select mortality table. S say. In the case of with profits policies. the sum assured may be increased by additions. The corresponding force of interest per annum is δ = log(1 + i) and the present value of 1 due at time t years is v t = (1 + i)−t = e−δt (3. on the death of a given life (the life assured) at any future time. which we shall discuss later. If the benefits (and premiums) are completely specified in money terms in the contract. on the death of (x) at any future date. i.2. Policies under which payments depend on the death or survival of more than one life may also be issued. One also encounters unitlinked policies. i per annum.2) 37 . The present value of this benefit (assumed for the moment to be payable immediately on death) is Z = g(T ) = Sv T (3. payable immediately on the death of (40). a life now aged x + t who was selected t years ago) we replace x by [x] + t. The M.2.f.4) When S = 1 .a.T.1.2. No.T. a person who has just been selected) aged x.P. we replace x by [x]. = 10. Solution.2. is called the mean (or expected) present value (M. E(Z). Thus.38 CHAPTER 3.P. and for an “[x] + t” life (i.) of the whole life assurance benefit.P. of v T ∞ = 0 v t t p[x] µ[x]+t dt (3. 198. 000 × 0. 12-Males. of the sum of £S payable immediately on the death of (x) is therefore ¯ S Ax Example 3. Find the M.V. according to E.P.7) . ¯ M. No.3) where f (t) is the p.L. hence ∞ E(Z) = S 0 v t t px µx+t dt (3.V. 000 A40 on E. ASSURANCES The mean of this variable. and is ∞ E(Z) = 0 g(t)f (t) dt (3.V.d.10 Select tables For a select life (i. of T (for t > 0).P. we write ¯ A[x] = M. letting T = T ([x]) denote the future lifetime of “[x]”.e.P. we write ∞ E(Z) = 0 ¯ v t t px µx+t dt = Ax (3.6) If T = T ([x] + t) = future lifetime of “[x] + t” .V.31981 from Tables = £3.V.2. 000 .L. 4% interest = 10.5) where A stands for “assurance” and the bar indicates that the sum assured is payable immediately on the death of (x).e. we have ¯ A[x]+t = M.2. of a whole life assurance of £10.V. 12-Males with interest at 4% p.2. of v T ∞ = 0 v r r p[x]+t µ[x]+t+r dr (3. 4 The variance of the present value of benefits We recall that Z = g(T ) = Sv T is the present value of S due immediately on the death of (x).g. Mx etc.1) .2) (3.3. Define Dx = v x l x ¯ Cx = ¯ Mx = t=0 1 0 ∞ (3. e. We have var(Z) = E(Z 2 ) − [E(Z)] ∞ 2 2 = 0 ¯ S 2 v 2t t px µx+t dt − S Ax ∞ 0 = S2 ¯ (v ∗ )t t px µx+t dt − Ax 2 (where v ∗ = v 2 ) ¯ ¯∗ = S 2 Ax − Ax 2 (3. .3.032% ). ¯ Ax = = so that ∞ t=0 ¯ Cx+t 3. . a modern computer can easily compute Ax directly at any rate of interest required. and ¯ that the mean of Z is S Ax . In view of their importance in the Tables for Actuarial Examinations we shall look at them in detail. (In fact.3.4.1) (3. What is the variance of Z? Answer.3.3) v x+t lx+t µx+t dt ¯ Cx+t Now ¯ Ax = v x+t lx+t µx+t dt v x lx ¯ ¯ ¯ Cx + Cx+1 + Cx+2 + .4) Dx Dx We shall consider commutation functions again in relation to temporary and deferred assurances. COMMUTATION FUNCTIONS 39 3. Commutation functions are numerical devices (developed by Griffith Davies and others) which allow the calculation of certain common assurance (and annuity) values at a specified rate of interest from ¯ a small number of columns.3 Commutation functions ¯ Functions such as Ax may easily be evaluated by numerical integration. etc. At this point we consider only whole life assurances payable immediately on death. = Dx 1 0 ∞ 0 r+1 (on writing r v x+t lx+t µx+t dt = v x+r+u lx+r+u µx+r+u du by the change of variable u = t − r) ¯ Mx (3. labelled Dx . 4.3.3. and are often tabulated ¯ directly at various rates of interest. 4 and 3. found by taking the square root of var(Z).e.6) An application. Jensen’s inequality Let g(t) be convex.3).4) 2 (3. Let g(t) = e−δt .4. Zi = present value of benefit under ith policy = Sv T (i) with T (i) = future lifetime of ith life.V’s even if the lives are not independent. for i = 1.5 may be generalised to cover the case when the sums assured and/or ages of the lives are different (see exercise 3.4. . .5) We remark that the mean present value of a group of policies is the sum of their separate M. which is the case of g (t) ≥ 0 for all t > 0. the variance of Z is n Var(Z) = i=1 n Var(Zi ) ¯∗ ¯ S 2 Ax − Ax 2 2 = i=1 ¯ ¯∗ = nS 2 Ax − Ax and so the standard deviation of the total present value is √ S n ¯x ¯ A∗ − Ax (3. (3. 1 1 = v2 = ∗ 1+i (1 + i)2 CHAPTER 3. the total present value of the assurance benefits is Z = Z1 + Z2 + . We have g (t) = δ 2 e−δt ≥ 0 for all t > 0 .4. . of course.4.P. n. δ ∗ . . + Zn where. If we consider a block of n identical whole life policies on independent lives aged x.3) The standard deviation of Z is. Formulae 3. corresponding to i∗ is such that v ∗ = e−δ = v 2 = e−2δ δ ∗ = 2δ ∗ i.40 where the rate of interest i∗ is such that v∗ = i. Since the variables {T (i) } are assumed to be independent.e. where δ > 0. .4. ASSURANCES i∗ = 2i + i2 (3.2) Note that the force of interest. 2. .4.4. Jensen’s inequality shows that E[g(T )] ≥ g[E(T )] (3. 4. .T. 4% p. E(v T ) ≥ v E(T ) but E(T ) = ex and ¯ E(v T ) = Ax so we have shown that ¯ Ax ≥ v ex ◦ ◦ ◦ 41 (3.12-Males. Example 3.1) = SAx . It is now convenient to use the random variable K = [T ] = the integer part of T which has discrete probabilities P r{K = k} = k |qx The present value of the assurance is now Z = g(K) = Sv K+1 which has mean ∞ (k = 0. .7) That is.4. No. so which is less than e50 = 22.7 for x = 50 on the basis of E.a. Suppose that the death benefit S is payable at the end of the year of death (years being measured from the date of issue of the policy).68 v e50 = 0. 2. at any positive rate of interest.5. Verify formula 3.5. ¯ Ax .3. ) E[g(K)] = k=0 Sv k+1 k |qx (3. ASSURANCES PAYABLE AT THE END OF THE YEAR OF DEATH.5 Assurances payable at the end of the year of death. 1. interest.L.44429 ◦ 3. ◦ Solution.41085 ¯ A50 = 0.1. . v ex (which is the present value of £1 on the assumption that (x) dies when he reaches his expectation of life) is less than or equal to the mean present value.4. By Jensen’s inequality. 8) which follows from the mean value theorem for integrals.1. ASSURANCES Ax = k=0 v k+1 k |qx (3.3) (3.4) My = k=0 Cy+k and find that Ax = = ∞ k=0 v x+k+1 dx+k Dx (3.3) and (3. on average.2) We define the commutation functions Cy = v y+1 dy ∞ (3.5.5) Mx Dx Example 3.3. Show that the variance of Z (= Sv K+1 ) is S 2 Ax − (Ax ) where ∗ ∗ 2 (3. We thus have the approximate relationship ¯ Ax (1 + i) 2 Ax 1 (3.5. Var(Z) =E(Z 2 ) − [E(Z)]2 ∞ =S 2 (v ∗ )k+1 k |qx − (Ax )2 k=0 =S 2 A∗ − (Ax )2 x ¯ The relationship between Ax and Ax .5.7) This may be confirmed by establishing the following approximate relationship ¯ Cx (1 + i) 2 Cx 1 (3. the year between two consecutive policy anniversaries). Solution.4) now show that ¯ Mx (1 + i) 2 Mx 1 (3.5.5.5.42 where ∞ CHAPTER 3. Formulae (3. benefits payable at the end of the year of death will be received.5.5. Assuming that deaths occur on average mid-way through each policy year (i.5.e.9) . 6 months later than those immediately payable on death.6) indicates at the rate of interest 2i + i2 .5. 5. .5. since we have U.10) Theorem 3.5.11) δ ∞ ∞ ¯ ¯ Proof. ) Now. ). and hence ¯ Mx ¯ Ax = Dx 1 (1 + i) 2 Mx = (1 + i) 2 Ax Dx 1 43 (3. . 1.1. 1. of D.D. 2. It is sufficient (since Ax = Dx and Ax = [ k=0 Cx+k ]/ Dx ) to show that k=0 Cx+k i ¯ Cy = Cy δ for y = x + k (k = 0. between ages y and y + 1. .5. if i is small we have i δ 1 1+ i 2 and (1 + i) 2 1 1 1+ i 2 so we sometimes find the following approximations (which are suitable only when i is small): 1 ¯ Ax = (1 + i)Ax 2 1 ¯ (3. . ASSURANCES PAYABLE AT THE END OF THE YEAR OF DEATH. By the mathematics of finance. 2. .12) Cx = (1 + i)Cx 2 1 ¯ Mx = (1 + i)Mx 2 . ly+t µy+t = dy Therefore ¯ Cy = 0 1 (0 ≤ t < 1) v y+t ly+t µy+t dt 1 0 = dy v y = dy v y v t dt 1−v δ i = dy v y+1 δ i = Cy δ Note. If there is a uniform distribution of deaths between the ages x + k and x + k + 1 (for k = 0.3.5. . we have i ¯ Ax = Ax (3. 6.1) This may be evaluated by numerical integration or by commutation functions. we find that (as one would expect from general reasoning) ¯ A(m) −→ Ax x Corresponding to formula 3. the sum assured is payable at the end of the month of death. we have the useful approximation Ax ¯ v 1/(2m) Ax (3. ASSURANCES 3. . which we usually write as n years. Suppose that £1 is payable at the end of the 1/m year (measured from the issue date) following the death of (x).6. Thus ¯1 Ax:n = n 0 v t t px µx+t dt (3. on the death of (x) within the term of the policy. .7.3) (3. i. The position is illustrated in Figure 3. The function Ax is not often used in practice.6.time (years) Figure 3. say. for example. n−1 ¯ Cx+t ¯1 Ax:n = t=0 Dx = ¯ ¯ Mx − Mx+n Dx (3. in which ∗ indicates the time when (x) dies.7. .2) where the factor of v 1/(2m) allows for interest for 1/(2m) year.1: The mean present value of this benefit is ∞ A(m) = x r=0 v r+1 m r 1 m|m qx (3. r m .7 .44 CHAPTER 3. the “1” indicating that the status (x) must fail before the status n (a fixed period of n years) for a payment to be due (and that the payment is due when this event occurs). . which is the average “delay” in (m) receiving death benefit. . 3.6. if m = 12 .5.7 Temporary and deferred assurances A term (or temporary) assurance contract pays a sum assured of £S. If the benefit is payable immediately on death. the present value (as a random variable) is Z = g(T ) = Sv T 0 if T ≤ n if T > n ¯1 The mean of this variable is written as S Ax:n .1) As m → ∞ . death occurs here | 0 | 1 m payment is made here J J ^ | (r+1) m | 2 m | .6. .e.1 below. . .2) .6 Assurances payable at the end of the 1/m of a year of death. 8) ¯ Cx+t = = t=n Dx ¯ Mx+n Dx (3.7. The corresponding M.6) (3.7.7. n−1 Cx+t 1 Ax:n = t=0 Dx = Mx − Mx+n Dx (3. If the benefit is payable immediately on death of (x) after the deferred period has elapsed.) In terms of commutation functions.’s are written as n | Ax ¯ = v n n px Ax+n ¯ = v n n px Ax+n (3.7.7. the present value is Z = g(T ) = Sv T 0 if T > n if T < n if the sum assured is payable immediately on death.9) .5) A deferred assurance provides a sum of £S (say) on the death of (x) if this occurs after a certain period. TEMPORARY AND DEFERRED ASSURANCES 45 (provided that n is an integer) If the sum assured is payable at the end of year of death (if this occurs within the n year term).3) (It is assumed that n is an integer. the period of deferment).7. we have the approximations ¯1 Ax:n 1 (1 + i) 2 Ax:n 1 i 1 A δ x :n (3. called the deferred period (or.V. we have the present value Sv K+1 if K < n Z = g(K) = 0 if K ≥ n 1 The M. more correctly.7) and n |Ax We have the formulae n | Ax ¯ ∞ = n ∞ v t t px µx+t dt (3.4) ¯ In view of the relationship between Cx and Cx . or Z = g(K) = Sv K+1 0 if K ≥ n if K < n if the sum assured is payable at the end of year of death. of this benefit is written as S Ax:n .V.7.3.P.7. and we have n−1 1 Ax:n = t=0 v t+1 t |qx (3.P. which we usually write as n years. we see that the M.7.7. the value is obtained by multiplying the M.7. n Ex 1 ¯ 1 = Ax:n = Ax:n = v n n px In terms of commutation functions we have the useful result that n Ex = Dx+n Dx (3.12) n |Ax δ The evaluation of term and deferred assurance functions is often simplified by the observation that a whole life policy may be thought of as a combination of a term assurance and a deferred assurance.14) 3.e.8 Pure endowments and endowment assurances We now consider a pure endowment of £1 of term n years.13) (3.V. i. is written as n Ex 1 or Ax:n vn 0 if T ≥ n if T < n ¯ 1 or Ax:n (3. This M. a policy providing the sum of £1 at time n years if (x) is then alive. is v n n px (since P r{T ≥ n} = n px ). of a deferred assurance may be ¯ written as the product of n Ex and Ax+n or Ax+n .6) and (3.P. of the assurance at the “vesting date” (when the life attains age x + n) by a pure endowment factor.7.V.P.8.2) Returning to the formulae (3. In this policy.P.P. we have the approximations: i (3. The present value of this benefit is Z= and hence the M.e.46 and ∞ n |Ax CHAPTER 3. ASSURANCES = t=n ∞ v t+1 t| qx Cx+t (3.7. We finally consider an endowment assurance of term n years with sum assured £S.7. so that n | Ax ¯ (1 + i) 2 n |Ax 1 and Ax ¯ Ax = = 1 Ax:n + 1 ¯ Ax:n + n |Ax n | Ax ¯ (3. the sum assured is payable if (x) dies within n years or on maturity of the contract at time .V.7) .7.10) = = t=n Dx Mx+n Dx (3.V. That is.1) i.11) ¯ In view of the relationship between Cx and Cx .8. a. For example.8.8.8.4) Similarly. C[x]+t = v x+t+1 d[x]+t Example 3. the present value of the benefits is Z = g(T ) = ¯ The mean of Z is written S Ax:n . Thus 1 1 Ax:n = Ax:n + Ax:n 1 = Ax:n + n Ex Mx − Mx+n + Dx+n = Dx Sv K+1 Sv n if K < n if K ≥ n (3. if the death benefit is payable at the end of the year of death.8.5) (3. On the basis of A1967-70 select mortality and 4% p. and in compound interest functions.8) Select tables The adjustments to the formulae when a select table is used are straightforward: one need only replace ‘x + t’ by ‘[x] + t’ (or ‘[x]’ if the life assured has just been selected) and dispense with [ ] if the duration since selection is at least equal to the select period.V.8. It follows that an endowment assurance is a combination of a term assurance and a pure endowment (of the same term). interest.6) In view of the fact that the term Dx+n occurs in formulae (3.6). is written as S Ax:n . calculate the mean present value of each of the following assurance benefits for a life aged 30: . If we assume that the death benefit is payable immediately on death.8.4) and (3.3. the present value is Z = g(K) = and the M. if death occurs within n years.8.7) (3.1.8. PURE ENDOWMENTS AND ENDOWMENT ASSURANCES 47 n years.8.3) Sv T Sv n if T < n if T ≥ n (3. it is not correct to write ¯ Ax:n (1 + i) 2 Ax:n 1 A COMMON MISTAKE Correct relationships are (for example): ¯ Ax:n and ¯ Ax:n (1 + i) 2 (Mx − Mx+n ) + Dx+n Dx 1 1 1 (1 + i) 2 Ax:n + Ax:n 1 (3.P.Thus ¯ Ax:n = = = ¯1 ¯ 1 Ax:n + Ax:n ¯1 Ax:n + n Ex ¯ ¯ Mx − Mx+n + Dx+n Dx (3. whichever occurs first.8. s years. 9 Varying assurances Suppose that a contract provides the sum of β(t) immediately on the death of (x) at time t years. (i) ¯ 10.48 (i) (ii) (iii) CHAPTER 3.56 (iv) 100. 012.V.04) 2 × 1.2) . 070.9. 000. A 20-year endowment assurance for £50. 360.87 1 (ii) 50.1) ¯¯ If β(t) = t for all t > 0 . the M.90 + 50. 934. payable immediately on death. 000 (M40 − M50 ) = £1. with the death benefit payable immediately on death. 000.04) 2 A[30] = £1. if death occurs between ages 40 and 50 exactly. i. ASSURANCES A whole life assurance for £10.90 D[30] (iii) (1. A 20-year term assurance for £50.P. 000A[30] = 10.9. payable at the end of the year of death. 000. 012.91 D[30] 3. 000.P. ¯¯ (I A)x = 0 ∞ tv t t px µx+t dt (3. 000(1. 1 (iv) [Use the factor (1 + i) 2 for accelerating payments from end of year of death to the moment of death. is written as (I A)x .e. payable at the end of the year of death. 000 (M[30] − M50 ) = £1. A deferred temporary assurance for £100. is ∞ 0 (T > 0) v t β(t)t px µx+t dt (3.] Solution. The present value of this benefit is Z = g(T ) = β(T ) v T and hence the M. 000 1 D50 D[30] = £23.V. 9.5) (I A)x − Ax 2 We now consider the case when a death benefit of β(k) in year k is payable at the end of the year of death. . .9) . is written as (I A)x . .3. is ∞ (K = 0.V. .P.7) In terms of commutation functions we have ∞ (k + 1)Cx+k (IA)x = k=0 Dx Mx + Mx+1 + · · · = Dx Rx = Dx (3.9.9.V.3) ¯ Rx = k=0 ¯ Mx+k 1 2 (3. 1.8) where ∞ Rx = k=0 Mx+k (3. .9. the M.9. . is written (IA)x .) β(k + 1)v k+1 k |qx k=0 (3.V.9.4) less when β(t) = t than in the present case.9. where [t] indicates the integer part of t. and it may easily be shown that ∞ ¯ (t + 1)Cx+t ¯ (I A)x = = = = where ∞ t=0 Dx ¯ ¯ ¯ ¯ Cx + Cx+1 + · · · + Cx+1 + Cx+2 + · · · + · · · ¯ ¯ Mx + Mx+1 + · · · Dx ¯ Rx Dx Dx (3. the M. 2.P.9. . 2.P. VARYING ASSURANCES 49 ¯ If β(t) = [t] + 1 . so we have ∞ (IA)x = k=0 (k + 1)v k+1 k |qx (3. The present value is Z = g(K) = β(K + 1) v K+1 and hence the M. we have In view of the fact that the average benefit is the approximation ¯¯ (I A)x 1¯ ¯ (3.6) When β(k) = k for k = 1. In with profits contracts. the usual system for sharing profits is by means of additions to the basic benefit (sum assured.) in the form of bonuses.9. . and (3) the two-tier (or supercompound) bonus system.K. Nowadays one may also encounter unitised with profit policies.W..11) We may also define similar symbols for increasing temporary and endowment policies. These are not the same as unit-linked policies: the maturity proceeds of the latter are linked to the performance of a unit trust or internal fund of the life office.P.g. In this book we shall study reversionary bonuses only. etc. while for U.. ASSURANCES i Rx δ (1 + i) 2 Rx 1 (3. These may be either reversionary bonuses.10 Valuing the benefits under with profits policies So far we have studied without profits contracts. pension. policies the office credits each policy with internal units which are subject to certain guarantees. for which the premiums and benefits are completely fixed (in money terms). There are 3 systems of adding bonuses in common use: (1) the simple bonus system. e. or terminal bonuses.12) = and Rx − Rx+n − nMx+n Dx 1 (IA)x:n = (IA)x:n + n · n Ex (3. which are usually added annually and are not subsequently reduced. we find that ¯ Rx and ¯ (I A)x i (IA)x δ (1 + i) 2 (IA)x 1 1 CHAPTER 3.10) (3. the premiums are higher but the policyholder has the right to a share of the profits of the office. which are paid only on death or maturity (and possibly on surrender) and are not guaranteed to continue. n−1 1 (IA)x:n = k=0 n−1 (k + 1)v k+1 k |qx (k + 1)Cx+k = = k=0 Dx [Cx + 2Cx+1 + · · · ] − [Cx+n + 2Cx+n+1 + · · · ] − n [Cx+n + Cx+n+1 + · · · ] Dx (3. In the U.9.13) 3.9. (2) the compound bonus system. etc.50 ¯ In view of the approximations Cx (1 + i) 2 Cx .9. the benefits on maturity are normally equal to those on death in the final year. the formula used to calculate new bonus addition is new bonus = S × rate of bonus p.3.a. the benefits will be in year 1 in year 2 (3. Consider a whole life with profits policy issued t years ago to a life then aged x.7) Assuming a bonus rate of b p. the formula for bonus additions is basic sum assured new bonus = plus bonuses × rate of bonus p. and suppose that a bonus is about to be added.V.4) (3) In the case of an n-year endowment assurance with profits.a. and so on.10. giving a total benefit in year k of S + B + kbS (3. we have t = 0 and B = 0.10. The total sum assured is thus S + B + bS in year 1 from the present time.10.V. bonuses are calculated only on the basic sum assured (B.).10. (2) If the benefits are payable immediately on death. the M. which we shall denote by S. Suppose that bonuses vest annually in advance. S + B + 2bS in year 2 from the present time. In this system.8) .A.S.10. (applying now and for the forseeable future) as b (say). already vesting (S + B)(1 + b) (S + B)(1 + b) and so on. the M.a.6) (2) Compound bonuses In this system. giving (S + B)(1 + b)k in year k 2 (3.3) (3. (3.1) To value the benefits we must estimate the rate of simple bonus p.10.10.10.P. is ¯ ¯ (S + B)Ax+t + bS(I A)x+t (3.10. applies now and for the forseeable future. let the bonuses already vested be B.5) (3.a. of the benefits is ∞ [S + B + (k + 1)bS] v k+1 k |qx+t k=0 =(S + B)Ax+t + bS(IA)x+t Notes. (1) At the inception of the policy.P. VALUING THE BENEFITS UNDER WITH PROFITS POLICIES 51 (1) Simple bonuses. This leads to the following formulae for the value of benefits: at end of year of death immediately on death : : (S + B)Ax+t:n−t + bS(IA)x+t:n−t ¯ ¯ (S + B)Ax+t:n−t + bS(I A)x+t:n−t (3.2) If the benefits are payable at the end of the year of death. V. (3) Supercompound (two-tier) bonuses The formula is bonus rate p.a.V. the corresponding formulae are (S + B)A∗ x+t:n−t and 1 1 (S + B) (1 + i) 2 A∗ x+t:n−t + A∗ x+t:n−t 1 1 (3.10.A.10.S.15) .12): only the term assurance part is “accelerated” (2) At inception.10. in respect of the B.10) If the benefits are payable immediately on death.a.10.A.A.10.13).10. their M. new bonus =(B.13) Notes. in respect vesting of bonuses already vesting (3.12) (3. we observe that the pure endowment part is as in formula (3. is approximately (S + B)(1 + i) 2 A∗ x+t NOT (S + B)A∗ x+t For endowment assurances.14) Let us assume that the bonus rate p.S.10. giving the conjecture B(k + 1) = B(1 + b)k+1 + aS × sk+1 b (3.11) (3. we have t = 0 and B = 0. and let us define B(0) = B. (1) In formula (3.9) i−b 1+b (3. their M. ASSURANCES If the benefits are payable at the end of the year of death.S. + bonus rate bonuses already p.) in respect of B.a. is ∞ (S + B)(1 + b)k+1 v k+1 k |qx+t k=0 ∞ =(S + B) =(S + where A∗ is at rate of interest k=0 B)A∗ x+t 1+b 1+i k+1 k |qx+t (3. We have B(1) = B(1 + b) + aS B(2) = B(1 + b)2 + aS[1 + (1 + b)] and so on.10. Let B(k) = total bonus in year k (from the present time).10.P. and previous bonus additions are a and b respectively (now and for the forseeable future).P.52 CHAPTER 3. the M.V. (3) The “ordinary” compound bonus system is a special case of the two-tier system with a = b. the M. We have B(k + 2) = [aS + bB(k + 1)] + B(k + 1) = (1 + b) (1 + b)k+1 B + aS (1 + b)sk+1 b + 1 = (1 + b)k+2 B + aS sk+2 b so the result is also true for B(k + 2).P.16) Notes. we have t = 0 and B = 0.V.10.) .10. It follows that.18) (3.10.19) 1 a 1 1 S + B (1 + i) 2 A∗ x+t:n−t + A∗ x+t:n−t b (Note that in formula (3.17) b b (2) At inception of the policy.10. (4) For an endowment assurance the formulae are: at end of year of death immediately on death : S 1− : S 1− a a Ax+t:n−t + S + B A∗ x+t:n−t b b a ¯ Ax+t:n−t b + (3. of a whole life assurance is ∞ [S + B(k + 1)]v k+1 k |qx+t k=0 (assuming death benefits are payable at end of the year of death) ∞ = k=0 ∞ S + aS S 1− (1 + b)k+1 − 1 + B(1 + b)k+1 b k |qx+t a a + S + B (1 + b)k+1 b b k=0 a a =S 1 − Ax+t + S + B A∗ x+t b b = where A∗ is at rate x+t i−b 1+b k |qx+t (3.10. Now assume formula (3.10.10. VALUING THE BENEFITS UNDER WITH PROFITS POLICIES Proof by induction. as required. is approximately 1 a a ¯ Ax+t + S 1− S + B (1 + i) 2 A∗ x+t (3.19) the term assurance benefits are “accelerated” but the pure endowment parts are the same as in formula (3.18). Since B(1) = (aS + bB) + B = B(1 + b) + aS 53 the result is true for k = 0.3. in this bonus system.P.15) is true for B(k + 1). (1) If the sum assured is payable immediately on death. k being some value.10. a life assured may elect that compound reversionary bonuses should be added to the policy instead of simple reversionary bonuses.1.11. A company issues 20-year endowment assurances each with a basic sum assured of £1. if earlier. i−b 1+b . 000 to male lives aged 45.11 Guaranteed bonus policies As their name suggests. Derive an expression from which can be calculated the guaranteed rate of compound bonus the company should offer so that the value of the benefits at the outset is the same. as they provide additions to the basic benefit at rates which are fixed at the outset and do not depend on future the experience of the life office. these are not with profits contracts.25% of the current sum assured vest on payment of each annual premium. ASSURANCES 3. Solution. The sum assured and added bonuses will be payable at maturity. or at the end of the year of death. and hence b may be found. But many of the formulae used to value these benefits are the same as those we have derived in the previous section.54 CHAPTER 3.0225(IA)45:20 ] = 1000A∗ at rate i∗ = 45:20 This may be solved for i∗ . Alternatively. Example 3. Guaranteed simple reversionary bonuses at the rate of 2. Let b be the annual rate of guaranteed compound bonus. The equation of value for b is: 1000 [A45:20 + 0. at the outset of a policy. . (i) (ii) ¯ ¯ Show that E(Z) = S1 Ax1 + · · · + Sn Axn on Table B with interest at rate i p. Let Z be the total present value of a contract providing the sum of £Si immediately on death of (xi ).L.12.3.7 A life aged exactly 60 wishes to arrange for a payment to be made to a charity in 10 years’ time. interest. The n lives are subject to the same non-select mortality table.985. E. If he is still alive at that date the payment will be £1000. 4%. i = 0. and the interest is taken to be fixed at rate i p. Assuming an effective interest rate of 6% per annum and . . 4%. 4%.98. n.1 Evaluate (a) A60 and (b) A60 on the bases: (i) (ii) 3.5 A life aged 50 who is subject to the mortality of the A1967-70 Select table. Table B. interest.a. .6 What are the random variables (in terms of K = curtate future lifetime of (x)) whose means are represented by the following symbols? (i) (ii) (iii) n Ex 1 Ax:n 1 Ax:n 3. 4% p. 4% p. . p61 = 0. interest.12 . on A1967-70 Ultimate.6 . If he dies before the payment date.a. . on English Life Table No. EXERCISES 55 Exercises ¯ 3. Show that Ax = vqx + vpx Ax+1 Given that p60 = 0.3 Consider n lives now aged x1 . evaluate A61 and A60 3. regarded as a random variable.12 Males. show that n 2 ¯ ¯ Var(Z) = Sj [A∗ − (Ax )2 ] x j j j=1 where ∗ refers to an interest rate of 2i + i2 p.2 (i) (ii) A1967-70 ultimate. on A1967-70 Ultimate. 3. No. the amount given will be £500. 4%. xn respectively. Assuming further that the future lifetimes T (xi ) of the lives are independent variables.Males.T. i = 1. 3.a. . .05 and A62 = 0. (i) (ii) Write down the present value of the benefits under this contract.a. calculate the mean and the variance of the present value of the benefits available under this contract. Assuming an effective rate of interest of 5% per annum. on A1967-70 Ultimate.4 Using commutation functions or otherwise calculate the values of the following: (i) (ii) (iii) (iv) (v) A[40]:10 1 A30:20 ¯1 A30:20 ¯ A30:20 ¯ A30:20 on A1967-70. effects a pure endowment policy with a term of 20 years for a sum assured of £10. 3. 2.000. 4% p. .a.a. x2 . .58.06.8 (Difficult) You are given that (i) (ii) (iii) (iv) 1000 (IA)50 =4. 1 1000 A50:1 =5.56 CHAPTER 3. 3.996.12-Males.75 . 1000 A51 = 249. calculate the standard deviation of the present value of the liability. ASSURANCES mortality according to ELT No.05 and i =0. Calculate 1000(IA)51 . 04) 2 A60 1 = (1. we have Var(Z) = n 2 Sj var(v Tj ) j=1 n = j=1 2 ¯ ¯ Sj [A∗ j − (Axj )2 ] x 3.9552 − 1767.04) 2 0.985 · 0.55749 3.57905) 1.05 = 0.5555 = = 0.58317 from tables 1 ¯ (b) A60 = (1.3 (i) We have n Z= j=1 Sj v T j so n n E(Z) = j=1 Sj E(v Tj ) = j=1 ¯ Sj Axj (ii) By independence of Txj . SOLUTIONS 57 Solutions 3.57185 ∞ 3.2 (i) Ax = k=0 k |qx v k+1 ∞ = qx v + k=1 px ∞ k−1 |qx+1 v k+1 = qx v + px v k=0 k |qx+1 v k+1 (ii) A61 A62 = qx v + px vAx+1 1 = (0.67812 D[40] 6981.015 + 0.4 (i) (ii) M[40] − M50 + D50 1904.310 .020550 D30 10433.98 · 0.52750 = 0.8595 − 1767.51726 = 0.1 (i) (a) ¯ A60 = (1 + 0.04)− 2 A60 = 0.05 = 1 (0.3.13.6) 1.0607 = = 0.5555 + 4597.51726 from tables 1 (b) A60 (ii) ¯ (a) A60 = 0.5977 M30 − M50 1981.57905 = 0.02 + 0. )2 ] − [E(P.5 (i) P. 828.48 E[(P.33v 10 at 6% = £128.d.39 ∴ s.7510 p60 ] − (mean)2 = 770. 000.05−40 = 10.6(i) 0 v K+1 if K ≥ n if K < n vn 0 if K ≥ n if K < n 10.58 (iii) (iv) ¯1 A30:20 1 (1 + i) 2 A30:20 = 0. 000 = 10. 10 p60 500 with prob.615 . 811.V.207 Var(X) = 10002 10 p60 + 5002 (1 − 10 p60 ) − (mean)2 1 = 1. 681. 968 ⇒ V ar(P. 0002 × 1.725539 × 1.V. 000v 20 0 if T ≥ 20 if T < 20 Let X payment to be made at time 10 years.46157 D30 1 A30:20 (v) ¯1 A30:20 (1 + i) 1 2 = (1 + i)− 2 1 ¯ ¯ M30 − M50 = 0.) = (20 p[50] × v 20 + (1 − 20 p[50] ) × 0)10. of X = 230.03286 D30 3. X= 1000 with prob. 000[0. 000v 20 )2 = 0.6 (i) Z= (ii) Z= (iii) 3.)]2 = 20 p[50] (10.842 3.05−20 = 2734.)]2 = 2. 587 = 1.d.33 Present value is v 10 (X) and so has s.25 + 0. 000[10 p60 + (1 − 10 p60 )] − (mean)2 4 = 1.725539 × 10.020956 1 CHAPTER 3.207)2 = 53.V.) = E[(P.V. (1 − 10 p60 ) ∴ E(X) = 100010 p60 + 500(1 − 10 p60 ) = 500(1 + 10 p60 ) = £847. 230. ASSURANCES D50 ¯ ¯1 A30:20 = A30:20 + = 0. 000 × 0. 305. (ii) E(P.V.7 Same as 3.1 − (847. = where T = future lifetime of (50).V. 000. 051. 07305 ∴ 1. 073 . 000(IA)51 = 5.24905 + (IA)51 ] ∴ (IA)51 = 5.93782 D50 ∴ 4.00558 + 0.00558 59 Now vq50 D51 ∴ = v − vq50 = 0.8 (IA)50 = C50 + 2C51 + · · · (C51 + C52 + · · · ) + (C51 + 2C52 + · · · ) 1 = A50:1 + D50 D50 D51 1 = A50:1 + (A51 + (IA)51 ) D50 1 = A50:1 = 0.93782[0. SOLUTIONS 3.99675 = 0.3.13. 60 CHAPTER 3. ASSURANCES . i.e. expressed as a random variable.1. if (x) is then alive. is g(T ) = aT for T > 0 ¯ This variable has mean ax .1) ax = E(¯T ) = ¯ a An alternative definition is: 0 at t px µx+t dt ¯ (4. ¯ ∞ (4.Chapter 4 ANNUITIES 4. as follows: ∞ 0 at ¯ t px µx+t dt t at = ¯ So v r dr 0 d d (−t px ) = (t qx ) dt dt = t px µx+t d (¯ ) = v t a dt t 61 .1. i.2) ∞ ax = ¯ 0 vt interest (or discounting) factor t px dt (4.e. The present value of this benefit.3) probability of survival of (x) for t years To show that these definitions are equivalent.a.1. the rate of payment at time t years is £1 p. for life. we may use integration by parts.1 Annuities payable continuously Let (x) be entitled to £1 p. payable “continuously”.a. which gives ¯ 1 = δ¯x + Ax a The variance of the present value of an annuity. Express Var(¯T ) in terms of ax and a∗ .1. Var(¯T ) = a = = 1 − 2δ(¯∗ ) − (1 − δ¯x )2 ax a δ2 1 − 2δ(¯∗ ) − 1 − δ 2 (¯x )2 + 2δ¯x ax a a 2 δ 1 ∗ 2(¯x − ax ) − δ(¯x )2 a ¯ a δ ¯ A∗ = 1 − (2δ)¯∗ since force of interest is 2δ ax x . the corresponding force of interest is δ ∗ = 2δ.e.1. δ We may also argue as follows.v. by formula (4.1.4) which follows from: ∞ ax ¯ = 0 ∞ at t px µx+t dt ¯ 1 − vt δ t px µx+t = 0 dt 1 ¯ (1 − Ax ). a formula connecting assurance and annuity values): ¯ Ax = 1 − δ¯x a (4.5) where ∗ indicates “at the rate of interest 2i + i2 p.a.62 CHAPTER 4. Since aT = ¯ we have Var(¯T ) = a = 1 Var(v T ) δ2 ¯x ¯x A∗ − A2 δ2 1 − vT δ (4. = 1 = δ¯T a + vT (for all T > 0) Capital p. so. By the Mathematics of Finance. of p. ¯ ¯x so some writers use 2 Ax rather than A∗ .”. Example 4.5).1.v.1. a ¯ ¯x Solution. invested return of capital of interest Take expected values of each side. ANNUITIES ∞ 0 at t px µx+t dt = at (−t px ) ¯ ¯ ∞ ∞ 0 ∞ + 0 v t t px dt = 0 v t t px dt Note the following conversion relationship (i. the p. Note that ¨ ax = 1 + ax ¨ VITAL! (4. ) d .2. Show that ax = ¯ Solution. (The former is called an annuity-due.2. so ∞ 0 ¯ Nx = from which we obtain ¯ Nx = Dx v x+r lx+r dr ∞ 0 v t t px dt = ax . ¯ 4. 2 . .4. is payable (i) annually in advance. .8) v x+t lx+t dt ¯ Dx+t Example 4. 1.4) But aK+1 = ¨ 1 − v K+1 (K = 0.2. . The means are denoted by ax and ax respectively.1. ¯ Nx = ∞ ¯ Nx Dx ¯ Dx+t (4.1. . ANNUITIES PAYABLE ANNUALLY Commutation Functions.3) (since the annuitant gets extra immediate payment of 1 in case (i)).7) (4.2) (4.) In case (i).1. Now ax =E aK+1 ¨ ¨ ∞ = k=0 ak+1 k |qx ¨ (4. it is g(K) = aK = aK+1 − 1 ¨ where we take a0 = 0.. Define Dx = v x lx ¯ Dx = ¯ Nx = t=0 1 0 ∞ 63 (as stated in Chapter 3) (4.1. (ii) annually in arrear.2.6) (4.a.9) t=0 ¯ but (on change of variable) Dx+t = t t+1 v x+r lx+r dr for t = 0. of the benefit is g(K) = aK+1 ¨ (4. 2.1) In case (ii). 1. .1.2.2 Annuities payable annually Now consider the cases when an annuity of £1 p.2.v. ) + v 2 (dx+2 + .7) (4.2. lx ∞ v t t px . . t=0 as required. lx (dx + dx+1 + dx+2 + . v t t px (2) By the definition. . ANNUITIES so ax =E ¨ 1 − v K+1 1 − E(v K+1 ) = d d 1 − Ax = d (4.7).8) since ax = ax − 1 then gives (4.1. .64 CHAPTER 4... ) + . .2.2. . ∞ ax ¨ = k=0 ak+1 k |qx ¨ a1 ¨ dx lx + a2 ¨ dx+1 lx + a3 ¨ dx+2 lx + . . = = = = = 1 dx + (1 + v)dx+1 + (1 + v + v 2 )dx+2 + . ) + v(dx+1 + dx+2 + . lx lx + vlx+1 + v 2 lx+2 + .8) = It is sufficient to prove (4. . . .. We note the following results: ¨ ∞ (4.2.5) We thus have the conversion relationship Ax = 1 − d¨x a Important formulae for ax and ax . . .2.. . ¨ Two Proofs (1) Regard the annuity as sum of pure endowments due at times 0. . Thus ax ¨ = = t=0 0 Ex ∞ + 1 Ex + 2 Ex .2.6) ax ax ¨ = t=1 ∞ v t t px v t t px t=0 (4. Evaluation of ax by Commutation Functions Define ¨ ∞ Nx = t=0 Dx+t . .2. . . .4. If n is an integer.10) 4. THERE ARE NO SUCH THINGS AS a 1 ¯ Commutation Functions. ¨ We use the formula Var aK+1 ¨ = Var = 1 − v K+1 d 1 Var v K+1 d2 A∗ − (Ax )2 = x 2 d where ∗ indicates the rate of interest i∗ = 2i + i2 p. (that is. .) The p. ) Dx The variance of aK+1 .2) = 0 by integration by parts.n} ¯ if T ≥ n (4.3. whichever is earlier. payable continuously for at most n years. and suppose that (x) is entitled to £1 p. use ax = ax − 1 or ¨ ax = Nx+1 (since Nx+1 = Dx+1 + Dx+2 + . . .3.2. v x lx Dx + Dx+1 + .1) ax:n = ¯ 0 n at t px µx+t dt + an ¯ ¯ v t t px dt t px µx+t n dt (4.2.v. (Check this!) Note. . is g(T ) = The mean of g(T ) is n ∞ aT ¯ an ¯ if T < n = amin{T.a. Dx Nx Dx 65 Remark If we require ax .a.9) (4.3 Temporary annuities Let us first consider the “continuous payments” case. .3. ax:n = ¯ x:n and a ¯ x:n 1 !! n 0 ¯ ¯ v x+t lx+t dt Nx − Nx+n = xl v x Dx (4.3) . payments cease when x dies or after n years. TEMPORARY ANNUITIES We obtain ax ¨ = = = v x lx + v x+1 lx+1 + .3. Note For an annuity payable in arrear. we use the result that Var (aK ) = Var aK+1 − 1 ¨ = Var aK+1 ¨ (4. 8) Dx+1 + Dx+2 + · · · + Dx+n Nx+1 − Nx+n+1 = Dx Dx (4. the sum of the values of n pure endowments each for £1) = Hence ax:n = ¨ Nx − Nx+n Dx (4. prove the conversion relationships (i) (ii) Ax:n = 1 − d¨x:n .3. we obtain ax:n = mean of ¨ n−1 aK+1 ¨ an ¨ if K < n = E amin{K+1.” Example 4.6) (b) ax:n and a[x]:n are tabulated for certain values of x+n in the A1967-70 section of “Formulae ¨ ¨ and Tables.4) = k=0 ak+1 k |qx + an n px ¨ ¨ which may be shown to be equal to 1 + vpx + v 2 2 px + · · · + v n−1 n−1 px (i.7) aK an if K < n = E amin{K. By expressing Ax:n and ax:n as expectations of appropriate random variables.1. and are limited to at most n payments.n} if K ≥ n (4.5) Dx + Dx+1 + · · · + Dx+n−1 Dx If the payments are made annually in arrear. ANNUITIES If payments are made annually in advance.3.3. we obtain ax:n = mean of leading to ax:n = Note that (a) ax:n = 1 + ax:n−1 ¨ (4.3.. or ¨ otherwise.e. and a ¯ Ax:n = 1 − δ¯x:n .n} ¨ if K ≥ n (4. a .66 CHAPTER 4.3.3. 4.V. Select Tables The modifications needed for select tables are straightforward. Thus a m |¯x = M. Note that ax = ax:n + n |¨x ¨ ¨ a and a m |¨x = Dx+m Dx Nx+m Dx+m = Nx+m Dx (4. beginning in m years’ time It is often best to evaluate m |¯x by the formula a a m |¯x = Dx+m Dx pure endowment factor Similarly.4. of annuity of 1 p. .4.3) × ax+m ¯ (4.4) (4.n} .4 Deferred annuities These are annuities which commence in m (say) years’ time.B.n} = ¯ 1 − v min{T.4.1 below.4. and are illustrated in Table 4.1) tinuously.4.5) = Dx+m Dx ax+m ¨ (4. to (x). δ 1 − v min{K+1. a m |¨x and so on. DEFERRED ANNUITIES Proof.n} . payable con(4.n} = ¨ Take expected values to obtain ax:n = ¨ which gives the required result. amin{K+1. in Finance.a. Pensions are (essentially) deferred annuities. (ii) Take expected values in the equation amin{T.P. d 1 − Ax:n d 67 4.4.4. provided that the annuitant is then active. (i) By Maths.2) annuity factor at age x + m N. Select Mortality Table (select period 2 years) 4. Use the Euler-Maclaurin formula to deduce an approximate formula for ax in terms ¯ of ax ¨ . ANNUITIES Type of Annuity Symbol In terms of t p[x] ∞ In terms of Commutation Functions N[x]+1 D[x] N[x] D[x] N[x]+1 − Nx+n+1 D[x] N[x] − Nx+n D[x] Nx+m+1 D[x] Nx+m D[x] (n ≥ 1) Immediate a[x] t=1 ∞ v t t p[x] v t t p[x] t=0 n Annuity-due a[x] ¨ Temporary (n years) Temporary annuity-due (n years’ payments) a[x]:n t=1 n−1 v t t p[x] a[x]:n ¨ t=0 v t t p[x] ∞ (n ≥ 2) Deferred (m years) Deferred annuity-due (m years) m |a[x] t=m+1 ∞ v t t p[x] v t t p[x] t=m (m ≥ 1) a m |¨[x] (m ≥ 2) Table 4.5.1) Woolhouse’s formula may be then deduced: 1 m ∞ f t=0 t m ∞ f (t) − t=0 m−1 2m f (0) + m2 − 1 12m2 f (0) (4.5.) Example 4.5.1. Life aged x.68 CHAPTER 4.4.5 Annuities payable m times per annum We require the following formula from numerical analysis: The Euler-Maclaurin formula: ∞ f (t) dt 0 1 1 f (t) − f (0) + f (0) 2 12 t=0 ∞ (4.2) (We have assumed that f (t) → 0 and f (t) → 0 as t → ∞.1: Expression for Values of Single-Life Curtate Annuities. By E. giving ex (m) (m) ◦ ◦ ex + 1 1 − µx 2 12 1 2 (4.5. ∴ f (0) = 1 and f (0) = −(δ + µx ). Find an approximate formula for ax ¨ (m) in terms of ax . Let f (t) = v t t px = exp − f (t) = −(δ + µx+t ) exp − ax ¯ t (δ 0 t (δ 0 69 + µx+r ) dr . Set i = 0 in example 4.10) ax − ¨ m−1 2m − m2 − 1 12m2 (µx + δ) (4. giving ax − ¨ ax ¯ ax − ¨ 1 1 = ax + 2 2 ◦ (4.5.5.11) .5.4.5.1. Solution. Apply Woolhouse’s formula to f (t) = v t t px . (4.8) Example 4.6) The symbols ax and ax refer to the expected present values of an annuity of 1 per annum payable ¨ monthly in advance and monthly in arrear respectively.5) ex + (4.5.4) Example 4.9) − 1 (ignoring the final term) m (4.2.5. ¨ Solution.5. a(m) = a(m) − ¨x x = ax − ¨ = ax + 1 m m−1 2m m−1 2m ax − ¨ m−1 2m (4. Hence + µx+r ) dr .7) = t=0 t 1 m v t px m m (4.5. This gives a(m) ¨x In practice one usually uses a(m) ¨x Note. Find an approximate formula for ex in terms of ex .5.3) 1 1 1 1 − (µx + δ) or ax + − (µx + δ) 2 12 2 12 In practice the final term is usually ignored.5.5. ANNUITIES PAYABLE M TIMES PER ANNUM Solution. Thus ∞ a(m) ¨x and a(m) x = t=0 ∞ t 1 m v t px m m (4.5.-M. This gives ex The final term is usually omitted.3. 1. .14) 4. where t = 0. .a. If we have U. on death at time 1 This final payment varies from 0 to m and hence is on average about “complete” or “with final proportion”. i. monthly in arrear with a final payment immediately on death at time r since the last full payment. etc.6.D.P.12) (m) m−1 2m (4.V.P. x + 1 to x + 2. of D in each of the age-ranges x to x + 1.13) 1− (4.6. of £1 p. and 0 ≤ r < m 1 2m .5.2) More accurate formulae for ax .e. 2. we have the exact formulae ax = ax + ◦ i−δ δ2 Ax = ax + i−δ iδ ¯ Ax (4. . payable mthly in advance to (x) ¨ for at most n years = a(m) − n |¨(m) ¨x ax Dx+n (m) a ¨ = a(m) − ¨x Dx x+n m−1 Dx+n ax − ¨ − ax+n − ¨ 2m Dx m−1 Dx+n = ax:n − ¨ × 1− 2m Dx CARE! DO NOT OMIT THIS TERM We also have a(m)n = ax:n + x: m−1 2m Dx+n Dx (4.a. giving ax ◦ ◦ ◦ (m) ◦ 1¯ ax + Ax 2 (4. ANNUITIES Temporary mthly annuities Define ax:n = M.5.5.. 1 m Suppose that (x) is to receive 1 p. The final payment is shown below: Hence (using the formula for a varying assurance) .3) Proof.6 Complete annuities (or “annuities with final proportion”) t + r.70 CHAPTER 4. and we define This annuity is called ax ◦ (m) = M.V.1) When m = 1 we may write ax =ax .6. of complete annuity 1 ¯ a(m) + Ax x 2m (4. ) 4.. The present value of these payments (as a random variable depending on the future lifetime T of (x)) is (by the Mathematics of Finance) T g(T ) = 0 v t b(t) dt (4.2) .D. . firstly..1) The mean present value is thus ∞ E[g(T )] = 0 g(t)t px µx+t dt (4.7.D.3) follows from Ax = δ Ax (which holds by U.7 Varying annuities Suppose.4.7. that payments are made continuously so long as (x) survives at the rate b(t) p. VARYING ANNUITIES final payment (on death of (x) ) 6 1 .a. at time t.7. 1 2 3 4 . .6.. ax = ax + t=0 ◦ ∞ v t t px qx+t 0 1 rv r dr ¯a = (I¯)1 = ∞ a1 −v ¯ δ = ax + t=1 v t+1 t |qx (1 + i) s1 − 1 ¯ δ i−δ δ2 a1 − v ¯ δ (1 + i) − 1 − 1 /δ δ = ax + Ax = ax + Ax = ax + Ax i ¯ The second part of (4.time (years) ax = ax + t=0 ◦ ∞ v t t px 0 1 rv r r px+t µx+t+r dr value at time t of death benefit in year t + 1 pure endowment factor to age x + t By U. of D.D. 71 0 . ¯a Find an expression for (I¯)x in terms of an integral.V. ANNUITIES Using the following important formula (obtained from integration by parts): ∞ t ∞ ∞ u(t) 0 0 v(r) dr dt = 0 v(t) t u(r) dr dt (∗) we obtain the alternative expression ∞ E[g(T )] = 0 vt interest factor b(t) rate of payment at time t t px dt (4. of an increasing annuity to (x) in which the rate of payment at time t is t. t ≥ 0. of annuity to (x) with payment at rate t p.V. ¯a (I¯)x 1 1 tv t t px − f (0) + f (0) 2 12 t=0 ∞ d But f (0) = 0 and f (0) = [v t t px + t dt (v t t px )]t=0 = 1. let and v(t) = v t b(t) u(t) = t px µx+t so that t ∞ u(r) dr = t px then use formula (∗) ¯a Example 4. . so ∞ ¯a (I¯)x t=1 tv t t px + = (Ia)x + 1 12 (4. ¯a Solution.4) An approximation.72 CHAPTER 4..P.. By Euler-Maclaurin.7. Let f (t) = tv t t px .5) 1 (see definition of (Ia)x below) 12 Now suppose that 1 in year 1 b(t) = [t] + 1 = 2 in year 2 . Let b(t) = t. This gives g(T ) = (I¯)T .7.7.3) probability of survival To prove this. (t ≥ 0).P. at time t ∞ = 0 tv t t px dt (4. and we have ¯a (I¯)x = M.7. Define (I¯)x = M.a.1. 8) since the last payment is made at time K. 1. (If b(0) > 0. etc.V. } lx 1 = {g(0)[dx + dx+1 + . lx = ∞ E[g(K)] = t=0 b(t) vt t px (4.6) ¯ Sx = t=0 ¯ Nx+t (4.P. . the benefit at time t years being b(t) (t = 0.7) Now suppose that benefits are paid annually so long as (x) survives. . . . . . of the varying annuity is thus ∞ E[g(K)] = k=0 g(k)k |qx 1 {g(0)dx + g(1)dx+1 + . . . . . .9) benefit interest at time t factor survival factor This formula forms the basis of “spreadsheet” calculations for pension schemes. ¯ a a ¯x + Nx+1 + .7. [This may also be found by summing the pure endowments] . .7.P. The present value of the benefits (regarded as a random variable) is K g(K) = t=0 v t b(t) (4.7. VARYING ANNUITIES The M.7. 2.V.4. ] +[g(2) − g(1)][dx+2 + dx+3 + . ] + . .7. The M. } 1 = b(0)lx + vb(1)lx+1 + v 2 b(2)lx+2 + . ¯ N = Dx ¯ Sx = Dx where ∞ (4. this is a variable annuity-due). . ). . . . . ] lx + [g(1) − g(0)][dx+1 + dx+2 + . . of the corresponding annuity is written (I¯)x . Thus a ∞ 73 (I¯)x = a 0 ([t] + 1)v t t px dt ∞ ∞ ∞ = 0 + 1 + 2 v t t px dt = ax + 1 |¯x + 2 |¯x + . . ANNUITIES Applications Let b(t) = t + 1 (t = 0. . 1.7.7. . ) + (Dx+1 + Dx+2 + .2. ∞ ∞ (Ia)x = t=1 tv t px = t=1 t t Dx+t Dx (4. .P.7. Dx (Dx + Dx+1 + .V. . . .V.11) Similarly. . . .12) Sx+1 = Dx Example 4. when b(t) = t (t = 1. Show that (Ia)x = (I¨)x − ax a ¨ Solution. . is defined as (I¨)x . ) + . . 2. .P. . .74 CHAPTER 4. ) we get the M.7. so we have a ∞ (I¨)x = a = (t + 1)v t t px t=0 ∞ (t + 1)Dx+t /Dx t=0 = Dx + 2Dx+1 + . . 2. The M. ). ∞ (Ia)x = t=0 ∞ tv t t px (t + 1 − 1)v t t px t=0 ∞ ∞ = = t=0 (t + 1)v t t px − t=0 v t t px =(I¨)x − ax a ¨ Temporary increasing annuities .10) where ∞ Sx = t=0 Nx+t (4. = Dx Sx = Dx (4. = Dx Nx + Nx+1 + . ) − Dx Sx+1 − Sx+n+1 − nNx+n+1 = Dx n x:n Also.7. . . . .V. viz a Sx+m − Sx+m+n − nNx+m+n Dx ¯¯ + IA (1) (2) (3) m| (I¨)x:n = a Conversion relationships for increasing annuities and assurances ¯a ax = δ I¯ ¯ x x ax = d (I¨)x + (IA)x ¨ a ¯ ax = δ (I¯) + I A ¨ a x x Proof To show (1). 2.7. . . ) (Dx+n+1 + 2Dx+n+2 + . . . (Ia)x:n = Dx+1 + 2Dx+2 + · · · + nDx+n Dx (Dx+1 + 2Dx+2 + . .P. .7. take expected values on each side of the maths of finance formula ¯a aT = δ I¯ ¯ T + T vT .13) Similarly. . of payments of t + 1 at time t a (t = 0. . ) = − Dx Dx (nDx+n + nDx+n+1 + . VARYING ANNUITIES Define (I¨)x:n = M. ) − Dx Sx − Sx+n − nNx+n = Dx (4. ) − = Dx Dx (nDx+n+1 + nDx+n+2 + . 1. ) (Dx+n + 2Dx+n+1 + .14) (I¯)x:n = a Note that (for example) m| ¯ ¯ ¯ Sx − Sx+n − nNx+n Dx Dx+m Dx (I¨)x+m:n a (I¨)x:n = a which enables us to express m| (I¨)x:n in terms of commutation functions. n − 1). . . provided that (x) is then alive n−1 n−1 75 = t=0 (t + 1)v t t px = t=0 (t + 1)Dx+t /Dx = Dx + 2Dx+1 + · · · + (n − 1)Dx+n−2 + nDx+n−1 Dx (Dx + 2Dx+1 + . . ¯a I¯ = 0 tv t t px dt (4.4. . 0 ≤ r < 1) and observing that n+r a (I¯)t = (I¯)n + a v s (n + 1) ds 1 − vr δ n an − nv n ¨ + (n + 1)v n = δ a ¨ − (n + 1)v t = n+1 δ Note. . . in year 2. 3. t = 0 v r ([r] + 1) dr (∗∗) holds for t = 0.a. .6. . 2 p. ¯a ¯¯ ax:n = δ(I¯)x:n + (I A)x:n ¯ .6.a. we need to use the formula (I¯)t = a where a[t]+1 − ([t] + 1)v t ¨ δ for all t ≥ 0 CHAPTER 4. take expected values on each side of aK+1 = d (I¨)K+1 + (K + 1)v K+1 ¨ a To show (3). 2. formula 3. in year a 1. There are also conversion relationships for temporary increasing annuities.76 To show (2). It may be proved for general t ≥ 0 by letting t = n + r ( n integer.v. of an annuity-certain payable continuously at rate 1 p. by McCutcheon and Scott. ANNUITIES (∗∗) (I¯)t = p.g. e. . . . ceasing at time t exactly. 1. 4 Using the A1967-70 table with 4% p.5 (i) (ii) (iii) Find the present value of a deferred annuity of £1000 p. a40:30 .739 lx+1 90.509. ¨ ¨ ¨ a a 4. and ax+1 . EXERCISES 77 Exercises 4. a[39]+1:30 . which has a select period of 1 year and is such that q[x] = 0.6qx for each x.a.6 According to a certain mortality table. . and. p[x] and ax+1 .a.a. a71 at 10% interest = 5.094 85. 4. interest a45 = 15. 5 |(Ia)40:25 4. to a man aged 40.) As above. if he is then alive.132 88. On a certain select mortality table the select period is one year.2 Given that ax = 20.641. and continue thereafter annually for life. a70 at 10% interest = 5. Payments commence on his 60th birthday. Find a[70] at 10% interest.a.874 84.3 (i) (ii) Write down an expression for ax in terms of v. find the values of n Ex and ax:n . interest. Age x 55 56 l[x] 90.636 89. interest find the values of a[40]:30 . As above.8. 4 % p. Express a[x] in terms of v.4. but A1967-70 ultimate mortality to exact age 60 and a(55) males ultimate mortality above exact age 60. px .032 89. ¨ 4. Given that q[x] = . ax:n = 18 and ax+n = 8.1 The following is an extract from a life table with a select period of 1 year. Basis: A1967-70 ultimate mortality.586 age x + 1 56 57 58 59 60 61 Evaluate a56:5 and a[56]:5 at 5 % per annum interest. 5 |¨[40]:25 .6qx and that at 4 1 % p.151 87.449. but select mortality (at entry. ¨ ¨ 4.719 and 4 a46 = 15. find the value of a[45] (at the same rate of interest). (I¨)40 . 3 (i) (ii) ax = vpx (1 + ax+1 ) Note that with a one-year select period a[45] = vp[45] a46 ¨ = (1.0073864 Then q[45] = 0.468 = 12.509p[45] .e.949 directly tabulated = 276.960 directly tabulated = 16. Now a45 = vp45 a46 ¨ i.953 = 16. We must determine p[45] from the given data.1 l56 + vl57 + · · · + v 4 l60 = 4.2 ax = ax:n + n Ex ax+n =⇒ n Ex = 1 4 ax:n = ax:n − n Ex + 1 = 18.337 5 |(Ia)40:25 . p45 = Hence q45 = 0.75 ¨ 4.7658 (1 + i)a45 15. ANNUITIES Solutions 4.509 (a) 4.451 l56 l[56] + vl57 + · · · + v 4 l60 = = 4.719 = 1.0425)−1 16.9926136 a46 ¨ 16.0425 × = .0044318 Hence p[45] = 1 − q[45] = 0.6q45 = 0.342 = 122.4 a[40]:30 ¨ a[39]+1:30 = (N[39]+1 − N70 )/D[39]+1 ¨ a40:30 ¨ (I¨)40 = S40 /D40 a a 5 |¨[40]:25 = (N45 − N70 )/D[40] = (S46 − S71 − 25N71 )/D40 = 16.463 l[56] a56:5 = ¨ a[56]:5 ¨ 4.9955682 From (a) we then get a[45] = 15.78 CHAPTER 4. 08.6 a70 = v(1 − q70 )¨71 . 4.N60 /D40 = 5130. SOLUTIONS 79 4.625 = ¨ 5160. where the factor D60 is evaluated on A1967-70 mortality and the factor a60 D40 ∗ a60 is evaluated on a(55) males mortality. say £5130 1000.40873 × 12. say £5134 D60 1000. Hence q[70] = 0.5 (i) (ii) (iii) 1000. D40 .N60 /D[40] = 5133.022918. so 1 − q70 = 0.730 a .9. say £5160.22. and q[70] = a v(1 − q[70] )¨71 = 5. The value is thus 1000 × .68.96218.6q70 = 0.¨∗ .4. ANNUITIES .80 CHAPTER 4. V. The frequency of regular payments may be. Solution. (5. that is. we replace equation (5.2) where B = M.1.1. for example.1 Principles of premium calculations Premiums calculated without an allowance for expenses are called net premiums. Let Z = g(T ) = P. on the expiry of the term of a temporary (term) assurance policy.1) Example 5.V. and the maximum number of premiums payable may be limited to (for example) 20. Give a formula for P in terms of annuity functions. we shall assume that the equivalence principle (5. when the policy matures. A life assurance policy may be issued with (a) a single premium payable at the date of issue. although in some cases the use of “conservative” assumptions regarding mortality. until the death of (x). assuming that the equivalence principle applies. of profit on the contract.1. of course. In the absence of information to the contrary. or when there is no longer any possibility of future benefits: for example.a. interest and expenses means that the office has an implicit margin of expected profit. and in some cases for a profit to the office. of profit to office on this contract = P aT − v T ¯ 81 .Chapter 5 PREMIUMS 5.1.1) by E(Z) = B (5. Office premiums are usually calculated with an explicit allowance for expenses. whilst premiums which are actually charged are called office or gross premiums. Consider a whole life assurance of 1 payable immediately on the death of (x). Premiums should. which may be stated as follows: E(Z) = 0 where Z = present value of profit to the life office on the contract In some cases an explicit loading for profit is included in the calculation.1. and suppose that premiums are payable continuously at rate P p.P. or (b) regular premiums payable in advance and usually (but not always) of level amount. yearly or monthly.1) is to be used. Premiums are usually calculated by the equivalence principle.1. The rate of interest is usually taken to be fixed (not random). cease on the death of the assured life. however. but the standard symbols Px . The general symbols P.P. of profit to the office (if any) In example 5.V. P (Ax ) .e. at least for straightforward policies.V. When benefits are payable at end of year of death we may (optionally) shorten the notation as follows: (e.g. refer to a sum assured of £1. or equations of value. of premiums =M. Premiums are assumed to continue for as long as the contract can provide benefits.1) and (5. i. whichever occurs first) A = x:20 ax:10 ¨ 3. P may be ¯ ¯ used for any sum assured. P ) for the premium (single or regular). t P (m) ( ).g. we write t P ( ).) P (Ax ) = Px . The symbols P ( ). of expenses (if any) + M. If the policy is complicated it is best to just use the symbol P (or P . If premiums are limited to. PREMIUMS ¯ using the conversion relationship Ax = 1 − δ¯x a The equations (5. the ¯ term of the benefits). P . the equation of value is ¯ P ax = Ax ¯ (5. =1 (payable at end of year of death or on survival for 20 years.E (¯T ) = E v T a ¯ P ax = Ax ¯ ¯ Ax 1 P = = −δ ax ¯ ax ¯ CHAPTER 5.P. we have E{P aT − v T } = 0 ¯ ∴ ∴ ∴ P.1.A.2) may be expressed as equations of payments.).2 Notation for premiums The International Actuarial Notation (see Formulae and Tables for Actuarial Examinations) should be used. limited to at most 10 payments (i.3) 5. P (m) indicate that payments are made annually in advance.e. at most t years’ payments (where t < the term of contract.1. or P ( ).e. payments cease on death or after 10 payments are made. P.P. 10 P (Ax:20 ) = net annual premium for 20-year EA with S. i. continuously thly and m in advance respectively. . of benefits + M.1. n years for a n-year contract. 2.e. e. i. etc.82 Since we require that E{g(T )} = 0.P. P . provided of course that the life assured is ¯ still alive.1.V. or P (m) ( ) indicate the level net annual premium for the benefit indicated in the brackets. M. Some of the more important rules of the International Notation are the following: ¯ 1.V. t P ( ).1. THE VARIANCE OF THE PRESENT VALUE OF THE PROFIT ON A POLICY.1 relates to a mortality table with select period 2 years.3 The variance of the present value of the profit on a policy. Some offices employ a policy fee system.000 may be quoted as “£10 per £1. for example the office annual premium for a policy with a sum assured for £20.4.1. Consider n-year endowment assurance without profits with sum assured £1 issued to a select life aged x.3.) Note that g(T ) = P 1 − vT − vT δ P P +δ = − vT δ δ P +δ δ P +δ δ 2 ∴ Var[g(T )] = = Var(v T ) 2 ¯x A∗ at rate 2i + i2 ¯ − Ax 2 (5. 5. 83 Table 5. 5. There may also be expenses of payment of benefits (especially for pensions and annuities) and the expenses of maintaining records of policies with continuing benefits after premiums have ceased.1) Note the idea of “collecting” all the terms involving v T before taking the variance: a similar technique may be used for endowment assurance policies (but not generally). g(T ) = p. giving £20 × 10 + £15 = £215.4 Premiums allowing for expenses Most expenses may be classified as either: (a) initial (incurred at the outset only) .5. Example 5.v.000 plus a policy fee of £15”. . and refers to policies in which premiums are payable annually in advance. P . whereby a fixed addition of (say) £15 is added to the annual premium. or (b) renewal (incurred on the payment of later premiums).3. of profit to office on contract = P aT − v T ¯ Consider again the policy of Example 5.1. E[g(T )] = 0. Expenses may also be divided into: (i) (ii) commission payments. Expenses are e per premium payment (including the first) plus I at issue date (so the total initial expense is I + e ). We have (We need not assume the equivalence principle holds: if it does. and other costs.2. This system reflects the fact that certain administrative costs do not depend on the size of the benefit.1. for example A1967-70. Find formulae for the level office annual premium. 1: Expressions for Annual Premiums. When n < 2 or t < 2. Select Mortality Table (select period 2 years) Note.2. PREMIUMS Type of Assurance Symbol 1 A[x]:n a[x]:n ¨ Net Annual Premium payable throughout duration of contract In terms of A and a ¨ M[x] − Mx+n N[x] − Nx+n Mx+m − Mx+m+n N[x] − Nx+m+n M[x] N[x] Mx+m N[x] A[x]:n a[x]:n ¨ M [x] − Mx+n + Dx+n N[x] − Nx+n tP Net Annual Premium Limited to t payments Symbol In terms of A and a ¨ 1 t P [x]:n 1 A[x]:n a[x]:t ¨ 1 m |A[x]:n a[x]:t ¨ 1 m |A[x]:n In terms of commutation Functions (n ≥ 2) In terms of commutation Functions (n ≥ 2 and t ≥ 2) Temporary Deferred Temporary P A[x] a[x] ¨ m |A[x] a[x] ¨ 1 m |A[x]:n m |A[x]:n a[x]:m+n ¨ 1 1 P [x]:n M[x] − Mx+n N[x] − Nx+t Mx+m − Mx+m+n N[x] − Nx+t t P[x] Whole-life Deferred Whole-life P (m |A[x] ) Endowment Assurance P[x]:n P[x] A[x] a[x]:t ¨ t P (m |A[x] ) m |A[x] a[x]:t ¨ M[x] N[x] − Nx+t Mx+m N[x] − Nx+t t P[x]:n A[x]:n a[x]:t ¨ M[x] − Mx+n + Dx+n N[x] − Nx+t Table 5.CHAPTER 5. 84 . these formulae should be suitably modified. say.1.5. PREMIUMS FOR WITH PROFITS POLICIES Solution. The basic sum assured and bonuses are payable immediately on the man’s death.5. Example 5.e.11)) By linear interp. ∴ Value of benefits = £4033.32804. Suppose further that e = kP proportion k of the office premium): we have (1 − k)P a[x]:n = A[x]:n + I ¨ A[x]:n + I ∴ P = (1 − k)¨[x]:n a = 1 I P[x]:n + (1 − k) a[x]:n ¨ 85 (5.4. 000.4.10. as illustrated in the next example. In modern conditions there is often an explicit allowance for possible future bonus rates for with profits (or “participating”) policies.. This leads to (1 − k)P a[x]:n = A[x]:n + cP ¨ A[x]:n ∴ P = (1 − k)¨[x]:n − c a (For example.2) Sometimes I is a proportion of the first premium.96% 1 1 (see formula (3. 000(1 + i) 2 A∗ [45] where ∗ is at rate i − 0. the value of A[45] is 0. payable for at most 20 years. premiums for policies with the right to participate in the profits of the life office were calculated on conservative assumptions but without an explicit allowance for bonus declarations. Value of benefits 12. A∗ [45] 0. Let annual premium be P .34587).) 5.1) (i.01)2 v 2 1 |q[45] + · · · ] = 12.5. k = 2 1 % and c = 47 1 % if expenses are 50% of the first premium and 2 1 % of all 2 2 2 subsequent premiums .01 1 + 0. The policy has half-yearly premiums. cP . The basic sum assured is £12.01 3. Calculate the half-yearly premium on the following basis: (Note: at 3. 000(1 + i) 2 [1.5 Premiums for with profits policies In early days of life assurance. The office assumes that at the start of each policy year there will be bonus additions at the rate of 1% per annum compound. A with-profits whole-life assurance is about to be issued to a man aged 45. renewal expenses are a (5. Solution.75% per annum interest. The equation of value is P a[x]:n = A[x]:n + e¨[x]:n + I ¨ a where P = level annual premium.69 .01vq[45] + (1. the equation of value for P is: Dx+n 1 P ax:n = 1000 ¨ + P (IA)x:n (5. the term P (IA)x:n in equation (5.I.58 5.6. {A∗ x:n − Ax:n } [j/(1 + j)] i−j 1+j Notes. formula (5.W. 1.a.1) Dx survival benefit return of premiums on death This may be solved for P .16P a[45]:1 ¨ (2) 0. In practice. abbreviated to R. If i (the rate of interest used to discount premiums and benefits in the equation of value) equals j. If there are no expenses. mortality (within n years) may be ignored giving the compound interest equation P an = 1000v n ¨ . 2.6 (i) (ii) Return of premium problems £1. Now suppose that the premiums are returned with compound interest at rate j p. such policies are often described as “ Return With Interest” (R.86 Equation of value is 0.6.79 = 0.96P a[45]:20 ¨ (2) 12.). if (x) dies within n years. a policy issued to (x) with level annual premiums. and a return of all premiums paid. providing: Assume firstly that the premiums are returned without interest (sometimes described as “Return No Interest”.I.1) must be replaced by n−1 v k+1 k |qx k=0 P sk+1 j ¨ death benefit in year k + 1 n−1 =P k=0 (1 + j)k+1 − 1 k+1 v k |qx dj (5. 000 on survival for n years.69 P = £361.N.98753 CHAPTER 5. as an example. P . PREMIUMS + 200 + 4033. at the end of the year of death. If the premiums are returned immediately on death. replace (IA)x:n by (I A)x:n .2) should be multi1 1 1 plied by (1 + i) 2 /(1 + j) 2 = (1 + i∗ ) 2 . If the death benefit is paid 1 at the end of the year of death. If the death benefit is payable immediately on death. Consider.6.6.362 ∴ Hence the half-yearly premium is £180. 1 ¯ 1 Note. some of these problems may be simplified: if there are no expenses.).2) = where ∗ indicates the rate of interest P 1 1 . Let C be available at age 65.1.02)(R[45] − R65 − 20M65 ) D65 1000[76. The equation of value is 1000¨[45]:20 = C a ∴ C= D65 ¯ 1 + 1000(I A)[45]:20 D[45] 1000 (N[45] − N65 ) − (1. 000 are payable annually in advance during the deferred period.7 Annuities with guarantees Annuities are sometimes sold with the provision that payments will certainly continue until their total equals the purchase price.a.. Ignoring expenses. B say (or possibly some proportion of this purchase price. 258 2144.5. ANNUITIES WITH GUARANTEES 87 This result may best be proved by reference to reserves. Let f (B) = aB + B |¯x − B ¯ a B (5. find the cash sum at age 65. In respect of a life now aged 45.1) has a unique solution Proof. the premiums paid are returned immediately without interest.) Let us consider an annuity of £1 p. Equation (5. It follows that .7.a. 407. payable continuously with this guarantee to a life aged x at the issue date. given that the office uses the following basis: A1967-70 select 4% p. A life office sells policies each providing a cash sum at age 65. f (B) → −∞ (as ax → ¯ ¯ f (B) = 0 has a unique solution 1 δ and a B |¯x → 0).1713 5. 722. interest expenses are ignored Solution. which we shall discuss later.6.7.02 × 7. Premiums of £1.7.24] = = £32. as B → ∞.1.1) (B ≥ 0) v t t px dt − B = 0 v t dt + ∞ B Hence f (B) = v B − v B t px − 1 = v B B qx − 1 <0 for all B > 0 Now f (0) = ax > 0 and. the equation of value for B is B = aB + B |¯x ¯ a Theorem 5.7 − 1. On the death of the policyholder during the period of deferment. Example 5.7. 000 f (n) = an + n |¯65 − 1. The annuity payments are guaranteed to continue until the total payments reach £20.88 CHAPTER 5.2) subject to the condition n − 1 < B ≤ n. if the guarantee consists of a payment of the balance of 85% of the purchase price over the total annuity instalments received. . Let n be the guarantee period. solve n 0 n (1) (2) nZ = 20. f (n) → −∞. If the annuity instalments are paid annually in arrear we must solve the equation B = an + n |ax (5.) This is solved by trial and error. and prove that this equation has a unique solution.7. as n → ∞. which lies between 6 and 7 years.1. A man aged 65 buys a guaranteed annuity payable continuously for a purchase price of £32. ¯ f (n) = d dn v t dt + v t t p65 dt − 1.7. 000/n in (1) to obtain an + n |¯65 = ¯ a i.6129 = v n n p65 − 1. v n n p65 < 1 = v n − v n p65 − 1. The office issuing the contract uses the following basis: A1967-70 ultimate. Example 5. 4% p. Another type of guaranteed annuity is that in which the balance (if any) of the purchase price (or a proportion of it) over the total annuity payments received is paid immediately on the death of the annuitant. 000 32.7. 000. expenses are ignored. For example. ∴ f (n) decreases.6129 < 0.85B − t]v t t px µx+t dt (5. equation (5. for n > 0. The purchase price of an annuity with this guarantee may be considerably larger than for an ordinary annuity. (Here n is the guarantee period.3) 0 where n = 0. We have the equation 32. Give an equation for n. so f (n) = 0 has a unique solution. 258 n = 1.e.1) would be replaced by n B = aB + ¯ [0.6129n since. which must be an integer. and.85B (the time when the annuity payments equal 85% of the purchase price).6129n 20.6129n = 0 ¯ a ∞ n Now f (0) = a65 > 0. interest. Solution. PREMIUMS Note. 258.7. 258 = Z(¯n + n |¯65 ) a a subject to Replace Z by 20. Let Z be the annual rate of payment of the annuity.a. P.V. especially if the F. if he dies within n years.1 below.171 89 5. and possibly obtain the same benefits more cheaply by effecting a new policy.8.8. Case 1.The benefits made m in arrear. payments are at rate £B per annum.B. for the balance of the term.I. if premiums are payable for the full n-year term.8. As in case 2.I. of the family income benefit is B a(m) − a(m)n (5.8.2903 2. the policy may have a negative reserve at the later durations (see later discussion of reserves.6093 × 8. 1 As payments are received on average 2m year earlier than in case 2.6774 2.) Consider the combination of a F.5) Premiums. 144. is ¯ B (¯n − ax:n ) a thly (5.5.B.8 Family income benefits A family income benefit of term n years is a series of instalments payable from the date of death of the assured life. and an mthly temporary annuity of term n years.n} ¯ ¯ and hence their M.V.34626 + >0 Hence 6 < n < 7 1. but with payments beginning immediately on death. In practice.V.P. These are found by an equation of value in the usual way. . a danger that. payable in (m) arrear.The benefits are payable continuously Suppose that F. The present value of the benefits is B(¯n − aT ) if T < n a ¯ 0 if T ≥ n The P.4) n x: Case 3. the effect of expenses and other factors may make this possibility hardly profitable.) This means that the policyholder may be able to lapse the contract leaving the office with a loss. and the life assured is aged x at the issue date. however. the M.V.1) (5.8.112 − 9.8. The possibility of lapse option may be avoided by making the premium paying term shorter than the benefit term. There is. of the benefits may thus be written in the form B an − amin{T. 289.2) (5. FAMILY INCOME BENEFITS Try n = 7 f (n) = 6.I. It is clear that their total present value is Ban no matter when (x) dies.7567 × 7.8. is part of a more general assurance contract.772 − 11.B. 144.171 1. Hence the M. is approximately B(1 + i) 2m a(m) − a(m)n n x: 1 (5. 401. as in Example 5. It may be considered as a decreasing term assurance in which the benefit on death is an annuity-certain for the balance of the term.3) Case 2. beginning at the end of the 1/m year of death (measured from the issue date.12136 + <0 Try n = 6 f (n) = 5.P. 97P a35:25 = 2000A25 + 3600(1 + i) 24 a(12) − a(12)25 ¨ 25 35: P = 2000 × 0.90 CHAPTER 5. .8527) 0. Solution.1). 0. Ten years ago a woman aged exactly 35 effected an assurance policy by level annual premiums payable for a maximum of 25 years. if this occurs within 25 years.1. Table 3. The final payment is made in the month ending 25 years after the issue date.97 × 13. The policy provided the following benefits: (i) (ii) a whole life assurance benefit of £2.282 = £96. The similarity arises from the fact that the loan outstanding after the tth payment has been made is equal to the value of the future loan instalments (see Mc Cutcheon and Scott. Let P be the annual premium. a family income benefit of term 25 years.12187 + 3600(1.8. 000 payable at the end of the year of death. although the death benefits are payable in one sum rather than in instalments over balance of the term. Calculate the annual premium on the basis given below: A1967-70 6% interest expenses are 3% of all premiums.8.1312 − 12.06) 24 (13. “An Introduction to the Mathematics of Finance”. beginning immediately on death. with payments of £300 per month. PREMIUMS Example 5.93 1 1 Mortgage protection policies are similar to those for family income benefits. payable at the end of the year of death or on survival to end of the term. 5. if death occurs within the term of the policy.a. (i) Let g(T ) be the present value of the profit to the life office.5.Males. are payable annually in advance. each for a term of 25 years. Interest: 6% p. and that expenses are negligible.a. payable continuously during the term. 5.000 (payable at the end of the year of death or on survival to the end of the term). (a) Write down an expression for g(T ).a. interest expenses are 10% of all office premiums A life office sells immediate annuities. EXERCISES 91 Exercises 5. Level annual premiums are payable throughout the term. payable continuously for the term of the policy. 4% p. in respect of an n-year without profits endowment assurance to (x) with sum assured (payable immediately on death if this occurs within n years) and premium P per annum.1 A life aged 40 effects a 25-year without profits endowment assurance policy with a sum assured of £50. with additional initial expenses of 1% of the sum assured. Assuming that the mortality of annuitants does follow this table. Level premiums are payable annually in advance throughout the term of the policy or until earlier death of the life assured. issued to a woman aged 55. has a sum assured of £50. Level premiums. at the issue date. Calculate the annual premium for each policy. Expenses: none 5. limited to at most 3 years’ payments. (c) For what value of P is the mean of g(T ) equal to zero? An office issues a block of 400 without profits endowment assurances. The office’s premium basis is: A1967-1970 ultimate. that investments will earn 4% per annum. The sum assured is payable at the end of the year of death. to lives aged exactly 35.3 A 5-year temporary assurance.2 An office issues a large number of 25-year without-profit endowment assurances on lives aged exactly 40. and (2) the variance of g(T ). Basis: A1967-1970 select mortality 4% p. The sum assured under each policy is £10. interest.000. P . (b) Derive expressions for (1) the mean. and the sum assured of each policy is £10. 4% p.a. 000 and the premium is £260 per annum. Calculate the annual premium.5 (ii) . find the probability that the office will make a profit on the sale of an annuity payable continuously to a life aged 55. Expenses are ignored in all calculations. The sum assured is payable immediately on death.000 in the first year. 12 .000 each year. interest with no expenses as the premium basis. using English Life Table No. using the following premium basis Mortality: A1967-70 Ultimate. expenses are 5% of each annual premium including the first.4 5. Calculate the level premium. reducing by £10.9. PREMIUMS Assuming that the office will earn 4% interest per annum. Premiums are calculated according to the following basis: Mortality: English Life Table No.12-Males Interest: 4% p.15646 on A1967-70 ultimate 8.16% 5.a. The benefits under this contract are as follows: On death before age 60: an immediate lump sum of £1. 000 On survival to age 60: an annuity of £500 p. and (b) the standard deviation of the present value of this profit. find (a) the mean present value of the profit to the office on the block of policies. ¯ A35:25 = 0.. Expenses: Nil Calculate the annual premium.6 A life office issued a certain policy to a life aged 40.a. . payable continuously for the remaining lifetime of the policyholder. and that expenses may be ignored. Level annual premiums are payable continuously until age 60 or earlier death.92 CHAPTER 5. that the future lifetime of the lives may be described statistically in terms of the A1967-70 ultimate table. 70 5. of annuity.081 = 50000 × 0.9P ¨ ∴P = [N[55] − N58 ] D[55] 10000{5C[55] + 4C[55]+1 + 3C57 + 2C58 + C59 } 0.P.5.434 (or 43. SOLUTIONS 93 Solutions 5.95¨40:25 a = £276.P. of premiums = M.a.5 (i) (a) g(T ) = .9P a[55]:3 = 0. The purchase price is a55 and the office will make a profit if death ¯ occurs before time t.06 0.P. ¯ ¯ log A55 log A55 = = 16. of benefits P a40:25 = 50000A40:25 at A67-70 ultimate.V.9[N[55] − N58 ] 10000 × 371.95P a40:25 ¨ 10000A40:25 + 100 ∴P = 0.1 Equation of value is: M. of premiums less expenses is 0. of assurance benefits is (from first principles) 10000 1 1 5C[55] + 4C[55]+1 + 3C57 + 2C58 + C59 [= 60000A[55]:5 − 10000(IA)[55]:5 ] D[55] M. M. 1 − vt = a55 . 6% p.a.3 Let the annual premium be P .511 = = £395.2 Let the annual premium be P .25955 (see page 67 of tables) =⇒ P = £992.99 log v −δ The probability of making a profit is thus l71. ¯ δ t= so t= log [1 − δ¯55 ] a log v i.P.99 l55+t =1− = 0.V.e.09 5. We have 100 + 10000A40:25 = 0.10.4 Consider £1 p.V. where a55 = at at 4% ¯ ¯ That is.9 × 10449 5.interest ¨ =⇒ P × 13.V.4%) t q55 = 1 − l55 l55 (by interpolation) P aT − Sv T ¯ P an − Sv n ¯ if T < n if T ≥ n 5. 81 = £161. E[g(T )] = 0 when P = ax:n ¯ (ii) (a) First consider 1 policy.261 .P.39043 1 2 10. 000A40:20 + 500 ¯ ∴ P = ¯ N60 D40 95.542 − using a35:25 ¯ a35 − ¨ − = 136. 098 5. ¯ E[g(T )] = 260¯35:25 − 10.16% ¯ + 10. of profit is 2 260 ¯ 8.v. variance of p.60 ∴ For 400 policies.662915 × 104 )2 × 0. 000A35:25 a = 260 ×15.94 (b) (1) (2) ¯ E[g(T )] = P ax:n − S Ax:n ¯ Write g(T ) = P 1 − h(T ) − Sh(T ) δ vT vn if T < n if T ≥ n CHAPTER 5. 866 √ ∴ s. for 400 policies = 20 × 1.84 + 2. 000[ 1 − δ¯35:25 ] a 0.004024415 = 1.6 Let P = annual premium. 000 A35:25 − ( A35:25 )2 δ 0. PREMIUMS where h(T ) = P − δ = P + S h(T ) δ 2 ∴ Var[g(T )] = P +S δ P +S δ Var[h(T )] 2 = ¯x:n ¯ A∗ − Ax:n 2 where ∗ indicates the rate of interest 2i + i2 p. ¯1 P a40:20 = 1.V.004024415 = £21.12 13.662915 × 104 × 0.d.39043 D60 a D35 (¨60 − 1) 2 = (1. ¯ S Ax:n (c) By (b)(1). M.15646 0. 640 (b) For 1 policy.a. 040. of profit = £54. 112. 2) (6.2. and some practical considerations are beyond our present scope. by reference to future cashflows) is defined as tV = E(L) = M. the reserve of a contract may be considered as an accumulation of past premiums less expenses and the cost of death claims (and other benefits).2. or policy values.1 What are reserves? Reserves. etc. (3) (4) for inclusion in statutory returns to the Department of Trade and Industry (or other supervisory bodies) for the purpose of demonstrating the office’s solvency.e.V.g. Reserves are required for various purposes. of future premiums 95 (6. i. for internal office calculations to decide the bonus rates of with-profits policies. Define the NET LIABILITY or PROSPECTIVE LOSS of the office in respect of this policy to be the random variable L = present value of future benefits + present value of future expenses − present value of future premiums (6. e.P.2 Prospective reserves Consider a life assurance policy issued t years ago to a life then aged x.V.1) The reserve or policy value of the contract (calculated prospectively. (1) to pay surrender values (or transfer values in a pension fund).3) .P. the distribution of profits to shareholders. (2) to work out the revised premium or sum assured if a policy is altered or converted to another type. The bases used to calculate “reserves” for each of these purposes may be different.Chapter 6 RESERVES 6. 6. of future benefits and expenses − M. are sums of money held by financial institutions such as life offices and pension funds to cover the difference between the present value of future liabilities (including expenses) and the present value of future premiums or contributions.2. Alternatively. The general rule is therefore that. A negative reserve should not normally be held because the life office is thereby treating the contract as an asset: if the policy is discontinued there is no way in which the policyholder can be made to pay money to the office! Similarly. The premiums valued are also calculated on this basis. tV = net premium reserve = M. If these bases agree (or are assumed to agree) and there are no expenses.V of future benefits − M.P. the reserve t V is calculated just before receipt of any premium then due: the reserve just after payment of this premium is tV +P −e (6. and a rate of interest. This may or may not agree with the premium basis. The policy was issued t years ago by level annual premiums payable continuously throughout life. and the premiums valued are calculated on the reserve basis). By formula (6.M.5) where P is the premium paid at duration t years and e is the expense then incurred.V of future premiums (6.2.V. .P. the discussion of family income benefit policies in Section 5.3 Net premium reserves These are reserves calculated without allowance for expenses.8). payable immediately on the death of (x).4). Find a formula for the net premium reserve t V (on a given mortality and interest basis). this should be replaced by zero. at least for some policies and at early durations t. which we shall consider in the next section.3.1) where the valuation assurance and annuity factors are calculated on the specified mortality and interest basis. if a formula gives a negative value of t V. RESERVES = M.2. interest and expense assumptions used to evaluate t V are known as the reserving basis. of future benefits .96 If expenses may be ignored.1. 6. Are negative reserves allowable? Certain formulae may give negative values of t V. and on the assumption that the premium and the reserve bases agree.P.P.V. Consider a whole life policy with sum assured £1 without profits. the reserves stated in Statutory Returns should not be negative. although negative reserves may be permissible in certain internal office calculations.4) The mortality. the actual premium basis is different. By convention.3. Policies should in general be designed so that negative reserves do not arise (cf. Example 6. we obtain net premium reserves. (In some cases. we have tV CHAPTER 6.2. of future premiums (6. A net premium reserve basis is unambiguously specified by (i) (ii) a mortality table. We have L = v U − P aU ¯ ¯ ¯ where U = future lifetime of (x + t). etc.e. Let L be the net liability (a random variable).3. The general symbol t V may ¯ ¯ be used for any sum assured and any reserve. 97 Notation. but t Vx . with the addition of “t” to indicate the duration. monthly. etc. t V (Ax ). i. say). If premiums are limited to (say) h years of payment. Similar considerations apply if premiums are payable half-yearly.3.) The net premium reserve just after receipt of this premium is of course t Vx + Px .3. CONVENTIONS (1) The symbols t Vx . The International Notation for net premium reserves is similar to that for premiums. If select mortality tables are used. whether net premium or not.6. etc..1: Net Premium Reserves (premiums payable continuously) . It appears that the convention n V = 0 is used in profit testing but not elsewhere. (2) By formula (6. the symbol h is placed above the duration t.3.1) with t = 0. In the case of n-year endowment assurances or pure endowments. as shown in Table 6. NET PREMIUM RESERVES Solution. we write t V[x] . whether to assume that the sum assured has or has not already been paid. it is not always clear whether to take n V as zero or the sum assured (S.1. Type of Policy Whole Life Assurance n-year Term Assurance n-year Endowment Assurance h-Payment Years Whole Life Assurance h-Payment Years n-year Endowment Assurance n-year Pure Endowment Notation t V(Ax ) Prospective Formula for Reserve ¯ ¯ ¯ a Ax+t − P (Ax )¯x+t ¯ ¯ ¯ ¯1 t V(Ax:n t V(Ax:n ) ) ¯ 1 ¯ ¯1 Ax+t:n−t − P (Ax:n )¯x+t:n−t a ¯ ¯ ¯ Ax+t:n−t − P (Ax:n )¯x+t:n−t a ¯ ¯ ¯ a Ax+t − h P(Ax )¯x+t:h−t ¯x+t A t<h t≥h ¯ ¯ h¯ ¯ t V (Ax ) h¯ ¯ t V (Ax:n ) ¯ ¯ ¯ Ax+t:n−t − h P(Ax:n )¯x+t:h−t a ¯ Ax+t:n−t 1 1 a ¯ Ax+t:n−t − P (Ax:n )¯x+t:n−t t<h t≥h 1 ¯ t V(Ax:n ) Table 6. etc. etc. refer to net premium reserves for a policy with a sum assured of £1. ¯ ¯ ¯ a ∴ t V = E(L) = Ax+t − P (Ax )¯x+t . refer to the reserves just before payment of the premium due at time t (if a premium is then payable. we have 0 V = 0 for all net premium reserves. and P = P (Ax ).. 3. (i) (ii) What is the International Notation for the reserve of Example 6. Solution. The net premium reserve at integer duration t is just before premium is paid t Vx Vx + Px just after premium is paid t Now consider duration r + k.3.2.2) Reserves at non-integer durations Let us consider (for example) a whole life non-profit policy with sum assured £1 payable at the end of the year of death.98 Type of Policy Whole Life Assurance Endowment Assurance Temporary Assurance Pure Endowment CHAPTER 6. (i) (ii) tV tV ¯ (Ax ) ¯ ¯ (Ax ) = Ax+t − P (Ax )¯x+t ¯ ¯ ¯ ¯ a = 1 − δ¯x+t − ( a =1− ax+t ¯ ax ¯ 1 − δ)¯x+t a ax ¯ using conversion relationships (6.2: Net Premium Reserves (premiums payable annually) Example 6.3. The reserve interpolation between r Vx + Px and r+1 Vx . 0 < k < 1. r+k Vx r+k Vx is estimated by linear (6.e. Px . integer r. effected by (x) by level annual premiums.3) (1 − k)(r Vx + Px ) + k r+1 Vx (0 < k < 1) [The actual value is v 1−k [1−k qx+r+k + (1 − 1−k qx+r+k )Ax+r+1 ] − Px v 1−k (1 − 1−k qx+r+k )¨x+r+1 a .3.3. RESERVES Symbol t Vx Prospective Policy Value Ax+t − Px ax+t ¨ Ax+t:n−t − Px:n ax+t:n−t ¨ 1 1 Ax+t:n−t − P x:n ax+t:n−t ¨ Retrospective Policy Value Dx 1 Px ax:t − Ax:t ¨ Dx+t Dx 1 Px:n ax:t − Ax:t ¨ Dx+t Dx 1 1 P x:n ax:t − Ax:t ¨ Dx+t Dx [P 1 a ] ¨ Dx+t x:n x:t t Vx:n 1 t V x:n 1 t V x:n 1 1 ¨ Ax+t:n−t − P x:n ax+t:n−t Table 6.1 ? Express this reserve in terms of annuity functions. i. 6. and 1 r+ m V . which is obtained by accumulating the premiums less expenses and the cost of benefits. say. used in . Interest is assumed to be earned at the rate i p. e. and sharing the money out among the survivors. but these are seldom employed. and then dividing the hypothetical funds among the survivors.g. Such a surrender value is related to the retrospective reserve of the contract. RETROSPECTIVE RESERVES 99 which is complicated to evaluate. The annual premium. and the reserve at duration t (where t is an integer multiple of m ) is tV (m) ¯ = 1. without random fluctuations). A1967-70 ultimate. dx+1 in the second policy year. 6% p.4. is 1 ¯ P (m) (Ax ) = P . at least in the early years of the policy. The (prospective) net premium reserve (on a specified mortality and interest basis. with sum assured £1 payable at the end of the year of death. with lx+t survivors at time t.) To simplify matters let us suppose that the sum assured is payable immediately on death: this avoids the calculation of awkward assurance factors at non-integer ages.e. In practice.6. ¯ ¯ ¯ a (Ax ) = Ax+t − P (12) (Ax )¨x+t (4) ¯1 ¯ 1 ¯1 (Ax:10 ) = Ax+t:10−t − P (4) (Ax:10 )¨x+t:10−t a (4) tV tV (12) (12) if t is an integer multiple of 1 12 and 1 4 respectively. Consider a policy providing £1. which are very common in practice. assuming that the contract will not be surrendered. there are dx deaths in the first policy year. for example. e. issued t years ago by level annual premiums to a life then aged x. interest) just before payment of the premium now due is t Vx = Ax+t − Px ax+t ¨ The corresponding retrospective reserve is found by accumulating the funds of (say) lx identical policies until time t.a.000 immediately on the death of (x). of a hypothetical large group of identical policies whose mortality follows the office’s tables exactly (i. and in that event will. payable mthly in advance. effected by mthly premiums during the lifetime of (x). 000Ax+t − P ax+t ¨ just before payment of the premium then due. a whole-life policy. 1 m The reserve at duration t = r + k m (where r is an integer multiple of P m and 0 < k < 1) is usually estimated by linear interpolation between r V + NOTE One may use standard symbols.g. and so on. Since mortality is assumed to follow the table exactly.4 Retrospective reserves So far we have calculated the reserve of a policy by referring to future cashflows. Let us again assume that there are no expenses and that the premium and reserve bases agree. the policyholder may wish to surrender the contract. Consider. probably expect a surrender value related to the accumulation of his premiums less expenses and the cost of life assurance cover. or tV + P m just after payment of this premium. Reserves when premiums are payable mthly Suppose that premiums are payable mthly in advance (e.] Similar formulae apply for other classes of business with premiums payable annually in advance. m=12.a. which corresponds to monthly premiums.g. Proof.V.P. RESERVES the premium and the reserving bases. in duration t years the first t years If there are no expenses.P. We illustrate the argument by means of the whole life policy discussed earlier in this section.V. so we have proved the ¨ desired result. the prospective and the retrospective reserves of a given policy are equal. less benefits in Dx+t duration t years first t years (6. we have M. (at issue date) of retrospective Dx premiums. including situations in which there are expenses.4. The retrospective and prospective reserves at policy duration t are VR = and VP = Ax+t − Px ax+t ¨ respectively.100 CHAPTER 6. we have retrospective Dx M.2 above.1) (6. If the premium and the reserving bases agree. . we find that retrospective Dx 1 reserve at [Px ax:t − Ax:t ] ¨ = Dx+t duration t years More generally. Theorem 6.2) Some examples of formulae for retrospective reserves are given in Table 6.) ¨ ¨ a Dx 1 [Px ax:t − Ax:t ] ¨ Dx+t But Px ax − Ax = 0. This argument may easily be extended to cover other policies.3. reserve at = Dx+t of benefits and expenses. so the accumulated funds at time t are [lx Px (1 + i)t + lx+1 Px (1 + i)t−1 + · · · + lx+t−1 Px (1 + i)] − [dx (1 + i)t−1 + dx+1 (1 + i)t−2 + · · · + dx+t−1 ] 1 =lx (1 + i)t [Px ax:t − Ax:t ] ¨ On division by lx+t (the assumed number of survivors at policy duration t).4.V.P.4. (at issue date) of reserve at = premiums. minus M. Hence Dx+t 1 (VR − VP ) = Px ax:t − Ax:t ¨ Dx Dx+t Dx+t − Ax+t + Px ax+t ¨ Dx Dx = Px ax − Ax ¨ 1 (using the facts that ax = ax:t + t |¨x and Ax = Ax:t + t |Ax .1. by the equation of value for the level annual premium. however. if the premium and reserving bases agree (or are assumed to agree in reserve calculations).6. Give formulae for the following net premium reserves in terms of other monetary functions: (i) (ii) t Vx:n tV .3) (ii) tV ¯ ¯ (Ax:n ) = 1 − ax+t:n−t ¯ ax:n ¯ (6.4. the prospective method is more often used. Show that ax+t:n−t ¨ ax:n ¨ (i) t Vx:n =1− (6. Example 6. one need not specify whether a prospective or retrospective reserve is required. Solution. and (b) the retrospective method.4. ¯ (Ax:n ). ¯ by (a) the prospective method. (i) (a) Ax+t:n−t − Px:n ax+t:n−t ¨ (b) (ii) Dx 1 {Px:n ax:t − Ax:t } ¨ Dx+t ¯ ¯ ¯ (a) Ax+t:n−t − P (Ax:n )¯x+t:n−t a (b) Dx ¯ ¯ ¯1 {P (Ax:n )¯x:t − Ax:t } a Dx+t Example 6.4) Solution.2.4.4. RETROSPECTIVE RESERVES 101 It follows that. In practice.1.4. (i) t Vx:n = Ax+t:n−t − Px:n ax+t:n−t ¨ 1 = 1 − d¨x+t:n−t − ( a − d)¨x+t:n−t a ax:n ¨ a ¨ = 1 − x+t:n−t ax:n ¨ . 3) are called prospective gross premium reserves. the two methods will normally give different results.) If the premium and reserve bases do not agree. (In practice. of future premiums multiplied by the factor (1 − k) (6. of future benefits − M. however.1) Example 6. The reserves obtained by the formula (6. which are no longer assumed to be the same. an interest rate. together with the reserving basis. We begin by considering prospective gross premium reserves. An important feature of gross premium reserves are the facts that the prospective reserve at inception is not normally equal to zero. interest: 4% p. whereas those obtained by a retrospective accumulation of premiums less expenses and the cost of life assurance benefits are called retrospective gross premium reserves.4 above.P. interest.5 Gross premium valuations and asset shares. If the allowance for future expenses is a proportion k of future premiums (including any due now).1. We now allow for the possibility of expenses in the premium and reserving bases.) . and the retrospective gross premium reserve of an n-year policy just after it matures is not generally equal to zero. Consider the whole life policy discussed in section 6.102 (ii) tV CHAPTER 6. The premium basis (or the premiums themselves) must be specified. and suppose that (i) (ii) the premium basis is A1967-70 ultimate. RESERVES ¯ ¯ ¯ ¯ (Ax:n ) = A ¯ a x+t:n−t − P (Ax:n )¯x+t:n−t 1 = 1 − δ¯x+t:n−t − ( a − δ)¯x+t:n−t a ax:n ¯ a ¯ = 1 − x+t:n−t ax:n ¯ 6.a. 6% p.P. expenses: 5% of all future premiums.5. and the allowance for future expenses. the formula for the prospective reserve is tV = M.V. Find a formula for the reserve at duration t years (just before receipt of the premium then due.V. which gives (i) (ii) (iii) a mortality table. and reserves are to be calculated prospectively by the gross premium method on the following basis: mortality: A1967-70 ultimate.a. with no allowance for expenses. the phrase “gross premium reserve” usually refers to prospective reserves only..2.5. 1. As will be stated in the next chapter. Solution The annual premium is P = Px so the formula for the reserve is tV 103 on A1967-70 ultimate.5. one must specify the allowance for past expenses.5.3) . in the policy of example 6.5.a. are on A1967-70 ultimate at 4% p. and Dx /Dx+t . the death cost is Dt+1 =q[x]+t (St+1 − t+1 V) (which is negative if St+1 < t+1 V).5. GROSS PREMIUM VALUATIONS AND ASSET SHARES. one allowed for expenses of 10% of all past premiums. before payment of the premium then due.5.95P ax+t ¨ where the factors Ax+t and ax+t are on A1967-70 ultimate. In order to illustrate the calculations. 4% interest. let us consider an n-year policy issued t years ago to (x). = premium payable at the start of policy year t. If. By substituting formula (6. = rate of interest earned by the office in policy year t.) Asset shares The asset share of a life assurance policy is a retrospective gross premium reserve calculated on the basis of the mortality. 6% p. = actual expenses incurred at the start of policy year t.6. ¨ If a retrospective gross premium reserve is required.5. Dt+1 0≤t≤n−1 (6. (6. and the following relation holds: t+1 V = (t V + Pt+1 − et+1 )(1 + it+1 ) − death cost. the formula for the retrospective gross premium reserve would be tV = Dx 1 0. Let t V denote the asset share at duration t of the policy. interest (or whatever mortality and interest basis is to be used. = observed mortality rate among lives aged x + t who entered assurance at age x. we may obtain an expression for t+1 V in terms of t V and other known quantities. we must consider the given policy’s share of the cost of making the asset share (at the end of the policy year) up to the sum assured. In practice. however. for those policies which became claims in the year. We clearly have 0 V = 0. 2.5. St+1 .a.3) in formula (6. one may perhaps replace Dt+1 by a suitable approximate value and use formula (6.2) directly. Since the proportion of policies in force at the start of the year which became claims was q[x]+t . Notes 1. the quantity St+1 − t+1 V is called the death strain at risk of the policy in year t + 1.9P ax:t − Ax:t ¨ Dx+t where the annuity and assurance factors. and define St Pt et it q[x]+t = sum assured in year t (payable at end of year of death). = Ax+t − 0.2) To calculate the “death cost”. interest and expenses actually experienced by the office.2). 6. by Lidstone’s theorem. Its mean is defined as the reserve t V. The office must also allow for random fluctuations from the mean numbers of deaths (especially when there is a small number of policies with very large sums assured or death strains at risk) and consider the exchange or reinsurance of large risks. and (at least in the early years of the contract) will expect a surrender value related to the accumulation of his past premiums less expenses. RESERVES 3. In practice. Asset shares are particularly important for with profit policies. Var(L). however.2. but we make some remarks. Similar arguments lead to the “net premium method” for with-profit policies (provided that terminal bonuses are allowed for separately): cf. as their surrender and maturity values are not fixed in advance and the asset share gives one method of deciding “fair” surrender and maturity values. i. The above discussion refers mainly to the calculation of reserves for solvency testing. also involved. It would at first seem obvious that a prospective gross premium valuation should be used. How may this best be calculated? . the bases being perhaps different (reflecting expected future conditions and actual past conditions respectively. alterations to policies.9 below.6 The variance of L We return to the definition of the random variable L given by formula (6. In the unlikely event that mortality . since the premium value is unrelated to the actual premium being charged) may have the property that the corresponding reserves are comparable with. either by choice or because this is required by the supervisory authorities: such a basis (though apparently artificial. etc. interest and expenses follow a certain valuation basis exactly. This suggests that the office should hold reserves equal to the greater of (i) (ii) a prospective premium reserve.7 below) of the reserves (subject to limits on I). as this represents the mean of the present value of the net liability. a net premium valuation at a low rate of interest may be used. in case Zillmerisation is seen as a sign of financial weakness. but in certain problems we also require its variance. Net premium versus gross premium reserves A full description of the advantages and disadvantages of the various approaches is outside the scope of this course. and a retrospective gross premium reserve (or asset share). The asset share at policy duration n years may of course differ from the maturity value. but reserves are also used for internal purposes: deciding upon bonus levels. 4. A number of other considerations are. 5.104 CHAPTER 6. a retrospective gross premium reserve. and we shall not pursue the discussion of asset shares. For these purposes the office may use other bases.1). The reason for the use of a low rate of interest in net premium reserves is that. net premium reserves normally increase as the rate of interest falls. but many offices prefer to publish unmodified net premium reserves. the more “scientific” reserves given above. The supervisory authorities may be prepared to allow Zillmerisation (see Section 6.) The position is further complicated by the possible existence of guaranteed surrender values and the difficulty of estimating future conditions. or generally greater than. leading to a profit or loss to the office at that time. But this assumes that the policy will not be surrendered: the policyholder may do this at any time. the asset share of a policy will equal the retrospective gross premium reserve on that basis.e. Section 6. the average of the net liabilities is N ¯ L= j=1 Lj /N (6.2) where ∗ indicates a rate of interest of 2i + i2 p. The random variable L is defined as L = prospective loss on contract (at duration t) = Sv U − P aU ¯ (6. We now consider a number of life assurance policies. who is now aged x + t.a. they are of the same sum assured. The total reserves for the group of policies is N E[Lj ] = V1 + V2 + · · · + VN j=1 (6. etc.6. with premiums of P p. It is sometimes also assumed that N is so N large that the distribution of j=1 ¯ Lj or L is approximately normal (by the central limit theorem and related results).5) which has mean E(L) and variance Var(L)/N . N N Var j=1 Lj = j=1 Var(Lj ) (6.6. respectively. These results are sometimes used in solvency calculations. and expenses are ignored.6. labelled from j = 1 to j = N . . If we now suppose that all the policies are identical (i.e. and were issued at the same date to N different lives all of the same age).6.a.6.6. so we express L in terms of v U .4) where Var(Lj ) may be calculated by (say) formula (6. type and duration. the net liability and the reserve of the jth policy. THE VARIANCE OF L 105 Let us consider a policy providing £S immediately on the death of a life aged x at issue. although this may only be accurate for very large values of N .1) where U is the future lifetime of the policyholder. payable continuously throughout life. If we further assume that the lives assured are independent. That is. where L is the net liability of a given policy. ¯ L = Sv U − P = from which we obtain Var(L) = = S+ S+ P δ P δ 2 S+ P δ 1 − vU δ P vU − δ Var(v U ) 2 ¯x+t ¯ A∗ − Ax+t 2 (6.6.6. Now we cannot use the formula Var(L) = Var(Sv U ) + Var(P aU ) ¯ wrong! (because Sv U and P aU are not independent).3) where Lj and Vj are.2). This technique may also be used for certain other types of contracts. tV Z = M.106 CHAPTER 6. just before payment of the premium then due.7 Zillmerised reserves Let us now suppose that the premium basis is the same as the reserving basis (or is assumed to be the same for the purpose of calculating reserves. The premium and reserving basis includes the following allowances for expenses: expenses of e on the payment of each premium. In certain important classes of policy.1.1) where t V denotes the corresponding net premium reserve. RESERVES 6.) There are level premiums of P . payable at the end of the year of death (if this occurs within the term of the policy. who was aged x at the issue date. Let us consider an n-year endowment assurance for a sum assured of £1 without profits. however. The reserve for the above policy is tV Z = (1 + I)t V − I (6. with additional initial expenses of I (so the total initial expense is I + e). the retrospective and prospective methods give the same results. “Zillmerised net premium reserve” or sometimes “modified net premium reserve”. Theorem 6.P.4.3) t Vx:n =1− ax+t:n−t ¨ ax:n ¨ . The reserve is required at duration t. i. but reserves may not be “net premium reserves” because of the effect of expenses.7.V.e.) As stated earlier. the reserve is obtainable by means of a simple adjustment to the net premium reserve: the corresponding reserve is called a “Zillmerised reserve”. tV = t Vx:n =Ax+t:n−t − Px:n ax+t:n−t ¨ a ¨ =1 − x+t:n−t ax:n ¨ Proof. of future benefits and expenses less premiums = Ax+t:n−t − (P − e)¨x+t:n−t a = Ax+t:n−t − I + Ax:n ax:n ¨ ax+t:n−t ˙ I ax:n ¨ ax+t:n−t ˙ = Ax+t:n−t − Px:n ax+t:n−t − ¨ (where Px:n is the net annual premium) = t Vx:n − I ax+t:n−t ¨ ax:n ¨ Now we use the fact that (by formula 6. We first note that the annual premium P is such that P ax:n = Ax:n + e¨x:n + I ¨ a I + Ax:n ∴P −e= ax:n ¨ Now consider the reserve at duration t by the prospective method. payable annually in advance for n years or until earlier death of the policyholder.7. Calculate (i) (ii) the annual premium.e. payable at the end of the year of death (or on survival for 20 years). the reserve per £1000 sum assured of a whole life without profits policy might be quoted as 1020t Vx − 20 (corresponding to I= 2% ). The sum assured is £10. sometimes used in practice for all policies and all durations. and . 3. a policy should not be treated as an asset to the office.000. When the duration t is short (e. Note that formula (6. If the sum assured is not £1. we naturally multiply formula (6.7. (It is. expenses are 3% of all office premiums (including the first) with additional initial expenses of 1. ZILLMERISED RESERVES to give the formula tV Z 107 = t Vx:n − I [1 − t Vx:n ] = (1 + I)t Vx:n − I as required. and the reserve. interest.1) gives Z = −I 0V which is correct if the additional initial expense I is thought of as being disbursed before the first premium is received: if it is not.1.7. which should. 6% p.5% of the sum assured. The basis for premiums and reserves is: A1967-70 ultimate. however.7.1) if the sum assured is payable ¯ ¯ immediately on death and premiums are payable continuously. Example 6. one should write 0V Z =0 (as is always assumed in profit-testing: see later. i. as a rule. In this case. (a) just before receipt of the premium now due.a.7.) 2. be replaced by zero. and premiums are payable annually in advance.7. The case of a whole life policy is similarly dealt with (put n = ∞ in the above formulae.6.) 4.) 5. t = 1) Zillmer’s formula may give a negative reserve. A similar argument may be used to establish formula (6.1) by the sum assured: for example. Ten years ago life office issued a 20-year endowment assurance without profits to (35). Notes 1. Zillmer’s formula does not in general hold for other classes of policy.g. t V = t V(Ax:n ). nor when the premium due at time t has been paid. 49 Interpolate: i − 0.5)) l55 a55 ¨ l35 Example 6. RESERVES = £288.49 − 11. The reserve is (a) 10000[1.a. (i) The office premium is P = 10000 A35:20 + 0.108 (b) just after receipt of the premium now due. i s10 ¨ 0.7.015] = £3497 (b) £3497 + 0.2.11 − 11.1 were to surrender his policy (before paying the premium now due) and the office grants a surrender value equal to the reserve. so we use these formulae: a35:20 = a35 − v 20 ¨ ¨ = 11.03 0.44% at rate i=yield p.04 12.03 ∴i 12. as defined in that question.2.01510 V35:20 − 0.015 0.97 × £288.81 0. .e.81 3. i. Suppose that the policyholder of example 6.7.67 Note that tables at 6% interest are limited.11 ¨ We use the compound interest tables.81 = 0.03 11.04 − 0.76 = £3777 (see formula (6.67¨10 = 3497 s ∴ s10 = 12.9969 and A35:20 = 1 − d¨35:20 a (ii) One may use Zillmer’s formula in this case (or other methods).97 a35:20 ¨ CHAPTER 6. The equation of value is 288. Solution.44 12. What yield per annum would the policyholder have obtained? Solution. which we shall write as t V F P T . Similar formulae hold if the sum assured is payable immediately on death. The use of full preliminary term reserves was suggested by T. when premiums are payable continuously or monthly. reserve? We observe that.T. the reserve just before payment of the second annual premium is zero. and for certain other types of policy..8. 1V FPT = Dx 1 (P − e)¨x:1 − I − Ax:1 a Dx+1 =0 1 P = e + I + Ax:1 from which we obtain (6. Sprague in 1870. and the 1 cost of the first year’s life assurance cover.1) That is.8 Full preliminary term reserves. Ax:1 . and suppose that the reserve for a whole life or endowment assurance policy is such that 1V =0 (6. we naturally multiply these expressions by the sum assured.6. we may set n = ∞ to obtain tV FPT If the sum assured is not £1.8. We must have 1V a = Ax+1:n−1 − (P − e)¨x+1:n−1 = 0.8.4).2) (6. I + e. 109 6. but we do not pursue this topic. by the retrospective method. in which the allowance for total initial expenses may be smaller than that of formula (6.4) This states that the first premium is exactly sufficient to pay the initial expenses. Let us again suppose that the sum assured is £1. Other actuaries (particularly in North America) have devised formulae for “modified full preliminary term reserves”. .P.8. and that expenses are as in the previous section. B.8.8. FULL PRELIMINARY TERM RESERVES. We again assume that the premium and reserving bases agree. payable at the end of the year of death. What level of initial expenses leads to a F.3) In the case of a whole life policy. We have shown that tV FPT = t−1 Vx+1:n−1 = t−1 Vx+1 (6. etc. Hence P −e= Hence tV Ax+1:n−1 = Px+1:n−1 ax+1:n−1 ¨ ¨ = Ax+t:n−t − Px+1:n−1 ax+t:n−t = A(x+1)+(t−1):(n−1)−(t−1) − Px+1:n−1 a(x+1)+(t−1):(n−1)−(t−1) ¨ = t−1 Vx+1:n−1 = the net premium reserve for the corresponding policy on a life aged x + 1 at entry with term n − 1 years at duration t − 1 This reserve is called the Full Preliminary Term reserve. a. using A1967-70 ultimate mortality and interest at 3%p. and that the premium now due has not yet been paid.9.3) where P is the annual premium for the corresponding non-profit contract.9.110 CHAPTER 6. and (3) the asset share method. i.9. plus the mean present value of the bonuses already declared. Reversionary bonuses are added on the payment of each premium. 3 the office has declared a compound reversionary bonus of 3 4 % per annum vesting in advance on st each 1 January from outset.) The principal methods used for calculating reserves for with profits policies are: (1) the net premium method.9 Reserves for with-profits policies We shall deal with “traditional” with profits policies only. at the end of the year of death or on maturity. There were no lapses or surrenders during 1993 or 1994. For this group of policies. . The basic sum assured under each contract was £20. with attaching bonuses.e.) We suppose that the policy was issued with level annual premiums.) Let us consider an n-year endowment assurance issued t years ago to a life then aged x with basic sum assured S (payable. we have tV WP ¯ = (S + B)Ax+t:n−t − P ax+t:n−t ¨ ¯ S Ax:n ¯ = SP (Ax:n ) ax:n ¨ (6.9. (1) The net premium method. In the case of the endowment assurance policy discussed above. If the sum assured is payable immediately on death. RESERVES 6. 000. the reserve is given by the formulae: tV WP = St Vx:n + BAx+t:n−t ¨ = (S + B)Ax+t:n−t − SPx:n ax+t:n−t (6. Level annual premiums are payable in advance throughout the term. etc. P = (6. plus any bonuses. The rationale of this method is that the additional premiums for a with profits policy (relative to the corresponding without profits policy) are considered to have “earned” the bonus B so far declared.9. and we denote the total bonus added to date by B (hence the total death benefit in the year just ending is S + B. Among the group of policies there were 2 deaths during 1993 and 4 deaths during 1994. (2) the bonus reserve (or gross premium) method. and the basic sum assured. Example 6. Calculate the total reserves which the office held for the group of policies on 31st December 1994. so an additional reserve is required to cover the value of these bonuses.2) where all the actuarial functions are calculated on the given mortality and interest basis.1.4) Similar formulae may be given for whole life policies and for policies with monthly premiums. The reserve is taken to be the net premium reserve for corresponding non profit policy. and ignore terminal bonuses (although these may in practice be very important. is payable on maturity or immediately on earlier death.1) (6. The office values the policies by a net premium method. On 1st January 1993 a life office issued a with-profit assurance endowment policy with a term of 10 years to each of 100 male lives then aged exactly 50. 79562 = £443. 1 1 . We require 94×reserve of one policy. 000 50:10 = 1. in the case of the endowment assurance policy discussed above. 000.5) tV x+t:n−t where ∗ indicates the rate of interest (i − b)/(1 + b). 528.0375)2 − 1] (i.03) 2 (0. As might be expected.0375)2 A52:8 − P a52:8 ¨ per policy where P = net premium for corresponding non-profit policy ¯ A = 20.68419 = 0.e. Similar formulae may be derived for other policies. according to the system of bonus declarations used by the office. RESERVES FOR WITH-PROFITS POLICIES 111 Solution.03) 2 (A50:10 − v 10 10 p50 ) + v 10 10 p50 = 0. The value of the benefits is therefore calculated by the formulae given in Section 3. a formula for the reserve is WP = (S + B)A∗ − (1 − k)P ax+t:n−t ¨ (6.e.94 a50:10 ¨ [where we have used ¯ A50:10 1 1 (A50:10 )(1. it is assumed that future bonuses will be at the rate b per annum on the compound system and a proportion k of future premiums will be absorbed in expenses. 000(1.03) 2 + A50:10 1 1 = (1. 760.03) 2 (A52:8 − v 8 8 p52 ) + v 8 8 p52 = (1. reserves are calculated on the assumption that reversionary bonuses will be declared at certain annual rates.75219 ] ¯ ∴ Reserves needed = 94 × [21.73329 = 0.10 above.06142) + 0.125 A52:8 − 1. 760. If. ¯ reserve = 20.068 + 0. and in the early years of a contract the method may give reserves which are quite unrelated to the accumulation of premiums. B = 20. ¯ ¯ reserve = S[2 V(A50:10 )] + B A52:8 per policy where S = 20. As the name suggests. bonuses already vesting) i.9. 000[(1.9. this method is sometimes referred to as the gross premium method for with profits policies. the reserves calculated by the bonus reserve method depend greatly on the assured future levels of bonus. Using the net premium method for with profits policies. 417 (2) The bonus reserve method. and the reserve is then found by subtracting the value of the office premiums less projected future expenses. Since the premium valued is the actual (or office) premium for the policy.94¨52:8 ] a = (1.6. the reserves given by the asset share method do not depend greatly on past bonus declarations. the formulae are almost unchanged for with profit policies. the only difference being the fact that the sum assured in each policy year includes the vested bonuses. This has been described (for non-profit policies) in Section 6. A complete description of the advantages and disadvantages of these methods of calculating reserves is beyond the scope of this book. . Since these only affect the death cost.112 CHAPTER 6.5 above. RESERVES (3) The asset share method. 06942 = 0. 4 6.a.1 (i) Express t Vx in terms of Ax and Ax+t . assuming that the premium now due has been paid. given that 1 − Ax+2n = Ax+2n − Ax+n = Ax+n − Ax Calculate 1 V40 (ii) given that P40 = . p40 = .6 A whole life assurance with sum assured £100. show that your expressions are equal. Zillmerised reserves. payable at the end of the year of death (or on survival for 20 years). The basis for calculating reserves for this policy is as follows: net premium method: A1967-70 ultimate.) 6. all premiums paid will be returned without interest at the end of the year of death.06.10. Estimate the policy value at duration 28 1 by interpolation. The policy has annual premiums payable throughout life. In the event of death within the n years. the reserve.02. and premiums are payable annually in advance.2 Consider an n-year pure endowment policy. find P x:n . Using commutation functions.11430 20 10 Vx 1 P x:10 10 Vx Calculate . The sum assured is £10000.3 (Difficult) 1 1 Given that Px = 0. .4 Ten years ago a life office issued a 20-year endowment assurance without profits to (35). interest. Obtain expressions for the reserve at duration t (i) (ii) prospectively and retrospectively. (Assume that the premium due at time t has not yet been paid. and P x:n = 0. and ignoring expenses. EXERCISES 113 Exercises 6. 6.6. (Difficult) You are given: (i) (ii) (iii) (iv) Px = 0. calculate (i) (ii) 6.99647 and i = . 5% p. n Vx = 0. with sum assured 1 and with annual premiums payable throughout the duration of the policy.01536.7 Describe the following terms briefly: (i) (ii) net premium reserves. or otherwise.01212 20 Px = 0. issued to (x).05.01508 = 0. Using A1967-70 ultimate 6%.25. Hence. 6. find the values of n Vx and n Vx+n .5 the annual premium.000 payable at the end of the year of death was purchased by a life aged 30. 6. with level premiums payable annually in advance throughout the term and with a sum assured of £1 payable at the end of the year of death. RESERVES in each case giving suitable formulae in respect of a whole life assurance issued t years ago to a life then aged x. . CHAPTER 6. and gross premium reserves.114 (iii) (iv) Full Preliminary Term reserves. Assume that the premium now due has not yet been paid. 6.11. SOLUTIONS 115 Solutions 6.1 (i) t Vx =1− ax+t ¨ d¨x − d¨x+t a a Ax+t − Ax = = ax ¨ d¨x a 1 − Ax The given information implies that Ax+n = 2Ax+2n − 1 and Ax = 2Ax+n − Ax+2n = 3Ax+2n − 2 Hence n Vx = Ax+n − Ax 1 − Ax+2n 1 = = 1 − Ax 3 − 3Ax+2n 3 1 − Ax+2n 1 Ax+2n − Ax+n = = 1 − Ax+n 2 − 2Ax+2n 2 l40 P40 (1 + i) − q40 = = 0.01264 l41 p40 and n Vx+n = (ii) Retrospective method gives 1 V40 = [P40 − vq40 ](1 + i) 6.2 The equation of value for the annual premium P is: P ax:n = n Ex + P (Rx − Rx+n − nMx+n )/Dx ¨ or P (Nx − Nx+n ) = Dx+n + P (Rx − Rx+n − nMx+n ) The expressions for the reserves are as follows: Retrospective: tV (∗) = [P (Nx − Nx+t ) − P (Rx − Rx+t − tMx+t )] /Dx+t (∗∗) Prospective: tV 1 1 = n−t Ex+t + tP Ax+t:n−t + P (IA)x+t:n−t − P ax+t:n−t ¨ = [Dx+n + tP (Mx+t − Mx+n ) + P (Rx+t − Rx+n − [n − t]Mx+n ) − P (Nx+t − Nx+n )] /Dx+t The expressions (∗∗) and (∗ ∗ ∗) are equal if and only if P Nx − P Rx = Dx+n − tP Mx+n − P Rx+n − P [n − t]Mx+n + P Nx+n i.e. if and only if Dx+n + P (Rx − Rx+n − nMx+n ) = P (Nx − Nx+n ) which is true by (∗). 6.3 We know that n Vx (∗ ∗ ∗) =1− ax+n ¨ = 0.06, ax ¨ so ax+n ¨ = 0.94 ax ¨ 116 1 ax:n ¨ − ax − ax:n ¨ ¨ 1 = ax ¨ ax ax:n ¨ ¨ ¨ n Ex ax+n = ax ax:n ¨ ¨ 1 = 0.94P x:n = 0.235 1 ax:n ¨ −d − 1 −d ax ¨ = 1 ax:n ¨ CHAPTER 6. RESERVES Px:n − Px = so − 1 = 0.235 ax ¨ Px:n = 0.255 1 1 P x:n = Px:n − P x:n = 0.005 6.4 (i) The annual premium is 10000P35:20 = 10000 since a35:20 = a35 − v 20 ¨ ¨ (ii) The reserve is 1000010 V35:20 = 10000 1 − since a45:10 = a45 − v 10 ¨ ¨ a45:10 ¨ a35:20 ¨ = £3593 1 a35:20 ¨ −d = £267.51 l55 a55 = 11.997 ¨ l35 l55 a55 = 7.687 ¨ l45 ∴ Reserve after premium is paid = £3593 + £267.51 = £3861 (to nearest £1) 6.5 The prospective method does not work. By the retrospective method, (reminder: 20 10 Vx refers to limited payments policy) Dx [20 Px ax:10 − Px ax:10 ] ¨ ¨ Dx+10 Dx ax:10 ¨ Dx+10 (since benefits are the same under both policies) 20 10 Vx − 10 Vx = = (20 Px − Px ) = 20 Px − Px 1 P x:10 = 0.04264 ∴ 20 Vx = 0.15694 10 6.6 28 V 29 V = 100, 000 = 100, 000 29 V, 28 V30 29 V30 = 33, 101 = 34, 754 Interpolate between 28 V + P and P30 = 724.00. This gives 28 1 V 4 where P is the annual premium, i.e. P = 100, 000 1 3 (33, 101 + 724) + (34, 754) 4 4 = £34, 057 = 6.11. SOLUTIONS 6.7 (i) Net premium reserve = reserve calculated ignoring expenses and assuming premium basis agrees with reserving basis = t Vx = Ax+t − Px ax+t ¨ in example given (ii) Zillmerised reserve = (1 + I)t V − I per unit sum assured 117 (where t V = net prem. reserve, I = additional initial expenses per £1 sum assured) = (1 + I)t Vx − I in example given. (iii) F.P.T. reserve = net premium reserve for corresponding policy of term n − 1, entry age x + 1, duration t − 1 = = t−1 Vx+1:n−1 t−1 Vx+1 per unit sum assured for E.A. in example given value of future gross premiums less expenses. (iv) Gross prem. reserve = value of future benefits − = Ax+t − (1 − k)P ax+t ¨ in example, where P = office A.P. and k = propn. of premiums allowed for future expenses. 118 CHAPTER 6. RESERVES Chapter 7 APPLICATIONS OF RESERVES 7.1 Surrender values Most life assurance policies (except term assurances) provide for a sum of money known as a surrender value to be paid to the holder if the payment of premiums is discontinued and the contract terminated. If the scale of surrender values is fixed in advance, the policy is said to have guaranteed surrender values. For example, a whole life contract (with level annual premiums payable throughout life) may be issued with guaranteed surrender values equal to the full preliminary term reserve, i.e. (SV )t = S.t−1 Vx+1 (7.1.1) where x is the age at inception of the policy, t is the duration, (SV)t is the surrender value, S is the sum assured, and t Vx is calculated on a specified mortality and interest basis. (It is assumed that the premium due at time t is unpaid.) More usually, however, surrender values are not guaranteed, although there may be guaranteed minimum surrender values, as is the case for Industrial Assurance policies in the U.K. (i.e. those in which the premiums are collected regularly from the home of the policyholder.) Surrender values are normally related in a simple way to the reserve of the contract, but the method of calculating the reserve for this purpose may be different from those used for other purposes, e.g. statutory returns to the Department of Trade and Industry. Surrender value scales must take several factors into account: (i) (ii) (iii) (iv) (v) the experience of the office since the policy was issued, the expected future experience, consistency between different classes of business, competition between life offices (and possibly with a market for “second hand” life policies), and guaranteed minimum surrender values (if applicable.) A full discussion of all points involved is beyond our present scope. In some cases surrender values are calculated from paid up policy values, which we discuss in the next section. Transfer values in pension schemes correspond to surrender values of life assurance policies: a transfer value is, however, payable only to another pension scheme and not to the scheme member directly. 119 120 CHAPTER 7. APPLICATIONS OF RESERVES 7.2 Paid-up policy values Instead of surrendering his policy, a policyholder who wishes to stop paying the premiums may ask for the contract to continue in force with reduced benefits (payable when the original benefits were to be paid.) The reduced benefit is called the paid-up sum assured (P.U.S.A.). If the original sum assured on death differs from that on maturity, the paid-up benefits will normally be in the same ratio. The general symbol t W denotes the paid-up sum assured at duration t years, assuming that the premium then due is unpaid. If expenses are ignored and all calculations are conducted on the same mortality and interest basis, net premium paid-up policy values may be defined, using the notation t Wx , t Wx:n , etc. For example, t Wx = the paid-up sum assured at duration t in respect of a whole-life assurance issued to a life then aged x by level annual premiums, with sum assured £1 payable at the end of the year of death (7.2.1) Suppose that, at the date of discontinuance of premiums of this whole life policy, the life office wishes to make the mean net liability for the altered contract the same as that of the original contract. This leads to the equation t Vx = t Wx Ax+t (7.2.2) from which we obtain t Wx = t Vx Ax+t (7.2.3) Similar calculations hold for other policies, e.g. t Wx:n = t Vx:n Ax+t:n−t (7.2.4) More generally, we obtain the following general formula for the P.U.S.A.: tW = tV A (7.2.5) ¯ where t V is the reserve or surrender value and A is an assurance factor, such as Ax+t or Ax+t:n−t . (We note that no expenses have been allowed for at or after the date of alteration: if such an allowance is desired, the formulae should be modified.) In the important case of endowment assurances, the proportionate rule is sometimes used in practice, i.e. t S (7.2.6) tW = n where S is the original sum assured, t is the total number of premiums actually paid and n is the number originally payable. Surrender values of endowment assurances (or other policies) are sometimes found from the P.U.S.A. by the formula (SV )t = (P.U.S.A.)A (7.2.7) where A is the appropriate assurance factor at age x + t. In particular, if the proportionate rule for endowment assurances is used, the surrender value at duration t years is sometimes taken as t SAx+t:n−t n (7.2.8) 7.3. ALTERATIONS AND CONVERSIONS 121 if the sum assured is payable at the end of the year of death, if this occurs within the balance of the term, n − t years, or t ¯ S Ax+t:n−t (7.2.9) n if the sum assured is payable immediately on death within this period. Example 7.2.1. (i) Four years ago a life then aged 35 effected a 25-year without profits endowment assurance by annual premiums for a sum assured of £50,000 payable at the end of the year of death or on survival for 25 years. He surrendered the policy just before paying the premium now due. The life office calculated the premiums for this policy on the basis of A1967-70 select at 4% interest with the following expense loadings: 2.5% of each premium, including the first, plus a further initial expense of 1% of the sum assured. The office also calculates reserves on this basis, and surrender values are equal to 95% of reserves. (a) Calculate the office annual premium. (b) Calculate the surrender value of the policy. (ii) Suppose that the policyholder of (i) had made his policy paid-up instead of surrendering it, and that the office calculates paid-up values by the proportionate rule. Calculate the yield per annum (to the nearest 0.1%) that he would have obtained on his entire transaction, assuming that he survives to maturity date. Solution. (i) (a) Let A.P. be P . We have 0.975P a[35]:25 = 50000(A[35]:25 + 0.01) ¨ ∴ P = £1289.10 (b) Reserve = 50000A39:21 − 0.975P a39:21 ¨ = £4790.00 ∴ S.V. = 0.95 × reserve = £4550.50 (ii) PUSA=50000 × 4/25 = 8000. Solve i.e. By trials and interpolation, i P s4 = 8000v 21 ¨ (1 + i)21 s4 = 6.20588 ¨ 1.9%. 7.3 Alterations and conversions Life offices are frequently asked to alter a policy; for example, a policyholder with an endowment assurance maturing at age 65 may request that the policy should be changed to an endowment 122 CHAPTER 7. APPLICATIONS OF RESERVES assurance maturing at age 60. The policyholder might also request his policy to be converted to another class of business: for example, from a whole life assurance to an endowment assurance. The usual rule when carrying out these calculations is to equate the reserves before and after conversion. If V1 and V2 denote the reserves before and after the conversion respectively, the rule may be stated as follows: V1 = V2 (7.3.1) If there are expenses of C for the conversion itself, the formula becomes V1 − C = V2 (7.3.2) The bases used to calculate V1 and V2 may in practice be different, and V2 must be calculated prospectively by the gross premium method. To show that formulae (7.3.1) and (7.3.2) hold, we argue as follows. At the date of conversion, the policyholder’s original policy is worth V1 on immediate surrender. (In practice, however, the reserve V1 used in conversion calculations may exceed the surrender value, as the office is not losing business on a conversion or alteration.) Imagine that V1 is applied as a special single premium towards the new contract. The equation of value of the new contract is thus V1 + M.P.V. of future premiums (including any due now) = M.P.V. of future benefits + future expenses + C Rearranging this equation gives: V1 − C = M.P.V. of future benefits + expenses − premiums = V2 The office then solves this equation for the unknown quantity (usually the new sum assured or premium.) An example of the use of formula (7.3.1) has already been encountered in equation (7.2.2) above. Example 7.3.1. Ten years ago, a man now aged 40 effected a whole of life assurance for a sum assured of £10000, payable at the end of the year of death, by level annual premiums. The premiums were calculated using A1967-70 ultimate 4% and an allowance for expenses of 50% of the first year’s premium and 5% of each subsequent premium. Immediately before payment of the eleventh annual premium the man requests that the policy be converted into an endowment assurance maturing on his sixtieth birthday, with the annual premium remaining unaltered. Calculate his revised sum assured using a full preliminary term reserve (on A1967-70 ultimate, 4% interest) for finding the reserve of the original policy, and A1967-70 ultimate, 4% interest, expenses of 5% of premiums for calculating the reserve of the new policy. Solution. Annual premium is P , where 0.95P a30 = 10000A30 + 0.45P . ∴ P = £97.13. ¨ V1 = Reserve of original policy = 10000 9 V31 = £949.85 1− a40 ¨ a31 ¨ 7.3. ALTERATIONS AND CONVERSIONS Equation of value at date of conversion is reserve before conversion = (prosp.) reserve after conversion ∴ 949.85 = SA40:20 − 0.95 × 97.13 a40:20 ¨ allowing 5% of premiums for expenses ∴ S = £4717 123 Example 7.3.2. An office issued a ten-year endowment assurance to a man aged exactly 45. The monthly premium was £25 during the first five years increasing to £50 thereafter. The sum assured, payable immediately on death, was calculated using A1967-70 select 4%, allowing for expenses of £1 per month with additional initial expenses of 2 1 % the sum assured. 2 After five years the man requested that the premium remain at £25 per month for the next five years, with the death benefit staying unaltered. Calculate the reduced sum payable on survival using the premium basis and allowing for a £30 alteration charge. Solution. Let S =original sum assured. 300¨(12):10 + 300 a [45] D50 (12) ¯ = S 0.025 + A[45]:10 a ¨ D[45] 50 :5 ∴ S = £4871 (using a(12):10 = 8.154, a(12):5 = 4.489, ¨ [45] ¨ 50 D50 ¯ = 0.8093, A[45]:10 = 0.6808 ) D[45] + 12¨(12):10 a [45] Let S.A. on survival be reduced to S . Equate reserves before and after conversion [allowing for cost of conversion]: ¯ 4871A50:5 − (600 − 12)¨(12):5 − 30 a 50 D55 ¯1 − (300 − 12)¨(12):5 a 50 = 4871A50:5 + S D50 D55 ∴ (4871 − S ) = 300¨(12):5 + 30 a 50 D50 ∴ S = £3144 Example 7.3.3. In a certain country, the Universities operate the following superannuation scheme for academic staff, of whom there are two grades, Lecturer and Professor. Lecturers retire at age 65 with a lump sum of £30, 000, while Professors retire at age 70 with a pension of £12, 000 per annum payable monthly in advance. Staff pay for their benefits by means of level annual premiums, payable annually in advance until retirement or earlier death. Premiums are returned without interest on death before retirement, and withdrawals may be ignored. The employers do not contribute to the scheme. Death benefits are paid at end of year of death. The benefits are provided by policies issued by a life office which uses the following basis for all calculations: A1967-70 ultimate 124 4% per annum interest expenses are ignored. A Professor cannot be demoted to Lecturer. (i) (ii) (iii) CHAPTER 7. APPLICATIONS OF RESERVES Calculate the annual premium payable by a Lecturer on entry to the scheme at 25. Calculate the reserve, just before payment of the premium now due, for a Lecturer aged 45 with 20 years’ service. The Lecturer of (ii) is now promoted to Professor. Calculate his or her new annual premium, the first due at once. Solution. (i) Let the annual premium be P . The equation of value is 1 1 P a25:40 = 30000A25:40 + P (IA)25:40 ¨ =⇒ P = £274.44 (ii) V1 = 30000 D65 1 1 ¨ + 20P A45:20 + P (IA)45:20 − P a45:20 D45 = £8536 (iii) Let the new premium be P . We have, V1 = V2 That is, 8536 = value of future benefits − future premiums for new contract D70 (12) 1 1 + 20P A45:25 + P (IA)45:25 − P a45:25 a ¨ ¨ = 12000 D45 70 =⇒ P = £1525.87 7.4 The actual and expected death strains Consider N identical policies on lives aged x at entry and now aged x + t, and assume that the premiums now due are unpaid. The death strain at risk for a given contract in the coming policy year is S − t+1 V (7.4.1) where S denotes the sum payable on death (assumed payable at the end of the policy year) and t+1 V denotes the reserve at the end of the policy year. Since the life office may be considered to be holding the reserve t+1 V for each policy in force at time t + 1 in an imaginary “bank account” for the policyholder, the death strain at risk measures the extra cash required if he dies in policy year t + 1. Let θx+t denote the number of deaths in the coming year. Note that (in the notation used for mortality investigations) N = Ex+t = the q-type exposed to risk at age x + t (since each life is exposed to the risk of death for the whole policy year) 7.5. MORTALITY PROFIT AND LOSS Assuming independence of the lives we have θx+t ∼ binomial(N, qx+t ) which has mean N qx+t . Now consider the random variable A.D.S. = actual = θx+t (S − t+1 V) death strain The mean of this variable (even without the assumption of independence of the lives) is EDS = expected = N qx+t (S − t+1 V) death strain 125 (7.4.2) (7.4.3) Note. Once the number of deaths θx+t , is known (at the end of the policy year), ADS ceases to be a random variable: it is then a realisation of the variable. The position may be generalised to cover cases in which policies considered are not identical, provided that their premiums are due on the same date. The actual death strain is ADS = death claims in policy year t + 1 (S − t+1 V) (7.4.4) and the expected death strain is EDS = all policies in force at the start of policy year t + 1 qx+t (S − t+1 V) (7.4.5) Note. The sum assured S may depend on the policy year, t + 1. 7.5 Mortality profit and loss Consider a block of business in force at the start of policy year t + 1, the premiums now due being unpaid, and suppose that the life office holds funds equal to the total reserves for this block of business. Let us (i) ignore expenses, (ii) suppose that the premium and reserving bases agree, and (iii) assume that interest will be earned in policy year t + 1 at the valuation rate, i p.a. The mortality profit in policy year t + 1 is defined as the funds at the end of the year less the money needed to pay death claims and to set up reserves for the survivors. Hence mortality profit = all in force at the start of year t + 1 (t V + P )(1 + i) − death S− survivors claims t+1 V (7.5.1) Hence Notes. Under the present assumptions.5.5. the mortality profit to the life office in respect of this group of policies. Equation (7. a certain life office sold 100 10-year without profits endowment assurance policies to lives then aged 50.3) which may be thought of as saying that the accumulated funds at the end of the year.2).D.5.S. mortality profit = EDS − ADS as required. (t V + P )(1 + i) . minus A. and the sum assured under each contract is £30000.5. by equation (7.S. is equal to E.1. t + 1.4) may be rearranged to give (t V + P )(1 + i) = t+1 V + qx+t (S − t+1 V) On summing over all policies in force at the start of the year. Proof. mortality profit in policy year t + 1 = E. the difference between the actual funds at the end of the policy year and the funds required to pay claims and set up reserves. 2.5.6) Hence.4) (7. plus the expected cost of setting up reserves for the survivors.g. (1 − qx+t )t+1 V.5. we find that (t V + P )(1 + i) = all in force at the start of year all in force at the start of year t+1 V (7.D. The sum assured S and the premium P may depend on the policy year. must provide: (i) (ii) the expected cost of death claims.S.7) Example 7.S. 1.D. Equation (7. Sqx+t .5. − A.. i.5..2) Theorem 7.5. payable on maturity or at the end of the year of previous death. On 1st January 1988.126 mortality profit = at the start of year CHAPTER 7. Premiums are payable annually in advance. (t Vx + Px )(1 + i) = qx+t + (1 − qx+t )t+1 Vx (7.5.1.e.4) may be verified directly for particular policy types: e. APPLICATIONS OF RESERVES (t V + P )(1 + i) − deaths (S − t+1 V) − at the start of year t+1 V all in force all in force (7. that is.D. We have equation (t V + P )(1 + i) = Sqx+t + (1 − qx+t )t+1 V (7. The office calculates premiums and reserves on the following basis: . for a whole life policy with sum assured £1.5. Consider a particular policy in the group.5) + EDS (7. MORTALITY PROFIT AND LOSS mortality: A1967-70 ultimate interest: 5% per annum 127 expenses: nil All 100 policies were still in force on 1st January 1994.94 = 30000 − 300007 V50:10 = 30000 Hence mortality profit = E. Solution. The office calculates premiums. Basis: mortality: A1967-70 ultimate interest: 4% per annum expenses: nil. Solution.1. During 1993 sums assured of £50.5. Sums assured in force on 1. with annual premiums payable in advance throughout the term and sums assured payable on maturity or at the end of the year of death. Calculate the profit or loss from mortality for this group of policies for the year ending 31st December 1993. and maintains net premium reserves. on the basis given below.000.5.000 remained in force.2.S. and 2 policyholders died during 1994. Death strain at risk for each policy at 31.000 became death claims. at 31 December 1994. = (100q56 − 2) × 10753 = −11378 (Loss of £11378) a57:3 ¨ a50:10 ¨ = 10753 Example 7.1993 = £4050000 ∴ Expected death strain = q59 (4050000)[1 − 15 V45:20 ] = q59 (4050000) a60:5 ¨ a45:20 ¨ Actual death strain = 50000(1 − 15 V45:20 ) = 50000 a60:5 ¨ a45:20 ¨ .D. Calculate the mortality profit or loss to the office during 1994 in respect of this business.12.D. and there were no other exits in 1993.7.S. total sums assured of £4. At 31st December 1993. − A. On the 1st January 1979 a life office issued a number of 20-year non-profit endowment assurance policies. to lives then aged exactly 45. (It is assumed that the actual expenses are as allowed for in the basis used for premiums and reserves. If the premium basis differs from the reserving basis. is one half of the sum assured that will be paid if the policyholder survives to age 65.488 Some generalisations 1.S.a. where e is the expense incurred at the beginning of policy year t + 1.3.D.489 13. interest. given that the following reserving basis is used: prospective method. which we do not discuss here. Solution.88 the office’s funds were equal to its reserves.5. Then q55 [Sj − {Sj (A56:9 + 9 E56 ) − Pj a56:9 }] EDS = ¨ all j . one obtains expense profits or losses. APPLICATIONS OF RESERVES ∴ mortality profit = E.) 2. − A. Details of the policies in force on 31 December 1988 are as follows: Exact age 55 Total sums assured payable on death (£) 3500000 Total annual premiums (£) 250000 The claims in 1989 were on policies with the following total sums assured and premiums: Total sums assured payable on death (£) 50000 Total annual premiums (£) 4000 Assuming that at 31.3) remains correct. Let EDS and ADS denote the expected death strain and the actual death strain in 1989.8) remains true (with mortality. If expenses are included in the calculations.01299373 − 50000) = £874 4. payable at the end of the year of death during the term. = (4050000 × 0. If not.) This follows from the fact that equation (7.128 CHAPTER 7. gross premium reserve using A1967-70 ult. The equation corresponding to (7.5. no expenses.) Example 7. The sum assured..4) is (t V + P − e)(1 + i) = Sqx+t + (1 − qx+t )t+1 V (7.5. we need only alter t V + P to t V + P − e. calculate the mortality profit or loss in 1989.5. interest and expenses according to the reserving basis and P on the premium basis. For several years an insurance company has issued a large number of special endowment assurances.5.5. 4% p. Each policy matures at exact age 65 and is effected by annual premiums payable on each January 1st throughout the term. formula (7.12.8) and formula (7.D.S.3) remains true provided that reserves are calculated by a gross premium method (either retrospective or prospective. 25P . Assuming as before that the funds at the end of policy year t equal the reserves for the business under consideration.000 after 10 years and to £30.7.1) Example 7.000 was issued to a life then aged 40 exactly. Interest profits. 7. The sum assured. The office’s basis for premiums and reserves is A1967-70 ultimate 4% per annum interest expenses are ignored (i) Calculate the initial annual premium P .223. we shall give certain straightforward results on interest and surrender profits here. i. and (b) 31st December 1990.e. Thus ADS = Claims [Sj − {Sj (A56:9 + 9 E56 ) − Pj a56:9 }] ¨ Sj ) − ( Claims Claims =( Sj )(A56:9 + 9 E56 ) + ( Claims Pj )¨56:9 a Pj = 40000) = 13082 (since Claims Sj = 50000 and Claims The mortality profit is thus = 5859−13082=−7223.6.a.a. On 1st January 1978 a special twenty-year endowment assurance policy for an initial sum assured of £20. the interest profit is [ all in force at the start of policy year t + 1 (t V + P − e)](i − i) (7. The annual premium was P for the first ten years. and thereafter increased to 1. i.1. in policy year t + 1.000 on maturity.e. and the reserve for the contract at (a) 31st December 1989. which is payable at the end of the year of death or on survival to maturity date. OTHER SOURCES OF PROFIT AND LOSS where the summation is over all policies in force at the start of the year. Suppose that the office earns interest at a rate i p.6.6 Other sources of profit and loss Although some of these topics may also be considered under the heading of “profit-testing”. increased to £25. EDS = q55 ( Sj ) − ( Sj )(A56:9 + 9 E56 ) + ( Sj = 3500000 and all j all j 129 Pj )¨56:9 a Pj = 250000) = q55 (694048) (since = 5859 The actual death strain is obtained by a summation of the death strains at risk over those policies which become claims.6. a loss of £7. . the interest rate in the reserving basis being i p. D.D. (i) Let initial premium be P . The office found that it made neither profit nor loss in 1990 in respect of the policies of (ii) above.25P ) = 2494 ∴ i = 0.04) − 25000q52 ] 1 − q52 = £15772 (this may also be calculated directly) (ii) E. APPLICATIONS OF RESERVES Find the profit or loss from mortality for the year ending 31st December 1990 for a group of such policies.S. given that one policyholder dies in 1990 and 120 policies remained in force at 31st December 1990.25N50 − 1.) surrenders (7.2510 |¨40:10 ] = 20000A40:20 + 500010 |A40:10 + 5000 a a 20000M40 + 5000M50 − 25000M60 + 30000D60 N40 + 0. The surrender profit is (t+1 V − S.00 D60 D40 ∴P = (a) 12 V = 25000A52:8 + 5000 = £14085 D60 − 1.14% Surrender profits. P [¨40:20 + 0.04)[121](12 V + 1.0414.) (iii) Solution.S. = 121q52 (25000 − 13 V) = £6734 A.2) .130 (ii) CHAPTER 7.6. What rate of interest did it earn in 1990? (Expenses are negligible. The interest profit is (i − 0.25P a52:8 ¨ D52 (b) 13 V = [(12 V + 1. or 4.25N60 = £907. = 25000 − 13 V = £9228 ∴ mortality profit = 6374 − 9228 = −£2494 (iii) Let interest rate earned be i . There were no withdrawals in 1990. all effected on 1st January 1978 by lives then aged 40 exactly.25P )(1.V. Assume that surrenders may take place only at the end of a policy year (just before payment of any premium then due). 6.6. Premiums are payable annually in advance during the term of the policies and death claims are paid on 31st December each year. interest Ignore expenses. is the surrender value paid. On 31st December 1987 ten endowment assurance policies and five pure endowment policies were surrendered.6. OTHER SOURCES OF PROFIT AND LOSS where S. Reserve per policy at end of 1987 = 100011 V40:20 = 1000 1 − a51:9 ¨ a40:20 ¨ 7531 = 1000 1 − 13.15 = 26202 A.V.7.S.3) Example 7. Calculate the profit or loss arising during the year 1987.15 ∴ E.15 = 32829 . without payment of any sum assured. interest Surrender values: Net premium reserve on A1967-70 ultimate at 6% p. On 1st January 1977 an office issued a block of without profits endowment assurance policies and a block of without profits pure endowment policies to men aged exactly 40.a. showing separately the contributions from mortality and from surrender if the office uses the following basis: Premiums and reserves: A1967-70 ultimate at 4% p. Solution. = 1000q50 × 547. = 60 × 547. Summary of profit and loss Profit or loss for year = accumulation of funds at end of year − cost of death claims − cost of surrenders − cost of setting up reserves for remaining policies = Mortality profit + interest profit + surrender profit 131 (7.85 ∴ Death strain at risk per policy in force at start of 1987 is 1000 − 452.764 = 452. Each policy was for a term of 20 years and a sum assured of £1000. On 1st January 1987 there were 1000 endowment assurance policies and 5000 pure endowment policies in force. by the deaths of the assured lives.S.D. Endowment Assurances. Let us consider each type of policy separately. During the year to 31st December 1987 sixty endowment assurance policies became claims by death and twenty pure endowment policies were terminated.2.85 = 547.a.D. = 100011 V40:20 = 1000 1 − at 6 % = 407.55 ∴ Surrender profit is 5(425. = −1678 S.S.D.132 CHAPTER 7.55) = −10189 A.a.08 = 425.D.S.25 = 23. interest ¨ l51 = 546.55 per policy. = −20 × 425. Define P = net premium at 6% p. Annual premium is P = Reserve per policy at end of 1987 is P N40 − N51 D51 = 29.00 Summary Mortality profit = −£8305 Surrender profit = £670 Total profit = −£7635 (a loss of £7.55-381. = −6627 Surrender profit=10(Reserve-S.55)=220.52 11.a.D.87 − 165.V.029 11.S. − A.V.D.635) .55 = −8511 ∴ mortality profit = E.D.70 Pure Endowments.V. − A. APPLICATIONS OF RESERVES ∴ mortality profit = E. interest = 1000v 20 l60 l40 a40:20 ¨ = 279. It is easier to find the reserve prospectively.55 D60 1000 D40 a40:20 ¨ = 29.S.88 7.70 132002 − 68970 4399.70 Death strain at risk is −425. E.871 ∴ Surrender profit = 449. where S.).32 = 381.S.= net premium reserve at 6%. = 5000q50 (−425.S.871 ∴ Reserve = 1000v 9 l60 − P a51:9 at 6% p.D. expenses: Nil (i) (ii) (iii) Show that the annual premium for each of these policies is £242. on the premium basis. On 31 December 1988 there were 8567 policies still in force.4 On 1 January 1991. assuming that the premium then due is unpaid. In 1995. Premiums were payable annually in advance throughout life. the policyholder requests that the policy be converted to a without profits endowment assurance for the same sum assured.000.27.1 Explain briefly how a life office calculates the new sum assured or annual premium on the alteration or conversion of an existing policy. Expenses of alteration may be ignored. a total of 10. all premiums paid are returned without interest. On 1 January 1995. each with sum assured £1000 (payable at the end of the year of death. Immediately before payment of the 15th premium. EXERCISES 133 Exercises 7. a life office issued a number of identical special endowment assurances to lives then aged 45. using the following basis for all calculations: mortality: A1967-70 select interest: 4% p. on 1 January 1996 for one of these policies. a sum assured of £1000 is payable at the end of the year of death and. if this occurs within the term) to a group of lives all then aged 25. in addition. Find the revised annual premium.2 A life office issued a whole life without profits policy to a woman aged 30. 7. and the sum assured was payable at the end of the year of death. with expenses of 4% of each premium. Each policy had a term of 15 years and level annual premiums were payable throughout the term. The premium basis for these policies was as follows: mortality: A1967-70 select interest: 4% p. On death before the end of the term. 7. if before age 60. expenses: none 7.3 On 1 January 1972 an office issued a large number of 40-year endowment assurance policies. and the office gave surrender values equal to 95% of the reserve on the premium basis. payable on maturity at age 60 or at the end of the year of death. a sum assured of £5000 is payable.a. the expected death strain and the mortality profit or loss for the business in 1989. Calculate the profit or loss to the office in 1995 from this block of business in respect of (a) mortality .000 of these policies were still in force. and . and the life office earned 4% interest on its investments.7. During 1989 there were 13 deaths among the policyholders.7. Expenses were negligible in 1995. 47 of these lives died. The surrenders all took place at the end of 1995.a. The office calculates premiums and maintains reserves on the basis of A1967-70 ultimate mortality and 4% per annum interest. Calculate the reserve. with sum assured of £110. ten policies were surrendered. Find the actual death strain. On survival to the end of the term. decreasing by £10. £90. payable at the end of the year of death. The premium and reserve basis for this policy is as follows: mortality: A1967-70 Ultimate interest: 4% p.000 each year. .134 (b) surrenders.000 if the life dies in the second year. is £100. Level annual premiums are payable throughout the term of the policy. CHAPTER 7. APPLICATIONS OF RESERVES 7. and so on. The sum assured.a. expenses: Nil (i) (ii) Calculate the annual premium for the policy.5 A life office has issued a 5-year decreasing temporary assurance to a life aged 60. Write down a relationship between the reserve at integer duration t and the reserve at duration t + 1.000 if the life dies in the first year. ¨ V1 = 110. be P .04 Reserve before conversion = V1 = 110. per policy in force at the start of the year. SOLUTIONS 135 Solutions 7.52−9.3 On 31/12/88 the duration is 17 years. 825 7.2 Let P = original A.P. 923 0.4)=£1. 000A44:16 − 0. = 110000 P30 1 − 0. therefore. 923 Let new A.1 The office calculates V1 .97)=£716. we have: total expected death strain = 8567 × 1. a death strain at risk of (1000-283.14. The expected death strain per policy is 716.54703 − 16. The office then sets up the equation V1 − C = V2 where C = cost of alteration (if any) and V2 = (prospective) reserve of new policy.31105 = 11.234.4 The 1989 mortality profit is therefore (11. 000 × . the reserve for the original policy at the date of conversion. 00014 V30 = 110.926.8. 308.7.03.e. 234.P. We solve for P in the equation V1 = V2 . The reserve at the end of 1989 (per unit sum assured) therefore equals ¨ a 18 V[25]:40 = 1 − a43:22 /¨[25]:40 = . Solve for unknown quantity (usually a sum assured or premium. For the block of business.777 = £3.03 × q42 = 1.31105.52 total actual death strain = 13 × 716.) 7. . 000 1 − = 16. i.28397 Hence in 1989 we have.96 × 11. 000A44 − 0.308.03 = 9.96P a44:16 a44 ¨ a30 ¨ ∴ P = 110.96P a44 ¨ = 110. CHAPTER 7.A.R.05 × 5 V = £675. 798 (loss of £3. be P .5 (i) Let the Annual Premium be P .60855) ∴ P = £242. 000q49 × DSAR = £36.P.50271) D[45] 1 + P (IA) [45]:15 (= 0. 7. . in 1995 in respect of 1 policy = (1000 + 5P ) − 5 V = £861. 000t).03 (ii) (t V + P )(1.798) (b) Surrender profit per policy = 0.56732) ¨ + 4000 D60 (= 0.250) = 1.97 EDS = 10. 000t) + (1 − q60+t )t+1 V (since the sum assured on death in year t + 1 is 100.S. 000 − 10.136 7. 000A[45]:15 (= 0. 714 ADS = 47 × DSAR = £40. 349 D60 1 1 + 5P A50:10 + P (IA)50:10 − P a50:10 ¨ D50 (iii) (a) D. 000(IA)60:5 ¨ which gives P = £1. APPLICATIONS OF RESERVES P a[45]:15 (= 11. 513 ∴ mortality profit = EDS − ADS = −£3.4 (i) Let A.04) = q60+t (100.05 × reserve ∴ Total surrender profit = 10 × 0. 000A60:5 − 10. 1 1 P a60:5 = 110.27 (ii) 5V = 1000A50:10 + 4000 = £1. 347. 000 − 10. 1) µ∗ [x]+r dr (µ[x]+r + k) dr (8. Those accepted on special terms are considered to be subject to an extra mortality risk due to a health impairment.Chapter 8 EXTRA RISKS 8.2. (If a non-select table is used. (Females are sometimes treated as males 5. years younger.1 Introduction It is usual in practice for a life office to accept the majority of proposers for life assurance policies on “normal” (or “standard”) terms. a dangerous occupation or a dangerous spare-time activity (such as motor racing. 8. but shall assume that the addition to the normal rates or force of mortality is known. say.2 A constant addition to the force of mortality Let µ[x]+t refer to the force of mortality at duration t years for a life who was selected for assurance to age x on the office’s normal terms.e. although this rule is very inaccurate at certain young ages. Some offices make extensive use of “rating up”: that is impaired lives are treated as normal lives a number of years older than actual age. i. the remainder being either declined or accepted on special terms. we shall not go into this. omit [ ].) Smokers are sometimes treated as non-smokers 6 years older. and let µ∗ [x]+t be the corresponding force of mortality for a life accepted on special terms. We now assume that there is k > 0 such that: µ∗ [x]+t = µ[x]+t + k Hence ∗ t p[x] = exp − t 0 t (0 ≤ t ≤ n) (8.2.) Let n years be the duration of the policy.) The assessment of extra risks is usually carried out by experienced underwriters using a “numerical rating system”.2) = exp − 0 = e−kt t p[x] 137 . they are accepted but with higher premiums or reduced benefits. 019048 = δ. Solution. 000 1 1 − = £108.019048 to the force of mortality for the next 15 years.2.5) This rule does not apply to assurances and premiums. Extra annual premium is ∗ 10.2. payable at the end of the year of death.06 ¨ But d is still at 4% interest.04 ¨ .000.138 Also. aged 50.2.1.06 ¨ a50:15 . EXTRA RISKS = v t t p∗ [x] = e−(δ+k)t t p[x] = e−δ t t p[x] (8. Let ∗ indicate special mortality basis.51 a50:15 .2. or on survival for 15 years. Since an annuity factor is a sum or integral of pure endowment factors. as in the following example. a∗ = a[x] ¨[x] ¨ Example 8.4) and hence i = eδ − 1 = eδ+k − 1.06 = a50:15 0.04 + 0.3) where δ =δ+k (8. A certain life office uses the following basis for calculating premiums for assurance policies on lives accepted at normal rates: A1967-70 ultimate 4% interest expenses are ignored A certain proposer. The sum assured is £10. Hence the extra premium is 10. but one may sometimes use conversion relationships to express these in terms of annuity factors. for a 15-year endowment assurance without profits by annual premiums is considered by the office to be subject to the mortality of the office’s normal table with an addition of 0. Calculate the addition to the normal annual premium. ∗ t E[x] CHAPTER 8. we have (for example) on normal mortality but with force of interest δ = δ + k (8. 000[P50:15 − P50:15 ] Note that ∗ P50:15 = 1 − d by a conversion relationship a∗ ¨50:15 1 − d since δ. 5 lx i.a.5 = lx+t e−kt = 0.t p∗ dt (by integration by parts) x t. e.a.V.2. A VARIABLE ADDITION TO THE FORCE OF MORTALITY 139 In some problems we must evaluate assurance factors directly.P. To find the median future lifetime of the impaired life.2.8.3)) = 0 t px e dt (8.3 A variable addition to the force of mortality µ∗ [x]+t = µ[x]+t + ξ(t) (0 ≤ t ≤ n) (8.1) Let us now suppose that there is ξ(t) ≥ 0 such that .3.7) = ax ¯ Note also that at force of interest k p. we solve the equation ∗ t qx (8.e.a. E[(T ∗ )2 ] = 2 0 ∞ t. ∗ t px 8.g.3.8) ∞ =2 0 ¯a = 2(I¯)x Hence ¯a Var(T ∗ ) = 2(I¯)x − (¯x )2 at force of interest k p.2.t px e−kt dt at force of interest k p. of an increasing temporary assurance for an impaired life aged x at entry n = 0 n tv t t p∗ µ∗ dt x x+t tv t (t px e−kt )(µx+t + k) dt t (v )t t px (µx+t = 0 n = 0 + k) dt v is at rate i (with δ = δ + k) ¯¯ 1 = (I A)x:n at rate of interest i at rate of interest i (8.6) ¯a + k(I¯)x:n A formula for the expectation of life of an impaired life We have e∗ = E(T ∗ ) = x ∞ o ∞ 0 −kt ∗ t px dt (by formula (1. ¯¯ 1 (I A)∗ x:n = M.9) = 0. a Remark. (8. without the use of conversion relationships.4.2. expenses: 2% of the single premium.1104 × 0.a. The addition is 0. 000 0 v t t p∗ µ∗ 75 75+t dt 15 ∗ µ75 + v 15 15 p∗ µ∗ 75 90 2 = 50.3.020 at age 90.005 + 0. if this occurs within 15 years.005t + 0.001t ∴ t p∗ = t p75 exp − 75 t 0 (0 ≤ t ≤ 15) (0. A certain impaired life aged 75 experiences mortality according to a(55) males ultimate with an addition to the force of mortality.005 at age 75.1. 15 0. as in the next example. (i) (ii) Find a formula for the probability that this life will be alive at age 75 + t years (0 ≤ t ≤ 15).140 We obtain the formula ∗ t p[x] CHAPTER 8. increasing linearly to 0.005 + 0.4 Rating up In practice. EXTRA RISKS t = t p[x] exp − ξ(r) dr 0 (8.98P = 50. the single premium for a temporary assurance of £50. 000 × 8. .) This allows the life office to base all its premiums on one set of tables.2) t In some cases 0 ξ(r) dr may be calculated analytically.06550 + 0. (i) µ∗ 75+t = µ75+t + 0. on the following basis: mortality: as described above interest: 10% p.5 [0.3.23939 × 0. 000 × 7.2636] ∴ P = £27. Example 8. Solution. impaired lives are often considered to experience the mortality of a “normal” life r years older (where r depends on the severity of the impairment. by approximate integration using the trapezoidal rule (not repeated). Estimate.0005t2 ) (ii) Let single premium be P .001r) dr = t p75 exp −(0. 730 50.000 payable immediately on death of (75). Suppose that the normal mortality table follows Gompertz’ law. The impaired life finds that the level annual premium for this policy is the same for a policy providing identical benefits for a life his age accepted at normal rates. so we have the approximate rule r= log(1 + θ) log 1.1) for some constant.4 = 4. r log 1. θ. if θ = 0.4. This life requires a contract providing the following benefits: (i) (ii) £30. (This formula is usually suitable for “medical” impairments.4.4. but it may be sufficiently accurate in practice since the value of θ may in general be estimated only rather approximately.4.2) (0 ≤ t ≤ n) In practice c is usually about 1. r= log(1 + θ) log c (8.8. aged 50.) Hence µ∗ = (1 + θ)Bcx+t x+t = Bc(x+r)+t where cr = 1 + θ.08 For practical purposes this value of r would be rounded to 4 or 5 years. if within 10 years.4. . Example 8.1.37 log 1. A certain life office’s premium basis for lives accepted at normal rates is as follows: Mortality: A1967-70 ultimate Interest: 4% per annum Expenses: none A certain impaired life.08 (8. This rule is not precise.3) For example.e. 000 + S).) Suppose that the impairment is such that µ∗ = (1 + θ)µx+t x+t (0 ≤ t ≤ n) (8. so there are constants B. the sum of £(30. RATING UP 141 A practical method of finding “r” is as follows.08. on survival for 10 years. i. Find S. c such that µx+t = Bcx+t (0 ≤ t ≤ n) (This is usually a fairly accurate representation of mortality for all ages over about 30. for accident risks it is generally better to add a constant to the normal force of mortality.4 (corresponding to a 40% increase in the force of mortality). is considered to be subject to the office’s normal mortality rates for a life 10 years older. 000 at the end of the year of death. A life office calculates annual premiums for without profits endowment assurances using the following basis: Mortality: Interest: Initial expenses: A1967-70 select 4% 50% of the first premium Renewal expenses: 5% of each premium after the first (i) Calculate the annual premium payable for a 25 year endowment assurance taken out by a male life aged exactly 45 for a sum assured of £50. or (b) diminishing in arithmetic progression to zero in the final policy year. Such reductions in the death benefits are called “debts” or “liens”. 809 8.5. 000(0. A man aged exactly 45 effects a policy identical to that in (i) above but rated up 7 years.142 Solution. (ii) .1.068168 = £25. normal premium = 30.5 Debts Instead of paying higher premiums. a proposer accepted on special terms may elect to pay the normal premiums with reduced death benefits. 000P60:10 + S Equate these to give S= = a50:10 ¨ a60:10 ¨ 30. the life office offers to charge the standard premium but to impose a level debt.075689 − 0. 000(P60:10 − P50:10 ) D60 D50 a50:10 ¨ − D70 D60 a60:10 ¨ 30. (Note that a debt is a reduction in the death benefit only: the maturity benefit is not reduced. EXTRA RISKS = 30. (b) Alternatively. payable throughout the term of the policy. and may be either (a) a level amount. 000A50:10 + S D60 D50 a50:10 ¨ D60 D50 D70 D60 CHAPTER 8. 000P50:10 + S “impaired life” = 30. We first give an example involving a level debt. The death benefit is payable at the end of the year of death. Calculate the amount of the debt. (a) Calculate the level extra premium.08339) 0.08986 − 0.) Example 8.000. 752. the deduction from the sum assured) such that the debt reduces uniformly each year to nil in the last policy year. 0. a bar should be placed over the A symbols.95P a[45]:25 = 50. 000A[52]:25 − DA[52]:25 + 0... D in year n − 1 0 in year n Consequently the M.5. The equation of value is 1 0.95P a[52]:25 = 50.V. Let P be premium.. Level monthly premiums are payable in advance throughout the term or until earlier death. A life office issues 20-year without-profit endowment assurance policies providing a sum assured of £8.. 649 143 Now suppose that we are dealing with an n-year endowment assurance... the debts run as follows: (n − 1)D in year 1 (n − 2)D in year 2 .e..8.91 (b) Let death benefit be reduced by D. DEBTS Solution..P. Assuming that the ordinary premium appropriate to this life’s actual age is payable.. (i) Let A..000 on maturity or at the end of the year of earlier death. that is. Example 8.5. calculate the initial amount of the debt (i. of the debts. 0..5... 000A[45]:25 + 0. A certain impaired life aged 45 is considered by the office to have the mortality rates of a “normal” (select) life 5 years older.41 (ii) (a) Consider a “normal” (select) life aged 52.45P ¨ ∴ P = £1.32 ∴ Extra premium = £259.45P ¨ ∴ P = 1. be P . is 1 1 D{nA∗ [x]:n − (IA)∗ [x]:n } (8.. and that the debt diminishes in arithmetical progression to zero in the final policy year..1) where ∗ indicates impaired lives’ mortality..95P a[52]:25 = 50. 492. Basis: Mortality of “normal” lives: A1967-70 select Interest: 4% . 000A[52]:25 + 0.) The calculations are illustrated in the next example. at the issue date.2.45P ¨ ∴ D = £12.P.. the death benefits are taken to be payable at the end of the year of death (if these benefits are payable immediately on death. 12 − 296.507 − 0.95P a[45]:20 − 0.45833 1 − D[50]+1 D[50] D70 D[50] 20(M[50] − M70 ) − (R[50] − R70 − 20M70 ) D[50] = = 3973.5606 − 0. Let P be “normal” monthly premium.62253] − 0.14 (£26. The equation of value for D is 1 1 0.25 1 − 11 24 1− D[45]+1 D[45] = = 3844.98111] 1 4581. The equation of value for P is 12 ¨ 0.66887] + 78.24541 = £312.25[1 − 0. 896.45833 × 0.25P a[45]:1 = 8000A[45]:20 ¨ P = 8000 × [0. EXTRA RISKS Solution. 844.95[13.289 − 9360.71 .087 − 0.035[0.95 × a[50]:20 − 0.14¨[50]:20 − 0.01 per month) Let the initial debt be 19D.45833 × 0.45833 1 − ¨ +0.08 12.14¨[50]:1 = 8000A[50]:20 − 20DA[50]:20 − D(IA)[50]:20 a a (12) (12) D= 8000 × 0.95 a[45]:20 − ¨ 11 24 (12) (12) 1− D65 D[45] − 0.53[13.25 × 312.322 [20 × 758.49664 − 312.898 = 205.09 1.08 {0.45833 × 0.48051] .14 × 0.22] 259.14 1 − 0.2672 ∴ Initial Debt is 19D = £3.95 × 312.25 × 312.04014]} 3.144 expenses: 30% of premiums in first year 5% of later premiums CHAPTER 8. 0025 at age 60.8. will die between 20 and 30 years from the present time. and think of an approximate relationship between (a) the level debt and (b) the initial debt when debts decrease linearly to zero.6. increasing linearly to 0. Calculate this level extra premium.0005 at age 40. The explorer is about to undertake a hazardous expedition which will last three years. The life office quotes a level extra premium payable throughout the term. Can you envisage circumstances under which an office could offer an impaired life who wishes to pay the “normal” premiums a level debt but not a diminishing debt? Hint Consider a life who is very severely impaired.009569 to the normal force of mortality. For lives accepted at normal rate. The life office quotes a level extra premium payable for the first three years. 8. the mortality of a life 8 years older.2 3% per annum none An impaired life aged exactly 55 wishes to effect a without profit endowment assurance for a sum assured of £1. a constant addition of 0.000 payable at the end of 10 years or at the end of the year of earlier death.1 An explorer aged exactly 57 has just made a proposal to a life office for a whole life assurance with a sum assured of £10.4 (i) . but after three years will experience normal mortality. level annual premiums are payable until death under this policy. The addition is 0. will die within 30 years. Basis: normal mortality: A1967-70 ultimate interest: expenses: 3% per annum none 8.02871 to the normal force of mortality. The life office estimates that during these three years the explorer will experience a constant addition of 0. EXERCISES 145 Exercises 8. Find the probability that an impaired life aged exactly 40 (i) (ii) (iii) will die within 20 years. Special terms are offered on the assumption that the life will experience mortality which can be represented by: (a) for the first five years. and (b) for the remaining five years. at which level the addition remains constant. Calculate this level extra premium on the following basis: normal mortality: A1967-70 ultimate interest: expenses: 8.3 A group of impaired lives now aged 40 experience mortality according to A1967-70 ultimate with an addition to the force of mortality.000 payable at the end of the year of death. Level annual premiums are payable throughout the term of the policy. 5 The mortality of a certain impaired life aged exactly 60 may be represented as follows: (1) at ages up to exact age 65. mortality follows that of a “normal” (ultimate) life 4 years older. Calculate the level debt which would be deducted from the normal death benefit. and (2) at ages between exact age 65 and exact age 70. and suggests that the office should make the normal death benefits subject to a debt which decreases in arithmetical progression to zero in the final year. for a 20-year endowment assurance by annual premiums with sum assured £40. with an additional initial expense of 2 1% of the sum assured (before deducting any debt. 000 payable on maturity or at the end of the year of earlier death. Calculate the initial debt. (b) Suppose that the proposer also asks your office to quote a level debt. aged 40.146 (ii) CHAPTER 8.a. The life wishes. however. to pay the “normal” level annual premium for his age. which is that of A1967-70 select. interest expenses of 6% of all premiums.a. Your office’s basis for calculating premiums for “normal” lives is A1967-70 ultimate 4% p. there is a constant addition of 0.009569 to the “normal” force of mortality. expenses: 2 1 % of all premiums. 8. EXTRA RISKS A certain proposer. (a) The proposer asks to pay the annual premium for a “normal” life aged 40. Suppose that this life requires a 10-year endowment assurance without profits with sum assured £10.000 without profits (payable at the end of the year of death) is considered by your office to be subject to the mortality of the normal table with an addition of 10 years to the age. Calculate the level debt.) . using the basis given below: mortality: as given above interest: 4% p. 3207 a57:3 ¨ PNORMAL = 1000P55:10 = £85.804 ¨57:3 a∗ 3 |¨57 = 1.677 ∴ P ∗ = £91.06−3 (1 − A60 ¨ l57 = 0. Then a57 × P = 10. Then. 000A∗ − P (¨∗ a57 57 = 256.2 a57:3 + 3 |¨∗ ) 10.0005 + 0.0001t (0 ≤ t ≤ 20) µ40+t + 0.1 Let P be the normal annual premium. µ∗ to impaired lives.95 8.79 so extra annual premium = £5..84 P ∗ = 1000 1 a∗ ¨55:10 − d at 4% ∗ a∗ ¨55:10 = a∗ ¨55:5 + [5 E55 ]¨∗ a60:5 l60 = a55:5 at 5% + v 5 ¨ at 5% [¨68:5 at 4%] a l55 = 7. SOLUTIONS 147 Solutions 8.41 at 3% Let E be the extra premium charged for 3 years.7.0025 (t > 20) . the equation of value will be (* denotes impaired mortality): (P + E)¨∗ a57:3 + P 3 |¨∗ = 10. 000A57 ¨ P 377.59572 at 3%) Hence E= 8.3 Let µx refer to A67-70 ult. x µ∗ 40+t = µ40+t + 0.076 ∗ 1 A∗ = A∗ 57:3 + 3 E57 A60 at 3% 57 ∗ ∗ = A∗ − 3 E57 + 3 E57 A60 at 3% 57:3 l60 = 1 − d3% a57:3 6% − 1.8.06−3 l60 a60 ¨ l57 at 3% = 11. 000A∗ a57 57 where ¨ a∗ 3% = a57:3 6% = 2. 89 × 13.17 D= = £12. 575.A. 000P40:20 = 1454.148 (i) ∗ 20 p40 20 CHAPTER 8.e.94 If initial debt is 19D.0025] l60 = 0.4 (i) 0.895579 × 0. we have the equation of value 1 1 0. > level debt (in cases of serious impairment) 40.49818 − 0.04074 = 1. 929. debt is one-half of the initial debt).0005 + 0.16819 8.16819 − 2. 000 × 0. 000A50:20 − D{20A50:20 − (IA)50:20 } a i.89 0.45 . exceed the “normal” sum assured on death.94 × 1454. The level debt must be (very roughly) about one-half of the initial debt (since the average dim.86911 × 10 p∗ 60 l70 = 0. 000 A[60]:10 + 100 ¨ 7. 930) 1 1 1 (b) Replace D{20A50:20 − (IA)50:20 } in above by the term DA50:20 .P.5 Let P be the normal A. 0. EXTRA RISKS = exp − 0 (µ40+t + 0.69677 ∴ P = £919.26 Normal A.94 (£29.01 + 0. of course.1309 (ii) ∗ 30 p40 = 0.884 0. so we may have cases in which: initial debt (when debts diminish) (ii) (a) > S.86911 × × exp[−10 × 0.89)¨50:20 = 40. 392 0.975P a[60]:10 = 10.66656 ∴ Ans = 0.0001t) dt 20 0 1 = 20 p40 exp − 0.047 D= 20 × 0.P. = ∴ Initial debt = £29.02)} l40 = 0.3334 (iii) 8.0001t2 ) 2 l60 = × exp {−(0.3334−0.1309=0.86911 ∴ ans = 0.94(1454.2025 The debt cannot.0005t + (0. to give 2084.9704455 = 0. 40. 5 a∗ ¨[60]:10 = a[60]:5 0.04 ] l65 l74 l[60] l69 = 0.230719 = £1.05 v.04 ¨ 4.230719 ∴D= 402.05 + v0.708835 − [v.452 = 7.7.975P a∗ ¨[60]:10 = 10. 0.570 Let level debt be D.8. SOLUTIONS Let ∗ indicate special mortality.05 ¨ 149 l65 l[60] a69:5 0.29496 − 1 D A∗ [60]:10 + 100 1 − d¨∗ a[60]:10 = 0.478114 = 0.708835 − 0.708835 5 5 0. 000 7.119 0. 743 .570 A∗ [60]:10 0.72603 4. 150 CHAPTER 8. EXTRA RISKS . (In a mutual life office there are no shareholders. that is. and the vector of profits (indexed by the policy year. in each policy year in two stages. (We assume here that the age at entry to assurance is x. the sale of a with-profits policy may be considered to be an “investment” on the part of the office’s shareholders.) The corresponding vector of profits (after allowing for reserves) is called the profit signature.1 Principles of profit-testing In this book we shall consider “conventional” life assurance business only. 151 . The resulting net cash flow is called the in force net cash flow. The estimated net cash flow per policy sold in policy year t is called the initial net cash flow for that year. of a given contract is calculated on the assumption that the policy is still in force at the start of that year. which is written as {σt }. which is written as {(PROt )}. though the position is complicated by the fact that future profits are shared between the with-profits policyholders and the shareholders. t−1 px . (4) Random fluctuations in mortality are ignored. Stage B. and the expected profit. and the profit after allowing for appropriate reserves. t) is called the profit vector. so premiums less expenses are accumulated to the end of the appropriate policy year at the rate of interest which the office assumes it will earn on the life funds. not unit-linked policies (which are considered in part D1. and is found by multiplying the “in force” net cash flow by the probability that the policy is still in force at the start of the tth policy year. (3) All cash flows are considered at the end of the policy year. (2) The office calculates the expected net cash flow. each policy is considered as if it were one of a large number of identical contracts whose experience exactly follows the mortality table used by the office in its projections.) The main new ideas in profit-testing are as follows: (1) The sale of a without-profits policy is considered by the life office as an “investment” on the part of the shareholders and/or with-profits policyholders.) Similarly. The net cash flow in each policy year. and that a non-select mortality table is used.Chapter 9 PROFIT-TESTING 9. Stage A. which is that which. We now define. Basis 3: the experience basis.e.e. An endowment assurance policy. the net cash flow in year t per policy sold. the office assumes.2 Cash flow calculations We consider an n-year “generalised endowment assurance”. In some cases all three of these bases agree. In general.3) prob. that policy will still be in force at time t − 1.e.2. is issued to a life aged 55.2.2) (9. for 1 ≤ t ≤ n. (CF)t = the “in force” expected net cash flow in policy year t We clearly have (CF)t = (Pt − et )(1 + i) − accumulation of premium less expenses to end of year qx+t−1 Dt − px+t−1 St expected cost of death benefits expected cost of survival benefits (9. Dt = death benefit in policy year t (paid at the end of the policy year). The symbols qx+t . while Pt is calculated using Basis 1. term five years and level annual premiums. i.152 CHAPTER 9. The annual premium is calculated on the following basis: . with sum assured £5000. i = rate of interest p. Basis 2: the reserving basis. at start of policy year t Example 9. t−1 px . refer to the mortality table which the life office assumes will apply to the policyholders under consideration. St = survival benefit at the end of policy year t (St is usually zero for t < n). will be experienced.1. payable at time t − 1 years from the issue date). etc.2. and use the following notation: Pt = premium payable in policy year t (at the start of that year. is found from the important equation: “initial” net cash flow in year t = t−1 px (CF)t (9. i and qx+t given above all refer to Basis 3. Note.a. issued to (x). three distinct bases are used in profit-testing: Basis 1: the premium basis. et = expenses assumed to be incurred in policy year t (at the start of that year). i. which is used only to calculate the premiums. i. or two of them agree.2. which the life office assumes it will earn on its life funds. PROFIT-TESTING 9. Note that the symbols et .1) The “initial” expected net cash flow in year t. which is used only to calculate the reserves. 9.2. CASH FLOW CALCULATIONS Mortality: Interest: A1967-70 ultimate 6% per annum 153 Initial expenses: £250 Renewal expenses (associated with the payment of the second and each subsequent premium): £42 at the time of payment of the second premium, increasing thereafter by 5% per annum (compound). The death benefit is payable at the end of the year of death. (i) (ii) Show that the annual premium is £948.74 Assume that in calculating cash flows and profit signature for the policy the office uses the premium basis. On this basis determine for each year of the policy’s duration (a) the ‘in force’ expected cash flow, and (b) the ‘initial’ expected cash flow. Solution. (i) Let the annual premium be P . Then 4 P a55:5 = 5000A55:5 + 250 + ¨ t=1 42(1.05)t−1 t p55 v t (where all functions are on A1967-70 ultimate, 6%) , from which it follows that P = 5000 × 0.75171 + 250 + 152.62 4.386 = £948.74 154 (ii) We draw up the following table: (1) Year t 1 2 3 4 5 (2) Premium Pt 948.74 948.74 948.74 948.74 948.74 (7) In force expected cash flow (CF)t 698.46 914.04 906.42 898.13 −4045.87 (3) Expenses et 250.00 42.00 44.10 46.30 48.62 (8) Prob. in force t−1 p55 CHAPTER 9. PROFIT-TESTING (4) Interest .06(Pt − et ) 41.92 54.40 54.28 54.14 54.01 (9) Initial expected cash flow (5) Death Cost 5000q54+t 42.20 47.10 52.50 58.45 64.95 (6) Survival Cost St p54+t 0.00 0.00 0.00 0.00 4935.05 1.00000 .99156 .98222 .97191 .96054 698.46 906.33 890.30 872.90 −3886.22 Column(7) = (2) − (3) + (4) − (5) − (6) Column(9) = (7) × (8), where column (8) gives the probability that the policy will be in force at the start of year t. 9.3. THE PROFIT VECTOR AND THE PROFIT SIGNATURE 155 9.3 The profit vector and the profit signature We now suppose that the office maintains suitable reserves t V at the end of each policy year. We assume that 0 V = 0 and n V = 0 (i.e., any survival benefit payable when the contract matures at time n years has been paid.) Corresponding to the “in force” and “initial” net cash flows, we have the profit vector and profit signature; that is, (PRO)t = cash which will, according to office’s projections, be transferred from the life fund to the with-profits policyholders/shareholders at the end of policy year t, per policy in force at the start of that year σt = as for (PRO)t , but per policy sold The relationship between these quantities is therefore σt = t−1 px (PRO)t Calculation of (PRO)t We observe that (PRO)t = t−1 V(1 (9.3.1) (9.3.2) (9.3.3) + i) + (CF)t “in force” net cash flow in year t − px+t−1 .t V money needed for reserves at end of the year (9.3.4) accum. of reserves until end of year (since the chance that a given policy in force at the start of the policy year remains in force at the end is px+t−1 .) This formula is sometimes written in the form (PRO)t = (CF)t + it−1 V interest on reserve where (IR)t = px+t−1 .t V − t−1 V = money needed at end of the year for reserves, less money held at the start We may then calculate {σt } by formula (9.3.3). (9.3.6) − (IR)t increase in reserves required (9.3.5) Example 9.3.1. Consider the policy described in example 9.2.1. Assuming that the premium and experience basis are as described in that question and that the reserves held are as follows: 0 V=0 156 1 V=919 2 V=1,876 3 V=2,873 4 V=3,914 5 V=0 CHAPTER 9. PROFIT-TESTING Calculate the profit vector and profit signature. Solution. We draw up the following table: (1) Year t 1 2 3 4 5 (2) Reserve at start of year t−1 V (3) Survival prob. p54+t .99156 .99058 .98950 .98831 .98701 (8) Prob. in force t−1 p55 0.00 919.00 1876.00 2873.00 3914.00 (7) In force profit for year (PRO)t −212.78 29.85 52.15 75.26 102.97 (4) Increase in reserve (IR)t 911.24 939.33 966.83 995.25 −3914.00 (9) Profit signature σt −212.78 29.60 51.22 73.15 98.91 (5) Interest on reserve .06t−1 V 0.00 55.14 112.56 172.38 234.84 (6) In force expected cash flow (CF)t 698.46 914.04 906.42 898.13 −4045.87 1.00000 .99156 .98222 .97191 .96054 Column(4) = p54+t .t V − t−1 V Column(7) = (6) + (5) − (4) Column(9) = (7) × (8), where column (8) is the probability that the policy will be in force at the start of year t. 9.4. THE ASSESSMENT OF PROFITS 157 Example 9.3.2. A life office issues a 3-year without profits endowment assurance policy to a life aged 62. The sum assured of £1, 500 is payable on maturity or at end of the year of death, if within 3 years, and there are level annual premiums of £472.50 payable in advance. The office uses the following “experience” basis: mortality: interest: initial expenses: A1967-70 ultimate 6% per annum £20 renewal expenses: £5 at the beginning of the second and third policy years. The office’s reserve basis is as follows: net premium method, using A1967-70 ultimate mortality and 3% p.a. interest. Determine the profit signature of this policy. Solution. We first work out the “in force” net cash flows, (CF)t . (1) (2) t Prem. Exp. (3) (4) Interest Death cost = 0.06((1) − (2)) = 15, 000q61+t 27.15 28.05 28.05 (5) (6) Maturity (CF)t = (1) + (2) Cost +(3) − (4) − (5) 453.03 466.07 −1,004.45 1 472.50 2 472.50 3 472.50 20 5 5 26.62 0 29.48 0 ← − − 1, 500 − − − −−− −−→ Now we find the profit vector, (PRO)t , and the profit signature, σt . (1) Reserve at start of year t 1 2 3 0 476.58 975.71 (2) Interest on Reserve = 0.06(1) 0 28.60 58.54 (3) p61+t (4) (IR)t (5) (PRO)t = (CF)t + (2) − (4) 0.98225 468.12 0.98035 466.07 0.97826 −975.71 −15.09 14.71 29.80 1 0.98225 0.96295 (6) t−1 p62 (7) σt = (5) × (6) −15.09 14.45 28.70 by net premium method do not use office premium in valuation 9.4 The assessment of profits The profit signature {σt } can be assessed in one or more of the following ways. 158 CHAPTER 9. PROFIT-TESTING (1) One could work out the Internal Rate of Return (or yield) by solving the equation n v t σt = 0 t=1 (9.4.1) (the internal rate of return, i0 , being the solution of this equation.) (2) The shareholders may value the net profits at a certain rate of interest, j per annum. This rate is called the Risk Discount Rate, and may reflect uncertainties in {σt }, with j normally higher than i. The net present value of the profits is thus n NPV(j) = t=1 v t σt at rate j (9.4.2) (3) The Profit Margin is defined as n.p.v. of profits , both at some rate of interest, im , say n.p.v. of premiums n v t σt = n−1t=1 , at rate im t t=0 Pt+1t px v (im is often equal to the risk discount rate, j.) (9.4.3) Example 9.4.1. A life office issues a three-year non-profit endowment assurance policy to a man aged 30. The sum assured is £60,000 on maturity or at the end of the year of earlier death. Level premiums of £19,000 are payable annually in advance. The office maintains reserves as follows: policy year 1 2 reserve at end of policy year £19,000 £38,000 The office expects that its life funds will earn interest at 7% p.a. over the next 3 years. The office expects expenses to be as follows: initial expenses: 10% of the first year’s premium, renewal expenses: 2% of later premiums. Mortality is expected to follow A1967-70 ultimate. Calculate (i) (ii) the profit signature; the net present value at the issue date of the profit to the office, using a risk discount rate of 10% p.a. 999346 = 2.883 −40.620 18. 000q30+t−1 mat.22 18. (CF)t .975 −38. This follows from the formula (PRO)t = t−1 V(1 + i) + (CF)t − px+t−1 .620 (1) (Pt − et )(1 + i) 18. 000p30+t−1 ] We now work out the profit vector.5 Some theoretical results about {σt } σt = (PRO)t = 0 for all t Theorem 9.297 19. Theorem 9.258 19. 623 9. 237 583 × 0. t 1 2 3 (Pt − et ) 17.077 (2) i.000 0 Hence profit signature is: −729 2.t V = 0 which is merely a restatement of the equation (t−1 V + Pt − et )(1 + i) = qx+t−1 Dt + px+t−1 St + px+t−1 .000 38. 238 × 0. If all three bases defined in section 9.t V which holds (cf chapter 7) since all three bases agree.238 583 159 t 1 2 3 tV 19.t V − t−1 V 18. SOME THEORETICAL RESULTS ABOUT {σT } Solution.cost (in final year only) = 60.258 40. If the premium basis and experience basis agree.833 60.26 19.923 (2) (CF)t death + mat. Proof.a.923 19. earned by the office’s life funds.t−1 V 0 1330 2660 (3) p30+t−1 .2 agree.1.5. the internal rate of return earned by the office on a sale of a policy is equal to the rate i p.100 18.5.000 −40. costs = (1) − (2) 39. .2. 237v 2 + 582v 3 at 10% = £1.077 [ death cost = 60. (1) (CF)t 18.5.000 (4) (PRO)t = (1) + (2) − (3) −729 2. per policy in force at the start of the policy year. (i) We work out the net cash flows.987 18.998676 = 582 (ii) −729v + 2.9. R. Hence n n n v t σt = t=1 t=1 n−1 v t t−1 px (1 + i)t−1 V − t=1 n v t t px . .5. We must show that n CHAPTER 9.a. We shall now assume that surrenders may occur.R. We may assume that wn = 0. What is the I. n − 1).r V − t=1 v t t px .t V = r=1 v r r px .R. . associated with the profit signature of example 9. that a profit at time t policy in force per surrender at time t − 1 at that time will be surrendered at time t (9.6 Withdrawals So far we have ignored the possibility of surrender. on the sale of a policy in excess of the rate of interest it believes it will earn on the life funds. PROFIT-TESTING v t σt = 0 at rate i p. except that there are additional profits due to the surrender of (px+t−1 )wt [t V − (SV)t ] prob.6.1) .t−1 V(1 + i) − t px .R. 2.1.a. If the life office wishes to obtain an I.R.160 Proof. Since the premium and reserving bases agree. using the fact that 0 V = n V = 0 Corollary.R. . Let (SV)t denote surrender value at time t. must be i = 6% p. .t V] But n v t t−1 px (CF)t = 0 t=1 as this is merely a restatement of the equation of value for calculating premiums. 9. t=1 Now we have n n v σt = t=1 t=1 t v t [t−1 px (CF)t + t−1 px . Let wt denote the chance that a policy will be surrendered at the end of year t (t = 1. We observe that (PRO)t is as before. the I.1? Solution.3. Example 9. since the policyholder will receive the maturity benefits (if any) at that time.t V = 0. but only at the end of a policy year (just before payment of the premium then due). premiums must be higher than those calculated on the “experience” basis. 3. .952 . 2. 3.6.71 54.9.26 + 0. 873 = 72.99 79. {σt } is not the same as {σt } because the factors {t−1 px } are not the same as {t−1 px }. Solution. 5% of the surviving policyholders will surrender.96054 × . but now assume that at each of the durations 1.99156 × . . .6.87 31. t 1 2 3 4 5 (PRO)t −211.3.05 × . Even if (PRO)t = (PRO)t for t = 1.2. withdrawal rates depend to a great extent on economic and commercial factors which are less easy to forecast accurately. (1 − wt−1 ) σt = revised profit signature = t−1 px (PRO)t (9.02 × 1.4) Example 9.13 102. 914 = 102.15 + 0.93 80.6.99156 × . 876 = 52.87 48.75 65.97 + 0 = −211.87 29.05 × .6.1. to allow for (i) (ii) the change from (PRO) to (PRO) .954 Hence we have: σt −211.6. Consider the policy of Example 9. . . . and the changed probability of the policy being in force at time t − 1. We work out the revised profit vector.85 + 0. .R. which are fairly predictable by actuaries. (PRO)t : t 1 2 3 4 5 (PRO)t + p55+t−1 (0.05 × . The I.71 54.97 t−1 px 1 .R.4 years. 2.99058 × .05)(0.97191 × .56 Notes 1.98950 × . allowing for withdrawals may be found by solving the equation n v t σt = 0 t=1 (9.13 102. n − 1.2) We now have t−1 px = probability that policy is in force at time t − 1 = t−1 px (1 − w1 )(1 − w2 ) . Find the new profit signature.78 + 0.3) Hence (9.6.953 .1. 161 (9.5) 2.97 * when t = 5. is 98% of the office’s reserve. .02 × 2.02 × t V)∗ −212.87 31.02 × 919 = 29.02 × 3.V.98222 × . (PRO)t = (PRO)t . . .05 × . . WITHDRAWALS Hence (PRO)t = profit vector at time t. We must also adjust σt .99 79. Unlike mortality rates. and that the S. n.95 .98831 × . allowing for surrenders = (PRO)t + px+t−1 wt [t V − (SV)t ] Note that (PRO)t = (PRO) if t V = (SV)t for t = 1. A life office issues a 5-year with-profits endowment assurance policy to a life aged exactly 60. These formulae may be easily modified if the policy is altered. Simple reversionary bonuses vest at the start of each year.7. The actual year-end profits emerging in the n years (maximum) of the policy are as follows: (1) If the policy is no longer in force at the beginning of policy year t. and expenses and bonuses will follow the premium basis. and (t−1 V + Pt − et )(1 + it ) − St − t V if the life survives the year. (2) If the policy is still in force at the beginning of the year. .627.1) Example 9. the profit is (t−1 V + Pt − et )(1 + it ) − Dt if the life dies during the year. assuming that the office will earn interest at 7% per annum on its assets.7.1. The premium is calculated according to the following basis: mortality: A1967-70 select interest: 4% per annum simple reversionary bonuses at the rate of 4% per annum are assumed initial expenses: 60% of the first premium renewal expenses: 5% of each premium after the first (i) (ii) Show that the premium is equal to £2. and use the symbols it and et to denote the actual interest rate earned by the life funds in year t. Level premiums are payable annually is advance throughout the term of the policy. Calculate the profit signature for this policy.162 CHAPTER 9. (9. mortality follows the A1967-70 ultimate table. of course. The policy has a basic sum assured of £10. Let us suppose here that there are no withdrawals. The office holds net premium reserves using a rate of interest of 3% per annum and A1967-70 ultimate mortality . the profit emerging in that year is zero. including the first. PROFIT-TESTING 9.7 The actual emergence of profits The actual profits emerging from the sale of a given policy depend. and the year in which such an event occurs.7.2) (9. as in the following example.000 payable at the end of the year of death or at the maturity date. on whether the policyholder dies (or surrenders the policy) during the term. and the actual expenses incurred at the beginning of that policy year. Solution. including declared bonuses.P.V.02 (ii) The reserves held are: 1V R[60] − R65 − 5M65 + 5D65 D[60] ¨ = 10.88 Hence P = £2.7563P ¨ M. 000 1 − 61:4 a60:5 ¨ = 2191.67 + 1200A63:2 = 6.P.7.60 + 400A61:4 at 3% interest a62:3 ¨ a60:5 ¨ a62:3 ¨ 3 V = 10.95P a[60]:5 − 0.55P = 3.31 + 1600A64:1 = 9. 000 × 0. with no entitlement to further bonuses. 862.9. of premiums less expenses is 0. 000A[60]:5 + 400(IA)[60]:5 = 10. Expenses followed the premium assumptions up to the alteration date.V. M. 000P60:5 a61:4 a ¨ = 10. calculate the actual year-end profit earned on the policy.17 . The paid up sum assured was 60% of the benefits immediately before the alteration. and bonuses in the first three years followed the premium assumptions. THE ACTUAL EMERGENCE OF PROFITS (iii) 163 Immediately before the fourth premium was due (before the fourth bonus declaration) the policy was made paid-up. Interest was earned on the life funds was at 6% per annum over the period of the contract. 400A61:4 − 10. The policyholder survived to the maturity date. 000 1 − + 800A62:3 = 4. 475. 000 1 − a60:5 ¨ 2V = 10. For each of the five years of the policy term.82565 + 400 = 9867. (i) Let P be the annual premium. 366. 627. and no expenses were incurred after the policy was made paid-up. 000 1 − a60:5 ¨ a64:1 ¨ 4 V = 10. of benefits is 10. (5) = cost of death and maturity claims (including bonuses) Col. 504.24 418.67 6862.60 4475.02 2627.35 131.71 438.02 2627.25∗ 0 Survival benefits (5) 0 0 0 0 6720 Profit Year Pt − et (1) 1050.43p60 .21 131.35 (4) 73.08 0 Profit per policy in force at start of year (7) −1185.37.000 CHAPTER 9.17 Interest Cost of Claims (5) 150. (6) = p60+t−1 .67 0 0 t 2 3 4 5 (3) 63.84 527. . (3) = 0.06[(1) + (2)] Col.25.31 9366.71.11 487.60 4475.80 227.35 131.56 328.25 Interest Reserve at end tV (4) 2191.02 602.164 This gives the following table: Year Premiums Expenses Reserve t−1 V at start (3) 0 2191.10 172.67 2495.31 6524. the sum assured after alteration is 0.28 411.71 ∗6720A64:1 at 3% = 6524.00 6740.6[10. PROFIT-TESTING Cost of Reserves at End (6) 2159. 432. 000 + 3 × 400] = 6720.74 492.81 2495.07[(1) − (2) + (3)] Col. (6) = (1) + (2) + (3) − (4) − (5) .33 Col.35 131.67 6862.51 9182.71.99 655.67 6862. 483.10.174 p60 ) = (−1185. 574.43 520.02 2627. (4) = 0. 692.17 t 1 2 3 4 5 (1) 2627.06 830.05 281.99 12.t V Col.74 391. and the office calculates reserves by the net premium method. Col.95 198.38) (iii) We draw up the following table: Reserve at start t−1 V (2) 0 2191. 646.31 749.46 (6) −1077.80 195.60 4475.02 (2) 1576.97 4404.31.31 6524. . (7) = (1) − (2) + (3) + (4) − (5) − (6) Hence the profit signature for the contract is (−1185.02 2627.97 692. At each of these times.8. calculate the surplus accruing to the office at the end of each of the first three policy years. minus a surrender penalty of £40. so profits arise only from surrenders. For each policy.6973 = 1. per policy sold. Bonuses will be declared annually at a rate of 3% of the basic sum assured.4681 = 2. Level premiums are payable annually in advance. interest and expenses followed these assumptions. on the sale of this contract. 30). Note that (PRO)t = 0. Hint.) Calculate the annual premium (P ) required for the shareholders/with profits policyholders to achieve an internal rate of return of 12% p. which the office transferred to the surplus account. but there were surrenders just before payment of the premiums due at durations 1. 3.8672 = v = 0. 9.2 A life office issues a 5-year guaranteed bonus endowment assurance policy to a life aged 60. A certain life office sells assurance policies with term 3 years to lives aged 70. Given that the mortality of the policyholder is expected to follow A1967-70 ultimate. the profit vector is estimated to be (−50.1 (i) (ii) In the context of profit-testing.a. The sum assured. EXERCISES 165 Exercises 9. 2. (i) (ii) For each policy year. 9. and had a sum assured of £40.a. Interest on premiums and reserves is expected to be earned at an effective rate of 8% p.000. . 3% p. For each duration t = 1. 30. (b) the net present value of the profit to the office on the basis of a risk discount rate of 8% per annum. Expenses of 40% of the first year’s premiums and 5% of subsequent years’ premiums will be incurred. explain the difference between the “profit vector” and the “profit signature”. with attaching bonuses.9. payable on survival or at the end of the year of death.a. is payable at the end of the year of death or at maturity. using the basis A1967-70 ult at 3% p. Each policy was effected by annual premiums. The office holds net premium reserves.a. and were given a surrender value equal to the office’s reserve. 3% of the surviving policyholders surrendered their contracts.000. Mortality is expected to follow A1967-70 ultimate. given the following values on A1967-70 ultimate. calculate the total sum assured. or otherwise. Bonuses vest at the start of each policy year.5% of the sum assured. with basic sum assured £30. 4 years.2 and 3 years. The office’s premium basis was A1967-70 ultimate 5% interest expenses of 2% of all premiums with additional initial expenses of 0. calculate (a) the profit signature per policy sold.3 Ten years ago a life office issued a large block of 10-year without profits endowment assurances to lives then aged 30. which was calculated on the premium basis.1853 = 3. Withdrawals may be ignored. interest: (IA)60:5 (IA)61:4 (IA)62:3 (IA)63:2 (IA)64:1 (iii) = 4. By using a profit-testing approach. It was found that mortality .97087 (Difficult. calculate the reserve (t V) immediately before payment of the premium then due. a. calculate the net present value of the profit signature on the following assumptions: premium basis: mortality: A1967-70 ultimate interest: 6% p. . Expenses are expected to be as follows: initial expenses: £10 renewal expenses: £2 incurred at the beginning of the 2nd and each subsequent policy year. net premium method using A1967-70 ultimate.a. 4% p. PROFIT-TESTING 9. 000. payable at the end of the year of death. In respect of a policy with sum assured £10. The sum assured is £15. Calculate the office premium on each of the following reserving bases: (i) The office holds zero reserves at each year-end. expenses: 3% of all premiums. The risk discount rate used by the office is 15% per annum. expenses: mortality: risk discount rate: 3% of office premiums A1967-70 ultimate 10% p. and that mortality follows the A1967-70 ultimate table. 000. 9. Level premiums are payable annually in advance.a.166 CHAPTER 9.4 Your office is considering the issue of 3-year annual-premium endowment assurance policies without profits to lives aged 62. The office calculates the annual premium by requiring that the net present value of the expected profit on each policy is equal to 20% of one office premium. interest reserve basis: rate of interest to be earned in life fund: 8% p. It is assumed that interest of 7% per annum will be earned on the life funds.5 A life office issues 3-year term assurance policy to a man aged exactly 59. payable at the end of the year of death (if within 3 years) or on maturity. (ii) The office sets up a reserve at each year-end (except the last) equal to 80% of one office premium.a. 90 = £0. 353 12.72 + 21.700 4 33. = σ1 v + σ2 v 2 + σ3 v 3 at 8% = −50v + 28.2 Note: In (ii). 900A61:4 + 900(IA)61:4 − P1 a61:4 31. that policy is in force at time t − 1 (a) σt = t−1 p70 (PRO)t . 60064:1 + 900(IA)64:1 − P1 a64:1 ¨ = = = = 6.83 σ3 = 30 = 27. whilst the profit signature refers to profits per policy sold.59v 3 = −46. 000 × 0.V. 3% (ii) This leads to the table (note that 0 V = 5 V = 0): t 1 2 3 4 (iii) tV ¨ 30.A. 983 We first work out (CF)t (in terms of P ): . 1 30.59 (b) N.500 The formula is ¨ t V = (30.1 (i) The profit vector refers to expected profits per policy in force at start of year. SOLUTIONS 167 Solutions 9. we need the increasing assurance factors (A65+t:5−t and a65+t:5−t are directly ¨ tabulated) (i) policy year t total S.900 2 31. 000 + 900t)A60+t:5−t + 900(IA)60+t:5−t − P1 a60+t:5−t where P1 = net premium on A67-70 ult. Relationship is σt = (PRO)t × prob.P. 947 19.572 on A67-70 ult.32 9. 811 26. (ii) so σ1 = −50 σ2 = 30 l71 l70 l72 l70 = 28.86682 + 900 × 4. 80062:3 + 900(IA)62:3 − P1 a62:3 ¨ 32. 70063:2 + 900(IA)63:2 − P1 a63:2 ¨ 33.600 5 34.9. 3% = 30.30 + 24.83v 2 + 27. 000A60:5 + 900(IA)60:5 a60:5 ¨ 30.1853 = = £6511.800 3 32.9.67 4. is σt = (1. and 5P 34.36 1584. 500 (3)cash flow (CF)t 0.026P − 5358 v t t−1 p60 (PRO)t = 0 at 12% interest i.96 1.08) 0.3 .026P − 34.026P 1.39 1.2)0..76 6512.66 2158.026P − 509.168 (1)prs.t V − t−1 V 0 6261.08t−1 V (IR)t = p59+t .3 Profits arise only in respect of surrenders at times 1. 488 ∴ P = £6.2p30+t−1 surrender penalty The probability of being in force at the start of year t is t−1 t−1 p30 = t−1 p30 (1 − 0. per policy sold.42 1. 596 (Rough check: right order of magnitude:.93385)v 5 =0 ∴ 3.026P − 6057)(0.P = P1 .2.64 −26983.31 508.e.23 1.96 1.39 0 34500 since sum paid on death or survival We now work out (PRO)t : t 1 2 3 4 5 (1)interest on reserve (2)increase in reserve 0.026P − 5358)(0.969785)v 3 +(1.026P − 509.2.026P 1.648P 1.2577P = 21. 9. PROFIT-TESTING (2)death costs (3)survival costs q59+t Dt p59+t St 445.648P − 6707)v +(1.23 0 580.026P − 6388 1.026P − 660. less exp.026P − 5717 1.026P CHAPTER 9.026P − 6388)(0.24 6386.97t−1 t p30 t=1.648P − 445.026P − 6057 1. 2.67 1035.03) (40) = 1.42 1.026P − 5717)(0.648P − 445.00 We have the equation 5 t=1 t 1 2 3 4 5 (1)−(2)−(3) (CF)t 0.23 1. The profit at time t years (t = 1.952572)v 4 +(1. 3) per policy in force at start of year is (PRO)t = (1 − q30+t−1 )(0.026P − 34.026P − 580.3.88 6641.648P − 6707 1.985568)v 2 +(1. 500 (1)-(2)+(3) (PRO)t 0. with interest (P − et )(1. 500 (maturity benefit)).42 0 660.03) ∴ Expected surplus arising at time t.026P 1.96 0 509.39 1.026P − 660.026P − 580. (0. 000 = −6. = 0.06 = 124.2(0. 472.98225 0.30263 = 0.97 × 0.V.59 + 517.972 × 0.86 + 251.2 × ×0.t−1 V − [p62+t−1 .38 3. 000 1 − = 3. 5. 0001 V62:3 . 000 × 0.71 . 000q63 = 3.97)t−1 t p30 = σt 1. 090.2 × 0. 071.519 3V =0 Net cash flows (ignoring reserves) t 1 2 3 (CF)t 0.9.96294 σt 0.06 10.9.86 0.29 ∴ N.08) − 10.30 249.146.t V − t−1 V] tV 1 3. 1 V = 10. 119. 071. 146.4 1.97 0. 090. 731.472.P.29v 3 at 10% = £288.38v + 122.97 = £3. 731.91 0.16 1.66.88 t−1 p62 1 0. 199.99935 = £1.90 Reserves: 0 V = 0.51 3 0 −6.97P (1. 090.38 2 6.08) − 10.04 = 10.38 2V a63:2 ¨ a62:3 ¨ at 4% = 10.99868 = £1.66 = 258. SOLUTIONS t 1 2 3 9.91 + 0 − 3.53 = 0.71 − 3.2 × 0.138 6.59 Profit vector and signature t (PRO)t = (CF)t + i.30v 2 + 249. 472.20 1.08) − 10.81 + 6.38 122. 000q62 = 3.97P (1.97P (1. 000P62:3 .99798 = £1.13 169 Let P = annual premium P = 10. 000 1 − = 6.66 3. 15−1 + 1.7 0.2P = vσ1 + v 2 σ2 + v 3 σ3 at 15% ∴ P = £233.74 1 2 3 P−LO P−2 P−2 0.8P 0 0.926 P−242.8P 0.74 We solve the equation: 0.28039P−205.15−2 + 1.07P−205.90 .34 0.15−3 496.49 240. be P (1) (2) Interest = 0.79 1.07 × (1) (3) Expected death costs (4) (CF)t = (PRO)t = (1) + (2) − (3) Premiums less expenses (5) t−1 p59 1 0.07P − 0.056P 0.2133 (3) (CF)t (ii) (1) Year Reserve at start of year (2) Interest on reserve = 0.056P 1.28039P−205.20 1.8P 0. PROFIT-TESTING Let A.07 × 1.63 1.07P−218.5 (i) Year CHAPTER 9.14 0.0409P−235.t V − t−1 V (6) (PRO)t = = (2) + (3) − (5) 1 2 3 0 0.61 1.8P 0 0.87354P−235.0409 × 1.P.76 1.63 1.07P − 0.2 + 1.8 0.96 = £224.13755P−218.0561P−215.14 194.53 = 2.61 1.7896 −0.07P−205.12277P−215.07P−218.63 1.61 1.15−1 + 215.07P−242.15−3 −0.(PRO)t 0.07P − 0.61 × 1.170 9.79 × 1.0561 × 1.987006 0.2P = vσ1 + v 2 σ2 + v 3 σ3 at 15% ∴ P = 205.07 × (1) (4) Reserve at end of year (5) Increase in reserve = p59+t−1 .972761 (6) σt = (4) × (5) 1.34 We solve the equation: 0.74 × 1.61 1.91 216.15−2 + 235.07P−242.61 1.34 (7) σt = t−1 p59 .01155 −0.07 P−205. 2) = 0 lx+t dt lx ◦ (as it may be shown that limt→∞ (t.1. Note that ex = 0 ◦ ∞ ◦ We recall that t.t px ) = 0 when ex is finite.1.Chapter 10 STATIONARY POPULATIONS 10.) We recall also that ex = curtate expectation of life at age x = E(K) where K is the integer part of T .3) and. including fractions.1 Some Definitions ex = complete expectation of life at age x = E(T ) where T refers to the future lifetime of (x) in years. We find that ex = t=1 ∞ t px (10. ex ex − 171 ◦ 1 2 .1) Integrating by parts gives ex = [t(−t px )]∞ + 0 ∞ ◦ ∞ 0 t px dt = 0 ∞ t px dt (10.1. using the Euler–Maclaurin formula.t px µx+t dt (10. ) In any timeinterval of length t years. so the rate at which lives attain age x is klx per annum.5) = k.1.1.t.1. Also. Figure 10. By integrating over all ages greater than or equal to x. STATIONARY POPULATIONS Tx = 0 ∞ lx+t dt ly dy x = from which we obtain (on setting y = x + t) ex = = and hence ◦ ∞ ly x dy lx Tx lx Tx = lx ex ◦ (10.1. the number of lives attaining a specified exact age x is klx . we can see that ∞ Total population aged ≥ x = x kly dy (10.Tx The position is illustrated in Figure (10.1: a stationary population (It is nearly always advisable to draw a diagram in stationary population questions. Note that (if lives are considered at all ages) there will be kl0 births each year. klx lives ‘flow’ continuously over exact age x each year.6) .4) A stationary population is an idealised large population such that (in an old-fashioned mathematical notation) the number of lives in the population between ages x and x + dx is always equal to klx dx where k is called the scaling factor and lx is according to some life table. the number of deaths each year between ages x1 and x2 (x1 < x2 ) is k(lx1 − lx2 ) The number of lives in the population between any two ages x1 and x2 (x1 < x2 ) is k(Tx1 − Tx2 ) (10.1).e.172 Define ∞ CHAPTER 10.1. i.1.7) (10. SOME DEFINITIONS 173 Example 10.12 .Males. and English Life No. .1. there are no withdrawals. 115.500. 172.1. (b) the number of staff who retire each year. The total number of staff is k(T20 − T60 ) = 500 ◦ ◦ (l20 e20 −l60 e60 ) l20 = 19.10.T50 = 11. (b) Number who retire each year = kl60 = 500 l20 l60 = 410 (c) Number of pensioners = kT60 = 500 l60 l20 e60 ◦ = 6. What is the annual number of entrants ? Solution One knows that k. find: (a) the size of the staff. 500 ◦ = 507. A large company has for many years maintained a staff in a stationary condition by recruiting 500 annual entrants at exact age 20. Example 10. If the staff retire at age 60. They are subject to the mortality of English Life Table No. The membership of a certain learned society is stationary at 11.1.05 or 507. Members enter only at exact age 50.2.1. 12 – Males mortality is experienced. (c) the number of pensioners. and there are no withdrawals. uniformly over the year. 500 e50 ◦ kl50 e50 = 11. Solution (a) We have kl20 = 500. 500 So The annual number of entrants is kl50 = 11. 10) =ex − lx+n ex+n lx .t px ]n + 0 n dt + n.174 Some more symbols We let CHAPTER 10.t px µx+t dt + n.n px (by integration by parts) = 0 t px dt Hence ex:n = ◦ Tx − Tx+n lx lx+n ◦ ◦ ex+n =ex − lx (10. STATIONARY POPULATIONS Lx = the number of lives between ages x and x + 1 in a stationary population with scaling factor k = 1.d. t px µx+t for 0 < t < n.n px n 0 t px = [−t. and there is discrete probability n px that T ∗ = n. Hence E(T ∗ ) = 0 n t.8) We may use the approximations Lx Another definition is lx+ 1 2 ex:n = average lifetime lived by (x) between ages x and x + n = E(T ∗ ) where the random variable T ∗ is defined as follows: T∗ = T n if (x) dies within n years if (x) lives for n years. + lx+n lx (10.1.. = Tx − Tx+1 1 = 0 lx+t dt 1 (lx + lx+1 ) 2 (10..1. ◦ The variable T ∗ has p.) ¯ Similarly.9) (Notice that ax:n =ex:n if i = 0 . ex:n = average number of complete years lived by (x) between ages x and x + n n ◦ = t=1 t px = lx+1 + lx+2 + .f.1. 2.2 Define The Central Death Rate h mx = the central death rate between ages x and x + h lx − lx+h = x+h ly dy x number of deaths each year between ages x and x + h = population aged between x and x + h in a stationary population mx = dx Lx (10. You are given that lx = l1 1 − Find e1:10 . The average age at death of those who die between ages x and x + n may be worked out as follows.11) Note that this result refers to a randomly-chosen life aged x: we need not assume that there is a stationary population.n px x + E(T ∗ |T ∗ < n) ex:n −n. giving . We have E(T ∗ ) = E(T ∗ |T ∗ < n)P r{T ∗ < n} + E(T ∗ |T ∗ ≥ n)P r{T ∗ ≥ n} so that ex:n = E(T ∗ |T ∗ < n)n q x + n.3.1) If h = 1.1. THE CENTRAL DEATH RATE Example 10. 10. it is omitted.1.10.n px =x + n qx Tx − Tx+n − nlx+n =x + lx − lx+n ◦ ◦ Hence the average age at death of those who die between ages x and x + n is (10.2.2097. Solution ◦ 10 0 11 ◦ 175 log x 9 log 10 (1 ≤ x ≤ 11) e1:10 = lt+1 dt = l1 11 1 lx dx l1 log x = 1− dx 9 log 10 1 1 x=11 = 10 − [x log x − x]x=1 9 log 10 11 log 11 − 10 = 10 − 9 log 10 = 9. STATIONARY POPULATIONS 1 l µ dt 0 x+t x+t 1 l dt 0 x+t µx+ 1 2 (10. does not hold. then iF = B − C Note: In case (b). (10.2) (10. etc.1) .1) 2 0 (as the trapezoidal rule is exactly correct. lx+t is linear for 0 ≤ t ≤ 1.3.2.4 Stationary Funds Consider a pension scheme. (b) If income and outgo are received and paid only at the end of each year..3 Relationships Between mx and qx Assume that there is a Uniform Distribution of Deaths (U. F denotes the fund at the start of the year. are such that the interest on them pays B − C.D. 1 Since lx+1 = lx − dx . C are rates of payment per annum). then the funds will also remain constant. There are two cases: (a) If income and outgo are received and paid continuously. We have 1 1 Lx = lx+t dt = (lx + lx+1 ) (10.3.D.e.2) 10. not just an approximation).4) If U.2) dx lx − 1 dx 2 qx = 1 − 1 qx 2 We may rearrange this equation to get qx = mx 1 + 1 mx 2 (10. these results may be used as approximations. If the funds. then δ. such that B = annual benefit outgo for scheme (plus expenses.D.D. F . i. 10.3.) between ages x and x + 1.4.4. if any) and C = annual contribution income are constant. we have Lx = lx − dx 2 Hence mx = = dx Lx (10.3.176 Note that mx = CHAPTER 10.3) (10.F = B − C (B. 000 1 + e25 − + 10. Hence δ. 000P35 = £56. 716 − 22. Each year for many years a life office has issued 10. Therefore Fund = 18992591 δ = £484. 000 1 l35 × (1 + e35 ) × 56. ignoring expenses. and one quarter effect a 25-year without profits endowment assurance for twice that sum assured. 634.29 l25 10. 769. 125 = 18. 057 = £22.30 l25 4 l35 = 328. 249. N25 −N Whole life premium = 5. 723.29.4. Solution 35 The term assurance premium = 5. The office calculates premiums on A1967-70 ultimate 4%.000 temporary assurance policies each with a term of ten years and a sum assured of £5. Annual rate of payment of claims is 10000 5000 l25 − l35 l25 + l35 l25 1 1 × 5000 + × 10. All premiums are payable annually in advance and death claims are paid at the end of the year of death. 731. 065 + 14.30. .F = 41.85.1. 000 × 1 + e35 − (1 + e60 ) × 245. Endowment assurance premium = 10. 125. If the office’s experience follows this basis.4. 000 M25 −M35 = £3. 992. 000 3 4 = £41. calculate the size of the fund held for these contracts. One third of those who survive to age 35 then effect a without profits whole life policy for the same sum assured as the term policy.85 l25 3 l35 1 l60 + 10.000 to lives aged 25 exactly. Policies are issued uniformly throughout the year. 000P35:25 = £245.04. 731. 003 + 7.10. 000. STATIONARY FUNDS 177 Example 10. 716. 723. Annual premium income is l35 (1 + e35 ) × 3. 591 where i = 0. and those sick or unemployed between ages 15 and 65 also receive this amount.000 births per year. All teachers enter the profession at age 21 and retire at age 60. under which all employed men between ages 15 and 65 must contribute a fixed sum every week. 4 of those attaining age 60 retire immediately and the survivors retire at age 65. the same sum being also payable by their employers. STATIONARY POPULATIONS Exercises 10.Males. all remaining employees are moved out of the division. All employees are assessed at age 35. the mortality of males follows English Life Table No.3 For many years a company has recruited. Staff experience mortality according to English Life Table No. Find the annual number of new teachers recruited. 10. (b) The government of the country has decided to introduce a social security plan. Men over age 65 will receive a pension of 100 units of currency per week. (ii) The country’s teacher training colleges are such that a constant flow of new entrants to the profession is maintained. (i) Find the number of pupils at any given time. 10. spread uniformly over the year.178 CHAPTER 10.2 A certain country’s school system provides education for all children between the ages of 5 and 16 exactly. The population of the country is subject to the mortality of English Life Table No. Teachers are recruited uniformly over the year. 12 .000 male births per annum. If it may be assumed that at any time 95% of men between ages 15 and 65 are employed. uniformly distributed over the year. Basis: English Life Table No.4 The male population of a certain country has been stationary for many years. 10. At the end of one year in the job. 12 . Employees leave the company only through death or retirement. 200 employees on their 20th birthdays and a fixed number of additional employees on their 25th birthdays. The country’s population is stationary.12 .000. there being no withdrawals. Mortality has followed English Life Table No. calculate the weekly contribution payable by each employed worker.Males. . Of the 1 remainder. and the total number of employees is 10. (a) Calculate the size of the country’s male population at any time. At age 40. uniformly over each year. there being 100.5 A large manufacturing company has for many years staffed one of its divisions by the recruitment. there being 100. One third of those reaching age 30 leave immediately.Males. uniformly over each year. 12 . and one third of employees reaching their 60th birthdays retire on that date.Males.000 staff at exact age 20. while the remaining 5% are sick or unemployed. and 20% are dismissed. and the ratio of pupils to teachers is 20 to 1. 12 . Find the total number of new recruits each year. The scheme is to be financed on a pay-as-you-go basis (no fund is built up) and administrative costs are to be ignored. and migration may be ignored. (ii) the number of deaths in service each year. of 1. The only other decrement is death.Males 10.1 The staff of a large company is maintained as a stationary population by 500 new entrants each year at exact age 20. Death is the only other reason for leaving the division. new staff are examined for suitability. Calculate (i) the number of staff. Calculate the number of staff in the division. Employees may retire on their 60th or 65th birthdays. and 50% are immediately moved out of the division. 000 each year.q[x] (i) Express p[x] in terms of px . the following relationships apply at age x: 1 l[x] = (lx + lx+1 ) 2 (0 ≤ t ≤ 1) t q[x] = t. All remaining civil servants retire when they reach age 60 or die in service before this age. Mortality. ◦ ◦ . No policies terminate except by death or maturity. 12 .Males. if death occurs before age 65.000. (ii) Express e[x] in terms of ex and px . and (iv) the size of the fund which has been built up. and will do so for the indefinite future.6 In a certain country.7 In a mortality table with a one-year select period.000. Of those who reach exact age 40. plus 25% of office premiums in the first year. (Birthdays are assumed to occur uniformly over the calendar year). The sums assured issued each year have always been £5. This is to be done by immediately lowering the retirement age of civil servants. (iii) Express e[x] in terms of ex+1 and px . 10. 10% obtain employment in private companies and leave the civil service immediately. all civil servants are recruited on their 25th birthdays.5. and civil servants are recruited at the rate of 3. Premiums are calculated on the following basis: mortality: English Life Table No.10. Civil servants have for many years experienced the mortality of English Life Table No. (ii) the annual rate of payment of maturity claims. assuming that there are no other changes. interest and expenses have always been in accordance with this basis. (iii) the annual rate of payment of death claims. calculate (i) the annual rate of gross premium income. (i) How many civil servants are in service at any given date? (ii) The government of the country has just decided to reduce the size of the civil service by 10%. EXERCISES 179 10. Find the new retirement age (to the nearest month). In respect of this business. Premiums are payable continuously and the sum assured is payable immediately on death.8 For many years a life office has issued a steady flow of 10-year endowment assurances without profits to lives aged 55. (iv) Express m[x] in terms of px . 12 – Males interest: 4% expenses: 5% of all office premiums. The population of civil servants has been stationary for many years. spread uniformly over the year. 10. 066.81 3. 431 teachers per annum. 000 3 3 3 3 where k = 200 . Let annual number of recruits be c.1 (i) kl20 = 500.000 As l0 = 100.00053436 Hence number recruited each year = 200 + cl25 = 251 10. 320. 409 1 (ii) Number of teachers = 20 × 1. Then the number of employees is 1 2 1 2 k T20 − T60 − T65 + c T25 − T60 − T65 = 10. 10. 000 − 8157.4 (a) kl0 = 100. . 10. k = 1. 447. Then the equation is c (T21 − T60 ) = 53. STATIONARY POPULATIONS Solutions 10. Number of pupils = k(T5 − T16 ) = l5 e5 −l16 e16 = 1. 483 = 0. 066. 000. Let cl25 be the number who enter at age 25 per annum. 1 1 1 Number of Staff = k T20 − T30 − T60 − T65 3 6 2 500 1 1 1 ◦ ◦ ◦ ◦ = l20 e20 − l30 e30 − l60 e60 − l65 e65 l20 3 6 2 = 15. 409 = 53. 320 l21 ◦ ◦ 500 l20 1 1 1 l30 + l60 + l65 3 6 2 Hence c = 1. 361 (ii) Number of deaths is 500 − = 89. l20 Solving this for c gives c= 10. 000 Hence k = 1.3 kl20 = 200 employees enter at age 20 per annum.180 CHAPTER 10.2 (i) kl0 = number of births each year = 100. 4T35 − 0.9T60 ) 3000 ◦ ◦ ◦ = (l25 e25 −0. SOLUTIONS 181 Total male population = kT0 = 100. 000 Number of staff = k(T20 − T21 ) + 0. 242 l25 Hence Tx = 1.8k(T21 − T35 ) + 0.4T40 ) l20 = 14. Number of contributors = 0.2T21 − 0. 735 and T65 c = 12. 455 giving c= ◦ T15 = 5. 516. 10.10. Hence 2c × (number of contributors) = 100 × (no.1l40 e40 −0. 524. Then 3000 (T25 − 0. Let the employee s contribution be c per week. 571 So using linear interpolation gives x 55.9Tx ) = 85.95(T15 − T65 ) = 818. 242 Let x be the new age of retirement.9l60 e60 ) l25 = 94713 (ii) Number of civil servants required is 0. 352. 060 10. of beneficiaries) Therefore 100[T65 + 0.05k(T15 − T65 ). 333 T56 = 1.6.1T40 − 0.91 55 years and 11 months . 602.6 (i) kl25 = 3. 000 (b) Number of pensioners = kT65 Number aged between 15 and 65 who receive benefit = 0. 475 T55 = 1. 809.4k(T35 − T40 ) 1000 = (T20 − 0.95k(T15 − T65 ). 000 e0 = 6.05(T15 − T65 )] 2 × 0.5 kl20 = number of staff recruited per annum = 1.13 units. 000 Number of civil servants is k(T25 − 0.1T40 − 0.9 × 94713 = 85. 8 (i) Let P be the office annual premium per £1 sum assured.000. 000. so the scaling factor is ◦ 5. .25P a55:1 ¯ ¯ Hence P = 0.. 000 e55 − = £45.100663. The total sums assured in force at any given time are 5. 854. 000 .) l55 l65 ◦ e65 l55 Total sums assured = 5. ex+1 2 1 1 + 3px 2px ◦ ex+1 = + 2 1 + px 1 + px (iv) m[x] = = = = 1 l µ 0 [x]+t [x]+t 1 l dt 0 [x]+t [ l[x]+t dt + dt l[x] − lx+1 1 2 (l[x] + lx+1 ) 1 2 (1 1 − p[x] + p[x] ) (using part (i)) 2(1 − px ) 1 + 3px 10. l[x] p[x] lx = · ex = · ex = l[x] px 1 l[x] 1 ∞ 2 1 + px ex (iii) e[x] = = ◦ [ 0 1 l[x]+t dt] ∞ lx+t dt] l[x] 0 1 1 1 ◦ = [ (l[x] + lx+1 ) + lx+1 .. 000 (T55 − T65 ) l55 (£5.182 10. 000. We have 0. 618.q[x] .000 “flows across” age 55. l[x]+t is linear for 0 ≤ t ≤ 1) 1 ◦ = (1 + p[x] ) + p[x] .95P a55:10 = A55:10 + 0. STATIONARY POPULATIONS lx+1 = l[x] 1 2 (lx lx+1 2px = 1 + px + lx+1 ) e[x] = lx+1 + lx+2 + . ex+1 ] l[x] 2 (as t q[x] = t. 000.7 (i) p[x] = (ii) CHAPTER 10. 592. 985. 000. 592.6. 014. We have δF = 5. 592. 130 + 0. 130 p. 130 (iv) Annual rate of payment of benefits = £5. 000. 000 + 355. 870 = £1. 000 e55 − = £355.05 × 4. 000. 000 − 3. SOLUTIONS Hence rate of gross premium income = £4. 000. 000 (final figures are unreliable. 209 Let funds be F . 130 Hence F = £19.100663 . 000 l65 = £3. 000. (ii) 5.a. 456.) ◦ l56 ◦ e56 l55 × 0. 985.25 5.10. 000 Annual rate of payment of expenses = 5% of gross premium income + 25% of gross premium income in first policy year = 0. 209 − 4. 870 l55 183 (iii) 5. 184 CHAPTER 10. STATIONARY POPULATIONS . 1. (y) will survive for t years at least} = t pI · t pII y x (by independence of T1 . or business partners.e.1. subject to the mortality of Tables I and II respectively.) That is. etc. T2 ) (11. Table I is a(55) males ultimate. But in practice independence is normally assumed. i. Define I II t pxy = P r{both (x). T is the “joint future lifetime” of (x). Table II is a(55) females ultimate. Remark If (x) and (y) are a married couple.3) Note A common error is to write I II t q xy = t q I · t q II x y If Table I = Table II. 185 . we may omit the superscripts. for example. so (for example) t pxy = t px · t py on the table under consideration.2) (11.1. refers to a male life aged x and a female life aged y (the male life comes first. T2 is questionable: consider.Chapter 11 JOINT-LIFE FUNCTIONS 11. Note In the a(55) tables. the assumption of independence of T1 .1) and I II t q xy = P r{T ≤ t} = the distribution function of T =1− I II t pxy (11. Let T = min{T1 . the time until the first of them dies. That is. a car accident in which both lives are killed. where T1 = future lifetime of (x) and T2 = future lifetime of (y). T2 }. (y). the notation t pxy .1 Joint-Life Mortality Tables Consider 2 independent lives aged x and y respectively. 1.1. t pxy = lx+t:y+t lxy Note that t q xy = 1 − t pxy = 1 − t px . JOINT-LIFE FUNCTIONS From now on we will omit “I. Given lxy = 1. II” (unless needed. of two lives aged x and y. q Example 11. only one will survive for 10 years.t py = 1 − (1 − t q x )(1 − t q y ) = t q x + t q y − t q x .4) Using equation 11. Solution 10 px lx+10:y = 0.5) = µx + µy (since h qx h → µx .1136. 000 lx+10:y = 960 lx:y+10 = 920 calculate the probability that. The force of mortality is defined as µxy = lim = lim h qxy h qx h→0+ h h→0+ h + h qy h − h qx h · h qy h h (11.1. II respectively). α2 are the youngest ages in tables I.1. where α1 .92 10 py = lxy = Required probability =10 px (1 −10 py ) +10 py (1 −10 px ) = 0.t q y (11. By equation 11.1.186 CHAPTER 11.) Define lxy = lx ly (x ≥ α1 .1.96 lx:y lx:y+10 = 0. . y ≥ α2 .3. hhy → µy as h → 0+ ).1. t q xy = lxy − lx+t:y+t lxy and dxy = lxy − lx+1:y+1 . 6)).d.5) The probability density function (pdf ) of T. Note that K = min{K1 .f. is always equal to 1. t t px t py t = exp[− 0 t µx+r dr] · exp[− 0 µy+r dr] = exp[− 0 t (µx+r + µy+r ) dr] µx+r:y+r dr] 0 = exp[− by (11.1. .1. of T = F (t) d = [1 − t pxy ] dt d =− exp − dt t (for t > 0) t µx+r:y+r dr 0 (from (11. JOINT-LIFE MORTALITY TABLES Theorem t pxy 187 t = exp(− 0 µx+r:y+r dr) (11. we must have ∞ 0 t pxy µx+r:y+r t pxy µx+t:y+t 0 (t > 0) (t < 0) (11.f.11.6) Proof.1.1).7) ∞ dr = 0 t pxy (µx+t + µy+t ) dt =1 since the integral of a p. Now f (t) = p.1. From (11. y). K2 }. (using Chain Rule) = exp − 0 µx+r:y+r dr µx+t:y+t Thus f (t) = In particular.1. (When t < 0. The discrete variable K = integer part of T This is the number of complete years to be lived in the future by the “joint life status” of (x. the distribution function of T is zero). Note that F (t) = P r{T ≤ t} = t q xy is the distribution function of T .d.1. 2.l73 = l[59]+1:[69]+1 = = 0. JOINT-LIFE FUNCTIONS where K1 = integer part of T1 . . Define l[x]+r:[y]+u = l[x]+r · l[y]+u which leads to t p[x]+r:[y]+u = l[x]+r+t:[y]+u+t l[x]+r:[y]+u Example 11.3 Extensions to More than 2 Lives We extend the definitions of T . y + u respectively who were selected r.t py . z respectively. u years ago respectively).856. Solution 3 p[59]+1:[69]+1 l[59]+4:[69]+4 l[59]+1:[69]+1 l63 . We also define µ[x]+r:[y]+u = lim h→0+ h q[x]+r:[y]+u h = µ[x]+r + µ[y]+u The p.188 CHAPTER 11.2 Select Tables We replace (x).f. This variable is discrete. T2 . (y) by [x] + r. T3 respectively (assumed to be independent). On A67 − 70 calculate 3 p[59]+1:[69]+1 . whose future lifetimes are T1 . [y] + u) is f (t) = t p[x]+r:[y]+u . of T = future lifetime of ([x] + r.t pz lx+t:y+t:z+t = lxyz . y.8) 11.d.. t pxy . 11. 1. K2 = integer part of T2 .1.) lxy (11. . t pxyz = P r{(x). etc. as follows: consider 3 lives aged x. [y] + u (referring to lives aged x + r. with probabilities P r{K = k} = P r{1st death occurs between times k and k + 1} = k |qxy =k pxy −k+1 pxy (by an argument similar to that for a single life) lx+k:y+k − lx+k+1:y+k+1 = lxy dx+k:y+k = (k = 0.µ[x]+r+t:[y]+u+t 0 (t ≥ 0) (t < 0). (y) and (z) all survive for at least t years} = t px . 2 .1. 5 0 t pxxxx (4µx+t ) dt. y. h→0 µxyz = lim+ h qxyz f (t) = t pxyz µx+t:y+t:z+t = t pxyz (µx+t + µy+t + µz+t ) (t ≥ 0) 0 (t < 0) Example 11. use the formula P r{(x) survives but (y). = 0. The probability that exactly one of three lives aged x will survive for n years is 27 times the probability that all three will die within n years.3. T3 } has p. (z) die} +P r{(z) survives but (x).9.3.3.d.2. Example 11.1. (y) die} (note that these events are mutually exclusive) = t px (1 − t py )(1 − t pz ) + t py (1 − t px )(1 − t pz ) + t pz (1 − t px )(1 − t py ) Similar calculations apply to P r{exactly one life dies} and other possibilities. α2 . Find the probabilities that (a) at least two will survive n years (b) at least one will die within n years Solution (a) 3n px (1 −n px )2 = 27(1 −n px )3 thus n px = 9(1 −n px ) hence n px = 0. in particular = µx + µy + µz h and T = min{T1 . α2 . (z) die} +P r{(y) survives but (x).f.11. T2 . To find the probability that exactly one life out of three survives for t years. α3 (α1 . α3 being the youngest ages in tables I. III). Evaluate 0 ∞ t pxxxx µx+t:x+t dt Solution ∞ 0 t pxxxx µx+t:x+t ∞ dt = 0. .5. Other functions follow similarly. II.1. Also P r{at least 1 dies within t years} = t q xyz = 1 − P r{all 3 survive} = 1 − t pxyz = 1 − t px t py t pz = 1 − (1 − t q x )(1 − t q y )(1 − t q z ). z ≥ α1 . so n qx = 0. EXTENSIONS TO MORE THAN 2 LIVES 189 where lxyz = lx ly lz for all x. 1) We also have exy = curtate joint expectation of life of (x.190 CHAPTER 11.9)3 = 0. where T = min{T1 . y) = E(K) where K = integer part of T ∞ = t=1 t pxy (11. Hence exy = 0 ◦ ∞ t. . JOINT-LIFE FUNCTIONS By the binomial distribution. similar to the single-life case).t pxy . we get exy = 0 ◦ ∞ t pxy dt (11.4 Define The Joint Expectation of Life ◦ exy = the complete joint expectation of life for the pair aged (x.t pxy dt − 0 (Use integration by parts.t pxy µx+t:y+t dt − (exy )2 ∞ 2 t pxy dt ◦ = 0 2t.4.271 11.972 (b) P r{at least one will die} = 1 − (n px )3 = 1 − (0.4. T2 } as before.2) The Euler–Maclaurin formula gives the approximation exy The variance of T is calculated as follows: Var(T ) = E(T 2 ) − [E(T )]2 ∞ exy − ◦ 1 2 = 0 ∞ t2 . y) = E(T ).µx+t:y+t dt Proceeding as in single-life case. the probability of at least two survivors = 3 (n px )2 (1 −n px ) + (n px )3 2 = 3(n px )2 − 2(n px )3 = 0. using integration by parts. 2 Evaluation of annuities using commutation functions.v. MONETARY FUNCTIONS 191 11.5.p.e.1) Note also that (the proof is similar to single-life case. i. See “International Actuarial Notation” in “Formulae and Tables for Actuarial Examinations”.2) Other joint-life annuity functions are ∞ axy = t=1 v t .) of an annuity ¯ of £1 per annum (p.) payable continuously so long as both (x).5 Monetary Functions Notation is very similar to single-life case (changing ‘x’ to ‘xy’). = 0 (11. m−1 2m m−1 − (1 − n pxy v n ) 2m a(m) ¨xy axy:n ¨ The Euler–MacLaurin formula gives axy ¯ (m) axy − ¨ axy:n ¨ axy − ¨ 1 2 1 axy + . and the only functions available in the examination . of an annuity of £1 p.11. payable at the ¨ start of each year so long as both (x).p. Commutation functions are not often used. (y) are alive = E[¨K+1 ] where K = integer part of T a ∞ = t=0 v t t pxy .5. (y) are alive = E[¯T ] where T = future lifetime of joint-life status a ∞ = 0 ∞ at| .t pxy µx+t:y+t dt ¯¯ v t t pxy dt 1 1+i ).t pxy = axy − 1 ¨ axy:n = axy −n pxy v n ax+n:y+n ¨ ¨ ¨ Woolhouse’s approximation for annuities payable mthly hold for joint lives. (11.5.v.a. (a) Annuities We define axy = mean present value (m.a. note that v = axy = m. 4) Nxy = t=0 Dx+t:y+t From these we get n Exy = m.5. Dxy lx+t:y+t lxy = Note: In A67 − 70 we only find Dxx and Nxx (i.e.5.3) (11. of £1 payable immediately on the failure of the joint-life status = E(v T ) where T = future lifetime of joint-life status ∞ = 0 v t . equal ages only. (y) are both alive = v n . Note that axy = E ¯ so we have the “conversion” relationship ¯ Axy = 1 − δ¯xy . of pure endowment of £1 payable at time n if (x).p.v.n pxy Dx+n:y+n = Dxy and ∞ axy = ¨ t=0 vt Nxy .5.192 tables are CHAPTER 11.5) This can be evaluated numerically by approximate integration.p.v. JOINT-LIFE FUNCTIONS Dxy = v 2 (x+y) lxy ∞ 1 (11. a We also have Axy = E[v K+1 ] = t=0 ∞ 1 − vT δ = ¯ 1 − Axy δ v t+1 t |qxy (11.) (b) Assurances ¯ Axy = m.5.6) and it will now be shown that Axy = 1 − d¨xy a Proof axy = E[¨K+1 ] = E ¨ a = 1 (1 − Axy ) d 1 − v K+1 d .t pxy µx+t:y+t dt (11. 1. (y) if within n :n xy ¯ Example 11. . ¯ Axy 1 ¯ ¯ P (Axy ) = = −δ axy ¯ axy ¯ If there are expenses. ¯ = Axy:n − v n ·n pxy :n xy = (1 − δ¯xy:n ) − v n · a lx+n ly+n · lx ly 1 lx+n ly+n 1 1 − δ(axy + ) − v n · · 1 − δ(ax+n:y+n + ) . The general symbols P and P may be used for any net or gross premium.5. in the International Actuarial Notation. Show how one can evaluate A Solution.. e.5. calculate office premiums as for single-life policies by setting up an equation of value.g.11. MONETARY FUNCTIONS So Axy = 1 − d. Method 1: ¯ A 1 1 on the a(55) tables (male/female. 1 193 ¯ (1 + i)− 2 Axy 1 to indicate that a sum assumed is payable on the first death of (x).) :n xy (1 + i) 2 A :n 1 2 1 1 1 xy xy :n = (1 + i) [Axy:n − v n ·n pxy ] = (1 + i) 2 Now use axy:n = axy − v n ¨ ¨ and finally use axy ¨ Method 2: ¯ A 1 (as in single-life case). If there are no expenses we obtain net premiums. 1 − d¨xy:n − v n a lx+n ly+n · lx ly lx+n ly+n · ax+n:y+n ¨ lx ly = axy + 1.¨xy a An approximation which can often be used is Axy ¯ Write A years. 2 lx ly 2 (c) Premiums We employ equations of value in the same way as for single-life policies. as follows: a61:69 1 [a60:68 + a62:68 + a60:70 + a62:70 ]. :t xy (Axy ) = 1 − ax+t:y+t ¯ axy ¯ using conversion relationships. The sum assured is payable immediately on the first death. . ¯ ¯ ¯ a = Ax+t:y+t − P (Axy )¯x+t:y+t (using prospective method). The policy has annual premiums. Two lives. 4 Example 11.2. ¨ ¨ axxx . aged 50 and 60. effect a 10-year temporary joint life assurance with sum assured £60. Note: The formula for retrospective reserve in this example is Dxy Dx+t:y+t Also. e..194 (d) Reserves Calculated as for single-life policies. Estimate the value of the integral by the following form of Simpson’s rule: 50 : 60:10 10 f (t) dt = 0 10 · [f (0) + 4f (5) + f (10)] 6 (ii) Hence.a. payable immediately on the first death of two lives aged x and y at the issue date.000. We use the general symbol t V for the reserve at duration t.5. a[xx] (= a[x]:[x] ).a. payable during the joint lifetime of both lives for at most 10 years. JOINT-LIFE FUNCTIONS ¯ (Axy ) = net premium reserve at duration t years for ¯ the joint-life assurance of £1. affected by annual premiums payable continuously during the joint-lifetime. ¨ Nxx . 6% and 8% interest. Basis: A1967 − 70 ultimate mortality 4% p.g. one may show that tV [P axy:t − A ¯ 1 ]. estimate the values of A50:60:10 and a50:60:10 . tV CHAPTER 11. or otherwise. Exam tables (i) A67-70 gives: axx . (ii) a(55) gives axy at 2-year intervals for male aged x and female aged y (x. ¨ axxxx ¨ all at 4% p. and calculate the annual ¨ premium for the policy. ¯ (i) Express A 1 as an integral. Interpolation is needed to get (for example) a61:69 . interest. at 4%. provided that this occurs within 10 years of the issue date. interest No expenses. just before payment of any premium then due. Dxx . y ≥ 60 only). 2) giving t pxy = t px + t py − t px · t py t q xy and = t qx · t qy . T2 }. Note that t q xy is the distribution function of the variable Tmax = max{T1 .04)− 2 .6 Define. (y) dies within t years}. 768.a.1) (11. d ¯ A 1 P = 60000 50 : 60:10 a50:60:10 ¨ = £1. we may write t pxy = t px · t py (11.24 11. By elementary probability theory. If they are independent.70792. LAST SURVIVOR PROBABILITIES (TWO LIVES ONLY) Solution ¯ (i) A 1 = 50 : 60:10 195 10 0 v t .f.2238. 50 : 60:10 (ii) A50:60:10 = A 1 50 : 60:10 1 + v 10 l60 l70 · l50 l60 + v 10 l70 l50 1 ¯ (1. t pxy = t px + t py − t pxy = t q x + t q y − t q xy and hence t q xy These formulae are true even if (x).6.11.d. ¯ A Simpson’s rule gives an approximate answer of 1 = 0. of Tmax = time to death of last survivor is f (t) = d d d d (t q ) = (t q x ) + (t q y ) − (t q xy ) dt xy dt dt dt . (y) will be alive in t years’ time} = 1 − t q xy .6. 1 − A50:60:10 = = 7. is = 0. (y) are not independent.594.A 50 : 60:10 a50:60:10 ¨ The premium P p.t p50:60 (µ50+t + µ60+t ) dt.6. Last Survivor Probabilities (two lives only) t pxy = P r{at least 1 of (x). where t q xy = P r{last survivor of (x). and the p. T2 } + v max{T1 .196 Assume independence of T1 . We get: CHAPTER 11.7 Last Survivor Monetary Functions (a) Annuities Define.d. On multiplying the equation t pxy = t px + t py − t pxy by v t .T2 } .1) a(m) ¨xy = a(m) ¨x + a(m) ¨y − We may also define temporary last survivor annuity functions. of an annuity of £1 p.t py (µx+t + µy+t ) t pxy Note This is not often used in practice. 11.v. e.f. F (t) = the distribution function of Tmax . JOINT-LIFE FUNCTIONS f (t) = t px µx+t + t py µy+t − t px t py (µx+t + µy+t ) (t > 0) Now the “force of mortality” has to be defined as follows: µxy (t) = hazard rate at time t = where f (t) = the p. T2 . axy = m.t pxy v t [t px + t py − t pxy ] t=0 = ¨ ¨ = ax:n + ay:n − axy:n ¨ Note: these formulae hold even without the assumption of independence of the lives.7. This gives µxy (t) = t px µx+t f (t) 1 − F (t) + t py µy+t − t px . one obtains the following relationships for annuity functions ¨ ¨ ¨ axy = ax + ay − axy ¨ axy = ax + ay − axy ¯ ¯ ¯ ¯ (m) axy ¨ (11.p. for example. of Tmax .g. payable annually ¨ in advance so long as at least one life is alive. and summing or integrating over t. n−1 axy:n = ¨ t=0 n−1 v t . (b) Assurances Taking expected values on each side of the relationship v T1 + v T2 = v min{T1 .a. ¯ ¯ ¯ = Ax + Ay − Axy Similarly.p. of a last survivor assurance of £1 payable immediately on the death of (x).3) Axy = Ax + Ay − Axy = (1 − d¨x ) + (1 − d¨y ) − (1 − d¨xy ) a a a = 1 − d¨xy a ¯ a and. Example 11.11.7.v.7.2) (11.7.K2 +1} + v max{K1 +1. (y) Axy = axy ¨ Using the conversion relationships given above. using the International Actuarial Notation.7.K2 +1} and hence Axy = Ax + Ay − Axy (which is still true even when the lives are not independent. when the sum assured is payable at the end of the year of death. (y).20246 (c) Premiums. we have v K1 +1 + v K2 +1 = v min{K1 +1. similarly.052] = 0. Evaluate A40:40 on A1967 − 70 ultimate.894 − 17. For example.) Note also the conversion relationships 197 (11.1. We may use an equation of value to calculate net or gross premiums. the following equations can be derived: ¯ Axy 1 ¯ ¯ P (Axy ) = = −δ axy ¯ axy ¯ and Pxy = 1 − d. axy ¨ . 4% interest. Solution A40:40 = 1 − d¨40:40 a = 1 − d[2¨40 − a40:40 ] a ¨ = 1 − d[2 × 18. Axy = 1 − δ¯xy . LAST SURVIVOR MONETARY FUNCTIONS one obtains ¯ Axy = m. etc. we have Pxy = net annual premium payable annually in advance for a policy providing £1 at the end of the year of the death of the second to die of (x). t px (1 − t py ) t py (1 − t px ) . one may value the policy as a ‘weighted average’ of t V 1 . Proof t V1 = S (1 − d¨x+t:y+t ) − Pxy ax+t:y+t a ¨ = S 1 − (Pxy + d)¨x+t:y+t a a ¨ ¨ < S [1 − (Pxy + d)¨x+t ] (as ax+t < ax+t:y+t ) = S [(1 − d¨x+t ) − Pxy ax+t ] a ¨ ¨ = S [Ax+t − Pxy ax+t ] = t V 2 ( and similarly for t V3 ) If it is not known whether one or both of (x). depends on whether or not both lives are still alive at time t. with weights proportional to the probabilities t pxy .8 Reserves for Last Survivor Assurances Consider a whole-life assurance providing £S at the end of the year of death of the last survivor of (x). Finally. Let t V1 = reserve assuming both lives are still alive t V2 = reserve assuming (x) now alive but (y) has died t V3 = reserve assuming (y) now alive but (x) has died We find that t V1 t V2 = S Ax+t:y+t − Pxy ax+t:y+t ¨ = S [Ax+t − Pxy ax+t ] ¨ ¨ t V3 = S [Ay+t − Pxy ay+t ] Theorem When the first life dies. Then Annual premium. Therefore the weighted average reserve is V = t pxy t pxy · tV 1 + − t py ) t py (1 − t px ) · tV 2 + · tV 3 pxy t t pxy i. the prospective reserve will increase (from t V1 to t V2 or t V3 ). respectively t pxy t pxy t px (1 in order for them to add up to 1. We ignore expenses and assume that the premium and reserve bases are the same.e. t V 2 and t V 3 . t px (1 − t py ) and t py (1 − t px ) respectively. :t xy .198 CHAPTER 11. P = SPxy = S 1 −d axy ¨ The value of the prospective reserve at duration t. JOINT-LIFE FUNCTIONS 11. (y).t pxy 1 . just before payment of the premium then due. V = 1 S Pxy axy:t − A ¨ v t . it can be shown that the retrospective reserve is equal to the weighted average reserve. Hence the weights are t pxy t pxy . (y) survive at time t (and in practice this may be the case since one death does not result in a claim or a change in the premium). on the premium basis. the man died.96 − 1. be P .975 × 10.549 24 24 65:60 am − ¨ = 1 − d¨ m a 65:60 1 f = 0. Premiums were payable monthly in advance so long as at least one of the couple survived.0075 − = 10.55 . 773. (ii) Just after the payment of the first premium. 000 × (1.15.18463 10.5% of all office premiums .11. 0. 000 immediately on the death of the second to die of a man aged 65 and his wife aged 60. and a f a60 60 Hence Reserve = £694. payable monthly. A last-survivor policy provided £10. Solution (i) Let annual premium.942 24 11 = 9. RESERVES FOR LAST SURVIVOR ASSURANCES 199 Example 11. The office which issued the policy used the following basis: mortality : a(55)ultimate. (i) Find the monthly premium. expenses : 2. males/females as appropriate interest : 8% per annum .82 (using Af = 1 − d.975P a f 60 = 2.18463 Hence P = = £186. 000(1 + i) 2 A m 1 65:60 65:60 f am ¨ (12) 65:60 (12) 65 (12) 60 f f am ¨ 65:60 65 f − 11 = 7. (ii) Reserve = 10000(1 + i) 2 Af − 0.¨f . (12) 60 1 (12) 60 = af + 60 11 ) 24 .08) 2 × 0.8.1. Find the office’s prospective reserve just after this event.836 af − ¨ 24 60 11 11 am f − ¨ = 11.55 0.975P a m ¨ (12) f = 10.228 24 am ¨ af ¨ Hence a m ¨ Am (12) 65:60 65:60 f 11 = 7. 476.8.549 Therefore monthly premium = £15. Show that. and give a corresponding result for case (ii). if the office uses the following basis for premiums and reserves. (a) all three will survive one year (b) at least one will survive one year (c) exactly two will survive one year (d) at least two will survive one year 11.6 The probability that at least one of three lives aged 60 will survive to age 65 is eight times the probability that exactly one will survive to age 65. find the probability that. who is 4 years older than himself. or otherwise. 1 expenses: 3% of all office premiums including the first. of three lives (x). 11. (y) and (z). (y) and (z). in each case at a given rate of interest i p.200 CHAPTER 11.8 12 years ago a man then aged 48 effected a without profits whole life assurance for £10.7 (i) Define t pxy and show that t pxy = t px + t py − t pxy (ii) Hence.a. 11.a. The premium now due is unpaid. Assuming that the 3 lives are independent and subject to the same table of mortality.4. 11.(I¨)xy ¨ a 11. ceasing on the first death. with additional initial expenses of 1 2 % of the sum assured. . of the lives (x).6.) 11. T2 }) using P r{K = k} = k |qxy . and pz the probabilities that.000 (payable at the end of the year of death) by annual premiums. ¯xy ¯ var(L) = A∗ − (Axy )2 where ∗ indicates a rate of interest of 2i + i2 p.3 Express in terms of px . (a) none will survive n years (b) exactly one will survive n years (c) at least one will survive n years. mortality: A1967-70 ultimate. rated down 4 years for female lives. interest: 4% per annum. n py = 0.1 Given that n px = 0.4 Derive the formula exy = E(K) = t=1 ∞ t pxy where K = integer part of T (T = min{T1 . 11.. Calculate the revised office annual premium.2 Prove that (IA)xy = axy − d. py . show that axy = ax + ay − axy ¨ ¨ ¨ ¨ 11. providing that the sum assured does not increase. JOINT-LIFE FUNCTIONS Exercises 11. find the probability that exactly one life will survive to age 65.3.5 Evaluate A75:75 on the basis of A1967 − 70 ultimate at 4% interest. He now wishes to alter the policy so that the same sum assured will be payable at the end of the year of the first death of himself and his wife. and (ii) the second death. in case (i). (This additional initial expense is not charged again on the conversion of an existing policy.9 Consider the random variable L equal to the present value of £1 payable immediately on (i) the first death of (x) and (y). n pz = 0. 436 = 1 − 0.04) 2 [1 − d.04) 2 A75 − (1. 1 1 .4) + 0. pz (b) 1 − (1 − px )(1 − py )(1 − pz ) (c) px py (1 − pz ) + px pz (1 − py ) + py pz (1 − px ) (d) px py (1 − pz ) + px pz (1 − py ) + py pz (1 − px ) + px py pz 11.1 (a) (b) (1 − n px )(1 − n py )(1 − n pz ) n px (1 = 0. py .7)(0.4 exy = curtate joint expectation of life ∞ = k=1 k.6 × 0.4(0.10.64676.4 = 0.168 − n py )(1 − n pz ) + n py (1 − n px )(1 − n pz ) + n pz (1 − n px )(1 − n py ) = 0.k |qxy 1 lxy ∞ = = k.dx+k:y+k k=1 1 [dx+1:y+1 + 2dx+2:y+2 + 3dx+3:y+3 + · · · ] lxy 1 = [(dx+1:y+1 + dx+2:y+2 + · · · ) + (dx+2:y+2 + dx+3:y+3 + · · · ) + · · · ] lxy 1 = [lx+1:y+1 + lx+2:y+2 + · · · ] lxy = = t=1 1 lxy ∞ ∞ lx+t:y+t t=1 t pxy 11.7 × 0. y) (I¨)xy = E (I¨)K+1 a a =E = aK+1 − (K + 1)v K+1 ¨ d (as (IA)xy = E[(K + 1)v K+1 ]) 1 [¨xy − (IA)xy ] a d Hence (IA)xy = axy − d(I¨)xy .6(0.6) (c) 1 − (1 − n px )(1 − n py )(1 − n pz ) 11.¨75:75 ] a = 0. = 0.5 ¯ ¯ ¯ A75:75 = 2A75 − A75:75 2(1.168 = 0.4) + 0.11.3 (a) px .7)(0.832.3(0.2 Let K = curtate future lifetime of (x. SOLUTIONS 201 Solutions 11. ¨ a 11.6)(0. 015(1 − ) − 0.97P a60:60 ¨ = 10.202 CHAPTER 11.51 11. T2 }.¨60:60 ) − 0. Equation: V1 = 10.76862 (ignoring root > 1) = 46 Hence probability of exactly one survivor = 3p(1 − p)2 = 0. 000[1.015] (using Zillmer’s formula). 000(1 − d.6 Let 5 p60 = p. 000A60:60 − 0.015] a48 ¨ = £2.01512 V48 − 0. L = v T where T = min{T1 .6447P Hence P = £397. 341.97P a60:60 a ¨ = 6175. . We have var (L) = E(L2 ) − [E(L)]2 ¯ = E[(v ∗ )T ] − (Axy )2 ¯xy ¯ = A∗ − (Axy )2 .8 = ax + ay − axy ¨ ¨ ¨ V1 = reserve of original policy = 10. a60 ¨ = 10.12345 p= 45 ± 11. We have (using binomial distribution) 1 − (1 − p)3 = 8 × 3p(1 − p)2 Hence 3p − 3p2 + p3 = 24p(1 − p)2 3 − 3p + p2 = 24 − 48p + 24p2 So 23p2 − 45p + 21 = 0 (−45)2 − 4 × 23 × 21 46 √ 45 ± 93 = 0.89 (there is no need to find the original premium) Let P be the revised annual premium.7 (i) t pxy = P r{ the last survivor of (x) and (y) survives for t years} = P r{(x) survives for t years but (y) dies} + P r{(y) survives for t years but (x) dies} + P r{(x) and (y) both survive for t years} = t px (1 − t py ) + t py (1 − t px ) + t px t py = t px + t py − t px t py (ii) ∞ axy = ¨ t=0 ∞ t pxy v t = t=0 ∞ v t (t px + t py − t px t py ) ∞ ∞ = t=0 v t t px + t=0 v t t py − t=0 v t t pxy 11.77 − 9.9 In case (i). 000[1. JOINT-LIFE FUNCTIONS 11. SOLUTIONS The corresponding result for case (ii) is ¯xy ¯ var (L) = A∗ − (Axy )2 . 203 .11.10. 204 CHAPTER 11. JOINT-LIFE FUNCTIONS . (y). giving q1 = Note also that ∞ ∞ q1 xy xy 0 t pxy µx+t dt = 0 t pxy µx+t t t q1 x:x dt = P r{(x) dies before (y)} r pxx µx+r If x = y. and before (y)} i.Chapter 12 CONTINGENT ASSURANCES 12. it is easy to make the necessary adjustments in the formulae.e.2) .1) If t = 1 we may omit it. we clearly have = 0 dr = 1 t r pxx µx+r:x+r dr 2 0 1 = t q xx 2 205 (12. (x) dies within t years and (y) dies after this event (not necessarily within t years).1. Let T1 = future lifetime of (x) with pdf f1 (t1 ) = t1 px µx+t1 T2 = future lifetime of (y) with pdf f2 (t2 ) = t2 py µy+t2 Then t q1 xy xy (t1 > 0) (t2 > 0) = P r{T1 ≤ t and T1 ≤ T2 } = = 0 t t1 0<t1 ≤t t1 ≤t2 t ∞ f1 (t1 )f2 (t2 ) dt1 dt2 t1 px µx+t1 · t2 py µy+t2 dt2 dt1 = 0 t t1 px µx+t1 r pxy µx+r · t1 py dt1 dr 1 = 0 (12.1.1 Contingent Probabilities Suppose we have 2 independent lives (x). (If these tables are different.) Define t q1 = P r{(x) dies within t years. subject to the same mortality table. 206 CHAPTER 12.) 12.H.3) Note also that ∞ q1 xy + ∞q xy 1 = ∞ q xy = 1 One may also define deferred contingent probabilities.S. CONTINGENT ASSURANCES By general reasoning. and we find that t |q 1 x:y = t pxy . . the following result holds: t q1 xy + tq xy 1 = t q xy (12.2 Contingent Assurances Consider a benefit of £1 payable immediately on the death of (x) if this occurs before the death of (y).1. as expected by general reasoning. are the probabilities of mutually exclusive events.V.1. The M. t qx = t q1 + t q2 xy xy (12.2. we have t t q2 xy = 0 (1 − r py )r px µx+r dr xy = t qx − t q1 Thus.∞ q x+t:y+t 1 Consideration of the second death = P r{(x) dies after (y) and within t years} xy By calculations similar to those for t q1 .4) which follows from P r{T1 ≤ t} = P r{T1 ≤ t and T1 ≤ T2 } + P r{T1 ≤ t and T1 > T2 } (the two terms on the R. (at a specified rate of interest) is: v T1 ¯ A1 = E xy 0 = = 0 ∞ t1 ≤t2 ∞ t1 if T1 ≤ T2 if T1 > T2 v t1 (t1 px µx+t1 )(t2 py µy+t2 ) dt2 dt1 ∞ t1 px µx+t1 t1 t2 py µy+t2 v dt2 dt1 = 0 ∞ v t1 t1 px µx+t1 (t1 py ) dt1 v t t pxy µx+t dt (12.P.1) = 0 This is often evaluated by approximate integration. t q2 xy t |∞ q 1 x:y = t pxy .q x+t:y+t 1 and We also have the definition. 2. CONTINGENT ASSURANCES 207 Example 12. by approximate integration. (b) It has been suggested that a policy providing the above benefit should be issued by annual premiums ceasing on the death of the female life or after 15 years. the value of a contingent assurance of £100. A very accurate approximation to the value is not expected.v. so that the integral is over a range of 48 years.2.2. we also have (12. Using Simpson’s rule ¯ A1 50:60 = 2 [f (0) + 4f (6) + 2f (12) + 4f (18) + 2f (24) 4f (30) + 2f (36) + 4f (42) + f (48)] = 0.1546. and that of the male follows A1967-70 ultimate. whichever is earlier.] Solution ¯ A1 ∞ 50:60 = 0 48 v t t p50 µ50+t t p60 dt (at 4% interest) f (t) dt = 0 where f (t) = v t t p50:60 µ50+t . Break this into 4 sub-intervals and use Simpson’s rule over each. (a) Estimate. by approximate integration.2. provided a male aged 50 is still alive and provided her death occurs within 15 years.4) Axy = A 1 + A 1 xy xy Note also that Example 12. The interest rate is 7 2 % per annum and expenses are ignored. of £1 payable immediately on death of (x) if this occurs before the death of (y) and within n years n x:y:n = 0 v t t pxy µx+t dt (12. Temporary contingent assurance functions A1 = m.p.12.1. x) will be the first to die). State with reasons whether you agree with this suggestion. Note. . other than issuing the contract by a single premium. ¯ We may therefore evaluate contingent assurances on 2 equal ages from tables of Axx . the value of ¯ A1 50:60 [Assume l108 = 0. As expected by general reasoning.2.000 payable immediately on the death of a female aged 60.3) xx 2 (since there is a 50% chance that the first “x” of (x. suggest a more suitable premiumpaying term.2. Using A1967-70 mortality and 4% interest estimate.2. If you do not.2) 1¯ ¯ A 1 = Axx (12. The mortality of the female follows a(55) ultimate 1 (females). v.208 Solution CHAPTER 12.3. CONTINGENT ASSURANCES ¯ (a) Value = 100. the values of (i) A 1 . e.6) ¯ ¯ ¯ Example 12. ¯xy A2 = Hence ¯xy ¯ ¯xy Ax = A1 + A2 as expected. We may also encounter temporary contingent assurances payable on the second death.p. 000A m = 100. the m. equals n 0 ∞ 0 v t t px µx+t (1 − t py ) dt (12. (12. 657 (b) No: premiums must also cease on death of (50). 000 50:60:15 15 v t t p50:60 µ60+t 0 1 f dt Evaluate integral by (for example) the three-eighths rule: 15 f (t) dt 0 15 [f (0) + 3f (5) + 3f (10) + f (15)] 8 = 0. Hence value = £12. if earlier. of £1 payable immediately on death xy of (x) if this occurs after the death of (y).v. ¯xy By a similar argument to that for A1 . provided that (x) dies after (y). Define ¯ A 2 =m.g. We may also encounter contingent assurances payable on the second death. (ii) A 2 and (iii) A 1 60:60 60:60 60:60:10 .2.5) v t t px µx+t (1 − t py ) dt x:n = A1 Note also that − A1 xy:n 1¯ ¯xx A2 = Axx 2 since there is a fifty per cent chance that the first “x” will be the last survivor.2.12657. Restrict premium-paying period so that premiums cease on first death or after 15 years. otherwise they may still be payable with no prospect of claim: negative prospective reserve.p. Using the fact that A1 + A 1 = Axy . of £1 payable immediately on death of (x) within n years. find on A1967 − 70 ultimate at 4% interest xy xy ¯ ¯ ¯ .2. etc. xy 1 (12.15513 = The symbols A1 . etc.: xy Axy = A 1 + A xy xy xy xy 1 Ax = A1 + A2 A1 A2 xx xx 1 = Axx 2 1 = Axx 2 The variance of the present value of a contingent assurance Let Z = present value of £1 payable on death of (x).2.12. xy ¯ These are the same as A1 . this has mean ¯ E(Z) = A1 xy . etc.8) Note the following results. An exact expression is xy ∞ A1 = xy t=0 v t+1 t pxy q x+t:y+t 1 (12.04) 2 [1 − d¨60:60 ] a 0.2. we may use the approximations A1 xy ¯ (1 + i)− 2 A1 .2. CONTINGENT ASSURANCES Solution. if this occurs before the death of (y) = v T1 0 if T1 ≤ T2 if T1 > T2 where v = 1 1+i As we have shown above.04) 2 A60:60 − A70:70 2 D60:60 = 0.7) In practice. etc. but with the benefit payable at the end of the year of death. (i) ¯ A 60:60 1 209 ¯ = 0..5A60:60 0.31491 1 (ii) Use A 60:60 2 = A60 − A 60:60 1 to obtain ¯ A 2 60:60 = 0. similar to those given above for A 1 .5(1.21259 (iii) ¯ A 60:60:10 1 1¯ 1 A 2 60 : 60:10 1 1 D70:70 (1. Hence the reserves are as follows: (i) (ii) 10 V = 1000A40 − P a40 if only A is alive = 211. provided A dies after B. the annual premium.210 The variance of Z is CHAPTER 12.9) 12.00327) = 3. Reserves are usually calculated prospectively. Two lives (A and B) are both aged 30.3 Premiums and Reserves for Contingent Assurances Premiums are calculated by the usual equation of value (including expenses if necessary). Calculate.3. making allowance for any deaths which have already occurred. Find the policy value (on the premium basis) after 10 years (before the premium then due is paid) if (i) only A is then alive. payable during the lifetime of A.49 ¨ 10 V = 1000A 2 40:40 . to provide an insurance of £1000 payable at the end of the year of death of A. CONTINGENT ASSURANCES E(Z 2 ) − [E(Z)]2 where E(Z 2 ) = E =E (v 2 )T1 0 (v ∗ )T1 0 xy if T1 ≤ T2 if T1 > T2 if T1 ≤ T2 if T1 > T2 where v ∗ = 1 with i∗ = i2 + 2i 1 + i∗ ¯1 = A∗ at rate of interest i2 + 2i (or force of interest 2δ) Hence ¯1 ¯ Var(Z) = A∗ − (A1 )2 xy xy (12. Solution A2 30:30 = A30 − A 1 30:30 1 1 = A30 − A30:30 = (1 − d¨30 ) − (1 − d¨30:30 ) a a 2 2 1 1 = − d(¨30 − a30:30 ) a ¨ 2 2 Hence A2 premium = 30:30 a30 ¨ · 1000 = 1000 1 1 a30:30 ¨ −d 1− 2¨30 a 2 a30 ¨ = 1000. (ii) both lives are then alive. on the basis of A1967 − 70 ultimate mortality and 4% p.1.a.56 ¨ − P a40 if both are alive = 39. Example 12. interest.(0.2. say.27 = P. (See Example 12. as this may lead to a lapse option against the office.) 12. An absolute reversion is a reversion in which the sum is paid under all circumstances. A PRACTICAL APPLICATION – THE PURCHASE OF REVERSIONS 211 Lapse options One should avoid having a negative prospective reserve at any duration.000.The life office issuing the policy would examine the health (and any other risk factors) of (50) very carefully. 000A 50:80 1 In practice. provided that he is then alive. and will “plug the gap” with an insurance policy (which pays out on the death of (80) if she dies second.1.a. 000 A80 − A ¯ = 1. (Note that the purchaser might assume different tables for the lives).12. 000. 000..4. 000A 2 50:80 ¯ Hence purchase price = 1. one should ensure that premiums cease as soon as there is no possibility of future benefits. . 000A 50:80 1 50:80 2 (as found before) Note:. Brown and to his mother. The interest is contingent since it requires (50) to be alive when (80) dies.) Suppose that the life office issuing the insurance policy uses the same mortality and interest basis as the purchaser. 000A80 − cost of premium for insurance policy ¯ ¯ = 1. 000. 000.000 on the death of his mother (aged 80). ¯ Single premium to buy insurance = 1.4. J.P. and assumes that a certain mortality table applies to both J.4 A Practical Application – The Purchase of Reversions Definitions A reversion is a contract providing a sum of money payable on the death of a certain life. Brown(50) will receive £1. the purchaser will want to be sure of getting the money when (80) dies. whilst in a contingent reversion the sum is paid only if certain other lives are (or are not) then alive.2. = 1. Example 12.V.2 above. He wishes to sell his “interest”. In particular. 000. (y). i p. and also ignores expenses. What is it worth? Solution Suppose that a purchaser uses a certain rate of interest. Hence ¯ M. (y) having died first} ∞ = 0 t px µx+t (1 xz xyz − t py )t pz dt (12.5. (y).2) = ∞q 1 − ∞q 1 ∞q 2 xyz = P r{(x) dies second of (x). but we do not pursue this topic. (z)} = ∞q 2 xy z 1 + ∞q 2 xyz 1 (since either (y) or (z) dies first) xyz = ∞ q 1 + ∞ q 1 − 2∞ q 1 xy xz using formula (12.2) Similar definitions and relationships apply for contingent assurances. CONTINGENT ASSURANCES 12.5.5 Extension to Three Lives ∞q 1 xyz We briefly mention the following symbols and formulae: = P r{(x) dies first of (x). (z)} ∞ = 0 ∞q 2 xy z 1 t px µx+t · t py · t pz dt (12.212 CHAPTER 12.g. A2 = m.1) = P r{(x) dies second. . (y). e. of £1 payable immediately on the death of (x) if he dies second =A 1 + A 1 − 2A 1 xy xz xyz xyz More complicated functions may be defined and evaluated.5.v.p. 000 immediately on the death of (x) if she dies before (y) is to be issued by a life office to a group of trustees. if (a) both Smith and Jones are alive. 12.12. (iii) Write down (but do NOT evaluate) formulae for the reserve at duration 10 years (immediately before payment of the premium then due) on the premium basis. estimate the single premium payable by Adams. with an additional initial expense of £100.2 Adams (aged 40) and Brown (aged 50) are two business partners. Using Simpson’s rule. a further £5. Adams wishes to provide for the sum of £80.000 at the end of the year of death of the first to die of these two lives. axx and the rate of interest. if Smith is the second to die. (i) Ignoring expenses.000 will be payable at the end of the year of his death. 12. The policy is to have annual premiums payable during the joint lifetime of Smith and Jones. ¯ ¯ Axx = 2Ax ¯ ¯ ¯ Axx = 2Ax − A1 ¯ ¯ ¯ Axx = A1 + A2 xx xx xx xx 12.) 12. (ii) The trustees suggest that level annual premiums should be payable in advance until the death of the last survivor of (x) and (y).4 (i) Express A2 in terms of ax . and effects a policy providing this benefit by single premium. D. Axx = 2A1 B.6.000 to be paid immediately on Brown’s death if Brown predeceases him within ten years. In addition. The life office issuing the contract employs the following basis: Mortality (both lives) : A1967 − 70 ultimate Interest : 6% Expenses: 2% of the single premium. EXERCISES 213 Exercises 12. Calculate the annual premium on the following basis: A1967 − 70 ultimate mortality 4% interest expenses are 5% of all premiums. ¨ ¨ xx ∞ q1 74:84 on the basis of A1967 − 70 ultimate mortality. C. (ii) Smith and Jones are both aged 60. . Break this into 3 sub-intervals and use Simpson’s Rule over each. or otherwise.5 A policy providing the sum of £100.3 Estimate the value of (Assume l108 = 0. so that the integral is over a range of 24 years. and (b) Jones has died but Smith is alive.1 Which (if any) of the following statements are correct? ¯ ¯ A. A life office has been asked to issue a special joint-life assurance policy providing £10. write down an expression for the single premium in terms of an integral. Obtain an expression.) 12. 12. An interest rate of 7. but a person aged exactly 45 is not. CONTINGENT ASSURANCES (a) Ignoring expenses.214 CHAPTER 12. (i) Assuming that the two lives are independent.000 immediately on the death of a woman now aged 80 within 15 years. by the trapezoidal rule or another suitable rule for approximate integration. the position of chief is given to someone from outside the previous chief’s family. who are then permanently debarred from becoming chief. . and may be succeeded only by a person aged from 36 to 45. aged exactly 37 and 33 respectively.7 Define the following functions in words. the single premium for a temporary contingent assurance of £50.5% p. give a formula for the annual premium. n q1 only. in terms of quantities of the form n px . The present chief is aged exactly 47 and has two brothers.9 Your life office has been asked to quote a single premium for a contingent assurance policy providing £300. for the probability that (33) will become chief. (i) ∞ q1 xy xy x:y:n ¯ (ii) A2 ¯ (iii) A1 12. provided that at the date of her death a man now aged 60 has died.) (iii) Would you subject the male life to stringent underwriting procedures? Give brief reasons for your answer. (You are NOT required to carry out the evaluation. (ii) State a suitable non-repeated rule of approximate integration for evaluating this integral.000 payable immediately on the death of Mrs Smith (aged 60). is to be used. provided that this event occurs within 5 years and that her husband (aged 50) is alive at the date of her death. (b) Would you advise the life office to issue the policy with premiums payable as suggested? Give reasons for your answer. Mrs Smith is subject to the mortality of a(55) ultimate (females) and Mr Smith is subject to the mortality of A1967 − 70 ultimate. and allowance is to be made for expenses of 6% of the single premium.) The customs of the tribe require that a chief’s successor be his oldest eligible brother. and give an expression for each of them in terms of an integral. (Note A very accurate answer is not expected.8 The chief of a certain tribe holds that office until age 50 or earlier death. (A person aged exactly 36 is eligible. The chief and his brothers may be regarded as independent lives subject to the mortality of a given table. if there is no eligible brother.a. write down a formula for the single premium in terms of an integral.a(55) ultimate (females) interest : 8% per annum expenses : 10% of the single premium. Your office uses the following basis: mortality : males .a(55)ultimate (males) females . xy 12.6 Estimate. 943)] 1.5 × 9.95P axx = 10.1 Only A is correct. 000 Putting x = 60 gives 0. xx 2 1 = (1 − d¨x ) − (1 − d¨xx ) a a 2 1 i 1 = − (¨x − axx ) a ¨ 2 1+i 2 (ii) Let the annual premium be P. and set x = 60.95P · (9.04 =6175.98 0 80000 10 [f (0) + 4 · f (5) + f (10)] using Simpson’s rule 0.5 − (12.2 Let single premium be P .4 xx (i) A2 = (or use A2 xx 1 1 Axx = Ax − Axx 2 2 1 = Ax − A1 = Ax − Axx ).17 7318. 000A2 + 100 ¨ xx = 10.943) − 100 =10.75 9.02P at 6% interest = 80000 10 f (t) dt where f (t) = v t t p50 µ50+t .04 0. 000[1 − d¨xx ] a + 5. 000Axx + 5.551 − 0.17 = £774.04 + 5. 12.44585 1 1 − d(¨x − axx ) + 100 a ¨ 2 2 Hence P = .12. 000[0.77 + 1042.943 1.2907.7. ∞ 12. 000 1 − 0.t p84 4 [f (0) + 4f (4) + 2f (8) + 4f (12) + 2f (16) + 4f (20) + f (24)] 3 = 0. 686. SOLUTIONS 215 Solutions 12.3 ∞ q1 74:84 = 0 24 t p74 µ74+t . P = 80000 0 10 v t t p50 µ50+t · t p40 dt + 0.40 = 7218.t p84 dt = 0 f (t) dt where f (t) = t p74 µ74+t . 0.t p40 0.98 6 = £4.04 · 9. 12. of a temporary contingent assurance of £1 payable immediately on death of (x). CONTINGENT ASSURANCES (iii) (a) Reserve = 10.t p m dt 50 0. because premiums should not continue after first death.94 = P r{(x) will die before (y)} ∞ = 0 t pxy µx+t dt (ii) ¯ A2 x:y = m. 219 0. of a contingent assurance of £1 payable immediately on death of (x). 000A1f = 50. 000A − 0.041726 Hence P 12. There is thus a lapse option on the first death.6 Let single premium = P . 000A70:70 + 5.94P = 50.t pxy µx+t dt . 0 m 60:50:5 5 v t t p f µ f . 000 ∞ 0 70:70 2 v t t pxy µx+t dt 100000 ∞ 0 (ii) (a) Annual premium.216 CHAPTER 12.t p m 50 60 60+t 0 1 at 7 % interest 2 dt 5 vt t p f µ 60 60+t f . axy ¨ (b) No.v. n = 0 v t .p.p.7 (i) ∞ q1 xy 50000 × 0. If (x) dies first. and if (y) dies first there is no possibility of benefit and the policy be lapsed.041726 = £2. the benefit is paid and the policy may be lapsed. and before (y) dies. 12. 000 By the trapezoidal rule.v.5 (i) 100. ∞ = 0 v t t px µx+t (1 − t py ) dt (iii) ¯ A1 x:y:n = m. 000A70 12. P = v t t pxy µx+t dt .95P a70:70 ¨ (b) Reserve = 5. provided this occurs after the death of (y). provided that this occurs within n years. Then ¯ 0. e. (x1 ) having died. Note.7.g. participation in a dangerous sport.8 Let chief = (y) = (47). 2 (iii) Yes. 217 (2) (x1 ) becomes chief (by succeeding (y) on his death or “retirement”) and dies after time 3 and before time 12. . (x2 ) = (33).90 (ii) The three-eighths rule would be suitable.9 q 12. since it avoids evaluation of the integrand when t = 71. If (60) dies soon. (2) whether he has any occupational or other risks. (x2 ) becomes chief if and only if (1) he survives for 3 years and (y) “retires” then. The sum assured is large enough to justify the costs of a medical examination. brothers (x1 ) = (37).9 x1 +3:x2 +3 300. The office must check: (1) the health of (60). leaving (x2 ) alive} 12 = 3 9 t px1 µx1 +t · t px2 dt · r+3 px2 dr dr = 0 r+3 px1 µx1 +r+3 9 = 3 px1 x2 0 r px1 +3:x2 +3 µx1 +r+3 1 = 3 px1 · 3 px2 . Probability of event (1) is 3 py (1 − 3 px1 )3 px2 Probability of event (2) is P r{(x1 ) dies between times 3 and 12. 000 (i) annual premium = 15 0 vt t p f µ 80 80+t f (1 − t p m ) dt 60 0.12. there is a high chance that (80) will die before age 95 and so give rise to a claim. SOLUTIONS 12. leaving (x2 ) alive. CONTINGENT ASSURANCES .218 CHAPTER 12. p. = t px µx+t (t > 0).1.Chapter 13 REVERSIONARY ANNUITIES 13. The present value of this reversionary annuity is Z= where T U = future lifetime of (x). If (x) and (y)’s mortality rates follow different tables. this should be indicated. = future lifetime of (y). with p.v.f. ¯ aU − aT ¯ 0 if U > T if U ≤ T Note. of Z ¯ = u>t ∞ (au − at )(u py µy+u )(t px µx+t ) du dt ¯ ∞ t = 0 (¯u − at )u py µy+u du a ¯ t px µx+t dt (13. Observe that Z + amin(T. payable continuously to (y) after the death of (x).a.1) Theorem ax|y = ay − axy ¯ ¯ ¯ Proof It is easiest to proceed indirectly. as follows.U ) = ¯ (¯U − aT ) + aT a ¯ ¯ aU ¯ if U > T if U ≤ T = aU (13.1.f.2) Taking expected values on both sides gives E(Z) + axy = ay ¯ ¯ Another important formula ∞ ax|y = ¯ 0 v t t px µx+tt py ay+t dt ¯ 219 .d. with p. Define ax|y = m.d. = u py µy+u (u > 0).1 Reversionary Annuities Payable Continuously Consider an annuity of £1 p. REVERSIONARY ANNUITIES Proof ∞ ax|y = ¯ = 0 v t t py (1 − t px ) dt f (t)g (t) dt ∞ 0 where f (t) = 1 − t px = t q x (which is such that f (t) = t px µx+t ) and ∞ g(t) = −t |¯y = − t v r r py dr (which is such that g (t) = v t t py ) a Using integration by parts ax|y = [f (t)g(t)]∞ − ¯ 0 ∞ f (t)g(t) dt 0 ∞ 0 = [(−t |¯y )t q x ]∞ + a 0 ∞ a t px µx+t (t |¯y ) dt dt.1.656 (using interpolation) 13.v.p. the first payment being at the end of the m year (m) (measured from the issue date) following the death of (x).1. = 0 t ¯ t px µx+t v t py ay+t Evaluation of reversionary annuities Using the Euler–Maclaurin formula we have ax|y = ay − axy ¯ ¯ ¯ 1 1 (ay + ) − (axy + ) 2 2 = ay − axy = ax|y Example 13. a f − am 69 65:69 f = 7. Calculate an appoximate value of a m ¯ Solution am ¯ 65|69 f 65|69 f on the a(55) tables at 8% interest. and use the symbol ax|y to refer to this case.2 Reversionary Annuities Payable Annually or mthly 1 We first assume that payments are made mthly .533 − 5. Using the equation ax|y + a(m) = a(m) xy y it can be seen that ax|y = m.220 CHAPTER 13. of reversionary annuity of £1 p.a. payable monthly to (y) after the death of (x) = a(m) − a(m) y xy (m) (m) .877 = 1. a.v.2) In practice this gives results similar to (13.) begins immediately on the death of 1 (x). Suppose now that a mthly reversionary annuity (of £1 p. only at this stage and suppose that a male employee retires at age 65.3. m .2. 13. = Death In Service) and/or on death after retirement (D.R.R. is approximately 1 (m) (1 + i) 2m ax|y (13. is payable on the death of (65) only if he was married to the same woman at retirement.v. = Death After Retirement) We consider D. giving ax|y = ay − axy By Woolhouse’s formula ax|y (m) 221 m−1 m−1 ) − (axy + ) 2m 2m = ay − axy = ax|y .2. (ay + An alternative approach is to regard this reversionary annuity as a collection of “pure endow1 2 ments” (payable at times m . at age 65 per married man is am ¯ 65|y f (assuming pension is payable continuously) .e.a. Since the payments begin on average 2m year earlier than in the case discussed previously.A.2.A. it may be omitted.1).p. In this case.I. m py t = = t=1 (m) ay − 1 m a(m) xy (as before). Now consider 2 different cases:Case 1 A widow’s pension of £1 p. · · · ) which are paid if (x) has died and (y) is alive.1) An exact formula is ∞ 0 ∞ 0 v t t px µx+t·t py ay+t dt ¨ v t t pxy µx+t ay+t + ¯ 1 2m dt (m) =¯x|y + a 1 ¯ A1 2m xy 1 ¯ ax|y + A1 2m xy (13.S. WIDOW’S (OR SPOUSE’S) PENSION ON DEATH AFTER RETIREMENT When m = 1. the m.13. i.3 Widow’s (or Spouse’s) Pension on Death after Retirement Many pension schemes provide a spouse’s pension on the death of the member in service (D. ax|y = t=1 (m) ∞ 1 t t v m P r{(x) has died but (y) is alive at time } m m ∞ t t v m (1 − m px ).p. the m. of widow’s annuity to the widow of a ¨ man dying at age 65 + t. Suppose.3. Case 2 (now more common) The widow’s pension is payable to any widow: we use the “collective” approach. A life office sells “personal pensions” policies under which the benefits for men on retirement at age 65 consist of: (a) a member’s pension (payable monthly in advance for 5 years certain and for life thereafter). and it is assumed that.p.a. Note. is payable monthly in advance.1. The possibility of divorce of men aged over 65 may be ignored.222 where CHAPTER 13.2) where a65−d+t = m. The m.3.p. that the widow’s pension of £1 p. beginning immediately on death of her husband. the fund available to purchase pension at age 65 will be £100.v. We ignore the possibility of divorce (or assume that the ex-wife still gets pension). beginning immediately on the death of the member) of half the member’s pension. and post-retirement marriages do not give rise to spouse’s pension. where d = age difference between husband and wife (approximately 3 years in practice) Hence the value of widow’s pension for each member retiring at age 65 (marital status unknown) is h65 a m ¯ 65|65−d f (13.1) where hx = the probability that a man aged x is married. REVERSIONARY ANNUITIES y = average age of wife of member aged 65 = 65 − d. Suppose that the life office uses the following basis to calculate the amount of pension which may be purchased at retirement in respect of a given fund: mortality of males: a(55) males ultimate mortality of females: a(55) females ultimate interest: 8% per annum expenses: 1% of the fund (at age 65) . if widow is assumed to be d years younger than husband (12)f Example 13. The member’s contributions are invested in certain unitised with-profits funds. and (for men married at age 65 only) (b) a spouse’s pension (payable monthly in advance. for a man retiring at age 65 (marital status then unknown) is ∞ 0 v t t pm µm h65+t a ¨ 65 65+t (12) 65−d+t f dt (13.v.000.3. in respect of a certain Mr Brown’s policy. for example. and (ii) he is assumed to be married at age 65. (i) Let P be annual pension. 008.1782P Hence P = £12.9 v 3 2 ·3 |q m af ¨ 65 (12) 1 65−d+3 2 ¨ + v 4 2 ·4 |q m af 66 1 (12) 65−d+4 1 2 + ··· (assuming that the man dies on average half-way through each year of age).99 × 100.90 for t ≥ 3.13. but married men must buy this.1637 + 0. 000 = P (¨5 a = P( (12) +5 |¨65 ) on males’ table a + v5 l70 11 a70 − ¨ ) l65 24 (12) a d(12) 5 i = P (4. Calculate Mr Brown’s expected monthly pension if (i) he is assumed to be single at age 65.85 for 0 ≤ t < 1 0. 105.89v 2 2 2 |q m a 65 1 1 + 0. and his wife is 3 years younger. so monthly pension = £1. 0. 553.2026) Hence P = £10. payable monthly in advance. . 000 Notes (1) There may be rules to try to exclude benefits for widows of “deathbed marriages”. WIDOW’S (OR SPOUSE’S) PENSION ON DEATH AFTER RETIREMENT 223 Men who are not married at age 65 need not buy spouse’s pension. The integral in (13. i. there may be an “actuarial reduction” (see later).86621 × 6. In the U.1782 + 1. (5) h65+t is sometimes assumed to be “piecewise continuous”.87v 1 2 1 |q m a 65 (12) 1 65−d+2 2 f 1 (12) 1 65−d+1 2 f ¨ + 0. e.75 (ii) The equation of value is now 1 1 P (8. (2) There may be problems if the man was married more than once (and his ex-wives are still alive). the last wife receives all the pension.87 for 1 ≤ t < 2 h65+t = 0.3.g.08) 24 a f − a m f 2 62 65:62 P (8.1782) + P (1.e. 0.8097) = 8. (4) Widow’s pensions might cease on remarriage. but this rule is no longer common.89 for 2 ≤ t < 3 0.K. ∞ 0 v t t pm µ 65 65+t m (h65+t )¨ a (12) 65−d+t f dt m 0. so monthly pension = £879.85v 2 q65 a ¨ 1 (12) 65−d+ 1 2 f ¨ + 0. Solution.3.2 ) may be evaluated approximately by a sum.68058 × 0. (3) If a wife is very much younger than her husband.42 99. ) Let the widow aged y get £R p.I. (This rule may also apply to widow’s D. in a pension policy or scheme.¨ a (12) x−10 f af ¨ . the rules state that a reduction applies to the normal widow’s pension if wife is more than 10 years younger than her husband. for each £1 p.a. leaving widow aged y (y < x − 10).4 Actuarial Reduction Factors Suppose that. of course. pensions). Then (Full widow’s pension). reduced. REVERSIONARY ANNUITIES 13.224 CHAPTER 13. of ‘normal’ widow’s pension.a.R¨ f a This gives a ¨ R = actuarial reduction factor = (12) x−10 (12) y f (12) y = (Full widow’s pension). The value of this widow’s pension is calculated to be the same (actuarially speaking) as for a widow aged x − 10 (whose pension is not. Suppose a member dies aged x.S. life table and compound interest functions.500 per annum during the lifetime of the wife.000 per annum.000 and £20. 13. EXERCISES 225 Exercises 13.000 per annum until the man’s death. Under this policy. Obtain an expression for the mean present value of the benefits in terms of joint-life and single. Obtain an expression for the present value of this annuity in terms of single and joint-life annuity factors. (b) if the wife dies first. A 2 and ay|x . (c) payments at the rate of £3.1 (i) Define the following symbols in words. Hence write down a relationship involving A 1 . commencing immediately on the first death and ceasing immediately on the second death. Calculate the annual premium if the office uses the basis given below: Mortality males: a(55) males ultimate females: a(55) females ultimate 20% of the first premium 5% of each premium after the first 6% per annum. On the first death of the couple. The conditions of payment are: (a) so long as both survive the rate of payment will be £3. payable yearly in arrear.2 A special life policy on 2 lives aged x and y respectively provides cash sums of £10.4 A special annuity. the survivor will receive an annuity of £10. In addition.3 An office issues a policy on the lives of a woman aged 60 and her husband aged 64.000 per annum will be paid continuously. payable weekly.) of (1) and (2) are equal. the first payment being at the end of the year of his death.000 per annum will continue for six years certain after the death of the husband. and will be reduced thereafter to £1. beginning immediately on the first death. and give a formula in terms of an integral for each of them: (a) A xy xy 1 (b) A 2 (c) ay|x (ii) Consider the following sets of payments: (1) £1 immediately on the death of (y) if (y) dies before (x). plus £1 immediately on the death of (x) if this occurs after that of (y). payable continuously to (x) after the death of (y). Expenses: Interest: 13. the rate of payment will be £2. Assume that the same (nonselect) table of mortality is appropriate for the two lives.000 per annum. .13.life annuity functions and the force of interest.000 immediately on the first and second deaths respectively. is effected on the lives of a man aged x and his wife aged y. Prove that the present values (at force of interest δ p. level premiums are payable annually in advance for 20 years or until the first death of the couple. xy xy 13.a. and (2) an income of £δ p.a. an annuity at the rate of £1. Ignore expenses. if earlier.5. payable continuously. provided that his wife has already died. The office issuing the contract uses the following basis: mortality : A1967-70 ultimate interest : 4% per annum expenses are ignored. 10% of all premiums Ignore the possibility of divorce.5 A single-premium policy provides the following benefits to a husband and wife each aged 40. (1) An annuity of £5. Calculate the annual premium on the undernoted basis: Males’ Mortality: Females’ Mortality: Interest: Expenses: a(55) males ultimate a(55) females ultimate 8% p. and (2) a reversionary annuity of £5. Level premiums are payable annually in advance until the first death.226 CHAPTER 13. .000 p. payable continuously throughout the lifetime of the surviving spouse after the death of the first. 13. (2) A return of half the single premium without interest immediately on the death of the husband within 25 years. and continuing so long as either husband or wife is alive. Calculate the single premium.a. REVERSIONARY ANNUITIES 13.000 payable immediately on the first death.6 A husband and wife. or on his survival for 25 years.a.000 per annum. effect a policy under which the benefits are (1) a lump sum of £10. commencing on the husband’s death within 25 years. aged 70 and 64 respectively. 6. 000 a ¯ a 13. 000(2 − 2δ¯x + 2 − 2δ¯y − 1 + δ¯xy ) + 1000(¯x + ay − 2¯xy ) a a a a ¯ a = (1000 − 20.95 × 8.) to give income so long as (x) lives (assuming (y) dies first.v. if this occurs before that of (x). 000(2Ax + 2Ay − Axy ) + 1000(¯x + ay − 2¯xy ) a ¯ a = 10.3 Let annual premium be P .15P = 10000[¯ m a 64:60:20 64|60 f +af ¯ f 60|64 m ] 64 64:60 f = 10000[¯ f − a m a ¯ 60 60 64:60 ¯ + am − am ¯ 64:60 f ] 10000[a f + a m − 2a m 64 ] = 42230 lf · lf am ¨ 64:60:20 f = am ¨ 64:60 f − v 20 lm lm 84 64 80 · am ¨ 60 84:80 f = 8. of £1 payable immediately on the death of (y).2 Benefit = 10. ∞ A2 = xy 0 v t t px µx+t (1 − t py ) dt (c) The m.13.p.5709 − 0. the capital (£1) is paid immediately.5709 42230 = £5.v.a. 000δ − 2000)¯xy + 30.15 13.p. Payments (2) thus have the same present value as payment (1) (both being random variables depending on the future lifetimes of (x) and (y)).95P a m f ¨ − 0. if after that of (y). 0. On the death of (x).¯y|x + A 2 xy xy 13. 000δ)(¯x + ay ) + (10. 000(Axy + 2Ax + 2Ay − 2Axy ) + 1000(¯x + ay − 2¯xy ) a ¯ a = 10. Hence P = (b) value of benefit = 2000(ax − axy ) .a.4 (a) value of benefit = 3000axy . Suppose it is invested (at force of interest δ p. Take means of these present values to get a A 1 = δ. 000Axy + 1000(¯x|y + ay|x ) a ¯ ¯ = 10.v. pf £1 payable immediately on the death of (x) if this occurs after that of (y). 284 . ∞ A xy 1 = 0 v t t pxy µy+t dt (b) The m. SOLUTIONS 227 Solutions 13. payable continuously to (x) after the death of (y). 000Axy + 20. Then 0. otherwise there is no payment (1)).1 (i) (a) The m.p. of a reversionary annuity of £1 p. ∞ ay|x = (other expressions also possible) 0 v t t py µy+t · t px · ax+t dt (ii) Consider payment (1). (b) husband survives for 25 years. (b).) In case (a).a.V.5 × 0.¨40:40:25 − a 2 D40 2 D40:40 P (0. is v 25 25 pm 40 =v 25 25 p m 40 25 p f a m f 40 65:65 40 65 + (1 − 25 p f )a m 40 65:65 f 65 · 25 p f (¯ f − a m a ¯ ∞ 40 )+v f 25 ¯ 25 p m a m 40 65 a ¯ f =v 25 25 p m · 25 p f · 40 0 vt t p m 65:65 65+t 65+t µ m ¯ dt + v 25 25 p m a m 40 65 Therefore total M.842 + 3. (c) to find total value of the annuity. for 6 years) beginning at end of year of death of (x)..P.17511) = 0.P.094993 − 0.a. is 25 0 v t t pm µ 40 m pf a f ¯ 40+t·t 40 40+t dt In case (b). is 0 ∞ vt t p m + =¯ m a 40|40 f 40:40 40+t 40+t ¯ v 25 25 p m a m 40 65 f µ m a ¯ f ¯ dt + v 25 25 p m a m 40 65 Benefit (1) = 5000(¯40|40 ) + 5000v 25 25 p40 a65 a ¯ = 5000(¯40 − a40:40 ) + 5000v 25 25 p40 a65 a ¯ ¯ 1 5000[¨40 − a40:40 + v 25 25 p40 (¨65 − )] a ¨ a 2 = 5000(1.a.V. P 25 t v t p40 µ40+t (1 − t p40 ) dt 2 0 P [A 1 −A1 ] 40:40:25 2 40:25 P 1 [A 1 − A 1 ] 2 40:25 2 40 : 40:25 1 D65 1 D65:65 P (1.V.228 CHAPTER 13. Consider benefit (1) with payment at rate £1 p. M..5 Let single premium be P .P.1418) = 24. (Indicate “m. 919. REVERSIONARY ANNUITIES (c) value of benefit is value of (1) an annuity-certain (of 3000 p. payable at times t (t ≥ 7) if (y) alive and (x) dead 6 years previously ∞ = 3000¨6 · Ax + 1500 a t=7 v t t py (1 − t−6 px ) = 3000¨6 (1 − d¨x ) + 1500v 6 6 py (ay+6 − ax:y+6 ) a a Add (a).003719P 2 Benefit (2) = = = = = .04) 2 A40:25 − − 1 − d.f” to clarify which life is which. plus (2) 1500 p. 13. There are two cases: (a) husband dies within 25 years. M. 000[1 − δ(a m 70:64 1 + )] + 5000[a m + a f − 2a m f ] 70 2 64 70:64 = 5. SOLUTIONS Hence equation of value is P = 24919 + 0. 5.6. 000A m f 229 70:64 + 5000(¯ m a 70:64 f f 70|64 f +af ¯ 70 64|70 m ) 70:64 f = 10. 017 .9184P .003719P Therefore P = £25.85 + 18.85 = £4. 13.9P a m ¨ Hence P = 23773.9184 f 70:64 = 5. . 012.6 Let annual premium be P . 323.85 Value of premiums less expenses = 0. Value of benefits = 10.13. 773. 450 = 23. 000(1 − δ¯ m a a ) + 5000(¯ m + a f − 2¯ m a ¯ 64 ) 10. REVERSIONARY ANNUITIES .230 CHAPTER 13. Unit prices are quoted at two levels. into which are paid deductions from the premiums for expenses and fund management charges. the bid price and the offer price.). and from which it pays the actual office expenses and death guarantee costs. the benefits are usually equal to the value of the units (at bid price). On death.2 Mechanics of the Unit Fund Unit prices are quoted at 2 prices.K. subject to a minimum death benefit. the office holds a balancing account. called the Sterling Fund (or Sterling Reserves). The bid price is the “real” price and all valuation calculations use the bid price. We use the following notation: 231 . Define 1−λ= Bid Price Offer Price where λ is called the Bid/Offer Spread (λ being perhaps 0. In addition. 14.05). The proceeds on maturity (at policy duration n) or earlier surrender are usually equal to the bid value of the units. equities. the “Bid Price” (at which the units can be sold) and the “Offer Price” (at which the units must be bought). The assets underlying each policyholder’s units form the Unit Fund of the policy. The office may also transfer profits/losses from the sterling reserves to the shareholders or withprofits policyholders (who may be considered to be “investing” in the sale of the unit-linked policies).1 Unit-Linked Policies Most of the money paid in premiums by the policyholders is used to purchase “units”: that is. the value of which may be calculated from the unit price and the number of units held. money is placed in a unitised investment of some kind (U.Chapter 14 PROFIT TESTING FOR UNIT-LINKED POLICIES 14. which is an artificial higher price. property. etc. They may not take money from the unit fund as this belongs entirely to the policyholders. The bid price of the units is 95% of the offer price..1.. usually a percentage of the value of the unit fund. (14..2. The units are subject to an annual management charge of 0. The annual premium is £1. we have ct = m [Ft−1 + (1 − λ)at Pt ] (1 + iu ) and thus Ft = (1 − m) [Ft−1 + (1 − λ)at Pt ] (1 + iu ) Note. Hence Pt − (1 − λ)at Pt may be transferred to the sterling reserves as a deduction for expenses. . The accumulation of the unit fund Let iu be the assumed rate of growth per annum of the unit fund. Define ct = the fund management charge in year t = a charge made at the end of year t. 2.2) (14. but before payment of any premium then due) asssuming that the policy is still in force.3) (14. according to the office’s projections.2. PROFIT TESTING FOR UNIT-LINKED POLICIES Pt = the office premium in year t (t = 1.232 CHAPTER 14. n) (which is actually paid at time t − 1) at = the allocation proportion in year t = the proportion of the premium Pt which is allocated to buying units (at the offer price) The cost of allocation in year t is the money actually used by the office to buy units.1) Example 14. which is transferred to the sterling fund.75% of the bid value of the fund at the end of each policy year. A life office issues a large block of 3-year unit-linked endowment assurances under which 80% of the first year’s premium and 101% of subsequent premiums are invested in units at the offer price. that is (1 − λ)at Pt . . Define F0 = 0 Then Ft = [Ft−1 + (1 − λ)at Pt ] (1 + iu ) − ct Suppose that the fund management charge is a proportion m of the unit fund.2. Calculate the bid value of the units at the end of each year. In some cases ct may be a fixed sum rather than a proportion of the fund.2.000 and unit prices are assumed to grow at 9% per annum. Define Ft =the value of the unit fund at time t (after deduction of the fund management charge. is given by t−1 px .2) Maturity Bonuses In some policies. 14.2)) − px+n−1 × (Maturity Bonus) (14..146.14.a.47 3. and St = the guaranteed minimum death benefit in year t.40 1. or a proportion of the bid value of the fund at maturity. This money must come from the sterling fund.3.927. Define (SCF )t = the expected net cash flow in the sterling fund in year t per policy in force at the start of the year = the “in force” net cash flow The “initial” expected net cash flow in the sterling fund in year t.3) .C.3.(SCF )t (t = 1.3.09) × (2) (4) = 0.80 (4) (5) F. 6.19 1.0075 × (3) (5) = (3) − (4).942. in the sterling reserves.3 The Sterling Fund (or Sterling Reserves) et = projected expenses for office in year t (payable at the start of year t) ct = fund management charge (as before) (DG)t = death guarantee cost in year t = qx+t−1 (St − Ft ) 0 if St > Ft if St ≤ Ft Money is assumed to earn interest at rate is p.1) The formula for (SCF )t is (SCF )t = [Pt − (1 − λ)at Pt − et ](1 + is ) + ct − (DG)t (14. that is the expected net cash flow per policy sold.M.. .97 (3) Fund at end of year before F.3. Define where x = the age of the policyholder at the start of the policy.M.123.3.21 14. THE STERLING FUND (OR STERLING RESERVES) Solution (1) Cost of Allocation t Pt (1 − λ)at 1 2 3 760.C.50 959.781..886.04 3. random variations being ignored.20 (3) = (1.00 959. 2. so we must adjust (SCF )n as follows: (SCF )n = “normal” (SCF )n (as in (14. We suppose that the policy is one of a large number of similar unit-linked contracts on lives whose mortality follows a specified table.00 1. n) (14.57 23. 828.50 233 (2) Fund brought forward from start Ft−1 + Pt (1 − λ)at 760.60 Ft 822.69 2. there may be a maturity bonus in the form of a fixed sum. 3. calculate the profit signature.e. 2. payable at the end of the year of death..3.5) where (IR)t = increase in reserves = px+t−1 ·t V −t−1 V Therefore (P RO)t = (SCF )t + (1 + is )t−1 V − px+t−1 ·t V In both cases the profit signature is found by equation 4.. Renewal expenses are expected to be £20. The profit vector is (P RO)t = (SCF )t + is ·t−1 V − (IR)t (14. is the greater of twice the annual premium and the bid value of the units. The calculations are very similar to those for conventional profit-testing. PROFIT TESTING FOR UNIT-LINKED POLICIES The Profit Vector and Profit Signature Define (P RO)t = the profit vector = the expected net profit to the office in year t per policy in force at the start of the year σt = the profit signature = the expected net profit to the office in year t per policy sold =t−1 px .1.3.3.. .1.4. Assuming that the office holds zero sterling reserves at the end of each policy year. the death benefit. σt =t−1 px · (P RO)t (14.1) Suppose that. i. Ignore the possibility of withdrawal..2. in addition to the information in example 14.234 CHAPTER 14. n Case 2 Suppose now that the office wishes to maintain sterling reserves of t V at the end of year t (t = 1. (continued from Example 14.. The mortality rate at each age is assumed to be 0. we have (P RO)t = (SCF )t for t = 1. The office expects to incur initial expenses on these policies of 20% of the first premium. n − 1).(P RO)t (14.6) Example 14.2.003. Case 1 If there is no need (or desire) to maintain sterling reserves at the end of each policy year.3.. 2.4) There are 2 cases to consider. . payable at the beginning of each policy year after the first. Sterling reserves are assumed to earn interest at 4% per annum. It is assumed that 0 V = n V = 0. . 21 14.28 35.57 23.22 0 (5) FMC ct 6.997)t−1 (9) = (7) × (8) 1 0.4 The Assessment of Profits The profit signature {σt } can be assessed in one or more of the following ways.32 21.56 44.p. (1) One could work out the Internal Rate of Return (or yield) by solving n v t σt = 0 where v = t=1 1 1+j (the internal rate of return. (3) The Profit Margin is defined as n. at rate im . and may reflect uncertainties in {σt }. j. both at some rate of interest.60 (6) In Force cash flow (SCF )t 44. j per annum.v.60 21. of premiums n v t σt = n−1 t=1 .p. being the solution of this equation.28 35.67 44.50 235 (2) Expenses et 200 20 20 (3) Accumulation of Sterling Fund in year 41.04) (4) = 0.28 35.997 0.) (2) The shareholders may value the net profits at a certain rate of interest. The net present value of the profits is thus n N P V (j) = t=1 v t σt at rate j.67 44. of profits .50 40.53 0. THE ASSESSMENT OF PROFITS Solution (1) Premium less cost of allocation t Pt − Pt (1 − λ)at 1 2 3 240.4.00 40. This rate is called the Risk Discount Rate.v.92 (8) t−1 px t 1 2 3 (9) Profit signature σt 44. with j normally higher than is .003(2000 − Ft ) (6) = (3) − (4) + (5) (7) Profit Vector (P RO)t 44.92 (3) = [(1) − (2)] × (1.65 (7) = (6) (as no reserves held) (8) = (0.32 (4) Death Guarantee Costs (DG)t 3.994 14. t t=0 Pt+1 · t px v .14. im say n. 62 Notes 1 at rate 4 % 2 .65v 3 = 103.1 at a risk discount rate of 10%. Solution 3 n. Let X be the sum needed at time 1 to pay the negative cash flows at times 2. as there may not be enough money to cover the “negative profits” at times 2. PROFIT TESTING FOR UNIT-LINKED POLICIES Example 14.46. 14.a.56v 2 + 44. 3 and 4.76.5 Zeroisation of the Profit Signature Suppose that the profit signature.29 0.62 It may be considered undesirable for the shareholders to take a profit of £26.4. 2 Then X = 8.e.46 = £11. is of the following form: 26. σt . (for example).46 Note X is the amount required at time 1. Find the net present value of the profit signature in example 4. The shareholders may still take a profit of £0.22 −8. Hence the shareholders may take a profit at time 1 of 26. with no provision for sterling reserves at the end of each year (i.5.22 at time 1.62 at time 5.3.18 −2.76 0 (σt ) = 0 0 0.22 − 14. 3 and 4. and hence the negative cash flows are discounted to time 1 (not the start of the policy).v.05v + 5.29v 3 = £14. Thus the profit taken by the shareholders in year 1 should be reduced in order to avoid negative sterling fund profits in years 2. t V = 0 for all t). of profit = t=1 v t σt at 10% = 44.05 (14. There will now be no need for capital injections (from the shareholders to the Sterling Reserves) at times 2. 3 and 4 when the Sterling Reserves earn interest at rate is = 4 1 % p. 3 and 4.1.p. The zeroised profit signature {σt } is thus 11.58v + 35.1) (σt ) = −5.236 CHAPTER 14.18v 2 + 2. This process is called Zeroisation. Now assume that the surrender of a policy may occur only at the end of a policy year (just before payment of the premium then due).01 for t = 1.951 (6) tV 14.5.6.. and supposing that qx+t−1 = 0. Thus Funds needed at time t per policy sold = t V × P r{ policy is still in force at time t}. n − 1) Assume that wn = 0 as the policyholder will receive the maturity benefit at that time.6 Withdrawals So far we have ignored the possibility of surrender. Example 14. Using the profit signature (14. Solution (1) Original profit signature t σt 1 2 3 4 5 26.1.26 0 0 Observe that column (4) gives the sterling fund per policy sold. WITHDRAWALS (1) This process may be carried out even if σ1 < 0.22 -8.14.46 -8. and the probability of being in force at time t is t px . .18 -2.. 3.05 -5.18 -2. . t V . (2) We may calculate the revised profit vector using the equation (P RO)t = σt t−1 px 237 (3) We may also calculate the sterling reserves. the bid value of the policy’s units. 2.5.99 0.29 0.61 7.980 0. that policy is in force t px 0. calculate the sterling reserves in each year that are implied by zeroisation of the profit signature.19 0 0 (5) Prob.1).06 2.62 (2) Zeroised profit signature σt 11. implied by zeroisation. The chance that a policy in force at the start of year t is surrendered at the end of year t is therefore. 5.05 -5. of (3) at is 14.29 0 (6) = (4) (5) (4) Accum. 2.46 7.970 0..961 0.62 (3) Remainder of net-cash flow σt − σt 14.20 2. Define wt = the probability that a policy will be surrendered at the end of year t (t = 1. 14. in some cases minus a surrender penalty of (say) £10 or 1% of Ft .76 0 0 0 0. 4. px+t−1 · wt Define (SV )t = the surrender value at time t which is usually equal to Ft . The withdrawal benefit is 98% of the bid value of the units. then obviously (P RO)t = (P RO)t + px+t−1 · wt [Ft − (SV )t ]. If t V = 0 for all t. 2% of the surviving policyholders withdraw. A fund management charge of 0. Mortality is assumed to follow A1967-70 ultimate. σt . The profit signature..6. the probability of the policy being in force at time t should be changed from t px to t px (1 − w1 )(1 − w2 ). after deducting the management charge. A life office issues a large number of 3-year unit-linked endowment policies to men aged 65.(1 − wt ) Example 14. The death benefit.1) (14. allowing for withdrawals is (P RO)t = (P RO)t + px+t−1 · wt [Ft − (SV )t +t V ] where (P RO)t is as before. The office incurs expenses of £100 at the start of the first year and £20 at the start of each of the second and third years.2) (14. must also be adjusted to allow for surrenders. we have σt =t−1 px · (P RO)t where t−1 px (14. and hence (P RO)t = (P RO)t .1. 80% of the first premium and 105% of each subsequent premium is invested in units at the offer price. before payment of any benefits then due. PROFIT TESTING FOR UNIT-LINKED POLICIES The revised profit vector.238 CHAPTER 14. Note that if there is no surrender penalty (i. σt will still differ from σt as the probability of the policy still being in force in each year will be different.5% of the bid value of the policyholder’s fund is deducted at the end of each policy year.6. (SV )t = Ft ). under each of which level annual premiums of £1.3) = P r{ policy is in force at time t − 1.6. The profit signature allowing for withdrawals... which is payable at the end of the year of death.6.e. is £3. When calculating the reserves. (P RO)t = (P RO)t .000 or the bid value of the units if greater. allowing for withdrawals} = t−1 px (1 − w1 )(1 − w2 ). {σt }.. It is assumed that. . may be zeroised in the same way as before.(1 − wt−1 ) 1 if t ≥ 2 if t = 1 Note Even if (SV )t = Ft . (P RO)t . at the end of each of the first two policy years. implied by zeroisation. when t > 1. the bid price being 95% of the offer price. t V . There is a bid/offer spread in unit values.000 are paid. The maturity value is equal to the bid value of the units. at end of year t Pt (1 − λ)at Pt (1 − λ)at + Ft−1 of charge ct Ft 1 2 3 We now Policy Year t 1 2 3 (P RO)t 99. Hence profit taken at time 1 is 100.00 813.91245 -2.40 -18.50 2920. WITHDRAWALS 239 (a) Assuming that the growth in the unit value is 7% p.67 1923.57 0 σt F.02Ft ) 0.75 calculate (P RO)t = (SCF )t and then find (P RO)t + px+t−1 wt (0. Policy Cost of Funds brought Funds on end Unit fund Year allocation forward from start before deduction F.63 In force cash flow (SCF )t 99.20 4.10v + 2.10 0 -2. calculate the profit emerging at the end of each policy year per policy sold.00 760. costs (DG)t -52.48 = £64.82 -37.66 .45 -2.C.92 where 760.09 9. less cost of allocation Pt − Pt (1 − λ)at 240 2.66 t−1 px =t−1 px (1 − w1 )..M. G.a.50 1806. (b) Let reduction in profit at time 1 be X.a.(1 − wt−1 ) with w3 = 0.48.38 15.45 -2.67 15.07 9.14.50 2.63 1933.M.63 3109. assuming it remains in force.50 px+t−1 wt (0.C. Ft .02Ft ).92 0. Solution (a) We first work out the unit fund. Expenses et 100 20 20 (P RO)t Accumulation in Sterling Fund at 6% interest 148. (b) Calculate the revised profit emerging at the end of each year if the office takes a smaller profit in year 1 in order to ensure that the profit emerging in the second and third policy years is zero.95645 -35.92 Prem.42 997. and that the office holds unit reserves equal to the bid value of units and zero Sterling Reserves at the end of each year. where x = 65.07 809.32 100.75 -36. per policy sold.13 997.55 t−1 px D.14 0.14 − 35.65 -28. ct 4.66v 2 at is = 6% = 35. Then X = 35. Sterling Reserves are assumed to earn interest at 6% p.6.82 -37.70 0..14 1 100.92 3125.55 -18. The allocation proportion is 90% in year 1 and 97% thereafter. the policy pays the higher of £10.12 −111. incurred at the start of the second and the third years 2% per annum. (iii) the profit vector of a policy. and .1 A life office issues a three-year unit-linked endowment policy to a life aged exactly 60. define the following terms briefly: (i) the Unit Fund.000 and the bid value of units allocated to the policy. the profit vector per policy sold. Ignore the possibility of surrenders. 14. Construct tables to show the following: (i) the growth of the unit fund. ignoring any need to maintain Sterling Reserves at the end of each year.64 before receipt of the premium due at time 2 years. taken at the end of each year before payment of any benefits 4% per annum 6% 10% per annum. is as follows (£): 191. In each case. A bonus of 2% of the (bid) value of the unit fund is payable at maturity. (b) A life office issues a large number of identical 4-year annual premium unit-linked endowment assurances to lives aged 65. and Sterling Reserves earn interest at 5% per annum.28 10. ignoring withdrawals and assuming that no Sterling reserves are maintained at the end of each year. PROFIT TESTING FOR UNIT-LINKED POLICIES Exercises 14.000. payable at the start of each year.2 (a) In the context of profit-testing of unit-linked business. The annual premium is £2.45 −3. (iv) the profit signature of a policy. indicate clearly how you calculate your table entries.95 The office’s mortality basis is A1967-70 ultimate.48 before receipt of the premium due at time 1 year and £78. (ii) the Sterling Reserves. (iii) the profit signature after taking into account sterling reserves.240 CHAPTER 14. and (vi) zeroisation of Sterling Reserves. after deduction of the fund management charge. At the end of year of death during the term. The life office makes the following assumptions in projecting future cash flows: Mortality Initial expenses: Renewal expenses: Fund management charge: Sterling fund interest rate: Bid/offer spread: Unit fund growth rate: A1967-70 ultimate £300 £50. (v) the risk discount rate. According to the office’s calculations. assuming that no sterling reserves are held. given that the sterling reserves per policy are to be £36. (ii) the profit signature. Calculate (i) the profit signature per policy sold. it is customary to eliminate these negative values by setting up sterling reserves at the end of each year.75%. Management charges of 4 % of the bid value of the units are deducted at the end of each year (before payment of death and maturity claims). Describe briefly the technique (“zeroisation”) by which these reserves are calculated. It assumes that. EXERCISES (ii) the profit signature per policy sold if Sterling reserves are zeroised. The maturity value is the bid value of the units at maturity. Calculate (i) the revised profit signature per policy sold.000. is £1. that sterling reserves need not be maintained at the end of each year.100. . The allocation proportion is 70% for the first annual premium and 98% for all subsequent annual premiums. Using a profit testing analysis.7. per policy then in force.000 or the bid value of units if greater.01 per annum at each age and sterling reserves earn interest at 8% p. The death benefit. The sum assured. for a 3. (b) An office issues a 3-year unit-linked policy with a yearly premium of £500. and (iii) the net present value. . ignoring any need to maintain Sterling Reserves at the end of each year. 3% of the surviving policyholders will surrender (just before payment of the second and third annual premiums respectively.) Surrender values are equal to the value of the policyholder’s units (after deduction of fund management charges). Calculate the sterling reserves required at the end of each policy year. is the bid value of the units held. The bid price of units is 95% of 1 the offer price. 241 (c) The office now wishes to make an allowance for surrenders. Assume that the unit fund grows at 8% per annum (before deduction of management charges). of the revised profit signature per policy sold. with a surrender penalty of £10. 95% of each premium is invested in units at the offer price. at a risk discount rate of 15% per annum. 14. 14. (ii) the revised profit signature per policy sold if the Sterling Reserves are zeroised. given that the rate of mortality is 0.3 (a) If a profit test for a unit-linked policy reveals negative cash flows in the second or any subsequent policy year.a. In determining the sterling reserves necessary for the policy the office makes the following assumptions: .4 If a profit test for a unit-linked policy reveals negative cash flows in the second or any subsequent policy year. assuming that the Sterling Reserves are zeroised. payable at the end of the year of death. 14. it is customary to eliminate these negative values by setting up sterling reserves at the end of each year. subject to a guaranteed minimum death benefit of £2. at the end of the first and the second policy years. payable at the end of year of death or at the maturity date.50). (ii) the net present value at the issue date of the expected profit from one policy assuming a risk discount rate of 10% per annum. The mortality of policyholders follows A1967-70 ultimate and sterling reserves earn interest at 6% per annum during each policy year. and that the possibility of surrender may be ignored. For units the bid/offer spread is 5% and the annual rate of management charge is 0.14.5 An office issues a unit-linked endowment assurance with annual premium £400 and term five years to a life aged 60 who is subject to A1967-70 ultimate mortality. The office expects to incur expenses of £75 at the start of the first year and £25 at the start of each subsequent year.year policy for which the profit signature (with no allowance for sterling reserves at the end of each year) is (250. calculate for a life aged 60 at entry (i) the expected profit in each of the 3 years per policy in force at the beginning of the year. Suppose.a. (b) Hence determine the sterling reserves which should be held by the office to eliminate the sterling fund negative cash flows in the second and subsequent years of the policy’s duration. determine the resulting sterling fund profit vector and signature.242 CHAPTER 14.a.a. (i) Construct a table showing the growth of the unit fund over the duration of the policy and (ii) Construct a table showing the growth of the sterling fund in the absence of reserves. and (ii) construct a table showing the growth of the sterling fund in the absence of reserves. Growth rate for units: 7% p. compound from the outset of the policy. and that the inflationary growth rate for renewal expenses will be 4% p. (a) On this basis (i) construct a table showing the growth of the unit fund over the duration of the policy.a. however. (iii)Assuming that the office sets up the sterling fund reserves found above. that the growth rate for units will be 10% p.. that the sterling fund interest rate will be 6% p. Interest rate for sterling fund: 4% p. PROFIT TESTING FOR UNIT-LINKED POLICIES Initial expenses: £125 Renewal expenses (associated with the payment of the second and each subsequent premium) £20 increased by 7% p.a..a. . Find also the internal rate of return corresponding to the profit signature. (c) Consider the unit-linked policy described above. (from the outset of the policy). 46 (DG)t 118.96979 σt -108.89 (SCF )t -72. X = 108.10 (ii) Let X be the amount withheld at time 1 to cover negative cash flows at times 2 and 3.62 t−1 p60 1 0.97597 0.41 75. t 1 2 3 (SCF )t -72.95 77.98557 0.79 p59+t .46 112.20 4012.58 5755.2 (a) Simple definition of each term required.1 1861.41 74. Check definitions with text.40 176.77 -3.75 (iii) (P RO)t = (SCF )t + (1 + is )t−1 V −t V.54 68. (b) (i) t 1 2 3 4 (P RO)t 191.12v 2 at 5% interest = 106.02 146.60 1823.25 126.92226 σt 191.63 (Maturity cost in year 3 = 0.60 (1) Pt (1 − λ)at + Ft−1 1692 3647.77v + 3.10 150.95007 0.22 80.22 80.M.37 FMC 37.89 66.70 0 Hence σt = 0 10.83 (1.17 67.02(6204.40 (3) et 300 50 50 [(2) − (3)] × 1.83 t−1 p60 1 0.32 131.09 6204.C.14.28 10.94 81.12 -108.10 (c) (i) Notice that Ft − (SV )t = 10 for all t.34 6331.12 10.04)t−1 V 0 37.98557 0.63) × p62 .42 84.46 114.) t 1 2 3 (P RO)t -72.px+t−1 .8.04 8.46 114.96979 σt -72.07 14.54 68.26 F.83 Pt (1 − λ)at 1692 1823. 37.12 -111.00 97.25 126.69 (1) × 1.1 (i) t 1 2 3 (ii) t 1 2 3 (2) Pt − Pt (1 − λ)at 308 176.95 t−1 p65 1 0.45 -3.98 3932.63 Ft 1823.t V 35.46 131.54 68.63 Maturity Cost 0 0 121. SOLUTIONS 243 Solutions 14. Hence (P RO)t = (P RO)t + 10wt p64+t .46 114.38 0 (P RO)t -108. 94 Ft 486.98557 0.53 3.50v 4 at 15% interest = £82.95 10wt p64+t 0.70 29.29 0 0 (P RO)t 191.22 2. 14.23 -2.02v + 27.25 451.93 9.88 (ii) Retain X at time 1.12 -111. (b) (i) t 1 2 3 Pt (1 − λ)at 451.96979 σt -34.11 1 0. 88.02 2 27.03 for t = 1.38 1461.53v + 9.53 3. 2 0 for t = 3.23v 3 = £12.70 3 29.41 -105.95 t−1 p65 1 0.75 t−1 p60 [(2) − (3)] × 1.25 451.28 10.16 -3.244 CHAPTER 14.42 0 0 FMC 1.28 10.23v + 2.41 -111.02 27. 14.02 27.98 FMC 1.89392 0.25 (1) Pt (1 − λ)at + Ft−1 451.37 1577.30 28.53 0 Hence zeroised profit signature is 0 9.13 1009.75 48.30v 2 + 28.93v 2 at 5% interest = 102.94 (SCF )t = (P RO)t -34.06 -27.84.23 3 Hence net present value = t=1 σt v t at 10% interest = −34.17 25.94669 0.86775 σt 191.04 t 1 2 3 (2) Pt − Pt (1 − λ)at 48.11 (ii) t (P RO)t 1 -34.08 487.25 937.82 25. 4 X = 105.3 (a) Zeroisation is the process whereby the profit in the first year is reduced to pay for any future negative cash flows (as explained in text).45 -3.75 48. PROFIT TESTING FOR UNIT-LINKED POLICIES t 1 2 3 4 (P RO)t 191.22 2.35 1012.29 0.50 (iii) NPV = 88.4 250 σt = −100 −50 .17 (DG)t 7.41.50 Notice that wt = 0.84 1574.09 (3) et 75 25 25 (1) × 1. 30 -9.95257 0.54 0 0 t px tV 0.89 15.60 27.99 0.44 (DG)t 24.90 2110.46 114.5 (a) (i) t 1 2 3 4 5 (ii) t 1 2 3 4 5 t 1 2 3 4 5 (2) Pt − Pt (1 − λ)at 134 27.22 σt -13.40 22.22 1.89 1067.36 6.91 7.40 (1) Pt (1 − λ)at + Ft−1 266.05 1599.30 -9.83 47.04 9.30 -9.27 t−1 p60 Pt (1 − λ)at 266.00 654.87 1506.18 -1.46 46.09 16.45 4.93385 (b) Let X be the amount that has to be withheld at time 1 to pay for the negative cash flows at times 2 and 3.54 Hence the zeroised profit signature is σt = 0 0 (1) σt − σt 135.91 7.13 5.44 17.87 0 FMC 2.37 7.13 5.41 So the zeroised profit signature is −23.05 -1.49 695. Hence X = 9.57 12.13 (1) × 1.14.62 700.27 1 0.60 27.62 1611.85 7.81 2094.09 15.05v + 1.83 Ft 282.8.43 [(2) − (3)] × 1.970 136.89 3.40 372.40 372.50 26.30 0 245 t 1 2 3 σt 250 -100 -50 σt 114.980 0.90 24.47 1134.83 (SCF )t -13.60 27. Then X = 100v + 50v 2 at 8% interest = £135.96979 0.71 0 σt = 0 7.26 8.18 -1.09 16.09 15.45 1972.57 12.26 FMC 2. SOLUTIONS Let X be the sum retained in year 1 to cover the later negative cash flows.60 (P RO)t -13.73 1142.24 0 14.00 372.98557 0.44 17.46 -100 -50 Accumulation of (1) 135.26 8.21 (3) et 125 21.79 20.13 .07 284.40 372.85v 2 at 4% interest = £10. 78 0 t 1 2 3 4 5 t px 0.60 27.40 [(5) − (6)] × 1.00 27.60 27.85 0 0 CHAPTER 14.35 1.11v 4 + 20.62 1196. Hence j = 0. −23.98 2085.60 2.76 3 1.00 372.40 372.81 0 0 0 (P RO)t -23.04 12.58 1712.94 -7.65 Pt (1 − λ)at 266.47 9.41 1.62 1725.98557 0.t V Let the internal rate of return be j p.60v 2 + 2.a.44 14.147 = 14.00 20.06 9.60 (6) et 125.84 0 0 0 (1.11 12.94 2276.65 t−1 p60 1 0. PROFIT TESTING FOR UNIT-LINKED POLICIES Accumulation of (4) at 4% interest 10.56 1.65 This uses tV 10.47 9.94 2 -7.41 723.40 372.62 3.a.98557 0.21 6.34 (4)×1 · 1 292.56 1.26 5.246 (4) σt − σt 10.97 12.96979 0.67 (iii) t (SCF )t 1 -12.45 (DG)t 24.54 7.40 372.96979 0.41 4.95257 0.84 0 0 0 (c) (i) t 1 2 3 4 5 (ii) t 1 2 3 4 5 (5) Pt − Pt (1 − λ)at 134. .67 20.22 (P RO)t = (SCF )t + (1 + is )t−1 V − px+t−1 .95 0 0 p59+t · t V 10.76 1.63 22.60 729.41 1.e.05 -1.35v + 1.50 23.64 0 FMC 2.11 20.87173.60 27.93385 σt -23. Then j solves 5 σt v t = 0 where t=1 v= 1 .80 21.81 1096.02 1568.71 21.7% p.95257 - tV 10.00 662.11 4 12.35 1. 1+j i.20 Ft 290.19 5.41 -9. Solving this equation by trials and interpolation gives v = 0.04 12.22v 5 = 0.19 1.94 17.94 17.09 1205.33 5.19 5.40 Pt (1 − λ)at + Ft−1 266.20 (SCF )t -12.87 FMC 2.88 2293.97v 3 + 12.71 21.71 5 21.06)t−1 V 0 11.06 12. (In the above example. α and β.) =P r{T ≥ t} 247 . Define t (ap)x =the probability that (x) will “survive” for at least t years with respect to both modes of decrement. marriage.1 Introduction Consider a body of lives subject to two “modes of decrement”. whether he is then married or not (At this point it is assumed that all bachelors eventually marry.Chapter 15 MULTIPLE-DECREMENT TABLES 15.) Let T = min{T1 . he will not marry or leave the service within t years. from which men can leave by either mode α. T2 } =time until exit from the group of bachelor employees. whichever comes first. For example. by either mode α (marriage) or mode β (withdrawal from service). whether or not he is still an employee of the company and T2 =time until (x) leaves the service of the company. and let T1 =time to marriage of (x). of a bachelor employee aged x Note This is similar to the joint-life situation in Chapter 11. leaving the company (mortality being ignored). consider a group of bachelor employees of a large company. Take a bachelor employee aged x. or by mode β. MULTIPLE-DECREMENT TABLES =P r{T < t} =P r{(x) leaves before time t. by merely putting “a” in front of the various functions. either by mode α or mode β} In the example above t (aq)x = the probability that (x) gets married or leaves service (or does both these things) within t years We may proceed through the development of “life tables”. (x ≥ x0 . t≥0 . x ≥ x0 . (al)x . t ≥ 0) (al)x When t = 1 it may be omitted. For example. is constructed to be such that t (ap)x = (al)x+t . define the “force of exit” from the double-decrement table (by whichever mode occurs first) by (aµ)x = lim+ h→0 h (aq)x h t It follows as for ordinary life tables that t (ap)x = exp − 0 (aµ)x+r dr The probability density function of T is t (ap)x (aµ)x+t 0 and its distribution function is t t (aq)x . the function. t<0 = 0 r (ap)x (aµ)x+r dr (t ≥ 0) We also define (am)x =central rate of “total decrement” (modes α and β together) (ad)x = = (aL)x 1 (al)x+t (aµ)x+t 0 1 (al)x+t dt 0 dt .248 and t (aq)x CHAPTER 15. giving (aq)x = and t |(aq)x (ad)x (al)x − (al)x+1 = (al)x (al)x = (ad)x+t (al)x Also. 2 The Associated Single-Decrement Tables Each mode of decrement may be thought of as possessing its own “life table”. for α β example. whether exit by mode β happens or not} ( = P r{(x) does not marry within t years } in our example. x The functions.) β t px = P r{(x) does not exit by mode β before time t. denoted by. t ≥ 0 . lx or lx .2. THE ASSOCIATED SINGLE-DECREMENT TABLES 249 15. whether exit by mode β happens or not} = P r{T1 ≤ t} = 1 − t pα . whether exit by mode α happens or not} ( = P r{(x) does not leave service within t years} in our example. x Similarly β t qx = P r{T2 ≤ t} = 1 − t pβ . lx .3 The Relationships between the Multiple-Decrement Table and its Associated Single-Decrement Tables It is usually assumed that the variables T1 and T2 are independent. t q β are called the independent rates (or probabilities) of exit within t years x x at age x in their respective single-decrement tables.3. Thus α t px = P r{(x) does not exit by mode α before time t.15.) α t px Note that We also have = P r{T1 ≥ t} and t pβ = pr {T2 ≥ t}. This gives t (ap)x = P r{T1 ≥ t and T2 ≥ t} = P r{T1 ≥ t}P r{T2 ≥ t} = t pα t pβ x x (15. In particular t (ap)x = lβ lα (al)x+t = x+t · x+t β α (al)x lx lx for all x ≥ x0 .1) α β Hence we may construct the function (al)x from lx . t q α . 15. x α t qx = P r{(x) exits by mode α before time t. t1 > 0. MULTIPLE-DECREMENT TABLES α β (al)y = k. assuming independence of T1 and T2 . x x = t q α + t q β − t q α .t q β x x x x If t = 1 . respectively .h ax h α h qx = lim h→0+ h + lim β h qx h→0+ h + lim α h qx β h qx h→0+ h h h = µα + µβ (similar to joint-life argument). y = x0 + t to obtain CHAPTER 15.250 Put x = x0 .ly for all y ≥ x0 ( as k = 1) Some Important Formulae t (aq)x = 1 − t (ap)x = 1 − (1 − t q α )(1 − t q β ).3.lx0 .ly . then α β (al)y = ly .4 Define Dependent Rates of Exit α t (aq)x = P r{(x) exits by mode α within time t.’s f1 (t1 ) = t1 pα µα 1 . If we choose the radix (al)x0 to be equal to lx0 .ly (y ≥ x0 ) α β where k is a constant. x x+t f2 (t2 ) = So β β t2 px µx+t2 .d. We have α t (aq)x = P r{T1 < t and T1 < T2 } where T1 . α β α β (aq)x = qx + qx − qx qx (15. x x 15.f.2) Also (aµ)x = lim h (aq)x h→0+ = lim+ h→0 h β α β α h q x + h q x − h q x . this is the probability that (x) gets married within t years while still an employee of the company. . t2 > 0. exit by mode β not having previously occurred} In terms of our example. T2 have p. 1 · 30 30 3 5 = 0. (The word ‘dependent’ indicates that (aq)α depends not only on mode α but also on mode β).µx+r . it is known that µα = 0.4.1.t px dt. DEPENDENT RATES OF EXIT 251 α t (aq)x = = 0 t1 <t2 t1 <t ∞ t t1 t f1 (t1 ).05t [35 − (30 + r)]−1 dr 1 dr 5−r 0 5−t = exp[log(5 − r)]t = . (b) the probability that he will leave by mode α before age 32.2 pβ = e−0. For 30 ≤ x ≤ 32.4. (aq)α = x 0 1 β α α t px µx+t .t2 px µx+t2 ∞ t1 dt2 dt1 dt2 dt1 = 0 t α α t1 px µx+t1 β β t2 px µx+t2 = 0 α α β t1 px µx+t1 . 0 5 = exp − Thus 2 (ap)30 =2 pα .f2 (t2 ) dt1 dt2 α α β β t1 px µx+t1 . α and β.1) = 0 dr When t = 1. Solution (a) t pα = exp − 30 β t p30 = exp − t 0 t 0 t µα 30+r dr = e−0. (aq)β are called the dependent rates (or probabilities) of exit at age x by mode α. x Example 15.t1 px dt1 Thus t 0 t α t (aq)x = α α β r px . (aq)α . calculate (a) the probability that he will still be in the population at age 32.5429 . β respecx x tively.r px α r (ap)x µx+r dr (15.15.05 x and µβ = (35 − x)−1 x For an individual in this population at exact age 30.4. A population is subject to 2 modes of decrement. 05e−0.01 x mw = 0. for mode β) α = qx t − t2 β q 2 x 1 β α = qx (1 − qx ).05 0 e−0.1 ) + e−0. (a) lx+t is linear (0 ≤ t ≤ 1) (b) t q x = t. death (d) and withdrawal (w). the central rates of decrement at a certain age x are md = 0.2.qx ) dt t=1 t=0 (using U.1 ) (using integration by parts) 5 = 0.05t dt − 0.1 − 4(1 − e−0.qx (0 ≤ t ≤ 1) (c) t px µx+t = qx (0 ≤ t ≤ 1) Hence (aq)α = x = 0 α = qx 1 0 1 α α β t px .252 CHAPTER 15. the dependent rate of mortality would be reduced by approximately 0.2 x Show that if the central rate of withdrawal was doubled.00076.07645.µx+t . holds approximately in each single-decrement table.t px dt α qx (1 − t q β ) dt (using U. Example 15. of D.01 t.05t dt 2 = (1 − e−0. of D. Theorem If there is a Uniform Distribution of Decrements between ages x and x + 1 in each single-decrement table. MULTIPLE-DECREMENT TABLES (b) 2 (aq)α = 30 = 2 0 2 0 β α α t p30 µ30+t .e−0. of D.D.D.D.) for a life table. 2 Note It is often assumed that U.D. then 1 β α (aq)α = qx 1 − qx x 2 and 1 α β (aq)β = qx 1 − qx x 2 Proof Under a Uniform Distribution of Decrements (U. In a double-decrement table with 2 causes of decrement. so the above results may be used as approximations. .05t 2 5−t 5 = 0.4. of D.t p30 dt dt 2 0 0. for mode α) x 1 0 β (1 − t. e. qx from (aq)α .D.3333) = 0. α α (qx )2 + qx [(aq)β − (aq)α − 2] + 2(aq)α = 0.2 Hence (aq)d x i.4 = 0. i.18182 1 1 + 2 mw x 253 d qx w qx Hence (aq)d x 1 w d qx (1 − . x x x (i) (ii) α β This can be solved (rejecting any solution outside the range 0 and 1) to find qx .5.. of D.00076. 15.. α β Getting qx . and hence qx may be found from (i) or (ii). 1 + 0. we have shown that d qx (1 − 1 × 0.00829 2 1 β α (aq)α = qx (1 − qx ).qx ) = 0. md x = 0. (aq)β x x Assuming U.e.009950 1 1 + 2 md x mw x = 0. x 2 1 α β β (aq)x = qx (1 − qx ) 2 β Substituting qx = (aq)β x α 1− 1 qx 2 α from (ii) into (i) gives a quadratic equation for qx .4.D. x Then w qx 0. of D.5 We define Practical Construction of Multiple-Decrement Tables (ad)α =the expected number of exits by mode α x between ages x and x + 1 in a multiple-decrement table with (al)x lives at age x =(al)x (aq)α x Note also that (aq)x = (aq)α + (aq)β x x .00905 2 Now suppose mw is increased to 0.3333. PRACTICAL CONSTRUCTION OF MULTIPLE-DECREMENT TABLES Solution Assuming U. a reduction of 0.15. . x0 + 1. (For example..1. of D. qx (x = x0 . which can be omitted) (ad)x = (al)x (aq)x = (ad)α + (ad)β x x Also. d = death i = permanent disability . assuming x x U. Step 3 Calculate (ad)α0 = (al)x0 (aq)α0 x x and (ad)β0 = (al)x0 (aq)β0 x x Then find (al)x0 +1 = (al)x0 − [(ad)α0 + (ad)β0 ] x x = (al)x0 − (ad)x0 Repeat this procedure for x0 + 1.000).. (aq)β (x = x0 . (al)x0 may be 10.). x0 + 1. MULTIPLE-DECREMENT TABLES + t (aq)x = P r{T1 ≤ t and T1 ≤ T2 } + P r{T2 ≤ t and T2 ≤ T1 } = P r{min{T1 .5. α t |(aq)x = P r{(x) leaves by mode α in the multiple-decrement table between ages x + t and x + t + 1} = t (ap)x (aq)α x+t = (ad)α x+t (al)x Procedure There are 3 basic steps in the construction of a multiple-decrement table. β and hence (putting t = 1.) and hence evaluate (aq)α . .D.000 or 100. Step 2 α β Calculate qx . T2 } ≤ t} = t (aq)x . Example 15.254 To prove these results. x0 + 2 and so on. we note that α t (aq)x CHAPTER 15. A certain population is subject to 2 modes of decrement.. Step 1 Choose a radix (al)x0 where x0 = the youngest age considered. . 01 at each age x (60 ≤ x ≤ 62). and there is an independent rate of disablement of 0.571 95.01 Applications of Multiple-Decrement Tables 1.679 (ad)i x 993 968 942 0.01593 0. 62 (and include the number of survivors at age 63.554 1.01 0. calculate (a) The probability that a person aged exactly 60 is alive and not disabled at age 63.428 (ad)d x 1. 000 (for example) x 60 61 62 63 d qx i qx (aq)d x 0.049 92. Solution (a) 3 (ap)60 (al)63 (al)60 92428 = = 0. PRACTICAL CONSTRUCTION OF MULTIPLE-DECREMENT TABLES 255 The independent rates of mortality are in accordance with A67-70 ult. of D.5.01436 0. . as will be shown in the next chapter.) Solution Use U.01443 0.15. Construct a multiple-decrement table for ages x = 60.92428 100000 = (b) i 3 (aq)60 = (ad)i + (ad)i + (ad)i 60 61 62 (al)60 993 + 968 + 942 = 100. Financial calculations can be carried out using multiple-decrement tables. Example 15. The probabilities of various events can be calculated. (b) The probability that a person aged 60 becomes disabled within 3 years.01775 0. Using the multiple-decrement table constructed in example 15.01776 (aq)i x 0.2.000 97.D. to find d (aq)d = qx (1 − x 1 i ·q ) 2 x 1 d ·q ) 2 x and i (aq)i = qx (1 − x Choose a radix of (al)x0 = 100.00992 0.5. 61.00991 (al)x 100.1. 000 = 0.02903 2.00993 0.5..01 0.01601 0.436 1. 1)) Compare this to the corresponding independent central rate of decrement: .6. o (aµ)α x = lim h d α dh [ 0 t (ap)x µx+t d dh (h) α h (ap)x µx+h dt] h→0+ = lim+ 1 h→0 = µα (as lim h (ap)x = 1) x h→0+ This result is called the identity of the forces. MULTIPLE-DECREMENT TABLES 15.6 Further Formulae α h (aq)x The Identity of the forces Define (aµ)α = lim+ x h→0 h (15. Note that (aµ)x = lim+ h→0 h (aq)x = lim h→0+ h β (aq)α h h (aq)x x + h h = (aµ)α + (aµ)β x x = µα + µβ x x Central Rates of Decrement The dependent central rate of decrement by mode α at age x is given by (am)α = x = = (ad)α x (aL)x l (al)x+t (aµ)α x+t 0 1 (al)x+t dt 0 1 (al)x+t µα dt x+t 0 1 (al)x+t dt 0 α µx+ 1 2 dt (from (15. x x Then (aµ)α = lim+ x h→0 α h (aq)x h h (ap)x µα x+t 0 t = lim+ h→0 dt h Now.256 CHAPTER 15. by L’Hˆpital’s rule.6.1) Theorem (aµ)α = µα for all x x x Proof Assume that µα and µβ are continuous. 7 Generalization to 3 Modes of Decrement Suppose there are now 3 modes of decrement.t px dt (using U.t px . using d d qx = q[x] .D. By generalising the results for 2 modes. of D.qx )(1 − t.7.D.t px dt Theorem Under U.D. 2 3 Note also that the identity of the forces remains true.qx dt t=1 α = qx t − α = qx t2 β t3 β γ γ (qx + qx ) + qx .15. x x x . on β and γ) α qx (1 − t q β )(1 − t q γ ) dt x x 1 0 1 0 β γ (1 − t. GENERALIZATION TO 3 MODES OF DECREMENT 257 mα = x 1 α l µα dt 0 x+t x+t 1 α l dt 0 x+t α µx+ 1 2 Therefore it is normally assumed that (am)α x mα x Note We may also construct tables using select rates of decrement. on α) (using U. .qx ) dt β γ β γ 1 − t(qx + qx ) + t2 qx . β and γ. of D. of each mode of decrement in its single-decrement table 1 β 1 β γ α γ (aq)α = qx [1 − (qx + qx ) + qx · qx ] x 2 3 Proof (aq)α = x = 0 α = qx α = qx 1 0 1 α α β γ t px µx+t .qx 2 3 t=0 1 β γ 1 β γ 1 − (qx + qx ) + qx . qx+1 = q[x]+1 . of D. 15. for example.t px .. α. and (aµ)x = µα + µβ + µγ . it can be shown that α β γ t (ap)x = t px .qx ..t px .t px and (aq)α = x 0 1 γ β α α t px µx+t . 100.e. 000 15. Solution (a) Let d = death. i. It follows that µα is not defined for t = k and hence formulae such as x+t (aq)α = x 1 0 α β α t px .03 at age 51. However. the independent withdrawal rate is 0. p = promotion. of D. we find the values of (al)x .253 1 w 1 w p d p (aq)d = qx [1 − (qx + qx ) + qx .02 0.000 at age 50.01953 0.00728 0.03946 0.7. Assume mode β operates “smoothly” (i.258 CHAPTER 15. α and β operating between ages x and x + 1. x + k (0 ≤ k ≤ 1).t px µx+t dt .qx ]. 12 .02 at age 50 and 0.04 0.e. (ad)α . The members of a large company’s manual workforce are subject to three modes of decrement.00795 52 This table assumes U.00706 51 0. death.. d w p x qx qx qx (aq)d (aq)w (aq)p x x x 50 0. MULTIPLE-DECREMENT TABLES In the practical construction of a multiple-decrement table. and their independent promotion rate is 0.1.D. x (ad)β .02928 (ad)d x 706 742 (ad)w x 3946 3665 (ad)p x 1953 2735 (al)x 100. x 2 3 with similar formulae for withdrawal and promotion rates. µβ exists and is continuous) between ages x and x + 1.03 0. including the value of (al)52 . (a) Draw up a service table for manual workers from age 50 to age 51 with a radix of 100. That is 1 for t < k α t px = α 1 − qx for t > k. It is known that these workers’ independent rates of mortality are those of English Life Table No.000 93.04 at each age. (b) Calculate the probability that a life aged exactly 50 will gain promotion within 2 years.04688. 0. x mode α only operates at a particular age. w = withdrawal.00823 0. (b) p 2 (aq)50 = (ad)p + (ad)p 50 51 (al)50 1953 + 2735 = = 0.04 0.Males.03924 0. x x x Example 15.395 86. (ad)γ and use x x (al)x+1 = (al)x − [(ad)α + (ad)β + (ad)γ ]. withdrawal and promotion to supervisor.8 “Abnormal” Incidence of Decrement Suppose that there are 2 modes of decrement. for mode β is assumed α β (aq)α = qx (1 − k. there is a chance qx of exit by mode α.8.2) Special Cases 1.qx x α = qx (1 − k q β ) x If (as usually the case) U. so (aq)α = P r{(x) survives to age x + k under mode β only} x × P r{(x) leaves at age x + k by mode α. we consider exits by mode α to occur just before reaching age x + 1) α β (aq)α = qx (1 − qx ) x and β (aq)β = qx . x we may use (aq)β = (aq)x − (aq)α x x α β α β α β = [qx + qx − qx .e.8. Military personnel attend a training camp for an intensive course which involves 3 weeks of continual exercises. at age x + k.D. we argue that from age x to x + k. If k = 1. of D.15.qx ] − qx (1 − k.8.1) To obtain (aq)β . Then. If k = 1 . (i. If k = 0 α (aq)α = qx x and β α (aq)β = qx (1 − qx ) x 3.qx ) So β α (aq)β = qx [1 − (1 − k)qx ] x (15. mode β α operates by itself.8. provided he is not hospitalised because . we obtain the familiar formulae 2 α (aq)α = qx (1 − x 1 β ·q ) 2 x 1 α ·q ) 2 x and β (aq)β = qx (1 − x 2. 259 To deal with this “abnormal” mode of decrement.qx ) x (15. “ABNORMAL” INCIDENCE OF DECREMENT cannot be used. given survival until then} α = k pβ . x Example 15.1. Each soldier remains in the camp for exactly 3 weeks. 08731 775 75 68 2 0. f f h q0 is now 0.058 0.07285 1 0.. Calculate the revised numbers of soldiers successfully completing the course.102 0.058 0. .04175 632 37 26 3 570 Hence now the number who complete the course = 570. but all the other values of qx and qx are unchanged. being hospitalised and being failed. MULTIPLE-DECREMENT TABLES of injury or “failed” by one of the instructors and sent back to his base.102 0.092 3 0. f = failed. During weeks 2 and 3. Anyone sent to hospital or failed leaves the camp immediately and does not return. Assume a uniform distribution over each week of hospitalisation and failure. (ii) Some time later the course is altered so that in the first week the instructors test the soldiers only at the start of the sixth day. h f x qx qx (aq)h x 0 0.078 0. There are no other modes of decrement.09731 2 0.09731 0. the number who will be hospitalised and the number who will be failed. Solution (i) Consider “age” as time (in weeks) since entry to camp.260 CHAPTER 15.07443 0. Assume U.043 (i) Of a group of 1. testing takes the form of continuous assessment as previously. which gives 1 f h (aq)h = qx (1 − qx ) x 2 1 h f f and (aq)x = qx (1 − · qx ) 2 where h = hospitalised.043 0. k = 5 7 (test is at end of 5th day).000 soldiers who start the course.160.092 0.043 0.15109 1000 74 151 1 0. h f h f qx qx (aq)x (aq)x (al)x (ad)h (ad)f x x x 0 0.16 0.16.078 0. assuming nothing else changes.058 0.05675 0. (aq)f x 0.092 0.102 0.132 2 0. calculate the number who will successfully complete the course. The independent weekly Hospitalised Week through injury Being failed 1 0.12685 0.08731 0. so h h (aq)x = qx (1 − 2 f ·q ) 7 x 5 h ·q ) 7 x and f (aq)f = qx (1 − x for x = 0 only. of D.04175 (ad)h x 73 78 37 (ad)f x 127 70 27 (ii) Here.D.132 0.05675 3 Hence number who complete the rates of decrement are as follows: (al)x 1000 800 652 588 course = 588.078 0. and the independent weekly rate of being failed in week 1 alters to 0. with no change to the independent weekly rates of being failed. “ABNORMAL” INCIDENCE OF DECREMENT 261 Application to Profit-Testing Let α = withdrawal.15.qx+t−1 withdrawal where qx+t−1 was denoted by wt and mortality P r{ death occurs in policy year given it is in force at start of year} = qx+t−1 These formulae correspond exactly to those given in this section when k = 1. x Then the numbers of exits by modes β and γ between ages x and x + 1 can be worked out as if there were no other modes of decrement. It is assumed that withdrawals occur only at the end of a policy year.2) may be awkward to derive.8. These problems are best handled from first principles. . formulae similar to (15.1) and (15. say) operates at exact age x and the others (β and γ) operate uniformly over the year of age from x to x + 1. Often one mode (α.8. and (al)+ = number of lives at age x after mode α operates. That is.8. the following results were used: P r{ policy in force at time t − 1 will be surrendered at time t} mortality withdrawal = (1 − qx+t−1 ). Note however that (ad)β = (al)+ (aq)β x x x 1 γ β where (aq)β = qx (1 − qx ) x 2 Similar calculations apply to “abnormal” exits at the end of the year. Extension to 3 modes of decrement With 3 modes of decrement. mode α only operates at time k = 1. In chapter 14. This can be dealt with by defining (al)x = number of lives at age x before mode α operates. and β = mortality. 09 58 0.8.01299 0. of D.D.854 5.01124 0.343 65.01169 0.964 (al)+ x 80.” So construct a table of (aq)d and (aq)r (ignoring mode r ). 60 . the central rates of retirement are as follows: Central Rate Age of Retirement 57 0. Among the employees of a certain firm retirement may take place at or after age 57. x x x x 57 58 59 60 Note We may (al)x 100.01267 0. x x x 57 58 59 Now d qx r qx mr x 1 1+ 2 mr x (aq)d x (aq)r x 0.08 59 0.07692 0.07647 0.000 72.998 0 (ad)r x 6. Construct a service table for employees from age 57 to age 60.000 0 0 61. (ad)r = 20.000 72.04878 0. Leaving aside these retirements.01004 0. Then.2.05 There are no withdrawals or ill-health retirements after age 50. but is compulsory at age 60. to refer to retirement at exact age x.998 61. MULTIPLE-DECREMENT TABLES Example 15.04846 construct the multiple-decrement table. Solution Create a special mode of decrement.000 employees attaining age 57.343 65. both of which are approxi57 mately “U. noting that (ad)r = (al)+ (aq)r .532 3. 000.08612 0.08567 0. r . 20% of those attaining age 57 retire at once. 57 leaving (al)+ lives who are subject to ‘normal’ retirements and death.964 (ad)r x 20.198 (ad)d x 803 813 836 take (al)60 = (ad)r since all survivors at age 60 retire at once. with a radix of 100.262 CHAPTER 15. The independent rates of mortality of the employees are those of A1967-70 ultimate.01050 0. [Regard failing the test as a special mode of decrement ] (ii) How many people are recruited on each 1st July? . are exposed to the risk of injury.decrement table. EXERCISES 263 Exercises 15.051219 at age 18 and 0. widowed and divorced men are subject to the following independent q-type rates of decrement: mortality: English Life Table No. for whom remarriage is not permitted. ten per cent fail a final test of competence and are dismissed. The independent rates of mortality of wives follow a(55) ultimate (females). the end of former marriage.Males remarriage: rates depend on the age at.025 0. Of those who complete this period of training. and the duration since. construct a double-decrement table for married women from age 70 to age 72 inclusive. distinguishing between those about to take the final test of competence and those who pass it. Employees may also leave service voluntarily at any time. 12 .023 0.020 Calculate the probability that a man aged exactly 50 whose marriage has just ended will remarry within 2 years. The number of employees attaining age 21 each year is 500. at a relocation cost of £1. An employee who is injured is transferred to alternative work with a subsidiary company. the number of wives at age 73. 15. 15.15 for trainees and 0.9.4 A large industrial company recruits a constant number of school leavers aged exactly 18 years on 1 July each year.000 and assuming a uniform distribution of each mode of decrement in its associated single-decrement table. α2 be the modes of decrement in a double-decrement table.050030 at ages 19 and above. the dependent qtype rate of widowhood at each integer age x from 70 to 72 inclusive is twice the corresponding dependent q-type rate of mortality.1 Let α1 . workers undergo training for one year.042 duration 1 year 0.Males. the following table is an extract from these rates: exact age at end of former marriage 50 51 52 duration 0 year 0. The central rate of voluntary withdrawal from service is 0. Suppose that α1 is uniformly distributed over the year of age from x to x + 1 in its associated single. (i) Construct a service table covering the first 3 years of employment with the original company.050 0.15. x+t α α Find formulae for (aq)α1 and (aq)α2 in terms of qx 1 and qx 2 . (i) Using a radix of 100. x x 15.10 at each age for fully trained employees. Upon joining. and (b) be widowed within 3 years. including trainees. (ii) Find the probabilities that a wife now aged 70 will (a) be alive and married at age 73.2 For a certain group of married women. The mortality of all employees follows English Life Tables No. giving also the value of (al)73 .3 In a certain country. The occupation is hazardous and all workers.045 0.12 . The independent q-type rate of injury is 0.000. and µα2 = c for 0 ≤ t ≤ 1. 326 2. 72 x x where w = widowhood d = death.119 (ad)d x 2. Using U.509 4.02760 79119 100.05519 (ii)(a) 3 (ap)70 = (b) 3 (aq)w = 70 15.02559 0.D.02839 and (aq)w = 2(aq)d .264 CHAPTER 15.000 (aq)w x 0. so α1 α1 α1 t px µx+t = qx µα2 = c. (ad)w +(ad)w +(ad)w 70 71 72 (al)70 . so x+t α2 t px t for 0≤t≤1 = exp[− 0 1 0 t µα2 dt] = e−ct for 0 ≤ t ≤ 1 x+t (aq)α1 = x = α1 α1 α2 t px µx+t .02307 0.04989 0.1392.02494 0.761 (aq)d x 0.t px α qx 1 e−ct dt t=1 t=0 dt 0 1 α = qx 1 − e−ct c −c α (1 − e ) = qx 1 c α −qx 2 α1 = qx α log(1 − qx 2 ) α as c = − log(1 − qx 2 ) (aq)α2 = (aq)x − (aq)α1 x x α α α α α = [qx 1 + qx 2 − qx 1 qx 2 ] − qx 1 α −qx 2 α log(1 − qx 2 ) (from above) 15. of D.79119 = 0.260 79.000 93. 71.1 α1 is uniformly distributed.2 (i) (aq)w = 2(aq)d for x = 70.3 Denote d = death = 0. x x (al)x 100.02254 0. 1 d 1 w w d (aq)w = qx (1 − qx ) = 2qx (1 − qx ) = 2(aq)d x x 2 2 w Hence qx = d 2qx 1 d 1+ 2 qx d 1 d qx (1− 2 qx ) . MULTIPLE-DECREMENT TABLES Solutions 15.380 (ad)w x 4.04509 0.237 86.651 4.254 2. 1 d 1+ 2 qx Therefore (aq)d = x x 70 71 72 73 d qx 0. 308 91.193 (ad)d x 710 767 (ad)r x 4. 500 employees aged exactly 21.13588 0.00112 0.04762 (aq)w x 0.409 (ad)f x 0 816 0 (al)+ x 10.09280 Define (al)19 = number of people before test (at age 19 exactly) (al)+ = number of people who pass test.09280 0. SOLUTIONS r = remarriage r Notice that q51 = q[50]+1 . 000 (for example) x 18 19 20 21 (al)x 10.10.09524 0.000 7.301 (ad)d x 10 8 7 (ad)i x 476 349 300 (ad)w x 1359 681 585 Notice that (ad)f = (al)x − (al)+ x x (ii) 10.982 2.00117 0. x 18 19 20 d qx 0.04982 0.00728 0. .00710 0. injuries and withdrawals. 19 with (al)+ = 0.e.4 (i) Denote d = death w = withdrawal (excluding those who fail test) f = those who fail test i = injury Use w qx mw x 1 1 + 2 mw x and in view of U.000 entrants give 5. Hence 924 entrants are needed to get.04762 0.348 (ad)r + (ad)r 50 51 (al)50 = 0. on average. and treat the test as an “abnormal” mode of exit.02490 r 2 (aq)50 (al)x 100.15.025 (aq)d x 0. i.00823 r qx 0.339 6. duration 1 year from age 50 at end of former marriage. use formulae such as 1 w 1 w i d i (aq)d = qx [1 − (qx + qx ) + qx qx ] x 2 3 for deaths.D.001105 (aq)i x 0.04762 0.13953 0.05003 0.301 5.9(al)19 19 Let (al)18 = 10. of D.409 employees after 3 years. 265 x 50 51 d qx 0.0733.000 94.000 8.09524 (aq)d x 0.001016 0.00813 (aq)r x 0.05122 0.05003 w qx 0.050 0.00119 i qx 0. = 15.155 6.001087 0. 266 CHAPTER 15. MULTIPLE-DECREMENT TABLES . v. of premiums = m. on average.v.v. the annual premium or the sum assured payable on death) Note It should normally be assumed that exits occur. and there are no commutation functions (except for pensions.1 Principles One may value cash flows. Notation is best considered from first principles. If it is not.2 The Use of “Defective” Variables lim F (t) = k < 1 A defective random variable. let T = time of exit of (x) by mode α in a double-decrement table. and that premiums cease if a life exits by any mode of decrement.p. Stage 2 Write down an equation of value of the form m. Then 267 .Chapter 16 FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES 16. in the middle of each policy year. 16.) Then solve the equation of value for the unknown quantity (for example. T .p. the other mode of exit being β. which are covered later). calculate premiums and find reserves using the same ideas as are used when there is just one mode of decrement (death). of expenses (This assumes that the expected present value of the profits to the office is zero. The procedure is generally as follows: Stage 1 Construct a multiple-decrement table.p. is such that t→∞ where F (t) is the distribution function of T . For example. of benefits + m. add a suitable profit term to the equation. (A) Integrals The theory of ‘defective’ variables is used to value the m. T is a defective variable with p.t pα µα .1) . t > 0 f (t) = 0. of this benefit is n S 0 v t . The m.v.d.f. as a random variable. α t→∞ β t→∞ lim F (t) = lim t (aq)x < 1 A defective variable T may have a probability density function. α α β t px . β α α t px µx+t . FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES F (t) = P r{T ≤ t} α = t (aq)x t = 0 α α β r px µx+r .p. consider a benefit of £1 payable immediately on exit by mode α to a life now aged x.268CHAPTER 16.v. and suppose that there is a benefit of £S payable immediately on the exit of (x) by mode α within n years. where T = time to exit by mode α in the double-decrement table. We also consider the m.p.t pβ dt x x+t x (16.t px (t > 0). we still have results such as ∞ E[g(T )] = −∞ g(t)f (t) dt For example.3 Evaluation of Mean Present Values This can either be done by integrals or sums. In the above example.v.t pβ dt x x+t x Notice the similarities to the joint-life and contingent assurance functions in chapters 11 and 12. 16.t px . (i) Consider a double-decrement table with exits by mode α and mode β. is Z= Sv T 0 if T < n if T > n. The present value of this benefit. f (t) = F (t).p.µx+t . t<0 With defective variables.r px dr Since we have α ∞ (aq)x + ∞ (aq)x = 1. of this benefit is E[v T ] = 0 ∞ v t . Hence the m.p.v. of premium payments.3. of benefits on exit by a given mode of decrement.t pα µα . 2) n 0 (assuming 2 modes of decrement.t pβ dt x x (16. The mean present value is n P 0 vt · (al)x+t dt = P (al)x =P n 0 v t . (i) Consider a benefit of £S payable at the end of the year of exit of (x) by mode α within n years. The m. of the benefit is S v2 1 α α 3 (ad)x+1 1 (ad)x+n−1 (ad)α x + v2 + · · · + v n− 2 (al)x (al)x (al)x (ii) Consider an annual premium of £P per annum. The m. The m.p. payable continuously for at most n years while (x) remains a member of the group under consideration. Calculate the value of premiums of £P per annum payable continuously for at most n years while (x) remains a member of the group.t (ap)x dt v t .3. Solution µw = k for all t. death (d) and withdrawal (w).t pw dt x x e−δt .e−kt dt x =P 0 (as v t = e−δt where δ is the force of interest) n =P 0 e−(δ+k)t . mid-way through year.1) and (16. (al)x (al)x (al)x If the benefit is payable immediately.t pd . for at most n years.p.t pd dt x = P ax:n at force of interest δ = δ + k ¯ (B) Sums Sums can be used to value death and other benefits and premium payments either exactly (if financial transactions occur at the end of policy years) or by approximating integrals (16.v.3.p.3.16. of premiums = P 0 n v t . of this benefit is S v (ad)α (ad)α (ad)α x+1 x+n−1 x + v2 + · · · + vn . x n m.v. x+t Hence t pw = e−kt for all t. and there is a constant force of withdrawal of k per annum.3.v.t pα . is P 1+v (al)x+2 (al)x+n−1 (al)x+1 + v2 + · · · + v n−1 (al)x (al)x (al)x . then exits can be assumed to occur. EVALUATION OF MEAN PRESENT VALUES 269 (ii) Consider premiums of £P per annum.3. Suppose there are 2 modes of decrement.v.1.p. payable in advance while (x) is still a member of the group.2) if benefits are payable immediately or premiums are paid continuously. on average.t pd . α and β) Example 16. 3.8325P ) = 722. Benefits other than cash sums In some cases. The m.79 Therefore P = £268.p. for at most 3 years.79 P [(al)60 + v(al)61 + v 2 (al)62 ] (al)60 = 2.v. an annuity of £2. P .2) can be approximated to a sum.61 per annum. providing £20. Interest is at 4% per annum. etc. and (ii) immediately on death before age 60 while not permanently disabled. must often be found by interpolation. 2 2 Example 16. of benefits = m. Solution 20. m.3.p. is now 1 (al)x+ 1 3 (al)x+ 3 1 (al)x+n− 1 2 2 2 P v2 + v2 + · · · + v n− 2 (al)x (al)x (al)x Note (al)x+ 1 . expression (16.436 Suppose a 3-year term assurance is issued to a life aged 60.000 immediately on death.p.p. calculate the annual premium. payable in advance while a policyholder. treat the m.000 at the end of the year of death.v. on permanent disability or on reaching age 60.000 per annum payable weekly for life and £20.) Example 16. 000 [v.270CHAPTER 16. A multiple-decrement table referring to mortality (d) and withdrawal (w) from a life assurance contract is x (al)x (ad)w (ad)d x x 60 10. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES If the premiums are payable continuously.3.2.8325P. the benefits on exit by a certain mode of decrement consists not of a cash sum but an annuity or some other benefit.v.618 48 134 63 9. In such cases. £20. Calculate the office annual premium. of premiums = Hence P solves 0. If expenses consist of 5% of each premium.810 61 131 62 9.95 × (2. (The annuity will often have to be evaluated by interpolation if it begins mid-way through a year of age. payable weekly and ceasing on death.(ad)60 + v 2 (ad)61 + v 3 (ad)62 ] (al)60 = 722.000 64 126 61 9.000.v. for a life aged 58 if the office uses the following basis: .3. A life office issues policies to lives aged under 60 providing the following benefits: (i) on becoming permanently disabled before age 60. (al)x+ 3 . of the annuity (or other benefit) at the date of exit as if it were paid out at that time. 13.12 . 12 .L. 000¯ih 1 + 20. The value of premiums less expenses is 0. 000Aih 1 a58 58 2 2 (al)58 i 3 (ad) 59 ¯ + v2 2.T.01169 0.01295 0.975P v 2 1 (al)58 1 2 (al)58 + v2 3 1 (al)59 2 (al)58 − 50 (al)58 1 2 (al)59 1 2 1 [(al)58 + (al)59 ] = 99119 2 97310 Hence value of premiums less expenses is 1. 12 .00596 100. Expenses: 2 1 % of all office premiums.006 60 Value of benefits: 0. Consequently P = £470.34 + 348.381 1.Males = 0.006.006 59 0. Let P be the annual premium.918 ¯58 2 ¯ ¯ Aih 1 = A65 on E.34 (ad)58 (al)58 ¯ aih 1 = a65 on E. 59 2 So value of (i) = 180. 2 Permanent disability: a constant independent rate of 0.01165 0. EVALUATION OF MEAN PRESENT VALUES 271 Mortality: the independent rates of mortality of those not permanently disabled are those of A1967-70 ultimate.8422P − 50 = 468. d i x qx qx (aq)d (aq)i (al)x (ad)d (ad)i x x x x 58 0.Males = 8. assuming U. Interest: 4% per annum.587 ¯59 ¯ 2 ¯ ¯ Aih 1 = A66 = 0.3.8422P − 50.00596 0.65023 58 2 Similarly aih 1 = a66 = 8. 2 Solution Construct service table with d = death and i = disability.35.D.24 + 167.13 .16. 000Aih 1 a59 59 2 2 (al)58 1 1 (ii) 20. the permanently disabled are subject to the mortality of English Life Table No. Therefore 1. 000 v 2 For benefit (i) d 3 (ad) (ad)d 58 59 + v2 = 468.238 96.01299 0. .000 98.66323.L.165 1. of D. plus £50 at the issue date.89 = 348.T.Males with the age rated up by 6 1 years. 000¯ih 1 + 20.272 596 585 (i) v 2 (ad)i 58 ¯ 2. as covered in the Actuarial Subject A2. nor the rate of interest. 16. calculate the annual premium payable continuously by the newspaper to provide this “free” policy. Example 16. whereas an office premium (allowing for expenses) would be larger. A national newspaper recently advertised “free insurance for subscribers”.v. in which case the formulae in Section 16.42 per month. n m. Also notice that £0.000 would be paid immediately on accidental death (mode α) within n years. this will often be the force of mortality for a given life table. Example 16. such as motor racing.272CHAPTER 16.5 Extra Risks Treated as an Additional Mode of Decrement Suppose that in addition to “normal” mortality (mode α).t pβ dt x x+t x n 0 = 10000 × 0.4 Benefits on Death by a Particular Cause Sometimes a policy will provide death benefits if (x) dies from a particular cause. x+t When these are added together to give (aµ)x+t .5.t pα µα .4. Mode β may refer to: (a) Certain occupational hazards. nor the term of n years. Solution Let the annual premium be P . a m. the problem can be treated as an “extra risk” question.v. a group of lives are subject to certain additional hazards (mode β). of premiums = P ax:n Thus P = £5 per annum (or £0.1. Assume µα = 0. (b) Risks associated with a leisure activity.p.42 is a net premium. and x+t µβ = force of mortality from all other causes.) Note: Observe that P does not depend on the mortality table (for all causes).1.t pA67−70ult dt x = 5¯x:n on A67-70 ult. One may write µα = force of mortality from cause α.0005 v t .0005 for t ≥ 0 (for all x) and mortality x+t due to all causes follows A1967-70 ult. If the death benefit is paid on any cause of death. (c) Some medical conditions. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES 16. whereby a benefit of £10. A certain life office’s premium basis for policies accepted at normal rates is: . Assuming a given rate of interest and ignoring expenses.p. of benefits = 10000 0 v t .3 may be used. One may wish to calculate premiums for policies which pay out only on death from the “additional” cause. of this benefit is found by considering exit by mode β only.03922 (rate of interest = 4%) So δ = 0.000.04) [(1 − d. and offers to pay an additional single premium at the outset for this extra cover. expenses are ignored. (b) The proposer requests that. 4% interest. payable throughout the term of the policy. is £10. 000(1.05 = 5%.87 a[45]:20 0. of benefits is 10. The sum assured. 273 A proposer.78 (b) The m.p. which is payable immediately on death.04) 2 A∗ 1 1 2 [45]:20 1 = 10. the sum assured should be doubled.v. 000 = 10. in the event of death occurring as a result of the special occupational hazard. EXTRA RISKS TREATED AS AN ADDITIONAL MODE OF DECREMENT A1967-70 select.04) 2 [1 − d.p.05 ¨ = £167. Calculate this single premium.t p[45] (µ[45]+t + k) dt (where k = 0. i = eδ − 1 = 0. aged 45.¨∗ a[45]:20 ) − A∗ ] [45]:20 Now use the rule (as in extra risks) that the rate of interest (in annuity and pure endowment functions) may be altered to allow for the addition to the force of mortality. 102.16. for temporary assurance ceasing at age 65 is subject to an extra occupational hazard which is considered to be equivalent to an addition of 0.t pβ (µα [45] [45] [45]+t + µ[45]+t ) dt 20 = 10. β = extra occupational mortality Value of benefits is 20 10000 0 β v t t pα .5.05 −A = 1 ] [45]:20 0.05 l65 20 v ) l[45] 0. (a) Calculate the level annual premium. 000(1.87 a∗ ¨[45]:20 2102. Hence m.009569) where * indicates normal plus extra mortality 1 [45]:20 = 10.v.05 So annual premium = = 2102. 000(1.048790 and hence.009569 to the force of mortality at all ages.87 (using A 1 0.05 1 [45]:20 0. 000A 0 ¯∗ 1 v t e−kt . Solution (a) Consider α = normal mortality. a∗ ¨[45]:20 is at force of interest δ = δ + k where δ = log(1.¨[45]:20 a = £2. This gives a single .04) = 0. These problems may often be treated from first principles.t p50 µ50+t l52 l50+t dt (the expression in brackets giving survival as a foreman to age 52).69 × 12.1899 = £1.e−kt .009569) = 10.6. Manual workers between the ages of 50 and 60 are subject to 2 modes of decrement. rather than by the construction of a multi-state model. Example 16. which approximates to (ad)p l (ad)p l52 50 51 · 52 + (al)50 l50 1 (al)50 l51 1 2 2 (b) The probability is approximately (ad)p l 50 1 − 52 (al)50 l50 1 2 + (ad)p l 51 1 − 52 (ad)50 l51 1 2 . 000k¯[45]:20 a 95.t p[45] . 000 0 v t .05 95. 166.45 16. 000k¯∗ a[45]:20 e−δt . Find the probabilities that (a) A manual worker aged 50 will be alive and a foreman at age 52.1. etc... 000 = 0 10. d = death. (b) A manual worker aged 50 will be promoted to foreman but then die before age 52. Solution (a) The exact formula is 2 0 p p d t p50 . Foremen are subject to the mortality of another table.05 0.274CHAPTER 16.05 1 l65 20 − (1 − v ) 2 l[45] 0.µβ [45] [45] [45]+t dt 20 = 10. T .69 a[45]:20 ¨ 0.6 Calculations Involving a Change of State Often there is interest in the probabilities of survival for a life who leaves a multiple-decrement table and continues in another state.t pα .t pβ . FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES premium of 20 10. and p = promotion to foreman.k dt (k = 0. with functions lx . ) 16. The surrender value is to be equal to 0. EXERCISES 275 Exercises 16. The following multiple-decrement table is applicable to employees going overseas at exact age 47. The central rate of withdrawal of apprentices .415 15. and the benefit on withdrawal is zero. The employees serve an apprenticeship during the first two years of employment. i = ill-health retirement) x 47 48 49 50 (al)x 100.007 8.16.604 9. 16. subject to the mortality and withdrawal rates shown in the above table.. (b) on survival as an employee of the company to the end of the term.005 (ad)d x 853 616 478 (ad)i x 13. On the basis of rate of interest of 6% p. payable immediately on death. w = withdrawal.336 (ad)w x 24 21 15 (ad)d x 51 58 60 (i) What is the probability that a policyholder aged 51 will withdraw before attaining age 53? (ii) Suppose that a single-premium three-year term assurance contract is to be issued to a life aged 50. Those who undertake this assignment effect a 3-year policy providing the following benefits payable at the end of the year of claim: (a) on death or ill-health retirement during the term.059 11.7. 1 (c) on withdrawal from the company during the second year an amount equal to 1 4 times the annual premium. Assuming that surrenders in each policy year take place on average half-way through the year. x 50 51 52 (al)x 15. a lump sum.a..875 Using an interest rate of 4% p. and allowing for expenses of 10% of the single premium.3 An organization recruits new employees at exact age 20. these modes being referred to as ‘d’ and ‘w’ respectively.490 15.000 in advance. (Do NOT proceed to the evaluation of this premium.1 The following is an extract from a multiple-decrement table referring to mortality and withdrawal from certain life assurance contracts. the sum of £6.a.2 Employees of a certain company are given the opportunity of early retirement immediately after they complete a 3-year overseas assignment.468 1. The sum assured is £50. calculate the lump sum for an employee going overseas at exact age 47 who pays level annual premiums of £2.000.5 per cent of the single premium for each week between the date of withdrawal and the end of the term. No benefit is payable on withdrawal from the company in 2 the first year. write down an equation of value from which you could calculate the revised single premium.000. calculate the single premium for the above policy.393 31. (iii) Suppose now that the office issuing the policy of (ii) above wishes to introduce surrender values for these three-year contracts. (d = death.081 42.000 63. age.035 (ad)w x 23. and on withdrawal from the company during the third year an amount equal to 2 1 times the annual premium. Ignore expenses. The force of withdrawal of permanent staff is 0. 16. Remarriage: rates depend on the age at. Weekly benefits may be taken as payable continuously.05 per annum. rated up by 7 years 4% p. who retire at exact age 60.025 0. .a. apprentices join the permanent staff.000 on death in service to permanent staff.050 0. The employer pays an immediate benefit of £2. Both apprentices and permanent staff experience mortality according to A1967-70 ultimate. ceasing at age 65 or earlier death.a. Evaluate the single premium for this contract. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES at ages 20 and 21 is 0. will attain age 52 as a widower who has not remarried between ages 50 and 52. married men are subject to the following independent q-type rates of decrement: Mortality: A1967-70 ultimate Widowhood: wives of married men are subject to A1967-70 ultimate Divorce: 0.040 duration 1 year 0. (b) Calculate. The following basis is used to calculate the single premium for this contract: Independent rates of mortality of policyholders (not disabled): Force of disablement: Mortality of disabled lives: Rate of interest: Expenses are ignored.02 per annum at all ages Widowed and divorced men are subject to the following independent q-type rates of decrement: Mortality: English Life Table No.020 0.5 In a certain country. and the duration since. A1967-70 ultimate.023 0.045 0. 12 . 16. (a) Construct a double-decrement table for apprentices between ages 20 and 22 with a radix of 100. At exact age 22.Males.01 p. and withdrawals are uniformly spread over the year of age (in the single-decrement table for withdrawals).018 Calculate the probability that a married man aged 50. The death benefit (but not the withdrawal benefit) is paid to apprentices.276CHAPTER 16. the end of the former marriage. the value of the benefits in respect of a new employee aged 20.000 at age 20. A1967-70 select (at entry) 0. to those becoming permanently disabled before age 65.4 A life office sells policies to lives aged 63 which provide a benefit of £50 per week. the following table is an extract from these rates: age at end of former Marriage 1 50 2 1 51 2 52 1 2 53 1 2 duration 0 year 0. whose wife is also aged 50.0094787 per annum.000 on withdrawal and £8.042 0. using an interest rate of 5% per annum. v 3 ] 49 (al)48 (al)49 2 v+ v (al)47 (al)47 2000 w 2 (al)47 [1.16.005 × 1.5)] 50 51 52 (al)50 16.v = 279. 000 [v((ad)d + (ad)i ) + v 2 ((ad)d + (ad)i ) + v 3 ((ad)d + (ad)i )] 47 47 48 48 49 49 (al)47 = 1987.5(ad)w .009977 = 498.99 + 0.2 Let the lump sum be X.05 mw 20 w = 0. the surrender value in year 1 is on average 52. 000 (al)50 = 50.p. The equation of value is 0.04878 = q21 = 1.25(ad)48 .v. benefits: (a) 6.8. and so on. The average surrender value in year 2 is 52.29.27563(X) (al)47 + 2.005 + P [v 2 (ad)w (2. SOLUTIONS 277 Solutions 16. Then P solves 1 1 1 (ad)d v 2 + (ad)d v 1 2 + (ad)d v 2 2 50 51 52 0.3 (a) w q20 0.1 (i) w 2 (aq)51 (ad)w + (ad)w 51 52 (al)51 = 0.9P = 50.18 × 0. 000 × 0. premiums = 2000 1 + = 3996.18 × 0.5P .18 weeks in 1 year.025 1 + 1 mw 2 20 .18 × 0.p.92 (b) X.5) + v 1 2 (ad)w (1.53 16.9P = value of death benefit (as above) 1 1 1 52.005 × 2.5) + v 2 2 (ad)w (0.99 = 2266.002335 = (ii) Let single premium be P .v 3 (c) (al)50 = 0.27563(X) Hence X = £6276. m.07 So X solves 3996.86 Hence P = £554.5P . m.v. So. There are on average 52.99. at 6% interest (iii) Let P be the premium allowing for surrender values. 04878 (aq)d x 0.k dt 22 .04878 0. i −0.06 = 285.987 (ad)d x 835 1216 (ad)i x 991 971 m.4 Let d = death and i = disablement.325 (ad)d x 87 78 (ad)w x 4876 4634 (b) Consider value of benefits for employee attaining age 22. valued at age 20.06 − l60 38 v l22 0.k¯22:38 a 1 l60 38 = 2000 × 0. Let k = 0. Value of death benefit is 38 8000 0 v t e−kt . d d Note that q63 = q[63] and q64 = q[63]+1 on A67-70 tables.06) 2 A22:38 = 206.v.037 90. the double-decrement table is.278CHAPTER 16.174 95. x x 20 21 22 d qx 0.00995 0.04876 0.t pd .80 Value of withdrawal benefit is 2000 0 0.01 i = 0. q63 = q64 = 1 − e x 63 64 65 d qx 0.06 1 = 8000(1.01239 (aq)i x 0.64 Value of benefit in first 2 years.000 95.59 + v 2 [206.0094787 a22:38 − (1 − ¨ v ) 2 l22 0.00995 Also.000821 (aq)w x 0.00989 (al)x 100.04876 (al)x 100.03 16.00839 0.058269 22:38 0.00991 0.p.000889 0.0094787.80 + 285.000841 w qx 0.59 Hence value of benefits for new employee is (al)22 12.t pd µd 22 22+t dt 22:38 ¯ = 8000A∗1 ¯ = 8000A 1 at force of interest δ = δ + k = 0.000 98.06 = 2000.64] (al)20 = £416.06 38 v t e−kt . FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES 1 d w Using (aq)d = qx (1 − 2 qx ).00835 0.¯∗ 1 1 + v 1 2 (ad)i a∗ 1 1 63 a63 :1 64 ¯64 : (al)63 2 2 2 2 . benefits where 1 50 × 52.18 1 v 2 (ad)i .0.000868 0.00995 (aq)d x 0. is 1 8000 1 [v 2 (ad)d + v 1 2 (ad)d ] 20 21 (al)20 = 12.01245 i qx 0. 97064.97322 = 0.T.08. 12-Males) l50 1 2 2 (i) 1 attains age 52 without remarrying} 2 (ii) P r{ widower aged 50 = 0.004730 0. 16. but short cuts can be taken.009399 .96010] = 1.005154 51 50 1 attains age 52 without remarrying} 2 l52 1 = (1 − 0.8.48005 2 Hence single premium = £47.025) (using E.02 0.005309 A multi-decrement table can be used.045) = 0. β γ α (ap)50 = (1 − q50 )(1 − q50 )(1 − q50 ) = 0.92674 + 0.02 x 50 51 α qx 0.96010) = 0. P r{ widower aged 51 = 1 l52 (1 − × 0. So. SOLUTIONS 279 a∗ ¯ 1 63 2 :1 1 2 = a70 1 :1 1 on A67-70 ultimate ¯ 2 2 1 [¯ a + a71:1 ] ¯ 2 70:2 1 = [1.92674. l51 1 2 2 Hence required probability is (aq)β × (i) + (ap)50 (aq)β × (ii) 50 51 = 0.97322.L.004789 0.004730 × 0.005377 β qx 0.005377 (aq)β x 0.004789 0.4046 2 a∗ ¯ = a71 1 : 1 on A67-70 ultimate ¯ 2 2 64 1 :1 1 2 2 1 (0. P r{ married man aged 50 becomes widower between ages 51 and 52} = 1 |(aq)β = (ap)50 (aq)β = 0. Use formula of the form 1 β 1 β γ α γ (aq)α = qx [1 − (qx + qx ) + qx qx ] x 2 3 γ qx 0.050)(1 − × 0.005154 × 0.84904 + 0.5 Denote α = mortality of married men β = mortality of wives γ = divorce.16. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES .280CHAPTER 16. (b) Modernist .uses continuous-time stochastic processes with a finite number of states.1 Two Points of View There are essentially two approaches to the theory of multiple-decrement tables: (a) Traditionalist . In chapters 15 and 16. n = 3 in the following system.2 Kolmogorov’s Forward Equations Suppose there are n possible states.Chapter 17 MULTIPLE-STATE MODELS 17. This chapter shall briefly take a more modernist approach. and both produce the same practical results. x + t) =P r{(x) will be in state j at age x + t given that he is in state i at age x} Figure 17. For example.2.1: a double-decrement model 281 .uses the underlying single-decrement tables. which refers to a double-decrement table: Define pij (x. 17. and show the differences between the two methods. Both view-points have advantages and disadvantages. the traditional approach and notation was used. A development of the modernist approach for multiple-state models is covered in Actuarial Subject D2. x + t) = 1 − p12 (x. x + t) − p13 (x.2. x + t) = − p12 (x. are defined as µij (y) = lim+ h→0 CHAPTER 17. j ≤ n. µi (y) = i=j µij (y) = the force of transition from state i to any other state at age y Note If all these forces are constant. x + t) dt dt dt (17.282 Note that the initial conditions are pij (x. t ≥ 0) (17. x + t) = p11 (x.1) The three-state model of Figure 17. x + t) − p13 (x.2) .2.1 is such that µ1 (y) = µ12 (y) + µ13 (y) µ2 (y) = 0 µ3 (y) = 0 (as no one leaves states 2 or 3) so Kolmogorov’s Forward Equations give d p12 (x.2. x + t) and hence d d d p11 (x. y + h) h (i = j) Also. x) = The forces of transition. x + t) = p11 (x. it is called inhomogeneous. x + t) + ν=j (1 ≤ i. MULTIPLE-STATE MODELS 1 0 if i = j if i = j. this is a homogeneous chain. x + t)µ13 (x + t) dt Note also that p11 (x. x + t) = −µj (x + t)pij (x. x + t)µ12 (x + t) dt and d p13 (x. If the forces of transition vary with age.2. pij (y. Kolmogorov’s Forward Equations (for the general case in which there are n states) give the following system for each fixed x: d pij (x.4) (17. x + t) dt µνj (x + t)piν (x. µij (y).3) (17.2. 3) gives d [p12 (x. x + t)[µ12 (x + t) + µ13 (x + t)] dt = p11 (x.2.2. x + t)µ1 (x + t) dt (Equation (17.6) This corresponds exactly to a result derived in chapter 15. i. x + t) = exp − 0 µ12 (x + r) dr In “traditional” notation. x + t) = −p11 (x. LIFE TABLES AS STOCHASTIC PROCESSES Adding equations (17.2. d p11 (x.2.2.e. i. t t (ap)x = exp[− 0 t (aµ)x+r dr] (µα + µβ ) dr] x+r x+r = exp[− 0 where µα corresponds to µ12 (x + r). x + t) = exp[− 0 µ1 (x + r) dr] (17. x) = 1. x) = 0 gives t p12 (x.2) and (17. x + t)µ1 (x + t) And hence.2. x + t) + p13 (x. x + t)] = p11 (x. using equation (17. (17.2) and solving with the initial condition p12 (x.3. this is just the familiar result that t t px = exp − 0 µx+r dr . Using the initial condition p11 (x.e. we find that Kolmogorov’s Forward Equations may be solved to give t p11 (x. α t (aq)x t = 0 α r (ap)x µx+r dr 17. x + t) = 0 p11 (x.1) with i = j = 1.4).2.) 283 (17.2.5) can be solved to give the unique solution t p11 (x. x+r x+r Returning to the differential equation (17. x + r)µ12 (x + r) dr Again this corresponds to a “traditional” formula in chapter 15.5) This is a straightforward first-order differential equation.2.2.3 Life Tables as Stochastic Processes Life tables just contain one mode of decrement (death) and hence the continuous-time model is Note that µ12 (y) = the force of mortality at age y = µy in “traditional” notation Proceeding as in section 17.17. but with only one mode of decrement. and µβ corresponds to µ13 (x + r) .5) also follows directly from equation (17. That is.1: the life table model Remark In the stochastic processes approach.z−y py 17. and for the n states in the continuous-time model.z µy νy.B. recovery at age y and duration of sickness z. y)pkj (y. x + t) = P r{ a healthy life aged x will be sick at time t}. mortality of sick lives at age y and duration of sickness z The various forces of transition have been graduated using Permanent Health Insurance data.z = = = = the the the the force force force force of of of of sickness at age y.I. z) (17. In Figure 17.284 CHAPTER 17. as the forces of recovery and mortality of sick lives depend on .3.3. mortality of healthy lives at age y. Suppose one wishes to calculate probabilities such as p12 (x. and clearly p21 (x1 . MULTIPLE-STATE MODELS Figure 17.1 there are only 2 states (alive and dead).) has used the following multiple-state model for sickness rates: Define σy py.1) for all x ≤ y ≤ z. x2 ) = 0 for all x1 ≤ x2 and p22 (x1 . x2 ) = 1 for all x1 ≤ x2 Equation (17.1) with i = 1 and j = 1 is just a re-statement of the following axiom of the traditional life table: z−x px =y−x px .M. n pij (x.3. z) = k=1 pik (x. The existence of the parameter z complicates the multiple-state model: Kolmogorov’s Forward Equations have to be modified.4 Sickness Models The Continuous Mortality Investigation Bureau (C.3. it is assumed that the Chapman–Kolmogorov equations hold. 3. . t2 . Disadvantages: 1.. a healthy life (who has not been sick since age x)..4. A pictorial representation of this is as follows: In this set-up. one can work out P r{ a healthy life aged x will be sick at age x + t} = P r{ a life in state 1 at age x will be in state j at time t} j=2. Each term may be evaluated. This leads to “semi-Markov” processes which are more difficult to handle than the cases so far discussed. a sick life (who has been sick exactly once). . . Define T1 .4. 2.17. Tj ). + Tj > t} = a multiple integral (we omit the details) Advantages of this method: 1. but T1 + . tj ) of the variable (T1 .6...4. to be the times (in years) spent as 1. T2 .. That is. a healthy life (who has been sick exactly once)... one may wish to calculate P r{ a healthy life aged x will be sick at time t and has been sick for at least d weeks (d is the deferred period)} A possible simplification One possible way to simplify this model is to initially ignore mortality and to consider only forward transitions between “healthy” and “sick” states... and so on. 2. The deferred period may be easily handled by modifying the range of the multiple integrals. T3 .. + Tj−1 < t.. using the joint density function f (t1 .1: a sickness model duration of sickness as well as age.. SICKNESS MODELS 285 Figure 17. . The mathematics is not so difficult to follow. P r{ a life aged x in state 1 will be in state j at time t} =P r{T1 + ... T2 . For example. A further complication is caused by the need (in practical applications) to consider “deferred periods”. There can be difficulties in evaluating the multiple integrals for large values of j.. 4. MULTIPLE-STATE MODELS Figure 17. . the Manchester Unity System is often used to calculate sickness functions. this will be covered in Chapter 18.2: sick and healthy states In practice.286 CHAPTER 17. Define (h) zx = P r{(x) is entitled to sickness benefit for the time period until age x + h} It is supposed that. Suppose 1 that benefit is payable in advance over intervals of length h years. each year is divided into n intervals.18 0 ¯ t px zx+t dt (18. where h = n . We also define zx = the central rate of sickness at age x = 52. uniformly on any bounded age interval.18h)zx+jh Letting n → ∞ (or h → 0+ ) gives sx = expected cash paid out in sickness benefit 1 = 52. at age x is the probability that a life aged exactly x is “sick” (according ¯ to the rules of the scheme).Chapter 18 SICKNESS FUNCTIONS 18. zx . The expected cash to be paid in sickness benefit between ages x and x + 1 for a life now aged x is n−1 j=0 (h) jh px (52. Consider these periods to be of length h years.1.g. In practice.e. i. this amounts to £52. h→0+ (h) lim zx = zx ¯ If benefit is paid at the rate of £1 per week.1 Rates of Sickness The force of sickness. sickness benefit must be claimed for short periods (e. days or weeks) during which time the life is either “sick” or “not sick”.18 1 p z ¯ 0 t x x+t 1 p dt 0 t x dt (18.1.2) 287 .1) sx is called the annual rate of sickness at age x: it is the expected number of weeks of sickness between ages x and x + 1 for a life now aged x.18 per year on average. 3) (18.1) and (18. This time-period is often 6 months or a year. Note that zx = zx ¯13 ¯0/13 = P r{(x) is “sick” and has been sick for less than 13 weeks} Also. Fuller details of this experience can be found in the Appendix to this chapter. in a D1 policy. Define zx ¯m/n = P r{(x) is “sick” and has been sick for more than m weeks but not more than m + n weeks.4) (18. which refers to the length of the deferred period in weeks. The Tables give the following values: 104/all 52/52 26/26 13/13 13 . zx ¯13/13 = P r{(x) is “sick” with duration of sickness more than 13 weeks but less than 26 weeks}.18¯x+ 1 z 2 p zx x (18. (c) The Off-Period This is the minimum time that must elapse between 2 bouts of sickness in order for them not to be considered as the same bout of sickness for benefit calculations. 1 px zx+ 2 ¯ 1 2 52. For example.5) “Formulae and Tables for Actuarial Examinations” makes use of the Manchester Unity Experience 1893-97. SICKNESS FUNCTIONS Equations (18.} Similar modifications are used to define zx m/n and sx m/n . or 2 years}. zx zx . For example.1.1. D4. (b) The Waiting Period This is the time between joining a friendly society or sickness benefit scheme and being able to claim sickness benefits.288 CHAPTER 18.2) give the approximations: sx zx and hence sx 1 2 52.1. 104/all 26/26 52/52 13/13 all 13 + zx + zx zx = zx = zx + zx + zx Some More Definitions (a) The Deferred Period This is the time between falling sick and being able to claim sickness benefits. zx . zx Furthermore. etc. .1. Occupational Groups AHJ. one encounters the notation D1. Sickness by Duration Benefits may depend on the duration of sickness. a member must be sick for 1 week before being allowed to receive sickness benefit.1. In Permanent Health Insurance. zx ¯104/all = P r{(x) is sick with duration of sickness greater than 104 weeks.18. zx . t px zx+t dt ¯ (18.2.18¯x+t+ 1 ) 2 2 x 1 = t=0 v t+ 2 .t+ 1 px zx+t 2 2.2.v.t+ 1 px zx+t 2 1 (18. i.P.1) This can be approximated by a sum. subject to a certain mortality table. there may be a temptation for people to temporarily “recover” and then to start claiming benefit again at the higher initial rate.2. and suppose that sickness benefit will be payable at the rate of £1 (per week) during all sickness within the next T years. of a sickness benefit may be evaluated by direct evaluation of a sum. we may let T → ∞. The mean present value of the sickness benefit is thus nT −1 j=0 v jh jh px (52. by approximate integration or by commutation functions.18h)zx+jh Letting n → ∞ gives T (h) m.18 0 v t . VALUING SICKNESS BENEFITS 289 If the sickness benefit falls with duration of sickness.5) . T −1 m.t+ 1 p (52.18 0 v x+t lx+t zx+t dt ¯ (18.18 0 ∞ t=0 1 v t . the off-period is 1 year.v. as a continuation of the first spell. for benefit purposes. To prevent this. of sickness benefit t=0 T −1 z v t+ 2 .e. an off-period is specified so that if 2 spells of sickness are separated by less than the off-period.2. the later spell is treated.t px zx+t dt ¯ (18. Commutation functions Define 1 Hx = 52.2 Valuing Sickness Benefits Consider a life aged x. If the sickness benefit is payable for sickness of duration greater than m weeks but less than m + n m/n m/n and zx .2. weeks.3) (18.2.p. 18. If benefits do not fall as duration of sickness increases there is no need for an off-period rule. and suppose that sickness benefit is payable in advance on sickness during any of these short-age intervals.p. of sickness benefit = 52.p. respectively.2) Notes 1. In the Manchester Unity experience.V. of sickness benefit = 52.18. Consider the age-range from x to x + T to be split into nT short intervals. replace zx and zx by zx ¯ ¯ The M. If sickness benefits continue throughout life (although this is unusual as it is not easy to define “sickness” among the very old). giving ∞ m. each of length h years.v.4) v t+ 2 . If there is a waiting period of 6 months for sickness benefits. etc.12 . etc. SICKNESS FUNCTIONS 1 ¯ 52. We also define Kx = t=0 CHAPTER 18.p.2. Commutation functions are available in “Formulae and Tables” on the following basis only: • English Life Tables. of sickness benefit of £1 per week to (x) is Kx − K65 (18. J) Temporary and Deferred benefits If sickness benefit ceases at a certain age.2.290 This can be approximated to give Hx and so Hx where Dx = lx v x .6) Dx+ 1 zx 2 (18. we obtain Kx+n Dx rather than Kx Dx This is used in connection with waiting periods.v. are defined by + Kx + Kx + Kx Kx = Kx = Kx + Kx 13 replacing zx by zx . No.4)) = = Hx+t Dx (18.. say 65.7) ∞ Hx+t (18.8) so that the m.. replace Kx by Kx+ 1 (interpolate in the Tables.9) Kx Dx Note 104/all 52/52 26/26 13/13 13 13 all 13 .) 2 .p. of a sickness benefit of £1 per week for life to (x) is approximately ∞ t=0 Dx+t+ 1 2 Dx ∞ t=0 zx+t (from (18. then the m. as usual.Males • 4% interest • Manchester Unity 1893-97 (A.2. in the formulae for Hx and Kx .10) Dx If benefits are deferred for n years. H.2.2.18v x+ 2 lx+ 2 zx+ 1 2 1 (18.v.2. where Hx and Kx . of a sickness benefit of £1 per week payable up to age 65 on sickness lasting more than 13 weeks Since Kx Kx 13/all 13/all 13/13 is not given directly in the Tables.2. VALUING SICKNESS BENEFITS 291 Example 18. l31 1 . The value is 50 1 11 21 2 2 2 1 v0.1.2. one must use + Kx 26/26 = Kx + Kx 52/52 + Kx 104/all Example 18. There is no waiting period and the off-period is as in the Tables provided. Kx 13/all − K65 Dx 13/all = m.05 l30 1 z30 + v0. Find the single premium on each of the following bases: mortality: English Life Table No.p. and Hx replace Kx and Hx by Kx For example. The benefits are £50 per week during sickness within the next three years.Males interest: (i) 4% p. to give m.05 l32 2 z32 2 2 l30 1 The value of l30 1 .01 (ii) Calculate value of benefits from first principles.v. an annuity of £8 per week payable weekly in advance for as long as he survives.05 l33 1 z31 + v0.v.12 . m/n m/n respectively.05 If sickness benefit is payable for sickness lasting more than m weeks but less than m + n weeks. No. the sum of £4.p.18. (ii) 5% p.a. (c) on sickness. Solution (i) m.v. of benefits = 50 × 249682 95265 = £131. of benefits = 50 · K30 − K33 D30 1784760 − 1706624 = 50 29372 = £133. A friendly society issued a policy providing the following benefits to a man aged exactly 25 at entry: (a) on death at any time before age 60.2.p. l32 2 can be found by interpolation using English Life Tables.000 payable immediately.. an income benefit to be payable during sickness of £32 per week for the first 6 months reducing to £16 per week for the next 18 months and to £8 per week thereafter. sickness: Manchester Unity 1893-97 (AHJ) expenses: none. Sickness . A life office is proposing to issue 3-year sickness benefit policies to lives aged 30.12 2 2 Males. (b) on survival to age 60.2.a. 1) − K65 Dx 104/all (18.92 ¨ = 1855. which is correct.2) is usually optional. Occupation Group AHJ 4% p. with all benefits ceasing at age 65. mortality: sickness: interest: expenses: English Life Table No.05 26/78 104/all 104/all (b) 52.05 + 926. 18. formula (18.65 + 660. Premiums are payable monthly in advance for at most 35 years.3 Various Other Points (a) A possible adjustment Consider a policy providing a sickness benefit of £10 per week on sickness lasting more than 2 years for an entrant aged x.92 Let P = annual premium payable monthly. There is no waiting period.3.65 + 16 K25 26/78 26 26 K25 −K60 D25 −K60 D25 +8 K25 −K60 D25 = 660. Find the monthly premium.292 CHAPTER 18.3. none Solution Value of benefits: ¯ (a) 4000A 1 = 4000 25:35 M 25 −M 60 D25 = 268. and it is concluded that neither formula is exactly right. For example when x = 63. The society uses the following basis to calculate premiums.3. The only exception (in which (18.3.18 × 8 · (c) 32 N 60 D25 = 926.1) is not) occurs when the term of the policy is very short. . and are not waived during periods of sickness.3.2) is accurate and (18.a.Males Manchester Unity Sickness Experience 1893-97. SICKNESS FUNCTIONS benefit is not payable after age 60. One might want to adjust the formula 10 to 10 Kx+2 104/all Kx 104/all − K65 Dx 104/all (18. This point is discussed in the Appendix to this Chapter.2) since the benefit cannot be received in the first 2 years.62 (12) a25:35 = a25 − ¨ ¨ (12) D60 (12) a ¨ D25 60 1 D60 = (¯25 + ) − a 24 D25 (12) a60 + ¯ 1 24 = 18. Hence P = £100. so the use of adjusted version (18.2) gives zero.495.3.12 .36. Then the equation of value is P a25:35 = 268.33 and monthly premium = £8. p. VARIOUS OTHER POINTS 293 (b) Waiver of Premium Benefits Suppose that a policy has a weekly premium of £P but this is waived during sickness (or during sickness of a certain duration).12 (Males) sickness: Manchester Unity Sickness Experience 1893-97.9 × 52.1. Occupation Group AHJ. All benefits cease at age 65. one should adjust the equation of value appropriately. plus 10% of all premiums payable after the first year. as a general rule.v.31 (c) Reserves These are calculated prospectively or retrospectively (usually the former) in a way similar to that used for life policies. No. If this is not the case.3. interest: 4% p. There is no waiting period. one must specify how to find the reserve. This is handled by assuming that all premiums are paid by the policyholder.63 + 543. Solution Let the weekly premium be P .276P K38 13/all 13/all (ii) Setting (i) = (ii) gives P (713. Example 18. A friendly society issues sickness insurance policies which provide income during periods of sickness as follows: (a) £100 per week while a sickness has duration in excess of 13 weeks but less than 1 year. Calculate the weekly premium for a life aged 38 at entry. (b) £75 per week while a sickness has duration in excess of 1 year but less than 2 years. expenses: 50% of all premiums payable in the first year. of benefits and premiums waived is 100[ K38 13/39 (i) − K65 D38 13/39 ] + 75[ K38 52/52 − K65 D38 52/52 ] + 50[ K38 104/all − K65 D38 104/all ] + P[ − K65 ] D38 = 784. If the premium and reserve bases agree.18.L.p. Premiums are payable weekly until age 65 and are waived when sickness benefit is being paid.98 − 22. expenses will apply even when premiums are waived. If the bases differ. .3.00 + P × 22.a.18P a38:1 ¯ = 713. Basis: mortality: E. of (premiums-expenses) is 0.43 Hence P = £2. the prospective and retrospective reserves are equal.v. but there is an additional sickness benefit of £P per week.276 = 1595.98P m.18P a38:27 − 0.T.43 + 22.80 + 267. m.276) = 1595. (c) £50 per week while a sickness has duration in excess of 2 years. Also.4 × 52. zx . by an argument similar to that used for life policies. Pages 86-87 give 13 13/13 26/26 52/52 104/all Dx . Kx . ceasing at age 65 (ignoring expenses). zx . . Kx . Consider a policy issued to (x) providing a sickness benefit of £10 per week ceasing at age 65.294 CHAPTER 18. Pages 84-85 give 13 Kx Kx Kx Kx Kx . Dx Dx Dx Dx Dx 13/13 26/26 52/52 104/all . (b) Calculate the reserve at duration t years by (i) the prospective method. zx .3.18P ax:65−x = 10 ¯ (b) (i) t V = 10 Kx+t −K65 Dx+t Kx −K65 Dx − 52.18P ax:t − 10 Kx −Kx+t ¯ Dx Note that (i) and (ii) are equal.2. (a) Give a formula for the weekly premium. P . zx (= zx ) 2. Pages 82-83 give 13 13/13 26/26 52/52 104/all all zx . . Kx . SICKNESS FUNCTIONS Example 18.18P ax+t:65−x−t ¯ Dx (ii) t V = Dx+t 52. Kx . Solution (a) 52. on the premium basis. (d) Understanding the Tables for Actuarial Exams 1. . and (ii) the retrospective method. zx . Kx Dx 3. Kx . 510 = 23. EXERCISES 295 Exercises 18. calculate the percentage by which the normal weekly premium (i. 425 . only half the premium (including the extra premium for the waiver benefit) is waived. no expenses.4. The policy provides the following benefits: (a) on survival to exact age 65. or on survival until the end of the term. an annuity of £5. mortality: English Life Table No. Calculate the revised premium payable assuming that the alteration basis follows the premium basis above. 12 . a man then aged exactly 30 effected an insurance policy providing sickness benefits of £100 per week for the first six months of sickness.000 per annum payable monthly in advance. Note.129 = 24. If the man has been sick for 6 months or more when a premium falls due.18.2 A policy issued by a life office to a male life aged exactly 35 is subject to level weekly premiums ceasing at exact age 65. 405 = 1. Basis: English Life Table No. a return of all premiums paid (including those waived during sickness) together with compound interest at 4% per annum to the date of death. = 554. during the second year of the policy.12-Males.Males. Occupation Group AHJ. with benefit ceasing at age 60. The sum assured under the endowment policy is payable immediately on death.12-Males. interest 4% per annum. Using the basis given below. Occupation Group AHJ. Calculate the weekly premium payable to age 50 on the following basis: Mortality: English Life Table No.1 (i) In a combined sickness and mortality table Kx+1 zx Dx Dx+1 Estimate Kx . Expenses: 10% of each premium. Interest: 4% per annum. the premium for a policy without the waiver benefit) for a life aged exactly 30 at entry should be increased in order to provide the waiver benefit. and. Manchester Unity Sickness Experience 1893-97. £50 per week for the remainder of the first year and £30 per week thereafter. interest: 4% per annum. (b) on death before age 65.e. 18. (ii) An office offers an optional waiver of premium benefit on sickness of any duration in respect of a 25-year with or without profits endowment assurance policy with weekly premiums payable for 25 years or until earlier death. expenses: 15% of each extra premium for the waiver benefit. Sickness: Manchester Unity 1893-97. (b) The man now wishes to alter his policy so that premiums will in future be waived during all periods of sickness. Occupation Group AHJ. Expenses are incurred even when premiums are waived. There is a waiting period of 12 months for the waiver of premium benefit. Calculate the weekly premium. sickness: Manchester Unity 1893-97. the premium is waived. 18.3 (a) Ten years ago. There is no waiting period and the off periods are the same as those underlying the tables in Formulae and Tables for Actuarial Examinations. waived during sickness and ceasing at age 65 or the previous death of the member. Note Use Simpson’s rule for approximate integration.T. (a) Calculate the weekly contribution payable by a new member aged 35 on the basis given below.Males with an age-deduction of 5 years interest: 4% per annum sickness: Manchester Unity 1893-97 AHJ expenses: 5% of all contributions. b b−a a+b f (t) dt [f (a) + 4f ( ) + f (b)] 6 2 a 18. 12 . there is no benefit on the death of wives of marriages taking place after entry to the society. and (iv) during any spell of sickness of the member before age 65. a lump sum of £1.296 CHAPTER 18. including those waived during sickness Wives of members are taken as being of the same age as their husbands. immediately on the death of the member’s wife before her husband.L. There is a six-month waiting period for the sickness benefit (including the premium waiver) and the off-period may be assumed to be the same as that underlying the tables in “Formulae and Tables for Actuarial Examinations”.Males Manchester Unity Sickness Experience 1893-97.50 per week thereafter. No. a lump sum of £2.T. SICKNESS FUNCTIONS 18. and the possibility of divorce is to be ignored. for an entrant aged 25.000 plus a return of contributions (including any waived) without interest. (ii) during the first 10 years of membership. an annuity of £10 per week for life.000.Males mortality of wives: E. Contributions are payable by level weekly amounts until age 60. Contributions are waived during sickness. .5 A Friendly Society issues policies providing the following benefits: (i) A sickness benefit of £25 per week for the first 13 weeks of sickness and £12. £10 per week reducing to £6 per week after 3 months’ sickness. 12 . (ii) On death before age 60. The society provides the following benefits: (i) immediately on the death of the member at any age.4 A certain friendly society recruits only married men aged under 55. Basis: English Life Table No. There is no waiting period and the off-period is as in the Tables provided. £2. (iii) on survival of the member to age 65. i. No. The possibility of withdrawal is ignored. £500.12 . (a) Calculate the weekly contribution rate. Occupation Group AHJ interest 4% per annum.e.L.000. (iii) On survival to age 60. (b) Calculate the reserve (on the basis given below) to be held for this member five years after he joins the Society. benefit ceasing at age 60. (b) Calculate the reserve for a member aged 35 who joined at age 25. expenses are 5% of all premiums (including those waived). The basis for all calculations is: mortality of members: E. 4% per annum interest. and the off-period is as in the Manchester Unity sickness tables. using the following basis: English Life Table No.T.a. Each policy provides an income of £20 per week during any period of sickness which has lasted more than 6 months.6 A friendly society provides the following benefits: 297 (i) On sickness.) Calculate the reserve which should be held for a member aged 50. There is no waiting period. £40 per week for the first 26 weeks and £50 per week for the next 26 weeks. No sickness benefit is payable after age 65. Members contribute £1 per week. 12 .18.12 . EXERCISES 18. (ii) On attaining age 65 or immediately on earlier death.000.Males Interest: 4% p. ceasing at age 65 or earlier death.4. 18.7 On 1 January 1995 a friendly society had in force a large group of policies providing certain sickness and life assurance benefits. and off-periods are as in “Formulae and Tables for Actuarial Examinations”. These policies had all been issued 10 years ago to lives now aged 45.L. . The basis for the calculation of premiums and reserves is: Mortality: E. Sickness: Manchester Unity Experience 1893-97 (Occupation Group AHJ) Expenses: 20% of all premiums. but this is waived during periods of sickness (whether benefit is payable or not. (b) Calculate the reserve per policy in force on 1 January 1995.000 immediately on death before age 65. Premiums are waived during sickness (even when no benefit is payable). and the policies have level weekly premiums payable throughout the term of the policy. Manchester Unity Sickness Experience 1893-97 (AHJ).Males. the sum of £3. including those waived (a) Calculate the weekly premium for each of these policies. and a lump sum of £20. The benefits cease when the policyholders reach age 65. 18.55 + 203. Note One does not need to calculate the actual premium P in this example. .81 Therefore k = 0. 464 (ii) Let weekly premium for the basic policy be P .07k = (1 + k)20. Hence 690.P be the extra premium for the waiver benefit.11%. The equation of value is 1 (K31 − K32 ) + (K32 − K55 ) 52. The equation of value is 26 26 K30 − K60 26/26 26/26 52.298 CHAPTER 18.18 × 0. The equation of value is 52.3 (a) Let weekly premium be P .9P a30:20 = 100 ¯ + 50 K30 − K60 D30 + 30 K30 52/all − K60 D30 52/all Hence 640.8815P Hence P = £15.35 + 125. 18.t p35 µ35+t (52.¯35:30 = 5000 a D65 (12) + a ¨ D35 65 26/all 30 + P K35 − K65 0 26/all v t .zx + Kx+1 2 1 (Dx + Dx+1 )zx + Kx+1 2 = 581.85kP a30:25 = P (1 + k) 2 ¯ D30 (note the waiting period of 1 year).18 × 0.0311 = 3.18P st| ) dt ¯¯ /Dx (i) 30 Note that 0 v t .18P.1 (i) Kx = Hx + Kx+1 Dx+ 1 .18P 0 1 − vt δ ¯ −A1 t p35 µ35+t dt = 52. 994.25 per week.74 and so P = £3.40.51P = 1711.2 Let weekly premium be P .21 + 242.18P δ 30 q35 35:30 Hence (i) gives 869. SICKNESS FUNCTIONS Solutions 18.t p35 µ35+t (52.26P + 16.13P = 9. and let k.18P st| ) dt ¯¯ 30 = 52. 12.88 + 62. m.4 (a) Let weekly contribution be P .43) D35 D35 13 13 K35 − K65 +4 − 0.008638 = 4.95 × 52. 500 0 v t .115.18P a35:25 = 1.18¯35:30 a D35 = 537.54P = 1329.37 per week.18 × 10 + (6 + 1.853P ¯ Hence 927.18P (I A) 1 35:25 35:25 D60 + 2000 + (12.p.46 and so.28 +4 13 13 K25 −K65 D25 K25 −K65 D25 = 254.28 + 254.5 + P ) D35 13 13 K35 1 − K60 2 + 12.18 · (iv) (6 + P ) D65 ¯ D25 a65 = 693. SOLUTIONS (b) Let extra premium be E per week.319 (using Simpson’s Rule) (iii) 10 × 52. (b) 10 V ¯ N65 K35 − K65 ¯ = 2000A35 + 52.64 (ii) 10 299 say £0. 18.18 × 0. 18. The equation of value is ¯ ¯¯ 0.5 D35 Use.67E and so E = 0.43 × 52.43.t p20 µ20+t .38 + 1038. ¯¯ (I A) 1 and find K 35 1 2 35:25 K35 1 − K60 2 D (i) ¯ (I A) 1 35:25 1¯ − A1 2 35:25 and 13 K35 1 2 by linear interpolation.71 = £728. .95P a25:40 = 377.95 × 1.9E¯40:10 a D40 Therefore (E + 3. P = £1.25) × 13. The equation of value for E is K40 − K50 (E + 3.5. Hence revised premium is £3.18 + 270.319 + 693.18 × 0.5 (a) Let the weekly contribution be P . benefits: ¯ (i) 2000A25 = 377.t p25 dt 500 × 0.64 + 4.55 − 1180.25) = 52.18.853P Hence the equation of value is 52.23 + 31.v. 000A 1 ¯ + 52.077 = 382.23 + 31.28 The reserve may also be calculated by the retrospective method. 44 + 337.46P Hence P = £4.30P = 2472.53 − 1222.68 = £374.84 − 542.18P a35:30 = 20000A1 ¯ + 20 D35 35:30 +P K35 − K65 D35 This gives 695.06 + 59.5 40:20 K40 − K60 D40 13 13 K40 − K60 D40 (Prospectively) Using P = 1.8 × 52.35 + 113.25 18.5 + P )28. 26 26 K50 − K65 Reserve = 40 D50 ¯ + 3000 A1 + 50:15 + 50 K50 26/26 − K65 D50 26/26 + K50 − K65 D50 D65 − 52.99 + 768.066 + 165. SICKNESS FUNCTIONS Equation (i) gives 759.57 + (12.86 per week (b) 5V ¯ = (1000 + 5 × 52.88 = £2163.6 Using the prospective method.18P )A1 D60 + 2000 + (12.11 + 413.92 and so P = £1.18P a40:20 ¯ 40:20 ¯¯ + (52.60 .36 + 39.85 + 209.42 − 2285.300 CHAPTER 18.7 (a) Let the weekly premium be P .36 18. (b) Using the prospective method 10 V ¯ = 20000A1 + 20 K45 26/all 45:20 − K65 D45 26/all +P K45 − K65 D45 − 0.8 × 52.95 × 52.18¯50:15 a D50 = 681. this gives 10 V = 3176.27 per week.5 + P ) D40 − 0.27.08 + 1775.63 + 36. this gives 5 V = 142.91 = £1475.18 × 1.38P + 625.82 + 416.27 + 168.28 + 71. The equation of value is 26/all 26/all K35 − K65 ¯ 0.52P = 86.86)(I A)1 + 12.86.18P a45:20 ¯ With P = 4. an association of lodges with some measure of central control. zx . 1911). giving zx . The off-period assumed in the investigation . From 1911 to 1946. A part of their income was provided by a Government grant. (The statistical basis of the graduation of sickness rates is more complex than for mortality rates. in modern conditions: under the Friendly Societies Act 1992. and others concentrate on social activities. i. 3. G. and indeed prosper.H. of weeks of sickness at age x last birthday c Ex These rates were graduated by an adjusted-average formula to produce the published rates. zx . Manchester Unity (a large “affiliated order” friendly society.) There was (and to some extent still is) a wide variety of societies ranging in size and financial stability. A calendar year system was used. Since 1946 the State has provided welfare benefits (including sickness benefits) directly.e. usually dental and ophthalmic) benefits. it being found that sickness rates were lowest in the first group and increased as one moved from one group to the . Friendly societies existed largely to protect wage-earners in times of adversity. Watson. there being no State benefits before 1908 (unless one counts the Workmen’s Compensation Acts of 1897. and published in 1903. 4.5. zx was 1 year. zx .18. the principal published set of sickness rates were those derived from the experience in 1893-97 of the Independent Order of Oddfellows. Until fairly recently. BCD.I. either in return for national insurance contributions or under a means test. EF. sickness benefits for wage-earners were provided by Approved Societies (friendly societies or life offices approved under the National Insurance Act.) The rates were also subdivided by periods of 104/all 52/52 26/26 13/13 13 . they are able to conduct a greater variety of business.M. and they had to use a valuation basis specified by the Government Actuary. this being the actual off-period for most of the lodges. zx sickness. SOLUTIONS 301 Appendix: The Manchester Unity Experience 1893-97 1. and not all societies provided sickness benefits.I. The friendly societies have generally declined in importance. and we do not attempt a discussion. A certain number (particularly of centralized societies) continue. (Permanent Health Insurance). when the C. published statistics concerning P. lives being classified according to age nearest birthday at the start of the calendar year.) The members were all male. The unadjusted sickness rates are of the form zx = ˆ no. 2. Societies with favourable experience were able to provide additional (in practice. many have disappeared. which covered injuries at work. The Manchester Unity 1893-97 investigation was carried out by Alfred W. Members were also subdivided according to the following occupational groups: A: Agricultural Workers B: Outdoor Tradesmen and Labourers C: Railwaymen D: Seamen and Fishermen E: Quarry Workers F: Iron and Steel Workers G: Miners H: Rural workers not included in A-G J: Urban workers not included in A-G Sickness rates were calculated for the four groups AHJ. sickness rates should be “select”.. as used in Tables for Actuarial Examinations. excluding those in the more hazardous occupations. which are now valued as 10[Dx+1.) in which the exposed to risk included some recent entrants who could not claim “26/26” benefit because they had not been eligible for sickness 26/26 benefit for more than 6 months. It is therefore better to use the formula + Dx+1. The sickness rates of M.e. subject to a waiting period of 6 months. with separate tables for 3 geographical areas and for urban and rural areas. sickness rates were high because the jobs demanded a high level of physical fitness. (whole society) were found to be remarkably similar to those for employed men in the national insurance scheme in 1953-58. the “26/26” benefits.302 CHAPTER 18.5Dx+0.U.5. this represents the experience of wage-earners in 1893-97.) 5. Since the middle classes and well-to-do did not (in general) join friendly societies providing sickness benefits. except perhaps when benefits cease soon after the entry age. for example. (In some cases. so formula (ii) understates the expected value of the benefits. 7. This argument leads us to use sickness rates without deferment (except for the waiting period). Mortality investigations were also conducted.. Strictly speaking. sickness rates with a more modern mortality table (e.5 zx+2 + . But when comparing the actual weeks of sickness claim at the later durations in a given experience with that expected on the basis of Manchester Unity (say). zx+2 should be slightly increased. the uncertain definition of “sickness” and other factors are likely to be of much greater importance in the estimation of sickness rates.5 zx+1 + Dx+2. 12Males.. zx+1 should be slightly adjusted upwards relative 26/26 to that of M. since new entrants are not normally accepted unless they are reasonably healthy. they should depend on duration of membership as well as on attained age.5 zx 26/26 26/26 26/26 26/26 26/26 . But formula (ii) is not quite right. Similarly. one must bear in mind any differences in the proportions of members whose durations of membership are so short that they cannot claim. In practice. English Life Table No. The experience of occupation group AHJ is given in Tables for Actuarial Examinations.g. That is. i. an experience consisting mainly of recent entrants may be expected to have low rates of sickness at the later durations. Another point is illustrated by the example of a new entrant aged x who is to receive £20 per week for the first 26 weeks of sickness. The value of the benefits might be adjusted from 26 20Kx+ 1 + 10Kx+ 1 + 5Kx+ 1 2 2 2 26/26 52/all Dx to 26 20Kx+ 1 + 10Kx+1 + 5Kx+1. however. as in formula (i).5 zx+1 + .U.) 6. 10[0..5 2 (i) 26/26 52/all (ii) Dx to allow for the fact that no “26/26” benefit can be received before age x + 1 and no “52/all” benefit can be received before age x + 1.U. consider.] Dx Now zx+1 is based on an experience (M. but these have long been out of date. It is usual to combine the M. SICKNESS FUNCTIONS next.U. Thus. for example. £10 for the second 26 weeks and £5 per week for the remainder of sickness.] Dx since the “extra” first term allows approximately for the fact that the later terms are too low. Obviously.benefit schemes These are pension schemes whereby the pension and other benefits are set out in the rules of the scheme. 19.contribution schemes These are also called “money purchase” schemes.2 Valuation Principles Only defined-benefit schemes shall be considered.benefit schemes. which may vary from time to time according to actuarial advice. Remarks 1. a fixed percentage of salary) and (ii) the employer’s contributions. 1 in a typical U.contribution schemes. the annual pension depends on the rate of return obtained on the employer’s and member’s contributions and the terms on which pension may be purchased at retirement (unless pension is purchased directly when the contributions are received). the annual pension is 80 × the final salary per year of service. There may be various other benefits such as death benefits and a lump sum on retirement.K. Most schemes of this type provide benefits which depend on the ‘final salary’. the contributions of both employees and employer are fixed (often as percentages of the salary). For example. Defined .K. 2. (“Active” refers to a 303 . The pension benefit is what these combined contributions can buy. public sector final salary scheme. A “non-contributory” pension scheme is one in which the employees do not contribute (e. or (b) defined . usually after investment (either directly by the scheme or by life office contracts) until retirement age. civil service). U. and the mean present value of the future benefits and future contributions of an “active” member aged x will be calculated. Defined .Chapter 19 PENSION FUNDS 19. The benefits are paid for by a combination of (i) the employee’s contributions (e.1 General Introduction Pension schemes may be described (in broad terms) as either: (a) defined .g.g. An “insured” pension scheme is one in which benefits are secured by contracts with a life office. as in “Formulae and Tables in Actuarial Examinations”: lx = number of members at exact age x dx = number of deaths at age x last birthday ix = number of “ill-health” retirements at age x last birthday rx = number of “age” retirements at age x last birthday wx = number of withdrawals at age x last birthday Notes 1. We may also have to deal with benefits for people with “deferred” pensions in the scheme who no longer work for the company. 19. etc. and all members must retire at age 65 at the latest. Expenses are usually ignored. In the “Formulae and Tables”.1) sx = x sy dy = ¯ 0 sx+t dt ¯ (19. a salary scale is required to estimate future salaries from current salaries. reserve = mean present value of future benefits .4. the salary scale function. PENSION FUNDS member who has not yet retired). 3. i per annum. i is called the valuation interest rate.mean present value of future contributions (of both employee and employer) Notes 1. as they are paid for separately by the employer.4 Salary Scales Assume that salaries are revised continuously.2) . 5. or both. The reserve for each member is calculated prospectively.) 19. 2. ill-health retirement. “Age” retirements may be concentrated at an exact age (for example 60 or 65) or may be spread uniformly between ages 60 and 65. The rate of interest.4. so a mortality table for them is sometimes required (see later. Define {¯x }. A service table is required.304 CHAPTER 19. 4. If benefits or contributions depend on the employee’s salary. used in valuing the benefits and contributions is normally gross (free of tax). Members often have to retire by a certain age (often 65). Sometimes one may require to find the employer’s contribution rate by setting the reserve equal to zero at entry to the scheme of a new entrant.3 Service Tables The following notation is used. “Age” retirements are the retirements at or above the minimum normal retirement age (NRA) of the scheme. or for a group of new entrants. to be such that the salary rate per annum at age x + t of a s life now aged x with current salary rate £(SAL) per annum is (SAL) × Also define x+1 1 sx+t ¯ sx ¯ (19. That is. 3. This is a multiple-decrement table with various modes of decrement: death. “age” retirements occur only from age 60 onwards. 2. To calculate the value of benefits one may require separate mortality tables for “age” retirement pensioners and ill-health retirement pensioners. 4) Note To estimate s65 .192 in the coming year. (vi) the total amount earned over his last year of life.80 = £24. between ages x − 1 and x). (b) Calculate revised answers for (i) to (vi) assuming that the person aged exactly 25 is expected to earn £9.e.4. so we find sx by linear interpolation: ¯ ¯ sx ¯ 1 [sx−1 + sx ] 2 sx− 1 2 (19. Solution (a) (i) 9192 · s53 ¯ s25 ¯ (ii) (iii) 9192 s25 s64 ¯ 9192 s25 ¯ 2 9192 1 s24 1 s52 2 2 = 9192 × 4. If SAL refers to the expected earnings in the coming year (i.4. (iv) his annual salary rate at exact age 65.77 1.4. between ages x and x + 1).4.5) where SAL is the current salary rate (in £) earned by (x). adjust the denominator from sx to sx−1 . determine the average values of (v) his annual salary rate at the moment of death. If he dies at age 57 last birthday. s60 +s61 +s62 +s63 +s64 5 . is estimated as: (SAL) sx+t sx ¯ (19.19. one must use s64 1 ¯ ¯ 2 s64 and s63 1 ¯ 2 s63 and then employ linear extrapolation: see Example 19. adjust the denominator from sx to sx . 359 9192 s24 1 s64 = £27.e.2). interest. Assume that salaries are revised continuously.4. ¯ Example 19. We also find that: sx+t ¯ = assumed salary rate per annum sx ¯ at exact age x + t (SAL) (19. (a) Consider a person now aged exactly 25 whose annual salary rate is currently £9.1. ¯ 2. (iii) the average amount earned by him each year between exact ages 60 and 65.a.6) Adjustments 1. Estimate (i) his annual salary rate at exact age 53.4. The salary to be earned by (x) between ages x + t and x + t + 1. 576 £27. and use the pension table in “Formulae and Tables for Actuarial Exams” with 4% p. SALARY SCALES Using approximate integration in (19. we have sx sx+ 1 ¯ 2 1 [¯x + sx+1 ] s ¯ 2 305 (19. 198. given that he is an active member during this age-interval.4. (ii) his earnings between exact ages 64 and 65.1 below. If SAL refers to the earnings received in the past year (i.3) “Formulae and Tables” gives values of sx (not sx ).192.4. (b) contributions are at a fixed rate of £F per annum (payable continuously).1) (Note the summation is up to “64 − x”). (as previously defined) .962567. 544 (iv) £26.306 ¯ (iv) Annual salary rate at 64 is 9192 s64 s25 ¯ s64 1 ¯ CHAPTER 19. 775 Estimation of sx and sx ¯ sx is usually estimated by a product of 2 factors. The inflation factor is usually of the form (1 + e)x . 499) = £27. (iii) £26.5. one to allow for future inflationary increases and ¯ one to take account of “career progression”. PENSION FUNDS 27. 738. It is often assumed that i−e the “real” rate of interest (assuming wage inflation and price inflation are about the same) 2% (or some similar figure) where i is the valuation interest rate and future salary inflation is assumed to be e per annum. 19. 499 1 Annual salary rate at 64 2 is 9192 s252 27. 891 s25 ¯ (vi) 9192 s56 1 £25. (a) The mean present value of the future contributions (of employee. employer or both) at rate k% of salary for a member age x with current salary rate of £SAL per annum is k SAL · 100 sx ¯ This is usually approximated to k SAL · 100 sx ¯ 64−x 65−x 0 vt lx+t sx+t dt ¯ lx v t+ 2 t=0 1 lx+t+ 1 2 lx sx+t (using sx+ 2 ¯ 1 sx ) (19. 576 ¯ Hence by linear extrapolation the annual salary rate at age 65 is approximately 27. 576 − 27. 922 s24 1 2 s25 0. 576 + (27. s25 ¯ 2 (b) Adjusting the denominator from s25 to s25 results in multiplying each answer by a factor of ¯ s25 ¯ s25 This gives answers of: (i) £23. 447 (ii) £26. Now define Dx = v x lx .5 The Value of Future Contributions There are 2 main ways of calculating contributions: (a) contributions are at the rate of k% of salary. where e is the assumed annual rate of future salary escalation. 653 (v) On average he will die at exact age 57 1 . 180 (vi) £24. Hence answer is 2 9192 · s57 1 ¯ 2 s25 ¯ 9192 · s57 = £25. 618 (v) £24. change sx to sx−1 in the denominator. . THE VALUE OF FUTURE CONTRIBUTIONS 1 ¯ Dx = 0 Dx+t dt Dx+ 1 2 s ¯ ¯ Dx = sx Dx and 64−x ¯ s ¯ Nx = t=0 s Dx+t 1 2 [Dx 307 + Dx+1 ] One can now evaluate (19. Consider a life aged 35 with current salary rate £10.2) Example 19. change sx to sx in the denominator. 100 sx Dx ¯ So.p. Solution m. 224 = £9520. calculate the mean present value of the employee’s future contributions. of contributions is 5 10000 s ¯ · · N35 100 s35 D35 ¯ 500 × 417.lx+t+ 1 2 v x lx Dx+t+ 1 2 Dx s 1 sx+t k SAL 100 sx ¯ k SAL 100 sx Dx ¯ 64−x t=0 64−x sx+t ¯ Dx+t t=0 = k SAL s ¯ · N x.v.000 who contributes 5% of his salary to a pension scheme. ¯ ¯ 2. of future contributions is k SAL s ¯ · Nx 100 sx Dx ¯ Notes 1. the m.p. (19.v.1.5.08) × 7232 1 2 (2. If SAL refers to the past year.5. giving a mean present value of k SAL · 100 sx ¯ = 64−x t=0 v x+t+ 2 . + 3.5.19.5.98 (b) Suppose contributions are independent of salary and that these are a fixed annual sum of £F.1) by means of commutation functions. If SAL refers to the coming year. Using the Examination Tables. Employees contribute to a pension scheme at a rate of 6% of “pensionable salary”. 606.500. where pensionable salary is defined as “salary . of these contributions is k SAL · 100 sx ¯ s ¯ Nx Dx − k ·A 100 ¯ Nx Dx (19.p. 377.3. 096. If a life aged 35 contributes £500 each year to his pension scheme.5. Solution ¯ 15500 × 0.v. D35 Suppose now that contributions are k% of the “pensionable salary”. Example 19.p. Then the mean present value of future contributions is 65−x F 0 vt lx+t dt lx 1 64−x F t=0 64−x v t+ 2 lx+t+ 1 2 lx =F t=0 Dx+t+ 1 2 Dx 64−x F· =F ¯ where Nx = 64−x t=0 1 Dx ¯ Nx Dx ¯ Dx+t t=0 (19.5.v.308 CHAPTER 19.5.3) ¯ Dx+t .5 = £16.06 s29 1 · D30 D30 2 = 18. 771 m.v. 19. SAL. Assuming that the current salary rate. = 500 · ¯ N35 = £7.2. PENSION FUNDS payable continuously. Solution m. calculate the value of his future contributions.p.A”. (b) A fraction of the average salary per year of service. where “pensionable salary” equals actual annual salary rate less £2000 (to allow for other pension arrangements). Find the present value of the future contributions payable by a member aged exactly 30 whose current annual salary rate is £15.6 The Value of Pension Benefits Pension benefits are usually of one of the following forms: (a) A fixed pension of £P per year of service.7 − 1. the m.06 s ¯ N30 · N30 − 2000 × 0.5. A being a fixed amount. .4) Example 19. already exceeds A. in this case the F.P. is (t + 1 )P if retirement occurs (due to ill-health) at age x + t + 1 .v.P. is nP .7. including fractions pro rata. which is the pension earned in respect of past service with the employer.7 Fixed Pension Schemes Consider a fixed pension of £P per year of service. 2. Ill-health retirements The m.1) Define the following commutation functions: ia Cx = v x+ 2 ix ai 1 .).v.7.S.S. 3. is 64−x nP t=0 v t+ 2 1 ix+t i a ¯ 1 + P lx x+t+ 2 64−x v t+ 2 t=0 1 ix+t 1 (t + )¯i a 1 lx 2 x+t+ 2 (19. and is possibly subject to a guarantee that at least 5 years’ payments will be made. 2 2 So the m. Similarly. of the benefit is 64−x v t+ 2 t=0 1 1 ix+t (n + t + )P ai ¯x+t+ 1 2 lx 2 This can be divided into 2 terms: 1. the benefit on age retirement at age 65 is annual pension × ar ¯65 where r indicates age retirement mortality rates. ai refers to a life retiring at age x due to ill-health.S. for a life now aged x with n years’ past service. (12) (12) ar = a5 + 5 |¨65 ¯65 ¨ a on a suitable mortality table.p.). e.19. The benefits on the date of age retirement or ill-health retirement are valued by multiplying the annual pension by an appropriate annuity function. which is the pension which will be earned in future. 2 64−x ¯ ia Rx = t=0 ¯ ia Mx+t . For example. Ill-health and age retirements will be considered separately. ¯65 monthly in advance. the benefit may be payable in various ways. Years of service normally include fractions counting pro rata. The value of the Past Service Pension (P.S. FIXED PENSION SCHEMES (c) A fraction of the final salary per year of service. ¯x+ 2 1 64−x ia Mx = t=0 ia Cx+t ¯ ia Mx and 1 ia ia = Mx − Cx . 2.g.P. Ill-health retirement mortality is ¯x usually heavier than that for age retirements. The value of the Future Service Pension (F. 19. in this case the P.P. Although the symbol ar is usually used.p. 309 Notes 1. In this example. + C64 ) 2 1 ia ia ia + ( Cx+1 + Cx+2 + . + C64 ) 2 + . + M64 1 ia ia ia = ( Cx + Cx+1 + .v... but with a final term corresponding to age retirement at exact age 65.2) Age retirements The valuation of benefits caused by age retirements is very similar to that for ill-health retirements. + (64 − x + )C64 2 2 2 64−x = 1 ia (t + )Cx+t .7. of age retirement benefits is 64−x v t+ 2 t=0 1 rx+t lx 1 r65 (n + t + )P ar ¯x+t+ 1 + v 65−x (n + 65 − x)P ar ¯65 2 2 lx (19.310 It follows that ¯ ia Rx = Proof. 1 ia + C64 2 1 ia 1 ia 1 = Cx + 1 Cx+1 + .1). 2 t=0 These commutation functions can now be used to calculate (19.Mx + Rx Dx Hence the value of all ill-health pension benefits is P (19. is 64−x nP t=0 v x+t+ 2 ix+i ai ¯x+t+ 1 2 v x lx 1 Dx 64−x ia Cx+t t=0 1 = nP · = nP The value of the F.3) .7.P. PENSION FUNDS 64−x 1 ia (t + )Cx+t 2 t=0 ¯ ia ¯ ia ¯ ia ¯ ia Rx = Mx + Mx+1 + .. The m.S.p.. CHAPTER 19.... is 64−x ia Mx Dx P t=0 v x+t+ 2 ix+t 1 (t + )¯i a 1 xl v x 2 x+t+ 2 1 Dx 64−x 1 =P· =P 1 ia (t + )Cx+t 2 t=0 ¯ ia Rx Dx ia ¯ ia n..S. The value of the P.P.7... P.P.) 60 1 + × (total future salary).8. x < 65 ¯x+ ra 2 Cx = 65 v r65 ar ¯65 . 60 Suppose that the annual pension on retirement is . T.S.Mx (i+r)a ¯ (i+r)a + Rx Dx (19. terms.P.Mx + Rx . x = 65 ra Mx = 65−x t=0 ra Cx+t (note the summation is up to “65 − x”) 311 1 ra ra ¯ ra Mx = Mx − 2 Cx and 64−x 64−x ¯ ra Rx = t=0 ¯ ra Mx+t = 1 ra ra (t + )Cx+t + (65 − x)C65 2 t=0 (using a very similar proof to that for ill-health retirements). then the So if one defines Mx value of a fixed pension of £P per annum payable for any retirement is P n. this may be separated into the P.S. Dx ia ra = Mx + Mx . 19.7. is 1 nP · Dx = nP · The value of the F.S. and the F. Then the value of the P. and similarly for other commutation functions. Define the commutations 1 v x+ 2 rx ar 1 .P. AVERAGE SALARY SCHEMES Again.19.P.4) for a member aged x with n years’ past service.S.S.8 Average Salary Schemes 1 × (total salary in service of company) 60 1 = × (total past salary. is P· 1 Dx 64−x 1 1 (t + )v x+t+ 2 rx+t ar ¯x+t+ 1 + (65 − x)v 65 r65 ar ¯65 2 2 t=0 64−x ¯65 v x+t+ 2 rx+t ar ¯x+t+ 1 + v 65 r65 ar t=0 ra Mx 2 1 Dx =P· =P 1 Dx ¯ ra Rx Dx 64−x 1 ra ra (t + )Cx+t + (65 − x)C65 2 t=0 Hence the value of all age retirement pension benefits is P (i+r)a ra ¯ ra n. Mx m..2). Value of P. is SAL 60¯x Dx s = 64−x ¯ ia sx+t Mx+t t=0 64−x s t=0 SAL 60¯x Dx s ¯ ia M x+t SAL s ¯ ia = · Rx 60¯x Dx s Consider age retirements now.v. PENSION FUNDS Thus the benefits can be separated into the Past Service Pension and the Future Service Pension.8.P. The m. of benefits is therefore 1 60 = 64−x 1 2 is v t+ 2 t=0 1 1 ix+t SAL (sx + sx+1 + ..8.1) 60 Dx Value of F.v... (i+r)a T.8.S. + C64 ) 2 1 ia + . + C64 ) 2 1 ia ia ia + sx+1 ( Cx+1 + Cx+2 + .P. + s64 M64 x x+1 So the value of the F..p. Consider ill-health retirement first. using the functions defined in the previous section. The salary to be earned between ages x + t and x + t + estimated as 1 sx+t (SAL) 2 sx ¯ The m. + sx+t−1 + sx+t )¯i ax+t+ 1 2 lx sx ¯ 2 (19..P.p..2) can be re-written as 1 ia ia ia sx ( Cx + Cx+1 + .S..P. with all ‘i’ terms being replaced by ‘r’ terms and with an extra term relating to retirements at exactly age 65.. this means that a final term of SAL ra · (sx + sx+1 + .. of the future service pension is very similar to the summation in (19.8.. + s64 )C65 60¯x Dx s ..S.P.312 CHAPTER 19.S. + sx+t−1 + sx+t )Cx+t 2 t=0 s Define the commutation functions and s ¯ ia ¯ ia Mx = sx Mx 64−x s t=0 ¯ ia Rx = ¯ ia M x+t The ‘summation’ in (19.p..) so. 1 This is just the value of a fixed pension of 60 (T. of past service benefits = (19.v.s64 C 2 64 ¯ ia ¯ ¯ = sx M ia + sx+1 M ia + .S.2) SAL 60¯x Dx s 64−x 1 ia (sx + sx+1 + .. 0. Solution Let k% be the contribution rate. On defining the commutation functions 64−x s ¯ ra s ¯ ra ¯ ¯ Mx = sx M ra and s Rra = M x x t=0 x+t 313 and using a similar argument to that used for ill-health retirements. m In “Formulae and Tables”. 80 Suppose that the annual pension is given by the formula “Final salary” is usually defined as the average annual salary in the last m years of service. + sx−1 ).1) .S. is found to be SAL s ¯ ra · Rx 60¯x Dx s Thus the value of all benefits for an average salary scheme is TPS 60 Mx Dx (i+r)a + SAL 60¯x s s ¯ (i+r)a Rx Dx (19.P.. of benefits for a member aged x with n years past service is 64−x t=0 1 ix+t 1 SAL zx+t+ 1 (n + t + )v t+ 2 ai ¯ 1 2 80¯x s 2 lx x+t+ 2 (19.p. and hence zx = 1 (sx−3 + sx−2 + sx−1 ) 3 The method of valuing benefits is similar to that in section 19. calculate the appropriate contribution rate (as a percentage of salary) for a new entrant aged 20. Ill-health retirements The m. If contributions are paid entirely by the employer. FINAL SALARY SCHEMES has to be added.6s N20 19.9. k = ¯ (i+r)a R20 ¯ = 5.9 Final Salary Schemes 1 × “final salary” per year of service. m = 3. Then k must solve k SAL · 100 s20 ¯ s s ¯ N20 D20 = SAL 60¯20 s s ¯ (i+r)a R20 D20 Hence.v. the value of the F.19.8.8. A pension scheme provides each member who retires (whether for “age” or “ill1 health” reasons) with an annual pension of 60 th of his average annual income over a member’s entire service. for each year of service.1.9. allowing for salary factors. Define 1 zx = (sx−m + sx−m+1 + ..25%. Fractions of years of service are included when calculating the amount of pension payable.3) Example 19.7. following the usual steps. (19. The new commutation functions are z ra Cx = ra 1 zx+ 2 Cx ra z65 C65 if x < 65. 65−x z ra Mx = t=0 z z C ra (note the upper limit of 65 − x) x+t 1 ¯ ra Mx = z M ra − z C ra x 2 x 64−x z ¯ ra Rx = t=0 z ¯ ra M x+t We find that the m.p.2) Age retirements Again this is similar to the fixed pension case.9.1) becomes SAL 80¯x Dx s 64−x 1 ia (n + t + )z Cx+t 2 t=0 and.314 Define the commutation functions z ia ia Cx = zx+ 1 Cx . of benefits due to age retirements is SAL 80¯x Dx s = 64−x 1 ra ra (n + t + )z Cx+t + (n + 65 − x)z C65 2 t=0 (19.4) .7) Expression (19. x+t 1 ¯ ia Mx = z M ia − z C ia . we find that the m.z Mx + · 80¯x Dx s 80¯x Dx s This shows the terms for past and future service pensions separately. if x = 65. but with salary factors inserted. 2 64−x z ia Mx = t=0 z z CHAPTER 19. x 2 x z and z 64−x ¯ ia Rx = t=0 ¯ ia M x+t = 64−x 1 (t + )z C ia x+t 2 t=0 (note similarity to definitions in section 19.v. of the benefits is equal to ia ¯ ia SAL z Rx SAL n.9.z M (i+r)a + z Rx x 80¯x Dx s (19.3) ra ¯ ra SAL z Rx SAL n.9.z Mx + 80¯x Dx s 80¯x Dx s Thus the value of all pension benefits for a final salary scheme is SAL ¯ (i+r)a n.9. PENSION FUNDS C ia .p.v. as the latter refers to average if m = 1.9.9. Give separately the values of the past-service and future-service benefits.z Mx 2 in formula (19. 071 (SAL)i z ¯ (i+r)a 1 · i=1 · R45 100 s45 D45 ¯ 1 41000 ¯ ia ¯ ra (z R45 + z R45 ) 100 1 (s44 + s45 )D45 2 3 Value of F. If there is no past service. 3. Solution 1 (i+r)a i=1 ni (SAL)i z · M 45 100 s45 D45 ¯ ia 1 490000(z M45 + z M ra ) 45 1 100 (s44 + s45 )D45 2 3 Value of P. So one must substitute 2 1 (i+r)a (i+r)a (n − ). It may be the case that the number of years of service are restricted to. of 100 th of the average salary in the final 3 years before retirement for each year of service.z Mx + z Rx 80¯x Dx s 80 n. 4.4) (even when n = 0). FINAL SALARY SCHEMES 315 Example 19. (In calculations these points are sometimes ignored.v.P. If “final salary” means the salary rate at the date of retirement. 929.9.000 5 years For these members find the present value of a pension. payable on age-retirement or on ill-health 1 retirement. then we change zx to sx . = = 42. There are 2 cases to consider: . it is usually sufficiently accurate to adjust the n years’ past service to n − 1 . including fractions. Hence total benefits have m. say. of the pension benefit in a final salary scheme is now SAL (i+r)a ¯ (i+r)a − A n. assuming a latest retirement age of 65.Mx (i+r)a ¯ (i+r)a − Rx Dx 5. This will affect lives joining at ages under 25.) 2.p.S. Remarks 1.000 10 years C : £14.000. the factor zx should be adjusted to allow for the actual past salary progression. Also. the final salary must be redefined as the average over a shorter period. Benefits may depend on a “pensionable salary” equal to salary minus some fixed sum.19.v. If so. If retirement is possible within m years. then the m. zx = sx−1 . If fractions of a year are not included when calculating the pension value. But one must not change Rx to Rx salary schemes. Three members of a pension scheme whose age nearest birthday is 45 have the following annual rates of salary and exact numbers of years of past service: A : £15. ¯ s ¯ (i+r)a z ¯ (i+r)a . say £A.P. £72.z Mx for n. consider the member 2 as having “− 1 ” year past service.S. if someone retires with less than m years’ service.1. = = 29.000 20 years B : £12.p. Also. 40 for pension purposes. zero. now aged 47 exactly has 14 years of past service and expects to earn £80.000 over the coming year.S. upon retirement due to age between the ages of 60 and 65. = ¯ ra ¯ ra 80000 z R47 − z R63 · 45 s47 D47 (notice that £80. Define the commutation functions i Cx = ix v x+ 2 z i i 1 Cx = zx+ 2 Cx . service is restricted to a maximum of 30 years for pension purposes.S. Hence.10.v. then a lump sum of 3P on retirement has m.10 Lump Sums on Retirement Suppose there is a cash payment on retirement of 3 times the annual pension.6) 80 sx Dx ¯ 1 Example 19.2) Dx .2. Using the symbols defined in the Formulae and Tables for Actuarial Examinations. 19. then leave the P. the value of the F. i+r ¯ i+r n. changing the annuity factor in all commutation functions (both “age” or “ill-health”) to 1.P. A director.P.v. In the final salary scheme mentioned above. what is the expected present value of the future service pension on age retirement for this member? 2 30 Solution As the maximum pension is 3 = 45 of the final salary.S.9).9.9.5) (b) If n ≥ 40.p. .P. of the lump sum on retirement is 3· SAL ¯ i+r (n. Final salary is defined as salary in the 3 years prior to retirement.P. with a maximum of 3 of final salary. then restrict the past service to 40 years and make the F.1) If the annual pension is a fixed sum of £P per year of service. as n is 14.Mx + Rx 3P (19.z M i+r + z Rx ) x 80¯x Dx s (19. where the pension is 1 80 × final salary per year of service (as in section 19.10. and hence we put s47 in the denominator). An executive pension scheme provides a pension of 45 of final salary for each 2 year of scheme service. is altered to SAL 80¯x Dx s z ¯ (i+r)a −z R(i+r)a ¯ Rx x+40−n (19. unchanged but restrict the F. Hence the value of pension benefits is 40 SAL z (i+r)a · Mx (19.v.p. Using nearly the same arguments as before.S. of F. m.P.000 is the salary next year and not the salary rate. These functions are tabulated (on the same basis as in “Formulae and Tables”) in the Supplement to this book.9.S.p. 1 and so on.316 CHAPTER 19. the m. PENSION FUNDS (a) If n < 40. p.3) B· Withdrawals Similar formulae may be developed (replacing the ‘d’ with ‘w’) but withdrawal benefits are usually in the form of a return of contributions or a deferred pension. a spouse’s pension. are tabulated in the Supplement to this book.v. The benefit for a person aged x. with current salary rate SAL. 3. 64−x 2(SAL) t=0 v t+ 2 1 dx+t sx+t lx sx ¯ (19. £B.11. Note that d j d w w Mx corresponds to Mx when j = 0 (page v of the Supplement) and Mx corresponds to j Mx when j = 0 (page vi of Supplement). has m. then the m. .11 Death and Withdrawal Benefits The benefits on death in service usually consist of one or more of: 1.11. a fixed sum.11. is 64−x B t=0 v t+ 2 d Mx Dx 1 dx+t lx (19. 2. We begin by considering the first type of benefit. Mx . s d d Cx = sx Cx 64−x d Mx = t=0 64−x s d Mx = t=0 s d Cx+t 1 Cd x+t Expression (19. or a certain multiple of salary.1) can be written in the form 2 and hence the value of the benefit is 2 SAL sx D x ¯ 64−x d sx+t Cx+t t=0 d SAL s Mx · sx ¯ Dx (19. DEATH AND WITHDRAWAL BENEFITS 317 19.11.) Note d w The commutation functions Mx . Suppose the death benefit is 2× the salary rate at date of death.2) If the benefit on death is just a fixed amount.v.p. accumulated at a rate of interest j (where j may be zero).11.19. etc. (We shall discuss the formulae for valuing these benefits later. a return of the employee’s contributions.1) Define the commutation functions d Cx = v x+ 2 dx . v. Then k solves 0.p. 0. sj ¯w Rx . 000 s ¯ ¯ (i+r)a + 10. etc. 000 M40 · N 40 = × z R40 s40 D40 ¯ 80¯40 D40 s D40 19.01k · Thus.7 = 14827. Note w The values of j M x .11%. 000 10.p. of the return of contributions on withdrawal is (T P C)(1 + j)−x j w j ¯w Mx Rx +A· Dx Dx (19. the m. Suppose that the employees contribute at rate k% of salary. payable continuously. PENSION FUNDS Example 19.11. 64−x j w Mx = t=0 j 1 1 C x+t w 1 w w ¯w Mx = j M x − j C x 2 sj ¯ w j ¯w Mx = sx M x j 64−x sj ¯w Rx = t=0 1 ¯w (1 + j)−(x+t+ 2 )sj M x+t Similar functions for returns of contributions on death are defined by substituting ‘d’ for ‘w’. Then the m. d 10. are given for j = 0.12 Return of Contributions on Death or Withdrawal Suppose that the employee’s contributions are returned with compound interest at rate j per annum if the member leaves service.12.2) .000. The following additional commutation functions are defined: j w Cx = (1 + j)x+ 2 v x+ 2 wx .01k × 172628. final salary is the average annual salary in the 3 years prior to retirement. In addition the sum of £10. If all contributions are paid by the employer. Use the basis of “Formulae and Tables” (with the supplement).12.v. See the Appendix to this Chapter.318 CHAPTER 19. A company has a pension scheme which provides a pension upon retirement 1 of 80 th of the final salary per year of service.000 is paid on death in service of a member. If the employee’s contributions are fixed at £A per annum. Let TPC be the member’s total past contributions to the scheme.60 + 895.6 Hence k = 9. Solution Let k be the contribution rate per cent. find the contribution rate as a percentage of salary required for a new entrant aged 40 with a salary rate of £10.1. of the return of contributions on withdrawal for a member aged x is (T P C)(1 + j)−x · j w Mx k SAL sj ¯ w + · · Rx Dx 100 sx Dx ¯ (19.03 and j = 0 in the Supplement to this book.1) Proof. 000 p.12.e. You are actuary to a pension scheme which provides the following benefits: 1 (a) a pension of 80 × final salary for each year of service (including fractions pro rata) on retirement for “age” or “ill-health” reasons.000.19..12. and he asks for his transfer value to be used to give him “added years” of service. with salary at the rate of £20. Suppose that a life aged 45 joins the scheme.2) to (19.a.12. Find n. w ¯ Rw Mx +A· x Dx Dx (19.5) Dx 100 sx Dx ¯ d+w where j Mx = jMx + jMx ¯x ¯x ¯ d+w = sj Rd + sj Rw .) Note When j = 0 one may omit the ‘j’ from the commutation functions. and you calculate reserves prospectively.12.a. formulae (19.) d w Example 19. Furthermore. the prospective reserve at the entry date is zero.12. so if contributions are returned without interest their mean present value is (T P C) w Mx k SAL s ¯ w + · Rx Dx 100 sx Dx ¯ (19. 3. i.a. and sj Rx (Similar remarks apply to formulae (19. Find the employer’s contribution rate for a member joining at age 20. 1. bringing a transfer value of £10. The total contribution rate is assessed for each member separately. Your actuarial basis is that given in the “Formulae and Tables”. Find the reserve held for a member aged 45.000 per annum. one may adjust formula (19. (b) a lump sum on retirement of 3 times the annual pension. who joined the employer at age 20.1. when the member joins. . Final salary is the average annual salary in the 3 years before retirement. to credit him with n years of past service.12. in the event of death in service. RETURN OF CONTRIBUTIONS ON DEATH OR WITHDRAWAL 319 (see the Appendix to this Chapter. as expected.1) to (19. given that.1) to j d+w Mx k SAL sj ¯ d+w (T P C)(1 + j)−x + · · Rx (19. Note There are no withdrawals over age 45 in the Tables.4) If employee’s contributions are returned on death (with or without interest).12.3) if the employee contributes k% of salary. i.4) are changed by just substituting ‘d’ for ‘w’. and (T P C) if he contributes at the fixed rate of £A p. the transfer value will be returned with compound interest at 3% p.e.12. ignoring expenses. Members pay contributions at the rate of 2% of salary.4).12. 2. The member’s current salary rate is £15. and is the proportion of salary which will pay for the benefits. so we need not specify what happens to the transfer value in that event.12. Your pension scheme accepts transfer values from other pension schemes. if contributions are returned on either death or withdrawal. z Ri+r ¯ R20 20 sN ¯20 = 7. (a) Spouse’s D. of this benefit is 64−x v t+ 2 t=0 1 dx+t (12) 1 × (spouse’s pension p. 000 (0.z R20 ] 100 s20 D20 ¯ 80¯20 D20 s Thus k= 100 80 z ¯ (i+r)a + 3.1) . a double-decrement table (for death and remarriage) should be constructed and used in place of the mortality table for spouses.) pension. i.13 Spouse’s Benefits A pension scheme may provide the following benefits (among others): (a) a spouse’s death in service (D. Consider (for example) a male member aged x of a pension scheme providing an annuity payable (say) monthly in advance to his widow if he dies in service. 000 (i+r)a ¯ ¯ (i+r)a + 3(25.443.v.13.) 19. 3. and/or (b) a spouse’s death after retirement (D.03)−45 where j = 0.06 years (The transfer value is treated as if it were the T. The m.z M45 ) 80¯45 D45 s j d M45 + 10000(1. 2. 585. 000 = 1420.47.p. Then k SAL s ¯ SAL z ¯ (i+r)a ¯ i+r · N20 = [ R20 + 3. Reserve is 20.A.03 D45 Hence 10. average value of (husband’s age . If spouse’s pension ceases on remarriage. Let k be total contribution rate per cent for a new member aged 20.P.S. Let n years of past service be credited. PENSION FUNDS Solution 1.C.z M i+r +z Ri+r )] − 20.I.0744)s N45 ¯ [25.443% − 2% = 5.) pension.I. As in chapter 13.R.)¨ h a f lx x+t+ 2 x+t+ 1 −d 2 (19. So employer contributes at rate 7. we define hx = probability that a man is married at age x and d = the average age difference (in years) between husband and wife.n + 1394.443% of salary.320 CHAPTER 19.a.z M 45 +z R45 45 45 80¯45 D45 s s45 D45 ¯ = £59.S.42 Therefore n = 6.e. The equation of value is 10000 = 15000n z (i+r)a i+r ( M45 + 3.wife’s age). 1.A. consider a man retiring at age 65 in normal health. The spouse’s D.13. Assuming that spouse’s pensions are payable continuously. SPOUSE’S BENEFITS 321 The size of spouse’s pension may depend on the member’s salary at or near the date of death and the number of years of service. sx .R. and that the member’s current salary rate per annum is SAL.2) 65|65−d Case (ii) : (annual widow’s pension). between ages 60 and 65. Solution (a) Define sx .A.) (ii) Any widow may receive the pension.R. or at exact age 65.t p m µ 65 65+t m h65+t a ¯ 65−d+t f dt (19. Widows’ pensions on death after retirement are payable to any widow (not just to a widow who was married to the member when he retired. ∞ 0 v t .3) Example 19.) Widows’ pensions are payable monthly in advance and do not cease on remarriage. and that ill-health retirements may take place at any age between 35 and 60.13. A pension fund provides the following benefits for widows of male members: (a) on death of the member in service.13. a widow’s pension of annual amount equal to one-third of the member’s salary rate at the time of his death.13. Let ¯ zx = Let y = average age of wife of (x) =x−d Then value of benefit is SAL 3 64−x 1 (sx−3 + sx−2 + sx−1 ) 3 as before. a widow’s pension of 120 th of the member’s average salary in the 3 years immediately preceding retirement for each year of service. Commutation functions need not be developed. One must consider the following 2 cases: (i) Widow’s pension is only payable if the widow was married to the scheme member when he retired. and then allowing for survivorship and interest before retirement. the value at that age of the spouse’s D.h65 a m f ¯ (19. and 1 (b) on death of the member after age or ill-health retirement. Note. You may assume that a service table has already been constructed. (Post-retirement marriages do not count for benefit purposes. benefit is as follows: Case (i) : (annual widow’s pension). You may assume that all age-retirements take place at exact age 60. Develop formulae for valuing the widow’s benefits for a male member aged x with n years’ past service (including fractions).A. (b) Spouse’s D. v t+ 2 t=0 1 dx+t lx sx+t+ 1 ¯ 2 sx ¯ hx+t+ 1 a ¨ 2 (12) 1 y+ t+ 2 f .19. Commutation functions are complicated and are usually ignored. fractions of a year counting pro rata. For example. pension is valued by first considering the appropriate value at the retirement date of the member. hx as previously.R. of this withdrawal benefit for a life l age x is SAL 60 64−x v 65−x t=0 ˆ65 1 1 wx+t 1 zx+t+ 2 l (n + t + ) (1 + j)65−(x+t+ 2 ) · a65 ¯ ˆ lx 2 sx ¯ lx+t+ 1 2 (19. this pension may escalate between the date of leaving and age 65. 19.1) by commutation functions. ‘Final’ salary refers to the average salary in the 3 years before leaving service. vesting at age 65. m. note that l65 = r65 ). 65.14 Preserved Pensions on Leaving Service Suppose that (for example) “early leavers” get a deferred annual pension of 1 60 × final salary per year of service. suppose that escalation is at rate j per annum compound. x+t+ x+t+ 2 2 (ii) value = SAL 60−x r60 v 120 lx SAL 65−x r65 + v 120 lx z60 (n + 60 − x)aN H 60 sx ¯ z65 (n + 65 − x)aN H 65 sx ¯ (rx refers to retirements at exact ages 60.14.v.1) where a65 is the annuity factor.v. ¯ To evaluate (19. etc. one needs to define suitable ‘new’ commutation functions. (i) value = 64−x 1 SAL 1 rx+t zx+t+ 1 2 (n + t + )aN H 1 v t+ 2 120 t=60−x lx sx ¯ 2 x+t+ 2 where aN H 1 is as for aIH 1 . For age retirements.p.14. but with a mortality table for age retirements. there are terms for retirement (i) between ages 60 and 65.p.r pih 1 µih 1 +r hx+t+ 2 +r ay+t+ 1 +r dr ¨ x+t+ x+t+ 2 2 2 (12) (“ih” indicates that a mortality table for men retiring due to ill-health is employed). (ii) at exact ages 60 and 65. Also. That is.1) is equal to SAL 60¯x Dx s where z wa Cx 64−x 1 wa (n + t + )z Cx+t 2 t=0 = ˆ65 a65 v 65 l 1 wx zx+ 2 (1 + j)65−x− 2 ˆ 1 l 1 x+ 2 . In practice. with the pension possibly payable monthly. PENSION FUNDS SAL 120 59−x v t+ 2 t=0 1 ix+t lx 1 zx+t+ 2 sx ¯ 1 (n + t + )aIH 1 2 x+t+ 2 ∞ 0 1 v r . The m. one must specify a mortality table for the early leavers: let us use the notation ˆx for their life table.14.322 (b) Consider ill-health retirement first. = where aIH 1 = x+t+ 2 CHAPTER 19. the expression (19. 2) n. but is an example of “do-it-yourself” commutation functions.14. . x+t 1 ¯ wa Mx = z M wa − z C wa x 2 x and z 64−x z ¯ wa Rx = t=0 z ¯ wa M x+t (Note the similarities with the definitions used in previous sections.) The present value of the deferred pension is SAL wa ¯ wa (19. PRESERVED PENSIONS ON LEAVING SERVICE We now define 64−x z wa Mx 323 = t=0 z C wa .19.z Mx + z Rx 60¯x Dx s Note The notation used here is not standard.14. Employees contribute to the scheme at the rate of 3% of salary.000. 6% between ages 35 and 45. give formulae for valuing benefits (2). (You need not derive the formulae. (3) on death in service. and 7 1 % when aged 45 or over.000. payable continuously. (b) Hence or otherwise find the employer’s contribution rate for this new member. using suitable commutation functions.1 Contributions to a pension scheme by employees are made at a rate of 5% of salary when aged under 35. For each active member of the scheme there is recorded . (a) Calculate the contribution rate paid by the company. There is no benefit on withdrawal. Salaries are revised continuously. (b) A valuation of the fund is to be conducted. and is such that the prospective reserve for each new entrant is zero. (2) a lump sum on retirement of 3 times the annual pension. a lump sum of £30. The percentage is that which will exactly cover the cost of benefits for a new entrant to the fund at age 30 with an initial salary rate of £10. In addition.000 p. (4) on withdrawal from service. assuming the last retirement age is 65. find the value of each of the benefits (1). in terms of suitable commutation functions. (2).) (c) Hence find a formula for the employer’s contribution rate for a new member aged x and a starting salary rate of £10. Calculate the present 2 value of the future contributions payable by a member aged exactly 30 who in the past year has received a total salary of £12. PENSION FUNDS Exercises 19. All members who reach age 65 retire immediately. Contributions are payable continuously. (3) and (4) for a new entrant aged 45 with salary rate £10.) (b) In respect of a new entrant aged x with annual salary rate SAL. Expenses are ignored. Final pensionable salary is defined as the average annual salary in the three years immediately before retirement.324 CHAPTER 19. (ii)(a) Using the basis given in the pension fund section of the Formulae and Tables (and the supplement).2 A company pension scheme provides the following benefits for all members: (1) a pension on retirement (on grounds of ill-health or of age) of one-eightieth of final pensionable salary for each year of service (including fractions). a return of the employee’s contributions. (i) (a) Derive a formula. The company pays a constant percentage of all the members’ salaries into the pension fund. Expenses are negligible.000 per annum.718.health’ reasons) of amount equal to one per cent of the member’s total earnings throughout his service. in the event of a member dying in service there is payable at the time of death a lump sum of £30.3 The pension scheme of a certain company provides an annual pension on retirement (for ‘age’ or ‘ill.a. (You need not define the service table functions. for valuing benefit (1) above in respect of a new entrant aged x with annual salary rate SAL. accumulated at 3% per annum compound. 19. 19.000 per annum. The employer’s contribution rate is assessed for each member separately. and the employees do not contribute to the scheme. The pension is payable weekly. (3) and (4) above. a pension of 80 th of annual pensionable salary. member age current salary rate (£) 1 45 30. (ii) on withdrawal or death in service. (b) A new employee. Salaries are revised continuously. Pensionable salary is defined as salary less £4000. EXERCISES 325 (i) the age nearest birthday (which is regarded as the member’s exact age) at the valuation date.000 4 35 10.000 3 35 10. 19. . Calculate the employer’s contribution rate. the totals of (ii) and (iii) are recorded and the following is an extract from the data. Final salary is the average income over the last 3 years of service. that in the event of death in service a benefit is payable equal to the return without interest of both the member’s and the employer’s contributions. accumulated at 3% per annum compound interest.4 A pension scheme provides each member who retires (for any reason) with annual pension 1 equal to 60 × final salary per year of service. Assuming that equal contributions are payable by the member and his employer. 19. (ii) the annual salary rate at the valuation date. will exactly pay for the benefits. and (iii) the total past earnings in service (prior to the valuation date.100 25 Assuming that the basis of the Tables provided is appropriate.000 (a) The employer has decided to contribute the proportion of total salaries which.000. and that in the event of withdrawal from service a return without interest is made of the member’s contributions. find the liability at the valuation date for the benefits payable to the members aged 25.100 Total of annual salary rates for members aged x £ 70. aged 35 and with current salary rate £8. a return of the member’s contributions. is about to be hired.15.19. Age x No. for each year of future service including fractions. averaged over the previous three years. and determine whether the future contributions payable in respect of these members are more or less than sufficient to cover the benefits. You have decided to use the pension fund tables in Formulae and Tables for Actuarial Examinations as the basis for all calculations. which has just been established.000 2 45 20. Calculate the surplus or deficiency in the pension fund after this new member joins. The scheme provides the following benefits to employees: 1 (i) on retirement (for ill-health reasons or otherwise). Employees contribute 2% of salary to the scheme.) For each age. of members aged x 11 Total past earnings for members aged x £ 302. together with the employees’ contributions.5 You are consulting actuary to a small pension scheme. and fractions of a year of service are not included when calculating the pension. and contributions are made continuously. Details of the current membership are as follows. calculate the appropriate contribution rate payable by both the member and his employer in respect of a new entrant aged 40. 326 CHAPTER 19. PENSION FUNDS 19.6 It is desired to set up a pension scheme for the group of employees described below. For each member there is recorded his exact age, his exact length of service with the company, and his annual rate of salary. Past Service (Years) 4 6 5 3 1 Rate of Salary p.a. £ 8,800 8,500 8,500 8,600 8,600 Past Service (Years) 12 18 15 5 10 Rate of Salary p.a. £ 12,400 12,500 12,800 12,500 12,400 Past Service (Years) 25 15 10 20 5 Rate of Salary p.a. £ 15,000 14,000 13,000 13,200 13,000 Age Age Age 25 25 25 25 25 35 35 35 35 35 45 45 45 45 45 1 The scheme will provide pensions of 60 th of “pensionable salary” for each year of service (fractions of a year being included) and on death or withdrawal from service a return will be made of the member’s contributions with 3% compound interest. All members will contribute at the same rate and the employer will contribute the same amount as each member. The contribution rate will be such as to provide exactly the benefits for future service. The basis of “Formulae and Tables” is to be used. (a) Assuming that pensionable salary is the average annual earnings over the three-year period ending on the retirement date, calculate the contribution rate payable by each member. (b) Calculate also the total liability for past service benefits. The employer wishes to meet this liability by paying additional contributions proportional to future salary payments. At what rate should these additional contributions be made? (c) Immediately after the scheme is set up as described above the 45 year old member with 5 years of service withdraws. Is the position of the fund improved or worsened by this withdrawal? 19.7 The pension scheme of a large company provides the following benefits, among others: (1) on death in service of married members: a spouse’s pension of one-third of the member’s annual salary at the date of death; (2) on death after normal retirement of the member at age 65 or after ill-health retirement at an earlier age: a spouse’s pension of 1% of annual salary at the date of retirement for each year of scheme membership (including fractions of a year); no pension is payable if the marriage has taken place after the member’s retirement date. Assume that a service table has been constructed, and that the proportion of members married at exact y is hy . Assume further that spouses are the same age as members, and that a unisex mortality table is used for all calculations, with the age rated up by 5 years on ill-health retirement. You are given the age nearest birthday, current salary rate and years of past service (including fractions) of each member. Spouse’s pension is payable monthly in advance, beginning immediately on the death of the member. (i) Using the rate of interest i per annum, find formulae for the mean present value of each of the benefits (1) and (2) above for a member aged x (x < 65). You are NOT required to construct commutation functions. (ii) Suppose that the scheme’s rules are to be changed so that benefit (2) is to be payable to any surviving spouse. Show how to modify the formulae of (i) to accommodate this change. 19.16. SOLUTIONS 327 Solutions 19.1 12, 718 ¯ ¯ ¯ {0.05s N30 + 0.01s N35 + 0.015s N45 } s29 × D30 = 12, 718 × 1.30162 = £16, 554 19.2 (i)(a) SAL m.p.v. = 80¯x Dx s = SAL 80¯x s 64−x t=0 z ¯ (i+r)a Rx 1 1 ia ra (t + )z Cx+t + (t + )z Cx+t 2 2 ra + (65 − x)z C65 Dx (b)(2): 3.SAL 80¯x s · z ¯ i+r Rx Dx (3) 30000 · (4) 0.03 · d Mx Dx SAL sj ¯ w sx Dx Rx ¯ where j = 0.03 (c) Let employer’s contribution rate be k for this member. Then SAL s ¯ N x = above benefits. (k + 0.03) sx D x ¯ So k + 0.03 = 1 sN x ¯ 1 z ¯ (i+r)a 3 ¯ i+r d ¯w R + z Rx + 3¯x Mx + 0.03.sj Rx s 80 x 80 (ii)(a) Values of benefits are: (1) 13,088 (2) 3,642 (using Supplement) (3) 3,016 (4) 0 Hence total value of benefits = £19,746. (b) value of contributions = 142580(k + 0.03) 19746 Hence k + 0.03 = 142580 and hence k = 10.85%. 19.3 (a) Let k be the contribution rate. Then ¯ (i+r)a SAL s ¯ SAL s R30 Md k· · N30 = · + 30000 · 30 s30 D30 ¯ 100¯30 s D30 D30 where SAL = 10, 000. Hence k = s¯ R s30 ¯ 0.01 30 sN ¯30 s30 ¯ (i+r)a d + 3.M30 = 0.0535 = 5.35% 328 CHAPTER 19. PENSION FUNDS (b) Value of benefits for the group is (i+r)a ¯ (i+r)a 302100 M25 70100 s R25 Md · + · + 11 × 30000 25 100 D25 100¯25 s D25 D25 = 2977.37 + 54730.83 + 11811.67 = £69, 528 Value of future contributions is 70100 s ¯ N 25 s25 D25 ¯ = £76, 718. 0.0535 × Hence contributions are more than sufficient. 19.4 Let k% of salary be the contribution rate for both the member and the employer. Then if the member has a salary rate of £SAL, 2k SAL s ¯ SAL 1 (i+r)a z ¯ (i+r)a · N40 = R40 − z M 40 100 s40 D40 ¯ 60¯40 D40 s 2 k SAL s ¯ w 2k SAL s ¯ d · · R40 + · · R40 + 100 s40 D40 ¯ 100 s40 D40 ¯ Hence 6342.42k = 35513.07 + 13.74k + 562.35k. Therefore k = 6.16% 19.5 (a) Construct the following table, using j = 0.03: Member 1 2 3 4 age 45 45 35 35 salary 30,000 20,000 10,000 10,000 SAL 80¯x s · xx D 39,264 26,176 15,189 15,189 95,818 z ¯ R(i+r)a 4000 80 · xDx 3,999 3,999 3,498 3,498 14,994 ¯ R(i+r)a 0.02 SAL Dx sx ¯ x 899 599 549 549 2,596 sj ¯ Rd+w Let the employer contribute k% of salary, so m.p.v. of contributions is 50000 s ¯ 20000 s ¯ (k + 0.02) N 45 + · N35 s45 D45 ¯ s35 D35 ¯ = 1093716(k + 0.02) Therefore k + 0.02 = 95818−14994+2596 1093716 and hence k = 5.627%. (b) Reserve for new member is ¯ (i+r)a ¯ (i+r)a 8000 z R35 4000 R35 · − · 80¯35 s D35 80 D35 8000 sj ¯ d+w + 0.02 × . R35 s35 D35 ¯ 8000 ¯ − × 0.07627s N35 s35 D35 ¯ = 12, 151 − 3, 498 + 440 − 11, 617 = −2, 524. Hence there is now a surplus of £2,524. 19.6 (a) For each age x, Value of Past Service Pension = (n×SAL) 60¯x s · z (i+r)a Mx Dx 19.16. SOLUTIONS Value of Future Service Pension = Value of return of contributions = x 25 35 45 SAL 43,000 62,600 68,200 (n × SAL) 163,100 752,300 1,044,000 SAL 60¯x s SAL 100¯x s 329 · · ¯ (i+r)a Rx Dx sj ¯ d+w Rx (where Dx z j = 0.03) per 1% of salary P.S.P. 7,776 55,421 103,230 166,427 F.S.P. 76,783 126,780 119,012 322,575 return of contributions (per 1% of salary) 2,223.7 1,720.5 1,022.7 4,966.9 (1) The value of the contributions is, per 1% of contributions, ¯ SAL s Nx 100¯x Dx s x 25 35 45 Value of (1) 8,791.2 11,919.1 9,724.0 30,434.3 Let members’ contribution be k%. The equation of value is thus 2k × 30, 434.3 = 322, 575 + 4, 966.9k Hence k = 5.77%. (b) The past service liability is £166,427. Let the contribution rate per cent needed to pay for this be p. Then p solves, p × 30, 434.3 = 166, 427 Hence p = 5.468%. (c) The reserve for the withdrawn member is 13000 sj ¯ d+w 13, 000 z ¯ (i+r)a 5 × 13000 z (i+r)a + + 0.0577 × · R45 M 45 · R45 60¯45 D45 s 60¯45 D45 s s45 D45 ¯ 13000 s ¯ · N45 − (2 × 0.0577 + 0.05468) · s45 D45 ¯ = 30, 238 − 31, 526 = −1, 288 So the position of the pension fund has worsened by £1,288. 19.7 (i) Benefit (1) has value SAL 3 64−x 1 v t+ 2 t=0 dx+t sx+t (12) 1a h ¨ 1 lx x+t+ 2 x+t+ 2 sx ¯ Benefit (2): On normal retirement the value is 1 r65 SAL (12) · v 65−x .¯65 (n + 65 − x) × s h65 a65|65 (1 + i) 24 100¯x s lx On ill-health retirement, the value is 64−x 1 1 sx+t ix+t SAL 1 (12) ) (n + t + )[hx+t+ 1 ax+t+ 1 +5|x+t+ 1 (1 + i) 24 ] v t+ 2 ( 2 2 2 100 t=0 sx ¯ lx 2 330 (ii) Benefit (1): no change Benefit (2): For normal retirement change h65 a65|65 (1 + i) 24 to 0 ∞ (12) 1 CHAPTER 19. PENSION FUNDS v t .t p65 µ65+t h65+t a65+t dt ¨ (12) For ill-health retirement, change 1 (12) hx+t+ 1 ax+t+ 1 +5|x+t+ 1 (1 + i) 24 2 2 2 to ∞ 0 ¨ v r .r px+t+ 1 +5 µx+t+ 1 +5+r hx+t+ 1 +r ax+t+ 1 +r dr 2 2 2 2 (12) 19.16. SOLUTIONS 331 Appendix: Formulae for valuing a return of contributions Suppose employee’s contributions are to be returned on withdrawal with interest at rate j per annum compound. Consider a member aged x whose current salary rate is SAL and whose total past contributions, accumulated at rate j p.a., are TPC. Assume that the employee will in future contribute k% of salary. The value of the return of future contributions is k SAL 100 sx ¯ k = 100 64−x v t+ 2 t=0 1 wx+t 1 [sx (1 + j)t + sx+1 (1 + j)t−1 + ... + sx+t−1 (1 + j) + sx+t ] lx 2 1 1 1 1 {v 2 wx × sx + v 1 2 wx+1 [sx (1 + j) + sx+1 ] 2 2 1 1 + v 2 2 wx+2 [sx (1 + j)2 + sx+1 (1 + j) + sx+2 ] 2 + ... 1 1 + v 64 2 −x w64 [sx (1 + j)64−x + sx+1 (1 + j)64−x−1 + ... + s63 (1 + j) + s64 ]} 2 SAL sx lx ¯ Collecting the coefficients of sx , sx+1 , ... gives k 100 1 1 SAL 1 1 {sx [ v 2 wx + v 1 2 wx+1 (1 + j) + ... + v 64 2 −x w64 (1 + j)64−x ] sx lx ¯ 2 1 1 1 1 + sx+1 [ v 1 2 wx+1 + v 2 2 wx+2 (1 + j) + ... + v 64 2 −x w64 (1 + j)64−x−1 ] 2 1 1 + ... + s64 [ v 64 2 −x w64 ]} 2 Define j w Cx = (1 + j)x+ 2 v x+ 2 wx 64−x j w Mx = t=0 j 1 1 C x+t w 1 w w ¯w Mx = j M x − j C x 2 sj ¯ w ¯w Mx = sx .j Mx j 64−x sj ¯w Rx = t=0 1 ¯w (1 + j)−(x+t+ 2 ) .sj Mx+t The value of the return of future contributions can be written as k 100 SAL sx v x l x ¯ {sx (1 + j)−(x+ 2 ) 1 1 1 1 1 1 (v(1 + j))x+ 2 wx + (v(1 + j))x+1 2 wx+1 + ... + (v(1 + j))64 2 w64 2 + sx+1 (1 + j)−(x+1 2 ) + ... + s64 (1 + j)−64 2 = k 100 SAL sx D x ¯ ¯w .sj Rx 1 1 1 1 (v(1 + j))x+1 2 wx+1 + (v(1 + j))x+2 2 wx+2 + ... + (v(1 + j))64 2 1 1 (v(1 + j))64 2 w64 } 2 f rac12 w64 but are instead £A each year. with the salary scale function taken as 1. This gives a present value of j ¯w Rx A· Dx where j 64−x ¯w Rx = t=0 ¯w (1 + j)−(x+t+ 2 ) . payable continuously. The value of the return of future contributions is obtained in a similar way to the previous derivation.j Mx+t . PENSION FUNDS The value of the return of past contributions with interest at rate j is 64−x (T P C) t=0 v t+ 2 (1 + j)t+ 2 64−x 1 1 wx+t lx 1 = (T P C) Dx (1 + j)−x (v(1 + j))x+t+ 2 wx+t t=0 j w Mx Dx = (T P C)(1 + j)−x Hence the total value is (T P C)(1 + j)−x j w Mx k SAL sj ¯ w + Rx Dx 100 sx Dx ¯ Suppose now that the employee’s contributions do not depend on salary.332 CHAPTER 19. 1 . One should ensure that an approximate time-scheme (say. such as the correct use of actuarial tables and commutation functions. it is essential to get off to a good start. by being in the right frame of mind and picking the most suitable questions to attempt first. when one is too tired to think out new ideas but can still do (or check) arithmetic. It is sometimes necessary to omit (ii) or even (i) under extreme time-pressure. and arithmetic.Appendix A Some notes on examination technique Each candidate should spend some time going over his or her notes. enabling the candidate to learn the most important facts and then have a “revision aid”. Routine arithmetic (or even looking up the tables) may sometimes be left till near the end. it is satisfying to give a complete answer. one should tick the questions about which one is most confident . 333 .) In actuarial examinations (particularly in life contingencies) it is important to separate the basic ideas or “rationale” of the solution from (i) (ii) technical aspects. 15 minutes per 10-mark question) is adhered to.especially bookwork. Leave the less familiar questions until later. On first seeing the examination paper. making a note of subjects and formulae of particular importance. (He or she should ask a friend to choose a card at random. For this purpose a set of postcards/computer cards is useful: each topic may be summarised on one postcard. and this should be attempted if time is not short. within reasonable limits. As in musical performances. On the other hand. but one should leave space to return to the question later if time permits. the textbook and (especially) past examination papers. to give the title and to ask for a description of the topic. One should keep the left-hand pages of the examination book free for rough working and arithmetic. 5.5 are estimated by linear interpolation. Remember that for a(55) there are two tables . If only limited tables are available. 334 .5 2 l50. This saves time. e. Remember to make use of the functions maturing at ages 60. it is nearly always best to avoid the use of commutation functions.5 µ50. suppose that one is asked to evaluate A50. 65.5 at 10% interest on A1967-70 ultimate.) lx Dx 3. ax:n at 6% = ax − v n ¨ ¨ lx+n ax+n ¨ lx (There is no point writing v n lx+n Dx+n as since Dx and Dx+n are not given at 6% in the tables. For other interest rates one most use the more limited tables provided. Avoid excessive reliance on commutation functions.5 where µ50.5 l53 µ53 µ50. but also in a(55)) that you distinguish between “select” and “ultimate” tables and use the correct rate of interest. Be careful (especially in A1967-70.5 and l50.) Assume that males and females are subject to the table of the appropriate sex (unless the question states otherwise.5:2. etc.Appendix B Some technical points about the tables used in examinations 1. In theoretical/statistical questions.5 + v 2.g. using suitable approximations.male and female (each with a select period of 1 year. In the A1967-70 section of “Formulae and Tables” commutation functions are given only at 4% interest. For ¯ 1 example. This may be done by the trapezoidal rule as follows: 2.5+t dt 2. one must proceed directly. 4.5 0 v t t p50. 2.) 6. 3. 11. 10. = Ax P.V. ax:n = 1 + ax:n ¨ µx = t px var(v T ) ax:n = 1 + ax:n−1 ¨ µx = − lx lx t lx lx t = exp 0 µx+r dr t px = exp − 0 µx+r dr ax:n = Nx − Nx+n Dx ax:n = Nx+1 − Nx+n+1 Dx 335 . 4. 9. Mean present value = v T ¯ Present value (as random variable) = Ax var(P aT − v T ) ¯ = var(P aT ) − var(v T ) ¯ or var(P aT ) + var(v ) ¯ T ¯ M. ¯ Ax:n (1 + i) 2 Ax:n 1 CORRECT ¯ Ax:n 1 1 (1 + i) 2 Ax:n + Ax:n 1 (1 + i) 2 (Mx − Mx+n ) + Dx+n = Dx 2.V. = var − P +δ δ vT + P δ = 8. = v T var P (1 − v T ) − vT δ P +δ δ 2 7. 5.P.Appendix C Some common mistakes WRONG 1. ax:n t do not exist t 0 cr dr = cr or cr+1 r+1 t 0 0 t cr dr = cr log c t (Note that cr = er log c ) 0 6. 0 1 ax ¯ a(m)n ¨ x: i ax δ i ax:n d(m) ax ¯ a(m)n ¨ x: ax + 1 2 or ax − ¨ 1 2 ax:n − ¨ m−1 2m 1− Dx+n Dx 1 1 ax:n . Mid-point rule b f (x)dx a (b − a)f a+b 2 2. Three-eighths rule b f (x)dx a b−a 8 f (a) + 3f 2a + b 3 + 3f a + 2b 3 + f (b) 336 . Trapezoidal rule b f (x)dx a b−a 2 [f (a) + f (b)] 3.Appendix D Some formulae for numerical integration 1. Simpson’s rule b f (x)dx a b−a 6 f (a) + 4f a+b 2 + f (b) 4. 336 . 3 11554.2 16670.6 2.8 1.1 17546.3 14.1 186.3 4010.0 7550.4 755.9 6269.4 904.8 879.0 0.1 2586.1 654.7 10731.8 Rx 6835.2 2.4 13.9 31040.9 21101.7 5.8 0.2 5.1 19.2 5136.0 21.8 893.0 0.1 56.3 147.7 2.1 9.2 2.2 430.9 Ill-health retirement functions 337 .8 2086.2 6048.3 58.7 28327.5 112.3 1609.0 18428.6 13.3 13228.0 5.9 0.6 1020.7 808.2 132.9 103.4 39.3 796.5 z i TABLES FOR LUMP SUM BENEFITS ON RETIREMENT 0.1 1456.8 360.0 0.0 0.4 904.0 0.1 208.0 i Mx 188.4 904.4 9.8 2417.5 1.6 853.5 1765.4 4630.8 33.8 465.5 6646.7 10.3 828.0 11.0 0.6 14935.1 2931.4 902.4 904.7 845.5 119.7 148.8 3826.0 2.5 889.1 24709.5 25614.2 54.9 409.5 14078.5 174.6 151.7 15.1 3.8 126.0 0.6 751.0 0.3 5327.8 68.9 904.9 25.0 4196.0 0.1 21.8 180.4 904.1 15799.2 13.0 0.3 620.8 188.5 837.8 188.7 10.3 50.5 5702.4 4.5 684.8 1.2 160.8 188.1 29231.7 632.8 188.4 904.4 709.2 8.0 0.8 188.2 181.0 0.age x 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 SUPPLEMENT 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 i Cx 0.7 6457.8 582.0 165.6 4758.0 0.4 904.0 2.9 3.3 5.2 223.2 93.5 790.4 40.9 26518.2 138.0 43.7 183.5 38.3 19315.3 29.7 12387.4 6.3 3323.6 111.4 12.2 81.3 867.0 2.7 154.4 2.6 2758.7 95.7 23805.2 10.8 188.8 5.6 3961.6 i z i Cx 0.8 188.4 3107.7 2721.4 4947.3 18.9 4.0 0.6 1.3 22902.6 2250.1 884.9 8.2 22.8 9115.0 6.0 0.6 860.5 732.4 1924.8 539.7 157.8 188.2 168.7 87.1 0.2 900.4 4.5 170.7 32849.2 176.9 885.4 904.9 7.3 2.3 5891.1 188.3 184.0 818.0 i Mx 904.2 782.8 4570.0 178.0 0.6 163.2 4.5 80.7 8.3 27422.6 516.9 2160.4 172.8 1.3 9917.6 289.5 1186.4 z Rx 33753.9 768.8 188.8 1646.8 187.0 0.9 20206.8 3644.0 0.9 1161.4 1.4 904.2 1306.4 112.5 4383.0 897.9 187.3 3463.4 2.9 217.0 873.2 71.3 6790.3 3.5 8326.4 904.3 17.5 30136.0 5325.8 5514.7 7.5 3284.4 143.1 6080.3 31944.4 311.0 0.5 489.0 22000.8 188.3 70.3 6.2 2. 0 4988.0 1524.2 55003.0 0.2 0.0 0.5 45859.0 r Mx 1524.0 0.8 77.5 4712.9 30619.0 Rx 65671.6 234413.7 210267.7 8048.0 1524.0 0.0 0.0 1524.7 8048.7 1107.6 4397.7 649.0 0.7 8048.2 16904.0 1524.0 0.7 8048.0 0.7 8048.0 0.0 0.4 50431.0 1524.8 113683.2 19952.7 8048.0 1524.7 8048.0 0.4 194169.7 8048.0 0.0 0.7 8048.0 0.1 178072.1 59575.3 242461.9 65391.5 44335.5 6236.0 0.0 378.7 8048.0 0.0 0.3 51955.7 8048.0 1952.0 0.7 8048.0 0.0 0.0 0.0 1524.8 z r TABLES FOR LUMP SUM BENEFITS ON RETIREMENT 0.0 0.0 0.4 3691.2 129780.2 56527.7 8048.2 105634.2 290753.7 186121.7 8048.7 8048.0 62623.1 153926.1 61099.0 1524.0 0.0 0.5 212.0 1524.0 0.9 649.4 933.0 1524.8 688.0 1524.5 121731.9 29096.3 93.1 21476.0 0.0 0.0 1524.5 3484.3 218315.8 32143.6 49294.9 89537.7 8048.4 10808.7 8048.0 0.6 258558.0 0.0 1524.1 8048.0 226364.0 820.2 58051.0 0.3 13856.0 0.0 0.7 8048.0 1524.6 41287.0 0.5 3899.3 15380.0 0.7 8048.0 18075.7 1461.0 0.0 0.3 12332.7 z Rx 347094.0 1524.0 0.2 1524.0 0.0 0.1 339045.5 97585.7 8048.7 8048.7 8048.4 9284.0 202218.7 8048.0 1524.7 8048.5 7760.0 27572.0 0.0 0.0 0.8 137829.0 0.0 0.9 250510.0 0.6 73439.0 0.0 0.0 1524.4 48907.7 8048.7 8048.0 0.7 8048.1 23000.9 274656.7 6096.4 7840.4 591.0 0.7 8048.8 161975.0 1524.0 1524.0 498.0 1524.0 0.5 306850.0 0.8 33667.0 1524.0 0.0 0.7 8048.0 1524.0 0.4 170023.9 2338.0 0.0 0.7 8048.0 1524.1 314899.7 8048.7 8048.0 1524.2 266607.0 1524.0 1524.7 726.7 38239.0 0.0 0.8 298802.7 36715.5 282704.0 24524.0 0.2 81488.0 0.3 57342.7 8048.7 8048.0 64147.0 1524.0 0.0 0.0 0.0 41245.SUPPLEMENT age x 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 r Cx 0.5 145877.0 1524.0 1145.0 0.6 3377.0 0.0 1524.7 8048.7 8048.3 53479.0 26048.8 35191.0 0.0 1524.2 18428.4 47383.0 1524.7 8048.0 0.0 1524.0 1524.0 r Mx 8048.4 112.7 8048.7 25148.7 8048.0 1524.0 0.0 0.0 0.0 1524.0 0.6 3484.0 415.6 39763.1 r z r Cx 0.0 0.6 42811.8 322948.7 12533.1 Age retirement functions 338 .0 0.0 0.0 1524.0 0.4 330996.3 33196.0 1524.0 1524.0 1524. 94 63315.64 490.29 79.71 2861.46 5291.80 1177.89 13941.34 24305.dx d 64−x d (2) j Mx = t=0 j C x+t j d d 64 j d y=x C y 3108.29 33.33 147.85 1804.09 43.5) j = y=x (1 + j) .04 j = 0.45 225.42 76568.76 12500.03 x = 20 25 30 35 40 45 50 55 60 61 62 63 64 j d Mx sj Rx Rx d d 64 64 −(y+.02 243.36 784.80 108.67 49478.15 17508.(1 + i)−(x+.70 181.59 2524.20 8. d (1) j Cx = (1 + j)x+.5) .sj M y 31393.97 8479.18 2878.85 (3) j M x = j M x − (4) sj d d d M x = sx · M x 1 2 j · jCx d d 339 .57 2700.45 6005.SUPPLEMENT Functions for valuing (at interest rate i per annum) refunds on death of contributions with compound interest at interest rate j per annum i = . M y = y=x (1 + j)−(y+.59 2338.11 350.67 Notes.14 36324.41 655.12 2120.17 1286.15 1333.12 423.5) .5 .98 23700.46 87740. 90 7831.98 2399.12 46.39 3281.97 1049.25 5520.53 1875.70 140.85 1425.17 38299.26 284. M y = y=x (1 + j)−(y+.(1 + i)−(x+.28 3732.18 31.04 j = 0.wx w 64−x w (2) j Mx = t=0 j C x+t w w w 1 (3) j M x = j M x − 2 · j C x w w (4) sj M x = sx · j M x j w w 43547.20 24.32 13393.08 42216.22 1608.58 6.5) j = y=x (1 + j) .97 666.43 353.5 .19 340 .95 388.SUPPLEMENT Functions for valuing (at interest rate i per annum) refunds on withdrawal of contributions with compound interest at interest rate j per annum i = .5) .16 8041.53 2391.03 x = 20 25 30 35 36 37 38 39 40 41 42 43 44 j 64 j w y=x C y w Mx sj Rx Rx w w 64 64 −(y+.97 98759.16 207941.75 2334.sj M y 119414.67 Notes.23 1426.49 1063.03 92.50 10746.82 482.89 767.50 19045.95 3006. w (1) j Cx = (1 + j)x+.63 204.95 124.5) .20 735. sj M y 18575.(1 + i)−(x+.17 756.40 53.85 674.98 527.38 50789. M y = y=x (1 + j)−(y+.43 6483.54 104.21 178.59 2467.83 36247.75 56093.5) .62 466.38 326.16 594.45 4303.72 34.62 142.88 44005.49 19889.91 227.94 11769.80 8965.57 33.11 412.05 14932.07 16.07 404.75 77.08 76.69 1071.5) .42 1229.18 (3) j M x = j M x − (4) sj d d d M x = sx · M x 1 2 j · jCx d d 341 . d (1) j Cx = (1 + j)x+.67 Notes.5) j = y=x (1 + j) .10 12048.59 8.dx d 64−x d (2) j Mx = t=0 j C x+t j d d 64 j d y=x C y 792.57 5469.5 .09 43.33 232.42 28117.04 j=0 x = 20 25 30 35 40 45 50 55 60 61 62 63 64 j d Mx sj Rx Rx d d 64 64 −(y+.SUPPLEMENT Functions for valuing (at interest rate i per annum) refunds on death of contributions with compound interest at interest rate j per annum i = . 82 1516.52 345.20 1009. M y = y=x (1 + j)−(y+.01 745.81 580.59 637.99 3051.5 .96 374.60 87256.SUPPLEMENT Functions for valuing (at interest rate i per annum) refunds on withdrawal of contributions with compound interest at interest rate j per annum i = .08 7847.05 137.wx w 64−x w (2) j Mx = t=0 j C x+t w w w 1 (3) j M x = j M x − 2 · j C x w w (4) sj M x = sx · j M x j w w 20579.04 j=0 x = 20 25 30 35 36 37 38 39 40 41 42 43 44 j 64 j w y=x C y w Mx sj Rx Rx w w 64 64 −(y+.58 123. w (1) j Cx = (1 + j)x+.sj M y 103387.77 1374.17 180950.67 Notes.(1 + i)−(x+.57 2288.5) .14 6.29 2916.89 37164.61 311.65 3546.55 214.94 31.57 432.96 971.53 758.5) .54 2185.24 5208.5) j = y=x (1 + j) .75 12076.20 24.46 79.40 342 .41 199.53 38.10 10002.18 34605.20 90.36 12.93 7337.