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Faculty of ActuariesInstitute of Actuaries EXAMINATION 26 April 2010 (am) Subject CT5 — Contingencies Core Technical Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. Enter all the candidate and examination details as requested on the front of your answer booklet. 2. You must not start writing your answers in the booklet until instructed to do so by the supervisor. 3. Mark allocations are shown in brackets. 4. Attempt all 14 questions, beginning your answer to each question on a separate sheet. 5. Candidates should show calculations where this is appropriate. Graph paper is NOT required for this paper. AT THE END OF THE EXAMINATION Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this question paper. In addition to this paper you should have available the 2002 edition of the Formulae and Tables and your own electronic calculator from the approved list. CT5 A2010 © Faculty of Actuaries © Institute of Actuaries 1 Explain what the following represent: (a) l[ x ]+ r (b) n|m q x (c) dx [3] 2 Define spurious selection, giving two distinct examples. 3 Calculate the standardised mortality ratio for the population of Urbania using the following data: Age 60 61 62 Standard Population Population Deaths 2,500,000 26,170 2,400,000 29,531 2,200,000 32,542 [3] Urbania Population Deaths 10,000 130 12,000 145 11,000 173 [3] 4 A life insurance company offers an increasing term assurance that provides a benefit payable at the end of the year of death of 10,000 in the first year, increasing by 100 on each policy anniversary. Calculate the single premium for a five year policy issued to a life aged 50 exact. Basis: Rate of interest Mortality Expenses 4% per annum AM92 Select Nil [4] 5 A population is subject to the force of mortality μx = e0.0002x−1. Calculate the probability that a life now aged 20 exact: (i) survives to age 70 exact (ii) dies between ages 60 exact and 70 exact CT5 A2010—2 [2] [3] [Total 5] 6 You are provided with the following extract from a life table: x 50 51 52 lx 99,813 97,702 95,046 Calculate 0.75p50.5 using two different methods. 7 [5] A company is about to establish a pension scheme that will provide an age retirement benefit of n/60ths of final pensionable salary where n is total number of years of service. Final pensionable salary is the average salary in the three years before retirement. An employee who will become a member of the pension scheme is currently aged 55 exact has and will be granted exactly 20 years of past service. The employee’s salary in the year before the valuation date was £40,000. (i) (ii) Calculate the present value of benefits for this member (including future service). [3] Calculate the contribution required to fund this benefit as a percentage of future salaries. [3] Basis: Pension Scheme from the Formulae and Tables for Actuarial Examinations [Total 6] 8 100 graduates aged 21 exact decide to place the sum of £1 per week into a fund to be shared on their retirement at age 66 exact. (i) Show that each surviving member can expect to receive on retirement a fund of approximately £7,240. [4] Basis: Rate of interest Mortality 4% per annum AM92 Ultimate One of the survivors uses the accumulated fund to buy a weekly annuity payable for 10 years certain. After 10 years the annuity is payable at two-thirds of the initial level for the rest of life. (ii) Calculate the weekly amount of the annuity on the basis used in part (i). [2] [Total 6] CT5 A2010—3 PLEASE TURN OVER 9 A life insurance company models the experience of its pension scheme contracts using the following three-state model: Active(A) μx 10 σx Dead (D) Retired (R) μx (i) Derive the dependent probability of a life currently Active and aged x retiring in the year of age x to (x + 1) in terms of the transition intensities. [2] (ii) Derive a formula for the independent probability of a life currently Active and aged x retiring in the year of age x to (x + 1) using the dependent probabilities. [4] [Total 6] The decrement table extract below is based on the historical experience of a very large multinational company’s workforce. Age (x) Number of employees (al ) x 40 41 42 10,000 9,855 9,684 Deaths (ad ) dx 25 27 Withdrawals (ad ) wx 120 144 Recent changes in working conditions have resulted in an estimate that the annual independent rate of withdrawal is now 75% of that previously used. Calculate a revised table assuming no changes to the independent death rates, stating your results to one decimal place. [7] 11 Thiele’s differential equation for the policy value at duration t (t > 0), tVx , of an immediate life annuity payable continuously at a rate of £1 per annum from age x is: ∂ t V x = μ x +t × t V x − 1 + δ× t V x ∂t (i) Derive this result algebraically showing all the steps in your working. (ii) Explain this result by general reasoning. CT5 A2010—4 [5] [3] [Total 8] 12 On 1 January 2005, a life insurance company issued 1,000 10-year term assurance policies to lives aged 55 exact. For each policy, the sum assured is £50,000 for the first five years and £25,000 thereafter. The sum assured is payable immediately on death and level annual premiums are payable in advance throughout the term of this policy or until earlier death. The company uses the following basis for calculating premiums and reserves: Mortality Interest Expenses AM92 Select 4% per annum Nil (i) Calculate the net premium retrospective reserve per policy as at 31 December 2009. [6] (ii) (a) Give an explanation of your numerical answer to (i) above. (b) Describe the main disadvantage to the insurance company of issuing this policy. (c) Give examples of how the terms of the policy could be altered so as to remove this disadvantage. [3] There were, in total, 20 deaths during the years 2005 to 2008 inclusive and a further 8 deaths in 2009. (iii) CT5 A2010—5 Calculate the total mortality profit or loss to the company during 2009. [3] [Total 12] PLEASE TURN OVER 13 A life insurance company issues a 3-year unit-linked endowment assurance policy to a male life aged 45 exact. Level premiums of £4,000 per annum are payable yearly in advance throughout the term of the policy or until earlier death. 95% of the premium is allocated to units in the first policy year, 100% in the second and 105% in the third. A policy fee of £50 is deducted from the bid value of units at the start of each year. The units are subject to a bid-offer spread of 5% on purchase. An annual management charge of 1.75% of the bid value of units is deducted at the end of each policy year. Management charges are deducted from the unit fund before death, surrender and maturity benefits are paid. If the policyholder dies during the term of the policy, the death benefit of 125% of the bid value of the units is payable at the end of the policy year of death. On maturity, 100% of the bid value of the units is payable. The policyholder may surrender the policy only at the end of the first and second policy years. On surrender, the bid value of the units less a surrender penalty is payable at the end of the policy year of exit. The surrender penalty is £1,000 at the end of the first policy year and £500 at the end of the second policy year. The company uses the following assumptions in carrying out profit tests of this contract: Rate of growth on assets in the unit fund 5.5% per annum in year 1 5.25% per annum in year 2 5.0% per annum in year 3 Rate of interest on non-unit fund cash flows 4.0% per annum Mortality AM92 Select Initial expenses £200 Renewal expenses £50 per annum on the second and third premium dates Initial commission 15% of first premium Renewal commission 2.0% of the second and third years’ premiums Rate of expense inflation 2.0% per annum Risk discount rate 7.0% per annum For renewal expenses, the amount quoted is at outset and the increases due to inflation start immediately. In addition, you should assume that at the end of the first and second policy years, 12% and 6% respectively of all policies still in force then surrender immediately. (i) Calculate the profit margin for the policy. (ii) Calculate the expected present value of profit for the policy if the company assumed that there were no surrenders at the end of each of the first and second policy years. [3] [Total 16] CT5 A2010—6 [13] 92308% of the sum assured.000. £250 on maturity Future reversionary bonus: 1. (ii) Calculate the gross prospective policy reserve at the end of that policy year immediately before the premium then due. the actual past bonus additions to the policy have been £145.5% of each quarterly premium £90 at the start of each policy year £1. Policy reserving basis: Mortality: Interest: Bonus loading: Renewal commission: Renewal expenses: Claim expense: AM92 Ultimate 4% per annum 4% of the sum assured and attaching bonuses.14 A life insurance company issues a 30-year with profits endowment assurance policy to a life aged 35 exact. Pricing basis: Mortality: Interest: Initial commission: Initial expenses: Renewal commission: AM92 Select 6% per annum 100% of the first quarterly premium £250 paid at policy commencement date 2. £500 on maturity [6] [Total 16] END OF PAPER CT5 A2010—7 .e.000 on death. The sum assured of £100. compounded and vesting at the end of each policy year (i. (i) Show that the quarterly premium payable in advance throughout the term of the policy or until earlier death is approximately £616. the death benefit does not include any bonus relating to the policy year of death) [10] At the end of the 25th policy year. compounded and vesting at the end of each policy year 2.5% of each quarterly premium from the start of the second policy year Renewal expenses: £45 at the start of the second and subsequent policy years Claim expense: £500 on death.000 plus declared reversionary bonuses are payable on survival to the end of the term or immediately on death if earlier. © Faculty of Actuaries © Institute of Actuaries . R D Muckart Chairman of the Board of Examiners July 2010 Comments These are given in italics at the end of each question.Faculty of Actuaries Institute of Actuaries EXAMINERS’ REPORT April 2010 Examinations Subject CT5 — Contingencies Core Technical Introduction The attached subject report has been written by the Principal Examiner with the aim of helping candidates. They have however given credit for any alternative approach or interpretation which they consider to be reasonable. The questions and comments are based around Core Reading as the interpretation of the syllabus to which the examiners are working. 95% Question generally answered well. Question generally answered well.000 = 415 SMR = 448/415 = 107. To that degree Time Selection may be spurious. (iii) The number of lives that die between x and (x + 1) out of l x lives alive at x.010468 × 10. 3 The Standardised mortality ratio is the ratio of actual deaths in the population divided by the expected number of deaths in the population if the population experienced standard mortality.500.000 + 0.000 = 0. However withdrawers from the scheme may be having an effect as their mortality could be different.0104680 Age 61: 29.531 / 2.0147918 Expected number of deaths for Urbania = 0.170 / 2. Page 2 .0147918 × 11. So Region is spurious and being confounded with occupation.000 = 0.400. Actual number of deaths for Urbania = 130+145+173 = 448 Mortality rates in standard population are: Age 60: 26. Question generally answered well.0123046 × 12. For example mortality differences by region may be put down to the actual class structure of the region itself whereas a differing varying mix of occupations region by region could be having a major effect.542 / 2. 2 Spurious selection occurs when mortality differences ascribed to groups are formed by factors which are not the true causes of these differences.000 + 0. (ii) The probability that a life age x will die between age x + n and x + n + m. Credit was given for a wide range of valid examples.0123046 Age 62: 32.200.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report 1 (i) The number of lives still alive at age x + r out of lx lives alive at age x subject to select mortality. Another example might be in a company pension scheme which might be showing a significant change in mortality experience which could be viewed as change over time.000 = 0. 38950) 9706.900(0.0002 S − 1)ds ) ∫ ⎛ ⎡e0.0002( x +t ) − e0.30 Many students answered the question well.5639 − v5 (5*0.32868 − v5 9557.38950 + 8.0002 x ⎤ ⎞ ⎦ +t⎟ = exp ⎜ − ⎣ ⎜ ⎟ 0.000 as the multiplier before the temporary assurance function rather then 9. The most common error was the use of 10.900.0002 S x +t e ds + ds ) x x (e0.0977 9557.57976) ⎟ 9706.96 + 4.900( A[50] − v5 5 p[50] A55 ) + 100(( IA)[50] − v5 5 p[50] (5 A55 + ( IA)55 )) = 9.0002 ⎝ ⎠ = 0.34 = 137. 000 − 100) A[50]:5 + 100( IA)1[50]:5 = 9.6362 Page 3 . 5 t px = exp(− ∫ x +t x +t x μ s ds ) = exp(− ∫ x = exp(− ∫ x +t 0.0002 ⎝ ⎠ (i) Probability = ⎛ ⎡e0.0977 ⎝ ⎠ = 132.8179 ⎛ ⎞ +100* ⎜ 8.0002 x 20 ⎤ ⎞ ⎣ ⎦ ⎜ = exp − + 50 ⎟ ⎜ ⎟ 0.0002 x 70 − e0.8179 *0.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report 4 1 EPV = (10. 5*0.027561 = 0.0002 x 20 ⎤ ⎞⎛ ⎛ ⎡e0.978850 * (1 − 0.8773) = 0.702 / 99.989368* 0. 000 60 s54 D55 Page 4 . This method was however credited – solution not published as not in CT5 7 (i) Age retirement benefit z ra ra (20 z M 55 + R55 ) 1 40. 1 − exp − + (70 − 60) ⎟ ⎜ ⎟ ⎜⎜ ⎜ ⎟ ⎟⎟ 0.027561 0.5 p50 Constant force of mortality μt = –ln(pt) μ50 = –ln(0.25 p51 =e −0.25* (1 − 0.0002 x 60 − e0.978850) = 0.5* (1 − 0.5 p50 * 0.25*0.725 x(1 − 0.982574 Generally answered well.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report (ii) Probability = This is the probability that the life survives to 60 and then dies between 60 and 70 40 p20 (1 − 10 p60 ) ⎛ ⎡e0.972815)) = = = 0.0002 x 70 − e0.0889 This question was answered poorly overall.978850 p51 = 95.25(1 − p51 )) 0.972815 Uniform distribution of deaths p50 0.021377 μ51= –ln(0.702 = 0.021377 * e −0.978850)) 0.0002 ⎝ ⎠⎝ ⎝ ⎠⎠ = 0.993133 = 0.0002 x 60 ⎤ ⎞⎞ ⎣ ⎦ ⎣ ⎦ ⎜ ⎜ ⎟ ⎜ ⎟ = exp − + (60 − 20) .5(1 − p50 )) (1 − 0. It was an unusual representation of the μ x function but other than that was a straight forward probability and integration question. A limited number of students used the Balducci Assumption as one of their answers. This is not in the CT5 Course whilst the above 2 methods clearly are.046 / 97. 6 p50 = 97.972815) = 0.0002 0.982588 (1 − 0.25 p51 p50 (1 − 0.813 = 0. 026 + 963.202 9976.389 = 173.17805* 8821.045 + .777 therefore fund = ( 52*1.6199 ⎞ a21:45 = a21:45 − *(1 − v 45 * l66 / l21 ) = a21:45 − * ⎜1 − 0.584 / 261.615 9. Also students often did not include the past service benefits in the final contribution rate believing the final result would have been too high (the question however was quite specific on providing past benefits). 8 (i) Fund = 52* 1. N 55 s54 D55 = K . 66. 000.2612 = 21.3909 ⎠ = a21:45 − 0.000 x 88. Most common error was the wrong sx function.3909 ) Page 5 . Also some students included early retirement calculations which were not asked for.000 60 9.3% Most students answered reasonably well.745*1.04(66−21) a21:45 45 p21 1 1 ⎛ 8695.389 = 261.40.6199 9976.e.17120* ⎟ 2 2 ⎝ 9976.584 (ii) Contributions s K 40.745*1.3909 ⇒ a21:45 = 20.868 i.0445 (20.869) 40. 240 8695.42539 a21:45 = a21:44 + v 44 * l65 / l21 = 21.777) = 7.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report = 1 (20 *128.868K Therefore K = 173. a76 ) 3 ⎡ (1 − v10 ) ⎤ 6589. 365.5) ⎥ = 52 P ⎢ + 2 * 0.25/7) would have been an acceptable alternative to 52 as the multiplier which will of course have adjusted the answer slightly.26P P = 12.04) ⎦ = 52 P [8.675564 * 3 8695.618]] = 566. Then EPV of annuity at 66 is 52 P ( a10 + 2 * v10 10 p66 .e.26P Therefore pension is given by 7.6199 ⎣ ln(1.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report (ii) Let annuity be £P per week.79 Many students struggled with this question and indeed a large number did not attempt it.272 + 2. As will be seen from the solution above the actuarial mathematics involved are relatively straightforward. Note that 52.9258 (8. 240 = 566.18 (i.169 − 0. 9 (i) We are looking to derive (aq) rx in terms of σx and μx Use the Kolmogorov equations (assuming the transition intensities are constant across a year age): ∂ r − ( σ+μ )t t ( aq ) x = σe ∂t σ (aq) rx = (1 − e−( σ+μ ) ) (σ + μ ) (ii) Similarly (aq) dx = μ (1 − e −(σ+μ ) ) (σ + μ) Note that: 1 − ((aq) rx + (aq) dx ) = e −(σ+μ ) ⇒ σ + μ = − log(1 − ((aq) rx + (aq) dx )) Page 6 . the question did not specify that constant forces must be assumed. and so would merit full marks. So. deaths occur on average at age x + ½ . then qxr = 1 − ⎡1 − ((aq) rx ⎣ + (aq) dx ) ⎤ ⎦ ( aq ) rx (( aq )rx + ( aq )dx ) In general this was poorly answered with most students making a limited inroad to the question. The following alternatives could be valid: (1) Assume dependent decrements are uniformly distributed over the year of age With this assumption. so: qxr = (ad ) rx + 12 (ad ) dx × qxr (al ) x = (aq ) rx + 12 (aq) dx × qxr ⇒ qxr = (aq) rx 1 − 12 (aq) dx (This is covered by the Core Reading in Unit 8 Section 10. For part (ii) a solution is only possible if some assumption is made.3. the above expression will turn into the answer in the above solution. a valid alternative to part (i) is: ⎡ t ⎤ = ∫ t (ap ) x σ x +t dt = ∫ exp ⎢ − ∫ ( μ x + r + σ x + r ) dr ⎥ σ x +t dt 0 0 ⎣⎢ 0 ⎦⎥ 1 (aq ) rx 1 This makes no assumptions and provides an answer in the form asked for in the question. If constant forces are assumed.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report So (aq ) rx = σ (− log(1 − ((aq ) rx + (aq ) dx ))) ((aq ) rx + (aq ) dx ) this can be rearranged to show −σ = (aq ) rx (aq ) rx log(1 − ((aq ) rx + (aq ) dx )) + (aq ) dx Given that: qxr = 1 − e −σ . However.1.) (2) Assume independent decrements are uniformly distributed over the year of age Page 7 . 00900 4 ⎝ 2 ⎠ 3 ⎛ 1 ⎞ w (aq ) 41 = .) Whilst a full description has been given above to assist students.00274 .00125) = .00276 w w q40 =.01200/(1–.00250 .00252 and q41 = .01201 and q41 = . in reality those who successfully attempted this question did assume constant forces.00250/(1–.00137) = .00276* ⎜ 1 − * *.00731) = .000 9.00274/(1–.00251 ⎝ 2 4 ⎠ ⎛ 1 3 ⎞ (aq ) d41 = .006) = . (This is covered by the Core Reading Unit 8 Section 10.01201⎟ = .855 9.01463 ⎟ = .01463 Adjusting for the 75% multiplier of independent withdrawal decrements: ⎛ 1 3 ⎞ (aq ) d40 = .01096 4 ⎝ 2 ⎠ Page 8 .Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report This leads to two simultaneous equations: q xd = (aq ) dx and 1 − ½q xr q xr = (aq ) rx 1 − ½q xd which results in a quadratic equation in qxr .00274 ⎝ 2 4 ⎠ 3 ⎛ 1 ⎞ w (aq ) 40 = . 10 First calculate (aq) dx and (aq ) wx Age (x) 40 41 42 Number of employees (al ) x 10.01463* * ⎜1 − *.00276 ⎟ = .01200 .01461 From this table and relationship 1 1 qxd = (aq) dx / (1 − *(aq) wx ) and qxw = (aq ) wx / (1 − *(aq ) dx ) 2 2 d w Calculate qx and qx d d q40 =.00252 ⎟ = .1.6.01461/(1–.684 (aq) dx (aq ) wx .01201* * ⎜1 − *.00252* ⎜1 − * *. Many failed to do so and were penalised accordingly.9*. 11 (i) Policy value at duration t of an immediate annuity payable continuously at a rate of £1 per annum and secured by a single premium at age x is given by: ∞ tVx = ax +t = ∫ e −δs s px +t ds 0 ∞ ∞ 0 0 ∂ ∂ ∂ ∂ ⇒ tVx = ax +t = ∫ e−δs s px +t ds = ∫ e −δs s px +t ds ∂t ∂t ∂t ∂t 1 ∂ ∂ ∂ × s px +t = ln( s px +t ) = (ln l x +t + s − ln l x +t ) = −μ x +t + s + μ x +t ∂t ∂t s p x +t ∂t ⇒ ∂ s p x +t = s p x +t ( −μ x +t + s + μ x +t ) ∂t ∞ ∂ ⇒ tVx = ∫ e −δs s px +t (μ x +t − μ x +t + s )ds ∂t 0 ∞ = μ x +t × ax +t − ∫ e −δs s px +t × μ x +t + s ds 0 ∞ ⎧⎪ ∞ ⎪⎫ = μ x +t × ax +t − ⎨ ⎡ −e −δs s px +t ⎤ − δ ∫ e−δs s px +t ds ⎬ ⎣ ⎦0 ⎪⎭ 0 ⎩⎪ = μ x +t × ax +t − 1 + δ× ax +t Page 9 . Because of the limited effect on the answer from the original table students were asked to show the result to 1 decimal place.00900=90.000 9.884.9 9749.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report Using the above data the Table can now be reconstructed Age (x) Number of employees (al ) x 40 41 42 10.884.0 so full credit was given for this answer also.1 10000*.00274)=27.01096=108.3 It should be noted that if more decimal places are used in the aq factors then the deaths at 40 become 25.884.00251)=25.1 9.5 Deaths Withdrawals (ad ) dx (ad ) wx (10000*.0 (9.9*. 46 8.52786) ⎟ = 25.4564) ⎟ ⎜ ⎟ 9545.38879 − 0.38879 − 0.228 Page 10 . 000 A[55]:10 + 25. 495.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report = μ x +t × t Vx − 1 + δ× t Vx (ii) Consider a short time interval (t.38879 − 0.39 Therefore P = 2118.82193 × 9287. t + dt) −1× dt = annuity payments made in time interval (t.9929 × 0.000 × 1.9929 ⎝ ⎠ = 25.10 × ( (0.2612 ⎛ ⎞ ⎜ (0. t + dt) then equation implies: t + dtV − tV = μ x +t × t Vx × dt − 1× dt + δ× t Vx × dt + o(dt ) where μ x +t × t Vx × dt = reserve released as a result of deaths in time interval (t.32953) + (0.38879 − 0. 12 (i) Annual premium P for the term assurance policy is given by: P = 1 1 25. 000 A[55]:5 a[55]:10 where 1 1 25.39 = 257. 000 A[55]:5 ( = 25. 000 A[55]:10 + 25. t + dt) In general very poorly answered on what was a standard bookwork question.000 × (1 + i )1/2 × ( A[55] − v10 10 p[55] A65 ) + ( A[55] − v 5 5 p[55] A60 ) ) 8821.36496) ) = 2118.67556 × 9545. t + dt) δ× t Vx × dt = interest earned on reserve over time interval (t.2164 × 0.019804 × ⎜ ⎟ ⎜ + (0. Possible alterations to policy structure Collect premiums more quickly by shortening premium payment term or make premiums larger in earlier years.9929 × [257. with size of debt equal to negative reserve. 042 = 980 × 0. a mortality loss) Quite reasonably answered by the well prepared student.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report Net Premium Retrospective Reserves at the end of the fifth policy year is given by: (1 + i )5 × = 1. 042 = 350. Disadvantage – if policy lapsed during the first 5 years (and possibly longer). 042 = 400.46 × 4.71 (ii) Explanation – more cover provided in the first 5 years than is paid for by the premiums in those years. explicitly stated so and then did the easier prospective calculation full marks were given. smaller in later years Change the pattern of benefits to reduce benefits in first 5 years and increase them in last 5 years.336 = -£50. 000 A[55]:5 ⎣ ⎦ 9545.00714 × 50. If the student recognised this.182 (i. Page 11 .59 − 50. Hence policyholder “in debt” at time 5.336 Total Mortality Profit = 350.21665 × l[55] l60 1 ⎤ × ⎡ Pa[55]:5 − 50.154 ADS = 8 × 50.38879 − 0.36496)] 9287.154 – 400. (iii) Mortality Profit = EDS – ADS Death strain at risk = 50.000 – (–42) = 50. the company will suffer a loss which is not possible to recover from the policyholder.000 × 1. No credit was given for a prospective calculation without explanation. In (i) it should be noted that in this case the retrospective and prospective reserves are equal.2164 = −41.e.019804 × (0.042 EDS = (1000 − 20) × q59 × 50. 074 yr 2 3690.0% 2.0% 4.000 50.0% 100.0% Allocation % (1st yr) Allocation % (2nd yr) Allocation % (3rd yr) B/O spread Management charge 95.768 12001.0% 5.001201 0.000 210.11986 0.5% 5.641 4200.001802 (aq) sx 0.000 390.00000 (ap) 0.001557 0.00 x 45 46 47 (aq) dx 0.037 7693.001201 0.727 3690.824920 Unit fund (per policy at start of year) value of units at start of year Alloc B/O policy fee Interest management charge value of units at year end yr 1 0.0% 1.0% 5.000 190.06 0.800 65.0% Multiple decrement table: x 45 46 47 q xd 0.641 yr 3 7693.998198 t −1 ( ap ) 1.000 200.00 7.12 0.938536 0.25% 5.000 581.000 195.604 137.554 Page 12 .0% 105.000 50.000000 0.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report 13 Annual premium Risk discount rate Interest on investments (1st yr) Interest on investments (2nd yr) Interest on investments (3rd yr) £4000.75% Surrender penalty (1st yr) Surrender penalty (2nd yr) £1000 £500 Policy Fee £50 % prem 15.878943 0.878943 0.000 50.0% Interest on non-unit funds Death benefit (% of bid value of units) 125% Initial expense Renewal expense Expense inflation (i) £ 200 50 2.001557 0.001802 qxs 0.000 3800.074 4000.682 213.05991 0. of profit 0.020 –2.781.769 2882.802.000 –14. 133. Then: EPV of premiums: 4 Pa(4) [35]:30 @ 6% = 56.781. of premiums profit Margin (ii) yr 2 50.926 probability in force discount factor 1 0.000 132.400 65.000 190.000 200.856 –189.768 5. Substantial credit was given to students who showed how they would tackle this question even if they did not complete all the arithmetical calculations involved.069 10167.000 4. The most common error was to ignore dependent decrements.v. 257.878943 0. 133.904+ 108.000 210.407 0.727 1.000 131. 14 (i) Let P be the quarterly premium.953 287.1408 P where (4) a[35]:30 = a[35]:30 − 3 1 − 30 p[35]v 30 8 ( ) Page 13 .461 0.760 137.31% Revised profit vector (–309.000 133.837 1.873439 yr 3 –150.v.643 = 44.824920 0.816298 133.995 29.755 3285.515 + 224.108 119.934579 expected p. 257.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report Cash flows (per policy at start of year) unallocated premium + pol fee B/O spread expenses Interest man charge extra death benefit surrender penalty end of year cashflow yr 1 250.000 expected p.492.032 Again most well prepared students made a good attempt at this question.093) Revised PVFNP = –289.037 2.000 800.280 premium signature 4000.881 213.461) Revised profit signature (–309. . 000v30 30 p[35] @ i′ × (1.5 × A[35]:30 (1 + b) = 100.9151 = 4.5 (1 + b)0.352 Page 14 .30832 × 8821. 000(1 + b)30 v 30 30 p[35] = 100.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report 3 ⎛ 8821.9151 ⎠ = 14.352 − ⎜ 1 − × 0..97857 + 45 × 13. + 1 29 q[35] (1 + b) 29 v 29.9151 ⎠ ⎝ +100. 742. 234.025 × 4 Pa[35]:1 + 45 ⎡ a[35]:30 − 1⎤ ⎣ ⎦ 1 +500 A[35]:30 + 250v30 30 p[35] = P + 250 + 0.5 × q[35] (1 + b)v + q[35] (1 + b) 2 v 2 + .1408 P − 0.06)0.06)0.5 ) +100.5 ⎛ 8821.2612 ⎞ = 14. + q[35] (1 + b)30 v 30 1 29 (1 + b)0.17411⎟ 8 ⎝ 9892.2612 9892.30832 × ⎟ (1 + b) 9892.0192308 = ( 100.5 + . 492..32187 − 0.06 − 1 = 0.706 where i′ = 1.594 + 27.06)0. 000 1 @ i′ + 100.5 + q[35] (1 + b)v1.04 1+ b EPV of expenses (at 6%) (4) (4) = P + 250 + 0. 000 × (1 + b)30 v30 30 p[35] where b = 0. 000(q[35]v 0.0352 EPV of benefits: 100.025 × 4 P × 0.025 × 4 Pa[35]:30 − 0.5 ) +100..025 × 56. 000 (1.2612 ⎞ × ⎜ 0.112 = 32. 000 × (1. 000 × 0. 5 ⎜ 0.04 = 245.04 −1 = 0 and i′′ = 1.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report 8821.30566 P + 906.55 − ⎜1 − 0.97857 8 ⎝ 9892.2164 × 0.9151 ⎠ 9892.82499 − 0. 234.5 1 × A60:5 @ i′′ + 245.9151 ⎝ = 2.000 (1 + i )1/2 A 1 @ i′′ + 245.1408P = 32.028 = £615.30566 P + 906.040.5 ⇒ A160:5 @ i′′ = ∑ d60+t 0 l60 = 465.17411× 9892.025 × 4 Pa(4) + 90a60:5 − 4 Pa(4) 60:5 60:5 (1 + b) 60:5 +1000 A160:5 + 500v5 5 p60 = 245. 000 × 0.4678 60:5 8⎝ l60 ⎠ 8⎝ 9287. 000 (1.975 × 4 Pa(4) 60:5 l60 ⎛ l ⎞ l +1000 ×1.2612 ⎞ where a(4) = a60:5 − ⎜ 1 − v5 × 65 ⎟ = 4.2164 ⎠ 4 1.78069 ) + 500 × 0.60 53.5 ( 0.040.04) 0. 000 (1.78069 Page 15 .4678 +1000 ×1.82193 × ⎟ = 4.2612 ⎛ +500 ×1.706 + 2. 000 × v5 l65 @ i′′ + 90a60:5 − 0.2612 ⎞ 8821.060.18763 − 0.8351 Gross prospective policy value (calculated at 4%) is given by: V prospective = 245.05017 9287.55 − 0.17411× ⎟ + 250 × 0.322 ⇒P= (ii) 33.9434 ⎟ = 0.05017 + 245.141.5 ⎜ A60:5 − v5 65 ⎟ + 500v5 65 l60 ⎠ l60 ⎝ l ⎞ 3⎛ 3⎛ 8821.9551 = 0.9151 ⎠ Equation of value gives: 56.60 × 4.94983 + 90 × 4.04)0.322 where a(4) [35]:1 = a[35]:1 − ( 3 1 − p[35]v 8 ) 3 ⎛ 9887.000 × v 5 5 p60 @ i′′ + 0.2069 ⎞ = 1 − ⎜1 − × 0.975 × 4 × 615. 35 + 409.954 + 232.880 Part (i) answered reasonably well.345 = 234.Subject CT5 (Financial Mathematics Core Technical) — April 2010 — Examiners’ Report = 12.5 − 10.177 + 390. Students had more problems with (ii) END OF EXAMINERS’ REPORT Page 16 . 726.473 + 45. 708. 052. Candidates should show calculations where this is appropriate. Graph paper is NOT required for this paper. 2. Mark allocations are shown in brackets. 5. In addition to this paper you should have available the 2002 edition of the Formulae and Tables and your own electronic calculator from the approved list. with any additional sheets firmly attached. 3. Attempt all 14 questions. CT5 S2010 © Faculty of Actuaries © Institute of Actuaries . 4. and this question paper.Faculty of Actuaries Institute of Actuaries EXAMINATION 6 October 2010 (am) Subject CT5 — Contingencies Core Technical Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. You must not start writing your answers in the booklet until instructed to do so by the supervisor. beginning your answer to each question on a separate sheet. AT THE END OF THE EXAMINATION Hand in BOTH your answer booklet. Enter all the candidate and examination details as requested on the front of your answer booklet. Basis: AM92 Ultimate [3] 3 Calculate the single premium payable for a temporary reversionary annuity of £12. Basis: Rate of interest Mortality of male life Mortality of female life Expenses 4% per annum PMA92C20 PFA92C20 Nil [4] 4 A gymnasium offers membership for a three-year period at a fixed fee of £240 per annum payable monthly in advance. Monthly premiums cease immediately on the death of the member.75 using the Uniform Distribution of Deaths assumption.1 Calculate: (a) 20|10 q[45] (b) 30 p[45]:[50] Basis: AM92 Select [3] 2 Calculate 0.000 per annum payable monthly in arrear to a female life currently aged 55 exact on the death of a male life currently aged 50 exact.5p45. Calculate the expected present value of membership fees if the gymnasium sells 120 memberships: Basis: Rate of interest Rate of mortality Probability of renewal Expenses 6% per annum 1% per annum 80% at each anniversary Nil [5] CT5 S2010—2 . The contract may only be cancelled at a renewal anniversary. No payment is made after 20 years from the date of purchase. .. Give a formula for the expected cashflows between the 66th and 67th birthdays as a result of entitlement from this past service. Normal retirement age is 65 and age retirement is only permitted between ages 60 and 65 exact.01 throughout for the life aged 30 now μ = 0..000.1. A member of the pension scheme currently aged 45 exact has 12 years of service and their salary in the year before the valuation date was £25. CT5 S2010—3 [6] PLEASE TURN OVER . Calculate the monthly premium for this policy using the following basis: Mortality Interest Expenses AM92 Select 4% per annum Nil [6] 8 Describe the causal factors that explain observed differences in mortality and morbidity. 000 × (1 + 0.02 throughout for the life aged 40 now δ = 4% per annum [6] 7 A life insurance company issues a 10-year term assurance policy to a life aged 55 exact... Level premiums are payable monthly in advance throughout the term of the policy or until earlier death.. The sum assured which is payable immediately on death is given by the formula: 50.5 A pension scheme provides an age retirement benefit of n/80ths of final pensionable salary where n is total number of years of service.1t ) t = 0. Final pensionable salary is the average salary in the three years before retirement.. [5] 6 Calculate: (a) A30:40 (b) a30:40:20 Basis: μ = 0.9 where t denotes the curtate duration in years since the inception of the policy.. 2. [3] The company sets up reserves in order to zeroise future negative cash flows.9 The actuary advising a pension scheme has decided that the independent mortality in the standard table for pension schemes (PEN) from page 142 of the Formulae and Tables for Actuarial Examinations is no longer appropriate for that pension scheme. (ii) (iii) Calculate the net present value of the profits after zeroisation using a risk discount rate of 6% per annum. assuming that the revised independent mortality rate at that age is 80% of the previous independent mortality rate.1 −32. The rate of interest earned on non-unit reserves is 2.2 −43. The following non-unit cash flows. (i) Show that the annual internal rate of return is 6%. Calculate the revised row of the service table for age 61.3.5% per annum. CT5 S2010—4 [3] [1] [Total 7] . Comment on the results obtained in (i) and (ii) above. [7] 10 Define the following terms.5 Assume that the annual mortality rate for the male life is constant at 1% at all ages. are obtained at the end of each year t per policy in force at the start of the year t: Year t NUCFt 1 2 3 4 −50.2.4). NUCFt (t = 1.1 145. giving formulae and defining all notation used: (a) (b) Crude mortality rate Indirectly standardised mortality rate [7] 11 A life insurance company issues a four-year unit-linked policy to a male life. Level premiums are payable annually in advance under the policy until age 65 or earlier death. defining symbols where necessary. (i) Give an expression for the gross future loss random variable under the policy at the outset.000 and attaching bonuses are payable immediately on death. The company declares simple reversionary bonuses at the start of each year including the first year and the bonus entitlement on the policy is earned immediately the bonus is declared.12 A life insurance company issued a with profits whole life policy to a life aged 40 exact on 1 January 2000.750. Bonuses declared to date total £13. the basic sum assured of £50. the policy is still in force.5% per annum simple £300 £25 at the start of the second and subsequent policy years while the policy is in force £250 [4] On 31 December 2009. (iii) Calculate the gross premium prospective reserve for the policy as at 31 December 2009 using the following assumptions: Mortality Interest Bonus loading Renewal expenses Claim expenses CT5 S2010—5 AM92 Ultimate 4% per annum 3% per annum simple £35 at the start of each policy year while the policy is in force £250 [4] [Total 12] PLEASE TURN OVER . [4] (ii) Calculate the annual premium using the following assumptions: Mortality Interest Bonus loading Initial expenses Renewal expenses Claim expenses AM92 Select 6% per annum 2. Under the policy. in respect of a policy where both the male and the female life died during the year. Premiums under each policy are payable annually in advance while at least one of the lives is alive.000 policies for the calendar year 2009. a life insurance company issued 10. [10] [Total 14] CT5 S2010—6 .13 On 1 January 2009.000 joint life whole life assurance policies to couples. there were 20 males and 10 females who died during the year. Under each policy. Each couple comprised one male life aged 60 exact and one female life aged 55 exact when the policy commenced. The life insurance company uses the following basis for calculating premiums and net premium reserves: Mortality Interest Expenses (i) PMA92C20 for the male PFA92C20 for the female 4% per annum Nil Calculate the annual premium payable under each policy. there was one claim for death benefit. In addition. a sum assured of £100. [4] During the calendar year 2009. (ii) Calculate the mortality profit or loss for the group of 10.000 is payable immediately on the death of the second of the lives to die. 500 payable on maturity or at the end of the year of death if earlier. The company uses the following assumptions in carrying out profit tests of this contract: Rate of interest on cash flows 4% per annum Mortality AM92 Select Surrenders 10% of all policies still in force at the end of each of the first. The sum assured is £21. Premiums of £5.14 A life insurance company issues four-year without profits endowment assurance policies to male lives aged 56 exact. second and third policy years Initial expenses £600 Renewal expenses £45 per annum on the second and subsequent premium dates Risk discount rate 6% per annum Calculate the expected profit margin for this contract. The company holds net premium reserves for these policies. calculated using AM92 Ultimate mortality and interest of 4% per annum. END OF PAPER CT5 S2010—7 [15] . Surrenders occur only at the end of a year immediately before a premium is paid.000 are payable annually in advance throughout the term of the policy. The surrender value is 70% of the net premium reserve calculated at the time the surrender value is payable. They have however given credit for any alternative approach or interpretation which they consider to be reasonable. The questions and comments are based around Core Reading as the interpretation of the syllabus to which the examiners are working. T J Birse Chairman of the Board of Examiners December 2010 © Institute and Faculty of Actuaries .INSTITUTE AND FACULTY OF ACTUARIES EXAMINERS’ REPORT September 2010 examinations Subject CT5 — Contingencies Core Technical Introduction The attached subject report has been written by the Principal Examiner with the aim of helping candidates. 000 ⎜ ⎢(18.0977 Question generally done well.75 = . However many students did not appreciate the split in line 1 above and attempted to apply formula directly. Page 2 )) ⎤ 24 ⎦ .821.792 − 13 − ⎢ 16.623 9941.380951 l[45] l[50] 9.955 ⎛⎡ ⎤ 10.001465) = .134 ⎡ ⎤⎞ 8.000367) *(1 − .75* q45 ) = .210 − 13 ) − v ( 24 24 9917.923 ⎣ = 12.4604 = = 0.000406) = .2612 − 6.001465 / (1 − .1673) / 9.933 − 13 )⎥ = 12.623 ⎦ ⎝⎣ 12 × 1.099)) = 2.000 × a (12) − a (12) 55:20 50:55:20 20 55 20 55 75 50:55 20 20 p50:55 (ä 70:75 − 13 20 ( ) ( ) 8784.25* q46 = .1673 5.5 p45.25* q45 / (1 − .25 p45.955 9238.25*.000 ( ⎡( ä − 13 ) − v p ( ä − 13 )⎤ − ⎡( ä − 13 ) − v 24 24 ⎦ ⎣ 24 ⎣ 8784.25 p46 = .5 p45. 2 .25 q45.0837 = 0. 3 Value of Single Premium is: ( ) = 12.201) − (16.25 q46 Hence .879.879.0837 9. However question generally done well by well prepared students. 798.000406 .999227 In general question done well.668 − 4.75 . 266.75*.75 = (1 − . 798.75 * .Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report 1 20|10 q[45] (a) (b) = (l65 − l75 ) / l[45] = (8. 706.198212 30 p[45]:[50] = l75 l80 6.000((17.367 − 3.25*.000367 by UDD = .001622 = .909 − 13 − v 20 24 ⎥⎦ ⎟⎠ 24 9917.388 Many students struggled with how to break down the monthly annuity functions into those which could then utilise the Tables. 06) = 0.792 / 1.Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report 4 The value of 1 per annum payable monthly for 1 year is ä (12) = äx(12) − v.792 * 0.96973 × (1 + 0. Page 3 . p x ) x:1 Where äx:1 = 1 Therefore ä (12) = 1 − 11/ 24(1 − 0.99 /1.5 ⎦⎥ ⎣ 80 s44 l45 ( rl )65 ⎦ ⎣⎢ t =15 80 s44 Question done very poorly.6273 The value is therefore 120 × 240 × 0.06)2 ) = 64.06 + 0.5 r45+ t ( rl )66+ t ⎤ ⎡ 12 z65 r65 ( rl )66 ⎤ 25000 ⎢ ∑ ⎥ + 25000 ⎢ ⎥ l45 ( rl )45+ t + 0. Many students attempted to use annuity functions whereas the question sought was a pure cash flow one.6273 / (1.99*0. 5 The formula is: ⎡ 19 12 z45+ t + 0.792 = 0.388 This question was overall done very poorly with few students realising that the key element to the calculation involved a one year annuity due payable monthly.792 Year 3 = 0. px äx(12) +1 = äx:1 − 11/ 24(1 − v.8 = 0.96973 x:1 The probability of reaching the beginning of each year is : Year 1 = 1 Year 2 = 0. Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report 6 (a) ∞ A______ = ∫ e−.04t {e −.01t (1 − e −.02t ) *.01 + e −.02t (1 − e−.01t ) *.02}dt 0 30:40 ∞ = ∫ {.01*(e −.05t − e −.07t ) + .02*(e −.06t − e−.07t )}dt 0 ∞ = ∫ (.01*e−.05t + .02* e −.06t − .03* e−.07t )dt 0 ∞ ⎡ .01 −.05t .02 −.06t .03 −.07t ⎤ *e *e = ⎢− *e − + ⎥ .06 .07 ⎣ .05 ⎦0 = (1/ 5 + 1/ 3 − 3 / 7) = .10476 a30:40:20 = ∫ (b) =∫ 20 −.04t e 0 * e−.01t * e −.02t dt 20 −.07t e dt 0 20 ⎡ 1 −.07t ⎤ = ⎢− e ⎥⎦ ⎣ .07 0 = (1/ .07) − e −1.4 / .07) = 10.763 Question generally done well. 7 Let P be the monthly premium. Then equating expected present value of premiums and benefits gives: 12 Pa(12) [55]:10 1 = 45000 A[55]:10 + 5000( IA)1[55]:10 where a(12) [55]:10 = a[55]:10 − ( ( ) 11 8821.2612 ⎞ ⎛ 1 − v10 × 10 p[55] = 8.228 − 0.458 ⎜1 − .67556 × ⎟ = 8.056 24 9545.9929 ⎠ ⎝ ) 1 A[55]:10 = 1.040.5 A[55]:10 − v10 × 10 p[55] = 1.040.5 ( 0.68354 − 0.62427 ) = 0.06044 ( IA)1[55]:10 = 1.040.5 (( IA) [55] − v10 × 10 p[55] × ( IA )65 − 10v10 × 10 p[55] × A65 = 1.040.5 (8.58908 − 0.62427 × 7.89442 − 10 × 0.62427 × 0.52786) = 0.3728 45000 × 0.06044 + 5000 × 0.3728 ⇒ 12 P = = 568.99 8.056 ⇒ P = £47.42 In general question done well by well prepared students. Page 4 ) Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report 8 Occupation – either because of environmental or lifestyle factors mortality may be directly affected. Occupations may also have health barriers to entry, e.g. airline pilots Nutrition – poor quality nutrition increases morbidity and hence mortality Housing – standard of housing (reflecting poverty) increases morbidity Climate – climate can influence morbidity and may also be linked to natural disaster Education – linked to occupation but better education can reduce morbidity, e.g. by reducing smoking Genetics – there is genetic evidence of a predisposition to contracting certain illnesses, even if this has no predictive capability A straightforward bookwork question generally done well although not all students captured the full range. All valid examples not shown above were credited. Students who misunderstood the question and tried to answer using Class, Time, Temporary Initial Selection were given no credit. 9 Use the formula q xα = (aq )αx (1 − 0.5((aq ) −α x )) to derive the independent probabilities: q xd q ix q xr = = = (aq ) dx (1 − 0.5((aq ) −x d )) = (50 / 6548) = 0.00809 (1 − 0.5*((219 + 516) / 6548)) (aq )ix = (219 / 6548) = 0.03496 (1 − 0.5*((50 + 516) / 6548)) (aq ) rx = (516 / 6548) = 0.080455 (1 − 0.5*((50 + 219) / 6548)) (1 − 0.5((aq )−x i )) (1 − 0.5((aq ) −x r )) Then the revised qxd = 80% *0.00809 = 0.006472 then use the formula 1 1 (aq)αx = qxα (1 − (qβx + ...) + (qβx .qxγ + ...) − ...) 2 3 Page 5 Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report to derive dependent probabilities: 1 1 (aq)dx = qxd (1 − (qix + qxr ) + (qix .qxr )) = 0.0061046 2 3 1 1 (aq)ix = qix (1 − (qxd + qxr ) + (qxd .qxr )) = 0.0334465 2 3 1 1 (aq) rx = qxr (1 − ( qxd + qix ) + ( qxd .qix )) = 0.0787948 2 3 The resulting service table is: lx dx ix rx 6,548 40 219 516 This question was done poorly. Many students appeared not to remember the derivation process for multiple decrements etc. Some students wrote down the final table without showing intermediate working. This gained only a proportion of the marks. 10 (a) Crude mortality rate = actual deaths / total exposed to risk = ∑ Exc,t mx,t x ∑ Exc,t x where Exc,t is central exposed to risk in population between age x and x+t mx,t is central rate of mortality in population between age x and x+t (b) Indirectly standardised mortality rate ∑ s Exc,t s mx,t x = ∑ s Exc,t x ∑ Exc,t s mx,t x ∑ Exc,t mx,t x Page 6 9801 0.1 (iii) As expected.unit cash flow losses have been accelerated and the risk discount rate is greater than the accumulation rate.1 ⇒ 1V = 72.9703 –50.01 0.99 0.1 = 31.3 revised cash flow in year 1 = −50.8 = 0. 11 Year t qx px t −1 p x NUCFt Profit Signature 1 2 3 4 0.025 × 1.99 0.99 0. the NPV after zeroisation is smaller because the emergence of the non.2 –42.6 = −121. Page 7 .01 0.8 and NPV of profit = –121.99 1 0.t s is central exposed to risk in standard population between age x and x+t mx.06 + 111.01 0.1 –32.2 − 71.2 − px × 1V = −50.5v3 + 141.1 145.0 − 26.0 ⇒ IRR = 6% (ii) 2V 1V = 32.01 0.2v − 42.4 + 111.99 0.5 –50.2 –43.3 1.4 − 38. In Part (ii) many students failed to develop the formulae properly although they realised the effect in (iii).t is central rate of mortality in standard population between age x and x+t This question generally done well.5 141. Parts (i) and (iii) done well generally.025 − px × 2V = 43.7v 2 − 31.Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report s c Ex.2 (i) PV of profit @ 6% = −50.7 –31. Other symbol notation was accepted provided it was consistent and properly defined.2v 4 = −47.8/1.8 = -3. 000 ⎡⎣1 + b ( K 40 + 1) ⎤⎦ vT40 + ( I − e) + eaK 40 +1 + fvT40 − Pamin( K 40 +1.560 + 13. 000 × 1.500 ×1.5 × 0.060.32907 + 1.25) Note: select functions also acceptable where b I e is the annual rate of bonus is the initial expense is the annual renewal expense payable in the 2nd and subsequent years f is the claim expense P is the gross annual premium K40(T40) is the curtate (complete) random future lifetime of a life currently aged 40 (ii) The annual premium P is given by Pa[40]:25 = 50.35 ⇒ P = £901.Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report 12 (i) The gross future loss random variable is 50. 250 ( IA ) + 300 + 25(a[40] − 1) [ 40] ⇒ P × 13.253 = 21.79 × a50:15 = 64.79 ×11. 477.85489 +300 + 25(15. 093.84 = £25.040. 250 ×1.060.5 × 3. 033.12296 + 1.065 + 300 + 362.54 − 10.444 − 901.500 ( IA ) 50 + 35a50 − 901.29 = 50. In (i) credit also given if the formulae included a limited term on the expense element although in reality this is unlikely.040.147.79 (iii) The required reserve is 64.196 + 610.29 P = 6361.5 × 0. Page 8 .5 × 8. 250 × 1. 250 A[ 40] + 1.494 − 1) ⇒ 13. 000 A50 + 1.55929 +35 × 17.32 In general question done well by well prepared students.402 + 4961. 04 ⎞ 100000 ×1.04 ⎞ 100000 ×1.086) = 0.Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report 13 (i) Let P be the annual premium.040.5 × ⎜1 − ×15.756 = 19.83 ×17.83 ×15.038462 × 19.086 60m :55 f = 1. 420. Then equating expected present value of premiums and benefits gives: Pa 60m :55 f where a 60m :55 f A = 100000 A 60m :55 f = a60m + a55 f − a60m:55 f = 15.756 ⎠ • Where the male life is alive only: 100000 A61m − Pa61m ⎛ 0.040.254 = 20475.040.917 − 14.5 × (1 − 0.01 ⎝ 15.12 ⎝ 1.210 − 14.632 + 18.040.356 ⎞ = 100000 ×1.040.2711804 ∴ P × 19.040.210 − 14.94 ⎝ 1.5 × ⎜1 − ×17.632 + 18.2711804 ⇒ P = £1.917 ⎟ − 1420. (ii) Reserves at the end of the first policy year: • Where both lives are alive: ⎛ a m f 100000 ×1.5 × A 60m :55 f = 1.04 ⎠ Page 9 .83 .254 ⎟ − 1420.04 ⎠ • Where the female life is alive only: 100000 A56 f − Pa56 f ⎛ 0.917 = 6247.5 × ⎜1 − ⎟ = 1448.254 + 17.040.5 × ⎜1 − 61 :56 ⎜ a m f 60 :55 ⎝ ⎞ ⎟ ⎟ ⎠ ⎛ 15.5 × (1 − d × a 60m :55 f ) = 1.086 = 100000 * 0. 02 = −£68.02 Hence overall total mortality profit = −97954.01 = (10.01) = (10. Only limited partial credit was given if students used only joint life situations.040.94 − 1448.169. 000 × 0.99 (b) Males only die during 2009 = 20 actual deaths (and therefore we need to change reserve from joint life to female only surviving). 000 × q60m × q55 f − 1 × 100000 ×1. Mortality Profit ( ) = 10.12 − 1448.001046 − 10 ) × (19027. 000 × p55 f × q60m − 20 × ( 6247.002451 − 20 ) × ( 4799. a mortality loss Part (i) generally done well.95 + 8265.99 + 21520.001046 − 1) × (100532.95 (c) Females only die during 2009 = 10 actual deaths (and therefore we need to change reserve from joint life to male only surviving).002451× 0.Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report Mortality Profit = Expected Death Strain – Actual Death Strain (a) Both lives die during 2009 = 1 actual claim. 000 × 0.93) = 8265. Mortality Profit ) ( ( ) = 10. Mortality Profit ( ) = 10. Page 10 .998954 × 0.997549 × 0.38 ) = −97954. Part (ii) was challenging and few students realised the full implications of “reserve change” on 1st death. 000 × p60m × q55 f − 10 × ( 20475.e.01) = (10.11) = 21520.5 − 1448. 000 × 0.01 i. 1 0.23 4319.00 0.0 0.005507 0.38 1096.47 275.07 0.007140 0.0 0.91 24.09963 0.992860 1.000000 0.09945 0.00 200.31 715.745 = 1− 1.02 4174.09936 0.792094 –338.42 3920.717685 d s Probability in force (ap)[56]+t −1 = (1 − q[56] +t −1 ) × (1 − q[56]+t −1 ) The calculations of the profit vector.870 = 0.00000 0.745 Multiple decrement table: T d q[56] + t −1 1 2 3 4 0.40 136.802525 0.20 198.00 45.45 118.05 630.36 –358.006352 0.53 3816.005507 0.1 0.72 –15759.82 514.00 45.475 –16346.20 198.93 411.943396 0.896632 0.007140 s q[56] + t −1 (aq) d[56]+t −1 (aq) s[56]+t −1 (ap)[56]+t −1 t −1 ( ap )[56] 0.00 21346.895044 0.57 153.0 = 0.47797 3.955 = 0.89663 0.Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report 14 Reserves required on the policy per unit sum assured are: 0V56:4 = 1− a56:4 1V56:4 = 1− a57:3 2V56:4 = 1− 3V56:4 = 1− a56:4 a56:4 a58:2 a56:4 a59:1 a56:4 =0 = 1− 2.20 80.78 345.894283 0.63 1.96 346.890000 0. profit signature and NPV are set out in the table below: Policy year Premium 1 2 3 4 5000 5000 5000 5000 Expenses Interest Death claim Maturity claim Surrender claim In force cash flow 600.003742 0.51 0.73298 3.00 198.1 0.80 Policy year Increase in reserves Interest on reserves Profit vector Cum probability of survival Discount factor NPV profit 1 2 3 4 4504.63 Page 11 .839619 0.00 45.24 Total NPV = 308.006352 0.00 176.80253 0.00 4145.71768 0.84 42.49 350.745 = 1− 1.896632 0.23364 3.1613 0.00 0.003742 0. 00 5000.0% 2 3 4 5000.v. Significant credit was awarded in such situation.80253 0. of premium signature => expected p.71768 0.839619 4229.Subject CT5 (Contingencies Core Technical) — September 2010 — Examiners’ Report The calculations of the premium signature and profit margin are set out in the table below: Policy year Premium probability in force discount factor 1 5000.89663 0.v.00000 5000.00000 1.00 5000.22 3012.890000 0.943396 0.40 3571. of premiums 15813.000 p.00 0.00 1.91 Many well prepared students were able to outline the process required without being totally accurate on the calculation. Many students failed to appreciate the multiple decrement element. END OF EXAMINERS’ REPORT Page 12 .53 profit margin = 2. INSTITUTE AND FACULTY OF ACTUARIES EXAMINATION 26 April 2011 (am) Subject CT5 — Contingencies Core Technical Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. Enter all the candidate and examination details as requested on the front of your answer booklet. 2. You must not start writing your answers in the booklet until instructed to do so by the supervisor. 3. Mark allocations are shown in brackets. 4. Attempt all 13 questions, beginning your answer to each question on a separate sheet. 5. Candidates should show calculations where this is appropriate. Graph paper is NOT required for this paper. AT THE END OF THE EXAMINATION Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this question paper. In addition to this paper you should have available the 2002 edition of the Formulae and Tables and your own electronic calculator from the approved list. CT5 A2011 © Institute and Faculty of Actuaries 1 Give a different example of selection shown by each of the following mortality tables: (a) (b) (c) ELT15 PMA92 AM92 [3] 2 Calculate: (a) (b) 10|5 q60 (c) s65:10 23 p65 Basis: Mortality Rate of interest 3 PMA92C20 4% per annum [4] Calculate ( Ia ) x Basis: μ x = 0.02 for all x δ = 4% per annum [4] 4 Outline the benefits that are usually provided by a pension scheme on retirement due to ill health. [5] 5 A pension scheme uses the following model to calculate probabilities, where the transition intensities are μ = 0.05 and ρ = 0.08. ρ Active Retired μ μ Dead Calculate: (a) (b) the dependent probability of retirement the independent probability of death from active service using the Kolmogorov equations. CT5 A2011—2 [5] 6 (i) Define uniform distribution of deaths [2] (ii) Using the method in (i) above calculate 1.25q65.5 [4] Basis: Mortality ELT15(Males) [Total 6] 7 Explain how education influences morbidity. [6] 8 A life insurance company issues a with profits whole life assurance policy to a life aged 40 exact. The sum assured of £100,000 plus declared reversionary bonuses are payable immediately on death. Level premiums are payable annually in advance to age 65 or until earlier death. A simple bonus, expressed as a percentage of the sum assured, is added to the policy at the start of each year (i.e. the death benefit includes the bonus relating to the policy year of death). The following basis is used to price this policy: Mortality AM92 Select Rate of Interest 4% per annum Initial expenses £300 plus 50% of the first annual premium, incurred at the policy commencement date Renewal commission 2.5% of each premium from the start of the second policy year Claim expense £350 at termination of the contract Using the principle of equivalence, calculate the level simple bonus rate that can be supported each year on this policy if the annual premium is £3,212. [6] CT5 A2011—3 PLEASE TURN OVER is payable at the end of the policy year of death. PFA92C20 (female life) Rate of interest 4% per annum Expenses Nil 10 Calculate the expected present value and variance of the present value of an endowment assurance of 1 payable at the end of the year of death for a life aged 40 exact. Premiums are payable for a period of five years.000 immediately on first death. CT5 A2011—4 . If the policyholder dies during the term of the policy. The policy pays a sum assured of £100. monthly in advance.5% of the bid value of units is deducted at the end of each policy year before death. In the first policy year 40% of the premium is allocated to units. Basis: Mortality Rate of interest Expenses 11 [7] AM92 Select 4% per annum Nil [8] A life insurance company issues a 4-year unit-linked endowment policy to a life aged 61 exact under which level premiums of £2. The percentage is based on the policy year of surrender as follows: Policy year % of total premiums payable as a surrender value 1 2 3 4 0 25 50 75 On maturity. surrender and maturity benefits are paid. The unit prices are subject to a bid-offer spread of 5%. An annual management charge of 0. Basis: Mortality PMA92C20 (male life).9 A male life aged 52 exact and a female life aged 50 exact take out a whole life assurance policy.500 are payable yearly in advance throughout the term of the policy or until earlier death. in which case a value equal to a fixed percentage of the total premiums paid on the policy is payable at the end of the policy year of surrender. The policyholder may surrender the policy. 105% of the bid value of units is payable. the death benefit of £10. while in the second and subsequent policy years 110% of the premium is allocated to units.000 or the bid value of the units. with a term of 15 years. whichever is higher. Calculate the monthly premium payable. [3] (ii) Construct tables showing the growth of the unit fund and the non-unit fund. Include all commissions in the non-unit fund.25% per annum Rate of interest on non-unit fund cash-flows 3.5% of the second and subsequent years’ premiums Risk discount rate 5. [3] [Total 13] CT5 A2011—5 PLEASE TURN OVER . [7] (iii) Calculate the profit margin for this policy on the assumption that the company does not zeroise future expected negative cashflows.5% per annum (i) Construct a multiple decrement table for this policy assuming that there is a uniform distribution of both decrements over each year of age in the single decrement table.The company uses the following assumptions in carrying out profit tests of this contract: Rate of growth on assets in the unit fund 4.5% per annum Independent rate of mortality AM92 Select Independent rate of surrender 6% per annum Initial expenses £325 Renewal expenses £74 per annum on the second and subsequent premium dates Initial commission 10% of first premium Renewal commission 2. inflating at 4% per annum compound. The initial sum assured was £75.5% of each premium from the start of the second policy year Renewal expenses £75 per annum. at the start of the second and subsequent policy years (the renewal expense quoted is as at the start of the policy and the increases due to inflation start immediately) Claim expense £300 on termination (the claim expense is fixed over the duration of the policy) (i) Show that the monthly premium for the policy is approximately £56. a life insurance company issued a 25-year term assurance policy to a life then aged 40 exact.12 On 1 April 1988. (ii) Calculate the gross premium prospective reserve as at 31 March 2011. The sum assured is payable immediately on death and level monthly premiums are payable in advance throughout the term of the policy or until earlier death. The company uses the following basis for calculating premiums and reserves: Mortality AM92 Select Rate of interest 4% per annum Initial commission 50% of the total premium payable in the first policy year Initial expenses £400 paid at the policy commencement date Renewal commission 2.000 which increased by 4% per annum compound at the beginning of the second and each subsequent policy year. [6] [Total 16] CT5 A2011—6 [10] . 000 term assurance policies to male lives then aged 40 exact and 20. During the first ten years. On 1 January 2000. including formulae. there were 22 actual deaths from the term assurance policies and 36 actual deaths from the endowment assurance policies. (a) Calculate the death strain at risk for each type of policy during 2010.000 endowment assurance policies to male lives then aged 35 exact.000 The death benefit under each type of policy is payable at the end of year of death. During 2010. Basis: Mortality AM92 Ultimate Rate of interest 4% per annum Expenses Nil [11] [Total 17] END OF PAPER CT5 A2011—7 . the following expressions assuming that the sum assured is payable at the end of the year of death: • • • (ii) death strain at risk expected death strain actual death strain [6] A life insurance company issues the following policies: • • 25-year term assurances with a sum assured of £200. the company sold 10. (b) Calculate the total mortality profit or loss to the office in the year 2010. there were 145 actual deaths from the term assurance policies written and 232 actual deaths from the endowment assurance policies written. premiums are payable annually in advance. (c) Comment on the results obtained in (b) above.13 (i) Explain. For each type of policy. Assume that there were no lapses/withdrawals on each type of policy during the first eleven years.000 25-year endowment assurances with a sum assured of £100. They have however given credit for any alternative approach or interpretation which they consider to be reasonable.INSTITUTE AND FACULTY OF ACTUARIES EXAMINERS’ REPORT April 2011 examinations Subject CT5 — Contingencies Core Technical Introduction The attached subject report has been written by the Principal Examiner with the aim of helping candidates. T J Birse Chairman of the Board of Examiners July 2011 © Institute and Faculty of Actuaries . The questions and comments are based around Core Reading as the interpretation of the syllabus to which the examiners are working. Only minimal credit was given for this..8662 × 0. 647.06) = 294..06 + 3( e −.131 = (a) 23 p65 (b) 10|5 q60 (c) s65:10 = = (1 + i )10 a65:10 10 p65 = (1 + i )10 (a65 − v10 10 p65 a75 ) 10 p65 ( (1.67556 × 0.04−10 ) × 8.19 This question was not done well..456) = 1.... 405.160) = 70 75 = = 0.)a 1 at force of interest 6% = (1/(1 − e −.06 ))2 × ((1 − e −. April 2011 1 (a) (b) (c) Time selection – because it is based on a period of three calendar years Class selection – applies only to male pensioners Temporary initial selection – as there are select rates Other valid answers acceptable This question was generally done well.970591 = 286. Instead most attempted ∞ to compute ( Ia ) x = ∫ tvt t px dt .. 2 l88 3534.456) / 0..666 − 0.160 9.87120 × 9.48024 × (13.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.366307 l65 9647.04)10 (13. 3 1 2 3 ( Ia ) x = ∫ v t t p x dt + 2 ∫ v t t p x dt + 3∫ v t t p x dt + .87120 = 13.666 − (1.04 *e 1 −. Hence ( Ia ) x = (1 + 2e −0.797 (8. although the payment ā is continuous... 0 Now vp x = e −.06 ) / .054 = = 0. 647..160 9.134 − 8405. However some students did not supply different selection types for each part and this was penalised. 0 Page 2 ..06 )3 + ..06 )2 + 4( e −.797 (l − l ) (9238.797) ) × 9. 405.02 2 =e −.764 This question was generally done well for parts (a) and (b) but students struggled more with part (c).06 throughout. The majority of students failed to realise that the increasing function I was not continuous.084771 l60 9826.. pensionable service is usually more generous than under age retirement with years beyond those served in the scheme being credited to the member e. Other valid points were credited. 5 The Kolmogorov equations in this case are: δ r − (μ+ρ)t t (aq ) x = ρe δt δ d − (μ+ρ)t t (aq ) x = μe δt For the case where t = 1 the solution for the dependent probability of retirement is: ( aq) rx = ρ (1 − e − (μ+ρ ) ) ρ+μ Hence the dependent probability of retirement is 0.07502 (aq) rx = The formula for the independent probability of death is qxd = 1 − e −μ Hence the independent probability of death is: qxd = 1 − e −0.04877 Generally this question was completed satisfactorily by well prepared students. actual pensionable service subject to a minimum of 20 years.08 (1 − e −(0. However. Benefits are usually related to salary at the date of ill-health retirement in similar ways to age retirement benefits.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.08) ) 0.05+0. April 2011 4 Schemes usually allow members to retire on grounds of ill-health and receive a pension benefit after a minimum length of scheme service.g.05 = 0. Page 3 . Generally this bookwork question was done well. A lump sum may be payable on retirement and a spouse pension on death after retirement.05 = 0. or pensionable service that would have been completed by normal retirement age.08 + 0. 5 p65. Shows in: • • • • • • • Increased income Better diet Increased exercise Better health care Reduced alcohol and tobacco consumption Lower levels of illicit drug use Safer sexual practices Some effects are direct (e.5 × 0.5 × 0.96753 = 0.5 ) = (1 − (0. (ii) We have 1.75 p66 = (1 − 0. drug use).02447))) = 0.98761 × 0.5 = 1 − 0.75 × q66 = 1 − 0.5 q65. which reduces morbidity.03247 A straightforward question that was generally done well.5q65 / (1 − 0.25 p65.5 × 0. such as public health awareness campaigns. exercise) Students generally scored on a range of points but in most cases did not write enough of them to gain all the marks.75 × 0. some are indirect (e.5 = 0. Page 4 .5 p65.5q65 ))) by UDD = (1 − ((0.97967 = 0.75 p66 = 1 − 0.25 p65.25 p65.97967 Hence 1. Students who mentioned over indulgence risks for the better educated were given credit. Education includes formal and informal processes.qx (alternatively t pxμ x +t is constant).Subject CT5 (Contingencies Core Technical) — Examiners’ Report.98761 0.5 0. 7 Education influences the awareness of a healthy lifestyle.02711 = 0. April 2011 6 (i) The definition of the uniform distribution of deaths (UDD) is s qx = s.g.25 q65.5 = 1 − 1.g.02447) / (1 − 0.96753 ⇒ 1.5 = 0.75 q66 = 1 − 0. 025) = (100.5423 + 8. 000 + 350) × (1. a simple bonus rate of 3% per annum Generally done well although some students treated b as not vesting in the first year.04) × a52:50 ) = 101. 9 Value of benefits using premium conversion 100.85 There was an anomaly in this question in that it was not fully clear that the premium paying period ceased on 1st death within the 5 year period.115.e.23041 + 1. interest is (where P = 3.244 × 9.04 ) 0.3823 = 627.000 × (1.9564b + 1.e. April 2011 8 Let b be the simple bonus rate (expressed as a percentage of the sum assured).04 / 1.100 ⎛ v5 × l57:55 ⎞ = (0.458 = 15.295 − 0.348.3823P Therefore: P = 34.837 − 0. 000b × (1.04)1/2 × A52:50 = 100.6179 = 23.458 = 16. This was credited.025) = (100.5 P P(.9564 i.81491 Hence 12 Pä (12) 52:50:5 = 12 P(16.89 / 54.81491 × 15.000 A52:50 = 100. Page 5 .4 × (1 − 0.558 − 0.295) = 34.212): P(.833.00 ⇒b= 8.143.906 24.952.196 × 9.04 ) 0. ignoring the joint life contingency.697) ⎜ ⎟ l 52:50 ⎠ ⎝ = 0.82193 × 9.880. Then the equation of value at 4% p.95835 + 300 + 0.837 (12) ä57:55 = ä57:55 − 11/ 24 = 15.0384615 × 17. 000 + 350) A[40] + 1.115.975a[40]:25 + 0.930.0756 = 3.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.000 × (1.579.a.04)1/2 × (1 − (0. 000b( IA)[40] + 300 + 0.5 × 7.143.917.887 + 0.100) = 54.5 P ⇒ 49.623) / (9.975 × 15.980. Even though the vast majority of students who completed this question used the above solution a small minority used 12Pa5(12) i.5 × 0.89 Value of monthly premium of P 12 Pä (12) 52:50:5 ⎛ (12) ⎛ 5 l57:55 ⎞ × ä (12) ⎞ = 12 P ⎜ ä52:50 −⎜v × l52:50 ⎟⎠ 57:55 ⎟⎠ ⎝ ⎝ (12) ä52:50 = ä52:50 − 11/ 24 = 17. 3036 ⎝ ⎠ = 0. Question done well by well prepared students.23041 − 0.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.55526 × × 0.55919 Variance = 2 A [40]:15 −( A [40]:15 )2 2 1 A [40]:15 = 2 A[40]:15 + 2 A[40]:115 = 2 A[40] − (v 2 )15 15 p [40] 2 A55 + (v 2 )15 15 p [40] 9.20977 + 0.06775 − 0.557.854.557.8179 9. Page 6 .559192 = 0.854.557.05318 + 0.3036 9.30832 × × 0.854.55526 × 9.8179 ⎛ ⎞ = 0.23041 − ⎜ 0.3036 ⎝ ⎠ = 0.000917 Note answers are sensitive to number of decimal places used.31361 − 0.854.30832 × 9.29904 = 0. Many students failed to realise that the endowment function needed to be split into the term and pure endowment portions.8179 9.17785 ⎟ + 0.557.31361 So variance = 0.3036 9. April 2011 None the less many students struggled with this question 10 Expected present value is A [40]:15 where 1 A [40]:15 = A[40]:15 + A[40]:115 = A[40] − v15 15 p [40] A55 + v15 15 p [40] 9.53855 = 0.38950 ⎟ + 0.8179 ⎛ ⎞ = 0.06775 − ⎜ 0. 012716 0.750.75 3.0% 2.536.05966 0.000000 0.07 6. April 2011 11 Summary of assumptions: Annual premium Risk discount rate Interest on investments Interest on sterling provisions Minimum death benefit 2.930886 0.933953 0.750.006433 0.00 40.08 2.42 2.5% Total 575 136.05981 0.929337 0.50 269.00 137.000.00 £ 325 74 Initial expense Renewal expense (i) (ii) Allocation % (1st yr) Allocation % (2nd yr +) Man charge B/O spread 40% 110% 0.05971 0.37 4.807969 Unit fund (per policy at start of year) value of units at start of year alloc B/O interest management charge value of units at year end yr 1 yr 2 yr 3 yr 4 0.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.732.91 18.869404 0.46 Page 7 .00 137.012335 0.00 137.15 2.009405 0.000.0% Maturity benefit 105% % prm 10.009696 0.06 x (aq) dx (aq) sx (ap) t −1 ( ap ) 61 62 63 64 0.25% 3.95 985.5% 5.05962 0.928047 1.750.006240 0.65 33.08 3.5% 10.06 0.933953 0.5 Multiple decrement table: x qxd qxs 61 62 63 64 0.15 6.732.73 47.50 152.00 5.011004 0.581.92 9.581.42 985.5% 4.06 0.50 390.00 50.011344 0.06 0.500.00 1. 79 1.996 P where (12) a[40]:25 = a[40]:25 − = 15.20 0.37512 ⎟ ⎜1 − 24 ⎝ 9854.933953 0.94 0.00 34.v.72 18.869404 0.386.947867 0.952.76 –250.00 575.36 –250.887 − = 15. Then: EPV of premiums: 12 Pa(12) [40]:25 @ 4% = 186.12 4.00 –93.00 –149.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.72 –121. April 2011 Non-unit fund (per policy at start of year) yr 1 yr 2 yr 3 yr 4 unallocated premium B/O spread expenses/commission interest man charge extra death benefit extra surrender benefit extra maturity benefit end of year cashflow 1.07 37.03 premium signature expected p.00 1.41 442.00% Credit was given to students who showed good understanding of the processes involved even if the calculations were not correct.72 47. Generally well prepared students did this question quite well.213.v.50 –8. 12 (i) Let P be the monthly premium.00 50.00 137.92 5.50 –8.50 –8.16 1. of profit 419.50 136.19 (iii) expected p.62 probability in force discount factor 1 0.00 137.898452 0.72 33.851614 0.75 58.71 –250.016.583 Page 8 ( 11 1 − 25 p[40]v 25 24 ) 11 ⎛ 8821.00 137.25 –58.95 56.807217 2.51 –536.2612 ⎞ × 0.91 0.50 136.807969 0.3036 ⎠ .62 –168. of premiums profit margin 2.720.15 5.50 136.500.95 –148.500.00 8. 000 × (1 + i )0.025 ×12 Pa[40]:25 − 0.6880 where i/ = 1.5 + q[40] (1 + b)v1.5 ×12 P + 400 + 0. + l64 ) = e[40] − l64 e64 l[40] 8934..025 ×12 Pa[40]:1 + 75 ⎡ a[40]:25 − 1⎤ ⎣ ⎦ 1 +300 A[40]:25 = 6 P + 400 + 0. + 1 24 q[40] (1 + b) 24 v 24.5 (1.8771 ×17. 000 ⎛ 8821.00 i.982025 + 75 × 23..5 75.96154 ⎟ = 0.025 × 12 P ×15. 000 × (1 + i )0.3803P + 2161.27542 +300 × 0.05422 = 6 P + 400 + 4.5 1 ⎡⎣ A[40] − 25 p [40] v 25 A65 ⎤⎦ @ i / × A[40]:25 @ i/ = (1 + b) (1 + b) = 75.04 = 75.04) ⎝ 9854.27541 9854.3036 ⎠ = 7709.5 + .9218 where ( ) 11 1 − p[40]v [40]:1 24 11 ⎛ 9846.04 − 1 = 0.5384 ⎞ = 1 − ⎜1 − × 0.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.071 − 1 l[40] (l [40]+1 + .2612 ⎞ × ⎜1 − × 1× 1⎟ 0..025 ×12 P × 0.e.583 − 0.982025 24 ⎝ 9854.5 ) where b = 0.6749 P − 0.421 = 23.3036 ⎠ a(12) = a[40]:1 − @0% −1 = a[40]:25 = 39. 000(q[40]v 0.3036 Page 9 .. April 2011 EPV of benefits: 75. i / = 0% 1+ b EPV of expenses (at 4% unless otherwise stated (12) (12) @0% = 0.6558 + 16..266 = 10.2946 P + 1745. 5106 = £3.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.2612 ⎤ ⎡ = 1. 1. i. 463. the present value at time t + 1 of all benefits payable on death during the year t + 1) over the end of year provision.98058 × [ 0.96154] + 0.37512 × = 0.854 [1 + (1.96154] − 12 × 55.92456 × ⎟ = 1.04) 23 × v 0.04) 23 [1 + (1.012716 × 0.3803P + 2161. 2…) is the excess of the sum assured (i.89 ×1.04) × p63q64 × v ] + 300v 0.89 ×1.335.011344 + 0.6880 + 10.38896 − 0.89 176.040. EDS for year t + 1 = q ( S − t +1V ) Page 10 .011344 + (1.90629 where a(12) = a63:2 − 63:2 11 ⎛ 11 ⎛ 8821. 000 × (1.04) × 0.5 A[40]:25 = 1.996 P = 7709.5 ⎢ 0.3036 ⎥⎦ ⎣ Equation of value gives: 186.853. 278.e.0628 + 6.5 ⎢ A[40]:25 − v 25 65 ⎥ A[40]:25 l[40] ⎦⎥ ⎣⎢ 8821. DSAR for year t + 1 = S − t +1V The “expected death strain” for year t + 1 (t = 0.90629 +184.98058 × [ 0. In part (i) although it was commonly recognised that a resultant rate of interest of 0% emerged students did not often seem to know how to progress from there.012716 × 0.8932 + 31.040.66 × 0.6157 Gross prospective policy value at t = 23 (calculated at 4%) is given by: (ii) V prospective = 75.951 − ⎜1 − 0.6098 ⇒P= = £55.988656 × 0. April 2011 ⎡ l ⎤ 1 1 = 1.025 ×12 Pa(12) 63:2 +75 × (1.040.6104 − 1.988656 × 0.04) × 0.5 [ q63 + p63q64v ] + 0.04) p63v ] − 12 Pa(12) 63:2 = 184.9218 9871.e.90629 24 ⎝ 24 ⎝ 9037. 2…) is the amount that the life insurance company expects to pay extra to the end of year provision for the policy.2612 ⎞ 2 l65 ⎞ ⎜ 1 − v × ⎟ = 1.96154] +300 × 0.988656 × 0.9628 + 367. 1.3973 ⎠ l63 ⎠ = 4.05422 9854. 13 (i) The death strain at risk for a policy for year t + 1 (t = 0.02 This question was generally not done well especially part (ii).5 [ q63 + (1.025 ×12 × 55.e. i. 37512 × PTA = 8.05334 = 671.38907 − 0.e. (ii) (a) Annual premium for endowment assurance with £100. 1.58393 − 2.000 × 0.891.38359 = 2.62 15.000 sum assured given by: 11V EA = 100.000 sum assured given by: P TA = 1 200.000 A40:25 a40:25 1 where A40:25 = A40:25 − v 25 25 p40 = 0.2863 200.40 a35:25 16. 000 100.393.8 = 32.884 Reserves at the end of the 11th year: – for endowment assurance with £100.38907 − 0.393.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.393.856.40 × 10.05334 9. 2…) is the observed value at t+1 of the death strain random variable i. April 2011 The “actual death strain” for year t + 1 (t = 0.000 sum assured given by: Page 11 .2612 = 0.000 × A46:14 − P EAa46:14 = 100. ADS for year t + 1 = ( S − t +1V ) if the life died in the year t to t+1 = 0 if the life survived to t + 1 Note: Full credit given if definition of death strain is given for a block of policies rather than for a single policy as per above.027 Annual premium for term assurance with £200.818 = 58.0 − 25.000 sum assured given by: P EA = 100.000 × 0.821.33573 = 0.2 – for term assurance with £200. 000 × A35:25 = × 0.501. 57748 × 11V TA 8.538.06301 9.000 – 5.e.183.501.2 = 67.8 = 2.002508 × 194. 602. the actual number of deaths for the endowment assurances is approximately 25% higher than expected.5 (i.8 = 19768 × 0.8 mortality profit = –475. 498.6 = 9. sums at risk are: Endowment assurance: DSAR = 100.855 × . 246.3 ADS = 36 × 67. Further investigation would be required to determine reasons for poor mortality experience for the endowment assurances. a loss) For term assurance EDS = 9. 498.52583 = 0.4 (c) Although there is an overall mortality profit in 2010.58884 − 0. which is a concern. there may have been limited underwriting requirements applied to this type of contract when they were written. 429. total mortality profit = 528.4 = 194.577. 498.7149 = 200. Many students did not attempt (c) or at best gave a somewhat sketchy answer. April 2011 11V TA 1 = 200. 773.355.8 = 1.6 = 4.0 − 7. Generally (a) was done well.6 Mortality profit = EDS – ADS For endowment assurance EDS = 19768 × q45 × 67.001465 × 67.06301 − 671. 000 A51:14 − PTA a51:14 1 where A51:14 = A51:14 − v14 14 p51 = 0.6 = 5.9 Hence.577.821. The most common error in (b) was to assume reserves at 10 years rather than 11.5 = £53.956.4 Therefore.g. 000 × 0.69 = 12.9 – 475.1 ADS = 22 × 194. e.183.809.2 mortality profit = 528.707. 280. 422.577. On the whole well prepared students coped with (b) well.179.000 − 32.62 × 10.855 × q50 × 194.498.58884 − 0.6 = 4.422.8 Term assurance: (b) DSAR = 200.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.954.577.2612 = 0. 687. END OF EXAMINERS’ REPORT Page 12 .538. In addition to this paper you should have available the 2002 edition of the Formulae and Tables and your own electronic calculator from the approved list. 2. Mark allocations are shown in brackets. with any additional sheets firmly attached. 5. 4. beginning your answer to each question on a separate sheet. Graph paper is NOT required for this paper. CT5 S2011 © Institute and Faculty of Actuaries . and this question paper. Attempt all 14 questions.INSTITUTE AND FACULTY OF ACTUARIES EXAMINATION 4 October 2011 (am) Subject CT5 — Contingencies Core Technical Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. Enter all the candidate and examination details as requested on the front of your answer booklet. You must not start writing your answers in the booklet until instructed to do so by the supervisor. Candidates should show calculations where this is appropriate. 3. AT THE END OF THE EXAMINATION Hand in BOTH your answer booklet. (b) Calculate A1x: y on the following assumptions: [4] μ x = 0.03 for all y δ = 4% per annum CT5 S2011—2 [5] . the following relationships are true for all ages: 0.5 q75. Calculate the expected value and variance of this contract.5 = 0. AM92 [3] In a special mortality table with a select period of one year.1 Calculate: (a) 10|1 q[50] (b) 10 p[60]+1 a(12) [40]:20 (c) Basis: Mortality Rate of interest 2 Calculate 0.5 q[ x ] = 0.5 q[ x ]+ 0.25qx 0.45qx Express p[x] in terms of px .02 for all x μ y = 0. Basis: Mortality Rate of interest 5 AM92 Ultimate 4% per annum (a) Write down the random variable form of A1x: y .25 Basis: Mortality 3 AM92 6% per annum using the assumption of a constant force of mortality. 4 [3] [3] A term assurance contract with a term of 20 years pays a sum assured of 1 immediately on death to a life now aged 30 exact. 500 15. Basis: Mortality PMA92C20 (male life). (b) Hence or otherwise derive the standardised mortality ratio and the indirectly standardised mortality rate. [6] Members of a pension scheme are subject to three decrements: (a) Deaths . PFA92C20 (female life) Rate of interest 4% per annum Expenses Nil 8 The following data is extracted from a population census: Age 20–29 30–39 40–49 50–64 9 [6] All Professions Population Deaths 120. whilst other decrements act uniformly across the year of age. [6] CT5 S2011—3 PLEASE TURN OVER .6 Explain why it is necessary to have different mortality tables for different classes of lives. Age retirements are assumed to take place on the attainment of the exact age. Calculate the expected present value of this annuity.000 14.000 156.000 256 458 502 600 Profession A Population Deaths 12.025) (c) Age retirement .with an independent decrement rate of 0.000 30 40 50 60 (a) Calculate the area comparability factor for Profession A using the data for All Professions as the standard population.01 at age 50 exact increasing by 0.with an independent decrement rate of 0.000 178. Payments commence on the first death and continue for 5 years after the second death. Calculate the probability that a member currently aged 59 exact will retire at age 62 exact.with independent decrement rates that are assumed to follow ELT15(Males) (b) Ill-health retirement .2 at each age from 60 to 64 all exact.005 for each additional year of age (so the ill-health independent decrement at age 53 exact is 0.000 123. [6] 7 A special joint life last survivor annuity of £10.000 16.000 per annum is payable continuously in respect of a male and female life each aged 60 exact. [2] [Total 7] A pension scheme provides a pension on retirement of 1% of final pensionable salary for each completed year of pensionable service. Basis: Mortality AM92 Select Rate of interest 4% per annum (b) 11 [5] Suggest two reasons why a life insurance company might use the super compound method of adding bonuses to with profits policies. Final pensionable salary is defined as the average salary in the last three years before retirement. [8] CT5 S2011—4 . pensionable service is calculated as service that would have been completed by the normal retirement age of 65. The super compound method of adding bonuses to the policy is used as follows: • each year there is a simple bonus of 2. Define all symbols used. without using commutation functions. as opposed to the compound method. The sum assured is £150.000 in the year after entry to the scheme. for the present value of the benefits for a new member age 30 with salary of £20.10 (i) Five years ago a with profits whole life assurance policy was sold to a life then aged 30 exact.5% on the sum assured • and an additional bonus of 5% on all existing bonuses (excluding the simple bonus relating to that policy year) Assume that bonuses vest at the start of each policy year and that the actual past bonus additions have followed the assumptions stated above. Derive an expression. Calculate the net premium policy value just before payment of the 6th premium. On retirement due to ill-health.000 payable at the end of year of death and premiums are payable annually in advance throughout life. 000 payable on maturity or at the end of the year of death if earlier.5% per annum [5] [Total 9] A life insurance company issues a 3-year without profits endowment assurance policy to a male life aged 57 exact for a sum assured of £15. The surrender value payable is 75% of total premiums paid on the contract at the time the surrender value is payable. –334. Assume that at the end of the first and second policy years.05.82) (ii) Determine the net present value of the profits of this policy. 10% and 5% respectively of all policies still in force at that time then surrender. The office holds net premium reserves for these policies. (iii) Calculate the revised net present value of profits and comment on your answer. –292. –933. Surrenders occur only at the end of a year immediately before a premium is paid.89.5% per annum 7. assuming that the company sets up reserves in order to zeroise future negative expected cash flows on the policy. Premiums of £4. [4] A four-year unit-linked policy issued by a life insurance company to a life aged 56 exact has the following profit vector: (1525. (ii) Calculate the internal rate of return for this contract. Basis: Mortality Rate of interest on non-unit fund cash flows Risk discount rate 13 AM92 Ultimate 4.12 (i) List the main features of a unit-linked policy.08. calculated using AM92 Ultimate mortality and interest of 4% per annum.700 are payable annually in advance throughout the term of the policy. The company uses the following assumptions in carrying out profit tests of this contract: Rate of interest on cash flows and Reserves Mortality Initial expenses Renewal expenses Risk discount rate 5% per annum AM92 Select 10% of the annual premium £65 per annum on the second and subsequent premium dates 7% per annum (i) Calculate the net present value of profits for this contract. [4] [Total 16] CT5 S2011—5 PLEASE TURN OVER . [10] [2] The company weakens the reserving basis by assuming that net premium reserves for these policies are now calculated using AM92 Ultimate mortality and interest of 6% per annum. During 2010. [4] There were 385 policies in force on 1 January 2010. Premiums are payable annually in advance. there were three actual deaths.5% of each premium from the start of the second policy year (i) Write down the recursive relationship between the gross premium reserves at successive durations of these policies. (iii) Calculate the profit or loss made by the company from both mortality and interest in respect of these policies for the year 2010 based on the formula stated in (i) above.000 at maturity.803.000 at the end of the year of earlier death to lives then aged 35 exact. [4] (ii) Show that the annual premium for each policy is approximately £1.14 On 1 January 2001. defining all symbols used. The company uses the following basis for calculating premiums and reserves: Mortality Interest Initial commission Initial expenses Renewal expenses AM92 Select 4% per annum 50% of the premium payable in the first policy year £300 paid at policy commencement date 2. a life insurance company issued a number of 30-year endowment assurance policies that pay £100. actual interest earned by the company was 5% and expenses were as expected. [10] [Total 18] END OF PAPER CT5 S2011—6 . or £50. the questions set.INSTITUTE AND FACULTY OF ACTUARIES EXAMINERS’ REPORT September 2011 examinations Subject CT5 — Contingencies Core Technical Purpose of Examiners’ Reports The Examiners’ Report is written by the Principal Examiner with the aim of helping candidates. This is much more than a model solution – it would be impossible to write down all the points in the report in the time allowed for the question. T J Birse Chairman of the Board of Examiners December 2011 © Institute and Faculty of Actuaries . will generally be based on Core Reading. Other valid approaches are always given appropriate credit. and also those who have previously failed the subject. the Examiners are not required to examine the content of Core Reading. For numerical questions the Examiners’ preferred approach to the solution is reproduced in this report. Notwithstanding that. both those who are sitting the examination for the first time and who are using past papers as a revision aid. and particularly the open-ended questions in the later subjects. this report contains all the points for which the Examiners awarded marks. this is also noted in the report. which is designed to interpret the syllabus. The Examiners are charged by Council with examining the published syllabus. Although Examiners have access to the Core Reading. and the following comments. For essay-style questions. where there is a commonly used alternative approach. Questions that were done less well were 7. In the case of descriptive answers credit is also given where appropriate to different valid points made which do not appear in the solutions below. Comments on the September 2011 paper The general performance was slightly worse than in April 2011 but well-prepared candidates scored well across the whole paper. Students should note that for long questions a reasonable level of credit is given if they can describe the right procedures although to score well reasonable accurate numerical calculation is necessary. September 2011 General comments on Subject CT5 CT5 introduces the fundamental building blocks that stand behind all life insurance and pensions actuarial work. 11 and 14(iii) and here more commentary is given to students to assist with further revision. a different notation system produced by a student to that used by examiners is acceptable provided it is used consistently. is relevant and is properly defined and used in the answer. Page 2 .Subject CT5 (Contingencies Core Technical) — Examiners’ Report. Most of the short questions were very straightforward where an answer could be produced quickly and this is where many successful candidates scored particularly well. In questions where definitions of symbols and then formulae are requested. 9. Credit is given to students who produce alternative viable numerical solutions. 10. 5 ) = (1 − 0.55 + .20 l[50] l[60]+1 9.5 q[ x]+0.95795) = 0.0977 = 7.5 = (1 − 0.45(1 − p x )) = (0.85285 v 20l60 )) l[40] = 12.5 p[ x ]+0.4125 + 0.25qx ) *(1 − 0.04296dt 75.25 = e = e −0.4508 = 0.04296 75. September 2011 1 d 60 = 74.45 px ) = 0.95795 = e −μ where μ is the constant force Hence μ = − ln(0.00768 9.45qx ) = (1 − 0.1673 = 0.97875 Hence 0. 706.6568 = 0.589.5 p75.879.2164 / 9.25 = 1 − 0. 287.1125 p x2 This question was done reasonably well but many students failed to make the connection in line 1.25 = 0.3036) = 11.5 q[ x] )*(1 − 0.5 q75.5 ) .02125 Again done well.Subject CT5 (Contingencies Core Technical) — Examiners’ Report. 3 p[ x ] = 0. 2 We have: 1 p75 = 6.9258 / 6.25 0.5020 (a) 10|1 q[50] = (b) 10 p[60]+1 = l71 (c) a(12) = (a[40]:20 − 11/ 24 × (1 − [40].75 Hence −∫ 0.75 + .854.676 Straightforward question generally done well.5 p[ x ] * 0.3118 × 9.5 p75.854.25(1 − p x )) *(1 − 0.25 px ) *(. 209.02148 = 0.000 − 11/ 24 × (1 − 0.475 p x + 0. Credit was given to those students who jumped straight to the solution of (1 − ( 1 p75 )0. . 0728 / 9. The most common error was to forget to use continuous functions which was penalised as one of the key attributes being tested was to see if students could work out the 1.13065)) × 1.e .0728 / 9.04 factor for the variance.04t ⋅ t p ⋅μ x +t dt 0 xy ∞ ∞ 0 0 = ∫ e −.04 = 0.a simultaneous death where it could be argued either that Z =0 or is undefined.45639 × (9.03528 − (0.09 = 0. Page 4 .02)dt = 0. Part (a) comes directly from Core Reading but there is some debate about the situation where Tx = Ty i. Part (b) was generally well done. September 2011 4 1 The expected value is A 30:20 This equals ( A30 − ( v 20 × (l50 / l30 ) × A50 )) × (1.712. The examiners decided to accept all these alternative situations.09t dt = 0.013532 = 0.04) = (0.2094) × 0.02 / 0.02t ⋅ e−.008814 This question was done reasonably well.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.008997 − 0. 1 (b) A x: y ∞ = ∫ e −.16023 − (0.20829 × (9.22222 In part(a)many students did not appreciate what a random variable form was.008997 Variance = 0.32907)) × 1.04t ⋅ e −.019804 = 0.925. 5 (a) ⎧⎪viTx if Tx ≤ Ty Z =⎨ if Tx > Ty ⎪⎩0 where i is the valuation rate of interest.02∫ e−.2094) × 0.925.03t ⋅ (0.712.01353 The variance equals 2 1 A30:20 2 − ( A1 30:20 )2 2 20 2 2 1 A30:20 = (( A30 − ( v ) × (l50 / l30 ) × A50 )) × (1.04)1/2 = (0. Well prepared students answered this question well.g. • For this reason separate mortality tables are usually constructed for groups which are expected to be heterogeneous. It would have very restricted uses. This can manifest itself as class selection e. all the lives to whom the table applies follow the same stochastic model of mortality represented by the rates in the table.g. but they are tabulated in an efficient (space saving) way. or as time selection e. but combined after the end of the select period. and the remainder are homogeneous e. separate tables for males and females. September 2011 6 • When a life table is constructed it is assumed to reflect the mortality experience of a homogeneous group of lives i. Such a table could only be used to model mortality in a group with the same mixture. • If a life table is constructed for a heterogeneous group then the mortality experience will depend on the exact mixture of lives with different experiences that has been used to construct the table. . This means that the table can be used to model the mortality experience of a homogeneous group of lives which is suspected to have a similar experience. the experience after the end of the select period for life assurance policyholders.Subject CT5 (Contingencies Core Technical) — Examiners’ Report. separate tables for males in England and Wales in 1980–82 (ELT14) and 1990–92 (ELT15). In fact there are separate (homogeneous) mortality tables for each age at selection. However many did not get to the heart of the homogeneity discussion and went off on tangents regarding various forms of selection.g. whole life and term assurance policyholders. • Sometimes only parts of the mortality experience are heterogeneous e. the experience during the initial select period for life assurance policyholders. In such cases the tables are separate (different) during the select period.e. annuitants and pensioners.g. 667 38. 000 × a5 × A60:60 f m a60:60 = a60 + a60 − a60:60 = (15. 000 × (17.042 = 0.487 68.000 577.973 Indirectly standardised mortality rate = 1. Page 6 30 40 50 60 180 Total Expected deaths 185.000 16.30603 Therefore (1 − v5 ) × 0.694 Ax: y = (1 − δax: y ) = 1 − ln(1.632 − 0.042 = 0.816 12. 000(a60:60 − a60:60 ) + 10.500 26.000 178.Subject CT5 (Contingencies Core Technical) — Examiners’ Report. 000 × δ = 41.694 = 0.090 − 0.30603 EPV = 10.694 − (14.003062 577.5) + (16.000 156. September 2011 7 EPV is 10.000 123. 000 180 Straightforward with no issues and generally well done.042 1. 8 Age 20–29 30–39 40–49 50–64 Total (a) All professions Population Deaths 120.04) ×17.894 = 54.5)) + 10.816 185.000 256 458 502 600 1.934 Many students struggled here with the second term in the equation in the 2nd line and did not appreciate how to mix a continuous assurance factor with an annuity.595 51.042 Area comparability factor = (b) Population Profession A Deaths Expected deaths .090 − 0.000 57.5) − (14.042 = 0.5) = 17.652 − 0.500 15.500 Standardised mortality ratio = 180/185.000 14.293 185.978 577. 000 57.816 185.040 + 13. 509569 = 0.06 0.750.750 3.750 – 187.054658 0. In fact this question can be solved with the same answer without using multiple decrements and the few students who realised this were given credit.01560 qix 0.000 150.2 * 0.721.97 20.687.5qβx ) Age (aq) dx (aq)ix 59 60 61 0.933254 Probability of retiring at age 60 = 0.692038 Probability of retiring at age 61 = 0.19% This question was not done well overall.01392 0.2 * 0.821.50 11.186651 Probability of reaching 61 = 0.692038 * (1-0.064493 Probability of reaching 60 = (1 − 0.065 The dependent decrements are calculated as: (aq )αx = qxα (1 − 0.055 0.000 150.013502-0.09 808.8 * 0.38 591.8 * 0.750 3. September 2011 9 Age retirement can be ignored in constructing the dependent decrements.000 150.012088 − 0.000 3. we have: Year SA b1 b2 0 1 2 3 4 5 150.015093 0.000 150.750 3. The following rates are required: Age q xd 59 60 61 0.12 .012088 0.01243 0.2 * 0.138408 Probability of reaching 62 = 0.059582) = 0.692038 = 0. 10 (a) At the end of the 5th policy year.00 7.013502 0.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.15 ∑ b 3.750 3.933254 * (1 − 0.509569 Probability of retiring at age 62 = 0.059582 0.162.000 150.015093 − 0.88 16.054658) = 0. Students struggled to follow through the logical sequences.064493) = 0.101914 Overall required probability thus 10.933254 = 0.50 384. 711.81 21.810.Subject CT5 (Contingencies Core Technical) — Examiners’ Report. 721. This question was also done poorly overall.5a30 + t + 0. relative to other methods.5a30 + (65 − 30) z65r65v 35a65 ⎟⎟ / s30l30 t + + 0. 000 A[30] a[30] = 150. A very large number of students attempted to construct a complex “net premium” from the existing bonus flow where the question was only seeking the normal net premium method. 11 Retirement other than ill-health: ⎛ 65− 30 −1 * * ⎞ 0. This method rewards longer standing policyholders and discourages surrenders.89 − 23.5r30 + t v t + 0. net premium reserve at end of 5th policy year is given by: 5V (b) = (150.837 Therefore.01 × 20.81× 21.58 The sum assured and bonuses increase more slowly than under other methods for the same ultimate benefit. 000 × 0.12 × 0.01 × 20.003 = 32.721.5 / s30l30 .16011 = 1099.000 × (65 − 30) × 65− 30 −1 ∑ t =0 Page 8 * z30+ t + 0.099.099.5i30+ t v t + 0. enabling the office to retain surplus for longer.000 × ⎜ ∑ tz30 + t + 0. September 2011 If net premium denoted by P then P= 150. Part (b) was done better.12) A35 − Pa35 = 170.5 ⎜ ⎝ t =0 ⎠ Retirement due to ill-health: 0. 000 + 20.31 = £9.19219 − 1. minimum death benefit) • Unit-linked contracts are generally endowment assurance and whole of life contracts . rx . September 2011 Where a*x is the annuity value at age x including any contingent spouse pensions ix .g.Subject CT5 (Contingencies Core Technical) — Examiners’ Report. fund management charge) and may be varied after notice of change given. lx are values from a multiple decrement table at age x sx is the salary index for age x where s x +1 / sx is the ratio of salary in the year beginning age x + 1 to salary in the year beginning age x zx (sx−3 + sx−2 + sx−1)/3 Other schemes were accepted but overall very few students managed to derive a full answer in this question. which are priced regularly (usually daily) • Policyholder receives the value of the units allocated to their own policy • Benefits are directly linked to the value of the underlying investments • Unallocated premiums are directed to the company’s non-unit fund • Bid/offer spread is used to help cover expenses and contribute towards profit • Charges are made from the unit account periodically to cover expenses and benefits (i. 12 (i) • Allocated premiums are invested in a fund(s) chosen by the policyholder which purchases a number of units within that fund(s) • Each investment fund is divided into units.e. • Unit-linked contracts may offer guaranteed benefits (e. 870 Multiple decrement table: T d q[57] + t −1 s q[57] + t −1 ( aq ) d[57]+t −1 ( aq ) s[57]+t −1 (ap)[57]+t −1 t −1 (ap)[57] 1 2 3 0. 0.13 = 138.77 Revised profit vector becomes (138.870 = 1− 1.000000 0.992860 1.006180 0. 0.99435 p56 = 0.651568 2.89 − p56 × 1.13 The revised cash flow for year 1 will become: 1.007140 0.394.045 3V = 2V × 1. 0) and Net present value of profits = 138.004171 0.318815 2.82 = 893.05 0.09 This question was generally done well.955 = 0.049691 0.045 − p58 × 3V = 292.007140 0.99365 p57 = 0.099583 0.006180 0.77.896246 0.944129 0.075) = 129.896246 0.08 ⇒ 1V = 1.000000 0.00 0. 13 (i) Reserves required on the policy per unit sum assured are: 0V57:3 = 1− a57:3 1V57:3 = 1− a58:2 2V57:3 = 1− a59:1 a57:3 a57:3 a57:3 =0 = 1− 1.129.77/(1.394.61 1.Subject CT5 (Contingencies Core Technical) — Examiners’ Report. September 2011 (ii) To calculate the expected reserves at the end of each year we have (utilising the end of year cashflow figures): p58 = 0.846172 d s Probability in force (ap)[56]+t −1 = (1 − q[56] +t −1 ) × (1 − q[56]+t −1 ) Page 10 .004171 0.17 1V ×1.99497 933.0 = 0.10 0.525.05 ⇒ 2V = 1.045 − p57 × 2V = 334. 68 Interest 211.02(1 + i ) −2 + 109.15(1 + i ) −1 + 195.57 92.07 Total NPV profit = 18.00 0.89625 0.12 then LHS of equation = 2.91 4423. profit signature and NPV are set out in the table below: Policy Year Premium Expenses 1 2 3 4700 4700 4700 470.84617 0.52 0.11 488.95 Death claim Maturity claim 62.25 Cum probability of survival Discount factor NPV Profit 1. September 2011 The calculations of the profit vector.0 = 0.00 14892.00 65.00 In force cash flow 4027.74 Therefore IRR is 13% (iii) The revised reserves required on the policy per unit sum assured are: 0V57:3 = 1− 1V57:3 = 1 − 2V57:3 = 1 − a57:3 a57:3 a58:2 a57:3 a59:1 a57:3 =0 = 1− 1.02 0.00 239.05 4445.03 350.75 Profit vector −258.75 231.14 then LHS of equation = −2.15 217.00 Policy year Increase in reserves Interest on reserves 1 2 3 4286.90 Surrender claim 351.15 (ii) IRR is determined by solving the following equation for i: −258.645012 2.312389 2.13 then LHS of equation = −0.26 170.70 107.11(1 + i ) −3 = 0 If i = 0.24 −9773.00000 0.34 89.64 If i = 0.817 = 1− 1.937 = 0.10 0.60 128.10 If i = 0.93458 0.87344 0.50 231.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.00 65.73 −10133.81630 −241.817 . Note that it is possible to solve (ii) using a quadratic equation process.74 Total NPV profit = 21.76 −171.89625 0. In general well prepared students made a reasonable attempt with this question.78 −9675.02 The NPV of profit increases slightly if the reserving basis is weakened. September 2011 And the revised cashflows become: Policy year 1 2 3 Increase in reserves Interest on reserves Revised Profit vector Cum probability of survival Discount factor NPV profit 4199.66 4448.24 25.18 0.87344 0.69 1.76 209. 14 (i) Formula is (t V + P − e) × (1 + i ) = qx +t × S + px +t × t +1V Definitions: tV = gross premium reserve at time t qx +t / p x +t = probability that a life aged x+t dies within /survives one year on premium/valuation basis P = office premium e = initial/renewal expense incurred at start of policy year i = rate of interest in premium/valuation basis S = sum assured payable at end of year of death Page 12 .81630 −160. Credit was given to students who showed they understood the processes even if not all the arithmetical calculations were correct.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.79 17.93458 0.52 163. as a result of the surplus emerging being brought forward and the fact that the risk discount rate is greater than the interest rate being earned on reserves.29 483.00000 0.84617 0.00 234. 46998 − 0.2329 = 2.39443 − 0.27492 = 0.821.02 The gross premium prospective reserve per policy at the end of 2010 is given by: 10 V PRO = 50.Subject CT5 (Contingencies Core Technical) — Examiners’ Report. 000 A[35]:30 + 100. 000 × 0. 000 A144:21 + 100. 443.05815 9.2329 ⇒9 V PRO = 50.41075 = 0.975 P ×14.907.975 Pa44:21 where A144:21 = A44:21 − v 21 21 p44 = 0.39443 = 0.5 P + 0.6313 P= 30. 000 A145:20 + 100.30832 × 8.05815 + 100.46998 − 0.32187 − 0.975 Pa45:20 where A145:20 = A45:20 − v 20 20 p45 = 0.814.50 + 39. 000v 21 21 p44 − 0. 000v 20 20 p45 − 0.71552 (iii) The gross premium prospective reserve per policy at the end of 2009 is given by: 9V PRO = 50.329.43883 × 8.021.3359 a44:21 = 14.5 = 1.821.892.45258 − 0. Then equation of value is: ( ) 1 Pa[35]:30 = 50.27492 + 300 + 0.2612 = 0.04695 9.2612 = 0.6313 ⇒ 17.05923 9.821. 000 × 0.04695 + 100.45258 − 0.025P ×16.025 P a[35]:30 − 1 where 1 A[35]:30 = A[35]:30 − v30 30 p[35] = 0.139.00 − 25.3123 .803. 000v30 30 p[35] + 300 + 0. September 2011 (ii) Let P be the annual premium.08 16.32187 − 0.6313P = 50.5 P + 0.45639 × 8. 000 × 0.48 = 17.2612 = 0.801. 000 × 0.9151 a[35]:30 = 17. 423.23 – 1.75 ADS = 3 × DSAR = 3 × 30.810.189. a mortality loss) Interest profit = 385 × (17.145.23 (i.549.83 i. a combined mortality and interest loss of £1.e.975 P ×13. 000 × 0.7805 = 2.05923 + 100.34 = 30. 715.975 ×1.329.05 − 3 × 50.803. In part (iii) most well prepared students were able to derive the mortality profit but most struggled with the interest portion.329.08) × (0.568. 000 − 19. END OF EXAMINERS’ REPORT Page 14 .620.66 = 15.34 = 7.75 − 90. 423.568.88 = −1.41075 − 0.189.145.66 EDS = 385 × q44 × DSAR = 385 × 0. 620.e.02 + 0.524.489.00 − 24.189.98 Therefore Mortality profit = EDS – ADS = 15.66 = 90. If a student got the combined total correct but then did not split up the content it was decided to give full credit. 000 − (385 − 3) ×19.83 which can be split between mortality profit and interest profit separately as follows: DSAR = 50.961.02 + 0.975 × 1.04 Alternatively: Interest profit = 75. 000 × 0.001327 × 30.40 (the small discrepancy with the figure for interest profit above is due to figures being used from the Actuarial Tables with only a limited number of decimal places) Part (i) and (ii) were done well.810. 000 − 7. 226.05 − 150.810.04) = 73.803.08 ) ×1. September 2011 a45:20 = 13.7805 ⇒10 V PRO = 50.16 = 19.83 = 73.05 – 0.567.34 Combined mortality and interest profit = 385 × (17.Subject CT5 (Contingencies Core Technical) — Examiners’ Report.620.50 + 41.929.075.98 = −75. with any additional sheets firmly attached. 5. beginning your answer to each question on a separate sheet. You must not start writing your answers in the booklet until instructed to do so by the supervisor. Graph paper is NOT required for this paper. and this question paper. CT5 A2012 © Institute and Faculty of Actuaries .INSTITUTE AND FACULTY OF ACTUARIES EXAMINATION 23 April 2012 (am) Subject CT5 – Contingencies Core Technical Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. Attempt all 15 questions. 2. In addition to this paper you should have available the 2002 edition of the Formulae and Tables and your own electronic calculator from the approved list. 4. 3. AT THE END OF THE EXAMINATION Hand in BOTH your answer booklet. Mark allocations are shown in brackets. Candidates should show calculations where this is appropriate. Enter all the candidate and examination details as requested on the front of your answer booklet. 950 0.1 (a) Define 4 5 q[60]+1 in words. A level premium of £3. the death benefit payable at the end of year of death is a return of premiums paid without interest. For a policy in force at the start of the 12th policy year. Calculate the expected present value of the contract. Basis: Mortality 2 AM92 [3] Under a policy issued by a life insurance company. (b) Calculate its value. PFA92C20 (female life) 4% per annum Nil [4] CT5 A2012–2 . 3 [3] Calculate: (a) a50:15 (b) 1 ( IA)50:15 Basis: Mortality Rate of interest 4 AM92 6% per annum [4] A joint life assurance contract provides a death benefit of £100.000 payable immediately on the second death of two lives.000 is payable annually in advance throughout the term of the policy. you are given the following information: Reserve at the start of the policy year Reserve at the end of the policy year per survivor Probability of death during the policy year Expenses incurred at the start of the policy year Rate of interest earned £25.130 £28. a male life currently aged 60 exact and a female life currently aged 55 exact.03 £90 4% per annum Reserves given above are immediately before payment of the premium due. Basis: Mortality Rate of interest Expenses PMA92C20 (male life). Calculate the profit/loss expected to emerge at the end of the 12th policy year per policy in force at the start of that year. 000 on death. Basis: AM92 Ultimate [4] 7 Describe the benefits typically provided by a salary-related pension scheme for active members on age retirement.25 using the assumption of a constant force of mortality and the value derived in (a) above. 5. probability of death at each age) 2. −4. [2] (ii) Give one example of each category in part (i) and indicate the manner in which it is usually allowed for in the calculation of premiums.5% per annum [4] 6 (a) Calculate the constant force of mortality applicable to a life aged between 67 and 68 exact. (a) (b) Construct a multiple state transition model for these policies.5 A 10-year unit-linked policy has the following profit vector: (−40. [6] CT5 A2012–3 PLEASE TURN OVER . 8. [4] [Total 6] 10 [6] An insurance company writes policies that provides benefits of £1. 25. −12. [6] 8 Explain the impact of occupation on mortality and morbidity. 30) Determine the revised profit vector if reserves are set up to zeroise future negative cash flows on the following basis: Mortality Interest 0.5 q67. −1. 20.000 in the event of becoming disabled due to accident and £10.e. −6. 9 (i) List the main categories of expenses incurred by life insurance companies. Give a formula for the expected present value of the benefits. (b) Calculate the value of 0.5% per annum (i. 000 Number of deaths in sub-population (ii) 12 1. compounded and vesting at the end of each policy year (i.000 80.250 Calculate the Standardised Mortality Ratio using ELT15 (Males) as the mortality rate for the standard population. The sum assured of £85.000 95. [3] [Total 7] An endowment assurance contract with a term of 10 years pays a sum assured of £100. Calculate the expected present value and variance of this contract.92308% of the sum assured.03 throughout 5% per annum [8] A life insurance company issues a 40-year with profit endowment assurance policy to a life aged 20 exact. [10] . the death benefit does not include any bonus relating to the policy year of death). Basis: Mortality Interest Initial commission Initial expenses Renewal commission Renewal expenses CT5 A2012–4 AM92 Select 6% per annum 480% of the first monthly premium £325 2. The renewal expense is assumed to increase by £5 per annum from the start of the third policy year. The company assumes that future annual bonuses will be declared at a rate of 1. Calculate the monthly premium payable in advance throughout the term of the policy.5% of each monthly premium excluding the first £75 per annum at the start of the second and subsequent policy years.11 (i) State the advantages and disadvantages of using crude mortality rates and directly standardised mortality rates as the comparison measure of mortality in two or more different populations [4] You are given the following data in respect of a sub-population: Age Population 50 55 60 100.000 on survival for 10 years. Basis: Mortality Rate of interest 13 µx = 0.e.000 immediately on death and a sum of £50.000 plus declared reversionary bonuses is payable on survival to the end of the term or immediately on death if earlier. 000 in the twentieth and final policy year.000 in the first policy year. which is payable at the end of the year of death. [4] 625 policies were in force at the start of the 10th policy year and 3 policyholders died during that policy year.000 each year until the benefit is £10.000 in the second policy year thereafter reducing by £10. The death benefit. [6] (iii) Comment briefly on the results obtained in part (ii) above. £190. is £200. (i) Show that the annual net premium for each policy is approximately equal to £204 using the basis below. The company calculates its reserves on a net premium basis and negative reserves are permitted. Premiums on the policies are payable annually in advance for 20 years or until earlier death. [2] Basis: Mortality Interest Expenses AM92 Ultimate 4% per annum Nil [Total 12]  CT5 A2012–5 PLEASE TURN OVER .14 A life insurance company issues 20-year decreasing term assurance policies to single lives aged 40 exact. (ii) Calculate the mortality profit or loss to the life insurance company during the 10th policy year using the basis below. 000 payable at the end of year of death.5% of the second and third years’ premiums 6% per annum (i) Write down the gross future loss random variable at the outset of the policy. [4] (iii) Derive the office premium using a discounted cash flow projection.15 A life insurance company issues a three-year term assurance policy to a male life aged 57 exact under which level premiums are payable annually in advance throughout the term of the policy or until earlier death. assuming no withdrawals and using the same profit criterion as in part (ii). [2] [Total 17] END OF PAPER CT5 A2012–6 . increasing the risk discount rate to 8% per annum. The sum assured is £150. setting the expected value of the gross future loss random variable to zero. The company uses the following assumptions to calculate the premium for this policy: Rate of interest on cash flows Mortality Initial expenses Renewal expenses Initial commission Renewal commission Risk discount rate 6% per annum AM92 Select £350 £50 per annum on the second and third premium dates 15% of first premium 2. [6] (iv) Without further calculation explain the effect of: (a) setting up reserves within the calculation of part (iii). (b) having set up the reserves in part (a). [5] (ii) Calculate the office premium using assurance and annuity functions. For essay-style questions. will generally be based on Core Reading. This is much more than a model solution – it would be impossible to write down all the points in the report in the time allowed for the question. For numerical questions the Examiners’ preferred approach to the solution is reproduced in this report. and the following comments. the Examiners are not required to examine the content of Core Reading. the questions set. Other valid approaches are always given appropriate credit. The Examiners are charged by Council with examining the published syllabus. both those who are sitting the examination for the first time and who are using past papers as a revision aid. T J Birse Chairman of the Board of Examiners July 2012 © Institute and Faculty of Actuaries . this is also noted in the report. Notwithstanding that. which is designed to interpret the syllabus. and particularly the open-ended questions in the later subjects. where there is a commonly used alternative approach. this report contains all the points for which the Examiners awarded marks. Although Examiners have access to the Core Reading. and also those who have previously failed the subject.INSTITUTE AND FACULTY OF ACTUARIES EXAMINERS’ REPORT April 2012 examinations Subject CT5 – Contingencies Core Technical Introduction The Examiners’ Report is written by the Principal Examiner with the aim of helping candidates. Subject CT5 (Contingencies) – April 2012 – Examiners’ Report General comments on Subject CT5 CT5 introduces the fundamental building blocks that stand behind all life insurance and pensions actuarial work. 12. In the case of descriptive answers credit is also given where appropriate to different valid points made which do not appear in the solutions below. Students should note that for long questions a reasonable level of credit is given if they can describe the right procedures although to score well reasonable accurate numerical calculation is necessary. 10. Questions that were done less well were 2. is relevant and is properly defined and used in the answer. Comments on the April 2012 paper The general performance was better this session than in recent diets and many students scored well with a very pleasing increase in the number passing. Most of the short questions were very straightforward where an answer could be produced quickly and this is where many successful candidates scored particularly well. Page 2 . In questions where definitions of symbols and then formulae are requested. 13 and 15(i) and (iii) and here more commentary is given to students to assist with further revision. a different notation system produced by a student to that used by examiners is acceptable provided it is used consistently. Credit is given to students who produce alternative viable numerical solutions. Subject CT5 (Contingencies) – April 2012 – Examiners’ Report 1 (a) 4|5 q[60]+1 is the probability that a life now aged 61 exact who entered the selection period 1 year ago will die between the ages of 65 and 70 both exact (b) 4 | 5 q[60] + 1 = l 65 − l 70 / l [60]+1 = (8821.2612 × (10. wholly incorrect.0544) / 9209.569 − 1) 9.130 + 3.40177) 9.821.6568 = 0. 712.417 (b) ( IA) 1 50:15 ⎛l ⎞ = ( IA)50 − v15 ⎜ 65 ⎟ (( IA)65 + 15 A65 ) ⎝ l 50 ⎠ 8.03 − (1 − 0.84555 − 0. It was also not on many occasions appreciated that the accumulation minus the benefit costs gave the profit.000 Profit emerging per policy in force at the start of the year is: (11V + P − e) × (1 + i ) − (qx +t × S ) − px +t × 12V = (25.0728 = 9.0833 ( ) A gentle starter generally done well 2 The death benefit in year 12 is £36. 000 × 0.2612 − 8054.950 = £0.04 − 36.03) × 28.41727 × (5.41727 × 8.10 This question overall caused problems and students sometimes had only a vague recall of the iterative formula in line 3.2612 = 4.0728 = 0.712.47329   In (a) a surprising number of students thought that the required function could be derived from the a due function for the same term minus 1 which is. 000 − 90) ×1. Otherwise the question was generally well done.50985 + 15 × 0. 3 (a) ⎛ l 65 ⎞ a 50:15 = a 50 − v15 ⎜⎜ ⎟⎟ ( a 65) l ⎝ 50 ⎠ = (14.044 − 1) − 0. The most common error was to forget expenses and the survival factor before the closing reserve.821. Page 3 . of course. 995 × 3V ) = 6. 8.5)) = £27104 Generally well done.976 3V = 1. 0. 3 and 2 are: 4 = 3.756 − 0. 30) Generally well done by well prepared students who were able to recall the techniques involved.025 1 (12 + .995 × 2V ) = 18.218 => revised profit vector: (−58.039221 × (15. 0.801 2V = 1.309 1V = 1. 20.118 Revised cash flow in policy year 1 = −40 − 0.22.632 + 18.04)(a60 + a55 − a60:55 )) = 100000 × (1 − ln(1. Other methods such as multiplying non continuous functions by (1.025 5V = Revised cash flow in policy year 5 = 5 − 0.025 1 = 0.902 1. 4.995 × 5V = 1.210 − 14. 0.04)([a60 − 1 / 2] + [a55 − 1 / 2] − [a60:55 − 1 / 2])) = 100000 × (1 − 0. 5 The reserves required at the beginning of policy years 6.12.995 × 1V = −58. 0. Page 4 . 25. 1.Subject CT5 (Contingencies) – April 2012 – Examiners’ Report 4 JJJJJK where 60 relates to the male life and 55 the female life.04)1/2 to obtain the continuous one were quite acceptable.025 1 (6 + . The value is 100000A60:55 JJJJJK = 100000 × ( A + A − A 100000 A60:55 60 55 60:55 ) = 100000 × (1 − ln(1. 75 ∫ 67. although some credit was given. Other relevant comments were credited. e. Page 5 . Annual salary averaged over a few years before retirement (final average salary) 3. However some students ignored the instruction to use (a) to get (b) choosing the more direct route. Annual salary averaged the whole of service (career average salary) Pensions are commonly increased in payment to offset the effect of inflation. Pensions may also be paid for an initial guarantee period like 5 years. ⇒ μ = − ln( p67 ) = − ln(1 − q67 ) = − ln(0. 1/80 is described as the accrual rate. The examiners penalised this approach.Subject CT5 (Contingencies) – April 2012 – Examiners’ Report 6 (a) p67 = exp(− ∫ 68 67 μ dx) where μ is the constant force.25 = 1 − exp( − 67. Pension for each year of service is usually related to pensionable salary.g. as defined in the scheme rules.5 × μ ) = 1 − exp( −0.017985 (b) Using the constant force assumption: 0. Age retirement benefits may be provided on early or late retirement. Some benefit may be in the form of cash. 7 Pension schemes usually have a fixed Normal Pension Age (NPA).5 p 67.5 × 0. complete years of membership. for example 1/80ths of pensionable salary for each year of service. No credit was given for any discussion on ill-health retirement as this was not required from the question.25 μ dx) = 1 − exp( −0. Pensionable salary can be defined as: 1. sometimes by converting pension to cash. Pension usually depends on pensionable service at retirement. There can be a spouse’s pension for married pensioners which is often a percentage of the main pension on the member’s prior death. Many students scored reasonable marks.982176) ⇒ μ = 0.25 = 1 − 0.008952 Generally well done.5 q 67. Salary at retirement (final salary) 2.017985) = 0. Processing proposal and issuing policy (allowed for on a per policy basis) or. this effect can be produced without formal checks. chemicals. the occupation may involve exposure to harmful substances e. and this permits them to adopt a particular lifestyle e.g. Generally well done and credit was given for any other relevant points.Subject CT5 (Contingencies) – April 2012 – Examiners’ Report 8 Occupation can have several direct and indirect effects on mortality and morbidity. This will inflate the mortality rates of newspaper sellers. bus drivers have a sedentary and stressful occupation while bus conductors are more active and less stressed. Some work environments e. content and pattern of diet. quality of housing. or to potentially dangerous situations e. Investment Expense (charged as a deduction from investment funds). However. A person’s occupation largely determines their income. Much of this is moderated by health and safety at work regulations. This may be accentuated by health checks made on appointment or by the need to pass regular health checks e. The environment may be rural or urban. e.g. This effect can be positive and negative e. Marketing (allowed for on a per policy basis on estimated volumes) Renewal Expense Administration (allowed for on a per policy per annum basis with allowance for inflation) or. give exposure to a less healthy lifestyle. former miners who have left the mining industry as a result of ill-health and then chosen to sell newspapers.g. Some occupations by their very nature attract more healthy or unhealthy workers. Occupation determines a person’s environment for 40 or more hours each week.g. Some occupations are more healthy by their very nature e. over indulgence. Commission (allowed for directly and usually premium related) or. Claim Expense Calculation and payment of benefit (allowed for on a per policy per annum basis with allowance for inflation) Page 6 .g. Commission (allowed for directly and usually premium related) or.g. working at heights. publicans.g.g. airline pilots.  9 (i) Initial Expense Renewal Expense Claim Expense Overhead Expense (ii) Initial Expense Underwriting (allowed for on a per policy basis although medical expenses might be sum assured related) or. The resultant formulae were.ν x +t )dt 0 ∞ Value of disablement benefit = 1. Other relevant examples were credited however. on the whole poorly done.Subject CT5 (Contingencies) – April 2012 – Examiners’ Report Overhead Expense Central services e. 10 The multiple state transition model is: σx   a = able   i = disabled ρx   µx d = dead   νx   Define the force of interest δ ∞ Value of death benefit = 10. IT. 11 (i) Crude Mortality Rate Advantage – do not need population and deaths split by age Disadvantage – differences in age structure between populations will be confounded Directly Standardised Mortality Rate Advantage – only reflects differences in mortality rates Page 7 . legal (allowed for on a per policy per annum basis with allowance for inflation) ] Many students did not give a full answer referring only to direct and indirect expenses for which only partial credit was given. This was accepted so long as the assumptions were stated.σ x +t )dt 0 Generally the diagram was completed satisfactorily. 000 ∫ e−δt ( t pxaa . premises.g. however.μ x +t + t pxai . Also many did not give the full number of distinctly different examples. Many students took the view that returning to the able state from the disabled one was impossible and thus omitted the return arrow. 000 ∫ e−δt ( t pxaa . 07879 ⎣ 0 0.03)t ) × .25% and ln(1.45480 The expected value is thus 100000 × 0.01392 = 2335 SMR = 1250 / 2335 = 0.2758)] =0.03 ⎢ − ⎥⎦ 0.03∫ = 1 x:n A 10 0 10 ⎡ exp( −0.00797+80000*0.00464+95000*0.16949 0.03∫ = 2 1 x:n A 10 0 10 ⎡ exp(−0.535 A straightforward question generally done well by well prepared students.20759 0.20759 + 50000 × 0. Many students failed to realise that the integration process was the same as for the expected value with the exception of building in the 10.7879) = 0.07879t )dt = 0. However by contrast the part relating to the variance was done poorly.45480 = £43499 For the variance the rate used is (1.03 [1 − exp( −0.07879 = exp(−0.12758 = exp(−1.27921 − (43499)2 = (22378)2 2 2 The part relating to the expected value was generally done well.12758 ⎣ 0 0.Subject CT5 (Contingencies) – April 2012 – Examiners’ Report Disadvantage – requires age specific mortality rates for the observed population (ii) SMR = actual deaths / expected deaths Expected deaths = 100000*0.16949 + ( 50000 ) × 0. 12 1 x :n The expected value is 100000 Ax1:n + 50000 A Ax1:n = ∫ 10 0 exp(( − ln(1. Hence: 2 1 Ax:n = 0.03dt = 0.07879t ) ⎤ exp( −0. Page 8 .05)2−1=10.2758) = 0.7879)] =0.03 ⎢ − ⎥⎦ 0.05) − .03 [1 − exp(−1.12758t ) ⎤ exp(−0.27921 The variance is thus (100000 ) × 0.1025)=0.25% interest rate. Some students struggled to find the distinctive advantages and disadvantage given above.09758.12758t )dt = 0. 5 + q[20] (1 + b)v1.2164 ⎞ = 15..801 − 0.550 = 18.461 Page 9 .801 − ⎜1 − 0. 000 × (1.09722 × ⎟ = 15. 000(1 + b) 40 v 40 40 p[20] 85.19383 = 0.21746 − 0.5 × A[20]:40 @ i′ + 85.2164 = 0. 028.2432 ⎠ (12) a[20]:40 = a[20]:40 − EPV of benefits: ( 85.19383 (1 + b) = 2. 000 × = ( ) (1.911 + 16. 000 × 0..0192308 = 85.06 − 1 = 0..06)0. 000v 40 40 p[20] @ i′ (1 + b) where i′ = and 1.21746 − 0.2432 EPV of benefits 85..02363 9980.5 + (1 + b) 40 40 p[20]v 40 ) where b = 0.4169 = 15.06)0.Subject CT5 (Contingencies) – April 2012 – Examiners’ Report 13 Let P be the monthly premium. + 1 39 q[20] (1 + b)39 v39. + q[20] (1 + b) 40 v 40 1 39 (1 + b) +85. 000 q[20]v 0.02363 + 85.6092 P where 11 1 − v 40 40 p[20] ) ( 24 11 ⎛ 9287.5 = × 0.5 q[20] (1 + b)v + q[20] (1 + b) 2 v 2 + . 000 1 × (1. 475.5 + .20829 × 9287.04 1+ b 1 A[20]:40 @ i′ = A[20]:40 − v 40 40 p[20] = 0.06)0. 504. Then: EPV of premiums: 12 Pa(12) [20]:40 @ 6% = 184.3841 24 ⎝ 9980. 891 + 113.801 + 5 × 208.8P + 325 + 0.Subject CT5 (Contingencies) – April 2012 – Examiners’ Report EPV of expenses (12) = 4.025P + 65 × ⎡ a[20]:40 − 1⎤ + 5 × ⎡( Ia)[20]:40 − 1⎤ ⎣ ⎦ ⎣ ⎦ = 9.09722 × 9287.219 The difficult part of this question was related to the EPV of Expenses and most students failed to complete this complex part. 000 A40:20 − 10.516] 9980.6092 P = 18.504.2863 ( IA)140:20 = ( IA)40 − v 20 20 p40 [ 20 A60 + ( IA)60 ] = 7.43544 Page 10 . 14  (i) Annual net premium for the decreasing term assurance is given by: P = 1 210.95699 − 0.90 175.895 where ( Ia)[20]:40 = ( Ia)[20] − v 40 40 p[20] ⎡⎣ 40a60 + ( Ia)60 ⎤⎦ = 262. The rest of the question was however generally reasonably done by well prepared students.833. 000( IA)140:20 a40:20 1 = A40:20 − v 20 20 p40 where A40:20 = 0.461 + 9.2164 = 0.025 ×12 × Pa[20]:40 − 0.666 − 0.895 ⇒P= 20. 328.356 = £118.46433 − 0.43004 = 0.45640 + 8.3902 P + 325 + 65 × 14.03429 9856.43004 × [ 20 × 0.46433 − 0.2164 × [ 40 ×11.328.36234 ] = 0.2432 = 262.300 = 209.666 − 53.366 Equation of value gives 184.3902 P + 2.366 = 9.45639 × and 9287.3902 P + 2. 20875 10V = 110.2164 = 0. sum at risk per policy in the 10th policy year is: DSAR = 110.64601× [10 × 0.45640 + 8.021.03429 − 10.e.50 Mortality profit = EDS – ADS EDS = 625 × q49 ×110.86 ADS = 3 ×110.Subject CT5 (Contingencies) – April 2012 – Examiners’ Report P = (ii) 210.03423 9712.000 – (−21. 021.36234 ] = 0.0728 and ( IA)150:10 = ( IA)50 − v10 10 p50 [10 A60 + ( IA)60 ] = 8.67556 × 9287.50 ) = 110.965. mortality profit = − 175.03423 − 10.30 − 2. 087. 000 A50:10 − 10. 000 × 0. 765.36 (i. 021.64601 = 0. 064. 098. 000 × 0.55929 − 0. 021.002241×110. 000 × 0. a loss) Page 11 .e.50 = 625 × 0.43544 = 204.68024 − 0. 699. 000( IA)150:10 − P a50:10 where 1 A50:10 = A50:10 − v10 10 p50 = 0.50 Therefore.927 Reserve at the end of the 10th policy year given by: 10V 1 = 110.50 = 154. 000 × 0.50 i.50 = 330.314 = 3.30 = −21.50 − 1.68024 − 0.20875 − 204.39 × 8.39 13. 975a3) − 0. 000 × 0. 000v K[57] +1 + 350 + 50aK [57] − P(0. a mortality experience investigation would need to consider a longer time period and ideally.82502 = 0.125⎤⎦ where Α1[57]:3 = Α[57]:3 − v 3 3 p[57] = 0. a larger number of policies to determine whether actual mortality experience is heavier than expected.Subject CT5 (Contingencies) – April 2012 – Examiners’ Report (iii) The death strain at risk per policy in the 10th policy year for this decreasing term assurance is very large (approximately equal to the sum assured payable in the event of death).5938 and a[57]:3 = 2. The actual number of deaths during the 10th policy year (at 3) is approximately double that expected (at 1. 044. 000 A[57]:3 + 350 + 50 ⎡⎣ a[57]:3 − 1⎤⎦ = P ⎡⎣ 0. 15 (i) Gross future loss random variable = 150.83962 × 9287.995829 0.125) if K[57] ≥ 3 (ii) E (Gross future loss random variable) = 0 1 ⇒ 150. However.6245 (iii) Mortality table: x 57 58 59 Page 12 t 1 2 3 q[ x ]+t −1 p[ x]+t −1 t −1 p[ x ] 0.989675 .0 = 1.006180 0.975a[57]:3 − 0.77 2.01534 9451.84036 − 0.820 ⇒ 150.007140 0.0 + 350 + 91.6245 P ⇒P= 2.975aK [57] +1) − 0.000000 0. This was a straightforward question of its type.004171 0.992860 1.84036 − 0.993820 0.301.01534 + 350 + 50 × 1.125) if K[57] < 3 350 + 50a2 − P (0.4) which accounts for the mortality loss. Question generally done well by well prepared student.995829 0.820 = 2.2164 = 0. NPV falls.890000 0.8588P − 933.025P + 50 0. Most students struggled with part (i) but well prepared ones completed parts (ii) and (iv) satisfactorily.0585P − 3 1071.0292P − 975.95 2.989 Therefore: 3 ∑ NPV = 0 =2.240 0.395 0. the premium would need to be increased to satisfy the same profit criterion.995829 1.15P + 350 0.65 2 P 0.00 1.000000 0.051P − 21 625.94340 0.791 = 1.0228P − 1112.0335P−1124.0335P−980.6248P − 2742.9010P − 996.83962 0. Therefore. 044.00 1.6248 which is consistent with the premium calculated in (ii) above (allowing for rounding) (iv) (a) profit is deferred but as the earned interest rate is equal to the risk discount rate.791 ⇒ P = 1 2742.025P + 50 0.0585P − 3 927.85P − 940. there is no change to the NPV or premium (b) profit is deferred and because the risk discount rate is greater than the earned interest rate.Subject CT5 (Contingencies) – April 2012 – Examiners’ Report Cash flows (per policy at start of year) assuming annual premium is denoted by P: Year Premium Expenses Interest Claim Profit vector Cumulative probability of survival Profit signature Discount factor NPV of profit 1 P 0.901P − 996. Part (iii) caused students great difficulties as often occurs with this approach and many even failed to realise that the answers to (ii) and (iii) should numerically be the same within rounding. END OF EXAMINERS’ REPORT Page 13 .650 0.65 0.9160P − 868.912 0.00 3 P 0.989675 1.562 0.00 1. You must not start writing your answers in the booklet until instructed to do so by the supervisor. Graph paper is NOT required for this paper. Candidates should show calculations where this is appropriate. 5. CT5 S2012 © Institute and Faculty of Actuaries . 4. beginning your answer to each question on a separate sheet. 3. Attempt all 15 questions. 2. In addition to this paper you should have available the 2002 edition of the Formulae and Tables and your own electronic calculator from the approved list. AT THE END OF THE EXAMINATION Hand in BOTH your answer booklet. Enter all the candidate and examination details as requested on the front of your answer booklet.INSTITUTE AND FACULTY OF ACTUARIES EXAMINATION 2 October 2012 (am) Subject CT5 – Contingencies Core Technical Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. with any additional sheets firmly attached. Mark allocations are shown in brackets. and this question paper. Write down an expression for the net future loss random variable at outset for this policy defining all symbols that are used. under which the sum assured S and any attaching bonuses. (b) Explain what is measured by the Area Comparability Factor by considering the ratio of the numerator to the denominator. Compound bonuses are added annually in advance. 4 Calculate 3 p55. [4] 6 A life insurance company issues a with profit whole life assurance policy to a life aged 40 exact.1 Calculate: (a) 12 p43 (b) 10|5 q55 (c) a45:10 Basis: Mortality Rate of interest AM92 6% per annum [3] 2 Give three different forms of selection that would be expected in a group of lives purchasing immediate annuities with an example of each.t x ∑ Exc.t x s Exc. [4] CT5 S2012–2 .t x ∑ ∑ Exc.t s mx. Premiums are payable annually in advance ceasing at exact age 85 or on earlier death. [3] 3 Explain how nutrition affects mortality and morbidity.75 using the assumption of Uniform Distribution of Deaths. are payable immediately on death. [4] Basis: Mortality 5 ELT15 (Females) [4] The Area Comparability Factor is defined as: F = ∑ s Exc.t s mx.t x (a) Define the notation used. [3] [Total 7] A pension scheme provides a lump sum benefit to members on reaching retirement at age 65 equal to one month’s pensionable salary for each complete year of service. Calculate the total mortality profit or loss to the life insurance company during 2011 assuming the company calculates net premium reserves on the following basis: Mortality Interest Expenses AM92 Select 4% per annum Nil [4] 8 Examine the column of dx shown in the English Life Table No. There were 3521 pure endowment policies still in force on 1 January 2011 and 8 policyholders died during 2011.000. 15 (Males) in the Formulae and Tables for Examinations (Pages 68–69).000 payable on maturity. a life insurance company sold a large number of 30-year pure endowment policies to lives then aged 35 exact. Describe the key characteristics of this mortality table using the data to illustrate your points. Premiums are payable annually in advance throughout the term of the policy. Pensionable salary is defined as average annual salary in the last two years before retirement.7 On 1 January 2007. with salary in the next year of £20. [6] 9 (i) Explain what is meant by the following in the context of life insurance policies: (a) (b) (ii) 10 gross premium prospective reserve gross premium retrospective reserve [4] State the conditions necessary for gross premium prospective and gross premium retrospective reserves to be equal. The sum assured under each policy is £125. Basis: Pension Scheme tables in the Formulae and Tables for Examinations Interest 4% per annum [8] CT5 S2012–3 PLEASE TURN OVER . Calculate the cost of this benefit as a percentage of salary for a new member of the scheme aged 35 exact. Calculate the expected present value of this annuity.03 for all y Force of interest 5% per annum Expenses Nil CT5 S2012–4 [10] . • At the end of the guarantee period if only one life has survived the annuity reduces to two-thirds of the initial level and continues at this reduced level until the second life dies.02 for all x Life y: µy = 0. Under this policy. At the end of 25 years a sum of £50. • At the end of the guarantee period if both lives are still surviving the annuity continues at the same level until one life dies at which time it reduces to twothirds of the initial level and continues at this reduced level until the second life dies.000 is paid immediately on the second death within the 25 year term. Basis: Mortality Life x: µx = 0. Basis: Mortality Rate of interest Expenses 12 PMA92C20 (male life).000 is paid to each survivor. Calculate the annual premium paid continuously under this policy assuming this is paid throughout the term or until the second death if earlier.11 A special joint life annuity of £500 per week is payable in arrear in respect of a male life aged 65 exact and a female life aged 62 exact. a sum assured of £100. PFA92C20 (female life) 4% per annum Nil [8] A life insurance company issues a special endowment assurance policy for a 25 year term to two lives x and y. • At the end of the guarantee period if both lives have previously died then the annuity ceases. The annuity has the following features: • The annuity is guaranteed in any event for the first 5 years at the level of £500 per week. The sum assured is £75. the death benefit does not include any bonus relating to the policy year of death).13 A life insurance company issues a with profit whole life assurance policy to a life aged 55 exact.5% of the monthly premiums Bonuses Simple bonus of 2.e.5% of the second and subsequent monthly premiums Simple bonus of 2.5% of basic sum assured per annum [4] [Total 10] CT5 S2012–5 PLEASE TURN OVER . Level premiums are payable monthly in advance ceasing on the policyholder’s death or on reaching age 85 if earlier.000 together with any attaching bonuses and is payable immediately on death. [6] (ii) Calculate the gross prospective policy value at the end of the 30th policy year given that the total actual past bonus additions to the policy have followed the assumptions stated in the premium basis above (including the bonus just vested). Simple annual bonuses are added at the end of each policy year (i.0% of basic sum assured per annum (i) Calculate the monthly premium for this policy. Policy value basis: Mortality AM92 Ultimate Interest 4% per annum Expenses Renewal Claim £80 at the start of each policy year and payable until death £250 on death Commission Renewal 2. The company calculates the premium on the following basis: Mortality AM92 Select Interest 4% per annum Expenses Initial Renewal Claim Commission Initial Renewal Bonuses £275 £65 at the start of the second and subsequent policy years and payable until death £200 on death 75% of the total premium payable in the first policy year 2. 023744 .049361 . 15  [10] A life insurance company issues a three-year unit-linked endowment assurance policy to a male life aged 45 exact.098727 . a return of 50% of total premiums paid.024680 0 .000 are payable annually in advance throughout the term of the policy or until earlier claim. Premiums of £14.000. 100% in the second and 105% in the third Policy fee: £35 is deducted from the bid value of units at the start of each policy year Death benefit: 150% of the bid value of the units is payable at the end of the policy year of death Maturity benefit: 100% of the bid value of the units is payable CT5 S2012–6 . The company uses the following basis to profit test this contract: Interest earned on cash flows Expenses Reserves 3% per annum 5% of each premium paid Ignore The company has also calculated the following dependent rates of mortality.000636 . The main features of the contract are: Premiums: £3. surrender and redundancy which are used to profit test this contract: Year t d (aq )[30] +t −1 s (aq)30 +t −1 r (aq)30 +t −1 1 2 3 4 .14  A life insurance company issues a four-year policy to a male life aged 30 exact that offers the following benefits: • On death during the term of the policy or on survival to the end of the term.000447 . • On surrender during the term of the policy. a return of 100% of total premiums paid.000548 . The death. a sum of £60. surrender and redundancy benefits are payable immediately on claim.000 per annum are payable yearly in advance throughout the term of the policy or until earlier death Allocation rates: 75% of premium is allocated to units in the first policy year.024368 . • On redundancy during the term of the policy.024680 0 Calculate the expected profit margin to the company on this policy using a risk discount rate of 5% per annum. The contract ceases on payment of any claim.000602 . The company uses the following assumptions in carrying out profit tests of this contract: Rate of growth on assets in the unit fund 5.5% per annum Initial commission Renewal commission Rate of expense inflation Risk discount rate For renewal expenses.Bid-offer spread: 5% Annual management charge: 1. the amount quoted is at outset.5% of the second and subsequent years’ premiums 2. second year of the contract.5% of the bid value of units is deducted at the end of each policy year (management charges are deducted from the unit fund before death and maturity benefits are paid). (i) Calculate the non-unit fund cash flows in each year of the contract and hence the expected present value of profit assuming that the policyholder dies in the third year of the contract.0% per annum 6. [9] (ii) Calculate the expected present value of profit for the policy if the policyholder dies in the: (a) (b) first year of the contract.0% per annum in year 1 4.0% per annum Mortality Withdrawals Initial expenses Renewal expenses AM92 Select None £275 £80 per annum on the second and subsequent premium dates 20% of first premium 2.0% per annum in year 3 Rate of interest on non-unit fund cash flows 3. [4] (iii) Hence calculate the expected present value of the contract allowing for the possibility that the policyholder survives to the end of the contract. and the increases due to inflation start immediately.5% per annum in year 2 4. [2] [Total 15] END OF PAPER CT5 S2012–7 . particularly the open-ended questions in the later subjects. For essay-style questions. The Examiners are charged by Council with examining the published syllabus. which is designed to interpret the syllabus. The Examiners have access to the Core Reading. the report may contain more points than the Examiners will expect from a solution that scores full marks. other valid approaches are given appropriate credit. D C Bowie Chairman of the Board of Examiners December 2012 © Institute and Faculty of Actuaries . For numerical questions the Examiners’ preferred approach to the solution is reproduced in this report.INSTITUTE AND FACULTY OF ACTUARIES EXAMINERS’ REPORT September 2012 examinations Subject CT5 – Contingencies Core Technical Introduction The Examiners’ Report is written by the Principal Examiner with the aim of helping candidates. both those who are sitting the examination for the first time and using past papers as a revision aid and also those who have previously failed the subject. and will generally base questions around it but are not required to examine the content of Core Reading specifically or exclusively. accurate numerical calculation is necessary. Questions that were done less well were 6. Credit is given to students who produce alternative viable numerical solutions. Page 2 .Subject CT5 (Contingencies) – September 2012 – Examiners’ Report General comments on Subject CT5 CT5 introduces the fundamental building blocks that stand behind all life insurance and pensions actuarial work. However most of the short questions were very straightforward and this is where many successful candidates scored particularly well. to score well reasonable. more commentary on these questions is given in this report to assist candidates with further revision. 11. however. and 15(ii) and (iii). In questions where definitions of symbols and then formulae are requested. In the case of descriptive answers credit is also given to alternative valid points which do not appear in the solutions below. Comments on the September 2012 paper The general performance was lower than average this session. 12. is relevant and is properly defined and used in the answer. a different notation system produced by a student to that used by examiners is acceptable provided it is used consistently. Students should note that for long questions a reasonable level of credit is given if they can describe the right procedures. 8179 ⎛ ⎞ = 14.g.8179 = = 0.g.g.g.8179 9557.Subject CT5 (Contingencies) – September 2012 – Examiners’ Report 1 = l55 9557. Excessive or inappropriate (e.057 ⎟ 9801.2612 − 8054. • Time Selection – mortality rates vary over time (in annuities generally improves). Many candidates gave a reasonable answer but there was a tendency to overlook the obesity risk in the second paragraph.55839 × ×13. Also. alcohol consumption.850 − ⎜ 0. Inappropriate nutrition may be the result of economic factors e.2060 (a) 12 p43 (b) 10|5 q55 (c) ⎛ ⎞ ⎛l ⎞ a45:10 = a45 − ⎜ v10 × ⎜ 55 ⎟ × a55 ⎟ at 6% ⎜ ⎟ ⎝ l45 ⎠ ⎝ ⎠ = l65 − l70 8821.97269 l43 9826. • Adverse Selection – lives in better health than average may be more likely to purchase an annuity Generally well done-other plausible types and examples were credited. heart disease.0544 = = 0. hypertension) leading to increased morbidity and mortality. too much fat) eating can lead to obesity and an increased risk of associated diseases (e. Page 3 . males and females have different mortality/longevity characteristics. lack of income to buy appropriate foods or the result of a lack of health and personal education resulting in poor nutritional choices.g. social and cultural factors encourage or discourage the eating of certain foods e.740 Generally well done 2 • Class selection – e.3123 ⎝ ⎠ = 7. 3  Nutrition has an important influence on morbidity and in the longer term on mortality. Poor quality nutrition can increase the risk of contracting many diseases and hinder recovery from sickness.08027 l55 9557. 45) = annual rate of future bonus = annual net premium = curtate future lifetime of a life aged 40 exact = complete future lifetime of a life aged 40 exact Generally not done well.00475 ⎠ ⎠ = 0. s m x .t Central exposed to risk in population being studied between ages x and x + t. F is therefore is a measure of variation between population age structures. Central exposed to risk for a standard population between x and x + t.t s Exc. Page 4 . Central rate of mortality either observed or from a life table in population being studied for ages x to x + t.75 × p56 × p57 ×0.98274 Generally well done.99408 × 0.Subject CT5 (Contingencies) – September 2012 – Examiners’ Report 4 3 p55.99505 = 0.75 × 0. (b) The area comparability factor (F) is the ratio of the mortality rates in the standard population weighted by the age structure distribution of the standard population to the mortality rates in the standard population weighted by the age structure distribution of the observed population.99881× 0.t Central rate of mortality either observed or from a life table in standard mx.25 × 0.t population for ages x to x + t.75 = 0.75 q58 ) ⎛ ⎛ 0.25 q55.25 p55. 5 (a) Definitions of three terms are: Exc. It is often the case that candidates have difficulties in setting out the random variable expressions. Many candidates gave formulae that were not required. Also many did not give a complete answer.00660) ⎝ ⎝ 1 − 0.99408 × (1 − 0.75 ) × (1 − q56 ) × (1 − q57 ) × (1 −0.00475 ⎞ ⎞ = ⎜1 − ⎜ ⎟ ⎟ × 0. 6 The net future loss random variable is given by: S (1 + b) K 40 +1 v b P K40 T40 T40 − Pamin( K 40 +1.99469 × 0.99469 × 0.75 p58 = (1 −0.75 × 0. 9151 = 1949.006.631 8821. 000v 25 25 p40 − Pa40:25 = 125.2612 − P ×15. 714.966. This is often attributed to a rise in accidental deaths during young adulthood. • There is a distinct “hump” in the deaths at ages around 18–25.13 ⇒5 V = 41. reaching a peak at about age 80. • Mortality then falls dramatically during the first few years of life and is at lowest around ages 8–10. 000 × 0.00087 × DSAR = −33.98 = 11.00 − 30. and is called the “accident hump”.30832 × 17.2863 where annual net premium for the policy is given by P= 125.048.02 = −88.16 Profit = EDS – ADS=54.37512 × 8821.Subject CT5 (Contingencies) – September 2012 – Examiners’ Report 7 Reserve at the end of the 5th policy year is given by: 5V = 125. Generally well done but many candidates did not score all available marks.75 Generally done fine by well-prepared candidates 8 • Mortality just after birth (“infant mortality”) is very high. 000v30 30 p[35] a[35]:30 = 125.   • The number of deaths at higher ages falls again (even though the mortality rate qx continues to increase) since the probabilities of surviving to these ages are small.02 EDS = 3521q39 × DSAR = 3521× .2612 9892.006. 000 × 0. 006. Page 5 .959.884 9856.02 DSAR = 0 5V = −11. • From middle age onwards there is a steep increase in mortality.333.41 ADS = −8 × 11. 000. (b) The gross premium retrospective policy reserve is the expected accumulation of past gross premiums received. Page 6 .64 This question was generally poorly answered despite being a relatively straightforward question. 000 ⎜ ⎟ ( 65 − 35 ) 18866 ⎝ 12 ⎠ 11.K %. expenses) using the equivalence principle Generally done well but many answers were incomplete on a standard bookwork question. s N 35 s D35 = 20. and • the gross premium is that determined on the original basis (mortality.151 + 11. 502. 000 ⎜ ⎟ ( 65 − 35 ) l35 ⎝ 12 ⎠ s63 + s64 2 s35 3757 ( 65−35) ⎛ 1 ⎞ = v 20.655 = 5185  Assume value of contributions is K% of salary Value of contributions of K% of salary 20. and • the expenses valued are the same as those used to determine the original gross premium.K % 31. The main issue was in understanding how the benefit value arose. 000. less expected expenses and benefits including bonuses included in past claims.328 2 6.Subject CT5 (Contingencies) – September 2012 – Examiners’ Report 9 (i) (ii) (a) The gross premium prospective policy reserve is the expected present value of future benefits (including declared bonus and an allowance for future bonus if applicable) and future expenses less the expected present value of future gross premiums.836 = 316.816 Therefore K = 1.K %. Gross premium retrospective and prospective reserves will be equal if: • the mortality and interest rate basis used is the same as used to determine the gross premium at the date of issue of the policy.   10 Value of benefit: r65 ( 65−35) ⎛1⎞ v 20. interest. 090. The question was generally done poorly and many candidates failed to realise that a weekly annuity could be closely approximated by a continuous one.18 = 26090 p. Working initially for a unit annualised payment: PV = a5 + v5 +v5 m l70 m l65 × f l67 ⎛2 ⎞ __________ 1 × × a ⎜ m ):67( f ) + × a70( m ):67( f ) ⎟ 70( f 3 ⎠ l62 ⎝ 3 f ⎞ f m ⎛ m l67 l70 ⎛2 ⎞ 5 ⎛ l70 ⎞ l67 ⎛ 2 ⎞ × − × a + v − × ⎜1 f ⎟ ⎜ 70( m) ⎟ ⎜⎜1 m ⎟⎟ f × ⎜ a67( f ) ⎟ at 4% m ⎜ ⎟ l65 ⎝ l62 ⎠ ⎝ 3 ⎠ ⎠ ⎝ l65 ⎠ l62 ⎝ 3 a5 = (i / δ) × a5 = 1.611 3 = 4.1176+0.82193 × 0.a.97973 × ⎜ ×14.Subject CT5 (Contingencies) – September 2012 – Examiners’ Report 11 Because the annuity is payable weekly this can reasonably be represented by continuous annuity functions.062 + 13.733 __________ a70( m ):67( f ) = 11.611 − 9.940 ⎛ 1 ⎛2 ⎞⎞ PV = 4.5 = 11.3103 = 15. Once this is done the question is then a relatively simple calculation of annuity functions.04246 × 0.1498 The annualised benefit is 500 × 52.95754 × 0.019869 × 4. (NB 52 acceptable) So PV = 26090 ×15.258 The key to this question is to break down carefully the component parts of the annuity.733 = 14.5 = 9.062 3 2 +0.95754.733 ⎟ ⎟ 3 ⎝3 ⎠⎠ ⎝ 2 +0.5403 + ⎜ 0.611.5403+10.97973 9804.940 + × 9. Page 7 . a70( m ):67( f ) = a70( m ):67( f ) − 0.95754 × 0. a67( f ) = a67( f ) − 0.82193 × 0.82193 × 0.1498 = £395.5403 m l70 m l65 9238.062.1816+0.483 = 0.173 a70( m ) = a70( m ) − 0.134 = = 0.5 = 13.797 f l67 f l62 = 9605.02027 × ×11.97973 × ×13.4518 = 4. 9647. 08t − e −.07t + .1t ⎤ ⎫⎪ 0.1 ⎦⎥ ⎪ ⎢ ⎪⎩ ⎣⎢ ⎦⎥ 0 ⎣⎢ ⎣ 0 0 ⎭ ⎧⎪⎛ 2 3 1 ⎞ ⎛ 2e −1.08t ⎤ 25 ⎡ −.8208)} = 13.07t + e −.03e−.07t ⎤ 25 ⎡ ⎡ .75 3e−2 e −2.02t e {e (1 − e −.02t (1 − e −.5 ⎞ ⎫⎪ = 100000 ⎨⎜ + − ⎟ − ⎜ + − ⎟⎬ ⎜ 8 2 ⎟⎠ ⎪⎭ ⎪⎩⎝ 7 8 2 ⎠ ⎝ 7 = 100000{(0.75 ) + 50000e −.75 + 50000e −.1 ⎟⎠ ⎭⎪ ⎝ ⎠ ⎝ ⎩⎪ = P{(14.7 + 6766.75 (1 − e −.05t {e −.1t )dt 25 25 ⎧⎡ −.1 ⎥⎦ 0 ⎪⎭ ⎧⎪⎛ 1 1 1 ⎞ ⎛ e −1.5 − 10) − (2.08 ⎥⎦ 0 ⎢⎣ 0.2857 + 12.5 × e −.02e = 100000 ⎨ ⎢ − ⎥ + ⎢− ⎥ − ⎢− ⎥ ⎬ .03t ) + e −.75 e−2 e −2.1t )dt ⎧ ⎡ −.75 + 50000e−2 = 8688.07 .5 )) = 50000e−1.08 .6917 − 0.03t (1 − e−.08 .5 ⎞ ⎫⎪ = P ⎨⎜ + − ⎟−⎜ + − ⎟⎬ ⎜ .05075 − .03}dt 0 = 100000 ∫ 25 0 (.08 ⎦⎥ .02 + e −.1t ⎤ 25 ⎫ −e −e ⎪ −e ⎪ = P ⎨⎢ ⎥ +⎢ ⎥ −⎢ ⎥ ⎬ ⎪⎩ ⎢⎣ 0.04965 + .5 (1 − e −.03t (1 − e −.5) − (.05t }dt 0 = P∫ 25 0 (e −.07t ⎤ 25 ⎡ −.07 ⎥⎦ 0 ⎢⎣ 0.08t ⎤ ⎪ 0.Subject CT5 (Contingencies) – September 2012 – Examiners’ Report 12 The value of the death benefit is: 25 100000 ∫ e −.05e −.02t ) × .05t −.02e −.05e −.375 − 0.02t ) + e −.25 × (100000e −.07 .8 = 15456 say The value of annualised premium P is: P∫ 25 −.4825 + 1.04104)} = 10135 The value of the survival benefits are: e −1.03e−.08t − .432P Page 8 .28571 + 0.1 .03t ) × .07 . 355 P = 42.58908] +275 + 65 ×14.900 ⇒ (0.02 S ( IA)[55] +275 + 65(a[55] − 1) + 0.98 × 75.75 ×12 P ⇒ 0.62] + 275 + 967.69 Page 9 .04)0.5 [ 73.5 [28. As in Question 11 the key is to organise the component parts logically. 000 + 200) A[55] + 0. 360.025)P = (1.23 13.02 × 75.30832 × ⎜ 5.Subject CT5 (Contingencies) – September 2012 – Examiners’ Report So P = 10135 + 15456 = £1905.046 + 275 + 967.500 × 8.432 Many well prepared candidates made a very good attempt at this difficult question but in general terms it was done quite poorly. 13 (i) If the monthly premium and sum assured are denoted by P and S respectively then: 0.04)0.025 P = (0. 700 × 0.883.9929 ⎝ 24 ⎠ ⎝ = 15.9 + 0.915 ⇒ P = £263.2479 ⎛ 11 ⎞ ⎛ = ⎜15.915 + 9 P ⇒ 165.38879 + 1.891 − ⎟ − . 653. 000( IA)[55] ⎤⎦ +275 + 65(a[55] − 1) + 9 P where a(12) [55]:30 (12) (12) = a[55] − v 3030 p[55]a85 11 ⎞ 11 ⎞ ⎛ ⎛ = ⎜ a[55] − ⎟ − v3030 p[55] ⎜ a85 − ⎟ 24 ⎠ 24 ⎠ ⎝ ⎝ 11 ⎞ 3385.5 ⎡⎣ (0.98S + 200) A[55] + 0.333 − ⎟ 24 ⎠ 9545.355 P = (1.891 + 9 P ⇒ 174.823 + 12.975 × 12 Pa(12) [55]:30 + 0.433 − 0.025 P = (1.975 × 12 × 14.04)0.533 = 14.975 × 12 Pa(12) [55]:30 + 0. 04 12271.025S ( IA)85 + 80a85 where B = 30 × 0.024680 0 .57 => Cum probability of survival 1.72 8475.44 Page 10 .84 Profit Vector 12633.024368 .333 = 104.35 -46301.771370 Cash flows: Year t 1 2 3 4 Premium Expenses Interest P E on P-E 14000.5 (118.822702 NPV of Profit signature 12031.000636 .950038 .0 .389. 000 = 45.04)0.73 0 0 Maturity Claim 0 0 0 59961.000 × (1 + .37 701. 14 We have the following multiple decrement table: Year t d (aq )[30] + t −1 s (aq)30 +t −1 r (aq)30 +t −1 (ap)[ x ]+t −1 1 2 3 4 .049361 .51 + 426.375 × 0.35 -46301.863838) = 45.000548 .15 Generally part (i) was done well. 000 ⇒ V prospective = (1.024680 0 .00 399.863838 .00 700.000000 .5 {( 0.098727 .72 -29383.04)0.925723 .00 14000.907029 + .60 .71 -35715.000602 .66 526.771370 Profit signature Discount factor 12633. 000( IA)85 } + 80a85 = (1.811935 .875 × 4.333.877082 × .00 Death Surrender Redundancy Claim Claim Claim 27.04 12271.00 700.00 399.00 14000.31 Total NPV of profit = 886.000447 .877082 .40856) + 80 × 5.975S + B + 250 ) A85 + 0. 000 + 45.82 12084.811935 .877082 .7949 + 1.00 700.46 9762.82 12084.999364 t −1 (ap)[30] 1.00 14000.04 10763.99 38. surrender and redundancy costs Year t Profit Vector 1 2 3 4 12633.38 337.00 1051.00 399.023744 . 000 + 250 ) A85 + 0.40 9811.00 700.64 = £104.907029 .975 × 75.816.771370 × .811935 × .34 692.00 399.025 × 75. Very few candidates successfully completed part (ii) as is often the case with prospective reserve calculations.22 701.46 36.57 Note: allowance for ½ year interest roll up is included in death.37 33.952381 + .Subject CT5 (Contingencies) – September 2012 – Examiners’ Report (ii) Gross prospective policy value (calculated at 4%) is given by: V prospective = ( 0.952381 .59 NPV of premium = 14.877082 .02 × 75. 00 6.0% 5.001557 0.5% Allocation % (1st yr) Allocation % (2nd yr) Allocation % (3rd yr) B/O spread Management charge Policy Fee 75.333.733 126.828 3150.001802 0.000 150.500 35.114 2174.000 323.000 105. 15 Annual premium Risk discount rate Interest on investments (1st yr) Interest on investments (2nd yr) Interest on investments (3rd yr) Interest on non-unit funds Death benefit (% of bid value of units) Initial expense Renewal expense Expense inflation £3000. Very few got to the final answer however.44 = 1.998799 0.Subject CT5 (Contingencies) – September 2012 – Examiners’ Report Therefore.5% 5.50 35.0% 20.528 78.998198 1.001201 0.997244 Unit fund (per policy at start of year) yr 1 value of units at start of year Alloc B/O policy fee interest management charge value of units at year end 0.000 224.805 Page 11 .256 8290.0% 4.0% 100.000 2250.511 yr 2 2174.0% 2. profit margin = 886.5% 4.000 35.96% Credit was given for correct data items and many well prepared candidates scored a reasonable proportion of the marks available.511 3000.998443 0.0% 3.5% £35 Mortality table: X q[ x ]+t −1 p[ x]+t −1 t −1 p[ x ] 45 46 47 0.000000 0.000 157.828 yr 3 5135.000 112.211 5135.61/45.0% 1.998799 0.0% 105.125 33.0% 150% £ % premium 275 80 2. 852 + 78.418 If policyholder dies in the 3rd year of contract.805 ) = −4138.891 (ii) (a) If policyholder dies in the 1st year of contract.600 0. non unit cash flow at end of 1st year is: yr 1 = ( 785 + 112.464 yr 3 −115.5 − 158.967 × v × q[45] = −968.998443 × 0.472 + 126.852 78.289 × v − 2460.451 => expected present value of these cash flows is given by: ⎡56.746 − 3425.998799 × 0.854 − 2169.256 7.232 − 3.511) = −1030.472 126.163 (b) If policyholder dies in the 2nd year of contract.351 => expected present value of these cash flows is given by: ⎡56.000 156.500 158.351× v3 ⎤ × p[45] × p[45]+1 × q47 ⎣ ⎦ = [52.984 yr 2 35.289 yr 2 = ( 35 + 150 − 156.5 × 8290. non unit cash flows at end of each year are: yr 1 == 56.211 − 0.289 × v + 107.211 3.6 + 0.000 157.289 (derived above) yr 2 = ( 35 + 150 − 156.000 150.5 × 2174.675 33.967 × 0.828 ) = −2460.256 − 0.500 875.000 112.114 ) = 56.291 Page 12 .232 −3.854 + 94.001201 = −1.5 − 875 + 0.470 −0.967 => expected present value of this cash flow is given by: −1030. non unit cash flows at end of each year are: yr 1 = ( 785 + 112.463 × v 2 − 4138.852 + 78.114 − 0.6 + 0.998799 × 0.001802 = −5.463 yr 3 = ( −115 + 157.114 1.675 + 33.5 × 5135.451× v 2 ⎤ × p[45] × q[45]+1 ⎣ ⎦ = [52.211) = 107.306 54.000 0.001557 = −3.279] × 0.Subject CT5 (Contingencies) – September 2012 – Examiners’ Report Cash flows (per policy at start of year) yr 1 unallocated premium + pol fee b/o spread expenses interest man charge extra death benefit profit vector (i) 785.5 − 875 + 0.675 + 33.930] × 0.998 103. 998198 = 152. Part credit was given in (i) for correctly calculating the data items and well prepared candidates scored a fair proportion of the marks here.998443 × 0.854 + 94.163 – 3.39 Candidates generally found this question difficult particularly parts (ii) and (ii).838] × 0.256 ) = 7.052 => expected present value of these cash flows is given by: ⎡56.5 − 158.232 − 3.737 = 142.052 × v3 ⎤ × 3 p[45] ⎣ ⎦ = [52.Subject CT5 (Contingencies) – September 2012 – Examiners’ Report (iii) If policyholder survives until end of contract.891 + 152.746 + 5.463 (derived above) yr 3 = ( −115 + 157.289 (derived above) yr 2 = 107. non unit cash flows at end of each year are: yr 1 == 56.998799 × 0.472 + 126.291 − 5.737 Expected present value of policy is therefore = −1.463 × v 2 + 7. END OF EXAMINERS’ REPORT Page 13 .289 × v + 107.