SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETYEXAM C CONSTRUCTION AND EVALUATION OF ACTUARIAL MODELS EXAM C SAMPLE QUESTIONS Copyright 2008 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study note are taken from past SOA/CAS examinations. C-09-08 PRINTED IN U.S.A. 1. You are given: (i) Losses follow a loglogistic distribution with cumulative distribution function: bx / θg F b xg = 1+ bx / θ g γ γ (ii) The sample of losses is: 10 35 80 86 90 120 158 180 200 210 1500 Calculate the estimate of θ by percentile matching, using the 40th and 80th empirically smoothed percentile estimates. (A) (B) (C) (D) (E) Less than 77 At least 77, but less than 87 At least 87, but less than 97 At least 97, but less than 107 At least 107 2. You are given: (i) The number of claims has a Poisson distribution. (ii) (iii) Claim sizes have a Pareto distribution with parameters θ = 0.5 and α = 6 . The number of claims and claim sizes are independent. The observed pure premium should be within 2% of the expected pure premium 90% of the time. (iv) Determine the expected number of claims needed for full credibility. (A) (B) (C) (D) (E) Less than 7,000 At least 7,000, but less than 10,000 At least 10,000, but less than 13,000 At least 13,000, but less than 16,000 At least 16,000 C-09-08 -1- 3. You study five lives to estimate the time from the onset of a disease to death. The times to death are: 2 3 3 3 7 Using a triangular kernel with bandwidth 2, estimate the density function at 2.5. (A) (B) (C) (D) (E) 8/40 12/40 14/40 16/40 17/40 4. You are given: (i) Losses follow a Single-parameter Pareto distribution with density function: f ( x ) = (α +1) , x α x >1, 0< α < ∞ (ii) A random sample of size five produced three losses with values 3, 6 and 14, and two losses exceeding 25. Determine the maximum likelihood estimate of α . (A) (B) (C) (D) (E) 0.25 0.30 0.34 0.38 0.42 C-09-08 -2- 1 At least 1.7 6. For a sample of dental claims x1.5. Calculate E X ∧ 500 for the fitted distribution. You are given: The annual number of claims for a policyholder has a binomial distribution with (i) probability function: ⎛ 2⎞ 2− x p ( x q ) = ⎜ ⎟ q x (1 − q ) . but less than 175 At least 175. . but less than 1. 1. 0 < q < 1 This policyholder had one claim in each of Years 1 and 2.3. x = 0.574. x2 .802 Claims are assumed to follow a lognormal distribution with parameters μ and σ . 2 ⎝ x⎠ (ii) The prior distribution is: π ( q ) = 4q 3 . (A) (B) (C) (D) (E) Less than 125 At least 125. x10 . . you are given: (i) (ii) (iii) ∑ xi = 3860 and ∑ xi2 = 4. Determine the Bayesian estimate of the number of claims in Year 3. but less than 275 At least 275 -3- C-09-08 . but less than 1. μ and σ are estimated using the method of moments.5 At least 1. . (A) (B) (C) (D) (E) Less than 1.1. . but less than 1.7 At least 1.3 At least 1. but less than 225 At least 225.5. losses in Year 1 were 5. but less than 3 At least 3. 11. θ >1 bg θ Calculate Bühlmann’s k for aggregate losses. but less than 4 At least 4 9. Claim counts and claim sizes are independent. 0< x<∞ 2 ( x +θ ) For half of the company’s policies θ = 1 . The prior distribution has probability density function: 5 π θ = 6 . DELETED DELETED You are given: (i) Losses on a company’s insurance policies follow a Pareto distribution with probability density function: f (xθ ) = (ii) θ . (A) Less than 1 (B) (C) (D) (E) At least 1. DELETED You are given: Claim counts follow a Poisson distribution with mean θ . C-09-08 -4- . but less than 2 At least 2. given θ .7. (i) (ii) (iii) (iv) Claim sizes follow an exponential distribution with mean 10θ . 8. while for the other half θ = 3 . 10. For a randomly selected policy. C-09-08 -5- .27 12.11 0.3512 0. (A) (B) (C) (D) (E) 655 670 687 703 719 13. (A) (B) (C) (D) (E) 0.19 0. You are given total claims for two policyholders: Year Policyholder X Y 1 730 655 2 800 650 3 650 625 4 700 750 Using the nonparametric empirical Bayes method. A particular line of business has three types of claims.3744 Number of Claims in Current Year 112 180 138 You test the null hypothesis that the probability of each type of claim in the current year is the same as the historical probability.21 0. determine the Bühlmann credibility premium for Policyholder Y.2744 0.Determine the posterior probability that losses for this policy in Year 2 will exceed 8.15 0. The historical probability and the number of claims for each type in the current year are: Type A B C Historical Probability 0. but less than 12 At least 12 14. The information associated with the maximum likelihood estimator of a parameter θ is 4n . (A) (B) (C) (D) (E) Less than 9 At least 9. 1 1 4 (A) (B) (C) (D) (E) 2n n n 8n 16n C-09-08 -6- . Calculate the asymptotic variance of the maximum likelihood estimator of 2θ . where n is the number of observations. but less than 10 At least 10.Calculate the chi-square goodness-of-fit test statistic. but less than 11 At least 11. (A) (B) (C) (D) (E) 0.450 0.0080 0. For a survival study with censored and truncated data. you are given: Time (t) 1 2 3 4 5 Number at Risk at Time t 30 27 32 25 20 Failures at Time t 5 9 6 5 4 16. given survival past Time 1. is 3 q1 . The probability of failing at or before Time 4. You are given: (i) (ii) (iii) The probability that an insured will have at least one loss during any year is p.0091 0.500 0. Calculate Greenwood’s approximation of the variance of 3 q1 . Use the following information for questions 21 and 22. The prior distribution for p is uniform on [ 0.5] .0067 0.0.0073 0. Determine the posterior probability that the insured will have at least one loss during Year 9.15. (A) (B) (C) (D) (E) 0.550 0.0105 C-09-08 -7- .475 0.625 16-17. An insured is observed for 8 years and has at least one loss every year. but less than 10.07) (0.44.56. Determine the Bühlmann credibility estimate of the second claim amount from the same risk.5 0.800 At least 10.89) (0.800 C-09-08 -8- .31. A claim of 250 is observed. (A) (B) (C) (D) (E) Less than 10.99) (0. (A) (B) (C) (D) (E) (0.400 At least 10. 0.54) (0.39.79) bg 18. based on the Nelson-Aalen estimate. 0. 0.2 Probability of Claim Amount for Risk 2 0.2 0.600. Calculate the 95% log-transformed confidence interval for H 3 .600 At least 10.000 (ii) Probability of Claim Amount for Risk 1 0.3 0. 1.1 Risk 1 is twice as likely to be observed as Risk 2. 1.200.7 0.30. You are given: (i) Two risks have the following severity distributions: Amount of Claim 250 2. but less than 10.400.500 60.17. but less than 10.200 At least 10. ∑ 2 Determine the bootstrap approximation to the mean square error of g. 5 and 9. (A) (B) (C) (D) (E) 1 2 4 8 16 C-09-08 -9- .…. X2) = 1 ( X i − X )2 . [θ θ σ σ 0< x<∞ (ii) (iii) θ >σ ∑ xi = 150 and ∑ xi2 = 5000 Estimate θ by matching the first two sample moments to the corresponding population quantities. You are given a sample of two values.19. (A) (B) (C) (D) (E) 9 10 15 20 21 20. x10 is drawn from a distribution with probability density function: 1 2 1 exp(− x ) + 1 exp(− x )]. x2 . You estimate Var(X) using the estimator g(X1. You are given: (i) A sample x1 . (iii) ∑ ln ( xi + 7.9 17.92. The natural logarithm of the likelihood function evaluated at the maximum likelihood estimates is −817. The claim frequencies of different insureds are independent.8 .21. C-09-08 . (A) (B) (C) (D) (E) 16. You are given: (i) (ii) (iii) The number of claims incurred in a month by any insured has a Poisson distribution with mean λ . You are given: (i) (ii) The maximum likelihood estimates are α = 1.6.7 16.0 22.10 - .8) = 607.3 17.5 and θ = 7.64 Determine the result of the test. The prior distribution is gamma with probability density function: 6 100λ ) e −100λ ( f (λ ) = 120λ (iv) Month 1 2 3 4 Number of Insureds 100 150 200 300 Number of Claims 6 8 11 ? Determine the Bühlmann-Straub credibility estimate of the number of claims in Month 4.4 and θ = 7.6 18. You fit a Pareto distribution to a sample of 200 claim amounts and use the likelihood ratio test to test the hypothesis that α = 1. 0 6.005 significance level.6 C-09-08 .θ .005 level. Reject at the 0. Reject at the 0.050 significance level.11 - .025 significance level. Do not reject at the 0.010 level.8 7.050 significance level. For a sample of 15 losses. 2] (2.4 6.2 7. ∞ ) (ii) Observed Number of Losses 5 5 5 Losses follow the uniform distribution on 0. Reject at the 0. where E j is the expected number of losses in the jth interval and O j is the observed number of losses in the jth interval. but not at the 0.(A) (B) (C) (D) (E) Reject at the 0. but not at the 0. b g 2 Estimate θ by minimizing the function ∑ 3 ( E j − Oj ) Oj j =1 . you are given: (i) Interval (0. (A) (B) (C) (D) (E) 6.025 level. 5] (5. 23.010 significance level. but not at the 0. 12 - .63 0.54 0. πθ = bg Determine the posterior probability that θ is greater than 0. 0<θ <1 2 A randomly chosen insured is observed to have exactly one claim.67 0.58 0.72 25. You are given: (i) (ii) The probability that an insured will have exactly one claim is θ . The prior distribution of θ has probability density function: 3 θ.24. The distribution of accidents for 84 randomly selected policies is as follows: Number of Accidents 0 1 2 3 4 5 6 Total Number of Policies 32 26 12 7 4 2 1 84 Which of the following models best represents these data? C-09-08 . (A) (B) (C) (D) (E) 0.60. One claim from a high-hazard risk is observed. You are given: (i) (ii) (iii) (iv) (v) (vi) Low-hazard risks have an exponential claim size distribution with mean θ . High-hazard risks have an exponential claim size distribution with mean 3θ . 2 and 3.(A) (B) (C) (D) (E) Negative binomial Discrete uniform Poisson Binomial Either Poisson or Binomial 26.13 - . No claims from low-hazard risks are observed. Determine the maximum likelihood estimate of θ . Three claims from medium-hazard risks are observed. of size 15. of sizes 1. Medium-hazard risks have an exponential claim size distribution with mean 2θ . (A) (B) (C) (D) (E) 1 2 3 4 5 C-09-08 . C-09-08 . You are given: Claim Size (X) b100. 200 b50. 50 b0.27. You are given: (i) (ii) X partial = pure premium calculated from partially credible data μ = E⎡ ⎣ X partial ⎤ ⎦ Fluctuations are limited to ± k μ of the mean with probability P Z = credibility factor (iii) (iv) Which of the following is equal to P? (A) (B) (C) (D) (E) Pr ⎡ ⎣ μ − k μ ≤ X partial ≤ μ + k μ ⎤ ⎦ Pr ⎡ ⎣ Z μ − k ≤ Z X partial ≤ Z μ +k ⎤ ⎦ Pr ⎡ ⎣ Z μ − μ ≤ Z X partial ≤ Z μ +μ ⎤ ⎦ Pr ⎡ ⎣1 − k ≤ Z X partial + (1 − Z ) μ ≤ 1 + k ⎤ ⎦ Pr ⎡ ⎣ μ − k μ ≤ Z X partial + (1 − Z ) μ ≤ μ + k μ ⎤ ⎦ 28. 25 Number of Claims 25 28 15 6 Assume a uniform distribution of claim sizes within each interval.14 - . 100 b25. 40 0.1 0. At least 200.4 One randomly chosen risk has three claims during Years 1-6.15 - .33 0.22 0.2 0. but less than 400 At least 400. Determine the posterior probability of a claim for this risk in Year 7.46 C-09-08 .1 Annual Claim Probability 0.28 0. but less than 500 At least 500 29. (A) (B) (C) (D) (E) 0.Estimate E X 2 − E X ∧ 150 (A) (B) (C) (D) (E) Less than 200 c h b g 2 . (i) (ii) Type of Risk I II III Prior Probability 0.7 0. but less than 300 At least 300. You are given: Each risk has at most one claim each year.2 0. a (i) new policy was added. (ii) (iii) Policies are surrendered only at the end of a policy year. (A) (B) (C) (D) (E) 358 371 384 390 396 . F What is the value of n? 8 (A) (B) (C) (D) (E) 9 10 11 12 31. r j .542.30. meaning that the risk set. is always equal to 100. The number of policies surrendered at the end of each policy year was observed to be: 1 at the end of the 1st policy year 2 at the end of the 2nd policy year 3 at the end of the 3rd policy year n at the end of the nth policy year (iv) The Nelson-Aalen empirical estimate of the cumulative distribution function at time n. ˆ ( n) . You are given the following claim data for automobile policies: 200 255 295 320 360 420 440 490 500 520 1020 Calculate the smoothed empirical estimate of the 45th percentile. You are given the following about 100 insurance policies in a study of time to policy surrender: The study was designed in such a way that for every policy that was surrendered.16 - C-09-08 . is 0. was 23/132. H(t). (A) (B) (C) (D) (E) 0.37 0. You are given: (i) The number of claims made by an individual insured in a year has a Poisson distribution with mean λ. Three claims are observed in Year 1.35 1. In a study of claim payment times.35 0.43 33.41 1.36 1.40 1.41 0.17 - . immediately following the second paid claim.32. Determine the Nelson-Aalen estimate of the cumulative hazard function. The Nelson-Aalen estimate of the cumulative hazard function. At most one claim was paid at any one time. H(t). Using Bühlmann credibility. (ii) The prior distribution for λ is gamma with parameters α = 1 and θ = 1.39 0. and no claims are observed in Year 2. (A) (B) (C) (D) (E) 1. immediately following the fourth paid claim. estimate the number of claims in Year 3. you are given: (i) (ii) (iii) The data were not truncated or censored.2 .43 C-09-08 . 50 3. where β is unknown and r is known.34. You wish to estimate β based on n observations.75 C-09-08 . (A) (B) (C) (D) (E) 0. You are given the following information about a credibility model: First Observation 1 2 3 Unconditional Probability 1/3 1/3 1/3 Bayesian Estimate of Second Observation 1.75 1.00 Determine the Bühlmann credibility estimate of the second observation.25 1. where x is the mean of these observations. Determine the maximum likelihood estimate of β . The number of claims follows a negative binomial distribution with parameters β and r.50 1.18 - . (A) x r2 x r (B) (C) (D) (E) x rx r2x 35. given that the first observation is 1.00 1.50 1. 7 At least 0.695. bg b g Determine S t0 .779 bg 37.762 (C) 0. Determine the maximum likelihood estimate of α .36.843 . but less than 0. (A) (B) (C) (D) (E) Less than 0.9 C-09-08 .769 (E) 0. A random sample of three claims from a dental insurance plan is given below: 225 525 950 Claims are assumed to follow a Pareto distribution with parameters θ = 150 and α . 0. For a survival study.8. bg The Product-Limit estimator S t0 is used to construct confidence intervals for bg (ii) The 95% log-transformed confidence interval for S t0 is 0.758 (B) 0.765 (D) 0.7.6.19 - .9 At least 0. (A) 0.6 At least 0. but less than 0.8 At least 0. you are given: (i) S t0 . but less than 0. You are given the following information about a commercial auto liability book of business: (i) (ii) (iii) Each insured’s claim count has a Poisson distribution with mean λ .60 ∑ c Xi − X h i =1 2 = 3. (A) (B) (C) (D) (E) Less than 0. gamma distribution with α = 15 Individual claim size amounts are independent and exponentially distributed with mean 5000. calculate the Bühlmann credibility factor for an individual policyholder.80 At least 0.74 At least 0.2 .74. An insurer has data on losses for four policyholders for 7 years. You are given: ∑ ∑ d X ij − X i i i =1 j =1 4 4 7 2 = 33.38. but less than 0.83 39.80.90. but less than 0. The loss from the i th policyholder for year j is X ij .77 At least 0. C-09-08 . but less than 0. and θ = 0.77.83 At least 0.30 Using nonparametric empirical Bayes estimation. The full credibility standard is for aggregate losses to be within 5% of the expected with probability 0. where λ has a .20 - . The prior distribution for σ 2 is exponential with mean 1. The mean of the exponential distribution is estimated using the method of moments.Using classical credibility. (A) (B) (C) (D) (E) 0. determine the expected number of claims required for full credibility. C-09-08 . You are given: (i) Annual claim frequency for an individual policyholder has mean λ and variance σ 2 . (ii) (iii) The prior distribution for λ is uniform on the interval [0. You are given: (i) A sample of claim payments is: 29 (ii) (iii) 64 90 135 182 Claim sizes are assumed to follow an exponential distribution.14 0. Using Bühlmann credibility.25.19 0. estimate the number of claims in Year 2 for the selected policyholder.16 0. Calculate the value of the Kolmogorov-Smirnov test statistic.5.21 - .5].25 0. 1. A policyholder is selected at random and observed to have no claims in Year 1. (A) (B) (C) (D) (E) 2165 2381 3514 7216 7938 40.27 41. (A) (B) (C) (D) (E) 0.94 42. claim sizes follow a Pareto distribution with parameters α = 2 and θ . Calculate the posterior probability that Θ exceeds 2.22 - .33 0. DELETED You are given: (i) The prior distribution of the parameter Θ has probability density function: πθ = (ii) bg 1 Given Θ = θ .56 0.42 0. (A) (B) (C) (D) (E) 0. θ2 .64 C-09-08 . 43.83 0.71 0. 1<θ < ∞ A claim of 3 is observed.65 0.50 0.58 0. X. but less than 2100 At least 2100. θ > 500 . x h . but less than 2400 At least 2400 45. Two claims. are observed. x h = 3G H θ JK Calculate the Bayesian premium. θ > 600 f cθ x .θ . ∞ ) Frequency 7 6 7 Calculate the maximum likelihood estimate of θ . You are given: (i) (ii) The amount of a claim. A random sample of 20 losses is distributed as follows: Loss Range [0. Ec X x . 3 1 2 4 3 1 2 C-09-08 .44. You are given: (i) (ii) Losses follow an exponential distribution with mean θ . 2000] (2000.23 - . but less than 2250 At least 2250. 1000] (1000. is uniformly distributed on the interval 0. The prior density of θ is π θ = bg 500 θ2 . You calculate the posterior distribution as: F 600 I . (A) (B) (C) (D) (E) Less than 1950 At least 1950. x1 = 400 and x2 = 600 . calculate the probability that the loss on a policy exceeds 8.(A) (B) (C) (D) (E) 450 500 550 600 650 46. The claim payments on a sample of ten policies are: 2 3 3 5 5 + 6 7 7 + 9 10 + + indicates that the loss exceeded the policy limit Using the Product-Limit estimator. (A) (B) (C) (D) (E) 0.20 0.25 0. You are given the following observed claim frequency data collected over a period of 365 days: Number of Claims per Day 0 1 2 3 4+ Observed Number of Days 50 122 101 92 0 Fit a Poisson distribution to the above data. using the method of maximum likelihood.30 0. by number of claims per day.36 0.24 - . into four groups: 0 1 2 3+ C-09-08 . Regroup the data.40 47. Determine the result of the chi-square test.005 level.4 0.Apply the chi-square goodness-of-fit test to evaluate the null hypothesis that the claims follow a Poisson distribution. Reject at the 0.79 0. but not at the 0. Reject at the 0.86 0. (A) (B) (C) (D) (E) Reject at the 0.1 0. You are given the following joint distribution: Θ X 0 0 1 2 0. Reject at the 0.75 0. but not at the 0.010 level.005 significance level.82 0.010 significance level. 48. but not at the 0.025 significance level.25 - .1 0.1 1 0.1 For a given value of Θ and a sample of size 10 for X: ∑ xi = 10 i =1 10 Determine the Bühlmann credibility premium. Do not reject at the 0.2 0.050 significance level.025 level.050 significance level. (A) (B) (C) (D) (E) 0.89 C-09-08 . You are given four classes of insureds.49 –0.1 3 0. If five insureds are selected at random from the same class.08 0. when n = 4. You are given: x 0 0. μ . 2 Calculate the bias of Sn . σ 2 . and four insureds are selected at random from the class.5 0.24 –0.8 0.49.26 - . The total number of claims is two.72 –0.1 Pr[X = x] The method of moments is used to estimate the population mean.5 0.9 0.5 1 0.00 50. (A) (B) (C) (D) (E) –0. respectively. by X and S 2 n ∑( X = i −X) 2 n . each of whom may have zero or one claim.1 0.9 A class is selected at random (with probability ¼). with the following probabilities: Class I II III IV Number of Claims 0 0.3 2 0. and variance. estimate the total number of claims using Bühlmann-Straub credibility.2 0. C-09-08 .1 1 0. (B) or (C) 53. 52.2 2.0 2.8 51.(A) (B) (C) (D) (E) 2.27 - .4 2. DELETED With the bootstrapping technique. Determine the variance of the aggregate loss. (B) or (C) None of (A). C-09-08 .6 2. You are given: Number of Claims 0 1 2 Probability 1 3 5 5 Claim Size 25 150 50 200 Probability 1 3 2 2 3 3 1 5 1 3 Claim sizes are independent. the underlying distribution function is estimated by which of the following? (A) (B) (C) (D) (E) The empirical distribution function A normal distribution function A parametric distribution function selected by the modeler Any of (A). 500 12.510 15.(A) (B) (C) (D) (E) 4. A random sample of losses is distributed as follows: Loss Range (0 – 100] (100 – 200] (200 – 400] (400 – 750] (750 – 1000] (1000 – 1500] Total Estimate θ by matching at the 80th percentile.612 54.100 10.050 8. (A) (B) (C) (D) (E) 249 253 257 260 263 Number of Losses 32 21 27 16 2 2 100 C-09-08 .28 - . You are given: (i) (ii) Losses follow an exponential distribution with mean θ . 55. You are given: Class 1 2 3 Number of Insureds 3000 2000 1000 Claim Count Probabilities 0 1 3 1 1 1 3 6 2 1 2 1 3 3 6 3 0 1 2 6 3 4 0 0 1 6 0 0 0 A randomly selected insured has one claim in Year 1. Determine the expected number of claims in Year 2 for that insured. (A) (B) (C) (D) (E) 1.00 1.25 1.33 1.67 1.75 56. You are given the following information about a group of policies: Claim Payment 5 15 60 100 500 500 Policy Limit 50 50 100 100 500 1000 Determine the likelihood function. (A) f(50) f(50) f(100) f(100) f(500) f(1000) 29 (B) (C) (D) (E) f(50) f(50) f(100) f(100) f(500) f(1000) / [1-F(1000)] f(5) f(15) f(60) f(100) f(500) f(500) f(5) f(15) f(60) f(100) f(500) f(500) / [1-F(1000)] f(5) f(15) f(60) [1-F(100)] [1-F(500)] f(500) 57. You are given: Claim Size 0-25 25-50 50-100 100-200 Number of Claims 30 32 20 8 Assume a uniform distribution of claim sizes within each interval. Estimate the second raw moment of the claim size distribution. (A) (B) (C) (D) (E) Less than 3300 At least 3300, but less than 3500 At least 3500, but less than 3700 At least 3700, but less than 3900 At least 3900 58. You are given: 30 (i) (ii) The number of claims per auto insured follows a Poisson distribution with mean λ. The prior distribution for λ has the following probability density function: b500λ g e f bλ g = λΓb50g (iii) Number of claims Number of autos insured 50 −500 λ A company observes the following claims experience: Year 1 75 600 Year 2 210 900 The company expects to insure 1100 autos in Year 3. Determine the expected number of claims in Year 3. (A) (B) (C) (D) (E) 178 184 193 209 224 59. The graph below shows a p-p plot of a fitted distribution compared to a sample. 31 75 32 . You are given the following information about six coins: Coin 1–4 5 6 Probability of Heads 0. The tails of the fitted distribution are too thick on the left and on the right.50 0. The tails of the fitted distribution are too thin on the left and on the right.25 0. 60. and the fitted distribution has less probability around the median than the sample.Fitted Sample Which of the following is true? (A) (B) (C) (D) (E) The tails of the fitted distribution are too thick on the left and on the right. and the fitted distribution has less probability around the median than the sample. too thin on the right. and the fitted distribution has more probability around the median than the sample. and the fitted distribution has more probability around the median than the sample. and the fitted distribution has less probability around the median than the sample. The tail of the fitted distribution is too thick on the left. The tails of the fitted distribution are too thin on the left and on the right. An insurer writes a large book of home warranty policies.54 0.0.59 0. Determine the mean of the fitted distribution.1 l q l q Determine E X 5 S using Bayesian analysis.A coin is selected at random and then flipped repeatedly.52 0. The following sequence is obtained: S = X 1 . X 4 = 11 . X 3 . You observe the following five ground-up claims from a data set that is truncated from below at 100: 125 150 165 175 250 You fit a ground-up exponential distribution using maximum likelihood estimation. (A) (B) (C) (D) (E) 0.63 c h 61. . You are given the following information regarding claims filed by insureds against these policies: 33 . X 2 . where “1” indicates heads and “0” indicates tails.56 0. (A) (B) (C) (D) (E) 73 100 125 156 173 62. X i denotes the outcome of the ith flip. 08 0. The variance of the individual insured claim probabilities is 0.17 0.19 34 . The probability of a claim varies by insured.(i) (ii) (iii) (iv) (v) A maximum of one claim may be filed per year. The prior distribution of θ is π θ = (ii) bg 500 θ2 . The probability of a claim for each insured remains constant over time. Determine the probability that the next claim will exceed 550. and the claims experience for each insured is independent of every other insured. θ > 500 .04 0. Determine the Bühlmann credibility estimate for the expected number of claims the selected insured will file over the next 5 years. (iii) Two independent claims of 400 and 600 are observed. 64. DELETED For a group of insureds.10. (A) 0. An insured selected at random is found to have filed 0 claims over the past 10 years.01. The overall probability of a claim being filed by a randomly selected insured in a year is 0. you are given: (i) The amount of a claim is uniformly distributed but will not exceed a certain unknown limit θ . (A) (B) (C) (D) (E) 0.25 63.22 0. To estimate E X . (iv) Yi1. X 4 and X 5 with the following results: 1 2 3 4 5 You want the standard deviation of the estimator of E X to be less than 0.0 and θ = 1000 .25 0. X 2 .31 65.. . Using classical credibility theory. You are given the following information about a general liability book of business comprised of 2500 insureds: (i) (ii) X i = ∑ Yij is a random variable representing the annual loss of the ith insured. determine the partial credibility of the annual loss experience for this book of business.50 0. (iii) following a Pareto distribution with α = 3..34 0.28 0... . Yi 2 .. j =1 Ni N1. (A) (B) (C) (D) (E) 0.05.22 0.2. you have simulated X 1 . X 3 . 35 .53 66.. N 2 .(B) (C) (D) (E) 0. N 2500 are independent and identically distributed random variables following a negative binomial distribution with parameters r = 2 and β = 0.42 0.47 0. YiN i are independent and identically distributed random variables The full credibility standard is to be within 5% of the expected aggregate losses 90% of the time. you are given: (i) Country A Country B Sj rj Sj rj ti 1 20 200 15 100 36 . (A) (B) (C) (D) (E) Less than 150 At least 150. For a mortality study of insurance applicants in two countries..045 0. Yi1. Z. j =1 Ni N1. .826 0. . YiN i are independent random variables distributed according to a Pareto distribution with α = 3. (iii) (iv) Unknown parameter r has an exponential distribution with mean 2. N100 are independent random variables distributed according to a negative binomial distribution with parameters r (unknown) and β = 0. N 2 .. but less than 900 At least 900 67. You are given the following information about a book of business comprised of 100 insureds: (i) (ii) X i = ∑ Yij is a random variable representing the annual loss of the i th insured.905 68. (A) (B) (C) (D) (E) 0.. Yi 2 .Estimate the total number of simulations needed.... Determine the Bühlmann credibility factor.2. for the book of business. but less than 400 At least 400. but less than 650 At least 650.500 0.000 0.0 and θ = 1000 . 06 0. You fit an exponential distribution to the following data: 1000 1400 5300 7400 7600 Determine the coefficient of variation of the maximum likelihood estimate of the mean. (A) (B) (C) (D) (E) 0. ti . (A) (B) (C) (D) 0. Deaths.08 0.10 bg bg 69. S bt g is the Product-Limit estimate of S bt g based on the data for all study i −1 i T i r j is the number at risk over the period ti −1 .33 0.2 3 4 (ii) 54 14 22 180 126 112 20 20 10 85 65 45 (iii) bt .45 0. t g are assumed to occur at t . S j .07 0. bg bg Determine S T 4 − S B 4 . (iv) S B t is the Product-Limit estimate of S t based on the data for study participants in Country B. θ. during the period b g participants.70 1.09 0.00 37 . 8 2.5 0.0 1.06 Pleasure Use Expected Claim Claims Variance 1.0 0.19 0.(E) 1.5 2. You are investigating insurance fraud that manifests itself through claimants who file claims with respect to auto accidents with which they were not involved. You are given the following information on claim frequency of automobile accidents for individual drivers: Rural Urban Total Business Use Expected Claim Claims Variance 1.21 70.05 0. Your evidence consists of a distribution of the observed number of claimants per accident and a standard distribution for accidents on which fraud is known to be absent.12 You are also given: (i) (ii) Each driver’s claims experience is independent of every other driver’s. The two distributions are summarized below: Number of Claimants per Accident Standard Probability 38 Observed Number of Accidents .8 1. Determine the Bühlmann credibility factor for a single driver.17 0. (A) (B) (C) (D) (E) 0.3 1.5 2.09 0.0 1.0 1. There are an equal number of business and pleasure use drivers.27 71. 005 significance level.01 1. 72.25 .5 39 . Reject at the 0.010 level.010 significance level. The expected value of the process variance is 8000. The following experience is observed for a randomly selected policyholder: Year 1 2 3 Average Loss per Employee 15 10 5 Number of Employees 800 600 400 Determine the Bühlmann-Straub credibility premium per employee for this policyholder.050 significance level. Reject at the 0. The overall average loss per employee for all policyholders is 20. but not at the 0. but not at the 0.1 2 3 4 5 6+ Total 0.025 level.11 .005 level. Do not reject at the 0.025 significance level.050 significance level. The variance of the hypothetical means is 40. (A) Less than 10.35 . (A) (B) (C) (D) (E) Reject at the 0. Reject at the 0.04 .00 235 335 250 111 47 22 1000 Determine the result of a chi-square test of the null hypothesis that there is no fraud in the observed accidents.24 . but not at the 0. You are given the following data on large business policyholders: (i) (ii) (iii) (iv) (v) Losses for each employee of a given policyholder are independent and have a common mean and variance. 67 0. Estimate.80 74.000. 75.000] (15.000-30.77 0.74 0.70 0.5 73.5. but less than 11.5 At least 13.000] Number of Claims in Interval 1 1 4 Number of Claims Censored in Interval 2 2 0 Assume that claim sizes and censorship points are uniformly distributed within each interval. DELETED You are given: (i) Claim amounts follow a shifted exponential distribution with probability density function: 1 f x = e− b x −δ g/θ . δ < x < ∞ bg θ 40 . using the life table methodology.000] (30.5. but less than 13. You are given the following information about a group of 10 claims: Claim Size Interval (0-15.5.(B) (C) (D) (E) At least 10. (A) (B) (C) (D) (E) 0.000-45.5 At least 12. but less than 12.5 At least 11. the probability that a claim exceeds 30. 81 0. X 2 ..(ii) A random sample of claim amounts X1.78 0. (A) (B) (C) (D) (E) 0.5 4.0 3.0 i = 100 and ∑X 2 i = 1306 Estimate δ using the method of moments. X10 : 5 5 5 6 8 9 11 12 16 23 (iii) ∑X 3.. You are given: (i) (ii) The annual number of claims for each policyholder follows a Poisson distribution with mean θ . The distribution of θ across all policyholders has probability density function: f θ = θ e −θ .0 4. Determine the posterior probability that this same policyholder will have at least one claim this year.70 0.. θ > 0 bg (iii) z 0 ∞ θ e − nθ dθ = 1 n2 A randomly selected policyholder is known to have had at least one claim last year. (A) (B) (C) (D) (E) 76.5 5..75 0.86 41 . 77. 2. 2000 Calculate the expected size of the next claim using Bühlmann credibility. A survival study gave (1. (A) (B) (C) (D) (E) (0. 3. (A) (B) (C) (D) (E) 1025 1063 1115 1181 1266 42 . 2. 1075.55) as the 95% linear confidence interval for the cumulative hazard function H t0 . 2. 2. bg Calculate the 95% log-transformed confidence interval for H t0 .84.58. has mean μ and variance 500.63.60) (1.49. 0. The random variable μ has a mean of 1000 and variance of 50.94) (0.60) bg 78.68.50) (1. The following three claims were observed: 750.68. You are given: (i) (ii) (iii) Claim size.34) (1. X. (A) (B) (C) (D) (E) F pe .000 JK GH 100 10.F pe .000 JK F pe . from a two point mixture.2 −1 −0.000 JK F pe + b1 − pge I + F pe + b1 − pge I GH 100 10.000 with probability 1 − p .3. b1 − pge I . Calculate the simulated value of Y . b1 − pge I + F pe .69 in that order.01 −20 −0. You simulate the mixing variable where low values correspond to the exponential distribution.79. 1 are 0.2 −1 −0. DELETED You wish to simulate a value.2 80. 81. Then you simulate the value of Y . Y is exponentially distributed with mean 0. b1 − pge I GH 100 10.000 JK H 100 10.F pe + b1 − pge I GH 100 10. (A) 0.000 JK −1 −0.000 JK F pe + b1 − pge I .5.01 −20 −0. G GH 100 10. Losses of 100 and 2000 are observed. With probability 0. Losses come from a mixture of an exponential distribution with mean 100 with probability p and an exponential distribution with mean 10.01 −20 −0.000 JK GH 100 10.7.000 JK F e + e I + b1 − pg. Your uniform random numbers from 0. where low random numbers correspond to low values of Y .000 JK GH 100 10. Y is uniformly distributed on −3.25 and 0.2 −1 −0.2 −1 −0. b1 − pge I GH 100 10. With probability 0. 3 .000 JK GH 100 10.01 −20 −0.19 43 . Determine the likelihood function of p.01 −20 −0. Y .F e + e I p. be independent random variables following the uniform distribution on (0.77 0.V2 . n = 1. N follows the distribution: Pr( N = n) = 0. i = 1.9(0.30 v2 0.(B) (C) (D) (E) 0.52 v4 0.1) n −1 .05 0. You use the inverse transformation method with U to simulate N and Vi to simulate X i with small values of random numbers corresponding to small values of N and X i .… Let U and V1 .38 0. (A) (B) (C) (D) (E) 0 36 72 108 144 44 . N is the random variable for the number of accidents in a single year.. X i follows the distribution: g ( xi ) = 0. You are given the following random numbers for the first simulation: u v1 0..… X i is the random variable for the claim amount of the ith accident.01xi .59 0.01 e−0. 2.22 v3 0.95 82. xi > 0.46 Calculate the total amount of claims during the year for the first simulation.. 2. 1). E( B ) = 100 Calculate c. 5000]. no treasure will be found. where large random numbers from the uniform distribution on [0. The outcome. 1] correspond to large outcomes. The rewards are high: with probability 0. A health plan implements an incentive to physicians to control hospitalization under which the physicians will be paid a bonus B equal to c times the amount by which total hospital claims are under 400 ( 0 ≤ c ≤ 1) . is uniformly distributed on [1000. Your random numbers for the first two trials are 0. You use the inverse transformation method to simulate the outcome. (A) (B) 0.75 and 0. if treasure is found.85.83. Calculate the average of the outcomes of these first two trials. (A) (B) (C) (D) (E) 0 1000 2000 3000 4000 84.48 45 . The effect the incentive plan will have on underlying hospital claims is modeled by assuming that the new total hospital claims will follow a two-parameter Pareto distribution with α = 2 and θ = 300 .80. and thus the outcome is 0.44 0.20 treasure will be found. You are the consulting actuary to a group of venture capitalists financing a search for pirate gold. It’s a risky undertaking: with probability 0. (A) (B) (C) (D) (E) 80 90 100 110 120 86. The number of maintenance calls and the costs of the maintenance calls are all mutually independent.52 0.56 0. Using the normal approximation for the distribution of the aggregate maintenance costs. calculate the minimum number of computers needed to avoid purchasing a maintenance contract.(C) (D) (E) 0. The cost for a maintenance call has mean 80 and standard deviation 200. Computer maintenance costs for a department are modeled as follows: (i) (ii) (iii) The distribution of the number of maintenance calls each machine will need in a year is Poisson with mean 3. Aggregate losses for a portfolio of policies are modeled as follows: (i) (ii) The number of losses before any coverage modifications follows a Poisson distribution with mean λ . The severity of each loss before any coverage modifications is uniformly distributed between 0 and b.60 85. 46 . The department must buy a maintenance contract to cover repairs if there is at least a 10% probability that aggregate maintenance costs in a given year will exceed 120% of the expected costs. 24 47 .20 0. The modified claim amount is uniformly distributed on the interval ⎡ ⎣ 0. x 120 Calculate the loss elimination ratio for an ordinary deductible of 20. The insurer models its claims with modified frequency and severity distributions.000 0 80 Loss amount.The insurer would like to model the impact of imposing an ordinary deductible. on each loss and reimbursing only a percentage.010 0. It is assumed that the coverage modifications will not affect the loss distribution. c ( b − d ) ⎤ ⎦.008 f(x ) 0. c ( 0 < c ≤ 1) . d ( 0 < d < b ) . (A) (B) (C) λ λc λ λ d b b−d b b−d b (D) (E) λc 87. The graph of the density function for losses is: 0. Determine the mean of the modified frequency distribution.006 0.002 0.012 0. (A) (B) 0.004 0. of each loss in excess of the deductible. (C) (D) (E) 0. Compute the loss elimination ratio for the coming year.28 0. A towing company provides all towing services to members of the City Automobile Club. You are given: Towing Distance Towing Cost Frequency 0-9.36 88.99 miles 30+ miles (i) (ii) (iii) 80 100 160 50% 40% 10% The automobile owner must pay 10% of the cost and the remainder is paid by the City Automobile Club. You are given: (i) Losses follow an exponential distribution with the same mean in all years.32 0.000 in any given year. 48 . (ii) (iii) The loss elimination ratio this year is 70%. The ordinary deductible for the coming year is 4/3 of the current deductible.99 miles 10-29. calculate the probability that the City Automobile Club pays more than 90. The number of towings has a Poisson distribution with mean of 1000 per year. (A) (B) (C) (D) (E) 3% 10% 50% 90% 97% 89. The number of towings and the costs of individual towings are all mutually independent. Using the normal approximation for the distribution of aggregate towing costs. The number of auto vandalism claims reported per month at Sunny Daze Insurance Company (SDIC) has mean 110 and variance 750. calculate the probability that SDIC’s aggregate auto vandalism losses reported for a month will be less than 100.20 0. Individual losses have mean 1101 and standard deviation 70.31 91.31 0. They have concluded that the number of windshield claims filed per year per driver follows the Poisson distribution with parameter λ .24 0. The number of claims and the amounts of individual losses are independent. Calculate the probability that a driver selected at random will file no more than 1 windshield claim next year.24 0.19 0. (A) (B) (C) 0. Using the normal approximation. (A) (B) (C) (D) (E) 0.15 0.000.(A) (B) (C) (D) (E) 70% 75% 80% 85% 90% 90. Actuaries have modeled auto windshield claim frequencies.36 49 . where λ follows the gamma distribution with mean 3 and variance 3. calculate the probability that a player starting with 15.000 will have at least 15. and the amount of each prescription is 40. At the beginning of each round of a game of chance the player pays 12. are modeled assuming the number of claims has a geometric distribution with mean 4. Using the normal approximation. The player then rolls N dice and wins an amount equal to the total of the numbers showing on the N dice.(D) (E) 0.01 0.5.06 0.49 92. (A) (B) (C) (D) (E) 0.12 50 . All dice have 6 sides and are fair. The player then rolls one die with outcome N.000 after 1000 rounds. Calculate E S − 100 (A) (B) (C) (D) (E) 60 82 92 114 146 b g + . 93. S. Prescription drug losses.09 0.39 0.04 0. 51 . (A) (B) (C) (D) (E) 0. Hunt’s bonus will be a percentage of his earned premium equal to 15% of the difference between his ratio and 60%. If the ratio is less than 60%.130 0.33 0.135 0. X is a discrete random variable with a probability function which is a member of the (a.27 0.35 bg 96. x = 1.31 0.125 0.3.94. b g Calculate Fs 3 . S is the aggregate claim amount in the period. (A) (B) (C) (D) (E) 0. Hunt’s annual earned premium is 800. The number of claims in a period has a geometric distribution with mean 4. The number of claims and the claim amounts are independent. Calculate P X = 3 .0) class of distributions.4. The amount of each claim X follows P X = x = 0. Insurance agent Hunt N. You are given: (i) P( X = 0) = P ( X = 1) = 0.25 .b.25 (ii) P( X = 2) = 01875 .29 0. Quotum will receive no annual bonus if the ratio of incurred losses to earned premiums for his book of business is 60% or more for the year.140 b g 95.120 0.000.2. but less than 22. Calculate the expected total claim payment after these changes.000.000 52 .25 0.000 17.000 At least 20.000 and α = 2.000 29. You decide to impose a per claim deductible of 100.000 At least 22.000.000. Calculate the expected value of Hunt’s bonus. with θ = 500.000 35. A group dental policy has a negative binomial claim count distribution with mean 300 and variance 800. but less than 20. Ground-up severity is given by the following table: Severity 40 80 120 200 Probability 0.000 At least 24.000 97.25 You expect severity to increase 50% with no change in frequency. but less than 24. (A) (B) (C) (D) (E) 13.25 0.000 (B) (C) (D) (E) At least 18.Incurred losses are distributed according to the Pareto distribution. (A) Less than 18.25 0.000 24. or 3. For a certain company.00 2.96 53 . 2. calculate the probability that you will pay your employees a bonus next month.36 2. Calculate the expected claim payments for this insurance policy.16 0. You own a fancy light bulb factory. The boxes have varying numbers of light bulbs in them. Assuming independence and using the normal approximation. (A) (B) (C) (D) (E) 2.31 99.98. subject to an annual aggregate deductible of 2.23 0. each with probability 1/3. losses follow a Poisson frequency distribution with mean 2 per year. the entire box is destroyed. and of each other.81 2.27 0. (A) (B) (C) (D) (E) 0.45 2. Your workforce is a bit clumsy – they keep dropping boxes of light bulbs. You are given: Expected number of boxes dropped per month: Variance of the number of boxes dropped per month: Expected value per box: Variance of the value per box: 50 100 200 400 You pay your employees a bonus if the value of light bulbs destroyed in a month is less than 8000.19 0. Loss amounts are independent of the number of losses. and the amount of a loss is 1. and when dropped. An insurance policy covers all losses in a year. 100.8e −0.0 0. Calculate the expected payment under this policy for one claim. has the following characteristics: x F x bg E X∧x 0 91 153 331 b g 0 100 200 0. (A) (B) (C) (D) (E) 250 300 350 400 450 54 .02 x − 0. The unlimited severity distribution for claim amounts under an auto liability insurance policy is given by the cumulative distribution: F x = 1 − 0. The random variable for a loss. bg x≥0 The insurance policy pays amounts up to a limit of 1000 per claim. (A) (B) (C) (D) (E) 57 108 166 205 240 101. X.6 1000 1.0 Calculate the mean excess loss for a deductible of 100.2e −0.2 0.001x . 59 55 . You are given: (i) (ii) (iii) Combined revenue for the two factories is 3.3 0. WidgetsRUs will pay a dividend equal to the profit.2 0.47 0. retained major repair costs.4 0. less the sum of insurance premiums.1 0 1 2 3 (iv) At each factory.51 0. The insurance premium is 110% of the expected claims. (A) (B) (C) (D) (E) 0. All other expenses are 15% of revenues. and all other expenses. Major repair costs at the factories are independent.55 0. if it is positive.43 0.102. The distribution of major repair costs for each factory is k Prob (k) 0. It buys insurance to protect itself against major repair costs. WidgetsRUs owns two factories. Profit equals revenues. (v) Calculate the expected dividend. the insurance policy pays the major repair costs in excess of that factory’s ordinary deductible of 1. (ii) (iii) (iv) He then generates x from the Poisson distribution with mean λ . The expected lifetime of a watch is 8 years. (A) (B) (C) (D) (E) 0.44 0. then generating x from the Poisson distribution with mean λ . (A) (B) (C) (D) (E) 75 100 125 150 175 56 . Calculate the probability that the lifetime of a watch is at least 6 years.61 0. For watches produced by a certain manufacturer: (i) (ii) Lifetimes follow a single-parameter Pareto distribution with α > 1 and θ = 4.103. The repetitions are mutually independent.67 104.56 0. He repeats the process 999 more times: first generating a value λ . He generates 1000 values of the random variable X as follows: He generates the observed value λ from the gamma distribution with α = 2 and (i) θ = 1 (hence with mean 2 and variance 2). Glen is practicing his simulation skills. Calculate the expected number of times that his simulated value of X is 3.50 0. The number of eggs released by each salmon has a distribution with mean of 5 and variance of 5. A dam is proposed for a river which is currently used for salmon breeding.2 and variance 0.40 You have 106. An actuary for an automobile insurance company determines that the distribution of the annual number of claims for an insured chosen at random is modeled by the negative binomial distribution with mean 0. Using the normal approximation for the aggregate number of eggs released. (A) (B) (C) (D) (E) 0. The number of salmon going through the dam each hour it is open and the numbers of eggs released by the salmon are independent.20 0. determine the least number of whole hours the dam should be left open so the probability that 10.000 eggs will be released is greater than 95%.35 0.105.25 0. modeled: (i) (ii) (iii) For each hour the dam is opened the number of salmon that will pass through and reach the breeding grounds has a distribution with mean 100 and variance 900. Calculate the variance of this gamma distribution.4. (A) (B) (C) (D) (E) 20 23 26 29 32 57 .30 0. The number of claims for each individual insured has a Poisson distribution and the means of these Poisson distributions are gamma distributed over the population of insureds. …. Determine p 4 .03 2.4 1 0.09 2.00 2.09 0.1 The stop-loss insurance has a deductible of 1 for the group. k b g k = 1.2 3 0. you are given the recursion relation pk = bg 2 * p k −1 .06 2. For a stop-loss insurance on a three person group: (i) (ii) Loss amounts are independent.12 108. (A) (B) (C) (D) (E) 0. (A) (B) (C) (D) (E) 2.11 bg 58 .107.08 0. The distribution of loss amount for each person is: Loss Amount Probability 0 0. For a discrete probability distribution. (iii) Calculate the net stop-loss premium.3 2 0.10 0.07 0. 2. 109.7 0.2 0. (A) (B) (C) (D) (E) 306 316 416 510 518 59 . and prize amounts. X. regardless of vehicle type. Loss amounts. have exponential distribution with θ = 200.1 b g 0 100 1000 Your budget for prizes equals the expected prizes plus the standard deviation of prizes. N. The expected number of losses is 20. two modifications are to be made: a certain type of vehicle will not be insured. (A) (B) (C) (D) (E) 1600 1940 2520 3200 3880 110. (ii) a deductible of 100 per loss will be imposed. A company insures a fleet of vehicles. Calculate the expected aggregate amount paid by the insurer after the modifications.8 0. Calculate your budget.2 b g x Pr X = x 0. The number of prizes. In order to reduce the cost of the insurance. It is estimated that this will (i) reduce loss frequency by 20%. Aggregate losses have a compound Poisson distribution. You are the producer of a television quiz show that gives cash prizes. have the following distributions: n 1 2 Pr N = n 0. respectively. g 1 − Φ 0. made on an insurance portfolio follows the following distribution: n Pr(N=n) 0 2 3 0. Calculate the variance in the total number of claimants.8 and 0.1 If a claim occurs. the benefit is 0 or 10 with probability 0. 60 . respectively. (A) (B) (C) (D) (E) 20 25 30 35 40 112. g 1 − Φb316 . physicians volunteer their time on a daily basis to provide care to those who are not eligible to obtain care otherwise.93g 1 − Φb313 . The number of claims.2 0. The number of patients that can be served by a given physician has a Poisson distribution with mean 30.111. or 3 claimants with probabilities 2 . The number of accidents follows a Poisson distribution with mean 12. 3 . (A) (B) (C) (D) (E) b g 1 − Φb0.2. using the normal approximation with continuity correction. 2. N. 6 . In a clinic. 1 1 1 Each accident generates 1.7 0. Determine the probability that 120 or more patients can be served in a day at the clinic. The number of physicians who volunteer in any day is uniformly distributed on the integers 1 through 5.68 113.72g 1 − Φb0. 05 0.61 0.5].71 0. Calculate the variance of the amount paid by the insurance company for one claim.66 0.000 1. (A) (B) (C) (D) (E) 0.The number of claims and the benefit for each claim are independent.81 The mean 115.76 0.000 990.12 114.09 0. of the Poisson distribution is uniformly distributed over the interval [0. A claim count distribution can be expressed as a mixed Poisson distribution. An insurance company will pay the amount of each claim in excess of a deductible of 100. Calculate the probability that there are 2 or more claims. (A) (B) (C) (D) (E) 0.07 0.02 0. (A) (B) (C) (D) (E) 810.000 900. including the possibility that the amount paid is 0.000. A claim severity distribution is exponential with mean 1000.000 61 .000 860. Calculate the probability that aggregate benefits will exceed expected benefits by more than 2 standard deviations. 116. No bonus is paid if total claims exceed 500. Total hospital claims for the health plan are now modeled by a new Pareto distribution with α = 2 and θ = K . The health plan begins to provide financial incentives to physicians by paying a bonus of 50% of the amount by which total hospital claims are less than 500.337 0 Discharges from the hospital are uniformly distributed between the durations shown in the table. Total hospital claims for a health plan were previously modeled by a two-parameter Pareto distribution with α = 2 and θ = 500 .739 161. (A) (B) (C) (D) (E) 250 300 350 400 450 117.384 5. you are given: (i) Duration In Days 0 5 10 15 20 25 30 35 40 (ii) Number of Patients Remaining Hospitalized 4. for a patient who has been hospitalized for 21 days.386. 62 . in days.349 1.554 486.801 53. Calculate the mean residual time remaining hospitalized.000 1. Calculate K. For an industry-wide study of patients admitted to hospitals for treatment of cardiovascular illness in 1998. The expected claims plus the expected bonus under the revised model equals expected claims under the previous model.488 17.461. 3 118. For an individual over 65: (i) (ii) (iii) The number of pharmacy claims is a Poisson random variable with mean 25. The amount of each pharmacy claim is uniformly distributed between 5 and 95. g 1 − Φb2.9 5.(A) (B) (C) (D) (E) 4.3 5. 1 − Φ 133 .33g 63 .66g 1 − Φb3. The amounts of the claims and the number of claims are mutually independent.4 4. Determine the probability that aggregate claims for this individual will exceed 2000 using the normal approximation. (A) (B) (C) (D) (E) b g 1 − Φb166 .33g 1 − Φb2.8 6. 000 6.000. Each year. F b xg = 1 − G H x + 2000K 2 (iii) (iv) Reinsurance will pay the excess of each loss over 3000.300.200. x > 0.08 64 .06 1. (A) (B) (C) (D) (E) 1.500.04 1. equal to 110% of the expected losses covered by the reinsurance.000 3.119-120 Use the following information for questions 119 and 120. In the year 2001 individual losses have a Pareto distribution with F 2000 IJ . the reinsurer is paid a ceded premium. Calculate C2001. (v) (vi) Individual losses increase 5% each year due to inflation.000.07 1. 119. Calculate C2002 / C2001 . An insurer has excess-of-loss reinsurance on auto insurance.600. Cyear .05 1.400. You are given: (i) (ii) Total expected losses in the year 2001 are 10.000 120. The frequency distribution does not change.000 4. (A) (B) (C) (D) (E) 2.000 5. X N . 0. 4.3. 0. 3. N. X 2 . Then you repeat another N. and so on until you have performed the desired number of simulations. 2.5. Claim amounts are uniformly distributed on {1.7. 0. and 0.8.7.1. X 1. 0. All simulations use the inverse transform method. Your random numbers from (0. its claim amounts. You simulate the number of claims. 0. (v) (vi) (vii) Calculate the aggregate claim amount associated with your third simulated value of N. You are simulating a compound claims distribution: (i) (ii) (iii) (iv) The number of claims. and are independent of the number of claims. N.1. When the simulated number of claims is 0. 1) are 0.5. 0.… . Claim amounts are independent. is binomial with m = 3 and mean 1. 5} . 0.1.9. with low random numbers corresponding to few claims or small claim amounts. (A) (B) (C) (D) (E) 3 5 7 9 11 65 . DELETED 122.3. then the amounts of each of those claims. you do not simulate any claim amounts. 0.121. The insured then pays 5% of the remaining costs. The insured then pays 25% of the costs between 250 and 2250. The insured pays 100% of the costs above 2250 until the insured has paid 3600 in total.6 and m = 8. Determine the expected annual plan payment. 9000). The number of calories of a scientist is uniformly distributed on (7000. Calculate the probability that two or more scientists are eaten and exactly two of those eaten have at least 8000 calories each. The numbers of calories of scientists eaten are independent. 66 . Annual prescription drug costs are modeled by a two-parameter Pareto distribution with θ = 2000 and α = 2 . and are independent of the number of scientists eaten. A prescription drug plan pays annual drug costs for an insured member subject to the following provisions: (i) (ii) (iii) (iv) The insured pays 100% of costs up to the ordinary annual deductible of 250.123. For a tyrannosaur with a taste for scientists: (i) (ii) (iii) The number of scientists eaten has a binomial distribution with q = 0. (A) (B) (C) (D) (E) 1120 1140 1160 1180 1200 124. 27 0.(A) (B) (C) (D) (E) 0.45 67 .30 0.25 0. Calculate the normal approximation to the probability that the total of claim amounts exceeds 18. 1) (0.39 0.43 0.23 0. Type of Claim I II Poisson Parameter λ for Number of Claims 12 4 For each type.35 125. the number of claims follows a Poisson distribution and the amount of each claim is uniformly distributed as follows: Range of Each Claim Amount (0.41 0. 5) The numbers of claims of the two types are independent and the claim amounts and claim numbers are independent. Two types of insurance claims are made to an insurance company. (A) (B) (C) (D) (E) 0.37 0. (A) (B) (C) (D) (E) 5/9 5/8 2/3 3/4 4/5 68 . An insurance for the losses has an ordinary deductible of 5 per loss. Losses in 2003 follow a two-parameter Pareto distribution with α = 2 and θ = 5. The size of each loss has a two-parameter Pareto distribution with θ = 10 and α = 2. Losses in 2004 are uniformly 20% higher than in 2003. An insurance covers each loss subject to an ordinary deductible of 10. (A) (B) (C) (D) (E) 8 13 18 23 28 127.5 .126. Calculate the expected value of the aggregate annual payments for this insurance. Calculate the Loss Elimination Ratio in 2004. The number of annual losses has a Poisson distribution with a mean of 5. 128. DELETED 129. DELETED 130. Bob is a carnival operator of a game in which a player receives a prize worth W = 2 N (i) (ii) N has a Poisson distribution with mean Λ . if the player has N successes, N = 0, 1, 2, 3,… Bob models the probability of success for a player as follows: Λ has a uniform distribution on the interval (0, 4). Calculate E [W ] . (A) (B) (C) (D) (E) 5 7 9 11 13 69 131. You are simulating the gain/loss from insurance where: (i) (ii) (iii) (iv) (v) Claim occurrences follow a Poisson process with λ = 2 / 3 per year. Each claim amount is 1, 2 or 3 with p(1) = 0.25, Claim occurrences and amounts are independent. The annual premium equals expected annual claims plus 1.8 times the standard deviation of annual claims. i=0 p(2) = 0.25, and p(3) = 0.50 . You use 0.25, 0.40, 0.60, and 0.80 from the unit interval and the inversion method to simulate time between claims. You use 0.30, 0.60, 0.20, and 0.70 from the unit interval and the inversion method to simulate claim size. Calculate the gain or loss from the insurer’s viewpoint during the first 2 years from this simulation. (A) (B) (C) (D) (E) loss of 5 loss of 4 0 gain of 4 gain of 5 70 132. Annual dental claims are modeled as a compound Poisson process where the number of claims has mean 2 and the loss amounts have a two-parameter Pareto distribution with θ = 500 and α = 2 . An insurance pays 80% of the first 750 of annual losses and 100% of annual losses in excess of 750. You simulate the number of claims and loss amounts using the inverse transform method with small random numbers corresponding to small numbers of claims or small loss amounts. The random number to simulate the number of claims is 0.8. The random numbers to simulate loss amounts are 0.60, 0.25, 0.70, 0.10 and 0.80. Calculate the total simulated insurance claims for one year. (A) (B) (C) (D) (E) 294 625 631 646 658 133. You are given: (i) The annual number of claims for an insured has probability function: ⎛3⎞ 3− x p ( x ) = ⎜ ⎟ q x (1 − q ) , x = 0, 1, 2, 3 ⎝ x⎠ (ii) The prior density is π (q ) = 2q , 0 < q < 1. A randomly chosen insured has zero claims in Year 1. Using Bühlmann credibility, estimate the number of claims in Year 2 for the selected insured. - 71 - 3 234.72 - .7 246. (A) (B) (C) (D) (E) 219.33 1.50 1.00 1.(A) (B) (C) (D) (E) 0.8 .50 134.33 0. You are given the following random sample of 13 claim amounts: 99 133 175 216 250 277 651 698 735 745 791 906 947 Determine the smoothed empirical estimate of the 35th percentile.6 256.4 231. 3 1.73 - .65.5 1.55 At least 0.55.6 − 1.60 At least 0. but less than 0. S10 (1.6).5 1. (A) (B) (C) (D) (E) Less than 0.1 2. For observation i of a survival study: • • • di is the left truncation point xi is the observed value if not right censored ui is the observed value if right censored You are given: Observation (i) 1 2 3 4 5 6 7 8 9 10 di 0 0 0 0 0 0 0 1.7 − 2. but less than 0.3 Determine the Kaplan-Meier Product-Limit estimate.1 − ui − 1.135.9 − 1.6 xi 0.2 − 1.65 At least 0. but less than 0.7 − − 2.70 At least 0.60.70 .5 − − 1. 800 137. (A) (B) (C) (D) (E) 4.81 0.1 4. 0 < x < 1 .200 At least 10.95 You fit a distribution with the following density function to the data: f x = p + 1 x p .200. You are given the following three observations: 0. You are given: (i) Two classes of policyholders have the following severity distributions: Claim Amount 250 2.2 Probability of Claim Amount for Class 2 0.0 4.1 Class 1 has twice as many claims as Class 2.74 0. Determine the Bayesian estimate of the expected value of a second claim from the same policyholder.2 0. (A) (B) (C) (D) (E) Less than 10.3 0.136.500 60.2 4. A claim of 250 is observed. but less than 10.400. p > −1 bg b g Determine the maximum likelihood estimate of p.400 At least 10.000 (ii) Probability of Claim Amount for Class 1 0.74 - .4 .5 0.7 0.600. but less than 10.800 At least 10.3 4. but less than 10.600 At least 10. 7 Number of Claims 1 0. 1 or 2 claims. q) model with the following requirements: (i) (ii) The mean of the fitted model equals the sample mean.1 0. as shown below: Year 1 2 Number of Insureds 20 30 Number of Claims 7 10 Determine the Bühlmann-Straub credibility estimate of the number of claims in Year 3 for 35 insureds from the same class.138.1 A class is chosen at random.1 0.9 0.1 0. (A) (B) (C) (D) (E) 2 3 4 5 6 139.75 - . Members of three classes of insureds can have 0.0 0.8 0. . Determine the smallest estimate of m that satisfies these requirements. and varying numbers of insureds from that class are observed over 2 years. You are given the following sample of claim counts: 0 0 1 2 2 You fit a binomial(m. The 33rd percentile of the fitted model equals the smoothed empirical 33rd percentile of the sample.2 2 0. with the following probabilities: Class I II III 0 0. 76 - .1 11.450 7.498 0. Calculate the chi-square goodness-of-fit statistic.800 25.500 3.9 11.810 30.930 0.(A) (B) (C) (D) (E) 10.800 3.90 12.390 230 2.000 2. but less than 10 At least 10.55 4.6 10.876 0.920 4. but less than 16 At least 16 .200 140 2.000 4.000 1.800 15.000 600 3.100 3.750 560 2.81 7.910 12.16 500 0. (A) (B) (C) (D) (E) Less than 7 At least 7.000 You test the hypothesis that auto claims follow a continuous distribution F(x) with the following percentiles: x F(x) 310 0.95 You group the data using the largest number of groups such that the expected number of claims in each group is at least 5.000 35. You are given the following random sample of 30 auto claims: 54 2.500 7.300 1.000 1.6 140.4 11.800 55.27 2. but less than 13 At least 13.498 0.580 11.000 1.870 32. 9 2. (A) (B) (C) (D) (E) 1.700) is a 95% log-transformed confidence interval for the cumulative hazard rate function at time t.141.56 0.5 1. The interval (0. Determine k.61 142.50 0.53 0. 0.59 0.575. Determine the value of the Nelson-Aalen estimate of S(t). The prior density is: π (θ ) = (iii) e −θ .77 - . where the cumulative hazard rate function is estimated using the Nelson-Aalen estimator. You are given: (i) (ii) The number of claims observed in a 1-year period has a Poisson distribution with mean θ.7 1.3 .357. (A) (B) (C) (D) (E) 0. 0 <θ < k 1 − e− k The unconditional probability of observing zero claims in 1 year is 0.1 2. 78 - . but less than 5 At least 5. The parameters of the inverse Pareto distribution F ( x) = ⎡ ⎣ x / ( x + θ )⎤ ⎦ τ are to be estimated using the method of moments based on the following data: 15 45 140 250 560 1340 Estimate θ by matching kth moments with k = −1 and k = −2 . You are given the following simulations from the sample: Simulation 1 2 3 4 5 6 7 8 9 10 600 1500 1500 600 600 600 1500 1500 300 600 Claim Amounts 600 300 300 600 300 600 1500 300 600 600 1500 1500 600 300 1500 1500 1500 1500 300 600 Determine the bootstrap approximation to the mean square error of the estimate. but less than 50 At least 50 144. (A) (B) (C) (D) (E) Less than 1 At least 1. A sample of claim amounts is {300. 600. but less than 25 At least 25.143. .125. the loss elimination ratio for a deductible of 100 per claim is estimated to be 0. By applying the deductible to this sample. 1500}. 021 0.6.6 At least 0. for Company III.000 50 Year 2 50.003 0.000 100 ? ? 150.79 - .2 At least 0.081 145.000 300 150.(A) (B) (C) (D) (E) 0. You are given the following commercial automobile policy experience: Company Losses Number of Automobiles Losses Number of Automobiles Losses Number of Automobiles I II III Year 1 50.2. but less than 0.8 At least 0. but less than 0.054 0.4 At least 0.000 500 ? ? Year 3 ? ? 150.010 0.000 200 150. (A) (B) (C) (D) (E) Less than 0.000 150 Determine the nonparametric empirical Bayes credibility factor.8 .4. but less than 0. Z. ym denote independent random samples of losses from Region 1 and Region 2. 4. Single-parameter Pareto distributions with θ = 1 .0.0. 4.. From a population having distribution function F.7.. You intend to calculate the maximum likelihood estimate of α for Region 1.3..50 0.53 0.. x2 .4. Which of the following equations must be solved? (A) n α n − ∑ ln ( xi ) = 0 − ∑ ln ( xi ) + m (α + 2 ) 2∑ ln ( yi ) − =0 2 3α (α + 2 ) (B) α n (C) α n − ∑ ln ( xi ) + 2 ln ( yi ) 2m − ∑ =0 2 3α (α + 2 ) (α + 2 ) 6 ln ( yi ) 2m − ∑ =0 2 α (α + 2 ) (α + 2 ) 6 ln ( yi ) 3m − ∑ =0 2 α (3 − α ) (3 − α ) (D) α n − ∑ ln ( xi ) + (E) α − ∑ ln ( xi ) + 147.0. 4.5 times the expected value of losses in Region 1.. are used to model losses in these regions.7.. y2 . 4.31 0. (A) (B) (C) (D) (E) 0. Let x1. 3.7 Calculate the kernel density estimate of F(4). 3. you are given the following sample: 2. Past experience indicates that the expected value of losses in Region 2 is 1..3. 4. xn and y1.. respectively.41 0.80 - . using the data from both regions. but different values of α . using the uniform kernel with bandwidth 1.146.63 . 148. You are given: (i) The number of claims has probability function: ⎛ m⎞ m− x p ( x ) = ⎜ ⎟ q x (1 − q ) , ⎝x ⎠ (ii) (iii) x = 0, 1, 2,…, m The actual number of claims must be within 1% of the expected number of claims with probability 0.95. The expected number of claims for full credibility is 34,574. Determine q. (A) (B) (C) (D) (E) 0.05 0.10 0.20 0.40 0.80 149. If the proposed model is appropriate, which of the following tends to zero as the sample size goes to infinity? (A) (B) (C) (D) (E) Kolmogorov-Smirnov test statistic Anderson-Darling test statistic Chi-square goodness-of-fit test statistic Schwarz Bayesian adjustment None of (A), (B), (C) or (D) - 81 - 150. You are given: (i) (ii) (iii) Losses are uniformly distributed on ( 0, θ ) with θ > 150. The policy limit is 150. A sample of payments is: 14, 33, 72, 94, 120, 135, 150, 150 Estimate θ by matching the average sample payment to the expected payment per loss. (A) (B) (C) (D) (E) 192 196 200 204 208 151. You are given: (i) (ii) (iii) (iv) A portfolio of independent risks is divided into two classes. Each class contains the same number of risks. For each risk in Class 1, the number of claims per year follows a Poisson distribution with mean 5. For each risk in Class 2, the number of claims per year follows a binomial distribution with m = 8 and q = 0.55 . A randomly selected risk has three claims in Year 1, r claims in Year 2 and four claims in Year 3. (v) The Bühlmann credibility estimate for the number of claims in Year 4 for this risk is 4.6019. Determine r . - 82 - (A) (B) (C) (D) (E) 1 2 3 4 5 152. You are given: (i) A sample of losses is: 600 (ii) (iii) 700 900 No information is available about losses of 500 or less. Losses are assumed to follow an exponential distribution with mean θ . Determine the maximum likelihood estimate of θ . (A) (B) (C) (D) (E) 233 400 500 733 1233 - 83 - μ .51 0.20 7.66 35. (viii) The prior distribution has joint probability density function: f ( λ . You are given: (v) (vi) (vii) Claim counts follow a Poisson distribution with mean λ .82 3.84 - . 0 < μ < 1 . .49 0.62 12.153. Claim sizes follow a lognormal distribution with parameters μ and σ .66 1. but less than 8 At least 8 155. Claim counts and claim sizes are independent.24 You use the method of percentile matching at the 40th and 80th percentiles to fit an Inverse Weibull distribution to these data. but less than 4 At least 4.71 5. 0 < σ < 1 Calculate Bühlmann’s k for aggregate losses. DELETED 154. but less than 6 At least 6. Determine the estimate of θ . (A) (B) (C) (D) (E) Less than 2 At least 2. σ ) = 2σ . 0 < λ < 1 . You are given the following data: 0. For this policyholder. Determine the maximum likelihood estimate of λ .35.00 1.65 156. but less than 1.(A) (B) (C) (D) (E) Less than 1. The prior density is π (q) = q3 .8. Observations other than 0 and 1 have been deleted from the data. . 0.07 A randomly selected policyholder has one claim in Year 1 and zero claims in Year 2.50 0.35 At least 1.7 < q < 0.55 At least 1. The probability of a claim for a policyholder during a year is q .50 157. 0. You are given: (i) (ii) (iii) In a portfolio of risks.55.85 - .65 At least 1.8.6 < q < 0. determine the posterior probability that 0.45. (A) (B) (C) (D) (E) 0. You are given: (i) (ii) (iii) The number of claims follows a Poisson distribution with mean λ .75 1. each policyholder can have at most one claim per year.25 1. The data contain an equal number of observations of 0 and 1. but less than 1.45 At least 1. but less than 1. 22. 92.4. 22.5 At least 0.22 0.6 158. 22.4 At least 0. but less than 0. 51. but less than 0. 230 (ii) (iii) ˆ ( x) is the Nelson-Aalen empirical estimate of the cumulative hazard rate function. Calculate H 2 1 (A) (B) (C) (D) (E) 0. ˆ (75) − H ˆ (75) . but less than 0.6 At least 0.33 0.00 0.(A) (B) (C) (D) (E) Less than 0. You are given: (i) The following is a sample of 15 losses: 11.44 . 120.5. 92. H 1 ˆ ( x) is the maximum likelihood estimate of the cumulative hazard rate function H 2 under the assumption that the sample is drawn from an exponential distribution. 69.3 At least 0. 36.11 0. 69.86 - .3. 161. 69. 161. 2 0.1 0.35.7 1. you are given: (i) (ii) The number of claims for each policyholder has a conditional Poisson distribution. For a portfolio of motorcycle insurance policyholders. but less than 0. but less than 0. for Year 2.40 At least 0.35 At least 0. but less than 0.3 You test the hypothesis that the probability density function is: f ( x) = 4 (1 + x ) 5 . For Year 1.87 - .159.40.30. (A) (B) (C) (D) (E) Less than 0. You are given a random sample of observations: 0.45 At least 0.5 0. . the following data are observed: Number of Claims 0 1 2 3 4 Total Number of Policyholders 2000 600 300 80 20 3000 Determine the credibility factor.30 At least 0.45 160. x>0 Calculate the Kolmogorov-Smirnov test statistic. Z. X.(A) (B) (C) (D) (E) Less than 0. (A) (B) (C) (D) (E) 150 175 200 225 250 .05. A consistent estimator is also unbiased.05 At least 0. 162. but less than 0. A loss.15. For an unbiased estimator. but less than 0.35 161. Which of the following statements is true? (A) (B) (C) (D) (E) A uniformly minimum variance unbiased estimator is an estimator such that no other estimator has a smaller variance.15 At least 0.35 At least 0. You are given: 5 E⎡ ⎣ X − 100 X > 100 ⎤ ⎦ = 3E⎡ ⎣ X − 50 X > 50 ⎤ ⎦ Calculate E ⎡ ⎣ X − 150 X > 150 ⎤ ⎦. One computational advantage of using mean squared error is that it is not a function of the true value of the parameter.25 At least 0. but less than 0.25. An estimator is consistent whenever the variance of the estimator approaches zero as the sample size increases to infinity. follows a 2-parameter Pareto distribution with α = 2 and unspecified parameter θ . the mean squared error is always equal to the variance.88 - . B’s Latin class have a normal distribution with mean θ and standard deviation equal to 8.00 164. However.88 0. (A) (B) (C) (D) (E) 0.500 81. Each year.000 . Calculate the variance of the aggregate payments of the insurance.77 0. For a collective risk model the number of losses. B chooses a student at random and pays the student 1 times the student’s score.163. The common distribution of the individual losses has the following characteristics: (i) (ii) (iii) (iv) E [ X ] = 70 E [ X ∧ 30] = 25 Pr ( X > 30 ) = 0.75 2 ⎤ E⎡ ⎣ X X > 30 ⎦ = 9000 An insurance covers aggregate losses subject to an ordinary deductible of 30 per loss. has a Poisson distribution with λ = 20 .000 67. N.92 1. then there is no payment.500 108. given that there is a payment. The scores on the final exam in Ms.89 - .000 94. if the student fails the exam (score ≤ 65 ). Calculate the conditional probability that the payment is less than 90. Ms.85 0. (A) (B) (C) (D) (E) 54. θ is a random variable with a normal distribution with mean equal to 75 and standard deviation equal to 6. 35 0. (A) (B) (C) (D) (E) 0. 2. The common distribution of the individual losses is: x 1 2 fx ( x) 0.5 for k = 1. For a collective risk model: (i) (ii) The number of losses has a Poisson distribution with λ = 2 .89 0.06 0.29 0.40 .90 - .6 0.13 0. A discrete probability distribution has the following properties: (i) (ii) ⎛ 1⎞ pk = c ⎜1 + ⎟ pk −1 ⎝ k⎠ p0 = 0.94 166.74 0.84 0.79 0. (A) (B) (C) (D) (E) 0.… Calculate c.165.4 An insurance covers aggregate losses subject to a deductible of 3. Calculate the expected aggregate payments of the insurance. The repair costs for boats in a marina have the following characteristics: Boat type Power boats Sailboats Luxury yachts Number of boats 100 300 50 Probability that repair is needed 0. and 0.000 168.000.000 At most one repair is required per boat each year. Calculate Y. Calculate Var Y P . ( ) .91 - . The insurance has an ordinary deductible of 150 per loss.000 2.3 0.2. For an insurance: (i) (ii) (iii) Losses can be 100.000 210. 0. equal to the aggregate mean repair costs plus the standard deviation of the aggregate repair costs.6.6 Mean of repair cost given a repair 300 1000 5000 Variance of repair cost given a repair 10. Y.2.1 0.000 230.000 240. 200 or 300 with respective probabilities 0.000 400. Y P is the claim payment per payment random variable.167. (A) (B) (C) (D) (E) 200.000 220. The marina budgets an amount. The distribution of a loss. Calculate Pr ( X ≤ 200 ) .82 0.2.8. With probability 0.(A) (B) (C) (D) (E) 1500 1875 2250 2625 3000 169.92 - . X has a two-parameter Pareto distribution with α = 2 and θ = 100 . The distribution of the population and the mean number of colds are as follows: Proportion of population Children Adult Non-Smokers Adult Smokers 0. is a two-point mixture: With probability 0.76 0. (A) (B) (C) (D) (E) 0. X has a two-parameter Pareto distribution with α = 4 and θ = 3000 .60 0. (i) (ii) X .88 170.10 Mean number of colds 3 1 4 .85 0. In a certain town the number of common colds an individual will get in a year follows a Poisson distribution that depends on the individual’s age and smoking status.30 0.79 0. S: (i) (ii) The number of losses has a negative binomial distribution with mean 3 and variance 3.16 0. The common distribution of the independent individual loss amounts is uniform from 0 to 20.28 171. (A) (B) (C) (D) (E) 61 63 65 67 69 .12 0.93 - . Calculate the 95th percentile of the distribution of S as approximated by the normal distribution.24 0.Calculate the conditional probability that a person with exactly 3 common colds in a year is an adult smoker. For aggregate losses.6.20 0. (A) (B) (C) (D) (E) 0. 05 significance level.172.10 1.36 n 0.025 0.7 0. that the probability density function for the population is: f ( x) = 4 (1 + x ) 5 .22 n 0.94 - .48 1.05 significance level. . Reject H 0 at the 0. Reject H 0 at the 0.63 n n Determine the result of the test.3 You use the Kolmogorov-Smirnov test for testing the null hypothesis. Reject H 0 at the 0.025 significance level.05 1. x>0 (iii) Critical values for the Kolmogorov-Smirnov test are: Level of Significance Critical Value 0. (A) (B) (C) (D) (E) Do not reject H 0 at the 0.025 significance level. Reject H 0 at the 0.01 significance level.9 1. You are given: (i) A random sample of five observations from a population is: 0.1 1.10 significance level.01 1. but not at the 0.10 significance level. but not at the 0.01 significance level. but not at the 0. H 0 .2 (ii) 0. Determine the expected number of claims needed for aggregate losses to be within 10% of expected aggregate losses with 95% probability. but less than 2400 At least 2400 . You are given: (iv) The number of claims follows a negative binomial distribution with parameters r and β = 3.95 - . but less than 1600 At least 1600.2 (v) The number of claims is independent of the severity of claims.173. Claim severity has the following distribution: Claim Size 1 10 100 (iii) Probability 0.4 0. (A) (B) (C) (D) (E) Less than 1200 At least 1200. but less than 2000 At least 2000.4 0. 58 At least 0. tk = time of the k th death ˆ (t ) = 39 .56. but less than 0.62 . (A) (B) (C) (D) (E) Less than 0.60. You are given: (vi) (vii) (viii) (ix) A mortality study covers n lives.58.60 At least 0. A Nelson-Aalen estimate of the cumulative hazard rate function is H 2 380 Determine the Kaplan-Meier product-limit estimate of the survival function at time t9 . None were censored and no two deaths occurred at the same time. but less than 0.56 At least 0. but less than 0.62 At least 0.96 - .174. Three observed values of the random variable X are: 1 1 4 You estimate the third central moment of X using the estimator: g ( X1.0 At least 3.5 At least 3.0. but less than 3. X 3 ) = 3 1 Xi − X ) ( ∑ 3 Determine the bootstrap estimate of the mean-squared error of g. (A) (B) (C) (D) (E) Less than 3.5.97 - . but less than 4. X 2 .175. but less than 4.0 At least 4.5 At least 4.5 .0. x ≥ 0 Uniform on [1. x ≥ 1 F(x) = x / (1 + x).2 0.2 0.0 0. 100] Exponential with mean 10 Normal with mean 40 and standard deviation 40 (A) (B) (C) (D) (E) . You are given the following p-p plot: 1.176.0 0.4 0.8 1.25 .0 0.4 0.6 0.98 - .6 0.0 Fn(x) The plot is based on the sample: 1 2 3 15 30 50 51 99 100 Determine the fitted model underlying the p-p plot.8 F(x) 0. F(x) = 1 – x − 0. 7. but less than 0. You are given: (i) (ii) Claims are conditionally independent and identically Poisson distributed with mean Θ . but less than 0.8 At least 0.6 At least 0.7 At least 0.99 - . The prior distribution function of Θ is: ⎛ 1 ⎞ F (θ ) = 1 − ⎜ ⎟ ⎝ 1+θ ⎠ Five claims are observed.9 2.6 . (A) (B) (C) (D) (E) Less than 0. Determine the Bühlmann credibility factor.6.177.9 At least 0. θ >0 178. but less than 0. DELETED .8. (A) (B) (C) (D) (E) 0.15 0.07 0.22 . Use the following information for questions 179 and 180.11 0.19 180.04 0. A random sample of size two has a mean time of 6. (A) (B) (C) (D) (E) 0. Use the delta method to approximate the variance of the maximum likelihood estimator of FY (10) .12 0. 179.179-180. The time to an accident follows an exponential distribution. Let Y denote the mean of a new sample of size two.16 0.19 0. Determine the maximum likelihood estimate of Pr (Y > 10 ) .08 0.100 - . determine k. Using Bühlmann’s credibility for aggregate losses.101 - . You are given: (i) (ii) (iii) (iv) (v) (vi) The number of claims in a year for a selected risk follows a Poisson distribution with mean λ .181. The prior distribution of λ is exponential with mean 1. (A) (B) (C) (D) (E) 1 4/3 2 3 4 . The prior distribution of θ is Poisson with mean 1. λ and θ are independent. A priori. The severity of claims for the selected risk follows an exponential distribution with mean θ . The number of claims is independent of the severity of claims. 0. (B) and (C) . (B) or (C) All of (A).20 u2 = 0. Thus you would like to use a claim count model that has its mean equal to its variance.09 Calculate the average of the simulated values. Determine which of the following models may also be appropriate.03. (A) (B) (C) (D) (E) 1 3 1 5 3 7 3 3 183. An obvious choice is the Poisson distribution. The deaths are independent.182. (A) (B) (C) (D) (E) A mixture of two binomial distributions with different means A mixture of two Poisson distributions with different means A mixture of two negative binomial distributions with different means None of (A). Do this three times using: u1 = 0.03 u3 = 0. You are given claim count data for which the sample mean is roughly equal to the sample variance. A company insures 100 people age 65. The annual probability of death for each person is Use the inversion method to simulate the number of deaths in a year.102 - . 1.437) .4) e − λ / 6 + (0.9 10. (A) (B) (C) (D) (E) (0.5). 1 6 1 − λ /12 e .248.266) (0.545) (0. The prior distribution of λ has probability density function: π ( λ ) = (0.283. The observed data are as follows: Time of First Claim Number of Claims 1 2 2 1 3 2 4 2 5 1 6 2 7 2 Using the Nelson-Aalen estimator.6 9. (A) (B) (C) (D) (E) 9. 12 λ >0 Determine the Bayesian expected number of claims for the insured in Year 2. 1. 1.402) (0. 1.103 - .206. 1.6) Ten claims are observed for an insured in Year 1. Twelve policyholders were monitored from the starting date of the policy to the time of first claim.314.8 9.184.189.0 185.7 9. calculate the 95% linear confidence interval for the cumulative hazard rate function H(4. You are given: (i) (ii) Annual claim frequencies follow a Poisson distribution with mean λ.361) (0. A randomly selected policy had x claims in Year 1.104 - .2 0. you are given: (i) (ii) E[ X ] = θ . The prior distribution of β has the Pareto density function π (β ) = ( β + 1)(α +1) α . You are given: (i) (ii) The annual number of claims on a given policy has a geometric distribution with parameter β. For the random variable X. where α is a known constant greater than 2. k +1 (iv) ⎤ MSE θˆ (θ ) = 2 ⎡ ⎣ bias θˆ (θ ) ⎦ Determine k. (A) (B) (C) (D) (E) 0.186. . θ >0 Var ( X ) = θ2 25 k >0 2 (iii) θˆ = k X.5 2 5 25 187. 0< β <∞ . (C) or (D) is true. (B). using sample data to estimate the parameters of the distribution tends to decrease the probability of a Type II error. The Anderson-Darling test tends to place more emphasis on a good fit in the middle rather than in the tails of the distribution. the critical value for the chi-square goodness-of-fit test becomes larger with increased sample size. The Kolmogorov-Smirnov test can be used on individual or grouped data.105 - . Which of the following statements is true? (A) For a null hypothesis that the population follows a particular distribution. (B) (C) (D) (E) . None of (A). For a given number of cells. (A) 1 α −1 (α − 1) x (B) (C) (D) α x + 1 α (α − 1) x +1 α x +1 α −1 (E) 188.Determine the Bühlmann credibility estimate of the number of claims for the selected policy in Year 2. DELETED 189. 70.68. and the probability distribution of Θ are as follows: Number of claims Probability 0 2θ 1 2 1 − 3θ θ 0.72 At least 1. the conditional probability of the annual number of claims given Θ = θ . but less than 1. (A) (B) (C) (D) (E) Less than 1.05 0.30 0. For a particular policy.74 At least 1.106 - .74 .68 At least 1.80 θ Probability Two claims are observed in Year 1.20 Calculate the Bühlmann credibility estimate of the number of claims in Year 2.190. but less than 1.72.70 At least 1. 0. but less than 1. 00 3. x2 = 3) .75 4.107 - . (A) (B) (C) (D) (E) 3. Determine E ( Λ x1 = 5. The prior distribution of Λ is gamma with probability density function: (2λ )5 e−2λ f (λ ) = .00 192.25 3.50 3. You are given the kernel: ⎧2 2 ⎪π 1 − ( x − y) . You are given: (i) (ii) The annual number of claims for a policyholder follows a Poisson distribution with mean Λ . . ⎪ ⎩ You are also given the following random sample: 1 3 3 5 y −1 ≤ x ≤ y +1 otherwise Determine which of the following graphs shows the shape of the kernel density estimator.191. 24λ λ >0 An insured is selected at random and observed to have x1 = 5 claims during Year 1 and x2 = 3 claims during Year 2. ⎪ ky (x) = ⎨ ⎪ 0. (B) (A) (D) (C) (E) .108 - . 000 90 200 Year 3 15. (A) (B) (C) (D) (E) Less than 250 At least 250.56 . but less than 280 At least 280.27 34.000 50 200 16.52 0. You are given: Group Total Claims Number in Group Average Total Claims Number in Group Average Total Claims Number in Group Average 1 Year 1 Year 2 10. calculate the limited expected value at 500.000 100 160 18.000 190 178. You are also given a Use the nonparametric empirical Bayes method to estimate the credibility factor for Group 1. but less than 310 At least 310.95 59. The following claim data were generated from a Pareto distribution: 130 20 350 218 1822 Using the method of moments to estimate the parameters of a Pareto distribution.54 0.03.193.000 110 227.000 60 250 Total 25.000 300 196. but less than 340 At least 340 194.67 2 ˆ = 651.50 0.48 0.109 - . (A) (B) (C) (D) (E) 0. 47 0.10.000 and 6.000 .1. .3.000 50.40 0.110 - .000 .000 .5.000 .000 25.25.000.50.50 196.000 10.000 over 100.000 20.45 0.000 .000 .000 3.000 Number of Claims 16 22 25 18 10 5 3 1 Use the ogive to estimate the probability that a randomly chosen claim is between 2.100.000 1.000 5.000 ∞ Past experience indicates that these losses follow a Pareto distribution with parameters α and θ = 10.000 10.195. (A) (B) (C) (D) (E) 0.000 . Determine the maximum likelihood estimate of α .36 0.000 400 Number of Losses 3 3 4 6 4 Deductible 200 0 0 0 300 Policy Limit ∞ 10. You are given the following 20 bodily injury losses (before the deductible is applied): Loss 750 200 300 >10. You are given the following information regarding claim sizes for 100 claims: Claim Size 0 . 0 At least 4.111 - . Each policyholder was insured for the entire 2-year period.443 . but less than 3.0. You are given: (i) During a 2-year period. but less than 5.0 At least 2.0 At least 3.0 197.0 At least 5.403 0. A randomly selected policyholder had one claim over the 2-year period.387 0.(A) (B) (C) (D) (E) Less than 2.0. 100 policies had the following claims experience: Total Claims in Years 1 and 2 0 1 2 3 4 (ii) (iii) Number of Policies 50 30 15 4 1 The number of claims per year follows a Poisson distribution.393 0.380 0. Using semiparametric empirical Bayes estimation.0. determine the Bühlmann estimate for the number of claims in Year 3 for the same policyholder. (A) (B) (C) (D) (E) 0. but less than 4. 0 to 4. but less than 2100 At least 2100. x>0 A sample of four claims is: 130 240 300 540 The values of two additional claims are known to exceed 1000. you are given: (i) (ii) The expected number of claims in a year for these risks ranges from 1. For the same risk. For five types of risks.0.2 .198. Determine the maximum likelihood estimate of θ.2. During Year 1. (A) (B) (C) (D) (E) Less than 300 At least 300. but less than 1200 At least 1200. n claims are observed for a randomly selected risk.. but less than 3000 At least 3000 200. .9.112 - . . The number of claims follows a Poisson distribution for each risk. Personal auto property damage claims in a certain region are known to follow the Weibull distribution: F ( x) = 1− e − x (θ ) 0. DELETED 199..1. both Bayes and Bühlmann credibility estimates of the number of claims in Year 2 are calculated for n = 0. Which graph represents these estimates? . 113 - .(A) (B) (C) (D) (E) . but is not rejected at the 0. (A) (B) (C) (D) (E) The hypothesis is not rejected at the 0.05 significance level. but is not rejected at the 0.201. You test the hypothesis that a given set of data comes from a known distribution with distribution function F(x).130 0.035 0.025 significance level.01 significance level.10 significance level. Determine the result of the test.025 significance level.630 0. The hypothesis is rejected at the 0.10 significance level. The hypothesis is rejected at the 0.05 significance level. The following data were collected: Interval x<2 2≤x<5 5≤x<7 7≤x<8 8≤x Total F ( xi ) 0. You test the hypothesis using the chi-square goodness-of-fit test. but is not rejected at the 0.000 Number of Observations 5 42 137 66 50 300 where xi is the upper endpoint of each interval. The hypothesis is rejected at the 0.01 significance level.114 - .830 1. . The hypothesis is rejected at the 0. 75 .115 - .0808 0.2 . (A) (B) (C) (D) (E) 3. A randomly selected policy had two claims in Year 1.6 3. You use simulation to generate severities.4207 Using these numbers and the inversion method.7881 0.6 and σ = 0. The following are six uniform (0.4 3. but less than 340 At least 340.8 4.6179 0. calculate the average payment per claim for a contract with a policy limit of 400. (A) (B) (C) (D) (E) Less than 300 At least 300. and the remaining two-thirds have β = 5. One-third of the policies have β = 2. but less than 360 At least 360 203. Unlimited claim severities for a warranty product follow the lognormal distribution with parameters μ = 5.202.4602 0.0 4. You are given: (i) (ii) The annual number of claims on a given policy has the geometric distribution with parameter β. 1) random numbers: 0.9452 0. but less than 320 At least 320. Calculate the Bayesian expected number of claims for the selected policy in Year 2. 250 0. and the mean number of sessions received for residents receiving therapy are displayed in the following table: State # 1 2 3 Number in state 400 300 200 Probability of needing therapy 0. The length of time. calculate the probability that more than 3000 physical therapy sessions will be required for the given month.116 - .2 0.204. Calculate the probability that a person drawn at random from this population will remember an actuarial statistic less than 1 2 year. Temporarily in a Health Center (State #2) or Permanently in a Health Center (State #3). Transitions between states occur at the end of the month.5 0. in years. If a resident receives physical therapy. Y has a gamma distribution with α = θ = 2 . the probability of needing physical therapy in the month. In a CCRC.125 0.3 Mean number of visits 2 15 9 Using the normal approximation for the aggregate distribution. The numbers of sessions received are independent. .875 205. In a certain population. The number in each state at the beginning of a given month. the number of sessions that the resident receives in a month has a geometric distribution with a mean which depends on the state in which the resident begins the month. residents start each month in one of the following three states: Independent Living (State #1). (A) (B) (C) (D) (E) 0.500 0. that a person will remember an actuarial statistic is modeled by an exponential distribution with mean 1Y .750 0. (A) (B) (C) (D) (E) 0.0 . the number of projects that require you to work overtime has a geometric x 5 10 20 f ( x) distribution with β = 2 .5 18.21 0.5 The number of projects and number of overtime hours are independent.50 206. the distribution of the number of overtime hours in the week is the following: 0.2 28.1 26.34 0.8 22.3 0.27 0. (A) (B) (C) (D) (E) 18. For each project.2 0. Calculate the expected number of overtime hours for which you will get paid in the week. You will get paid for overtime hours in excess of 15 hours in the week.117 - .42 0. In a given week. (ii) The secondary distribution is the Poisson with probability generating function λ z −1 P( z) = e ( ) . .118 - . Calculate λ .4 208. (A) (B) (C) (D) (E) 2.2 3.02 x 0 < x < 10 fX ( x) = ⎨ elsewhere ⎩0 (ii) (iii) The insurance has an ordinary deductible of 4 per loss. −2 (iii) The probability of no claims equals 0. P ⎤ Calculate E ⎡ ⎣Y ⎦ .0 3. An actuary has created a compound claims frequency model with the following properties: (i) The primary distribution is the negative binomial with probability generating function P( z) = ⎡ ⎣1 − 3 ( z − 1) ⎤ ⎦ .3 3.207. Y P is the claim payment per payment random variable. For an insurance: (i) Losses have density function ⎧0.9 3.067. 119 - .2 times the expected claims.1 times the expected reinsured claims. Pi . equals 1.1 0.(A) (B) (C) (D) (E) 0.52 0.66 . The risk is reinsured with a deductible that stays the same in each year. (A) (B) (C) (D) (E) 0. the premium in year i.6 2.1 In 2006 209.4 1.58 0.46 0.7 3. equals 1. R 2005 P 2005 = 0. An insurance on the risk has a deductible of 600 in each year. losses inflate by 20%. Ri .55 Calculate R 2006 P2006 .55 0. the reinsurance premium in year i. In 2005 a risk has a two-parameter Pareto distribution with α = 2 and θ = 3000 . 97 1. A vaccine which will eliminate Disease 1 and costs 0. Each life within a group medical expense policy has loss amounts which follow a compound Poisson process with λ = 0.16 .06 . the probability that it is for Disease 1 is Loss amount distributions have the following parameters: Mean per loss 5 10 Standard Deviation per loss 50 20 1 . Given a loss.03 1.00 1.15 per person has been discovered. Define: A = the aggregate premium assuming that no one obtains the vaccine. (A) (B) (C) (D) (E) 0. and B = the aggregate premium assuming that everyone obtains the vaccine and the cost of the vaccine is a covered loss.94 0.24. Calculate A/B.210. 16 Disease 1 Other diseases Premiums for a group of 100 independent lives are set at a level such that the probability (using the normal approximation to the distribution for aggregate losses) that aggregate losses for the group will exceed aggregate premiums for the group is 0.120 - . This distribution is replaced with a spliced model whose density function: (i) (ii) (iii) is uniform over [0. (A) (B) (C) (D) (E) 36 48 72 96 120 . There is an ordinary deductible of 4 per loss. 3] is proportional to the initial modeled density function after 3 years is continuous Calculate the probability of failure in the first 3 years under the revised distribution. Loss amounts are uniformly distributed on (0. An actuary for a medical device manufacturer initially models the failure time for a particular device with an exponential distribution with mean 4 years.121 - .211. Loss amounts and the number of losses are mutually independent.51 212.47 0.43 0. (A) (B) (C) (D) (E) 0.49 0. For an insurance: (i) (ii) (iii) (iv) The number of losses per year has a Poisson distribution with λ = 10 .45 0. Calculate the variance of aggregate payments in a year. 10). 3 0.4 8.2 Each claim amount has a Poisson distribution with mean 3.213. (A) (B) (C) (D) (E) 4. and The number of claims and claim amounts are mutually independent.4 .122 - .1 0. For an insurance portfolio: (i) The number of claims has the probability distribution n 0 1 2 3 (ii) (iii) pn 0.2 12.8 6.0 10.4 0. Calculate the variance of aggregate claims. 05 At least 0. the credibility estimate for the number of claims in Year 2 is 0.00 At least 1.02 At least 0. but less than 0.214. the credibility estimate for the number of claims in Year 3 is 0.05 -123- . For policyholders with an average of 2 claims per year in Year 1 and Year 2.75 At least 0. but less than 0.00.04.03.50 At least 0. Determine θ . You are given: (i) (ii) (iii) (iv) The conditional distribution of the number of claims per policyholder is Poisson with mean λ . but less than 1.75. but less than 0. (A) (B) (C) (D) (E) Less than 0. For policyholders with 1 claim in Year 1.03 At least 0.15.02.50.25 At least 1. (A) (B) (C) (D) (E) Less than 0. but less than 0.25 215.04 At least 0. A portfolio of policies has produced the following claims: 100 100 100 200 300 300 300 400 500 600 Determine the empirical estimate of H(300).20. but less than 1. The variable λ has a gamma distribution with parameters α and θ . but less than 220 At least 220 217. ^ ^ -124- . 25% of the claim amounts in the sample exceed 500. but less than 210 At least 210.216. For a portfolio of policies. but less than 200 At least 200. ^ ^ Determine the absolute difference between S 1 (1250) and S 2 (1250) . Estimate θ by percentile matching. (A) (B) (C) (D) (E) Less than 190 At least 190. S 2 (1250) is the maximum likelihood estimate of S(1250) under the assumption that the losses follow an exponential distribution. you are given: (i) (ii) There is no deductible and the policy limit varies by policy. (iii) (iv) S 1 (1250) is the product-limit estimate of S(1250). A random sample of claims has been drawn from a Burr distribution with known parameter α = 1 and unknown parameters θ and γ . A sample of ten claims is: 350 350 500 500 500+ 1000 1000+ 1000+ 1200 1500 where the symbol + indicates that the loss exceeds the policy limit. You are given: (i) (ii) 75% of the claim amounts in the sample exceed 100. 05 0.03 0.00 0.5 -125- . (A) (B) (C) (D) (E) Less than 4.0 At least 4. One other value exceeds 4.0.0 At least 5. Calculate the maximum likelihood estimate of θ.5 At least 4. but less than 5.5 At least 5.5. but less than 5. but less than 4.0.07 0. The random variable X has survival function: SX ( x) = (θ θ4 2 + x2 ) 2 Two values of X are observed to be 2 and 4.(A) (B) (C) (D) (E) 0.09 218. 10 in that order.13 . as needed.219. -126- . you are given: (i) The annual claim amount on a policy has probability density function: f ( x|θ ) = (ii) 2x . The number of claims is simulated using u = 0. (A) (B) (C) (D) (E) 0.56 220. 0 < x <θ θ2 The prior distribution of θ has density function: π (θ ) = 4θ 3 .1 in Year 1.50 0. Determine the Bühlmann credibility estimate of the claim amount for the selected policy in Year 2. u2 = 0. The number of claims has a Poisson distribution with λ = 4 . The claim amounts are simulated using u1 = 0. Determine the total losses. For a portfolio of policies.05 .45 0. 0 < θ < 1 (iii) A randomly selected policy had claim amount 0.43 0. Total losses for a group of insured motorcyclists are simulated using the aggregate loss model and the inversion method. The amount of each claim has an exponential distribution with mean 1000.53 0.95 and u3 = 0. (B). You are given: (i) The sample: 1 (ii) 2 3 3 3 3 3 3 3 3 ˆ ( x) is the kernel density estimator of the distribution function using a uniform F 1 kernel with bandwidth 1. ˆ ( x) = F Determine which of the following intervals has F 1 2 (A) (B) (C) (D) (E) 0 < x <1 1< x < 2 2< x<3 3< x< 4 None of (A).(A) (B) (C) (D) (E) 0 51 2996 3047 3152 221. (iii) ˆ ( x) for all x in the interval. ˆ ( x) is the kernel density estimator of the distribution function using a triangular F 2 kernel with bandwidth 1. (C) or (D) -127- . 05 significance level. The hypothesis is rejected at the 0. but is not rejected at the 0. but is not rejected at the 0. Any interval in which the expected number is less than one is combined with the previous interval.025 significance level. The number of work days missed is given below: Number of Days of Work Missed 0 1 2 3 or more Total Total Number of Days Missed Number of Workers 818 153 25 4 1000 230 The chi-square goodness-of-fit test is used to test the hypothesis that the number of work days missed follows a Poisson distribution where: (i) (ii) The Poisson parameter is estimated by the average number of work days missed.222. The hypothesis is rejected at the 0. (A) (B) (C) (D) (E) The hypothesis is not rejected at the 0. Determine the results of the test.10 significance level. 1000 workers insured under a workers compensation policy were observed for one year.10 significance level.05 significance level. -128- . The hypothesis is rejected at the 0. but is not rejected at the 0.01 significance level.025 significance level.01 significance level. The hypothesis is rejected at the 0. The maximum likelihood estimates are μ = 6.88 224.84 and σ = 149 (ii) (iii) ˆ and σ ˆ is: The covariance matrix of μ LM0. Determine the credibility factor for Year 3 using the nonparametric empirical Bayes method.000 25 Year 2 14.88 At least 0. but less than 0.0444 N0 (iv) ∂ F −φ ( z ) = ∂μ σ 0 0. (A) (B) (C) (D) (E) Less than 0. .0222 OP Q bg The partial derivatives of the lognormal cumulative distribution function are: and ∂ F −z × φ z = σ ∂σ -129- .000 30 The estimate of the variance of the hypothetical means is 254.73 At least 0. .78.78 At least 0. You are given the following data: Year 1 Total Losses Number of Policyholders 12. but less than 0.223.73.83 At least 0. You are given: (i) Fifty claims have been observed from a lognormal distribution with unknown parameters μ and σ .83. DELETED 225. but less than 0. 85 -130- . PH] Determine PL .79 0.(v) An approximate 95% confidence interval for the probability that the next claim will be less than or equal to 5000 is: [PL.73 0.82 0.76 0. (A) (B) (C) (D) (E) 0. 3. but less than 1.1 At least 1.3 At least 1.10 0. (A) (B) (C) (D) (E) Less than 1.2 At least 1.20 θ θ Probability One claim was observed in Year 1.2. such that the number of simulations N is sufficiently large to be at least 95 percent confident of estimating F(1500) correctly within 1 percent. For a particular policy.4 227.80 1 2 1 − 3θ 0.1. but less than 1. Let P represent the number of simulated values less than 1500. but less than 1. and the probability distribution of Θ are as follows: Number of Claims Probability 0 2θ 0.30 0. (A) (B) (C) (D) (E) N = 2000 N = 3000 N = 3500 N = 4000 N = 4500 P = 1890 P = 2500 P = 3100 P = 3630 P = 4020 -131- .4 At least 1. the conditional probability of the annual number of claims given Θ = θ . Determine which of the following could be values of N and P. Calculate the Bayesian estimate of the expected number of claims for Year 2.226. You simulate observations from a specific distribution F(x). A random sample of size n is drawn from a distribution with probability density function: f ( x) = θ (θ + x) 2 .4128 and H ( y4 ) = 0.014448 ∧ ^ ∧ ^ Determine the number of deaths at y4 . 0 < x < ∞.5691 ^ ^ (iii) The estimated variances of the estimates in (ii) are: Var[ H ( y3 )] = 0. you are given: (i) (ii) Deaths occurred at times y1 < y 2 < … < y 9 . For a survival study. -132- . (A) (B) (C) (D) (E) 2 3 4 5 6 229.009565 and Var[ H ( y4 )] = 0.228. The Nelson-Aalen estimates of the cumulative hazard function at y 3 and y 4 are: H ( y3 ) = 0. θ > 0 Determine the asymptotic variance of the maximum likelihood estimator of θ . The Bühlmann credibility estimate of the number of claims for the same risk in Year 2 is 11.983. For a portfolio of independent risks. Determine x. the number of claims for each risk in a year follows a Poisson distribution with means given in the following table: Mean Number of Claims per Risk 1 Class Number of Risks 900 1 2 3 10 20 90 10 You observe x claims in Year 1 for a randomly selected risk. (A) (B) (C) (D) (E) 13 14 15 16 17 -133- .(A) 3θ 2 n 1 3nθ 2 (B) (C) 3 nθ 2 n 3θ 2 1 3θ 2 (D) (E) 230. determine the symmetric linear 95% confidence interval for S(5).01 -134- .00 30. 0.87 4.… The parameters μi vary in such a way that there is an annual inflation rate of 10% for losses. 1.00 55.75) (0.283.32. 0. 0. determine the method of moments estimate of μ3 . 3.71 63. for i = 1.23.73) (0. (ii) (iii) σ i = σ .267) as the symmetric linear 95% confidence interval for H(5). 0.26. (A) (B) (C) (D) (E) (0. Using the delta method. The following is a sample of seven losses: Year 1: Year 2: 20 30 40 40 50 90 (iv) 120 Using trended losses.31.28.69) (0.231. A survival study gave (0. You are given: (i) Losses on a certain warranty product in Year i follow a lognormal distribution with parameters μi and σ i . 0. (A) (B) (C) (D) (E) 3.72) (0.80) 232. 2. but less than 0. the maximum likelihood estimate of θ is 816. DELETED 235. ∑ ln ( f ( xi ) ) = −33.7 234. but less than 0. but less than 0. You are given: (i) A random sample of losses from a Weibull distribution is: 595 (ii) (iii) (iv) 700 789 799 1109 At the maximum likelihood estimates of θ and τ . You use the likelihood ratio test to test the hypothesis H0 :τ = 2 H1 : τ ≠ 2 -135- . The number of insureds is constant over time for each territory.4.6. (A) (B) (C) (D) (E) Less than 0. You are given: (i) A region is comprised of three territories. Determine the Bühlmann-Straub empirical Bayes estimate of the credibility factor Z for Territory A.5 At least 0.4 At least 0.05 .7.233.7 At least 0. When τ = 2 .5. Each insured in a territory has the same expected claim frequency.6 At least 0. Claims experience for Year 1 is as follows: Territory A B C (ii) (iii) (iv) Number of Insureds 10 20 30 Number of Claims 4 5 3 The number of claims for each insured each year has a Poisson distribution. 01 level of significance.025 level of significance.5. and variance.…. Calculate Cov X i . Xn ..4. conditional on Θ.01 level of significance. but less than 2.5 At least 2.025 level of significance. but not at the 0.5 At least 0...5 At least −0. (A) (B) (C) (D) (E) Less than −0. but not at the 0. Reject H 0 at the 0. n ( ) j = 1. μ (θ ) = E ( X j Θ = θ ) . You are given: (i) (ii) j = 1.. but not at the 0. The Bühlmann credibility assigned for estimating X5 based on X1.. losses X1.5 . v (θ ) = Var X j Θ = θ . but less than 0. are independently and identically distributed with mean. 2. Reject H 0 at the 0. Reject H 0 at the 0.5. X j .05 level of significance..10 level of significance. X4 is Z = 0.5 At least 1. The expected value of the process variance is known to be 8. (A) (B) Do not reject H 0 at the 0..5 ( ) -136- . but less than 1. (C) (D) (E) 236. Reject H 0 at the 0. n ..05 level of significance.…. For each policyholder.10 level of significance. i ≠ j . 2.Determine the result of the test. (A) (B) (C) (D) (E) Less than 630 At least 630.1) random numbers: 0.9941 0.8686 0. but less than 730 At least 730. You use simulation to estimate claim payments for a number of contracts with different deductibles. calculate the average payment per loss for a contract with a deductible of 100.6 and 0. but less than 680 At least 680.6217 0. Losses for a warranty product follow the lognormal distribution with underlying normal mean and standard deviation of 5.75 respectively. but less than 780 At least 780 238. The following are four uniform (0. Calculate the mean-squared error of X 2 as an estimator of θ 2 . The random variable X has the exponential distribution with mean θ .237.0485 Using these numbers and the inversion method. 20 θ 4 21 θ 4 22 θ 4 23 θ 4 24 θ 4 (A) (B) (C) (D) (E) -137- . A Poisson distribution is fitted to the data using the method of moments.12 -138- . Calculate P − Q . Let Q = probability of zero claims using the fitted Poisson model.09 0. (A) (B) (C) (D) (E) 0.239.00 0. Let P = probability of zero claims using the fitted geometric model.06 0.03 0. You are given the following data for the number of claims during a one-year period: Number of Claims 0 1 2 3 4 5+ Total Number of Policies 157 66 19 4 2 0 248 A geometric distribution is fitted to the data using maximum likelihood estimation. During Year 1.240.15.60 -139- .45 At least 0. the following data are observed for 8000 policyholders: Number of Claims 0 1 2 3 4 5+ Number of Policyholders 5000 2100 750 100 50 0 A randomly selected policyholder had one claim in Year 1. Determine the semiparametric empirical Bayes estimate of the number of claims in Year 2 for the same policyholder. but less than 0. For a group of auto policyholders. (A) (B) (C) (D) (E) Less than 0.45. but less than 0.30.60 At least 0. but less than 0. you are given: (i) (ii) The number of claims for each policyholder has a conditional Poisson distribution.15 At least 0.30 At least 0. (A) (B) (C) (D) (E) − 0. For each year. You are given: (i) The following are observed claim amounts: 400 (ii) (iii) 1000 1600 3000 5000 5400 6200 An exponential distribution with θ = 3300 is hypothesized for the data.07 0.5 A randomly selected policyholder had two claims in Year 1 and two claims in Year 2. determine the Bayesian estimate of the expected number of claims in Year 3. Determine ( s − t ) − D ( 3000 ) . Let (s.10 0. The goodness of fit is to be assessed by a p-p plot and a D(x) plot.00 0. 0.12 − 0.039 0.2 < q < 0.90 − q q π (q) = q2 . each policyholder can have at most two claims per year.07 0. t) be the coordinates of the p-p plot for a claim amount of 3000.12 242. -140- . For this insured. the distribution of the number of claims is: Number of Claims 0 1 2 (iii) The prior density is: Probability 0. You are given: (i) (ii) In a portfolio of risks.241. 000) [5.60 At least 1.30.000) [2. (A) (B) (C) (D) (E) Less than 65 At least 65. 500) [500. ∞ ) Number of Claims 200 110 x y ? ? ? You are also given the following values taken from the ogive: F500(1500) = 0.000. 25. but less than 1.000) [25. but less than 1.(A) (B) (C) (D) (E) Less than 1.000) [1.000. but less than 70 At least 70.50 At least 1. 5.839 Determine y.50. For 500 claims.30 At least 1.000) [10.000. 1.689 F500(3500) = 0. 2.60 243. but less than 1.000. but less than 80 At least 80 -141- .40 At least 1. but less than 75 At least 75.000. 10.40. you are given the following distribution: Claim Size [0. when the parameters of the distribution in the null hypothesis are estimated from the data. (B). (C). The number of claims and claim sizes are independent. (C) or (D) is false. the probability of rejecting the null hypothesis decreases. For the Kolmogorov-Smirnov test. the critical value for right censored data should be smaller than the critical value for uncensored data. (B). Which of statements (A).142 - . The full credibility standard has been selected so that actual aggregate losses will be within 10% of expected aggregate losses 95% of the time. (C) (D) (E) 245. Claim sizes follow a gamma distribution with parameters α (unknown) and θ = 10.000 . but less than 450 At least 450. The Anderson-Darling test does not work for grouped data. None of (A). but less than 500 At least 500 The expected number of claims required for full credibility cannot be determined from the information given. For the Kolmogorov-Smirnov test. You are given: (i) (ii) The number of claims follows a Poisson distribution. and (D) is false? (A) (B) The chi-square goodness-of-fit test works best when the expected number of observations varies widely from interval to interval. (iii) (iv) Using limited fluctuation (classical) credibility. (A) (B) (C) (D) (E) Less than 400 At least 400. determine the expected number of claims required for full credibility.244. . 25 Poisson with λ = 0.7 At least 0. (A) (B) (C) (D) (E) Less than 2.6. but less than 0. but less than 0.50 Poisson with λ = 1.8 247.5 At least 3.8 At least 3. but less than 3.5.2 At least 3.246.4 At least 0. Calculate the estimate of γ.2. An insurance company sells three types of policies with the following characteristics: Type of Policy I II III Proportion of Total Policies 5% 20% 75% Annual Claim Frequency Poisson with λ = 0. (A) (B) (C) (D) (E) Less than 0.5. but less than 3.5 At least 0. A random sample of 15 losses is: 195 255 270 280 350 360 365 380 415 450 490 550 575 590 615 (iii) The parameters γ and θ are estimated by percentile matching using the smoothed empirical estimates of the 30th and 65th percentiles. determine the Bayesian estimate of the expected number of claims in Year 5. but less than 3.143 - . You are given: (i) (ii) Losses follow a Burr distribution with α = 2. but less than 0. For the same policyholder.4.9 At least 2.00 A randomly selected policyholder is observed to have a total of one claim for Year 1 through Year 4.9.7 .6 At least 0. 5 At least 1.312 0. You are given a random sample of 10 claims consisting of two claims of 400. (a) (b) For the number of losses use the random number 0.7537 0. You are given: (i) The cumulative distribution for the annual number of losses for a policyholder is: n 0 1 2 3 4 5 (ii) (iii) FN ( n ) 0. For loss amounts use the random numbers: 0.0.0.7654.5152 0.5 249.5.5 At least 2.144 - . but less than 1. seven claims of 800. The inversion method is used to simulate the number of losses and loss amounts for a policyholder.656 0. and one claim of 1600.2738 0.6481 0. (A) (B) (C) (D) (E) Less than 1.773 0. but less than 2. but less than 2.248.125 0.0 At least 1.855 The loss amounts follow the Weibull distribution with θ = 200 and τ = 2. Determine the empirical skewness coefficient. There is a deductible of 150 for each claim subject to an annual maximum out-ofpocket of 500 per policy.500 0.0 At least 2.3153 . but less than 26 At least 26 . (A) (B) (C) (D) (E) 106. Losses follow an inverse exponential with distribution function F ( x ) = e−θ / x . x > 0 Calculate the maximum likelihood estimate of the population mode.32 224.Use the random numbers in order and only as needed. Based on the simulation.05 250. You have observed the following three loss amounts: 186 91 66 Seven other amounts are known to be less than or equal to 60.53 520.145 - . calculate the insurer’s aggregate payments for this policyholder.93 161. (A) (B) (C) (D) (E) Less than 11 At least 11. but less than 16 At least 16. but less than 21 At least 21.44 347. 146 - . but less than 2150 At least 2150 . Calculate the Bühlmann credibility estimate of the loss on the selected policy in Year 4 based on the data for Years 1. you are given: (i) (ii) (iii) (iv) (v) (vi) The annual loss on an individual policy follows a gamma distribution with parameters α = 4 and θ.251. but less than 2050 At least 2050. The loss on the selected policy in Year 3 was 2763. the Bühlmann credibility estimate of the loss on the selected policy in Year 4 is 1800. A randomly selected policy had losses of 1400 in Year 1 and 1900 in Year 2. (A) (B) (C) (D) (E) Less than 1850 At least 1850. After the estimate in (v) was calculated. Loss data for Year 3 was misfiled and unavailable. The prior distribution of θ has mean 600. the data for Year 3 was located. For a group of policies. Based on the data in (iii). 2 and 3. but less than 1950 At least 1950. X 1 . ˆ (11).048 0.016 0. identically distributed Bernoulli random variables with parameter q. b = 99.252.147 - .… . Determine the smallest value of m such that the mean of the marginal distribution of Sm is greater than or equal to 50.031 0.064 0. X 2 .075 253. You are given: (i) For Q = q. Determine Greenwood’s approximation to the variance of the product-limit estimate S (A) (B) (C) (D) (E) 0. The following is a sample of 10 payments: 4 4 5+ 5+ 5+ 8 10+ 10+ 12 15 where + indicates that a loss exceeded the policy limit. Sm = X 1 + X 2 + + Xm (ii) (iii) The prior distribution of Q is beta with a = 1. and θ = 1 . X m are independent. (A) (B) (C) (D) (E) 1082 2164 3246 4950 5000 . You are given: (i) (ii) (iii) A portfolio consists of 100 identically and independently distributed risks. the following loss experience is observed: Number of Claims 0 1 2 3 Total Number of Risks 90 7 2 1 100 Determine the Bayesian expected number of claims for the portfolio in Year 2. (A) (B) (C) (D) (E) 8 10 11 12 14 . The prior distribution of λ is: (50λ ) 4 e−50 λ π (λ ) = .148 - . The number of claims for each risk follows a Poisson distribution with mean λ .254. λ >0 6λ During Year 1. (A) (B) (C) (D) (E) 944 1299 1559 1844 2213 256.781 −7. You are planning a simulation to estimate the mean of a non-negative random variable. known that the population standard deviation is 20% larger than the population mean.730 .255.750 −6.000 insurance policies is: Number of Claims per Policy 0 1 2 or more (ii) Number of Policies 5000 5000 0 You fit a binomial model with parameters m and q using the method of maximum likelihood. It is Use the central limit theorem to estimate the smallest number of trials needed so that you will be at least 95% confident that the simulated mean is within 5% of the population mean. Determine the maximum value of the loglikelihood function when m = 2.931 −6. (A) (B) (C) (D) (E) −10.397 −7.149 - . You are given: (i) The distribution of the number of claims per policy during a one-year period for 10. 55 At least 0.60 At least 0. You are given: (i) Over a three-year period. (A) (B) (C) (D) (E) Less than 0.257.55.60.150 - .70 . but less than 0.70 At least 0. but less than 0. but less than 0.65. Determine the semiparametric empirical Bayes estimate of the claim frequency per vehicle for Insured A in Year 4.65 At least 0. the following claim experience was observed for two insureds who own delivery vans: Year Insured A 1 2 3 B (ii) Number of Vehicles Number of Claims Number of Vehicles Number of Claims 2 1 N/A N/A 2 1 3 2 1 0 2 3 The number of claims for each insured each year follows a Poisson distribution. 258. Determine the estimated coefficient of variation of the estimator of the Poisson parameter. (A) (B) (C) (D) (E) 100 200 300 400 500 259. You use the method of maximum likelihood to fit a Poisson model. . You are given: (i) A hospital liability policy has experienced the following numbers of claims over a 10-year period: 10 (ii) (iii) 2 4 0 6 2 4 5 4 2 Numbers of claims are independent from year to year.15 The Nelson-Åalen Estimate H 300 100 400 X Determine X.151 - . For the data set 200 you are given: (i) (ii) (iii) (iv) k=4 s2 = 1 r4 = 1 ˆ (410) > 2. You are given: (i) (ii) (iii) Claim sizes follow an exponential distribution with mean θ .0 .00 260.152 - .26 1. but less than 6.10 0. For 80% of the policies.8 At least 5. A randomly selected policy had one claim in Year 1 of size 5. θ = 2 .2 At least 6. θ = 8.(A) (B) (C) (D) (E) 0.6.22 0. Calculate the Bayesian expected claim size for this policy in Year 2. (A) (B) (C) (D) (E) Less than 5.16 0. For 20% of the policies.6 At least 6.8. but less than 6.0 At least 7. but less than 7.2. Determine the Kaplan-Meier multiple-decrement estimate of the number of individuals in Group A that survive to be at least 40 years old.100 0.182 0. For a double-decrement study.261.600 1.153 - . you are given: (i) The following survival data for individuals affected by both decrements (1) and (2): j 0 1 2 3 (ii) (iii) (iv) 2 q j′ ( ) = 0. (A) (B) (C) (D) (E) 343 664 736 816 861 .05 for all j cj 0 20 40 60 T q (j ) 0. Group A is affected by only decrement (1).000 Group A consists of 1000 individuals observed at age 0. Three light bulbs burn out at times 5. (A) (B) (C) (D) (E) Less than 10 At least 10. and 13 hours. there are 5 working light bulbs. 9. The maximum likelihood estimate of ω is 29. but less than 12 At least 12. ω ) .262. while the remaining light bulbs are still working at time 4 + p hours. Determine p. You are given: (i) (ii) (iii) (iv) (v) At time 4 hours. but less than 16 At least 16 . The 5 bulbs are observed for p more hours. but less than 14 At least 14. The distribution of failure times is uniform on ( 0.154 - . but less than 315 At least 315 . Γ ( 2 ) = 1 (v) Month 1 2 3 Number of Insureds 100 150 250 Number of Claims 10 11 14 Determine the Bühlmann-Straub credibility estimate of the number of claims in the next 12 months for 300 insureds.77245. The claim frequencies of different insureds are independent. Γ (1.88623. The prior distribution of λ is Weibull with θ = 0.155 - . but less than 275 At least 275. (A) (B) (C) (D) (E) Less than 255 At least 255. Some values of the gamma function are Γ ( 0. but less than 295 At least 295.263. You are given: (i) (ii) (iii) (iv) The number of claims incurred in a month by any insured follows a Poisson distribution with mean λ .5 ) = 1. Γ (1) = 1.1 and τ = 2 .5 ) = 0. 5 At least 0.1 At least 0. but less than 0. but less than 0.156 - .264. 5000 and 7500 in (i) reflect right-censored observations and that the remaining observed values are uncensored. You are given: (i) The following data set: 2500 2500 2500 3617 3662 4517 5000 5000 6010 6932 7500 7500 (ii) H1 (7000) is the Nelson-Åalen estimate of the cumulative hazard rate function calculated under the assumption that all of the observations in (i) are uncensored. but less than 0. H 2 (7000) is the Nelson-Åalen estimate of the cumulative hazard rate function calculated under the assumption that all occurrences of the values 2500. (A) (B) (C) (D) (E) Less than 0.7 ^ ^ .3. ^ ^ (iii) Calculate| H1 (7000) − H 2 (7000) |.1.3 At least 0.7 At least 0.5. 157 - .1210 0.000.6179 Using the simulated values. is modeled by y = 0. Five models are fitted to a sample of n = 260 observations with the following results: Model Number of Parameters Loglikelihood I 1 − 414 II 2 − 412 III 3 − 411 IV 4 − 409 V 6 − 409 Determine the model favored by the Schwarz Bayesian criterion.000 266. For a warranty product you are given: (i) (ii) Paid losses follow the lognormal distribution with μ = 13.801 x 0.8238 0.000. but less than 500. y.494 .2877 0. calculate the empirical estimate of the average unpaid losses for purchase year 2005. The ratio of estimated unpaid losses to paid losses.1) random numbers: 0.000.000 At least 600. . but less than 600. (A) (B) (C) (D) (E) Less than 300.265.747 x where x = 2006 − contract purchase year The inversion method is used to simulate four paid losses with the following four uniform (0.000 At least 300.294 and σ = 0.000 At least 500. but less than 400.851 e− 0.000 At least 400. The Bayesian estimate of this risk’s expected number of claims in Year 2 is 2.158 - . but less than 2. Determine the Bühlmann credibility estimate of the expected number of claims for this risk in Year 2. For 25% of the risks. For 75% of the risks.3 At least 2.(A) (B) (C) (D) (E) I II III IV V 267. A randomly selected risk had r claims in Year 1. λ = 1 . (A) (B) (C) (D) (E) Less than 1. λ = 3 .7.3.9.98. You are given: (i) (ii) (iii) The annual number of claims for an individual risk follows a Poisson distribution with mean λ . but less than 2.1 .1 At least 3.9 At least 1.7 At least 2. but less than 3. The random variables X 1 . You are given the following ages at time of death for 10 individuals: 25 30 35 35 37 39 45 47 49 55 Using a uniform kernel with bandwidth b = 10. (A) ⎛ n +1⎞ 2 ⎜ ⎟θ ⎝ n ⎠ ⎛ n +1⎞ 2 ⎜ 2 ⎟θ ⎝ n ⎠ (B) (C) θ2 n (D) (E) θ2 n θ2 .400 0.439 0.… .268.159 - . determine the kernel density estimate of the probability of survival to age 40. x≥0 θ 2 ⎤ Determine E ⎡ ⎣X ⎦ .417 0. X n are independent and identically distributed with probability density function f ( x) = e− x /θ . X 2 . (A) (B) (C) (D) (E) 0.485 269.377 0. but less than 1.40.40 At least 0.00 271.33 0.80 At least 0. (A) (B) (C) (D) (E) Less than 0. but less than 0. (A) (B) (C) (D) (E) 0.00 At least 1. You are given: (i) (ii) The number of claims made by an individual in any given year has a binomial distribution with parameters m = 4 and q. (iii) 0 < q < 1.80. DELETED 272.50 0. Two claims are made in a given year. Three individual policyholders have the following claim amounts over four years: Policyholder Year 1 Year 2 Year 3 Year 4 X 2 3 3 4 Y 5 5 4 6 Z 5 5 3 3 Using the nonparametric empirical Bayes procedure.67 0. but less than 0. The prior distribution of q has probability density function π (q) = 6q(1 − q). calculate the estimated variance of the hypothetical means.160 - .60 At least 0. Determine the mode of the posterior distribution of q.270.83 .17 0.60. Determine k. The number of claims follows a Poisson distribution.161 - . where claim severity has probability density function: f ( x) = 1 . (A) (B) (C) (D) (E) 720 960 2160 2667 2880 274. determine the expected number of claims necessary to obtain full credibility under the new standard.575.000 Using limited fluctuation credibility. you are given the following: Time 3 5 6 10 Number of Deaths 1 3 5 7 Number at Risk 50 49 k 21 You are also told that the Nelson-Åalen estimate of the survival function at time 10 is 0. A company has determined that the limited fluctuation full credibility standard is 2000 claims if: (i) (ii) The total number of claims is to be within 3% of the true value with probability p. .273.000 0 ≤ x ≤ 10. 10. The standard is changed so that the total cost of claims is to be within 5% of the true value with probability p. For a mortality study with right censored data. (A) (B) (C) (D) (E) 28 31 36 44 46 275. A dental benefit is designed so that a deductible of 100 is applied to annual dental charges. The reimbursement to the insured is 80% of the remaining dental charges subject to an annual maximum reimbursement of 1000. You are given: (i) (ii) The annual dental charges for each insured are exponentially distributed with mean 1000. Use the following uniform (0, 1) random numbers and the inversion method to generate four values of annual dental charges: 0.30 0.92 0.70 0.08 Calculate the average annual reimbursement for this simulation. (A) (B) (C) (D) (E) 522 696 757 947 1042 - 162 - 276. For a group of policies, you are given: (i) Losses follow the distribution function F ( x ) = 1 − θ / x, (ii) θ < x<∞. A sample of 20 losses resulted in the following: Interval x ≤ 10 10 < x ≤ 25 Number of Losses 9 6 5 x > 25 Calculate the maximum likelihood estimate of θ. (A) (B) (C) (D) (E) 5.00 5.50 5.75 6.00 6.25 277. You are given: (i) (ii) Loss payments for a group health policy follow an exponential distribution with unknown mean. A sample of losses is: 100 200 400 800 1400 3100 Use the delta method to approximate the variance of the maximum likelihood estimator of S (1500 ) . - 163 - (A) (B) (C) (D) (E) 0.019 0.025 0.032 0.039 0.045 278. You are given: (i) A random sample of payments from a portfolio of policies resulted in the following: Interval (0, 50] (50, 150] (150, 250] (250, 500] (500, 1000] (1000, ∞ ) Total (ii) Number of Policies 36 x y 84 80 0 n Two values of the ogive constructed from the data in (i) are: Fn ( 90 ) = 0.21, and Fn ( 210 ) = 0.51 Calculate x. (A) (B) (C) (D) (E) 120 145 170 195 220 - 164 - (A) (B) (C) (D) (E) 6 7 8 9 10 .165 - . (A) (B) (C) (D) (E) 37 39 43 47 49 280.03. Calculate k.6 0.4 where k > 5 The expected cost of an aggregate stop-loss insurance subject to a deductible of 5 is 28. 100 < x ⎪ ⎩ 1 An insurance pays 80% of the amount of the loss in excess of an ordinary deductible of 20. given that a payment has been made. Loss amounts have the distribution function ⎧( x /100 )2 . A compound Poisson claim distribution has λ = 5 and individual claim amounts distributed as follows: x 5 k fX ( x) 0. Calculate the conditional expected claim payment. subject to a maximum payment of 60 per loss. 0 ≤ x ≤ 100 ⎪ F ( x) = ⎨ .279. 4% 10.0% 11.43% − 1. For a random variable X which has a normal distribution with mean μ and standard deviation σ . subject to a minimum guaranteed crediting rate of 3%. you are given: (i) (ii) (iii) All deposits are credited with 75% of the annual equity index return.9% 9.43% μ = 8% 0. you are given the following limited expected values: E [ X ∧ 3% ] μ = 6% σ = 12% σ = 16% − 0.19% E [ X ∧ 4% ] μ = 6% σ = 12% σ = 16% 0.281.99% μ = 8% 0. For a special investment product. The annual equity index return is normally distributed with a mean of 8% and a standard deviation of 16%.31% − 1.7% 11.6% .58% Calculate the expected annual crediting rate.166 - .95% − 0. (A) (B) (C) (D) (E) 8.15% − 1. and γ = 1 . An insurance pays 100 per doctor visit beginning with the 4th visit per family. Aggregate losses are modeled as follows: (i) (ii) The number of losses has a Poisson distribution with λ = 3 . The annual number of doctor visits for each individual in a family of 4 has a geometric distribution with mean 1. (A) (B) (C) (D) (E) 12 14 16 18 20 283. The amount of each loss has a Burr (Burr Type XII.282. The annual numbers of visits for the family members are mutually independent. (A) (B) (C) (D) (E) 320 323 326 329 332 . θ = 2 .5.167 - . (iii) Calculate the variance of aggregate losses. Calculate the expected payments per year for this family. The number of losses and the amounts of the losses are mutually independent. Singh-Maddala) distribution with α = 3. 400 49. so that the expected insurance cost remains the same. Your annual budget for persons appearing on the show equals 10 times the expected number of persons plus 10 times the standard deviation of the number of persons.800 47.600 44.37 0.284.42 285.000 (B) (C) (D) (E) . The number of members of an accepted club has a distribution with mean 20 and variance 20. (A) (B) (C) (D) (E) 0. An alternative insurance replaces the deductible with coinsurance α . The probability of a club being accepted is 0.20. Club acceptances and the numbers of club members are mutually independent.168 - .32 0.27 0. 1000 actuarial clubs audition for the show.200 45. which is the proportion of the loss paid by the insurance. (A) 42. You are the producer for the television show Actuarial Idol.22 0. Calculate α . Each year. Calculate your annual budget for persons appearing on the show. A risk has a loss amount which has a Poisson distribution with mean 3. An insurance covers the risk with an ordinary deductible of 2. Michael is a professional stuntman who performs dangerous motorcycle jumps at extreme sports events around the world. Michael pays 20% of repair costs between 1000 and 6000 each year. Michael pays 100% of the annual repair costs above 6000 until Michael has paid 10.000 in out-of-pocket repair costs each year. Calculate the expected annual insurance reimbursement. Michael pays 10% of the remaining repair costs each year. The annual cost of repairs to his motorcycle is modeled by a two parameter Pareto distribution with θ = 5000 and α = 2 . An insurance reimburses Michael’s motorcycle repair costs subject to the following provisions: (i) (ii) (iii) (iv) Michael pays an annual ordinary deductible of 1000 each year.286. (A) (B) (C) (D) (E) 2300 2500 2700 2900 3100 .169 - . 125 p 0.125 p 2 0. The claim amounts are uniformly distributed on the interval (0. Using the normal approximation for aggregate losses. 8). calculate the premium such that the probability that aggregate losses will exceed the premium is 5%.125 p -170- .5 and m = 2 . N has a binomial distribution with q = 0. (A) (B) (C) (D) (E) 500 520 540 560 580 288.375 + 0.125 p 2 0. The random variable N has a mixed distribution: (i) (ii) With probability p. With probability 1 − p .375 + 0.287. Which of the following is a correct expression for Prob ( N = 2 ) ? (A) (B) (C) (D) (E) 0.5 and m = 4 . N has a binomial distribution with q = 0. The number of claims and claim amounts are mutually independent. For an aggregate loss distribution S: (i) (ii) (iii) The number of claims has a negative binomial distribution with r = 16 and β = 6 .125 p 2 0.375 − 0.375 − 0. 038 0.04 Calculate the probability that aggregate claims will be exactly 600.022 0.060 0. A compound Poisson distribution has λ = 5 and claim amount distribution as follows: x 100 500 1000 p ( x) 0. (A) (B) (C) (D) (E) 0.80 0.16 0.289.070 -171- .049 0.