Exam C, Fall 2006 PRELIMINARY ANSWER KEY

Exam C, Fall 2006 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 E 19 B 2 D 20 D 3 B 21 A 4 C 22 A 5 A 23 E 6 D 24 E 7 B 25 D 8 C 26 A 9 E 27 C 10 D 28 C 11 E 29 C 12 B 30 B 13 C 31 C 14 A 32 A 15 B 33 B 16 E 34 A 17 D 35 A 18 D

**BEGINNING OF EXAMINATION** 1. You are given: (i) Losses follow a Burr distribution with α = 2. (ii) 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 30 th and 65 th percentiles. Calculate the estimate of γ. (A) Less than 2.9 (B) At least 2.9, but less than 3.2 (C) At least 3.2, but less than 3.5 (D) At least 3.5, but less than 3.8 (E) At least 3.8 Exam C: Fall 2006-1 - GO ON TO NEXT PAGE

2. An insurance company sells three types of policies with the following characteristics: Type of Policy Proportion of Total Annual Claim Frequency Policies I 5% Poisson with λ = 0.25 II 20% Poisson with λ = 0.50 III 75% Poisson with λ = 1.00 A randomly selected policyholder is observed to have a total of one claim for Year 1 through Year 4. For the same policyholder, determine the Bayesian estimate of the expected number of claims in Year 5. (A) Less than 0.4 (B) At least 0.4, but less than 0.5 (C) At least 0.5, but less than 0.6 (D) At least 0.6, but less than 0.7 (E) At least 0.7 Exam C: Fall 2006-2 - GO ON TO NEXT PAGE

3. You are given a random sample of 10 claims consisting of two claims of 400, seven claims of 800, and one claim of 1600. Determine the empirical skewness coefficient. (A) Less than 1.0 (B) At least 1.0, but less than 1.5 (C) At least 1.5, but less than 2.0 (D) At least 2.0, but less than 2.5 (E) At least 2.5 Exam C: Fall 2006-3 - GO ON TO NEXT PAGE

4. You are given: (i) The cumulative distribution for the annual number of losses for a policyholder is: n F ( n ) 0 0.125 1 0.312 2 0.500 3 0.656 4 0.773 5 0.855 M M (ii) The loss amounts follow the Weibull distribution with θ = 200 and τ = 2. N (iii) There is a deductible of 150 for each claim subject to an annual maximum out-ofpocket of 500 per policy. The inversion method is used to simulate the number of losses and loss amounts for a policyholder. (a) For the number of losses use the random number 0.7654. (b) For loss amounts use the random numbers: 0.2738 0.5152 0.7537 0.6481 0.3153 Use the random numbers in order and only as needed. Based on the simulation, calculate the insurer s aggregate payments for this policyholder. (A) 106.93 (B) 161.32 (C) 224.44 (D) 347.53 (E) 520.05 Exam C: Fall 2006-4 - GO ON TO NEXT PAGE

5. You have observed the following three loss amounts: 186 91 66 Seven other amounts are known to be less than or equal to 60. Losses follow an inverse exponential with distribution function θ / ( ) F x = e, x> 0 x Calculate the maximum likelihood estimate of the population mode. (A) Less than 11 (B) At least 11, but less than 16 (C) At least 16, but less than 21 (D) At least 21, but less than 26 (E) At least 26 Exam C: Fall 2006-5 - GO ON TO NEXT PAGE

6. For a group of policies, you are given: (i) The annual loss on an individual policy follows a gamma distribution with parameters α = 4 and θ. (ii) The prior distribution of θ has mean 600. (iii) A randomly selected policy had losses of 1400 in Year 1 and 1900 in Year 2. (iv) (v) (vi) Loss data for Year 3 was misfiled and unavailable. Based on the data in (iii), the Bühlmann credibility estimate of the loss on the selected policy in Year 4 is 1800. After the estimate in (v) was calculated, the data for Year 3 was located. The loss on the selected policy in Year 3 was 2763. Calculate the Bühlmann credibility estimate of the loss on the selected policy in Year 4 based on the data for Years 1, 2 and 3. (A) Less than 1850 (B) At least 1850, but less than 1950 (C) At least 1950, but less than 2050 (D) At least 2050, but less than 2150 (E) At least 2150 Exam C: Fall 2006-6 - GO ON TO NEXT PAGE

7. 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. Determine Greenwood s approximation to the variance of the product-limit estimate Ŝ (11). (A) 0.016 (B) 0.031 (C) 0.048 (D) 0.064 (E) 0.075 Exam C: Fall 2006-7 - GO ON TO NEXT PAGE

8. Determine f ( 3) using the second degree polynomial that interpolates the points: (2, 25) (4, 20) (5, 30) (A) Less than 15 (B) At least 15, but less than 18 (C) At least 18, but less than 21 (D) At least 21, but less than 23 (E) At least 23 Exam C: Fall 2006-8 - GO ON TO NEXT PAGE

9. You are given: (i) For Q = q, X1, X2, K, Xm are independent, identically distributed Bernoulli random variables with parameter q. (ii) Sm = X1+ X2 + L + X m (iii) The prior distribution of Q is beta with a= 1, b= 99, and θ = 1. Determine the smallest value of m such that the mean of the marginal distribution of S m is greater than or equal to 50. (A) 1082 (B) 2164 (C) 3246 (D) 4950 (E) 5000 Exam C: Fall 2006-9 - GO ON TO NEXT PAGE

10. You are given: (i) A portfolio consists of 100 identically and independently distributed risks. (ii) The number of claims for each risk follows a Poisson distribution with mean λ. (iii) The prior distribution of λ is: (50 ) 4 50 λ λ e πλ ( ) =, λ> 0 6λ During Year 1, the following loss experience is observed: Number of Claims Number of Risks 0 90 1 7 2 2 3 1 Total 100 Determine the Bayesian expected number of claims for the portfolio in Year 2. (A) 8 (B) 10 (C) 11 (D) 12 (E) 14 Exam C: Fall 2006-10 - GO ON TO NEXT PAGE

11. You are planning a simulation to estimate the mean of a non-negative random variable. It is known that the population standard deviation is 20% larger than the population mean. 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. (A) 944 (B) 1299 (C) 1559 (D) 1844 (E) 2213 Exam C: Fall 2006-11 - GO ON TO NEXT PAGE

12. You are given: (i) The distribution of the number of claims per policy during a one-year period for 10,000 insurance policies is: Number of Claims per Policy Number of Policies 0 5000 1 5000 2 or more 0 (ii) You fit a binomial model with parameters m and q using the method of maximum likelihood. Determine the maximum value of the loglikelihood function when m = 2. (A) 10,397 (B) 7,781 (C) 7,750 (D) 6,931 (E) 6,730 Exam C: Fall 2006-12 - GO ON TO NEXT PAGE

13. You are given: (i) Over a three-year period, the following claim experience was observed for two insureds who own delivery vans: Year Insured 1 2 3 A Number of Vehicles 2 2 1 Number of Claims 1 1 0 B Number of Vehicles N/A 3 2 Number of Claims N/A 2 3 (ii) The number of claims for each insured each year follows a Poisson distribution. Determine the semiparametric empirical Bayes estimate of the claim frequency per vehicle for Insured A in Year 4. (A) Less than 0.55 (B) At least 0.55, but less than 0.60 (C) At least 0.60, but less than 0.65 (D) At least 0.65, but less than 0.70 (E) At least 0.70 Exam C: Fall 2006-13 - GO ON TO NEXT PAGE

14. For the data set you are given: (i) k = 4 (ii) s 2 = 1 (iii) r 4 = 1 200 300 100 400 X (iv) The Nelson-Åalen Estimate H ˆ (410) > 2.15 Determine X. (A) 100 (B) 200 (C) 300 (D) 400 (E) 500 Exam C: Fall 2006-14 - GO ON TO NEXT PAGE

15. You are given: (i) A hospital liability policy has experienced the following numbers of claims over a 10-year period: 10 2 4 0 6 2 4 5 4 2 (ii) (iii) Numbers of claims are independent from year to year. You use the method of maximum likelihood to fit a Poisson model. Determine the estimated coefficient of variation of the estimator of the Poisson parameter. (A) 0.10 (B) 0.16 (C) 0.22 (D) 0.26 (E) 1.00 Exam C: Fall 2006-15 - GO ON TO NEXT PAGE

16. You are given: (i) Claim sizes follow an exponential distribution with mean θ. (ii) For 80% of the policies, θ = 8. (iii) For 20% of the policies, θ = 2. A randomly selected policy had one claim in Year 1 of size 5. Calculate the Bayesian expected claim size for this policy in Year 2. (A) Less than 5.8 (B) At least 5.8, but less than 6.2 (C) At least 6.2, but less than 6.6 (D) At least 6.6, but less than 7.0 (E) At least 7.0 Exam C: Fall 2006-16 - GO ON TO NEXT PAGE

17. For a double-decrement study, you are given: (i) The following survival data for individuals affected by both decrements (1) and (2): j c j ( T ) q j 0 0 0.100 1 20 0.182 2 40 0.600 3 60 1.000 (ii) ( 2) q = 0.05 for all j j (iii) Group A consists of 1000 individuals observed at age 0. (iv) Group A is affected by only decrement (1). Determine the Kaplan-Meier multiple-decrement estimate of the number of individuals in Group A that survive to be at least 40 years old. (A) 343 (B) 664 (C) 736 (D) 816 (E) 861 Exam C: Fall 2006-17 - GO ON TO NEXT PAGE

18. You are given: (i) (ii) (iii) At time 4 hours, there are 5 working light bulbs. The 5 bulbs are observed for p more hours. Three light bulbs burn out at times 5, 9, and 13 hours, while the remaining light bulbs are still working at time 4 + p hours. (iv) The distribution of failure times is uniform on ( 0,ω ). (v) The maximum likelihood estimate of ω is 29. Determine p. (A) Less than 10 (B) At least 10, but less than 12 (C) At least 12, but less than 14 (D) At least 14, but less than 16 (E) At least 16 Exam C: Fall 2006-18 - GO ON TO NEXT PAGE

19. You are given: (i) (ii) The number of claims incurred in a month by any insured follows a Poisson distribution with mean λ. The claim frequencies of different insureds are independent. (iii) The prior distribution of λ is Weibull with θ = 0.1 and τ = 2. (iv) Some values of the gamma function are Γ ( 0.5) = 1.77245, Γ ( 1) = 1, Γ ( 1.5) = 0.88623, Γ ( 2) = 1 (v) Month Number of Insureds Number of Claims 1 100 10 2 150 11 3 250 14 Determine the Bühlmann-Straub credibility estimate of the number of claims in the next 12 months for 300 insureds. (A) Less than 255 (B) At least 255, but less than 275 (C) At least 275, but less than 295 (D) At least 295, but less than 315 (E) At least 315 Exam C: Fall 2006-19 - GO ON TO NEXT PAGE

20. You are given: (i) The following data set: 2500 2500 2500 3617 3662 4517 5000 5000 6010 6932 7500 7500 (ii) (iii) ^ H 1 (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. ^ 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, 5000 and 7500 in (i) reflect right-censored observations and that the remaining observed values are uncensored. Calculate H ^ (7000) H ^ (7000). 1 2 (A) Less than 0.1 (B) At least 0.1, but less than 0.3 (C) At least 0.3, but less than 0.5 (D) At least 0.5, but less than 0.7 (E) At least 0.7 Exam C: Fall 2006-20 - GO ON TO NEXT PAGE

21. For a warranty product you are given: (i) Paid losses follow the lognormal distribution with μ = 13.294 and σ = 0.494. (ii) The ratio of estimated unpaid losses to paid losses, y, is modeled by where 0.851 0.747x y = 0.801x e x = 2006 contract purchase year The inversion method is used to simulate four paid losses with the following four uniform (0,1) random numbers: 0.2877 0.1210 0.8238 0.6179 Using the simulated values, calculate the empirical estimate of the average unpaid losses for purchase year 2005. (A) Less than 300,000 (B) At least 300,000, but less than 400,000 (C) At least 400,000, but less than 500,000 (D) At least 500,000, but less than 600,000 (E) At least 600,000 Exam C: Fall 2006-21 - GO ON TO NEXT PAGE

22. 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. (A) (B) (C) (D) (E) I II III IV V Exam C: Fall 2006-22 - GO ON TO NEXT PAGE

23. You are given: (i) The annual number of claims for an individual risk follows a Poisson distribution with mean λ. (ii) For 75% of the risks, λ = 1. (iii) For 25% of the risks, λ = 3. A randomly selected risk had r claims in Year 1. The Bayesian estimate of this risk s expected number of claims in Year 2 is 2.98. Determine the Bühlmann credibility estimate of the expected number of claims for this risk in Year 2. (A) Less than 1.9 (B) At least 1.9, but less than 2.3 (C) At least 2.3, but less than 2.7 (D) At least 2.7, but less than 3.1 (E) At least 3.1 Exam C: Fall 2006-23 - GO ON TO NEXT PAGE

24. 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, determine the kernel density estimate of the probability of survival to age 40. (A) 0.377 (B) 0.400 (C) 0.417 (D) 0.439 (E) 0.485 Exam C: Fall 2006-24 - GO ON TO NEXT PAGE

25. The following is a natural cubic spline passing through the points (0, 3), (1, 2), (3, 6): f ( x) 3 ( ) ( 1 ) 3 2 x+ 2 x x 3, 0 1 = 3 2 3 ( )( ) ( 1 2 + 2 x 1 4)( x 1 ), 1 x 3 Using the method of extrapolation as given in the Loss Models text, determine f ( 4. ) (A) 7.0 (B) 8.0 (C) 8.8 (D) 9.0 (E) 10.0 Exam C: Fall 2006-25 - GO ON TO NEXT PAGE

26. The random variables X1, X2, K, Xn are independent and identically distributed with probability density function x / θ e f( x) =, x 0 θ Determine 2 E X. (A) (B) (C) (D) (E) n + 1 2 θ n n + 1 2 θ 2 n 2 θ n 2 θ n 2 θ Exam C: Fall 2006-26 - GO ON TO NEXT PAGE

27. 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, calculate the estimated variance of the hypothetical means. (A) Less than 0.40 (B) At least 0.40, but less than 0.60 (C) At least 0.60, but less than 0.80 (D) At least 0.80, but less than 1.00 (E) At least 1.00 Exam C: Fall 2006-27 - GO ON TO NEXT PAGE

28. You are given: (i) (ii) (iii) A Cox proportional hazards model was used to compare the fuel economies of traditional and hybrid cars. A single covariate z was used with z = 0 for a traditional car and z = 1 for a hybrid car. The following are sample values of miles per gallon for the two types of car: Traditional: 22 25 28 33 39 Hybrid: 27 31 35 42 45 (iv) The partial maximum likelihood estimate of the coefficient β is 1. Calculate the estimate of the baseline cumulative hazard function ( ) H 0 32 using an analog of the Nelson-Åalen estimator which is appropriate for proportional hazard models. (A) Less than 0.7 (B) At least 0.7, but less than 0.9 (C) At least 0.9, but less than 1.1 (D) At least 1.1, but less than 1.3 (E) At least 1.3, but less than 1.5 Exam C: Fall 2006-28 - GO ON TO NEXT PAGE

29. 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. The prior distribution of q has probability density function π ( q) = 6 q(1 q), 0< q< 1. (iii) Two claims are made in a given year. Determine the mode of the posterior distribution of q. (A) 0.17 (B) 0.33 (C) 0.50 (D) 0.67 (E) 0.83 Exam C: Fall 2006-29 - GO ON TO NEXT PAGE

30. A company has determined that the limited fluctuation full credibility standard is 2000 claims if: (i) The total number of claims is to be within 3% of the true value with probability p. (ii) The number of claims follows a Poisson distribution. The standard is changed so that the total cost of claims is to be within 5% of the true value with probability p, where claim severity has probability density function: 1 f ( x ) =, 0 x 10, 000 10,000 Using limited fluctuation credibility, determine the expected number of claims necessary to obtain full credibility under the new standard. (A) 720 (B) 960 (C) 2160 (D) 2667 (E) 2880 Exam C: Fall 2006-30 - GO ON TO NEXT PAGE

31. For a mortality study with right censored data, you are given the following: Time Number of Deaths Number at Risk 3 1 50 5 3 49 6 5 k 10 7 21 You are also told that the Nelson-Åalen estimate of the survival function at time 10 is 0.575. Determine k. (A) 28 (B) 31 (C) 36 (D) 44 (E) 46 Exam C: Fall 2006-31 - GO ON TO NEXT PAGE

32. 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) 522 (B) 696 (C) 757 (D) 947 (E) 1042 Exam C: Fall 2006-32 - GO ON TO NEXT PAGE

33. For a group of policies, you are given: (i) Losses follow the distribution function F( x) = 1 θ / x, θ < x<. (ii) A sample of 20 losses resulted in the following: Interval Number of Losses x 10 9 10 < x 25 6 x > 25 5 Calculate the maximum likelihood estimate of θ. (A) 5.00 (B) 5.50 (C) 5.75 (D) 6.00 (E) 6.25 Exam C: Fall 2006-33 - GO ON TO NEXT PAGE

34. 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. ( ) (A) 0.019 (B) 0.025 (C) 0.032 (D) 0.039 (E) 0.045 Exam C: Fall 2006-34 - GO ON TO NEXT PAGE

35. You are given: (i) A random sample of payments from a portfolio of policies resulted in the following: Interval Number of Policies (0, 50] 36 (50, 150] x (150, 250] y (250, 500] 84 (500, 1000] 80 (1000, ) 0 Total n (ii) Two values of the ogive constructed from the data in (i) are: F ( 90) = 0.21, and ( ) n F 210 = 0.51 n Calculate x. (A) 120 (B) 145 (C) 170 (D) 195 (E) 220 **END OF EXAMINATION** Exam C: Fall 2006-35 - STOP

FALL 2006 EXAM C SOLUTIONS Question #1 Key: E With n + 1 = 16, we need the 0.3(16) = 4.8 and 0.65(16) = 10.4 smallest observations. They are 0.2(280) + 0.8(350) = 336 and 0.6(450) + 0.4(490) = 466. The equations to solve are: γ 2 γ 2 θ θ 0.3 = 1 and 0.65 1 γ γ = γ γ θ + 336 θ + 466 1/2 γ 1/2 γ (0.7) = 1 + (336 / θ ) and (0.35) = 1 + (466 / θ ) -1/2 γ (0.7) 1 (336 / θ ) = 1/2 γ (0.35) 1 (466 / θ ) γ 0.282814 = (336 / 466) ln(0.282814) = γ ln(336 / 466) γ = 3.8614. Question #2 Key: D Let E be the even of having 1 claim in the first four years. In four years, the total number of claims is Poisson(4λ). 1 Pr( E TypeI) Pr( TypeI) e (0.05) 0.01839 Pr( Type I E) = = = = 0.14427 Pr( E) Pr( E) Pr( E) 2 e (2)(0.2) 0.05413 Pr( Type II E) = = = 0.42465 Pr( E) Pr( E) 4 e (4)(0.75) 0.05495 Pr( Type III E) = = = 0.43108 Pr( E) Pr( E) Note : Pr( E) = 0.01839 +.05413 +.05495 = 0.12747 The Bayesian estimate of the number of claims in Year 5 is: 0.14427(0.25) + 0.42465(0.5) + 0.43108(1) = 0.67947.

Question #3 Key: B The sample mean is 0.2(400) + 0.7(800) + 0.1(1600) = 800. 2 2 2 The sample variance is 0.2(400 800) + 0.7(800 800) + 0.1(1600 800) = 96, 000. The sample third central moment is 3 3 3 0.2(400 800) + 0.7(800 800) + 0.1(1600 800) = 38, 400, 000. The skewness coefficient is 1.5 38,400,000 / 96,000 = 1.29. Question #4 Key: C Because 0.656 < 0.7654 < 0.773, the simulated number of losses is 4. To simulate a loss by inversion, use F( x) = 1 e 1 u = e ln(1 u) = ( x / θ ) x = θ ( ln(1 u)) 1 2 3 4 τ ( x / θ ) τ ( x / θ ) u = 0.2738, x u = 0.5152, x u = 0.7537, x u = 0.6481, x 1 2 3 4 = u τ 1/ τ = 200( ln(1 u)) = 113.12 = 170.18 = 236.75 = 204.39 1/ 2 With a deductible of 150, the first loss produces no payments and 113.12 toward the 500 limit. The second loss produces a payment of 20.18 and the insured is now out-of-pocket 263.12. The third loss produces a payment of 86.75 and the insured is out 413.12. The deductible on the fourth loss is then 86.88 for a payment of 204.29 86.88 = 117.51. The total paid by the insurer is 20.18 + 86.75 + 117.51 = 224.44.

Question #5 Key: A 2 / x The density function is f ( x) = θ x e θ and the likelihood function is 2 θ/186 2 θ /91 2 θ / 66 θ / 60 7 L( θ) = θ(186 ) e θ(91 ) e θ(66 ) e ( e ) 3 0.148184θ θ e l( θ) = ln L( θ) = 3ln( θ) 0.148184θ 1 l ( θ) = 3θ 0.148184 = 0 θ = 3/ 0.148184 = 20.25. The mode is θ / 2 = 20.25 / 2 = 10.125. Question #6 Key: D We have μ( θ) = 4 θ and μ = 4E( θ) = 4(600) = 2400. The average loss for Years 1 and 2 is 1650 and so 1800 = Z(1650) + (1 Z)(2400) which gives Z = 0.8. Because there were two years, Z = 0.8 = 2 /(2 + k) which gives k = 0.5. For three years, the revised value is Z = 3/(3+ 0.5) = 6/7 and the revised credibility estimate (using the new sample mean of 2021), (6 / 7)(2021) + (1/ 7)(2400) = 2075.14. Question #7 Key: B The uncensored observations are 4 and 8 (values beyond 11 are not needed). The two r values are 10 and 5 and the two s values are 2 and 1. The Kaplan-Meier estimate is S ˆ(11) = (8/10)(4 / 5) = 0.64 and Greenwood s estimate is 2 2 1 (0.64) + = 0.03072. 10(8) 5(4)

Question #8 Key: C There are two ways to approach this problem. One is LaGrange s formula: (3 4)(3 5) (3 2)(3 5) (3 2)(3 4) f (3) = 25 + 20 + 30 = 18.33. (2 4)(2 5) (4 2)(4 5) (5 2)(5 4) 2 Or, if the equation is f ( x) = a+ bx+ cx then three equations must be satisfied: 25 = a+ 2b+ 4c 20 = a+ 4b+ 16c 30 = a+ 5b+ 25c The solutions is a = 63.3333, b = -27.5, and c = 4.1667. The answer is 63.3333 27.5(3) + 4.1667(9) = 18.33. Question #9 Key: E S Q~ bin( m, Q ) and Q~ beta (1,99). Then m 1 E( Sm) = E[ E( Sm Q)] = E( mq) = m = 0.01m. For the mean to be at least 50, m must be at 1+ 99 least 5,000. Question #10 Key: D The posterior distribution is 4 50λ λ 90 λ 7 2 λ 2 3 λ λ e 17 150λ πλ ( data) ( e ) ( λe ) ( λe ) ( λe ) = λ e which is a gamma distribution λ with parameters 18 and 1/150. For one risk, the estimated value is the mean, 18/150. For 100 risks it is 100(18)/150 = 12. Alternatively, The prior distribution is gamma with α = 4 and β = 50. The posterior will be continue to be gamma, with α / = α + no. of claims = 4 + 14 = 18 and β / = β + no. of exposures = 50 + 100 = 150. Mean of the posterior = α / β = 18/150 = 0.12. Expected number of claims for the portfolio = 0.12 (100) = 12.

Question #11 Key: E 0.95 = Pr(0.95μ < X < 1.05 μ) 2 2 ~ ( μσ, / = 1.44 μ / ) X N n n 0.95μ μ 1.05μ μ 0.95 = Pr < Z < 1.2 μ/ n 1.2 μ/ n 0.95 = Pr( 0.05 n/1.2 < Z < 0.05 n/1.2) 0.05 n /1.2 = 1.96 n = 2212.76. Question #12 Key: B 5000 5000 2 2 Lq ( ) = (1 q) q(1 q) = 2 q (1 q) 0 1 lq ( ) = 5000 ln(2) + 5000 ln( q) + 15000 ln(1 q) l q = q q = 1 1 ( ) 5000 15000(1 ) 0 2 5000 5000 15000 qˆ = 0.25 l(0.25) = 5000 ln(2) + 5000 ln(0.25) + 15000 ln(0.75) = 7780.97. Question #13 Key: C The estimate of the overall mean, μ, is the sample mean, per vehicle, which is 7/10 = 0.7. With the Poisson assumption, this is also the estimate of v = EPV. The means for the two insureds are 2/5 = 0.4 and 5/5 = 1.0. The estimate of a is the usual non-parametric estimate, 2 2 5(0.4 0.7) + 5(1.0 0.7) (2 1)(0.7) VHM = aˆ = = 0.04 1 10 (25 + 25) 10 (The above formula: Loss Models page 596, Herzog page 116, Dean page 25) Then, k = 0.7/0.04 = 17.5 and so Z = 5/(5+17.5) = 2/9. The estimate for insured A is (2/9)(0.4) + (7/9)(0.7) = 0.6333.

Question #14 Key: A Item (i) indicates that X must one of the four given values. Item (ii) indicates that X cannot be 200 Item (iii) indicates that X cannot be 400. First assume X = 100. Then the values of r are 5, 3, 2, and 1 and the values of s are 2, 1, 1, and 2 1 1 1 1. Then H ˆ (410) = + + + = 2.23 and thus the answer is 100. As a check, if X = 300, the r 5 3 2 1 1 1 2 1 values are 5, 4, 3, and 1 and the s values are 1, 1, 2, and 1. Then, H ˆ (410) = + + + = 2.12. 5 4 3 1 Question #15 Key: B The estimator of the Poisson parameter is the sample mean. Then, E( ˆ λ) = E( X) = λ Var( ˆ λ) = Var( X ) = λ/ n cv.. = λ / n/ λ = 1/ nλ It is estimated by 1/ nλ = 1/ 39 = 0.1601. Question #16 Key: E Pr( X1 = 5 θ = 8) Pr( θ = 8) Pr ( θ = 8 X1 = 5) = Pr( X1 = 5 θ = 8) Pr( θ = 8) + Pr( X1 = 5 θ = 2) Pr( θ = 2) 5(0.125) 0.125 e (0.8) = = 0.867035. 5(0.125) 5(0.5) 0.125 e (0.8) + 0.5 e (0.2) Then, EX ( X= 5) = E( θ X= 5) = 0.867035(8) + 0.132965(2) = 7.202. 2 1 1 Question #17 Key: D ( T) ( T) ( T) (1) 1 q 1 q q 0.05 = 1 = 1 = (2) ( T ) (1) (2) We have q = 1 (1 q )(1 q ) and so q 1 q 1 0.05 0.95 (1) (1) q = 0.05/ 0.95 = 0.05263, q = 0.132 / 0.95 = 0.1389, and 20 0 20 20. Then, (1) 40 p 0 = 0.9474(0.8611) = 0.8158. Out of 1000 at age 0, 816 are expected to survive to age 40.

Question #18 Key: D 2 111 ω 4 p 2 ωωω ω ( ω 4 p) L( ω) = = 5 5 ω 4 ( ω 4) ω l( ω) = 2ln( ω 4 p) 5ln( ω 4) 2 5 l ( ω) = = 0 ω 4 p ω 4 2 5 0 = l (29) = 25 p 25 p = 15. The denominator in the likelihood function is S(4) to the power of five to reflect the fact that it is known that each observation is greater than 4. Question #19 Key: B μ( λ) = v( λ) = λ μ = v= E( λ) = 0.1 Γ (1 + 1/ 2) = 0.088623 VHM = 2 2 a = Var( λ) = (0.1) Γ (1 + 2 / 2) 0.088623 = 0.002146 500 Z = = 0.92371. 500 + 0.088623/ 0.002146 The estimate for one insured for one month is 0.92371(35/ 500) + 0.07629(0.088623) = 0.07142. For 300 insureds for 12 months it is (300)(12)(0.07142) = 257.11. Question #20 Key: D With no censoring the r values are 12, 9, 8, 7, 6, 4, and 3 and the s values are 3, 1, 1, 1, 2, 1, 1 (the two values at 7500 are not needed). Then, 3 1 1 1 2 1 1 H ˆ 1 (7000) = + + + + + + = 1.5456. 12 9 8 7 6 4 3 With censoring, there are only five uncensored values with r values of 9, 8, 7, 4, and 3 and all five s values are 1. Then, 1 1 1 1 1 H ˆ 2 (7000) = + + + + = 0.9623. The absolute difference is 0.5833. 9 8 7 4 3

Question #21 Key: A 1 The simulated paid loss is exp[0.494 Φ ( u) + 13.294] where Φ 1 ( u) is the inverse of the standard normal distribution function. The four simulated paid losses are 450,161, 330,041, 939,798, and 688,451 for an average of 602,113. The multiplier for unpaid losses is 0.851 0.747(2006 2005) 0.801(2006 2005) e = 0.3795 and the answer is 0.3795(602,113) = 228,502 Question #22 Key: A The deduction to get the SBC is ( r/ 2)ln( n) = ( r/ 2)ln(260) = 2.78r where r is the number of parameters. The SBC values are then -416.78, -417.56, -419.34, -420.12, and -425.68. The largest value is the first one, so model I is to be selected. Question #23 Key: E Pr( X1 = r λ = 1)Pr( λ = 1) Pr ( λ = 1 X1 = r) = Pr( X1 = r λ = 1)Pr( λ = 1) + Pr( X1 = r λ = 3) Pr( λ = 3) 1 e (0.75) r! 0.2759 = =. 1 3 r r e e 3 0.2759 + 0.1245(3 ) (0.75) + (0.25) r! r! Then, r 0.2759 0.1245(3 ) 2.98 = (1) + (3) r r 0.2759 + 0.1245(3 ) 0.2759 + 0.1245(3 ) r 0.2759 + 0.3735(3 ) = r. 0.2759 + 0.1245(3 ) Rearrange to obtain r r 0.82218 + 0.037103(3 ) = 0.2759 + 0.03735(3 ) r 0.54628 = 0.00025(3 ) r = 7. Because the risks are Poisson, (μ = EPV, a = VHM): μ = v= E( λ) = 0.75(1) + 0.25(3) = 1.5 a= Var( λ) = 0.75(1) + 0.25(9) 2.25 = 0.75 1 Z = = 1/3 1+ 1.5/ 0.75 and the estimate is (1/3)(7) + (2/3)(1.5) = 3.33.

Question #24 Key: E The uniform kernel spreads the probability of 0.1 to 10 units each side of an observation. So the observation at 25 contributes a density of 0.005 from 15 to 35, contributing nothing to survival past age 40. The same applies to the point at 30. For the next 7 points: 35 contributes probability from 25 to 45 for 5(0.005) = 0.025 above age 40. 35 contributes probability from 25 to 45 for 5(0.005) = 0.025 above age 40. 37 contributes probability from 27 to 47 for 7(0.005) = 0.035 above age 40. 39 contributes probability from 29 to 49 for 9(0.005) = 0.045 above age 40. 45 contributes probability from 35 to 55 for 15(0.005) = 0.075 above age 40. 47 contributes probability from 37 to 57 for 17(0.005) = 0.085 above age 40. 49 contributes probability from 39 to 59 for 19(0.005) = 0.095 above age 40. The observation at 55 contributes all 0.1 of probability. The total is 0.485. Question #25 Key: D f (3) = 2 + (3/ 2)(4) (1/ 4)(8) = 6 f (3) = (3/ 2)(2)(2) (1/ 4)(3)(4) = 3 f(4) = f(3) + (4 3) f (3) = 6+ 1(3) = 9. Question #26 Key: A X ~ Exp( θ ) n i= 1 X i ( ) ~ Γ( n, θ ) X ~ Γ( n, θ / n) ( θ / ) ( )( 1) ( 1) θ /. 2 2 2 E X = n n n+ = n+ n The second line follows because an exponential distribution is a gamma distribution with α = 1 and the sum of independent gamma random variables is gamma with the α parameters added. The third line follows because the gamma distribution is a scale distribution. Multiplying by 1/n retains the gamma distribution with the θ parameter multiplied by 1/n.

Question #27 Key: C The sample means are 3, 5, and 4 and the overall mean is 4. Then, 1+ 0+ 0+ 1+ 0+ 0+ 1+ 1+ 1+ 1+ 1+ 1 8 vˆ = = 3(4 1) 9 2 2 2 (3 4) + (5 4) + (4 4) 8 / 9 7 aˆ = = = 0.78. 3 1 4 9 Question #28 Key: C The ordered values are: 22t, 25t, 27h, 28t, 31h, 33t, 35h, 39t, 42h, and 45h where t is a traditional car and h is a hybrid car. The s values are all 1 because there are no duplicate values. The c values are 1 for 1 traditional cars and e for hybrid cars. Then 1 1 1 1 1 Hˆ 0 (32) = + + + + = 1 1 1 1 1 1.0358. 5+ 5e 4+ 5e 3+ 5e 3+ 4e 2+ 4e Question #29 Key: C 2 2 3 3 π ( q 2) = 6 q (1 q) 6 q(1 q) q (1 q) The mode can be determined by setting the derivative equal to zero. 2 3 3 2 π ( q 2) 3 q (1 q) 3 q (1 q) = 0 (1 q) q= 0 q = 0.5. Question #30 Key: B For the severity distribution the mean is 5,000 and the variance is 10,000 2 /12. For credibility based on accuracy with regard to the number of claims, 2 z 2 2000 =, z = 1.8 0.03 where z is the appropriate value from the standard normal distribution. For credibility based on accuracy with regard to the total cost of claims, the number of claims needed is 2 2 z 10000 /12 2 1+ 960. 2 = 0.05 5000

Question #31 Key: C Hˆ ˆ(10) = e (10) = 0.575 1 3 5 7 S Hˆ (10) = ln(0.575) = 0.5534 = + + +. 50 49 k 12 The solution is k = 36. Question #32 Key: A The annual dental charges are simulated from x /1000 u = 1 e x = 1000ln(1 u). The four simulated values are 356.67, 2525.73, 1203.97, and 83.38. The reimbursements are 205.34 (80% of 256.67), 1000 (the maximum), 883.18 (80% of 1103.97), and 0. The total is 2088.52 and the average is 522.13. Question #33 Key: B 9 6 5 θ θ θ θ L( θ ) = 1 (10 θ) θ 10 10 25 25 l( θ) = 9ln(10 θ) + 11ln( θ) 9 11 l ( θ ) = + = 0 10 θ θ 11(10 θ) = 9θ 110 = 20θ θ = 110 / 20 = 5.5. 9 11

Question #34 Key: A The maximum likelihood estimate is ˆ θ = x = 1000. The quantity to be estimated is 2 S( θ ) = exp( 1500 / θ ) and S ( θ ) = 1500θ exp( 1500 / θ). For the delta method, 2 Var[ S( ˆ θ)] [ S ( ˆ θ)] Var( ˆ θ) 2 2 2 = [1500(1000) exp( 1500/1000)] (1000 / 6) = 0.01867. 2 This is based on Var( ˆ θ) = Var( X ) = Var( X )/ n = θ / n. Question #35 Key: A Based on the information given 36 0.4x 0.21 = + n n 36 x 0.6y 0.51 = + + n n n n= 200 + x+ y. Then, 0.21(200 + x+ y) = 36 + 0.4x 0.51(200 + x + y) = 36 + x+ 0.6y and these linear equations can be solved for x = 119.37.

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS NOVEMBER 2006 SOA EXAM C/CAS 4 SOLUTIONS 1. The sample is of size 15. We assign smoothed percentiles to the sample points: "*& Þ Þ Þ #)! $&! Þ Þ Þ %&! %*! Þ Þ Þ " % & "! "" "' ÞÞÞ "' "' ÞÞÞ "' "' ÞÞÞ Þ!'#& Þ Þ Þ Þ#&! Þ$! Þ$"#& Þ Þ Þ Þ'#& Þ'&! Þ')(& Þ Þ Þ "' % %Þ) & "! "!Þ% "" Since 4.8 is 80% of the way from 4 to 5, the smoothed empirical estimate of the 30th percentile is 80% of the way from 280 to 350, which is 336. Since 10.5 is 40% of the way from 10 to 11, the smoothed empirical estimate of the 65th percentile is 40% of the way from 450 to 490, which is 466. # " The cdf of the Burr distribution with α œ# is JÐBÑœ"Ò"ÐBÎ) Ñ Ó #. Applying the percentile matching method, we use the estimated percentiles in the cdf to get # # " " J Ð$$'Ñ œ " Ò"Ð$$'Î) Ñ Ó œ Þ$! J Ð%''Ñ œ " Ò"Ð%''Î) Ñ Ó # and # œ Þ'&. After a little algebraic juggling, these equations become # " # " Ð$$'Î) Ñ œ È " œ Þ"*&##* and Ð%''Î) Ñ œ " œ Þ'*!$!* Þ( È. Þ$& %'' # Dividing the second equation by the first results in ( $$' ) œ $Þ&$&*, and 68 $Þ&$&* # œ %'' œ $Þ)'. Answer: E 68 Ð Ñ $$' 2. Denote by R the number of claims in one year, and A is the annual claim frequency. The prior distribution of A is T ÐA œ!þ#&ñ œ Þ!& ß T ÐA œ!þ&!ñ œ Þ#! ß and T ÐA œ "!!ÞÑ œ Þ(&. The Bayesian estimate is % % IÒR l R œ "Ó œ IÒR la œ!þ#&ó T ÐA œ!þ#& l R œ "Ñ & 3 & 3 3œ" 3œ" % % IÒR la œ!þ&!ó T ÐA œ!þ&! l R œ "Ñ IÒR la œ "Þ!!Ó T ÐA œ "Þ!! l R œ "Ñ & 3 & 3 3œ" 3œ" % % % A 3 A 3 A 3. 3œ" 3œ" 3œ" œ ÐÞ#&ÑT Ð œ Þ#& l R œ "Ñ ÐÞ#ÑT Ð œ Þ& l R œ "Ñ ÐÞ(&ÑT Ð œ " l R œ "Ñ We find the posterior probabilities as follows, and note that if R is Poisson with mean -, then % R is Poisson with mean %-. 3œ" 3 Given: Given: T ÐA œ!þ#&ñ œ Þ!& T ÐA œ!þ&!ñ œ Þ#! T ÐA œ "!!ÞÑ œ Þ(& % % % TÐR œ "la œ Þ#&Ñ TÐR œ "la œ Þ&Ñ TÐR œ "la œ "Ñ 3 3 3 3œ" 3œ" 3œ" " # % œ/ œ#/ œ%/ Ì Ì Ì % % T ÐR œ " A œ Þ#&Ñ T ÐR œ " A œ Þ&Ñ 3 3 3œ" 3œ" " # % œþ!&/ œþ%/ œ$/ Ì S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS % % % TÐR œ"ñœtðr œ" A œþ#&ñtðr œ" A œþ&ñ 3 3 3 3œ" 3œ" 3œ" % 3 " # % 3œ" T Ð R œ " A œ "Ñ œ Þ!&/ Þ%/ $/. % T ÐA œ Þ#& l Þ!&/ R3 œ " Ñ œ œ Þ"%%#*& ß Þ!&/ " Þ%/ # $/ % 3œ" % 3 3œ" % 3 3œ" Þ%/ T ÐA œ Þ& l R œ " Ñ œ œ Þ%#%''& ß Þ!&/ " Þ%/ # $/ % $/ T ÐA œ " l R œ " Ñ œ œ Þ%$"!%" Þ!&/ Þ%/ $/ # " Ì % " # %. The Bayesian estimate is ÐÞ#&ÑÐÞ"%%#*&Ñ ÐÞ&ÑÐÞ%#%''&Ñ Ð"ÑÐÞ%$"!%"Ñ œ Þ'(*. Answer: D IÒÐ\ Ñ Ó 3. The skewness coefficient is. ÐZ +<Ò\ÓÑ$Î# #Ð%!!Ñ(Ð)!!Ñ"'!! The empirical estimate of. is \ œ "! œ )!!. The empirical estimate of Z+<Ò\Óis " # " # # # "! ÒDÐ\ 3 \Ñ Ó œ "! Ò#Ð%!! )!!Ñ (Ð)!! )!!Ñ Ð"'!! )!!Ñ Ó œ *'ß!!!. The empirical estimate of IÒÐ\. Ñ $ Ó is " " "! ÒDÐ\ 3 \Ñ $ Ó œ "! ÒDÐ\ 3 )!!Ñ $ Ó " $ $ $ œ "! Ò#Ð%!! )!!Ñ (Ð)!! )!!Ñ Ð"'!! )!!Ñ Ó œ $)ß %!!. $)ß%!!ß!!! The estimated skewness coefficient is œ "Þ#*". Answer: B Ð*'ß!!!Ñ $Î#. $ 4. To simulate the number of claims from uniform number?, we find 8 so that JRÐ8 "Ñ Ÿ? JR Ð8Ñ. From the given cdf for R, and? œ ('&%, we have JRÐ$Ñ œ Þ'&' Ÿ Þ('&% Þ(($ œ JR Ð%Ñ. The simulated number of claims is R œ %. The cdf of the Weibull distribution is ÐBÎ) Ñ J\ ÐBÑœ"/ 7 ÐBÎ#!!Ñ B Î%!ß!!! œ"/ # œ"/ #. Given uniform Ð!ß "Ñ number?, the simulated value of B is the solution of the equation # B Î%!ß!!! "Î#? œ " /, or equivalently, B œ Ò %!ß!!! 68Ð"?ÑÓ. The simulated values of the loss amounts are: from? œ Þ#($), the simulated B is ""$Þ"$, claim amount is 0 after deductible of 150 ; from? œ Þ&"&#, the simulated B is "(!Þ"), claim amount is 20.18 after deductible of 150; from? œ Þ(&$(, the simulated B is #$'Þ(&, claim amount is 86.75 after deductible of 150; from? œ Þ'%)", the simulated B is #!%Þ$*, claim amount is 117.52 after deductible of 86.87. The deductible is 50 on the 4th claim to bring the total policyholder out of pocket to the maximum of 500 (113.13 out-of-pocket for first claim, 150 out-of-pocket for each of the next 2 claims and 86.87 out-of-pocket for the 4th claim to bring the total out-of-pocket to 500). Insurer's aggregate payment is #!Þ") )'Þ(& ""(Þ&# œ ##%Þ%& Þ Answer: C S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS 5. From the table of distributions, for the inverse exponential distribution with parameter ), the ) )/ mode is #. The pdf of the inverse exponential is 0ÐBÑ œ ÎB ) B#, and the log of the pdf is ) " " 68 0ÐBÑ œ 68 ) B # 68 B, and the derivative of the log of the pdf is ) B. The log of the cdf is 68 J ÐBÑ œ ) B and the derivative of the log of the cdf is. ".) 68 J ÐBÑ œ B. The likelihood function is P œ 0Ð")'Ñ 0Ð*"Ñ 0Ð''Ñ ÒJ Ð'!ÑÓ (, the loglikelihood is 68 P œ 68 0Ð")'Ñ 68 0Ð*"Ñ 68 0Ð''Ñ ( 68 J Ð'!Ñ. The derivative of the loglikelihood is. " " " " " " ".) 68P œ Ð ) ")' ÑÐ ) *" ÑÐ ) '' Ñ(Ð '! Ñ.. Setting.) 68P œ! and solving for ) results in the mle s) œ #!Þ#&. #!Þ#& The mle of the mode is # œ "!Þ". Answer: A 6. The annual loss will be \. We are told that the conditional distribution of \ given ) is gamma with α œ% and ). The hypothetical mean is IÒ\l) Óœα) œ% ), # # and the process variance is Z+<Ò\l) Óœα) œ% ). The expected hypothetical mean is. œ IÒ IÒ\l) Ó Ó œ IÒ% ) Ó œ %IÒ) Ó œ %Ð'!!Ñ œ #%!! (since we are told that the prior distribution of ) has mean 600). The Buhlmann credibility premium based on Years 1, 2 and 3 is ^\ Ð" ^Ñ., "%!!"*!!#('$ $ where \ œ $ œ #!#", and. œ #%!!, and ^ œ @ $. + # # @œexpected process variance œiò% ) Óœ%IÒ) Ó, and + œ variance of hypothetical mean œ Z +<Ò% ) Ó œ "'Z +<Ò) Ó. From the Buhlmann credibility premium based on Years 1 and 2, we have "%!!"*!! ")!! œ ^\ Ð" ^Ñ., where \ œ # œ "'&!,. œ #%!! (as above), # and ^ œ. Therefore, ")!! œ "'&!^ #%!!Ð" ^Ñ œ #%!! (&!^. # @ + Solving for ^results in ^œþ)œ #, so that @ @ # + œþ&. + Then, going back to the premium based on Years 1, 2 and 3, we have ^œ $ œ $ œ ' @ $ $Þ& (, + ' ' and the Buhlmann credibility premium is Ð ( ÑÐ#!#"Ñ Ð" ( ÑÐ#%!!Ñ œ #!(&. Answer: D 7. The loss Ÿ 11 that is nearest to 11 is 8, so WÐ""Ñ s œ WÐ)Ñ s. We assume that there is no truncation since there is no indication that there is any truncation. The first payment amount is C" œ%, and there are < " œ"! at risk and = " œ# losses of that amount. The second payment amount is C# œ ), and there are < # œ & at risk (only count losses above the last censoring point), and there is = # œ " loss of that amount. The product-limit estimate is WÐ)Ñ s # " "' œ Ð" ÑÐ" Ñ œ WÐ)Ñ s "! & #&. The Greenwood approximation to the variance of is ÒWÐ)ÑÓ s # = " = # "' # # " Ò < Ð< = Ñ < Ð< = Ñ Ó œ Ð #& Ñ Ò "!Ð"!#Ñ &Ð&"Ñ Ó œ Þ!$!(#. " " " # # # Answer: B S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS 8. The interpolating polynomial can be found by the Lagrange method. With points ÐB! ß C! Ñ ß ÐB" ß C" Ñ ß ÐB# ß C# Ñ, the 2nd degree interpolating polynomial is ÐBB" ÑÐBB# Ñ ÐBB! ÑÐBB" Ñ ÐBB! ÑÐBB" Ñ TÐBÑ œ C! ÐB B ÑÐB B Ñ C " ÐB B ÑÐB B Ñ C #! "! # "! " # ÐB# B! ÑÐB# B" Ñ. ÐB%ÑÐB&Ñ ÐB#ÑÐB&Ñ ÐB#ÑÐB%Ñ This is Ð#&Ñ Ð#%ÑÐ#&Ñ Ð#!Ñ Ð%#ÑÐ%&Ñ Ð$!Ñ Ð&#ÑÐ&%Ñ Þ Ð$%ÑÐ$&Ñ Ð$#ÑÐ$&Ñ Ð$#ÑÐ$%Ñ && Then T Ð$Ñ œ Ð#&Ñ Ð#%ÑÐ#&Ñ Ð#!Ñ Ð%#ÑÐ%&Ñ Ð$!Ñ Ð&#ÑÐ&%Ñ œ $. Answer: C 9. The Bernoulli distribution is a 0,1 distribution, with TÐ\ œ "Ñ œ ; and TÐ\ œ!ñ œ ";. W7 has a binomial distribution with parameters 7 and ;. Using the double expectation rule, we have IÒW7Ó œ IÒ IÒW7lUÓ Ó. We know that IÒW7lU œ ;Ó œ 7;, since W7 has a binomial distribution with parameters 7 and ;. Then, IÒW7ÓœIÒ7UÓœ7IÒUÓ. The pdf of Uis > Ð"**Ñ "" **" **x *) *) > Ð"Ñ > Ð**Ñ ; Ð";Ñ œ!x *)x Ð";Ñ œ**ð";ñ. The expected value of ; is ' " *) ' " *)!; **Ð" ;Ñ.; œ **! Ò" Ð" ;ÑÓÐ" ;Ñ.; œ **Ò' " *) Ð" ;Ñ.; ' " ** " " "!! Ð" ;Ñ.;Ó œ **Ò ** "!! Ó œ "!!. 7 Then, IÒW7Ó œ 7IÒUÓ œ "!!. In order for this to be at least 50, we must have 7 &!!!. Answer: E 10. The prior distribution of - is gamma with α œ% and ) œþ!#. The model distribution is Poisson with a mean of -. There are a total of 8 œ "!! observations ( B" ß B# ß ÞÞÞß B"!! ) in Year 1, and the total number of claims is DB3 œ (Ð"Ñ #Ð#Ñ "Ð$Ñ œ "%. The gamma-prior/poisson-model combination results in a posterior distribution which is also w gamma, with updated parameters, α œ αdb3 œ %"% œ ") and ) w ) Þ!# Þ!# œ 8" ) œ "!!ÐÞ!#Ñ" œ $. The Bayesian premium for one risk in Year 2 is IÒ\ "!" lb" ß B# ß ÞÞÞß B"!! Ó œ '! IÒ\ "!" l-ó 1Ð-lB" ß B# ß ÞÞÞß B"!! Ñ.- œ '! - 1Ð-lB" ß B# ß ÞÞÞß B"!! Ñ.- œ mean of posterior distribution œ α) w w Þ!# œ Ð")ÑÐ $ Ñ œ Þ"#. The expected number of claims for the portfolio of 100 risks is "!! Þ"# œ "#. Alternatively, once we know α and ) from the prior gamma distribution of -, and 8 and DB 3, for the gamma/poisson combination, the predictive distribution of \ 8" lb" ß ÞÞÞß B8 is negative ) binomial with <œαd B3 and " œ 8" ), ) so that IÒ\ 8" lb" ß ÞÞÞß B8Ó œ < " œ Ðα DB3ÑÐ 8" ) Ñ, as above. Answer: D S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS 11. We want T Òl\ IÐ\Ñl Þ!&IÐ\ÑÓ Þ*&. According to the central limit theorem, \ has a distribution which is approximately normal. Z+<Ð\Ñ Also, IÐ\Ñ œ IÐ\Ñ and Z +<Ð\Ñ œ 8, where 8 is the number of items in the sample. T Òl\ IÐ\Ñl Þ!&IÐ\ÑÓ can be "standardized" to be written in the form \IÐ\Ñ Þ!&IÐ\Ñ Þ!&IÐ\Ñ TÒl È l Ó œ T Òl^l Ó ^ Z+<Ð\Ñ ÈZ+<Ð\Ñ È, where is standard normal. Z+<Ð\ÑÎ8 Þ!&IÐ\Ñ In order for this probability to be at least.95, we must have È "Þ*'. Z+<Ð\ÑÎ8 Þ!& È8 We are given that ÈZ +<Ð\Ñ œ "Þ#IÐ\Ñ, so that the inequality becomes "Þ# "Þ*', which becomes 8 ##"$. Answer: E 12. When 7 is given, the maximum likelihood estimate of ; is found from 7; œ \. \â\ " "!ß!!! There are 10,000 policies, so \œ "!ß!!!, where each \ has a binomial distribution with 7 œ #. From the given data set, D\ 3 œ &!!!, so \ œ Þ&. \ Since 7œ#, we get the mle for ; to be s;œ # œþ#&. The likelihood function for a discrete random variable is the product of the probability function values at all the sample points. For the binomial with 7œ# and ;œþ#&, we # have 0Ð!Ñ œ T Ð\ œ!ñ œ ÐÞ(&Ñ œ Þ&'#&, 0Ð"Ñ œ T Ð\ œ "Ñ œ #ÐÞ#&ÑÐÞ(&Ñ œ Þ$(& # and 0Ð#Ñ œ T Ð\ œ #Ñ œ ÐÞ#&Ñ œ Þ!'#&. &!!! &!!! The likelihood function for the 10,000 data points is ÐÞ&'#&Ñ ÐÞ$(&Ñ, since there are 5000 '0's and 5000 '1's. The loglikelihood is &!!! 68ÐÞ&'#&Ñ &!!! 68ÐÞ$(&Ñ œ (ß ()". Answer: B 13. There are 7 œ 5 Type A vehicles observed with ]" ] # ] $ ] % ] & œ #. The estimate for the number of claims for a Type A vehicle in Year 4 is ^] Ð" ^Ñ., # & where ]œ & œþ%, ^œ & @, and. œið\ñ. +. is estimate as \ based on the entire data set of Type A and Type B, which is ""#$ ##"$# œþ(. # From the data for Type A, we have hypothetical mean IÒ\lEÓ œ & œ Þ% and for Type B we & have hypothetical IÒ\lFÓ œ & œ "Þ Since \ is Poisson for each Type, we also have Z +<Ò\lEÓ œ Þ% and Z +<Ò\lFÓ œ ", so that the expected process variance is @ œ Þ(, the same as.. We do not have individual data values for all of the vehicles, but we do have \ E œ Þ% based on 7E œ & observations and \ F œ " based on 7F œ & observations. We use the nonparametric < " estimate of +, s+ œ Ò # < 73Ð\ 3 \Ñ @Ð<"ÑÓ s. In this case < œ # insureds. " 7 7# 3œ" 7 3œ" 3 " s+ œ œ " Ò &ÐÞ% Þ(Ñ &Ð" Þ(Ñ Þ(Ð# "Ñ Ó œ Þ!% "! Ð& # & # Ñ "! # #. & Then, ^ œ œ Þ#### The semiparametric empirical Bayes estimate of the claim frequency & Þ( Þ!% for a vehicle of Type A in Year 4 is ÐÞ####ÑÐÞ%Ñ ÐÞ((()ÑÐÞ(Ñ œ Þ'$$. Answer: C S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS 14. 5œ% means that there are 4 distinct sample values but there are & data points, so \ must be "!!ß #!!ß $!! or %!!. LÐ%"!Ñ s œ LÐ%!!Ñ s since there are no deaths after time 400. Since < % œ ", there is only 1 at risk and 1 death at time 400, so = % œ ". Exactly one of = " and = $ is 1 and the other is 2. By trial and error: - if = " œ " and = $ œ #, then \ œ $!! and we have LÐ%!!Ñ s " " # " œ & % $ " œ #Þ"# #Þ"&,so this is not correct. Therefore, \ œ "!!, and = " œ #, s $ œ ", and LÐ%!!Ñ s # " " œ & $ # " œ #Þ#$. Answer: A 15. The mle of the Poisson parameter is \. The mean of the estimator is Z+<Ð\Ñ - IÐ\Ñ œ IÐ\Ñ œ -, and the variance of the estimator is Z +<Ð\Ñ œ 8 œ "!. È Z+<Ð\Ñ È-Î"! " The coefficient of variation of the estimator is œ œ È. IÐ\Ñ - "!- " The mle of - is \œ$þ*, so the estimated coefficient of variation is œþ"'. È"!Ð$Þ*Ñ Answer: B 16. IÒ\ # l\ " œ &Ó œ IÒ\ # l) œ )Ó T Ð) œ )l\" œ &Ñ IÒ\ # l) œ #Ó T Ð) œ #l\" œ &Ñ œ Ð)Ñ T Ð) œ )l\ œ &Ñ Ð#Ñ T Ð) œ #l\ œ &Ñ " ". We find the posterior probabilities TÐ) œ)l\ " œ&ñ and TÐ) œ#l\ " œ&ñ. " The pdf of the exponential distribution with mean ) is. ) /BÎ) TÐ) œ )Ñ œ Þ) TÐ) œ #Ñ œ Þ# " &Î) " &Î# TÐ\ " œ&l) œ)ñœ ) / TÐ\ " œ&l) œ#ñœ # / Ì Ì " &Î) " &Î# T Ð\ " œ & ) œ )Ñ œ Ð ) / ÑÐÞ)Ñ T Ð\ " œ & ) œ #Ñ œ Ð # / ÑÐÞ#Ñ Þ'#& #Þ& œþ"/ œþ"/ Ì TÐ\ " œ&ñœtð\ " œ& ) œ)ñtð\ " œ& ) œ#ñ Þ'#& #Þ& œ Þ"/ Þ"/ Ì T Ð) œ) \ œ&ñ Þ'#& " Þ"/ TÐ) œ)l\ " œ&ñœ T Ð\ œ&ñ œ Þ"/ Þ'#& Þ"/ #Þ& œþ)'(, " T Ð) œ# \ œ&ñ #Þ& " Þ"/ and TÐ) œ#l\ " œ&ñœ T Ð\ œ&ñ œ Þ"/ Þ'#& Þ"/ #Þ& œþ"$$. IÒ\ # l\ " œ &Ó œ Ð)ÑÐÞ)'(Ñ Ð#ÑÐÞ"$$Ñ œ (Þ#. " Note that we have used the notation TÐ\ " œ #Ñ, but since \ is exponential, it is actually a density and not a probability. Answer: E S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS ÐX Ñ 4 17. ; is the probability that a person at age - departs due to some decrement by time -. 4 4" ÐX Ñ wð"ñ wð#ñ 4 4 4 For a two decrement model ; œ " Ð" ; ÑÐ" ; Ñ. Using these equations, we get ÐX Ñ wð"ñ wð#ñ wð"ñ wð"ñ Þ" œ ;! œ " Ð" ;! ÑÐ" ;! Ñ œ " Ð" ;! ÑÐ" Þ!&Ñ p " ;! œ Þ*%($')ß ÐX Ñ wð"ñ wð"ñ Þ")# œ ;" œ " Ð" ;" ÑÐ" Þ!&Ñ p " ;" œ Þ)'"!&$. Since Group A is affected only by decrement 1, the survival probability to age 40 for Group A is wð"ñ wð"ñ Ð" ;! ÑÐ" ;" Ñ œ Þ)"&(, and the expected number of survivors to age 40 from 1000 Group A individuals observed at 0 is "!!!ÐÞ)"&(Ñ œ )"'. Answer: D 18. To formulate the likelihood function we must first note that since observation of the bulbs begins at 4 hours, all data is conditional given that X%. There are 3 known burnout times, so these would be included as conditional density in the likelihood function, and the 2 bulbs that survive : more hours would be included as conditional survival probabilities. The likelihood function is P œ 0Ð&lX %Ñ 0Ð*lX %Ñ 0Ð"$lX %Ñ ÒT ÐX % :lx %ÑÓ #. " = % For the uniform distribution on Ð!ß = Ñ, 0Ð>Ñ œ =, T ÐX %Ñ œ =, "Î= " = %: so 0Ð>lX%Ñœ Ð= %ÑÎ= œ = %, and TÐX%:lX%Ñœ = %. # " $ = %: # Ð= %:Ñ The likelihood function is PœÐ= % Ñ Ð = % Ñ œ Ð= %Ñ&. # & Then, 68Pœ#68Ð= %:Ñ&68Ð= %Ñ, and. = 68Pœ = %: = %.. The mle of = occurs where.= 68 P œ!, so substituting = œ #* results in # & #*%: #*% œ!, from which we get : œ "&. Answer: D 19. This Buhlmann-Straub situation can be considered an ordinary Buhlmann model with 8 œ &!! observations. We will find ^\ Ð" ^Ñ., which will be the Buhlmann credibility estimate of the expected number of claims for one insured for one month. The expected number of claims for 300 insureds for 12 months will be $!! "# times as large. "!"""% \œ &!! œþ!( for the given data. \ is the number of claims for one insured for one month. The hypothetical mean is IÒ\l-Ó œ - and the process variance is Z +<Ò\l-Ó œ -. The first and second moments of the Weibull distribution are " IÒ- Ó œ )> Ð" 7 Ñ œ ÐÞ"Ñ > Ð"Þ&Ñ œ Þ!))'#$ # # # # and IÒ- Óœ) > Ð" 7 ÑœÐÞ"Ñ > Ð#ÑœÞ!". Then,. œ expected hypothetical mean œ IÒ IÒ\l-Ó Ó œ IÒ-Ó œ Þ!))'#$ ß @ œ expected process variance œ IÒ Z +<Ò\l-Ó Ó œ IÒ-Ó œ Þ!))'#$, and +œvariance of hypothetical mean œz+<òiò\l-óó œ Z +<Ò-Ó œ IÒ- # Ó ÐIÒ-ÓÑ # œ Þ!" ÐÞ!))'#$Ñ # œ Þ!!#"%&*'%. &!! &!! Then, ^ œ @ &!! œ Þ!))'#$ œ Þ*#$(!', + &!! Þ!!#"%&'*% and the credibility estimate for the number of claims for one month for the insured is ÐÞ*#$(!'ÑÐÞ!(Ñ ÐÞ!('#*%ÑÐÞ!))'#$Ñ œ Þ!("%#". The estimate for the expected number of claims for the next year for 300 insureds is $!! "# ÐÞ!("%#"Ñ œ #&(. Answer: B S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS 20. Ls $ " " " # " " " Ð(!!!Ñ œ "# * ) ( ' % $ œ "Þ&%&'$&. Ls " " " " " # Ð(!!!Ñ œ * ) ( % $ œ Þ*&#$!#. Note that since the three observations at 2500 and the two at 6000 become censored, we lose $ "# and # from, so we lose a total of $ # ' Ls " Ð(!!!Ñ "# ' œ Þ&)$. The censoring at 7500 is irrelevant since the estimate at 7000 is based on deaths up to 7000 but not at 7500. Answer: D 68 > "$Þ#*% 21. The cdf of the lognormal distribution is JÐ>Ñ œ FÐ!Þ%*% ÑÞ The four simulated losses are: 68 > "$Þ#*% 68 > "$Þ#*%? œ Þ#)(( œ FÐ Þ&'Ñ œ FÐ!Þ%*% Ñ p Þ&' œ!þ%*% p > œ %&!ß "'", 68 > "$Þ#*% 68 > "$Þ#*%? œ Þ"#"! œ FÐ "Þ"(Ñ œ FÐ!Þ%*% Ñ p "Þ"( œ!þ%*% p > œ $$$ß!%", 68 > "$Þ#*% 68 > "$Þ#*%? œ!þ)#$) œ FÐÞ*$Ñ œ FÐ!Þ%*% Ñ p Þ*$ œ!þ%*% p > œ *$*ß (*), 68 > "$Þ#*% 68 > "$Þ#*%? œ!þ'"(* œ FÐÞ$Ñ œ FÐ!Þ%*% Ñ p Þ$ œ!þ%*% p > œ '))ß %&". Þ(%( For losses based on contract year 2005, the ratio C is Þ)!"/ œ Þ$(*&. The estimate of the average unpaid losses for 2005 is %&!ß"'"$$$ß!%"*$*ß(*)'))ß%&" ÐÞ$(*&ÑÐ % Ñ œ ##)ß ()'. Answer: A < 22. According to the Schwarz Bayesian criterion, we compare j # 688 for each estimated model, where j is the maximized loglikelihood, < is the number of parameters estimated and 8 is < the number of data points. The model favored is the one with the largest value of j # 688. We get the following values " # Model I: %"% # 68 #'! œ %"'Þ(), Model II: %"# # 68 #'! œ %"(Þ&', $ % Model III: %"" # 68 #'! œ %"*Þ$%, Model IV: %!* # 68 #'! œ %#!Þ"#, ' Model V: %!* # 68 #'! œ %#&Þ'). Model I has the largest value. Answer: A 23. Year 1 gives us one observed value of \ (annual number of claims) of <. The Bayesian estimate for the second year is IÒ\ # l\ " œ<óœiò\ # l- œ"ó TÐ- œ"l\ " œ<ñiò\ # l- œ$ó TÐ- œ$l\ " œ<ñ œ Ð"Ñ T Ð- œ "l\" œ <Ñ Ð$Ñ T Ð- œ $l\ " œ <Ñ œ #Þ*). If we denote TÐ- œ"l\ œ<ñœ-, then TÐ- œ$l\ œ<ñœ"-, and then " " - " -$Ð"-Ñœ#Þ*) gives us -œtð œ"l\ œ<ñœþ!". We can formulate TÐ\ " œ <Ñas T Ð\ " œ <Ñ œ T Ð\ " œ <l- œ "Ñ T Ð- œ "Ñ T Ð\ " œ <l- œ $Ñ T Ð- œ $Ñ " $ < / / $ œ Ð <x Ñ ÐÞ(&Ñ Ð <x Ñ ÐÞ#&Ñ. Then T Ð-œ" \ " œ<ñ T Ð\ " œ<l-œ"ñ T Ð-œ"Ñ TÐ- œ"l\ " œ<ñœ TÐ\ œ<ñ œ " TÐ\ " œ<ñ " Ð/ Î<xÑÐÞ(&Ñ $ œ œ œ Þ!" / " / $ $ < Ð <x Ñ ÐÞ(&ÑÐ <x Ñ ÐÞ#&Ñ $/ # $ <. < It follows that $ œ #"*%Þ&&, and < œ (. S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS The Buhlmann credibility estimate is ^\ Ð" ^Ñ.. " \œ<œ( and 8œ" observed value, ^œ. " @ + Since \l- is Poisson with mean -, we have that the hypothetical mean is IÒ\l-Ó œ - and the process variance is Z+<Ò\l-Óœ-. Then,. œ expected hypothetical mean œ IÒ-Ó œ Ð"ÑÐÞ(&Ñ Ð$ÑÐÞ#&Ñ œ "Þ&, @ œ expected process variance œ IÒ-Ó œ Ð"ÑÐÞ(&Ñ Ð$ÑÐÞ#&Ñ œ "Þ&, and # + œ variance of hypothetical mean œ Ð$ "Ñ ÐÞ(&ÑÐÞ#&Ñ œ Þ(&. " " " # Then, ^ œ "Þ& œ, and the Buhlmann credibility estimate is Ð ÑÐ(Ñ Ð ÑÐ"Þ&Ñ œ $Þ$$. " $ $ $ Þ(& Answer: E 24. There are 9 distinct observed values, The empirical probabilities are :ÐCÑ œ Þ" for all C- values except :Ð$&Ñ œ Þ#. We wish to estimate T ÐX %!Ñ œ " J Ð%!Ñ. With bandwidth,œ"!, the interval for the uniform kernel at observed value Cis from C"! to * C "!. The kernel density estimate of J Ð%!Ñ is Js Ð%!Ñ œ :ÐC Ñ O Ð%!Ñ. 3œ" 3 C 3 For any C3 that is Ÿ $!, the interval around that C3 is totally to the left of B œ %!, so JC3 Ð%!Ñœ" for that C3; this applies to Cœ#& and $!. For any C 3 &!, the interval around that C3 is totally to the right of Bœ%!, so JC Ð%!Ñœ! for that C3; this applies to Cœ&&. 3 For any C3 for which C3 "! Ÿ %! Ÿ C3 "!, JC3Ð%!Ñ œ Ò%! ÐC3 "!ÑÓÐÞ!&Ñ " " (this is the are of the rectangle whose base is from C3 "! to 40 with height #, œ #! œ Þ!&ÑÞ The kernel density estimate of J Ð%!Ñ is Js Ð%!Ñ œ ÐÞ"ÑÐ"Ñ ÐÞ"ÑÐ"Ñ ÐÞ#ÑÒ%! Ð$& "!ÑÓÐÞ!&Ñ ÐÞ"ÑÒ%! Ð$( "!Ñ %! Ð$* "!Ñ %! Ð%& "!Ñ %! Ð%( "!Ñ %! Ð%* "!ÑÓÐÞ!&Ñ! œ Þ&"&. The estimate of WÐ%!Ñ is " Þ&"& œ Þ%)&. Answer: E 25. The method of extrapolation for a natural cubic spline uses a straight line extrapolation from the right endpoint of the data set. The line has the same slope as the spline at the right w $ # w endpoint. 0ÐBÑœ$ÐB"ÑÐ% ÑÐB"Ñ for "ŸBŸ$, so 0Ð$Ñœ$. 0Ð$Ñ œ ', so the equation of the extrapolation line starting at B œ $ is ' $ÐB $Ñ œ $B $. The extrapolated value of 0Ð%Ñ is $Ð%Ñ $ œ *. Answer: D 26. \ has an exponential distribution with mean ), so Z+<Ò\Ó œ ) #. Z+<Ò\Ó For a sample of size 8 and any distribution, it is always true that Z+<Ò\Ó œ 8, ) and IÒ\Ó œ IÒ\Ó. Therefore, in this case, Z +<Ò\Ó œ # 8 and IÒ\Ó œ ). From the definition of variance, we know that # # # # Z +<Ò\Ó œ IÒ\ Ó ÐIÒ\ÓÑ, so that IÒ\ Ó œ Z +<Ò\Ó ÐIÒ\ÓÑ. # ) It follows that IÒ\ Ó œ # # 8" # 8 ) œ Ð 8 Ñ) for the exponential distribution. Answer: A S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS 27. There are <œ$ policyholders, and 8œ% observations for each policyholder. We use the equal sample size version of nonparametric empirical Bayes estimation. \" œ $ß \# œ &ß \ $ œ %ß and \ œ %. % " s@ " œ # " # # # # # %" Ð\ "4 \ " Ñ œ $ ÒÐ#$Ñ Ð$$Ñ Ð$$Ñ Ð%$Ñ Óœ $. 4œ" # % s@@@ " s# s $ ) # $, and $ $, so that $ *. $ " # s @ " # # # )Î* $" 3 % # %. 3œ" s@ œ s@ œ s@ œ œ s+ œ Ð\ \Ñ œ ÒÐ$ %Ñ Ð& %Ñ Ð% %Ñ Ó œ Þ(() Answer: C 28. The numbers at risk for each covariate class at each data point Ÿ$# are: ## #& #( #) $" D œ!: & % $ $ # D œ ": & & & % % There is a single observation at each data point. " " " " " The estimate of L! Ð$#Ñ is &&/" %&/" $&/ " $%/ " #%/". Substituting " œ " into this expression gives a value of 1.036. Answer: C 29. The prior distribution is a beta distribution (with ) œ" ), with +œ# and,œ#. The Bayesian combination of a beta prior distribution for ; with parameters + and,, and a binomial model distribution for \ with parameters 7 and ;, and with 8 observed values B" ß ÞÞÞß B 8 results in a posterior distribution for ; which is also beta, with parameters w w + œ +DB 3 and, œ,78db3 The given beta prior has +œ",,œ", and the binomial model distribution has 7œ%, and there is 8œ" (one year) observation with Bœ#. The posterior distribution is beta with w w + œ##œ% and, œ#ð%ñð"ñ#œ%. The pdf of the posterior distribution is > Ð%%Ñ $ $ > Ð%Ñ > Ð%Ñ ; Ð";Ñ,!;". The mode of the posterior is point at which the posterior pdf is maximized. This occurs where log pdf is maximized. Log pdf of posterior is 68 "%! $68 ; 68Ð" ;Ñ, and the derivative is $ $ ; ";. Setting the derivative to 0 and solving for ; results in ; œ Þ&. This could have been anticipated from the form of the posterior cdf which is symmetric on the interval Ð!ß "Ñ. Answer: C 30. R (number of claims) is Poisson and we are told that the standard for full credibility for R based on total number of claims (not total number of exposures of R) is 2000. This standard is D: Ð Þ!$ Ñ # Z+<ÒRÓ IÒR Ó œ #!!!. Since R is Poisson, we have Z +<ÒRÓ œ IÒRÓ, so that D: # ÐÞ!$ Ñ œ #!!!. Note that D: is the standard normal value such that T Ð D: ^ D: Ñ œ :. Note also that in the first part of this problem we are applying limited fluctuation credulity to the claim number random variable R. In the second part of the problem, we are applying credibility to the compound aggregate claims distribution W (with frequency R and severity \ that is uniform from 0 to 10,000). The standard applied to W has "closeness" parameter 5% and the same probability :. S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS The standard is based on expected number of claims (not expected exposures of W), and that D: # Z+<ÒWÓ # standard is ÐÞ!& Ñ ÐIÒWÓÑ# IÒRÓ. Since R is Poisson, Z +<ÒWÓ œ IÒRÓ IÒ\ Ó, and using IÒWÓ œ IÒRÓ IÒ\Ó, the standard becomes D # : # IÒ\ Ó D: # # ÐÞ!& Ñ ÐIÒ\ÓÑ#. From ÐÞ!$ Ñ œ #!!! we get D : œ "Þ), and from the uniform distribution for # # "!ß!!! \ we get IÒ\Ó œ &!!!, IÒ\ Ó œ $. The expected number of claims needed for full D # # : # IÒ\ Ó "Þ) "!ß!!! Î$ credibility of W under this standard is ÐÞ!& Ñ ÐIÒ\ÓÑ# œ ÐÞ!&Ñ# Ð&!!!Ñ# œ *'!. Answer: B LÐ"!Ñ s 31. The Nelson-Aalen estimate of WÐ"!Ñ is /, where LÐ"!Ñ s is the Nelson-Aalen estimate LÐ"!Ñ s of LÐ"!Ñ. We are given that / œ Þ&(&, so that LÐ"!Ñ s œ Þ&&$%. From the data set, we have LÐ"!Ñ s " $ & ( œ &! %* 5 #" œ Þ&&$%, from which we get 5 œ $'. Answer: C Þ!!"B 32. The cdf of the exponential distribution with mean 1000 is JÐBÑ œ "/, for B!. The simulated dental charges, and related reimbursements are: Þ!!"B? œ!þ$! œ " / p B œ $&'Þ'(, 257.67.8 œ 206 is reimbursed, Þ!!"B? œ!þ*# œ " / p B œ #&#&Þ($, 2425.73.8 œ 1941, so 1000 is reimbursed, Þ!!"B? œ!þ(! œ " / p B œ "#!$Þ*(, 1103.97.8 œ 883 is reimbursed, Þ!!"B? œ!þ!) œ " / p B œ )$Þ$), 0 is reimbursed. #!'"!!!))$! Average annual reimbursement is % œ &##. Answer: A 33. The likelihood function is the product of probabilities for each interval. ) "!) The probability for the interval B Ÿ "! is T Ð\ Ÿ "!Ñ œ J Ð"!Ñ œ " "! œ "!. ) ) The probability for the interval "! B Ÿ #& is J Ð#&Ñ J Ð"!Ñ œ "! #& œ Þ!' ). ) The probability for the interval B #& is " J Ð#&Ñ œ #& œ Þ!% ). The likelihood function is * ' & "!) * ' & * "" P œ ÒJ Ð"!ÑÓ ÒJ Ð#&Ñ J Ð"!ÑÓ Ò" J Ð#&ÑÓ œ Ð "! Ñ ÐÞ!' ) Ñ ÐÞ!% ) Ñ œ -Ð"! ) Ñ ), * ' & where - œ ÐÞ"Ñ ÐÞ!'Ñ ÐÞ!%Ñ. The log of Pis 68Pœ68- *68Ð"!) Ñ ""68), and the derivative is. * "" "".) 68 P œ "!) ). Setting this equal to 0 and solving for ) results in the mle ) œ #. Answer: B S. Broverman, 2006 www.sambroverman.com

NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS 34. For the exponential distribution with mean ), WÐ"&!!Ñ œ / "&!!Î). If s) is the mle of ), then the mle of WÐ"&!!Ñ is / "&!!Î s), and according to the delta method, the. "&!!Î ) # variance pf this estimate is Ð / Ñ Z+<Òs.) ) Ó. For the exponential distribution, the mle of ) is \ œ "!!!, and the variance is Z+<Ò\Ó ) Z +<Ò\Ó œ œ # 8 œ ' s œ "!!! s "!!! 8 8. Since and ), the estimated variance of ) is # '.. "&!!Î) "&!!Î) "&!!.) / œ / Ð )# Ñ. Applying the delta method, the variance of the estimate of # "&!!Î ) "&!! # ) WÐ"&!!Ñ is Ò/ Ð )# ÑÓ 8, which is estimated to be # "&!!Î"!!! "&!! # "!!! Ò/ Ð"!!! # ÑÓ ' œ Þ!")(. Answer: A 35. Using linearity (uniform distribution) within each interval, we have $'Þ%B J8Ð*!Ñ œ 8 œ Þ#", so that $' Þ%B œ Þ#"8. $'BÞ%C Similarly, J8Ð#"!Ñ œ 8 œ Þ&", so that $' B Þ'C œ Þ&"8. The total number of losses is 8œ$'BC)%)!œ#!!BCœ8. The first two equations become $' Þ%B œ Þ#"Ð#!! B CÑ œ %# Þ#"B Þ#"C and $' B Þ'C œ Þ&"Ð#!! B CÑ œ "!# Þ&"B Þ&"C. Solving these last two equations for B and C results in B œ "#! and C œ )!. Answer: A S. Broverman, 2006 www.sambroverman.com