Exam C, Fall 2006 PRELIMINARY ANSWER KEY


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1 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
2 **BEGINNING OF EXAMINATION** 1. You are given: (i) Losses follow a Burr distribution with α = 2. (ii) A random sample of 15 losses is: (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 GO ON TO NEXT PAGE
3 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 GO ON TO NEXT PAGE
4 3. You are given a random sample of 10 claims consisting of two claims of 400, seven claims of 800, and one claim of 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 GO ON TO NEXT PAGE
5 4. You are given: (i) The cumulative distribution for the annual number of losses for a policyholder is: n F ( n ) 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 outofpocket 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 (b) For loss amounts use the random numbers: Use the random numbers in order and only as needed. Based on the simulation, calculate the insurer s aggregate payments for this policyholder. (A) (B) (C) (D) (E) Exam C: Fall GO ON TO NEXT PAGE
6 5. You have observed the following three loss amounts: 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 GO ON TO NEXT PAGE
7 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 After the estimate in (v) was calculated, the data for Year 3 was located. The loss on the selected policy in Year 3 was 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 GO ON TO NEXT PAGE
8 7. The following is a sample of 10 payments: where + indicates that a loss exceeded the policy limit. Determine Greenwood s approximation to the variance of the productlimit estimate Ŝ (11). (A) (B) (C) (D) (E) Exam C: Fall GO ON TO NEXT PAGE
9 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 GO ON TO NEXT PAGE
10 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 GO ON TO NEXT PAGE
11 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 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 GO ON TO NEXT PAGE
12 11. You are planning a simulation to estimate the mean of a nonnegative 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 GO ON TO NEXT PAGE
13 12. You are given: (i) The distribution of the number of claims per policy during a oneyear period for 10,000 insurance policies is: Number of Claims per Policy Number of Policies 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 GO ON TO NEXT PAGE
14 13. You are given: (i) Over a threeyear period, the following claim experience was observed for two insureds who own delivery vans: Year Insured A Number of Vehicles Number of Claims 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 GO ON TO NEXT PAGE
15 14. For the data set you are given: (i) k = 4 (ii) s 2 = 1 (iii) r 4 = X (iv) The NelsonÅalen Estimate H ˆ (410) > 2.15 Determine X. (A) 100 (B) 200 (C) 300 (D) 400 (E) 500 Exam C: Fall GO ON TO NEXT PAGE
16 15. You are given: (i) A hospital liability policy has experienced the following numbers of claims over a 10year period: (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 GO ON TO NEXT PAGE
17 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 GO ON TO NEXT PAGE
18 17. For a doubledecrement study, you are given: (i) The following survival data for individuals affected by both decrements (1) and (2): j c j ( T ) q j (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 KaplanMeier multipledecrement 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 GO ON TO NEXT PAGE
19 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 GO ON TO NEXT PAGE
20 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) = , Γ ( 1) = 1, Γ ( 1.5) = , Γ ( 2) = 1 (v) Month Number of Insureds Number of Claims Determine the BühlmannStraub 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 GO ON TO NEXT PAGE
21 20. You are given: (i) The following data set: (ii) (iii) ^ H 1 (7000) is the NelsonÅalen 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 NelsonÅalen estimate of the cumulative hazard rate function calculated under the assumption that all occurrences of the values 2500, 5000 and 7500 in (i) reflect rightcensored 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 GO ON TO NEXT PAGE
22 21. For a warranty product you are given: (i) Paid losses follow the lognormal distribution with μ = and σ = (ii) The ratio of estimated unpaid losses to paid losses, y, is modeled by where x 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: Using the simulated values, calculate the empirical estimate of the average unpaid losses for purchase year (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 GO ON TO NEXT PAGE
23 22. Five models are fitted to a sample of n = 260 observations with the following results: Model Number of Parameters Loglikelihood I II III IV V Determine the model favored by the Schwarz Bayesian criterion. (A) (B) (C) (D) (E) I II III IV V Exam C: Fall GO ON TO NEXT PAGE
24 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 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 GO ON TO NEXT PAGE
25 24. You are given the following ages at time of death for 10 individuals: Using a uniform kernel with bandwidth b = 10, determine the kernel density estimate of the probability of survival to age 40. (A) (B) (C) (D) (E) Exam C: Fall GO ON TO NEXT PAGE
26 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 = ( )( ) ( 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 GO ON TO NEXT PAGE
27 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 θ n n θ 2 n 2 θ n 2 θ n 2 θ Exam C: Fall GO ON TO NEXT PAGE
28 27. Three individual policyholders have the following claim amounts over four years: Policyholder Year 1 Year 2 Year 3 Year 4 X Y Z 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 GO ON TO NEXT PAGE
29 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: Hybrid: (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 NelsonÅalen 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 GO ON TO NEXT PAGE
30 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 GO ON TO NEXT PAGE
31 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 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 GO ON TO NEXT PAGE
32 31. For a mortality study with right censored data, you are given the following: Time Number of Deaths Number at Risk k You are also told that the NelsonÅalen estimate of the survival function at time 10 is Determine k. (A) 28 (B) 31 (C) 36 (D) 44 (E) 46 Exam C: Fall GO ON TO NEXT PAGE
33 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 You are given: (i) (ii) The annual dental charges for each insured are exponentially distributed with mean Use the following uniform (0, 1) random numbers and the inversion method to generate four values of annual dental charges: Calculate the average annual reimbursement for this simulation. (A) 522 (B) 696 (C) 757 (D) 947 (E) 1042 Exam C: Fall GO ON TO NEXT PAGE
34 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 < 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 GO ON TO NEXT PAGE
35 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: Use the delta method to approximate the variance of the maximum likelihood estimator of S ( ) (A) (B) (C) (D) (E) Exam C: Fall GO ON TO NEXT PAGE
36 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 STOP
37 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 γ γ = γ γ θ θ /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 = (336 / 466) ln( ) = γ ln(336 / 466) γ = 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) Pr( Type I E) = = = = Pr( E) Pr( E) Pr( E) 2 e (2)(0.2) Pr( Type II E) = = = Pr( E) Pr( E) 4 e (4)(0.75) Pr( Type III E) = = = Pr( E) Pr( E) Note : Pr( E) = = The Bayesian estimate of the number of claims in Year 5 is: (0.25) (0.5) (1) =
38 Question #3 Key: B The sample mean is 0.2(400) + 0.7(800) + 0.1(1600) = The sample variance is 0.2( ) + 0.7( ) + 0.1( ) = 96, 000. The sample third central moment is ( ) + 0.7( ) + 0.1( ) = 38, 400, 000. The skewness coefficient is ,400,000 / 96,000 = Question #4 Key: C Because < < 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)) τ ( x / θ ) τ ( x / θ ) u = , x u = , x u = , x u = , x = u τ 1/ τ = 200( ln(1 u)) = = = = / 2 With a deductible of 150, the first loss produces no payments and toward the 500 limit. The second loss produces a payment of and the insured is now outofpocket The third loss produces a payment of and the insured is out The deductible on the fourth loss is then for a payment of = The total paid by the insurer is =
39 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 ) θ θ e l( θ) = ln L( θ) = 3ln( θ) θ 1 l ( θ) = 3θ = 0 θ = 3/ = The mode is θ / 2 = / 2 = Question #6 Key: D We have μ( θ) = 4 θ and μ = 4E( θ) = 4(600) = 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) = 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 KaplanMeier estimate is S ˆ(11) = (8/10)(4 / 5) = 0.64 and Greenwood s estimate is (0.64) + = (8) 5(4)
40 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) = = (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 = , b = 27.5, and c = The answer is (3) (9) = 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 least 5,000. Question #10 Key: D The posterior distribution is 4 50λ λ 90 λ 7 2 λ 2 3 λ λ e λ 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 = = 18 and β / = β + no. of exposures = = 150. Mean of the posterior = α / β = 18/150 = Expected number of claims for the portfolio = 0.12 (100) = 12.
41 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 = Question #12 Key: B Lq ( ) = (1 q) q(1 q) = 2 q (1 q) 0 1 lq ( ) = 5000 ln(2) ln( q) ln(1 q) l q = q q = 1 1 ( ) (1 ) qˆ = 0.25 l(0.25) = 5000 ln(2) ln(0.25) ln(0.75) = 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 nonparametric estimate, 2 2 5( ) + 5( ) (2 1)(0.7) VHM = aˆ = = ( ) 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) =
42 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 Then H ˆ (410) = = 2.23 and thus the answer is 100. As a check, if X = 300, the r values are 5, 4, 3, and 1 and the s values are 1, 1, 2, and 1. Then, H ˆ (410) = = 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 = 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) e (0.8) = = (0.125) 5(0.5) e (0.8) e (0.2) Then, EX ( X= 5) = E( θ X= 5) = (8) (2) = 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) (1) q = 0.05/ 0.95 = , q = / 0.95 = , and Then, (1) 40 p 0 = (0.8611) = Out of 1000 at age 0, 816 are expected to survive to age 40.
43 Question #18 Key: D 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 = 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) = VHM = 2 2 a = Var( λ) = (0.1) Γ (1 + 2 / 2) = Z = = / The estimate for one insured for one month is (35/ 500) ( ) = For 300 insureds for 12 months it is (300)(12)( ) = 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, H ˆ 1 (7000) = = 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, H ˆ 2 (7000) = = The absolute difference is
44 Question #21 Key: A 1 The simulated paid loss is exp[0.494 Φ ( u) ] 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 ( ) 0.801( ) e = and the answer is (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 , , , , and 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! = =. 1 3 r r e e (3 ) (0.75) + (0.25) r! r! Then, r (3 ) 2.98 = (1) + (3) r r (3 ) (3 ) r (3 ) = r (3 ) Rearrange to obtain r r (3 ) = (3 ) r = (3 ) r = 7. Because the risks are Poisson, (μ = EPV, a = VHM): μ = v= E( λ) = 0.75(1) (3) = 1.5 a= Var( λ) = 0.75(1) (9) 2.25 = Z = = 1/ / 0.75 and the estimate is (1/3)(7) + (2/3)(1.5) = 3.33.
45 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 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) = above age contributes probability from 25 to 45 for 5(0.005) = above age contributes probability from 27 to 47 for 7(0.005) = above age contributes probability from 29 to 49 for 9(0.005) = above age contributes probability from 35 to 55 for 15(0.005) = above age contributes probability from 37 to 57 for 17(0.005) = above age contributes probability from 39 to 59 for 19(0.005) = above age 40. The observation at 55 contributes all 0.1 of probability. The total is 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 / 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.
46 Question #27 Key: C The sample means are 3, 5, and 4 and the overall mean is 4. Then, vˆ = = 3(4 1) (3 4) + (5 4) + (4 4) 8 / 9 7 aˆ = = = 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 Hˆ 0 (32) = = e 4+ 5e 3+ 5e 3+ 4e 2+ 4e Question #29 Key: C 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 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 =, z = 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 / =
47 Question #31 Key: C Hˆ ˆ(10) = e (10) = S Hˆ (10) = ln(0.575) = = 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 , , , and The reimbursements are (80% of ), 1000 (the maximum), (80% of ), and 0. The total is and the average is Question #33 Key: B θ θ θ θ L( θ ) = 1 (10 θ) θ l( θ) = 9ln(10 θ) + 11ln( θ) 9 11 l ( θ ) = + = 0 10 θ θ 11(10 θ) = 9θ 110 = 20θ θ = 110 / 20 =
48 Question #34 Key: A The maximum likelihood estimate is ˆ θ = x = 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) = [1500(1000) exp( 1500/1000)] (1000 / 6) = This is based on Var( ˆ θ) = Var( X ) = Var( X )/ n = θ / n. Question #35 Key: A Based on the information given x 0.21 = + n n 36 x 0.6y 0.51 = + + n n n n= x+ y. Then, 0.21(200 + x+ y) = x 0.51(200 + x + y) = 36 + x+ 0.6y and these linear equations can be solved for x =
49 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 œ" 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,
50 NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS % % % TÐR œ"ñœtðr œ" A œþ#&ñtðr œ" A œþ&ñ œ" 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 after deductible of 150; from? œ Þ(&$(, the simulated B is #$'Þ(&, claim amount is after deductible of 150; from? œ Þ'%)", the simulated B is #!%Þ$*, claim amount is after deductible of The deductible is 50 on the 4th claim to bring the total policyholder out of pocket to the maximum of 500 ( outofpocket for first claim, 150 outofpocket for each of the next 2 claims and outofpocket for the 4th claim to bring the total outofpocket to 500). Insurer's aggregate payment is #!Þ") )'Þ(& ""(Þ&# œ ##%Þ%& Þ Answer: C S. Broverman,
51 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 ^ $. + # 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 #, # + 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 productlimit 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,
52 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 gammaprior/poissonmodel 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\ "!" ló 1ÐlB" ß B# ß ÞÞÞß B"!! Ñ. œ '!  1ÐlB" 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,
53 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 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 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,
54 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#$. Answer: A 15. The mle of the Poisson parameter is \. The mean of the estimator is Z+<Ð\Ñ  IÐ\Ñ œ IÐ\Ñ œ , and the variance of the estimator is Z +<Ð\Ñ œ 8 œ "!. È Z+<Ð\Ñ ÈÎ"! " The coefficient of variation of the estimator is œ œ È. IÐ\Ñ  "! " The mle of  is \œ$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,
55 NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS ÐX Ñ ; is the probability that a person at age  departs due to some decrement by time " ÐX Ñ wð"ñ wð#ñ 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 BuhlmannStraub 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 $!! "# times as large. "!"""% \œ &!! œþ!( for the given data. \ is the number of claims for one insured for one month. The hypothetical mean is IÒ\lÓ œ  and the process variance is Z +<Ò\lÓ œ . The first and second moments of the Weibull distribution are " IÒ Ó œ )> Ð" 7 Ñ œ ÐÞ"Ñ > Ð"Þ&Ñ œ Þ!))'#$ # # # # and IÒ Óœ) > Ð" 7 ÑœÐÞ"Ñ > Ð#ÑœÞ!". Then,. œ expected hypothetical mean œ IÒ IÒ\lÓ Ó œ IÒÓ œ Þ!))'#$ œ expected process variance œ IÒ Z +<Ò\lÓ Ó œ IÒÓ œ Þ!))'#$, and +œvariance of hypothetical mean œz+<òiò\lóó œ Z +<ÒÓ œ IÒ # Ó ÐIÒÓÑ # œ Þ!" ÐÞ!))'#$Ñ # œ Þ!!#"%&*'%. &!! &!! Then, ^ &!! 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,
56 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&)$. The censoring at 7500 is irrelevant since the estimate at 7000 is based on deaths up to 7000 but not at Answer: D 68 > "$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 Ðœ" \ " œ<ñ T Ð\ " œ<lœ"ñ T Ðœ"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,
57 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\lÓ œ  and the process variance is Z+<Ò\lÓœ. Then,. œ expected hypothetical mean œ IÒÓ œ Ð"ÑÐÞ(&Ñ Ð$ÑÐÞ#&Ñ œ œ expected process variance œ IÒÓ œ Ð"ÑÐÞ(&Ñ Ð$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,
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