Course 4 Examination Questions And Illustrative Solutions Novemer 000
1. You fit an invertile first-order moving average model to a time series. The lag-one sample autocorrelation coefficient is 0.35. Determine an initial guess for θ, the moving average parameter. (A) 0. (B) 0.4 (C) 0.6 (D) 0.8 (E) 1.0
. The following data have een collected for a large insured: Numer Of Average Claim Year Claims Size 1 100 10,000 00 1,500 Inflation increases the size of all claims y 10% per year. A Pareto distriution with parameters α distriution. = 3 and θ is used to model the claim size Estimate θ for Year 3 using the method of moments. (A),500 (B) 3,333 (C) 4,000 (D) 5,850 (E) 6,400 3
3. You are given the following for a dental insurer: (i) (ii) (iii) Claim counts for individual insureds follow a Poisson distriution. Half of the insureds are expected to have.0 claims per year. The other half of the insureds are expected to have 4.0 claims per year. A randomly selected insured has made 4 claims in each of the first two policy years. Determine the Bayesian estimate of this insured s claim count in the next (third) policy year. (A) 3. (B) 3.4 (C) 3.6 (D) 3.8 (E) 4.0 4
4. You are studying the length of time attorneys are involved in settling odily injury lawsuits. T represents the numer of months from the time an attorney is assigned such a case to the time the case is settled. Nine cases were oserved during the study period, two of which were not settled at the conclusion of the study. For those two cases, the time spent up to the conclusion of the study, 4 months and 6 months, was recorded instead. The oserved values of T for the other seven cases are as follows: 1 3 3 5 8 8 9 Estimate Pr[3 T 5 ] using the Product-Limit estimator. (A) 0.13 (B) 0. (C) 0.36 (D) 0.40 (E) 0.44 5
5. You are investigating the relationship etween per capita consumption of natural gas and the price of natural gas. You gathered data from 0 cities and constructed the following model: You have determined: Y = α + βx + ε, where Y is per capita consumption, X is the price, and ε is a normal random error term. α$ = 138. 561 β$ = 1104. X = 90, 048 i Y = 116, 058 i x = X X = 10, 668 i i y = Y Y = 0, 838 i c c d i ε$ = Y Y$ = 7, 83 i i i h h i Determine the shortest 95% confidence interval for β. (A). 1, 0. 1 g g g g g (B) 1. 9, 0. 3 (C) 1. 7, 0. 5 (D) 1. 5, 0. 7 (E) 1. 3, 0. 9 6
6. You have oserved the following claim severities: 11.0 15. 18.0 1.0 5.8 You fit the following proaility density function to the data: g F HG g I K J > > f x = 1 x 1 x x x 0 0 π exp µ,, µ Determine the maximum likelihood estimate of µ. (A) Less than 17 (B) At least 17, ut less than 18 (C) At least 18, ut less than 19 (D) At least 19, ut less than 0 (E) At least 0 7
7. The following information comes from a study of roeries of convenience stores over the course of a year: (i) X i is the numer of roeries of the i th store, with i = 1,,..., 500. (ii) X i = 50 (iii) X i = 0 (iv) The numer of roeries of a given store during the year is assumed to e Poisson distriuted with an unknown mean that varies y store. Determine the semiparametric empirical Bayes estimate of the expected numer of roeries next year of a store that reported no roeries during the studied year. (A) Less than 0.0 (B) At least 0.0, ut less than 0.04 (C) At least 0.04, ut less than 0.06 (D) At least 0.06, ut less than 0.08 (E) At least 0.08 8
8. For a study of the excess mortality of smokers, you are given: (i) (ii) (iii) 100 smokers are oserved for a period of three years. The hazard rate h t g for smokers is modeled as htg = βtgθ tg, where θtg is the g is the factor indicating excess mortality. reference hazard rate for nonsmokers, and β t The tale elow summarizes the information aout the study: Time Interval Value Of θtg Over The Interval Numer Of Smokers At The Beginning Of The Interval Numer Of Deaths Over The Interval (0,1] 0.0 100 3 (1,] 0.04 97 3 (,3] 0.06 94 4 (iv) Deaths during each interval are assumed to occur at the end of the interval. Calculate B $ 3g, the estimate of the cumulative relative excess mortality at t = 3. (A) Less than (B) At least, ut less than 3 (C) At least 3, ut less than 4 (D) At least 4, ut less than 5 (E) At least 5 9
9. You are using a three-point moving average to forecast values of a time series. The last three recorded values of the time series are as follows: y y y 98 99 100 = = 100 99 = 101 Determine y$ $ 105 y104, the difference etween the five-step-ahead and the four-step-ahead forecasted values. (A) 0. (B) 0.04 (C) 0.00 (D) 0.04 (E) 0. 10
10. You are given: (i) Sample size = 100 (ii) (iii) The negative loglikelihoods associated with five models are: Model Numer Of Parameters Negative Loglikelihood Generalized Pareto 3 19.1 Burr 3 19. Pareto 1. Lognormal 1.4 Inverse Exponential 1 4. F I HG K J. The form of the penalty function is rln n π Which of the following is the est model, using the Schwartz Bayesian Criterion? (A) (B) (C) (D) (E) Generalized Pareto Burr Pareto Lognormal Inverse Exponential 11
11. For a risk, you are given: (i) The numer of claims during a single year follows a Bernoulli distriution with mean p. (ii) The prior distriution for p is uniform on the interval [0,1]. (iii) (iv) The claims experience is oserved for a numer of years. The Bayesian premium is calculated as 1/5 ased on the oserved claims. Which of the following oserved claims data could have yielded this calculation? (A) (B) (C) (D) (E) 0 claims during 3 years 0 claims during 4 years 0 claims during 5 years 1 claim during 4 years 1 claim during 5 years 1
1. You are given the following linear regression results: t Actual Fitted 1 77.0 77.6 69.9 70.6 3 73. 70.9 4 7.7 7.7 5 66.1 67.1 Determine the estimated lag 1 serial correlation coefficient after one iteration of the Cochrane- Orcutt procedure. (A) 0.3 (B) 0. (C) 0.1 (D) 0.0 (E) 0.1 13
13. A sample of ten oservations comes from a parametric family f x, y; θ 1, θ g with loglikelihood function 10 1 g i i 1 g 1 1 1 i= 1 ln L θ, θ = ln f x, y ; θ, θ = 5. θ 3θ θ θ + 5θ + θ + k, where k is a constant. Determine the estimated covariance matrix of the maximum likelihood estimator, (A) (B) (C) (D) (E) L NM L NM L NM L NM L NM O QP O QP O QP O QP 0. 5 03. 0. 3 0. 0 30 30 50 0. 03. 03. 05. 5 3 3 3 3 5 O QP θ ˆ 1. θ ˆ 14
14. For an insurance portfolio, you are given: (i) (ii) (iii) For each individual insured, the numer of claims follows a Poisson distriution. The mean claim count varies y insured, and the distriution of mean claim counts follows a gamma distriution. For a random sample of 1000 insureds, the oserved claim counts are as follows: Numer Of Claims, n 0 1 3 4 5 Numer Of Insureds, f n 51 307 13 41 11 6 nf n = 750 n f = 1494 n (iv) Claim sizes follow a Pareto distriution with mean 1500 and variance 6,750,000. (v) (vi) Claim sizes and claim counts are independent. The full crediility standard is to e within 5% of the expected aggregate loss 95% of the time. Determine the minimum numer of insureds needed for the aggregate loss to e fully credile. (A) Less than 8300 (B) At least 8300, ut less than 8400 (C) At least 8400, ut less than 8500 (D) At least 8500, ut less than 8600 (E) At least 8600 15
15.-16. Use the following information for questions 15 and 16. Survival times are availale for four insureds, two from Class A and two from Class B. The two from Class A died at times t = 1 and t = 9. The two from Class B died at times t = and t = 4. 15. For question 15 only, you are also given: A proportional hazards model is used to model the difference in hazard rates etween Class A (Z=0) and Class B (Z=1). No other covariates are used. The (partial) maximum likelihood estimate of the parameter β is. Determine the value of Breslow s estimator of the cumulative hazard rate for Class A at time t = 3. (A) (B) (C) (D) (E) 1 7 1 1 1 e + + e + 1 1 e e + + e + 1 e e e + + e + 1 16
15.-16. (Repeated for convenience) Use the following information for questions 15 and 16. Survival times are availale for four insureds, two from Class A and two from Class B. The two from Class A died at times t = 1 and t = 9. The two from Class B died at times t = and t = 4. 16. For question 16 only, you are also given: Nonparametric Empirical Bayes estimation is used to estimate the mean survival time for each class. Uniased estimators of the expected value of the process variance and the variance of the hypothetical means are used. Estimate Z, the Bühlmann crediility factor. (A) 0 (B) /19 (C) 4/1 (D) 8/5 (E) 1 17
17. You are given the following information aout an ARMA(1,1) model: φ θ 1 1 = 03. = 0. 4 Calculate ρ. (A) 0.10 (B) 0.03 (C) 0.03 (D) 0.04 (E) 0.10 18
18. A jewelry store has otained two separate insurance policies that together provide full coverage. You are given: (i) The average ground-up loss is 11,100. (ii) Policy A has an ordinary deductile of 5,000 with no policy limit. (iii) Under policy A, the expected amount paid per loss is 6,500. (iv) Under policy A, the expected amount paid per payment is 10,000. (v) Policy B has no deductile and a policy limit of 5,000. Given that a loss has occurred, determine the proaility that the payment under policy B is 5,000. (A) Less than 0.3 (B) At least 0.3, ut less than 0.4 (C) At least 0.4, ut less than 0.5 (D) At least 0.5, ut less than 0.6 (E) At least 0.6 19
19. For a portfolio of independent risks, you are given: (i) The risks are divided into two classes, Class A and Class B. (ii) Equal numers of risks are in Class A and Class B. (iii) For each risk, the proaility of having exactly 1 claim during the year is 0% and the proaility of having 0 claims is 80%. (iv) All claims for Class A are of size. (v) All claims for Class B are of size c,an unknown ut fixed quantity. One risk is chosen at random, and the total loss for one year for that risk is oserved. You wish to estimate the expected loss for that same risk in the following year. Determine the limit of the Bühlmann crediility factor as c goes to infinity. (A) 0 (B) 1/9 (C) 4/5 (D) 8/9 (E) 1 0
0. Fifteen cancer patients were oserved from the time of diagnosis until the earlier of death or 36 months from diagnosis. Deaths occurred during the study as follows: The Nelson-Aalen estimate H ~ 35g is 1.5641. Time In Months Numer Of Since Diagnosis Deaths 15 0 3 4 30 d 34 36 1 Calculate the Aalen estimate of the variance of H ~ 35g. (A) Less than 0.10 (B) At least 0.10, ut less than 0.15 (C) At least 0.15, ut less than 0.0 (D) At least 0.0, ut less than 0.5 (E) At least 0.5 1
1. You are given the following two regression models, each ased on a different population of data: Model A: Yi = A1 + AX i + A3 X3i + ε i where i = 1,,..., 30 Model B: Yj = B1 + BX j + B3 X3j + ε j where j = 1,,..., 50 You assume that the variances of the two models are equal and pool the data into one model: Model G: Yp = G1 + GX p + G3 X3p + ε p where p = 1,,..., 80 You calculate R model and the error sum of squares, denoted as ESS model, for all three models. Which of the following is the F statistic for testing the hypothesis that Model A is identical to Model B? (A) (B) (C) (D) (E) F F F F F 3, 74 6, 77 6, 74 3, 74 6, 77 = = = = = d d ESSG ESSA ESSB ESS + ESS 74 A ESSG ESSA ESSB ESS + ESS 77 A ESSG ESSA ESSB ESS + ESS 74 A G A B R R R d R A + RB i / G A B R R R d R A + RB i / i 74 i B / 77 B B g g g 3 / 6 / / / g g g / / / 3 6 6
. You are given the following information aout a random sample: (i) The sample size equals five. (ii) The sample is from a Weiull distriution with τ =. (iii) Two of the sample oservations are known to exceed 50, and the remaining three oservations are 0, 30 and 45. Calculate the maximum likelihood estimate of θ. (A) Less than 40 (B) At least 40, ut less than 45 (C) At least 45, ut less than 50 (D) At least 50, ut less than 55 (E) At least 55 3
3. You are given: (i) The parameter Λ has an inverse gamma distriution with proaility density function: g λ λ 4 10/ λ g = 500 e, λ > 0 (ii) The size of a claim has an exponential distriution with proaility density function: c h 1 x/ λ f x Λ = λ = λ e, x > 0, λ > 0 For a single insured, two claims were oserved that totaled 50. Determine the expected value of the next claim from the same insured. (A) 5 (B) 1 (C) 15 (D) 0 (E) 5 4
4. The Product-Limit estimator was used to estimate the survival function for a set of lifetime data: t i d i Y i $ S t g σ s tg 1 1 0 0.950 0.006 3 1 19 0.900 0.0056 7 1 18 0.850 0.0088 8 1 15 0.793 0.0136 9 1 14 0.737 0.0191 11 1 1 0.675 0.067 15 1 6 0.563 0.0600 Confidence Coefficients c. 05aL, aug For 95% Equal Proaility Confidence Bands a L a U 0.08 0.10 0.1 0.40.8055.7666.7309 0.4.8178.7801.7456 0.44.895.7931.7597 0.46.8408.8055.773 0.48.8517.8174.786 0.50.863.890.7987 0.5.876.840.8108 0.54.886.8511.86 0.56.894.8618.8341 Determine the confidence interval for S (5) within the 95% equal proaility linear confidence and for S (t) over the range 3 t 15. (A) (0.6, 1.00) (B) (0.69, 1.00) (C) (0.71, 1.00) (D) (0.73, 1.00) (E) (0.77, 1.00) 5
5. You are given the following information aout an AR() model: ρ 1 = 0.5 ρ = 0.4 Determine φ 1. (A) 0.1 (B) 0. (C) 0.3 (D) 0.4 (E) 0.5 6
6. You are given a random sample of two values from a distriution function F: 1 3 g You estimate θ F X X = + X 1. = Var Xg using the estimator g X1, X = Xi X g c h i= 1, where Determine the ootstrap approximation to the mean square error. (A) 0.0 (B) 0.5 (C) 1.0 (D).0 (E).5 7
7. You are given the following information on towing losses for two classes of insureds, adults and youths: Exposures Year Adult Youth Total 1996 1997 1998 1999 Total 000 1000 1000 1000 5000 450 50 175 15 1000 450 150 1175 115 6000 Pure Premium Year Adult Youth Total 1996 1997 1998 1999 Weighted average 0 5 6 4 3 15 15 1 10.755 4.400 7.340 3.667 4.167 You are also given that the estimated variance of the hypothetical means is 17.15. Determine the nonparametric empirical Bayes crediility premium for the youth class, using the method that preserves total losses. (A) Less than 5 (B) At least 5, ut less than 6 (C) At least 6, ut less than 7 (D) At least 7, ut less than 8 (E) At least 8 8
8. Prior to oserving any claims, you elieved that claim sizes followed a Pareto distriution with parameters θ = 10 and α = 1, or 3, with each value eing equally likely. You then oserve one claim of 0 for a randomly selected risk. Determine the posterior proaility that the next claim for this risk will e greater than 30. (A) 0.06 (B) 0.11 (C) 0.15 (D) 0.19 (E) 0.5 9
9. Prior to oserving any claims, you elieved that the survival function for claim sizes followed a Dirichlet distriution with parameter function: g 4 F H G I K J α t, = 10 10 + t You then oserve one claim of 0 for a randomly selected risk. Determine the posterior proaility that the next claim for this risk will e greater than 30. (A) 0.0 (B) 0.5 (C) 0.30 (D) 0.35 (E) 0.40 30
30. You are given the following information aout an MA() model: µ θ θ 1 σ ε = = 10. 03. = 0. = 40. Determine the width of the forecast confidence interval of radius two standard deviations around a forecast of y t five time periods ahead. (A) 8.5 (B) 8.8 (C) 9.8 (D) 17.0 (E) 19.6 31
31. You are given: (i) y = β x + ε (ii) i i i Var ε i xi g F = H G I K J i x i y i 1 1 8 5 3 3 3 4 4 4 Determine the weighted least squares estimate of β. (A) 0.4 (B) 0.9 (C) 1.4 (D).0 (E).6 3
3. You are given the following for a sample of five oservations from a ivariate distriution: (i) x 1 4 5 6 y 4 3 6 4 (ii) x = 36., y = 38. A is the covariance of the empirical distriution F e as defined y these five oservations. B is the maximum possile covariance of an empirical distriution with identical marginal distriutions to F e. Determine B A. (A) 0.9 (B) 1.0 (C) 1.1 (D) 1. (E) 1.3 33
33. A car manufacturer is testing the aility of safety devices to limit damages in car accidents. You are given: (i) (ii) (iii) (iv) (v) (vi) A test car has either front air ags or side air ags (ut not oth), each type eing equally likely. The test car will e driven into either a wall or a lake, with each accident type eing equally likely. The manufacturer randomly selects 1,, 3 or 4 crash test dummies to put into a car with front air ags. The manufacturer randomly selects or 4 crash test dummies to put into a car with side air ags. Each crash test dummy in a wall-impact accident suffers damage randomly equal to either 0.5 or 1, with damage to each dummy eing independent of damage to the others. Each crash test dummy in a lake-impact accident suffers damage randomly equal to either 1 or, with damage to each dummy eing independent of damage to the others. One test car is selected at random, and a test accident produces total damage of 1. Determine the expected value of the total damage for the next test accident, given that the kind of safety device (front or side air ags) and accident type (wall or lake) remain the same. (A).44 (B).46 (C).5 (D).63 (E) 3.09 34
34. Phil and Sylvia are competitors in the light ul usiness. Sylvia advertises that her light uls urn twice as long as Phil s. You were ale to test 0 of Phil s uls and 10 of Sylvia s. You assumed that the distriution of the lifetime (in hours) of a light ul is exponential, and separately estimated Phil s parameter as θ $ P = 1000 and Sylvia s parameter as θ $ S = 1500 using maximum likelihood estimation. Determine θ *, the maximum likelihood estimate of θ P restricted y Sylvia s claim that θ = θ. S P (A) Less than 900 (B) At least 900, ut less than 950 (C) At least 950, ut less than 1000 (D) At least 1000, ut less than 1050 (E) At least 1050 35
35. You are analyzing a large set of oservations from a population. The true underlying model is: y = 01. t z + ε You fit a two-variale model to the oservations, otaining: y = 03. t + ε You are given: * t = 0 t = 16 z = 0 z = 9 Estimate the correlation coefficient etween z and t. (A) 0.7 (B) 0.6 (C) 0.5 (D) 0.4 (E) 0.3 36
36. You study the onset of a disease that includes oth left and right censoring, using Turnull s modification of the Product-Limit estimator. You are given: j Time t j Numer Left Censored c j Numer Of Events d j 4 13 c 4 7 5 14 6 15 0 7 16 8 >16 0 In the first iteration of Turnull s algorithm, you calculate: j i 4 5 6 7 1 0.066 0.048 0.038 0.035 0.0 0.145 0.116 0.107 3 0.33 0.3 0.184 0.171 4 0.409 0.93 0.33 0.16 5 0.8 0.4 0.08 6 0.05 0.190 7 0.07 p ij d$ 4 = 30. 063 Determine c 4. (A) 5 (B) 6 (C) 7 (D) 8 (E) 9 37
37. Data on 8 home sales yield the fitted model: Y$ = 439. + 038. X 00009. X + 014718. X 6. 68X 069. X 3 4 5 6 where Y = sales price of home X = taxes X 3 = size of lot X 4 = square feet of living space X 5 = numer of rooms X 6 = age in years You are given that the estimated variance-covariance matrix (lower-triangular portion) of the variales in the regression is: Y X X 3 X 4 X 5 X 6 Y 0,041.4 X 36,909.0 80,964. X 3 9,66.6 439,511.8 5,93,16.9 X 4 71,479. 19,03.9 907,497.1 300,11.4 X 5 17. 44.5 1,589.3 53.5 1.3 X 6 585.4 1,40.5 1,877.4 1,343.4 0. 190.9 $ * β j is the standardized regression coefficient associated with X j. Which of the following is correct? (A) β$ > β$ > β$ * * * 3 4 (B) β$ > β$ > β$ * * * 4 3 (C) β$ > β$ > β$ * * * 3 4 (D) β$ > β$ > β$ * * * 4 3 (E) β$ > β$ > β$ * * * 4 3 38
38. An insurance company writes a ook of usiness that contains several classes of policyholders. You are given: (i) The average claim frequency for a policyholder over the entire ook is 0.45. (ii) The variance of the hypothetical means is 0.370. (iii) The expected value of the process variance is 1.793. One class of policyholders is selected at random from the ook. Nine policyholders are selected at random from this class and are oserved to have produced a total of seven claims. Five additional policyholders are selected at random from the same class. Determine the Bühlmann crediility estimate for the total numer of claims for these five policyholders. (A).5 (B).8 (C) 3.0 (D) 3.3 (E) 3.9 39
39. You are given the following information aout a study of individual claims: (i) 0 th percentile = 18.5 (ii) 80 th percentile = 35.80 Parameters µ and σ of a lognormal distriution are estimated using percentile matching. Determine the proaility that a claim is greater than 30 using the fitted lognormal distriution. (A) 0.34 (B) 0.36 (C) 0.38 (D) 0.40 (E) 0.4 40
40. You are given two random walk models. These models are identical in every respect, except that one includes a known positive drift parameter and the other does not include a drift parameter. Which of the following statements aout these random walk models is incorrect? (A) (B) (C) (D) For the random walk without drift, all forecasted values from time T are equal. For the random walk without drift, the standard error of the forecast from time T increases as the forecast horizon increases. For the random walk with drift, the forecasted values from time T will increase linearly as the forecast horizon increases. For the random walk with drift, the standard error of a forecasted value from time T is equal to the standard error of the corresponding forecasted value for the random walk without drift. (E) For the random walk with drift, the standard error of the forecast from time T increases or decreases, depending on the drift parameter, as the forecast horizon increases. Question #1 θ Solve ρ = 035. = 1 1 + θ which yields the equation 0 35. θ θ + 0. 35 = 0. The two solutions of this quadratic equation are 0.408 and.449. To e invertile, the parameter must e less than one in asolute value, so the answer is 0.408. (B) Question # First inflate the claims. Then there are 100 averaging 10,000 (1.1) = 1,100 for a total of 1,10,000 and 00 averaging 1,500 (1.1) = 13,750 for a total of,750,000. The grand total is 3,960,000 on 300 claims for an average of 13,00. The mean of this Pareto distriution is θ. Setting this equal to 13,00 yields θ $ = 6, 400. (E) 41
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #3 From Bayes Theorem: c h c h g c h g c h g c h g Pr 4, 4 λ = Pr λ = Pr λ = 4, 4 = Pr 4, 4λ = Pr λ = + Pr 4, 4 λ = 4 Pr λ = 4 4 e / 4! 05. = 4 4 4 ce / 4! h 05. g + ce 4 / 4! h 05. g = 017578. Because E Xg = λ, the expected next claim count is the posterior expected value of λ which is 0.17578() + 0.84(4) = 3.648. (C) 4
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #4 The data reveal death times of 1, 3 ( deaths), 5, 8 ( deaths), and 9 and withdrawal times of 4 and 6. The Product-Limit estimate produces a discrete distriution with proaility at each death time. The proaility of dying at time 3 is given y (8/9)(/8) = /9. The proaility of dying at time 5 is given y (8/9)(6/8)(1/5) = /15. The total proaility of dying etween times 3 and 5 inclusive is /9 + /15 = 16/45 = 0.356. (C) Question #5 The following facts are needed: s 783 = = 43511., t18,. 05 = 101.. 18 The confidence interval is 1104 ± 101 43511... or 1104. ± 0. 44. 10, 668 (D) Question #6 L 1 Lµ g exp µ g. µ g µ g µ g. µ NM g 11 1 304. 15 1 36 18 1 4 1 1 516. 58 d 1 ln Lµ g = µ g +. µ g + µ g+ µ g +. µ g dµ 11 11 1 15. 15 1 18 18 1 1 1 1 58. 58 O QP and setting this equal to zero yields 5 0. 98633µ = 0 for µ $ = 16. 74. (A) 43
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #7 50 v$ = x = = 01. and 500 a$ = s 01. = 0 500 01. 499 g 01. = 033086.. Then $ 1 Z = = 076791.. 01. 1+ 033086. The crediility estimate is then 0. 76791 0 + 0309. 01. = 0. 0309. g g (B) Question #8 g g g g Q 1 = 0. 0 100 =. Q( ) = 004. 97 =. 38 g Q 3 = 006. 94 =. 444 Then B$ 3 g = + + = 4 89 3 3 4.. 38. 444.. (D) 44
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #9 100+ 99 + 101 y$ 101 = = 100, 3 99 + 101+ 100 y$ 10 = = 100, 3 101+ 100+ 100 y$ 103 = = 100. 33, 3 100+ 100+ 100. 33 y$ 104 = = 10011., 3 100+ 100. 33+ 10011. y$ 105 = = 10015.. 3 Then y$ $ 105 y104 = 0. 04. (D) Question #10 n The score is ln L r lnf H G I K J = ln L. 7673r. π The scores are: Generalized Pareto: 191. 3 7673. = 7. 40 g g Burr: 19. 3. 7673 = 7. 50 Pareto: 1.. 7673 = 6. 73 Lognormal: 14.. 7673 = 6. 93 Inverse exponential: 4.. 7673= 697. g g The highest value is for the Pareto. (C) 45
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #11 The posterior distriution is π c h c h g g x n x pn, x f x n, p π p p 1 p 1, where n is the numer of years and x is the numer of claims. This is a eta distriution. The Bayesian estimate is the posterior mean, which is x + 1 n+. The only comination among the five answer choices that produces 0. is x = 0 and n = 3. (A) Question #1 The residuals are 0. 6, 0. 7,. 3, 0, 1. The estimate of the lag-one serial correlation coefficient is n t= n t= ε$ ε$ t ε$ t 1 t 1 0. 4 161. + 0+ 0 =.. 036. + 0. 49 + 59. + 0 = 01938 (B) Question #13 θ 1 ln L = 5θ 1 3θ + 5, ln L = 3θ 1 θ +, ln L = 5, ln L =, ln L = 3. θ θ θ θ θ L NM O The information matrix is 5 3 3 QP and the covariance matrix is its inverse: 3 3 5 1 L NM O QP. 1 (E) 46
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #14 g g Y g g g Y g Y g g ξ = E X = E N θ = 075. 1500 = 115. σ = Var X = E N σ + Var N θ = 0. 75 6, 750, 000 + 0. 9343 1500 = 7, 160, 473. 196. 0. 05 7, 160, 473 115 n F H G I K J = 8, 694. (E) Question #15 g g g 1 1 W 1, = + e, W, = 1+ e, H$ 0 3 =. e + + e + 1 (C) Question #16 1 v$ = + + + =. 1 1 5 9 5 3 4 3 17 g g g g g g g 1 a$ = + =. <. 1 5 4 3 4 17 65 0 When this happens, the convention is to set Z $ = 0. (A) 47
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #17 ρ 1 φ1θ 1gφ1 θ 1g φ1 0 3 1 01. g 0. 3 0. = 4 g =. 1+ θ 1 φ1θ 1 1+ 016. 04. = 009.. (B) Question #18 6, 500 10, 000 =, F 5, 000 = 0. 35, Pr X 5, 000 = 0. 65. 1 F 5, 000 g g g (E) Question #19 g For class A, the mean is 0. = 04. and the variance is 0. 016. = 0. 64. For class B, the mean is 0. c and the variance is 0. c 0. 04c = 016. c. g g Then, µ = 05. 0. 4 + 05. 0. c = 0. + 01. c. g g and Also, a = 05. 04. 0. 01. c + 05. 0. c 0. 01. c = 004. 004. c + 0. 01c g c h v = 05. 0. 64 + 05. 016. c = 03. + 008. c. g 1 Then, Z = 0. 3 + 0. 08c 1+ 004. 004. c + 001. c 0. 04 0. 04c + 0. 01c = 036. 0. 04c + 0. 09c. As c goes to infinity, Z goes to 1/9. (B) 48
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #0 g = + + + + = 15641.. ~ H 35 15 3 13 d 10 8 8 d This leads to d + = 1 or d 16d + 48 = 0. 8 8 d The solution is d = 4. Then the estimated variance is 15 3 4 + + + + = 0. 341. 13 10 8 4 (D) Question #1 The F statistic, with 3 degrees of freedom in the numerator (the numer of restrictions), and 74 degrees of freedom in the denominator (the numer of degrees of freedom in the unrestricted regression), is ESSG ESSA ESSB / 3. ESS + ESS / 74 A B g g (A) Question # θ θ g g g g g g g g g θ θ / / / / L θ = f 0 f 30 f 45 1 F 50 θ e 0 θ e 30 θ e 45 e 50. This is equal to θ e 6 835/ θ Setting it equal to zero yields θ $ = 5. 68. 3 and the derivative of its logarithm is 6 / θ + 16650 / θ. (D) 49
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #3 c 1 h = 1 g x x e x 1/ λ 1 x/ λ 4 10/ λ 6 60/ λ λ, λ λ e λ e λ e. This is an inverse gamma distriution with α = 5 and θ = 60. The expected value of this random variale is 60/4 = 15. The next claim has an exponential distriution with mean λ, which itself has a mean of 15. (C) Question #4 g g g g g 0 00056. 0 006. al = = 01007., au = = 0545., c 01 0545 = 854 1 + 0 00056 1 + 0. 05.,.... 0 006. g The lower confidence limit is 0. 9. 854 00056. 0. 9 = 071.. The upper limit is 1.09, which is taken as 1 (the maximum possile value). (C) Question #5 ρ = φ + φ ρ, ρ = φ ρ + φ. 1 1 1 1 1 This yields 05. = φ + 05. φ, 0. 4 = 05. φ + φ. The solution is φ 1 = 04.. 1 1 (D) 50
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #6 The variance from the sample is θ F e There are four possile ootstrap samples: g MSE Sample g x1, x 1,1 0 1 1,3 1 3,1 1 3,3 0 1 And so the average MSE is 1. g g g = 05. 1 + 3 = 1. (C) Question #7 g g g g g g g g /,. $v = 000 0 3 + 1000 5 3 + 1000 6 3 + 1000 4 3 + 450 15 10 + 50 10 + 175 15 10 + 15 1 10 6 = 1 9 Then Z$ 5000., $ 1000 = = 087447 Z = = 05815.. 5000+ 19 / 17. 15 1000+ 19 / 17. 15 1 g g 087447. 3 + 05815. 10 Also, µ $ = 087447. + 05815. = 57976.. g g For the youth class, the estimate is 05815. 10 + 041785. 57976. = 8. 44. (E) 51
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #8 f f f c c c h h g h g 0α = 1 = 10 / 30 = 0. 011111, 3 0α = = 10 / 30 = 0. 007407, 3 4 0α = 3 = 3 10 / 30 = 0. 003704. The total of these three proailities is 0.0 and thus the posterior proailities are 1/, 1/3, and 1/6. The following tale yields the posterior expectation: α Posterior Proaility Pr X > 30α 1 1/ 10 / 40 = 0. 5 1/3 10 40 / g = 0. 065 3 1/6 10 40 3 / 0. 01565 The weighted average is 0.1484. c g = h (C) Question #9 g g g ~ α 30, + 0 4 10 / 40 S D 30 = = α 0, + 1 4+ 1 g = 0. (A) Question #30 g + + = The width of the confidence interval is 4 1 0. 3 0. 85.. (A) 5
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #31 The weighted least squares estimate is xi / σ igyi / σ ig F 8 10 9 16 F 1 4 9 = + + x / σ 05. 1. 5 4 0. 5 1. 5 i i g I HG K J I + + + HG K J = 4 16 1. 65. (E) Question #3 To maximize the covariance, sort the y values from smallest to largest. The pairs are then (1,), (,3), (4,4), (5,4), and (6,6). + 6 + 16 + 0 + 36g The expected product is = 16. 5 g The covariance is 16 36. 38. =. 3. For the original data, the covariance is The difference is 1.. g g 4 + 4 + 1 + 30+ 4 5 36. 38. = 11.. (D) 53
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #33 Let S = side airag, F = front airag, W = wall-impact, L = lake-impact, and the numer is the numer of crash test dummies. The first tale yields the posterior proaility of each configuration given that there was total damage of 1. Configuration Pro (damage = 1) Prior pro Product Posterior* SW 0.5 1/8 1/3 /7 SW4 0.00 1/8 0 0 SL 0.00 1/8 0 0 SL4 0.00 1/8 0 0 FW1 0.50 1/16 1/3 /7 FW 0.5 1/16 1/64 1/7 FW3 0.00 1/16 0 0 FW4 0.00 1/16 0 0 FL1 0.50 1/16 1/3 /7 FL 0.00 1/16 0 0 FL3 0.00 1/16 0 0 FL4 0.00 1/16 0 0 *The posterior proaility is the product divided y the sum of the items in the product column, 7/64. For the next accident, the posterior proailities are /7 for SW, 3/7 for FW, and /7 for FL. For an SW configuration, the expected next damage is 15. + 3 =. 5 (ased on the average of the expectations for and 4 dummies). For FW, it is For FL, it is 075. + 15. +. 5+ 3 4 15. + 3+ 45. + 6 4 g g = 375.. = 1875.. g g g g g g The posterior mean is / 7. 5 + 3/ 7 1875. + / 7 3. 75 =. 5. (C) 54
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #34 Because the parameter estimate is the sample mean, the sample means are 1000 and 1500 for Phil and Sylvia respectively. The likelihood function is g 0 1 xi j L θ = θ e θ e θ e i= 1 10 / θ 1 y / θ 30 0, 000/ θ 15, 000 / θ j= 1 g ecause the totals for Phil and Sylvia are 0,000 and 15,000 respectively. Setting the derivative of the logarithm of the likelihood equal to zero gives the equation 30 7, 500 * 7, 500 + = 0, θ = = 917. θ θ 30 (B) Question #35 g g g g,, 3. 0. 3 = 01. 1 Cov t z Cov t z, 0. =, Cov t, z =. Var t 16 / n n Also, Var z g = 9. g Then, Corr t, z n = 3. / n 16/ ng9 / ng = 067.. (E) Question #36 g g g g d$ = d + c p, 30063. = 7 + c 0. 409 + 093. + 0 0. 33 + 016., c = 5. 4 4 i i4 4 4 (A) 55
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #37 The formula is: β$ * j = β$ s j s Y X j, $ *. β = 038. 80964 = 0478., 00414. $ *. β 3 = 0. 0009 59316 9 = 0004., 00414. $ *. β 4 = 014718. 300114 = 0570.. 0041. 4 Then $ * $ * $ * β > β > β. 4 3 (D) Question #38 9 Z = = 0. 65. 9 + 1793. / 0. 37 g g The estimate for one policyholder is 0. 65 7 / 9 + 0. 35 0. 45 = 0. 6543. For 5 policyholders, it is 50. 6543g = 37.. (D) 56
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS Question #39 ln 185. µ The two equations to solve are 084. = σ and ln 358. µ 084. =. σ The solutions are σ $ = 0. 4 and µ $ = 341.. F I HG K J = = = ln 30 341. Then Pr X > 30g = 1 Φ 1 Φ04. g 1 06554. 03446.. 04. (A) Question #40 The forecast error always increases with the forecast horizon, so E is false. (E) 57
COURSE 4 EXAM NOVEMBER 000 ILLUSTRATIVE SOLUTIONS ANSWER KEY Question Numer Answer Question Numer Answer 1 B 1 A E D 3 C 3 C 4 C 4 C 5 D 5 D 6 A 6 C 7 B 7 E 8 D 8 C 9 D 9 A 10 C 30 A 11 A 31 E 1 B 3 D 13 E 33 C 14 E 34 B 15 C 35 E 16 A 36 A 17 B 37 D 18 E 38 D 19 B 39 A 0 D 40 E 58