Practice Exam 1. Loss Amount Number of Losses

Transcription

1 Practice Exam 1 1. You are given the following data on loss sizes: An ogive is used as a model for loss sizes. Determine the fitted median. Loss Amount Number of Losses (A) 2000 (B) 2200 (C) 2500 (D) 3000 (E) In a mortality study on 5 individuals, death times were originally thought to be 1, 2, 3, 4, 5. It then turned out that one of these five observations was a censored observation rather than an actual death. Determine the observation time of the five for which turning it into a censored observation would result in the lowest variance of the Nelson-Åalen estimator of H(4). (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 3. For an insurance coverage, the number of claims per year follows a Poisson distribution. Claim size follows a Pareto distribution with α = 3. The methods of classical credibility are used to determine premiums. The standard for full credibility is that actual aggregate claims be within 5% of expected aggregate claims 95% of the time. Based on this standard, 10,000 exposure units are needed for full credibility, where an exposure unit is a year of experience for a single insured. Determine the expected number of claims per year. (A) Less than 0.45 (B) At least 0.45, but less than 0.50 (C) At least 0.50, but less than 0.55 (D) At least 0.55, but less than 0.60 (E) At least 0.60 (A) (B) (C) (D) (E) 4. Which of the following statements is true? If data grouped into 7 groups are fitted to an inverse Pareto, the chi-square test of goodness of fit will have 5 degrees of freedom. The Kolmogorov-Smirnov statistic may be used to test the fit of a discrete distribution. The Kolmogorov-Smirnov statistic does not require adjustment for estimated parameters. The critical values of the Kolmogorov-Smirnov statistic do not vary with sample size. The critical values of the Anderson-Darling statistic do not vary with sample size. 425 Problems continue on the next page...

2 426 PART VI. PRACTICE EXAMS USE THE FOLLOWING INFORMATION FOR QUESTIONS 5 AND 6 There are 2 classes of insureds. In Class A, the number of claims per year has a Poisson distribution with mean 0.1 and claim size has an exponential distribution with mean 500. In Class B, the number of claims per year has a Poisson distribution with mean 0.2 and claim size has an exponential distribution with mean 250. Each class has the same number of insureds. An insured selected at random submits 2 claims in one year. Claim sizes are 200 and Calculate the probability that the insured is in class A. (A) 0.01 (B) 0.04 (C) 0.15 (D) 0.19 (E) Calculate the Bühlmann estimate of the aggregate losses for this insured in the following year. (A) Less than 42 (B) At least 42, but less than 47 (C) At least 47, but less than 52 (D) At least 52, but less than 57 (E) At least An insured s number of claims per year follows a Poisson distribution with mean Λ. Λ varies in accordance with a gamma distribution with α = 20, θ = You have the following information on the number of claims made by an insured in the past 6 years: 0, 0, 1, 0, 0, 1 Calculate the predictive variance of the number of claims per year for this insured. (A) Less than 0.18 (B) At least 0.18, but less than 0.19 (C) At least 0.19, but less than 0.20 (D) At least 0.20, but less than 0.21 (E) At least You are running a simulation, and want the estimated mean to be within 2% of the correct value 95% of the time. You estimate that 50,806 runs would be needed to achieve this result. Determine the coefficient of variation of the distribution you are simulating. (A) Less than 2.0 (B) At least 2.0, but less than 2.2 (C) At least 2.2, but less than 2.4 (D) At least 2.4, but less than 2.6 (E) At least 2.6 Problems continue on the next page...

3 PRACTICE EXAM An insurance coverage covers two types of insureds, A and B. There are an equal number of insureds in each class. Number of claims and claim sizes for insureds in each class have the following distributions: Number of claims Size of claims (Pareto parameters) A B A B α θ Within each class, claim size and number of claims are independent. Calculate the Bühlmann credibility to assign to 2 years of data. (A) 0.01 (B) 0.02 (C) 0.03 (D) 0.04 (E) An auto collision coverage is sold with deductibles of 500 and You have the following information for total loss size (including the deductible) on 86 claims: Deductible 1000 Deductible 500 Loss size Number of losses Loss size Number of losses Over Over You fit an exponential distribution to total loss size using maximum likelihood. Determine the fitted average total loss size for policies with a 500 deductible. (A) 671 (B) 707 (C) 935 (D) 1171 (E) You are given the following data from a 2-year mortality study. Duration Entries Withdrawals Deaths j d j u j x j c Withdrawals and new entries are assumed to occur uniformly. Based on this data, q 1 is estimated as Determine c. (A) 26 (B) 32 (C) 35 (D) 38 (E) 41 Problems continue on the next page...

4 428 PART VI. PRACTICE EXAMS 12. Estimated mortality rates are as follows: x 1000q x To determine 1000q x at intermediate ages, polynomial interpolation is used. An updated mortality study changed the estimated 1000q x at age 65 to 18. changed. Calculate the reduction in the interpolated value of 1000q 70 that results. (A) 2.00 (B) 2.75 (C) 3.25 (D) 3.75 (E) 4.00 No other rates were 13. The number of claims per year on an insurance coverage has a binomial distribution with parameters m = 2 and Q. Q varies by insured and is distributed according to the following density function: f(q) = 42q(1 q) 5 0 q 1 An insured submits 1 claim in 4 years. Calculate the posterior probability that for this insured, q is less than (A) 0.52 (B) 0.65 (C) 0.70 (D) 0.76 (E) You simulate a random variable with probability density function { 2x 1 x 0 f(x) = 0 otherwise using the inversion method. You use the following numbers random numbers from the uniform distribution on [0, 1]: Calculate the mean of the simulated observations (A) (B) (C) (D) (E) In a mortality study, there are 8 lives that survive 3 or more years. 3 deaths occur in the fourth year. Each death occurs at a different duration within the fourth year. No lives drop out of the study during the fourth year. The estimate of S(3), using the Nelson-Åalen estimator, is Calculate the Nelson-Åalen estimate of S(4). (A) 0.52 (B) 0.54 (C) 0.56 (D) 0.58 (E) 0.59 Problems continue on the next page...

5 PRACTICE EXAM You are given a sample of 5 claims: 2, 3, 4, x 1, x 2 with x 2 > x 1. This sample is fitted to a Pareto distribution using the method of moments. The resulting parameter estimates are ˆα = 47.71, ˆθ = Determine x 1. (A) 6.0 (B) 6.6 (C) 7.0 (D) 7.6 (E) The number of claims per year on a policy follows a Poisson distribution with parameter Λ. Λ has a uniform distribution on (0, 2). An insured submits 5 claims in one year. Calculate the Bühlmann credibility estimate of the number of claims for the following year. (A) 1.6 (B) 1.8 (C) 2.0 (D) 2.5 (E) For a group dental coverage, you have the following three years of experience from a covered group: There will be 120 members in the next year. Number of members Number of Aggregate Year in group claims claims , , ,000 The number of claims per member in any year follows a binomial distribution with parameters m = 4 and q. q is the same for all members in the group, but varies over groups, and is distributed uniformly over (0.45, 0.55). Claim size follows a gamma distribution with parameters α = 10, γ = 30. Calculate the Bühlmann-Straub estimate of aggregate claims in the next year. (A) Less than 55,300 (B) At least 55,300, but less than 55,500 (C) At least 55,500, but less than 55,700 (D) At least 55,700, but less than 55,900 (E) At least 55,900 Problems continue on the next page...

6 430 PART VI. PRACTICE EXAMS 19. There are two types of insured. The first type s claim sizes, before application of deductible, are distributed according to a Pareto with parameters α = 1, θ = The second type s claim sizes, before application of deductible, are distributed according to a Pareto with parameters α = 2, θ = An insured is equally likely to be of either type. You examine experience on an insured selected at random. The insured submitted 1 claim of size 500 (before application of the deductible) on a policy with deductible 250 and 1 claim of size 1500 (before application of the deductible) on a policy with deductible 500. Calculate the probability that the insured is of the first type. (A) 0.41 (B) 0.47 (C) 0.59 (D) 0.72 (E) A study is performed on the amount of time on unemployment. The records of 10 individuals are examined. 7 of the individuals are not on unemployment at the time of the study. The following is the number of weeks they were on unemployment: 5, 8, 10, 11, 17, 20, 26 3 individuals are still on unemployment at the time of the study. They have been unemployed for the following number of weeks: 5, 20, 26 Let T be the amount of time on unemployment. Using the Kaplan-Meier estimator with exponential extrapolation past the last study time, estimate Pr(20 T 30). (A) 0.17 (B) 0.21 (C) 0.28 (D) 0.32 (E) In a mortality study, you are given Time Risk set Number of deaths The Kaplan-Meier product limit estimator is used to estimate the survival function. Determine the width of a 95% log-transformed confidence interval for S(7). (A) (B) (C) (D) (E) Problems continue on the next page...

7 PRACTICE EXAM The distribution of auto insurance policyholders by number of claims submitted in the last year is as follows: Number of claims Number of insureds Total 100 The number of claims for each insured is assumed to follow a Poisson distribution. Use semi-parametric empirical Bayes estimation methods, with unbiased estimators for the variance of the hypothetical mean and the expected value of the process variance, to calculate the expected number of claims in the next year for a policyholder with 2 claims in the last year. (A) Less than 0.52 (B) At least 0.52, but less than 0.57 (C) At least 0.57, but less than 0.62 (D) At least 0.62, but less than 0.67 (E) At least You have the following experience for mortality in 3 classes of insureds. There are no withdrawals. Class 1 Class 2 Class 3 Z Number in class Deaths at time y Deaths at time y A Cox model with covariate Z is used for this population. β is estimated as 0.1. Calculate the Breslow estimate of the partial loglikelihood of this data. (A) 8.7 (B) 8.4 (C) 8.1 (D) 7.8 (E) You are given the following claims data from an insurance coverage with claims limit 10,000: 1000, 2000, 2000, 2000, 4000, 5000, 5000 There are 3 claims for amounts over 10,000 which are censored at 10,000. You fit this experience to an exponential distribution with parameter θ = 6,000. Calculate the Kolmogorov-Smirnov statistic for this fit. (A) 0.11 (B) 0.13 (C) 0.15 (D) 0.18 (E) The natural interpolating spline f(x) is passed through the points (0, 0), (1, 2) and (3, 0). Calculate f(2). (A) 1 (B) 4 3 (C) 3 2 (D) 5 3 (E) 7 4 Problems continue on the next page...

8 432 PART VI. PRACTICE EXAMS 26. For an insurance coverage, you observe the following claims sizes: 400, 1100, 1100, 3000, 8000 You fit the loss distribution to a lognormal with µ = 7 using maximum likelihood. Determine the mean of the fitted distribution. (A) Less than 2000 (B) At least 2000, but less than 2500 (C) At least 2500, but less than 3000 (D) At least 3000, but less than 3500 (E) At least In a mortality study performed on 5 lives, ages at death were 70, 72, 74, 75, 75 Estimate S(75) using kernel smoothing with a uniform kernel with bandwidth 4. (A) 0.2 (B) 0.3 (C) 0.4 (D) 0.5 (E) In a mortality study, y 1 is the first time at which deaths occurred. The 95% confidence interval for S(y 1 ), using the Kaplan-Meier estimator and the Greenwood approximation of variance, is (0.8241, ). Determine the number of deaths at time y 1. (A) 2 (B) 3 (C) 4 (D) 5 (E) For an insurance coverage, the number of claims per year follows a Poisson distribution with mean θ. The size of each claim follows an exponential distribution with mean 1000θ. You are examining one year of experience for four randomly selected policyholders, whose claims are as follows: You use maximum likelihood to estimate θ. Policyholder #1 2000, 4000, 4000, 7000 Policyholder # Policyholder #3 2000, 3000 Policyholder #4 1000, 4000, 5000 Determine the variance of aggregate losses based on the fitted distribution. (A) Less than 48,000,000 (B) At least 48,000,000, but less than 50,000,000 (C) At least 50,000,000, but less than 52,000,000 (D) At least 52,000,000, but less than 54,000,000 (E) At least 54,000,000 Problems continue on the next page...

9 PRACTICE EXAM You are going to the supermarket to buy bread. You select the brand of bread as follows: (i) (ii) (iii) (iv) If Brand A is on sale for 1.00 per pound, you will buy Brand A. The probability that Brand A is on sale is If Brand A is not on sale, you may buy Brand A, Brand B, or Brand C, with respective probabilities 0.25, 0.25, Brand A is 2.00 per pound when it is not on sale. Brand B is 1.00 per pound and Brand C is 1.25 per pound. You will buy 1, 2, 3, or 4 pounds of a single brand of bread, with equal probability. Given that you spent exactly 4.00, what is the probability that you bought 4 pounds of bread? (A) 0.50 (B) 0.67 (C) 0.70 (D) 0.75 (E) Past data on aggregate losses for two group policyholders is given in the following table. Group Year 1 Year 2 A Total losses Number of members B Total losses Number of members Calculate the credibility factor used for Group A s experience using non-parametric empirical Bayes estimation methods. (A) Less than 0.40 (B) At least 0.40, but less than 0.45 (C) At least 0.45, but less than 0.50 (D) At least 0.50, but less than 0.55 (E) At least You study the relationship of gender and duration of illness to survival after radiation treatment for cancer. (i) (ii) You model the relationship with a two-parameter Cox proportional hazards model. Gender is coded with an indicator variable z 1 = 1 for males. Duration is coded with the variable z 2 and is measured in years. The resulting partial likelihood estimates of the coefficients are: ˆβ 1 = 0.1 ˆβ 2 = 0.01 (iii) The covariance matrix of ˆβ 1 and ˆβ 2 is given by ( 0.2 ) Construct a 95% confidence interval for the relative risk of a male who has had the disease for 5 years to a female who has had the disease for 2 years. (A) (0.54, 1.73) (B) (0.59, 1.84) (C) (0.59, 1.95) (D) (0.63, 2.06) (E) (0.84, 1.69) Problems continue on the next page...

10 434 PART VI. PRACTICE EXAMS 33. For an insurance coverage, claim size follows a Pareto distribution with parameters α = 4 and θ. θ varies by insured and follows a normal distribution with µ = 3 and σ = 1. Determine the Bühlmann credibility to be assigned to a single claim. (A) 0.05 (B) 0.07 (C) 0.10 (D) 0.14 (E) Observations of a random variable X are fitted to a Pareto distribution using maximum likelihood. The estimated parameters are ˆα = 3, ˆθ = 600. The information matrix for (ˆα, ˆθ) is ( ) Calculate the width of a 95% confidence interval for the mean of X. (A) 59 (B) 72 (C) 98 (D) 144 (E) For a group dental coverage, aggregate losses per insured have a mean of 400 and a variance of 200,000. The number of claims per insured follows a negative binomial distribution with r = 2, β = 1. You are to use classical credibility methods to evaluate the credibility of experience of this group. The credibility standard is that aggregate losses should be within 5% of expected 99% of the time. Determine the number of expected claims needed for 25% credibility. (A) 207 (B) 415 (C) 829 (D) 1659 (E) A claims adjustment facility adjusts all claims for amounts less than or equal to 10,000. Claims for amounts greater than 10,000 are handled elsewhere. In 2002, the claims handled by this facility fell into the following ranges: Size of Claim Number of Claims Less than The claims are fitted to a parametric distribution using maximum likelihood. Which of the following is the correct form for the likelihood function of this experience? (A) [F (1000)] 100 [F (5000) F (1000)] 75 [F (10,000) F (5000)] 25 (B) (C) (D) (E) [F (1000)] 100 [F (5000) F (1000)] 75 [F (10,000) F (5000)] 25 [F (10,000)] 200 [F (1000)] 100 [F (5000) F (1000)] 75 [F (10,000) F (5000)] 25 [1 F (10,000)] 200 [1 F (1000)] 100 [F (5000) F (1000)] 75 [F (10,000) F (5000)] 25 [F (10,000)] 200 [1 F (1000)] 100 [F (5000) F (1000)] 75 [F (10,000) F (5000)] 25 [1 F (10,000)] 200 Problems continue on the next page...

11 PRACTICE EXAM An insurance coverage has a deductible of 500 and a claims limit of 10,000. You are given the following claims experience: (i) 10 payments below 9500 having average size (ii) 6 payments of The ground-up loss distribution is fitted to an exponential using maximum likelihood. Determine the fitted mean of the ground-up distribution. (A) 3688 (B) 7200 (C) 7700 (D) 8200 (E) For a sample from an exponential distribution, which of the following statements is false? (A) (B) (C) (D) (E) If the sample has size 2, the sample median is an unbiased estimator of the population median. If the sample has size 2, the sample median is an unbiased estimator of the population mean. If the sample has size 3, the sample mean is an unbiased estimator of the population mean. If the sample has size 3, 1.2 times the sample median is an unbiased estimator of the population mean. The sample mean is a consistent estimator of the population mean. 39. The median of a sample is 5. The sample is fitted to a mixture of two exponential distributions with means 3 and x > 3, using percentile matching to determine the weights to assign to each exponential. Which of the following is the range of values for x for which percentile matching works? (A) 3 < x < (B) 3 < x < (C) < x < (D) x > (E) x > The number of claims for an insured follows a Poisson distribution with mean λ. λ varies by insured according to a gamma distribution with α = 3, θ = 0.1. No claims are submitted for n years. Determine n such that the expected number of claims in the following year is 0.2. (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 Solutions to the above questions begin on page 635.

12 Appendix B Answers to the Practice Exams Answer Key for Practice Exam 1 1 A 11 B 21 D 31 D 2 D 12 D 22 E 32 D 3 E 13 D 23 A 33 A 4 E 14 A 24 D 34 A 5 D 15 C 25 E 35 B 6 C 16 C 26 A 36 B 7 D 17 C 27 B 37 C 8 C 18 D 28 E 38 A 9 A 19 A 29 E 39 E 10 E 20 D 30 C 40 B Practice Exam 1 1. [Section 2.2] The distribution function at 1000, F (1000), is , and F (5000) = 12. By definition, the median is the point m such that F (m) = 0.5. The ogive linearly interpolates between 1000 and Thus we solve the equation m = m 1000 = 1 (4000) = m = 2000 (A) 2. [Lesson 6] The variance is 1. To minimize the variance, maximize the r ri 2 i s. The r i s for the first 3 deaths are 3 out of {5,4,3,2}. By making 4 the censored observation, the r i s are 5, 4, and 3. (Making 5 the censored observation leads to a sum of 4 terms instead of 3, with the first 3 terms being the same as if 4 is censored, so it does not lead to minimal variance). (D) 3. [Lesson 27] The credibility formula in terms of expected number of claims, formula (27.2), requires 1+CVs 2, or the second moment divided by the first moment squared of the severity distribution. For a Pareto with α = 3, this is 2θ 2 (2)(1) ( θ 2 = 4 2) The expected number of claims needed for full credibility is then ( ) 2(4) = , so we have = 10,000λ λ = (E) 635

13 636 PRACTICE EXAM 1, ANSWERS TO QUESTIONS [Lessons 18, 19, 20] (A) is false because the number of degrees of freedom is n 1 minus the number of parameters estimated. Here n = 7 and the inverse Pareto has 2 parameters, so there are 4 degrees of freedom. (B) is false, as indicated on page 159. (C) is false, as indicated on page 159, where it says that the indicated critical values only work when the distribution is completely specified, not when parameters have been estimated. (D) is false; in fact, the critical values get divided by n. (E) is true. 5. [Lesson 30] The likelihood that an insured in class A submits 2 claims for 200 and 400 is the product of the Poisson probability of 2 claims and the 2 densities under the exponential distribution with mean 500, or ( ) ( ) ( ) 0.1 f(200, 400 A) = e e 200/500 e 400/500 = ! and similarly for an insured in class B: ( ) ( ) ( ) 0.2 f(200, 400 A) = e e 200/250 e 400/250 = ! Since the classes are of equal size, these are the relative probabilities of the two classes. Then the probability that the insured is in class A is Pr(A 200, 400) = = (D) [Lesson 37] Since the hypothetical means are the same in both classes (0.1)(500) and (0.2)(250) are 50, a = 0 and there is no credibility. Thus, 50 is the credibility premium. (C) 7. [Lesson 32] The prior parameters are α = 20 and γ = 1 θ = 100. The posterior parameters are α = 20+2 = 22 and γ = = 106. The posterior negative binomial has r = α and β = θ = 1 γ, and its variance is ( )( ) rβ(1 + β) = 22 = (D) [Lesson 45] A 95% confidence interval is constructed by adding and subtracting 1.96 times the standard deviation. If σ 2 is the variance of the distribution, then the variance of the mean of n observations is σ2 σ n, and the standard deviation is n. If µ is the mean of the distribution, the required accuracy will be achieved if 1.96σ n = 0.02µ 1.96σ = 0.02 n = ,806 = µ σ µ = = 2.3 (C) 1.96

14 PRACTICE EXAM 1, ANSWERS TO QUESTIONS [Lesson 37] Let µ A be the hypothetical mean for A, v A the process variance for A, and use the same notation with subscripts B for B. For process variance, we will use the compound variance formula. µ A = 0.1(25) = 2.5 v A = 0.1( ) (25 2 ) = µ B = 0.2(30) = 6 v B = 0.2( ) (30 2 ) = 684 ( ) 1 a = (6 2.5) 2 = v = 1 ( ) = For 2 years of experience, the credibility is Z = na na + v = (2)(3.0625) = (A) (2)(3.0625) [Lesson 11] The likelihood function (either using the fact that the exponential is memoryless, or else writing it all out and canceling out the denominators) is Logging and differentiating: Multiply through by L(θ) = (1 e (1000/θ)) 20 (e (1000/θ)) 10 (1 e (500/θ)) 32 (e (500/θ)) 24 ( l(θ) = 20 ln 1 e 1000/θ) ( + 32 ln 1 e 500/θ) 10, ,000 θ dl dθ = 20,000e 1000/θ θ 2 (1 e 1000/θ ) + 16,000e 500/θ θ 2 (1 e 500/θ ) + 22,000 θ 2 = 0 θ2 1000, and set x = e 500/θ. We obtain 22 20x2 1 x 2 16x 1 x = x 2 20x 2 16x 16x 2 = 0 58x x 22 = 0 x = (22)(58) 116 θ = 500 ln x = = An easier way to do this would be to make the substitution x = e 500/θ right in L(θ), and to immediately recognize that 1 x 2 = (1 x)(1 + x). This would avoid the confusing differentiation step: l(θ) = 20 ln(1 x 2 ) + 44 ln x + 32 ln(1 x) = 20 ln(1 x) + 20 ln(1 + x) + 44 ln x + 32 ln(1 x) = 52 ln(1 x) + 20 ln(1 + x) + 44 ln x dl dθ = 52 1 x x + 44 x = 0 52(x)(1 + x) + 20(x)(1 x) + 44(1 x) 2 = 0 52x 2 52x 20x x x 2 = 0

15 638 PRACTICE EXAM 1, ANSWERS TO QUESTIONS x x 44 = 0 which is double the quadratic above, and leads to the same solution for θ. Using the fact that the exponential distribution is memoryless, the average total loss size for a 500 deductible is = (E) 11. [Lesson 8] The conditional probability of death at the second duration (note that 0 is the first duration and 1 is the second), q 1, is estimated by s2 r 2, number of deaths over the risk set in duration 2. Duration 2 starts with = 867 lives, and since withdrawals and new entries occur uniformly, we add half the new entries and subtract half the withdrawals to arrive at r 2 = ( ) = Then 0.03 = ˆq 1 = c 1067 c = 32 (B) 12. [Lesson 22] By equation (22.1), the factor from 65 is (x 45)(x 55)(x 75) (65 45)(65 55)(65 75) For x = 70, and applying this to a reduction of 4 (from 22 to 18), we get (70 45)(70 55)(70 75) 1875 (4) = (4) = 3.75 (65 45)(65 55)(65 75) 2000 (D) 13. [Lesson 34] The prior distribution is a beta distribution with a = 2, b = 6. In general, a is 1 more than the exponent of q and b is 1 more than the exponent of 1 q. The number of claims is binomial, which means that 2 claims are possible each year. Of the 8 possible claims in 4 years, you received 1 and didn t receive 7. Thus a = = 3 and b = = 13 are the parameters of the posterior beta. The density function for the posterior beta is π(q x) = Γ(3 + 13) Γ(3)Γ(13) q2 (1 q) 12 = 1365q 2 (1 q) 12 since 15! 2 = We must integrate this function from 0 to 0.25 to obtain Pr(Q < 0.25). It is easier to integrate if we change the variable, by setting q = 1 q. Then we have 2!12! = (15)(14)(13) 1 Pr(Q < 0.25) = 1365 = (1 q ) 2 q 12 dq (q 12 2q 13 + q 14 )dq = q q q = 1365( ) = (D)

16 PRACTICE EXAM 1, ANSWERS TO QUESTIONS [Lesson 44] The distribution function is Inverting, F (x) = x 1 2udu = u 2 x 1 = 1 x2 u = 1 x 2 1 u = x 2 x = 1 u It is necessary to use the negative square root, since the simulated observation must be between 1 and 0. So 15. [Lesson 5] We back out Ĥ(3): x 1 = x 2 = = x 3 = = x 4 = = = (A) 4 Ĥ(3) = ln Ŝ(3) = ln Then we sum up si r i = 1 r i for the 3 deaths that occur between time 3 and 4: Ĥ(4) = ln = Ŝ(4) = e = (C) 16. [Lesson 9] We write the moment equations for the first and second moments: x 1 + x 2 = θ 5 α 1 ( ) x 1 + x 2 = 5 = x x 2 2 2θ 2 = 5 (α 1)(α 2) [ 2( x x 2 2 ] ) 2 = 5 = 654 (46.71)(45.71) We use the first equation to solve for x 2 in terms of x 1, and plug that into the second equation and solve. x 2 = 31 x x (31 x 1 ) 2 = x x = 654

17 640 PRACTICE EXAM 1, ANSWERS TO QUESTIONS Since x 2 is higher than x 1, x 1 = 7. (C) 2x x = 0 x x = 0 x 1 = 31 ± (168) 2 31 ± 17 = = 7, [Lesson 37] The hypothetical mean is Λ. The expected hypothetical mean µ = E(Λ) = 1 (the mean of the uniform distribution). The process variance is Λ. The expected process variance v, or the expected value of Λ, is 1. The variance of the hypothetical mean is Var(Λ). For a uniform distribution on (0, θ), the variance is θ2 12, so the variance is a = 1 3. The Bühlmann k is therefore 1 1/3 = 3. Z = = The credibility premium is 1 4 (5) (1) = 2. (C) 18. [Section 39.1] The hypothetical mean is 4qαθ = 1200q. The overall mean is E[1200q] = 600. The variance of q is = , since the variance of a uniform distribution is the range squared divided by 12. The variance of the hypothetical means is a = Var(1200q) = = 1200 The process variance, by the compound variance formula (26.2) on page 229 is The expected value of q 2 is The expected value of the process variance is P V = 4qαθ 2 + 4q(1 q)(αθ) 2 = 36,000q + 360,000q 360,000q 2 E(q 2 ) = Var(q) + E(q) 2 = = v = E(396,000q) E(360,000q 2 ) ( ) 601 = 198, , = 17,700 The exposure unit is a member-year; the number of claims is irrelevant. The total exposure units is = 305. The credibility factor is Z = na na+v = 305(1200) 305(1200)+17,700 = The observed aggregate losses is 40, , ,000 x = = The Bühlmann-Straub prediction of next year s aggregate claims is 120[ ( ) + ( )(600)] = 55,862 (D)

18 PRACTICE EXAM 1, ANSWERS TO QUESTIONS [Lesson 30] The density of a claim of 500 with a deductible of 250 from the first insured is 1000 f(500) 1 F (250) = The density of a claim of 1500 with a deductible of 500 from the first insured is 1000 f(1500) 1 F (500) = Multiplying these together (both types of insureds are equally likely, so we can ignore the 1/2 factor when calculating relative likelihoods) we get (1500)( = ) The density of a claim of 500 with a deductible of 250 from the second insured is 2(2000 f(500) 1 F (250) = 2500 ( The density of a claim of 1500 with a deductible of 500 from the second insured is ) ) 2 The product of these two densities is 2(2000 f(1500) 1 F (500) = 3500 ( ) ) ( ) 2500( = ) The posterior probability that the insured is of the first type is then = (A) [Lesson 5 and section 4.2] First we calculate Ŝ(26). y i r i s i S 10 (y i ) So Ŝ(26) = To extrapolate, we exponentiate Ŝ(26) to the 26 power, as discussed in example 5.4 on page 40: Ŝ(30) = /26 = Pr(20 T 30) = S(20 ) S(30), since the lower endpoint is included. But S(20 ) = S(17) = So the answer is = (D)

19 642 PRACTICE EXAM 1, ANSWERS TO QUESTIONS [Lesson 6, page 49] First we estimate S(7) using the product-limit estimator, equation (5.1) on page 37: S 100 (7) = ( )( ) = Then we use the Greenwood formula, equation (6.1) on page 48 to estimate the variance of S 100 (7): [ ] Var[S 100 (7)] = [S 100 (7)] 2 2 (98)(100) + 1 = (95)(94) Then we use equation (6.3) to construct a log-transformed confidence interval. ( ) 1.96 V U = exp = exp( ) = S ln S (S 1/U, S U ) = (0.9089, ) The width is = (D) 22. [Lesson 43] We estimate µ, v, and a: ˆµ = ˆv = x = 22(1) + 6(2) + 2(3) 100 = = 0.4 We will calculate s 2, the unbiased sample variance, by calculating the second moment, subtracting the square of the sample mean (which gets us the empirical variance) and then multiplying by n n 1 to turn it into the sample variance. â + ˆv = s 2 = [ 22(1 2 ) + 6(2 2 ) + 2(3 2 ] ) x = ( ) = â = = Z = a a + v = P C = (1.6) = 0.68 (E) [Lesson 14] We use the expression (14.1), with the proportionality constants c = e βz. [ e 0.1 ] [ e 0.4 ] ln (5 + 3e e 0.3 ) 2 + ln (4 + 2e e 0.3 = (A) ) 24. [Lesson 18] We set up a table for the empirical and fitted functions. Note that we do not know the empirical function at 10,000 or higher due to the claims limit. x F n (x ) F n (x) F (x) Largest difference e 1/6 = e 1/3 = e 2/3 = e 5/6 = , e 5/3 = Inspection indicates that the maximum difference occurs at 2000 and is (D)

20 PRACTICE EXAM 1, ANSWERS TO QUESTIONS [Lesson 23] Using equation (23.3), we have that u 1, 6 times the change in slope, is 6(( 1) 2) = 18 and g 1 = 2(3 0) = 6, so 6m 1 = 18 making m 1 = 3. Then using equations (23.6), we have a 1 = 2 c 1 = 3 2 d 1 = 0 ( 3) = 3 6(2) 12 = 1 4 b 1 = (22 ) 1 4 (23 ) 2 As indicated in that lesson, you don t have to memorize the formula for b, since you can back it out once you have the other coefficients. Then = 1 f(2) = = 7 4 (E) 26. [Lesson 11] See the discussion of transformations and the lognormal example 11.3 on 91, and the paragraph before the example. We will log each of the claim sizes and fit them to a normal distribution. You may happen to know that for a normal distribution, the MLE s of µ and σ are independent, so given µ, the MLE for σ will be the same as if µ were not given. Moreover, the MLE for σ for a normal distribution is the sample variance divided by n (rather than by n 1). If you know these two facts, you can calculate the MLE for σ on the spot. If not, it is not too hard to derive. The likelihood function (omitting the constant 2π) is (where we let x i = the log of the claim size) We calculate L(σ) = 1 σ 5 5 i=1 l(σ) = 5 ln σ e (x i 7) 2 2σ 2 5 (x i 7) 2 i=1 2σ 2 dl dσ = 5 5 σ + i=1 (x i 7) 2 σ 3 = 0 5 σ 2 i=1 = (x i 7) 2 5 σ 2 = (ln 400 7)2 + 2(ln ) 2 + (ln ) 2 + (ln ) 2 5 The mean of the lognormal is = exp ( µ + σ 2 /2 ) = exp( /2) = (A) 27. [Lesson 7] The kernel survival function for a uniform kernel is a straight line from 1 to 0, starting at the observation point minus the bandwidth and ending at the observation point plus the bandwidth. From the perspective of 74, we reverse orientation; the kernel survival for 74 increases as

21 644 PRACTICE EXAM 1, ANSWERS TO QUESTIONS the observation increases. Therefore, the kernels are 0 at 70, 1 8 at 72 (which is 1 8 of the way from 71 to 79), 3 8 at 74 (which is 3 8 of the way from 71 to 79), and 1 2 at 75 (which is 1 2 of the way from 71 to 79). Each point has a weight of 1 n = 1 5. We therefore have: [ ( )] (2) = 0.30 (B) [Lesson 6] 0.9, the mean, is s r = 1.96 Var[Ŝ(y 1)] = Var[Ŝ(y 1)] = Var[Ŝ(y 1)] The study is a complete study through time y 1 ; even if there were withdrawals between time 0 and time y 1, we could treat the study as if it started at the time of the last withdrawal before y 1. Hence the Greenwood approximation is the same as the empirical variance, and we can use equation (3.1) on page 20. ( s1 ) ( ) r 1 1 s1 r = = 0.9 r 1 ( 1 = ) r 1 r 1 r 1 = = 60 s 1 = (1 0.9)(60) = 6 ( 1 s ) 1 r 1 r1 2 = 0.09 r [Lesson 11] The likelihood function is the product of θni θ e n i! (E) for the number of claims n i for the 4 individuals, times the product of θ e xi/1000θ for each of the 10 claim sizes x i. These get multiplied together to form the likelihood function. We have ni = = 10 and xi = = θ 1000θ θ If we ignore the constants, the likelihood function is: L(θ) = e 4θ θ 10 1 θ 10 e 36/θ l(θ) = 4θ 36 θ

22 PRACTICE EXAM 1, ANSWERS TO QUESTIONS dl 36 = 4 + dθ θ 2 = 0 θ = 3 To complete the problem, use equation (26.2), or better, since number of claims is Poisson, equation (26.3). Let S be aggregate losses. Using either formula, we obtain Var(S) = 3[2( )] = 54,000,000 for the fitted distribution. (E) 30. [Lesson 30] There are 3 ways to spend 4.00: 1. 4 pounds of Brand A on sale, probability (0.25)(0.25) pounds of Brand A not on sale, probability (0.75)(0.25)(0.25) pounds of Brand B, probability (0.75)(0.25)(0.25). We need only the relative probabilities, so we can factor out the (0.25)(0.25). So we are looking at weights 1 on the first possibility and 0.75 on each of the second and third possibilities. Then the probability of the first and the third vs. the second is = 0.70 (C) [Subsection ] We apply formulas (42.2) and (42.3) x 1 = = x 2 = = x = = = 22 ) 2 ( ˆv = 40 ( ) 2 ( ) 2 ( ) = [ â = 90( ( ) 22)2 + 60( )2 ( )(1)] = = ˆk = ˆv â = = 73.2 Z 1 = 90 = (D) 90 + k 32. [Section 15.1] The estimated difference of the logarithms of the hazard rates is 0.1(1) (3) = The variance is Var( ˆβ ˆβ 2 ) = Var( ˆβ 1 ) + Var( ˆβ 2 ) + 2 Cov( ˆβ 1, ˆβ 2 ) = (3 2 ) 2(0.03)(3) = We exponentiate the estimated confidence interval for the logarithm of the relative hazard rate to obtain a confidence interval for the relative hazard rate. exp(0.13 ± ) = , (D)

23 646 PRACTICE EXAM 1, ANSWERS TO QUESTIONS [Lesson 38] For aggregate losses, the mean given θ is θ 3 2θ 2 9. Then and the variance given θ is 2θ2 6 ( θ 3) 2 = v = 2 9 E[θ2 ] = 2 9 (32 + 1) = 20 9 a = 1 9 Var(θ) = 1 9 Z = a a + v = 1 21 (A) 34. [Section 12.1] We invert the information matrix to obtain the covariance matrix. The determinant of the matrix is (100)(9) (1)(1) = 899, so the inverse matrix is ( ) θ The mean of a Pareto is g(α, θ) = α 1. We use equation (12.1) to calculate the variance. In the 2-variable case, the equation reduces to ( ) 2 ( ) 2 ( )( ) g g g g Var[g(α, θ)] = Var(α) + Var(θ) + 2 Cov(α, θ) α θ α θ so we have g α = θ (α 1) 2 = = 150 g θ = 1 α 1 = 1 2 [ Var(estimated mean) = 1 ( ) 2 ( ) ] 1 1 9( 150) ( 1)( 150) = The width of a 95% confidence interval is then 2(1.96)( ) = (A) 35. [Lesson 29] We use the general formula, (27.1) on page 236, to determine the exposures needed for credibility of aggregate losses, since the severity mean and variance are not given. ( 2 ( ) 2 ( ) σ ,000 e F = n 0 = µ) = (1.25) = The number of expected claims needed for full credibility is βr = For 25% credibility, we need ( )( ) = claims. (B) 36. [Lesson 11] The claims are truncated, not censored, at 10,000. The probability of seeing any claim is F (10,000). Any likelihood developed before considering this condition must be divided by this condition. The likelihood of each of the 100 claims less than 1000, if not for the condition, is F (1000). The conditional likelihood, conditional on the claim being below 10,000, is F (1000) F (10,000). The likelihood of each of the 75 claims between 1000 and 5000, if not for the condition, is F (5000) F (1000). The conditional likelihood is F (5000) F (1000) F (10,000).

24 PRACTICE EXAM 1, ANSWERS TO QUESTIONS The likelihood of each of the 25 claims between 5000 and 10,000, if not for the condition, is F (10,000) F (5000). The conditional likelihood is F (10,000) F (5000) F (10,000). Multiplying all these 200 likelihoods together we get answer (B). 37. [Lesson 11] See the exponential shortcut in example 11.2 on page 91, and the paragraph before the example. The exposure for the payments below 9500, x i, is x i, because the deductible is already taken into account in the payment size,so the sum of these exposures is 10(2000) = 20,000. The exposure for each of the 6 payments of 9500 is 9500 for the same reason. So the total exposure is 10(2000) + 6(9500) = 77,000. Dividing by the number of uncensored claims, 10, the answer is 77, = (C) Naturally, you could do the problem from first principles too. The likelihood function is f(xi+500) for the payments x i (after the deductible) below 9500, and Then ( 1 ) θ e [ 10 L(θ) = 10 i=1 (xi+500) 6(10,000)] ( ) e ( 500/θ)16 1 = θ ( 10 ) 1 = θ 10 e 77,000/θ l(θ) = 10 ln θ 77,000 θ dl dθ = 10 θ + 77,000 θ 2 = 0 θ = 7700 (C) 1 F (10,000) 1 F (500) e [ 10(2000) 10(500) 6(10,000)+16(500)]/θ 1 F (500) for the truncated payments of 38. [Lesson 1] The sample mean is an unbiased estimator of the population mean, and if the population variance is finite (as it is if it has an exponential distribution), the sample mean is a consistent estimator of the population mean. (C) and (E) are therefore true. For a sample of size 2, the sample median is the sample mean, so (B) is true. (D) is proved in Loss Models. That leaves (A). (A) is false, because (B) is true and the median of an exponential is not the mean. In fact, it is the mean times ln 2. So the sample median, which is an unbiased estimator of the mean, and therefore has an expected value of θ, does not have expected value θ ln 2, the value of the median. 39. [Lesson 10] For a mixture F is the weighted average of the F s of the individual distribution. The median of the mixture F is then the number m such that wf 1 (x) + (1 w)f 2 (x) = 0.5 where w is the weight. Here, it is more convenient to use survival functions. (The median is the number m such that S(m) = 0.5) m = 5. We have: we 5/3 + (1 w)e 5/x = 0.5 w(e 5/3 e 5/x ) = 0.5 e 5/x w = 0.5 e 5/x e 5/3 e 5/x

25 648 PRACTICE EXAM 1, ANSWER TO QUESTION 40 In order for this procedure to work, w must be between 0 and 1. Note that since x > 3, 5 3 < 5 x, so the denominator is negative. For w > 0, we need For w < 1, we need 0.5 e 5/x < 0 e 5/x > 0.5 5/x > ln 0.5 5/x < ln 2 x > 5 ln 2 = e 5/x 0.5 < e 5/x e 5/3 0.5 > e 5/3 = and this is always true. So percentile matching works when x > (E) 40. [Lesson 32] The credibility expectation, by equation (32.1) is 3 + n i=1 x i = n where x i is the number of claims in year i and n is the number of years. Here, x i = 0 for all i, so we have n = 0.2 n = 5 (B)