# SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM C CONSTRUCTION AND EVALUATION OF ACTUARIAL MODELS EXAM C SAMPLE QUESTIONS

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1 SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM C CONSTRUCTION AND EVALUATION OF ACTUARIAL MODELS EXAM C SAMPLE QUESTIONS Copyright 005 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study note are taken from past SOA/CAS examinations. C PRINTED IN U.S.A.

2 . You are given: (i) (ii) Losses follow a loglogistic distribution with cumulative distribution function: γ x Fx bg b / θg = γ + x / θ The sample of losses is: b g Calculate the estimate of θ by percentile matching, using the 40 th and 80 th empirically smoothed percentile estimates. (A) Less than 77 (B) At least 77, but less than 87 (C) At least 87, but less than 97 (D) At least 97, but less than 07 (E) At least 07. You are given: (i) The number of claims has a Poisson distribution. (ii) Claim sizes have a Pareto distribution with parameters θ = 0.5 and α = 6. (iii) The number of claims and claim sizes are independent. (iv) The observed pure premium should be within % of the expected pure premium 90% of the time. Determine the expected number of claims needed for full credibility. (A) Less than 7,000 (B) At least 7,000, but less than 0,000 (C) At least 0,000, but less than 3,000 (D) At least 3,000, but less than 6,000

3 (E) At least 6, You study five lives to estimate the time from the onset of a disease to death. The times to death are: Using a triangular kernel with bandwidth, estimate the density function at.5. (A) 8/40 (B) /40 (C) 4/40 (D) 6/40 (E) 7/40 4. You are given: (i) Losses follow a Single-parameter Pareto distribution with density function: f ( x α ) = ( ) x, x >, 0 < α < α + (ii) A random sample of size five produced three losses with values 3, 6 and 4, and two losses exceeding 5. Determine the maximum likelihood estimate of α. (A) 0.5 (B) 0.30 (C) 0.34 (D) 0.38 (E) 0.4 3

4 5. You are given: (i) The annual number of claims for a policyholder has a binomial distribution with probability function: x ( ) ( ) x p x q = q q, x = 0,, x (ii) The prior distribution is: ( ) 3 π q = 4 q, 0< q< This policyholder had one claim in each of Years and. Determine the Bayesian estimate of the number of claims in Year 3. (A) Less than. (B) At least., but less than.3 (C) At least.3, but less than.5 (D) At least.5, but less than.7 (E) At least.7 6. For a sample of dental claims x, x,..., x 0, you are given: (i) x = 3860 and x = 4,574,80 i i (ii) Claims are assumed to follow a lognormal distribution with parameters µ and σ. (iii) µ and σ are estimated using the method of moments. Calculate E X 500 for the fitted distribution. (A) Less than 5 (B) At least 5, but less than 75 (C) At least 75, but less than 5 (D) At least 5, but less than 75 (E) At least 75 4

5 7. Two independent samples are combined yielding the following ranks: Sample I:,, 3, 4, 7, 9, 3, 9, 0 Sample II: 5, 6, 8, 0,,, 4, 5, 6, 7, 8 You test the null hypothesis that the two samples are from the same continuous distribution. The variance of the rank sum statistic is: ( + m+ ) nm n Using the classical approximation for the two-tailed rank sum test, determine the p-value. (A) 0.05 (B) 0.0 (C) 0.05 (D) 0.0 (E) You are given: (i) Claim counts follow a Poisson distribution with mean θ. (ii) Claim sizes follow an exponential distribution with mean 0θ. (iii) Claim counts and claim sizes are independent, given θ. (iv) The prior distribution has probability density function: bg= 5 6, θ > πθ θ Calculate Bühlmann s k for aggregate losses. (A) Less than (B) At least, but less than (C) At least, but less than 3 (D) At least 3, but less than 4 (E) At least 4 5

6 9. You are given: (i) A survival study uses a Cox proportional hazards model with covariates Z and Z, each taking the value 0 or. (ii) The maximum partial likelihood estimate of the coefficient vector is: β, β = 07., 00. e j b g b g =. (iii) The baseline survival function at time t 0 is estimated as St Estimate St 0 (A) 0.34 (B) 0.49 (C) 0.65 (D) 0.74 (E) 0.84 bg for a subject with covariate values Z Z = =. 0. You are given: (i) (ii) Z and Z are independent N(0,) random variables. a, b, c, d, e, f are constants. (iii) Y = a+ bz + cz and X = d + ez + f Z Determine E( Y X ). (A) a (B) a+ ( b+ c)( X d) (C) a+ bbe+ cfgb X dg + b + g/ d + i (D) a be cf e f X b g d ib g (E) a+ be+ cf / e + f X d 6

7 . You are given: (i) Losses on a company s insurance policies follow a Pareto distribution with probability density function: f x θ = < x < ( θ ) ( x + θ ), 0 (ii) For half of the company s policies θ =, while for the other half θ = 3. For a randomly selected policy, losses in Year were 5. Determine the posterior probability that losses for this policy in Year will exceed 8. (A) 0. (B) 0.5 (C) 0.9 (D) 0. (E) 0.7. You are given total claims for two policyholders: Year Policyholder 3 4 X Y Using the nonparametric empirical Bayes method, determine the Bühlmann credibility premium for Policyholder Y. (A) 655 (B) 670 (C) 687 (D) 703 (E) 79 7

8 3. A particular line of business has three types of claims. The historical probability and the number of claims for each type in the current year are: Type Historical Number of Claims Probability in Current Year A B C You test the null hypothesis that the probability of each type of claim in the current year is the same as the historical probability. Calculate the chi-square goodness-of-fit test statistic. (A) Less than 9 (B) At least 9, but less than 0 (C) At least 0, but less than (D) At least, but less than (E) At least 4. The information associated with the maximum likelihood estimator of a parameter θ is 4n, where n is the number of observations. Calculate the asymptotic variance of the maximum likelihood estimator of θ. (A) n (B) n (C) (D) (E) 4 n 8n 6n 8

9 5. You are given: (i) The probability that an insured will have at least one loss during any year is p. (ii) The prior distribution for p is uniform on [ 0,0.5 ]. (iii) An insured is observed for 8 years and has at least one loss every year. Determine the posterior probability that the insured will have at least one loss during Year 9. (A) (B) (C) (D) (E) Use the following information for questions and. For a survival study with censored and truncated data, you are given: Time (t) Number at Risk at Time t Failures at Time t The probability of failing at or before Time 4, given survival past Time, is 3 Calculate Greenwood s approximation of the variance of 3 q. (A) (B) (C) (D) (E) q. 9

10 7. Calculate the 95% log-transformed confidence interval for Hb3g, based on the Nelson-Aalen estimate. (A) (0.30, 0.89) (B) (0.3,.54) (C) (0.39, 0.99) (D) (0.44,.07) (E) (0.56, 0.79) 8. You are given: (i) Two risks have the following severity distributions: Amount of Claim Probability of Claim Amount for Risk Probability of Claim Amount for Risk , , (ii) Risk is twice as likely to be observed as Risk. A claim of 50 is observed. Determine the Bühlmann credibility estimate of the second claim amount from the same risk. (A) Less than 0,00 (B) At least 0,00, but less than 0,400 (C) At least 0,400, but less than 0,600 (D) At least 0,600, but less than 0,800 (E) At least 0,800 0

11 9. You are given: (i) A sample x, x,, x is drawn from a distribution with probability density function: (ii) θ > σ 0 (iii) xi = 50 and xi = 5000 [ ] exp( x ) + exp( x ), 0 < θ θ σ σ x < Estimate θ by matching the first two sample moments to the corresponding population quantities. (A) 9 (B) 0 (C) 5 (D) 0 (E) 0. You are given a sample of two values, 5 and 9. You estimate Var(X) using the estimator g(x, X ) = ( X X ). i Determine the bootstrap approximation to the mean square error of g. (A) (B) (C) 4 (D) 8 (E) 6

12 . You are given: (i) (ii) (iii) The number of claims incurred in a month by any insured has a Poisson distribution with mean λ. The claim frequencies of different insureds are independent. The prior distribution is gamma with probability density function: f ( λ ) = ( ) λ e 0λ (iv) Month Number of Insureds Number of Claims ? λ Determine the Bühlmann-Straub credibility estimate of the number of claims in Month 4. (A) 6.7 (B) 6.9 (C) 7.3 (D) 7.6 (E) 8.0. You fit a Pareto distribution to a sample of 00 claim amounts and use the likelihood ratio test to test the hypothesis that α =.5 and θ = 7.8. You are given: (i) The maximum likelihood estimates are α =.4 and θ = 7.6. (ii) The natural logarithm of the likelihood function evaluated at the maximum likelihood estimates is (iii) ( ) ln x = i Determine the result of the test.

13 (A) (B) (C) (D) (E) Reject at the significance level. Reject at the 0.00 significance level, but not at the level. Reject at the 0.05 significance level, but not at the 0.00 level. Reject at the significance level, but not at the 0.05 level. Do not reject at the significance level. 3. For a sample of 5 losses, you are given: (i) Observed Number of Interval Losses (0, ] 5 (, 5] 5 (5, ) 5 (ii) Losses follow the uniform distribution on b0,θg. Estimate θ by minimizing the function losses in the jth interval and (A) 6.0 (B) 6.4 (C) 6.8 (D) 7. (E) j= ( E ) j Oj, where E j is the expected number of O j O j is the observed number of losses in the jth interval. 3

14 4. You are given: (i) The probability that an insured will have exactly one claim is θ. (ii) The prior distribution of θ has probability density function: 3 πθ bg= θ, 0< θ< A randomly chosen insured is observed to have exactly one claim. Determine the posterior probability that θ is greater than (A) 0.54 (B) 0.58 (C) 0.63 (D) 0.67 (E) The distribution of accidents for 84 randomly selected policies is as follows: Number of Accidents Number of Policies Total 84 Which of the following models best represents these data? (A) Negative binomial 4

15 (B) (C) (D) (E) Discrete uniform Poisson Binomial Either Poisson or Binomial 6. You are given: (i) Low-hazard risks have an exponential claim size distribution with mean θ. (ii) Medium-hazard risks have an exponential claim size distribution with mean θ. (iii) High-hazard risks have an exponential claim size distribution with mean 3θ. (iv) No claims from low-hazard risks are observed. (v) Three claims from medium-hazard risks are observed, of sizes, and 3. (vi) One claim from a high-hazard risk is observed, of size 5. Determine the maximum likelihood estimate of θ. (A) (B) (C) 3 (D) 4 (E) 5 7. You are given: (i) (ii) (iii) (iv) X partial = pure premium calculated from partially credible data = E X partial µ Fluctuations are limited to ± k µ of the mean with probability P Z = credibility factor 5

16 Which of the following is equal to P? (A) (B) (C) Pr µ kµ Xpartial µ + kµ Pr Zµ k ZXpartial Zµ + k Pr Zµ µ ZXpartial Zµ + µ (D) ( ) Pr k ZXpartial + Z µ + k (E) Pr ( ) µ kµ ZXpartial + Z µ µ + kµ 8. You are given: Claim Size (X) Number of Claims b 05, 5 b 5, 50 8 b 50, 00 5 b 00, 00 6 Assume a uniform distribution of claim sizes within each interval. Estimate Ec X h EbX 50g. (A) Less than 00 (B) At least 00, but less than 300 (C) At least 300, but less than 400 (D) At least 400, but less than 500 (E) At least 500 6

17 9. You are given: (i) Each risk has at most one claim each year. (ii) Type of Risk Prior Probability Annual Claim Probability I II III One randomly chosen risk has three claims during Years -6. Determine the posterior probability of a claim for this risk in Year 7. (A) 0. (B) 0.8 (C) 0.33 (D) 0.40 (E) You are given the following about 00 insurance policies in a study of time to policy surrender: (i) The study was designed in such a way that for every policy that was surrendered, a new policy was added, meaning that the risk set, r j, is always equal to 00. (ii) (iii) Policies are surrendered only at the end of a policy year. The number of policies surrendered at the end of each policy year was observed to be: at the end of the st policy year at the end of the nd policy year 3 at the end of the 3 rd policy year n at the end of the n th policy year (iv) The Nelson-Aalen empirical estimate of the cumulative distribution function at time n, F ˆ ( n ), is What is the value of n? 7

18 (A) 8 (B) 9 (C) 0 (D) (E) 3. You are given the following claim data for automobile policies: Calculate the smoothed empirical estimate of the 45th percentile. (A) 358 (B) 37 (C) 384 (D) 390 (E) You are given: (i) The number of claims made by an individual insured in a year has a Poisson distribution with mean λ. (ii) The prior distribution for λ is gamma with parameters α = and θ =.. Three claims are observed in Year, and no claims are observed in Year. Using Bühlmann credibility, estimate the number of claims in Year 3. (A).35 (B).36 (C).40 (D).4 (E).43 8

19 33. In a study of claim payment times, you are given: (i) (ii) (iii) The data were not truncated or censored. At most one claim was paid at any one time. The Nelson-Aalen estimate of the cumulative hazard function, H(t), immediately following the second paid claim, was 3/3. Determine the Nelson-Aalen estimate of the cumulative hazard function, H(t), immediately following the fourth paid claim. (A) 0.35 (B) 0.37 (C) 0.39 (D) 0.4 (E) The number of claims follows a negative binomial distribution with parameters β and r, where β is unknown and r is known. You wish to estimate β based on n observations, where x is the mean of these observations. Determine the maximum likelihood estimate of β. x (A) r (B) (C) (D) (E) x r x rx rx 9

20 35. You are given the following information about a credibility model: First Observation Unconditional Probability Bayesian Estimate of Second Observation /3.50 / / Determine the Bühlmann credibility estimate of the second observation, given that the first observation is. (A) 0.75 (B).00 (C).5 (D).50 (E) For a survival study, you are given: (i) The Product-Limit estimator St 0 St bg. 0 b g is used to construct confidence intervals for (ii) The 95% log-transformed confidence interval for St bg 0 is b0.695, 0.843g. Determine St b 0 g. (A) (B) 0.76 (C) (D) (E)

21 37. A random sample of three claims from a dental insurance plan is given below: Claims are assumed to follow a Pareto distribution with parameters θ = 50 and α. Determine the maximum likelihood estimate of α. (A) Less than 0.6 (B) At least 0.6, but less than 0.7 (C) At least 0.7, but less than 0.8 (D) At least 0.8, but less than 0.9 (E) At least An insurer has data on losses for four policyholders for 7 years. The loss from the i th policyholder for year j is X ij. You are given: 4 7 dxij Xii = i= j= 4 cxi Xh = i= 3.30 Using nonparametric empirical Bayes estimation, calculate the Bühlmann credibility factor for an individual policyholder. (A) Less than 0.74 (B) At least 0.74, but less than 0.77 (C) At least 0.77, but less than 0.80 (D) At least 0.80, but less than 0.83 (E) At least 0.83

22 39. You are given the following information about a commercial auto liability book of business: (i) (ii) (iii) Each insured s claim count has a Poisson distribution with mean λ, where λ has a gamma distribution with α = 5. and θ = 0.. Individual claim size amounts are independent and exponentially distributed with mean The full credibility standard is for aggregate losses to be within 5% of the expected with probability Using classical credibility, determine the expected number of claims required for full credibility. (A) 65 (B) 38 (C) 354 (D) 76 (E) You are given: (i) A sample of claim payments is: (ii) (iii) Claim sizes are assumed to follow an exponential distribution. The mean of the exponential distribution is estimated using the method of moments. Calculate the value of the Kolmogorov-Smirnov test statistic. (A) 0.4 (B) 0.6 (C) 0.9 (D) 0.5 (E) 0.7

23 4. You are given: (i) Annual claim frequency for an individual policyholder has mean λ and variance σ. (ii) The prior distribution for λ is uniform on the interval [0.5,.5]. (iii) The prior distribution for σ is exponential with mean.5. A policyholder is selected at random and observed to have no claims in Year. Using Bühlmann credibility, estimate the number of claims in Year for the selected policyholder. (A) 0.56 (B) 0.65 (C) 0.7 (D) 0.83 (E) You study the time between accidents and reports of claims. The study was terminated at time 3. You are given: Time of Accident Time between Accident and Claim Report Number of Reported Claims Use the Product-Limit estimator to estimate the conditional probability that the time between accident and claim report is less than, given that it does not exceed 3. 3

24 (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 You are given: (i) The prior distribution of the parameter Θ has probability density function: πθ bg= θ < θ < (ii) Given Θ=θ, claim sizes follow a Pareto distribution with parameters α = and θ. A claim of 3 is observed. Calculate the posterior probability that Θ exceeds. (A) 0.33 (B) 0.4 (C) 0.50 (D) 0.58 (E) You are given: (i) Losses follow an exponential distribution with mean θ. (ii) A random sample of 0 losses is distributed as follows: Loss Range Frequency [0, 000] 7 (000, 000] 6 (000, ) 7 4

25 Calculate the maximum likelihood estimate of θ. (A) Less than 950 (B) At least 950, but less than 00 (C) At least 00, but less than 50 (D) At least 50, but less than 400 (E) At least You are given: (i) The amount of a claim, X, is uniformly distributed on the interval 0,θ. (ii) The prior density of θ π θ 500 θ 500 θ,. is bg= > Two claims, x = 400 and x = 600, are observed. You calculate the posterior distribution as: f 3 c h F = H G I K J > θ x, x 3 600, θ θ Calculate the Bayesian premium, Ec X3 x, xh. (A) 450 (B) 500 (C) 550 (D) 600 (E) 650 5

26 46. The claim payments on a sample of ten policies are: indicates that the loss exceeded the policy limit Using the Product-Limit estimator, calculate the probability that the loss on a policy exceeds 8. (A) 0.0 (B) 0.5 (C) 0.30 (D) 0.36 (E) You are given the following observed claim frequency data collected over a period of 365 days: Number of Claims per Day Observed Number of Days Fit a Poisson distribution to the above data, using the method of maximum likelihood. Regroup the data, by number of claims per day, into four groups: 0 3+ Apply the chi-square goodness-of-fit test to evaluate the null hypothesis that the claims follow a Poisson distribution. Determine the result of the chi-square test. 6

27 (A) (B) (C) (D) (E) Reject at the significance level. Reject at the 0.00 significance level, but not at the level. Reject at the 0.05 significance level, but not at the 0.00 level. Reject at the significance level, but not at the 0.05 level. Do not reject at the significance level. 48. You are given the following joint distribution: Θ X For a given value of Θ and a sample of size 0 for X: 0 x i = i= 0 Determine the Bühlmann credibility premium. (A) 0.75 (B) 0.79 (C) 0.8 (D) 0.86 (E)

28 49. You are given: x 0 3 Pr[X = x] The method of moments is used to estimate the population mean, µ, and variance, σ, by X and S i n Calculate the bias of (A) 0.7 (B) 0.49 (C) 0.4 (D) 0.08 (E) 0.00 ( X X ) =, respectively. n S n, when n = You are given four classes of insureds, each of whom may have zero or one claim, with the following probabilities: Class Number of Claims 0 I II III IV A class is selected at random (with probability ¼), and four insureds are selected at random from the class. The total number of claims is two. If five insureds are selected at random from the same class, estimate the total number of claims using Bühlmann-Straub credibility. (A).0 (B). 8

29 (C).4 (D).6 (E).8 5. The following results were obtained from a survival study, using the Product-Limit estimator: t St b g VSt b g Determine the lower limit of the 95% linear confidence interval for x 0.75, the 75 th percentile of the survival distribution. (A) 3 (B) 36 (C) 50 (D) 54 (E) 56 9

30 5. With the bootstrapping technique, the underlying distribution function is estimated by which of the following? (A) (B) (C) (D) (E) The empirical distribution function A normal distribution function A parametric distribution function selected by the modeler Any of (A), (B) or (C) None of (A), (B) or (C) 53. You are given: Number of Claims Probability Claim Size Probability Claim sizes are independent. Determine the variance of the aggregate loss. (A) 4,050 (B) 8,00 (C) 0,500 (D),50 (E) 5,6 30

31 54. You are given: (i) Losses follow an exponential distribution with mean θ. (ii) A random sample of losses is distributed as follows: Loss Range Number of Losses (0 00] 3 (00 00] (00 400] 7 ( ] 6 ( ] ( ] Total 00 Estimate θ by matching at the 80 th percentile. (A) 49 (B) 53 (C) 57 (D) 60 (E) 63 3

32 55. You are given: Class Number of Insureds Claim Count Probabilities A randomly selected insured has one claim in Year. Determine the expected number of claims in Year for that insured. (A).00 (B).5 (C).33 (D).67 (E) You are given the following information about a group of policies: Claim Payment Policy Limit Determine the likelihood function. 3

33 (A) (B) (C) (D) (E) f(50) f(50) f(00) f(00) f(500) f(000) f(50) f(50) f(00) f(00) f(500) f(000) / [-F(000)] f(5) f(5) f(60) f(00) f(500) f(500) f(5) f(5) f(60) f(00) f(500) f(500) / [-F(000)] f(5) f(5) f(60) [-F(00)] [-F(500)] f(500) 57. You are given: Claim Size Number of Claims Assume a uniform distribution of claim sizes within each interval. Estimate the second raw moment of the claim size distribution. (A) Less than 3300 (B) At least 3300, but less than 3500 (C) At least 3500, but less than 3700 (D) At least 3700, but less than 3900 (E) At least

34 58. You are given: (i) The number of claims per auto insured follows a Poisson distribution with mean λ. (ii) (iii) The prior distribution for λ has the following probability density function: λ e fbλg b500 g = λγ b 50 g λ A company observes the following claims experience: Year Year Number of claims 75 0 Number of autos insured The company expects to insure 00 autos in Year 3. Determine the expected number of claims in Year 3. (A) 78 (B) 84 (C) 93 (D) 09 (E) 4 34

35 59. The graph below shows a q-q plot of a fitted distribution compared to a sample. Fitted Which of the following is true? Sample (A) (B) (C) (D) (E) The tails of the fitted distribution are too thick on the left and on the right, and the fitted distribution has less probability around the median than the sample. The tails of the fitted distribution are too thick on the left and on the right, and the fitted distribution has more probability around the median than the sample. The tails of the fitted distribution are too thin on the left and on the right, and the fitted distribution has less probability around the median than the sample. The tails of the fitted distribution are too thin on the left and on the right, and the fitted distribution has more probability around the median than the sample. The tail of the fitted distribution is too thick on the left, too thin on the right, and the fitted distribution has less probability around the median than the sample. 60. You are given the following information about six coins: Coin Probability of Heads

36 A coin is selected at random and then flipped repeatedly. X i denotes the outcome of the ith flip, where indicates heads and 0 indicates tails. The following sequence is obtained: Determine Ec X5 Sh using Bayesian analysis. (A) 0.5 (B) 0.54 (C) 0.56 (D) 0.59 (E) 0.63 l 3 4q l0q S = X, X, X, X =,,, 6. You observe the following five ground-up claims from a data set that is truncated from below at 00: You fit a ground-up exponential distribution using maximum likelihood estimation. Determine the mean of the fitted distribution. (A) 73 (B) 00 (C) 5 (D) 56 (F) 73 36

37 6. An insurer writes a large book of home warranty policies. You are given the following information regarding claims filed by insureds against these policies: (i) (ii) (iii) (iv) A maximum of one claim may be filed per year. The probability of a claim varies by insured, and the claims experience for each insured is independent of every other insured. The probability of a claim for each insured remains constant over time. The overall probability of a claim being filed by a randomly selected insured in a year is 0.0. (v) The variance of the individual insured claim probabilities is 0.0. An insured selected at random is found to have filed 0 claims over the past 0 years. Determine the Bühlmann credibility estimate for the expected number of claims the selected insured will file over the next 5 years. (A) 0.04 (B) 0.08 (C) 0.7 (D) 0. (E) A study of the time to first claim includes only policies issued during 996 through 998 on which claims occurred by the end of 999. The table below summarizes the information about the 50 policies included in the study: Number of Policies Year of Time to First Claim Issue year years 3 years

38 Use the Product-Limit estimator to estimate the conditional probability that the first claim on a policy occurs less than years after issue given that the claim occurs no later than 3 years after issue. (A) Less than 0.0 (B) At least 0.0, but less than 0.5 (C) At least 0.5, but less than 0.30 (D) At least 0.30, but less than 0.35 (E) At least For a group of insureds, you are given: (i) (ii) (iii) The amount of a claim is uniformly distributed but will not exceed a certain unknown limit θ. The prior distribution of θ is π θ 500 θ 500 θ,. bg= > Two independent claims of 400 and 600 are observed. Determine the probability that the next claim will exceed 550. (A) 0.9 (B) 0. (C) 0.5 (D) 0.8 (E)

39 65. You are given the following information about a general liability book of business comprised of 500 insureds: (i) X = Y i N i j= ij is a random variable representing the annual loss of the i th insured. (ii) N, N,..., N500 are independent and identically distributed random variables following a negative binomial distribution with parameters r = and β = 0.. (iii) Yi, Yi,..., YiN i are independent and identically distributed random variables following a Pareto distribution with α = 30. and θ = 000. (iv) The full credibility standard is to be within 5% of the expected aggregate losses 90% of the time. Using classical credibility theory, determine the partial credibility of the annual loss experience for this book of business. (A) 0.34 (B) 0.4 (C) 0.47 (D) 0.50 (E) To estimate E X, you have simulated X, X, X3, X4 and X5with the following results: You want the standard deviation of the estimator of E X to be less than Estimate the total number of simulations needed. (A) Less than 50 (B) At least 50, but less than 400 (C) At least 400, but less than 650 (D) At least 650, but less than 900 (E) At least

40 67. You are given the following information about a book of business comprised of 00 insureds: i N i (i) X = Y is a random variable representing the annual loss of the i th insured. j= ij (ii) N, N,..., N00are independent random variables distributed according to a negative binomial distribution with parameters r (unknown) and β = 0.. (iii) Unknown parameter r has an exponential distribution with mean. (iv) Yi, Yi,..., YiN i are independent random variables distributed according to a Pareto distribution with α = 30. and θ = 000. Determine the Bühlmann credibility factor, Z, for the book of business. (A) (B) (C) (D) 0.86 (E) For a mortality study of insurance applicants in two countries, you are given: (i) Country A Country B t i d i Y i θ i d i Y i θ i (ii) Y i is the number at risk over the period bti, tig. Deaths during the period bti, tig are assumed to occur at t i. (iii) θ i is the reference hazard rate over the period ti, tig. Within a country, θ i is the same for all study participants. b 40

41 (iv) S T btg is the Product-Limit estimate of St b g based on the data for all study participants. (v) S B btg is the Product-Limit estimate of St b g based on the data for study participants in Country B. T B Determine S b4g S b4g. (A) 0.06 (B) 0.07 (C) 0.08 (D) 0.09 (E) You fit an exponential distribution to the following data: Determine the coefficient of variation of the maximum likelihood estimate of the mean, θ. (A) 0.33 (B) 0.45 (C) 0.70 (D).00 (E). 4

42 70. You are given the following information on claim frequency of automobile accidents for individual drivers: Business Use Pleasure Use Expected Claims Claim Variance Expected Claims Claim Variance Rural Urban Total You are also given: (i) (ii) Each driver s claims experience is independent of every other driver s. There are an equal number of business and pleasure use drivers. Determine the Bühlmann credibility factor for a single driver. (A) 0.05 (B) 0.09 (C) 0.7 (D) 0.9 (E) 0.7 4

43 7. You are investigating insurance fraud that manifests itself through claimants who file claims with respect to auto accidents with which they were not involved. Your evidence consists of a distribution of the observed number of claimants per accident and a standard distribution for accidents on which fraud is known to be absent. The two distributions are summarized below: Number of Claimants per Accident Standard Probability Observed Number of Accidents Total Determine the result of a chi-square test of the null hypothesis that there is no fraud in the observed accidents. (A) (B) (C) (D) (E) Reject at the significance level. Reject at the 0.00 significance level, but not at the level. Reject at the 0.05 significance level, but not at the 0.00 level. Reject at the significance level, but not at the 0.05 level. Do not reject at the significance level. 7. You are given the following data on large business policyholders: (i) Losses for each employee of a given policyholder are independent and have a common mean and variance. (ii) The overall average loss per employee for all policyholders is 0. (iii) The variance of the hypothetical means is 40. (iv) The expected value of the process variance is (v) The following experience is observed for a randomly selected policyholder: 43

44 Year Average Loss per Number of Employee Employees Determine the Bühlmann-Straub credibility premium per employee for this policyholder. (A) Less than 0.5 (B) At least 0.5, but less than.5 (C) At least.5, but less than.5 (D) At least.5, but less than 3.5 (E) At least You are given the following information about a group of 0 claims: Claim Size Interval Number of Claims in Interval Number of Claims Censored in Interval (0-5,000] (5,000-30,000] (30,000-45,000] 4 0 Assume that claim sizes and censorship points are uniformly distributed within each interval. Estimate, using the life table methodology, the probability that a claim exceeds 30,000. (A) 0.67 (B) 0.70 (C) 0.74 (D) 0.77 (E)

45 74. You are making credibility estimates for regional rating factors. You observe that the Bühlmann-Straub nonparametric empirical Bayes method can be applied, with rating factor playing the role of pure premium. X ij denotes the rating factor for region i and year j, where i = 3,, and j = 34,,,. Corresponding to each rating factor is the number of reported claims, m ij, measuring exposure. You are given: i mi = m 4 j= ij X i = m i 4 j= mx ij ij v = m X X i 4 3 j= ijd ij ii mi Xi X c h Determine the credibility estimate of the rating factor for region using the method that preserves i= (A).3 (B).33 (C).35 (D).37 3 (E).39 mx i i. 45

46 75. You are given: (i) Claim amounts follow a shifted exponential distribution with probability density function: x fbg x = eb δg/ θ, δ < x < θ (ii) A random sample of claim amounts X, X,..., X0: (iii) X i = 00 and X i = 306 Estimate δ using the method of moments. (A) 3.0 (B) 3.5 (C) 4.0 (D) 4.5 (E) You are given: (i) (ii) The annual number of claims for each policyholder follows a Poisson distribution with mean θ. The distribution of θ across all policyholders has probability density function: f θ θ θ b g = e, θ > 0 z nθ (iii) θe dθ = n 0 A randomly selected policyholder is known to have had at least one claim last year. Determine the posterior probability that this same policyholder will have at least one claim this year. 46

47 (A) 0.70 (B) 0.75 (C) 0.78 (D) 0.8 (E) A survival study gave (.63,.55) as the 95% linear confidence interval for the cumulative hazard function Ht 0 bg. Calculate the 95% log-transformed confidence interval for Ht bg. 0 (A) (0.49, 0.94) (B) (0.84, 3.34) (C) (.58,.60) (D) (.68,.50) (E) (.68,.60) 78. You are given: (i) Claim size, X, has mean µ and variance 500. (ii) The random variable µ has a mean of 000 and variance of 50. (iii) The following three claims were observed: 750, 075, 000 Calculate the expected size of the next claim using Bühlmann credibility. (A) 05 (B) 063 (C) 5 (D) 8 (E) 66 47

48 79. Losses come from a mixture of an exponential distribution with mean 00 with probability p and an exponential distribution with mean 0,000 with probability p. Losses of 00 and 000 are observed. Determine the likelihood function of p. (A) (B) (C) (D) F HG F HG F HG F HG b g I F b g KJ HG b g I F b g KJ + HG b g I F +. b g + 0, 000 KJ HG 00, b g I F b + KJ + + g 0, 000 HG 00, pe. pe. pe. pe 00 0, , pe. pe pe. pe 00 0, , pe pe pe pe pe pe pe pe F HG I b g F HG (E) p e e e e. + p., KJ , 000 I KJ I KJ I KJ I KJ I KJ 80. A fund is established by collecting an amount P from each of 00 independent lives age 70. The fund will pay the following benefits: 0, payable at the end of the year of death, for those who die before age 7, or P, payable at age 7, to those who survive. You are given: (i) Mortality follows the Illustrative Life Table. (ii) i =

49 For this question only, you are also given: The number of claims in the first year is simulated from the binomial distribution using the inverse transform method (where smaller random numbers correspond to fewer deaths). The random number for the first trial, generated using the uniform distribution on [0, ], is 0.8. Calculate the simulated claim amount. (A) 0 (B) 0 (C) 0 (D) 30 (E) You wish to simulate a value, Y, from a two point mixture. With probability 0.3, Y is exponentially distributed with mean 0.5. With probability 0.7, Y is uniformly distributed on 3, 3. You simulate the mixing variable where low values correspond to the exponential distribution. Then you simulate the value of Y, where low random numbers correspond to low values of Y. Your uniform random numbers from 0, are 0.5 and 0.69 in that order. Calculate the simulated value of Y. (A) 0.9 (B) 0.38 (C) 0.59 (D) 0.77 (E)

50 8. N is the random variable for the number of accidents in a single year. N follows the distribution: Pr( N n) 0.9(0.) n = =, n =,, X i is the random variable for the claim amount of the ith accident. distribution: X i follows the 0.0 ( ) x i g x = 0.0 e, x > 0, i=,, i Let U and V, V,... be independent random variables following the uniform distribution on (0, ). You use the inverse transformation method with U to simulate N and V i to simulate X i with small values of random numbers corresponding to small values of N and X i. You are given the following random numbers for the first simulation: u v v v 3 v 4 i Calculate the total amount of claims during the year for the first simulation. (A) 0 (B) 36 (C) 7 (D) 08 (E) You are the consulting actuary to a group of venture capitalists financing a search for pirate gold. It s a risky undertaking: with probability 0.80, no treasure will be found, and thus the outcome is 0. The rewards are high: with probability 0.0 treasure will be found. The outcome, if treasure is found, is uniformly distributed on [000, 5000]. You use the inverse transformation method to simulate the outcome, where large random numbers from the uniform distribution on [0, ] correspond to large outcomes. 50

51 Your random numbers for the first two trials are 0.75 and Calculate the average of the outcomes of these first two trials. (A) 0 (B) 000 (C) 000 (D) 3000 (E)

52 SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM C CONSTRUCTION AND EVALUATION OF ACTUARIAL MODELS EXAM C SAMPLE SOLUTIONS Copyright 005 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study note are taken from past SOA/CAS examinations. C PRINTED IN U.S.A.

53 Question # Key: E The 40 th percentile is the.4() = 4.8 th smallest observation. By interpolation it is.(86) +.8(90) = 89.. The 80 th percentile is the.8() = 9.6 th smallest observation. By interpolation it is.4(00) +.6(0) = 06. The equations to solve are γ γ (89. / θ) (06 / θ).4 = and.8 = γ + (89. / θ ) + (06 / θ ) γ. Solving each for the parenthetical expression gives γ γ = (89. / θ ) and 4 = (06 / θ ). 3 Taking the ratio of the second equation to the first gives 6 = (06 / 89.) γ which leads to γ = ln(6) / ln(06 / 89.) =.407. Then / = 06 /θ for θ = Question # Key: E.645 Var( X ) The standard for full credibility is + where X is the claim size.0 EX ( ) (.5) variable. For the Pareto variable, EX ( ) =.5/ 5 =. and Var( X ) = (.) =.05. 5(4) Then the standard is + = 6,93 claims..0. Question #3 Key: B The kernel is a triangle with a base of 4 and a height at the middle of 0.5 (so the area is ). The length of the base is twice the bandwidth. Any observation within of.5 will contribute to the estimate. For the observation at, when the triangle is centered at, the height of the triangle at.5 is.375 (it is one-quarter the way from to the end of the triangle at 4 and so the height is one-quarter the way from 0.5 to 0). Similarly the points at 3 are also 0.5 away and so the height of the associated triangle is also.375. Each triangle height is weighted by the empirical probability at the associated point. So the estimate at.5 is (/5)(3/8) + (3/5)(3/8) + (/5)(0) = /40.

54 Question #4 Key: A x The distribution function is F ( x ) = αt α dt = t α = x α. The likelihood function is L= f(3) f(6) f(4)[ F(5)] α α α α = α3 α6 α4 (5 ) 3 α α [3(6)(4)(65)]. Taking logs, differentiating, setting equal to zero, and solving: ln L = 3lnα α ln57,500 plus a constant (ln L) = 3α ln57,500 = 0 ˆ α = 3/ ln57,500 =.507. x Question #5 Key: C π( q,) p( q) p( q) π( q) = q( q) q( q)4 q q ( q) q ( q) dq= /68, π ( q,) = 68 q ( q). 3 5 The expected number of claims in a year is EX ( q) = qand so the Bayesian estimate is 5 (,) = (68) ( ) = 4 / 3. 0 E q q q q dq The answer can be obtained without integrals by recognizing that the posterior distribution of q is beta with a = 6 and b = 3. The posterior mean is Eq (,) = a/( a+ b) = 6/9= /3. The posterior mean of q is then 4/3. Question #6 Key: D For the method of moments estimate, µ +.5σ µ + σ 386 = e, 457,480. = e = µ +.5 σ, = µ + σ ˆ µ = , ˆ σ =.8. Then

55 (.8) ln ln EX ( 500) = e Φ Φ.8.8 = 386 Φ(.853) + 500[ Φ(.7739)] = 386(.3877) + 500(.95) = 59. Note-these calculations use exact normal probabilities. Rounding and using the normal table that accompanies the exam will produce a different numerical answer but the same letter answer. Question #7 Key: D Because the values are already ranked, the test statistic is immediately calculated as the sum of the given values for Sample I: R = = 78. The other needed values are n = 9 and m =, the two sample sizes. The mean is n(n+m+)/ = 94.5 and the variance is nm(n+m+)/ = The test statistic is Z = ( ) / 73.5 =.5. The p-value is twice (because it is a two-tailed test) the probability of being more extreme than the test statistic, p= Pr( Z <.5) =.0. Question #8 Key: C Let N be the Poisson claim count variable, let X be the claim size variable, and let S be the aggregate loss variable. µθ ( ) = ES ( θ) = EN ( θ) EX ( θ) = θ0θ= 0θ 3 v( θ) = Var( S θ) = E( N θ) E( X θ) = θ00θ = 00θ 6 d θ θ θ θ 6 θ θ θ θ µ = E(0 θ ) = 0 θ (5 θ ) θ = 50 / 3 EPV = E(00 ) = 00 (5 ) d = 500 VHM = Var(0 ) = (0 ) (5 ) d (50 / 3) =. k = 500 /. =.5. Question #9 Key: A c = exp(.7() +.0()) = Then St ˆ z = Sˆ t = =. c.4843 ( 0; ) 0( 0)

56 Question #0 Key: E Y and X are linear combinations of the same two normal random variables, so they are bivariate normal. Thus E(Y X) = E(Y) + [Cov(Y,X)/Var(X)][X E(X)]. From the definitions of Y and X, E(Y) = a, E(X) = d, Var(X) = e + f, and Cov(Y,X) = be + cf. Question # Key: D f (5 θ = ) Pr( θ = ) Pr( θ = X = 5) = f(5 θ = ) Pr( θ = ) + f(5 θ = 3) Pr( θ = 3) (/ 36)(/ ) = = 6 / 43 (/ 36)(/ ) + (3/ 64)(/ ) Pr( X > 8 X = 5) = Pr( X > 8 θ = )Pr( θ = X = 5) + Pr( X > 8 θ = 3)Pr( θ = 3 X = 5) = (/ 9)(6 / 43) + (3/)(7 / 43) =.6. For the last line, is used. 8 Pr( X > 8 θ ) = θ( x+ θ) dx= θ(8 + θ) Question # Key: C The sample mean for X is 70 and for Y is 670. The mean of all 8 observations is 695. (730 70) + (800 70) + (650 70) + (700 70) + ( ) + ( ) + (65 670) + ( ) vˆ = = 3475 (4 ) (70 695) + ( ) 3475 aˆ = = kˆ = 3475/ 38.5 = 9.48 ˆ 4 Z = = P =.305(670) +.695(695) = c Question #3 Key: B There are 430 observations. The expected counts are 430(.744) = 7.99, 430(.35) =5.0, 430(.3744) = The test statistic is

57 ( 7.99) (80 5.0) ( ) + + = Question #4 Key: B From the information, the asymptotic variance of ˆ θ is /4n. Then Var( ˆ θ) = 4 Var( ˆ θ) = 4(/4 n) = / n. Note that the delta method is not needed for this problem, although using it leads to the same answer. Question #5 Key: A The posterior probability density is 8 8 π( p,,,,,,,) Pr(,,,,,,, p) π( p) p () p. π p ( p,,,,,,,) = = = 9(.5 ) p pdp (.5 ) / = = 0 9 = π p9(.5 ) p dp 9(.5 )(.5 ) / Pr( X,,,,,,,) Pr( X p) ( p,,,,,,,) dp = = = p Question #6 Key: A p ˆ = =. Greenwood s approximation is = (7) 6(3) 0(5)

58 Question #7 Key: D Ĥ (3) = 5/30 + 9/7 + 6/3 = Var ˆ ( H ˆ (3)) = 5/(30) + 9/(7) + 6/(3) = The 95% log-transformed confidence interval is: H ˆ (3) U, where U = exp ± = exp( ± ).6875 The confidence interval is: [ exp( ), exp( )] = [0.443,.067]. Question #8 Key: D The means are.5(50) +.3(,500) +.(60,000) =,875 and.7(50) +.(,500) +.(60,000) = 6,675 for risks and respectively. The variances are.5(50) +.3(,500) +.(60,000),875 = 556,40,65 and.7(50) +.(,500) +.(60,000) 6,675 = 36,738,5 respectively. The overall mean is (/3)(,875) + (/3)(6,675) = 0, and so EPV = (/3)(556,40,65) + (/3)(36,738,5) = 476,339,79 and VHM = (/3)(,875) + (/3)(6,675) 0, = 8,54,. Then, k = 476,339,79/8,54, = and Z = /( ) = The credibility estimate is.0767(50) (0,808.33) = 0,6. Question #9 Key: D The first two sample moments are 5 and 500, and the first two population moments are EX ( ) =.5( θ + σ ) and EX ( ) =.5(θ + σ ) = θ + σ. These can be obtained either through integration or by recognizing the density function as a two-point mixture of exponential densities. The equations to solve are 30 = θ + σ and 500 = θ + σ. From the first equation, σ = 30 θ and substituting into the second equation gives 500 = θ + (30 θ) = θ 60θ The quadratic equation has two solutions, 0 and 0. Because θ > σ the answer is 0.

59 Question #0 Key: D There are four possible samples, (5,5), (5,9), (9,5), and (9,9). For each, the estimator g must be calculated. The values are 0, 4, 4, and 0 respectively. Assuming a population in which the values 5 and 9 each occur with probability.5, the population variance is.5(5 7) +.5(9 7) = 4. The mean square error is approximated as.5[(0 4) + (4 4) + (4 4) + (0 4) ] = 8. Question # Key: B From the Poisson distribution, µ ( λ) = λ and v( λ) = λ. Then, = E( ) = 6 /00 =.06, EPV = E( ) =.06, VHM = Var( ) = 6 /00 =.0006 where the µ λ λ λ various moments are evaluated from the gamma distribution. Then, k =.06 /.0006 = 00 and Z = 450 /( ) = 9 / where the 450 is the total number of insureds contributing experience. The credibility estimate of the expected number of claims for one insured in month 4 is (9/)(5/450) + (/)(.06) = For 300 insureds the expected number of claims is 300( ) = 6.9. Question # Key: C The likelihood function is L( αθ, ) = αθ α 00 and its logarithm is j= α ( x j + θ ) + 00 i= l ( α, θ ) = 00ln( α ) + 00 α ln( θ ) ( α + ) ln( x + θ ). When evaluated at the hypothesized values of.5 and 7.8, the loglikelhood is The test statistic is ( ) = 7.7. With two degrees of freedom (0 free parameters in the null hypothesis versus in the alternative), the test statistic falls between the 97.5 th percentile (7.38) and the 99 th percentile (9.). i

60 Question #3 Key: E Assume that θ > 5. Then the expected counts for the three intervals are 5( / θ ) = 30 / θ,5(3/ θ) = 45/ θ, and 5( θ 5) / θ = 5 75/ θ respectively. The quantity to minimize is (30 5) (45 5) (5 75 5) θ + θ + θ 5. Differentiating (and ignoring the coefficient of /5) gives the equation 3 (30θ 5)30θ (45θ 5)45θ + (0 75 θ )75θ = 0. Multiplying through by θ and dividing by reduces the equation to (30 5 θ )30 (45 5 θ)45 + (0θ 75)75 = θ = 0 for a solution of ˆ θ = 8550 /5 = 7.6. Question #4 Key: E π ( θ ) θ(.5 θ ) θ.5.5. The required constant is the reciprocal of.5 and so π ( θ ) =.5θ. The requested probability is Pr( θ >.6 ) =.5 θ dθ = θ =.6 = θ dθ = θ /.5 =.4 0 Question #5 Key: A k / k k kn n Positive slope implies that the negative binomial distribution is a good choice. Alternatively, the sample mean and variance are.6 and.93 respectively. With the variance substantially exceeding the mean, the negative binomial model is again supported.

61 Question #6 Key: B /( θ ) /( θ) 3/( θ) 5/(3 θ) 8/ θ e e e e e The likelihood function is =. The loglikelihood 4 θ θ θ 3θ 4θ function is ln 4 4 ln( θ ) 8/ θ. Differentiating with respect to θ and setting the result equal to 0 yields θ + θ = which produces ˆ θ =. Question #7 Key: E The absolute difference of the credibility estimate from its expected value is to be less than or equal to kµ (with probability P). That is, [ ZX partial + ( Z) M] [ Zµ + ( Z) M] kµ kµ ZX partial Zµ kµ. Adding µ to all three sides produces answer choice (E). Question #8 Key: C In general, E( X ) E[( X 50) ] = x f ( x) dx x f ( x) dx 50 f ( x) dx = ( x 50 ) f ( x) dx Assuming a uniform distribution, the density function over the interval from 00 to 00 is 6/7400 (the probability of 6/74 assigned to the interval divided by the width of the interval). The answer is x 6 ( x 50 ) dx= 50 x = Question #9 Key: B The probabilities are from a binomial distribution with 6 trials. Three successes were observed Pr(3 I) = (.) (.9) =.0458, Pr(3 II) = (.) (.8) =.089, Pr(3 III) = (.4) (.6) =

62 The probability of observing three successes is.7(.0458) +.(.089) +.(.7648) = The three posterior probabilities are:.7(.0458).(.089).(.7648) Pr(I 3) = =.887, Pr(II 3) = =.3008, Pr(III 3) = = The posterior probability of a claim is then.(.887) +.(.3008) +.4(.50975) =.833. Question #30 Key: E ˆ ˆ H.54 ( ) ( n = Fn= e ), Hn ˆ( ) =.78. The Nelson-Aalen estimate is the sum of successive s/r values. From the problem statement, r = 00 at all surrender times while the s-values follow the pattern,, 3,. Then, n n( n+ ).78 = = and the solution is n = Question # 3 Answer: C g = [(.45)] = [5.4] = 5; h= = 0.4. ˆ π =.6 x +.4 x =.6(360) +.4(40) = (5) (6) Question # 3 Answer: D N is distributed Poisson(λ) µ = E( λ) = αθ = (.) =.. v= E λ = a= Var λ = αθ = =. 5 k = = ; Z = = / 6 7 Thus, the estimate for Year 3 is 5 (.5) + (.) = ( ).; ( ) (.).44. Note that a Bayesian approach produces the same answer. Question # 33 Answer: C

63 At the time of the second failure, ˆ 3 Ht ( ) = n. n + n = 3 = At the time of the fourth failure, ˆ Ht ( ) = = Question # 34 Answer: B The likelihood is: n x j rr ( + ) ( r+ x ) n j β r+ x j j= x j!( + β ) j= x r x j j L = β ( + β). The loglikelihood is: n l = xjln β ( r+ xj)ln( + β) j= n xj r+ xj l = = 0 j= β + β n 0 = x ( + β ) ( r+ x ) β = x rnβ j j j j= j= 0 = nx rnβ; ˆ β = x / r. n Question # 35 Answer: C The Bühlmann credibility estimate is Zx+ ( Z ) µ where x is the first observation. The Bühlmann estimate is the least squares approximation to the Bayesian estimate. Therefore, Z and µ must be selected to minimize + µ + + µ + + µ [ Z ( Z).5] [ Z ( Z).5] [3 Z ( Z) 3]. Setting partial derivatives equal to zero will give the values. However, it should be clear that µ is the average of the Bayesian estimates, that is,

64 µ = ( ) =. 3 The derivative with respect to Z is (deleting the coefficients of /3): ( Z +.5)( ) + (.5)(0) + ( Z )() = 0 Z =.75. The answer is.75() +.5() =.5. Question # 36 Answer: E / The confidence interval is ˆ θ ˆ θ ( St ( 0), St ( 0) ). Taking logarithms of both endpoints gives the two equations ln.695 = = ln St ˆ( 0) θ ln.843 =.7079 = θ ln St ˆ( ). 0 Multiplying the two equations gives.064 = [ln St ˆ( 0)] ln St ˆ( 0) =.498 St ˆ( 0 ) = The negative square root is required in order to make the answer fall in the interval (0,). Question # 37 Answer: B The likelihood is: α α α α50 α50 α50 L = (50 + 5) ( ) ( ) 3 3α α 50 =. α + (375i675i00) α+ α+ α+

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