Solutions to the Fall 2017 CAS Exam S

Transcription

1 Solutions to the Fall 2017 CAS Exam S (Incorporating the Final CAS Answer Key) There were 45 questions in total, of equal value, on this 4 hour exam. There was a 15 minute reading period in addition to the 4 hours. The Exam S is copyright 2017 by the Casualty Actuarial Society. The exam is available from the CAS. The solutions and comments are solely the responsibility of the author. CAS Exam S prepared by Howard C. Mahler, FCAS Copyright 2017 by Howard C. Mahler. The following questions are not on the syllabus of MAS-1: 25 to 30. Howard Mahler hmahler@mac.com

2 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 1 1. You are given the following information: Coins are tossed into a fountain according to a Poisson process at a rate of one every two minutes. The coin denominations are independently distributed as follows: Coin Denomination Probability Penny 0.50 Nickel 0.20 Dime 0.20 Quarter 0.10 Calculate the probability that two Quarters will be tossed into the fountain before four non-quarter coins. A. Less than 0.06 B. At least 0.06, but less than 0.07 C. At least 0.07 but less than 0.08 D. At least 0.08, but less than 0.09 E. At least D. The probability of a quarter is 0.1 and the probability if a non-quarter is 0.9. We require that at least 2 of the first 6-1 = 5 coins be quarters. The number of coins out of the first five that are quarters is Binomial with parameters q = 0.1 and m = 5. Thus the desired probability is: 5 2 (0.12)(0.93) (0.13)(0.92) (0.1 4 )(0.9) = Alternately, the chance that n claims from the first process occur before m claims from the second process is F(m-1) for a Negative Binomial Distribution with parameters r = n and β = λ 2 / λ 1. In this case, we want F(4-1) = F(3) for a Negative Binomial with parameters r = 2 and β = 0.9/0.1 = 9. F(3) = 1/ (2) 9/ {(2)(3)/2} 9 2 / {(2)(3)(4)/6} 9 3 /10 5 = Comment: The given rate of the Poisson process is irrelevant.

3 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 2 2. You are given the following information: A reinsurer classifies three types of events as catastrophes: hurricanes, earthquakes and wildfires. Each type of catastrophe follows a homogeneous Poisson process with the following rates of occurrence: Hurricanes: 5 per year Earthquakes: 1 per 5 years Wildfires: 1 per year Calculate the probability that 2 or more catastrophes occur in a six-month period. A. Less than 0.80 B. At least 0.80, but less than 0.81 C. At least 0.81, but less than 0.82 D. At least 0.82, but less than 0.83 E. At least C. The number of catastrophes over 6 months is Poisson. The expected number of catastrophes over 6 months (one half year) is: (1/2)(5) + (1/2)(1/5) + (1/2)(1) = 3.1. Prob[at least 2 catastrophes] = 1 - e e -3.1 =

4 3. You are given: A Poisson process N has a rate function: λ(t) = 3t 2. You've already observed 50 events by time t = 2.1. Calculate the conditional probability, Pr[N(3)=68 I N(2.1)=50]. A. Less than 5% B. At least 5%, but less than 10% C. At least 10%, but less than 15% D. At least 15%, but less than 20% E. At least 20% 3. B. What happens after 2.1 time is independent of what happened up to time 2.1. Thus we need the probability of exactly = 18 events from time 2.1 to 3. The number of events from time 2.1 to 3 is Poisson with mean: 3 3 λ(t) dt = 3t 2 dt Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 3 = t 3 ] t = 3 = = t = 2.1 f(18) = λ 18 e λ / 18! = e / 18! = 9.34%.

5 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 4 4. You are given the following information: Taxicabs leave a hotel with a group of passengers at a Poisson rate of 10 per hour. The number of people in each group taking a cab is independent and has the following probability: Number of people Probability Calculate the probability that at least 800 people leave the hotel in a cab during a 48-hour period using the normal approximation. A. Less than B. At least 0.200, but less than C. At least 0.210, but less than D. At least 0.220, but less than E. At least C. Mean severity is: (0.6)(1) + (0.25)(2) + (0.1)(3) + (0.05)(4) = 1.6. Second moment of severity is: (0.6)(1 2 ) + (0.25)(2 2 ) + (0.1)(3 2 ) + (0.05)(4 2 ) = 3.3. The mean aggregate is: (48)(10)(1.6) = 768. The variance of the aggregate is: (48)(10)(3.3) = Thus using the continuity correction, the probability of at least 800 people is approximately: Φ[ ] = 1 - Φ[0.7915] = 21.4%. 1584

6 5. You are given: The failure (hazard) rate function, λ(t) λ(t) = α t + θ ; for t > 0 The density function, f(t) αθ f(t) = α (t + θ) α + 1; for t > 0 α = 3 θ = 5 Calculate S(15), the probability a life age 0 lives to age 15. A. Less than B. At least 0.020, but less than C. At least 0.040, but less than D. At least 0.060, but less than E. At least A. The given density is that of a Pareto Distribution. Thus F(t) = 1 - S(15) = 5 3 (5 +15) 3 = 1.56%. Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 5 θ α (t + θ) α. Alternately, H(15) = λ(t) dt = t + 5 dt 0 0 = 3ln[t + 5] ] t = 15 t = 0 = 3 ln[20] - 3 ln[5] = S(15) = exp[-h(15)] = e = 1.56%. Alternately, λ(t) = f(t)/s(t). S(15) = f(15) / λ(15) = (3)(53 )/ / 20 = 1.56%.

7 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 6 6. A customer calls a customer service center with two servers. Both servers are busy at the time of the call, therefore the customer is placed in a queue that is currently empty. The next available server will handle this customer's call. You are given: Service times are exponentially distributed and independent. Server 1 is more experienced, so she handles calls in 5 minutes on average. Server 2 is less experienced, so she handles calls in 7 minutes on average. Calculate the expected total waiting plus service time, in minutes, until this customer is finished. A. Less than 7 B. At least 7, but less than 8 C. At least 8, but less than 9 D. At least 9, but less than 10 E. At least C. The combined service process has a rate of: 1/5 + 1/7 = 12/35. Thus the customer's average wait to be served is: 35/12 = /(5 + 7) = 7/12 chance that the customer gets served by the first server, and 5/(5+7) = 5/12 chance the customer gets served by the second server. Thus the average time the customer is being served is: (5)(7/12) + (7)(5/12) = 70/12 = Thus the expected total waiting plus service time is: = minutes. Comment: See Example 5.8 in An Introduction to Probability Models by Ross. In general with n servers, the average time until the customer is done being served is: (n+1) / (sum of their failure rates). In this case: (2+1) / (1/5 + 1/7) = 8.75.

8 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 7 7. A produce vendor must decide how many oranges to order per week in order to maximize her profit. You are given: The cost of oranges to the vendor is 0.50 per orange. The vendor is able to sell oranges at 1.00 per orange. Quantity demanded follows an exponential distribution with a θ of per week. Any inventory left over at the end of the week will rot and is worthless. There is no penalty to the vendor if she cannot meet all of the demand. Calculate the optimal amount of oranges the vendor should purchase in a week to maximize profit. A. Less than 650 B. At least 650 but less than 675 C. At least 675 but less than 700 D. At least 700 but less than 725 E. At least A and C. (See Comment) If the vendor buys y oranges and the demand is x, then the profit is: Min[x. y] y. Thus the expected profit is: E[X y] y. For an Exponential with mean θ, the limited expected value is: E[X y] = θ (1 - e -y/θ ). Then the expected profit is: (1 - e -y/0.001 ) - 0.5y. In order to maximize the expected profit, set the partial derivative with respect to θ equal to zero: 0 = e -1000y y = ln(2)/1000 = Alternately, assume the exam question intended to say a rate of and thus a mean of Then the expected profit is: 1000 (1 - e -y/1000 ) - 0.5y. In order to maximize the expected profit, set the partial derivative with respect to θ equal to zero: 0 = e -y/ y = 1000 ln(2) = 693. Comment: The CAS accepted either choice A or C. As written this exam question makes no sense; if on average the demand were per week, then there is no reason for the vendor to buy any oranges. Probably this exam question meant to say hazard rate of λ of per week. See Example 5.5 in An Introduction to Probability Models by Ross.

9 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 8 8. A 3-out-of-50 system is placed in parallel with a 48-out-of-50-system to form a combined system. Calculate the number of minimal path sets for the combined system. A. Fewer than 20,000 B. At least 20,000, but fewer than 30,000 C. At least 30,000, but fewer than 40,000 D. At least 40,000, but fewer than 50,000 E. At least 50, B. Since the two systems are in parallel, the combined systems fails if and only if both systems fail. The first system fails if fewer than three components operate. The second system fails if fewer than 48 components operate. A minimal path set is such that the system functions if all of the components of the set function, but the system would not function if any component in the set fails. So any set of three components of the first system would be a minimal path set for the combined system. Similarly, any set of 48 components of the second system would be a minimal path set for the combined system. Thus the number of minimal path sets is: = 19, = 20, Comment; If instead the two systems were put in series to form a combined system, then the number of minimal path sets would have been: = (19,600) (1225) = 24,010,

10 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 9 9. You are given: A 999-out-of-1000 system of independent identically distributed exponential components. The mean lifetime of each component is θ = 5. Calculate the expected system lifetime. A. Less than 0.05 B. At least 0.05, but less than 0.10 C. At least 0.10, but less than 1.00 D. At least 1.00, but less than E. At least A. A 999-out-of-1000 system fails when fewer than 999 components function. Thus it fails when the 2nd component fails. We want the expected value of the second failure out of 1000 Exponentials with θ = 5. The expected lifetime is: θ (1/ /999) = (5) (1/ /999) = Comment: For a k out of n system, if each component follows an Exponential distribution with mean n 1 θ, then the expected lifetime of the system is θ. i k

11 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following information: System X is a series system with two components. System Y is a parallel system with two components. All components are independent and function for an amount of time, uniformly distributed over (0, 1). Calculate the absolute value of the difference in expected system lifetimes between system X and system Y. A. Less than 0.10 B. At least 0.10, but less than 0.20 C. At least 0.20, but less than 0.30 D. At least 0.30, but less than 0.40 E. At least D. The lifetime of System X is the expected value of a minimum of two uniform distributions on (0, 1): 1/(2+1) = 1/3. The lifetime of System Y is the expected value of a maximum of two uniform distributions on (0, 1) is: 2/(2+1) = 2/3. 2/3-1/3 = 1/3 = Comment: For the uniform distribution, the order statistics are equally spaced over the support, including room on the left below the minimum and room on the right above the maximum.

12 11. You are given the following transition matrix for a Markov chain with two states 0, 1: A = At time t = 0, the Markov chain is in state 0. Calculate the expected number of steps needed to return to state 0. A. Fewer than 10 B. At least 10, but fewer than 30 C. At least 30, but fewer than 50 D. At least 50, but fewer than 70 E. At least E. The balance equations are πa = π: 0.25 π π 1 = π π π 1 = π 1. Also π 0 + π 1 = 1. Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 11 Therefore, 0.25 π (1 - π 0 ) = π 0. π 0 = 1/76. π 1 = 75/76. The expected number of steps needed to return to state 0 is 1/π 0 = 76.

13 12. You are given: A branching process has an initial population consisting of 16 individuals. There are three states: o State 0 for zero offspring o State 1 for one offspring o State 2 for two offspring P 0 = 2/5, P 1 = 1/10, P 2 = 1/2 Calculate the probability that the population will die out. A. Less than 0.02 B. At least 0.02, but less than 0.03 C. At least 0.03, but less than 0.04 D. At least 0.04, but less than 0.05 E. At least B. The mean is: (2/5)(0) + (1/10)(1) + (1/2)(2) = 1.1 > 1. Thus we want the smallest root of π 0 = P i π i 0 : i π 0 = 2/5 + (1/10)π 0 + (1/2)π 0 2. Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 12 π π = 0. π 0 = 1.8 ± (4)(1)(0.8) (2)(1) Thus the probability of a single individual dying out is π 0 = 1 or 0.8. The probability of an initial population of 16 dying out is: = Comment: If the mean were 1, then the population always dies out.

14 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following information about a homogeneous Markov chain: Planes move among three states servicing flights: o State 0: On-time o State 1: Delayed o State 2: Canceled P = A plane is currently in the state Canceled. Calculate the probability that the plane will be in the state Canceled after two transitions. A. Less than 0.07 B. At least 0.07, but less than 0.09 C. At least 0.09, but less than 0.11 D. At least 0.11, but less than 0.13 E. At least E. (0, 0, 1) P = (0.8, 0.2, 0). (0.8, 0.2, 0) P = (0.78, 0.08, 0.14).

15 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following information about a Markov chain with a transition probability matrix: P = The three states are 0, 1, and 2. Calculate the long-run proportion of time in State 2. A. Less than B. At least 0.200, but less than C. At least 0.220, but less than D. At least 0.240, but less than E. At least A. The balance equations are πp = π: 0.6 π π π 2 = π π π 1 = π 1. π 1 = (3/7) π π π 2 = π 2. π 2 = (1/3) π 0. Also π 0 + π 1 + π 2 = 1. π 0 = 1 / (1 + 3/7 + 1/3) = 21/37. π 1 = 9/37. π 2 = 7/37 =

16 15. A life annuity has the following values: ä 20 = ä 21 = The interest rate used is The force of mortality is constant between the ages of 20 and 30. From a sample of 500 independent twenty-year-olds, you wish to calculate a 95% confidence interval for the number of people who will die within five years implied by the annuity values above, using a normal approximation. Calculate the upper bound of this confidence interval. A. Less than 60.0 B. At least 60.0, but less than 65.0 C. At least 65.0, but less than 70.0 D. At least 70.0, but less than 75.0 E. At least B. ä x = 1 + v p x ä x = 1 + (1/1.05) p 20 ( ). p 20 = For a constant force of mortality, the survival function is: S(t) = e λt. e λ = λ = Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 15 Thus the probability of living 5 years is: exp[-(5)( )] = Thus the number of people who die within five years is Binomial with m = 500 and q = = The mean is: (500)( ) = The variance is: (500)( )( ) = Thus a 95% confidence interval is: ± = (35.11, 60.95). Comment: Since the force of mortality is constant, 5 p 20 = (p 20 ) 5 = =

17 16. You are given the following information about a machine warranty: A warranty will provide payment at the end of the year in which the machine fails. The annual interest rate, i, is The number of functioning machines at the start of the year and the number failing during the year are denoted by letters I and d respectively. Age x I x d x Payment Calculate the actuarial present value of the warranty payment per unit at time 0. A. Less than 100 B. At least 100, but less than 120 C. At least 120, but less than 140 D. At least 140, but less than 160 E. At least E. v = 1/1.05. d 0 = = d 1-50 = 50. d 1 = 200. Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 16 APV = 200 v v v v =

18 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following information: 0 θ 1 X has the probability distribution: X Pr(X) 3θ/4 θ/4 2(1-θ)/3 (1-θ)/3 You observe the following 36 values from this distribution: X Obs Calculate the maximum likelihood estimate of the parameter θ. A. Less than 0.20 B. At least 0.20, but less than 0.40 C. At least 0.40, but less than 0.60 D. At least 0.60, but less than 0.80 E. At least C. The loglikelihood is: 7 ln[3θ/4] + 12 ln[θ/4] + 13 ln[2(1-θ)/3] + 4 ln[(1-θ)/3] = 19 ln[θ] + 17 ln[1-θ] + constants. Set the partial derivative of the loglikelihood with respect to θ equal to zero: 19/θ - 17/(1-θ) = 0. 19(1-θ) = 17θ. θ = 19/36 = Comment: Note that the given probabilities sum to one, as they should.

19 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following information: X has the probability mass function: p(x = x α) = e -α / 2 α x 2 x, where α > 0 and X 0,1,2,3,... x! You observe the following 7 values from this distribution: 0, 0, 1, 1, 3, 4, 5 Calculate the maximum likelihood estimate of the parameter α. A. Less than 0.5 B. At least 0.5, but less than 1.5 C. At least 1.5, but less than 2.5 D. At least 2.5, but less than 3.5 E. At least E. A Poisson Distribution with mean α/2. Maximum likelihood is equal to the method of moments: α/2 = ( )/7 = 2. ^α = (2)(2) = 4.

20 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page ^α n is an estimator of a parameter α for a sample size of n. You are given: E[ ^α n ] = Var[ ^α n ] = 7 n n (n-1) 2n - 1 The true value of α is 7. for n = 2,3,4,... for n = 2,3,4,... A list of potentially true statements about ^α n is given below. I. ^α n is an asymptotically unbiased estimator of α. II. The mean squared error of ^α 8 is greater than the variance of the estimator of ^α n for n 2. III. ^α n is a consistent estimator of α. Determine which of the above statements are true. A. None are true B. I and II only C. I and III only D. II and III only E. The answer is not given by (A), (B), (C), or (D)

21 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page B. and E. (See Comment) E[ ^α n ] goes to 7 as n approaches infinity, the true value of α. Thus ^α n is an asymptotically unbiased estimator of α. Statement I is true. Var[ ^α n ] goes to 70 as n approaches infinity, rather than going to zero. Thus we can not show whether or not the estimator is consistent. Statement III is not true. MSE[ ^α 8 ] = VAR[ ^α 8 ] + Bias[ ^α 8 ] 2 = (140)(7)/15 + {(7)(8)/12-7} 2 = Var[ ^α n ] = 140 (n-1) 2n - 1 = 70 2n - 2 2n - 1 < 70 < MSE[ ^α 8 ]. Statement II is true. Comment: The CAS allowed either choice B or E. There is not enough information to conclude whether or not ^α n is a consistent estimator of α. Thus Statement III is neither true nor false. While I think therefore choice B is better, there is also some justification for choosing E. An asymptotically unbiased estimator, whose variance goes to zero as the number of data points goes to infinity, is consistent. However, an asymptotically unbiased estimator, whose variance does not go to zero as the number of data points goes to infinity, may or may not be consistent. Therefore, based on the given information, Statement III is neither true nor false. While I think B is the best choice, choice E also has some merit. A Pareto Distribution with α = 1.5 has a finite mean but not a finite variance. Nevertheless, in the case of a sample from a Pareto Distribution with α = 1.5, the sample mean is still a consistent estimator of the mean, even though this estimator has an infinite variance.

22 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following information about a random sample of size 20 from a normal distribution: A two-sided 95% confidence interval for the distribution variance is estimated, with an equal probability in each tail. The width of the confidence interval is 150. Calculate the unbiased sample variance for the 20 random draws. A. Less than 25 B. At least 25, but less than 50 C. At least 50, but less than 75 D. At least 75, but less than 100 E. At least D. Let S 2 be the sample variance. (20-1) S 2 / σ 2 follows a Chi-Square Distribution with 20-1 = 19 degrees of freedom. The 2.5 th percentile of the Chi-Square Distribution with 19 degrees of freedom is Therefore we have: 19S 2 / σ 2 > σ 2 < 2.132S 2. The 97.5 th percentile of the Chi-Square Distribution with 19 degrees of freedom is Therefore we have: 19S 2 / σ 2 < σ 2 < 0.578S 2. Thus the width of the 95% confidence interval for the variance is: 2.132S S 2 = 150. S 2 = 96.5.

23 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page Let X be a single observation from the distribution: F(x) = 1 - exp[- x 2 ]. θ You are testing the null hypothesis H 0 : θ = y against the alternative hypothesis H 1 : θ = 10y. The null hypothesis is rejected if X > k. The probability of a Type I error is 5.0%. Calculate the probability of a Type II error. A. Less than 4.0% B. At least 4.0%, but less than 8.0% C. At least 8.0%, but less than 12.0% D. At least 12.0%, but less than 16.0% E. At least 16.0% 21. A. Prob[X > k θ = y] = 5% = exp[-(k/y) 2 ]. k = y. The probability of a Type II error is: 1 - Prob[x > k θ = 10y] = 1 - exp[ y 10y 2 ] = 2.95%. Comment: A Weibull Distribution with τ = 2.

24 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the claim activity for a group of 5,000 policyholders in years 2015 and 2016, as shown in the table below: Group 2015 Claims 2016 Claims Policyholders I 0 0 3,500 II 0 >0 900 III > IV >0 >0 100 Your null hypothesis is that claim activity in year 2015 is independent of claim activity in The alternative hypothesis is that claim activity in 2015 is not independent of claim activity in Calculate the p-value of this test using the Neyman-Pearson Lemma. A. Less than 0.5% B. At least 0.5%, but less than 1.0% C. At least 1.0%, but less than 2.5% D. At least 2.5% but less than 5.0% E. At least 5.0%

25 22. D. The grid of observeds: Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 24 No claim 2016 Claims in 2016 Total No claim Claims in Total The grid of expecteds: No claim 2016 Claims in 2016 Total No claim Claims in Total For example in the first row and column: (row total) (column total) / (overall total) = (4400)(4000) / 5000 = The Chi-Square Statistic is: ( ) 2 / ( ) 2 / ( ) 2 / ( ) 2 / 120 = The degrees of freedom are: (2-1)(2-1) = 1. Looking in the table, 3.84 < < Thus the p-value is between 5% and 2.5%. Comment: Similar to CAS3L, 5/10, Q.23. I could not find anywhere in the syllabus readings that it states that the usual one-dimensional or two-dimensional Chi-Square Tests are the result of applying the Neyman-Pearson Lemma. The Neyman-Pearson Lemma is based on two specific sets of parameters, one set for the null and one set for the alternative hypothesis. This does not seem to be the case here; the alternative hypothesis is that activity in 2015 is not independent of claim activity in 2016 which covers lots of different forms of dependence.

26 23. Let X 1 and X 2 be a random sample from a distribution with density function: f(x) = θ xθ - 1, 0 < x < 1, where θ > 0. 0, otherwise The null hypothesis H 0 : θ = 4 is tested against the alternative hypothesis H 1 : θ = 3 using the statistic Y = max(x 1, X 2 ) Calculate the critical value given the probability of a Type I error is A. Less than 0.30 B. At least 0.30, but less than 0.40 C. At least 0.40, but less than 0.50 D. At least 0.50, but less than 0.60 E. At least C. A Beta Distribution with b = 1 and what is usually a is here called θ. The mean is θ/(θ+1); thus when the alternative is true the mean is smaller than when the null is true. Therefore, we reject when Y = max(x 1, X 2 ) is small. For θ = 4, f(x) = 4x 3. F(x) = x 4. Y < y, if X 1 < y and X 2 < y. Prob[Y < y] = (y 4 ) 2 = y 8. In order to have the probability of a Type I error be : Prob[Y < k θ = 4] = k 8 = k = Comment: Similar to 2, 5/92, Q. 22. Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 25

27 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following information: Data follows a Poisson distribution with mean λ. Under the null hypothesis H 0 : λ = 5, you observe one value from the distribution and reject the null hypothesis if that observation is either 0 or 1. The probability of committing a Type I error in this situation is α. Assuming that the alternative hypothesis is H 1 : λ = 1.5, under the same rejection rule as above the probability of a Type II error is β. Calculate β - α. A. Less than 0.15 B. At least 0.15, but less than 0.25 C. At least 0.25, but less than 0.35 D. At least 0.35, but less than 0.45 E. At least D. Prob[0 or 1 claim λ = 5] = e e -5 = = probability of Type I error = α. Prob[0 or 1 claim λ = 1.5] = e e -1.5 = Probability of Type II error = = = β. β - α = =

28 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following information: X has the following probability density function: f(x) = 1 7 e - x 7 X > 0 You draw a sample of size 30 from this distribution. Calculate the probability that the two smallest values in your sample are both smaller than 0.5. A. Less than 0.2 B. At least 0.2, but less than 0.4 C. At least 0.4, but less than 0.6 D. At least 0.6, but less than 0.8 E. At least D. F(0.5) = 1 - e -0.5/7 = S(0.5) = e -0.5/7 = Prob[0 values < 0.5] + Prob[1 value < 0.5] = (30)( )( ) = Prob[two smallest values < 0.5] = Prob[at least 2 values < 0.5] = =

29 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page Losses were separated into two types of claims and analyzed. The values of 12 randomly selected claims are provided in the following table: Claim Type W Claim Type X Loss Amount Rank Loss Amount Rank 100, , , , , , , , , , , , H 0 : Median Type W = Median Type X H 1 : Median Type W Median Type X Calculate the smallest value for a for which H 0 can be rejected using the Mann-Whitney U Test. A. Less than B. At least 0.010, but less than C. At least 0.200, but less than D. At least 0.400, but less than E. At least C. The sum of the ranks for the first sample W is: = 25. U = n 1 n 2 + (n 1 )(n 1 +1)/2 - (sum of the ranks for the first sample of size n 1 ) = (5)(7) + (5)(5+1)/2-25 = 25. In order to use the table, subtract from n 1 n 2 : (5)(7) - 25 = 10. From the table, with n 1 = 5 and n 2 = 7, there is probability in one tail. For this two-sided test, the p-value is: (2)(0.134) = Alternately, count up the number of elements in X prior to each element of W: U = = 10. Proceed as before.

30 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You have calculated the Pearson correlation coefficient, Spearman's Rho, and Kendall's Tau using the following five observations: Determine the correct statement. A. Pearson's correlation coefficient < Kendall's Tau < Spearman's Rho B. Pearson's correlation coefficient < Spearman's Rho < Kendall's Tau C. Pearson's correlation coefficient < Kendall's Tau = Spearman's Rho D. Spearman's Rho = Kendall's Tau < Pearson's correlation coefficient E. Kendall's Tau < Pearson's correlation coefficient < Spearman's Rho

31 27. C. From the given graph, the data is: (1,1), (2,3), (3,4), (4,5), (5,10). X = 3. Y = 4.6. Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 30 (1-3)(1-4.6) + (2-3)(3-4.6) + (3-3)(4-4.6) + (4-3)(5-4.6) + (5-3)(10-4.6) 5 (1-3) 2 + (2-3) 2 + (3-3) 2 + (4-3) 2 + (5-3) 2 5 = 2. (1-4.6) 2 + (3-4.6) 2 + (4-4.6) 2 + (5-4.6) 2 + (10-4.6) 2 Thus, Pearson's correlation coefficient is: = = The ranks are: (1,1), (2,2), (3,3), (4,4), (5,5). Spearman's Rho is the correlation of these ranks: 1. To get Kendallʼs Tau, the first step is to order the first elements of the pairs from smallest to largest. Then list the resulting ranks of the second elements. Compute the number concordant and discordant. The number concordant listed in a row is the number of ranks below it in the column that are greater than the given rank. The number discordant listed in a row is the number of ranks below it in the column that are less than the given rank. X Y Rank of Y Concordant Discordant Total Total τ = C - D C + D = = 1. Comment: One can compute Pearson's correlation coefficient using the stat functions of the calculator. When the ranks of X match the ranks of Y, then both Spearman's Rho and Kendallʼs Tau are one. = 4.

32 28. You are given the following information: The probability that a randomly selected person has a specific rare disease is If they have the disease, they will have a positive test 90% of the time. If they do not have the disease, they will have a positive test 1 % of the time. You gather a sample of 10 people who all have positive test results. Calculate the probability that no one in your sample actually has the disease. 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 E. Use Bayes Theorem. For an individual, Prob[healthy positive result] = Prob[healthy] Prob[positive healthy] Prob[healthy] Prob[positive healthy] + Prob[sick] Prob[positive sick] = ,000 1% ,000 1% + 1 = 10,000 90% Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page = ,000. Probability that all ten in the sample do not have the disease is: = Alternately, assume a total of 1 million people. 100 have the disease and the rest do not. Of the 100 with disease, 90 test positive. Of the 999,900 without disease, 9999 test positive. Thus of those who test positive, the portion that do not have the disease is: 9999 / ( ) = Proceed as before.

33 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following information: Annual claim frequency follows a Poisson distribution with mean λ. λ follows a gamma distribution with mean 3 and variance 5. You observe 4 policies with 0, 0, 2, and 3 claims in the previous year. Calculate the mean of the posterior distribution of λ. A. Less than 1.3 B. At least 1.3, but less than 1.8 C. At least 1.8, but less than 2.3 D. At least 2.3, but less than 2.8 E. At least B or E. (See Comment) A Gamma-Poisson. αθ = 3 and αθ 2 = 5. θ = 5/3 and α = 9/5. αʼ = α + C = 9/ = 34/5. 1/θʼ = 1/θ + E = 3/5 + 4 = 23/5. θʼ = 5/23. Posterior mean = αʼθʼ = (34/5) (5/23) = 34/23 = Alternately, for Buhlmann Credibility, K = 1/θ = 3/5. Z = 4/(4 + 3/5) = 20/23. Estimated future frequency is: (20/23)(5/4) + (1-20/23)(3) = Comment: The CAS allowed either choice B or E. If one assumed you observed 4 policies with 4 different lambdas than the policy in which we are interested, then this would not affect our prior estimate of lambda. The posterior mean would be the same as the prior mean of 3, which corresponds to choice E. It would have been better if the question said, you observe 4 policies all with the same lambda or you observe a policy for 4 years and there were 0, 0, 2, and 3 claims. An assumption in the Gamma-Poisson is that the observations are all drawn from Poisson distributions with the same lambda.

34 30. For a large portfolio of insurance contracts, suppose θ represents the proportion of policies for which there were claims made within one year. The value of θ is a random variable with the prior distribution expressed as: p(θ) = Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 33 Γ(a+ b) Γ(a) Γ(b) θa-1 (1- θ) b-1 ; for 0 < θ < 1 You choose a and b such that: The prior estimate of the mean of θ is 10% The posterior mean of θ gives twice as much weight to the observed average from a sample of 500 policies as the prior mean. Calculate the value of a implied for the prior distribution. A. Less than 22 B. At least 22, but less than 30 C. At least 30, but less than 38 D. At least 38, but less than 46 E. At least B. 10% = prior mean = mean of prior Beta Distribution = a/(a+b). The posterior mean of θ gives twice as much weight to the observed average as to the prior mean. Z = 2/3. 2/3 = 500 / (500 + K). K = 250. For the Beta-Bernoulli, for Buhlmann Credibility, K = a + b. 250 = a + b. a = (250)(10%) = 25. b = 225.

35 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following output from a GLM to estimate an insured's Pure Premium: Response variable Pure Premium Response distribution Gamma Link Log Parameter df β^ Intercept Risk Group 2 Group Group Group Territory Code 2 Region Region Region Calculate the predicted Pure Premium for an insured in Risk Group 3 from Region 2. Less than 250 At least 250, but less than 275 At least 275, but less than 300 At least 300, but less than 325 At least D. exp[ ] = e 5.75 =

36 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page Given a family of distributions where the variance is related to the mean through a power function: Var[Y] = a E[Y] p One can characterize members of the exponential family of distributions using this formula. You are given the following statements on the value of p for a given distribution: I. Normal (Gaussian) distribution, p = 0 II. Compound Poisson-gamma distribution, 1 < p < 2 III. Inverse Gaussian distribution, p = -1 Determine which of the above statements are correct. A. I only B. I and II only C. I and III only D. II and III only E. The answer is not given by (A), (B), (C), or (D) 32. B. For the Inverse Gaussian distribution, p = 3, thus Statement III is not true. Comment: See the subsection on Tweedie Distributions in the CAS Study Note Generalized Linear Models, by Larsen.

37 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page A distribution belongs to the exponential family if it can be written in the canonical form: f(y; θ) = exp[a(y)b(θ) + c(θ) + d(y)] Determine the values of a(y), b(θ), c(θ), and d(y) for the Poisson distribution. A. a(y) = y, b(θ) = In(θ), c(θ) = θ, and d(y) = In(y!) B. a(y) = -y, b(θ) = -In(θ), c(θ) = θ, and d(y) = In(y!) C. a(y) = y, b(θ) = In(θ), c(θ) = θ, and d(y) = 1/(y!) D. a(y) = y, b(θ) = In(θ), c(θ) = -θ, and d(y) = -In(y!) E. The answer is not given by (A), (B), (C), or (D) 33. D. For the Poisson with mean θ, f(y) = e θ θ y / y!. Therefore, ln f(y) = y ln(θ) - ln(y!) - θ. Therefore, the Poisson distribution is a member of a (linear) exponential family of the discrete type, with: a(y) = y, b(θ) = ln(θ), c(θ) = -θ, and d(y) = -ln(y!). Comment: In the canonical form: a(y) = y.

38 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page For an ordinary linear regression with 5 parameters and 50 observations, you are given: The total sum of squares, S yy = 996. The unbiased estimate for the constant variance, σ 2, is s 2 = Calculate the coefficient of determination. A. Less than 0.65 B. At least 0.65, but less than 0.75 C. At least 0.75, but less than 0.85 D. At least 0.85, but less than 0.95 E. At least D. (Regression SS) / (50-5) = s 2 = RSS = R 2 = 1 - RSS/ TSS = / 996 = Comment: I assume 5 parameters means 4 slopes plus an intercept.

39 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page Consider the following 2 models, which were fit to the same 30 observations using ordinary least squares: Model 1 Model 2 Response variable Y Response variable Y Response distribution Normal Response distribution Normal Link Identity Link Identity SS Total SS Total SS Error 2781 SS Error 2104 Parameter β^ d.f. Parameter β^ d.f. Intercept Intercept X X X X X X You test H 0 : β 3 = β 4 = 0 against the alternative hypothesis that at least one of β 3, β 4 0. Calculate the smallest significance level at which you reject H 0. A. Less than 0.01 B. At least 0.01, but less than 0.02 C. At least 0.02, but less than 0.05 D. At least 0.05, but less than 0.10 E. At least C. F = ( ) / / ( ) = 4.022, with 2 and 25 degrees of freedom. Looking in the table, < < Thus the p-value is between 2% and 5%. Comment: F = (ESS R - ESS UR ) / q, where p is the number of slopes fit in the unrestricted model, ESS UR / (N - p- 1) q is the dimension of the restriction, and ESS are the error sum of squares for the restricted and unrestricted models.

40 36. You are analyzing a dataset and have fit a multiple regression with 12 continuous explanatory variables and one intercept. You are given the following: Sum of square errors = 618 The 10 th diagonal entry of the hat matrix, h 10,10 = 0.35 The 10 th residual value = ε^10 = 9.5 Cook's Distance = D 10 = 5.0 Calculate the number of data points this model was fit to. A. Fewer than 200 B. At least 200, but fewer than 400 C. At least 400, but fewer than 600 D. At least 600, but fewer than 800 E. At least C. Let N be the unknown number of observed points. Then the residual standard error is: RSE = σ^ = The standardized residuals are: Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 39 ε^ i RSE 1 h ii. 618 N = 618 N Thus the tenth standardized residual is: 9.5 RSE 0.65 = / RSE. D i = (i th standardized residual) 2 Thus, 5.0 = D 10 = ( / RSE) 2 hii, where p is the number of fitted slopes. (p+ 1) (1 - h ii ) 0.35 (12 + 1) (1-0.35) = / RSE2. RSE 2 = N - 13 = N = 550.

41 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following residual Q-Q plot for a fitted GLM: I. The distribution of residuals is skewed to the left. II. The residuals are serially correlated. III. Observations 33 and 101 are influential points. Determine which of the above statements can be concluded to be true from the above chart. A. I only B. II only C. III only D. I, II and III E. The answer is not given by (A), (B), (C), or (D)

42 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page A. The Normal Distribution has zero skewness, since it is symmetric around its mean. The lefthand tail for the distribution of residuals has larger absolute values than for the Normal Distribution. (The points in the lefthand part of the graph are below the comparison line.) The righthand tail for the distribution of residuals has smaller absolute values than for the Normal Distribution. (The points in the righthand part of the graph are below the comparison line.) Thus the distribution of the residuals has a lighter righthand tail than a Normal Distribution and a lighter righthand tail than the Normal Distribution. Thus the distribution of residual has a skewness less than that of the Normal and thus a negative skewness. Thus the distribution of residual is skewed to the left. This is not a useful graph for determining whether or not the residuals are serially correlated. An influential point has a large absolute value of the standardized residual and a large leverage. While observations 33 and 101 fit the first criterion, there is no way to determine from this graph whether or not they have high leverage. Comment: See Figure 6.1 in Dobson and Barnett. Q-Q plots are discussed in Mahlerʼs Guide to Statistics. To determine whether or not there is serial correlation, one could graph ε^ i versus ε^ i+1. An influential data point has a disproportionate impact on the results of the model. A large absolute value of a standardized residual indicates an outlier; a rule of thumb is an absolute value greater than 2 or 3. A high leverage point has unusual value(s) of the predictor variable(s).

43 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page A sample of 100 is taken from the following distribution: f(y i ) = y i e - (y i / θ) θ 2 You are told that 100 y i = 2, i=1 You are provided with a plot of the log likelihood as a function of θ for this data. The method of scoring is used to estimate θ from the sample. Assume θ (1) = 5. Calculate θ (2). A. Less than or equal to 2.0 B. Greater than 2.0 but less than or equal to 6.5 C. Greater than 6.5, but less than or equal to 11.0 D. Greater than 11.0, but less than or equal to 15.5 E. Greater than 15.5

44 38. C. ln f(y) = ln(y i ) - y i /θ - 2 ln(θ). ln[f(y)] θ = y i /θ 2 ln[f(y - 2/θ. U = i )] θ = /θ2 - (100)(2/θ). i For θ (1) = 5, the score is: U (1) ln[f(y = i )] θ = /52 - (100)(2/5) = i 2 ln[f(y)] θ 2 = -2y i /θ 3 + 2/θ 2. Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 43 E[ 2 ln[f(y)] θ 2 ] = -2E[Y]/θ 3 + 2/θ 2 = -2(2θ)/θ 3 + 2/θ 2 = -2/θ 2. For θ (1) = 5, τ (1) = Information = -n E[ 2 ln[f(y)] θ 2 ] = -(100) (-2/5 2 ) = 8. θ (2) = θ (1) + U (1) / τ (1) = /8 = Comment: A Gamma Distribution with α = 2. In this case, maximum likelihood is equal to the Method of Moments: αθ = 2 θ = Y = /100 = θ^ = , consistent with where the loglikelihood is maximized in the given graph. In this case, the Method of Scoring converged in one iteration. Based on θ^ = , θ (2) should be move towards from 5; this eliminates choice A. Other than that, I did not find the graph helpful in answering this question. Uʼ = 2 ln[f(y i )] θ i 2 = -( )(2)/θ3 + (100)(2/θ2). For θ (1) = 5, Uʼ(1) = -( )(2)/5 3 + (100)(2/5 2 ) = Thus using instead the Newton-Raphson Method: θ (2) = θ (1) - U (1) / Uʼ(1) = / = The Newton-Raphson iterations are: , , , , ,

45 39. A set of n observations, y 1, y 2,..., y n, are assumed to be exponentially distributed, f(y i ) = exp[-y i /θ i ] / θ i, i = 1, 2,..., n A GLM is fit to the data with the following model specification: ln(θ i ) = β 0 + β 1 x 1i + β 2 x 2i + β 3 x 2i 2 + β 4 x 3i The vector of observed responses, Y and the design matrix, X, are given for the first four observations as well as the vector of all estimated parameters, β^ : Y = Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page X = β^ = The first four rows and columns of the hat matrix, H, are: H = Calculate the standardized residual for the second observation, r 2 A. Less than 1 B. At least 1, but less than 2 C. At least 2, but less than 3 D. At least 3 but less than 4 E. At least 4

46 39. D. Assume that they are asking for the standardized Pearson residual. Reading the values of the predictor variables for the 2nd observed point in the 2nd row of X: y^ 2 = exp[ (-0.27)(0) + (-0.67)(0.6) + (0.16)(0.36) + (0.91)(1)] = e = ε^2 = y 2 - y^ 2 = = Since Y follows an Exponential Distribution: StdDev[Y i ] = Y i. Get the Pearson Residual by dividing by the estimate of StdDev[Y 2 ] = y^ 2: / = Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 45 To get the standardized Pearson Residual, divide by 1 - h 22 : / = Comment: Dobson and Barnett uses r Pi for the i th standardized Pearson residual. One possibility would be to fit a GLM assuming a Gamma Distribution; one would not usually assume its special case an Exponential Distribution (Gamma with α = 1.)

47 40. You are fitting a Poisson regression model of the form: E(Y i ) = β 1 + β 2 x i Maximum likelihood estimates of the beta coefficients are obtained using iterative weighted least squares procedure. You are given three matrixes: W is the weight matrix. X is the design matrix. Z has the beta values from the prior iteration applied to the explanatory variables as well as the correction term. From the estimates of the first iteration, the following matrices are calculated using the matrixes as defined above: (X T WX) (1) = [(X T WX) (1) ] = (X T Wz) (1) = Calculate β 2 (2), the estimate for β 2 on the second iteration. A. Less than 0 B. At least 0, but less than 0.40 C. At least 0.40, but less than 0.80 D. At least 0.80, but less than 1.20 E. At least 1.20 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page B. The updated estimate of the betas is: (Xʼ W X) Xʼ W z = = The estimate for β 2 on the second iteration is

48 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page A GLM has been fit to a data set with 1,000 observations with the following model form: E(y i ) = g -1 (β 0 + β 1 X 1i + β 2 X 2i + β 3 X 3i ) The standardized residuals are plotted against the values of the variable represented by X 3 in the model. The plot is shown below. Determine the best alternate model parameterization based on the residual plot above. A. E(y i ) = g -1 (β 0 + β 1 X 1i + β 2 X 2i + β 3 X 3i + β 4 X 3i 2 ) B. E(y i ) = g(β 0 + β 1 X 1i + β 2 X 2i + β 3 X 3i ) C. E(y i ) = g -1 (β 0 + β 1 X 1i + β 2 X 2i + β 4 X 2i 2 + β 3 X 3i ) D. E(y i 2 ) = g -1 (β 0 + β 1 X 1i + β 2 X 2i + β 3 X 3i ) E. E[ln(y i )] = g -1 (β 0 + β 1 X 1i + β 2 X 2i + β 3 X 3i ) 41. A. The plot has larger residuals for both small and big values of X 3, indicates a nonlinear relationship of g(y) with X 3. There appears to be a quadratic effect of X 3. Adding to the model a term involving X 3 2 should help. Comment: Not adequately covered in Dobson and Barnett; see Figures 2.3 and 2.4 and page 35. One could instead examine Partial Residual Plots, not on the syllabus.

49 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page A practitioner built a GLM to predict claim frequency. She used a Poisson error structure with a log link. You are given the following information regarding the model summary statistics: The model has 5 parameters. The likelihood ratio chi-squared statistic C is Determine the best statement of what we can we conclude from the value of C. A. At least one of the non-intercept coefficients is likely to be non-zero using a p-value of.05, though we cannot conclude that more than one of the non-intercept coefficients are non-zero. B. The model does not fit well, when the parsimony of the model is taken into account. C. The model fit is significantly better with 5 parameters compared to 4 using a p-value of.05. D. Each of the coefficients is likely to be non-zero. E. The model fits well, even when the parsimony of the model is taken into account. 42. A. H 0 : The minimal model, in others words all of the slopes are zero: β 1 = = β p = 0. H 1 : Not H 0, in other words that at least one of the slopes is non-zero. C (approximately) follows a Chi-Square Distribution with degrees of freedom equal to the number of parameters in the fitted model minus one = 5-1 = 4 d.f. Since the 2.5% critical value is 11.14, and > 11.14, we reject the null hypothesis at 2.5%. Thus using a p-value of 2.5% (as well as 5%) we conclude that at least one of the non-intercept coefficients is likely to be non-zero. Comment: We testing whether all of the slopes are zero. It would require further tests to determine whether individual slopes or subsets of slopes are zero. This statistic does not separately take into account the parsimony of the model, in other words how many parameters the model uses. If instead they intended that the model has 5 slopes plus an intercept, then the likelihood ratio chi-squared statistic C has 5 d.f. The 5% critical value is Since > 11.07, we reject the null hypothesis at 5%.

50 43. You are given the following information from a time series: t X t Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page 49 x 4.0 Calculate the sample lag 4 autocorrelation. A. Less than B. At least -0.30, but less than C. At least -0.10, but less than 0.10 D. At least 0.10, but less than 0.30 E. At least C. c 0 = (4-4)2 + (3.5-4) 2 + (2.5-4) 2 + (5.5-4) 2 + (4.5-4) 2 + (4-4) 2 6 = 5/6. c 4 = (4-4)(4.5-4) + (3.5-4)(4-4) 6 = 0. r 4 = c 4 /c 0 = 0. Comment: r 1 = -0.15, r 2 = -0.30, and r 3 = Using the electronic calculator, the sample variance of all of the given data is: S 2 = 1. c 0 = S 2 (n-1)/n = (1)(5/6) = 5/6.

51 Solutions Fall 2017 CAS Exam S, HCM 12/9/17, Page You are given the following statements about stationarity: I. Linear models for time series are stationary when they include functions of time. II. All moving average processes are stationary. III. All random walk processes are non-stationary. Determine which of the above statements are true. A. None are true B. I and II only C. I and III only D. II and III only E. The answer is not given by (A), (B), (C), or (D) 44. D. If the linear model includes functions of time, then the mean and variance depend on time. Thus the time series is not stationary. Statement I is not true. Since it is the sum of a finite number of white noises, each of which is stationary, MA(q) is stationary. Thus Statement II is true. The variance of a random walk increases over time, thus it is not stationary. The covariances are a function of time, another reason it is not stationary. Thus Statement III is true. Comment: A stationary time series model has its mean, variance, autocovariances, and autocorrelations all independent of time. For Statement I, see Section of Cowpertwait and Metcalfe. For Statement II, see Section of Cowpertwait and Metcalfe. For Statement III, see Section of Cowpertwait and Metcalfe.