Errata and updates for ASM Exam C/Exam 4 Manual (Sixth Edition) sorted by date

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1 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date 1 Errata and updates for ASM Exam C/Exam 4 Manual (Sixth Edition) sorted by date Please note that the SOA has announced that when using the normal distribution table provided on the exam, you are to round, not interpolate. To calculate Φ(0.2314), for example, use Φ(0.23). The manual uses interpolated values, but do not do this on the exam. For practice exam 5 questions 6 and 16, make the changes noted below. Practice exam 5 question 32 is defective in that none of the 5 answer choices is correct. [5/4/2009] On page 195, the solution to exercise is incorrect. The correct solution is: For each insured, the Poisson parameter over two years is Λ = 2λ. Since E[Λ] = 2 E[λ] and Var(Λ) = 4 Var(λ), the parameter Λ follows a gamma distribution with mean 1 and variance 2. Let N be the number of losses over the two-year period. Then E[N] = E [ E[N Λ] ] = E[Λ] = 1 and the variance of N is Var(N) = E [ Var(N Λ) ] + Var ( E[N Λ] ) = E[Λ] + Var(Λ) = = 3 For 1500 insureds, the aggregate mean is 1500 and the aggregate variance is 1500(3) = We make a continuity correction and check the probability that a normal distribution with these parameters is greater than : ( ) Pr(N > 1600) = 1 Φ 4500 = 1 Φ(1.50) = = [4/2/2009] On page 164, the paragraph after Example 12A up to the end of the lesson are incorrect. Replace them with the following: The same parameter that gets multiplied by v in the (a, b, 0) class gets multiplied by v in the (a, b, 1) class. p 0 is then the balancing item, 1 k=1 p k. The textbook gives formulas for p M 0 in all cases (Table 8.3). Rather than memorizing the table, use the following formula: ( ) 1 p 1 p M 0 = (1 p M 0 ) 0 1 p 0 where asterisks indicate distributions with revised parameters. This formula works even when the unmodified distribution is improper (so that unmodified probabilities are negative or greater than 1), as in the ETNB family. This is illustrated in the following example: Example 12B Frequency of claims per year follows a zero-modified negative binomial distribution with r = 0.5, β = 1, and p M 0 = 0.7. Claim size follows a Pareto with α = 1, θ = 1000, and is independent of claim frequency. A deductible of 500 is imposed. Calculate the probability of no claims payments in a year. Answer: The probability of a payment given a claim is the Pareto survival function at 500: S(500) = θ θ = = 2 3

2 2 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date The revised negative binomial parameters are r = 0.5, β = 2/3. By the equation above: 1 p M 0 = 0.3 ( ) r ( ) p 0 = 1 = 1 = β 2 ( ) p 0 = 1 = /3 ( ) p M 0 = 0.3 = p M 0 = = [1/27/2009]On page 1130, in the solution to question 16, on the first displayed line, change K/S 0 to S 0 /K. [11/2/2008]On page 1036, in question 6, add the word annual on the first line before number of claims. On the third line, change number of claims to number of years of experience. [11/2/2008]On page 1128, in the solution to question 9, on the first line, change sis to is. On the third displayed line, change the denominator of the second integral from k to x. [11/2/2008]On page 1131, in the solution to question 18, on the third displayed line, replace E[HM 2 ] with E [ (N α) 2], where N is the number of claims. [10/19/2008]On page 1097, in the solution to question 35, on the 5th line, there should be a radical over the second Var(X) Var(Y ): ρ Var(X) Var(Y ) Var(X) Var(Y ). [10/17/2008]On page 233, in the solution to exercise 15.4, on the 5th displayed line, add 1 + after the left parenthesis: Pr(S 20) = 1 ( )e 5 = [10/11/2008]On page 1010, in question 13, change Pr(3 < X < x X < 10) to Pr(3 < X < x X < 10). [10/11/2008]On page 1076, in the solution to question 13, on the third and fourth displayed lines, change dq to dq. [10/11/2008]On page 1090, in the solution to question 13, on the third line, change = 0.25 to = Change Pr(3 < X < x X < 10) to Pr(3 < X < x X < 10) in the four places it appears. [9/28/2008]On page 22, in the solution to exercise 1.23, on the 7th line, change X can t be negative to Y can t be negative. [9/28/2008]On page 38, on the first line of the solution to exercise 2.17, change 3x 2/3 to 1 3 x 2/3. On the fourth line, delete l in S(x lλ). [9/28/2008]On page 132, on the second line, change must by to must be. [9/28/2008]On page 192, some of the notation is inaccurate. Replace the paragraph starting with Let U be claim size and the following 3 displayed lines with Let U be claim size. We would like to calculate Var(U) = Var I ( EU [U I] ) + E I [ VarU (U I) ]

3 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date 3 ( Let s calculate Var I EU [U I] ). [ E I EU [U I] ] = E I [200, 1000, 100, 1500] = 0.2(200) + 0.3(1,000) + 0.4(100) + 0.1(1,500) = 530 [ E I EU [U I] 2] = E I [200 2, , 100 2, ] = 0.2(200 2 ) + 0.3(1,000 2 ) + 0.4(100 2 ) + 0.1(1,500 2 ) = 537,000 ( Var I EU [U I] ) [ = E I EU [U I] 2] [ E I EU [U I] ] 2 = 537, = 256,100 [9/28/2008]On page 1119, in the solution to question 22, on the fifth line, change must by to must be. [9/13/2008]On page 678, in the solution to exercise 45.27, in the table, on the second line, the number under Side Wall should be 2 instead of 1. [9/8/2008] On page 245, make the following corrections to the solution to exercise 16.5: On the fourth line, the equation for p k should be p k = e (k 0.5)/2 e (k+0.5)/2. On the first displayed line, add an equals sign: E[X] = k=0 ( kp k = e 1/4 e 3/4 + 2 e 3/4 e 5/4) ( + 3 e 5/4 e 7/4) + Six lines from the bottom of the page, replace with On the second-to-last line, change E[X 1.6] to E[S 1.6]. [9/8/2008] On page 453, in exercise 29.41, the column Years of Disability should be (1, 2) instead of (0, 1) and [2, ) instead of [1, ). [9/7/2008] On page 142, on the 5th line of the answer to Example 10B, replace Then c... through the end of the paragraph with Then the ratio of each modified probability to the (a, b, 0) probability is By the definition of c as 1 p M 0, [9/7/2008] On page 971, on the third displayed line, change ln x to ln Q α. c 1 p 0, so c 1 p 0 = pm 1 p1 = = 8 5. [9/7/2008] On the last 3 lines of page 971, Q α is the α quantile of a standard normal distribution, not of Z. [8/14/2008]On page 365, in the second line of the answer to Example 25B, replace r 1 with r 0. [8/6/2008] On page 83, the paragraph after Definition 1, there are 2 typos in the first 3 sentences. The following replacement is clearer: To calculate E[S t S t < K], we carry out two steps. The first step is to integrate the lognormal random variable S t /S 0 over its probability density function from 0 to K/S 0, and multiply it by S 0. The result of S 0 times the integral is the partial expectation of S t. [8/6/2008] On page 85, on the last line before the exercises, a t is missing from the first exponent: S 0 e (α δ)t. [8/5/2008] On pages , the solution to exercise 14.3 starting with The Poisson rate is incorrect. The correct solution starting from that point is

4 4 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date Let N be the number of coins picked up in half an hour. Then Pr(N = 0) = 1 11 Pr(N = 1) = 1 11 λ=0 ( ) λ 10 e 0.5λ 11 ) λ a geometric series = 1 ( e 0.5 = 1 ( ) /11e 0.5 = ( ) λ λ e 0.5λ 11 λ=0 ( ) λ 10 λ 11e 0.5 = 1 22 λ=1 = 1 ( 22 10/11e 0.5 (1 10/11e 0.5 ) 2 ) = Pr(N 2) = = [8/5/2008] On page 224, on the 4th line, change Pr(S 2.8) to Pr(S > 2.8). [8/2/2008] On page 89, in the second paragraph, change the first sentence to A franchise deductible d means that if the loss is less than or equal d, nothing is paid, but if the loss is higher than d, the full amount is paid. [8/2/2008] On page 366, in the first displayed line on the page, q (τ) j in the exponent. and p (τ) j should have only one set of parentheses [8/2/2008] On page 439, on the second line of the paragraph under Weibull distribution, the first word should be then instead of the. [7/19/2008]On the last line of page 224, the formula is incorrect. The upper bound of the sum should be one lower, and a parenthesis is missing after the last S. The corrected formula is u 1 E[S d] = hs(hj) + (d hu)s(hu) j=0 where u = d/h 1. If u 1 < 0 (for example if d = 1 and h = 2), the sum is empty and only the second term is used. [7/10/2008]On page 470, delete the footnote, and replace the last complete paragraph on the page with the following: 1 The estimate for µ is based purely on the first and last observations. Thus it is not improved by more frequent observations, only by a longer time period. As a result it is not very accurate. We can see this mathematically. If the time from the first to last observation is t and the period is broken up into n periods, the mean of the lognormal for each interval is µt n and the variance for each interval is σ2 t n. The estimator for µt n, the sample mean of the logs, has variance equal to the variance of the distribution divided by n, or σ2 t n. Thus the variance of the estimate of µ is ( )( n 2 σ ) 2 t 2 t 2 n = σ 2 2 t, which does not depend on n. 1 I thank Ken Burton for pointing this out to me.

5 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date 5 For example, if the period is one year and we knew that the standard deviation σ = 0.3 the standard deviation of the estimator for µ, would be 0.3 and there would be a 95% probability of the sample mean being in an interval of width 2(1.96)(0.3) = centered at µ. This is a huge interval. [7/7/2008] On page 120, in exercise 8.16, add after You are given: : X has a continuous distribution. Some values for X s distribution function and limited expected values are given in the following table:. [4/24/2008]On page 1086, on the first line of the page, change 1 w 10 to 1 w 5. [4/23/2008]On page 1160, in the solution to question 8, on the first line of the page, the right hand side should be e 1 ( ). [4/17/2008]On page 81, replace the second line of Section 6.3 with α be the expected continuously compounded rate of return on a stock, [4/16/2008]On page 1040, change the last line of question 16 to Given that the option pays off, determine the expected value of S 1. [4/14/2008]On page 1123, in the solution to question 37, on the last line of the page, change the right hand side to e 2y1/θ. [4/11/2008]On page 698, in the solution to exercise 46.21, on the second line change = to -. On the seventh line change 3 to 2, and on the eighth lines change the first 3 to 2. [4/10/2008]On page 851, in the solution to exercise 57.2, on the 3rd line, delete the coefficient 2 of ( ) [4/2/2008] On page 502, in the solution to exercise 32.10(ii), on the second line, replace the numerator 1145 with [3/26/2008]On page 393, on the 5th displayed line of the solution to exercise 26.26, delete the coefficient 2 before θ 2 2. [3/26/2008]On page 415, on the first line, delete at. [3/26/2008]On page 780, in the solution to exercise 52.32, on the third line, change = 3 to = 2 3. [3/26/2008]On page 784, in the solution to exercise 52.45, on the last line, change the first to +. [3/26/2008]On page 810, in the solution to exercise 53.38, on the last line of the page, add E before [ e σ2]. [3/16/2008]On page 393, in the solution to exercise 26.25, the second to last displayed line should be split into two lines: 2θ (50 3θ 1) 2 = 400 5θ θ = 0 3 On the last line of the solution to exercise 26.25, add the word not between would and result. [3/16/2008]Replace the paragraph before the heading Coverage of this material... with the following: On pre-2000 exams, Poisson frequency was virtually the only case tested on; as a result, this lesson has lots of old exam questions. Between 2000 and 2004, however, questions on non-poisson frequency, discussed in the next lesson, were more frequent. At this point, it is unclear whether non-poisson frequency is still on the syllabus, so I once again expect most of the limited fluctuation credibility questions to be based on Poisson frequency. [3/8/2008] On page 605, on the first displayed line, the last denominator is missing a factorial: n i=1 x!. [3/1/2008] On page 312, in the solution to exercise 21.23, the final answer should be instead of

6 6 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date [2/25/2008]On page 931, one line after the second displayed formula, change variances to standard deviations. The third displayed formula should be ln(s ti ) = ln(s 0i ) + (α i 0.5σ 2 i )t + σ i t Z(i) [2/24/2008]On page 470.2, on the fifth line, change to [2/24/2008]On page 912, in exercise 61.17(ii), delete 0 from the set {1,2,3,4,5}. [2/16/2008]On page 64, on the second line, replace it if with if it. [2/16/2008]On page 111, in the solution to exercise 7.42, on the third displayed line, replace x2 3 [2/16/2008]On page 127, in the solution to exercise 8.17, on the first line, replace 2002 with with x3 3. [2/16/2008]On page 136, in the solution to exercise 9.4, on the third displayed line, in the expression (500 + K), K should be capitalized. [2/16/2008]On page 157, in exercise 11.10(i), replace a with α. [2/16/2008]On page 232, in the solution to exercise 15.1, on the first displayed line, replace with [2/9/2008] On page 106, in the solution to exercise 7.20, on the first displayed line, replace X θ with X d. [2/7/2008] On page 127, in the solution to exercise 8.13, on the first line, put parentheses around X 10,000. [2/7/2008] On page 208, in the solution to exercise 13.57, the final answer is 990,944. [2/7/2008] On page 874, 3 lines from the end, delete the parenthetical remark (on which they almost... ). This is now a frequent exam topic. [2/4/2008] On page 153, in the solution to exercise 10.24, on the second line, replace p n with pn p n 1. [1/30/2008]On page 969, at the very end of the second paragraph in Section 67.1, change Q α to CTE α. [1/30/2008]On page 980, on the second line, delete 1. Also, the negative sign of the exponent should be outside the parentheses. Thus the line should read S(x) = e (x/θ)τ [1/28/2008]On page 946, on the last line of the answer to Example 64B, replace with and 42.5% with 42.6%. [1/7/2008] On page 366, 4 lines from the bottom of the page, replace q (d) 2 with q (w) 2. [12/26/2007]On page 84, on the fourth line, change S to S 0. [12/26/2007]On page 367, on the fourth line, after ( ), replace with 0.l [12/26/2007]On page 375, in the displayed line of Example 26E, replace F (x) with f(x). [12/4/2007]On page 81, on the second displayed line, 2 t s are missing from the last expression, which should read N ( (α δ 0.5σ 2 )t, σ 2 t ) [12/4/2007]On page 181, exercise is the same as exercise 3.3. [12/4/2007]On page 321, exercise 22.1 is the same as exercise 4.1. [12/4/2007]On page 623, exercise 42.4 is the same as exercise

7 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date 7 [12/4/2007]On page 709, exercise is the same as exercise [12/4/2007]On page 710, exercise is the same as exercise [12/4/2007]On page 710, exercise is the same as exercise [12/2/2007]On page 51, 5 lines above Exercises, add the word payment between expected and per loss. [10/30/2007]On page 113, in the solution to exercise 7.49, on the second displayed line from the end, the second ( ) should not be squared. The line should read ( )( ) = (500/1.1) = [10/30/2007]On page 178, in exercise 13.4, on the first line, change than to that. [10/30/2007]On page 1137, the last 3 lines of the solution to question 32 is incorrect. Replace them with: We must also revise the probabilities of begin in Classes A and B to account for the conditioning. We use Bayes theorem to do this. Now we re ready to calculate the variance. Pr(X > 20 A) Pr(A) Pr(A X > 20) = Pr(X > 20 A) Pr(A) + Pr(X > 20 B) Pr(B) ( ) 3 10 Pr(X > 20 A) = S(20 A) = = ( ) 4 10 Pr(X > 20 B) = S(20 B) = = (1/8)(0.6) Pr(A X > 20) = (1/8)(0.6) + (1/16)(0.4) = 0.75 Pr(B X > 20) = 1 Pr(A X > 20) = = 0.25 Var(X X > 20) = E[Var(X I & X > 20)] + Var(E[X I & X > 20]) = [ (0.75)(300) + (0.25) ( ( 88 9)] 8 + (0.75)(0.25) = = [10/29/2007]On page 131, on 8th line from the bottom, add a right parenthesis at the end of the line. [10/29/2007]On page 738, 10 lines from the bottom, change than to that. [10/25/2007]On page 954, in exercise 65.2, on the last line, replace payoff with price. [10/25/2007]On page 956, in the solution to exercise 65.2, on the second displayed line, delete the negative sign before 0.6 in the second numerator. Replace the last three lines of the solution with The arithmetic average is = and the payoff is = Discounting this payoff 3 months, the value of the option in this run is e 0.01 (0.9516) = The geometric average is 3 ( )( )( ) = and the payoff is = Discounting this payoff 3 months, the value of the geometric option in this run is e 0.01 (0.9613) = The simulated value in this run of the arithmetic average option using the control variate method is ( ) = ) 2

8 8 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date [10/25/2007]On page 988, the last two lines of the page should read ( ) 5 = = The confidence interval is ± 1.645(0.4693) = (365.0, 366.5). [10/23/2007]On page 1098, in the solution to question 39, on the first line, change 0.2(10) = 2 to 0.1(20) = 2. [10/21/2007]On page 842, at the end of the first paragraph of subsection , replace the formula for x with r ni i=1 j=1 x = m ijx ij r ni i=1 j=1 m. ij [10/21/2007]On page 871, the solution to exercise is incorrect. The correct solution is For a geometric distribution, the hypothetical mean is µ(θ) = β and the process variance is v(θ) = β(1+β) = β+β 2. The expected hypothetical mean, expected process variance, and variance of hypothetical means are: µ = E θ [µ(θ)] = E[β] v = E θ [v(θ)] = E[β] + E[β 2 ] a = Var θ ( µ(θ) ) = Var(β) = E[β 2 ] E[β] 2 It then follows that a = v µ µ 2. We can estimate µ = x. By equation (57.3), ( ) 1 ( r r ) â = m m 1 m i ( x i x) 2 ˆv(r 1) Plugging in v = a + µ + µ 2, we get r i=1 â = m i( x i x) 2 (â ˆµ ˆµ 2 )(r 1) m m 1 r i=1 m2 i and letting D = m m 1 r i=1 m2 i, i=1 m 2 i i=1 r i=1 â = m i( x i x) 2 (r 1)(ˆµ ˆµ 2 ) D + r 1 In our exercise (in which r = 2 for 2 policyholders), m 1 = = 30 x 1 = 9 30 = 0.3 m 2 = = 20 x 2 = 2 20 = 0.1 D = = m i ( x i x) 2 = 30( ) ( ) 2 = 0.48 i=1 â = ( ) = ˆv = =

9 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date 9 The credibility factor for Policyholder A is Z = 30( ) 30( ) = [10/16/2007]On page 1120, in the solution to question 23, on the first displayed line, move the first θ into the exponent: e 0.3θ f Θ (θ)dθ [10/16/2007]On page 1123, in the solution to question 34, on the third line, replace the first 3 with 0.7. [10/16/2007]On page 1143, in the solution to question 6, on the second line, change the denominator of the fraction from n to 2n. [10/16/2007]On page 1159, in the solution to question 5, on the 4th line from the end, remove the exponent 2 from [10/14/2007]On page 176, on the first displayed line, change 52,500 to 52,250. [10/14/2007]On page 197, in the solution to exercise 13.19, on the fourth displayed line, change 1788λ to 1788 λ. [10/14/2007]On page 201, in the solution to exercise 13.27, on the fourth line, replace with [10/14/2007]On page 653, in the solution to exercise 44.15, on the second line from the end, Φ should be Φ 1. [10/14/2007]On page 719, in the solution to exercise 48.1, Φ(1.890) should be 1 Φ(1.890). [10/14/2007]On page 719, in the solution to exercise 48.5, on the second to last line, there should be a square root sign over 109,901 in the denominator. [10/13/2007]On page 1117, in the solution to question 16, on the 5th line of the page, change deductible of to deductible of [10/8/2007]On page 204, in the solution to exercise 13.40, in the last line, replace with and Φ(1.169) with Φ(1.164). The final answer is instead of [10/8/2007]On page 975, in the solution to exercise 67.2, on the second displayed line, change ln to twice. Also, one line above and below the displayed lines, change to e [10/3/2007]On page 160, in the solution to problem 8, on the last line, change Pr(N = 2) to Pr(N 2). [10/3/2007]On page 940, in the solution to exercise 63.1, on the 6th line, replace with [10/3/2007]On page 957, in the solution to exercise 65.5, on the first line, the second denominator should be instead of [9/26/2007]On page 292, in the solution to exercise 20.9, on the first displayed line, Pr(800 < X < 1000) should be replaced by number of losses between 800 and [9/26/2007]On page 921, in the solution to exercise 61.25, in the heading of the third column of the table, change 0.3u i to 0.3z i. [9/26/2007]On page 921, in the solution to exercise 61.26, change the heading of the last column of the table to e ni. [9/26/2007]On page 988, the solution to Example 68G is incorrect. The correct solution is Using the non-parametric method discussed above, 0.95(500) = 475. The variance of the binomial is 500(0.95)(0.05) = So we want to add and subtract = 8.017, which we round up to 9, to and from 475. The confidence interval is (L (466), L (484) ) = (350.3, 358.0).

10 10 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date [9/20/2007]On page 21, in the solution to exercise 1.19, on the 10th line of the page, change 3 2 (7500) to 2 3 (7500). [9/20/2007]On page 682, 9 lines from the bottom, change the exponent on θ from 13 to 14: = 13 14(2 13 1) θ14 [9/20/2007]On page 696, in the solution to exercise 46.16, 3 lines from the end, change a = 3b to a = 3c. [9/20/2007]On page 735, in the solution to exercise 50.7, on the last two lines replace twice with , and replace with [9/20/2007]On page 806, in the solution to exercise 53.23, on the second line, replace E[S] with Var(S). [9/20/2007]On page 830, in the solution to exercise 55.8 on the second line from the end, there should be an equals sign between v and E[θ]. [9/20/2007]On page 853, in the solution to exercise 57.8, two lines below the table (equation for ˆv), delete the 2 after the third summand. [9/6/2007] On page ix, 12 lines from the bottom, delete the word a before real exams. [9/6/2007] On page 398, the last line of the page, which is the last line of the solution to Example 27C, is incorrect. Replace it with: 20 ˆθ = ln 0.6 = = [9/3/2007] On page 642, in the solution to exercise 43.13, on the last line, replace n F with e F. [8/31/2007]On page 571, in the solution to exercise 39.2, on the 5th displayed line, the first numerator should be F (500) F (250) instead of F (500). [8/31/2007]On page 578, in the answer to Example 40G, on the 6th displayed line, replace N 3 with N > 3. [8/31/2007]On page 598, in the solution to exercise 40.15, on the first line, replace with [8/31/2007]On page 605, on the seventh line, replace λ 1 = λ 2 = 0.1 with λ 1 = 0.1, λ 2 = [8/31/2007]On page 610, in the solution to exercise 41.6, replace the 4th displayed line with 2 1 n i=1 l(θ) = n ln θ x i θ = n ln θ n x θ [8/30/2007]On page 241, on the 3rd and 6th lines of subsection , change x kh and x (k+1)h to x k and x k+1. This is the notation the textbook uses. [8/30/2007]On page 242, starting from the 9th line through the end of the solution to Example 16D, every superscript 4 should be replaced with a superscript 1. In other words, m 4 0 should be m 1 0 and m 4 1 should be m 1 1 in all places. [8/30/2007]On page 619, on the fifth line, the second sentence should start: The square of the coefficient of variation of the number of claims.... [8/30/2007]On page 822, in the solution to exercise 54.5, on the second displayed line (last line on the page), replace mu with µ. [8/28/2007]On page 551, in Figure 37.5(d), change the two 4 s in the caption to 40 s. [8/28/2007]On page 803, on the first displayed line of the solution to exercise 53.16, replace X 1 = 0 with X 1 = 340.

11 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date 11 [8/27/2007]On page 542, in the answer to Example 37C, on the 3rd line of 2, replace 920 with 9 20 [8/27/2007]On page 542, 11 lines from the bottom, change (3, 0.625) to (3, ). and 1120 with [8/26/2007]On page 519, in the solution to exercise 33.12, on the fourth displayed line, insert c into the last exponent so that the line reads L(θ, c) = )/θ c2 2 θ 2 e (12 θ 2 e c( )/θ [8/25/2007]On page 525, in the second line of the second paragraph, add the word be between now and denoted. [8/22/2007]On page 376, five and four lines before Section 26.2, a θ is missing on each line: 10θ θ + 14,500 = 0 θ 2 295θ = 0 [8/22/2007]On page 395, in the solution to exercise 26.31, on the second to last line, change greatest lower bound to least upper bound. [8/22/2007]On page 417, two lines from the bottom, delete the first θ: θ = 35θ [8/22/2007]On page 457, in the solution to exercise 29.19, on the second line, replace e ( 2 θ )2 with e ( t θ )2. [8/22/2007]On page 464, in the solution to exercise 29.40, change the equation for l(α) to l(α) = n ln α α ln x i + m ln 3α m ln(2 + α) 3α ln yi 2 + α [8/22/2007]On page 466, in the solution to exercise 29.46, on the fourth line, change q x to 0.6q x. [8/22/2007]On page 469, on the first line of the solution to Example 30A, change into to by. [8/19/2007]On the last line of page 242, change f i to f ih. (The textbook calls this f i, but it is actually the probability that the distribution is equal to ih, so f ih is consistent with the notation we ve previously used.) [8/19/2007]On page 246, in exercise 16.7, the severities must be modified, so the solution is incorrect. Replace the 6th line and later with We ll modify the negative binomial to have non-zero claims only by multiplying β by 1 f 0 ; (0.5)( ) = We must also modify the probabilities of severities to make them conditional on severity not being zero by dividing by 1 f 0 ; the modified f 1 is /( ) = Then ) r ) 2 = = Pr(S = 0) = ( β p 1 = p 0 ( rβ 1 + β ( ) ( ) 2( ) = ( ) = Pr(S = 1) = ( )( ) = Then F S (1) = = [8/19/2007]On pages , in the solution to exercise 16.8, replace the last three lines with ( )( ) ( ) F S (1) = 1 2 ( ) ( ) = =

12 12 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date The discretized estimate was 7.5% off. [8/19/2007]On page 472, in the solution to exercise 30.1, on the second line from the end, change to and delete the extra 0. in [8/16/2007]On page 188, in exercise 13.48, on the third line, change 1,500 to 1,300. [8/16/2007]On page 508, 3 lines from the bottom, change baseline smoker to baseline individual. [7/28/2007]On page 220, in the solution to exercise 14.14, on the fifth displayed line (the formula starting Pr(Y = 2)), replace numerator with Four lines from the end, replace with [7/28/2007]On page 236, in the solution to exercise 15.16, on the last line, replace with [7/28/2007]On page 297, in the table at the bottom of the page, s 4 = 1, not 2. [7/19/2007]On page 196, in the solution to exercise 13.16, change will to make on the second line, and change the final answer to The final answer given (0.0062) is what you would get without a continuity correction. [7/19/2007]On page 205, in the solution to exercise 13.43, on the last line, change the second expression to ( ) 100,000 96,000 1 Φ 9,760,000 [7/19/2007]On page 205, at the end of the solution to exercise 13.45, add the words making the final answer = [7/19/2007]On page 208, in the solution to exercise 13.56, on the last line, change to [7/19/2007]On page 211, on the first line, the reading from Loss Models should exclude subsection [7/19/2007]On page 240, on the fifth line of text, add the words as high after at least. In the third displayed line, the second summation sign is misplaced, and it should read n 1 x/θ xj F S (x) = p n 1 e j! n=0 j=0 [7/13/2007]On page 172, on the last line, change E[X 2 ] to E[X] 2. [7/8/2007] On page 142, on the third line, replace c = 1 pm 0 1 p 0 with c 1 p 0. [7/8/2007] On page 142, 4 lines after the answer to Example 10B, change r = 0 to r 0. [7/8/2007] On page 151, in the solution to exercise 10.16, on the second to last line, there should be a right parenthesis after the first e 5. [7/8/2007] On page 970, in the caption for Figure 67.1, delete 0.05 before Q [7/5/2007] On pages , exercises 16.1 and 16.2 are defective. Change the exercises to the following: 16.1 For a block of 50 policies: (i) Claim sizes on a coverage follow an exponential distribution with mean (ii) Claim counts for each policy follow a negative binomial distribution with parameters r = 0.02 and β = 25. (iii) Claim counts and claim sizes are independent.

13 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date 13 Determine the probability that aggregate claims are within standard deviations of the mean Claim sizes on a coverage follow an exponential distribution with mean lives are covered under the contract. Claim counts for each life follow a negative binomial distribution with mean 0.1 and variance 1.1. Claim counts and claim sizes are independent. A stop-loss reinsurance contract is available at 150% of expected claim cost. You are willing to pay 1500 for the contract. Determine the aggregate deductible needed to make the cost of the contract Change the solution to 16.1 to the following: For the block of 50 policies, claim counts follow a negative binomial with r = 50(0.02) = 1 and β = 25, or a geometric. The mean and variance are β = 25 and β(1 + β) = 650 respectively. Letting S be aggregate losses for the entire block, E[S] = 25(1500) = 37,500. By the compound variance formula, Var(S) = 25( ) + 650( ), and the standard deviation is σ = = 38,971. Then 37, (38,971) = 4686 and 37, (38,971) = 70,314, so we want the probability of 4686 β S 70,314. The aggregate distribution is a two-point mixture with weight (1+β) = on the exponential; the exponential has parameter θ(1 + β) = (26)(1500) = 39,000. F S (4686) = e 4686/39,000 = F S (70,314) = e 70,314/39,000 = So the probability of being in the interval is = If the normal approximation had been used, the probability would be 0.6 since Φ(0.842) = 0.8. Replace the first line of the solution to 16.2 with the following: For each life, rβ = 0.1 and rβ(1 + β) = 1.1, so β = 10 and r = For the group, r = 100(0.01) = 1 and β = 10, making the distribution geometric. The rest of the solution to 16.2 remains the same. [7/1/2007] In lesson 26 (starting on page 373), note that the tables for the Fall 2007 sitting will no longer have method of moments estimators. Thus all problems involving estimating parameters for the gamma, Pareto, and lognormal distributions must be done using first principles. This affects the solutions to Examples 26B and 26C, and the solutions to exercises 26.2, 26.3, 26.4, 26.5, 26.6, 26.11, 26.13, 26.15, 26.16, 26.17, and Replacement pages for subsections and for the solutions to the above exercises are at the end of these errata. (Ignore the different pagination.) [7/1/2007] On page 379, in exercise 26.5, change x 2 i /n to (x i x) 2 /n. [7/1/2007] On page 397, delete the first paragraph of Section The tables you get on the exam will not have p and q. [6/23/2007]On page 433, in the solution to exercise 28.17, on the fourth line, change survivor to survivors. [6/20/2007]On page 179, exercise 13.5 is a duplicate of exercise 3.6. [6/20/2007]On page 409, in the solution to exercise 27.14, on the second line, the second exponent α. [6/19/2007]On page 131, on the 3rd and 2nd lines from the bottom, change to [6/17/2007]On page 291, on the last line, change S07:30 to S07:40. θ 2θ k should have an

14 14 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date [6/15/2007]On page 115, replace the third paragraph with the following: Usually exam problems will tell you exactly what order to make the coverage modifications in. However, the policy limit is defined as the most that will be paid, and the maximum covered loss is defined as the claims limit of the policy, usually denoted with the letter u. If there is a deductible of 500, coinsurance of 80%, and a claims limit of 10,000, this means that the maximum covered loss is 10,000 and the policy limit is 0.8(10, ) = [6/15/2007]On page 129, in the solution to exercise 8.23, on the third displayed line, change 10 5 to [6/15/2007]On page 348, in the caption for subfigure 24.7(a), delete = k 70 (x). In the caption for subfigure 24.7(b), delete = K 70 (x). [6/12/2007]On page 10, on the last line, remove the exponent from [6/12/2007]On page 111, on the last line of the solution to exercise 7.43, eliminate the negative sign from the exponent on e.

15 Lesson 26 Method of Moments Reading: Loss Models 12.1 There is approximately one method of moments question on each exam. In the method of moments, to fit a k parameter distribution, you equate the first k sample moments with the first k moments of the distribution you re fitting. If k is 2 or greater, instead of matching the second moment, you may match the variance, but use the empirical distribution variance ( 1 n (xi x) 2 ), which uses division by n, not the sample variance ( 1 n 1 (xi x) 2 ), which uses division by n The method of moments for various distributions Exponential The exponential distribution has one parameter θ, which is its mean. To fit an exponential distribution using the method of moments, set θ equal to the sample mean. EXAMPLE 26A You are given a sample of 50 claims whose sum is 920. It is assumed that the underlying distribution of these claims is exponential. Determine the method of moments estimator of the loss elimination ratio at 10. ANSWER: The method of moments estimator of θ is x = = For an exponential distribution, the loss elimination ratio is E[X d] LER(d) = = 1 e d/θ E[X] In our case, LER(10) = 1 e 10/18.4 = Gamma A gamma distribution has two parameters, α and θ. Let X be the random variable having a gamma distribution. Then E[X] = αθ Var(X) = αθ 2 Equating these the the sample mean x and the biased sample variance ˆσ 2, αθ = x αθ 2 = ˆσ 2 Dividing the first line into the second, ˆθ = ˆσ2 x ˆα = x2 ˆσ 2 (26.1) Copyright 2007 ASM 371

16 METHOD OF MOMENTS EXAMPLE 26B You have the following sample of loss sizes: 3, 5, 8, 10, 15, 27, 38, 60 x i = 166 xi 2 = 6196 You are to fit the losses are assumed to a gamma distribution using the method of moments, matching the first 2 moments. Determine the estimate of θ. ANSWER: The sample mean is x = = The empirical variance is = By equation (26.1), ˆθ = = Pareto The following example shows how to calculate the method of moments estimators for a Pareto. EXAMPLE 26C You have the following sample of loss sizes: 3, 5, 8, 10, 20, 50, 100, 200 x i = 396 x 2 i = 53,098 You are to fit the losses are assumed to a Pareto distribution using the method of moments, matching the first 2 moments. Estimate e(20) using the fitted distribution. ANSWER: Matching first and second moments, θ α 1 = = θ 2 (α 1)(α 2) = 53,098 = By dividing the square of the first equation into the second we eliminate θ. 2(α 1) α 2 = = α 2 = α 2 = = ˆα = = Then we can calculate ˆθ by plugging ˆα into the equation of means. θ α 1 = 49.5 ˆθ = 49.5( ) = Therefore, e(20) = = (Actually, we did not have to calculate ˆθ in order to calculate e(20), since e(20) = E[X]+ 20 α 1 and the method of moments sets E[X] equal to the sample mean automatically, so we just needed ) Copyright 2007 ASM

17 METHOD OF MOMENTS [4-F04:24] You are given: (i) Losses are uniformly distributed on (0, θ) with θ > 150. (ii) The policy limit is 150. (iii) A sample of payments is: 14, 33, 72, 94, 120, 135, 150, 150 Estimate θ by matching the average sample payment to the expected payment per loss. (A) 192 (B) 196 (C) 200 (D) 204 (E) 208 Additional released exam questions: C-F05:21, CAS3-F06:3 Solutions For an exponential, set ˆθ equal to the sample mean ˆθ = X = 7 Pr(X > 500) = e ( 9000)(500) 7 = The first 2 raw sample moments are = We equate moments: m = = t = = 680,000 ˆθ ˆα 1 = m = 550 2ˆθ 2 ( ˆα 1)( ˆα 2) = t = 680, ( ˆα 1) = ˆα ˆα 2(680000) = 2(550 2 ) ˆα 2(550 2 ) ˆα = = ˆθ = 550( ˆα 1) = Either way, ( ) α θ Pr(X > 500) = θ ( ) = = Copyright 2007 ASM

18 EXERCISE SOLUTIONS FOR LESSON The first moment is m = 667, = , and the second moment is t = 20,354,674, = 203,546, Then, dividing the square of the first moment into the second moment, 2(α 1) α 2 = 203,546, = α 2 = α 2 = = ˆα = = ˆθ = m( ˆα 1) = ( ) = 11, We calculate the first and second raw sample moments m = = t = = 37, Dividing m 2 into t and equating to the distribution moments, 2(α 1) α 2 = 37,504.8 = α 2 = α 2 = = ˆα = ˆθ = m( ˆα 1) = To find the 95th percentile, we equate F(x) = 0.95, or 1 F(x) = ( ) α θ = 0.05 θ + x θ θ + x = α 0.05 (θ + x) α 0.05 = θ x = θ(1 α 0.05) α = ( ) = = (A) In order for the variance to exist, α must be greater than 2. When fitting with the method of moments, we divide the mean squared into the second moment to obtain 2(α 1) α 2 = t m = t Copyright 2007 ASM

19 METHOD OF MOMENTS α 2 = t The left hand side is greater than 2 since the fraction must be positive, so empirical variance t m 2 > 2(500 2 ) = = 250, The first two raw sample moments are ( ) m = = ( ) t = = 701, t > 2, implying t > 2(500 2 ). Then the Then 2(α 1) α 2 = t m = 701,401.6 = α 2 = α 2 = = ˆα = = ˆθ = m( ˆα 1) = (508)( ) = The limited expected value at 500 is E[X 500] = θ α 1 = ( ) α 1 θ θ ) = (C) 1 ( To bring this up with inflation, we multiply average claim size in year 1 by and in year 2 by 1.1. Then, average inflated claim size is 100(12,100) + 200(13,750) = 13, Setting the Pareto mean equal to this, θ 2 = 13,200 θ = 26,400 (E) This is a Pareto with θ = 1. Equating sample and distribution means, X = E[X] = 1 α 1 α = X = X + 1 X (E) Copyright 2007 ASM

20 EXERCISE SOLUTIONS FOR LESSON Only one parameter is estimated, so only one moment is matched: the mean. We will express the estimator in terms of the data. ˆθ 2 = So we must calculate the variance of the right hand side. Var(ˆθ) = i=1 ˆθ = i=1 i=1 X i X i Var(X i ) = Var(X i) X i has a Pareto distribution, and its variance is the can be read from the tables in the appendix. 2θ 2 ( ) 2 Var(X i ) = (α 1)(α 2) θ = 2θ2 α 1 2(1) θ2 4 = 3 4 θ2 Plugging this into the expression for Var(ˆθ), Var(ˆθ) = We equate the first 2 moments and solve for α and θ. ( ) ( ) θ2 = 0.01θ 2 (D) X = ˆαˆθ ˆα 1 = 30 E[ ˆX 2 ] = = ˆαˆθ 2 ˆα 2 = 1100 ( ˆα 1) 2 ˆα( ˆα 2) = ˆα 2 22 ˆα = 9 ˆα 2 18 ˆα ˆα 2 4 ˆα 9 = 0 ˆα = 4 ± = Notice that from the first moment equation. ˆθ = 30(2.3452/3.3452) = 21.03, which makes the fit impossible since data for a single-parameter Pareto may not be less than θ, and we are given the points 10 and 20. That s why θ for a single-parameter Pareto should be given in advance rather than fitted The mean of a single-parameter Pareto is αθ α 1. Here we have a single-parameter Pareto with θ = 1, so α α 1 = x α 1 = x As x, the fraction goes to 0 and ˆα 1. (C) ˆα = x 1 Copyright 2007 ASM

21 METHOD OF MOMENTS For the Nelson-Åalen estimator, the risk sets and events through time 15 are The Nelson-Åalen estimate is then y i r i s i Ĥ 1 (15) = = Notice that the large observation 131 has no effect on this estimate, but will affect the method of moments estimate. The sample mean is 2(11)+12+3(13)+2(14)+15+2(17)+2(19) = Setting this equal to the mean of a single αθ parameter Pareto, α 1, Then we get S (15) and then H 2 (15). Ŝ (15) = 10α α 1 = α α 1 = α = 344α α = 344 ( ) α θ = 15 ˆα = = ( ) ˆα 10 = 15 Ĥ 2 (15) = ln = The absolute difference is = Calculate the first two raw moments. ( ) = (60) x = = 60 6 t = (60 2 ) = ˆσ 2 = = Using formula (26.1), ˆα = = 10.8 (D) This is a gamma distribution with α = 2 and θ = c, so 2c = x = 8, c = 4 (E) Copyright 2007 ASM

22 EXERCISE SOLUTIONS FOR LESSON The first two raw sample moments are (200) x = = t = (200 2 ) = 43, ˆσ 2 = 43, = 22, By formula (26.1), ˆθ = ˆσ2 x = x = = t = = 977, ˆσ 2 = 977, = 27, ˆα = x2 ˆσ 2 = , = (D) x = = t = ˆσ 2 = 16,332, = 1,892,000 ˆθ = ˆσ2 x = 1,892,000 = (E) 3800 = 16,332, Let s calculate the sample mean. The sample variance is not needed since only one parameter is being estimated. When they don t specify which moments to use, always use the first k moments, where k is the number of parameters you re fitting (here k = 1). Then set E[X] equal to 4.2. X = = 4.2 β 2π = ˆβ = 2(4.2) = π (D) You may recognize this distribution as a beta distribution with a = α, b = 1, and the mean of a beta is. If not, it s not hard to integrate: a a+b = α α+1 E[X] = 1 0 αx α dx = α α + 1 Copyright 2007 ASM

23 EXERCISE SOLUTIONS FOR LESSON Not really a method of moments problem, but since it involves moments, I put the problem here. The mean is 1 2 (m 1 + m 2 ). The second moment is m m2 2. The square of the coefficient of variation is σ 2 µ = E(X2 ) E(X) 2 = E(X2 ) 2 E(X) 2 E(X) 1 2 so if we maximize the quotient of the second moment over the square of the first moment, we ll be done. That quotient is 4 m2 1 + m2 2 (m 1 + m 2 ) 2 We want to minimize the denominator. Since m 1 > 0 and m 2 > 0, (m 1 + m 2 ) 2 > m m2 2. On the other hand, they can be made arbitrarily close by making m 2 very small, close to zero. So the greatest lower bound of the quotient is 1. But then the answer is 4(1) 1 = 3. (C) Equating moments, µ + 0.5σ 2 = ln 386 = µ + 2σ 2 = ln 457,480.2 = σ 2 = ( ) = ˆµ = ( ) = ˆσ = = Now we use the formula for E[X 500]. The method of moments sets E[X] = x = 386. ( ) [ ( )] ln 500 µ σ 2 ln 500 µ E[X 500] = E[X]Φ + x 1 Φ σ σ ( ) [ ( )] ln ln = 386Φ Φ = 386Φ( ) + 500[1 Φ( )] = 386(0.3877) + 500( ) = (D) It is unusual to match negative moments, but for an inverse Pareto, there is little choice, since positive moments 1 and higher are not defined. In any case, you must follow the exam question s instructions, and default to using the first n moments for method of moments only when the question doesn t specify which moments to use. The sample 1 moment is (We ll use m and t for negative moments here, even though usually they mean positive moments.) m = 1 ( ) = The sample 2 moment is t = 1 6 ( ) = Let m be the sample 1 moment and t the sample 2 moment. We equate these with the fitted moments, using the formulas from the Loss Models Appendix: m = 1 θ(τ 1) Copyright 2007 ASM