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

Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page Errata and updates for ASM Exam C/Exam 4 Manual (Sixth Edition) sorted by page 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(0.234), 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 6, make the changes noted below. Practice exam 5 question 32 is defective in that none of the 5 answer choices is correct. [9/6/2007] On page ix, 2 lines from the bottom, delete the word a before real exams. [6/2/2007]On page 0, on the last line, remove the exponent from 5.76. [9/20/2007]On page 2, in the solution to exercise.9, on the 0th line of the page, change 3 2 (7500) to 2 3 (7500). [9/28/2008]On page 22, in the solution to exercise.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 2.7, change 3x 2/3 to 3 x 2/3. On the fourth line, delete l in S(x lλ). [2/2/2007]On page 5, 5 lines above Exercises, add the word payment between expected and per loss. [2/6/2008]On page 64, on the second line, replace it if with if it. [4/7/2008]On page 8, replace the second line of Section 6.3 with α be the expected continuously compounded rate of return on a stock, [2/4/2007]On page 8, 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 ) [8/6/2008] On page 83, the paragraph after Definition, 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. [2/26/2007]On page 84, on the fourth line, change S to S 0. [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. [8/2/2008] 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. [2/9/2008] On page 06, in the solution to exercise 7.20, on the first displayed line, replace X θ with X d. [2/6/2008]On page, in the solution to exercise 7.42, on the third displayed line, replace x2 x3 3 with 3. [6/2/2007]On page, on the last line of the solution to exercise 7.43, eliminate the negative sign from the exponent on e. [0/30/2007]On page 3, in the solution to exercise 7.49, on the second displayed line from the end, the second ( ) 500. should not be squared. The line should read ( )( ) = (500/.)2 28 500 + = 420.02 6000 33.

2 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page [6/5/2007]On page 5, 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 0,000, this means that the maximum covered loss is 0,000 and the policy limit is 0.8(0,000 500) = 7600. [7/7/2008] On page 20, in exercise 8.6, 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:. [2/7/2008] On page 27, in the solution to exercise 8.3, on the first line, put parentheses around X 0,000. [2/6/2008]On page 27, in the solution to exercise 8.7, on the first line, replace 2002 with 200. [6/5/2007]On page 29, in the solution to exercise 8.23, on the third displayed line, change 0 5 to 0 6. [0/29/2007]On page 3, on 8th line from the bottom, add a right parenthesis at the end of the line. [6/9/2007]On page 3, on the 3rd and 2nd lines from the bottom, change 234.48 to 243.48. [9/28/2008]On page 32, on the second line, change must by to must be. [2/6/2008]On page 36, in the solution to exercise 9.4, on the third displayed line, in the expression (500 + K), K should be capitalized. [7/8/2007] On page 42, on the third line, replace c = pm 0 p 0 with c p 0. [9/7/2008] On page 42, on the 5th line of the answer to Example 0B, 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 p M 0, c p 0, so [7/8/2007] On page 42, 4 lines after the answer to Example 0B, change r = 0 to r 0. c p 0 = pm p = 0.4 0.25 = 8 5. [7/8/2007] On page 5, in the solution to exercise 0.6, on the second to last line, there should be a right parenthesis after the first e 5. [2/4/2008] On page 53, in the solution to exercise 0.24, on the second line, replace p n with pn p n. [2/6/2008]On page 57, in exercise.0(i), replace a with α. [0/3/2007]On page 60, in the solution to problem 8, on the last line, change Pr(N = 2) to Pr(N 2). [4/2/2009] On page 64, the paragraph after Example 2A 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, ) class. p 0 is then the balancing item, k= 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: ( ) p p M 0 = ( p M 0 ) 0 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 ), as in the ETNB family. This is illustrated in the following example:

Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page 3 Example 2B Frequency of claims per year follows a zero-modified negative binomial distribution with r = 0.5, β =, and p M 0 = 0.7. Claim size follows a Pareto with α =, θ = 000, 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 + 500 = 000 500 = 2 3 The revised negative binomial parameters are r = 0.5, β = 2/3. By the equation above: p M 0 = 0.3 ( ) r ( ) 0.5 p 0 = = = 0.442 + β 2 ( ) 0.5 p 0 = = 0.290 5/3 ( ) 0.290 p M 0 = 0.3 = 0.208 0.442 [7/3/2007]On page 72, on the last line, change E[X 2 ] to E[X] 2. [0/4/2007]On page 76, on the first displayed line, change 52,500 to 52,250. p M 0 = 0.208 = 0.7892 [0/30/2007]On page 78, in exercise 3.4, on the first line, change than to that. [6/20/2007]On page 79, exercise 3.5 is a duplicate of exercise 3.6. [2/4/2007]On page 8, exercise 3.4 is the same as exercise 3.3. [8/6/2007]On page 88, in exercise 3.48, on the third line, change,500 to,300. [9/28/2008]On page 92, 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 Let s calculate Var I ( EU [U I] ). Var(U) = Var I ( EU [U I] ) + E I [ VarU (U I) ] E I [ EU [U I] ] = E I [200, 000, 00, 500] = 0.2(200) + 0.3(,000) + 0.4(00) + 0.(,500) = 530 E I [ EU [U I] 2] = E I [200 2, 000 2, 00 2, 500 2 ] = 0.2(200 2 ) + 0.3(,000 2 ) + 0.4(00 2 ) + 0.(,500 2 ) = 537,000 Var I ( EU [U I] ) = E I [ EU [U I] 2] E I [ EU [U I] ] 2 = 537,000 530 2 = 256,00 [5/4/2009] On page 95, the solution to exercise 3. is incorrect. The correct solution is:

4 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page 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 and variance 2. Let N be the number of losses over the two-year period. Then E[N] = E [ E[N Λ] ] = E[Λ] = and the variance of N is Var(N) = E [ Var(N Λ) ] + Var ( E[N Λ] ) = E[Λ] + Var(Λ) = + 2 = 3 For 500 insureds, the aggregate mean is 500 and the aggregate variance is 500(3) = 4500. We make a continuity correction and check the probability that a normal distribution with these parameters is greater than 600.5: ( ) 600.5 500 Pr(N > 600) = Φ 4500 = Φ(.50) = 0.9332 = 0.0668 [7/9/2007]On page 96, in the solution to exercise 3.6, change will to make on the second line, and change the final answer to 0.005785. The final answer given (0.0062) is what you would get without a continuity correction. [0/4/2007]On page 97, in the solution to exercise 3.9, on the fourth displayed line, change 788λ to 788 λ. [0/4/2007]On page 20, in the solution to exercise 3.27, on the fourth line, replace 0.274 with 0.274. [0/8/2007]On page 204, in the solution to exercise 3.40, in the last line, replace 250.5 with 249.5 and Φ(.69) with Φ(.64). The final answer is 0.23 instead of 0.2. [7/9/2007]On page 205, in the solution to exercise 3.43, on the last line, change the second expression to ( ) 00,000 96,000 Φ 9,760,000 [7/9/2007]On page 205, at the end of the solution to exercise 3.45, add the words making the final answer 0.9507 = 0.0493. [7/9/2007]On page 208, in the solution to exercise 3.56, on the last line, change 0.534 to 0.0534. [2/7/2008] On page 208, in the solution to exercise 3.57, the final answer is 990,944. [7/9/2007]On page 2, on the first line, the reading from Loss Models should exclude subsection 6.6. [8/5/2008] On pages 27 28, the solution to exercise 4.3 starting with The Poisson rate is incorrect. The correct solution starting from that point is

Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page 5 Let N be the number of coins picked up in half an hour. Then Pr(N = 0) = Pr(N = ) = λ=0 ( ) λ 0 e 0.5λ ) λ a geometric series = ( 0 e 0.5 = ( ) 0/e 0.5 = 0.20265 ( ) λ 0 0.5λ e 0.5λ λ=0 ( ) λ 0 λ e 0.5 = 22 λ= = ( 22 0/e 0.5 ( 0/e 0.5 ) 2 ) = 0.2454 Pr(N 2) = 0.20265 0.2454 = 0.6728 [7/28/2007]On page 220, in the solution to exercise 4.4, on the fifth displayed line (the formula starting Pr(Y = 2)), replace numerator 0.8394 with 0.08030. Four lines from the end, replace 0.797 with 0.707. [8/5/2008] On page 224, on the 4th line, change Pr(S 2.8) to Pr(S > 2.8). [7/9/2008]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 E[S d] = hs(hj) + (d hu)s(hu) j=0 where u = d/h. If u < 0 (for example if d = and h = 2), the sum is empty and only the second term is used. [2/6/2008]On page 232, in the solution to exercise 5., on the first displayed line, replace 0.35336 with 0.35335. [0/7/2008]On page 233, in the solution to exercise 5.4, on the 5th displayed line, add + after the left parenthesis: Pr(S 20) = ( + 2.5 + 4.625 + 6.3547)e 5 = 0.9024 [7/28/2007]On page 236, in the solution to exercise 5.6, on the last line, replace 0.6454 with 0.2698. [7/9/2007]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 x/θ xj F S (x) = p n e j! n=0 j=0 [8/30/2007]On page 24, on the 3rd and 6th lines of subsection 6.2.2, change x kh and x (k+)h to x k and x k+. This is the notation the textbook uses.

6 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page [8/30/2007]On page 242, starting from the 9th line through the end of the solution to Example 6D, every superscript 4 should be replaced with a superscript. In other words, m 4 0 should be m 0 and m 4 should be m in all places. [8/9/2007]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.) [7/5/2007] On pages 243 244, exercises 6. and 6.2 are defective. Change the exercises to the following: 6. For a block of 50 policies: (i) Claim sizes on a coverage follow an exponential distribution with mean 500. (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. Determine the probability that aggregate claims are within 0.842 standard deviations of the mean. 6.2 Claim sizes on a coverage follow an exponential distribution with mean 500. 00 lives are covered under the contract. Claim counts for each life follow a negative binomial distribution with mean 0. and variance.. Claim counts and claim sizes are independent. A stop-loss reinsurance contract is available at 50% of expected claim cost. You are willing to pay 500 for the contract. Determine the aggregate deductible needed to make the cost of the contract 500. Change the solution to 6. to the following: For the block of 50 policies, claim counts follow a negative binomial with r = 50(0.02) = and β = 25, or a geometric. The mean and variance are β = 25 and β( + β) = 650 respectively. Letting S be aggregate losses for the entire block, E[S] = 25(500) = 37,500. By the compound variance formula, Var(S) = 25(500 2 ) + 650(500 2 ), and the standard deviation is σ = 500 25 + 650 = 38,97. Then 37,500 0.842(38,97) = 4686 and 37,500 + 0.842(38,97) = 70,34, so we want the probability of 4686 β S 70,34. The aggregate distribution is a two-point mixture with weight (+β) = 25 26 on the exponential; the exponential has parameter θ( + β) = (26)(500) = 39,000. F S (4686) = 25 26 e 4686/39,000 = 0.473 F S (70,34) = 25 26 e 70,34/39,000 = 0.845 So the probability of being in the interval is 0.845 0.473 = 0.6942. 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 6.2 with the following: For each life, rβ = 0. and rβ( + β) =., so β = 0 and r = 0.0. For the group, r = 00(0.0) = and β = 0, making the distribution geometric. The rest of the solution to 6.2 remains the same. [9/8/2008] On page 245, make the following corrections to the solution to exercise 6.5: On the fourth line, the equation for p k should be p k = e (k 0.5)/2 e (k+0.5)/2.

Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page 7 On the first displayed line, add an equals sign: E[X] = k=0 ( kp k = e /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 0.0966745 with 0.22587. On the second-to-last line, change E[X.6] to E[S.6]. [8/9/2007]On page 246, in exercise 6.7, 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 f 0 ; (0.5)( 0.4959) = 0.42520. We must also modify the probabilities of severities to make them conditional on severity not being zero by dividing by f 0 ; the modified f is 0.2406/( 0.4959) = 0.28346. Then ( ) r ( ) 2 Pr(S = 0) = = = 0.49232 + β.42520 ( ) ( ) rβ 2(0.42520) p = p 0 = (0.49232) = 0.29376 + β.42520 Pr(S = ) = (0.28346)(0.29376) = 0.08327 Then F S () = 0.49232 + 0.08327 = 0.57559. [8/9/2007]On pages 246 247, in the solution to exercise 6.8, replace the last three lines with The discretized estimate was 7.5% off. ( )( ) 0.5 F S () = 2 (0.80074).5.5 = 0.35588 0.0874 = 0.53537 [6/7/2007]On page 29, on the last line, change S07:30 to S07:40. ( ) 2 0.5 (0.80074 + 0.7794).5 [9/26/2007]On page 292, in the solution to exercise 20.9, on the first displayed line, Pr(800 < X < 000) should be replaced by number of losses between 800 and 000. [7/28/2007]On page 297, in the table at the bottom of the page, s 4 =, not 2. [3//2008] On page 32, in the solution to exercise 2.23, the final answer should be 0.3556 instead of 0.3555. [2/4/2007]On page 32, exercise 22. is the same as exercise 4.. [6/5/2007]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). [8/4/2008]On page 365, in the second line of the answer to Example 25B, replace r with r 0. [8/2/2008] On page 366, in the first displayed line on the page, q (τ) j in the exponent. and p (τ) j [/7/2008] On page 366, 4 lines from the bottom of the page, replace q (d) 2 with q (w) 2. [2/26/2007]On page 367, on the fourth line, after (0.880064), replace 0.47573 with 0.l3548. should have only one set of parentheses

8 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page [7//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., 26.3, 26.5, 26.6, 26.7, and 26.32. Replacement pages for subsections 26.. 26..3 and for the solutions to the above exercises are at the end of these errata. (Ignore the different pagination.) [2/26/2007]On page 375, in the displayed line of Example 26E, replace F (x) with f(x). [8/22/2007]On page 376, five and four lines before Section 26.2, a θ is missing on each line: 0θ 2 2950θ + 4,500 = 0 θ 2 295θ + 450 = 0 [7//2007] On page 379, in exercise 26.5, change x 2 i /n to (x i x) 2 /n. [3/6/2008]On page 393, in the solution to exercise 26.25, the second to last displayed line should be split into two lines: 2θ 2 + 3 (50 3θ ) 2 = 400 5θ 2 00θ + 300 = 0 3 On the last line of the solution to exercise 26.25, add the word not between would and result. [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. [8/22/2007]On page 395, in the solution to exercise 26.3, on the second to last line, change greatest lower bound to least upper bound. [7//2007] On page 397, delete the first paragraph of Section 27.. The tables you get on the exam will not have p and q. [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 =.0446 ln 0.6 = 20 0.525689 = 38.0453 [6/20/2007]On page 409, in the solution to exercise 27.4, on the second line, the second exponent α. [3/26/2008]On page 45, on the first line, delete at. [8/22/2007]On page 47, two lines from the bottom, delete the first θ: 3.5 5θ = 35θ θ 2θ k should have an [6/23/2007]On page 433, in the solution to exercise 28.7, on the fourth line, change survivor to survivors. [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. [9/8/2008] On page 453, in exercise 29.4, the column Years of Disability should be (, 2) instead of (0, ) and [2, ) instead of [, ). [8/22/2007]On page 457, in the solution to exercise 29.9, on the second line, replace e ( 2 θ )2 with e ( t θ )2.

Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page 9 [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. [2/24/2008]On page 470.2, on the fifth line, change 0.0008 to 0.00008. [7/0/2008]On page 470, delete the footnote, and replace the last complete paragraph on the page with the following: 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. 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(.96)(0.3) =.76 centered at µ. This is a huge interval. [8/9/2007]On page 472, in the solution to exercise 30., on the second line from the end, change 0.0976 to 0.00976 and delete the extra 0. in 0.0.05084. [4/2/2008] On page 502, in the solution to exercise 32.0(ii), on the second line, replace the numerator 45 with 445. [8/6/2007]On page 508, 3 lines from the bottom, change baseline smoker to baseline individual. [8/26/2007]On page 59, in the solution to exercise 33.2, on the fourth displayed line, insert c into the last exponent so that the line reads L(θ, c) = +3 2 )/θ c2 2 θ 2 e (2 θ 2 e c(2 +4 2 )/θ [8/25/2007]On page 525, in the second line of the second paragraph, add the word be between now and denoted. [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, lines from the bottom, change (3, 0.625) to (3, 0.0625). [8/28/2007]On page 55, in Figure 37.5(d), change the two 4 s in the caption to 40 s. and 20 with 20. [8/3/2007]On page 57, 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). [8/3/2007]On page 578, in the answer to Example 40G, on the 6th displayed line, replace N 3 with N > 3. [8/3/2007]On page 598, in the solution to exercise 40.5, on the first line, replace 0.045 with 0.45. [3/8/2008] On page 605, on the first displayed line, the last denominator is missing a factorial: n i= x!. [8/3/2007]On page 605, on the seventh line, replace λ = λ 2 = 0. with λ = 0., λ 2 = 0.2. [8/3/2007]On page 60, in the solution to exercise 4.6, replace the 4th displayed line with n i= l(θ) = n ln θ x i θ I thank Ken Burton for pointing this out to me. = n ln θ n x θ

0 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page [8/30/2007]On page 69, on the fifth line, the second sentence should start: The square of the coefficient of variation of the number of claims.... [3/6/2008]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. [2/4/2007]On page 623, exercise 42.4 is the same as exercise 3.5. [9/3/2007] On page 642, in the solution to exercise 43.3, on the last line, replace n F with e F. [0/4/2007]On page 653, in the solution to exercise 44.5, on the second line from the end, Φ should be Φ. [9/3/2008]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. [9/20/2007]On page 682, 9 lines from the bottom, change the exponent on θ from 3 to 4: 2 3 = 4(2 3 ) θ4 [9/20/2007]On page 696, in the solution to exercise 46.6, 3 lines from the end, change a = 3b to a = 3c. [4//2008]On page 698, in the solution to exercise 46.2, 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. [2/4/2007]On page 709, exercise 47.22 is the same as exercise.3. [2/4/2007]On page 70, exercise 47.24 is the same as exercise.4. [2/4/2007]On page 70, exercise 47.25 is the same as exercise.6. [0/4/2007]On page 79, in the solution to exercise 48., Φ(.890) should be Φ(.890). [0/4/2007]On page 79, in the solution to exercise 48.5, on the second to last line, there should be a square root sign over 09,90 in the denominator. [9/20/2007]On page 735, in the solution to exercise 50.7, on the last two lines replace 2884.09 twice with 2886.43, and replace 0.8327 with 0.8324. [0/29/2007]On page 738, 0 lines from the bottom, change than to that. [3/26/2008]On page 780, in the solution to exercise 52.32, on the third line, change 2 2+ = 3 to 2 2+ = 2 3. [3/26/2008]On page 784, in the solution to exercise 52.45, on the last line, change the first to +. [8/28/2007]On page 803, on the first displayed line of the solution to exercise 53.6, replace X = 0 with X = 340. [9/20/2007]On page 806, in the solution to exercise 53.23, on the second line, replace E[S] with Var(S). [ [3/26/2008]On page 80, in the solution to exercise 53.38, on the last line of the page, add E before e σ2]. [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. [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].

Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page [0/2/2007]On page 842, at the end of the first paragraph of subsection 57.2., replace the formula for x with r ni i= j= x = m ijx ij r ni i= j= m. ij [4/0/2008]On page 85, in the solution to exercise 57.2, on the 3rd line, delete the coefficient 2 of ( 3 8) 2. [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. [0/2/2007]On page 87, the solution to exercise 58.3 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(+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), â = ( m m Plugging in v = a + µ + µ 2, we get and letting D = m m r i= m2 i, r i= m 2 i ) ( r ) m i ( x i x) 2 ˆv(r ) i= r i= â = m i( x i x) 2 (â ˆµ ˆµ 2 )(r ) m m r i= m2 i r i= â = m i( x i x) 2 (r )(ˆµ ˆµ 2 ) D + r In our exercise (in which r = 2 for 2 policyholders), m = 5 + 0 + 5 = 30 x = 9 30 = 0.3 m 2 = 0 + 0 = 20 x 2 = 2 20 = 0. D = 50 302 + 20 2 = 24 50 2 m i ( x i x) 2 = 30(0.3 0.22) 2 + 20(0. 0.22) 2 = 0.48 i= â = (0.48 0.22 0.222 ) = 0.008464 25 ˆv = 0.008464 + 0.22 + 0.22 2 = 0.276864

2 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page The credibility factor for Policyholder A is Z = 30(0.008464) 30(0.008464) + 0.276864 = 0.47839 [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. [2/24/2008]On page 92, in exercise 6.7(ii), delete 0 from the set {,2,3,4,5}. [9/26/2007]On page 92, in the solution to exercise 6.25, in the heading of the third column of the table, change 0.3u i to 0.3z i. [9/26/2007]On page 92, in the solution to exercise 6.26, change the heading of the last column of the table to e ni. [2/25/2008]On page 93, 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) [0/3/2007]On page 940, in the solution to exercise 63., on the 6th line, replace 3 3 2 with 5 2. [/28/2008]On page 946, on the last line of the answer to Example 64B, replace.282 with.22 and 42.5% with 42.6%. [0/25/2007]On page 954, in exercise 65.2, on the last line, replace payoff with price. [0/25/2007]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 38.5438+40.2789+38.3225 3 = 39.0484 and the payoff is 40 39.0484 = 0.956. Discounting this payoff 3 months, the value of the option in this run is e 0.0 (0.956) = 0.942. The geometric average is 3 (38.5438)(40.2789)(38.3225) = 39.0387 and the payoff is 40 39.0387 = 0.963. Discounting this payoff 3 months, the value of the geometric option in this run is e 0.0 (0.963) = 0.957. The simulated value in this run of the arithmetic average option using the control variate method is 0.942 + (0.65 0.957) = 0.64. [0/3/2007]On page 957, in the solution to exercise 65.5, on the first line, the second denominator should be 0.02 instead of 0.05. [/30/2008]On page 969, at the very end of the second paragraph in Section 67., change Q α to CTE α. [7/8/2007] On page 970, in the caption for Figure 67., delete 0.05 before Q 0.95. [9/7/2008] On page 97, on the third displayed line, change ln x to ln Q α. [9/7/2008] On the last 3 lines of page 97, Q α is the α quantile of a standard normal distribution, not of Z. [0/8/2007]On page 975, in the solution to exercise 67.2, on the second displayed line, change ln 7.563 to 7.563 twice. Also, one line above and below the displayed lines, change 7.563 to e 7.563. [/30/2008]On page 980, on the second line, delete. Also, the negative sign of the exponent should be outside the parentheses. Thus the line should read S(x) = e (x/θ)τ [9/26/2007]On page 988, the solution to Example 68G is incorrect. The correct solution is

Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page 3 Using the non-parametric method discussed above, 0.95(500) = 475. The variance of the binomial is 500(0.95)(0.05) = 23.75. So we want to add and subtract.645 23.75 = 8.07, which we round up to 9, to and from 475. The confidence interval is (L (466), L (484) ) = (350.3, 358.0). [0/25/2007]On page 988, the last two lines of the page should read 0.438 + 0.99(365.76 365.09) = 0.22026 = 0.4693 5 The confidence interval is 365.76 ±.645(0.4693) = (365.0, 366.5). [0//2008]On page 00, in question 3, change Pr(3 < X < x X < 0) to Pr(3 < X < x X < 0). [/2/2008]On page 036, 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. [4/6/2008]On page 040, change the last line of question 6 to Given that the option pays off, determine the expected value of S. [0//2008]On page 076, in the solution to question 3, on the third and fourth displayed lines, change dq to dq. [4/24/2008]On page 086, on the first line of the page, change w 0 to w 5. [0//2008]On page 090, in the solution to question 3, on the third line, change 0.8 0.6 0.8 0.6 0.2 = 0.25 to 0.8 0.2 = 0.333. Change Pr(3 < X < x X < 0) to Pr(3 < X < x X < 0) in the four places it appears. [0/9/2008]On page 097, 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 ). [0/23/2007]On page 098, in the solution to question 39, on the first line, change 0.2(0) = 2 to 0.(20) = 2. [0/3/2007]On page 7, in the solution to question 6, on the 5th line of the page, change deductible of 0000 to deductible of 000. [9/28/2008]On page 9, in the solution to question 22, on the fifth line, change must by to must be. [0/6/2007]On page 20, 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 [0/6/2007]On page 23, in the solution to question 34, on the third line, replace the first 3 with 0.7. [4/4/2008]On page 23, in the solution to question 37, on the last line of the page, change the right hand side to e 2y/θ. [/2/2008]On page 28, 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. [/27/2009]On page 30, in the solution to question 6, on the first displayed line, change K/S 0 to S 0 /K. [/2/2008]On page 3, in the solution to question 8, on the third displayed line, replace E[HM 2 ] with E [ (N α) 2], where N is the number of claims. [0/30/2007]On page 37, the last 3 lines of the solution to question 32 is incorrect. Replace them with:

4 Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page 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 0 Pr(X > 20 A) = S(20 A) = = 20 8 ( ) 4 0 Pr(X > 20 B) = S(20 B) = = 20 6 (/8)(0.6) Pr(A X > 20) = (/8)(0.6) + (/6)(0.4) = 0.75 Pr(B X > 20) = Pr(A X > 20) = 0.75 = 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) 30 26 2 3 = 8900 36 + 25 2 = 249.3056 [0/6/2007]On page 43, in the solution to question 6, on the second line, change the denominator of the fraction from n to 2n. [0/6/2007]On page 59, in the solution to question 5, on the 4th line from the end, remove the exponent 2 from 0.279 2. [4/23/2008]On page 60, in the solution to question 8, on the first line of the page, the right hand side should be e (0.0603375). ) 2

Lesson 26 Method of Moments Reading: Loss Models 2. 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 ( n (xi x) 2 ), which uses division by n, not the sample variance ( n (xi x) 2 ), which uses division by n. 26. The method of moments for various distributions 26.. 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 0. ANSWER: The method of moments estimator of θ is x = 920 50 = 8.4. For an exponential distribution, the loss elimination ratio is E[X d] LER(d) = = e d/θ E[X] In our case, LER(0) = e 0/8.4 = 0.493. 26..2 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.) Copyright 2007 ASM 37

372 26. METHOD OF MOMENTS EXAMPLE 26B You have the following sample of loss sizes: 3, 5, 8, 0, 5, 27, 38, 60 x i = 66 xi 2 = 696 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 = 66 8 = 20.75. The empirical variance is 696 8 20.75 2 = 343.9375. By equation (26.), ˆθ = 343.9375 = 6.5753. 20.75 26..3 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, 0, 20, 50, 00, 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 = 396 8 = 49.5 2θ 2 (α )(α 2) = 53,098 = 6637.25 8 By dividing the square of the first equation into the second we eliminate θ. 2(α ) α 2 = 6637.25 = 2.708805 49.5 2 2 + 2 α 2 = 2.708805 α 2 = 2.708805 2 = 0.354403 2 ˆα = 2 + 0.354403 = 4.8265 Then we can calculate ˆθ by plugging ˆα into the equation of means. θ α = 49.5 ˆθ = 49.5(4.8265 ) = 89.7 Therefore, e(20) = 20+89.7 4.8265 = 54.733. (Actually, we did not have to calculate ˆθ in order to calculate e(20), since e(20) = E[X]+ 20 α and the method of moments sets E[X] equal to the sample mean automatically, so we just needed 49.5 + 20 3.8265.) Copyright 2007 ASM

384 26. METHOD OF MOMENTS 26.34. [4-F04:24] You are given: (i) Losses are uniformly distributed on (0, θ) with θ > 50. (ii) The policy limit is 50. (iii) A sample of payments is: 4, 33, 72, 94, 20, 35, 50, 50 Estimate θ by matching the average sample payment to the expected payment per loss. (A) 92 (B) 96 (C) 200 (D) 204 (E) 208 Additional released exam questions: C-F05:2, CAS3-F06:3 Solutions 26.. For an exponential, set ˆθ equal to the sample mean. 00 + 200 + 500 + 000 + 500 + 2000 + 3700 ˆθ = X = 7 Pr(X > 500) = e ( 9000)(500) 7 = 0.6778 26.2. The first 2 raw sample moments are = 9000 7 We equate moments: 00 + 00 + 200 + 200 + 300 + 500 + 000 + 2000 m = = 550 8 t = 002 + 00 2 + 200 2 + 200 2 + 300 2 + 500 2 + 000 2 + 2000 2 8 = 680,000 ˆθ ˆα = m = 550 2ˆθ 2 ( ˆα )( ˆα 2) = t = 680,000 680000 2( ˆα ) = 550 2 ˆα 2 680000 ˆα 2(680000) = 2(550 2 ) ˆα 2(550 2 ) ˆα = 755000 75000 = 0.0667 ˆθ = 550( ˆα ) = 4986.667 Either way, ( ) α θ Pr(X > 500) = θ + 500 ( ) 0.0667 4986.667 = = 0.3822 5486.667 Copyright 2007 ASM

EXERCISE SOLUTIONS FOR LESSON 26 385 26.3. The first moment is m = 667,754 00 = 6677.54, and the second moment is t = 20,354,674,039 00 = 203,546,740.39. Then, dividing the square of the first moment into the second moment, 2(α ) α 2 = 203,546,740.39 = 4.56490 6677.54 2 2 + 2 α 2 = 4.56490 α 2 = 4.56490 2 =.28245 2 ˆα = 2 +.28245 = 2.779758 ˆθ = m( ˆα ) = 6677.54(2.779758 ) =,884.4 26.4. We calculate the first and second raw sample moments. 4 + 5 + 2 + 99 + 42 m = = 0 5 t = 42 + 5 2 + 2 2 + 99 2 + 42 2 = 37,504.8 5 Dividing m 2 into t and equating to the distribution moments, 2(α ) α 2 = 37,504.8 = 3.099570 0 2 2 + 2 α 2 = 3.099570 α 2 = 3.099570 2 = 0.549785 2 ˆα = 3.8889 ˆθ = m( ˆα ) = 30.0782 To find the 95th percentile, we equate F(x) = 0.95, or F(x) = 0.05. ( ) α θ = 0.05 θ + x θ θ + x = α 0.05 (θ + x) α 0.05 = θ x = θ( α 0.05) α 0.05 3.8889 0.05 = 0.45637 30.0782( 0.45637) = = 369.3659 (A) 0.45637 26.5. 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 ) α 2 = t m = t 2 500 2 Copyright 2007 ASM

386 26. METHOD OF MOMENTS 2 + 2 α 2 = t 500 2 The left hand side is greater than 2 since the fraction must be positive, so empirical variance t m 2 > 2(500 2 ) 500 2 = 500 2 = 250,000. 26.6. The first two raw sample moments are ( ) 30 + 20 + 350 + 28 + 822 m = = 508 5 ( 30 2 + 20 2 + 350 2 + 28 2 + 822 2 ) t = = 70,40.6 5 t 500 2 > 2, implying t > 2(500 2 ). Then the Then 2(α ) α 2 = t m = 70,40.6 = 2.77937 2 508 2 2 + 2 α 2 = 2.77937 α 2 = 2.77937 2 = 0.358968 2 ˆα = 2 + 0.358968 = 4.78576 ˆθ = m( ˆα ) = (508)(3.78576) = 923.67 The limited expected value at 500 is E[X 500] = θ α = 508 ( ) α θ θ + 500 ) 3.78576 = 296.223 (C) ( 923.67 2423.67 26.7. To bring this up with inflation, we multiply average claim size in year by. 2 and in year 2 by.. Then, average inflated claim size is 00(2,00) + 200(3,750) = 3,200 300 Setting the Pareto mean equal to this, θ 2 = 3,200 θ = 26,400 (E) 26.8. This is a Pareto with θ =. Equating sample and distribution means, X = E[X] = α α = + X = X + X (E) Copyright 2007 ASM

EXERCISE SOLUTIONS FOR LESSON 26 387 26.9. 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 = 300 300 So we must calculate the variance of the right hand side. Var(ˆθ) = 50 2 i= ˆθ = 300 50 300 i= i= X i X i Var(X i ) = 300 50 2 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 )(α 2) θ = 2θ2 α 2() θ2 4 = 3 4 θ2 Plugging this into the expression for Var(ˆθ), Var(ˆθ) = 26.0. We equate the first 2 moments and solve for α and θ. ( ) ( ) 300 3 50 2 4 θ2 = 0.0θ 2 (D) X = ˆαˆθ ˆα = 30 E[ ˆX 2 ] = 02 + 20 2 + 30 2 + 40 2 + 50 2 = 00 5 ˆαˆθ 2 ˆα 2 = 00 ( ˆα ) 2 ˆα( ˆα 2) = 9 ˆα 2 22 ˆα = 9 ˆα 2 8 ˆα + 9 2 ˆα 2 4 ˆα 9 = 0 ˆα = 4 ± 6 + 72 4 = 3.3452 Notice that from the first moment equation. ˆθ = 30(2.3452/3.3452) = 2.03, which makes the fit impossible since data for a single-parameter Pareto may not be less than θ, and we are given the points 0 and 20. That s why θ for a single-parameter Pareto should be given in advance rather than fitted. 26.. The mean of a single-parameter Pareto is αθ α. Here we have a single-parameter Pareto with θ =, so α α = x + α = x As x, the fraction goes to 0 and ˆα. (C) ˆα = + x Copyright 2007 ASM

388 26. METHOD OF MOMENTS 26.2. For the Nelson-Åalen estimator, the risk sets and events through time 5 are The Nelson-Åalen estimate is then y i r i s i 5 2 2 3 3 2 3 4 9 2 5 7 Ĥ (5) = 2 5 + 3 + 3 2 + 2 9 + 7 = 0.825336 Notice that the large observation 3 has no effect on this estimate, but will affect the method of moments estimate. The sample mean is 2()+2+3(3)+2(4)+5+2(7)+2(9)+25+3 5 = 344 5. Setting this equal to the mean of a single αθ parameter Pareto, α, Then we get S (5) and then H 2 (5). Ŝ (5) = 0α α = 344 5 α α = 344 50 50α = 344α 344 94α = 344 ( ) α θ = 5 ˆα = 344 94 =.77396 ( ) ˆα 0 = 5 Ĥ 2 (5) = ln 0.487254 = 0.78970 The absolute difference is 0.825336 0.78970 = 0.0637. 26.3. Calculate the first two raw moments. ( ).77396 2 = 0.48725 3 30 + 50 + 2(60) + 70 + 90 x = = 60 6 t = 302 + 50 2 + 2(60 2 ) + 70 2 + 90 2 = 3933 6 3 ˆσ 2 = 3933 3 602 = 333 3 Using formula (26.), ˆα = 602 333 3 = 0.8 (D) 26.4. This is a gamma distribution with α = 2 and θ = c, so 2c = x = 8, c = 4 (E) Copyright 2007 ASM

EXERCISE SOLUTIONS FOR LESSON 26 389 26.5. The first two raw sample moments are 0 + 20 + 40 + 80 + 00 + 2(200) + 500 x = = 43.75 8 t = 02 + 20 2 + 40 2 + 80 2 + 00 2 + 2(200 2 ) + 500 2 = 43,562.5 8 ˆσ 2 = 43,562.5 43.75 2 = 22,894.44 By formula (26.), 26.6. ˆθ = ˆσ2 x = 59.2935 000 + 850 + 750 + 00 + 250 + 900 x = = 975 6 t = 0002 + 850 2 + 750 2 + 00 2 + 250 2 + 900 2 = 977,96 2 6 3 ˆσ 2 = 977,96 2 3 9752 = 27,29 2 3 ˆα = x2 ˆσ 2 = 9752 27,29 2 3 = 34.832 (D) 26.7. 500 + 3500 + 800 + 4800 + 3900 + 6000 + 3800 + 5500 + 4200 + 3000 x = = 3800 0 t = 5002 + 3500 2 + 800 2 + 4800 2 + 3900 2 + 6000 2 + 3800 2 + 5500 2 + 4200 2 + 3000 2 0 ˆσ 2 = 6,332,000 3800 2 =,892,000 ˆθ = ˆσ2 x =,892,000 = 497.89 (E) 3800 = 6,332,000 26.8. 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 = ). Then set E[X] equal to 4.2. X = 4.9 +.8 + 3.4 + 6.9 + 4.0 5 = 4.2 β 2π = 4.2 2 ˆβ = 2(4.2) = 3.35 2π (D) 26.9. You may recognize this distribution as a beta distribution with a = α, b =, and the mean of a beta is. If not, it s not hard to integrate: a a+b = α α+ E[X] = 0 αx α dx = α α + Copyright 2007 ASM

EXERCISE SOLUTIONS FOR LESSON 26 393 26.3. Not really a method of moments problem, but since it involves moments, I put the problem here. The mean is 2 (m + m 2 ). The second moment is m 2 + 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) 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 + m2 2 (m + m 2 ) 2 We want to minimize the denominator. Since m > 0 and m 2 > 0, (m + m 2 ) 2 > m 2 + 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. But then the answer is 4() = 3. (C) 26.32. Equating moments, µ + 0.5σ 2 = ln 386 = 5.95584 2µ + 2σ 2 = ln 457,480.2 = 3.03349 σ 2 = 3.03349 2(5.95584) =.284 ˆµ = 5.95584 0.5(.284) = 5.39493 ˆσ =.284 =.0596 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 Φ σ σ ( ) [ ( )] ln 500 5.39493.0596 2 ln 500 5.39493 = 386Φ + 500 Φ.0596.0596 = 386Φ( 0.28526) + 500[ Φ(0.77389)] = 386(0.3877) + 500( 0.7805) = 259.4 (D) 26.33. It is unusual to match negative moments, but for an inverse Pareto, there is little choice, since positive moments 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 moment is (We ll use m and t for negative moments here, even though usually they mean positive moments.) m = ( 6 5 + 45 + 40 + 250 + 560 + ) = 0.07094 340 The sample 2 moment is t = 6 ( 5 + 2 45 + 2 40 + 2 250 + ) 2 560 + = 0.00083484 2 340 2 Let m be the sample moment and t the sample 2 moment. We equate these with the fitted moments, using the formulas from the Loss Models Appendix: m = θ(τ ) Copyright 2007 ASM