Student Solutions Manual to Accompany LOSS MODELS WILEY SERIES IN PROBABILITY AND STATISTICS Established by WALTER A. SHEWHART and SAMUEL S. WILKS Editors: David J. Balding, Noel A. C. Cressie, Garrett M. Fitzmaurice, Harvey Goldstein, Iain M. Johnstone, Geert Molenherghs, David W. Scott, Adrian E M. Smith, Ruey S. Tsay, Sanford We is berg Editors Emeriti: Vic Barnett, J. Stuart Hunter, Joseph B. Kadane, Jozej L. Teugels A complete list of the titles in this series appears at the end of this volume. Student Solutions Manual to Accompany LOSS MODELS From Data to Decisions Fourth Edition Stuart A. Klugman Society of Actuaries Harry H. Panjer University of Waterloo Gordon E.Willmot University of Waterloo SOCIETY OF ACTUARIES ©WILEY A JOHN WILEY & SONS, INC., PUBLICATION Copyright © 2012 by John Wiley & Sons, Inc. 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ISBN 978-1-118-31531-6 10 9 8 7 6 5 4 3 2 1 CONTENTS Introduction 2 Chapt er 2 solutions 2.1 3 Section 2.2 Chaplter 3 solutions 3 3 9 3.1 Section 3.1 9 3.2 Section 3.2 14 3.3 Section 3.3 15 3.4 Section 3.4 16 3.5 Section 3.5 20 v VI CONTENTS 4 Chapter 4 solutions 23 4.1 23 5 6 Section 4.2 Chapter 5 solutions 29 5.1 Section 5.2 29 5.2 Section 5.3 40 5.3 Section 5.4 41 Chapter 6 solutions 43 6.1 Section 6.1 43 6.2 Section 6.5 43 6.3 Section 6.6 44 Chapl :er 7 solutions 47 7.1 Section 7,1 47 7.2 Section 7.2 48 7.3 Section 7.3 50 7.4 Section 7.5 52 Chapter 8 solutions 55 8.1 Section 8.2 55 8.2 Section 8.3 57 8.3 Section 8.4 59 8.4 Section 8.5 60 8.5 Section 8.6 64 Chapter 9 solutions 67 9.1 Section 9.1 67 9.2 Section 9.2 68 9.3 Section 9.3 68 9.4 Section 9.4 78 9.5 Section 9.6 79 9.6 Section 9.7 85 9.7 Section 9.8 87 CONTENTS 10 11 Chapter 10 solutions 93 10.1 10.2 10.3 93 96 97 13 15 16 17 Section 11.2 Section 11.3 99 99 100 Chapter Chapt :er 12 solutions 105 12.1 12.2 12.3 12.4 105 111 114 116 Section Section Section Section 12.1 12.2 12.3 12.4 Chapter 13 solutions 119 Section Section Section Section Section 119 124 137 145 146 13.1 13.2 13.3 13.4 13.5 14 Section 10.2 Section 10.3 Section 10.4 Chapter 11 solutions 11.1 11.2 12 Vii 13.1 13.2 13.3 13.4 13.5 Chapter 14 147 14.1 147 Section 14.7 Chapter 15 151 15.1 15.2 151 161 Section 15.2 Section 15.3 Chapter 16 solutions 165 16.1 16.2 16.3 165 166 175 Section 16.3 Section 16.4 Section 16.5 Chapter 17 solutions 183 17.1 183 Section 17.7 viii 18 19 20 CONTENTS Chapter 18 187 18.1 187 Section 18.9 Chapter 19 219 19.1 219 Section 19.5 Chapter 20 solutions 227 20.1 20.2 20.3 20.4 227 228 229 230 Section Section Section Section 20.1 20.2 20.3 20.4 CHAPTER 1 INTRODUCTION The solutions presented in this manual reflect the authors' best attempt to provide insights and answers. While we have done our best to be complete and accurate, errors may occur and there may be more elegant solutions. Errata will be posted at the ftp site dedicated to the text and solutions manual: ftp://ftp.wiley.com/public/sci_tech_med/loss_models/ Should you find errors or would like to provide improved solutions, please send your comments to Stuart Klugman at sklugman@soa.org. Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Fourth Edition. By Stuart A. Klugman, Harry H. Panjcr, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. 1 CHAPTER 1 INTRODUCTION The solutions presented in this manual reflect the authors' best attempt to provide insights and answers. While we have done our best to be complete and accurate, errors may occur and there may be more elegant solutions. Errata will be posted at the ftp site dedicated to the text and solutions manual: ftp://ftp.wiley.com/public/sci_tech_med/loss_models/ Should you find errors or would like to provide improved solutions, please send your comments to Stuart Klugman at sklugman@soa.org. Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Fourth Edition. By Stuart A. Klugman, Harry H. Panjcr, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. 1 CHAPTER 2 CHAPTER 2 SOLUTIONS 2.1 SECTION 2.2 , r r n , c , , / O.Olx, 2.1 F5(x) = 1 - 5 8 (x) = | 0Q2x _ 01 * ^ = Fp// \ / ° /.(«) 5 (x) = | 0 0 2 ) M*) = $& 5 8 = < W 100 05 ' 5 0° << χx << 50 ' ?5 ΓΧ — , I. 75 - a; 0 < x < 50, < χ < ?5 50 50<x<75. 2.2 The requested plots follow. The triangular spike at zero in the density function for Model 4 indicates the 0.7 of discrete probability at zero. Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Fourth 3 Edition. By Stuart. A. Klugnmn, Harry H. PHiijor, Gordon E. Willmot Copyright ( ö 2012 John Wiley & Sons, Inc. CHAPTER 2 SOLUTIONS 0.6 F(x) 0.4 0.2 1 2 X 3 4 5 Distribution function for Model 3. 0.6 F(x) 0.4 0.2 100000 200000 x 300000 400000 500000 Distribution function for Model 4. 10 20 30 x 40 50 60 70 Distribution function for Model 5. SECTION 2.2 0.51 0.41 0.31 P(x) 0.21 0.Γ 1 X 3 4 Probability function for Model 3. 5e-06 100000 200000 x 300000 400000 500000 Density function for Model 4. 0.025 : 0.02: o.oi5: f(x) ■ o.oi: 0.005 - 10 20 30 x 40 50 60 Density function for Model 5. 70 6 CHAPTER 2 SOLUTIONS le-051 8e-06: 6e-06: h(x) ■ 4e-061 2e-06 ] 100000 200000 x 300000 400000 500000 Hazard rate for Model 4. 0.2 0.181 0.16 0.14 0.12 (x)0.1 0.08 i 0.061 0.04 0.02 10 20 30 x 40 50 60 70 Hazard rate for Model 5. 2.3 f'(x) = 4(1 + x 2 ) - 3 - 24x 2 (l + x 2 ) - 4 . Setting the derivative equal to zero and multiplying by (1 -f x 2 ) 4 give the equation 4(1 + x2) — 24x 2 = 0. This is equivalent to x2 = 1/5. The only positive solution is the mode of l/\/5· 2.4 The survival function can be recovered as 0.5 = 5(0.4) = = e e-fi'* -Ax-0.5e2x\°'1 _ e -0.4A-0.5e A +*2xdx '» () 8 - +0.5 Taking logarithms gives -0.693147 = -0.4Λ - 1.112770 -f 0.5, and thus A = 0.2009. SECTION 2.2 7 2.5 The ratio is / 10,000 \ 2 V10,000 + d / / 20,000 \ 2 \20,000+ cp) ( 20,000 + cP\2 \20,000 + 2(iJ 20,0002 + 40,000^ + d4 20,0002 + 80,000d + AcP' Prom observation or two applications of L'Höpital's rule, we see that the limit is infinity. SECTION 2.2 7 2.5 The ratio is / 10,000 \ 2 V10,000 + d / / 20,000 \ 2 \20,000+ cp) ( 20,000 + cP\2 \20,000 + 2(iJ 20,0002 + 40,000^ + d4 20,0002 + 80,000d + AcP' Prom observation or two applications of L'Höpital's rule, we see that the limit is infinity. CHAPTER 3 CHAPTER 3 SOLUTIONS 3.1 SECTION 3.1 /»OO OO 3.1 / = -OO (x - μ)3/(χ)άχ /4 -3^2M + 2/i3, OO / (x - = J — OO (x3 - 3χ2μ + 3χμ2 - μ3)/(χ)άχ μ)4/(χ)άχ -OO / O O (x 4 - 4χ 3 μ + 6 x V - 4χμ 3 + μ4)/(χ)άχ -OO = ^ 4 - V 3 / X + 6/X2M2-3/i4· 3.2 For Model 1, σ 2 = 3,333.33 - 502 = 833.33, σ = 28.8675. / 4 = Jo10" *3(0·01)<ίχ = 250,000, μ 3 = 0, Ί ι = 0. μ4 = / 0 100 χ 4 (0.01)(/χ = 20,000,000, μ 4 = 1,250,000, 7 2 = 1·8· Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Fourth Edition. By Stuart A. Klugman, Harry H. Panjer, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. 9 10 CHAPTER 3 SOLUTIONS For Model 2, σ 2 = 4,000,000 - Ι,ΟΟΟ2 - 3,000,000, σ - 1,732.05. μ'3 and μ4 are both infinite so the skewness and kurtosis are not defined. For Model 3, σ 2 = 2.25 - .93 2 = 1.3851, σ = 1.1769. μ!> = 0(0.5) + 1(0.25) + 8(0.12) + 27(0.08) + 64(0.05) = 6.57, μ 3 = 1.9012, 7 1 = 1.1663, μ\ = 0(0.5) + 1(0.25) + 16(0.12) + 81(0.08) + 256(0.05) = 21.45, μ4 = 6.4416, 7 2 = 3.3576. For Model 4, σ 2 = 6,000,000,000 - 30,0002 = 5,100,000,000, σ = 71,414. 3 3 00001 ^x = 1.8 χ 10 15 , μ>3 = 0 (0.7) + J T x (0.000003)eμ 3 = 1.314 χ 10**, 7ι = 3.6078. μ'Α = / 0 oo x 4 (0.000003)e- 00001:E (ia: - 7.2 χ 10 20 , μ4 = 5.3397 χ 10 20 , 7 2 = 20.5294. For Model 5, σ2 = 2,395.83 - 43.752 - 481.77, σ = 21.95. μ'3 = / ο 50 x 3 (0.01)dx + /5705 χ 3 (0.02)ώ = 142,578.125, μ3 = -4,394.53, 7 α = -0.4156. μ'Α - /05° x4{0.0l)dx + J5705 x4(0.02)da; - 8,867,187.5, μ4 = 439,758.30, 7 2 = 1.8947. 3.3 The standard deviation is the mean times the coefficient of variation, or 4, and so the variance is 16. From (3.3) the second raw moment is 16 4- 2 2 = 20. The third central moment is (using Exercise 3.1) 136 - 3(20)(2) + 2(2) 3 = 32. The skewness is the third central moment divided by the cube of the standard deviation, or 32/4 3 = 1/2. 3.4 For a gamma distribution the mean is αθ. The second raw moment is a(a + 1)#2, and so the variance is αθ2. The coefficient of variation is = a l'z = 1. Therefore a — 1. The third raw moment is a(a -f 1)(α + 2)θ = 6Θ . From Exercise 3.1, the third central moment is 60' — 3(2<92)<9 + 2(93 = 203 and the skewness is 2<93/(02)3/2 = 2. 3.5 For Model 1, _ />-».»■» 1 - O.Old 100 -d 2 For Model 2, f°° ( 2,000 \3 J d U + 2,000j ' 2,000 ci + 2,000 N3 2,000 + d 2 SECTION 3.1 11 For Model 3, 0.25(1 - d) + 0.12(2 - d) + 0.08(3 - d) + 0.05(4 - d) 0.5 = 1.86 - d, =< 0.12(2 - d) + 0.08(3 -d)+ 0.05(4 - d) 0 „ 0 0.25 ~ ^ 0.08(3-d) + 0 . 0 5 ( 4 - d ) = 3 3 8 4 6 _ d ; ^ d ' 2 < d < 3, U. l o O05(4-d)_ l 0.05 0 < d < 1, 3 < d < 4. ' ar Model 4, e J r?°0.3e- 000001 ^x W = " 0 . 3e -o.oooo lrf = 100 000 ' · The functions are straight lines for Models 1, 2, and 4. Model 1 has negative slope, Model 2 has positive slope, and Model 4 is horizontal. 3.6 For a uniform distribution on the interval from 0 to w, the density function is f(x) = 1/w. The mean residual life is e(d) = f™(x — d)w rW l dx Id w 2 (x - d) w 2w w —dd w {w - d)2 2(w - d) w—d l dx The equation becomes w - 30 100 - 30 + 4, with a solution of w — 108. 3.7 From the definition, e(A) S™(x-\)\-le-x'xdx /~A_1e-*Adar = λ. 12 CHAPTER 3 SOLUTIONS Ε(Χ) = / JO = [ xf(x)dx Jo xf(x)dx = / xf(x)dx Jo 4- / Jd d/(a;)da; + / Jd 4- d[l - F(d)} 4- e(d)S(d) (z - = E[X Ad}+ d)f(x)dx e(d)S(d). 3.9 For Model 1, from (3.8), E[lAw]= / Jo 4- M(1 - 0 . 0 1 M ) = M(1 - 0 . 0 0 5 M ) x(0.01)dx and from (3.10), 1QQ E [ X Λ u] = 50 - U - 0.01M) = M(1 - 0.005M). (l From (3.9), -oo 0.01M2 1 - O.Olxdx = M - ~ ' ^ ~ JO 2 Odx+ = M(1 - 0 . 0 0 5 M ) . For Model 2, from (3.8), r [ v L , J 3(2,000)3 Γ 7o (z + 2,000) 2,0003 , 4 (2, 000 4- M ) 3 4,000,000 -1000 " (2,000 4- u)2 and from (3.10), E[X Au] = 1,000 2,000 2,000 + u ( 2,000 4- u 1,000 4,000,000 (2,000+ M) 2 J From (3.9), -2,000 3 [ J 7o V 2,000 + * ; = 1,000 1 - 2(2,000 4- x)2 4,000,000 (2,000 + u)2 For Model 3, from (3.8), ( 0 ( 0 . 5 ) + M ( 0 . 5 ) = 0.5M, 0(0.5) 4-1(0.25) 4- M ( 0 . 2 5 ) - 0.25 4- 0.25M, E[X Aw] = 0(0.5) + 1(0.25) + 2(0.12) + M ( 0 . 1 3 ) = 0.49 4 - 0 . 1 3 M , 0(0.5) 4-1(0.25) 4- 2(0.12) 4- 3(0.08) + M ( 0 . 0 5 ) = 0.73 4- 0 . 0 5 M , 0 < M < 1, 1 < M < 2, 2 < M < 3, 3 < M < 4, SECTION 3.1 13 and from (3.10), E[X A u] = { 0.93 - (1.86 - M)(0.5) = 0.5u, 0.93 - (2.72 - M)(0.25) = 0.25 + 0.25M, \ 0.93 - (3.3846 - M)(0.13) = 0.49 + 0.13M, [ 0.93 - (4 - M)0(.05) - 0.73 + 0.05M, 0 < u < 1, 1 < u < 2, 2 < u < 3, 3 < u < 4. For Model 4, from (3.8), E[X Λ M] = Γ x ( 0 . 0 0 0 0 0 3 ) e - 0 0 0 0 0 1 ^ x + M(0.3)e-° o o o o l u Jo = 30,000[l-e-ooooolw], and from (3.10), E[X Au} = 30,000 - 100,000(0.3e-° oooolu ) = 30,000[1 - e- 0 0 0 0 0 l M ] 3.10 For a discrete distribution (which all empirical distributions are), the mean residual life function is 3(d) = Y,Xj>d(x3 - d )p(xj) EXj>dp(x3) When d is equal to a possible value of X, the function cannot be continuous because there is jump in the denominator but not in the numerator. For an exponential distribution, argue as in Exercise 3.7 to see that it is constant. For the Pareto distribution, e(d) E(X) - E(X A d) S{d) _θ a —1 Θ a-1 θ_ a —1 a-1 \θ+ά) \e+d) θ+d θ+ d Θ a-V which is increasing in d. Only the second statement is true. 3.11 Applying the formula from the solution to Exercise 3.10 gives 10,000 + 10,000 0.5-1 -40,000, 14 CHAPTER 3 SOLUTIONS which cannot be correct. Recall that the numerator of the mean residual life is E ( X ) — E ( X Λ d). However, when a < 1, the expected value is infinite and so is the mean residual life. 3 . 1 2 T h e right truncated variable is defined as Y = X given t h a t X < u. W h e n X > u, this variable is not defined. T h e fcth moment is E ( y } = F(u) = - ^ ) · 3 . 1 3 This is a single parameter Pareto distribution with parameters a — 2.5 and Θ = 1. T h e moments are μλ = 2.5/1.5 = 5/3 and μ2 = 2.5/.5 — ( 5 / 3 ) 2 = 20/9. The coefficient of variation is ^ 2 0 / 9 / ( 5 / 3 ) = 0.89443. 3 . 1 4 μ = 0.05(100) + 0.2(200) + 0.5(300) + 0.2(400) + 0.05(500) = 300. σ 2 - 0 . 0 5 ( - 2 0 0 ) 2 + 0 . 2 ( - 1 0 0 ) 2 + 0.5(0) 2 + 0.2(100) 2 + 0.05(20()) 2 = 8,000. μ 3 - 0 . 0 5 ( - 2 0 0 ) 3 + 0 . 2 ( - 1 0 0 ) 3 + 0.5(0) 3 H- 0.2(100) 3 + 0.05(200) 3 = 0. μ4 = 0 . 0 5 ( - 2 0 0 ) 4 + 0 . 2 ( - 1 0 0 ) 4 + 0 . 5 ( 0 ) 4 + 0 . 2 ( 1 0 0 ) 4 + 0 . 0 5 ( 2 0 0 ) 4 - 200,000,000. Skewness is ηλ = μ 3 / σ 3 = 0. Kurtosis is η2 = μ 4 / σ 4 = 200,000,000/8,000 2 = 3.125. 3 . 1 5 T h e Pareto mean residual life function is and so e x ( 2 0 ) / e x ( 0 ) - (2Θ -l· θ)/(θ + Θ) = 1.5. 3 . 1 6 Sample mean: 0.2(400) + 0.7(800) 4- 0.1(1,600) - 800. Sample variance: 0 . 2 ( - 4 0 0 ) 2 + 0.7(0) 2 + 0.1(800) 2 = 96,000. Sample third central moment: 0 . 2 ( - 4 0 0 ) 3 + 0.7(0) 3 + 0.1(800) 3 = 38,400,000. Skewness coefficient: 38,400,000/96,000 1 5 = 1.29. 3.2 SECTION 3.2 3 . 1 7 T h e pdf is f(x) = 2 x " 3 , x > 1. The mean is / ~ 2x~2dx = 2. T h e median is the solution to .5 = F(x) = 1 — a: - 2 , which is 1.4142. T h e mode is the value where the pdf is highest. Because the pdf is strictly decreasing, the mode is at its smallest value, 1. 3 . 1 8 For Model 2, solve p= l - ^ ^ g ^ - V and so πρ - and the requested percentiles are 519.84 and 1419.95. 2,000[(l-p)-1/3-l] SECTION 3 3 15 For Model 4, the distribution function j u m p s from 0 to 0.7 at zero and so 7Γ0.5 = 0. For percentile above 70, solve p = 1 - 0.3e~ 0OOOOl7r ^, and so πρ = -100,000 ln[(l - p ) / 0 . 3 ] and π 0 . 8 = 40,546.51. For Model 5, the distribution function has two specifications. From x = 0 to x — 50 it rises from 0.0 to 0.5, and so for percentiles at 50 or below, the equation to solve is p = 0.01π ρ for πρ — 100p. For 50 < x < 75, the distribution function rises from 0.5 to 1.0, and so for percentiles from 50 to 100 the equation to solve is p = 0.02π ρ — 0.5 for πρ — 50p + 25. T h e requested percentiles are 50 and 65. 3 . 1 9 T h e two percentiles imply 0.1 = 0.9 = 1- Θ β + 1 θ-k » + 50-3fcy Rearranging the equations and taking their ratio yield /6fl-3fc\Q \2e-k ) 0.9 0.1 Taking logarithms of both sides gives In 9 = a In 3 for a = In 9 / In 3 = 2. 3 . 2 0 T h e two percentiles imply 0.25 0.75 = = l-e-O.ooo/T, 1 - e ~(ioo,ooo/ ö r Subtracting and then taking logarithms of b o t h sides give In 0.75 = -(l,OOO/0) r , ln0.25 = -(1OO,OOO/0) T . Dividing the second equation by the first gives In 0.25 In 0.75 100 T . Finally, taking logarithms of both sides gives r In 100 = In [In 0.25/ In 0.75] for r - 0.3415. 3.3 SECTION 3.3 3 . 2 1 T h e sum has a g a m m a distribution with parameters a — 16 and Θ = 250. Then, P r ( S i 6 > 6,000) = 1 - Γ(16; 6,000/250) = 1 - Γ(16;24). From the 16 CHAPTER 3 SOLUTIONS Central Limit Theorem, the sum has an approximate normal distribution with mean αθ = 4,000 and variance αθ2 = 1,000,000 for a standard deviation of 1000. The probability of exceeding 6,000 is 1 - Φ[(6,000 - 4,000)/l,000] = 1 - Φ(2) = 0.0228. 3.22 A single claim has mean 8,000/(5/3) = 4,800 and variance 2(8,000) 2 /[(5/3)(2/3)] - 4,8002 = 92,160,000. The sum of 100 claims has mean 480,000 and variance 9,216,000,000, which is a standard deviation of 96,000. The probability of exceeding 600,000 is approximately 1 - Φ[(600,000 - 480,000)/96,000] = 1 - Φ(1.25) = 0.106. 3.23 The mean of the gamma distribution is 5(1,000) = 5,000 and the variance is 5(1,000)2 = 5,000,000. For 100 independent claims, the mean is 500,000 and the variance is 500,000,000 for a standard deviation of 22,360.68. The probability of total claims exceeding 525,000 is 1 - Φ[(525,000 - 500,000)/22,360.68] = 1 - Φ(1.118) = 0.13178. 3.24 The sum of 2,500 contracts has an approximate normal distribution with mean 2,500(1,300) = 3,250,000 and standard deviation V/275ÖÖ(400) = 20,000. The answer is Pr(X > 3,282,500) = Pr[Z > (3,282,500-3,250,000)/20,000] = Pr(Z > 1.625) = 0.052. 3.4 SECTION 3.4 3.25 While the Weibull distribution has all positive moments, for the inverse Weibull moments exist only for fc < r. Thus by this criterion, the inverse Weibull distribution has a heavier tail. With regard to the ratio of density functions, it is (with the inverse Weibull in the numerator and marking its parameters with asterisks) n-*fi*T* r θ x r r x π-τ*-1,>-(θ*/x)T* e rQ- x - e-^lQy The logarithm is (χ/θ)τ -(θ*/χ)τ* ocx-T-r\-(r/xV +(*/*)T. - ( τ + τ*)1ηχ. The middle term goes to zero, so the issue is the limit of (χ/θ)τ — (τ + τ*) Ιηχ, which is clearly infinite. With regard to the hazard rate, for the Weibull distribution we have SECTION 3.4 17 0.025 0.02 0.015 -Weibull - Inverse Weibull 0.01 4 0.005 23 Figure 3.1 25 Tails of a Weibull and inverse Weibull distribution. which is clearly increasing when r > 1, constant when r = 1, and decreasing when r < 1. For the inverse Weibull, h(x) = : 1 _ 77ΓΓΤΖ e-{9/xY oc 1 j.T+l[ e (0/*) T _ 1 ] ' The derivative of the denominator is (r + \)xT[e^^y - 1] + χτ+1β^χ^τθτ(-τ)χ-τ-\ and the limiting value of this expression is θτ > 0. Therefore, in the limit, the denominator is increasing and thus the hazard rate is decreasing. Figure 3.1 displays a portion of the density function for Weibull (r = 3, Θ = 10) and inverse Weibull (r = 4.4744, Θ = 7.4934) distributions with the same mean and variance. The heavier tail of the inverse Weibull distribution is clear. 3.26 Means: Gamma Lognormal Pareto 0.2(500) - 100. exp(3.70929 + 1.338562/2) = 99.9999. 150/(2.5 - 1) - 100. Second moments: Gamma Lognormal Pareto 5002(0.2)(1.2) = 60,000. exp[2(3.70929) 4- 2(1.33856)2] = 59,999.88. 1502(2)/[1.5(0.5)] = 60,000. 18 CHAPTER 3 SOLUTIONS Density functions: Gamma : 0.62851.T-°- 8 e -°· 002 '. (2π)- 1/2 (1.338566α:)~ 1 βχρ Lognormal 1 / I n s - 3.70929 V 1.338566 ) 688,919(x +150)~ 3 · 5 . Pareto The gamma and lognormal densities are equal when x = 2,221 while the lognormal and Pareto densities are equal when x — 9,678. Numerical evaluation indicates that the ordering is as expected. 3.27 For the Pareto distribution I e(x) S(x] Jo dy k (A) {θ + χ)α I (9 + y + x)-"dy Jo „ (Θ + y + x)- Q + 1 1 = (θ + χ -a + 1 0 + x (θ + χ)-η+1 (Θ + X a — 1 a —1 Thus e(x) is clearly increasing and e(x) > 0, indicating a heavy tail. The square of the coefficient of variation is 2Θ2 (a— l)(a — 2) Θ2 (o — 1)- a-2 - 1, which is greater than 1, also indicating a heavy tail. 3.28 My(t) etySx(y) °° Z"00 ety /■ Sx(y) dy — dy + E(X) t E(X) o . / o ~ E(X) E Mx(t) - 1 , Λ/.γ(ί) tE{X) tE{X) tE(X) / Jo e4t This result assumes lim^^^ etySx(y) — 0. An application of L'Hopital's rule shows that this is the same limit as (—t~l) lim^^^ ety fx(y). This limit must be zero, otherwise the integral defining Μχ(ί) will not converge. 3.29 (a) S(x) {\ + 2t2)e~2tdt = I = - -(l + t + i2)e-'2t\? (1 + x + x2)e~2x, x > 0 . SECTION 3.4 (b) = h(x) + x + x2) -AinS(x) = A^x)-^-\n(l ax ax ax 1+2X 21 + x + x2' = 19 (c) For y > 0, / Jy S(t)dt = / (1 + 1 + t2)e~2tdt Jy = - ( 1 + 1 + \t2)e~2X = (l + y + \y2)e~2\ Thus, (<0 e(x) = | ~ S(t)<ft _ 1 + x + 2^x 2 r o m a J,^~" =~~ -,1 ',+"Lx ',+ ^2x2 ^ () 5(ar) Jx anc * ( c )· (e) Using (b), lim /ι(χ) = *-^oo lim e(x) lim f 2 x—*oo Y x->oo 11 = 1 + 2x 1 + X + 1 2 lim h(x) χ—>οο (f) Using (d), we have e(x) = 1 Λ X 2 / ±x2 2 1 + x* + x 2 and so e'(x) v } = x 2 1 + x+ x t + \x2(l + 2x) _ -x(l (1 + x + x 2 ) 2 ——^ + x + x 2 ) + ±x 2 + x 3 (1 + x + x 2 ) 2 <0. (l + x + x2)2 ~ Also, from (b) and (e), h(0) = l,h(\) = ?> a n d M 0 0 ) = 2 · 3.30 (a) Integration by parts yields / /•DO (y-x)f(y)dy= «/Χ and hence S e (x) = j^° fe(y)dy follows. / /»OO 5(|/)dy, J X = — — /χ°°} S(y)dy, from which the result 20 CHAPTER 3 SOLUTIONS (b) It follows from (a) that E(X)Se(x) = / Jx yf(y)dy - x / Jx f(y)dy = / Jx yf(y)dy - xS(x), which gives the result by addition of xS(x) to both sides. (c) Because e(x) = E(X)Se(x)/S(x), from (b) yf(y)dy = S(x) J X ' , E(X)Se(x) S{x) S(x)[x + e(x)], and the first result follows by division of both sides by x + e(x). The inequality then follows from E(X) = J0°° yf(y)dy > f™ yf(y)dy. (d) Because e(0) = E(X), the result follows from the inequality in (c) with e(x) replaced by E(X) in the denominator. (e) As in (c), it follows from (b) that f.fl,i*.&i.i{>m + g } = M.){>ffl + f ! } , that is, 4- x e(x) from which the first result follows by solving for Se(x). The inequality then follows from the first result since E(X) = J 0 yf(y)dy > Jx yf(y)dy. 3.5 J™yf(y)dy = E(X)Se(x){-' e(x) SECTION 3.5 3.31 Denote the risk measures by p(X) = μχ 4- b x , p(Y) = βγ 4- kay, and p(X 4- Y) = Ρχ+γ 4- hax+γ. Note that μ χ + y = Ρχ 4- My and 0"x+y = < ö"x 4- cry 4- 2ρσχσγ σ 2 γ + (7y + 2σχσγ = (σχ 4- σγ)2, where p here is the correlation coefficient, not the risk measure, and must be less than or equal to one. Therefore, σχ+γ < σχ 4- σγ. Thus, p(X 4- Y) < μχ 4- pY 4- k(ax 4- a y ) - p(X) + p(Y), which establishes subadditivity. Because pcX — cpx and σ0χ = οσχ, p(cX) = cpx -\-kcax — cp(X), which establishes positive homogeneity. Because px+c = px 4- c and σχ+α = σχ, ρ(Χ 4- c) = μ γ 4- c 4- Α:σχ = p(X) + c, which establishes translation invariance. For the example given, note that X < Y for all possible pairs of outcomes. Also, ρχ = 3, pY = 4, σχ = y/3, σγ = 0. With k = 1, p(X) = 3 4- v ^ = 4.732, which is greater than p(Y) = 4 + 0 = 4, violating monotonicity. SECTION 3.5 21 3.32 The cdf is F(x) = 1 — exp(—x/0), and so the percentile solves p = 1 — exp(—πρ/θ) and the solution is VaRp(X) = πρ — —01n(l—p). Because the exponential distribution is memoryless, β(π ρ ) = 0 and so TVaR p (X) =VaR p (X)+ e(np) = πρ + θ. 3.33 The cdf is F(x) = 1 - [0/(0 + x)]a and so the percentile solves p = -1). From 1 - [0/(0 + πρ)]α and the solution is VsRp(X) = πρ = 0[(1 -p)~1^ (3.13) and Appendix A, TVaR p (X) - 7ΓΡ4- = ττρ 4- E(X) - E(X A πρ) 1 - F(np) _θ α —1 θ_ a—1 ν^+π ρ ; <*-1 α-1 7Γ ρ + 1-P Substitute 1 - ρ for [0/(0 + π ρ )] α to obtain TVaR p (X) = π ρ + = 7Γρ + α - 1 V 0 + 7Γ 0 + πρ α —1 3.34 First, obtain φ-!(0.90) = 1.28155, Φ" 1 (0.99) = 2.32635, and Φ" 1 (0.999) = 3.09023. Using πρ = μ + σΦ _ 1 (ρ), the results for the normal distribution are obtained. For the Pareto distribution, using 0 = 150 and a = 2.5 and the formulas in Example 3.17, VaR p (X) = 120 [(1 - p)~ 1 / 2 · 2 - 1 will yield the results. For the Weibull(50,0.5) distribution and the formulas in Appendix A, p = 1 — exp '-(ΞΕ) from which 0.5 VaR p pO = πρ = 50 { [ - ln(l -p)}2} which gives the results. , 22 CHAPTER 3 SOLUTIONS 3.35 Prom Exercise 3.34, VaRo.999pO = ττο.999 = 2,385.85. The mean is E(X) = 0Γ (1 + 1/r) = 50Γ (3) = 100. From Appendix A, Ε(ΧΛπ 0 .999) = 0r(l + l / r ) r [ l - f l / r ; ( ^ | £ i ) T ] Γ /7Γθ.999\ Τ Ί +π0,999 exp [- { θ ) J = 50Γ (3) Γ (3; 6.9078) + 2,385.85 exp (-6.9078) = 100Γ (3; 6.9078) +2,385.85exp (-6.9078). The value of Γ (3; 6.9078) can be obtained using the Excel function GAMMADIST(6.90775, 3, 1, TRUE). It is 0.968234. Therefore, E(X Λ πο.999) = 99.209. From these results using (3.13), TVaRo.999^0 = 2,385.85 + 1,000(100 - 99.209) = 3,176.63. (The answer is sensitive to the number of decimal places retained.) 3.36 For the exponential distribution, VaRo.95 (X) = 500 [- ln(0.05)] = 1,497.866. Therefore TVaRo.95 W = 1,997.866. From Example 3.19, for the Pareto distribution, VaRo.95 (X) = 1/3 1,000 ( 0 . 0 5 ) ~ - l Ε(ΧΛΤΓΟ.95) - 500 TVaRo. 95 PO = 1,714.4176 1,000 y 2,714.4176y/ 1,714.2+ 5 0 ° ^ 2 ' 1 0.05 4 0 432.140 = 3,071.63. 3.37 For x > x0 d E dx Wx>4 = IT yf{y)dy Έ S(x) χ χ ST yf(y)dv -/(*)] S(x) [S(x)Y XT yf(y)dy _ ί( ) = h(x) S(x) S{x) Because h(x) > 0 and e(x) > 0, the derivative must be nonnegative. CHAPTER 4 CHAPTER 4 SOLUTIONS 4.1 SECTION 4.2 4.1 Arguing as in the examples, FY(y) = Pr(X<y/c) Φ \ln(y/c)-v>] l σ J ["lny-(lnc + μ) φ 0" which indicates that Y has the lognormal distribution with parameters μ-hln c and σ. Because no parameter was multiplied by c, there is no scale parameter. To introduce a scale parameter, define the lognormal distribution function as F(x) = Φ ( l n x ~ l n t / ) . Note that the new parameter v is simply βμ. Then, Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugmaii, Harry H. Panjor, Gordon E. Willinot Copyright © 2012 John Wiley & Sons, Inc. Fourth23 24 CHAPTER 4 SOLUTIONS arguing as before, Pr(X < Fy(y) = Φ = Φ = Φ y/c) \n(y/c) — In v In ?/ — (In c -f In z/) In i/ — In ci/ demonstrating t h a t v is a scale parameter. 4.2 T h e following is not the only possible set of answers to this question. Model 1 is a uniform distribution on the interval 0 to 100 with parameters 0 and 100. It is also a beta distribution with parameters a = 1, b = 1, and 0 = 100. Model 2 is a Pareto distribution with parameters a = 3 and 0 = 2000. Model 3 would not normally be considered a parametric distribution. However, we could define a parametric discrete distribution with arbitrary probabilities at 0 , 1 , 2 , 3 , and 4 being the parameters. Conventional usage would not accept this as a parametric distribution. Similarly, Model 4 is not a standard parametric distribution, b u t we could define one as having arbitrary probability p at zero and an exponential distribution elsewhere. Model 5 could be from a parametric distribution with uniform probability from a to 6 and a different uniform probability from b to c. 4 . 3 For this year, PT(X > d) = 1 - F(d) = θ+d For next year, because Θ is a scale parameter, claims will have a Pareto distribution with parameters a = 2 and 1.06Ö. T h a t makes the probability (_Μ6θ_\2 T h 1.06(0+ rf)" 1.060+ d 1.123602 + 2.24720d + 1.1236d2 lim 1.123602 + 2.120d + d2 d—+oo 2.24720 + 2.2472d lim 2.120 + 2 d d—>οο 2.2472 lim d—»oo — — — = 1.1236. lim SECTION 4.2 25 4.4 The rath moment of a /c-point mixture distribution is v(Ym) = Jym[aifxl(y) + --- + akfxk(y)}dy = aiE(Yn + --- + akE(Yn- For this problem, the first moment is a-^— + (1 - a ) - ^ _ if a > 1. a -f1 a —I Similarly, the second moment is a l 20? ,., x 2(9? ., 7TT ^Τ + (1 - α ) τ ΤΓ" if a > 2. ν J (α-1)(α-2) (a + l)a 4.5 Using the results from Exercise 4.4, E(X) = JZi=i a*/4> a n d f° r the gamma distribution this becomes 5Ζΐ=ι a ^ a ^ . Similarly, for the second moment we have E(X 2 ) = Σί=ι αΦ2^ which, for the gamma distribution, becomes Σ ^ α ^ Ο ^ + Ι)^. 4.6 Parametric distribution families: It would be difficult to consider Model 1 as being from a parametric family (although the uniform distribution could be considered as a special case of the beta distribution). Model 2 is a Pareto distribution and as such is a member of the transformed beta family. As a stretch, Model 3 could be considered a member of a family that places probability (the parameters) on a given number of non-negative integers. Model 4 could be a member of the "exponential plus family," where the plus means the possibility of discrete probability at zero. Creating a family for Model 5 seems difficult. Variable-component mixture distributions: Only Model 5 seems to be a good candidate. It is a mixture of uniform distributions in which the component uniform distributions are on adjoining intervals. 4.7 For this mixture distribution, F(5,000) = = = 0,75φ p , 0 0 0 - 3 0 0 0 Λ α 2 5 φ ^5,000-4,000 1,000 j V 1,000 0.75Φ(2) + 0.25Φ(1) 0.75(0.9772) + 0.25(0.8413) = 0.9432. The probability of exceeding 5,000 is 1 - 0.9432 = 0.0568. 26 CHAPTER 4 SOLUTIONS 4.8 The distribution function of Z is F(z) = 0.5 11 — - 1-0.5 - 1 i + (z/yi^öö) 1 + 0.5 2 1 1 + 2/1,000 1,000 1,000 -0.5"'"1000 + * 1,000 + z2 0.5(1,0002 + l,000z + Ι,ΟΟΟ2 + 1,000z2) (1,000 + ζ 2 )(1,000 + ζ) The median is the solution to 0.5 = F(m) or (1,000 + m 2 )(l,000 + m) Ι,ΟΟΟ2 + 1,000m2 + 1,000m + m 3 m3 m = = = = 2(l,000) 2 + 1,000m + 1,000m2 2(1,000)2 + 1,000m + 1,000m2 1,0002 100. The distribution function of W is Fw(w) = PT(W < w) = P r ( l . l Z <w) = Pr(Z < w/1.1) = Fz(w/1A) 1 0.5 1 + 0.5 1 1 + (w/l.ly/TßÖÖ)2 l + z/1,100 This is a 50/50 mixture of a Burr distribution with parameters a = 1, 7 = 2, and Θ = 1.1^/1,000 and a Pareto distribution with parameters a = 1 and Θ - 1,100. 4.9 Right censoring creates a mixed distribution with discrete probability at the censoring point. Therefore, Z is matched with (c). X is similar to Model 5, which has a continuous distribution function but the density function has a jump at 2. Therefore, X is matched with (b). The sum of two continuous random variables will be continuous as well, in this case over the interval from 0 to 5. Therefore, Y is matched with (a). 4.10 The density function is the sum of five functions. They are (where it is understood that the function is zero where not defined) /i(x) f2(x) = = 0.03125, 1 < x < 5, 0.03125, 3 < x < 7, f3(x) f4(x) fc{x) = = - 0.09375, 4 < x < 8, 0.06250, 5 < x < 9, 0.03125, 8 < x < 12. SECTION 4.2 27 Adding the functions yields /{*) 0.03125, 0.06250, 0.15625, 0.18750, 0.15625, 0.09375, 0.03125, 1 < x < 3, 3 < x < 4, 4 < x < 5, 5 < x < 7, 7 < x < 8, 8 < x < 9, 9 < a : < 12 This is a mixture of seven uniform distributions, each being uniform over the indicated interval. The weight for mixing is the value of the density function multiplied by the width of the interval. 4.11 1/2" FY(y) = Φ = Φ μ y-ομ ^μ \y) ίθο \y 1/2" + exp I — 1 Φ + exp f2c6\^Φ \ομ ) y/c + μ (Bc_ μ \y y + ομ (Be ομ \y 1/2' 1/2" and so Y is inverse Gaussian with parameters ομ and c6. Because it is still inverse Gaussian, it is a scale family. Because both μ and Θ change, there is no scale parameter. 4.12 Fy(y) = Fx(y/c) = 1 — exp[— (y/c6)T], which is a Weibull distribution with parameters r and cd. SECTION 4.2 27 Adding the functions yields /{*) 0.03125, 0.06250, 0.15625, 0.18750, 0.15625, 0.09375, 0.03125, 1 < x < 3, 3 < x < 4, 4 < x < 5, 5 < x < 7, 7 < x < 8, 8 < x < 9, 9 < a : < 12 This is a mixture of seven uniform distributions, each being uniform over the indicated interval. The weight for mixing is the value of the density function multiplied by the width of the interval. 4.11 1/2" FY(y) = Φ = Φ μ y-ομ ^μ \y) ίθο \y 1/2" + exp I — 1 Φ + exp f2c6\^Φ \ομ ) y/c + μ (Bc_ μ \y y + ομ (Be ομ \y 1/2' 1/2" and so Y is inverse Gaussian with parameters ομ and c6. Because it is still inverse Gaussian, it is a scale family. Because both μ and Θ change, there is no scale parameter. 4.12 Fy(y) = Fx(y/c) = 1 — exp[— (y/c6)T], which is a Weibull distribution with parameters r and cd. CHAPTER 5 CHAPTER 5 SOLUTIONS 5.1 SECTION 5.2 5.1 FY(y) = 1 - (1 + y/9)~a = 1- (^)"· distribution. The pdf is fY(y) = dFY(y)/dy = This is the cdf of the Pareto {θ+^α+ι · 5.2 After three years, values are inflated by 1.331. Let X be the 1995 variable and Y = 1.331X be the 1998 variable. We want P r ( y > 500) = Pr(X > 500/1.331) - Pr(X > 376). From the given information we have Pr(X > 350) = 0.55 and Pr(X > 400) = 0.50. Therefore, the desired probability must be between these two values. 5.3 Inverse: FY(y) = 1 1 θ + y- 1 y + θ'1 Student Solutions Mmiual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugmaii, Harry H. Panjcr, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. Fourth29 30 CHAPTER 5 SOLUTIONS This is the inverse Pareto distribution with r = a and Θ = l/θ. Transformed: Fy(y) — 1 — \Q?T) lT 0 = 0' . . This is the Burr distribution with a — a, 7 = r, and Inverse transformed: Fy(y) = 1 - 1 - θ + ν~τ ντ + (θ~1/τΥ This is the inverse Burr distribution with r = a, 7 = r, and θ = Θ 5.4 FY(y) = 1 - flU (IT- !1//*) i + (ir70) 7 1 i + ür70)7 'T. w i + (</0)7' This is the loglogistic distribution with 7 unchanged and Θ = I/o. 5.5 F z (*) = Φ ln(*/0) - μ -Φ In z — In θ — μ which is the cdf of another lognormal distribution with μ — 1η#+μ and σ — σ. 5.6 The distribution function of Y is FY{y) = = = - P r ( y < y ) = Pr[ln(l + X / ö ) < j / ] Vi(\ + X/e<ey) Pr[X < 0 ( ^ - 1)] Θ 1<9 + 0 ( e ^ - l ) 1- l-e-ay. This is the distribution function of an exponential random variable with parameter 1/a. 5 . 7 X | 9 = 0haspdf fx\e(x\0) = τ[(χ/θγΓβχρ[-(χ/θ)η χΓ(α) and Θ has pdf /θ(β) τ[(δ/θ)ηββχΡ[-(δ/θγ} ΘΤ{β) SECTION 5.2 31 The mixture distribution has pdf ■lQ 1 e> ^Γ/"Α χΓ(α) J0 ητ^ίο\ ΘΤ(β) τ 2 χ τ α <ST/3 Γ(α + ß) χΓ(α)Γ(/3) τ ( ζ τ + <Γ)"+0 Γ(α + β)τχτα~ιδτβ τ Γ(α)Γ(/3)(χ + <Γ)"+0 ' which is a transformed beta pdf with 7 = r, r = a, a = ß, and Θ — δ. 5.8 The requested gamma distribution has a# = 1 and c*02 = 2 for a = 0.5 and Θ = 2. Then P r ( 7 V = 1) ί° io " ll = ^2°· W5Γ(0.5) n ^ d A Γ(0.5)>/2 .5)>/2 7o Z" 0 0 1 1.5Γ(0.5)>/3 Γ(1.5) 1.5Γ(0.5)Λ/3 0.5 = 0.19245. 1.5\/3 Line 3 follows from the substitution y — 1.5A. Line 5 follows from the gamma function identity Γ(1.5) = 0.5Γ(0.5). N has a negative binomial distribution, and its parameters can be determined by matching moments. In particular, we have E(N) = E[E(JV|A)] = E(A) = 1 and Var(iV) = E[Var(iV|A)] + Var[E(7V|A)] = Ε(Λ) + Var(A) = 1 + 2 = 3. 5.9 The hazard rate for an exponential distribution is h(x) = f(x)/S(x) = 0" 1 . Here Θ is the parameter of the exponential distribution, not the value from the exercise. But this means that the Θ in the exercise is the reciprocal of the exponential parameter, and thus the density function is to be written F(x) — 1 — e~9x. The unconditional distribution function is ,11 Fx(x) = (1-ε-"χ)0Λάθ / 0.1(0 + χ-1ε-θχ)\\1 1 + (β_ΠΧ β Χ) ά -" · 32 CHAPTER 5 SOLUTIONS Then, S*(0.5) = 1 - F x (0.5) = - ^ ( e - 5 5 - e"0-5) = 0.1205. 5.10 We have Pr(JV>2) = l-Fjv(l) = = 1- / (e-x+\e-x)0.2d\ Jo 1 - l-(l + \)e-x-e-x}0.2\l = l + 1.2e _5 + 0 . 2 e - 5 - 0 . 2 - 0 . 2 = 0.6094. 5.11 With probability p, pi = 0.52 = 0.25. With probability 1 - p, p2 = ©(Ο.δ) 4 = 0.375. Mixed probability is 0.25p + 0.375(1 -p) = 0.375 -0.125p. 5.12 It follows from (5.3) that fx{x) = -S'(x) = -M'A[-A{x)][-a(x)] = a(x)M'A[-A(x)}. Then nx[X > Sx(x) M A [-A(x)] · 5.13 It follows from Example 5.7 with a = 1 that Sx(x) = (1 + ΘχΊ)~1, which is a loglogistic distribution with the usual (i.e., in Appendix A) parameter θ replaced by 0 - 1 ' 7 . 5.14 (a) Clearly, a(x) > 0, and we have A(x)= px Jo a(t)dt=- n px 2 70 (1 + 0t)~Ut = Λ/1 + Θί\1 = y/l + θχ - 1. Because A(oo) = oo, it follows that /ΐχ|Λ(#|λ) — λα(χ) satisfies Ιΐχ\\(χ\λ) and J0 hx\h{x\\)dx = oo. > 0 (b) Using (a), we find that S X | A (x|A) = e - M ^ ^ - i ) . It is useful to note that this conditional survival function may itself be shown to be an exponential mixture with inverse Gaussian frailty. (c) It follows from Example 3.7 that MA(t) = (1 - t)~2(y and thus from (5.3) that X has a Pareto distribution with survival function Sx(x) = M\(l — yjl + θχ) = (1 + θχ)-α. SECTION 5.2 33 (d) The survival function Sx(x) = (1 + θχ)~α is also of the form given by (5.3) with mixed exponential survival function Sx\\(x\X) = e~Xx and gamma frailty with moment generating function M\(t) = (1 — 6t)~a, as in Example 3.7. 5.15 Write Fx[x) = 1- MA[-A(x)], and thus E(A)A(x) where MA(t) - 1 ίΕ(Λ) is the moment generating function of the equilibrium distribution of Λ, as is clear from Exercise 3.28. Therefore, Si(x) is again of the form given by (5.3), but with the distribution of Λ now given by the equilibrium pdf ΡΓ(Λ > λ)/Ε(Λ). Mi(t) 5.16 (a) ΜΑ.(ί) = J etX I e<*->A/A. (\)d\ f^WdX=1~Z7) all λ all λ MA(t - S) MA(-s) ' with the integral replaced by a sum if Λ is discrete. bM < > '^ = ^ Γ a n d thus Ε(Λ } M (0) Also, M'i(t) = M^{Zsy * = ^ = W0)- implying t h a t E Now, C c'A(t) = A(-S) = c'L(t) = (c) hx{x) M^y M'A(t) , and replacement of t by — s yields MA(t) Ε(Λ„). Similarly. M'Jjif) MA(t) \M'A(t) [MA(t) M'A{-s) c'A(s) ^ = M *» (0) = MA(-s) LMA(-S) and so 2 E ( A f ) - [ E ( A s ) ] 2 = Var(A s ). = ^ I n 5 x ( x ) = - £ l n MA\-A{x)\ = a(x)c'A[-A(x)}. = -±cA[-A{x)\ 34 CHAPTER 5 SOLUTIONS (d) d tix{x) = a'(x)c'A{-A(x)}+a(x) — = a'(x)c'A[-A(x)\ la(x)}2c'i[-A(x)}. - c'A[-A(x)} (e) Using (b) and (d), we have tix(x) - [α(χ)]2νΆτ[ΑΑ{χ)] = a\x)E[AA{x)] < 0 if a'(x) < 0 because E[A^ (;r )j > 0 and Var[A v4(;r )] > 0. But -^hX\A(x\X) λ α ' ( χ ) , and thus a'{x) < 0 when ^hx\A(x\\) < 0. = 5.17 Using the first definition of a spliced model, we have f , , _ [ }x{x) ~ T. 0 0 < x < 1,000 \ 7e-*/ , x> ι,οοο, where the coefficient r is a\ multiplied by the uniform density of 0.001 and the coefficient 7 is a^ multiplied by the scaled exponential coefficient. To ensure continuity we must have r = ^ g - 1 ' 0 0 0 / ^ Finally, to ensure t h a t the density integrates to 1, we have /•Ι,ΟΟΟ 1 7e- 1 000 ' /»OO θ Ίε-*' άχ = / Jo /"dx+ / Λ,οοο = l)0007e-1-000/e+7Öe-1'00°/e, which implies 7 = [(1,000+ 0 ) e - 1 , 0 0 0 / 0 ] - 1 . T h e final density, a one-parameter distribution, is fx(x) = { ( 0<x< 1,000 + 0 ' e-x/0 1,000, x > 1,000. (1,000 + 6>)β-^ 0 0 °/^ Figure 5.1 presents this density function for the value Θ = 1,000. 5.18 fY(y) 3 et/9 = exp(-|(lny)/0|)/20y. H o o = ¥*'*■ For x < 0, Fx(x) For x > 0 it is I + i f* exponentiation the two descriptions are Fy(y) FY{y) e = l-\e-^y' , e = jg Jl^e'^dt = -t/edi = i _ I e - / * . W i t h = \eXnyle, 0 < y < 1, and y>\. 5.19 F(ar) = / * it~Adt = 1 - x " 3 . y = 1.1X. P r ( y > 2.2) = 1 - Fy(2.2) = ( 2 . 2 / 1 . 1 ) " 3 = 0.125. FY{y) = 1 - (x/1.1)"3. SECTION 5.2 35 0.0006 ι 0.0005 0.0004 Uniform S 0.0003 \ Exponential j 0.0002 0.0001 0 (D 500 1000 1500 2000 X Figure 5.1 Continuous spliced density function. 5.20 (a) P r ( X " 1 < x) = P r ( X > l/x) fore, = P r ( X > l / x ) , and the pdf is, there- '"*<*> - έ Ρ Γ ( χ > ί) -W 1 X1 [θχ* 2πχ \X 1- _ μ χ \ exp θχ 2 i exp Θ ~2x (■ μχ ) X2) , x >0. (b) T h e inverse Gaussian pdf may be expressed as f(x) Thus, / 0 °° f(x)dx = θ_ 2π' θ/μ 1 5 Χ/^Ζ6 χ- · βΧρ[- θχ 2μ2 θ 2χ = 1 may be expressed as x x—,5expe x p ^θχ^ - - jθ\d x = ^ 2πe - ^ , f" (- T 36 CHAPTER 5 SOLUTIONS valid for Θ > 0 and μ > 0. Then 0t\X + t2X 1 /(*) 2π 2 π θ 2μ2 e e /"a;- 1 - 5 exp β ^ - β χ ρ ttl^-lf-^i ,χ ν θ 2μ, 2χ where θ* = θ - 2£2 and μ+ = M w / 2 ! ' 4 i · Therefore, if #* > 0 and μ* > 0, /»OO M(U,t2) = et,x+taX~'/(*)<** / 2π 2π (c) Μχ(ί) = βθ/μ Ι^1ε-θ./μ. θ-2ί2 θ-2ί2 0 exp ~μ~ μ θ θ-2ί2 exp θ - 2μΗχ y 6» - 2ί 2 θ-^(θ-2ί2)(θ-2μΗχ) μ Μ(ί,0) exp (d) 0»/^ Μ1/Χ(ί) V = - Μ / = exp »ίι-,/ι-¥< Μ(0,ί) 0 exp '-2ί 0-2ί = "-ϊ ι exp yίθ-^/θ(θ-2ί) Μ jp-v 1 -!. -1/2 where #ι = θ/μ2 and μχ = 1/μ. exp *,'-Η?· Thus M1/X(t) = MZl(t)Mz2(t) with Μζι (0 — (l — -§t) the mgf of a gamma distribution with a = 1/2 and Θ replaced by 2/Θ (in the notation of Example 3.7). Also, Mz2{t) is the mgf of an inverse Gaussian distribution with Θ replaced by θ\ — θ/μ2 and μ by μλ — Ι/μ, as is clear from (c). SECTION 5.2 (e) Η^ϊ Mz(t) = E { exp expl £ 2 μ2 ν 2t t^ and, therefore, ΪΜ[^,ί = e Mz(t) e 0 - ^(θ - 2ί) 2 Θ exp M 0-2t » Θ exp e-2t - 2i μ 6> - (6> - 2t) μ -1/2 '»-ϊ« the same gamma mgf discussed in (d). (f) First rewrite the mgf as i ( 0 s-wU?-' = exp Mx(t) Then, M^(i) = -(2<rp 2μ 2 - ί , (-Ι)Μχ(ί) ( - (ί) *(£-'Γ** ·>· Similarly, (0- I Λί£(«) + = *. 2μ 2 -ί) Μ χ (<) v( a)'"(^-)" ' ",+ ,4 Mx{t) α) · (έ-)" α)·(*-Γ 37 38 CHAPTER 5 SOLUTIONS and the result holds for k = 1, 2. Assuming that it holds for k > 2, differentiating again yields U 1 fc+l+3w £•-1 fc , v + fc + 2 + a n k + 2+ n 2μ 2 n=0 * Now, change the index of summation in the second sum to obtain M{xh+1)(t) /A-^„»±1=« / θ *, i* ST· (k + n-l)l JL fc-f 3n- 1 Λ~^~ fc + 1+ n Separate out the two terms when n = 0 and n = k and combine the others to obtain SECTION 5.2 = f(*+l) M^>(t) fc+1 fc+1 2 i\ Mx(t) Λ*±ι / 0 +Μχ(ί)^ ^ n=l xl 2 V (2fc-l)!/l\2fc^ , / Θ (fc + n - 1 ) ! / I (fc-n)!n! V2 v / 39 \~»+ fc+1 - n \ p - n) + 2n] ,v-' Μχ(ί) _fc±l±2i + r, n=0 fc+l+an fc+l+n (/c — n)!n! W and the kth derivative of Μχ (t) is as given by induction on n. Then fc-f-3n —fc— n G) · - & μ fc+n (261)"' and the result follows from E(Xk) = M{x]{0) (g) Clearly, κ Λτ-κ-' 20 p μ ' which implies from (f) that, for m = 1,2,..., the moment result may be stated Γ mfi ^ i^mi„ (θ\ ^^ m -x(f) 40 CHAPTER 5 SOLUTIONS which is also obviously true when m = 0. But, fix) 2π X 26 2/^xv x 2 2β X 2μ· and thus — e» / 2π J0 e *^ 2x —μ dx=\ e» Km_i or, equivalently, /•OO / Jo xn -2e~*^~'2*dx = 2μ7η-ΐΚπ Let a = 0 / ( 2 μ 2 ) , implying t h a t μ — y/0/(2a), and also t h a t θ/μ = 2μα y/2cS. 5.2 SECTION 5.3 5.21 Let r = a / 0 , and then in the Pareto distribution substitute τθ for a. T h e limiting distribution function is liml-' 0-+OO θ \X 1- + τθ 0 lim θ-^οο V X + 0 T h e limit of t h e logarithm is lim τ0[1η 0 - ln(x + 0)] 0—»οο = r lim 1 η 0 - 1 η ( χ + 0) -1 Θ—>oo 1 r r hm 0 " - ( x + 0)- 0-+oc —r lim —Ö—— -0" £0 0^oo X + I 2 -ra;. T h e second line and the final limit use L'Hopital's rule. T h e limit of the distribution function is then 1 — exp(—rx), an exponential distribution. 5.22 T h e generalized Pareto distribution is the transformed b e t a distribution with 7 = 1. T h e limiting distribution is then transformed g a m m a with r = 1, which is a g a m m a distribution. T h e g a m m a parameters are a = τ and 0 = ξ. SECTION 5.4 5.23 Hold a constant and let θτ1^ /(*) = Γ(α + -» ξ. Then let θ = ξτ'1^. 41 Then φχΐ*-1 ιτ Τ(α)Γ(τ)θ' (ί + χ'<θ-'ϊ)α+τ e-a-T(a + τ)α+τ-1(2π)1/2Ίχ-'τ-1 Γ(α)ε~τττ-1(2π)1/2ξ'ΪΤτ-τ(1 + χ-τξ~Ίτ)α+τ (1+1) α+τ-1 jx-^-1 Γ(α)τ-α-τξ'ι(-τ+α)ξ-Ίαχ-^τ+α)(1 + χΐΓΊτ)α+τ »0,+T-17a;-7a-l α-\-τ 'Ρ] Γ ( α ) Γ 7 α 1 + S£L 7ξ 7« Γ(α)χτα+ι β («Μ 7 5.3 SECTION 5.4 5.24 Prom Example 5.12, τ{θ) = - 1 / 0 and q(ß) = θα. Then, r'(0) = 1/02 and g'(0) = α 0 ° - 1 . From (5.9), μ(0) = , ' ,„. = . -2ηα ο.„. = αθ. r'(0)g(0) Θ'Η' With μ'(θ) = a and (5.10), Var(X) 0~ 2 r'(0) ' 5.25 For the mean, ln/(x;0) c>ln/(x;0) 8Θ In p(m, x ) + ror(0)x — m In g(0), mq'(e)~\ df(x-,e) i mr'(9)x 00 /(x;0) mg'(0) mr'(9)x /(*;*), = m dfjx-,θ) 00 - / m 0/(*;0) c£x «1/(0) | x/(x; θ)άχ - ^ p I /(x; θ)άχ 00 = E(X) = mq'{9) Qiß) ' ς'(θ)/[Γ'(θ)ς(θ)}=μ(θ). mr'(0)E(X) 42 CHAPTER 5 SOLUTIONS For the variance αί(χ;θ) ΘΘ 02ί(χ;θ) θθ 5— = πιν'(θ)[χ-μ(θ)]/(χ;θ), = mr"{fl)\x - μ(θ)}ί(χ; Θ) - τηΓ'(θ)μ'(θ)/(χ; 2 θ) 2 +τη[τ'(θ)} [χ-μ(θ)} ί(χ;θ), = I d2f 9) = 0- ^ dx Οθ -πιτ'(θ)μ'(θ) + m2[r'{ß)]2E{[X ror'(0)//(0)(l) - μ(θ)}2} + m2[r,(<9)]2Var(X), νχτ(Χ) = μ'(θ)/[πΐΓ'(θ)]. 5.26 (a) We prove the result by induction on m. Assume that Xj has a continuous distribution (if Xj is discrete, simply replace the integrals by sums in what follows). If m = 1, then S — X, and the result holds with p(x) = Pi(x). If true for m with p(x) — p^,(x), then for m 4- 1, S has pdf (by the convolution formula) of the form I p*m(s - x)erW>-*) n?=i QJV) = pm+1(x)erW* im+Λθ) erW* [ J J y ; ( s - ax x)pm+1(x)dx] which is of the desired form with Pm+i(s) = JQ Pm(s ~ x)Pm+i{x)dx. It is instructive to note that p(x) is clearly the convolution of the functions Pi{x),P2(x), · · - ,Pm(x)> (b) By (a), 5 has pf of the form If Xj has a continuous distribution and 5 has pdf / ( s ; #), then that of X is f\(x\9) = mf(mx',e) mp r * n (mx)e mr ^)* mv of the desired form with p(m,x) = mp*n(mx). If Xj is discrete, the pf of X is clearly of the desired form with p(ra, x) — p^h(mx). CHAPTER 6 CHAPTER 6 SOLUTIONS 6.1 b i SECTION 6.1 oo ΡΝ(Ζ) = Σ ν k=0 P'N{z) = P'NW P'NM 6.2 = = = oo ^ = Σν^χ"ζ fc=0 = ΜΝ{\ηζ) z-lM'N(\nz) MfN(0) = E(N) + z-2M^(lnz) -z-2M'N(\nz) 2 -E(iV) + E(iV ) = E[N(N - 1)]. SECTION 6.5 6.2 For Exercise 14.3, the values at k = 1,2,3 are 0.1000, 0.0994, and 0.1333, which are nearly constant. The Poisson distribution is recommended. For Exercise 14.5, the values at k = 1,2,3,4 are 0.1405, 0.2149, 0.6923, and 1.3333, which is increasing. The geometric/negative binomial is recommended (although the pattern looks more quadratic than linear). Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugman, Hairy H. Panjer, Gordon E. Willinot Copyright © 2012 John Wiley & Sons, Inc. FourthAZ 44 CHAPTER 6 SOLUTIONS 6.3 For the Poisson, λ > 0 and so it must be a = 0 and b > 0. For the binomial, m must be a positive integer and 0 < q < 1. This requires a < 0 and b > 0 provided — 6/a is an integer > 2. For the negative binomial, both r and β must be positive so a > 0 and 6 can be anything provided b/a > — 1. The pair a = — 1 and b = 1.5 cannot work because the binomial is the only possibility but — b/a = 1 . 5 , which is not an integer. For proof, let po be arbitrary. Then px = ( - 1 + 1·5/1)ρ = 0.5p and p2 = ( - 1 4- 1.5/2)(0.5p) = -0.125p < 0. 6.3 6 · 4 SECTION 6.6 Pk = Pk-i = Pk-2 = ß r-1 1 + /3 + k /c + r - l / c + r - 2 0 fc-l i8 Pi fc+r-1 fc ß β = Pk-i 14-/3 1+β ■/? /c + r - l / c + r - 2 ik i f e - 1 r + 1 2~" The factors will be positive (and thus pk will be positive) provided p\ > 0, β > 0, r > - 1 , and r ^ 0. To see that the probabilities sum to a finite amount, oo X^Pfc oo = Pi k=l ß + /?/ ΣΪ k=l lI v ^ Pifc=l Σι/ = = Pi ß + /? fe-l /c + r - l / c + r - 2 fc fe-l (l+/3)r+1 rß r-fl fc-1 fc + r - 1 /3 1 + /9 ■/3 fc + r - 1 fc The terms of the summand are the pf of the negative binomial distribution and so must sum to a number less than one (po is missing), and so the original sum must converge. 6.5 From the previous solution (with r = 0), ß i = Σ Ρ * = ^ Σ ( Ϊ+ß fc=l k=l ß fe=i = x fc-l +/ ι +β -In I 1 /3 Pi- ß 1+/ fc-l fc-lfc-2 fc fc- 1 SECTION 6.6 45 using the Taylor series expansion for ln(l — x). Thus ß Pi and Pk 6.6 P(z) = k ; (l + /3)ln(l + /3) _( ß Ϋ \l + ß) 1 feln(l + ßY * fV ß V1· l n (l + / 3 ) ^ V l + /?y *' _zß_ -In 1 ln(l + /3) l+ß m \l+ß-zß) ln(l + /3) ln[l-/?(z-l)] ln(l + /3) · 6.7 The pgf goes to 1 — (1 — z)~r as ß —> oo. The derivative with respect to z is —r(l — z)~r~1. The expected value is this derivative evaluated at z = 1, which is infinite due to the negative exponent. SECTION 6.6 45 using the Taylor series expansion for ln(l — x). Thus ß Pi and Pk 6.6 P(z) = k ; (l + /3)ln(l + /3) _( ß Ϋ \l + ß) 1 feln(l + ßY * fV ß V1· l n (l + / 3 ) ^ V l + /?y *' _zß_ -In 1 ln(l + /3) l+ß m \l+ß-zß) ln(l + /3) ln[l-/?(z-l)] ln(l + /3) · 6.7 The pgf goes to 1 — (1 — z)~r as ß —> oo. The derivative with respect to z is —r(l — z)~r~1. The expected value is this derivative evaluated at z = 1, which is infinite due to the negative exponent. CHAPTER 7 CHAPTER 7 SOLUTIONS 7.1 SECTION 7.1 7.1 Poisson: P(z) = e^ 2 " 1 *, B{z) = e 2 , Θ = \. Negative binomial: P(z) = [1 - ß(z - l)]" 7 ", B(z) = (1 - z)~r, Θ = ß. Binomial: P{z) = [1 + g(z - !)]"', B(z) = (1 + ^ ) m , Θ = q. 7.2 The geometric-geometric distribution has pgf PGG(Z) = (l-ZW-/^-!)]1 - /32(z - 1) 1-/32(1+ /?,)(*-!)" 1 -!}) - 1 The Bernoulli-geometric distribution has pgf PBGW = l + qill-ßiz-l)}-1-!} I - ß(l - q)(z - 1) l-ß(z-l) · Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugman, Harry H. Panjcr, Gordon E. Wilhiiot Copyright © 2012 John Wiley & Sons, Inc. FourthAl 48 CHAPTER 7 SOLUTIONS The ZM geometric distribution has pgf P ^ PZMG(Z) , [1-/?*(*-I)]"1-(1 + I T ^ι Po + ( l - f > o ) i = ! _ ( 1+ 1 Α«)-Ι l-(po/?*+po-l)(z-l) 1-/T(z-1) In PBG(Z), replace 1 — g with (1 + βλ)~λ and replace /? with β2(\ -f βλ) to see that it matches PGG(Z)· It is clear that the new parameters will stay within the allowed ranges. In PGG{Z), replace q with (1 — ρο)(1 + β*)/β* and replace β with β*. Some algebra leads to PZMG(Z)· 7.3 The binomial-geometric distribution has pgf PBG{Z) = {'^[i-ßl-i)-1}}"1 ri-^(i-g)(^-i)iro L ι-β(ζ-ΐ) J ' The negative binomial-geometric distribution has pgf PNBG(Z) 1 1 ßl [l-ß2(z-l) 1-02(1+&)(*-!) 1-β2(ζ-1) 1 ~ ß2(z ~ 1) L i - / 3 2 ( i + /?,)(*-i)J · In the binomial-geometric pgf, replace m with r, ß(l — q) with /? 2 , a n d ß with (1 + ßi)ß2 to obtain PNBG{Z)7.2 SECTION 7.2 7.4 = explf^Wz)-!]). ^=1 This is a compound distribution. The primary distribution is Poisson with parameter ^,7=1 X{. The secondary distribution has pgf Pz(z). SECTION 7.2 7-5 P(z) = ΣΓ=ο zkPk, P(1)(z) = ΣΓ=ο 49 k*k-lPk, oo Ρϋ)(«) £ fc(fc - 1) · · · (Ä - j + l)* f c -' P f c , ρΟ·)(ΐ) = fc=0 = f>(Ä-l)...(fc-.7 + l)pfc fc=0 E[iV(iV - 1) · · · (N - j + 1). P(z) pV\z) = exp{A[P 2 (z)-l]}, = μ = \P(z)P?\z), 1) p( (l) = p(l) *2 \P(l)m'1=\m'v pV\z) λ2Ρ(ζ) ß2~ß p(2)(l) = λ 2 (τηΊ) 2 + A(m2 - m'j). + Then μ'2 = λ (m'j ) 2 + λτη 2 and μ 2 = A*2 ~ M p(3>(*) = μ!ζ - 3μ2 4 2μ = \Ρ(ζ)Ρ?\ζ), λτη2. A 3 P( Ä )[P 2 ( 1 ) ]V3Ä 2 P( Ä )P 2 ( 1 ) («)P 2 W λ 3 (πι[) 3 4- 3A2mi (m 2 - mi) 4- A(m3 - 3m 2 4- 2m[). Then, μ3 = = = μ'3- 3μ2μ + 2μ 3 (^3 - 3μ 2 4- 2μ) 4- 3μ 2 - 2μ - 3μ'2μ 4- 2μ 3 λ 3 (mi ) 3 + 3A2mi (m 2 - mi) + A(mi> - 3m 2 + 2mi) +3[A 2 (m;) 2 4- \m'2] - 2Xm[ - 3Xm'1[X2(πι[)2 4- Xm'2] 4- 2A 3 (m;) 3 = λ3[Κ)3-3Κ)3 + 2Κ)3 = +A2[3m'1m2 - 3(m,;) 2 4- 3(m,;) 2 - 3m,; m 2 ] +A[m^ — 3m 2 4- 2m; 4 3m 2 — 2m;] Arn^. 7.6 For the binomial distribution, m[ = mq and m 2 = mq(l — q) 4- m2q2. For the third central moment, P^\z) = m(m - l)(m - 2)[1 4- q(z - l ) ] m - V and P(3>(1) = m(m - l)(m - 2)q3. Then, m'3 = m(m - l)(ra - 2)<73 4- 3mq 3mq2 4 3m2q2 - 2mq. 50 CHAPTER 7 SOLUTIONS μ = Χτη'λ — Xmq. <τ2 = \m'2 = X(mq — mq2 + m2q2) = Xmq(l - q + rag) = μ[1 4- g(ra - 1)]. μ3 = = = = Araij = Xmq(m2q2 — 3mq2 + 2g2 + 3 — 3q + 3ra — 2) Arag(3)(l - q + rag) - Arag(2) + Xmq(m2q2 - 3mq2 + 2q2) 3σ2 - 2μ + Xmq(m2q2 - 3mq2 + 2g2) 3σ 2 - 2μ + Arag(ra - l)(ra - 2)<?2. Now, τ η - 2 ( σ 2 - μ ) 2 ra — = μ ra ra — 1 - 2/i 2 [g(ra - l)] 2 , ow 1λχ 3 ^—^ — = (ra - 2)(ra - 1)λ7τιςΤ —1 μ and the relationship is shown. 7.3 SECTION 7.3 which is 7.7 For the compound distribution, the pgf is P(z) — PNB[PP(Z)\ exactly the same as for the mixed distribution because mixing distribution goes on the outside. 7.8 n n PN(z) = Y[PNt(z) i=\ = l[Pi[ex^% i=\ Then N is mixed Poisson. The mixing distribution has pgf P(z) = ΠΓ=ι Ρί(ζ)· 7.9 Because Θ has a scale parameter, we can write Θ = cX, where X has density (probability, if discrete) function fx(x). Then, for the mixed distribution (with formulas for the continuous case) Pk = Kk ^■ k\ je-x(,ekfx{e/c)c-'de je-^fx{x)dx. The parameter λ appears only as the product \c. Therefore, the mixed distribution does not depend on λ as a change in c will lead to the same distribution. 7.10 s Pk Jo SECTION 7.3 ° ° e - ¥ a2 (Θ + k\ a +1 v2 roo /c!(, fe!(o + 1) Λ 51 l)e-Q°de 3 fc!(Q + l ) 7 0 \a + l / T(fc + 2) + T(fe + 1) kl(a + l) + [ a + 1 g2[(fc + l ) / ( g + !) + !] dß k 2 (a l)fe+2 + T h e pgf of t h e mixing distribution is P(Z) = = / z0-?—(e + l)e-a<>de a JO + l rv2 i°° -^— / (0 + l)e-(a-lnz^de α + 1 Jo 0+ 1 fl + 1 -(α-\ηζ)θ a" a — Inz (a — lnz) 2 a + 1 v2 1 a 2 a + 1 a — lnz + (a — Inz) -(a-ln2)0 and so the pgf of the mixed distribution [obtained by replacing Inz with \(z - 1)] is a2 {Z) ~ f 1 α+ 1\α-λ(ζ-1) a a + 1 a - λ(ζ - 1) + 1 I [α-λ(ζ-Ι)]2/ +a+l [a- a X(z-l) which is a two-point mixture of negative binomial variables with identical values of β. Each is also a Poisson-logarithmic distribution with identical logarithmic secondary distributions. The equivalent distribution has a logarithmic secondary distribution with β = λ / α and a primary distribution that is a mixture of Poissons. The first Poisson has λ = 1η(1 + λ/α) and the second Poisson has λ = 2 ln(l + λ / α ) . To see that this is correct, note that the pgf is p z () v } = a f / λ \ ln[l - X(z - l ) / a ] 1 7 exp In 1 + , ^ w ( f a + 1 * \ \ a) 1η(1 + λ/α) J + βχρ 21η 1 + ln[l - \(z - 1)/«]] ln(l + λ/ a) / ^τϊ { ( έ) a + 1 In 1- Hz -1) a 1 a + 1 A(z-l) 52 7.4 CHAPTER 7 SOLUTIONS SECTION 7.5 7.11 (a) po = e~4 = 0.01832. a = 0,6 - 4 and then p1 = 4p0 = 0.07326 and P2 = (4/2)pi = 0.14653. (b) po = 5" 1 = 0.2. a = 4/5 = 0.8,6 = 0 and then a n d p 2 = (4/5)pi - 0 . 1 2 8 . Pl = (4/5)p 0 = 0.16 (c) po = (1 + 2)" 2 - 0.11111. a = 2/3,6 = (2 - l)(2/3) - 2/3 and then pi = (2/3 4- 2/3)po = 0.14815 and p 2 = (2/3 + 2/6)pi = 0.14815. (d) po = (1 - 0.5)8 = 0.00391. a = -0.5/(1 - 0.5) = - 1 , 6 = (8 + 1)(1) = 9 and then px = ( - 1 + 9)p 0 = 0.3125 and p 2 = ( - 1 + 9/2)pi = 0.10938. (e) po = 0, pi = 4/ln(5) = 0.49707. a = 4/5 = 0.8,6 = -0.8 and then P2 = (0.8 - 0.8/2)pi = 0.14472. (f) po = 0, pi = ^=£#^ = 0.72361. a = 4/5 = 0.8,6 = (-0.5 - 1)(0.8) = -1.2 and then p 2 = (0.8 - 1.2/2)pi = 0.14472. (g) The secondary probabilities were found in part (f). Then go = e - 2 = 0.13534, gi = \(1)(0.72361)(0.13534) - 0.19587, and 92 = |[1(0.723671)(0.19587) + 2(0.14472)(0.13534)] = 0.18091. (h) p ^ = 0.5 is given, pj = 1/5 - 0.2 and then p f = (1 - 0.5)0.2 = 0.1. From part (b), a = 0.8, 6 = 0 and then pif = (0.8)pf = 0.08. (i) The secondary Poisson distribution has /o = e _ 1 = 0.36788, f\ = 0.36788, and / 2 = 0.18394. Then, g0 = e" 4 ^" 0 · 3 6 7 8 8 ) = 0.07978. From the recursive formula, gi = \(1)(0.36788)(0.07978) = 0.11740, and g2 = |[1(0.36788)(0.11740) + 2(0.18394)(0.07978)] = 0.14508. (j) For the secondary ETNB distribution, / 0 = 0, fx = γ ^ τ ^ = 0.53333. With a = 6 - 1/3, f2 = (1/3 + l / 6 ) / i = 0.26667. Then, g0 = e~4 - 0.01832, 9\ = f (1)(0.53333)(0.01832) - 0.03908, and 92 = ^[1(0.53333)(0.03908) + 2(0.26667)(0.01832)] = 0.06123. (k) With /o = 0.5 (given) the other probabilities from part (j) must be multiplied by 0.5 to produce f\ = 0.26667 and / 2 = 0.13333. Then, go = SECTION 7.5 e -4(i-o.5) = 53 0.01832, <h = f (1)(0.26667)(0.01832) = 0.03908, and g2 = | [1(0.26667) (0.03908) + 2(0.13333) (0.01832)] = 0.06123. 7.12 Pk/Pk-i = (ns^) k. ¥" a + b/k for any choices of a, b, and p, and for all SECTION 7.5 e -4(i-o.5) = 53 0.01832, <h = f (1)(0.26667)(0.01832) = 0.03908, and g2 = | [1(0.26667) (0.03908) + 2(0.13333) (0.01832)] = 0.06123. 7.12 Pk/Pk-i = (ns^) k. ¥" a + b/k for any choices of a, b, and p, and for all CHAPTER 8 CHAPTER 8 SOLUTIONS 8.1 SECTION 8.2 8.1 For the excess loss variable, MV) = 0 · 0 0 0 0 °3 6 oooooU5.ooo) = O . O O O O l e — " , FY(y) = i-e-°-«™v. For the left censored and shifted variable, M 2 / ) " / 1 - 0.3e-° 0 5 = 0.71463, \ 0.000003e- 0 0 0 0 0 1 ^ + 5 ' 0 0 °), ϊγ\ν) - / 0.71463, \ 1 _ 0i3e -o.ooooi(i/+5,ooo) ? y = 0, j/>0, y = 0, ^ > 0, and it is interesting to note that the excess loss variable has an exponential distribution. Student Solutions Manual, to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Kluginan, Harry H. Panjer, Gordon E. Willniot Copyright © 2012 John Wiley & Sons, Inc. Fourth55 56 CHAPTER 8 SOLUTIONS 8.2 For the per-payment variable, r ( fy(y) x FY(y) 0.000003e~ 000001?/ m n m = Q>3e-o.ooooi(5,ooo) = Q= 1 - e-o.000010,-5,000)^ y > -o.ooooi(?,-5,ooo) QQQQle 5 ( m For the per-loss variable, / i \ fY{y) i7 I \ *Y\V) _ ~ _ ~ ! l ~ 0-3e-° 0 5 = 0.71463, y = 0, \ 0.000003e- 000001 ", y > 5,000, / 0.71463, 0 < y < 5,000, { ! _ 0.36-0.00001^ 2 / > 5 i 0 o o . 8.3 From Example 3.1 E(X) = 30,000 and from Exercise 3.9, E(X A 5,000) = 30,000[1 - e-o.ooooi(5.ooo)] = 1 4 6 3 12. 00001 5 000 ( · ' = 0.71463, and so for an ordinary Also F(5,000) = 1 - o.3e-° deductible the expected cost per loss is 28,536.88 and per payment is 100,000. For the franchise deductible, the expected costs are 28,536.88 + 5,000(0.28537) = 29,963.73 per loss and 100,000 + 5,000 = 105,000 per payment. 8.4 For risk 1, θ E(X)-E(X Ad) a —1 (a-l)(9 ° 1a — 1 + ' + *: k)"-1' The ratio is then Θ08" (0.8a - 1)(θ + k)0Sa~l <T I (a - 1)(θ + k)«-i I (6» + fc)a2'*(a-l) 02η θ (0.8α - 1) ' As k goes to infinity, the limit is infinity. 8.5 The expected cost per payment with the 10,000 deductible is E p Q - E(X Λ 10,000) _ 20,000 - 6,000 1 - F(10,000) ~ 1 - 0.60 " 35 ϋϋϋ ' · At the old deductible, 40% of losses become payments. The new deductible must have 20% of losses become payments and so the new deductible is 22,500. The expected cost per payment is E(X) - E(X Λ 22,500) _ 20,000 - 9,500 1 - F(22,500) 1 - 0.80 The increase is 17,500/35,000 = 50%. 52,500. SECTION 8.3 8.2 57 SECTION 8.3 8.6 From Exercise 8.3, the loss elimination ratio is (30,000 - 28,536.88)/30,000 = 0.0488. 8.7 With inflation at 10% we need E(X Λ 5,000/1.1) = 30,000[1 - e " 0 0 0 0 0 1 ^ 0 0 0 / 1 1 ) ] = 1333.11. After inflation, the expected cost per loss is 1.1(30,000-1333.11) = 31,533.58 an increase of 10.50%. For the per payment calculation we need ^(5,000/1.1) = l - O ^ e - 0 0 0 0 0 1 ^ ' 0 0 0 / 1 1 ) = 0.71333 for an expected cost of 110,000, an increase of exactly 10%. 8.8 E(X) = exp(7 + 2 2 /2) = exp(9) = 8103.08. The limited expected value is E(X Λ 2,000) In 2 , 0 0 0 - 7 - 2 2 ln2,000-7 = β9Φ = = 8,103.08Φ(-1.7) + 2,000[1 - Φ(0.3)] 8,103.08(0.0446) + 2,000(0.3821) - 1,125.60. 2,000 1 - Φ The loss elimination ratio is 1,125.60/8,103.08 = 0.139. With 20% inflation, the probability of exceeding the deductible is Pr(X > 2,000/1.2) /ln(2,000/1.2) Φ 2 Pr(1.2X > 2,000) V = = 1 - Φ(0.2093) 0.4171, and, therefore, 4.171 losses can be expected to produce payments. 8.9 The loss elimination ratio prior to inflation is E(X A 2k) E(X) k 2-1 2-1 \2k+k ) 2-1 2fc 2k+ k Because Θ is a scale parameter, inflation of 100% will double it to equal 2k. Repeating the preceding gives the new loss elimination ratio of 1- 2k 2k + 2k 1 2' 58 CHAPTER 8 SOLUTIONS 8.10 The original loss elimination ratio is 1,000(1 - e- 500 / 1 - 000 ) E(X A 500) E(X) ~~ 1,000 0.39347. Doubling it produces the equation 0.78694= l W l - ; o o d / 1 ' 0 0 0 ) = 1 _ ^ / , o o o . The solution is d = 1,546. 8.11 For the current year the expected cost per payment is E(X) - E(X A 15,000) _ 20,000 - 7,700 1-0.70 1-F(15,000) ~ After 50% inflation it is 1.5[E(X) - E(X Λ 15,000/1.5)] 1 - F(15,000/1.5) _ 1.5[E(X) - E(X A 10,000)] 1 - F(10,000) 1.5(20,000 - 6,000) 52,500. 1 - 0.60 8.12 The ratio desired is „6.9078+1.51742/2<f, / i n 10,000-6.9078-1.5174 2 \ ^ \ 1.5174 ) C E(X A 10,000) E(X ,ιηηηηΓι *■ /Ίηΐο,οοο-β.9θ7βΜ +10,000 [1 - Φ ^ '15174 )\ _ A 1,000) β 6.9078+1.5174 2 /2φ / i n 1,000-6.9078-1.5174 2 \ +1,000 [ ΐ - φ ( | η 1 ' 0 1 ° ^ 7 6 4 9 0 7 8 ) ] 6 8.059 φ ( 0 ) + 8059 ε xo^oQj! _ φ(ΐ,5ΐ74)] Φ(-1.5174) + 1,000[1 - Φ(0)] 3,162(0.5) + 10,000(0.0647) = 3.162. 3,162(0.0647) + 1,000(0.5) φ This year, the probability of exceeding 1,000 is Pr(X > 1,000) = 1 /i„i ,000-6.9078 \ = 0 5 W i t h 1 Q % i n f l a t i o i l ) t h e distribution is lognormal with parameters μ = 6.9078 + In 1.1 = 7.0031 and μ = 1.5174. The probability is 1 - Φ ('nl'01°577740031) = 1 - Φ(-0.0628) = 0.525, an increase of SECTION 8.4 59 5%. Alternatively, the original lognormal distribution could be used and then PT(X > 1,000/1.1) computed. 8.13 The desired quantity is the expected value of a right truncated variable. It is $1 °°°xf(x) F(1,000) E(X A 1,000) - 1,0QQ[1 - F(1,000)] E(X A 1,000) - 400 " F(1,000) ~ 0.6 = From the loss elimination ratio, E(X A 1,000) _ E(X A 1,000) E(X) ~ 2,000 and so E(X A 1,000) = 600, making the answer 200/0.6 = 333. 8.3 SECTION 8.4 8.14 From Exercise 3.9 we have E(X A 150,000) = 30,000[1 - e" 0 · 00001 ^ 50 ' 000 )] = 23,306.10. After 10% inflation the expected cost is 1.1E(X Λ 150,000/1.1) - 33,000[1 - e" 0 0 0 0 0 ^ 1 5 0 ' 0 0 0 / 1 1 )] = 24,560.94, for an increase of 5.38%. 8.15 From Exercise 3.10 we have θ+d 100 + d Therefore, the range is 100 to infinity. With 10% inflation, Θ becomes 110 and the mean residual life is ey(d) = 110 + d. The ratio is joo+d · ^ s ^ increases, the ratio decreases from 1.1 to 1. The mean residual life is ez(d) = fi°°{x - d)f(x)dx + Ζ~ ο (500 - ΓΤ^) d)f(x)dx This function is a quadratic function of d. It starts at 83.33, increases to a maximum of 150 at d = 200, and decreases to 0 when d = 500. The range is 0 to 150. 60 8.4 CHAPTER 8 SOLUTIONS SECTION 8.5 8.16 25 =E(X) = f™(l-x/w)dx Then /•50 = w/2 for w = 50. 5(10) = 1-10/50 = 0.. E(Y) = / (a;-10)(l/50)dx/0.8 = 20, E(y 2 ) = Var(F) = / (x - 10) 2 (l/50)d:r/0.8 = 533.33, Jio 533.33 - 202 = 133.33. 8.17 The bonus is B = 500,000(0.7 - L/500,000)/3 - (350,000 - L)/3 if positive. The bonus is positive provided L < 350,000. The expected bonus is E(B) = - / ,•350,000 (350,000 - l)fL(l)dl 3 Jo ^{350,000FL(350,000) - E(L Λ 350,000) {350,000FL (350,000) - E(L . +350,000(1 - FL(350,000)]} 600,000 \ 2 ' |35o,ooo - ««^22 950,000 ) . I = 56,556. 8.18 The quantity we seek is 1.1[E(X Λ 22/1.1) - E(X Λ 11/1.1)] 1.1(17.25-10) _ ~ 18.1 - 10.95 " ^ 1 1 5 ' E(X Λ 22) - E(X Λ 11) 8.19 This exercise asks for quantities on a per-loss basis. The expected value is E(X) - E(X A 100) - 1,000 - 1,000(1 - e" 100 / 1 ' 000 ) = 904.84. To obtain the second moment, we need Ε[(ΧΛ100) 2 ] = = = r-100 /»1UU \2 -100/1,000 / x20.001e-°-°°la!da: + (100)V Jo - e - 0 0 0 1 x ( a ; 2 + 2,000x + 2,000,000) |J°° + 9,048.37 9,357.68. SECTION 8.5 61 The second moment is E(X2) - E[(X Λ 100)2] - 200E(X) + 200E(X Λ 100) = 2(1,000)2 - 9,357.68 - 200(1,000) + 200(95.16) - 1,809,674.32, for a variance of 1,809,674.32 - 904.842 = 990,938.89. 8.20 Under the old plan the expected cost is 500/1 = 500. Under the new plan, the expected claim cost is K. The bonus is _ f 0.5(500-X), ~ { 0.5(500-500), X<500 X>500, which is 250 less a benefit with a limit of 500 and a coinsurance of 0.5. Therefore, E(B) 250 - 0.5E(X Λ 500) K2 K 250 The equation to solve is ~Ί + 2K + ι,οοο' 5 0 0 ^ + 250-f + ^ f l ^ and the solution is K = 354. 8.21 For year a, expected losses per claim are 2,000, and thus 5,000 claims are expected. Per loss, the reinsurer's expected cost is E(X) - E(X A 3,000) = 2,000-2,000 1 = 800, 2,000 2,000 + 3,000 and, therefore, the total reinsurance premium is 1.1(5,000)(800) = 4,400,000. For year b, there are still 5,000 claims expected. Per loss, the reinsurer's expected cost is 1.05[E(X) - E(X Λ 3,000/1.05)] J2,C 1= 1.05^2,000-2,000 = 864.706. 2,000 2,000 + 3,000/1.05 )]} 62 CHAPTER 8 SOLUTIONS and the total reinsurance premium is 1.1(5,000)(864.706) = 4,755,882. The ratio is 1.0809. 8.22 For this uniform distribution, pu E(XAu) = = = / J0 x(0.00002)dx + / /.50,000 Ju u(0.00002)dx O.OOOOlu2 + 0.00002u(50,000 - u) u- O.OOOOlix2. Prom Theorem 8.7, the expected payment per payment is E(X A 25,000) - E(X A 5,000) 1 - F(5000) 18,750 - 4,750 5,000 1 50.000 15,556. 8.23 This is a combination of franchise deductible and policy limit, so none of the results apply. From the definition of this policy, the expected cost per loss is /»ιυυ,υυυ / Λο,οοο xf(x)dx ,»100,000 + 100,000[1 - F(100,000)] Λ50,000 = / xf{x)dx + 100,000[1 - F(100,000)] - / xf(x)dx Jo Jo = E(X A 100,000) - E(X A 50,000) + 50,000[1 - F(50,000)]. Alternatively, it could be argued that this policy has two components. The first is an ordinary deductible of 50,000 and the second is a bonus payment of 50,000 whenever there is a payment. The first two terms in the preceding formula reflect the cost of the ordinary deductible and the third term is the extra cost of the bonus. Using the lognormal distribution, the answer is . / I n 100,000 - 10 - 1 . , ^ . In 100,000 - 10 ε 10 · 5 Φ ■— + 100,000 1 - Φ ' _ 6 ιο.5 φ / Ί η 5 0 , 0 0 0 - 1 0 - 1 1 = ε10·5Φ(0.513) + 100,000[1 - Φ(1.513)] - β 10 · 5 Φ(-0.180) = e 105 (0.6959) -h 100,000(0.0652) - e 105 (0.4285) - 16,231. 8.24 E ( r L ) = E ( X ) - E ( X Λ 30,000) = 10,000 - 10,000(1 - e" 3 ) = 497.87. E(YP) = 497.87/e" 3 = 10,000. SECTION 8.5 E[(F L ) 2 ] = = 63 E(X 2 ) - E[(X A 30,000)2] - 60,000E(X) +60, OOOE(X Λ 30,000) 2(10,000)2 - 10,0002(2)Γ(3; 3) - 30,000 2 e" 3 - 60,000(10,000) +60,000(9,502.13) 9,957,413.67. E[(y p ) 2 ] - 9,957,413.67/e" 3 - 200,000,000. Then Var(y L ) = 9,957,413.67497.872 = 9,709,539.14, and the cv is 9,709,539.141/2/497.87 = 6.259. For Yp, the variance is 100,000,000 and the cv is 1. 8.25 The density function over the respective intervals is 0.3/50 = 0.006, 0.36/50 - 0.0072, 0.18/100 = 0.0018, and 0.16/200 - 0.0008. The answer is /»50 / Jo /.100 Λ200 x2(0.006)dx + / x 2 (0.0072)dx+ / χ2(0.0018)ώζ Λο Λοο /»350 + / 7200 /»400 χ 2 (0.0008)ώτ+ / 7350 3502(0.0008)dx = 20,750. 8.26 The 10,000 payment by the insured is reached when losses hit 14,000. The breakpoints are at 1,000, 6,000, and 14,000. So the answer must be of the form aE{X A 1,000) + bE(X A 6,000) + cE(X A 14,000) + dE(X). For the insurance to pay 90% above 14,000, it must be that d = 0.9. To pay nothing from 6,000 to 14,000, c + d = 0, so c = -0.9. To pay 80% from 1,000 to 6,000, b + c + d = 0.8, so b = 0.8. Finally, to pay nothing below 1,000, a + b + c + d = 0, so a = —0.8. The answer is -0.8(833.33) + 0.8(2,727.27) - 0.9(3,684.21) + 0.9(5,000) = 2,699.36. 8.27 The expected payment is λ = 3. With the coinsurance, the expected payment is e*3. For the deductible, E(X A 2) = (0)e~ 3 + (l)(3e - 3 ) + 2(1 - e" 3 - 3e~ 3 ) = 1.751065. The expected cost is 3 - 1.751065 = 1.248935. Then, a = 1.248935/3 0.4163. 64 CHAPTER 8 SOLUTIONS 8.28 T h e maximum loss is 95. The expected payment per payment is n U.o E(X Λ 95) - E(X Λ 20) „,^v F(20) X QC = U.o- ? 5 _ 0.8 8.5 " ϊ θ ο ϋ ^ ~ Io° λ ~ 202 1 - 10.000 Wmdx 95^201 0 3 000) ζ =38.91. SECTION 8.6 8 . 2 9 (a) For frequency, the probability of a claim is 0.03, and for severity, the probability of a 10,000 claim is 1/3 and of a 20,000 claim is 2 / 3 . (b) P r ( X = 0) = 1/3 and P r ( X = 10,000) = 2 / 3 . (c) For frequency, the probability of a claim is 0.02, and the severity distribution places probability 1 at 10,000. 8.30 v = [1 - F ( l , 0 0 0 ) ] / [ 1 - F(500)] - 9/16. T h e original frequency distribution is P - E T N B with λ = 3, r = - 0 . 5 , and β = 2. T h e new frequency is P - Z M N B with λ = 3, r = - 0 . 5 , β = 2(9/16) = 9 / 8 , and p$ = 0-3 '+(i7/8)^-o(i7/8) r> = 0.37472. This is equivalent to a P - E T N B with λ = 3(1 - 0.37472) = 1.87584, β = 9 / 8 , and r = - 0 . 5 , which is P-IG with λ = 1.87584 and β = 9 / 8 . T h e new severity distribution has cdf F(x) (1,500+^ ) ' w h i c h is P a r e t o w i t h a = 2 a n d Θ = F T ( +^j~0^5Q0) = 1 - = 1,500. 8.31 T h e value of p$J will be negative and so there is no appropriate frequency distribution to describe effect of lowering the deductible. 8.32 We have PNP(Z) = ΡΝ(1-υ + υζ) - { l - ( l - v + vz)}-r r — (v — vz)~ -v'r{\-z)-r -v-r[l-PNL(z)} -v~r Therefore, + Pr(7V p = 0) = 1 - irrPNL(z). v~7\ SECTION 8.6 65 and Ρτ(Νρ = n) = [1 - Pr (Np = 0)] Pr (NL = n) ; n = 1,2,3,... . 8.33 Before the deductible, the expected number of losses is rß = 15. From the Weibull distribution, 5(200) = exp[-(200/1,000) 0 · 3 ] = 0.5395. The expected number of payments is 15(0.5395) — 8.0925. 8.34 5(20) = 1 - (20/100) 2 = 0.96. The expected number of claims without the deductible is 4/0.96 = 4.1667. SECTION 8.6 65 and Ρτ(Νρ = n) = [1 - Pr (Np = 0)] Pr (NL = n) ; n = 1,2,3,... . 8.33 Before the deductible, the expected number of losses is rß = 15. From the Weibull distribution, 5(200) = exp[-(200/1,000) 0 · 3 ] = 0.5395. The expected number of payments is 15(0.5395) — 8.0925. 8.34 5(20) = 1 - (20/100) 2 = 0.96. The expected number of claims without the deductible is 4/0.96 = 4.1667. CHAPTER 9 CHAPTER 9 SOLUTIONS 9.1 SECTION 9.1 9.1 The number of claims, AT, has a binomial distribution with m = number of policies and q — 0.1. The claim amount variables, X\,X<i,..., all are discrete with PT(XJ = 5,000) = 1 for all j . 9.2 (a) An individual model is best because each insured has a unique distribution. (b) A collective model is best because each malpractice claim has the same distribution and there are a random number of such claims, (c) Each family can be modeled with a collective model using a compound frequency distribution. There is a distribution for the number of family members and then each family member has a random number of claims. Student Solutions Manual to Accompany Loss Models: From. Data to Decisions, Edition. By Stuart A. Klugman, Harry H. Panjor, Gordon E. Willmot Copyright © 2012 John Wiley L· Sons, Inc. Fourth67 68 CHAPTER 9 SOLUTIONS 9.2 SECTION 9.2 9.3 E(iV) P^(z) P< x) (l) = = = P^(l), aQ(z)a~lQ^(z), aQW'-'LQWMota. 9.4 The Poisson and all compound distributions with a Poisson primary distribution have a pgf of the form P(z) = exp{X[P2(z) — 1]} = [<5(ζ)]λ, where Q(z) = exp[P 2 (z) - 1]. The negative binomial and geometric distributions and all compound distributions with a negative binomial or geometric primary distribution have P(z) = {1-β[Ρ2(ζ) - \}}-r = [Q(z)\r, where Q(z) = {1 - β[Ρ2(ζ) - l ] } " 1 . The same is true for the binomial distribution and binomial-X compound distributions with a = m and Q(z) = 1 + q[P2(z) — 1]. The zero-truncated and zero-modified distributions cannot be written in this form. 9.3 SECTION 9.3 9.5 To simplify writing the expressions, let Njp = μΝί = Nj = μΝί = XJP = Xj = Ε(Νή, Ε[(Ν-Ν1ρ)*], μχι=Ε(Χή, μΧί = Ε[(ΧΝ-Χ1ργ], and similarly for S. For the first moment, P^\z) so E(S) = Pg)(l) = P$\l)P$\l) = = P^)[Px(z)]Px1)(z), P$)[Px(l)]Plx\l) = (Nlp)(Xlp) = E(N)E(X). For the second moment use Pf(l) = = S2p-Slp = P%)[Px(z)]{Px1)(z)f + PlJ)lPx{z)}Px2)(z) 2 (N2p-Nlp)(Xpl) + Nlp(X2p-Xlp). Var(S) = = 52 = S2p - (Sip)2 = S2p - Sip + Sip - (Sip)2 (N2p - Nlp)(Xlp)2 + Nlp(X2p - Xlp) + (Nlp)(Xlp) -(Nlp)2(Xlp)2 Nlp{X2p - (Xlp) 2 ] + [N2p (Nlp)2](Xlp)2 (Nlp)(X2) + (N2)(Xlp)2. = = and SECTION 9.3 For the third moment use Pf = S3p-3S2p = = = P$){Px{z)}[Px1)(z)f +3P{2)[Px(z)}Px1)(z)Px2){z) + P^[Px(z)]P^\z) (N3p-3N2p + 2Nlp)(Xlp)3 +3(N2p - Nlp)(Xlp)(X2p - Xlp) +Nlp(X3p - 3X2p + 2Xlp). = S3 + 2Slp = S3p-3(S2p){Slp) + 2(Slp)3 S3p - 3S2p + 2S\p + 3 [52 + (51p) 2 ](l - 35lp) + 2(51p) 3 (N3p - 3N2p + 2Nlp)(Xlp)3 +3(N2p - Nlp)(Xlp)(X2p - Xlp) +Nlp(X3p - 3X2p + 2Xlp) +3{Nlp[X2p - (Xlp)2} +[N2p-(Nlp)2}(Xlp)2 +(Nlp)2(Xlp)2}[l 3(Nlp)(Xlp)] +2(Nlpf{Xlpf = 9 6 Nlp(X3) + 3(N2)(Xlp)(X2) + N3(Xlp)3. E(X) = 1,000 + 0.8(500) = 1,400. Var(X) = = = = = = Vax(X1) + Vai(X2)+2Cov(X1,X2) 5002 + 0.64(300)2 + 2(0.8) (100,000) 467,600. E{N)E(X) = 4(1,400) = 5,600. £(JV)VarpO+Var(JV)E(X) 2 4(467,600) + 4(1,400)2 = 9,710,400. E(5) Var(5) 9.7 E(5) = 15(5)(5) = 375. Var(5) = 15(5)(100/12) + 15(5)(6)(5) 2 = 11,875. StDev(5) = 108.97. The 95th percentile is 375 + 1.645(108.97) = 554.26. 9.8 Var(AT) Ε(λ) Var(A) Var(JV) = = = = E[Var(JV|A)] + Var[E(JV|A)] = Ε(λ) + Var(A), 0.25(5) + 0.25(3) + 0.5(2) = 3, 0.25(25) + 0.25(9) + 0.5(4) - 9 = 1.5, 4.5. 69 70 CHAPTER 9 SOLUTIONS 9.9 T h e calculations appear in Table 9.1. Results for Exercise 9.9. Table 9.1 X 0 1 2 3 4 5 6 h(x) h(x) 0.9 0.1 Mx) h+*(?) 0.45 0.32 0.21 0.02 0.5 0.3 0.2 0.25 0.25 0.25 0.25 fs(x) 0.1125 0.1925 0.2450 0.2500 0.1375 0.0575 0.0050 9 . 1 0 T h e calculations appear in Table 9.2. Table 9.2 X 0 1 2 3 4 5 6 7 h{x) h{x) 0.6 0.2 0.1 0.1 0.25 0.25 0.25 0.25 Results for Exercise 9.10. h+z{x) 0.150 0.200 0.225 0.250 0.100 0.050 0.025 h(x) 1-p P 0.06 = / 5 ( 5 ) - 0.1 - 0.05p, p = 0.8. 9 . 1 1 If all 10 do not have AIDS, E(5) - 10(1,000) - 10,000, Var(S) = 10(250,000) = 2,500,000, and so the premium is 10,000 + 0.1(2,500,000) 1 / 2 = 10,158. fs(x) 0.15p 0.15 + 0.05p 0.2 + 0.025p 0.225 + 0.025p 0.25-0.15p 0.1-0.05p 0.05 - .025p 0.025 - 0.025p SECTION 9.3 71 If the number with AIDS has the binomial distribution with m = 10 and q = 0.01, then, letting N be the number with AIDS, E(5) Var(S) = E[E(5|7V)] = E[70,000N + 1,000(10 - TV)] = = = 10,000 + 69,000[10(0.01)] 16,900, Var[E(S|iV)] + E[Var(S|iV)] Var[70,000iV + 1,000(10 - N)] +E[1,600,00(W + 250,000(10 - N)\ 69,0002[10(0.01)(0.99)] + 2,500,000 + 1,350,000[10(0.01)] 473,974,000, = = and so the premium is 16,900 + 0.1(473,974,000) ^ 2 - 19,077. The ratio is 10,158/19,077 = 0.532. 9.12 Let M be the random number of males and C be the number of cigarettes smoked. Then E(C) = E[E(C|M)] = E[6M + 3(8 - M)] = 3E(M) + 24. But M has the binomial distribution with mean 8(0.4) = 3.2, and so E(C) = 3(3.2) 4- 24 = 33.6. Var(C) = = = E[Var(C|Af)] 4- Vax[E(C|M)] E[64M + 31(8 - M)] + Var(3M + 24) 33E(M) + 248 + 9 Var(M) 33(3.2) + 9(8)(0.4)(0.6) = 370.88. The answer is 33.6 -h V370.88 = 52.86. 9.13 For insurer A, the group pays the net premium, so the expected total cost is just the expected total claims, that is, E(5) = 5. For insurer B, the cost to the group is 7 — D, where D is the dividend. We have f Ik - 5, \ 0, Then E(D) = / 0 (7fe - s)(0A)ds 7 - 2.45/c2, and so fc = 0.9035. S < 7fe, S > 7k. = 2.45fe2. We want 5 = E(7 - D) = 72 CHAPTER 9 SOLUTIONS 9.14 Let Θ be the underwriter's estimated mean. The underwriter computes the premium as /»DO 2E[(5 - 1.250)+] (s - 1.250)0 _ 1 β- Α / 0 ώ = 2 / = -2(s - 1.25θ)ε-^θ I CXD - 2θε-*Ιβ\ 11.250 = 2öe" 1 · 2 5 . Let μ be the true mean. Then Θ = 0.9/x. The true expected loss is /»OO E[(5 - 1.25(0.9)/x)+] = 2 / Λ.125μ (s - 1.125μ)μ- 1 6- β / μ ώ = / l e " 1 1 2 5 . The loading is 2(0.9)^·^ _1 = Q5885 μ6-1.125 9.15 A convenient relationship is, for discrete distributions on whole numbers, E[(X -d1)+] = E[(X - d)+] - 1 + F(d). For this problem, E[(X - 0)+] = E(X) = 4. E [ ( X - 1 ) + ] = 4 - 1 + 0 . 0 5 = 3.05, E [ ( X - 2 ) + ] = 3 . 0 5 - 1 + 0 . 1 1 = 2.16, E[(X - 3)+] = 2.16 - 1 + 0.36 = 1.52. Then, by linear interpolation, d = 2 + (2 - 2.16)/(1.52 - 2.16) = 2.25. 9.16 15 = J~ 0 [l - F(s)]ds, 10 = /~ 0 [1 - F(s)]ds, F(120) - F(80) = 0. Subtracting the second equality from the first yields 5 = f100 [1 — F(s)]ds, but over this range F(s) = F(80), and so /»120 />120 5= / [1 - F(s)]ds = / [1 - F(80)}ds = 20[1 - F(80)] Λοο Λοο and, therefore, F(80) = 0.75. E(A) E(B) = / ( | - 5θ) (0.01)dx = ( | r - δθχ^ (0.01) = 5 0 Α Γ 1 - 5 0 + 12.5/c. = / Jo rioo kx(0.01)dx= , 2 — (0.01) ^ o = 50fc. The solution to 50ÄT1 - 50 + 12.5/e = 50fc is k = 2/3. SECTION 9.3 73 9.18 E(X) = 440, F(30) = 0.3, f(x) = 0.01 for 0 < x < 30. E(benefits) = / /»30 Jo /»oo 20x(0.01)dx + / J30 /»OO [600 + 100(:r - 30)]/(x)da = 90 + / = 90 - 2,400[1 - F(30)] + 100 / Jo J30 + 100 / (-2,400)f(x)dx /»OO xf(x)dx J30 /»OO xf(x)dx /»30 -100 / Jo = 9 ·19 E(S) xf(x)dx /»30 90 - 2,400(0.7) + 100(440) - 100 / x(0.01)da; Jo 90 - 1,680 + 44,000 - 450 - 41,960. E(N)E(X) = [0(0.5) + 1(0.4) + 3(0.1)] [1(0.9) + 10(0.1)] 0.7(1.9) = 1.33. = = We require P r ( 5 > 3.99). Using the calculation in Table 9.3, P r ( 5 > 3.99) = 1 - 0.5 - 0.36 - 0.0729 = 0.0671. Table 9.3 Calculations for Exercise 9.19. X fx°(x) fx H*) 0 1 2 3 1 0 0 0 0 0.9 0 0 0 0 0.81 0 0 0 0 0.729 0.5 0.4 0 0.1 Pn Γχ\χ) Γχ\χ) fs(x) 0.5000 0.3600 0 0.0729 9.20 For 100 independent lives, E(S) = lOOmq and Var(S) = I00m2q(l-q) 250,000. The premium is lOOmq -f 500. For this particular group, E(5) Var(S) = = 97 (mq) + (Sm)q = lOOmq, 97m2q(l - q) + (3m) V 1 - q) 106m 2 ^(l -q)= 265,000, and the premium is lOOmq + 514.18. The difference is 14.78. 9.21 E(5) Var(S) = = 1(8,000)(0.025) + 2(8,000) (0.025) = 600, 1(8,000)(0.025)(0.975) +22(8,000)(0.025)(0.975) = 975. 74 CHAPTER 9 SOLUTIONS The cost of reinsurance is 0.03(2)(4,500) = 270. Pr(5 + 270 > 1,000) = Pr(5 > 730) '5-600 730-600 Pr > /975 /975 = 4.163 so K = 4.163. 9.22 E(Z) = / /•100 Λο 0.8(j/-10)(0.02)(l-O.Ol2/)dj/ /•100 3.016 / Jio 9.23 Pr(S > 100) = - 0 . 01y2 + I.ly-I0dy= 19.44. ] T Pr(7V - n) Pr(X* n > 100) n=0 = 0.5(0) +0.2 P r ( X > 100) +0.2Pr(X* 2 > 100) +0.1Pr(X* 3 > 100). Because X ~ 7V(100,9), X* 2 ~ N(200,18), and X* 3 ~ AT(300,27), and so Pr(S>100) 0.2Pr(Z>100-100Uo.2PrrZ>100-200 = Vis 100 - 300 +0.2 Pr [Z> \/27 0.2(0.5)+ 0.2(1)+ 0.1(1) = 0.4. 9.24 The calculations are in Table 9.4. The expected retained payment is 2,000(0.1) +3,000(0.15) +4,000(0.06) + 5,000(0.6275) = 4,027.5, and the total cost is 4,027.5 + 1,472 = 5,499.5. Table 9.4 /*(*) Pk Calculations for Exercise 9.24. Sx(x) fs(x) fx(x) 0 0 0 0 1 0 0 0 0 0 0 0.4 0.6 0 0.16 1/16 1/4 3/8 0.0625 0.0000 0 .1000 0.1500 0.0600 SECTION 9.3 75 9.25 In general, paying for days a through 6, the expected number of days is b J2(k - a + l)pk + (b - a 4-1)[1 - F(b)] k=a b k k=aj=a = έ Σ > + (&-«+ΐ)[ΐ-*χ*)] j=a k=j b = Yl[F(b) - F(j - 1)] + (6 - a + 1)[1 - F(b)} = ^[l-F(j-l)] = ^(0.8r1 (0.8)α~ι - (0.8)b .2 For the 4 through 10 policy, the expected number of days is (0.83—0.810)/0.2 = 2.02313. For the 4 through 17 policy, the expected number of days is (0.8 3 — 0.8 17 )/0.2 = 2.44741. This is a 21% increase. 9.26 oo / x3x~4dx = 3/2, x23x~4dx Var(S) = = 3, 3 - (3/2) 2 = 3/4. r (l+0)(3/2) = 3x~4dx l - [ ( l + 0)(3/2)]- 3 , and so Θ = 0.43629. 1.5+Ay^74 / = and so λ = 0.75568. 1-[1.5 + λΛ/3/4]-3, 3x~*dx 76 CHAPTER 9 SOLUTIONS 9.27 The answer is the sum of the following: 0.807?ioo pays 80% of all in excess of 100, 0.10i?noo pays an additional 10% in excess of 1,100, and 0.10i?2ioo pays an additional 10% in excess of 2,100. 9.28 PT(S = 0) = ΣΓ=οΡηΡΓ(£ η = 0) where Bn ~bin(n,0.25). Thus 00 Pr(5 p -2on ^ " ( 0 · 7 5 Γ = e - V · 5 = e" 1 / 2 . = 0) = ^ n=0 9.29 Let μ be the mean of the exponential distribution. Then, E(5) = = Ε[Ε(5|μ)] Ε(μ) 3,000,000. 9.30 E(N) = 0.1. Λ30 E(X) = / Jo = / r10 + / E{S) = = 100£/(£)d£ + 3,000Pr(T>30) 100ί(0.04)ώ+ / /■30 /.20 100i(0.035)di 100i(0.02)dt + 3,000(0.05) 200 + 525 + 500 + 150 = 1,350. 0.1(1,350) = 135. 9.31 Total claims are compound Poisson with λ = 5 and severity distribution fx(x) = 0AMx)+0.6f2(x) ( 0.4(0.001) + 0.6(0.005) = 0.0034, \ 0.4(0.001) + 0.6(0) = 0.0004, 0 < x < 200, 200 < x < 1,000. SECTION 9.3 77 Then, E[(X-100)+] = / (x -100) 200 / (x - 100)(0.0034)ώ JlOO /»OO /»oo + / = fx(x)dx J200 (x-100)(0.0004)d£ 177. 9.32 The calculations appear in Table 9.5. Table 9.5 Calculations for Exercise 9.32. 5 PT(S = S) d dPr(S = s) 0 1 2 3 0.031676 0.126705 0.232293 0.258104 0.351222 4 3 2 1 0 0.12671 0.38012 0.46459 0.25810 4+ Total 0 1.22952 9.33 The negative binomial probabilities are po = 0.026, p\ = 0.061, and p2 = 0.092, and the remaining probability is 0.821. The mean is 6, and so the total expected payments are 600. The expected policyholder payment is 0(0.026) + 100(0.061) + 200(0.092) + 300(0.821) = 270.80. The insurance company expected payment is 600 — 270.80 = 329.20. 9.34 Because the only two ways to get 600 in aggregate claims are one 100+500 and six 100s, their probabilities are —^-2(0.80)(0.16) = 0.02156 and — — (0.80)6 = 0.03833 for a total of 0.05989. The recursive formula can also be used. 9.35 The moments are E(N) = 16(6) - 96, Var(AT) = 16(6)(7) = 672, E(X) = 8/2 = 4. Var(X) = 8 2 /12 = 5.33, E(S) = 96(4) = 384, Var(S) = 96(5.33) + 672(4)2 = 11,264. The 95th percentile is, from the normal distribution, 384+ 1.645(ll,264)1/2 = 558.59. 78 CHAPTER 9 SOLUTIONS 9.36 The moments are (noting that X has a Pareto distribution) E(N) = 3, Var(iV) = 3, E(X) = 2/(3 - 1) = 1. Var(X) = 2 2 2/[(3 - 1)(3 - 2)] - l 2 = 3, Var(S) - 3(3) + 3(1) 2 - 12. 9.4 SECTION 9.4 9.37 Because this is a compound distribution defined on the nonnegative integers, we can use Theorem 7.4. With an appropriate adaptation of notation, PN{Px(z;ß)) = PN{Px(z);ß[l-fx(0)}}. So just replace ß by ß* = ß[l 9.38 (a) fq(x) fx(0)]. ßn £ = V oo ln(l + β) fs(x) — ;^n(l+/3)"ln(l+/?)0n(n-l)! 1 (b) χη-1£-χ/θ - = ß θ(1 + β)\ n=l i-x»-ie-«/e. η! xß x\n(l+ß) l·** [θ(1+β)\ e~x/9 v -χ/θ ' n=l 1 η\ xß xln(l + /3) \ exp*[θ(1 + β) exp -} — exp (-1) θ(1 + β) xln(l + /3) 9.39 To use (9.15) we need the binomial probabilities po = 0.63 = 0.216, Pi = 3(_0.6)20.4 = 0.432, p2 = 3(0.6)0.42 =_ 0.288, and p3 = 0.43 - 0.064. Then, P 0 = 0.432 + 0.288 + 0.064 - 0.784, ΡΎ = 0.288 -f 0.064 = 0.352, and P2 = 0.064. Then F 5 (300) = 1-e -300/100 0.89405. 0.784 ( 3 0 0 Λ 0 0 ) ° + 0 . 3 5 2 ( 3 0 0 / 1 0 0 ) 1 1! 0! (300/100) 2 +0.064 2! 9.40 For the three policies the means are 10,000(0.01) = 100, 20,000(0.02) = 400, and 40,000(0.03) = 1,200. Because the Poisson variance is equal to the SECTION 9.6 79 mean, the variances are 10,0002(0.01) = 1,000,000, 20,0002(0.02) = 8,000,000, and 40,000(0.03) = 48,000,000. The overall mean is 5,000(100) +3,000(400) + 1,000(1,200) = 2,900,000 and the variance is 5,000(l,000,000)+3,000(8,000,000)+ 1,000(48,000,000) = 7,7 x 10 10 . Total claims are the sum of 9,000 compound Poisson distributions which itself is a compound Poisson distribution with A = 5,000(0.01) + 3,000(0.02) + 1,000(0.03) = 140 and the severity distribution places probability 50/140 on 10,000, 60/140 on 20,000, and 30/140 on 40,000. Using the recursive formula with units of 10,000, 0) = e" -140 Pr(5 Pr(5 = Pr(5 = l) = ^ A e - 1 4 0 = 50e- 1 4 0 , 1 14 140 A 5 0 e- 1 4 0 + 2 6 -140 2) = i4 ' 14 .>-? A14l , 3 1 0 e - Pr(5 Pr(5 9.5 > o L 140 J £± - l,310e- 1 4 0 , + 2^-50e- 1 4 0 + 3 ^ - e - 1 4 0 = 24,196.67e- 140 , 14 14 3) = 1 - 25,557.67e~140 SECTION 9.6 9.41 Ak (fe+1>h mn x-kh-h f(x)dx -h Jkh kh t-κη MK+ijft = - rdx+ Tdx + (k + l){F[(k + Jo h J0 h E[X A(k + l)h] + (k + 1){1 - F[(k + l)h]} h l)h}-F(kh)} + τ Ε ( Χ Λ kh) - k[l - F(kh)} +(k + l){F[{k + = | E ( X Akh)-^E[X h a ,(fc.fe+1 m\ l)h}-F(kh)} = = < Jkh |E[X h >h X - kh h A(k + l)h] + l-F{kh), f{x)dx Λ (fc + l)h] - | Ε ( Χ Λ kh) - 1 + F[(k + l)h}. h 80 CHAPTER 9 SOLUTIONS For k = 1,2,... , m\-l+ml fk = = A kh) - \E[X \E(X ft ft +\nx ft ft = A{k- A (k + l)ft] 4- 1 - A kh) - \E\X \{2Έ(Χ h l)ft] - 1 + F(kh) A kh) - E[X A(k- F(kh) l)ft] - E[X A (k + l)h]}. Also, /o = m§ = 1 — E(X A h)/h. All of the rrij have nonnegative integrands and, therefore, all of the fk are nonnegative. To be a valid probability function, they must add to one, oo - 1 Λ + Σ Λ = oo 1-^Ε(ΧΛ/ι) + -^{Ε(ΧΛΑ:/ι)-Ε[ΧΛ(Α:-1)/ι]} fc=l fc=l i °° + - ] Γ { Ε ( Χ Λ /eft) - E[X Λ (fc 4- l)h}} k=\ = 1 - yE(X A h) + | E ( X Λ/ι) = 1, AI ft because both sums are telescoping. The mean of the discretized distribution is J T ftfc/fc = k=l ] T fc{E(X Λ /eft) - E[X A(k- l)h}} k=l OO + Σ kiE(X Λ kh ) - EiX Λ (fc + l)ft]} oo = E(X Λ ft) + 5^(jfc 4 1){E[X Λ (k + l)ft] - E(X Λ (jfeft)} oo + 5 3 ME(X Λ kh) - E[X A (k 4 l)h]} fc=l oo = E(X Ah) + J2{E[X A(k+l)h}-E{XA(kh)} k=\ = E(X) because E(X A oo) - E(X). 9.42 Assume x = 1. Then g0 = exp[-200(l - 0.76)] = exp(-48). recursive formula gives 9k = ^ ( 0 . 1 4 ^ _ ! 4- 0.10<?fe_2 4- 0.06^_3 + 0.12^_ 4 ) The SECTION 9.6 81 with gk = 0 for k < 0. Now use a spreadsheet to recursively compute probabilities until the probabilities sum to 0.05, which happens at k = 62. Then 62x = 4,000,000 for x = 64,516. The expected compensation is 200(0.14 + 0.10 + 0.06 + 0.12)(64,516) - 5,419,344. 9.43 (a) PN(z) = wP^z) + (1 ^ Ps(z) = w)P2(z). PN(PX(Z)) = wPi(Px(z)) = wPSl(z) + + (l-w)P2(Px(z)) (l-w)PS2(z), fs(x) = wfSl + (1 - w)fs2(x), Hence, first use (9.22) to compute fs1(x). the results. x = 0,1,2,... . Then take a weighted average of (c) Yes. Any distributions P\(z) and P2(z) using (9.22) can be used. 9.44 Prom (9.22), the recursion for the compound Poisson distribution, fs(0) = e~\ 5 5 fs(x) = -Y]yfx(y)fs(x x *-** - y)- Then /5(l) = 5/x(l)e-5, and so / χ ( 1 ) = \ since / s ( l ) = e - 5 . Then, fs(2) = | [ / x ( l ) / s ( l ) + 2/x(2)/ s (0)] = | [ I e - 5 + 2/x(2)e- 5 ] and, since /s(2) = §e~ 5 , we obtain fx(2) = | . 9.45 fs(7) = f [ l / x ( l ) / s ( 6 ) + 2 / x ( 2 ) / s ( 5 ) + 4 / x ( 4 ) / s ( 3 ) ] . Therefore, 0.041 = f [£/s(6) + §0.0271 + §0.0132] for / s ( 6 ) = 0.0365. 82 CHAPTER 9 SOLUTIONS 9.46 From (9.21) with /χ(0) = 0 and x replaced by z: M 1V1 (a + + l^)/χ(»)/5(«-ν) fs(z) = Σ ( α Zs M-l £ (a + bZ)fx(y)fs{z-v) »=i + (a + 6 ^ ) / x ( M ) / s ( 2 - M ) . Let z = x + M, /s(i + W) = Σ {a + b^-^)fx(y)fs(x + M-y) Rearrangement gives the result. (b) The maxmimum number of claims is m and the maximum claim amount is M. Thus, the maximum value of S is mM with probability [qfx(M)]m. This becomes the starting point for a backward recursion. 9.47 We are looking for 6 - E(5) - E(£>). 7 4 E(X) = 1(1) + f(2) = -A and E(5) = 1(2) = | . D /s(0) = Λ .5 -S, 0, S < 4.5 5>4.5. e" 2 , /s(l) = | 1 Κ 2 = Κ 2 > fs(2) = | ( l I l e - 2 + 2fe- 2 ) = f e - 2 , / S (3) = | ( l I f e - 2 + 2fIe-2) = I e - 2 , Λ(4) = I ( l | I e - 2 + 2ffe- 2 ) = i|e- 2 . Then E(D) = (4.5 + 3.5± + 2 . 5 f + 1.5f| + 0 . 5 § | ) e - 2 = 1.6411. The answer is 6 - 3.5 - 1.6411 = 0.8589. 9.48 For adults, the distribution is compound Poisson with λ = 3 and severity distribution with probabilities 0.4 and 0.6 on 1 and 2 (in units of 200). For SECTION 9.6 83 children it is λ = 2 and severity probabilities of 0.9 and 0.1. The sum of the two distributions is also compound Poisson, with λ = 5. The probability at 1 is [3(0.4) + 2(0.9)]/5 = 0.6 and the remaining probability is at 2. The initial aggregate probabilities are /s(0) = e-5, fs(l) = f l | e - 5 = 3e- 5 , / 5 (2) = |(l|3e-5 /s(3) = | ( l l f e - 5 + 2l3e-5) = fe- 5 , 2fe-5) = fe- 5 , + fs(4) = f(lff e -* + 2ffe-5) = ^ e - 5 . The probability of claims being 800 or less is the sum 1 + 3 + ^ + ^ + ^ ) e - 5 = 35.375e- 5 = 0.2384. 2 2 8 / 9.49 The aggregate distribution is 2 times a Poisson variable. The probabilities are P r ( 5 = 0) = e" 1 , Pr(5 = 2) = Pr(iV = 1) = e" 1 , P r ( 5 = 4) = Pr(AT = 2) = \e~\ E(D) = (6 - 0)e" 1 + (6 - 2)e~l + (6 - 4)±e~ 1 = l i e " 1 = 4.0467. 9.50 λ = 1 + 1 = 2, / χ ( 1 ) - [1(1) + l(0.5)]/2 = 0.75, / x ( 2 ) - 0.25. The calculations appear in Table 9.6. The answer is F£ 4 (6) = (81 +108+54)/256 = 243/256 - 0.94922. Table 9.6 Calculations for Exercise 9.50 X 0 1 2 3 4 5 6 fx°(x) 1 0 0 0 0 0 0 fxHx) 0 3/4 1/4 0 0 0 0 fx2(x) 0 0 9/16 6/16 1/16 0 0 fx3(x) 0 0 0 27/64 27/64 9/64 1/64 Γχ\χ) 0 0 0 0 81/256 108/256 54/256 9.51 56 = 29EpO, so E(X) = 56/29. 126 = 29£(X 2 ), so E(X 2 ) = 126/29. Let fi = Ρτ(Χ = Ϊ). Then there are three equations: / 1 + / 2 + /3 / i + 2 / 2 + 3/3 /i+4/2+9/3 = 1, = 56/29, = 126/29. 84 CHAPTER 9 SOLUTIONS The solution is f2 = 11/29. (Also, fx = 10/29 and / 3 - 8/29) and the expected count is 29(11/29) = 11. 9.52 Let fj = PT(X = j). 0.16 = Xfu k = 2λ/ 2 , 0.72 - 3λ/ 3 . Then /i = 0.16/λ and f3 = 0.24/λ, and so f2 = 1 - 0.16/λ - 0.24/λ. 1.68 = λ[1(0.16/λ) + 2(1 - 0.4/λ) + 3(0.24/λ)] - 0.08 + 2λ, and so λ = 0.8. 9.53 1 - F ( 6 ) = 0.04-0.02, F(6) = 0.98. 1 - F ( 4 ) = 0.20-0.10, F(4) = 0.90. P r ( 5 = 5 or 6) - F(6) - F(4) - 0.08. 9.54 For 1,500 lives, λ = 0.01(1,500) = 15. In units of 20, the severity distribution is Pr(X = 0) = 0.5 and Pr(X = x) = 0.1 for x = 1, 2,3,4, 5. Then E(X 2 ) = 0.5(0) + 0.1(1 + 4 + 9 + 16 + 25) = 5.5 and Var(S) = 15(5.5) - 82.5. In payment units it is 202(82.5) = 33,000. 9.55 Pr(iV = 2) = \ Pr(N = 2|class I) + f Pr(7V - 2|class II). Pr(7V - 2|class I) = Jo jrfßyildß = 0.068147. [Hint: Use the substitution y = ß/(l + /?)]. Pr(7V = 2|class II) - (0.25) 2 /(1.25) 3 = 0.032. Pr(7V = 2) = 0.25(0.068147) + 0.75(0.32) - 0.04104. 9.56 With E(iV) = 5 and E(X) = 5(0.6) + fc(0.4) - 3 + 0.4/c, E(5) = 15 + 2fc. With Pr(S = 0) = e~ 5 = 0.006738, the expected cost not covered by the insurance is 0(0.006738) + 5(0.993262) = 4.96631. Therefore, 28.03 = 15 + 2k - 4.96631 for k = 8.9982. 9.57 Using the recursive formula, Pr(S = 0) = e" 3 , Pr(5 = 1) = f (0.4)e~ 3 = 1.2e~3, P r ( 5 = 1) = f[0.4(1.2)e" 3 + 2(0.3)e" 3 ] = 1.62e"3, P r ( 5 = 3) = \[0.4(1.62)e-3+2(0.3)(1.2)e-3+3(0.2)e-3] = 1.968e"3. Pr(S < 3) = 5.788e~3 = 0.28817. 9.58 Using the recursive formula, Pr(S = 0) = (1 + 0.5)~ 6 = 0.08779. a = 0.5/(1 + 0.5) = 1/3, b = (6 - l)(l/3) = 5/3. SECTION 9.7 85 Pr(5 = 1) = (I + ^ Pr(5 = 2) = ( | + ^pj = 0.08545, = ( | + ^ ) (0.4)(0.08545) + Q + ^ ί Η ) (0.3)(0.07023) Pr(5 = 3) J (0.4)(0.08779) = 0.07023, (0.4)(0.07023) + ( | + W l \ + I i + ^ψ^- (0.3)(0.08779) ) (0.2)(0.08779) = 0.09593. Then, Pr(5 < 3) = 0.33940. 9.59 The result is 0.016. 9.60 The result is 0.055. Using a normal approximation with a mean of 250 and a variance 8,000, the probability is 0.047. 9.61 lfFx(x) = l-- e μχ ,χ > 0, then A-ft (*)-!-.-*. and for j = 1,2,3,. fi ·· 5 = = Fx IW>] i _ e~»h{i+h) - i -f .M = -Fx (1 - ßh\ β~μΗ) [('">] h e-» 0-2) „-ßh(j-l) (l-/o)(l-#^-\ ßh where φ = e~ . 9.6 SECTION 9.7 9.62 In this case, v = 1 - Fx(d) = e~^, and from (9.30), PNp(z) = B [0(1 -υ + υζ-1)] = Β [θν(ζ -1)] = Β [θε-μά(ζ - 1)] . Also, Yp = aZ, where Pr(Z >z) = Pr(X > z + d)/Pr(X > d) = e-rt*+*)/e-i* = e-"*. 86 CHAPTER 9 SOLUTIONS That is, FYp(y) = l - P r ( F F > y) = l - P r ( Z > y/a) = 1 - e~^. 9.63 (a) The cumulative distribution function of the individual losses is, from Appendix A, given by Fx(x) = 1 - (l + γ ^ ) e - " · ^ , x > 0. Also, E(X Ax) = 200r(3;^)+z[l-r(2;Ty].Then E(X Λ 175) E(X A 50) = = 200(0.25603) + 175(1 - 0.52212) = 134.835, 200(0.01439) + 50(1 - 0.09020) = 48.368, and the individual mean payment amount on a per-loss basis is E(YL) = E(X Λ 175) - E(X Λ 50) = 134.835 - 48.368 = 86.467. Similarly, for the second moment, E [(X Λ xf) = 60,000Γ (4; J L ) + , 2 [l - Γ ( 2 ; ^ ) ] , and, in particular, E [(X Λ 175)2] 2 E [(X A 50) ] - 60,000(0.10081) + 30,625(1 - 0.52212) = 20,683.645, - 60,000(0.00175) + 2,500(1 - 0.09020) = 2,379.587. Therefore, E[(rL)2] = = = E [{X A 175)2] - E [(X A 50)2] -100E(X Λ 175) + 100E(X Λ 50) 20,683.645 - 2,379.587 - 100(134.835) + 100(48.368) 9,657.358. Consequently, Var(F L ) = 9,657.358 - (86.467)2 = 2,180.816 = (46.70)2. For the negative binomially distributed number of losses, E(NL) = (2)(1.5) = 3 and Var(iVL) - (2)(1.5)(1 + 1.5) = 7.5. The mean of the aggregate payments is, therefore, E(5) - E(NL)E(YL) = (3)(86.467) = 259.401, and using Equation (9.9), the variance is Var(S) = E(7VL) Var(y L ) -f Var(7VL) [ E ( y L ) f = (3)(2,180.816) + (7.5)(86.467)2 = 62,616.514 = (250.233)2. SECTION 9 8 87 (b) The number of payments Np has a negative binomial distribution with r* = r = 2 and ß" = ß [1 - F x (50)] = (1.5)(1 - 0.09020) = 1.36469. (c) The maximum payment amount is 175 - 50 = 125. Thus FYp(y) = 1 for y > 125. For y < 125, FYp (y) = + 50) / P r ( X > 50) l-Pi(X>y (! + & ) « - * (d) /o = FYP(20)= 0.072105, /i /2 /3 = = = F y p (60) - Fyp (20) = 0.231664 - 0.072105 = 0.159559, F y p ( 1 0 0 ) - F y p ( 6 0 ) = 0.386868-0.231664 = 0.155204, Fyp(140) - Fyp (100) = 1 - 0.386868 = 0.613132. a = = = [1 + /?* (1 - 0.072105)]"2 = [1 + (1.36469)(1 - 0.072105)]" 2 0.194702 1.36469/2.36469 = 0.577112, 6 = (2 - 1) a = 0.577112. 9x = 3i g2 = = = = (e) 9o gz 9.7 9.64 Σΐ=ι (a + b%)fv9x-y n(tmiftQ'srf-.,V\f — : 7 = 0.602169 > 11 + - I fygx-y, i - aj0 £-j v XJ (0.602169) (2) (0.159559) (0.194702) = 0.037415, (0.602169) [(1.5)(0.159559)(0.037415) + (2)(0.155204)(0.194702)] 0.041786, 0.602169 [(4/3)(0.159559)(0.041786) + (5/3)(0.155204)(0.037415) +(2)(0.613132)(0.194702)] = 0.154953. SECTION 9.8 E and then E(IjBj\Ij) qj. Then n Ε W = Σ ^^) = j=i n Σ^^^Ι^)]' i=i = 0 with probability 1 — qj and = μ^ with probability Ε [ Ε ( / ^ · | / , ) ] = 0(1 - Qj) + μ^· = μ ^ , thus establishing (9.35). For the variance, n Var(S) = ^VaiiljBj) n = ^ V a r t E ^ B ^ J j ) ] +E[Var(J J -B i |J i )]. 88 CHAPTER 9 SOLUTIONS For the first term, Vav[E(IjB3\Ij)} = tf(l-qj) + v>qj-frjqj)* = 0 with probability 1 — qj and = σ | with For the second term, VS,T(IJBJ\IJ) probability qj. Then, E[V&T(IjBj\Ij)}=0(l-qj)+a2jqj. Inserting the two terms into the sum establishes (9.36). 9.65 Let 5 be the true random variable and let 5i, 52, and 53 be the three approximations. For the true variable, 2 J= l For all three approximations, Ε(5) = Χ;λΛ·, V a r ( 5 ) - ^ A ^ 2 . j=i i=i For the first approximation, with Aj = <7j it is clear that E(5i) = E(52), and because qj > qj{\ — qj), Var(5i) > Var(5). For the second approximation with Xj — — ln(l — qj), note that 2 " ln(l -Qj) = Qj + \ 3 + j + --'> Qj, and then it is clear t h a t E(52) > E ( 5 i ) . Because the variance also involves Xj, the same is true for the variance. For the third approximation, again using a Taylor series expansion, 2 Qj 3 4? 9? and, therefore, the quantities for this approximation are the highest of all. 9.66 For A, B E^5) = E(5) + 2SD(5) = 40(2) + 60(4) + 2^40(4) = 375.136. = = E^^jjvj] = Epjv + 60(10) + 4(100 - ΛΓ)] 400 - 2E(JV) = 400 - 2(40) = 320. SECTION 9.8 Var(S) = = 89 Var[E(S|]V)]+E[Var(S|iV)] Var(400 - 27V) + E[4AT + 10(100 - TV)] 4Var(AT)-6E(AT) + 1000 4(100)(0.4)(0.6) - 6(100)(0.4) 4-1,000 = 856. Therefore, A = 320 4- 2V856 - 378.515 and A/B = 1.009. 9.67 Premium per person is 1.1(1,000)[0.2(0.02) 4- 0.8(0.01)] = 13.20. With 30% smokers, E(5) Var(5) = = 1,000(0.3(0.02) 4- 0.7(0.01)] = 13, Ι,ΟΟΟ2 [0.3(0.02) (0.98) 4- 0.7(0.01)0(0.99)] = 12,810. With n policies, the probability of claims exceeding premium is Pr(5>13.2n) = Pr f z > ^ " ^ = Pr(Z > 0.0017671 y/n) = 0.20. Therefore, 0 . 0 0 1 7 6 7 1 ^ - 0.84162 for n = 226,836. 9.68 Let the policy being changed be the nth policy and let Fn(x) represent the distribution function using n policies. Then, originally, Fn(x) = 0.8F n _i(x) + 0.2F n _i(x - 1). Starting with x = 0: 0.40 = 0.58 = 0.64 0.69 = 0.70 = 0.78 = 0.80Fn_x(0) 0.80F n -i(l) 0.80Fn_i(2) 0.80Fn_!(3) 0.80Fn_i(4) 0.80F n-i(5) + 0.2(0), 4- 0.2(0.50), + 0.2(0.60), -l· 0.2(0.65), + 0.2(0.70), + 0.2(0.70), F n _i(0) = 0.50, F n - i ( l ) = 0.60, F n -i(2) = 0.65, F n _i(3) = 0.70, F n -i(4) = 0.70, F n -i(5) = 0.80. With the amount changed, F n (5) = 0.8F n _i(5) + 0.2F n _i(3) = 0.8(0.8) 40.2(0.7) = 0.78. 9.69 E(5) = 400(0.03)(5) + 300(0.07)(3) 4- 200(0.10)(2) = 163. For a single insured with claim probability q and exponential mean 0, Var(S) - E(N) Var(X) + Var(7V)E(X)2 = = qe2 + q(l-q)e2 q(2-q)e2. 90 CHAPTER 9 SOLUTIONS Var(S) = 400(0.03)(1.97)(25) +300(0.07)(1.93)(9) +200(0.10)(1.90)(4) 1,107.77. = P =E(S) + 1.645SD(S) = 163 + 1.645^1,107.77 = 217.75. 9.70 For one member, the mean is 0.7(160) + 0.2(600) + 0.5(240) = 352 and the variance is 0.7(4,900) + 0.21(160)2 + 0.2(20,000) + 0.16(600)2 +0.5(8,100) + 0.25(240)2 = 88,856 . For n members, the goal is 0.1 > " and thus, P r [ 5 > 1.15(352)n] / 1.15(352)n-352n\ P r ^Z> V88,856n ) ' 1.15(352)n-352« V88,856n > L28155 ' which yields n > 52.35. So 53 members are needed. 9.71 The mean is 100,000 = 0.2/c(3,500) + 0.6a/c(2,000) = (700 + l,200a)fc, and the variance is Var(S) = = 0.2(0.8)/c2(3,500) +0.6(0.4)(aA;)2(2,000) (560 + 480α 2 )/Λ Solving the first equation for k and substituting give Because the goal is to minimize the variance, constants can be removed, such as all the zeros, thus leaving 56 + 48a 2 (7+12a)2' SECTION 9.8 91 Taking the derivatives leads to a numerator of (the denominator is not important because the next step is to set the derivative equal to zero) (7 + 12α) 2 96α - (56 + 48α 2 )2(7 + 12α)12. Setting this expression equal to zero, dividing by 96, and rearranging lead to the quadratic equation 8 4 α 2 - 1 1 9 α - 9 8 = 0, and the only positive root is the solution a = 2. 9.72 λ = 500(0.01) + 500(0.02) = 15. fx(x) = 500(0.01)/15 = 1/3, fx(2x) = 2/3. E(X2) = (l/3)x 2 + (2/3)(2x) 2 = 3x 2 . Var(S) = 15(3x2) = 45z 2 = 4,500. x = 10. 9.73 All work is done in units of 100,000. The first group of 500 policies is not relevant. The others have amounts 1, 2, 1, and 1. E(S) Var(5) = = = = 500(0.02)(1) + 500(0.02)(2) + 300(0.1)(1) + 500(0.1)(1) 110, 500(0.02)(0.98)(1) + 500(0.02)(0.98)(4) +300(0.1)(0.9)(1) + 500(0.1)(0.9)(1) 121. (a) P = 110 + 1.645ν / Ϊ2Ϊ = 128.095. (b) μ + σ 2 /2 = In 110 = 4.70048. 2μ + 2σ 2 = ln(121 + HO2) = 9.41091. μ = 4.695505 and σ = 0.0997497. Ι η Ρ = 4.695505 + 1.645(0.0997497) = 4.859593, Ρ = 128.97. (c) αθ = 110, αθ2 = 121, θ = 121/110 = 1.1, α = 100. This is a chi-square distribution with 200 degrees of freedom. The 95th percentile is (using the Excel® 1 GAMMAINV function) 128.70. (d) λ = 500(0.02)+500(0.02)+300(0.10)+500(0.10) = 100. fx(l) = (10+30+ 50)/100 = 0.9 and fx(2) — 0.1. Using the recursive formula for the compound Poisson distribution, we find F 5 (128) = 0.94454 and F 5 (129) = 0.95320, and so, to be safe at the 5% level, a premium of 129 is required. 1 Excel® is cither a registered trademark or trademark of Microsoft Corporation in the United States and/or other countries. 92 CHAPTER 9 SOLUTIONS (e) One way to use the software is to note that S = X\ + 2X2 4- X3 where each X is binomial with m = 500, 500, and 800 and q = 0.02, 0.02, and 0.10. The results are FS(12S) = 0.94647 and F 5 (129) = 0.95635, and so, to be safe at the 5% level, a premium of 129 is required. 9.74 Prom (9.35), E(S) = 500(0.02)(500)+250(0.03)(750)+250(0.04)(l,000) = 20,625. Prom (9.36), Var(S) = - 500[0.02(500)2 + 0.02(0.98)(500)2] 250[0.03(750)2 + 0.03(0.97)(750)2] 250[0.04(1,000)2 + 0.04(0.96)(1,000)2] = 32,860,937.5 The compound Poisson model has λ - 500(0.02)+250(0.03)+250(0.04) = 27.5 and the claim amount distribution has density function, from (9.38), fv(r) JXK } = 1 _12_ J_p-*/500 , _I^__1 -s/750 ■ 1Q r -x/l,000 + + 27.5 500 27.5 750 27.5 1,000 This mixture of exponentials distribution has mean [10(500) + 7.5(750) + 10(l,000)]/27.5 = 750 and variance [10(2)(500)2 + 7.5(2)(750)2 + 10(2)(l,000) 2 ]/27.5 - 7502 = 653,409.09. Then E(5) Var(5) = - 27.5(750) = 20,625, 27.5(653,409.09) + 27.5(750)2 = 33,437,500. CHAPTER 10 CHAPTER 10 SOLUTIONS 10.1 SECTION 10.2 10.1 When three observations are taken without replacement, there are only four possible results. They are 1,3,5; 1,3,9; 1,5,9; and 3,5,9. The four sample means are 9/3, 13/3, 15/3, and 17/3. The expected value (each has probability 1/4) is 54/12 or 4.5, which equals the population mean. The four sample medians are 3, 3, 5, and 5. The expected value is 4, and so the median is biased. 10.2 E ( j ^ = E n (X 1 + . . . + X n ) -ipW-H + E(Xn)] (μ + · · · + μ) = μ. 10.3 For a sample of size 3 from a continuous distribution, the density function of the median is 6f(x)F(x)[l - F(x)}. For this exercise, F(x) = (χ-θ + 2)/4, Student Solutions Manual to Accompany Loss Models: From Data to Decisic Edition. By Stuart, A. Klugman, Harry H. Panjer, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. Fourth93 94 CHAPTER 10 SOLUTIONS Θ — 2 < x < Θ -{-2. The density function for the median is t ( \ ^ ^ χ ~ fmed(x) = 6(0.25) θ + 22~χ + θ and the expected value is χ(χ-θ + 2)(2-χ 64 J9-2 ^ j + θ), -y* + (6 - % 6_ [_£ 64 4 (6 - % 3 3 2 = 6 f\ 64 / (y + 0 ~ M ~ y dy ) + 4(0 - 2)ydy 4(fl-2)y2i ' 2 _6_ -64 + < 6 - ' > 6 4 + 32(0-2) 64 Θ, where the first line used the substitution y — x — θ Λ-2. 10.4 Because the mean of a Pareto distribution does not always exist, it is not reasonable to discuss unbiasedness or consistency. Had the problem been restricted to Pareto distributions with a > 1, then consistency can be established. It turns out that for the sample mean to be consistent, only the first moment needs to exist (the variance having a limit of zero is a sufficient, but not a necessary, condition for consistency). 10.5 The mean is unbiased, so its MSE is its variance. It is MqE m Var 42 P0 _4 MS£ m o a n (0) - — ^ — - j2(3j - g. The median is also unbiased. The variance is ρθ+2 Je-2 (x - θ)2(χ - Θ + 2)(2 - x + Θ) dx 64 64 J0 -y4+8y3-20y2 £ __6_ ~ 64 " 5 = 4/5, + + 16ydy 8(4) 4 _ 20(4) 3 4 3 + 16(4)2 2 and so the sample mean has the smaller MSE. _6_ r4 64 J0 (y - 2)2y(4 - y)dy SECTION 10.2 95 10.6 We have Var(0 c ) = - Var[w0A + (1 - w)0B] w2(160,000) + (1 - w)2(40,000) 200,000w2 - 80,000w + 40,000. The derivative is 400,000?/; — 80,000, and setting it equal to zero provides the solution, w = 0.2. 10.7 MSE - Var+bias 2 . 1 = Var+(0.2) 2 , Var = 0.96. 10.8 To be unbiased, m = E(Z) = a(0.8m) + ßm = (0.8a + ß)m, and so 1 = 0.8a + ß or ß = 1 - 0.8α. Then Var(Z) - a2m2 + /?21.5m2 = [a2 + (1 - 0.8a) 2 1.5]m 2 , which is minimized when a 2 + 1.5 — 2.4a + 0.96a 2 is minimized. This result occurs when 3.92a-2.4 = 0 or a = 0.6122. Then β = 1-0.8(0.6122) = 0.5102. 10.9 One way to solve this problem is to list the 20 possible samples of size 3 and assign probability 1/20 to each. The population mean is (1 + 1 + 2 + 3 + 5 + 10)/6 = l l / 3 . (a) The 20 sample means have an average of 11/3, and so the bias is zero. The variance of the sample means (dividing by 20 because this is the population of sample means) is 1.9778, which is also the MSE. (b) The 20 sample medians have an average of 2.7, and so the bias is 2.7 — 11/3 = -0.9667. The variance is 1.81 and the MSE is 2.7444. (c) The 20 sample midranges have an average of 4.15, and so the bias is 4.15 - 11/3 = 0.4833. The variance is 2.65 and the MSE is 2.8861. (d) E(aX (1 ) + bX(2) + cX (3 )) = 1.25a + 2.76 + 7.05c, where the expected values of the order statistics can be found by averaging the 20 values from the enumerated population. To be unbiased, the expected value must be 11/3, and so the restriction is 1.25a + 2.76 + 7.05c = 11/3. With this restriction, the MSE is minimized at a - 1.445337, b = 0.043733, and c = 0.247080 with an MSE of 1.620325. With no restrictions, the minimum is at a = 1.289870, b = 0.039029, and c = 0.220507 with an MSE of 1.446047 (and a bias of -0.3944). 96 CHAPTER 10 SOLUTIONS 10.10 bias(fli) = 165/75 - 2 - 0.2, Var(0i) = 375/75 - (165/75) 2 = 0.16, MSE(öi) - 0.16 + (0.2)2 - 0.2. bias(02) = 147/75 - 2 - -0.04, Var(#2) = 312/75 - (147/75) 2 = 0.3184, MSE(#2) = 0.3184 + (-0.04) 2 = 0.32. The relative efficiency is 0.2/0.32 = 0.625, or 0.32/0.20 = 1.6. 10.2 SECTION 10.3 10.11 (a) From the information given in the problem, we can begin with 0.95 = Pr(o < Χ/Θ < 6), where X/θ is known to have the gamma distribution with a = 50 and Θ — 0.02. This does not uniquely specify the endpoints. However, if 2.5% probability is allocated to each side, then a = 0.7422 and b = 1.2956. Inserting these values in the inequality, taking reciprocals, and multiplying through by X give 0.95 = Pr(0.7718X < Θ < 1.3473X). Inserting the sample mean gives an interval of 212.25 to 370.51. (b) The sample mean has E(X) = Θ and Var(X) = 6>2/50. Then, using 275 2 /50 as the approximate variance, 0.95 X = Pr (-1.96 < = Pr(-76.23 < X - Θ < 76.23). V < 1.96 ~,— ~ 275/^ ~ Inserting the sample mean of 275 gives the interval 275 ± 76.23, or 198.77 to 351.23. (c) Leaving the Θ in the variance alone, 0.95 = = = Pr f -1.96 < V X ~JL < 1.96 " Θ/VSÖ " Pr(-0.27726> < X - Θ < 0.2772Ö) Pr(X/1.2772 < Θ < X/0.7228) for an interval of 215.31 to 380.46. 10.12 The estimated probability of one or more claims is 400/2,000 = 0.2. For this binomial distribution, the variance is estimated as 0.2(0.8)/2,000 = 0.00008. The upper bound is 0.2 + 1.96^0.00008 = 0.21753. SECTION 10.4 10.3 97 SECTION 10.4 10.13 For the exact test, null should be rejected if X < c. The value of c comes from 0.05 = = Pr(X < c\0 = 325) Pr(X/325 < c/325), where X/325 has the gamma distribution with parameters 50 and 0.02. From that distribution, c/325 = 0.7793 for c = 253.27. Because the sample mean of 275 is not below this value, the null hypothesis is not rejected. The p-value is obtained from Pr(X < 275|0 = 325) = Pr(X/325 < 275/325). From the gamma distribution with parameters 50 and 0.02, this probability is 0.1353. For the normal approximation, 0.05 = Pr(X < c\0 = 325) Γ V " 325/vW ' Solving (c - 3 2 5 ) / ( 3 2 5 / Λ / 5 0 ) - -1.645 produces c = 249.39. Again, the null hypothesis is not rejected. For the p-value, Pr (z < V " 275 ~ ™ = -1.0879^ = 0.1383. 325/Λ/50 ) SECTION 10.4 10.3 97 SECTION 10.4 10.13 For the exact test, null should be rejected if X < c. The value of c comes from 0.05 = = Pr(X < c\0 = 325) Pr(X/325 < c/325), where X/325 has the gamma distribution with parameters 50 and 0.02. From that distribution, c/325 = 0.7793 for c = 253.27. Because the sample mean of 275 is not below this value, the null hypothesis is not rejected. The p-value is obtained from Pr(X < 275|0 = 325) = Pr(X/325 < 275/325). From the gamma distribution with parameters 50 and 0.02, this probability is 0.1353. For the normal approximation, 0.05 = Pr(X < c\0 = 325) Γ V " 325/vW ' Solving (c - 3 2 5 ) / ( 3 2 5 / Λ / 5 0 ) - -1.645 produces c = 249.39. Again, the null hypothesis is not rejected. For the p-value, Pr (z < V " 275 ~ ™ = -1.0879^ = 0.1383. 325/Λ/50 ) CHAPTER 11 CHAPTER 11 SOLUTIONS 11.1 SECTION 11.2 11.1 When all information is available, the calculations are in Table 11.1. As in Example 11.4, values apply from the current y-value to the next one. 11.2 (a) μ = £ > i / 3 5 = 204,900. μ'2 = Σ ζ 2 / 3 5 = 1-4134 x 10 11 . σ = 325,807. A3 = 1-70087 x 10 17 , As = 9.62339 x 10 16 , c = 325,807/204,900 = 1.590078, ηχ = 2.78257. (b) £„(500,000) = E j = i Vj + 5(500,000)]/35 = 153,139. E™ (500,000) = E j L j VJ + 5(500,000)2]/35 = 53,732,687,032. 11.3 A2 = A3 = 7! = μ\ = [2(2,000) + 6(4,000) + 12(6,000) + 10(8,000)]/30 = 6,000, [2(-4,000) 2 + 6(-2,000) 2 + 12(0)2 + 10(2,000)2]/30 = 3,200,000, [2(-4,000) 3 + 6(-2,000) 3 + 12(0)3 + 10(2,000)3]/30 = -3,200,000,000. -3,200,000,000/(3,200,000) 15 = -0.55902. Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugmtui, Harry H. Panjer, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. Fourth99 100 CHAPTER 11 SOLUTIONS Table 11.1 3 1 2 3 4 5 6 7 8 9 10 11 12 13 Vj 0.1 0.5 0.8 1.8 2.1 2.5 2.8 3.9 4.0 4.1 4.6 4.8 5.0 *F(x) = l-e Sj r3 1 1 1 2 1 1 1 2 1 1 2 2 14 30 29 28 27 25 24 23 22 20 19 18 16 14 H(x) Fso(x) 1. - 29/30 == 0.0333 ]L - 28/30 -= 0.0667 1L - 27/30 == 0.1000 ]L - 25/30 == 0.1667 ]L - 24/30 == 0.2000 JL - 23/30 == 0.2333 JL - 22/30 == 0.2667 ]L - 20/30 = 0.3333 ]L - 19/30 = 0.3667 ]L - 18/30 = 0.4000 L - 16/30 = 0.4667 :L - 14/30 = 0.5333 1 - 0/30 == 1.0000 F(xY 1/30 =-- 0.0333 0.0333 + 1/29 == 0.0678 0.0678 + 1/28 == 0.1035 0.1035 + 2/27 == 0.1776 0.1776 + 1/25 == 0.2176 0.2176 + 1/24 == 0.2593 0.2593 + 1/23 == 0.3027 0.3027 + 2/22 == 0.3937 0.3937+1/20 == 0.4437 0.4437+1/19 == 0.4963 0.4963 + 2/18 == 0.6074 0.6074 + 2/16 == 0.7324 0.7324+14/14 == 1.7324 0.0328 0.0656 0.0983 0.1627 0.1956 0.2284 0.2612 0.3254 0.3583 0.3912 0.4552 0.5192 0.8231 -H(x)m Table 11.2 Payment range Calculations for Exercise 11.4. Number of payments 0-25 25-50 50-75 75-100 100 150 150-250 250-500 500-1,000 1,000-2,000 2,000-4,000 11.2 Calculations for Exercise 11.1. 6 24 30 31 57 80 85 54 15 10 Ogive value 392 30 392 60 392 91 392 148 392 228 392 313 392 367 392 382 392 392 392 0.0153 0.0765 0.1531 = 0.2321 : 0.3776 = 0.5816 = 0.7985 = 0.9362 : 0.9745 : 1.0000 Histogram value 392(25) 24 39|25) 392(25) 31 392(25) 392(50) 80 392(100) 392(250) 392(500) 392(1000) 10 392(2000) 0.000612 0.002449 0.003061 : 0.003163 : 0.002908 : 0.002041 : 0.000867 = 0.000276 : 0.000038 : 0.000013 : SECTION 11.3 1 1 . 4 There are 392 observations and the calculations are in Table 11.2. For each interval, the ogive value is for the right-hand endpoint of the interval, while the histogram value is for the entire interval. Graphs of the ogive and histogram appear in Figures 11.1 and 11.2. SECTION 11.3 Figure 11.1 Figure 11.2 101 Ogive for Exercise 11.4. Histogram for Exercise 11.4. 11.5 (a) T h e ogive connects the points (0.5,0), (2.5,0.35), (8.5,0.65), (15.5,0.85), and (29.5,1). (b) T h e histogram has height 0.35/2 = 0.175 on the interval (0.5,2.5), height 0.3/6 = 0.05 on the interval (2.5,8.5), height 0.2/7 = 0.028571 on the interval (8.5,15.1), and height 0.15/14 = 0.010714 on the interval (15.5,29.5). 1 1 . 6 T h e plot appears in Figure 11.3. T h e points are the complements of the survival probabilities at the indicated times. 102 CHAPTER 11 SOLUTIONS 0.03 -, 0.025 0.02 - - -O- - - Refinances 0.015 ——A.—— Orininal m mm ■"■- ^ 0.01 ■■" v j i i y i i i a i 0.005 0<ψ ■ ■ - () -v i i 2 4 6 Years Figure 11.3 Ogive for mortgage lifetime for Exercise 11.6. Because one curve lies completely above the other, it appears possible that original issues have a shorter lifetime. 11.7 The heights of the histogram bars are, respectively, 0.5/2 = 0.25, 0.2/8 = 0.025, 0.2/90 = 0.00222, 0.1/900 = 0.000111. The histogram appears in Figure 11.4. 11.8 The empirical model places probability 1/n at each data point. Then Ε(ΧΛ2) = £ > , · ( 1 / 4 0 ) + Σ 2(1/40) = = (20 + 15)(l/40) + (14)(2)(l/40) 1.575. SECTION 11.3 Figure 11.4 103 Histogram for Exercise 11.7. 11.9 We have E(X A 7,000) = 1 2,000 2,000 + Σ *;+ Σ ^Ej<7,000 £ vXj<6,000 Σ 6,000<^<7,000 7,000 xj>7,000 Xj+ Σ 6,000 ajj> 6,000 (*i - 6,000) + J2 lfi0 Xj>7,000 °) J = E(X Λ 6,000) + [200,000 - 30(6,000) + 270(l,000)]/2,000 - 1,955. 11.10 Let n be the sample size. The equations are 36 x 0.6y „^ 36 0.4a: J Λ rn 0.21 = — + and 0.51 = — + - + — - . n n n n n Also, n = 200 + x + ?/. The equations can be rewritten as 0.21(200 + x + y) = 36 + 0.4x and 0.51(200 + x + y) = 36 + x + 0.6?/. The linear equations can be solved for x = 120. SECTION 11.3 Figure 11.4 103 Histogram for Exercise 11.7. 11.9 We have E(X A 7,000) = 1 2,000 2,000 + Σ *;+ Σ ^Ej<7,000 £ vXj<6,000 Σ 6,000<^<7,000 7,000 xj>7,000 Xj+ Σ 6,000 ajj> 6,000 (*i - 6,000) + J2 lfi0 Xj>7,000 °) J = E(X Λ 6,000) + [200,000 - 30(6,000) + 270(l,000)]/2,000 - 1,955. 11.10 Let n be the sample size. The equations are 36 x 0.6y „^ 36 0.4a: J Λ rn 0.21 = — + and 0.51 = — + - + — - . n n n n n Also, n = 200 + x + ?/. The equations can be rewritten as 0.21(200 + x + y) = 36 + 0.4x and 0.51(200 + x + y) = 36 + x + 0.6?/. The linear equations can be solved for x = 120. CHAPTER 12 CHAPTER 12 SOLUTIONS 12.1 SECTION 12.1 12.1 The calculations are in Tables 12.1 and 12.2. Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugman, Harry H. Panjer, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. Fourthl05 106 CHAPTER 12 SOLUTIONS Table 12.1 i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 di Ui 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Calculations for Exercise 12.1. Xi - 0.1 0.5 0.8 - 1.8 1.8 2.1 2.5 2.8 - 3.9 - 4.0 4.1 0.8 - 2.9 2.9 - 4.0 Table 12.2 - i 16 17 18 19-30 31 32 33 34 35 36 37 38 39 40 dt Ui 0 0 0 0 0.3 0.7 1.0 1.8 2.1 2.9 2.9 3.2 3.4 3.9 4.8 Xi - - 4.8 4.8 5.0 5.0 5.0 - 3.9 5.0 4.8 - 5.0 5.0 4.1 3.1 4.0 - - Further calculations for Exercise 12.1. Vj 1 2 3 4 5 6 7 8 9 10 11 12 0.1 0.5 0.8 1.8 2.1 2.5 2.8 3.9 4.0 4.1 4.8 5.0 3 17 30-0-0 31-1-0 32-2-0 33-3-1 34-5-1 35-6-1 35-7-1 39-8-4 40-10-4 40-11-6 40-12-7 40-15-8 30 or 0 + 30 - 0 - 0 = 30 : 30 or 30 + 1 - 1 - 0 = 30 30 or 30 + 1 - 1 - 0 = 30 : 29 or 30 + 1 - 1 - 1 = 29 = 28 or 29 + 1 - 2 - 0 = 28 : 28 or 28 + 1 - 1 - 0 = 28 27 or 28 + 0 - 1 - 0 = 27 : 27 or 27 + 4 - 1 - 3 = 27 : 26 or 27 + 1 - 2 - 0 = 26 : 23 or 26 + 0 - 1 - 2 = 23 : 21 or 23 + 0 - 1 - 1 = 21 : 17 or 2 1 + 0 - 3 - 1 = 17 SECTION 12.1 12.2 0 < £ <0.1, ^ 0.9667, 0.1 < t < 0.5, ΰ . 9 6 6 7 % 1 = 0.9344, 0.5 < t < 0.8, Q.9344% 1 = 0.9033, 0.8 < t < 1.8, 0.9033 2 929- 2 0.8410, 1.8 <t < 2 . 1 , Q . 8 4 1 028^ 1 = 0.8110, 2.1 < t < 2.5, 0.8110 2 828- 1 = 0.7820, 2.5 < t < 2.8, 0.7820 2 727- 1 = 0.7530, 2.8 < t < 3.9, 0.7530 27-2 27 0.6973, 3.9 < t < 4.0, 0 . 6 9 7 3 ^ = 0.6704, 4.0 < t < 4.1, 0 . 6 7 0 4 ^ = 0.6413, 4.1 < t < 4.8, 0 . 6 4 1 3 ^ = 0.5497, 4.8 < t < 5.0, = 11 Q . 5 4 9 7 17 ^ = 0, t > 5.0. 107 108 CHAPTER 12 SOLUTIONS 12.3 0 < t < 0.1, ( o, H(t) ^ = 0.0333 0.1 <t < 0 . 5 , 0.0333 4- ^ = 0.0667, 0.5 < t < 0.8, 0.0667 4- ^ = 0.1000, 0.8 < t < 1.8, 0.1000 4 - ^ = 0.1690, 1.8 < t < 2.1, 0.1690+^ = 0.2047, 2.1 < t < 2.5, 0.2047 + ^ = 0.2404, 2.5 < t < 2.8, 0.2404 4- 27 = 0.2774, 2.8 < t < 3.9, 0.2774 + |F = 0.3515, 3.9 < t < 4.0, 0.3515 + ^ = 0.3900, 4.0 <t < 4 . 1 , 0.3900 4- ± = 0.4334, 4.1 < t < 4.8, 0.4334 4- |[= 0.5763, 4.8 < t < 5.0, 0.5763 4- H = 1.5763, t > 5.0. e"° - 1, „-0.0333 -0.0667 -0.1000 e-0.1690 = e-0.2047 = S(t) e-0.2404 = e-0.2774 = 0.3515 _ 0.3900 _ 0.4334 _ 0.5763 _ -1.5763 0 <t < 0 . 1 , 0,9672, 0.1 < t < 0.5, 0.9355, 0.5 < t < 0.8, 9048, 8445, Q 8149, Q 7863, 0 ,7578, Q 7036, Q ,6771, Q ,6483, n .5620, .2076, 0 0.8 < t < 1.8, 1.8 < t < 2.1, 2.1 < t < 2.5, 2.5 < t < 2.8, 2.8 < t < 3.9, 3.9 < t < 4.0, 4.0<*<4.1, 4.1 < t < 4.8, 4.8 < t < 5.0, t > 5.0. 12.4 Using the raw data, the results are in Table 12.3. When the deductible and limit are imposed, the results are as in Table 12.4. Because 1,000 is a censoring point and not an observed loss value, there is no change in the survival function at 1,000. SECTION 12.1 Table 12.3 Value (x) 27 82 115 126 155 161 243 294 340 384 457 680 855 877 974 1,193 1,340 1,884 2,558 15,743 r s 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Table 12.4 Value (x) 115 126 155 161 243 294 340 384 457 680 855 877 974 109 Calculations for Exercise 12.4. SKM(X) 0.95 0.90 0.85 0.80 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 HNA(X) 0.0500 0.1026 0.1582 0.2170 0.2795 0.3462 0.4176 0.4945 0.5779 0.6688 0.7688 0.8799 1.0049 1.1477 1.3144 1.5144 1.7644 2.0977 2.5977 3.5977 SNA(X) 0.9512 0.9025 0.8537 0.8049 0.7562 0.7074 0.6586 0.6099 0.5611 0.5123 0.4636 0.4148 0.3661 0.3174 0.2686 0.2199 0.1713 0.1227 0.0744 0.0274 Further calculations for Exercise 12.4. r s SKM{X) 18 17 16 15 14 13 12 11 10 9 8 7 6 1 1 1 1 1 1 1 1 1 1 1 1 1 0.9444 0.8889 0.8333 0.7778 0.7222 0.6667 0.6111 0.5556 0.5000 0.4444 0.3889 0.3333 0.2778 HNA(x) 0.0556 0.1144 0.1769 0.2435 0.3150 0.3919 0.4752 0.5661 0.6661 0.7773 0.9023 1.0451 1.2118 SNA(X) 0.9459 0.8919 0.8379 0.7839 0.7298 0.6758 0.6218 0.5677 0.5137 0.4596 0.4056 0.3517 0.2977 12.5 Suppose the lapse was at time 1. The estimate of S(4) is (3/4) (2/3) (1/2) = 0.25. If it is at time 2, the estimate is (4/5)(2/3)(l/2)=0.27. If it is at time 3, the estimate is (4/5)(3/4)(1/2) = 0.3. If it is at time 4, the estimate is (4/5) (3/4) (2/3) = 0.4. If it is at time 5, the estimate is (4/5) (3/4) (2/3) (1/2) = 0.20. Therefore, the answer is time 5. 110 CHAPTER 12 SOLUTIONS Table 12.5 Age(i) #ds 0 1 2 3 4 5 7 300 #xs 20 30 #us r S(t) 6 300 ϋ-0.98 0 0 a 314 304 324 0.98§ff - 0.94879 0.94879||| = 0.91758 0 . 9 1 7 5 8 ^ ^ = 0.892 => a = 9 0.892^^ 45 b 9 10 12 13 Calculations for Exercise 12.7. 279-α - 2 7 0 35 6 15 244-a - b = 235 - b 12.6 # ( 1 2 ) = 7^ + 7^ + 7^ + 6 = ° · 6 5 · is 5(12) = e " 0 · 6 5 - 0.522. T h e 0 . 8 9 2 ^ ^ § | ^ = 0.856 =>b = 4 estimate of the survival function 1 2 . 7 T h e information may be organized as in Table 12.5 12.8 H(t10) S(ts) = e- - H(t9) 014307 = ^ = 0.077, n = 22. H(ts) = ± + ± + ^ = 0.14307, = 0.8667. 1 2 . 9 0.60 - 0 . 7 2 ^ , r4 = 12. 0.50 = 0 . 6 0 · ^ , r 5 = 6. W i t h two deaths at the fourth death time and the risk set decreasing by 6, there must have been four censored observations. 1 2 . 1 0 0.575 = 5(10) = e-"< 1 0 >. # ( 1 0 ) = - ln(0.575) = 0.5534 = ^ 4- ^ f + "Er ^ n e s o m t i o n is k = 36. + 1 2 . 1 1 W i t h no censoring, the r-values are 12, 9, 8, 7, 6, 4, and 3, and the .s values are 3, 1, 1, 1, 2, 1, and 1. Then #(7,000) = ^ + i + I + I + 2 + l + i = L5456> W i t h censoring, there are only five uncensored values with r values 9, 8, 7, 4, and 3, and all five s values are 1. Then # ( 7 , 0 0 0 ) =1+λ+λ+λ+±= °·9623· 12.12 ^ = ^ ^ = ^f = > r 4 = 13. T h e risk set at the fifth death time is the 13 from the previous death time less the 3 who died and less the 6 SECTION 12.2 who were censored. That leaves r 5 = 13 — 3 — 6 = 4. Then, %rr = *=Ρ => s5 = 2. 12.2 ' ° 0.5 111 r5 r5 s ^ = SECTION 12.2 12.13 To proceed, we need the estimated survival function at whole number durations. From Exercise 11.1, we have 5 30 (1) = 27/30, 5 30 (2) = 25/30, 5 30 (3) = 22/30, 5 30 (4) = 19/30, and 5 30 (5) = 14/30. Then the mortality estimates are q0 = 3/30, qx = 2/27, q2 = 3/25, fa = 3/22, q4 = 5/19, 5Po = 14/30. The six variances are Var(^o) Var(gi) Varfe) Var(g3) Var(g4) Var( 5 p 0 ) 3(27)/30 3 = 0.003, 2(25)/27 3 = 0.002540, 3(22)/25 3 - 0.004224, 3(19)/22 3 = 0.005353, 5(14)/19 3 = 0.010206, 14(16)/30 3 = 0.008296. = = = = = = The first and last variances are unconditional. 12.14 From the data set, there were 1,915 out of 94,935 that had two or more accidents. The estimated probability is 1,915/94,935 = 0.02017, and the estimated variance is 0.02017(0.97983)/94,935 = 2.08176 x 10" 7 . 12.15 From Exercise 12.13, 53o(3) = 22/30, and the direct estimate of its variance is 22(8)/30 3 = 0.0065185. Using Greenwood's formula, the estimated variance is 1 1 29(28) 28(27) 27(25) 1 1 1 \ 22(8) 25(24) 24(23) 23(22)/ 30 3 ' 30(29) \30J V30(i For 2^3, the point estimate is 8/22, and the direct estimate of the variance is 8(14)/22 3 = 0.010518. Greenwood's estimate is (M)'( 2 22(20) + 1 20(19) + 1 19(18) + 2 18(16) + 2 \ _ 8(14) 16(14) J ~~ 22 3 " 112 CHAPTER 12 SOLUTIONS 12.16 Prom Exercise 12.2, i>4o(3) = 0.7530, and Greenwood's formula provides a variance of 2 0.75302 f—j— + -J— + -J— + 30(29) + 2 ^ 30(29) + 28^7) 30(29) + 29(27) 2 ^ 6 ) ) =°·00571· Also, 2<?3 = (0.7530- 0.5497)/0.7530 = 0.26999, and the variance is estimated as °-730012 (2^5) + 26(W + 23(W + 2m) = °·00768· 12.17 Using the log-transformed method, P /l.96y«57l^ 1^0.753 In 0.753 ) u 4yjy1 ' ' The lower limit is 0.753 1 / 0 · 49991 = 0.56695, and the upper limit is 0.753 0 4 9 9 9 1 0.86778. 12.18 Prom Exercise 12.3, H(3) = 0.2774. The variance is estimated as The linear confidence interval is 0.2774 ± 1.96V0.0096342 or 0.0850 to 0.4698. The log-transformed interval requires U = exp 1.96x/0.0096342 0.2774 - 2.00074, or 0.49981. The lower limit is 0.2774(0.49981) = 0.13865 and the upper limit is 0.2774(2.00074) - 0.55501. 12.19 Without any distributional assumptions, the variance is estimated as (1/5) (4/5)/5 — 0.032. From the distributional assumption, the true value of 3<77 is [(8/15) - (5/15)]/(8/15) = 3/8, and the variance is (3/8)(5/8)/5 = 0.046875. The difference is -0.014875. 12.20 First, obtain the estimated survival probability as SECTION 12.2 113 Greenwood's formula gives ( 4032)2 °- + {vk) 8^6) + 4ö) + 6öiW) = °·00551· 12.21 The Nelson-Äalen estimates are the centers of the confidence intervals, which are 0.15 and 0.27121. Therefore, s i + i / r i + i = 0.12121. Prom the first confidence interval, the estimated variance is (0.07875/1.96)2 = 0.0016143, while for the second interval it is (0.11514/1.96)2 = 0.0034510 and, therefore, Si+i/rf^ = 0.0018367. Dividing the first result by the second gives r^+i = 66. The first equation then yields s^+i = 8. 12.22 Greenwood's estimate is 2 v = s2 V/ 50(48) + 4 45(41) + 8 \ (41 - c)(33 - c ) ) ' Then, 0.011467 = -L = 0.003001 + S2 ' (41-c)(33-c)' ( 4 1 - Χ 3 3 - ) = ÖÄ66= 9 4 5 · Solving the quadratic equation yields c — 6. 12.23 For the Nelson-Äalen estimate, 1.5641 = £ ( 3 5 ) = A + A + A + ^ + (1.5641 - 0.5641)8(8 - d) = d(8 - d) + 16, 0 = d2 - I6d + 48, d = 4. The variance is 2 Ϊ5^ + 3 Ϊ3* + 2 4 2 + + ϊο^ 8* 4^=0·23414· 12.24 The standard deviation is / 15 20 13 1/2 + + ττ^= 0.11983. -I 7^0 - -\ V1ÖÖ2 ^ 652 T 4Q2 ) - J^, 114 CHAPTER 12 SOLUTIONS Data for Exercise 12.28. Table 12.6 Vi P(VJ) 0.1 0.5 0.8 1.8 2.1 2.5 2.8 3.9 4.0 4.1 4.8 0.0333 0.0323 0.0311 0.0623 0.0300 0.0290 0.0290 0.0557 0.0269 0.0291 0.0916 1 2 . 2 5 T h e uncensored observations are 4 and 8, the two r-values are 10 and 5, and the two s-values are 2 and 1. T h e n 5(11) = y ^ | = 0.64. Greenwood's estimate is 1 2 (0.64) Ll0(8) + 5(4)J 0.03072. 1 2 . 2 6 At day 8 (the first uncensored time), r = 7 ( 8 - 1 ) and s = 1. At day 12, r = 5 and s = 2. T h e n # ( 1 2 ) = £ + § = 0.5429. Also, V a r [ # ( 1 2 ) ] = ^ + |> = 0.1004. T h e interval is 0.5429 ± 1.645\/0.1004, which is (0.0217,1.0641). 12.3 SECTION 12.3 1 2 . 2 7 In order for the mean to be equal to y, we must have θ/(α — 1) = y. Letting a be arbitrary (and greater t h a n 1), use a Pareto distribution with Θ — y(a — 1). This makes the kernel function ky(x) _ «K° - Pd"a + 1 [(a- I)?/+ x ] " 1 2 . 2 8 T h e d a t a points and probabilities can be taken from Exercise 12.2. They are given in Table 12.6 T h e probability at 5.0 is discrete and so should not be spread out by the kernel density estimator. Because of the value at 0.1, the largest available bandwidth is 0.1. Using this bandwidth and the triangular kernel produces the graph in Figure 12.1. This graph is clearly not satisfactory. T h e g a m m a kernel is not available because there would be positive probability at values greater t h a n 5. Your SECTION 12.3 Figure 12.1 115 Triangular kernel for Exercise 12.28. author tried to solve this by using the beta distribution. With Θ known to be 5, the mean (to be set equal to y) is 5a/(a + b). To have some smoothing control and to determine parameter values, the sum a + b was fixed. Using a value of 50 for the sum, the kernel is ky\x) — /x\1Qy Γ(50) Γ(10</)Γ(50 - lOj/) \ 5 / / V Χxν 5 0 - 1 0 2 / - 1 - I 5 / 1 -, 0 < x < 5 x and the resulting smoothed estimate appears in Figure 12.2 12.29 With a bandwidth of 60, the height of the kernel is 1/120. At a value of 100, the following data points contribute probability 1/20: 47, 75, and 156. Therefore, the height is 3(1/20)(1/120) = 1/800. 12.30 The uniform kernel spreads the probability of 0.1 to 10 units on either side of an observation. The observation at 25 contributes a density of 0.005 from 15 to 35, thus contributing nothing to survival past age 40. The same applies to the point at 30. The points at 35 each contribute probability from 25 to 45 and 0.25 of that probability is above 40. Together they contribute 2(0.25)(0.1) - 0.05. The point at 37 contributes (7/20)(0.1) = 0.035. The next four points contribute a total of (9/20+15/20+17/20+19/20)(0.1) = 0.3. The final point (at 55) contributes all its probability at points above 40 and so contributes 0.1 to the total, which is 0.05 + 0.035 + 0.3 + 0.1 = 0.485. 116 CHAPTER 12 SOLUTIONS 0.4 -r 0.35 \ 0.3 -I 0.25 \ g 0.2 J 0.15 \ 0.1 J 0.05 J 0J 0 1 2 3 4 5 6 x Figure 12.2 Beta kernel for Exercise 12.28. 1 2 . 3 1 (a) W i t h two of the five points below 150, the empirical estimate is 2 / 5 = 0.4. (b) T h e point at 82 has probability 0.2 spread from 32 to 132; all is below 150, so the contribution is 0.2. The point at 126 has probability from 76 to 176; the contribution below 150 is (74/100)(0.2) = 0.148. T h e point at 161 contributes (39/100)(0.2) = 0.078. T h e last two points contribute nothing, so the estimate is 0.2 + 0.148 + 0.078 = 0.426. (c) As in part (b), the first point contributes 0.2 and the last two points contribute nothing. T h e triangle has base 100 and area 0.2, so the height must be 0.004. For the point at 126, the probability excluded is the triangle from 150 to 176, which has base 26 and height at 150 of 0.004(26/50) = 0.00208. T h e area is 0.5(26)(0.00208) = 0.02704 and the contribution is 0.17296. For the point at 161, the area included is the triangle from 111 to 150, which has base 39 and height at 150 of 0.004(39/50) - 0.00312. T h e area is 0.5(39)(0.00312) = 0.06084. T h e estimate is 0.2 + 0.17296 + 0.06084 - 0.4338. 12.4 SECTION 12.4 1 2 . 3 2 T h e only change is the entries in the dj and w™ columns are swapped. T h e exposures then change. For example, at j = 2, the exposure is 28 + 0 + 3(1/2)-2(1/2) =28.5. SECTION 12.4 Kaplan-Meier calculations for Exercise 12.33. Table 12.7 x r 45.3 45.4 46.2 46.4 46.7 s S(x) 8 ]L 7 1L 8 ]L 6 1L 5 ]L 7/8 = 0.875 0.875(6/7) = 0.750(7/8) = 0.656(5/6) = 0.549(4/5) = Table 12.8 Cj PJ 250 500 0 6 12 14 12 12 5 4 2 1 0 1,000 2,750 3,000 3,500 5,250 5,500 6,000 10,250 10,500 117 nj 7 8 7 0 0 0 0 0 0 0 0 0.750 0.656 0.549 0.438 Calculations for Exercise 12.34. w* 0 0 1 1 0 1 1 1 0 1 0 dj 1 2 4 1 0 6 0 1 1 0 0 *r 1/7 2/14 4/19 1/14 0/12 6/12 0/5 1/4 1/2 0/1 - S(c3) 1.000 1.000(6/7) = 0.857 0.857(12/14) = 0.735 0.735(15/19) = 0.580 0.580(13/14) - 0.539 0.539(12/12) = 0.539 0.539(6/12) = 0.269 0.269(5/5) = 0.269 0.269(3/4) = 0.202 0.202(1/2) = 0.101 0.101(1/1) = 0.101 1 2 . 3 3 T h e Kaplan-Meier calculations appear in Table 12.7. T h e n q^ — 1 _ 0.750 - 0.250 and q46 = 1 - 0.438/0.750 - 0.416. For exact exposure at age 45, the first eight individuals contribute 1 + 1 + 0.3 + 1 + 0.4 + 1 + 0.4 4- 0.8 = 5.9 of exposure for qAb = 1 - e " 2 / 5 · 9 = 0.28751. For age 46, eight individuals contribute for an exposure of 0 . 7 + 1 + 1 + 1 + 0.3 + 0.2 + 0.4 + 0.9 = 5.5 for q46 = 1 - e ~ 3 / 5 · 5 = 0.42042. For actuarial exposure at age 45 the two deaths add 1.3 to the exposure for q45 = 2/7.2 = 0.27778. For age 46, the three deaths add 1.7 for q46 = 3/7.2 = 0.41667. 1 2 . 3 4 T h e calculations are in Table 12.8. T h e intervals were selected so t h a t all deductibles and limits are at the boundaries and so no assumption is needed about t h e intermediate values. This setup means t h a t no assumptions need to be made about the timing of entrants and withdrawals. W h e n working with the observations, the deductible must be added to t h e payment in order to produce the loss amount. T h e requested probability is 5(5,500)/5(500) = 0.269/0.857 = 0.314. 1 2 . 3 5 For the intervals, the n-values are 6, 6, 7, 0, 0, 0, and 0, the ^-values are 0, 1, 1, 1, 0, 0, and 0, and the d-values are 1, 2, 4, 7, 1, 1, and 0. the 118 CHAPTER 12 SOLUTIONS P-values are then 6, 6 - 0 - 1 + 6 = 11, 1 1 - 1 - 2 + 7 = 15, 1 5 - 1 - 4 + 0 = 10, 1 0 - 1 - 7 + 0 = 2, 2 - 0 - 1 + 0 = 1 , and 1 - 0 - 1 + 0 = 0. The estimates are 5(500) = 5/6 and 5(6,000) = (5/6)(9/ll)(ll/15)(3/10)(l/2) = 3/40, where both estimates are conditioned on survival to 250. The answer is the ratio (3/40)/(5/6) = 9/100 = 0.09. CHAPTER 13 CHAPTER 13 SOLUTIONS 13.1 SECTION 13.1 13.1 The mean of the data is μ[ = [27 + 82 + 115 + 126 + 155 + 161 + 243 + 13(250)]/20 = 207.95. The expected value of a single observation censored at 250 is Ε(ΧΛ250) = 0(1 - e _ 2 5 0 / ö ) . Setting the two equal and solving produce Θ = 657.26. 13.2 The equations to solve are βχρ(μ + σ 2 /2) exp(2/i 4- 2σ2) = = 1,424.4, 13,238,441.9. Taking logarithms yields μ + σ2/2 = 7.261506, 2μ + 2σ 2 = 16.398635. The solution of these two equations is μ = 6.323695 and σ 2 = 1.875623, and then σ = 1.369534. Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugmau, Harry H. Panjor, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. Fourth\\% 120 CHAPTER 13 SOLUTIONS 1 3 . 3 T h e two equations to solve are 0.2 = l-e-<5/">\ 0.8 = l-e-(12/"»T. Moving the 1 to the left-hand side and taking logarithms produces 0.22314 = (5/0)T, 1.60944 - (12/#)r. Dividing the second equation by the first equation produces 7.21269 = 2.4 T . Taking logarithms again produces 1.97584 - 0.87547T, and so f — 2.25689. Using the first equation, Θ = 5/(0.22314 1 / 2 · 2 5 6 8 9 ) = 9.71868. Then S(8) = e -(8/9.7i868) 2 - a5,i ^ 0.52490. = 1 3 . 4 T h e equations to solve are 0.5 = l-exp[-(10,000/6>)T], 0.9 = l-exp[-(100,000/#)r]. Then In 0.5 = -0.69315 = - ( 1 0 , 0 0 0 / # ) r , In 0.1 - - 2 . 3 0 2 5 9 = -(100,000/6>) r . Dividing the second equation by the first gives 3.32192 = 10r, In 3.32192 = 1.20054 = r l n l O = 2.30259r, f = 1.20054/2.30259 = 0.52139. Then 0.69315 = (lO,OOO/0) 0 · 52139 , 0.69315 1 / 0 · 5 2 1 3 9 = 0.49512 = 10,000/0, = 20,197. Θ SECTION 13.1 121 13.5 The two moment equations are 4 + 5 + 21 + 99 + 421 Θ 5 α-Γ 4 2 + 5 2 + 21 2 + 99 2 + 421 2 _ _ 202 5 ' ' (α-1)(α-2)' Dividing the second equation by the square of the first equation gives HO2 a —2 The solution is & = 3.8188. Prom the first equation, Θ = 110(2.8188) = 310.068. For the 95th percentile, , „nr Λ 0.95 = 1 — χ3 8188 310.068 ' ν310.068 + π 0 . 9 5 for 7Γο.95 = 369.37. 13.6 After inflation, the 100 claims from year 1 total 100(10,000)(1.1)2 = 1,210,000, and the 200 claims from year 2 total 200(12,500)(1.1) = 2,750,000. The average of the 300 inflated claims is 3,960,000/300 = 13,200. The moment equation is 13,200 - 0/(3 - 1), which yields Θ = 26,400. 13.7 The equations to solve are 0.2 = ^(18.25) = Φ 0.8 = F(35.8) = 'In 1 8 . 2 5 - μ ( $ ^ ^ The 20th and 80th percentiles of the normal distribution are —0.842 and 0.842, respectively. The equations become _0. 8 42 2 · 9 0 4 '^, σ _ o 3.578 -μ 0.842 -. σ Dividing the first equation by the second yields 2.904 - μ ~ 3.578 -μ The solution is μ — 3.241 and substituting in either equation yields σ = 0.4. The probability of exceeding 30 is Pr(X>30) = = 1 - F(30) = 1 - Φ = 1 - Φ(0.4) = 1 - 0.6554 = 0.3446. ( ^ ^ Ö T ^ ) = J ~ Φ(0,4) 122 CHAPTER 13 SOLUTIONS 13.8 For a mixture, the mean and second moment are a combination of the individual moments. The first two moments are E(X) E(X2) = = p(l) + ( l - p ) ( 1 0 ) = 1 0 - 9 p , p(2) + ( l - p ) ( 2 0 0 ) = 200-198p, Var(X) = 200 - 198p - (10 - 9p) 2 = 100 - 18p - Sip2 = 4. The only positive root of the quadratic equation is p = 0.983. 13.9 We need the 0.6(21) = 12.6th smallest observation. It is 0.4(38) + 0.6(39) = 38.6. 13.10 We need the 0.75(21) = 15.75th smallest observation. It is 0.25(13) + 0.75(14) = 13.75. 13.11 μ = 975, μ2 = 977,916§, σ 2 = 977,916§-975 2 = 27,291 §. The moment equations are 975 = αθ and 27,291 § = αθ2. The solutions are a = 34.8321 and Θ = 27.9915. 13.12 F(x) = (χ/θ)^/[1 + {χ/θγ). The equations are 0.2 = (100/<9)^/[l + (100/0)^] and 0.8 = (400/6>)7/[l + (400/0)^]. From the first equation 0.2 = 0.8(100/#) 7 or ΘΊ = 4(100) 7 . Insert this result in the second equation to get 0.8 = 4 ^ - 7 ( 1 + 4 7 " 1 ), and so 7 = 2 and then Θ = 200. 13.13 E(X) = Jo pxpdx = p/(l + p) = x. p = x/{l - x). 13.14 μ = 3,800 - αθ. μ'2 = 16,332,000, σ2 = 1,892,000 = αθ2. ά = 7.6321, θ = 497.89. 13.15 μ = 2,000 - βχρ(μ+σ 2 /2), μ'2 = 6,000,000 = βχρ(2μ+2σ 2 ). 7.690090 = μ 4- σ 2 /2 and 15.60727 = 2μ + 2σ 2 . The solutions are μ = 7.39817 and σ = 0.636761. Pr(X > 4,500) 13.16 μ = 4.2 = (ß/2)y/2n. = = 1 - Φ[(1η4,500 - 7.39817)/0.636761] 1-Φ(1.5919) = 0.056. ß = 3.35112. 13.17 X is Pareto, and so E(X) = 1,000/(a - 1) = x = 318.4. a = 4.141. SECTION 13.1 123 13.18 rß = 0.1001, rß(l + ß) = 0.1103 - 0.10012 = 0.10027999. 1 + ß = 1.0017981, ß = 0.0017981, r = 55.670. 13.19 rß = 0.166, rß(l + ß) = 0.252 - 0.1662 = 0.224444. 1 + ß = 1.352072, ß = 0.352072, r = 0.47149. 13.20 With τι+1 = 16, we need the 0.3(16) = 4.8 and 0.65(16) = 10.4 smallest observations. They are 0.2(280) + 0.8(350) = 336 and 0.6(450) + 0.4(490) = 466. The equations are 0.3 - 1- \2 ö7 ———— and 0.65 = 1 7 ' 0 +336V 1/2 = 1 + (336/0) and (0.35)~ ln(0.282814) (0.7)-1/2-! (0.19523) 1/3 ' 8614 = = = ln(336/466), 7 = 3.8614, (336/Ö) 38614 , 336/6», θ = 558.74. (0.7)- 7 1/2 iff1 \2 ν07+466τ, = 1 + (466/0)\ 13.21 With n + 1 = 17, we need the 0.2(17) = 3.4 and 0.7(17) = 11.9 smallest observations. They are 0.6(75) + 0.4(81) = 77.4 and 0.1(122) + 0.9(125) = 124.7. The equations are 0.2 - In 0.8 1.20397/0.22314 r Θ = = = = = l-e-(774/ö)Tand0.7=l-e-(124/7/^, 0.22314 - (77.4/0) T and - In 0.3 = 1.20397 = (124.7/0)7 5.39558 = (124.7/77.4) r , ln(5.39558)/ ln(124.7/77.4) = 3.53427, 77.4/0.22314 1/3 · 53427 = 118.32. 13.22 Shifting adds δ to the mean and median. The median of the unshifted distribution is the solution to 0.5 = S(m) = e~m/e for m = 01n(2). The equations to solve are 300 = Θ + δ and 240 = θ 1η(2) + δ. Subtracting the second equation from the first gives 60 = 0[1 — ln(2)] for Θ = 195.53. From the first equation, δ = 104.47. 124 CHAPTER 13 SOLUTIONS 13.2 SECTION 13.2 13.23 For the inverse exponential distribution, n Z(0) = n ^(Ιηθ-θχΤ1 - 2 1 η ^ ) = nlnΘ - ny0 - 2 ^ l n : ^ · , j=\ Ι'(θ) = j=\ ηθ~ι -ny, n 1 1 Θ = y~\ where y = - V — . n f-f x7· For Data Set B, we have Θ = 197.72 and the loglikelihood value is -159.78. Because the mean does not exist for the inverse exponential distribution, there is no traditional method-of-moments estimate available. However, it is possible to obtain a method-of-moments estimate using the negative first moment rather than the positive first moment. That is, equate the average reciprocal to the expected reciprocal: i£i-E(*-)-.-. n ^—: χΊ 3= 1 J This special method-of-moments estimate is identical to the maximum likelihood estimate. For the inverse gamma distribution with a — 2, n 1(0) = n ]Γ(21η6> - θχγ - 3 1 η ^ ) = 2nln0 - nyO - 3 ^ 1 η ^ · , 3=1 ϊ(θ) = 3=1 Ύ 2ηθ~ -ny,0 = 2/y. For Data Set B, Θ = 395.44 and the value of the loglikelihood function is — 169.07. The method-of-moments estimate solves the equation 1,424.4 = - ^ - = ~ , θ = 1,424.4, which differs from the maximum likelihood estimate. For the inverse gamma distribution with both parameters unknown, f(x\Q)= eae~9/x M P / x, ln/(x|0) = a In θ - θχ'1 - (a + 1) lna: - 1ηΓ(α). χ α + 1 1(a) The likelihood function must be maximized numerically. The answer is a — 0.70888 and Θ = 140.16 and the loglikelihood value is -158.88. The method- SECTION 13.2 125 Estimates for Exercise 13.25. Table 13.1 Model Original Exponential Gamma Inv. exponential Inv. gamma 9= a = Θ= ά = Censored 1,424.4 0.55616, θ = 2,561.1 197.72 0.70888, Θ = 140.16 Θ = 594.14 a = 1.5183, Θ = 295.69 Θ = 189.78 a = 0.41612, Θ = 86.290 of-moments estimate is t h e solution to t h e two equations 1,424.4 = 13,238,441.9 = Θ α-Ι' Θ* (α-1)(α-2)' Squaring the first equation and dividing it into the second equation give 6.52489 = a - 1 a-2' which leads to a = 2.181 and then Θ — 1,682.2. This result does not match the maximum likelihood estimate (which had to be the case because the mle produces a model t h a t does not have a mean). 1 3 . 2 4 For the inverse exponential distribution, the cdf is F(x) = e~Qlx. Numerical maximization yields Θ = 6,662.39 and the value of the loglikelihood function is —365.40. For the g a m m a distribution, the cdf requires numerical evaluation. In Excel® the function G A M M A D I S T ( x , a, #,true) can be used. T h e estimates are a = 0.37139 and Θ = 83,020. T h e value of the loglikelihood function is —360.50. For the inverse g a m m a distribution, the cdf is available in E x c e l ® as l - G A M M A D I S T ( l / z , a , l / 0 , t r u e ) . T h e estimates are a = 0.83556 and Θ = 5,113. T h e value of the loglikelihood function is -363.92. 1 3 . 2 5 In each case the likelihood function is / ( 2 7 ) / ( 8 2 ) · · · / ( 2 4 3 ) [ 1 - F ( 2 5 0 ) ] 1 3 . Table 13.1 provides the estimates for b o t h the original and censored d a t a sets. T h e censoring tended to disguise the t r u e n a t u r e of these numbers and, in general, had a large impact on the estimates. 1 3 . 2 6 T h e calculations are done as in Example 13.6, b u t with Θ unknown. T h e likelihood must be numerically maximized. For the shifted data, the estimates are a = 1.4521 and Θ = 907.98. T h e two expected costs are 907.98/0.4521 = 2,008 and 1,107.98/0.4521 = 2,451 for the 200 and 400 deductibles, respectively. For t h e unshifted data, the estimates are a = 1.4521 126 CHAPTER 13 SOLUTIONS Table 13.2 Probabilities for Exercise 13.29 Probability Event Observed Observed Observed Observed at at at at age age age age F(l)-F(0.4) 1-F(0.4) 35.4 and died 35.4 and survived 35 and died 35 and survived 1 1-0.4?/; 1-0.4?/; l-w F(l) = W 1-W and Θ = 707.98. T h e three expected costs are 707.98/0.4521 = 1,566, 2,008, and 2,451 for the 0, 200, and 400 deductibles, respectively. While it is always the case t h a t for the Pareto distribution the two approaches produce identical answers, t h a t will not be true in general. 1 3 . 2 7 Table 13.1 can be used. The only difference is t h a t observations t h a t were surrenders are now treated as x-values and deaths are treated as y-values. Observations t h a t ended at 5.0 continue to be treated as y-values. Once again there is no estimate for a Pareto model. T h e g a m m a parameter estimates are a = 1.229 and Θ = 6.452. 1 3 . 2 8 T h e contribution to the likelihood for the first five values (number of drivers having zero through four accidents) is unchanged. However, for the last seven drivers, the contribution is > 5)] 7 = [1 - p(0) - p ( l ) - p(2) - p(3) - p(4)] 7 , [PT(X and the maximum must be obtained numerically. T h e estimated values are λ = 0.16313 and q = 0.02039. These answers are similar to those for Example 13.8 because the probability of six or more accidents is so small. 1 3 . 2 9 There are four cases, with the likelihood function being the product of the probabilities for those cases raised to a power equal to the number of times each occurred. Table 13.2 provides the probabilities. T h e likelihood function is 0.6w 1-™ V \ 1-0 Aw 1-0.4W 8,, U2 ™14(l-w)16 and its logarithm is / = 141nw 4- 161n(l - w) - 101n(l - 0.4«;). T h e derivative is „ 14 I = w 16 4 + 1—w 1 — QAw SECTION 13.2 127 Set the derivative equal to zero and clear the denominators to produce the equation 0 = 14(1 - w)(l - OAw) - 16w(l - OAw) + Aw (I - w) 14-31.6w + 8w 2 , and the solution is w — (735 = 0.508 (the other root is greater than one and so cannot be the solution). 13.30 The survival function is f e"A,t' I e-2\y-(t-2)\^ cm*W- 0<t<2, t > 2 and the density function is }{t) - -b (t) - I λ2β_2λι_(ί_2)λ!>) t > 2 The likelihood function is L = _ /(1.7)S(1.5)S(2.6)/(3.3)5(3.5) λι6-1.7λΐ6-1.5λ,6-2λ,-0.6λ2λ26-2λι-1.3λ2(,-2λι-1.5λ2 The logarithm and partial derivatives are I = In λι + In λ 2 - 9.2λι - 3.4λ2, w - f-"-0· A2 0X2 and the solutions are λι = 0.10870 and λ 2 = 0.29412. 13.31 Let f(t) = w be the density function. For the eight lives that lived the full year, the contribution to the likelihood is Pr(T > 1) = 1 - w. For the one censored life, the contribution is Pr(T > 0.5) = 1 — 0.5u>. For the one death, the contribution is Pr(T < 1) — w. Then L InL dlnL dw = = _ (l-w)8(l-0.5w)w, 8ln(l - w) + ln(l - O.bw) + lnw, ___8 1— w 0.5 1 — O.bw 2. w' 128 CHAPTER 13 SOLUTIONS Setting the derivative equal to zero and clearing the denominators give 0 = - 8 w ( l - 0.5w) - 0.5w(l - w) + (1 - iu)(l - Q.bw). The only root of this quadratic that is less than one is w = 0.10557 = qx. 13.32 For the two lives that died, the contribution to the likelihood function is /(10), while for the eight lives that were censored, the contribution is 5(10). We have - -*<"-¥ΗΓ- /(*) L = InL = dlnL dk 0 = 31n(fc-10)-51n/c, 3 5 ■ ■ ( - -?)" /(10)aS(10)8=(^)2(l- 10\4 (fc-10) 3 " k ) * fc5 -o k-10 k "' 3k- 5(fc - 10) = 50 - Ik. ?herefor ■e, k = 25. 13.33 We have L = /(l,100)/(3,200)/(3,300)/(3,500)/(3,900)[5(4,000)] 495 _ ^ - l e - l , 1 0 0 / ( 9 ^ - 1 e -3,200/6>^-l e -3,300/6>^-l e -3,500/<9 χ InL dlnL M ^-1 -3,900/0 r -4,000/01495 _ ^-5-1,995,000/0 = -51n6>- — 5 1 Θ^ 1,995,000 Θ 1,995,000 Λ I : θ2 — 0 ' and the solution is Θ = 1,995,000/5 = 399,000. 13.34 For maximum likelihood, the contributions to the likelihood function are (where q denotes the constant value of the time-to-death density function) Event Survive to 36 Censored at 35.6 Die prior to 35.6 Die after 35.6 Contribution Pr(T > 1) = 1 - q Pr(T > 0.6) = 1 - 0.6q Pr(T < 0.6) = 0.6<? Pr(0.6 < T < 1) = OAq SECTION 13.2 129 Then, L In L dlnL dg = (1 - g) 7 2 (l - 0.6^) 15 (0.6^) 10 (0.4^) 3 oc (1 - q)72(l = 72 ln(l - q) + 15 ln(l - 0.6q) + 13 Inq, 72 9 13 _ 1 — </ 1 — 0.6g ^ 0.6q)15q13, The solution to the quadratic equation is q — 0.13911 = ^35. For the product-limit estimate, the risk set at time zero has 100 members. The first 10 observations are all deaths, and so the successive factors are (99/100), (98/99), . . . , (90/91) and, therefore, 5(0.6) = 90/100 - 0.9. The 15 censored observations reduce the risk set to 75. The next three deaths each reduce the risk set by one, and so 5(1) = 0.9(74/75)(73/74)(72/73) - 0.864. Then $35 = 1 - 0.864 = 0.136. 13.35 The density function is f(t) = —Sf(t) = 1/w. For actuary X, the likelihood function is /(l)/(»)/(4)/(4)S(5)-(I)4(l-i)-i-±. Setting the derivative equal to zero gives 0= 4 25 + - 5 , Aw = 25, w = 6.25. wö w° For actuary Y the likelihood function is / ( l ) / ( 3 ) / ( 4 ) / ( 4 ) / ( 6 ) = w-\ This function appears to be strictly decreasing and, therefore, is maximized at w = 0, an unsatisfactory answer. Most of the time the support of the random variable can be ignored, but not this time. In this case, f(t) = 1/w only for 0 < t < w and is zero otherwise. Therefore, the likelihood function is only w~5 when all the observed values are less than or equal to w, otherwise the function is zero. In other words, \ Γ 0, ^ < 6, T( L(w) = < _5 v ' [ ΰ; , w > ß6. This result makes sense because the likelihood that w is less than 6 should be zero. After all, such values of w are not possible, given the sampled values. This likelihood function is not continuous and, therefore, derivatives cannot be used to locate the maximum. Inspection quickly shows that the maximum occurs at w — 6. 130 CHAPTER 13 SOLUTIONS 13.36 The likelihood function is /(4)/(5)/(7)5(3 + r) w^w^w'^w - 3 - r)w~l 4 4 ™ ~ 5(3) " (w - 3) w~ 4 l(w) = \n(w-3-r)-4\n(w-3). ( } w-3-r (w - 3) 4 ' The derivative of the logarithm is i » = — \ w—ό—r —r w—ό Inserting the estimate of w and setting the derivative equal to zero yield 0 = 4 10.671 - r 10.67 = 10.67 - 4(10.67 - r) = -32.01 + 4r, = 8. r 13.37 The survival function is ( e-tx\ 0<£<5, β _ 5 λ ι _ ( ί _ 5 )λ^ β -5λ,-5λ 2 -(>-10)λ: % and the density function is ( 2 - 3 lnL(Ai,A 2 ,A 3 ) 0 ? 0<ί<5, Aae-^'-i*-5)^, 5<£<10, λ36-5λ1-5λ2-(ί-10)λ;% ^ ^ Λ^-βλ,^-ιολ,-δλ,^^θλ,^ολ,-ιΐλ, χ/6~5λι - >1 λιβ-^1, The likelihood function and its logarithm are £(Ai,A ,A ) 5 < * < 10, t -5λ2-5λ.-5\3 Ιηλι - 48λα + 21ηλ 2 - 40λ2 + 41ηΑ3 - 26λ 3 . The partial derivative with respect to λι is Aj"1 - 48 = 0 for Ai = 1/48. Similarly, A2 = 2/40 and A3 = 4/26. 13.38 The density function is the derivative f(x) — α 5 0 0 α χ _ α _ 1 . The likelihood function is L(a) = α 5 5 0 0 5 α ( Π χ ? · ) - α - \ and its logarithm is 1(a) = 5 In a-f (5a) In 5 0 0 - (α + 1)Σ1η^· = 5 Ina + 31.073a - 33.111(a + 1). SECTION 13.2 131 Setting the derivative equal to zero gives 0 = 5c*_1 - 2.038. The estimate is a = 5/2.038 = 2.45. 13.39 The coefficient (2πχ) - 1 / / 2 is not relevant because it does not involve μ. The logarithm of the likelihood function is m = -^(11^)2-3^(15-2-M)2-^(18-M)2-^(21-M)2-^(25.8-/x)2· The derivative is ,Ί,Λ 1 {μ) = 11-M , 15.2-μ -ΪΓ- + ^5^- + 18-μ -ϋ~ + 21 - μ + 25.8 - μ η9ο8βο„ 5 "ST" ^ 5 ^ - = - °· 2 9 8 6 3 μ · Setting the derivative equal to zero yields μ = 16.74. 13.40 The distribution and density function are F(x) = l - e - < * / e > a , f(x) = Θ ^ e - W \ The likelihood function is L(9) = /(20)/(30)/(45)[l-F(50)]2 e-(50/9) _ 2 ]2 ^-6 β -8,325/θ' 2 The logarithm and derivative are 1{θ) = Ι'(θ) = -61η6>-8,3256Τ 2 , -66» _1 + 16,6506»-3. Setting the derivative equal to zero yields Θ = (16,650/6) χ / 2 = 52.68. 13.41 For the exponential distribution, the maximum likelihood estimate is the sample mean, and so xp — 1000 and xs = 1500. The likelihood with the restriction is (using i to index observations from Phil's bulbs and j to index observations from Sylvia's bulbs) 132 CHAPTER 13 SOLUTIONS 10 20 L(r) = Yi^-'eM-xijnlivn-'eM-xj/zn 2= 1 \ 20 10 i=l j=l (6»*)- 3 O exp(-20ip/r-10x s /26l*) (0*)- 3 0 exp(-20,000/6>* - 7,500/6»*). = = Taking logarithms and differentiating yields - 3 O l n 0 * - 27,500/6»*, -30(6»*)" 1 +27,500(r)- 2 . 1(0*) = 1\θ*) = Setting the derivative equal to zero gives Θ = 27,500/30 = 916.67. 13.42 For the first part, L{ß) F(1,000) 62 [1 - F(1,000)] 38 = _ (I _ e-l,000/ö\62/ -l,000/ö\38 Let x = e- 1 » 000 / 0 . Then (l-z)6V8, 621n(l - x ) + 381nx, 62 38 l'{x) = x 1—x Setting the derivative equal to zero yields L(x) l(x) = = 0 x = = = -62x 4- 38(1 - x) 38 - lOOx, 0.38, and then Θ = -1,000/In 0.38 = 1,033.50. With additional information, 62 Up) 1{θ) Ι'(θ) 0 θ 62 [1 - F(1,000)] 38 = m ö ^ e - ^ / e ] e-38,ooo/ö = ^-62e-28?140/Öe-38,000/Ö _ = = = = -621n<9-66,140/0, - 6 2 / 0 + 66,14O/02 - 0, -620 4- 66,140, 66,140/62 = 1,066.77. 0-62-66,140/0 SECTION 13.2 Table 13.3 Likelihood contributions for Exercise 13.44. Observation Probability 1997-1 Pr(iV=l) Pr(iV = l)+Pr(AT=2) Pr(iV=2) Pr(iV=l)+Pr(iV=2) Pr(N=0) ΡΓ(ΛΓ=0)+ΡΓ(ΛΓ=1) Pr(JV = l) _ P r ( N = 0 ) + Pr(iV=l) Pr(iV=0) _ -, Pr(iV=0) — L 1997-2 1998-0 1998-1 1999-0 133 Loglikelihood (l-p)p 1 (l-p)p+(l-p)p2 1+p (I-P)P2 _ P (l-p)p+(l-p)p'2 1+P (1-p) _ 1 (l-p) + (l_p)p 1+p (I-P)P _ p (I-P)4-(I-P)P 1+P -31n(l+p) lnp — ln(l + p) -51n(l+p) 21np-21n(l+p) 0 13.43 The density function is f(x) = 0.5x-°-59-05e-^°\ The likelihood function and subsequent calculations are 10 10 = H 0.5xJ°-5e-°-5e-*i 0 . 5θ/ ) - 0 . 5 oc 0" 5 exp - 0 " 0 · 5 £ *,0.5 3=1 V 0- 5 exp(-488.976r 0 · 5 ), 1(0) = -51n0-488.976T 0 ' 5 , Ι'(θ) = -56>_1 + 244.4850 -1 · 5 = 0, = - 5 0 0 · 5 + 244.485, L{6) = 0 and so 0 = (244.485/5) 2 = 2,391. 13.44 Each observation has a uniquely determined conditional probability. The contribution to the loglikelihood is given in Table 13.3. The total is l(jp) = 3 lnp - l l l n ( l + p). Setting the derivative equal to zero gives 0 = l'(p) = 3p- 1 - 11(1 + p ) ~ \ and the solution is p = 3/8. 13.45 L = 2 η 6 » η ( Π ^ ) β χ ρ ( - έ » Σ ^ ) ) ' = n l n 2 + n l n 0 + £ l n x j - 0 £ > ? , = 0, 0 = η / Σ > 2 . Ι' = ηθ-1 -Σ,'ή 13.46 f(x) = ρχΡ~\ L = pn{X[xj)p~l, np~l + Y^\nXj = 0, p = — n/^2\nxj. I = nlnp + (p - l ^ l n a ^ · , I' = 13.47 (a) L = ( Π ^ ' ) α _ 1 e x p ( - Σ V ö ) [ Γ ( α ) ] _ n ί r ™ Q · ^ = dl/θθ ( a - l ) £ l n x j ; - 0 _ 1 £ Z J - η1ηΓ(α) - n a l n 0 = 81.61837(a - 1) - 38,OOO0_1 - 10 In Γ(α) - 10a In Θ. = 38,OOO0-2 - lOoiT 1 = 0, 134 CHAPTER 13 SOLUTIONS Θ = 38,000/(10 · 12) = 316.67. (b) I is maximized at a = 6.341 and Θ = 599.3 (with I - -86.835). 13.48 μ = ±Σ1ηχ 7 ; = 7.33429. σ 2 = ^ Σ ( 1 η ^ ) 2 - 7.334292 = 0.567405, σ = 0.753263. Ρ Γ ( Χ > 4,500) = 1 - Φ[(1η4,500 - 7.33429)/0.753263] = 1 - Φ(1.4305) = 0.076. 13.49 L = θ-ηβχρ(-Σ^/θ)θ~2Σχ3 = ° · O^Y^Xj/n. 13.50£ = ^ 1 0 I = -ηΐτιθ V = -ηθ~ι -ß-^Xj. + ( Π ^ > χ ρ [ - Σ ^ V = - 1 0 / Τ * 4- / Τ 3 Σ χ) = °· 0 = Λ / Σ Χ 2 / 1 0 = 3.20031. 13.51 / ( χ ) - a x " 0 " 1 . L - α ^ Π ^ Γ 0 " 1 Ζ' — η α - 1 — ^ZlnXj = 0. ά = n / ^ l n x j . / = η\ηα - (α + l J J Z l n x j . 1352 ' «·> - Κ Η έ - έ Η ΐ ) 5 ^ - ^ ' 1(0) = 91η(1Ο-0) + 111η(0), '<·> - -iöb + τ - 0 · 96» = 6» = 11(10-6»), 110/20 = 5.5. 13-53 — Γ(1Ι}) -LJ.1 (irülz£) 2 www = I'M = —\ 25-p w5 (i£zd) Z(w) 0 \ (w_4-p)2 / \ _ 21n(u; - 4 - p) - 51n(w - 4), w —4 —p 5 - = o, w —4 - w-äs^-s· = _Γ/ (w-4)5 10,p=15. ' SECTION 13.2 13.54 The density function is f(x) = 9x 2e L[ß) 135 θ χ Ι . The likelihood function is = ^(66- 2 )e- ö / 6 6 ^(91- 2 )e- ö / 9 1 ö(186- 2 )e- ö / 1 8 6 (e- 0 / 6 6 ) 7 OC β3β-0.148184βϊ 3 1 n ö - 0.148184(9, 36Γ 1 -0.148184 = 0, 3/0.148184 - 20.25. 1(θ) = Ι'(θ) = θ = The mode is 0/2 = 10.125. 13.55 L f(r.U\ /foil*») TT Q400 .Π (400 +a!j )- + _ /= i L(Q) = Π i- /(a) = 71na 4 7aln400 - (a 4-1) ]Tln(400 4- Xj) - 7aln0.8 /'(a) a = = 7 In a - 3.79a -47.29, 7 a " 1 - 3.79 - 0, 7/3.79 - 1.847. 13.56 L i I1 a = = = F ( ioo|e) αΊ7' \( 400 \°] [\ 400+100 ) J J=I α 5 λ 5 α [Π(λ + ^ · ) Γ α - \ 5 l n a 4 5aIn 1,000 - (a 4-1) £ l n ( l , 0 0 0 4 Xj), 5 a " 1 4 34.5388 - 35.8331 = 0, 3.8629. 13.57 (a) Three observations exceed 200. The empirical estimate is 3/20 = 0.15. (b) E(X) = 100α/(α - 1) = x = 154.5, ά - 154.5/54.5 = 2.835, Ρτ(Χ > 200) - (100/200) 2 · 835 = 0.140. (c) f(x) = αΙΟΟ^χ-"- 1 . L = α 2 0 1 0 0 2 0 α ( Π ^ ) ~ α _ 1 · I = 20Inα 4 20αIn 1 0 0 - (a + 1) ^ ΐ η ^ , Ι' = 20a" 1 4 20 In 1 0 0 - £ In Xj = 0, a = 2 0 / ( Σ > ^ - 20 In 100) = 20/(99.125 - 92.103) = 2.848. Pr(X > 200) = (100/200) 2 · 848 = 0.139. 136 CHAPTER 13 SOLUTIONS 13.58 The maximum likelihood estimate is Θ = 93.188. 13.59 (a) Σ (XJ Eh-^r - μ) 2 Σ(>ί-4)+Έ(4-*>+« μ2 Υ^ ί ) 4- — 1 Μ2Σ (b) . 2μ 2 2 /> η, 1 -V ü = de θ ~ 29 2L^ I τ-, ^ ) X η0 μ" χ. 9/3 26» Χ) 1\ - μ) - (χ - μ)22μ 25 μ 1 ^\Xj 1 θ^{ ηθ -μ22(χ ομ ^ θ l n Ö (χ - μγ 9/1.2 2μ: Ζ ^ 2 η η ^(Xj-μγ θ oc θη/2 exp L 1 2ημ + ηχ + 2 = 0, r ( x - A i ) 2 = 0, 2μ2ί η Σ(ί-ί)' 13.60 - τ τ ί --) oc 0 n / 2 exp Γ %XP θ^ (Xj - μ) 2 jixj-μ)2 2 mfi~ SECTION 13.3 ί(μ, θ) = - In θ - - ^2 ^—' ομ -^— = d£ öfl m j (xj ~ I1)2 + constant, 2^mj ·* m — Θ 2 . ™>j < 0, hence, maximum. nl 2 (9 λ-ι o 1 y-^ 2 1 ^-^ = — 2 n 0 = nj^m^-x) 2 ] -1 —ö2 Ö0 137 7T"~Ö < 0, hence, maximum. 2^ j=l n ^ j=l } n ί(θ) = Σΐηρ(ζ,·) + Γ(0)5>,·-η1η<?(0), Hiß) = ^ ) ^ χ , - η Therefore, *<*> r'(%(0) -*. But, <7'(0) = E(X) = μ(θ), r'(%(<?) and so μ(#) = x. 13.3 SECTION 13.3 13.62 In general, for the exponential distribution, Ι'(β) = 1"{θ) = Ε[1"(θ)] = Var(0) = -ηθ^+ηχθ-2, ηθ~ - 2ηχθ~ζ, ηθ~2 - 2ηθθ~3 = 2 -ηθ~2, θ2/η, where the third line follows from E(X) = E(X) = Θ. The estimated variance for Data Set B is Var(0) = l,424.4 2 /20 = 101,445.77 and the 95% confidence 138 CHAPTER 13 SOLUTIONS interval is 1,424.4±1.96(101,445.77)1/2, or 1,424.4±624.27. Note that in this particular case Theorem 13.5 gives the exact value of the variance, because Var(ö) = Var(X) - Var(X)/n - θ2/η. For the gamma distribution, 1(α,θ) = 021(α,θ) da2 2 θ 1{α,θ) δαδθ θ21(α,θ) δθ2 -^XjO"1 l)Y^]nxj Σι Θ1(α,θ) da dl(a,9) ΘΘ (a- ηΓ'(α) 1 i = i lnXj~~T(aY~ ^XjO'2 - - η 1 η Γ ( α ) - na In 0, ü ' -ηαθ~\ 3= 1 Γ(α)Γ"(α) - Γ'(α) 2 1 » = -ηθ~\ -2 ] Γ χάθ~3 + ηαθ'2 = -2ηχθ~3 + ηαθ'2. The first two second partial derivatives do not contain Xj, and so the expected value is equal to the indicated quantity. For the final second partial derivative, E(X) = E(X) = αθ. Therefore, n Ι(α,0) = no Γ(α)2 ηθ~ ηαθ The derivatives of the gamma function are available in some better computer packages, but are not available in Excel®. Using numerical derivatives of the gamma function yields 82.467 0.0078091 J(&,0) = 0.0078091 0.0000016958 and the covariance matrix is 0.021503 -99.0188 -99.0188 1,045,668 Numerical second derivatives of the likelihood function (using h\ = 0.00005 and h2 = 0.25) yield 1{ά,θ) 82.467 0.0078091 0.0078091 0.0000016959 SECTION 13.3 139 and covariance matrix Γ 0.021502 [ -99.0143 -99.0143 1 1,045,620 J * The confidence interval for a is 0.55616 ± 1.96(0.021502)1/2, or 0.55616 ± 0.28741, and for Θ is 2,561.1 ± 1.96(1,045,620)1/2, or 2,561.1 ± 2,004.2. 13.63 The density function is f(x\0) function is L{6) = = = 0 ~ \ 0 < x < Θ. The likelihood 0" n , 0<xi,...,xn 0, otherwise. <0, As a function of 0, the likelihood function is sometimes zero, and sometimes θ~η. In particular, it is θ~η only when Θ is greater than or equal to all the xs. Equivalently, we have L{9) — 0, Θ < m a x ( x i , . . . , x n ) = 0_n, Θ > max(xi,... ,xn). Therefore, the likelihood function is maximized at Θ = m a x ( # i , . . . , xn). Note that the calculus technique of setting the derivative equal to zero does not work here because the likelihood function is not continuous (and therefore not differentiable) at the maximum. From Examples 10.5 and 10.7 we know that this estimator is asymptotically unbiased and consistent, and we have its variance without recourse to Theorem 13.5. According to Theorem 13.5, we need 1(θ) = Z'(0) = /"(#) = — nln#, Θ > m a x ( x i , . . . , x n ) , — n # - 1 , Θ > m a x ( x i , . . . ,xn), 2 n 0 ~ , θ> m a x ( x i , . . . , x n ) . Then E[Z"(0)] = ηθ~2 because with regard to the random variables, ηθ~2 is a constant and, therefore, its expected value is itself. The information is then the negative of this number and must be negative. With regard to assumption (ii) of Theorem 13.5, !"Λ>=[-°-^->13.64 From Exercise 13.62 we have ά = 0.55616, Θ = 2,561.1, and covariance matrix Γ 0.021503 -99.0188 1 -99.0188 1,045,668 I ' 140 CHAPTER 13 SOLUTIONS The function to be estimated is g(a, θ) = αθ with partial derivatives of Θ and a. The approximated variance is [ 2,561.1 0.55616 ] 0.021503 -99.0188 2,561.1 0.55616 -99.0188 1,045,668 182,402. The confidence interval is 1,424.4 dz 1.96^/182,402, or 1,424.4 ± 837.1. 13.65 The partial derivatives of the mean are οβμ+σ 2 /2 θμ geß+<r2/2 da = β//+σ /2 = σβμ+σ*/2 = 134.458. = 123.017, The estimated variance is then [ 123.017 134.458 ] " 0.1195 0 0 0 0.0597 123.017 134.458 13.66 The first partial derivatives are da dl(a,ß) dß = - 5 α - 3β + 50, = -3α - 2 / 3 + 2. The second partial derivatives are d2l(a,ß) da2 2 d l(a,ß) dß2 2 d l(a,ß) dadß = -5, = -2, = -3, and so the information matrix is 5 3 3 2 The covariance matrix is the inverse of the information, or 2 -3 -3 5 2,887.73. SECTION 13.3 141 13.67 For the first case, the observed loglikelihood and its derivatives are 1 00 9 1 00( ö Z(0) = 6 2 1 n [ l - e - ' ° / ] + 381ne- ' V , i'(0) = -62e- 1 ' 0 0 0 / e (l,000/6' 2 ) I _ g-1,000/0 ~*~ 38,000 02 2 -62,0006"LMO/eg- + 38,0006>~2 - βδ,ΟΟΟβ" 1 ' 000 /^"· 2 I _ e -i,ooo/0 βδ,ΟΟΟβ1'000/0 - 100,000 ö2(ei.ooo/e _ i) 1»(Ω\ _02(ei.ooo/e _ i ^ o o O e ^ / ^ O O O i T 2 -^δ,ΟΟΟβ 1 - 000 /* - 100,000)^(e 1 - 0 0 0 ^ - 1) flV'O0O/el,000fl-2] 5/4^1,000/0 _ ! ) 2 Evaluating the second derivative at Θ = 1033.50 and changing the sign give Ι[θ) = 0.00005372. The reciprocal gives the variance estimate of 18,614. Similarly, for the case with more information 1(θ) = -621n0-66,1406T1, ϊ(θ) = - 6 2 0 _ 1 + 66,140ο -2 , 1"(θ) = 626»~2 - 132,2806Τ3, 1"{ 1,066.77) = -0.000054482. The variance estimate is the negative reciprocal, or 18,355.2 13.68 (a) f(x) = pxp-\ lnf(x) = lnp + ( p - l ) l n x , d2\nf(x)/dp2 I(p) = nE(p~2) = np~2, Var(p) = p2/n. (b) From Exercise 13.46, p = - η / Σ lnxj. The CI is p ± (c) μ = p/(l+p). The CI ispil+p)-1 p / ( l +p). δμ/dp = (1+p)-2. ± 1.96p(l +ρ)~2/^/η. = -p~2, IMp/y/n. Var(A) = (1 + ρ)-*ρ2/η. 13.69 (a) Inf(x) = -In6» - z/6>, <92ln/(x)/ö6>2 = 0 - 2 - 26Γ 3 :Ε, 1(0) = ηΕ(-θ~2 + 2Θ~3Χ) = ηθ~2, Var(0) = θ2/η. (b) From Exercise 13.49, 0 = x. The CI is x ± 1.965/v/n. (c) Var(X) = Θ2. dVax(X)/d0 = 2Θ. Var(X) = x 2 . 2 2 4 2 (2θ) θ /η = Αθ /η. The CI is x ± 1.96(2ä 2 )/v^. 2 Comparing the variances at the same Θ value would be more useful. Var[Vai(X)] = 142 CHAPTER 13 SOLUTIONS 1 3 . 7 0 In f(x) = - ( 1 / 2 ) 1 η ( 2 π ^ ) - χ 2 / ( 2 ^ ) , ο 2 1 η / ( χ ) / ο ^ 2 = ( 2 0 2 ) " 1 - x 2 ( 0 3 ) - 1 , and 7(0) = ηΕ[-(2θ2)-1 +Χ2(Θ3)'1} = η ( 2 β 2 ) " 1 since X - iV(O,0). Then MSE(0) = Var(0) = 2 0 2 / n = 20 2 /n = 8/40 = 0.2. 1 3 . 7 1 T h e maximum likelihood estimate is 0 = x = 1,000. Var(0) = Var(x) = 0 2 / 6 . T h e quantity to be estimated is 5(0) = e " 1 ' 5 0 0 / 0 , and then S , (0) = l , 5 O O 0 - 2 e - 1 ' 5 0 0 / 0 . From the delta method Var[5(0)] = [5'(0)] 2 Var(0) = [ l , 5 0 0 ( l , 0 0 0 ) - 2 e - 1 ' 5 0 0 / 1 ' 0 0 0 ] 2 ( l , 0 0 0 2 / 6 ) = 0.01867. T h e standard deviation is 0.13664 and with 5(1,000) = 0.22313, the confidence interval is 0.22313 ± 1.96(0.13664), or 0.22313 ± 0 . 2 6 7 8 1 . An alternative t h a t does not use the delta method is to start with a confidence interval for 0: 1,000 ± 1.96(l,000)/^6, which is 1,000 ± 800.17. P u t t i n g the endpoints into 5(0) produces the interval 0.00055 to 0.43463. 1 3 . 7 2 (a) L = F ( 2 ) [ l - F ( 2 ) ] 3 . F ( 2 ) = f^Xxe'^dx e " 4A . I = 1η(1-βλ = (1/4)1η(4/3). 4λ ) - 1 2 λ . dl/dX = (l-e- (b) Ρ α ( λ ) = 1 - e ~ 4 \ P 2 (A) = e " 4 \ 4A 1 )- 4e" 4A 2 -e -\x* = - 1 2 = 0. er P i (A) = 4e~ 4 A , Ρ^(λ) = 4X 0 = 1 = 3/4. -4e~4A. / ( λ ) - 4[16e~ 8 A /(l - e" 4 A ) + 1 6 e " 8 A / e - 4 A ] . Var(A) = {4[16(9/16)/(l/4) + 16(9/16)/(3/4)]}"1 - 1 / 1 9 2 . 1 3 . 7 3 T h e loglikelihood function is / = 81.61837(a—1)-38,OOO0~ 1 - 1 0 I n Γ ( α ) - 10a In 0. Also, a = 6.341 and 0 - 599.3. Using v = 4, we have 021(α,θ) da2 j_ — 1(6.3416341,599.3)-2/(6.341,599.3)+/(6.3403659,599.3) (0.0006341) 2 021(α,θ) Οαθθ j_ ~ 1(6.34131705,599.329965)-/(6.34131705,599.270035) -/(6.34068295,599.329965)+/(6.34068295.599.270035) (0.0006341)(.05993) = 0.0166861, _;_ — /(6.341,599.35993)-2/(6.341,599.3) + /(6.341,599.25007) — _ (0.05993) 2 ö 2 /(q,fl) dB2 7(ά,0) and its inverse is Var = 1.70790 0.0166861 0.0166861 0.000176536 7.64976 -723.055 -723.055 74,007.7 _ — _-, 1 7η7αη -'U'JU^ Π η η ΐ 7 Γ ^ Γ U.UUUl/ΟΟόΟ, n SECTION 13.3 143 The mean is αθ, and so the derivative vector is [ 599.3 6.341 1. The variance of αθ is estimated as 227,763 and a 95% CI is 3,800 ± 1.97^/227,763 = 3,800 ± 935. 13.74 a = 3.8629. 1η/(ζ) = 1ηα + α 1 η λ - ( α + 1)1η(λ + ζ). d 2 ln/(x)/<9a 2 —a - 2 . 7(a) = na~2. Var(a) = a 2 / n . Inserting the estimate gives 2.9844. Ε(ΧΛ 500) = /•500 / Jo :χα1,000 α (1,000 + +500 / xy^'dx x)-a-ldx αΙ,ΟΟΟ" (1,000 + 1,000 -(2/3)«1'500 a —1 v ' ' a —1 Evaluated at a, it is 239.88. The derivative with respect to a is 1,000 2 (a-1) "' V3; ( 1,500 (2\a 1,500, /2\ which is —39.428 when evaluated at a. The variance of the LEV estimator is (-39.4298) 2 (2.9844) = 5,639.45, and the CI is 239.88 ± 133.50. 13.75 (a) Let θ = μ/ΐχΐ + τ - 1 ) . From Appendix A, E(X) = ΟΓ(1 + τ " 1 ) = μ. (b) The density function is f(x) = exp { - [ r < 1+ ^ its logarithm is \nf(x) = Γ(1 + τ " 1 ) χ l)x T ]} ~x [ Γ(1+ ^ l)x ]\ + In r + r In m+r-1) + ( τ - l)lnx. + In r + r In m+r-1) + (r - 1) In Xj μ and The loglikelihood function is n , 1(μ) and its derivative is ew.±{jul±g*-z > , 144 CHAPTER 13 SOLUTIONS Setting the derivative equal to zero, moving the last term to the right-hand side, multiplying by μ, and dividing by r produce the equation Σ T-1)xj T(l+ μ 1 ra+r- ) J 1 ra+r- ) Σχΐ = η i=i n (13.1) ,,1/r and, finally, 1/r r(1+,-) Σ Ϊ ^: u=i (c) The second derivative of the loglikelihood function is ητ ~7& Erom(13.1), Σ ? = ι *; can be written as 1"{μ) = ~ ■τίτ + Ι ^ Ι + τ - ^ ν ^ Σ ^ · j=i ηλ'τμ\ Γ(1+τ-ΐ) and therefore the observed information - j - r i r + lJlXl + r - y A — nr "72" r ( r + l)n Γ2 2 — ηΎ'τμτ Γ(1 + τ - η r n ~2~> μ μ μ and the negative reciprocal provides the variance estimate. (d) The information requires the expected value of XT. From Appendix A, it is 6>ΤΓ(1 + 7) = θτ = [μ/Γ(1 + τ ~ 1 ) ] τ . Then Ε[/"(μ)1 — -τ(τ + 1)Γ(1 + τ-1Υμ-τ-*η\μ/Γ(1 μ τ(τ + \)η τ2η ητ 2 ' — μ* _ μ* μ = + τ-1)Υ Changing the sign, inverting, and substituting μ for μ produce the same estimated variance as in part (c). SECTION 13.4 145 (e) To obtain the distribution of μ, first obtain the distribution of Y = XT. We have SY(y) = Ρτ(Υ>ν) = Ρτ(Χτ>ν) = Pi(X>y1^) = exp I - = exp -^ 1 L f μ '- \Γ(ΐ + τ-ΐ)Υ J y \ H—μ—J r which is an exponential distribution with mean [μ/Γ(1+τ 1)]τ. Then 52?=i XT has a gamma distribution with parameters n and [μ/Γ(1 + r _ 1 ) ] T . Next look at ΈχΙ- μτ 3=1 Multiplying by a constant changes the scale parameter, so this variable has a gamma distribution with parameters n and 1. Now raise this expression to the 1/r power. Then has a transformed gamma distribution with parameters a = n, Θ = 1, and T = T. To create μ, this function must be multiplied by μ / η 1 / τ , which changes the scale parameter to θ = μ/η1/τ. Prom Appendix A, E(A) μΐΧη + τ - 1 ) nVrr(n) ' A similar argument provides the second moment and then a variance of - ^ VariW v a r W 13.4 Γ ( η + 2Τ-1) η2/τΓ(η) - "Tfr + T-12 )8 2 η /^Γ(η) ' SECTION 13.4 13.76 The likelihood function is L(9) = < Γ 2 ν 2 8 ' 4 8 8 / θ . Theni(Ö) = -2Ο1η(0)28,488/6». With Θ = 1,424.4, 10) = -165.23. A 95% confidence region solves 1(θ) = -165.23 - 1.92 = -167.15. The solutions are 946.87 and 2,285.07. Inserting the solutions in 5(200) = e _ 2 0 0 / e produces the interval 0.810 to 0.916. The symmetric interval was 0.816 to 0.922. 146 CHAPTER 13 SOLUTIONS 13.5 SECTION 13.5 13.77 For the Kaplan-Meier estimates, the variances are (by Greenwood's formula) Vär(<J45) = (0.750)2 Var(<746) = (0.584)2 1 8(7) 1 8(7) J_ + + = 0.02344, 7(6) J 1 1 - 0.03451. 6(5) + 5(4) For exact exposure, the deaths were 2 and 3 and the exposures 5.9 and 5.5 for the two ages. The estimated variances are Vär(<745) (1 - 0.28751) 2 —Ö = 0.02917, - (1 - 0.42042) 2 —^ = 0.03331. 5.5^ Var(£'46; The actuarial exposures are 7.2 for both ages. The estimated variances are v-w- Ϊ Var(q45) Var(g 46 ) = 0-28751(0.71249) ^7.2 = 0.02845, 0.42042(0.57958) 0.03384. 7.2 CHAPTER 14 CHAPTER 14 14.1 SECTION 14.7 14.1 (a) q = X/m, E(q) = E(X)/m b ( ) Var(<?) (c) I = = = = mq/m = q. Var(X)/m 2 = Var(X)/(nm 2 ) mq{\ — q)/(nm2) = q(\ — q)/(nm). \~] In ( \ + Xjlnq + (m — Xj) ln(l — q), n l ' = Σχά<1~λ ~ (jn-Xj)(l n -q)~\ J^-Xjq-Z-im-XjKl-q)-2, l" = j=i I(q) = E(-l") = n[mqq-2 + {m - mq)(l - q)-2} = nmlf'+a-?)- 1 ]. The reciprocal is (1 — q)q/{nm). Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugman, Harry H. Panje-r, Gordon E. Williiiot Copyright © 2012 John Wiley & Sons, Inc. FourthlAJ 148 CHAPTER 14 (d) q±za/2s/q{l-■q)/(nm). 1 ^ = Pr n 1 — Gi ί -za/s> < _ _J ~ Q_^_ \A(1 - q)l(nm) < Za/2 and so \q-q\ < ; nm(q — tf ' which implies Then, 1 V nm <z2a/2q(l-q). (nm + z^i2)q2 - (2nmq + ^/ 2 )<7 + nmq2 < 0. The boundaries of the CI are the roots of this quadratic: 2nmq + z2a/2 ± za/2\/l + Anmq(l - q) 2(nm + zl/2) 14.2 Because r = l,ß = X. Var(X) = Var(X)/n - β(1 + β)/η, n I = j=i = n ^ l n P r ( 7 V = x3) = ^ l n [ / 3 ^ ( l + β)"^'1] j=i ^ χ , · 1 η ( / 3 ) - ( χ , + 1)1η(1 + /3), n 3=1 E{1") = - η / 3 / Τ 2 + η(/3 + 1 ) ( 1 + / ? Γ 2 = η/[/3(1+/3)]. The reciprocal matches the true variance of the mle. 14.3 (a) The mle is the sample mean, [905+2(45) +3(2)]/10,000 - 0.1001, and the CI is 0.1001 ± 1.96^/0.1001/10,000 = 0.1001 ±0.0062, or (0.0939,0.1063). (b) The mle is the sample mean, 0.1001. The CI is 0.1001 ± 1.96χ/0.1001(1.1001)/10,000 = 0.1001 ± 0.0065, or (0.0936,0.1066). SECTION 14.7 149 (c) Numerical methods yield r = 56.1856 and ß = 0.00178159. (d) q = x/A = 0.025025. (e) 0.025025 ± 1.96>/0.025025(0.974975)/40,000 = 0.025025 ± 0.001531 and 2(10,000)(4)(0.025025) + 1.962 ± lMy/1 4- 4(10,000)(4)(0.025025(0.974975) 2[10,000(4)4-1.96 2 ] or 0.025071 ±0.001531. (f) The likelihood function increases as m —> oo and gf —> 0. 14.4 (a) The sample means are underinsured, 109/1,000 = 0.109, and insured, 57/1,000 = 0.057. (b) The Poisson parameter is the sum of the individual parameters, 0.109 40.057 = 0.166. 14.5 (a) λ is the sample mean, 0.166. (b) Let riij be the number observations of j counts from population i, where j = 0 , 1 , . . . and 2 = 1,2. The individual estimators are λ^ = Σ ^ = ο ^ η ΰ · The estimator for the sum is the sum of the estimators, which is λ = X}°L0 j(ji\j + ri2j), which is also the estimator from the combined sample. (c) β = 0.166. (d) Numerical methods yield r = 0.656060 and β = 0.253026. (e)q = 0.166/7 = 0.0237143. (f) The likelihood function increases as m —► oo and q —► 0. 14.6 (a) λ = 15.688. When writing the likelihood function, a typical term is (P4 +P5 4-p6) 47 \ a n ( l the likelihood must be numerically maximized. (b) β = 19.145. 150 CHAPTER 14 (c) f = 0.56418 and ß = 37.903. 1 4 . 7 T h e maximum likelihood estimate is the sample mean, which is 1. The variance of the sample mean is the variance of a single observation divided by the sample size. For the Poisson distribution, the variance equals the mean, so the estimated variance is 1/3,000. T h e confidence interval is 1 ± 1.645^/1/3,000 = 1 ± 0.030, which is 0.970 to 1.030. 14.8 The estimator is unbiased, so the mean is λ. The variance of the sample mean is the variance divided by the sample size, or X/n. The coefficient of variation is y/X/n/X — 1/ynX. T h e maximum likelihood estimate of A is the sample mean, 3.9. T h e estimated coefficient of variation is 1/^/10(3.9) = 0.1601. CHAPTER 15 CHAPTER 15 15.1 SECTION 15.2 1 5 1 Let f(v) = y$itmi+Ü>»' W = \nYy = 100e™ and dy = lOOe^dw. Thus, _ W ^ >~ In 100 = ln(F/100). Then 12(4.801121)12100ew _ 12(4.801121)12 13 100e™(0.195951 + w + lnl00) " (4.801121 + w ) 1 3 ' y > ' which is a Pareto density with a — 12 and Θ = 4.801121. 15.2 , , λ π(α|χ) oc α1010010(* ^ ^ (X α10^-1 οΡ-χβ-"/θ ^ ^ βχρΐ-^θ'1 - ΙΟΙηΙΟΟ + ^ l n ^ ) ] , which is a gamma distribution with parameters 10 + 7 and (θ~ — 10 In 100 + Σ ΐ η α ^ ) - 1 . The mean is aßayos = (1O + 7)(0 _ 1 - 10In 100 + ^ I n ^ ) " 1 . For the mle: I = /' = 101ηα + 10α1η100-(ο; + 1 ) ^ ; ΐ η χ ^ 10a" 1 + 10 In 1 0 0 - £ In x,; = 0, Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugmau, Harry H. Paiijor, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. FourthlSl 152 CHAPTER 15 for am\c — 10(^2m xj ~ 10 In 100) l. The two estimators are equal when 7 = 0 and Θ = oo, which corresponds to π(α) = a - 1 , an improper prior. n 15.3 (a) π(μ, σ|χ) oc σ exp -Σ^^Ψ^ (b) Let I = 1ηπ(μ,σ|χ) = — (n + 1) Ιησ — —σ Then 9/ d/z 2 TjOn^j — A*) 2 · 2 2 1η =^" Σ ( ^-^(- 1 ) = ° 2 and the solution is μ = ^ ^Z l n a ; j· Also, 81 £- = -(n + \)σ~ι + σ - 3 ^ ( l n a ; , · - μ)2 = 0, da and : (c) ^ϊΣ(ΐη^·-Α) 5 π(μ, σ|χ) oc exp ex = 1/2 Σ2 V 1 / In Xj; — μ £ 1 ημ 2 — 2μ V l n x 7 + ΥΎΐη:τ7·)5 Pl-2~^ ^2 — oc exp 1 μ2 - 2μμ + μ2 "2 ^ which is a normal pdf with mean μ and variance σ /η. The 95% HPD interval is μ db l.96a/y/n. 15.4 (a) π υ \ \*) ^ ρ200 Qß+l exp[-e-\\ + Zxj)} oc ^201+^ > which is an inverse gamma pdf with parameters 200 + ß and 30,000 4- λ. (b) £(2<9|x) - 2f^ll+_\- At β = A = 0, it is 2 ^ ^ - 301.51 while at β = 2 ? > 30,250 and λ = 250, it is 2 2 01 = 301.00. For the first case, the inverse gamma parameters are 200 and 30,000. For the lower limit, 0.025 - Pr(20 < a) = F{a/2) = 1 - Γ(200; 60,000/α) SECTION 15.2 153 for a = 262.41. Similarly, the upper limit is 346.34. With parameters 202 and 30,250, the interval is (262.14,345.51). (c) Var(20|x) = 4 Var(0|x) = 4 (30,000+A)2 _ /30,0 /30,000+λλ2 ^V 19' (199+j0)(198+0) 199+/3 I The two vari- ances are 459.13 and 452.99. The two CIs are 301.51±42.00 and 301.00±41.72. (d) I = -β'^Ο,ΟΟΟ - 200 In(9. V = 6>"230,000 - 2OO0"1 = 0. 0 = 150. For the variance, V = 0~"2 £>.,· - 2OO0~\ I" = - 2 0 " 3 ^Xj + 2OO0-2, E(-Z") = 20~3(2OO0) - 2OO0-2 = 2OO0"2 and so Var(0) = 02/2OO and Var(20) = 02/5O. An approximate CI is 300 ± 1.96(150)/y/EÖ = 300 ± 41.58. 15 5<a)/w ' - Ü^-v-^wwr^-*^ ίΚ\ Γ(ο + b) Γ(χ + α)Γ(Κ -x + b) \x)v(a)T(b) T(a + b + K) K\ T(x + a) T{K -x + b) T(a + b) x\(K-x)\ T(a) T(b) T(a + b + K) a(a + 1) · · · (x + a - 1) b(b + 1) - · ■ (K - x + b - 1) xl (K- x)\ K\ (a + b) ■ ■ ■ (K + a + b - 1) (-i)Tr6) (-aKb) Also, Ε(Χ\Θ) = ΚΘ, E{X) = Ε[Ε(Χ\Θ)} = Ε(ΚΘ) = K a a +b - 0 ) 6 - 1 , which is a beta density. There(b) π(0|χ) oc θΣχί(1 -θ)ΣΚ'-χ^θα~1{1 fore, the actual posterior distribution is with mean Ε(θ\ν) = a + b + Y^Kj' 154 CHAPTER 15 a)ß Jo 1 „, . ,w . o-l\-«-l - Γ ( α + 1)(χ + /Γ Γ(α)/?α α/Τ^/Τ'+χ)-"-1. = Using E(X|6>) = 6»"1, E(X) = Ε[Ε(Χ\Θ)}=Ε{Θ-1) , έ Τ - ν » / % _Γ(α-1)/Τ Γ(α)/Τ Γ(α)/Τ Jo β(α-ΐγ (b) π(0|χ) oc βηε-βΣχίθα-'ίβ-βΙβ = $n+r-1e-^^xi+ß '>, which is a gamma density. Therefore, the actual posterior distribution is π(0|χ) = Γ(η + α ) ( Σ ^ + / ? _ 1 ) with mean Ε(β|χ) = 15 • 7 ( a ) / ( x ) U+ a = f f{x\e)b{6)de x Jr(a)T(b)J0 y } r + x - 1\ Γ(α + b) T(r + α)Γ(& 4- x) x J Γ(α)Γ(6) Γ(Γ + a + 6 + x) T(r + x) Γ(α + 6) Γ(α + r)T(b + x) Γ(Γ)Χ! Γ(α)Γ(&) Γ(α + τ + 6 + χ ) ' (b) π(β|χ) α Π ^ Ι ^ ) Hence, π (#|x) is beta with parameters a* = a + nr, E(0|x) = a* a* + 6* a + nr a + nr + 6 4- Σ Xj SECTION 15.2 155 15.8 (a) /»oo / /(χ\θ)ο(θ)άθ = / Jo ß° / n oQ J^-e-^-^±-e^e-ßOde Γ(α) άθ V 2π r(« + j) (2π)ν2Γ(α)[ι(χ_μ)2+/?]^ r ( a + i) Γ (b) π(β|χ) α oc 1 ," [Π/(^Ίβ)]τ(β) ο " / 2 β χ ρ [ - | ^ ( α : , - μ ) 5 6, αa~- l1e -09 ^^•-v^·, where α * = α + §, ß* = ß + \ Y^(XJ - μ) 2 . Therefore, b(0|x) is gamma ( α % / η and E(g)x) = fr = ^ ° % _ μ ) ,. 15.9 (a) By the convolution formula, y=0 ^ o U A i + ey VI + ÖJ J/=0 te-y)! S e e (l+0)" The pf is of the form (5.6) with τ{θ) = Ιηβ, p(x) = £ * = 0 {")/(x - y)\, and q(9) = ee(l + e)N. (b) Theorem 15.18 applies, yielding π(θ) \q{e))-ke^kr^r'{e) ο(μ, k) β(μ, k) k l 0» - (l + 0)-Nke-kf) 156 CHAPTER 15 Because JQ π(θ)άθ = 1, c(M,fc) /»OO = / e»k-\i + e)-Nke-kede Jo pOO _ / ^ik-1^ ,^^k-Nk-\-l)-ßk-\e~kt^ JO = Γ(μ&)Φ(μΛ, μ/c - TVfc + 1, fc). 15.10 (a) Let N be Poisson(A). px(l-p)n-xe-x\n n\ n=x v m—nr ' X *,-\ °° \xe-^[(l-p)\}n P 1 — p J x\ ^—' (n — x)\ n=x ( P \ ' e - ^°°[ ( l - p ) A r ' x Λ-λ \l~pj P x\ n—0 ^ ^ 1 — p) e~pX(p\)x 6 x\ n\ [(l-p)\]xeV-rix a Poisson distribution with parameter p\. (b) Let N be binomial(ra, r). _ n\ *-^x\[n n=x \-p) v ,„_„. — x)\ ml n\(m — n)\ 7 v ^ ^ .Vx n=0 ' 1 - r . . /, ....... ... n!(m-n X ! SECTION 15.2 -1-pJ 2 - V K(1 - r'V»x\^ \ (V-&Y 1-r ) ψ ([1-P]r\n ^ \ l - r p ) - {&) 1-r ( l - r y n \l-rp) 1 (m-s)! (m-x)\ n\(m — n — x)! / 1— r m! (1 - r ) m 157 x x!(m — x)! \ 1 — rpy [1 — p\r \ ( 1 — r \ (m — x)! n\(m — n — x)! rp ) \ 1 — rp m—x Σ ^ n=0 m! (pr)x(l-rp)rn~x x!(m — x)! which is binomial (m,pr). (c) Let TV be negative binomial (r, /?). ^—' χ!(η — x)! \" / β 1 V r ( r + l)---(r + n ) :E / 1 \ \l + ß) r 1 °° l(i-p) x ! ^ 1 + /3 P γ ( i y 1 ^\β(ίΡ; /3(1-P) \l + ß) l-p) 1) ™! i + /V vi + /v x\^[ r(r + 1) · · · (r + n — 1) (n — x)! n+x 1+β r(r + 1) · · · (r + n -f x — 1) ( Pß Y( Σ n=0 n! 1 r^(i-p) I 1+0 Vr(r+ 1)···(Γ + Χ - 1 ) I n (r + x) · · · (r + n -f x — 1) n! and the summand is almost a negative binomial density with the term 158 CHAPTER 15 missing. Place this term in the sum so the sum is one and then divide by it to produce 1+ ß pß V ( 1 Y r(r + l ) - - - ( r + z - l ) χ} 1+PßJ \1+Pßj which is negative binomial with parameters r and pß. 15.11 Let D be the die. Then Pr(D = 2|2,3,4,l,4) P r ( 2 , 3 , 4 , l , 4 | D = 2)Pr(£> = 2) ~ Pr(2,3,4,1,4|D = 1) Pr(D = 1) + Pr(2,3,4,1, A\D == 2) Pr(£> == 2) _ 113 13 1 666662 3 I I I I I i λλ^ΙΆΙ 666662 1 "666662 o _ ° A' 15.12 (a) P r ( y = 0) = / J e-e(\)d0 = 1 - e" 1 = 0.63212 > 0.35. (b) P r ( y = 0) = Jo Θ2(1)άθ = 1/3 < 0.35. (c) P r ( y = 0) = / ^ ( l - 6>)2(l)cZ6> = 1/3 < 0.35. Only (a) is possible. 15.13 _ Pr(if = l/4|d = 1) Pr(d = 1\H = 1/4) Pr(if = 1/4) Pr(d = \\H = 1/4) Pr(H = 1/4) + Pr(d - \\H = 1/2) P r ( # = 1/2) 11 2 4 5 _ 45 + 2 5 Δ ό 15.14 The equations are αθ = 0.14 and αθ2 = 0.0004. The solution is a = 49 and 0 = 1/350. Prom Example 15.8, the posterior distribution is gamma with a = 49 + 110 - 159 and Θ = (l/350)/[620(l/350) + 1] = 1/970. The mean is 159/970 - 0.16392 and the variance is 159/970 2 = 0.00016899. 15.15 The posterior pdf is proportional to t e - t 5 i e - ' = t 2 e-* 6 , SECTION 15.2 159 which is a gamma distribution with a — 3 and ß = 1/6. The posterior pdf is n(t\x = 5) = l08t2e-6t. 15.16 Sm\Q ~ bin(m, Q) and Q ~ beta(l,99). Then, E(5 m ) - E[E(S m |Q)] = E(mQ) = ^ γ ^ = 0.01m. For the mean to be at least 50, m must be at least 5,000. 15.17 π(β\χ) oc f(x\ß)n(ß) = / T V ^ l O O / r V 1 0 ^ oc ß~*e-W+*)/ßm This is an inverse gamma distribution with a — 3 and # — 10 -f x. The posterior mean is θ/(α — 1) = (10 + x)/2. 15.18 The likelihood function is £(0) = ö -l e -HOO/0 e -l e -32OO/0 e -l e -33OO/0 ö -l e -35OO/e 0 -l e -39OO/0 xe -4000/01495 000/014 The posterior density is proportional to π(0|χ) oc ^ - V 1 · 9 9 5 · 0 0 0 ' · . This is an inverse gamma distribution with parameters 5 and 1,995,000. The posterior mean is 1,995,000/4 = 498,750. The maximum likelihood estimate divides by 5 rather than 4. 15.19 (a) ί(χ\θι,θ2) 6(0i|0 2 ) 6(02) ^2 / exp 02 2πσ exp 2 Γ(α)^ 2 6 ^2 a \2 //3 \ 2 160 CHAPTER 15 7Γ(0Ι,02|Χ) π(0ι|0 2 )π(0 2 ) OC ;=i OC Λη/2 e x ^2 5 4Σ 2 ^·-^) ,1/2 ^2 -/302 Θ2 P x exp \2 9<*-1 ,α+^ψ--! = Ö: x exp ^ - 0 2 π(βι|β 2 ,χ) //3 ^Κ^'^Σ^-·.)' oc ττ(0ι,0 2 |χ) oc exp exp ~ 2 which is normal with variance σ* = [^- + n#2] = θ, (\+ησ2\ which satisfies % = ^ + 6>2n:r. Then, u* = J-^J + ^ ^ . For the posterior distribution of 0 2 , 7r(#2|x) OC = Λ I n-r- d mean μ+, )άθλ [π(θ1,θ2\χ ff an 1 / βχρ {-|[(^) 2+ Σ(^-^) 2 ]}^· Now 52(XJ - Öi)2 = ^Z(xj - x) 2 + n (x - #i) 2 and, therefore, π(0 2 |χ) oc θ°+!ψ 1 expi-ff2 [/? + | £ ) ( * , ■ - ä ) 2 j } /•■Ι^) + η(α;-θι) ■döi. SECTION 15.3 161 To evaluate the integral, complete the square as follows: θ\ - μ \ 2 ,_ 2 -\-η(χ-θι) 1 2 θ2 2 '<?(£+-)-»-(£+«·) + £+■ = -1^(^+η) μ + ησ2χ\ [θ1 2 + 1 + ησ J #2 Τ ■ 22 ^ 2 (μ + ησ χ) 2 σ (1 + ησ2) Ö — ηχ σ2 T h e first term is a normal density and integrates to [^(l/^" 2 + n ) ] - 1 / 2 . T h e second term does not involve θ\ and so factors out of the integral. T h e posterior density contains #2 raised to the a + ^ — 1 power and an exponential term involving #2 multiplied by β + 2\Υΐ*ά-*) ^ 2 +\ 1 2 (μ -f ησ2χ)2 μ2 σ2(1 + ησ2) " ^ 2 η(χ-μ) 2 ^ ν ^ ηχ 2 2 ^ + 2 ( 1 + ησ ) ' This constitutes a g a m m a density with the desired parameters. (b) Because the mean of θ ι given ©2 and x does not depend on #2? it is also the mean of ©i given just x, which is μ*. T h e mean of ©2 given x is the ratio of the parameters, or a + n/2 1 β + 15.2 Σ { χ . . ^ 2 + n(x — μ ) 2 2(1 + η σ 2 ) SECTION 15.3 1 5 . 2 0 By Exercise 5.26(a), 5 | © has pf of the form fs\s(s\0) = n m ] n . Thus, π(θ\χ) oc β(χ\θ)π(θ) 7r(0)oC = 3=1 or π(θ\χ) oc exp x e r(e)EU n A^ ) m\ 162 CHAPTER 15 But π ( φ ) oc fS\e{8\0)ir{e) α ^ p f p also. 15.21 The posterior pdf is proportional to e-'0° β~ 0! θ e-20. = This is an exponential distribution. The pdf is n(6\y = 0) = 2e~29. 15.22 The posterior pdf is proportional to e θηΐ ^ 1! 0 _ /)2 -20 This is a gamma distribution with parameters 3 and 0.5. The pdf is n(0\y = l)^402e-20. 15.23 From Example 15.8, the posterior distribution is gamma with a = 50 + 177 - 227 and Θ = 0.002/[l,850(0.002) + 1] = 1/2,350. The mean is αθ = 0.096596 and the variance is αθ2 = 0.000041105. The coefficient of variation is λ / α ^ / ( α 0 ) = 1/^/5 = 0.066372. 15.24 The posterior pdf is proportional to ( ? V ( 1 " 0) 3-1 60(l - Θ) oc 02(1 - Θ)3. This is a beta distribution with pdf π(θ\τ = 1) = 6O02(1 - 6>)3. 15.25 The prior exponential distribution is also a gamma distribution with a = 1 and Θ — 2. From Example 15.8, the posterior distribution is gamma with a = l + 3 = 4 a n d # = 2/[l(2) + 1] = 2/3. The pdf is n{X\y = 3) = 27λ 36-3λ/2/32 15.26 (a) The posterior distribution is proportional to ζ)θ2(1 - θ)280θ3(1 - Θ)Α oc 0 5 (1 - Of which is a beta distribution. The pdf is π ( % = 2) = 27720 5 (l - 0) 5 . (b) The mean is 6/(6 + 6) = 0.5. SECTION 15.3 163 15.27 The posterior distribution is n(q\2) = 6<?2(1 -q)Hp{\ -q) oc q3(l -qf. The mode can be determined by setting the derivative equal to zero. 0 = 3q2(l — q)3 — 3g 3 (l — q)2, which is equivalent to q = 1 — q for q — 0.5. SECTION 15.3 163 15.27 The posterior distribution is n(q\2) = 6<?2(1 -q)Hp{\ -q) oc q3(l -qf. The mode can be determined by setting the derivative equal to zero. 0 = 3q2(l — q)3 — 3g 3 (l — q)2, which is equivalent to q = 1 — q for q — 0.5. CHAPTER 16 CHAPTER 16 SOLUTIONS 16.1 SECTION 16.3 16.1 For Data Set B truncated at 50, the maximum likelihood parameter estimates are f = 0.80990 and Θ = 675.25, leading to the graph in Figure 16.1. For Data Set B censored at 1,000, the estimates are f = 0.99984 and Θ = 718.00. The graph is in Figure 16.2. For Data Set C, the parameter estimates are f = 0.47936 and Θ — 11,976. The plot is given in Figure 16.3. 16.2 For Data Set B truncated at 50, the plot is given in Figure 16.4. For Data Set B censored at 1,000, the plot is given in Figure 16.5. 16.3 The plot for Data Set B truncated at 50 is given in Figure 16.6. For Data Set B censored at 1,000, the plot is given in Figure 16.7. Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugman, Harry H. Panjor, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. Fourthl65 166 CHAPTER 16 SOLUTIONS Weibull Fit 1 ' 0.90.80.7^ 0.6 - —Model I Emplrical| 0.4 0.3- 0.2 0.1 0 () 700 1400 2100 2800 3500 X Figure 16.1 Cdf plot for Data Set B truncated at 50. Weibull Fit 0.8 -I 0.70.6- *tr*^i 0.5 Model So.4- I Empirical 0 30.20.1 n . (3 200 1 1 1 400 600 800 1000 X Figure 16.2 16.2 Cdf plot for Data Set B censored at 1,000. SECTION 16.4 1 6 . 4 For D a t a Set B truncated at 50, the test statistic is 0.0887, while the critical value is unchanged from the example (0.3120). T h e null hypothesis is not rejected, and it is plausible t h a t the d a t a came from a Weibull population. For D a t a Set B censored at 1,000, the test statistic is 0.0991, while the critical value is 0.3041. T h e null hypothesis is not rejected. SECTION 16.4 167 Weibull Fit 0.00006 γ0.00005 \ 0.00004 j -Model - Empirical § 0.00003 J 0.00002 \ 0.00001 | oi 0 50000 100000 150000 200000 250000 300000 Figure 16.3 Pdf and histogram for Data Set C. Weibull Fit 0.1 -r 0.08 \ 0.06 \ 0.04 \ 0.02 \ 0 a -0.02 \ -0.04 -0.06 \ -0.08 ] -0.1 0 3500 Difference plot for Data Set B truncated at 50. Figure 16.4 16.5 T h e first step is to obtain the distribution function. It can be recognized as an inverse exponential distribution or the calculation done as /»DO pX F(x) = / J0 2y-2e-2/ydy= / J2/x 2 ( 2 / z ) - 2 e ~ z (2 z~2)dz /»OC - / J2/x e~zdz = e~2lx. In the first line, the substitution z — 2/y was used. T h e calculations are in Table 16.1. T h e test statistic is the maximum from the final column, or 0.168. 168 CHAPTER 16 SOLUTIONS Figure 16.5 Difference plot for Data Set B censored at 1,000. Weibull Fit 1 -1 • 0.90.8 0.7 ~ 0-6f0.50.4 0.3 0.2 0.1 ^»*\» ^—· · Π - () 0.2 0.4 0.6 0.8 I F„(x) Figure 16.6 p-p plot for Data Set B truncated at 50. 1 6 . 6 T h e distribution function is F(x)= I / 2(1+ y)-0dy= - ( 1 + y) -2\x l - ( l + x)"2. T h e calculations are in Table 16.2. T h e test statistic is the maximum from the final column, 0.189. 1 6 . 7 For D a t a Set B truncated at 50, the test statistic is 0.1631, which is less t h a n the critical value of 2.492, and the null hypothesis is not rejected. SECTION 16.4 169 Weibull Fit 1 0.9 0.80.7- * 0.6£- 0.5 0.40.3 0.2 0.1 0 F i g u r e 16.7 p-p plot for Data Set B censored at 1,000. Table 16.1 Calculations for Exercise 16.5. X F(x) Compare to Max difference 1 2 3 5 13 0.135 0.368 0.513 0.670 0.857 0, 0.2 0.2, 0.4 0.4, 0.6 0.6, 0.8 0.8, 1 0.135 0.168 0.113 0.130 0.143 Table 16.2 Calculations for Exercise 16.6. X F(x) Compare to Max difference 0.1 0.2 0.5 1.0 1.3 0.174 0.306 0.556 0.750 0.811 0, 0.2 0.2, 0.4 0.4, 0.6 0.6, 0.8 0.8, 1 0.174 0.106 0.156 0.150 0.189 For D a t a Set B censored at 1,000, the test statistic is 0.1712 and the null hypothesis is again not rejected. 16.8 T h e calculations for D a t a Set B truncated at 50 are in Table 16.3. T h e sum is 0.3615. W i t h three degrees of freedom, the critical value is 7.8147 and the Weibull model is not rejected. T h e p-value is 0.9481. T h e calculations for D a t a Set B censored at 1,000 are in the Table 16.4. T h e sum is 0.5947. W i t h two degrees of freedom, the critical value is 5.9915 and the Weibull model is not rejected. T h e p-value is 0.7428. 170 CHAPTER 16 SOLUTIONS Data Set B truncated at 50 for Exercise 16.8. Table 16.3 Range P Expected Observed x2 50-150 150-250 250-500 500-1,000 1,000-2,000 0.1599 0.1181 0.2064 0.2299 0.1842 0.1015 3.038 2.244 3.922 4.368 3.500 1.928 3 3 4 4 3 2 0.0005 0.2545 0.0015 0.0310 0.0713 0.0027 2,000-OQ Data Set B censored at 1,000 for Exercise 16.8. Table 16.4 Range P Expected Observed x2 0-150 150-250 250-500 500-1,000 l,000-oo 0.1886 0.1055 0.2076 0.2500 0.2484 3.772 2.110 4.151 4.999 4.968 4 3 4 4 5 0.0138 0.3754 0.0055 0.1997 0.0002 Table 16.5 Data Set C truncated at 7,500 for Exercise 16.8. Range P Expected Observed x2 7,500-17,500 17,500-32,500 32,500-67,500 67,500-125,000 125,000-300,000 300,000-oo 0.3299 0.2273 0.2178 0.1226 0.0818 0.0206 42.230 29.096 27.878 15.690 10.472 2.632 42 29 28 17 9 3 0.0013 0.0003 0.0005 0.1094 0.2070 0.0513 T h e calculations for D a t a Set C are in Table 16.5. T h e sum is 0.3698. W i t h three degrees of freedom, the critical value is 7.8147 and the Weibull model is not rejected. T h e /rvalue is 0.9464. 1 6 . 9 T h e calculations are in Table 16.6. For the test, there are three degrees of freedom (four groups less zero estimated parameters less one) and at a 5% significance level, the critical value is 7.81. T h e null hypothesis is accepted and, therefore, the d a t a may have come from a population with the given survival function. 171 SECTION 16.4 Calculations for Exercise 16.9 Table 16.6 Interval Observed Expected 0 to 1 1 to 2 2 to 3 3 to 4 Total 21 27 39 63 150 150F(1) = 150[F(2) 150[F(3) 150[F(4) 150 Table 16.7 No. of claims Observed 50 122 101 92 0 1 2 3 or more Chi-square 150(2/20) = 15 F(l)] = 150(4/20) = 30 F(2)] = 150(6/20) = 45 F(3)] = 150(8/20) = 60 §1 3Q = 2.40 0.30 If =0.80 60 = 0.15 3.65. Calculations for Exercise 16.10. Chi-square Expected 1 6438 365c" = 70.53 365(1.6438)e -1 · 6438 = 115.94 365(1.6438) 2 e- 1 6 4 3 8 /2 = 95.29 365 - 70.53 - 115.94 - 95.29 = 83.24 115.94 = 5.98 = 0.32 §05=0-34 § £ £ = 0.92 1 6 . 1 0 Either recall t h a t for a Poisson distribution the mle is the sample mean or derive it from L(X) = (β^γ°(Χβ-"γ 22 lnL(A) = 6001ηλ-365λ, 0 = 600λ_1-365, λ = 600/365 = 1.6438. / x2 λ\ /Xze~xx 101 / χ 3ö - λ X e~ λ * /χό 92 rsr \ 6 0 0 oc Λ e -365λ , For the goodness-of-fit test, the calculations are in Table 16.7. T h e last two groups were combined. T h e total is 7.56. There are two degrees of freedom (four groups less one estimated parameter less one). At a 2.5% significance level, the critical value is 7.38 and, therefore, the null hypothesis is rejected. T h e Poisson model is not appropriate. 1 6 . 1 1 W i t h 365 observations, the expected count for k accidents is 365Pr(AT = /c) 365e-°- 6 0.6 fc! T h e test statistic is calculated in Table 16.8. T h e total of the last column is the test statistic of 2.85. W i t h three degrees of freedom (four groups less one less zero estimated parameters), the critical value is 7.81 and the null hypothesis of a Poisson distribution cannot be rejected. 172 CHAPTER 16 SOLUTIONS Calculations for Exercise 16.11. Table 16.8 No. of accidents a Observed Expected Chi-square 209 111 33 7 3 2 200.32 120.19 36.06 7.21 1.08 0.14° 0.38 0.70 0.26 1.516 T h i s is 365 loss the sum of the other entries to reflect the expected for five or more accidents. T h e last three cells are grouped for an observed of 12 and an expected of 8.43. 16·12χ2 ^(0,-50)3 = 50 J'=l = 20 20 = 0.02 ^ Ο , 2 - 1 0 0 ^ Ο , + 2 0 ( 5 0 ) 2 j=l j=l 0.02(51,850 - 100(1,000) + 50,000] = 37. W i t h 19 degrees of freedom, the probability of exceeding 37 is 0.007935. 1 6 . 1 3 T h e density function and likelihood functions are αθα α20θ20α (χ + ey m=i(*;+*) ft+1 Then, 20 l(a, Θ) = 20 l n ( a ) + 20a ln(0) - ( a + 1) ^ \n(x3 + Θ). Under the null hypothesis, a = 2 and Θ = 3.1, the loglikelihood value is —58.7810. Under the alternative hypothesis, a = 2 and Θ — 7.0, and the loglikelihood value is —55.3307. Twice the difference is 6.901. There is one degree of freedom (no estimated parameters in the null hypothesis versus one in the alternative). T h e p-value is the probability of exceeding 6.901, which is 0.008615. 16.14 Π OC (nk + efc + l ) - - - e f e / nk\ k=\ /p»*(l+/3)-S("fc+efc). 1 1 + ßJ \l + ß T h e logarithm is ( l n / 3 ) ^ ] n , - [ l o g ( l + / 3 ) ] ^ K + e,), k=\ k=l SECTION 16.4 173 and setting the derivative equal to zero yields 6 6 k=\ k=\ ß'1 J > f c - (1 + ß)-1 ^2(nk + ek) = 0 for ß = Y?k=1 nk/YX=l ek = 0.09772. The expected number is Ek = ßek, which is exactly the same as for the Poisson model. Because the variance is ekß(l -f /3), the goodness-of-fit test statistic equals the Poisson test statistic divided by 1 + /?, or 6.19/1.09772 = 5.64. The geometric model is accepted. 16.15 The null hypothesis is that the data come from a gamma distribution with a = 1, that is, from an exponential distribution. The alternative hypothesis is that a has some other value. From Example 13.4, for the exponential distribution, Θ — 1,424.4 and the loglikelihood value is —165.230. For the gamma distribution, a = 0.55616 and Θ = 2,561.1. The loglikelihood at the maximum is L 0 = -162.293. The test statistic is 2(-162.293 4- 165.230) = 5.874. The p-value based on one degree of freedom is 0.0154, indicating there is considerable evidence to support the gamma model over the exponential model. 16.16 For the exponential distribution, Θ = 29,721 and the loglikelihood is -406.027. For the gamma distribution, a = 0.37139, Θ = 83,020, and the loglikelihood is —360.496. For the transformed gamma distribution, a = 3.02515, Θ = 489.97, f = 0.32781, and the value of the loglikelihood function is —357.535. The models can only be compared in pairs. For exponential (null) versus gamma (alternative), the test statistic is 91.061 and the p-value is essentially zero. The exponential model is convincingly rejected. For gamma (null) versus transformed gamma (alternative), the test statistic is 5.923 with a p-value of 0.015 and there is strong evidence for the transformed gamma model. Exponential versus transformed gamma could also be tested (using two degrees of freedom), but there is no point. 16.17 Poisson expected counts are 0 1 2 3 or more : 10,000(e-° 1001 ) = 9,047.47, : 10,000(0.1001e-° 1001 ) = 905.65, : 10,000(0.1001 2 e-° 1001 /2) = 45.33, : 10,000 - 9,047.47 - 905.65 - 45.33 - 1.55. The test statistic is (9,048 - 9,047.47)2/9,047.47 + (905 - 905.65)2/906.65 +(45 - 45.33) 2 /45.33 + (2 - 1.55)2/1.44 = 0.14. 174 CHAPTER 16 SOLUTIONS There are two degrees of freedom (four groups less one and less one estimated parameter) and so the 5% critical value is 5.99 and the null hypothesis (and therefore, the Poisson model) is accepted. Geometric expected counts are 9,090.08, 827.12, 75.26, and 7.54, and the test statistic is 23.77. W i t h two degrees of freedom, the model is rejected. Negative binomial expected counts are 9,048.28, 904.12, 45.97, and 1.63, and the test statistic is 0.11. W i t h one degree of freedom, the model is accepted. Binomial (m = 4) expected counts are 9,035.95, 927.71, and 36.34 (extra grouping is needed to keep the counts above 1), and the test statistic is 3.70. W i t h one degree of freedom, the model is accepted (critical value is 3.84). 1 6 . 1 8 Poisson expected counts are 847.05, 140.61, and 12.34 (grouping needed to keep expected counts above 1), and the test statistic is 5.56. W i t h one degree of freedom, the model is rejected. Geometric expected counts are 857.63, 122.10, 17.38, and 2.89, and the test statistic is 2.67. W i t h two degrees of freedom, the model is accepted. Negative binomial expected counts are 862.45, 114.26, 19.10, and 4.19, and the test statistic is 2.50. W i t h one degree of freedom, t h e model is accepted. Binomial (m = 7) expected counts are 845.34, 143.74, and 10.92, and the test statistic is 8.48. W i t h one degree of freedom, the model is rejected. 1 6 . 1 9 (a) To achieve a reasonable expected count, the first three rows are combined as well as the last two. T h e test statistic is 16,308. W i t h five degrees of freedom, the model is clearly rejected. (b) All nine rows can now be used. T h e test statistic is 146.84. W i t h seven degrees of freedom, the model is clearly rejected. (c) T h e test statistic is 30.16. W i t h six degrees of freedom, the model is clearly rejected. 1 6 . 2 0 x = 0.155140 and s2 = 0.179314. E(N) = λ ι λ 2 and Var(TV) = λ α ( λ 2 + λ | ) . Solving the equations yields λλ = 0.995630 and λ 2 = 0.155821. For the secondary distribution, fa = e - ° 1 5 5 8 2 1 ( 0 . 1 5 5 8 2 1 ) V j ! and then g0 = e x p [ - 0 . 9 9 5 6 3 ( l e -o.i5582i)] = 0.866185. Then, ^ 0.99563 . £ 9k = 2^ 1 JJjQk-j. For the goodness-of-fit test with two degrees of freedom, see Table 16.9, and the model is clearly rejected. SECTION 16.5 Table 16.9 Value Calculations for Exercise 16.20. Observed Expected Chi-square 103,704 14,075 1,766 255 53 103,814.9 13,782.0 1,988.6 238.9 28.6 0.12 6.23 24.92 1.09 20.82 0 1 2 3 4+ 175 Total 53.18 16.21 (a) (20,592-20,596.76)2 (2,651-2,631.03)2 , (297-318.37)2 20,596.76 ~τ~ 2,631.03 ~*~ 318.37 + (41-37.81) 2 (8-5.03) 2 37.81 "i" 5.03 0.0011 + 0.1516 + 1.4344 + 0.2691 + 1.7537 = 3.6099, df = 5 - 2 - 1 = 2, given a = 0.05, => χ* 0 0 5 = 5.99 > 3.6099 => fit is acceptable, or Pr(xL· l (2) > 3.6099) « 0.16 => fit is acceptable. (b) 0 Pi = . (r - l)/3 (a + 6 ) Ä « a + 6 (r + 1) £ i + /3 2 r +1 p2 pi f &\ = ^ = ^ = 5 S S = 0.12774, 1+β ρο 20,596.76 318.37 2,631.03 0.12101, 0.12774 = 1.05565 => f(2 - 1.05565) - 1.05565 0.12101 1.11786, a = - ^ = * H ™ =0.11427 1+/3 r 16.3 SECTION 16.5 16.22 These are discrete data from a discrete population, so the normal and gamma models are not appropriate. There are two ways to distinguish among the three discrete options. One is to look at successive values of kuk/rik-i· 176 CHAPTER 16 SOLUTIONS Table 16.10 Criterion Tests for Exercise 16.23. Exponential Weibull Trans, gam. B truncated at 50 K-S A-D x2 p- Value Loglikelihood SBC 0.1340 0.4292 1.4034 0.8436 -146.063 -147.535 0.0887 0.1631 0.3615 0.9481 -145.683 -148.628 0.0775 0.1649 0.5169 0.7723 -145.661 -150.078 B censored at 1,000 K-S A-D x2 p- Value Loglikelihood SBC 0.0991 0.1713 0.5951 0.8976 -113.647 -115.145 0.0991 0.1712 0.5947 0.7428 -113.647 -116.643 N/A N/A N/A N/A N/A N/A C K-S A-D x2 p- Value Loglikelihood SBC N/A N/A 61.913 lO" 1 2 -214.924 -217.350 They are 2.67, 2.33, 2.01, 1.67, decreasing, indicating a binomial to compute the sample mean and T h e variance is considerably less N/A N/A 0.3698 0.9464 -202.077 -206.929 N/A N/A 0.3148 0.8544 -202.077 -209.324 1.32, and 1.04. T h e sequence is linear and distribution is appropriate. An alternative is variance. They are 2 and 1.494, respectively. t h a n the mean, indicating a binomial model. 1 6 . 2 3 T h e various tests for the three d a t a sets produce the following results. For D a t a Set B truncated at 50, the estimates are a = 0.40982, f = 1.24069, and Θ — 1,642.31. For D a t a Set B censored at 1,000, there is no maximum likelihood estimate. For D a t a Set C, the maximum likelihood estimate is a = 4.50624, τ = 0.28154, and Θ = 71.6242. T h e results of the tests are in Table 16.10. For D a t a Set B truncated at 50, there is no reason to use a three-parameter distribution. For D a t a Set C, the transformed g a m m a distribution does not provide sufficient improvement to drop the Weibull as the model of choice. 1 6 . 2 4 T h e loglikelihood values for the two models are —385.9 for the Poisson and - 3 8 2 . 4 for the negative binomial. T h e test statistic is 2 ( - 3 8 2 . 4 + 3 8 5 . 9 ) = SECTION 16.5 177 7.0. There is one degree of freedom (two parameters minus one parameter), and so the critical value is 3.84. The null hypothesis is rejected, and so the data favors the negative binomial distribution. 16.25 The penalty function subtracts ln(100)/2 = 2.3 for each additional parameter. For the five models, the penalized loglikelihoods are: generalized Pareto: -219.1 - 6.9 - -226.0; Burr: -219.2 - 6.9 = -226.1; Pareto: -221.2 - 4.6 = -225.8; lognormal: -221.4 - 4.6 = -226.0; and inverse exponential: —224.3 — 2.3 = —226.6. The largest value is for the Pareto distribution. 16.26 The loglikelihood function for an exponential distribution is 1(θ) = In J p - V * ' / 0 = X > l n 0 - ^ = -ηΐηθ - ^ . Under the null hypothesis that Sylvia's mean is double that of Phil's, the maximum likelihood estimates of the mean are 916.67 for Phil and 1833.33 for Sylvia. The loglikelihood value is Un = = - 2 0 In 916.67 - 20,000/916.67 - 10 In 1,833.33 - 15,000/1833.33 -241.55. Under the alternative hypothesis of arbitrary parameters the loglikelihood value is /ait = - 2 0 In 1,000 - 20,000/1,000 - 10 In 1,500 - 15,000/1,500 - -241.29. The likelihood ratio test statistic is 2(-241.29 + 241.55) = 0.52, which is not significant (the critical value is 3.84 with one degree of freedom). To add a parameter, the SBC requires an improvement of ln(30)/2 = 1.70. Both procedures indicate that there is not sufficient evidence in the data to dispute Sylvia's claim. 16.27 The deduction to compute the SBC is (r/2) ln(n) = (r/2) ln(260) = 2.78r, where r is the number of parameters. The SBC values are —416.78, -417.56, -419.34, -420.12, and -425.68. The largest value is the first one, so that is the model to select. 16.28 Both the Poisson and negative binomial have acceptable p-values (0.93 and 0.74) with the Poisson favored. The Poisson has a higher loglikelihood value than the geometric. The negative binomial improves the loglikelihood by 0.01 over the Poisson, less than the 1.92 required by LRT or the 3.45 required by SBC. The Poisson is acceptable and preferred. 178 CHAPTER 16 SOLUTIONS Table 16.11 Calculations for Exercise 16.31. Model Parameters Poisson Geometric Negative binomial λ = 1.74128 β = 1.74128 f = .867043, β == 2.00830 NLL Chi-square df 2,532.86 2,217.71 2,216.07 1,080.80 170.72 165.57 5 7 6 1 6 . 2 9 Both the geometric and negative binomial have acceptable p-values (0.26 and 0.11), with the geometric favored. The geometric has a higher loglikelihood value t h a n the Poisson. T h e negative binomial improves the loglikelihood by 0.71 over the geometric, less t h a n the 1.92 required by LRT or the 2.30 required by SBC. T h e geometric model is acceptable and preferred. 1 6 . 3 0 T h e negative binomial distribution has the best p-value (0.000037) though it is clearly unacceptable. For the two one-parameter distributions, the geometric distribution is a better loglikelihood and SBC t h a n the Poisson. T h e negative binomial model improves the loglikelihood by 1132.25 — 1098.64 = 33.61, more t h a n the 1.92 required by the likelihood ratio test and is more t h a n the 0.5 In 503 = 3.11 required by the SBC. For the three models, the negative binomial is the best, but is not very good. It turns out t h a t zero-modified models should be used. 1 6 . 3 1 (a) For k = 1 , 2 , 3 , 4 , 5 , 6 , 7 , the values are 0.276, 2.315, 2.432, 2.891, 4.394, 2.828, and 4.268, which, if anything, are increasing. T h e negative binomial or geometric models may work well. (b) T h e values appear in Table 16.11. Because the sample variance exceeds the sample mean, there is no mle for the binomial distribution. (c) T h e geometric is better t h a n the Poisson by b o t h likelihood and chi-square measures. T h e negative binomial distribution is not an improvement over the geometric as the NLL decreases by only 1.64. W h e n doubled, 3.28 does not exceed the critical value of 3.841. T h e best choice is geometric, but it does not pass the goodness-of-fit test. 1 6 . 3 2 For each d a t a set and model, Table 16.12 first gives the negative loglikelihood and then the chi-square test statistic, degrees of freedom, and p-value. If there are not enough degrees of freedom to do the test, no p-value is given. For Exercise 14.3, the Poisson is the clear choice. It is the only oneparameter distribution acceptable by the goodness-of-fit test and no twoparameter distribution improves the NLL by more t h a n 0.03. SECTION 16.5 Table 16.12 179 Results for Exercise 16.32. Ex. 14.3 Ex. 14.5 E x . 14.6 Ex. 16.31 3,339.66 .00;1;.9773 488.241 5.56;1;.0184 3,578.58 16,308;4;0 2,532.86 1,081;5;0 Geometric 3,353.39 23.76;2;0 477.171 .28;1;.5987 1,132.25 146.84;7;0 2,217.71 170.72;7;0 Neg. bin. 3,339.65 .01;0 476.457 1.60;0 1,098.64 30.16;6;0 2,216.07 165.57;6;0 ZM Poisson 3,339.66 .00;0 480.638 1.76;0 2,388.37 900.42;3;0 2,134.59 37.49;6;0 ZM geometric 3,339.67 .00;0 476.855 1.12;0 1,083.56 1.52;6;.9581 2,200.56 135.43;6;0 ZM logarithmic 3,339.91 .03;0 475.175 .66;0 1,171.13 186.05;6;0 2,308.82 361.18;6;0 ZM neg. bin. 3,339.63 .00;-! 473.594 .05;0 1,083.47 1.32;5;.9331 2,132.42 28.43;5;0 Poisson For Exercise 14.5, the geometric is the best one-parameter distribution and is acceptable by the goodness-of-fit test. T h e best two-parameter distribution is the negative binomial, b u t the improvement in the NLL is only 0.714, which is not significant. (The test statistic with one degree of freedom is 1.428). T h e three-parameter ZM negative binomial improves t h e NLL by 3.577 over the geometric. This is significant (with two degrees of freedom). So an argument could be made for the ZM negative binomial, but the simpler geometric still looks to be a good choice. For Exercise 14.6, the best one-parameter distribution is the geometric, but it is not acceptable. T h e best two-parameter distribution is the ZM geometric, which does pass the goodness-of-fit test and has a much lower NLL. T h e ZM negative binomial lowers the NLL by 0.09 and is not a significant improvement. For Exercise 16.31, none of the distributions fit well. According to the NLL, the ZM negative binomial is the best choice, but it does not look very promising. 16.33(a) T h e mle is p = 3.0416 using numerical methods. (b) T h e test statistic is 785.18 and with three degrees of freedom the model is clearly not acceptable. 180 CHAPTER 16 SOLUTIONS Table 16.13 Results for Exercise 16.34. Ex. 14.3 Ex. 14.5 Ex. 14.6 Ex. 16.31 Poisson-Poisson 3,339.65 0.01;0 478.306 1.35;0 1,198.28 381.25;6;0 2,151.88 51.85;6;0 Polya-Aeppli 3,339.65 0.01;0 477.322 1.58;0 1,084.95 4.32;6;0.6335 2,183.48 105.95;6;0 Poisson-I.G. 3,339.65 0.01;0 475.241 1.30;0 1,174.82 206.08;6;0 2,265.34 262.74;6;0 Poisson-ETNB 3,339.65 0.01;-1 473.624 0.02;-l 1,083.56 1.52;5;0.9112 did not converge 1 6 . 3 4 Results appear in Table 16.13. T h e entries are the negative loglikelihood, the chi-square test statistic, degrees of freedom, and the p-value (if the degrees of freedom are positive). For Exercise 14.3, the Poisson cannot be topped. These four improve the loglikelihood by only 0.01 and all have more parameters. For Exercise 14.5, b o t h the Poisson-inverse Gaussian and P o i s s o n - E T N B improve the loglikelihood over the geometric. T h e improvements are 1.93 and 3.547. W h e n doubled, they are slightly (p-values of 0.04945 and 0.0288, respectively, with one and two degrees of freedom) significant. T h e goodnessof-fit test cannot be done. T h e geometric model, which easily passed the goodness-of-fit test, still looks good. For Exercise 14.6, none of the models improved the loglikelihood over the ZM geometric (although the P o i s s o n - E T N B , with one more parameter, tied). As well, the ZM geometric has the highest p-value and is clearly acceptable. For Exercise 16.31, none of the models have a superior loglikelihood versus the ZM negative binomial. Although this model is not acceptable, it is the best one from among the available choices. For the five 1 6 . 3 5 T h e coefficient discussed in the section is LH^Z^±MR^ d a t a sets, use t h e empirical estimates. For the last two sets, the final category is for some number or more. For these cases, estimate by assuming the observations were all at t h a t highest value. T h e five coefficient estimates are (a) -0.85689, (b) -1,817.27, (c) -5.47728, (d) 1.48726, and (e) - 0 . 4 2 1 2 5 . For all but D a t a Set (d), it appears t h a t a compound Poisson model will not be appropriate. For D a t a Set (d), it appears t h a t the Polya-Aeppli model will do well. Table 16.14 summarizes a variety of fits. T h e entries are the negative loglikelihood, the chi-square test statistic, the degrees of freedom, and the p- value. SECTION 16.5 Table 16.14 181 Results for Exercise 16.35. (a) (b) (c) (d) (e) 36,188.3 190.75;2;0 206.107 0.06;1;.8021 481.886 2.40;2;.3017 1,461.49 267.52;2;0 841.113 6.88;6;.3323 36,123.6 41.99;3;0 213.052 14.02;2;0 494.524 24.03;3;0 1,309.64 80.13;4;0 937.975 189.47;9;0 NB 36,104.1 0.09,1,.7631 did not converge 481.054 0.31;1;.5779 1,278.59 1.57;4;.8139 837.455 3.46;6;.7963 P-bin. ra = 2 36,106.9 2.38;1;.1228 did not converge 481.034 0.38;1;.5357 1,320.71 69.74;2;0 837.438 3.10;6;.7963 P-bin. m = 3 36,106.1 1.20;1;.2728 did not converge 481.034 0.35;1;.5515 1,303.00 63.10;3;0 837.442 3.19;6;.7853 PolyaAeppli 36,104.6 0.15;1;.6966 did not converge 481.044 0.32;1;.5731 1,280.59 6.52;4;.1638 837.452 3.37;6;.7615 Ney.-A 36,105.3 0.49;1;.4820 did not converge 481.037 0.33;1;.5642 1,288.56 24.14;3;0 837.448 3.28;6;.7736 P-iG 36,103.6 0.57;1;.4487 did not converge 481.079 0.31;1;.5761 1,281.90 8.75;4;.0676 837.455 3.63;6;.7260 P-ETNB 36,103.5 5.43;1;.0198 did not converge did not converge 1,278.58 1.54;3;.6740 did not converge Poisson Geometric (a) All two-parameter distributions are superior to the two one-parameter distributions. The best two-parameter distributions are negative binomial (best loglikelihood) and Poisson-inverse Gaussian (best p-value). The threeparameter Poisson-ETNB is not a significant improvement by the likelihood ratio test. The simpler negative binomial is an excellent choice. (b) Because the sample variance is less than the sample mean, mles exist only for the Poisson and geometric distributions. The Poisson is clearly acceptable by the goodness-of-fit test. (c) None of the two-parameter models are significant improvements over the Poisson according to the likelihood ratio test. Even though some have superior ^-values, the Poisson is acceptable and should be our choice. (d) The two-parameter models are better by the likelihood ratio test. The best is negative binomial on all measures. The Poisson-ETNB is not better, and so use the negative binomial, which passes the goodness-of-fit test. The 182 CHAPTER 16 SOLUTIONS moment analysis supported the Polya Aeppli, which was acceptable, but not as good as the negative binomial. (e) T h e two-parameter models are better by the likelihood ratio test. The best is Poisson-binomial with m = 2, though the simpler and more popular negative binomial is a close alternative. 1 6 . 3 6 Excel® solver reports the following mles to four decimals: p — 0.9312, λι = 0.1064, and A2 - 0.6560. The negative loglikelihood is 10,221.9. Rounding these numbers to two decimals produces a negative loglikelihood of 10,223.3 while Tröbliger's solution is superior at 10,222.1. A better two-decimal solution is (0.94,0.11,0.69), which gives 10,222.0. T h e negative binomial distribution was found to have a negative loglikelihood of 10,223.4. T h e extra parameter for the two-point mixture cannot be justified (using the likelihood ratio test). CHAPTER 17 CHAPTER 17 SOLUTIONS 17.1 1 7 1 SECTION 17.7 λ0 = E(A-) = E(X 2 ) = Var(X) = ηλ = (1.96/0.05) 2 = 1,536.64, ^ 0 0 100* - x2 - CLX '— όό ο 5 / 5,000 Jo 100 IQOa:2-*3, _ 2 1 [ ^ ^ 0 - ^ = 1,6665, Jo 1,666| - (33|) 2 = 555§, 1,536.64 1 + 'V555JP 33| 2Ί - 2,304.96. 2,305 claims are needed. 17.2 0.81 = nX/Xo. 0.64 = nX [λ0 ( l + 3.7647. aß = 100, /? = 26.5625. ^ ) = 0.81(1 + a " 11)\ - 1 Student Solutions Manual to Accompany Loss Models: Fivm Data to Decisions, Edition. By Stuart A. Klugman, Harry H. Panjer, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. a FourthlSZ 184 CHAPTER 17 SOLUTIONS 17.3 λ 0 = 1,082.41. μ = 600. Estimate the variance as σ2 = ° 2 + 752 +(~ 75 ) 2 = 75 2 . The standard for full credibility is l,082.41(75/600) 2 = 16.913. Z = ^3/16.913 = 0.4212. The credibility pure premium is 0.4212(475) + 0.5788(600) = 547.35. 17.4 μ = sßOy, σ2 — sßo\ + sß(\ 4- β)θγ; where s is used in place of the customary negative binomial parameter, r. Then T nAr τμ^/η 1.ο4ο — Q.Oby/nsß0Y — σ y/sßai+sßil + ßyf and so 1,082.41 = sßa . 2 Y " ^W/ r — — s - =-nsß nsß ,w 2 4 sß(l 4 β)θ H γ θ ±^ ' \ σ\ 4 Θ2Υ(1 4 ß) The standard for full credibility is nsß = 1,082.41 ( 1 4 ß 4 - f y Partial credibility is obtained by taking the square root of the ratio of the number of claims to the standard for full credibility. 17.5 λ 0 = (2.2414/0.03)2 = 5,582.08, and so 5,583 claims are required. (From the wording of the problem, λ = λο·) /»200,000 E(X) E 2 = l (* ) = Var(X) = /.200.000 / 2ÖPÖÖ^ = 100 ' 000 ' 2 ώ ^777^ = 200,000 13 333 333 i 333 > ' > ' >' ο3> 3,333,333,333^. The standard for full credibility is 1,082.41(1 + 3,333,333,333.33/10,000,000,000) = 1,443.21. Z = ^1,082/1,443.21 = 0.86586. 17.7 (1.645/0.06) 2 (1 + 7,500 2 /l,500 2 ) = 19,543.51, or 19,544 claims. 17.8 Z = ^6,000/19,543.51 = 0.55408. The credibility estimate is 0.55408(15,600,000) + 0.44592(16,500,000) = 16,001,328. SECTION 17.7 185 17.9 For the standard for estimating the number of claims, 800 = (y p /0.05) 2 , and so yp = y/2. /•ιυυ rioo i E(X) = / Jo 0.0002x(100-x)d£ = 3 3 - , 3 E(X2) = / Jo 0.0002x 2 (100-£)cte = 1,666-, 3 Var(X) ,400 9 = ΐ , 6 6 6 | - ( 3 3 ^ ) 2 = 555^. The standard for full credibility is ( v / 2/0.1) 2 [l + 555§/(33±) 2 ] - 300. 17.10 1,000 = (1.96/0.1) 2 (1 + C2). The solution is the coefficient of variation, 1.2661. 17.11 Z = ^10,000/17,500 = 0.75593. The credibility estimate is 0.75593(25,000,000) + 0.24407(20,000,000) - 23,779,650. 17.12 The standard for estimating claim numbers is (2.326/0.05) 2 = 2,164.11. For estimating the amount of claims, we have E(X) — f^° 2.bx~25dx = 5/3, E(X2) = J 1 0 0 2.5x- 1 5 dx - 5, and Var(X) - 5 - (5/3) 2 = 20/9. Then 2164.11 = (1.96/K) 2 [1 + (20/9)/(25/9)], and so K = 0.056527 17.13 E(X) = 0.5(1) + 0.3(2) 4- 0.2(10) = 3.1, E(X2) = 0.5(1) + 0.3(4) + 0.2(100) = 21.7, Var(X) = 21.7 - 3.1 2 = 12.09. The standard for full credibility is (1.645/. 1) 2 (1 + 12.09/3.12) = 611.04, and so 612 claims are needed. 17.14 3,415 - (1.96/fc)2(l + 4), k = 0.075, or 7.5%. 17.15 Z = ν/ίί/F, R = ZO + (1 - Z)P, Z = (R - P)/{0 n/F=(R-P)2/(0-P)2.n=^ff. - P) = vW^· 17.16 For the severity distribution, the mean is 5,000 and the variance is 10,0002/12. For credibility based on accuracy with regard to the number of claims, 2 000 ' =(äk) ' ^ 1 · 8 ' where z is the appropriate value from the standard normal distribution. For credibility based on accuracy with regard to the total cost of claims, the 186 CHAPTER 17 SOLUTIONS number of claims needed is * / 1 + io,ooo»/i2\ =m 2 0.052 V 5,000 CHAPTER 18 CHAPTER 18 18.1 SECTION 18.9 18.1 The conditional distribution of X given that Y = y is _ P r ( X = a:,Z = y - x ) F r ( X = x)P r (Z = y - x) _ ^ Γ — Pr(Y x\(y - x)\ \\i = y) -f X2J \y-x)\ (Ai+A 2 )»e \ λ ι + λ2, for x = 0 , 1 , 2 , . . . , 2/. This is a binomial distribution with parameters m = y and <? = λ ι / ( λ ι + λ2). Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugman, Harry H. Panjer, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. FourthW7 188 18.2 CHAPTER 18 f{x\y) = f(x,y) f(y) Pr(X = x, Z = y - x) Pr(Y = y) = ~ (ni+n2)pJ/(l -p)ni+n2-y ~ V a? / \y—x) ΎΤΤ' This is the hypergeometric distribution. 18.3 Using (18.3) and conditioning on N yields E(X) = E[E(X\N)} = E\NE(Yi)} = Ε[μγΝ] = μγΕ(Ν) = \μγ. For the variance, use (18.6) to obtain Var(X) = E[Var(X|JV)]+Var[E(X|7V)] = = = = = EfiVVar^)] + Var[ATE(yi)] Ε[σγΝ] + Υ&τ\μγΝ] σγΕ(Ν) + μγν&τ(Ν) \(μγ+σγ) AE[lf)]. 18.4(a) fx(0) = 0.3, / χ ( 1 ) = 0.4, fx(2) = 0.3. M O ) = 0.25, M l ) = 0-3, M 2 ) = 0-45. (b) The following array presents the values for x = 0,1,2: (c) / ( x | y = 0) = 0.2/0.25 = 4/5, 0/0.25 = 0, 0.05/0.25 = 1/5, f(x\Y = 1) = 0/0.3 = 0, 0.15/0.3 = 1.2, 0.15/0.3 = 1/2, f(x\Y = 2) = 0.1/0.45 = 2/9, 0.25/0.45 = 5/9, 0.1/0.45 = 2/9. Ε(ΛΤΚ = 0) = 0(4/5)+ 1(0)+2(1/5) = 2/5, E ( X | y = l) = 0(0)+ 1(1/2)+ 2(1/2) = 3/2, E(X\Y = 2) = 0(2/9) + 1(5/9) + 2(2/9) = 1, E(X2\Y = 0) = 0 ( 4 / 5 ) + 1(0)+4(1/5) = 4/5, E ( X 2 | F = 1 ) = 0(0)+ 1(1/2)+ 4(1/2) = 5/2, E ( X 2 | y = 2) = 0(2/9)+ 1(5/9)+ 4(2/9) = 13/9, V a r ( X | r = 0) = 4/5 - (2/5) 2 = 16/25, V a r ( X | r = 1) = 5/2 - (3/2) 2 = 1/4, Var(X|y = 2) = 13/9 - l 2 = 4/9. SECTION 18.9 (d) E(X) E[Var(X|Y)] Vai[E(X\Y)] Var(X) 189 (2/5)(0.25) + (3/2)(0.3) + 1(0.45) = 1, (16/25)(0.25)+ (l/4)(0.3) + (4/9)(0.45) = 0.435, (2/5) 2 (0.25) + (3/2) 2 (0.3) + 12(0.45) - l 2 = 0.165, 0.435 4- 0.165 = 0.6. = = = = 18.5 (a) f(x, y) oc exp { - « 1 2(1 -P2) Χ μι 2ρ( ~ χ-μ1\~ )(^)]) *?)]} βΧΡ {"2(Γ^) 2 Now a normal density Ν(μ, σ ) has pdf f(x) oc exp [— -^j(x2 fx\y(x\y) — 2μχ)] . Then 2 <* f(x, y) is N [ ^ +p^(y- μ2), (1 - Ρ )σ\j . (b) fx(x) = f(x,y)dy 2p « / exp {-2(i^7) + exp exp oc exp i (y- 1 ( x — μλ 2\ \dy σι μ2-ρ^(χ-^ιϊ °2 x / T ^ dy γ2πσ2\/1-p2 1 ( x — μλ 2 V σι Since the normal density is fx(x) [v-jH °2 1 ( x — μλ 2 V σι x / exp / = σ2 χ-μι\ o\ 1 J - exp — \ 1 ^/2πσ\ exp 2\ σι 2 (Χ-μ\ (^-^Y , in general, we have J ~Ν(μλ,σ\). 190 CHAPTER 18 (c) Suppose fx(x)fy(y) = /x,y(x,y). Then fx(x)fy(y) = fx,y{x,y) = fx\Y(x\y)fy(y)' Therefore, fx(x) = fx\y(x\y). From the results of (a) and (b),p = 0. Then Therefore, fXy(x,y) °1 = E(X) Var(X) 18.6 (a) x2 1 ocexp fx,y(x,y) / \ 2 OC °2 fX(x)fY(x). fx(x)fy(y). = = Ε[Ε(Χ|θι,θ2)] = Ε(θι) Ε[ν8τ(ΛΊΘι,θ2)]+ν8Γ[Ε(ΛΊΘι,θ2)] E(02) + Var(01). (b) / χ | θ ι , θ 2 ( χ | β ι , β 2 ) = ( 2 π ^ ) - 1 / 2 θ χ ρ f- i (χ - ^ ) fx(x) = J(2ne22y1/2exp J -II π(θ1,θ2)άθ1άθ·2 (2πθ:2 X - 1 / 2 exp ττι(θ1)π2(θ2)άθ1άθ2. We are also given, /y|e 2 (j/|ö 2 ) = ( 2 π ^ ) " 1 / 2 β χ ρ ( - ^ y 2 ) · Let Z = F + ©j. Then /ζ,Ιθ, Wfc) = ^ /κ|* 2 (2 - *ι |02) πι (0!)d0i and /z(*) = = J ίζ,\θΛζ\θ2)π2(θ2)άθ2 ί f /γ\θΑζ-θι\θ2)ττι(θ1)άθιπ2(θ2)άθ2 | | ( 2 π ο 2 ) - 1 / 2 exp j^mde 18.7 My) M*) e ιχ α r e~9ez -J5<*-'' =j π1(θ1)άθλπ2(θ2)άθ2 π(<9 - χ)άθ, SECTION 18.9 191 Let W = Y + Z. Then fw{w) = ^fy{y)fz(w-y) y=0 e-aay y y=0 -J ( α ε-( J +θ\α e-eQ{w-y) (w - y)\ §0(^)·(^Γ-«- + θΥ> ^ ( w \ wl J-—5^^π(^ + π(θ)άθ e-(° V(a ( a \ v ( θ Υ~υ + ey with the last line following because the sum contains binomial probabilities. Let r = a + Θ and so / = e~rrv _ t π(τ - a)dr w\ fx(x). 18.8 (a) 7r(9ij) = 1/6 for die i and spinner j . (b)(c) The calculations are in Table 18.1. Table 18.1 Calculations for Exercise 18.8(b) and (c). i 3 Pr(X = O|0ij) Pr(X = 3|0i,-) Pr(X == 8|0Ο·) 1 1 1 2 2 2 1 2 3 1 2 3 25/30 25/30 25/30 10/30 10/30 10/30 1/30 2/30 4/30 4/30 8/30 16/30 4/30 3/30 1/30 16/30 12/30 4/30 μ(θίά) 35/30 30/30 20/30 140/30 120/30 80/30 (d) Prpfi = 3 ) = (1 + 2 + 4 + 4 + 8 + 16)/(30 · 6) = 35/180. (e) The calculations are in Table 18.2. (f) The calculations are in Table 18.3. ν{θα) 6,725/900 5,400/900 2,600/900 12,200/900 10,800/900 5,600/900 192 CHAPTER 18 Calculations for Exercise 18.8(e). Table 18.2 i j 1 1 1 2 2 2 1 2 3 1 2 3 Pr(Xi = 3\θ13) ΡΓ(Θ = θία\Χι = 3) = 1/30 2/30 4/30 4/30 8/30 16/30 Table 18.3 ΡΓ(Χ|=3|0^)(1/6) 35/180 1/35 2/35 4/35 4/35 8/35 16/35 Calculations for Exercise 18.8(f). xi Pr(X 2 = x2\X1 = 3) = 0 3 8 - ^ [25(1) + 25(2) + 25(4) + 10(4) + 10(8) + 10(16)] = 455/1,050 M W l ) + 2(2) + 4(4) + 4(4) + 8(8) + 16(16)] = 357/1,050 M ά [4(1) + 3(2) + 1(4) + 16(4) + 12(8) + 4(16)] = 238/1,050 ΣΡΓ(Χ2 = = θί3)Ρν(θ = θα\Χχ = 3) Χ 2 |Θ (g) E(X2\X1 = 3) = 1 1 3Q 3^[35(1) + 30(2) + 20(4) + 140(4) + 120(8) + 80(16)] 2,975 1,050 (h) Ρτ(Χ2 = 0, Xi = 3) = ^ 25(1) t 900 10(4) 900 Pr(X 2 = 3|Jf! = 3) = ^ . 25(2) 25(4) 900 900 10(8) 900 Pr(X 2 = 8|Xi = 3) - 10(16) 900 455 5,400' - ^ 5,400* (i) Divide answers to (h) by PT(X\ = 3) = 35/180 to obtain the answers to (f). (j)E(X 2 |X 1 = 3) = 0 I f 5 + 3 1 f 5 + 8 ^050 (k) μ v = 1,050" 1 1 425 ^ ( 3 5 + 30 + 2 0 + 1 4 0 + 1 2 0 + 80) = — , = I TL· ( 5 ' 7 2 S + 5 > 4 o °+ 2 > 6 o °+ i 2 - 2 0 0 + i o > 8 0 0 + 5 » 6 °°) = ^ S » o yUU 5,4U0 » - δ4( 3 5 2 + 3 ο 2 + 2 0 2 + ι 4 θ 2 + ΐ 2 θ 2 + 8 0 2 '-(ιΙ) 2 -ϋδ^ SECTION 18.9 {1) Z = Pc = = = 193 1 > 4 2 5 A 4400 00 y 76,925/32,.,400 ) 0.228349. 0.228349(3) + 0.771651(425/180) 2.507. 43 (l+ 18.9 (a) 7r(0i) = 1/3, i = 1,2,3. (b)(c) The calculations appear in Table 18.4. Table 18.4 Calculations for Exercise 18.9(b) and (c). 1 2 3 0.1600 0.2800 0.3225 0.1750 0.0625 1.7 1.255 0.0625 0.0500 0.3350 0.1300 0.4225 2.8 1.480 0.2500 0.1500 0.3725 0.1050 0.1225 1.7 1.655 i Ρτ(Χ Pr(X Pv(X Pr(X Pr(X = = = = = O|0i) l\9i) 2\9i) 3|0») 4|0i) ß(0i) V(ßi) (d) Pr(Jfi = 2) = I (0.3225 + 0.335 + 0.3725) = 0.34333. (e) The calculations appear in Table 18.5. Table 18.5 i Calculations for Exercise 18.9(e). Pr(Xi = 2|0i) 1 2 3 0.3225 0.3350 0.3725 P r ( 6 = 0i|Xi = 2) = ΡΓ(Χ1=2|0,)(1/3) .34333 0.313107 0.325243 0.361650 (f) The calculations appear in Table 18.6. (g) E(X2|X!=2) = 1.7(0.313107)+ 2.8(0.325243)+ 1.7(0.361650) = 2.057767. 194 CHAPTER 18 Calculations for Exercise 18.9(f). Table 18.6 X2 Pr(X 2 = X2|Xi = 2) = Σ Ρ Γ ( Χ 2 = Χ2|θ = ö i ) P r ( 0 = 0i|Xi = 2) 0 1 2 3 4 0.16(0.313107) + 0.0625(0.325243) + 0.25(0.361650) = 0.160837 0.28(0.313107) + 0.05(0.325243) + 0.15(0.361650) = 0.158180 0.3225(0.313107) + 0.335(0.325243) + 0.3725(0.361650) = 0.344648 0.175(0.313107) + 0.13(0.325243) + 0.105(0.361650) = 0.135049 0.0625(0.313107) + 0.4225(0.325243) + 0.1225(0.361650) = 0.201286 (h) p r ( X 2 = 0, Xi = 2) - [0.16(0.3225) + 0.0625(0.335) + 0.25(0.3725)j/3 - 0.055221, Pr(X2 = 1, Xi = 2) = 0.054308. P r ( X 2 = 2, X a = 2) = 0.118329, P r ( X 2 = 3, X i = 2) = 0.046367. P r ( X 2 = 4, X i = 2) = 0.069108. (i) Divide answers to (h) by Ρτ(Χ\ (0(j) E(X2|Xi = 2 ) - — 2) = 0.343333 to obtain the answers to 0(0.160837) + 1(0.158180) + 2(0.344648) +3(0.135049) +4(0.201286) = 2.057767. (k) μ = ± ( 1 . 7 + 2 . 8 + 1.7) = 2.06667, v = ± ( 1 . 2 5 5 + 1.48+ 1.655) = 1.463333, 2 ±(1.7 + 2.8 2 + 1.72) - 2.06667 2 = 0.268889. 3 a (1) -1 / L463333V V .267779 y 0.155228, = = Pc = 0.155228(2) + 0.844772(2.06667) = 2.056321. (m) Table 18.4 becomes Table 18.7 and the quantities become μ — 1.033333, v = 0.731667, a = 0.067222. Z = 2+.73ΐββ7/0.067222 = 0.155228. Table 18.7 i 1 2 3 Calculations for Exercise 18.9(m). Ρν(Χ = 0\θι) Pr(X = l|ö0 Pr(X = 2|Ö0 0.40 0.25 0.50 0.35 0.10 0.15 0.25 0.65 0.35 μ(θι) 0.85 1.40 0.85 ν(θτ) 0.6275 0.7400 0.8275 SECTION 18.9 18.10 = 0.2(200) = 40, E(S\eA) E(S\9B) = = Ε(Ν\ΘΑ)Ε(Χ\ΘΑ) 0.7(100) = 70, Var(S|0 A ) Var (S\9B) μ3 vs = = = = = as = Ε(Ν\θΑ)νΆτ(Χ\θΑ)+νΒΐ(Ν\θΑ)[Ε(Χ\θΑ))2 0.2(400) + 0.2(40,000) = 8,800, 0.7(1500) + 0.3(10,000) = 4,050, | 4 0 + ±70 = 50, 18,800 3 " ' " " " T+ 3±4,050 = 7,216.67, k = ^ = 36.083, as Z = 4T4Ö83= 0 1 0 ' Pc = 2 _i__ ( νβτ[μ(0)] = |2/in2 402 + ±70 - 502 = 200, 0.10(125) + 0.90(50) = 57.50. 18.11 Let S denote total claims. Then μ8 = μ^μγ vs 195 = 0.1(100) = 10. E{E(AT|Ö 1 )Var(y|0 2 )+Var(iV|Öi)E[(y|Ö2)] 2 } +Ε[νΆι(Ν\θ1)]Ε{[Ε(Υ\θ2)}2} E[E(AT|0!)]E[Var(r|0 2 )] μΝνγ + νΝΕ{[Ε(Υ\θ2)}2}. = = = Since αγ = Var^ y (0)] = Ε{[μγ(θ)}2} - {Ε[μγ(θ)\}2, vs = μΝνγ + νΝ(αγ + μγ). Then αγ = Var[E(y|0 2 )] = Var(0 2 ) since F is exponentially distributed. But, again using the exponential distribution, Var(0 2 ) = E (Θ2) - [E(0 2 )] 2 = E[Var(y|0 2 )] - {E[E(y|0 2 )]} 2 = υγ - μ γ, from which αγ + μγ = νγ. Then vs = = μΝνγ+νΝ{υγ - μγ + μγ) = μΝυγ+νΝνγ 0.1(25,000) + 0.1(25,000) = 5,000. Also, as = = = = V a r ^ i , ^ ) ] = Ε{[μ3(θ1,θ2)}2} {Ε[μ3(θ1,θ2)}}2 Ε{[μΝ(θ1)}2}Ε{[(μγ(θ2)}2} - μ2Νμ2γ 2 (αΝ + μ%){αγ + μγ) - μ Νμγ [0.05 + (0.01)2](25,000) - (0.1) 2 (100) 2 = 1,400. Therefore, k = vs/as = 5,000/1,400 = 3.5714, and Pc = 0.4565 (ψ) + 0.5435(10) = 35.87. Ζ = 3 3+3 57U = 0.4565, 196 CHAPTER 18 18.12 (a) E(Xj) = Ε[Ε(Χ,·|Θ)] = Ε[/?,·μ(θ)] = /?,·Ε[μ(θ)] = Var(Xj-) = ßjß. EtVar(Xj|e)]+Var[E(Xj|e)] = Ε[τ^(θ) + rJ>jV(G)\ + V a r ^ M Q ) ] = TJ + ijjjv + /32a. CoviXi,^-) = = EiX.X^-EiX^EiXj) Ε[Ε(Χ,;Χ,·|θ)]-/3^μ2 = Ε[Ε(Χ,·|θ)Ε(Χ;|θ)]-&/3,.μ 2 = Ε{/3,;^.[μ(θ)]2}-^μ2 ^.(ΕΙΙΜΘ^Ι-ΙΕΙΜΘ)]} 2 ) = (b) The normal equations are E(Xn+1) n = α0 + Σά3Ε(Χ3)ι 3= 1 n Cov(X ? ; ,X n + 1 ) = ]Γ^·Οον(Χ?:,Χ,), where E(X n + 1 ) = /3 η+1 μ, E(Xj) = /^.μ and Cov(Xi,Xn+i) = βφη+λα, Cov(X?:, Xi) = Var (X*) = Tj + V ^ + ß j ß · On substitution, n and ßißn+la = a Σ jßißja + <*ι (Τ?; + V^) J= l = ) ßia + ®ί(Τί + V^)· [ßn+1 Hence, — β{α = ai(Ti + ψιν), yielding ά{ = —/3 ?: α (τ,; + ^ ι ; ) /i Then ζ= 1 μ i= \ and SECTION 18.9 197 Thus, ά άθ = ßn+Φ ~ Σ ^Ομ = # ^η+Ι^ - ά0α V J= l , 3= 1 which gives ffn+lM ^η+ιΜ α0 _ ßn+Φ 1 + am and a ßißn+1 1 + α?η τι + V ^ The credibility premium is ßn+Φ + ßn±\a -X, 1 4- am 1 + am J=i ' J + ^ ' V j=i E(Xn+l) 1 + am ffn+lQ Y^m3X l + amfijSv J E(Xn+1) [ ^ n + i a m ^ 1 4- am 1 + am ( l - Z ) E ( X n + 1 ) + Z/3 n + 1 X. 18.13 The posterior distribution is π(0|χ) oc π{θ) oc H [θ** (1 - 0)*'-*'] 0 a_1 (l - θγ-1 H f(Xj\9) 3=1 ßT,Xj+a-l J J=l Q _ ^^(Xj-x^+b-l 0a*(l-0)6% which is the kernel of the beta distribution with parameters α* = Σ Xj -f a and 6* = XX ^ ~~ xj)+ &· So 198 CHAPTER 18 E(X n + 1 |x) = / μη+1(θ)π(θ\χ)άθ Jo = f^^rhri!')/a-"i(i^)t·-1'« 70 Γ(α*)Γ(6*) - Γ(α»)Γ(6,) J0 Γ(α*+Κ)Γ(α* + 1)Γ(Κ) η+1 Γ(α*)Γ(ί>*)Γ(α* + 1 + 6») α κ * = Κ,η+\ Σχά+α ^Xj+α + Y^iKj -Xj) + b 53 -ftTj a a+b α Σ x j Σ -^j α+6 K + EKjE j+a + b a + b^Kj+a + bJ " Let Σ*0 -γ_ Σχί α _ ζ = Σ ^ /„f j ,+ ίo +■ . .ft'* =Σ^ ^ s · ' μ =α + &' ThenE(X„ + 1 |x) = Κη+λ[Ζ~Χ + {\-Ζ)μ\. (1 — Ζ)μ, the credibility premium. 18.14 (a) fXj\e(xj\e) = p(xj)erW*i/q(0), Normalizing, Ε(£^-|Χ) = ΖΧ + where p(x) = Γ(α + x)/[T(a)a:!], r(0) = ln(0) and q(0) = (1 - β)"°. Thus for 0 < Θ < 1, π(0) oc (1 - 6)akθμΜ{\/θ), which is the kernel of a beta pdf (Appendix A) with parameters a = μλ; and 6 = afc + 1 . As a > 0 and 6 > 0, the result follows. μ(0) - q'{9)l\r\0)q{e)} = αθ/(1 - Θ), and thus Ε Θ Μ >1 - j f £ j *(θ)άθ Γ(/χ/ί)Γ(α*; + 1) 7o which is infinite if a/c < 0 (orfc< 0 because a > 0). If a/; > 0, pf ίην, = ί Μ )J αΓ(μΑ; + afc + 1) Γ(μΑ: + l)r(afc) _ ~ Γ(μΛ)Γ(α*; + 1) T(ßk + ak + l) ~ μ' SECTION 18.9 199 Alternatively, because <Θ) kr'{6) = Γ(μΑ: + αΑ: + 1) kT(ßk)Y(ak + l) k K _ } k equals zero for Θ — 0 and 0 = 1 unless k < 0 (in which case it equals — oc for θ = 1), the result follows directly from (18.38). (c and d ( 1 Λ 1 ,Λ_ι *>-i$-[!(iV')]'·-E αο (1-0)2 w^S^/^-^· αΓ(μλ: + ak +_L) f1 kn 1) Jo As in (b), Ε[υ(θ)] = oc Ίϊ ak < 1, whereas if ak > 1, Ε[ν(θ)] mak .2, αΓ(μΑ; + ak + 1) Γ(μ& + 1)Γ(αΑ; - 1) Γ(μ*;)Γ(αΑ + 1) Γ(μ£ + ak) a(ßk + ak)μk ak(ak — 1) μ&(μ -f a) ak - 1 Similarly, [μ(<9)]2 - α2θ2/(1 - Θ)2 and which is infinite for ak < 1, whereas if ak > 1, Ε{[μ(θ)]^ α2Γ(μ& + afc + 1) Γ(μΛ + 2)r(afc - 1) T(ßk)T(ak + 1) Γ(μ& + afc + 1) 2 α (μ/ί)(μΑ;4-1) (ak)(ak - 1) 'μ/c + l a/i \afc-l, As fc > l / a > 0, Ε[μ(θ)] = μ from (b), and thus νβφ(θ)] = = Ε{[μ(θ)]2}-{Ε[μ(θ)]}2 (αμ)(μ& + 1) _ 2 ak — 1 ß2(ak) + αμ — ß2{ak — 1) ak — 1 M(Q + μ) afc — 1 ' 200 CHAPTER 18 Clearly, Ε[ν(θ)] = k V a r ^ ( 6 ) ] . Alternatively, because k > 1/a > Ο,π(0)l[kr'(ß)\ implying t h a t (18.52) holds. Furthermore, r'(0) = 0 for 0 = 0 and 0 = 1, l Γ(μΛ)Γ(α*; + 1) j which equals zero for 0 = 0 and 0 — 1 if k > 1/a, implying, in turn, that (18.52) reduces to k = Ε[υ(θ)}/ ν*ι[μ(θ)]. (d) π(0|χ) α |J}(l-0)a0^ with /e* = A; + n and μ„ = (μ/c + nx)/(k Qfik-l (1 - 0) c*k + n). As shown in (b), E ( X n + 1 | X = x) = Ε\μ(θ)\Χ = x] = μ, if /c* > 0 and is infinite if —1/a < k* < 0. But /e* = k + n, and /c* > 0 if and only if /c > —n. But /c* > — 1/a is the same as k > —n — 1/a, which is clearly satisfied because k > —1/a. Also, /e* < 0 is the same as k < —n. Therefore, if n < 1/a, it is possible to have —1/a < k < — n, or n — 1/a < /c* < 0, in which case the Bayesian premium is infinite. If k — n, then k* = 0, and π(0|χ) O C 0 ~ \ O < 0 < 1, which is impossible, t h a t is, there is no posterior pdf in this case. (e) If k > 1/a, then the credibility premium exists from (b), and the Bayesian premium is a linear function of the XjS from (d). Hence, they must be equal. 1 8 . 1 5 (a) T h e posterior distribution is π(0|χ) oc m Lj=i = exp(—m9xj) fo(0)r*exp(-0/xfc) W)Y fe(0)]-*-mnexp [-θ{τηΣχ3 This is the same form as the prior distribution, with k* — k + rrni and m μ* = Σ xj + M ran 4- k +μ*θ] · SECTION 18.9 201 The Bayesian premium is clearly given by (18.44), but with these new definitions offc*and /i^, because the derivation of (18.44) from (18.41) is completely analogous to that of (18.38) from (18.36). (b) Because the Bayesian premium is linear in the x η s as long as (18.45) holds, the credibility premium must be given by μ^ as defined in (a). (c) The inverse Gaussian distribution can be written as „ , / θ Υ (Ox θ θ\ Replace Θ with m and μ with (20) ""2 to obtain /(*) = ( £ 3 ) 4 «p [-"»«*+™(2β)*-^]· Now let p(m,x) = λ/^^χρ(-^),Γ(θ) that f(x) has the desired form. = 0, and q(0) = exp(-VW) to see 18.16 (a) They are true when Xi,X2,... represent values in successive years and r is an inflation factor for each year. (b) This is a special case of Exercise 18.12 with ßj = τ3\ Tj(9) = 0, ^ r2j /rrij for j = 1 , . . . , n. (c) From Exercise 18.12(b) n άο + Σ a X J J = (1- Ζ Ε Χ + ) ( η+ι) Ζτη+1Χ, J'=l where j=l rrij n I \r i=i and Z = \J—l rrij n / n i=i 3=1 Ση ra7 Ι + «Σ;=Ι^ TO fc +m v rar·? = 202 CHAPTER 18 Then, k k+ m E(Xn+1) + ^ - r " k+ m + 1 X k m v-^ m η+1 μ + k + mτ k + m *-^ m (d) This is the usual Bühlmann-Straub credibility premium updated with inflation to next year's dollars. (e) lnfx.\Q(xj\0) = ra^r ^Xjr(6) — rrij \nq(6) -f \np(xj, raj, r ) . a 0βίχ,\Θ(Χ^Θ) = q'{9) rrijT JXjr'(9) — rrij fx3\e(xj\0) d(0) mjr'WlT-Jxj μ(θ)}/ΧΑΘ(χι\θ). Integrate over all Xj and use Liebniz's rule to obtain 0 = rrijr'fflT-'EiXjie that is, = θ) - πιότ'(θ)μ(Θ)(1), Ε(^|θ) = τν(Θ). Also, d2 x e d(pfxM i\ ) = rnjr"(0)[r-'xJ-ß(e)]fXi\e{xj\e) -τηιΓ'(θ)μ'(θ)ΙΧΑθ(χ,\θ) +my(e)]2[r^xj -μ(θ)}2/Χ]ΙΘ(χί\θ) . Again integrating over all x yields 0 = 0 - τη^'(θ)μ'(θ) + m2T-2]{r'{e)]2VnT{Xj\Q = Θ), and solving for Var(X,|©) yields V a r ^ l © ) = ^ ' { f } = ΐϋίϊί«! (f) π(0|χ) oc π(θ) oc [9(6»)]-fc-mexp ßkr{e) + T-jr(e) ]T i=i = [q(e)}-h-mexp r{0) Ltfc + r - ^ i j=i = [q(6)]-k'e>1'k'r^r'(e), τ'(θ) SECTION 18.9 203 where /e* = k + m and j ßk + τ μ* Σ™=1 rrijXj k + m Thus, π ( 0 | χ ) is the same as (18.41), and from (18.44) Ε1μ(θ)|Χ = χ ] = μ . ^ + - ^ . But from (e) and (18.13), the Bayesian premium is E ( X n + 1 | X = x) = Εθ|χ=χ[Ε(Χη+1|θ)] = Εθ|χ=χ[τη+1μ(θ)] = τη+1Ε[/χ(θ)|Χ = χ]. 1 8 . 1 7 (a) T h e Poisson pgf of each X , is Ρχά{ζ\θ) = e"< 2-1 >, and so the pgf of S is Ps(z\&) = e n ö ( 2 - 1 ) 5 which is Poisson with mean ηθ. Hence, (n0)se- ί3\θ(8\θ) and so =1 {ηθ)°εΛ . - η β fs(s) -π(θ)άθ. (b) μ(0) = E(X|6>) = Θ and π ( 0 | χ ) = [Π^=ι f{Xj\0)] / ( X) = π(θ)άθ / f» π ( 0 ) / / ( ζ ) . We have T.X.J π(θ)άθ -ηθ. Therefore, ττ(0|χ) = θΕχ*ε-ηβπ(θ) ]θΣχ^-ηθπ{θ)άθ T h e Bayesian premium is E(Xn+1|x) = ίμ(θ)π(θ\χ )d9 ΙΘΣχ>+ιε-ηθπ(θ)άθ o oo /^c-»eir(i)i ^7(7^T^+1^"^(^ 5 / =f βΛβ-»βπ(ο)ίί0 o S + l /<;(.?+ 1) ™ /s(s) 204 CHAPTER 18 (c) 0 = ü l £ Z . /■ö-+«-ie-(«+^-,)ed0 s! Γ(α) 7 ns / T α T(g + g) s! Γ ( α ) ( η + 0 - 1 ) β + β r ( s + a) / 1 Γ(β + 1)Γ(α) \l + nßj Yfnß \l + nß where β^ = nß and r = a. This is a negative binomial distribution. 18.18 (a) Clearly, πθ|χ(0|χ) Θηχε-η°π{θ) oc oc θηχβ~ηθ (e-h~^9~^ From Exercise 5.20(g) with m = nx (a nonnegative integer because of the Poisson assumption) and Θ replaced by 7 yields the normalizing constant poo / θηχ-^€-α(η)θ-^άθ = 2 JO νΜφ Kn 2a(n) (b) /•OO /Xn+1|x(^n+l|x) = / χ η + Ι | θ ( α η + ΐ | 0 ) 7 Γ θ | 3 :(fl|x)d9 / JO 00 / 0 Jo 0Xn+] 6~θ ^n+l! 2α(η) 7 2 2K. 2a(n) 2(xn+1!)ffnä_i [V2a(n)7j Jo Again using Exercise 5.20(g) yields 4 ^ni-|e_a(n)0-^ ν^ϊΦ >d# SECTION 18.9 205 /x n + i|x(ffn+l| x ) x 2 4 Γ 7 n+l~ *^χη+1+ηχ-± [2α(η+1) y/2a(n + 1)7 ( α η + ι ! ) # η * - $ [χ/2ο:(η)7 7 Γ[2α(η+1) •^η + Ι 2 c*(n+l) 4 #* η+1 +η*-* [>/2α(η + 1)7J (χη+ι!)Κη5_ι (c) Let χ = ηθ, and so ö = x/n and άθ = (dx)/n. Jo s- Thus, Ln ^ n ' dx. But n ηny γ 2 π ( χ / η ) 3 exp \n/ 2πχ 3 exp 2(x/n) V μ ) n7 / x — ημ V 2x \ ημ / which is the pdf of the inverse Gaussian distribution with parameters 7 and μ replaced by 717 and ημ, respectively. As in Example 7.17, let μ* = ημ and β^ = ημ 2 /7, which implies that 717 — η2μ2/β* = μΙ/β*> Then n \nJ μ% 2πβ*χ* exp V^ßl exp μΐ (χ2ß*x V / \ 21 (χ-μ*)2' 2ß*x \ Therefore, S has a Poisson-inverse Gaussian distribution [or, equivalently, a compound Poisson distribution with ETNB (r = -\) secondary distribution]. with parameters μ* = ημ and β+ = η μ 2 / 7 , and the pf /s(s) can be computed recursively. As shown in Exercise 18.17(b), the Bayesian premium is then given by E(X n +l|X = x ) 1 + nx fs(l + nx) n fs{nx) 206 CHAPTER 18 1 8 . 1 9 T h e posterior distribution is π(0|χ) OC {Y\ ^=1 ex P = exp [{2υ)-\- oc exp 1 -i:^3-e? exp -2^-"> 2 + 2θηχ ~ ηθ2) - ( 2 α ) - 1 (0 2 - 2θμ + μ2)] Σx) έ + έ) +2 Η^ + £ Let p = ^ + ^ and <? = ^ -f ^ , and then note t h a t -ρθ2 + 2ς0 = -(ρι/2θ qp-^2f-q2p-1. - Then 7Γ(0|Χ) OC exp = exp -tf/H-qp-Wf 1 (0-qp- i \ 2 2 V(2p)~1/2 and the posterior distribution is normal with mean nx l· — V + a *= n 1 — — V + a ß*=QP and variance 1 /n 1 a* = — = - + 2p \ v a Then (18.10) implies t h a t X n + i | x is a mixture with Χ η + ι | θ having a normal distribution with mean Θ, and θ | χ is normal with mean μ* and variance a*. From Example 5.5, Xn-\-i |x is normally distributed with mean μ* and variance a* + v, t h a t is, Λ Υ „ + 1 | Χ ( ^ Π + Ι | Χ ) = [2π(α* + υ)] 7 exp 2(a* +v) , — OC < X ' n + i < OC. T h e Bayesian estimate is the mean of the predictive distribution μ + , which can be written as Zx + (1 — Ζ)μ, where Z = n/v n/v 4- 1/a na rm -f-1> n -f u / a ' Because the Bayesian estimate is a linear function of t h e data, it must be the Bühlmann estimate as well. To see this directly, μ(θ) = υ(θ) — v (not random), v = E(v) a = θ,μ = Ε(θ)=μ, Var(B) - a. ^ SECTION 18.9 207 thus indicating that the quantities in the question were chosen to align with the text. Then, k = v/a and Z — n/(n + k). 18.20 (a) Let Θ represent the selected urn. Then, / χ ( χ | θ = 1) = | , χ = 1,2,3,4 and / χ ( χ | θ - 2) = ±, x = 1,2,... ,6. Then, μ(1) = Ε(Χ|Θ = 1) = 2.5 and μ(2) - Ε(Χ\Θ = 2) = 3.5. For the Bayesian solution, the marginal probability of drawing a 4 is /x(4) - | x 4 + ^ x | - ^ 5 a n d the posterior probability for urn 1 is π(0 = 1|Χ = 4) and for urn 2 is / Χ | β (4|1)π(1) _ \ \ _ 3 _ /x(4) h24 5 π(0 = 2|Χ = 4) = 1 - § = §. The expected value of the next observation is E(X 2 |Xi = 4) = 2.5 ( | ) + 3.5 (§) = 2.9. (b) Using Bühlmann credibility, μ = ±(2.5+ 3.5) = 3, t,(l) = I ( l 2 + 2 2 + 3 2 + 4 2 ) - ( 2 . 5 ) 2 = 1.25, v(2) = ±(1 2 + · · · + 6 2 ) - ( 3 . 5 ) 2 = 2.917, υ = ±υ(1) + \υ{2) = 2.0835, a = Var[M(ö)] = E { [ ^ ) ] 2 } - { E K ö ) ] } 2 = i [ 2 . 5 2 + 3 . 5 2 ] - 3 2 = 0.25, 0.25 a - 0.1071352. 1 + 8.334 The credibility premium is Pc = Ζχ+{\-Ζ)μ 3.10709. 18.21(a) MW = = 0.1071352(4)4-0.8927648(3) = 0,ν(0) = 0, oo v = 3θ~3άθ = 1.5, / Ε(θ) = 1.5, /•OO α = Var(Ö) = / k = 1.5/0.75 = 2, Z = 2T2=°· 5 ' Pc = Ζθ~2άθ - 2.25 = 0.75, 0.5(10)+0.5(1.5) = 5.75. 208 CHAPTER 18 (b) Because the support of the prior distribution is Θ > 1, that is also the support of the posterior distribution. Therefore, the posterior distribution is not gamma. TT(0|JVI + N2 = 20) oc e" 2 ^ 2 0 *? - 4 = e " 2 ^ 1 6 . The required constant is i; β~2θθ16άθ = e" 2 1 17 2 ^ 4 ^ 17(16) 8 16! 2 17 l,179,501,863e" and the posterior distribution is π(θ\Ν1+Ν2 = 20) - e~ 2ö (9 16 /l,179,501,863e- 2 . The posterior mean is f™ee-29e16de 1,179,501,863e~2 8.5. The mean is actually slightly less than 8.5 (which would be the exact answer if integrals from zero to infinity were used to create a gamma posterior). 0.5 0.5+fc 18.22 Z 0.5, k = 0.5, Z - 3+0.5 = 6/7. 18.23 (a) Pr(Xx = 1|A) = 1|B) Pr(X! = \\C) Ρτ{Α\Χλ = 1) ΡΓ(*! ?τ{Β\Χχ = 1) Pr(C|Xi = 1) μ(Α) Ε(Χ2\Χ1 (b) μ(Β) μ(ϋ) = 1) μ a ν(Α) ν k Pc = = = = = = = = = 3(0.12)(0.9) = 0.027, 3(0.62)(0.4) = 0.432, 3(0.82)(0.2) = 0.384, 0.027/(0.027 + 0.432 + 0.384) = 27/843 = 9/281 144/281, 128/281, 3(0.9) = 2.7, 3(0.4) = 1.2, 3(0.2) = 0.6, = [9(2.7) + 144(1.2) + 128(0.6)]/281 = 0.97473. = (2.7 + 1.2 + 0.6)/3 = 1.5, = (2.72 + 1.22 + 0.6 2 )/3 - 2.25 = 0.78, = 3(0.9)(0.1) = 0.27, ν(Β) = 0.72, v(C) = 0.48, = (0.27 + 0.72 + 0.48)/3 = 0.49, = 49/78, Ζ = (1 + 49/78)- 1 = 78/127, = τ | ( 1 ) + τ | ( 1 · 5 ) = 1-19291. SECTION 18.9 18.24 (a) μ(λ) = λ) υ(λ) = λ> OO v = / Ε(λ) = 4/3, 4A_4dA = 4/3, OO / fc = Pc = 209 4Ä_3dÄ - 16/9 = 2/9, (4/3)/(2/9) = 6, Z = ^ A _ = l / 3 , (1/3)(1) + (2/3)(4/3) = 11/9. /•l (b) μ = v = Xd\ = l/2, Jo M=l/2, 2 / ' \ d\ k = - 1/4 = 1/12, Jo (l/2)/(l/12) = 6, Z = — Pc = (1/3)(1) + (2/3)(l/2) = 2/3. = 1/3, 18.25 μ(/ι) = h, μ = Ε(/ι) = 2, v(/i) = /i, v = E(ft) = 2, a = Var(ft) = 2, fc = 2/2 = 1, Z = i ^ i = 1/2. 18.26 (a) r ~ bin(3, 0), π(0) = 60(1 - 0). π(θ\Χ = 1) oc 30(1 - 0) 2 60(1 - 0) oc 0 2 (1 - 0) 3 , and so the posterior distribution is beta with parameters 3 and 4. Then the expected next observation is Ε(3Θ\Χ = 1) = 3(3/7) = 9/7. 00 μ{θ) = 30, υ(θ) = 30(1 - 0), μ = Ε(30) = 3 / 060(1 - θ)άθ = 1.5, Jo υ = Ε[30(1 - 0)] = 3 / 0(1 - 0)60(1 - θ)άθ = 0.6, Jo a = k Pc = = Var(30) = 9 f 02 60(1 - θ)άθ - 2.25 = 0.45, Jo 0.6/0.45 = 4/3, Z = (1 + 4 / 3 ) " 1 = 3/7, (3/7)(l) + (4/7)(1.5) = 9/7. 210 CHAPTER 18 18.27 (a) μ(Α) = 20, μ(Β) = 12, μ{0) = 10, υ{Α) = 416, υ{Β) = 288, v[C) = 308, μ = (20 + 12 + 10)/3 = 14, ν = (416 + 288 + 308)/3 = 337i, a = (202 + 122 + 1 0 2 ) / 3 - 1 4 2 = 1 8 | , k = 337±/18§ = 18-jL, Ζ = (1 + 1 8 ^ ) _ 1 = 14/267, P c = (14/267)(0) + (253/267)(14) = 13.2659. π(Α\Χ=0) = 2/(2 + 3 + 4) = 2/9, π(Β|Χ = 0) = 3/9, π{0\Χ = 0) = 4/9, Ε(Χ 2 |Χι = 0 ) = (2/9)20 + (3/9)12 + (4/9)10 = 12§. 18.28 (a) ΡΓ(ΛΓ = 0) = / 1 3 e- A (0.5)dA = (e" 1 - e~ 3 )/2 = 0.159046. (b) μ = υ a = Var(A) = / A2(0.5)dA - 4 = 1/3, k = 2/(1/3) = 6, Z = j Pc = (1/7)(1) + (6/7)(2) = 13/7. = Ε ( λ ) = ί A(0.5)dA = 2, ^ = 1/7, (c) 7r(A|Jfi = 1) =e-*A(.5)// 1 3 e- x A(.5)<t\ = e- A A/(2e- 1 E(A|X 1 18.29(a) = 1) ß(A) = v(A) = v{B) = μ = v = a = k = Z = 4e~3), = ή / e-x\2d\/(2e-1 = (5e _ 1 - 17e- 3 )/(2e- 1 - 4e'3) = 1.8505. -4e~3) (l/6)(4) = 2/3, μ(Β) = (5/6)(2) = 5/3, (l/6)(20) + (5/36)(16) = 50/9, (5/6)(5) + (5/36)(4) = 85/18, [(2/3) + (5/3)]/2 = 7/6, [(50/9) + (85/18)]/2 = 185/36, [(2/3) 2 + (4/3) 2 ]/2 - 49/36 = 1/4, (185/36)/(l/4) = 185/9, 41^579= 3 6 / 2 2 1 · SECTION 18.9 211 (b) (36/221)(0.25) + (185/221)(7/6) - 1,349/1,326 = 1.01735. 1 8 3 0 E(X 2 ) E(X2\X1 = 12) = = = = (l + 8 + 1 2 ) / 3 = 7 E[E(X 2 |X!)] [2.6 + 7.S + E(X2\X1 10.6. = 12)]/3. 18.31 (a) X ~ Poisson(A), π(λ) = e _ A / 2 / 2 . The posterior distribution with three claims is proportional to e - A A 3 e - A / 2 = λ e~ 1,5A , which is gamma with parameters 4 and 1/1.5. The mean is 4/1.5 = 2 | . (b) Μλ) μ a fc ^ = v(\) = A, = v = E(A) = 2, = Var(A) = 4, = 2/4 = 0.5, Z = = 9 i n L ^ = f, 8 2 ö(3) + ö(2) = ö = 2 ö - 18.32 (a) r ~ bin(3,0), π(0) = 28O03(1 - 0) 4 , which is beta(4,5). π(0|Χ = 2) oc 30 2 (1 - 0)28O03(1 - Θ)4 oc 0 5 (1 - 0) 5 , and so the posterior distribution is beta with parameters 6 and 6. Then the expected next observation is E(30|X = 2) = 3(6/12) = 1.5. (b) μ(θ) μ = = 30, ν(θ) = 30(1 - 0), Ε(30) = 3(4/9) = 4/3, ν = E[30(l - 0)] = 3 / 0(1 - 0)28O0J(1 - 0) 4 d0 ■f Jo = Var(30) = 9 ί 0228O03(1 - θ)4άθ - 16/9 = 2 , 5 2 0 « ^ - 1 6 / 9 = 2/9, k Z = = (2/3)/(2/9) = 3, ( l + 3 ) " 1 = l/4, Pr = (l/4)(2) + (3/4)(4/3) = 1.5. a 212 CHAPTER 18 μ(Α1) = ν(Α{) = μ = ν = α = k = 18.33(a) Ζ = 0.15, μ{Α2) = 0.05, 0.1275, ν(Α2) = 0.0475, (0.15 + 0.05)/2 = 0.1, (0.1275+ 0.0475)/2 = 0.0875, (0.152 + 0.05 2 )/2 - 0.1 2 = 0.0025, 0.0875/0.0025 = 35, 3Τ35=3/38· Thus, the estimated frequency is (3/38) (1/3) + (35/38) (0.1) = 9/76. == == μ == V == a -= k == 24, μ(Β2) = 34, 64, v{B2) = 84, (24 + 34)/2 = 29, (64 + 84)/2 = 74, (242 + 34 2 )/2 - 292 z --= ?—— = 25/99. 1 + 74/25 μ(Βι) v(B,) 25, 74/25, Thus, the estimated severity is (25/99) (20) + (74/99) (29) = 294/11. The estimated total is (9/76)(294/ll) = 1323/418 = 3.1651. (b) Information about the various spinner combinations is given in Table 18.i Table 18.8 Calculations for Exercise 18.33. Spinners AUB1 AUB2 A2,Bi A2,B2 μ 3.6 5.1 1.2 1.7 83.04 159.99 30.56 59.11 μ = v = a = (3.6 + 5.1 + 1.2 + 1.7)/4 = 2.9, (83.04 + 159.99 + 30.56 + 59.11)/4 = 83.175, (3.62 + 5.1 2 + 1.22 + 1.72)/4 - 2.92 = 2.415, k 83.175/2.415 = 34.441, Z = -—^—— = 0.080126. 1 3 + 34.441 = Thus, the estimated total is (0.080126)(20/3) + 0.919874(2.9) = 3.2018. SECTION 18.9 213 (c) For part (a), Pr(l|i4i) Pr(l|42) = = 3(0.15)(0.85)2 = 0.325125, 3(0.05)(0.95)2 = 0.135375, Pr(A 2 |l) = 1 - 0.706026 = 0.293974. Thus, the estimated frequency is 0.706026(0.15) + 0.293974(0.05) = 0.120603. Pr(20|Bi) Pr(20|B 2 ) = = 0.8, 0.3, Pr(ßi|20) K li ' Pr(S 2 |20) = _M^==8/ii, 0.8 + 0.3 ' 3/11. = Thus, the estimated severity is (8/ll)(24) + (3/ll)(34) = 26.727272. The estimated total is 0.120603(26.727272) = 3.2234. For part (b), Pr(0,20,0|4i,Bi) = (0.85)2(0.12) = 0.0867, ΡΓ(0,20,0|ΛΙ,Β2) = = = (0.85)2(0.045) = 0.0325125, (0.95)2(0.04) = 0.0361, (0.95)2(0.015) = 0.0135375. Ρ Γ (0,20,0μ 2 ,-Βι) Pr(0,20,0|A 2 ,ß 2 ) The posterior probabilities are 0.51347, 0.19255, 0.21380, 0.08017, and the estimated total is 0.51347(3.6) + 0.19255(5.1) + 0.21380(1.2) + 0.08017(1.7) = 3.2234. (d) Pr(A-1=0,...,X„_1=0|,4i,Bi) Pr(Xi=0,...,X„_i=0|i4i,B2) Pr(X1=0,...,Xn-1=0\A2,B1) ΡΓ(Λ-Ι=0,...,Χ„_Ι=0|Λ2,Β2) E(Xn\X1=0,...,Xn-i=0) = = = = (0.85)"- 1 , (0.85)n-\ (0.95)"- 1 , (0.95)"- 1 . (0.85)"- 1 (3.6) + (Q.85)"- 1 ^.!) + (Q.95)"- 1 ^·?) + (0.95)"-» (1-7) (0.85)"- 1 + (0.85)"" 1 + (0.95)"- 1 + (0.95)"- 1 _ 2.9 + 8.7(0.85/0.95)"- 1 2 + 2(0.85/0.95)"-! ' 214 CHAPTER 18 and the limit as n —» oc is 2.9/2 = 1.45. 18.34 PrOf = 0.12IA) - -== exp V ' J >/2^(0.03) ( 0 . 1 2 - 0 . 1 \2l 2(0.0009) 10.6483, PT(X = 0.12|£) = (X = 0.12|C) = 0 (actually, just very close to zero), so Pr(;4|X = 0.12) = 1. The Bayesian estimate is μ(Α) = 0.1. 18.35 Ε(Χ\Χλ = 4) = 2 = Z(4) + (1 - Z)(l), Z = 1/3 = ^ , k = 2 = v/a = 3/a, a = 1.5. 18.36 v = E{v) = 8, a = Var(^) = 4, fc = 8/4 = 2, Z = ^ = 0.6. 18.37 (a) /(j,) = / x-le-y^mx-'se-20^d\ Jo = 400 / Jo A-4e-(20+^/\u. Let 0 = (20 + y)/A, λ = (20 + y)/0, dA = -(20 + ν)/θ2άθ, and so /(j/) - 400 / (20 + y)-302e-e<W = 800(20 + j/)"" 3 , Jo which is Pareto with parameters 2 and 20, and so the mean is 20/(2 — 1) = 20. (b) μ(Α) = λ, v(X) = X . The distribution of λ is inverse gamma with a = 2 and 0 = 20. Then μ = E(A) = 20/(2 - 1) - 20 and v = Ε(λ 2 ), which does not exist. The Bühlmann estimate does not exist. (c) π(λ|15,25) oc A~ V 1 5 / A A _ V 2 5 / A 4 0 0 A - V 2 0 / A oc A" 5 e" 6 0 / A , which is inverse gamma with a = 4 and 0 = 60. The posterior mean is 60/(4 — 1) = 20. 1 8 ·38 μ(0) α υ k = = = = Z = 0, υ ( 0 ) = 0 ( 1 - 0 ) , Var(0) - 0.07, E ( 0 - 0 2 ) = E(0) - Vax(0) - [E(0)]2 0.25 - 0.07 - (0.25)2 = 0.1175, 0.1175/0.07=1.67857, 1 + 1.67857 = 0.37333. SECTION 18.9 215 18.39 (a) Means are 0, 2, 4, and 6 while the variances are all 9. Thus μ v = = (0 + 2 + 4 + 6)/4 = 3, 9, a = (0 + 4 + 16 + 36)/4 - 3 2 = 5, Z = ^ = 5/14 = 0.35714. (b)(i) v = 9, a = 20, Z = Y ^ Ö - 20/29 = 0.68966. (b)(ii) v = 3.24, a = 5, Z = 1+3 * 24/5 - 5/8.24 = 0.60680. (b)(iii) v = 9, a = (4 + 4-f 100-f 100)/4-36 - 16, Z = J^J^ = 16/25 = 0.64. ( b ) ( i v ) Z = 3 ^ - 1 5 / 2 4 = 0.625. (b)(v) a = 5, v = (9+9 + 2.25+2.25)/4 = 5.625, Z = 0.64. The answer is (i). 2+5 * 5 = 10/15.625 = 18.40 (a) Preliminary calculations are given in Table 18.9. Table 18.9 Calculations for Exercise 18.40. Risk 100 1,000 20,000 μ V 1 2 0.5 0.7 0.3 0.2 0.2 0.1 4,350 2,270 61,382,500 35,054,100 Pr(100|l) Pr = <1|100) = Expected μ (b) v a k 0.5, Pr(100|2) = 0.7, 0.5(2/3)(+2S(l/3) = 10/17 ' Pr(2|100) = 7/17. value is (10/17)(4350) + (7/17)(2,270) = 3,493.53. = (2/3)(4,350;) + (Γ/3)(2,270) = 3,656.33i, = (2/3)(61,382,500) + (1/3)(35,054,100) = 52,606,366.67, = (2/3)(4,350) 2 + (l/3)(2,270) 2 - 3,656.332 = 963,859.91, = 54.579, Z = 1/55.579 = 0.017992. Estimate is 0.017992(100) + 0.982008(3,656.33) = 3,592.34. 216 CHAPTER 18 18.41(a) ν(μ,λ) = = = = ν (b) = = = = = μ(μ,λ) a μ(2λ2) = 2μλ2, 2Ε(μλ 2 ) = 2 Ε ( μ ) ^ Γ ( λ ) + Ε ( λ ) 2 ] 2(0.1)(640,000 + 1,0002) 328,000. μλ, ν&τ(μ\) = Ε(μ'2λ2)-Ε(μ)2Ε(λ)2 [Var(M) + Ε(μ) 2 ] [Var(Ä) + Ε(λ) 2 ] - Ε(μ) 2 Ε(λ) 2 (0.0025 + 0.12)(640,000 + Ι,ΟΟΟ2) - 0.121,0002 10,500. 18.42 For the Bayesian estimate, Ρι·(λ = l|Xj = r) = Pr(Xi = r|Ä = 1) ΡΓ(Λ = 1) Pr(X! = r\X + 1) Pr(A = 1) + Ρτ(Χ1 = r | Ä = 3)Pr(Ä = 3) ^■(0.75) e_l( 0 .75) + ^21(0.25) 0.2759 0.2759 + 0.1245(3 r )' Then, 2 08 = 0-2759 0.2759 + 0.1245(3r)V 0.1245(3'·) 0.2759 + 0.1245(3r)V ; ' ; The solution is r = 7. Because the risks are Poisson, μ = υ = Ε(λ) = 0.75(1)+0.25(3) = 1.5, a = Var(A) = 0.75(1) + 0.25(9) - 2.25 = 0.75, Z = 1 + 1.5/0.75 = 1/3 ' and the estimate is (l/3)(7) + (2/3)(1.5) = 3.33. 18.43 For the Bayesian estimate, Pr(6> = 8|Xi = 5) Pr(Xi = 5|6> = 8) Pr(fl = 8) PT(X1 = 5|<9 = 8) Ρτ(θ = 8) + ΡΓ(ΛΊ = 5|0 = 2) Pr(6> = 2) (l/8)e- 5 / 8 (0.8) (l/8)e- /8( 0 .8) + (l/2)e- 5 / 2 (0.2) 5 0.867035. SECTION 18.9 217 Then, E(X 2 |Xi = 5) = Ε(0|ΛΊ = 5) = 0.867035(8) + 0.132965(2) = 7.202. For the Bühlmann estimate, μ = v = a = Z = 0.8(8) + 0.2(2) = 6.8, 0.8(64) + 0.2(4) = 52.8, 0.8(64) + 0.2(4) - 6.82 = 6.56, l + 5218/6.56=°·110512' and the estimate is 0.110512(5) + 0.889488(6.8) = 6.601. 18.44 The posterior distribution is π(λ|χ) oc ( β - λ ) 9 0 ( λ 6 - λ ) 7 ( λ ν λ ) 2 ( λ ν λ ) λ 3 ε - 5 0 λ = Ä 1 7 e - 1 5 0 \ This is a gamma distribution with a = 18 and Θ — 150. The estimate for one risk is the mean, 18(1/150) = 3/25 and for 100 risks, it is 300/25 - 12. Because a Poisson model with a gamma prior is a case where the Bayes and Bühlmann estimates are the same, the Bühlmann estimate is also 12. 18.45 We have μ = 0.6(2,000) + 0.3(3,000) + 0.1(4,000) = 2,500, v = 1,0002, a = 0.6(2,000)2 + 0.3(3,000)2 + 0.1(4,000)2 - 2,5002 - 450,000, 80 Z — ™ . ι,οοο,οοο ~ 0.97297, 80 + 450,000 X = 24,000 + 36,000 + 28,000 Λ η Λ η go -MOO, and the estimate is 0.97297(1,100) + 0.02703(2,500) = 1,137.84. 18.46 For one year, Trfolan) oc q^(l - qf-'^-^l - qf = ^ + - 1 ( 1 - q) 16-x which is a beta distribution with parameters x\ + a and 17 — X\. The mean is the Bayesian estimate of #, and 8 times that value is the estimate of the expected number of claims in year 2. Then, x\ = 2 implies 2.4344 = 8- 2 + a 17 + a 218 CHAPTER 18 for a = 5. For two years, *{q\xi =2,X2 = k) oc q\\ Then, - qfqk{\ - ς ) 8 " ^ 4 ( 1 - qf = q6+k{\ q)22~h. - 7 + A; 3.73333 = 8—^— for k = 7. 1 8 . 4 7 If p = 0, then the claims from successive years are uncorrelated, and hence the past d a t a x = (X\,..., X n ) are of no value in helping to predict X n + i so more reliance should be placed on μ. (Unlike most models in this chapter, here we get to know μ as opposed to only knowing a probability distribution concerning μ.) Conversely, if p = 1, then Xn+\ is a perfect linear function of x. Thus, no reliance need be placed on μ. 1 8 . 4 8 (a) E{[X n + 1 - 9 (X)] 2 } = E { [ X n + i - E ( X n + 1 |X) + E ( X n + 1 |X) - g(X)}2} 2 E{[Xn+1-E(Xn+1|X)] } +E{[E(Xn+1|X)-s(X)]2} +2E{[Xn+1 -E(Xn+i|X)][E(Xn+i|X) -^(X)]}. T h e third term is 2E{ [ X n + 1 - E ( X n + 1 | X ) ] [ E ( X n + 1 |X) - <7(X)]} = 2E(E{[Xn+1-E(Xn+1|X)][E(Xn+1|X)-^(X)]|X}) = 2E{ [ E ( X n + 1 |X) - E ( X n + 1 | X ) ] [ E ( X n + 1 |X) - ^(X)]} = 0, completing the proof. (b) T h e objective function is minimized when E { [ E ( X n + i | X ) — g(X)] 2 } is minimized. If g(X) is set equal to E ( X n + i | X ) , the expectation is of a random variable t h a t is identically zero, and so is zero. Because an expected square cannot be negative, this is the minimum. But this is the Bayesian premium. (c) Inserting a linear function, the mean-squared error to minimize is n E{[E(X n+1 |X) - a0 ~Σα3Χ3)2). But this is (18.19), which is minimized by the linear credibility premium. CHAPTER 19 CHAPTER 19 19.1 19.1 SECTION 19.5 Xi Vl V2 V3 V = 733±, X2 = 633i, X3 = 900, X = 755§, 2 2 2 = (16| + 66| + 83± )/2 = 5,833±, = = = (8| 2 + 33i 2 + 4 l | 2 ) / 2 = 1,458±, (02 + 502 + 502)/2 = 2,500, 3,263|, ä = i ( 2 2 f + 122| 2 + 144| 2 ) - 3,263§/3 = 17,060^, k = Z = 3,263§/17,060^ = 0.191316, 3/(3+ 0.191316) = 0.94005. The three estimates are 0.94005(733^) + 0.05995(755§) 0.94005(633^) + 0.05995(755§) 0.94005(900) + 0.05995(755|) = = = 734.67, 640.66, 891.34. Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart A. Klugnian, Harry H. Panjor, Gordon E. Willinot Copyright © 2012 John Wiley & Sons, Inc.. Fourt}i219 220 CHAPTER 19 19.2 χλ = 45,000/220 = 204.55, X2 X3 μ = = = 54,000/235 = 229.79, 91,000/505=180.20, χ = 190,000/960 = 197.91. a ~ [100(4.55)2 + 120(3.78)2 + 90(18.68)2 + 75(10.21)2 + 70(13.07)2 +150(6.87) 2 + 175(8.77)2 + 180(14.24) 2 /(1 + 2 + 2) 22,401, 220(204.55 - 197.91)2 + 235(229.79 - 197.91)2 +505(180.20 - 197.91)2 - 22,401(2) 960 - (2202 + 2352 + 505 2 )/960 ~ 617'54, k = 36.27, Zx = 0.8585, Z 2 = 0.8663, Z 3 = 0.9330. v = = The estimates are 0.8585(204.55) + 0.1415(197.91) 0.8663(229.79) + 0.1337(197.91) 0.9330(180.20) + 0.0670(197.91) = = = 203.61, 225.53, 181.39. Using the alternative method, A μ ~ _ 0.8585(204.55) + 0.8663(229.79) + 0.9330(180.20) 0.8585 + 0.8663 + 0.9330 ~ ' ' and the estimates are 0.8585(204.55) + 0.1415(204.32) 0.8663(229.79) + 0.1337(204.32) 0.9330(180.20) +0.0670(204.32) = = = 204.50, 226.37, 181.81. 19.3 X = 475, v = (02 + 752 + 75 2 )/2 = 5,625. With μ known to be 600, a = (475 - 600)2 - 5,625/3 = 13,750, k = 5,625/13,750 = 0.4091, Z = 3/3.4091 = 0.8800. The premium is 0.88(475) + 0.12(600) = 490. 19.4 (a) VsiiXij) = = (b) This follows from (19.6). ElVar^ieOl+Var^X^ie,)] Ε[ν(θί] + ν3χ[μ(θί)] = v + α. SECTION 19.5 221 (c) ΣΣ{χ^-χ? = ΣγΐίΧν-Xi + Xi-xf i=l j = \ i=\ j = l = Σ Σ Α - - **)2+2 Σ Σ( χ « - *i)(Xi - x) i=l j=l i=l j=l +£,£{*<-X? r t=l i = l n χ 2+2 = ΣΣ( «-*) r n Σ(*<-*)Σ(*«-**) +ηΣ(Χΐ-Χγ r i=l n i=\ j=\ i=l The middle term is zero because $Zi=iG^j — X{) = Σ ? = ι Xij — nX{ — 0. (d) E ^ΣΣ^-*) 2 nr r n 2 2 -^τ Σ Σ<*« ^) + ^ Σ ( ^ *) i = l j=l We know that ^ γ Y^=i(Xij — Xi)2 is an unbiased estimator of v(6i), and so the expected value of the first term is E E = E (Ε^τΣΣ(^-^)2ι^ 11 n-l Τ Σ^) i=l r(n — 1) nr — 1-v. For the second term note that Var(X^) — E[v(ßi)/n] -f Var^(©^)] = v/n H- a and ^τττ Σ ΐ = ι ( ^ ~~ X)2 ls a n unbiased estimator of v/n -f a. Then, for the 222 CHAPTER 19 second term, nr 1 f—' nr — 1 \n i=l I Then the total is r(n— 1) n(r — l)/v \ — -^υ + — zr (- +a) =v + a nr — 1 \n I nr — 1 n— 1 -a. nr — 1 (e) Unconditionally, all of the XjjS are assumed to have the same mean when in fact, they do not. They also have a conditional variance that is smaller, and so the variance from X is not as great as it appears from (b). 19.5 EfcbjiYj-Y)2} = Ε [ Σ Μ ^ - 7 + 7-?) 2 ] = E£>i(^-7) 2 + E M 7 - n 2 +2EW-7)(7-?)] = Σ bj(aj + a2/b3) + b Var(y) +2E [ 7 ZbjYj -Irri + byY-YY: E [7 ΣbjYj - &72 + frrY - Ϋ Var(r) b2 ! f2 ΣbjYj] Σήν&ΐ{Υ3)= = E{-ybY -Ιη2+ΙηΥ- = -E[6(y-7)2] = -ftVar(F). bjYj) b?2} ^{aJ+a2/b b2 E ^ + r2 E [ E W ' - ? ) 2 ] = EbM + ^/b^-bYariY) = Eaj(i>j-b2/b) + (n-l)a2. 19.6 X = 333/2,787 = 0.11948 = v. The sample variance is 447/2,787 0.119482 = 0.14611, and so a = 0.14611 - 0.11948 = 0.02663. Then, k = 0.11948/0.02663 = 4.4867 and Z = 1/5.4867 = 0.18226. The premiums are given in Table 19.1. SECTION 19.5 Table 19.1 Calculations for Exercise 19.6. No. of claims Premium 0 1 2 3 4 0.18226(0) + 0.18226(1) + 0.18226(2) + 0.18226(3) + 0.18226(4) + 0.81774(0.11948) 0.81774(0.11948) 0.81774(0.11948) 0.81774(0.11948) 0.81774(0.11948) = = = = = 0.09770 0.27996 0.46222 0.64448 0.82674 1 9 . 7 (a) See Appendix B. o, = V a r [ M ( © ) ] = V a r ( 6 ) , (b) ν - μ - μ v = Ε[ν(θ)]=Ε[θ(1 + θ)], μ = Ε[μ(θ)] = Ε ( θ ) , = Ε ( θ ) - Ε ( θ 2 ) - Ε ( θ ) - Ε ( θ ) 2 = V a r ( 6 ) = α. 2 (cj μ = Χ = 0.11948, α+ ΰ = 0.14611, a = ι) - 0.11948 - 0.11948 2 , = -0.133755. α-ν T h e solution is ä = 0.0061775 and v = 0.1399325. T h e n k = 0.1399325/0.0061775 = 22.652 and Z = 1/23.652 = 0.04228. T h e premiums are given in Table 19.2 Table 19.2 Calculations for Exercise 19.7. No. of claims Premium 0 1 2 3 4 0.04228(0) + 0.04228(1) + 0.04228(2) + 0.04228(3) + 0.04228(4) + 0.95772(0.11948) 0.95772(0.11948) 0.95772(0.11948) 0.95772(0.11948) 0.95772(0.11948) 19.8 /x,(Xi) = / Jo Π (mi^O^e-" 1 «"' j=1 - Ks^)i e-e^-'+rn,)et,de i v"t = = = = = 0.11443 0.15671 0.19899 0.24127 0.28355 223 224 CHAPTER 19 where U = X ^ l i Uj- T h e n the likelihood function is r 7=1 and the logarithm is 1(μ) = -r 1η(μ) - ]Γ(ί 7 ; + 1) 1η(μ -1 + m?;) and r ϊ(μ) = -τμ~ι - £(*< + ΐΚμ"1 + m,;)^-//- 2 ) = 0. T h e equation t o be solved is v^ ti + 1 > — ^-—' // νμ= 7= 1 19.9 (a) i=l j = l i=l j = l r n, 7= 1 j= l 7= 1 j = l r n, r = Σ Σ « (*« - xtf + Σ m *(* - *)2· m i=l j= l T h e middle term vanishes because ΣΤ=ι of X. 7=1 m ij(Xij~Xi) = 0 from the definition SECTION 19.5 (b) (TO-TO ^ T O 2 ) - 1 i=l ΣΣπιίό{Χίά i=l r rii m X ^ iJ - *i? - (r - 1)« t=i i = i (TO-TO ^ T O 2 ) - 1 i=l - Xf j=l "Σ Σ = 225 r nj t=l j=l -v^2(rii - 1) - ( r - l ) t ) i=l Also ra* = Σ ^ ^ ( ΐ - ^ ) ΣΙ=ι w< - 1 m-m-JEUrn? Σί=ιn* - * Then ,-i Σ[=ι ^t - 1 -1 m. i = l j=l ΣΖ=ιΣ%ιηα*{Χν-Χ) \i=l J 2 YJi=i ni - 1 19.10 The sample mean is 21/34 and the sample variance is 370/340 — (21/34) 2 = 817/1,156. Then, v = 21/34 and a = 817/1,156 - 21/34 = 103/1,156. k = (21/34)/(103/l,156) = 714/103 and Z = 1/(1 + 714/103) = 103/817. The estimate is (103/817) (2) + (714/817)(21/34) = 0.79192. 19.11 The sample means for the three policyholders are 3, 5, and 4, and the overall mean is μ — 4. The other estimates are v = a = Z = l2 + 0 2 + 0 2 + l2 + 0 2 + 0 2 + l2 + l2 + l2 + l2 + l2 + l2 3 ( 4 - 1) 2 . ) 2 + (4 - 4 ) 2 8/9 _ 7 ( 3 - -4) + ( 5 - 4 4 "9' 3 -- 1 4 7 4 + (8/9)/(7/9) 9- 9' The three estimates axe (7/9)(3)+(2/9)(4) = 29/9, (7/9)(5)+(2/9)(4) = 43/9, and (7/9) (4) + (2/9) (4) = 4. 226 CHAPTER 19 19.12 The estimate of the overall mean, μ, is the sample mean, per vehicle, which is 7/10 = 0.7. With the Poisson assumption, this is also the estimate of v. Then, - = 5(0.4 - 0.7)2 + 5(1.0 - 0.7)2 - (2 - 1)(0.7) _ io-(5'2 Z - 5 + 0.7/0.04 ~ 2/9 + 52) 10 · For insured A, the estimate is (2/9)(0.4) + (7/9)(0.7) = 0.6333, and for insured B, it is (2/9)(1.0) + (7/9)(0.7) = 0.7667. 19.13 46 34 ( ! ) +1 0 103 ( 2 ) + 5 ( 3 ) + 2 ( 4 ) - 0 . 8n 3oo , μ?, -_ v.-■ -- x =. - ... 2 _ - 46(-0.83) 2 + 34(0.17)2 + 13(1.17)2 + 5(2.17)2 + 2(3.17)2 _ - U.J5UÜ1, 9g ä = 0.95061-0.83 = 0.12061, Z - 1+0.83/0.12061 s (°) + =0 12688 · · The estimated number of claims in five years for this pohcyholder is 0.12688(3) + 0.87312(0.83) = 1.10533. For one year, the estimate is 1.10533/5 = 0.221. CHAPTER 20 CHAPTER 20 SOLUTIONS 20.1 SECTION 20.1 20.1 The first seven values of the cumulative distribution function for a Poisson(3) variable are 0.0498, 0.1991, 0.4232, 0.6472, 0.8153, 0.9161, and 0.9665. With 0.0498 < 0.1247 < 0.1991, the first simulated value is x = 1. With 0.9161 < 0.9321 < 0.9665, the second simulated value is x = 6. With 0.6472 < 0.6873 < 0.8153, the third simulated value is x = 4. 20.2 The cumulative distribution function is { 0.25x, 0.5, 0.1x + 0.1, For u = 0.2, solve 0.2 = 0.25a; for x = 0.8. from 2 to 4, so the second simulated value is 0.7 = 0.1x + 0.1 for x = 6. 0 < x < 2, 2 < x < 4, 4<x<9. The function is constant at 0.5 x = 4. For the third value, solve Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Edition. By Stuart. A. Klugman, Harry H. Panjcr, Gordon E. Willmot Copyright © 2012 John Wiley & Sons, Inc. Fourth227 228 CHAPTER 20 SOLUTIONS 20.2 SECTION 20.2 2 0 . 3 Because 0.372 < 0.4, the first value is from the Pareto distribution. Solve 0.693 = 1 - ( Y C ^ ) for x = 100 [(1 - 0 . 6 9 3 ) - 1 / 3 - l] = 48.24. Because 0.702 > 0.4, the second value is from the inverse Weibull distribution. Solve 0.284 = e-* 200 /** 2 for x = 2 0 0 [ - ln(0.284)]~ 1 / 2 = 178.26. 2 0 . 4 For the first year, the number who remain employees is bin(200,0.90) and the inversion method produces a simulated value of 175. T h e number alive but no longer employed is bin(25,0.08/0.10 = 0.80) and the simulated value is 22. T h e remaning 3 employees die during the year. For year 2, the number who remain employed is bin(175,0.90) and the simulated value is 155. T h e number of them who are alive but no longer employed is bin(20,0.80) and the simulated value is 17, leaving 3 deaths. For the 22 who began the year alive but no longer employed, the number who remain alive is bin(22,0.95) and the simulated value is 21. At the end of year 2 there are 155 alive and employed, 17 + 21 = 38 are alive but no longer employed, and 3 + 3 + 1 = 7 have died. T h e normal approximation produces the same values in all cases. Also note t h a t because the number in each state at time 1 was not required, an alternative is to determine the theoretical probabilities for time 2, given employed at time 0, as 0.81, 0.148, and 0.042 and then do a single simulation from this trinomial distribution. 20.5d = ln(0.3) = - 1 . 2 0 4 , c = - 3 ( - 1 . 2 0 4 ) = 3.612, λ0 = 3.612, λι = 3.612 - 1.204(1) = 2.408, sx = - ln(0.704)/2.408 = 0.1458, tx = 0.0427 + 0.1458 = 0.1885, A2 = 3.612 - 1.204(2) = 1.204, s2 = - ln(0.997)/1.204 = 0.0025, to = - ln(0.857)/3.612 = 0.0427, t2 = 0.1885 + 0.0025 = 0.1910. Because £ m _i = t2 < 2, the simulated value is m — 3. 20.6 xi w = = x2 = 2(0.108) - 1 = - 0 . 7 8 4 , 2(0.942) - 1 = 0.884, 2 (0.884) + ( - 0 . 7 8 4 ) 2 = 1.396 > 1, and so these values cannot be used. Moving to the next pair, X! w = = 2(0.217) - 1 = - 0 . 5 6 6 , 2 x2 = 2(0.841) - 1 = 0.682, 2 ( - 0 . 5 6 6 ) + (0.682) = 0.78548 < 1. These values can be used. Then, y = >/-21n(0.78548)/0.78548 = 0.78410, ζλ = -0.566(0.78410) = -0.44380, z2 = 0.682(0.78410) = 0.53476. SECTION 20.3 229 The simulated lognormal values are exp[5+1.5(—0.44380)] = 76.27 and exp[5+ 1.5(0.53476)] = 331.01. 20.3 SECTION 20.3 20.7 The requested probability is the same as the probability that at least a of the observations are at or below 7Γ0.9 and at most b are at or below πο.9. The number of such observations has a binomial distribution with a sample size of n and a success probability of Pr(X < πο.9) = 0.9. Let N be this binomial variable. We want 0.95 — Pr(a < N < b). Prom the central limit theorem, a - 0.9n b - 0.9n 0.95 = Pr \ ZJ "^ L λ/ο.θ9η ~ and a symmetric interval implies -1.96: ~ vömü a - 0.9n y/QMn' giving a = 0.9n — 1.96-^/0.9(0.l)n. Because a must be an integer, the result should be rounded down. A similar calculation can be done for b. 20.8 The mean and standard deviation are both 100. The analog of (20.1) is 0.02μ = 1.645. σ/y/n Substituting the mean and standard deviation produces n — (1.645/0.02) 2 = 6,766, where the answer has been rounded up for safety. For the probability at 200, the true value is F(200) = 1 - exp(-2) = 0.8647. The equation to solve is = 1.645 , °·02(0·8647) vO.8647(0.1353)/n for n = 1,059. When doing these simulations, the goal should be achieved 90% of the time. 20.9 The answer depends on the particular simulation. 20.10 The sample variance for the five observations is (1 - 3) 2 + (2 - 3) 2 + (3 - 3) 2 4- (4 - 3) 2 + (5 - 3) 2 _ , = 2.5. The estimate is the sample mean and its standard deviation is σ/y/n, which is approximated by ^/2.5/n. Setting this equal to the goal of 0.05 gives the answer, n = 1000. 230 CHAPTER 20 SOLUTIONS 20.4 SECTION 20.4 2 0 . 1 1 T h e inversion method requires a solution to u = Φ ( ^ Q Q Q 0 0 K where Φ(χ) is the standard normal cdf. x ~2^° For u — 0.5398, the equation to solve is = 0.1 for x = 15,200. After the deductible is applied, the insurer's cost for the first month is 5,200. T h e next equation is —1.2 = J ^ Q Q ° ° for x = 12,600 and a cost of 2,600. The third month's equation is - 3 . 0 = %*ooo°° for x — 9,000 and a cost of zero. T h e final month uses 0.8 = ^ Q Q Q for x = 16,600 and a cost of 6,600. The total cost is 14,400. 2 0 . 1 2 T h e equation to solve for the inversion method is u — Φ ( Q"^' 0 1 )· W h e n u = 0.1587, the equation is - 1 = 1 η ^ 0 1 for x = 0.99005, and the first year's price is 99.005. For the second year, solve 1.5 = ln Q"O2Q1 ^ o r x = 1.04081, and the price is 103.045. 2 0 . 1 3 For this binomial distribution, the probability of no claims is 0.03424, and so if 0 < u < 0.03424, the simulated value is 0. T h e probability of one claim is 0.11751, and so for 0.03424 < u < 0.15175, the simulated value is 1. T h e probability of two 2 claims is 0.19962, and so for 0.15175 < u < 0.35137. Because the value of 0.18 is in this interval, the simulated value is 2. 2 0 . 1 4 For claim times, the equation to solve is u = 1 — exp(—x/2) for x = —0.51n(l — u). The simulated times are 0.886, 0.388, and two more we don't need. T h e first claim is at time 0.886 and the second claim is after time 1, so there is only one claim in the simulated year. For the amount of t h a t claim, the equation to solve is 0.89 = 1 - ( γ ^ ^ : ) for x = 2,015. At the time of the first claim, the surplus is 2,000 + 2,200(0.886) - 2,015 = 1,934.2. Because ruin has not occurred, premiums continue to be collected for an ending surplus of 1,934.2 + 2,200(0.114) = 2,185. 2 0 . 1 5 T h e empirical distribution places probability 1/2 on each of the two points. T h e mean of this distribution is 2 and the variance of this distribution is 1. There are four possible bootstrap samples: (1,1), (3,1), (1,3), and (3,3). T h e value of the estimator for each is 0, 1, 1, and 0. T h e MSE is (1/4) [(0 l ) 2 + (1 - l ) 2 + (1 - l ) 2 + (0 - l) 2 ] = 0.5. 2 0 . 1 6 (a) T h e distribution function is F(x) = x / 1 0 and so is 0.2, 0.4, and 0.7 at the three sample value. These are to be compared with the empirical values of 0, 1/3, 2 / 3 , and 1. T h e maximum difference is 0.7 versus 1 for a test statistic of 0.3. SECTION 20.4 231 (b) Simulations by the authors produced an estimated p-value of 0.8882. 2 0 . 1 7 (a) T h e estimate is 4 ( 7 ) / 3 = 9.33. T h e estimated MSE is ( 2 8 / 3 ) 2 / 1 5 5.807. (b) There are 27 equally weighted bootstrap samples. For example, the sample 2, 2, 4 produces an estimate of 4 ( 4 ) / 3 — 5.33 and a contribution to the MSE of (5.33 - 8.0494) 2 = 7.3769. Averaging the 27 values produces an estimated MSE of 4.146. 1 0 For the first simulation, 0.6 0.8 the inverse normals are 2.1201 and —0.1181. T h e correlated normals are 2.1201 and 0.6(2.1201) + 0.8(-0.1181) - 1.1776. Next, apply the normal cdf to obtain the correlated uniform values 0.9830 and 0.8805. Finally, apply the Pareto and exponential inverse cdf functions to obtain the simulated loss of 28,891 and expenses of 850. T h e insured pays 500 of the loss, the insurer pays 10,000, and the reinsurer pays the remaining 18,391. T h e insurer's share of expenses is (10,000/28,391)(850) = 299 and the reinsurer pays 551. For the second simulation, the same calculations lead to a loss of 929 and expenses of 174. T h e total for the insured is 1,000, for the insurer is 10,000 + 299 + 429 + 174 = 10,902, and for the reinsurer is 18,391 + 551 = 18,942. 2 0 . 1 8 W i t h p = 0.6, the matrix L 2 0 . 1 9 T h e extra step requires a simulated g a m m a value with a = 3 and Θ = 1/3. W i t h u = 0.319, the simulated value is 0.6613. Dividing its square root into the simulated normals gives 2.1201/0.6613 1 / 2 = 2.6070 and 1.1776/0.6613 1 / 2 = 1.4480. Using the normal cdf to make them uniform and the Pareto and exponential inverse cdfs, respectively, gives a loss of 50,272 and expense of 1,043. For the second simulation, the loss is 710 and the expense is 155. After dividing them up, the insured pays 1,000, the insurer pays 10,000+210+210+155 = 10,575, and the reinsurer pays 39, 772+833 = 40,605. 2 0 . 2 0 T h e annual charges are simulated from u — 1 — e " * / 1 , 0 0 0 or x = - 1 , 0 0 0 l n ( l - u). T h e four simulated values are 356.67, 2,525.73, 1,203.97, and 83.38. T h e reimbursements are 205.34 (80% of 256.67), 1,000 (the maximum), 883.18 (80% of 1,103.97), and 0. T h e total is 2,088.52 and the average is 522.13. 2 0 . 2 1 T h e simulated paid loss is βχρ[0.494Φ _ 1 (ϋ) + 13.294]. T h e four simulated paid losses are 450,161, 330.041, 939,798, and 688,451, for an average of 232 CHAPTER 20 SOLUTIONS 602,113. T h e multiplier for unpaid losses is 0.801(2006 - 2 0 0 5 ) 0 · 8 5 1 β - 0 · 7 4 7 ( 2 0 0 6 - 2 0 0 5 ) = 0.3795 for an answer of 228,502. 2 0 . 2 2 0.95 = ΡΓ(0.95μ < X < 1.05μ) and X ~ Ν{μ,σ2/η Then, 0.95 = = = 1.44μ 2 /η). PrfOy■-* < z< l J%*f£\ V 1.2μ/ν/ή 1.2μ/^/η J Pr(-0.05>/n/1.2 < Z < 0 . 0 5 ^ / 1 . 2 , and therefore 0 . 0 5 ^ / 1 . 2 = 1.96 for n = 2,212.76. 2 0 . 2 3 F(300) = 1 - e " 3 0 0 / 1 0 0 = 0.9502. T h e variance of the estimate is 0.9502(0.0498)/n. T h e equation to solve is 0.9502(0.01) = 2 . 5 7 6 ^ « « . T h e solution is n = 3,477.81. 2 0 . 2 4 There are 4 possible b o o t s t r a p samples for the 2 fire losses and 4 for the wind losses, making 16 equally likely outcomes. There are 9 unique cases, as follows. T h e losses are presented as the first fire loss, second fire loss, first wind loss, second wind loss. Case 1: loss is ( 3 , 3 , 0 , 0 ) ; total is 6; eliminated is 0; fraction is 0; square error is (0 - 0.2) 2 = 0.04; probability 1/16. Case 2: loss is ( 3 , 3 , 0 , 3 ) ; total 9; eliminated 2; fraction 0.22; error 0.0005 probability 2/16 [includes ( 3 , 3 , 3 , 0 ) ] . Case 3: loss is ( 3 , 3 , 3 , 3); total 12; eliminated 4; fraction 0.33; error 0.0178 probability 1/16. Case 4: loss is ( 3 , 4 , 0 , 0 ) ; total 7; eliminated 0; fraction 0; error 0.04 probability 2/16. Case 5: loss is ( 3 , 4 , 0 , 3 ) ; total 10; eliminated 2; fraction 0.2; error 0 probability 4/16. Case 6: loss is ( 3 , 4 , 3 , 3 ) ; total 13; eliminated 4; fraction 0.3077; error 0.0116; probability 2/16. Case 7: loss is ( 4 , 4 , 0 , 0 ) ; total 8; eliminated 0; fraction 0; error 0.04; probability 1/16. Case 8: loss is ( 4 , 4 , 0 , 3 ) ; total 11; eliminated 2; fraction 0.1818; error 0.0003; probability 2/16. Case 9: loss is ( 4 , 4 , 3 , 3 ) ; total 14, eliminated 4; fraction 0.2857; error 0.0073; probability 1.16. T h e MSE is [1(0.04) + 2(0.0005) + · · ■ + 1(0.0073)]/16 = 0.0131. 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