Estimation, Evaluation, and Selection of Actuarial Models
Estimation, Evaluation, and Selection of Actuarial Models Estimation, Evaluation, and Selection of Actuarial Models
112 APPENDIX A. SOLUTIONS TO EXERCISES The logarithm and derivative are l(θ) = −6lnθ − 8325θ −2 l 0 (θ) = −6θ −1 +16, 650θ −3 . Setting the derivative equal to zero yields ˆθ =(16, 650/6) 1/2 =52.68. Exercise 45 For the exponential distribution, the maximum likelihood estimate is the sample mean and so ¯x P = 1000 and ¯x S =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) L(θ ∗ ) = ∝ 20Y i=1 (θ ∗ ) −1 exp(−x i /θ ∗ ) ⎛ (θ ∗ ) −30 exp ⎝− Taking logarithms and differentiating yields 20X i=1 10Y j=1 (2θ ∗ ) −1 exp(−x j /2θ ∗ ) ⎞ x i 10X θ ∗ − x j ⎠ 2θ ∗ j=1 = (θ ∗ ) −30 exp(−20¯x P /θ ∗ − 10¯x S /2θ ∗ ) = (θ ∗ ) −30 exp(−20, 000/θ ∗ − 7, 500/θ ∗ ). l(θ ∗ ) = −30 ln θ ∗ − 27, 500/θ ∗ l 0 (θ ∗ ) = −30(θ ∗ ) −1 +27, 500(θ ∗ ) −2 . Setting the derivative equal to zero gives ˆθ ∗ =27, 500/30 = 916.67. Exercise 46 For the first part, Let x = e −1000/θ .Then L(θ) = F (1000) 62 [1 − F (1000)] 38 = (1− e −1000/θ ) 62 (e −1000/θ ) 38 . L(x) = (1− x) 62 x 38 l(x) = 62ln(1− x)+38lnx l 0 (x) = − 62 1 − x + 38 x . Setting the derivative equal to zero yields and then ˆθ = −1000/ ln 0.38 = 1033.50. 0 = −62x + 38(1 − x) 0 = 38− 100x x = 0.38
113 With additional information ⎡ ⎤ ⎛ ⎞ 62Y Y62 L(θ) = ⎣ f(x j ) ⎦ [1 − F (1000)] 38 = ⎝ θ −1 e −x j/θ⎠ e −38,000/θ j=1 j=1 = θ −62 e −28,140/θ e −38,000/θ = θ −62 e −66,140/θ l(θ) = −62 ln θ − 66, 140/θ l 0 (θ) = −62/θ +66, 140/θ 2 =0 0 = −62θ +66, 140 ˆθ = 66, 140/62 = 1066.77. Exercise 47 The density function is f(x) =0.5x −0.5 θ −0.5 e −(x/θ)0.5 . The likelihood function and subsequent calculations are à Y10 L(θ) = θ −0.5 e −x0.5 j θ −0.5 ∝ θ −5 exp −θ −0.5 P 10 j=1 0.5x −0.5 j l(θ) = −5lnθ − 488.97θ −0.5 j=1 x 0.5 j ! = θ −5 exp(−488.97θ −0.5 ) l 0 (θ) = −5θ −1 +244.485θ −1.5 =0 0 = −5θ 0.5 +244.485 and so ˆθ = (244.485/5) 2 = 2391. Exercise 48 Each observation has a uniquely determined conditional probability. The contribution to the loglikelihood is given in the following table: Observation Probability Loglikelihood Pr(N=1) 1997-1 Pr(N=1)+Pr(N=2) = (1−p)p (1−p)p+(1−p)p = 1 2 1+p −3ln(1+p) Pr(N=2) 1997-2 Pr(N=1)+Pr(N=2) = (1−p)p 2 (1−p)p+(1−p)p = p 2 1+p ln p − ln(1 + p) Pr(N=0) 1998-0 Pr(N=0)+Pr(N=1) = (1−p) (1−p)+(1−p)p = 1+p 1 −5ln(1+p) Pr(N=1) 1998-1 Pr(N=0)+Pr(N=1) = (1−p)p (1−p)+(1−p)p = p 1+p 2lnp − 2ln(1+p) Pr(N=0) 1999-0 Pr(N=0) =1 0 The total is l(p) =3lnp − 11 ln(1 + p). Setting the derivative equal to zero gives 0=l 0 (p) = 3p −1 − 11(1 + p) −1 and the solution is ˆp =3/8. Exercise 49 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. Exercise 50 · ¸ E( ¯X) 1 = E n (X 1 + ···+ X n ) = 1 n [E(X 1)+···+ E(X n )] = 1 (µ + ···+ µ) =µ. n
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112 APPENDIX A. SOLUTIONS TO EXERCISES<br />
The logarithm <strong>and</strong> derivative are<br />
l(θ) = −6lnθ − 8325θ −2<br />
l 0 (θ) = −6θ −1 +16, 650θ −3 .<br />
Setting the derivative equal to zero yields ˆθ =(16, 650/6) 1/2 =52.68.<br />
Exercise 45 For the exponential distribution, the maximum likelihood estimate is the sample<br />
mean <strong>and</strong> so ¯x P = 1000 <strong>and</strong> ¯x S =1500. The likelihood with the restriction is (using i to index<br />
observations from Phil’s bulbs <strong>and</strong> j to index observations from Sylvia’s bulbs)<br />
L(θ ∗ ) =<br />
∝<br />
20Y<br />
i=1<br />
(θ ∗ ) −1 exp(−x i /θ ∗ )<br />
⎛<br />
(θ ∗ ) −30 exp ⎝−<br />
Taking logarithms <strong>and</strong> differentiating yields<br />
20X<br />
i=1<br />
10Y<br />
j=1<br />
(2θ ∗ ) −1 exp(−x j /2θ ∗ )<br />
⎞<br />
x i<br />
10X<br />
θ ∗ − x j<br />
⎠<br />
2θ ∗<br />
j=1<br />
= (θ ∗ ) −30 exp(−20¯x P /θ ∗ − 10¯x S /2θ ∗ )<br />
= (θ ∗ ) −30 exp(−20, 000/θ ∗ − 7, 500/θ ∗ ).<br />
l(θ ∗ ) = −30 ln θ ∗ − 27, 500/θ ∗<br />
l 0 (θ ∗ ) = −30(θ ∗ ) −1 +27, 500(θ ∗ ) −2 .<br />
Setting the derivative equal to zero gives ˆθ ∗ =27, 500/30 = 916.67.<br />
Exercise 46 For the first part,<br />
Let x = e −1000/θ .Then<br />
L(θ) = F (1000) 62 [1 − F (1000)] 38<br />
= (1− e −1000/θ ) 62 (e −1000/θ ) 38 .<br />
L(x) = (1− x) 62 x 38<br />
l(x) = 62ln(1− x)+38lnx<br />
l 0 (x) = − 62<br />
1 − x + 38 x .<br />
Setting the derivative equal to zero yields<br />
<strong>and</strong> then ˆθ = −1000/ ln 0.38 = 1033.50.<br />
0 = −62x + 38(1 − x)<br />
0 = 38− 100x<br />
x = 0.38
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