Estimation, Evaluation, and Selection of Actuarial Models
Estimation, Evaluation, and Selection of Actuarial Models Estimation, Evaluation, and Selection of Actuarial Models
116 APPENDIX A. SOLUTIONS TO EXERCISES Exercise 60 From Exercise 5, Ĥ(3) = 0.2774. The variance is estimated as dVar[Ĥ(3)] = 1 30 2 + 1 30 2 + 1 30 2 + 2 29 2 + 1 28 2 + 1 28 2 + 1 27 2 =0.0096342. The linear confidence interval is The log-transformed interval requires 0.2774 ± 1.96 √ 0.0096342 or 0.0850 to 0.4698. " # U =exp ± 1.96√ 0.0096342 =2.00074 or 0.49981. 0.2774 The lower limit is 0.2774(0.49981) = 0.13865 and the upper limit is 0.2774(2.00074) = 0.55501. Exercise 61 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 q 7 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. Exercise 62 First, obtain the estimated survival probability as Ŝ(4) = 12 56 20 54 15 80 25 60 =0.4032. Greenwood’s formula gives µ 3 (0.4032) 2 15(12) + 24 80(56) + 5 25(20) + 6 =0.00551. 60(54) Exercise 63 The Nelson-Åalen estimates are the centers of the confidence intervals, which are 0.15 and 0.27121. Therefore, s i+1 /r i+1 =0.12121. Fromthefirst 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 s i+1 /r 2 i+1 =0.0018367. Dividing the first result by the second gives r i+1 =66.The firstequationthenyieldss i+1 =8. Exercise 64 Greenwood’s estimate is µ 2 V = S 2 50(48) + 4 45(41) + 8 . (41 − c)(33 − c) Then, 0.011467 = V S 2 =0.003001 + 8 (41 − c)(33 − c) (41 − c)(33 − c) = 8/(0.008466) = 945. Solving the quadratic equation yields c =6.
117 Exercise 65 For the Nelson-Åalen estimate, 1.5641 = Ĥ(35) = 2 15 + 3 13 + 2 10 + d 8 + 2 8 − d (1.5641 − 0.5641)8(8 − d) = d(8 − d)+16 0 = d 2 − 16d +48 d = 4. The variance is 2 15 2 + 3 13 2 + 2 10 2 + 4 8 2 + 2 4 2 =0.23414. Exercise 66 The standard deviation is µ 15 100 2 + 20 65 2 + 13 1/2 40 2 =0.11983. Exercise 67 In general, for the exponential distribution, l 0 (θ) = −nθ −1 + n¯xθ −2 l 00 (θ) = nθ −2 − 2n¯xθ −3 E[l 00 (θ)] = nθ −2 − 2nθθ −3 = −nθ −2 Var(ˆθ) = θ 2 /n where the third line follows from E( ¯X) =E(X) =θ. The estimated variance for Data Set B is dVar(ˆθ) =1, 424.4 2 /20 = 101, 445.77 and the 95% confidence 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 3.29 gives the exact value of the variance. That is because For the gamma distribution, Var(ˆθ) =Var( ¯X) =Var(X)/n = θ 2 /n. l(α, θ) = (α − 1) ∂l(α, θ) ∂α ∂l(α, θ) ∂θ = = nX j=1 nX ln x j − j=1 ln x j − nΓ0 (α) Γ(α) nX x j θ −2 − nαθ −1 j=1 nX x j θ −1 − n ln Γ(α) − nα ln θ j=1 − n ln θ ∂ 2 l(α, θ) ∂α 2 = −n Γ(α)Γ00 (α) − Γ 0 (α) 2 Γ(α) 2 ∂ 2 l(α, θ) = −nθ −1 ∂α∂θ ∂ 2 l(α, θ) nX ∂θ 2 = −2 x j θ −3 + nαθ −2 = −2n¯xθ −3 + nαθ −2 . j=1
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116 APPENDIX A. SOLUTIONS TO EXERCISES<br />
Exercise 60 From Exercise 5, Ĥ(3) = 0.2774. The variance is estimated as<br />
dVar[Ĥ(3)] = 1<br />
30 2 + 1<br />
30 2 + 1<br />
30 2 + 2<br />
29 2 + 1<br />
28 2 + 1<br />
28 2 + 1<br />
27 2 =0.0096342.<br />
The linear confidence interval is<br />
The log-transformed interval requires<br />
0.2774 ± 1.96 √ 0.0096342 or 0.0850 to 0.4698.<br />
"<br />
#<br />
U =exp ± 1.96√ 0.0096342<br />
=2.00074 or 0.49981.<br />
0.2774<br />
The lower limit is 0.2774(0.49981) = 0.13865 <strong>and</strong> the upper limit is 0.2774(2.00074) = 0.55501.<br />
Exercise 61 Without any distributional assumptions, the variance is estimated as (1/5)(4/5)/5 =<br />
0.032. From the distributional assumption, the true value <strong>of</strong> 3 q 7 is [(8/15) − (5/15)]/(8/15) = 3/8<br />
<strong>and</strong> the variance is (3/8)(5/8)/5 =0.046875. The difference is −0.014875.<br />
Exercise 62 First, obtain the estimated survival probability as<br />
Ŝ(4) = 12 56 20 54<br />
15 80 25 60 =0.4032.<br />
Greenwood’s formula gives<br />
µ 3<br />
(0.4032) 2 15(12) + 24<br />
80(56) + 5<br />
25(20) + 6 <br />
=0.00551.<br />
60(54)<br />
Exercise 63 The Nelson-Åalen estimates are the centers <strong>of</strong> the confidence intervals, which are 0.15<br />
<strong>and</strong> 0.27121. Therefore, s i+1 /r i+1 =0.12121. Fromthefirst confidence interval, the estimated variance<br />
is (0.07875/1.96) 2 =0.0016143 while for the second interval it is (0.11514/1.96) 2 =0.0034510<br />
<strong>and</strong> therefore s i+1 /r 2 i+1 =0.0018367. Dividing the first result by the second gives r i+1 =66.The<br />
firstequationthenyieldss i+1 =8.<br />
Exercise 64 Greenwood’s estimate is<br />
µ 2<br />
V = S 2 50(48) + 4<br />
<br />
45(41) + 8<br />
.<br />
(41 − c)(33 − c)<br />
Then,<br />
0.011467 = V S 2 =0.003001 + 8<br />
(41 − c)(33 − c)<br />
(41 − c)(33 − c) = 8/(0.008466) = 945.<br />
Solving the quadratic equation yields c =6.