Estimation, Evaluation, and Selection of Actuarial Models

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114 APPENDIX A. SOLUTIONS TO EXERCISES Exercise 51 For a sample of size 3 from a continuous distribution, the density function of the median is 6f(x)F (x)[1 − F (x)]. For this exercise, F (x) =(x − θ +2)/4, θ − 2
115 The derivative is 400, 000w − 80, 000 and setting it equal to zero provides the solution, w =0.2. Exercise 55 To proceed, we need the estimated survival function at whole number durations. From Exercise 1, we have S 30 (1) = 27/30, S 30 (2) = 25/30, S 30 (3) = 22/30, S 30 (4) = 19/30, and S 30 (5) = 14/30. Then the mortality estimates are ˆq 0 =3/30, ˆq 1 =2/27, ˆq 2 =3/25, ˆq 3 =3/22, ˆq 4 =5/19, 5ˆp 0 =14/30. The six variances are Var(ˆq 0 ) = 3(27)/30 3 =0.003 Var(ˆq 1 ) = 2(25)/27 3 =0.002540 Var(ˆq 2 ) = 3(22)/25 3 =0.004224 Var(ˆq 3 ) = 3(19)/22 3 =0.005353 Var(ˆq 4 ) = 5(14)/19 3 =0.010206 Var( 5 ˆp 0 ) = 14(16)/30 3 =0.008296 The first and last variances are unconditional. Exercise 56 From the data set, there were 1,915 out of 94,935 that had 2 or more accidents. The estimated probability is 1, 915/94, 935 = 0.02017 andtheestimatedvarianceis 0.02017(0.97983)/94, 935 = 2.08176 × 10 −7 . Exercise 57 From Exercise 55, S 30 (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 µ 22 2 µ 1 30 30(29) + 1 29(28) + 1 28(27) + 2 27(25) + 1 25(24) + 1 24(23) + 1 = 22(8)/30 3 . 23(22) For 2ˆq 3 ,thepointestimateis8/22 and the direct estimate of the variance is 8(14)/22 3 =0.010518. Greenwood’s estimate is µ 14 2 µ 2 22 22(20) + 1 20(19) + 1 19(18) + 2 18(16) + 2 =8(14)/22 3 . 16(14) Exercise 58 From Exercise 4, S 40 (3) = 0.7530 and Greenwood’s formula provides a variance of µ 1 0.7530 2 30(29) + 1 30(29) + 1 30(29) + 2 29(27) + 1 28(27) + 1 28(27) + 1 =0.00571. 27(26) Also, 2ˆq 3 =(0.7530 − 0.5497)/0.7530 = 0.26999 and the variance is estimated as µ 2 0.73001 2 27(25) + 1 26(25) + 1 23(22) + 3 =0.00768. 21(18) Exercise 59 Using the log-transformed method, Ã 1.96 √ ! 0.00571 U =exp =0.49991. 0.753 ln 0.753 The lower limit is 0.753 1/0.49991 =0.56695 and the upper limit is 0.753 0.49991 =0.86778.
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Estimation, Evaluation, and Selection of Actuarial Models

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