Estimation, Evaluation, and Selection of Actuarial Models

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.