Estimation, Evaluation, and Selection of Actuarial Models

104 APPENDIX A. SOLUTIONS TO EXERCISES
Then Ŝ(8) = e−(8/9.71868)2.25689 =0.52490.
Exercise 22 The equations to solve are
Then
Dividing the second equation by the first gives
Then
0.5 = 1− exp[−(10, 000/θ) τ ]
0.9 = 1− exp[−(100, 000/θ) τ ].
ln 0.5 = −0.69315 = −(10, 000/θ) τ
ln 0.1 = −2.30259 = −(100, 000/θ) τ .
3.32192 = 10 τ
ln 3.32192 = 1.20054 = τ ln 10 = 2.30259τ
ˆτ = 1.20054/2.30259 = 0.52139.
0.69315 = (10, 000/θ) 0.52139
0.69315 1/0.52139 = 0.49512 = 10, 000/θ
ˆθ = 20, 197.
Exercise 23 The two moment equations are
4+5+21+99+421
5
4 2 +5 2 +21 2 +99 2 + 421 2
5
= 110 = θ
α − 1
2θ 2
= 37, 504.8 =
(α − 1)(α − 2) .
Dividing the second equation by the square of the first equation gives
37, 504.8
2(α − 1)
110 2 =3.0996 =
α − 2 .
The solution is ˆα =3.8188. From the first equation, ˆθ = 110(2.8188) = 310.068. For the 95th
percentile,
µ
310.068 3.8188
0.95 = 1 −
310.068 + π 0.95
for ˆπ 0.95 = 369.37.
Exercise 24 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 = θ/(3 − 1) which yields ˆθ =26, 400.