Estimation, Evaluation, and Selection of Actuarial Models

126 APPENDIX A. SOLUTIONS TO EXERCISES
Interval Observed Expected Chi-square
0to1 21 150F (1) = 150(2/20) = 15
6 2
15 =2.40
1to2 27 150[F (2) − F (1)] = 150(4/20) = 30
3 2
30 =0.30
2to3 39 150[F (3) − F (2)] = 150(6/20) = 45
6 2
45 =0.80
3to4 63 150[F (4) − F (3)] = 150(8/20) = 60
3 2
60 =0.15
Total 150 150 3.65.
For the test there are three degrees of freedom (four groups less zero estimated parameters less one)
and at a five percent significance level the critical value is 7.81. The null hypothesis is accepted
and therefore the data may have come from a population with the given survival function.
Exercise 84 Either recall that for a Poisson distribution the maximum likelihood estimator is the
sample mean, or derive it from
L(λ) =
³e −λ´50 ³ µ λe −λ´122 λ 2 e −λ 101 µ λ 3 e −λ 92
∝ λ 600 e −365λ
2
6
ln L(λ) = 600 ln λ − 365λ
0 = 600λ −1 − 365
ˆλ = 600/365 = 1.6438.
For the goodness-of-fit test,
No. of claims No. of obs. No. expected Chi-square
0 50 365e −1.6438 =70.53
20.53 2
70.53 =5.98
1 122 365(1.6438)e −1.6438 =115.94
6.06 2
115.94 =0.32
2 101 365(1.6438) 2 e −1.6438 /2=95.29
5.71 2
95.29 =0.34
3ormore 92 365 − 70.53 − 115.94 − 95.29 = 83.24
8.76 2
83.24 =0.92
The last two groups were combined. The total is 7.56. There are two degrees of freedom (4 groups
less 1 estimated parameter less 1). At a 2.5% significance level the critical value is 7.38 and therefore
the null hypothesis is rejected. The Poisson model is not appropriate.
Exercise 85 With 365 observations, the expected count for k accidents is
365 Pr(N = k) = 365e−0.6 0.6 k
.
k!
The test statistic is calculated as follows:
No. of accidents Observed Expected Chi-square
0 209 200.32 0.38
1 111 120.19 0.70
2 33 36.06 0.26
3 7 7.21 1.51**
4 3 1.08
5 2 0.14*
*This is 365 less the sum of the other entries to reflect the expected count for 5 or more accidents.
**The last three cells are grouped for an observed of 12 and an expected of 8.43.