Estimation, Evaluation, and Selection of Actuarial Models

120 APPENDIX A. SOLUTIONS TO EXERCISES
The covariance matrix is the inverse of the information, or
· ¸
2 −3
.
−3 5
Exercise 72 For the first case the observed loglikelihood and its derivatives are
l(θ) = 62ln[1− e −1000/θ ]+38lne −1000/θ
l 0 (θ) = −62e−1000/θ (1000/θ 2 )
1 − e −1000/θ +38, 000/θ 2
= −62, 000e−1000/θ θ −2 +38, 000θ −2 − 38, 000e −1000/θ θ −2
1 − e −1000/θ
= 38, 000e1000/θ − 100, 000
θ 2 (e 1000/θ − 1)
l 00 (θ) = −θ2 (e 1000/θ − 1)38, 000e 1000/θ 1000θ −2 − (38, 000e 1000/θ − 100, 000)[2θ(e 1000/θ − 1) − θ 2 e 1000/θ 1000θ −2 ]
θ 4 (e 1000/θ − 1) 2 .
Evaluating the second derivative at ˆθ = 1033.50 and changing the sign gives
The reciprocal gives the variance estimate of 18,614.
Similarly, for the case with more information
Î(θ) =0.00005372.
l(θ) = −62 ln θ − 66, 140θ −1
l 0 (θ) = −62θ −1 +66, 140θ −2
l 00 (θ) = 62θ −2 − 132, 280θ −3
l 00 (1066.77) = −0.000054482.
The variance estimate is the negative reciprocal, or 18,355. 1
Exercise 73 The null hypothesis is that the data come from a gamma distribution with α =1,
that is, from an exponential distribution. The alternative hypothesis is that α has some other value.
From Example 2.32, for the exponential distribution, ˆθ =1, 424.4 and the loglikelihood value is
−165.230. For the gamma distribution, ˆα =0.55616 and ˆθ =2, 561.1. The loglikelihood at the
maximum is L 0 = −162.293. The test statistic is 2(−162.293 + 165.230) = 5.874. The p-value
based on one degree of freedom is 0.0154 indicating there is considerable evidence to support the
gamma model over the exponential model.
Exercise 74 For the exponential distribution, ˆθ =29, 721 and the loglikelihood is −406.027. For
the gamma distribution, ˆα =0.37139, ˆθ =83, 020 and the loglikelihood is −360.496. For the
transformed gamma distribution, ˆα =3.02515, ˆθ =489.97, ˆτ =0.32781 and the value of the
loglikelihood function is −357.535. The models can only be compared in pairs. For exponential
(null) vs. gamma (alternative), the test statistic is 91.061 and the p-value is essentially zero. The
exponential model is convincingly rejected. For gamma (null) vs. transformed gamma (alternative),
the test statistic is 5.923 with a p-value of 0.015 and there is strong evidence for the transformed
1 Comparing the variances at the same ˆθ value would be more useful.