Estimation, Evaluation, and Selection of Actuarial Models

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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.
121 gamma model. Exponential vs. transformed gamma could also be tested (using two degrees of freedom), but there is no point. Exercise 75 For Data Set B truncated at 50 the maximum likelihood parameter estimates are ˆτ =0.80990 and ˆθ = 675.25 leading to the following graph. Weibull Fit F(x) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 700 1400 2100 2800 3500 Model Empirical x Cdf plot for Data Set B truncated at 50 For Data Set B censored at 1,000, the estimates are ˆτ =0.99984 and ˆθ = 718.00. The graph is presented below. Weibull Fit F(x) 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 200 400 600 800 1000 x Model Empirical Cdf plot for Data Set B censored at 1,000
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Estimation, Evaluation, and Selection of Actuarial Models

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