Estimation, Evaluation, and Selection of Actuarial Models

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128 APPENDIX A. SOLUTIONS TO EXERCISES Exercise 90 The loglikelihood function for an exponential distribution is l(θ) =ln nY θ −1 e −xi/θ = i=1 nX − ln θ − x i /θ = −n ln θ − n¯xθ −1 . i=1 Under the null hypothesis that Sylvia’s mean is double that of Phil’s, the maximum likelihood estimates of the mean are 916.67 for Phil and 1833.33 for Sylvia. The loglikelihood value is l null = −20 ln 916.67 − 20000/916.67 − 10 ln 1833.33 − 15000/1833.33 = −241.55. Under the alternative hypothesis of arbitrary parameters the loglikelihood value is l alt = −20 ln 1000 − 20, 000/1000 − 10 ln 1500 − 15, 000/1500 = −241.29. The likelihood ratio test statistic is 2(−241.29 + 241.55) = 0.52 which is not significant (the critical value is 3.84 with one degree of freedom). To add a parameter, the SBC requires an improvement of ln(30)/2 =1.70. Both procedures indicate that there is not sufficient evidence in the data to dispute Sylvia’s claim. Exercise 91 With separate estimates, all but the female-nonsmoker group had 1 surrender out of a risk set of 8 lives (For example, the female-smoker group had seven observations with left truncation at zero. The remaining three observations were left truncated at 0.3, 2.1, and 3.4. At the time of the first year surrender, 0.8, the risk set had eight members.). For those three groups, we have Ĥ(1) = 1/8, andŜ(1) = exp(−1/8) = 0.88250. For the female-nonsmoker group, there were no surrenders in the first year, so the estimate is Ŝ(1) = 1. For the proportional hazards models, let z 1 =1for males (and 0 for females) and let z 2 =1for smokers (and 0 for nonsmokers). The maximum likelihood estimates for the exponential are ˆβ 1 =0.438274, ˆβ 2 =0.378974, and ˆθ =13.7843. Letting the subscript indicate the z-values, we have Ŝ 00 (1) = exp(−1/13.7843) = 0.930023 Ŝ 01 (1) = exp(−e 0.378974 /13.7843) = 0.899448 Ŝ 10 (1) = exp(−e 0.438274 /13.7843) = 0.893643 Ŝ 11 (1) = exp(−e 0.438274+0.378974 /13.7843) = 0.848518. For the data-dependent approach, the two estimates are ˆβ 1 =0.461168 and ˆβ 2 =0.374517. The survival function estimates are Ŝ 00 (1) = Ŝ0(1) = 0.938855 Ŝ 01 (1) = 0.938855 exp(0.374517) =0.912327 Ŝ 10 (1) = 0.938855 exp(0.461168) =0.904781 Ŝ 11 (1) = 0.938855 exp(0.461168+0.374517) =0.864572. Exercise 92 For this model, S(t) =exp(−e βz t/θ) where z is the index of industrial production. This is an exponential distribution with a mean of θe −βz and a median of θe −βz ln 2. Thetwofacts imply that 0.2060 = θe −10β and 0.0411 = θe −25β ln 2. Dividing the first equation by the second equation (after dividing the second equation by ln 2) produces 3.47417 = e 15β for β =0.083024 and then θ =0.47254. Then, S 5 (1) = exp[−e 0.083024(5) /0.47254] = 0.04055.
129 Exercise 93 The relative risk of a male child subject versus a female adult subject is e β 1 /e β 2 = e β 1 −β 2 .Thevarianceofˆβ1 − ˆβ 2 can be estimated as £ 1 −1 ¤ · 0.36 0.10 0.10 0.20 ¸· 1 −1 ¸ =0.36 for a standard deviation of 0.6. Thus a 95% confidence interval is 0.25 + 0.45 ± 1.96(0.6) or (−0.476, 1.876). Exponentiating the endpoints provides the answer of (0.621, 6.527). Exercise 94 The first death time had one death so the numerator is 1. The denominator is the sum of the c-values for all four subjects or 1+1+e b + e b =2+2e b . The second death time also had one death and the denominator is the sum of the c-values for the three members of the risk set, 1+2e b .Thus,Ĥ(3) = 1 + 1 . 2+2e b 1+2e b Exercise 95 The contribution to the partial likelihood is the c-value for the one dying divided by the sum of the c-values for those in the risk set. Therefore L = e β 1 e β 2 e β 3 e β 1 + e β 2 + e β 3 e β 2 + e β 3 e β 3 = e β 1 +β 2 (e β 1 + e β 2 + e β 3)(e β 2 + e β 3) .
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Estimation, Evaluation, and Selection of Actuarial Models

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