Estimation, Evaluation, and Selection of Actuarial Models

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.