Estimation, Evaluation, and Selection of Actuarial Models

109
The only root of this quadratic that is less than one is w =0.10557 = ˆq x .
Exercise 36 For the two lives that died, the contribution to the likelihood function is f(10) while
for the eight lives that were censored, the contribution is S(10). Wehave
Therefore, ˆk =25.
Exercise 37 We have
f(t) = −S 0 (t) = 0.5
k
L = f(10) 2 S(10) 8 =
µ
1 − t −0.5
k
µ 0.5 2 µ
1 − 10
k k
ln L = 3ln(k − 10) − 5lnk
d ln L 3
=
dk k − 10 − 5 k =0
0 = 3k − 5(k − 10) = 50 − 2k.
−1 µ
1 − 10
k
L = f(1100)f(3200)f(3300)f(3500)f(3900)[S(4000)] 495
4
∝
(k − 10)3
k 5
= θ −1 e −1100/θ θ −1 e −3200/θ θ −1 e −3300/θ θ −1 e −3500/θ θ −1 e −3900/θ [e −4000/θ ] 495
= θ −5 e −1,995,000/θ
1, 995, 000
θ
= − 5 1, 995, 000
+
θ θ 2 =0
ln L = −5lnθ −
d ln L
dθ
and the solution is ˆθ =1, 995, 000/5 = 399, 000.
Exercise 38 For maximum likelihood, the contributions to the likelihood function are (where q
denotes the constant value of the time to death density function)
Then,
Event
Contribution
Survive to 36 Pr(T >1) = 1 − q
Censored at 35.6 Pr(T >0.6) = 1 − 0.6q
Die prior to 35.6 Pr(T ≤ 0.6) = 0.6q
Die after 35.6 Pr(0.6