Estimation, Evaluation, and Selection of Actuarial Models

110 APPENDIX A. SOLUTIONS TO EXERCISES
deaths each reduce the risk set by one and so Ŝ(1) = 0.9(74/75)(73/74)(72/73) = 0.864. Then
ˆq 35 =1− 0.864 = 0.136.
Exercise 39 The density function is f(t) =−S 0 (t) =1/w. For Actuary X, the likelihood function
is
µ 1 4 µ
f(1)f(3)f(4)f(4)S(5) = 1 − 5
= 1 w w w 4 − 5 w 5 .
Setting the derivative equal to zero gives
0=− 4 w 5 + 25 , 4w =25, ŵ =6.25.
w6 For Actuary Y the likelihood function is
f(1)f(3)f(4)f(4)f(6) = w −5 .
This function appears to be strictly decreasing and therefore is maximized at w =0, an unsatisfactory
answer. Most of the time the support of the random variable can be ignored, but not this time.
In this case f(t) =1/w only for 0 ≤ t ≤ w and is 0 otherwise. Therefore, the likelihood function is
only w −5 when all the observed values are less than or equal to w, otherwise the function is zero.
In other words,
½ 0, w < 6
L(w) =
w −5 , w ≥ 6.
This makes sense because the likelihood that w is less than 6 should be zero. After all, such values
of w are not possible, given the sampled values. This likelihood function is not continuous and
therefore derivatives cannot be used to locate the maximum. Inspection quickly shows that the
maximum occurs at ŵ =6.
Exercise 40 The likelihood function is
L(w) =
f(4)f(5)f(7)S(3 + r)
S(3) 4 = w−1 w −1 w −1 (w − 3 − r)w −1
(w − 3) 4 w −4 = w − 3 − r
(w − 3) 4
l(w) = ln(w − 3 − r) − 4ln(w − 3).
The derivative of the logarithm is
l 0 1
(w) =
w − 3 − r − 4
w − 3 .
Inserting the estimate of w and setting the derivative equal to zero yields
0 =
1
10.67 − r − 4
10.67
0 = 10.67 − 4(10.67 − r) =−32.01 + 4r
r = 8.
Exercise 41 The survival function is
⎧
⎨ e −tλ 1
,
0