Estimation, Evaluation, and Selection of Actuarial Models

100 APPENDIX A. SOLUTIONS TO EXERCISES
Exercise 8 Ĥ(12) = 2 15 + 1
12 + 1 10 + 2 6
e −0.65 =0.522.
=0.65. The estimate of the survival function is Ŝ(12) =
Exercise 9 The information may be organized as in the following table:
age(t) #ds #xs #us r Ŝ(t)
0 300
294
1 6 300
300 =0.98
2 20
3 10 314 0.98 304
314 =0.94879
4 30 10 304 0.94879 294
304 =0.91758
5 a 324 0.91758 324−a
324
=0.892 =⇒ a =9
7 45
9 b 279−a = 270 0.892 270−b
270
10 35
12 6 244−a − b =235− b 0.892 270−b 229−b
270 235−b
=0.856 =⇒ b =4
13 15
Exercise 10 Ĥ(t 10) − Ĥ(t 9)=
n−9 1 Ĥ(t 3 )=
22 1 + 21 1 + 20 1 3 )=
e −0.14307 =0.8667.
Exercise 11 0.60 = 0.72 r 4−2
r 4
, r 4 =12. 0.50 = 0.60 r 5−1
r 5
, r 5 =6. With two deaths at the fourth
death time and the risk set decreasing by 6, there must have been four censored observations.
Exercise 12 In order for the mean to be equal to y, wemusthaveθ/(α − 1) = y. Letting α be
arbitrary (and greater than 1), use a Pareto distribution with θ = y(α − 1). This makes the kernel
function
α[(α − 1)y]α
k y (x) =
[(α − 1)y + x] α+1 .
Exercise 13 The data points and probabilities can be taken from Exercise 4. They are:
y j p(y j )
0.1 0.0333
0.5 0.0323
0.8 0.0311
1.8 0.0623
2.1 0.0300
2.5 0.0290
2.8 0.0290
3.9 0.0557
4.0 0.0269
4.1 0.0291
4.8 0.0916
The probability at 5.0 is discrete and so should not be spread out by the kernel density estimator.
Because of the value at 0.1, the largest available bandwidth is 0.1. Using this bandwidth and the
triangular kernel produces the following graph.