Estimation, Evaluation, and Selection of Actuarial Models

Appendix A
Solutions to Exercises
Exercise 1 When all information is available, the calculations are as follows. As in Example 2.11,
values apply from the current y-value to the next one.
j y j s j r j F 30 (x) Ĥ(x) ˆF (x) =1− e
−Ĥ(x)
1 0.1 1 30 1 − 29
30 =0.0333 1
30
=0.0333 0.0328
2 0.5 1 29 1 − 28
30 =0.0667 0.0333 + 29 1 =0.0678 0.0656
3 0.8 1 28 1 − 27
30 =0.1000 0.0678 + 28 1 =0.1035 0.0983
4 1.8 2 27 1 − 25
30 =0.1667 0.1035 + 2 27
=0.1776 0.1627
5 2.1 1 25 1 − 24
30 =0.2000 0.1776 + 25 1 =0.2176 0.1956
6 2.5 1 24 1 − 23
30 =0.2333 0.2176 + 1 24
=0.2593 0.2284
7 2.8 1 23 1 − 22
30 =0.2667 0.2593 + 23 1 =0.3027 0.2612
8 3.9 2 22 1 − 20
30 =0.3333 0.3027 + 2 22
=0.3937 0.3254
9 4.0 1 20 1 − 19
30 =0.3667 0.3937 + 20 1 =0.4437 0.3583
10 4.1 1 19 1 − 18
30 =0.4000 0.4437 + 1 19
=0.4963 0.3912
11 4.6 2 18 1 − 16
30 =0.4667 0.4963 + 18 2 =0.6074 0.4552
12 4.8 2 16 1 − 14
30 =0.5333 0.6074 + 16 2 =0.7324 0.5192
13 5.0 14 14 1 −
30 0 14
=1.0000 0.7324 +
14
=1.7324 0.8231
Exercise 2 Thereare392observations. Foreachinterval,theogivevalueisfortheright-hand
endpoint of the interval while the histogram value is for the entire interval.
Payment range Number of payments Ogive value Histogram value
0—25 6
6
392
=0.0153
6
392(25) =0.000612
25—50 24
30
392 =0.0765 24
392(25) =0.002449
50—75 30
60
392 =0.1531 30
392(25) =0.003061
75—100 31
91
392 =0.2321 31
392(25) =0.003163
100—150 57
148
392 =0.3776 57
392(50) =0.002908
150—250 80
228
392 =0.5816 80
392(100) =0.002041
250—500 85
313
392 =0.7985 85
392(250) =0.000867
500—1000 54
367
392 =0.9362 54
392(500) =0.000276
1000—2000 15
382
392 =0.9745 15
392(1000) =0.000038
2000—4000 10
392
392 =1.0000 10
392(2000) =0.000013
95