Estimation, Evaluation, and Selection of Actuarial Models
Estimation, Evaluation, and Selection of Actuarial Models
Estimation, Evaluation, and Selection of Actuarial Models
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102 APPENDIX A. SOLUTIONS TO EXERCISES<br />
Exercise 14 With a b<strong>and</strong>width <strong>of</strong> 60 the height <strong>of</strong> the kernel is 1/120. At a value <strong>of</strong> 100, the<br />
following data points contribute probability 1/20 — 47, 75, <strong>and</strong> 156. Therefore, the height is<br />
3(1/20)(1/120) = 1/800.<br />
Exercise 15 The Kaplan-Meier estimate is equivalent to the empirical estimate based on a sample<br />
<strong>of</strong> size 18. The expected cost is<br />
15 + 26 + 55 + 61 + 143 + 194 + 240 + 284 + 357 + 580 + 755 + 777 + 874 + 5(900)<br />
18<br />
= 492.28.<br />
Exercise 16 From Example 2.22, the mean is estimated as 0.16313.<br />
estimated as<br />
81, 714(0) + 11, 306(1) + 1, 618(4) + 250(9) + 40(16) + 7(25)<br />
94, 935<br />
The estimated variance is<br />
0.21955 − 0.16313 2 =0.19294.<br />
The second moment is<br />
=0.21955.<br />
Exercise 17 The expected cost per loss is E(X ∧ 300, 000) − E(X ∧ 25, 000). The first quantity<br />
was calculated in Example 2.24 as 32,907.49. The second quantity requires evaluation at a point<br />
that is not a class boundary. Using the histogram as the empirical estimate <strong>of</strong> the density function,<br />
the general formula when c m−1
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