Estimation, Evaluation, and Selection of Actuarial Models

22 CHAPTER 2. MODEL ESTIMATION
Example 2.22 Use the empirical distribution in Example 2.6 to estimate the expected number of
accidents for one driver in one year.
For this discrete distribution the expected value is P 5
x=0
xp(x) which can be estimated by
5X
x=0
xp 94,935 (x) = 0 81,714
94,935 +111,306 1,618 250 40
+2 +3 +4
94,935 94,935 94,935 94,935 +5 7
94,935
= 15,487
94,935 =0.16313. ¤
When all of the observations are present (no grouping, censoring, or truncation), one view of an
empirical estimate is that it is calculated by doing to the data what you wish you could do to the
population. In the previous example, we wanted to calculate the population mean–the empirical
estimate calculates the sample mean.
Example 2.23 For Data Set B use the empirical distribution to estimate the probability that a loss
payment is greater than 150% of the mean payment.
The mean of the empirical distribution is the sample mean or (27+82+···+15,743)/20 = 1,424.4.
150% of that value is 2,136.6. In the sample, two of the observations exceed that value. The
empirical estimate is 2/20 = 0.1. ¤
Exercise 15 Consider Data Set B, but left truncated at 100 and right censored at 1000. Using the
Kaplan-Meier estimate from Exercise 6 estimate the expected cost per payment with a deductible of
100 and a maximum payment of 900.
Example 2.24 For Data Set C use the empirical distribution to estimate the expected premium
increase when raising the policy limit from 125,000 to 300,000.
In general, when the policy limit is u = c m (that is, the limit is a class boundary), the expected
cost per payment is
E(X ∧ u) =
=
Z u
0
mX
j=1
xf(x)dx + u[1 − F (u)]
Z cj
c j−1
x
n j
n(c j − c j−1 ) dx + c m
kX
j=m+1
mX n j c 2 j
=
− c2 kX
j−1
+ c m
n(c j − c j−1 ) 2
j=1
j=m+1
⎛
⎞
= 1 mX
⎝
c j + c j−1
kX
n j + c m n j
⎠ .
n 2
j=1
j=m+1
Z cj
c j−1
n j
n
n j
n(c j − c j−1 ) dx