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Estimation, Evaluation, and Selection of Actuarial Models

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3.3. VARIANCE AND CONFIDENCE INTERVALS 51<br />

It can be demonstrated that when there is no censoring or truncation, the two formulas will always<br />

produce the same answer. Recall that the development <strong>of</strong> Greenwood’s formula produced the<br />

variance only at death ages. The convention for non-death ages is to take the sum up to the last<br />

death age that is less than or equal to the age under consideration.<br />

With regard to 2ˆq 3 , arguing as in Example 3.17 produces an estimated (conditional) variance<br />

<strong>of</strong><br />

dVar( 2ˆq 3 )= (4/27)(23/27) = 92<br />

27 27 3 .<br />

For Greenwood’s formula, we first must note that we are estimating<br />

2q 3 =<br />

S(3) − S(5)<br />

S(3)<br />

=1− S(5)<br />

S(3) .<br />

As with the empirical estimate, all calculations must be done given the 27 people alive at duration<br />

3. Furthermore, the variance <strong>of</strong> 2ˆq 3 is the same as the variance <strong>of</strong> Ŝ(5) using only information from<br />

duration 3 <strong>and</strong> beyond. Starting from duration 3 there are three death times, 3.1, 4.0, <strong>and</strong> 4.8 with<br />

r 1 =27, r 2 =26, r 3 =25, s 1 =1, s 2 =1,<strong>and</strong>s 3 =2. Greenwood’s approximation is<br />

µ 23 2 µ 1<br />

27 27(26) + 1<br />

26(25) + 2 <br />

= 92<br />

25(23) 27 3 . ¤<br />

Exercise 57 Repeat the previous example, using time to surrender as the variable.<br />

Example 3.26 Repeat the previous example, this time using all 40 observations in Data Set D2<br />

<strong>and</strong> the incomplete information due to censoring <strong>and</strong> truncation.<br />

For this example, the direct empirical approach is not available. That is because it is unclear<br />

what the sample size is (it varies over time as subjects enter <strong>and</strong> leave due to truncation <strong>and</strong><br />

censoring). From Example 2.16, the relevant values within the first 3 years are r 1 =30, r 2 =26,<br />

s 1 =1,<strong>and</strong>s 2 =2.FromExample2.17,S 40 (3) = 0.8923. Then, Greenwood’s estimate is<br />

(0.8923) 2 µ 1<br />

30(29) + 2<br />

26(24)<br />

<br />

=0.0034671.<br />

An approximate 95% confidence interval can be constructed using the normal approximation. It is<br />

0.8923 ± 1.96 √ 0.0034671 = 0.8923 ± 0.1154<br />

which corresponds to the interval (0.7769, 1.0077). For small sample sizes, it is possible to create<br />

confidence intervals that admit values less than 0 or greater than 1.<br />

With regard to 2ˆq 3 , the relevant quantities are (starting at duration 3, but using the subscripts<br />

from the earlier examples for these data) r 3 =26, r 4 =26, r 5 =23, r 6 =21, s 3 =1, s 4 =2, s 5 =1,<br />

<strong>and</strong> s 6 =1. This gives an estimated variance <strong>of</strong><br />

µ 0.7215 2 µ 1<br />

0.8923 26(25) + 2<br />

26(24) + 1<br />

23(22) + 1 <br />

=0.0059502.<br />

21(20)<br />

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