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Estimation, Evaluation, and Selection of Actuarial Models

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52 CHAPTER 3. SAMPLING PROPERTIES OF ESTIMATORS<br />

Exercise 58 Repeat the previous example, using time to surrender as the variable. Interpret 2 q 3<br />

as the probability <strong>of</strong> surrendering before the five years expire.<br />

The previous example indicated that the usual method <strong>of</strong> constructing a confidence interval<br />

can lead to an unacceptable result. An alternative approach can be constructed as follows. Let<br />

Y =ln[− ln S n (t)]. Using the delta method (see Theorem 3.33 in the next Subsection), the variance<br />

<strong>of</strong> Y can be approximated as follows. The function <strong>of</strong> interest is g(x) = ln(− ln x). Its<br />

derivative is g 0 (x) = 1<br />

be approximated by<br />

− ln x<br />

−1<br />

x<br />

=(x ln x)−1 . According to the delta method, the variance <strong>of</strong> Y can<br />

{g 0 [E(S n (t))]} 2 Var[S n (t)] = Var[S n(t)]<br />

[S n (t)lnS n (t)] 2<br />

wherewehaveusedthefactthatS n (t) is an unbiased estimator <strong>of</strong> S(t). Then, an estimated 95%<br />

confidence interval for θ =ln[− ln S(t)] is<br />

q<br />

dVar[S n (t)]<br />

ln[− ln S n (t)] ± 1.96<br />

S n (t)lnS n (t) .<br />

Because S(t) =exp(−e θ ), putting each endpoint through this formula will provide a confidence<br />

interval for S(t). Fortheupperlimitwehave(whereˆv = Var[S d n (t)])<br />

n<br />

exp −e ln[− ln S n(t)]+1.96 √ˆv/[S o n (t)lnS n (t)]<br />

= exp<br />

n[ln o<br />

S n (t)]e 1.96√ˆv/[S n (t)lnS n (t)]<br />

"<br />

= S n (t) U 1.96 √ˆv<br />

#<br />

, U =exp<br />

.<br />

S n (t)lnS n (t)<br />

Similarly, the lower limit is S n (t) 1/U . This interval will always be inside the range zero to one <strong>and</strong><br />

is referred to as the log-transformed confidence interval.<br />

Example 3.27 Obtain the log-transformed confidence interval for S(3) as in Example 3.26.<br />

We have<br />

"<br />

1.96 √ #<br />

0.0034671<br />

U =exp<br />

=0.32142.<br />

0.8923 ln(0.8923)<br />

The lower limit <strong>of</strong> the interval is 0.8923 1/0.32142 =0.70150 <strong>and</strong> the upper limit is 0.8923 0.32142 =<br />

0.96404. ¤<br />

Exercise 59 Obtain the log-transformed confidence interval for S(3) in Exercise 58.<br />

Similar results are available for the Nelson-Åalen estimator. An intuitive derivation <strong>of</strong> a variance<br />

estimate proceeds as follows. As in the derivation for the Kaplan-Meier estimator, all results are<br />

obtained assuming the risk set numbers are known, not r<strong>and</strong>om. The number <strong>of</strong> deaths at death<br />

time t i has approximately a Poisson distribution 3 with parameter r i h(t i ) <strong>and</strong> so its variance is<br />

3 A binomial assumption (as was used for the Kaplan-Meier derivation) could also have been made. Similarly, a<br />

Poisson assumption could have been used for the Kaplan-Meier derivation. The formulas in this Note are the ones<br />

most commonly used.

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