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Estimation, Evaluation, and Selection of Actuarial Models

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

3.3 Variance <strong>and</strong> confidence intervals<br />

3.3.1 Introduction<br />

The examples in the previous Section were relatively simple. Problems <strong>of</strong> actuarial interest are<br />

<strong>of</strong>ten more complex. Complexity may be introduced via censoring or truncation or by having a<br />

goal that is not as easy to work with as a population mean. In this Section, accuracy measures for<br />

the estimators introduced in Chapter 2 will be introduced. These accuracy measures will then be<br />

used to provide interval estimates for the quantity <strong>of</strong> interest.<br />

3.3.2 Methods for empirical distributions<br />

Three cases will be covered. The first case is complete, individual data. That will be followed by<br />

an analysis <strong>of</strong> the grouped data case. The final case involves the effects <strong>of</strong> censoring or truncation.<br />

In Chapter 2, most <strong>of</strong> the time, the goal was estimation <strong>of</strong> the survival function. That case will<br />

be the only one covered in this Subsection. <strong>Estimation</strong> <strong>of</strong> related quantities will be covered in the<br />

following Subsection.<br />

When all <strong>of</strong> the information is available, analyzing the empirical estimate <strong>of</strong> the survival function<br />

is straightforward.<br />

Example 3.16 Demonstrate that the empirical estimator <strong>of</strong> the survival function is unbiased <strong>and</strong><br />

consistent.<br />

Recall that the empirical estimate <strong>of</strong> S(x) is S n (x) =Y/n where Y is the number <strong>of</strong> observations<br />

in the sample that are greater than x. ThenY must have a binomial distribution with parameters<br />

n <strong>and</strong> S(x). Then,<br />

E[S n (x)] = E(Y/n)=nS(x)/n = S(x),<br />

demonstrating that the estimator is unbiased. The variance is<br />

Var[S n (x)] = Var(Y/n)=<br />

S(x)[1 − S(x)]<br />

n<br />

which has a limit <strong>of</strong> zero, thus verifying consistency. ¤<br />

In order to make use <strong>of</strong> the result, the best we can do for the variance is estimate it. It is<br />

unlikely we know the value <strong>of</strong> S(x), because that is the quantity we are trying to estimate. The<br />

estimated variance is given by<br />

dVar[S n (x)] = S n(x)[1 − S n (x)]<br />

.<br />

n<br />

The same results hold for empirically estimated probabilities. Let p =Pr(a

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