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Estimation, Evaluation, and Selection of Actuarial Models

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

Suppose the value <strong>of</strong> x is between the boundaries c j−1 <strong>and</strong> c j .LetY be the number <strong>of</strong> observations<br />

at or below c j−1 <strong>and</strong> let Z be the number <strong>of</strong> observations above c j−1 <strong>and</strong> at or below c j .<br />

Then<br />

S n (x) =1− Y (c j − c j−1 )+Z(x − c j−1 )<br />

n(c j − c j−1 )<br />

<strong>and</strong><br />

E[S n (x)] = 1 − n[1 − S(c j−1)](c j − c j−1 )+n[S(c j−1 ) − S(c j )](x − c j−1 )<br />

n(c j − c j−1 )<br />

= S(c j−1 ) c j − x<br />

+ S(c j ) x − c j−1<br />

.<br />

c j − c j−1 c j − c j−1<br />

This estimator is biased (although it is an unbiased estimator <strong>of</strong> the true interpolated value). The<br />

variance is<br />

Var[S n (x)] = (c j − c j−1 ) 2 Var(Y )+(x − c j−1 ) 2 Var(Z)+2(c j − c j−1 )(x − c j−1 )Cov(Y,Z)<br />

[n(c j − c j−1 )] 2<br />

where Var(Y )=nS(c j−1 )[1 − S(c j−1 )], Var(Z) =n[S(c j−1 ) − S(c j )][1 − S(c j−1 )+S(c j )], <strong>and</strong><br />

Cov(Y,Z) =−n[1 − S(c j−1 )][S(c j−1 ) − S(c j )]. For the density estimate,<br />

f n (x) =<br />

Z<br />

n(c j − c j−1 )<br />

<strong>and</strong><br />

E[f n (x)] = S(c j−1) − S(c j )<br />

c j − c j−1<br />

which is biased for the true density function. The variance is<br />

Var[f n (x)] = [S(c j−1) − S(c j )][1 − S(c j−1 )+S(c j )]<br />

n(c j − c j−1 ) 2 .<br />

¤<br />

Example 3.20 For Data Set C estimate S(10,000), f(10,000) <strong>and</strong> the variance <strong>of</strong> your estimators.<br />

The point estimates are<br />

99(17,500 − 7,500) + 42(10,000 − 7,500)<br />

S 227 (10, 000) = 1 − =0.51762<br />

227(17,500 − 7,500)<br />

42<br />

f 227 (10,000) =<br />

227(17,500 − 7,500) =0.000018502.<br />

The estimated variances are<br />

<strong>and</strong><br />

dVar[S 227 (10,000)] =<br />

99<br />

10,0002<br />

227 227 128<br />

42<br />

+2,5002<br />

227 227 185<br />

99<br />

− 2(10,000)(2,500)<br />

227 227<br />

42<br />

227(10,000) 2 =0.00094713<br />

42 185<br />

227 227<br />

dVar[f 227 (10,000)] =<br />

227(10,000) 2 =6.6427 × 10−12 .

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