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Estimation, Evaluation, and Selection of Actuarial Models

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3.2. MEASURES OF QUALITY 39<br />

1 2 3 5 2 3 4 6<br />

3 4 5 7 5 6 7 9<br />

Combining the common values, the sample mean, usually denoted ¯X, has the following probability<br />

distribution.<br />

x p ¯X(x)<br />

1 1/16<br />

2 2/16<br />

3 3/16<br />

4 2/16<br />

5 3/16<br />

6 2/16<br />

7 2/16<br />

9 1/16<br />

The expected value <strong>of</strong> the estimator is<br />

E( ¯X) = [1(1) + 2(2) + 3(3) + 4(2) + 5(3) + 6(2) + 7(2) + 9(1)]/16 = 4.5<br />

<strong>and</strong> so the sample mean is an unbiased estimator <strong>of</strong> the population mean for this example. ¤<br />

Example 3.5 For Example 3.2 determine the bias <strong>of</strong> the sample mean <strong>and</strong> the sample median as<br />

estimators <strong>of</strong> the population mean.<br />

Thesamplemeanis ¯X =(X 1 + X 2 + X 3 )/3 where each X j represents one <strong>of</strong> the observations<br />

from the exponential population. Its expected value is<br />

µ <br />

E( ¯X)<br />

X1 + X 2 + X 3<br />

=E<br />

= 1 3<br />

3 [E(X 1)+E(X 2 )+E(X 3 )] = 1 (θ + θ + θ) =θ<br />

3<br />

<strong>and</strong> therefore the sample mean is an unbiased estimator <strong>of</strong> the population mean.<br />

Investigating the sample median is a bit more difficult. The distribution function <strong>of</strong> the middle<br />

<strong>of</strong> three observations can be found as follows, using Y as the r<strong>and</strong>om variable <strong>of</strong> interest <strong>and</strong> X as<br />

the r<strong>and</strong>om variable for an observation from the population.<br />

F Y (y) = Pr(Y ≤ y) =Pr(X 1 ,X 2 ,X 3 ≤ y)+Pr(X 1 ,X 2 ≤ y,X 3 >y)<br />

+Pr(X 1 ,X 3 ≤ y,X 2 >y)+Pr(X 2 ,X 3 ≤ y, X 1 >y)<br />

= F X (y) 3 +3F X (y) 2 [1 − F X (y)]<br />

= [1− e −y/θ ] 3 +3[1− e −y/θ ] 2 e −y/θ .<br />

The density function is<br />

f Y (y) =F 0 Y (y) = 6 θ<br />

³e −2y/θ − e −3y/θ´<br />

.<br />

The expected value <strong>of</strong> this estimator is<br />

E(Y |θ) =<br />

Z ∞<br />

0<br />

= 5θ<br />

6 .<br />

y 6 θ<br />

³e −2y/θ − e −3y/θ´<br />

dy

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