Estimation, Evaluation, and Selection of Actuarial Models

3.2. MEASURES OF QUALITY 43
When multiplied by 1.2, the sample median has second moment
Z ∞
E[(1.2Y ) 2 ] = 1.44 y 2 6 ³e −3y/θ´
−2y/θ − e dy
0 θ
= 1.44 6 · µ −θ
y 2
θ 2 e−2y/θ + θ µ θ
2
3 e−3y/θ − 2y
4 e−2y/θ − θ2
9 e−3y/θ
µ −θ
3
∞
+2
= 8.64
θ
8 e−2y/θ + θ3
27
¸¯¯¯¯
e−3y/θ
µ 2θ
3
8 − 2θ3 = 38θ2
27 25
for a variance of
38θ 2
25 − θ2 = 13θ2
25 > θ2
3 .
The sample mean has the smaller MSE regardless of the true value of θ. Therefore, for this problem,
it is a superior estimator of θ. ¤
Example 3.15 For the uniform distribution on the interval (0, θ) compare the MSE of the estimators
2 ¯X and n+1
n max(X 1,...,X n ). Also evaluate the MSE of max(X 1 ,...,X n ).
The first two estimators are unbiased, so it is sufficient to compare their variances. For twice
thesamplemean,
Var(2 ¯X) = 4 4θ2
Var(X) =
n 12n = θ2
3n .
For the adjusted maximum, the second moment is
0
E
" µn +1
n
Y n
2
#
=
(n +1)2 nθ 2
n 2 n +2 = (n +1)2 θ 2
(n +2)n
for a variance of
(n +1) 2 θ 2
(n +2)n − θ2 =
θ 2
n(n +2) .
Except for the case n = 1 (and then the two estimators are identical), the one based on the
maximum has the smaller MSE. The third estimator is biased. For it, the MSE is
nθ 2
(n +2)(n +1) 2 + µ nθ
n +1 − θ 2
=
2θ 2
(n +1)(n +2)
which is also larger than that for the adjusted maximum. ¤
Exercise 53 For the sample of size three in Exercise 51, compare the MSE of the sample mean
and median as estimates of θ.
Exercise 54 (*) You are given two independent estimators of an unknown quantity θ. For estimator
A, E(ˆθ A ) = 1000 and Var(ˆθ A ) = 160, 000, while for estimator B, E(ˆθ B ) = 1200 and
Var(ˆθ B )=40, 000. Estimator C is a weighted average, ˆθ C = wˆθ A +(1− w)ˆθ B . Determine the
value of w that minimizes Var(ˆθ C ).