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Unformatted text preview: d vehicle. Hence T = 12000 ⋅ X is a linear unbiased estimator of E (T ) .
ˆ
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Moreover, since var( X ) < var(µ) for any other linear unbiased estimator, µ , of
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E ( X ) implies that var(12000 ⋅ X ) < var(12000 ⋅ µ) , it follows by definition that
that T is the unique BLU estimate of E (T ) . In the present case, the desired
estimate is t = 12000 ⋅ x = 21,366 vehicles. For any estimator, L = a1 X 1 + a2 X 2 with a1 + a 2 = 1 , 2. E ( L) = a1 E ( X 1 ) + a2 E ( X 2 ) = (a1 + a2 ) µ = µ
So all three estimators are unbiased. This means that we need only compare their
variances to determine which is most efficient. But by the independence of the
random variables X 1 and X 2 :
2 2 L = a1 X 1 + a2 X 2 Þ var( L) = a1 var( X 1 ) + a2 var( X 2 )
2 2 = a1 (σ2 / 35) + a2 (σ2 /105) So we must have:
(i) var( L1 ) = σ2
105 11 91
σ2
)=
(ii) var( L2 ) = σ2 ( ⋅ + ⋅
16 35 16 105 140
11 11...
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 Fall '08
 ese302

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