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Unformatted text preview: ESTIMATION 165 Example 4.10.1 Let X = Y + N , where Y has the exponential distribution with parameter λ, and
N is Gaussian with mean 0 and variance σ 2 . Suppose the variables Y and N are independent, and
1
1
the parameters λ and σ 2 are known and strictly positive. (Recall that E [Y ] = λ and Var(Y ) = λ2 .)
(a) Find L∗ , the MSE linear estimator of Y given X , and also ﬁnd the resulting MSE.
(b) Find an unconstrained estimator of Y yielding a strictly smaller MSE than L∗ does.
Solution: (a) Since Y and N are independent, Cov(Y, N ) = 0. Therefore,
Cov(Y, X ) = Cov(Y, Y + N ) = Cov(Y, Y ) + Cov(Y, N ) = Var(Y )
Var(X ) = Var(Y + N ) = Var(Y ) + Var(N ). So, by (4.32),
L∗ (X ) = 1
1/λ2
+
λ 1/λ2 + σ 2 X− 1
λ = 1
1
+
λ 1 + λ2 σ 2 X− 1
λ = X + λσ 2
,
1 + λ2 σ 2 and by (4.34),
MSE for L∗ (X ) = 1
− (1/λ4 )/
λ2 1
+ σ2
λ2 = σ2
.
1 + λ2 σ 2 (b) Although Y is always nonnegative, the estimator L∗ (X ) can be negative. An estimator with
smaller MSE is Y = max{0, L∗ }, because (Y − Y )2 ≤ (Y − L∗...
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 Spring '08
 Zahrn
 The Land

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