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# Find p x 2y 1 y 172 chapter 4 jointly distributed

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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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