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Unformatted text preview: = E [XY ], so ∞ ∞ −∞ ∞ −∞ uvfX (u)fY |X (v |u)dvdu ρX,Y = E [XY ] = = ∞ vfY |X (v |u)dv du ufX (u) −∞ ∞ −∞ u2 fX (u)du = ρ. =ρ −∞ This proves (c). If ρ = 0 then ρX,Y = 0 so that X and Y are not independent. If ρ = 0 then fX,Y factors into the product of two single-variable normal pdfs, so X and Y are independent. This proves (d). By (4.33), L∗ (u) = ρu. Therefore, L∗ (u) = g ∗ (u), as claimed, proving (e). By (4.34), the MSE 2 for using L∗ is σe = 1 − ρ2 , so (f) follows from (4.39). 2 Example 4.11.3 Let X and Y be jointly Gaussian random variables with mean zero, σX = 5, 2 = 2, and Cov(X, Y ) = −1. Find P {X + 2Y ≥ 1}. σY 172 CHAPTER 4. JOINTLY DISTRIBUTED RANDOM VARIABLES Solution: Let Z = X + 2Y. Then Z is a linear combination of jointly Gaussian random variables, so Z itself is a Gaussian random variable. Also, E [Z ] = E [X ] + 2E [Y ] = 0 and 2 σZ = Cov(X + 2Y, X + 2Y ) = Cov(X, X ) + Cov(X, 2Y ) + Cov(2Y, X ) + Cov(2Y, 2Y ) 2...
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