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 singlevariable 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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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.
 Spring '08
 Zahrn
 The Land

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