hw6_sn - EE 178 Handout #18 Probabilistic Systems Analysis...

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Unformatted text preview: EE 178 Handout #18 Probabilistic Systems Analysis Tuesday, March 3, 2009 Homework #6 Solutions 1. (15 points) a. To find c we inspect the region R where the pdf f X,Y ( x,y ) is nonzero, as shown in Figure 1. The area of this region is 2—the area of the two triangles or half the area of the 2 × 2 square. Therefore 1 = integraldisplayintegraldisplay R cf X,Y ( x,y ) dxdy = c integraldisplayintegraldisplay R f X,Y ( x,y ) dxdy = 2 c ⇒ c = 1 2 .-1 1 y 1-1 x Figure 1: Joint pdf of X,Y b. It is obvious from the joint pdf f X,Y ( x,y ) that X and Y are not independent, since the support of the joint pdf is not a rectangle. Another proof: f X,Y (0 , 0) = 1 2 . f Y (0) = integraldisplay 1 2 dx = 0 . f X (0) = integraldisplay 1- 1 1 2 dx = 1 . Thus f X,Y (0 , 0) = 1 2 negationslash = 0 = f X (0) f Y (0). An even more general proof: For x ≥ f X ( x ) = integraldisplay- x- 1 1 2 dy + integraldisplay 1 x 1 2 dy = 1 − x, and for x < f X ( x ) = integraldisplay x- 1 1 2 dy + integraldisplay 1- x 1 2 dy = 1 + x. Thus, f X ( x ) = 1 − | x | for | x | ≤ 1. The marginal pdf of Y for y ≥ 0 is f Y ( y ) = integraldisplay y- y 1 2 dx = y, and for y < 0 is f Y ( y ) = integraldisplay- y y 1 2 dx = − y . Thus f Y ( y ) = | y | for | y | ≤ 1. We see that f X,Y ( x,y ) = 1 2 negationslash = (1 −| x | ) | y | = f X ( x ) f Y ( y ). Therefore X and Y are not independent. c. We compute the covariance Cov( X,Y ) = E( XY ) − E( X ) E( Y ). E( XY ) = integraldisplay 1- 1 integraldisplay | y |-| y | 1 2 xy dxdy = integraldisplay- 1 integraldisplay- y y 1 2 xy dxdy + integraldisplay 1 integraldisplay y- y 1 2 xy dxdy = integraldisplay- 1 1 4 yx 2 bracketleftbigg- y y dy + integraldisplay 1 1 4 yx 2 bracketleftbigg y- y dy = 0 . E( X ) = integraldisplay + ∞-∞ xf X ( x ) dx = integraldisplay 1- 1 x (1 − | x | ) dx = integraldisplay- 1 x (1 + x ) dx + integraldisplay 1 x (1 − x ) dx = parenleftbigg x 2 2 + x 3 3 parenrightbiggbracketleftbigg- 1 + parenleftbigg x 2 2 − x 3 3 parenrightbiggbracketleftbigg 1 = 0 ....
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hw6_sn - EE 178 Handout #18 Probabilistic Systems Analysis...

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