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# Sec7 - OR 3500/5500 Fall11 Section 7 Section 7 Problem 1 Go...

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OR 3500/5500, Fall’11, Section 7 Section 7 Problem 1 Go over homework 6. Problem 2 Verify Var( Z + Y ) = Var( Z ) + Var( Y ) + 2Cov( Z, Y ) . Problem 3 A continuous random vector ( Z, Y ) has a joint pdf given by f X,Y ( x, y ) = ( 3 x 0 x 1 , x - 1 y 1 - x 0 otherwise . 1. Find Cov( X, Y ). Are X and Y independent? 2. Find Var( X - 2 Y + 5). Problem 4 Suppose that n balls are placed into r boxes at random. For i = 1 , . . . , r , let X i = ( 1 if ith box is empty , 0 otherwise . 1. Compute E ( X i ). 2. Compute E ( X i X j ) for i 6 = j . 3. Let S be the number of empty boxes (i.e., S = r i =1 X i ). Find E ( S ) and Var( S ). 1

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OR 3500/5500, Fall’11, Section 7 Solutions: 1. Last homework 2. V ar ( Z + Y ) = ( E ( Z + Y )) 2 - E (( Z + Y ) 2 ) = ( EZ + EY ) 2 - E ( Z 2 + 2 ZY + Y 2 ) = ( EZ ) 2 - E ( Z 2 ) + ( EY ) 2 - E ( Y 2 ) + 2 EZEY - 2 EY Z = V ar ( Z ) + V ar ( Y ) - 2 Cov ( Y, Z ) 3. (a) For 0 x 1, f X ( x ) = R 1 - x x - 1 3 xdy = 6 x ( x - 1) For 0 y 1, f Y ( y ) = R 1 - y 0 3 xdx = 3 2 (1 - y ) 2 For - 1 y 0, f Y ( y ) = R 1+ y 0 3 xdx = 3 2 (1 + y ) 2 Hence for - 1 y 1, f Y ( y ) = 3 2 (1 - | y | ) 2 We should find EX , EY and EXY EX = Z 1 0 x 6 x (1 - x ) dx = 1 2 EY = Z 1 - 1 y 3 2 (1 - | y | ) 2 dy = 0 EXY = Z 1 0 Z 1 - x x - 1 xy 3 xdydx = Z 1 0 3 x 2 Z 1 - x x - 1 ydydx = 0 Hence Cov ( X, Y ) = EXY - EXEY = 0 but X and Y are not independent since f X,Y 6 = f X f Y (b) V ar ( X - 2 Y +5) = V ar ( X )+4 V ar ( Y ) - 4 Cov ( X, Y ) = V ar ( X )+4 V ar ( Y ) E ( X 2 ) = Z 1 0 x 2 6 x (1 - x ) dx = 3 10 E
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