# hw4sol - ORIE 3500/5500 Fall Term 2008 Assignment...

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ORIE 3500/5500 Fall Term 2008 Assignment 4-Solution 1. (a) For 1 i 6, E ( X i ) = P ( X i = 1) = (5 / 6) n . (b) For 1 i, j 6 and i 6 = j , E ( X i X j ) = P ( X i = 1 , X j = 1) = (4 / 6) n . (c) N = X 1 + ··· + X 6 . (d) E ( N ) = E ( X 1 ) + ··· + E ( X 6 ) = 6(5 / 6) n . (e) We have var ( N ) = var ( 6 X i =1 X i ) = 6 X i =1 var ( X i ) + 6 X i,j =1 , i 6 = j cov ( X i , X j ) . Now var ( X i ) = E ( X 2 i ) - [ E ( X i )] 2 = E ( X i ) - [ E ( X i )] 2 = (5 / 6) n (1 - (5 / 6) n ) and cpv ( X i , X j ) = E ( X i X j ) - E ( X i ) E ( X j ) = (4 / 6) n - (5 / 6) 2 n . So var ( N ) = 6(5 / 6) n (1 - (5 / 6) n ) + 30((4 / 6) n - (5 / 6) 2 n ) . 2. (a) We have 1 = Z -∞ f ( x ) dx = Z -∞ ce - 5 | x | dx = 2 c Z 0 e - 5 x dx = 2 c/ 5 c = 5 / 2 . (b) We have P ( X 6∈ ( - 1 , 1)) = 1 - Z 1 - 1 ce - 5 | x | dx = 1 - 2 c Z 1 0 e - 5 x dx = 1 - 2(5 / 2)(1 - e - 5 ) / 5 = e - 5 . (c) Since E ( | X | ) < , we have E ( X ) = R -∞ cxe - 5 | x | dx = 0, as the 1

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integrand is an odd function. Also using integration by parts
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## hw4sol - ORIE 3500/5500 Fall Term 2008 Assignment...

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