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# 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 = 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 i =1 X i ) = 6 i =1 var ( X i ) + 6 i,j =1 , i = 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 = -∞ f ( x ) dx = -∞ ce - 5 | x | dx = 2 c 0 e - 5 x dx = 2 c/ 5 c = 5 / 2 . (b) We have P ( X ( - 1 , 1)) = 1 - 1 - 1 ce - 5 | x | dx = 1 - 2 c 1 0 e - 5 x dx = 1 - 2(5 / 2)(1 - e - 5 ) / 5 = e - 5 . (c) Since E ( | X | ) < , we have E ( X ) = -∞ cxe - 5 | x | dx = 0, as the 1

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integrand is an odd function. Also using integration by parts E ( X 2 ) = -∞ cx 2 e - 5 | x | dx = 2 c 0 x 2 e - 5 x dx = 5 x 2 e - 5 x - 5 | 0 - 5 0 2 x e - 5 x - 5 dx = 2 0 xe - 5 x dx = 2 x e - 5 x - 5 | 0 - 2 0 e - 5 x - 5 dx = (2 / 5)(1 / 5) = 2 / 25 . So var ( X ) = E ( X 2 ) = 2 / 25. 3. (a) We see T = min { H, 1 / 2 } . So for 0 t < 1 / 2, P ( T t ) = P ( H t ) = t 0 2 xdx = t 2 , and for t 1 / 2, P ( T t ) = 1. So F T ( t ) = 0 if t < 0 t 2 if 0 t < 1 / 2 1 if t 1 / 2 . (b) We have E ( T ) = E (min { H, 1 / 2 } ) = 1 0 min { x, 1 / 2 } h ( x ) dx = 1 / 2 0 x · 2 xdx + 1 1 / 2 (1 / 2)2 xdx = (2 / 3)(1 / 2) 3 + (1 - (1 / 2) 2 ) / 2 = 11 / 24 . 4. (a) Mixed. (b) P (0 < X 1 . 5) = F (1 . 5) - F (0) = 1 / 2 - (1
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hw4sol - ORIE 3500/5500 Fall Term 2008 Assignment...

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