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hw8 - integrating R ∞(1(1-e-x n dx by parts 4.2.7 E 1 X 1...

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MATH 471 HW 8 Solutions Pengsheng Ji November 7, 2006 3.8.10 Let N be the number of successes. Then P ( N = k | X = n ) = C n,k p k (1 - p ) n - k for k n , so P ( N = k ) = X n = k P ( N = k | X = n ) P ( X = n ) = X n = k C n,k p k (1 - p ) n - ke - λ λ n n ! = ( λp ) k k ! e - λ X n = k [ λ (1 - p )] n - k ( n - k )! = ( λp ) k k ! e - λp . 3.9.17 The density f - Y ( y ) = λe λy , for y < 0, 0 otherwise so if z > 0, by Formula (7.2) we have f X - Y ( z ) = Z z f X ( x ) f - Y ( z - x ) dx = λ 2 e λx Z z e - 2 λx dx = λ 2 e λx . Since f Z ( - z ) = f Z ( z ), it follows that f Z ( z ) = λe - λ | z | / 2 . 4.1.23 (a) Using Exercise 4.1.20, E ( Y ) = Z 0 e - nx dx = 1 n . (b) The lack of memory property of the exponential implies EZ n = E ( Y + Z n - 1 , so EZ n = 1 /n + EZ n - 1 and EZ n = n k =1 1 /k . The recursion can also be derived by
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Unformatted text preview: integrating R ∞ (1-(1-e-x ) n ) dx by parts. 4.2.7 E ( 1 X + 1 ) = ∞ X n =0 1 n + 1 e-λ λ n n ! = e-λ λ-1 ∞ X n =0 λ n +1 ( n + 1)! = e-λ λ-1 ( e λ-1) 4.2.8 The generating function of X is r ( z ) = ∞ X n =1 z n (1-p ) n-1 p = pz 1-z (1-p ) . Then EX = r ( z ) = p (1-z + zp )-pz (-1 + p ) (1-z + zp ) 2 | z =1 = 1 /p....
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hw8 - integrating R ∞(1(1-e-x n dx by parts 4.2.7 E 1 X 1...

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