EE465-USC-Fall08-Assignment3_Solution

EE465-USC-Fall08-Assignment3_Solution - EE 465 : Homework 3...

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EE 465 : Homework 3 Solutions 1. a) Let A = { X > k } where k is a positive interger. F X ( x | A ) = P { X x | X > k } = P { X x,X > k } P { X > k } = P { k < X x } P { X > k } = F X ( x ) - F X ( k ) 1 - F X ( k ) If X is a geometric random variable, it has a CDF F X ( x ) = 1 - q x x =1, 2, ...q = 1 - p. where p is the success probability (hence q is the failure probability). Substituting, we have F X ( x | A ) = (1 - q x ) - (1 - q k ) 1 - (1 - q k ) = q k - q x q k = 1 - q x - k = 1 - (1 - p ) x - k for x > k 0 otherwise b) Let A = { X < k } where k is a positive interger. F X ( x | A ) = P { X x | X < k } = P { X x,X < k } P { X < k } P { X x,X < k } = P { X x } , x k - 1 P { X < k } x > k - 1 F X ( x | A ) = P { X < k } P { X < k } = 1 for X > k - 1 For x k - 1, F X ( x | A ) = P { X x } P { X < k } = F X ( x ) F X ( k - 1) = 1 - (1 - p ) x 1 - (1 - p ) k - 1 Therefore, F X ( x | X < k ) = 0 x 0 1
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This note was uploaded on 08/05/2009 for the course EE 465 taught by Professor Chow during the Fall '04 term at USC.

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EE465-USC-Fall08-Assignment3_Solution - EE 465 : Homework 3...

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