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Unformatted text preview: STA3007 Applied probability (0809) Solution to Assignment 8 Chapter 5 Problem 3.1 If W 1 > w 1 ,W 2 > w 2 , then there is no event occurring in the interval [0 ,w 1 ], and there may 0 or 1 event occurring in the interval ( w 1 ,w 2 ], which imply X ( w 1 ) = 0 ,X ( w 2 ) X ( w 1 ) = 0 or 1. It is obvious that X ( w 1 ) = 0 ,X ( w 2 ) X ( w 1 ) = 0 or 1 also imply W 1 > w 1 ,W 2 > w 2 . Pr( W 1 > w 1 ,W 2 > w 2 ) = Pr( X ( W 1 ) = 0 ,X ( w 2 ) X ( w 1 ) = 0 or 1) = Pr( X ( W 1 ) = 0)Pr( X ( w 2 ) X ( w 1 ) = 0 or 1) = exp( λw 1 )[1 + λ ( w 2 w 1 )] exp( λ ( w 2 w 1 )) = [1 + λ ( w 2 w 1 )] exp( λw 2 ) . Since Pr( W 1 > w 1 ,W 2 > w 2 ) = Z ∞ w 1 Z ∞ w 2 f ( x 1 ,x 2 ) dx 1 dx 2 , where f ( x 1 ,x 2 ) is the joint density function of W 1 ,W 2 , thus f ( w 1 ,w 2 ) = d dw 1 d dw 2 Z ∞ w 1 Z ∞ w 2 f ( x 1 ,x 2 ) dx 1 dx 2 = λ 2 exp( λw 2 ) for 0 < w 1 < w 2 . Problem 3.3 By changing variables according to S = W 1 , S 1 = W 2 W 1 , we have W 1 = S , W 2 = S + S 1...
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This note was uploaded on 02/16/2011 for the course ECON 1224 taught by Professor Afdsa during the Spring '11 term at CUHK.
 Spring '11
 afdsa

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