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# sol8 - STA3007 Applied probability(08-09 Chapter 5 Solution...

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STA3007 Applied probability (08-09) 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 ) = w 1 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 w 1 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 0 = W 1 , S 1 = W 2 - W 1 , we have W 1 = S 0 , W 2 = S 0 + S 1 , and the Jacobean matrix of this transformation ( w 1 , w 2 ) ( s 0 , s 1 ) = 1 1 0 1 Therefore, the joint distribution of the first two sojourn times is f ( s 0 , s 1 ) = f ( w 1 ( s 0 , s 1 ) , w 2 ( s 0 , s 1 )) det( ( w 1 , w 2 ) ( s 0 , s 1 ) ) = λ 2 exp( - λ ( s 0 + s 1 )) = λ exp( - λs 0 ) λ exp( -

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