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assignment5solution

assignment5solution - 10/11 THE UNIVERSITY OF HONG KONG...

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10/11 p. 1 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1301 Probability and Statistics I Assignment 5 Solution 1. (a) For any , the distribution function of 0 > x ( ) 1 , 0 ~ U X is if ( ) x x F X = 1 x and if . In terms of the indicator variable, it can be expressed as ( ) = x F X 1 > x 1 ( ) 1 1 > + = x x X I xI x F . Since X and Y are independent, we have t y t y y t y t X I I y t I I y t y t F y Y Y t X P < > + = + = = = 1 1 | . (b) Obviously, the support of the distribution of XY W = is ( ) 1 , 0 . For , 1 0 < < t ( ) ( ) = = Y t X P t XY P t F W = Y Y t X P E | ( ) + = < 1 0 dy y f I I y t Y t y t y ( ) ( ) + = t Y t Y dy y f dy y f y t 0 1 ( ) ( ) t F dy y f y t Y t Y + = 1 (c) From part (b), for , 1 0 < < t ( ) ( ) ( ) ( ) ( ) ( ) ( ) + = + + = 1 1 t Y Y Y t Y Y W y F d y t t F t F dy y f y t t F t F ( ) ( ) ( ) + + = 1 2 1 t Y t Y Y dy y F y t y F y t t F ( ) ( ) ( ) + = t Y Y Y Y dx x t F t F tF t F 1 1 (put x t y = ) + = 1 t Y dx x t F t (d) Since for , we can write ( ) 1 = y F Y 1 y t x t x Y Y I I x t F x t F x X X t Y P < + = = = | .
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