solution of assignment5

# solution of assignment5 - 09/10 THE UNIVERSITY OF HONG KONG...

This preview shows pages 1–3. Sign up to view the full content.

09/10 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 1 = x F X 1 > x ( ) 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 < + = = = | .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
09/10 p. 2 Therefore for 1 0 < < t , () ( ) = = X t Y P t XY P t F W = X X t Y P E | + = < 1 0 dx I I x t F t x t x Y + = t t Y dx dx x t F 0 1 t dx x t F
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 01/16/2012 for the course STAT 1301 taught by Professor Smslee during the Fall '08 term at HKU.

### Page1 / 7

solution of assignment5 - 09/10 THE UNIVERSITY OF HONG KONG...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online