STAT1301 (Assig5 ans)full

STAT1301 (Assig5 ans)full - 09/10 THE UNIVERSITY OF HONG...

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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 < + = = = | .
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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 t Y + = 1 (e) Using the expression in part (c), for 1 0 < < t , () 2 2 1 2 1 2 2 2 t t t t t x t t dx x t t t F t t W = + = + = + = . Therefore the cdf of is XY W = < < = 1 1 1 0 2 0 0 2 t t t t t t F W . 2. (a) For , if and only if ,... 2 , 1 = y y Y = y X y < 1 . Therefore the pmf of Y is given by () ( ) ( ) y X y P y Y P y p Y < = = = 1 ( ) ( ) 1 1 1 = y y e e y y e e = 1 ( ) 1 1 1 1 = e e y , ,... 2 , 1 = y Hence ( ) 1 1 ~ e Geometric Y . (b) Let , then the conditional distribution function of W given is given by 4 = X W 5 Y 4 | 4 5 | = = X w X P Y w W P w F W 4 4 4 + = X P w X P ( ) ( ) 4 4 4 1 1 1 1 + = e e e w , w e = 1 0 w Hence . 1 ~ | 5 Exp W Y
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09/10 p. 3 3. (a) () x n x p X p p x n x n p x f = 1 ! ! ! | | , n x ,..., 1 , 0 = ()() 1 1 1 Γ Γ + Γ = b a p p p b a b a p f , 1 0 < < p The joint pdf of X and p is ( ) ( ) p f p x f p x f p p X | , | = 1 1 1 ! ! ! + + Γ Γ + Γ = x b n x a p p b a b a x n x n , n x ,..., 1 , 0 = , . 1 0 < < p The marginal pmf of X is given by () ( ) ( ) + + Γ Γ + Γ = = 1 0 1 1 1 0 1 ! ! ! , dp p p b a b a x n x n dp p x f x f x b n x a X ( ) ( ) n b a x n b x a b a b a x n x n + + Γ + Γ + Γ Γ Γ + Γ = ! !
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STAT1301 (Assig5 ans)full - 09/10 THE UNIVERSITY OF HONG...

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