assignment 4 (ans)

assignment 4 (ans) - 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 STAT1801 Probability and Statistics: Foundations of Actuarial Science Assignment 4 Solution 1. () 4 1 3 y y Y e e y f + = , < < y . 2. 2 exp 1 exp b a x b b a x x f X + = , < < x . 3. (a) ( ) 1 1 e Geometric (b) . 1 Exp 4. (a) 2 1 2 ~ 1 , 0 ~ χ X W N X = 1 1 2 = = = Y W W X Y (inverse function exists) Hence 2 2 1 1 2 1 2 1 1 1 2 2 1 1 1 y e y dy dw y f y f y W Y Γ = = = y y 2 1 exp 2 1 2 3 π , 0 > y (a) Consider the distribution function of Y , for , we have 0 > y () ( ) ( ) ( ) ( )( y y y Z y P y Z P y Y P y F Y Φ Φ = = = = ) . Therefore the pdf of Y is () () () ( ) y y y F y f Y Y Φ + Φ = = ' ' ' 2 2 2 2 2 2 1 2 y y e e = × = , . 0 > y 5. (a) = 2 1 , , 2 1 u u f U U 2 1 2 1 3 2 1 1 u u u , 2 1 2 1 1 1 u u u < < , < < 2 2 u (c) < < < = 1 2 1 if 2 1 2 1 0 if 1 2 1 1 2 1 1 2 1 1 1 u u u u u f U (d) No
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10/11 p. 2 6. (a) () = v u f V U , , v u 2 2 1 , u v u 1 , < u 1 . (b) 2 log u u u f U = , ; < u 1 < < < = v v v v f V 1 if 2 1 1 0 if 2 1 2 7. X Y W = , X Y Z + = 2 W Z X = , 2 W Z Y + = (transformation is 1-1) < < < < + < < < < < z w w z w z y x 0 2 2 0 0 , 2 1 2 1 2 1 2 1 2 1 = = = z y w y z x w x J Joint pdf of W and Z is given by 2 1 2 2 exp 2 2 , 2 , + + = w z w z w z z w f Z W ( ) + + = 2 exp 2 w z w z w z , . < < < z w 0 8.
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assignment 4 (ans) - 10/11 THE UNIVERSITY OF HONG KONG...

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