assignment4solution

assignment4solution - 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 4 Solution 1. (a) () 36 1 2 , = = = i Y i X P for 6 , 5 , 4 , 3 , 2 , 1 = i 18 1 , = + = = j i Y i X P for 1 ,..., 2 , 1 = i j ; 6 , 5 , 4 , 3 , 2 , 1 = i (b) 36 , i i Y i X P = = = for 6 , 5 , 4 , 3 , 2 , 1 = i 36 1 , = = = j Y i X P for 6 ,..., 2 , 1 + + = i i j ; 6 , 5 , 4 , 3 , 2 , 1 = i (c) 36 1 , = = = i Y i X P for 6 , 5 , 4 , 3 , 2 , 1 = i 18 1 , = = = j Y i X P for 6 ,..., 2 , 1 + + = i i j ; 6 , 5 , 4 , 3 , 2 , 1 = i 2. , ( 2 1 1 , 2 2 2 1 1 x x p p x X x X P + = = = ) ,... 2 , 1 , 0 1 = x , ,... 2 , 1 , 0 2 = x 3. 10 1 , 2 1 = = = j N i N P , i j = 5 ,..., 1 , 4 , 3 , 2 , 1 = i . 4. (a) ( ) ( ) 35 . 0 25 . 0 1 . 0 1 , 1 0 , 0 = + = + = = p p Y X P ( ) 2 . 0 0 , 1 = = > p Y X P (b) ( ) ( ) ( ) 45 . 0 05 . 0 3 . 0 1 . 0 2 , 0 1 , 0 0 , 0 0 = + + = + + = = p p p X P ( ) ( ) ( ) 55 . 0 1 . 0 25 . 0 2 . 0 2 , 1 1 , 1 0 , 1 1 = + + = + + = = p p p X P Marginal pmf of X : = = = 1 55 . 0 0 45 . 0 x x x p X ( ) ( ) 3 . 0 2 . 0 1 . 0 0 , 1 0 , 0 0 = + = + = = p p Y P ( ) ( ) 55 . 0 25 . 0 3 . 0 1 , 1 1 , 0 1 = + = + = = p p Y P ( ) ( ) 15 . 0 1 . 0 05 . 0 2 , 1 2 , 0 2 = + = + = = p p Y P Marginal pmf of Y : = = = = 2 15 . 0 1 55 . 0 0 3 . 0 y y y y p Y (c) Possible values of Y X + : 0, 1, 2, 3 ( ) 1 . 0 0 , 0 0 = = = + p Y X P ( )( ) 5 . 0 2 . 0 3 . 0 0 , 1 1 , 0 1 = + = + = = + p p Y X P ( ) ( ) 3 . 0 25 . 0 05 . 0 1 , 1 2 , 0 2 = + = + = = + p p Y X P ( ) 1 . 0 2 , 1 3 = = = + p Y X P
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10/11 p. 2 The pmf of Y X W + = is . () = = = = = 3 1 . 0 2 3 . 0 1 5 . 0 0 1 . 0 w w w w w p W (d) 55 . 0 55 . 0 1 45 . 0 0 = × + × = X E ( ) 55 . 0 55 . 0 1 45 . 0 0 2 2 2 = × + × = X E ( ) ( ) 2475 . 0 55 . 0 55 . 0 2 2 2 = = = X E X E X Var 85 . 0 15 . 0 2 55 . 0 1 3 . 0 0 = × + × + × = Y E ( ) 15 . 1 15 . 0 2 55 . 0 1 3 . 0 0 2 2 2 2 = × + × + × = Y E () ( ) () ( ) 4275 . 0 85 . 0 15 . 1 2 2 2 = = = Y E Y E Y Var (e) 45 . 0 1 . 0 2 1 25 . 0 1 1 = × × + × × = XY E ( ) ( ) ( )() 0175 . 0 85 . 0 55 . 0 45 . 0 , = × = = Y E X E XY E Y X Cov () () 0538 . 0 4275 . 0 2475 . 0 0175 . 0 , , = = = Y Var X Var Y X Cov Y X Corr (f) The conditional pmf of Y given 1 = X is = = = = = = = = = 2 11 2 55 . 0 1 . 0 1 11 5 55 . 0 25 . 0 0 11 4 55 . 0 2 . 0 1 , 1 | y y y p y x p X y p X . 5. (a) , = otherwise 0 1 0 1 x x f X = otherwise 0 1 0 1 y y f Y (b) Yes (c) 16 1 π 6. (a) 8 1 (b) y Y e y y f = 3 6 1 , < < y 0 , x e x f x X + = 1 4 1 , < < x (c) 0 7. (a) Obviously,
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This note was uploaded on 01/16/2012 for the course STAT 1301 taught by Professor Smslee during the Fall '08 term at HKU.

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assignment4solution - 10/11 THE UNIVERSITY OF HONG KONG...

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