ass4 sol

# ass4 sol - 08/09 THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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08/09 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. for 1 . 0 , 2 1 = = = j N i N P i j = 5 ,..., 2 , 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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08/09 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) x e x f x X + = 1 4 1 , < < x , > = otherwise 0 0 6 3 y e y y f y Y (c) 0 7. (a) Obviously, cannot be factorized into ( y x cx y x f + = 2 2 , ) ( ) ( ) y h x g . Therefore X and Y are not independent. (b) () ( ) + = = 2 0 2 2 0 2 , dy xy x c dy y x f x f X x x c xy y x c + = + = 2 2 0 2 2 2 4 7 6 1 6 7 2 3 2 2 1 1 0 2 3 1 0 2 1 0 = = = + = + = c c x x c dx x x c dx x f X
08/09 p. 3 (c) () < < + = otherwise 0 1 0 2 7 6 2 x x x x f X (d) ∫∫ + = + = > 1 0 0 2 2 1 0 0 2 4 7 6 2 7 6 dx xy y x dydx xy x Y X P x x 56 15 56 15 14 15 1 0 4 1 0 3 = = = x dx x (e) < < + + = = otherwise 0 2 0 2 4 2 , | | y x y x x f y x f x y f X X Y (f) + = < < 2 1 0 0 2 2 7 6 2 1 y dxdy xy x Y X P + = 2 1 0 0 2 3 4 3 7 6 dy y x x y dy y = 2 1 0 3 2

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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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ass4 sol - 08/09 THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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