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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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