STAT1301 (Assig4 ans)full

STAT1301 (Assig4 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 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 2 1 1 1 1 , 2 2 2 1 1 x x x x p p p p p p x X x X P + = = = = , ,... 2 , 1 , 0 1 = x , ,... 2 , 1 , 0 2 = x 3. Just list out all the ten possible outcomes of 2 defectives out of 5, one can easily derive that 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
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09/10 p. 2 () ( ) 1 . 0 2 , 1 3 = = = + p Y X P 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) , ( 1 , 0 ~ U X ) () 1 , 0 ~ U Y = otherwise 0 1 0 1 x x f X , = otherwise 0 1 0 1 y y f Y (b) Yes, X and Y are independent. (c) For the point having a distance greater than 4 1 from the center of the square, the point must be located outside the circle having the same center as the square and radius 4 1 . Hence the probability is ( ) 16 1 1 4 1 1 square the of area circle the of area 1 2 π = = .
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09/10 p. 3 6. (a) () y y y y y y y y e y x e x e y dx e x y = = 3 3 2 2 2 3 4 3 ∫∫ = 0 3 0 2 2 3 4 dy e y dxdy e x y y y y y 8 4 3 4 = Γ = Hence 8 1 = c . (b) y y y y Y e y dx e x y y f = = 3 2 2 6 1 8 1 , < < y 0 (from part (a)) x e dy e x y x f x x y X + = = 1 4 1 8 1 2 2 , < < x (use integration by parts) (c) ( () ( ) 0 = = dx x xf X E X ( ) x xf X is an odd function ) 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 (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
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09/10 p. 4 (f) ∫∫ + = < < 2 1 0 0 2 2 7
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STAT1301 (Assig4 ans)full - 09/10 THE UNIVERSITY OF HONG...

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