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solution of assignment4

# solution of assignment4 - 09/10 THE UNIVERSITY OF HONG KONG...

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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 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) 0.35, 0.2 (b) , () = = = 1 55 . 0 0 45 . 0 x x x p X () = = = = 2 15 . 0 1 55 . 0 0 3 . 0 y y y y p Y (c) (d) 0.55, 0.85, 0.2475, 0.4275 () = = = = = + 3 1 . 0 2 3 . 0 1 5 . 0 0 1 . 0 w w w w w p Y X (e) –0.0175, –0.0538 (f) () = = = = = 2 11 2 1 11 5 0 11 4 1 | | y y y X y p X Y 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) No (b) 7 6 (c) () ( ) 1 0 , 2 7 6 2 < < + = x x x x f X (d) 56 15 (e) () 2 0 , 2 4 2 | | < < + + = y x y x x y f X Y (f) 656

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09/10 p. 2 8. (a) Since the supports of X and Y depends on each other, they are not independent. (b) marginal pdf of
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solution of assignment4 - 09/10 THE UNIVERSITY OF HONG KONG...

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