HW9 - Solution to HW9 5-18 They are not independent since...

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Solution to HW9: 5-18. They are not independent, since the joint pdf can not expressed as the product of two function related to x and y , respectively. [] () c x x dx x c dx x x x dx xy dydx y x c x x x x y x x 24 2 2 2 4 ) 2 ( ) ( 3 0 2 3 0 3 0 2 2 2 ) 2 ( 3 0 3 0 2 2 2 2 2 2 = + = + = + + = + = + ∫∫ + + + Therefore, c = 1/24. a) P(X < 1, Y < 2) equals the integral of ) , ( y x f XY over the following region. 0 0 1 2 2 x y Then, 10417 . 0 2 24 1 2 2 24 1 24 1 ) ( 24 1 ) 2 , 1 ( 1 0 2 2 3 0 2 3 1 0 1 0 2 2 2 3 2 2 = + = + = + = + = < < x x x y x x x dx x dx xy dydx y x Y X P b) P(1 < X < 2) equals the integral of ) , ( y x f XY over the following region. 0 0 1 2 2 x y

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∫∫ = + = + = + = + = < < + + 3 0 3 1 2 1 2 2 1 2 1 2 2 2 . 2 2 24 1 ) 2 4 ( 24 1 24 1 ) ( 24 1 ) 2 1 ( 2 x x dx x dx xy dydx y x X P x x y x x c) P(Y > 1) is the integral of fx y XY (,) over the following region. 9792 . 0 02083 . 0 1 2 1 2 1 2 24 1 1 2 3 2 1 24 1 1 ) 2 ( 24 1 1 ) ( 1 ) 1 ( 1 ) 1 ( 1 0 3 2 1 0 2 1 0 1 0 1 2 1 24 1 = = + = + = + = + = = > x x dx x x y xy dydx y x Y P Y P x x d) P(X < 2, Y < 2) is the integral of y XY (,) over the following region. 0 0 2 2 x y 6 1 ) ( 24 1 ) 2 , 2 ( 2 0 2 = + = < < x dydx y x x Y X P
e) ∫∫ = + = + = + = + = + + 3 0 3 0 2 3 2 3 0 3 0 2 2 2 2 8 15 3 4 24 1 ) 2 4 ( 24 1 24 1 ) ( 24 1 ) ( 2 x x dx x x dx y x dydx y x x X E x x xy x x f) 6094 . 0 8 15 20 2 3 4 4 3 24 1 8 15 ) 4 4 4 3 ( 24 1 8 15 24 1 8 15 ) ( 24 1 ) ( 2 3 0 5 2 3 4 2 3 0 4 2 3 2 3 0 3 0 2 2 3 2 2 2 2 2 = + + = + + = + = + = + + x x x x dx x x x x dx y x dydx y x x X V x x y x x x g) ) ( x f X is the integral of ) , ( y x f XY over the interval from x to x+2. That is, 12 1 6 24 1 ) ( 24 1 ) ( 2 2 2 2 + =

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HW9 - Solution to HW9 5-18 They are not independent since...

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