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Unformatted text preview: STAT 410 Fall 2009 Homework #3 (due Friday, September 18, by 3:00 p.m.) 1. Consider two continuous random variables X and Y with joint p.d.f. f X, Y ( x , y ) = < < < < otherwise , 81 2 2 K K y x y x a) Find the value of K so that f X, Y ( x , y ) is a valid joint p.d.f. 1 = ∫ ∫ K K dy dx y x 0 0 2 81 2 = 243 5 K . ⇒ K = 3 . b) Find P ( X > 3 Y ). P ( X > 3 Y ) = ∫ ∫ 3 3 2 81 2 dx dy y x x = ∫ 3 4 729 1 dx x = 15 1 . OR P ( X > 3 Y ) = ∫ ∫ 1 3 3 2 81 2 dy dx y x y = … = 15 1 . c) Are X and Y independent? Justify your answer. f X ( x ) = 2 9 1 x , 0 < x < 3, f Y ( y ) = y 9 2 , 0 < y < 3. f ( x , y ) = f X ( x ) ⋅ f Y ( y ). ⇒ X and Y are independent . d) Find Cov ( X, Y ). Since X and Y are independent, Cov ( X, Y ) = . 2. Let X and Y have the joint probability density function f X, Y ( x , y ) = < < < + otherwise 1 4 x y y x a) Find the marginal p.d.f.s f X ( x ) and f Y ( y ). f X ( x ) = ( 29 ∫ + x dy y x 4 = ( 29 2 2 x y y x + = 3 x 2 , 0 < x < 1. f Y ( y ) = ( 29 ∫ + 1 4 y dx y x = y y x x 1 2 4 2 + = 2 2 9 4 2 1 y y+ , 0 < y < 1. b) Are X and Y independent? Justify your answer. The support of ( X, Y ) is not a rectangle. ⇒ X and Y are NOT independent . OR f ( x , y ) ≠ f X ( x ) ⋅ f Y ( y ). ⇒ X and Y are NOT independent . c) Find Cov ( X, Y ). E ( X ) = ∫ ⋅ 1 2 3 dx x x = 4 3 . E ( X 2 ) = ∫ ⋅ 1 2 2 3 dx x x = 5 3 . E ( Y ) = E [ E ( Y  X ) ] = E ( 18 X 11 ) = 4 3 18 11 ⋅ = 24 11 . E ( X Y ) = E [ E ( X Y  X ) ] = E [ X E ( Y  X ) ] = E ( 18 X 11 2 ) = 5 3 18 11 ⋅ = 30 11 . Cov ( X, Y ) = E ( X Y ) – E ( X ) ⋅ E ( Y ) = 24 11 4 3 30 11 ⋅= 480 11 . 3. Suppose that the random variables X and Y have joint p.d.f. f ( x , y ) given by f ( x , y ) = C x 2 y , 0 < x < y , x + y < 2. a) Sketch the support of ( X , Y ). That is, sketch { 0 < x < y , x + y < 2 }. b) What must the value of C be so that f ( x , y ) is a valid joint p.d.f.? Must have ( 29 ∫ ∫ ∞ ∞ ∞ ∞dy dx y x f , = 1. ∫ ∫ 1 2 2 dx dy y x x x C = ∫ == 1 2 2 2 2 dx y x x y x y C = ( 29 [ ] ∫ 1 2 2 2 2 2 dx x x x C = ( 29 ∫1 3 2 2 2 dx x x C C = 1 4 3 2 3 2 x x C C = 6 C = 1. ⇒ C = 6 . c) Find P ( X + Y < 1 ). ∫ ∫ 5 . 1 2 6 dx dy y x x x = ( 29 ∫ == 5 . 1 2 2 3 dx y x x y x y = ( 29 [ ] ( 29 ∫5 . 2 2 2 1 3 dx x x x = ( 29 ∫5 . 3 2 6 3 dx x x = 5 . 4 3 2 3 x x = 4 3 2 1 2 3 2 1  = 32 3 8 1= 32 1 = 0.03125 ....
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This note was uploaded on 10/14/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Fall '08 term at University of Illinois at Urbana–Champaign.
 Fall '08
 AlexeiStepanov

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