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Unformatted text preview: Page 1 of 6 Exam 1 ( Version A â€“ white ) ( Answers ) 1. Let X and Y have the joint probability density function f X, Y ( x , y ) = & Â¡ & Â¢ Â£ < < < + otherwise 1 4 x y y x a) (4) Find f Y ( y ). f Y ( y ) = ( ) Â¤ + 1 4 y dx y x = y y x x 1 2 4 2 Â¥ Â¥ Â¦ Â§ Â¨ Â¨ Â© Âª + = 2 2 9 4 2 1 y y + , 0 < y < 1. b) (5) Find f Y  X ( y  x ). f X ( x ) = ( ) Â¤ + x dy y x 4 = ( ) 2 2 x y y x + = 3 x 2 , 0 < x < 1. f Y  X ( y  x ) = 2 3 4 x y x + , 0 < y < x , 0 < x < 1. f Y  X ( y  x ) is undefined for x < 0 or x > 1. c) (4) Find E ( Y  X ). E ( Y  X = x ) = Â¤ + â‹… x dy x y x y 2 3 4 = 3 2 2 3 4 2 3 1 x y y x x Â¥ Â¥ Â¦ Â§ Â¨ Â¨ Â© Âª + â‹… = 18 11 x , 0 < x < 1. E ( Y  X = x ) is undefined for x < 0 or x > 1. E ( Y  X ) = 18 X 11 . Page 2 of 6 2. (10) Let X and Y have the joint probability density function f X, Y ( x , y ) = & Â¡ & Â¢ Â£ < < < + otherwise 1 4 x y y x Let U = X Y and V = X. Find the joint probability density function of ( U, V ), f U, V ( u , v ). Sketch the support of ( U, V ). X = V, Y = V U . 0 < y Â¤ 0 < u , y < x Â¤ u < v 2 , x < 1 Â¤ v < 1. J = 1 1 2 v u v = v 1 .  J  = v 1 . f U, V ( u , v ) = f X, Y ( v , v u ) â‹…  J  = v v u v 1 4 â‹… Â¥ Â¦ Â§ Â¨ Â© Âª + = 2 4 1 v u + , 0 < v < 1, 0 < u < v 2 , f U, V ( u , v ) = 0 otherwise. Page 3 of 6 3. Alex goes to the bank during his lunch hour to deposit his paycheck. Suppose the time Alex goes to the bank during his lunch hour to deposit his paycheck....
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 Spring '08
 AlexeiStepanov
 Normal Distribution, Probability, Variance, Probability theory, Alex, probability density function

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