410Hw02ans - STAT 410 Fall 2011 Homework#2(due Friday September 9 by 3:00 p.m 1 Let X and Y have the joint p.d.f f X Y x y = C x 2 y 3 a 0 < x < 1

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STAT 410 Fall 2011 Homework #2 (due Friday, September 9, by 3:00 p.m.) 1. Let X and Y have the joint p.d.f. f X Y ( x , y ) = C x 2 y 3 , 0 < x < 1, 0 < y < x , zero elsewhere. a) What must the value of C be so that f X Y ( x , y ) is a valid joint p.d.f.? 1 0 0 3 2 dx dy y x x C = 1 0 4 4 dx x C = 20 C = 1. C = 20 . b) Find P ( X + Y < 1 ). y = x and y = 1 – x x = y 2 and x = 1 – y y = 2 1 5 - . P ( X + Y < 1 ) = - - 2 1 5 0 1 3 2 20 2 dy dx y x y y = ( ) - - - 2 1 5 0 9 3 3 3 20 1 3 20 dy y y y
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= - - - + - 2 1 5 0 9 6 5 4 3 3 20 3 20 20 20 3 20 dy y y y y y = 0 2 1 5 10 7 6 5 4 3 2 21 20 3 10 4 3 5 - - - + - y y y y y 0.030022. OR y < x and y = 1 – x x = 2 5 3 2 1 5 1 2 1 5 2 - = - - = - . P ( X + Y < 1 ) = - - - 1 2 5 3 1 3 2 20 1 dx dy y x x x = ( ) ( ) - - - - 1 2 5 3 4 2 4 1 5 5 1 dx x x x = ( ) - - + - + - - 1 2 5 3 6 5 4 3 2 5 20 25 20 5 1 dy x x x x x = 2 5 3 1 7 6 5 4 3 7 5 3 10 5 5 3 5 1 - - + - + - - x x x x x 0.030022. c) Let 0 < a < 1. Find P ( Y < a X ). P ( Y < a X ) = 1 0 0 3 2 20 dx dy y x x a = 1 0 6 4 5 dx x a = 4 7 5 a .
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d) Let 0 < a < 1. Find P ( X Y < a ). y = x and y = x a x = 3 2 a . P ( X Y < a ) = - 1 3 2 3 2 20 1 a x x a dx dy y x = - - 1 2 4 4 3 2 5 5 1 a dx x a x = 3 2 1 4 5 5 1 a x a x + - = 4 3 10 5 6 a a - .
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2. Let X and Y have the joint p.d.f. f X Y ( x , y ) = 20 x 2 y 3 , 0 < x < 1, 0 < y < x , zero elsewhere. a) Find f X ( x ). f X ( x ) = x dy y x 0 3 2 20 = 5 x 4 , 0 < x < 1. b) Find E ( X ). E ( X ) = 1 0 4 5 dx x x = 6 5 . c) Find f Y ( y ). f Y ( y ) = 1 3 2 2 20 y dx y x = ( ) 9 3 3 20 y y - , 0 < y < 1. d) Find E ( Y ). E ( Y ) = ( ) - 1 0 9 3 3 20 dy y y y = - 1 0 10 4 3 20 3 20 dy y y = 33 20 3 4 - = 11 8 . e) Find Cov ( X, Y ). E ( X Y ) = 1 0 0 3 2 20 dx dy y x y x x = 1 0 2 11 4 dx x = 13 8 . Cov ( X, Y ) = E ( X Y ) – E ( X ) E ( Y ) = 11 8 6 5 13 8 - = 858 8 0.009324.
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3. Suppose the joint probability density function of ( X , Y ) is ( ) = otherwise 0 1 0 , 2 x y y x C y x f a) Find the value of C that would make ( ) y x f , a valid probability density function. 1 =
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This note was uploaded on 01/26/2012 for the course STAT 410 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.

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410Hw02ans - STAT 410 Fall 2011 Homework#2(due Friday September 9 by 3:00 p.m 1 Let X and Y have the joint p.d.f f X Y x y = C x 2 y 3 a 0 < x < 1

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