Hw10ans - STAT 408 Spring 2009 Homework#10(due Friday April...

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Unformatted text preview: STAT 408 Spring 2009 Homework #10 (due Friday, April 10, by 3:00 p.m.) 1. Let X and Y have the joint probability density function x+4 y f X, Y ( x, y ) = 0 a) Find f Y ( y ). 0 < y < x <1 otherwise x2 1 1 9 f Y ( y ) = ( x + 4 y ) dx = +4xy = +4 y - y 2, 2 y 2 2 y 1 0 < y < 1. b) Find f Y | X ( y | x ). f X ( x ) = ( x + 4 y ) dy = x y + 2 y 2 0 x ( )x 0 = 3 x 2, 0 < x < 1. f Y|X( y | x ) = x+4y , 3x2 0 < y < x, 0 < x < 1. f Y | X ( y | x ) is undefined for x < 0 or x > 1. c) Find E ( Y | X ). x+4y 1 xy2 4y3 x 11 x E( Y | X = x ) = y dy = + = , 0 < x < 1. 2 2 2 0 18 3 3x 3x 0 E ( Y | X = x ) is undefined for x < 0 or x > 1. E( Y | X ) = 11 X . 18 x d) 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. e) Find Cov ( X, Y ). 1 E( X ) = 2 x 3 x dx = 0 1 3 . 4 E( X2) = 2 2 x 3 x dx = 0 3 . 5 11 X 11 3 11 ) = = . 18 18 4 24 E( Y ) = E[E(Y | X ) ] = E( E( X Y ) = E[E(X Y | X ) ] = E[ X E( Y | X ) ] = E( Cov ( X, Y ) = E ( X Y ) E ( X ) E ( Y ) = 11 X 2 11 3 11 ) = = . 18 18 5 30 11 3 11 11 - = . 30 4 24 480 2. a) b) 2 3 Let f ( x 1 , x 2 ) = 21 x1 x 2 , 0 < x 1 < x 2 < 1, zero elsewhere, be the joint pdf of X 1 and X 2 . Find the conditional mean and variance of X 1 , given X 2 = x 2 , 0 < x 2 < 1. Find the distribution of Y = E ( X 1 | X 2 ). That is, find the pdf of Y, f Y ( y ), or the cdf of Y, F Y ( y ). c) Determine E ( Y ) and Var ( Y ) and compare these to E ( X 1 ) and Var ( X 1 ), respectively. 3 f ( x 1 , x 2 ) = 21 x12 x 2 , 0 < x 1 < x 2 < 1. (a) 6 3 f 2 ( x 2 ) = 21 x12 x 2 dx1 = 7 x 2 , 0 x2 0 < x 2 < 1. 3 f 1 | 2 ( x 1 | x 2 ) = 3 x12 x 2 , 0 < x 1 < x 2 , 0 < x 2 < 1. x2 E( X1 | X2 = x2 ) = 0 2 3 x1 3 x1 x 2 dx1 = 3 x2 , 4 3 2 x2 , 5 0 < x 2 < 1. 2 E( X1 | X2 = x2 ) = x2 0 2 2 3 x1 3 x1 x 2 dx1 = 0 < x 2 < 1. 0 < x 2 < 1. Var ( X 1 | X 2 = x 2 ) = 3 2 9 3 2 2 x2 - x2 = x2 , 5 16 80 (b) Y = E( X1 | X2 ) = 3 X 2. 4 3 ( 0, 1 ) 0, 4 F2( x2) = 0 7 x2 1 x2 < 0 0 x2 < 1 1 x2 4 4 FY( y) = P ( Y y ) = P ( X2 y ) = y , 3 3 7 0<y< 3 . 4 4 fY( y) = 7 y 6, 3 7 0<y< 3 . 4 (c) 7 4 3 4 E ( Y ) = y 7 y 6 dy = 8 3 4 3 0 34 7 7 8 = 21 . 32 34 E( Y ) = 2 7 4 3 4 2 6 y 7 3 y dy = 9 3 4 0 7 441 7 - = . 16 1024 1024 7 7 9 = 7 . 16 Var ( Y ) = 3 f 1 ( x 1 ) = 21 x12 x 2 dx 2 = x12 1 - x14 , 4 x 1 1 21 ( ) 0 < x 1 < 1. 1 E( X1) = = 2 4 3 7 x1 4 x1 1 - x1 dx1 = 4 x1 - x1 dx1 0 0 21 ( ) 1 21 ( ) 21 21 21 - = . 16 32 32 1 2 E( X1 ) = 21 x12 x12 1 - x14 4 ( )dx1 1 = 0 21 4 8 x1 - x1 dx1 4 0 ( ) = 21 21 84 21 - = = . 20 36 180 45 21 441 1659 553 - = = . 45 1024 46080 15360 Var ( X 1 ) = E ( Y ) = E ( X 1 ). Var ( Y ) < Var ( X 1 ). 3. Let f X, Y ( x, y ) = 2, 0 < x < y, 0 < y < 1, zero elsewhere, be the joint pdf of X and Y. a) b) c) d) e) Find the marginal pdfs f X ( x ) and f Y ( y ). Find the conditional pdfs f X | Y ( x | y ) and f Y | X ( y | x ). Find the conditional means E ( X | Y = y ) and E ( Y | X = x ). Find the conditional variances Var ( X | Y = y ) and Var ( Y | X = x ). Find the correlation coefficient of X and Y, X Y . f ( x, y ) = 2, a) 0 < x < y < 1. f X ( x ) = 2 ( 1 x ), 0 < x < 1. f Y ( y ) = 2 y, f X | Y ( x | y ) = 1/y , f Y | X ( y | x ) = 1/( 1 x ) , y 0 < y < 1. b) 0 < x < y. x < y < 1. c) E( X |Y = y ) = 1 x y dx 0 1 = y 2 , 0 < y < 1. E(Y |X = x ) = 1 1+ x 1- x 2 y dx = = , 1- x 2 2 (1 - x ) x 0 < x < 1. d) E( X2|Y = y ) = y2 1 x 2 dx = , y 3 0 y 0 < y < 1. Var ( X | Y = y ) = y2 y2 3 - 2 = y2 12 , 0 < y < 1. 1 1+ x + x 2 1- x3 E( Y |X = x ) = y dx = = , 1- x 3 3 (1 - x ) x 2 1 2 0 < x < 1. Var ( Y | X = x ) = 1+ x + x 2 1+ x - 3 2 12 2 = 4 + 4x + 4x 2 - 3- 6x - 3x 2 12 0 < x < 1. 1- 2 x + x 2 = = 12 (1 - x ) 2 , [ X | Y = y has a Uniform distribution on ( 0, y ), Y | X = x has a Uniform distribution on ( x, 1 ). ] 1 e) E( X ) = x 2 ( 1 - x ) dx 1 0 = 1 0 2 1 = . 3 3 2 2 1 = . 3 4 6 E( X2) = 2 x 2 ( 1 - x ) dx = Var ( X ) = 1 1 - 6 3 2 = 1 . 18 2 E ( Y ) = y 2 y dy = . 3 0 Var ( Y ) = 1 E( Y ) = 2 1 0 2 y 2 y dy = 2 1 = . 4 2 1 2 - 2 3 2 = 1 . 18 y 1 1 E ( X Y ) = x y 2 dx dy = y 3 dy = . 4 0 0 0 1 Cov ( X, Y ) = 1 1 18 1 1 2 1 = . 4 3 3 36 = 1 . 2 = 36 1 18 4. Let X and Y have the pdf f ( x , y ) = 1, 0 < x < 1, 0 < y < 1, zero elsewhere. Find the cdf and pdf of the product Z = X Y. Hint: Find the cdf of Z, F Z ( z ) = P ( Z z ) = P ( X Y z ), 0 < z < 1, first. FZ( z ) = P( Z z ) = P( Y z X ) z 1 1 = 1d y d x + 0 0 z z x 1d y d x 0 = 1d x + x d x 0 z 0 < z < 1. 0 < z < 1. z 1 z = z z ln z, f Z ( z ) = F Z( z ) = ln z, ' 0 F Z ( z ) = z - z ln z 1 z0 0 < z <1 z 1 fZ( z ) = - ln z 0 0 < z <1 otherwise From the textbook: 4.2-8 (a), (b), (c), (d) 4.2-10 (a), (b) 4.2-12 (a), (b) 4.3-6 4.3-16 + (e) Compute P ( X + Y 1 ). Find the conditional mean E ( X | Y = y ). (f) X has a Uniform distribution on ( 0, 1 ): Y | X = x has a Uniform distribution on ( 0, x ): f X ( x ) = 1, 0 < x < 1. f Y | X ( y | x ) = 1/x , f ( x , y ) = 1/x , (e) P(X + Y 1) = 1 1 0 < y < x. 0 < y < x < 1. x 1 x dy 0 .5 1 - x dx = 2 x -1 0 .5 x 1 1 dx = 2 - dx x 0 .5 1 0 .5 = ( 2 x - ln x ) 1 = 1 + ln 0.5 = 1 ln 2. (f) f Y ( y ) = dx = ln y , y x f X | Y ( x | y ) = 1/x ln y , E( X |y) = - 1 1 0 < y < 1. y < x < 1. = - 1 x x ln y dx y 1- y ln y , 0 < y < 1. ...
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This note was uploaded on 04/27/2009 for the course STAT 420 taught by Professor Stepanov during the Spring '08 term at University of Illinois at Urbana–Champaign.

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