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### Hw03ans

Course: STAT 410, Spring 2008
School: UIllinois
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Word Count: 980

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410 Homework STAT #3 (due Friday, February 8, by 3:00 p.m.) Spring 2008 1. 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, a) Sketch the support of ( X , Y ). 0 &lt; x &lt; y, x + y &lt; 2. b) What must the value of C be so that f ( x, y ) is a valid joint p.d.f.? - - Must have f ( x, y ) dx dy = 1. 1 2- x 0 C x 2 y dy dx 1 0 x = C...

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410 Homework STAT #3 (due Friday, February 8, by 3:00 p.m.) Spring 2008 1. 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, a) Sketch the support of ( X , Y ). 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 f ( x, y ) dx dy = 1. 1 2- x 0 C x 2 y dy dx 1 0 x = C 2 x2 y2 y = 2-x dx y=x = 1 0 C 2 x 2 ( 2 - x )2 - x 2 [ ] dx = 1 0 ( 2 C x 2 - 2 C x 3 )dx 2C 3 C 4 x - x 3 2 1 = 0 = C 6 = 1. C = 6. c) Find P ( X + Y < 1 ). 0.5 1- x 0 6 x 2 y dy dx 0.5 0 x = (3 x 2 y 2 ) y y= =1 -x x dx = 0.5 0 (3 x 2 [ ( 1 - x ) 2 - x 2 ] )dx ( 3 x 2 - 6 x 3 )dx 3 4 x 2 - 3 1 2 2 0.5 0 4 = 0.5 0 = x3 - 1 2 3 = = 1 3 1 - = = 0.03125. 8 32 32 2. Suppose that the random variables X and Y have joint p.d.f. f ( x, y ) given by f ( x, y ) = 6 x 2 y, a) 0 < x < y, x + y < 2. Find the marginal probability density function for X. First, X can only take values in ( 0 , 1 ). fX( x) = f ( x, y ) dy = - 2- x 6 x 2 y dy = x (3 x 2 y 2 ) y y= =2 -x x 0 < x < 1. = 3x2 {( 2 - x ) 2 - x 2 } = 12 x 2 12 x 3 = 12 x 2 ( 1 x ), b) Find the marginal probability density function for Y. First, Y can only take values in ( 0 , 2 ). y 6 x 2 y dx 0 < y <1 fY( y) = f (x, y ) dx = - 0 2- y 0 6 x 2 y dx 1< y < 2 ( 2 x 3 y ) xx == 0y = 2 ( 2 x 3 y ) x = = -0 y x 0 < y <1 1< y < 2 2y4 = 0 < y <1 1< y < 2 2 y ( 2 - y )3 From the textbook: 2.1.6 FZ( z ) = P( Z z ) = P( Y z X ) = z 0 z-x 0 e- x - y d y d x z-x 0 = z 0 e- x e- y d y d x = z 0 e- x 1 - e- z + x d x e -x ( ) = z 0 dx- z 0 e- z d x = 1 - e-z - z e-z , z > 0. In particular, P ( Z 0 ) = 0, P ( Z 6 ) = 1 7 e 6 0.98265. f Z ( z ) = F Z( z ) = e z e z + z e z = z e z, ' 2.1.7 FZ( z ) = P( Z z ) = P( Y z X z > 0. ) = z 1 0 0 1d y d x + 1 z x 1d y d x z 1 0 = z 0 1d x + z z dx x 0 < z < 1. 0 < z < 1. = z z ln z, f Z ( z ) = F Z( z ) = ln z, ' 0 FZ( z ) = z - z ln z 1 z0 0 < z <1 z 1 fZ( z ) = - ln z 0 0 < z <1 otherwise 2.1.9 a) x2 x1 0 1 0 1 2 p1( x1 ) 7 12 5 12 2 12 2 12 4 12 3 12 2 12 5 12 2 12 1 12 3 12 p2( x2 ) b) P ( X 1 + X 2 = 1 ) = p ( 0, 1 ) + p ( 1, 0 ) = 3 2 5 + = . 12 12 12 2.1.10 f X ( x1 ) = 1 1 x1 x2 0 2 15 x1 x2 dx2 = 15 2 x1 2 x2 dx2 2 x1 x2 0 2 3 x1 dx1 1 = 15 2 2 x1 1 - x1 , 2 ( ) 0 < x1 < 1. f X ( x2 ) = 2 2 15 x1 x2 dx1 = 5 x2 1 2 1- 0 4 = 5 x2 , 0 < x2 < 1. P( X 1 + X 2 1 ) = 15 2 = x1 2 0 12 x1 x1 2 15 x1 x2 dx2 dx1 12 1- x1 x1 2 x2 dx2 dx1 = 15 2 x1 2 0 ( (1 - x1 ) 2 - x12 )dx1 12 = 15 2 x1 ( 1 - 2 x1 ) dx1 2 0 15 2 3 x1 - 15 x1 dy = 2 5 3 15 4 x - x1 2 1 4 12 0 = 12 = 0 5 15 5 - = . 16 64 64 2.1.12 x2 x1 1 2 1 2 p1( x1 ) 5 12 7 12 2 12 3 12 5 12 3 12 4 12 7 12 p2( x2 ) E( X1 ) = 1 E( X2 ) = 1 5 7 19 +2 = . 12 12 12 5 7 19 +2 = . 12 12 12 2 E( X1 ) = 1 2 E( X2 ) = 1 5 7 33 +4 = . 12 12 12 5 7 33 +4 = . 12 12 12 E( X1 X2 ) = 1 No, 2 6 4 30 +2 +4 = . 12 12 12 12 30 19 19 = E( X1 X2 ) E( X1 ) E( X2 ) = . 12 12 12 2 2 E( X1 2 6 X2 + 7 X1 X2 ) = 2 E( X1 ) 6 E( X2 ) + 7 E( X1 X2 ) = 2 19 33 30 50 6 +7 = . 12 12 12 12 2.1.14 M ( t 1, t 2 ) = 1 e t1 x1 + t 2 x 2 2 x1 =1 x 2 =1 x1 + x 2 = 1 e t1 x1 2 x1 =1 x1 =1 x1 1 e t 2 x2 2 x 2 =1 x2 = e t1 2 x1 e t2 2 x2 x 2 =1 Must have e t1 2 < 1, e t2 2 < 1 for the series to converge t1 < ln 2, t2 < ln 2. e t1 M ( t 1, t 2 ) = e t2 1- 2 2 1- e t1 2 e t2 2 = ( e t1 + t 2 , 2 - e t1 2 - e t 2 )( ) t1 < ln 2, t2 < ln 2. = M ( t 1 , t 2 ). M ( t 1, 0 ) M ( 0, t 2 ) = e t1 e t2 2 - e t1 2 - e t2 2.1.15 M ( t 1, t 2 ) = 0 0 x1 e t1 x1 + t 2 x 2 x1 e - x2 d x2 d x1 = x1 e t1 x1 x1 e t1 x1 x1 e t 2 x 2 e - x 2 d x2 d x1 e x2 ( t 2 -1 ) d x2 d x1 = 0 x1 Must have t2 < 1 for the integral to converge. M ( t 1, t 2 ) = 0 0 e x1 e t1 x1 x 2 ( t 2 -1 ) x1 t2 - 1 d x1 = x1 e t1 x1 e x1 ( t 2 -1 ) d x1 1 - t2 = 1 x1 e x1 ( t1 + t 2 -1 ) d x1 1 - t2 0 Must have t1 + t2 < 1 for the integral to converge. x1 e x1 ( t1 + t 2 -1 ) e x1 ( t1 + t 2 -1 ) 1 M ( t 1, t 2 ) = - 1 - t2 t1 + t 2 - 1 ( t1 + t 2 - 1 ) 2 = 0 1 ( 1 - t 2 ) ( 1 - t1 - t 2 ) 2 , t2 < 1, t1 + t2 < 1. 1 1 No, M ( t 1 , t 2 ) M ( t 1 , 0 ) M ( 0 , t 2 ) = ( 1 - t1 ) 2 (1 - t2 ) 3 . 2.3.2 2 f 1 | 2 ( x 1 | x 2 ) = c1 x1 x 2 , 0 < x 1 < x 2 , 0 < x 2 < 1. 4 f2( x2) = c2 x2 , 0 < x 2 < 1. 1 0 4 c 2 x 2 dx 2 = (a) 1 = - f 2 ( x 2 ) dx 2 = c2 5 . c 2 = 5. 1 = - f 1| 2 ( x1 | x 2 ) dx1 = x2 0 2 c1 x1 x 2 dx1 = c1 2 . c 1 = 2. (b) 2 f ( x 1 , x 2 ) = f 1 | 2 ( x 1 | x 2 ) f 2 ( x 2 ) = 10 x1 x 2 , 0 < x 1 < x 2 < 1. (c) 1 1 5 P < X1 < X2 = 4 2 8 | 12 = 14 2 x1 (5 8) 2 64 dx1 = 25 1 2 2 1 - 4 2 = 64 3 12 = . 25 16 25 12 12 (d) 1 1 P < X1 < 4 2 = 14 1 x1 2 10 x1 x 2 5 dx 2 dx1 = 10 3 x1 1 - x1 dx1 3 14 ( ) = 2 10 x1 x1 - 3 2 5 12 1 2 5 12 = 5 x1 - 2 x1 14 14 3 = ( ) = 1 5 1 5 1 - - + 3 4 16 16 512 1 449 449 = . 3 512 1536 ________________________________________________________________________ ________________________________________________________________________ If you are registered for 4 credit hours: 2.1.16 P( 2 X + 3 Y < 1 ) 12 ( to be handed in separately ) = 0 12 (1 - 2 x ) 0 6 ( 1 - x - y ) dy dx 2 3 = 0 (1 - 2 x ) 2 ( 1 - x )( 1 - 2 x ) - 3 5 14 8 - x + x 2 dx 3 3 3 dx 12 = = 0 = 5 7 1 13 - + = . 6 12 9 36 OR 12 0 P( 2 X + 3 Y < 1 ) = (1 - 2 x ) 0 6 ( 1 - x - y ) dy dx 3 let u = 1 x y, then d u = d y, 0 1 x, ( 1 2 x ) / 3 ( 2 x ) / 3 12 = 0 12 - (2 - x) 6 u du dx = 1- x 0 3 12 3 (1 - x ) 2 - ( 2 - x ) 2 dx 3 = 0 5 14 8 5 7 1 13 - x + x 2 dx = - + = . 3 3 3 6 12 9 36 E( X 2 ) = 1 0 1- x 0 x 2 6 ( 1 - x - y ) dy dx = 1 0 1 0 ( 6 x 2 (1 - x ) 2 - 3 x 2 (1 - x ) 2 )dx = 1- = 1 0 3 x ( 1 - x ) 2 dx 2 = ( 3 x 2 - 6 x 3 + 3 x 4 )dx = 1 0 6 3 + = 0.10. 4 5 E( X Y ) = 1 0 1- x 0 x y 6 ( 1 - x - y ) dy dx 1 0 ( 3 x (1 - x ) 3 - 2 x (1 - x ) 3 )dx = = 1 0 x ( 1 - x ) 3 dx = ( x - 3 x 2 + 3 x 3 - x 4 )dx 1 3 1 - 1 + - = 0.05. 2 4 5 E ( X Y + 2 X 2 ) = E ( X Y ) + 2 E ( X 2 ) = 0.25.
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