exam2sol - x y =0 (2 x + 2 y-4 xy ) dydx = Z 1 x =0 (-2 x 3...

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Exam2 Solution 1. (a) E ( X ) = Z 4 0 x 1 8 xdx = 8 3 (b) E ( X 2 ) = Z 4 0 x 2 1 8 xdx = 8 V ar ( X ) = E ( X 2 ) - ( EX ) 2 = 8 - ± 8 3 ² 2 = 8 9 . The standard deviation is p V ar ( X ) = q 8 9 = 2 2 3 . (c) The cdf is F ( x ) = x 2 16 , 0 < x < 4. The upper quartile is to solve x for F ( x ) = 3 4 , which is 2 3. The lower quartile is then 2. 2. E (2 X - 3 Y + 4) = 2 E ( X ) - 3 E ( Y ) + 4 = 2 × 2 - 3 × ( - 4) + 4 = 20 V ar (2 X - 3 Y + 4) = 2 2 V ar ( X ) + ( - 3) 2 V ar ( Y ) = 4 × 4 + 9 × 3 = 43 3. Let ¯ X be the average of 9 bricks. The mean is E ( ¯ X ) = μ = 1 . 2 kg , and the standard deviation σ ¯ X = σ n = . 1 9 = 1 30 . 4. (a) P (0 < X < 1 , 0 < Y < 0 . 5) = Z 1 x =0 Z 0 . 5 y =0 (2 x + 2 y - 4 xy ) dydx = Z 1 0 (2 xy + y 2 - 2 xy 2 ) | 0 . 5 y =0 dx = Z 1 0 ( x 2 + 1 4 ) dx = x 2 4 + x 4 ³ ³ ³ ³ 1 x =0 = 0 . 5 (b) f X ( x ) = Z 1 y =0 (2 x + 2 y - 4 xy ) dy = 1 , 0 < x < 1 . (c) f Y ( y ) = 1 , 0 < y < 1 . As f ( x, y ) 6 = f X ( x ) f Y ( y ), X and Y are not independent. (d) f X | Y = y ( x ) = f ( x, y ) f Y ( y ) = 2 x + 2 y - 4 xy, 0 < x < 1 . (e) cov ( X, Y ) = E ( XY ) - EXEY . EX = Z 1 0 xf X ( x ) dx = Z 1 0 xdx = 1 2 , EY = 1 2 E ( XY ) = Z 1 0 Z 1 0 xy (2 x + 2 y - 4 xy ) dxdy 1
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= Z 1 x =0 ±Z 1 y =0 (2 x 2 y + 2 xy 2 - 4 x 2 y 2 ) dy ² dx = Z 1 x =0 2 x 2 y 2 2 + 2 x y 3 3 - 4 x 2 y 3 3 ³ ³ ³ ³ 1 y =0 dx = Z 1 x =0 ( x 2 + 2 x 3 - 4 x 2 3 ) dx = x 3 3 + 2 3 x 2 2 - 4 3 x 3 3 ³ ³ ³ ³ 1 x =0 = 2 9 . Therefore, cov ( X, Y ) = 2 9 - 1 2 1 2 = - 1 36 . (f) P ( X + Y 1) = Z 1 x =0 Z 1 -
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Unformatted text preview: x y =0 (2 x + 2 y-4 xy ) dydx = Z 1 x =0 (-2 x 3 + 3 x 2-2 x + 1) dx = 1 2 . 5. (a) Find the cdf of Y , F Y ( y ) ±rst F Y ( y ) = P ( Y ≤ y ) = P (2 X + 3 ≤ y ) = P ( X ≤ y-3 2 ) = ± y-3 2 ² 2 The pdf f Y ( y ) = d dy F Y ( y ) = 1 2 ( y-3) , 3 < y < 5 . And EY = E (2 X + 3) = Z 1 (2 x + 3)2 xdx = 13 3 . (b) F Y ( y ) = P ( Y ≤ y ) = P ( e X ≤ y ) = P ( X ≤ log y ) = (log y ) 2 and the pdf of Y is f Y ( y ) = 2 log y y , 1 < y < e. and EY = Ee X = Z 1 e x 2 xdx = 2 2...
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This note was uploaded on 04/16/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

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exam2sol - x y =0 (2 x + 2 y-4 xy ) dydx = Z 1 x =0 (-2 x 3...

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