section25example

section25example - ≤ 1 . 2 8. Find Cov( X, Y ). Cov( X, Y...

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Example : Suppose the joint pdf f ( x, y ) = ( c ( x 2 + xy ) , 0 x 1 , 0 y 1 0 , otherwise 1. Determine c 1 = Z y = -∞ Z x = -∞ f ( x, y ) dxdy 1 = Z 1 y =0 ±Z 1 x =0 c ( x 2 + xy ) dx ² dy = Z 1 y =0 " c x 3 3 + x 2 y 2 !#³ ³ ³ ³ ³ 1 x =0 dy = Z 1 y =0 c ´ 1 3 + y 2 µ dy = c y 3 + y 2 4 ³ ³ ³ ³ 1 y =0 = 7 12 c That is c = 12 7 The joint pdf f ( x, y ) = ( 12 7 ( x 2 + xy ) , 0 x 1 , 0 y 1 0 , otherwise 2. Find P (0 X 1 2 , 0 Y 1) P (0 X 1 2 , 0 Y 1) = Z 1 y =0 Z 1 / 2 x =0 12 7 ( x 2 + xy ) dx ! dy = Z 1 y =0 12 7 x 3 3 + x 2 y 2 ³ ³ ³ ³ 1 / 2 x =0 dy = Z 1 y =0 12 7 ´ 1 24 + y 8 µ dy = ´ 1 14 y + 3 28 y 2 µ³ ³ ³ ³ 1 y =0 = 5 28 3. Find P ( Y X ) P ( Y X ) 1
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= Z 1 x =0 ±Z 1 y = x 12 7 ( x 2 + xy ) dy ² dx = Z 1 x =0 " 12 7 ³ x 2 y + 1 2 xy 2 ´µ µ µ µ 1 y = x # dx = Z 1 x =0 12 7 ±³ x 2 + 1 2 x ´ - ³ x 3 + 1 2 x 3 ´² dx = 12 7 x 3 3 + x 2 4 - 3 x 4 8 µ µ µ µ 1 x =0 = 5 14 4. Marginal pdf of X , f X ( x ). f X ( x ) = Z y = -∞ f ( x, y ) dy = Z 1 y =0 12 7 ( x 2 + xy ) dy = 12 7 ( x 2 + 1 2 x ) , 0 x 1 . And we can Fne E ( X ) by E ( X ) = Z x = -∞ xf X ( x ) dx = Z 1 x =0 x 12 7 ( x 2 + 1 2 x ) dx = ··· = 5 7 . 5. Marginal pdf of Y , f Y ( y ). f Y ( y ) = Z x = -∞ f ( x, y ) dx = Z 1 x =0 12 7 ( x 2 + xy ) dx = 12 7 ( 1 3 + 1 2 y ) , 0 y 1 . and E ( Y ) = Z y = -∞ yf Y ( y ) dy = Z 1 y =0 y 12 7 ( 1 3 + 1 2 y ) dy = ··· = 4 7 . 6. Conditional pdf of Y given X = x f Y | X = x ( y ) = f ( x, y ) f X ( x ) = 12 7 ( x 2 + xy ) 12 7 ( x 2 + 1 2 x ) = x + y x + 1 2 , 0 y 1 . And conditional pdf of Y given X = 1 2 is f Y | X = 1 2 ( y ) = 1 2 + y 1 2 + 1 2 = y + 1 2 , 0 y 1 . 7. Conditional pdf of X given Y = y f X | y = y ( x ) = f ( x, y ) f Y ( y ) = 12 7 ( x 2 + xy ) 12 7 ( 1 3 + 1 2 y ) = x 2 + xy 1 3 + 1 2 y , 0 x
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Unformatted text preview: ≤ 1 . 2 8. Find Cov( X, Y ). Cov( X, Y ) = E ( XY )-E ( X ) E ( Y ) . and E ( XY ) = Z ∞ y =-∞ Z ∞ x =-∞ xyf ( x, y ) dxdy = Z 1 y =0 ±Z 1 x =0 xy 12 7 ( x 2 + xy ) dx ² dy = Z 1 y =0 ± 12 7 Z 1 x =0 ( x 3 y + x 2 y 2 ) dx ² dy = Z 1 y =0 12 7 x 4 y 4 + x 3 y 2 3 !³ ³ ³ ³ ³ 1 x =0 dy = Z 1 y =0 12 7 y 4 + y 2 3 ! dy = 12 7 y 2 8 + y 3 9 !³ ³ ³ ³ ³ 1 y =0 = 17 42 That is, Cov( X, Y ) = E ( XY )-E ( X ) E ( Y ) = 17 42-5 7 × 4 7 =-. 0034 3...
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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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section25example - ≤ 1 . 2 8. Find Cov( X, Y ). Cov( X, Y...

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