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# 02_08ans - STAT 410 Examples for Spring 2008 Covariance of...

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STAT 410 Examples for 02/08/2008 Spring 2008 Covariance of X and Y σ XY = Cov ( X , Y ) = E [ ( X – μ X ) ( Y – μ Y ) ] = E ( X Y ) μ X μ Y (a) Cov ( X , X ) = Var ( X ) ; (b) Cov ( X , Y ) = Cov ( Y , X ) ; (c) Cov ( a X + b , Y ) = a Cov ( X , Y ) ; (d) Cov ( X + Y , W ) = Cov ( X , W ) + Cov ( Y , W ) . Cov ( a X + b Y , c X + d Y ) = a c Var ( X ) + ( a d + b c ) Cov ( X , Y ) + b d Var ( Y ) . Var ( a X + b Y ) = Cov ( a X + b Y , a X + b Y ) = a 2 Var ( X ) + 2 a b Cov ( X , Y ) + b 2 Var ( Y ) . 1. Find in terms of σ X 2 , σ Y 2 , and σ XY : a) Cov ( 2 X + 3 Y , X – 2 Y ) , Cov ( 2 X + 3 Y , X – 2 Y ) = 2 Var ( X ) – Cov ( X , Y ) – 6 Var ( Y ) . b) Var ( 2 X + 3 Y ) , Var ( 2 X + 3 Y ) = Cov ( 2 X + 3 Y , 2 X + 3 Y ) = 4 Var ( X ) + 12 Cov ( X , Y ) + 9 Var ( Y ) . c) Var ( X – 2 Y ) . Var ( X – 2 Y ) = Cov ( X – 2 Y , X – 2 Y ) = Var ( X ) – 4 Cov ( X , Y ) + 4 Var ( Y ) .

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2. Consider the following joint probability distribution p ( x , y ) of two random variables X and Y: y x 0 1 2 p X ( x ) Recall: 1 0.15 0.15 0 0.30 E ( X ) = 1.7 2 0.15 0.35 0.20 0.70 E ( Y ) = 0.9 p Y ( y ) 0.30 0.50 0.20 E ( X Y ) = 1.65 Find Cov ( X , Y ) = σ XY . Cov ( X , Y ) = 1.65 – 1.7 0.9 = 0.12 . 3. Let the joint probability density function for ( X , Y ) be ( ) ° ± ² + = otherwise 0 1 , 1 0 , 1 0 24 , y x y x y x y x f Recall: E ( X ) = 0.40, E ( Y ) = 0.40. Find Cov ( X , Y ) = σ XY . E ( X Y ) = ³ ³ ´ ´ µ · · ¸ ¹ - 1 0 1 0 24 dx dy y x y x x = ( ) ³ - 1 0 3 2 1 8 dx x x = ³ ´ µ · ¸ ¹ - + - 1 0 5 4 3 2 8 24 24 8 dx x x x x = 6 8 5 24 6 3 8 - + - =
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