06_22 - STAT 410 Examples for Summer 2009 2.4 Covariance...

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STAT 410 Examples for 06/22/2009 Summer 2009 2.4 Covariance and Correlation Coefficient 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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Correlation coefficient of X and Y ρ XY = Y X XY σ σ σ = ( 29 ( 29 ( 29 , Y Var X Var Y X Cov = - - Y Y , X X σ μ σ μ Y X Cov (a) 1 ρ XY 1; (b) ρ XY is either + 1 or – 1 if and only if X and Y are linear functions of one another. 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 ) 1 0.15 0.15 0 0.30 2 0.15 0.35 0.20 0.70 p Y ( y ) 0.30 0.50 0.20 1.00 Find Cov ( X , Y ) = σ XY and Corr ( X , Y ) = ρ XY . Recall:
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06_22 - STAT 410 Examples for Summer 2009 2.4 Covariance...

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