STAT 400 Lecture AL1 Answers for 4.2 Spring 2015 Dalpiaz 4.2 Covariance and Correlation Coefficient Covariance of X and Y XY = Cov ( X , Y ) = E [ ( X – X ) ( Y – Y ) ] = E ( X Y ) – X 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 ). 0. Find in terms of X2, Y2, and 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 ).Correlation coefficient of X and Y XY = Y XXY σσσ = , YVarXVarYXCov = YY XX σμσμ YXE (a) – 1 XY 1; (b) XY is either + 1 or – 1 if and only if X and Y are linear functions of one another. If random variables X and Y are independent, then E ( g ( X ) h ( Y ) ) = E ( g ( X ) ) E ( h ( Y ) ). Cov ( X, Y ) = XY = 0, Corr ( X , Y ) = XY = 0. 1. Consider the following joint probability distribution p ( x, y ) of two random variables X and Y: y Recall: E ( X ) = 1.75, E ( Y ) = 0.8, E ( X Y ) = 1.5. x 0 1 2 p X ( x ) 1 0.15 0.10 0 0.25 2 0.25 0.30 0.20 0.75 p Y ( y ) 0.40 0.40 0.20 1.00 Find Cov ( X , Y ) = XY and Corr ( X , Y ) = XY . Cov ( X , Y ) = XY = 1.5 – 1.75 0.8 = 0.10. E ( X 2 ) = 1 0.25 + 4 0.75 = 3.25. Var ( X ) = E ( X 2 ) – [ E ( X ) ] 2 = 3.25 – 1.75 2 = 0.1875. E ( Y 2 ) = 0 0.40 + 1 0.40 + 4 0.20 = 1.2. Var ( Y ) = E ( Y 2 ) – [ E ( Y ) ] 2 = 1.2 – 0.8 2 = 0.56.Corr ( X , Y ) = XY = 56.01875.010.0 0.3086.2. Let the joint probability density function for ( X , Y ) be otherwise01 ,10 ,10 60 ,2 yxyxyxyxf Find Cov ( X , Y ) = XY and Corr ( X , Y ) = XY . Recall: f X ( x ) = 2 2 1 30 xx , 0 < x < 1, E ( X ) = 21, f Y ( y ) = 3 1 20 yy , 0 < y < 1, E ( Y ) = 31, E ( X Y ) = 71. Cov ( X , Y ) = 312171 = 421. Var ( X ) = 2529. Var ( Y ) = 2528. XY = 25282529421 = 21 = 22 – 0.7071. 3. Let the joint probability density function for ( X , Y ) be otherwise010 ,10 ,yxyxyxf Find Cov ( X , Y ) = XY and Corr ( X , Y ) = XY . Recall: f X ( x ) = 21x, 0 < x < 1. f Y ( y ) = 21y, 0 < y < 1.E ( X Y ) = 10 10 dydxyxyx = 10 10 2 2 dydxyxyx = 10 2 23 dyyy = 31. Cov ( X , Y ) = 12712731 = 1441. XY = 14411144111441 = 111. 4. Let the joint probability density function for ( X , Y ) be otherwise00 ,101 12 ,2 yxxxyxfye Find Cov ( X , Y ) = XY and Corr ( X , Y ) = XY . Recall: f X ( x ) = 1 6 xx , 0 < x < 1. f Y ( y ) = ye2 2, y > 0. Since X and Y are independent, Cov ( X, Y ) = XY = 0, Corr ( X , Y ) = XY =
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