HW 8- solutions

HW 8- solutions - (c) To find the correlation, we evaluate...

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Unformatted text preview: (c) To find the correlation, we evaluate the product XY over all values of X and Y . Specifically, r X,Y = E [ XY ] = 4 summationdisplay x =1 x summationdisplay y =1 xyP X,Y ( x,y ) (8) = 1 4 + 2 8 + 3 12 + 4 16 + 4 8 + 6 12 + 8 16 + 9 12 + 12 16 + 16 16 (9) = 5 (10) (d) The covariance of X and Y is Cov[ X,Y ] = E [ XY ]- E [ X ] E [ Y ] = 5- (7 / 4)(10 / 4) = 10 / 16 (11) (e) The correlation coefficient is X,Y = Cov[ W,V ] radicalbig Var[ W ]Var[ V ] = 10 / 16 radicalbig (41 / 48)(5 / 4) . 605 (12) Problem 4.7.5 Solution For integers 0 x 5, the marginal PMF of X is P X ( x ) = summationdisplay y P X,Y ( x,y ) = x summationdisplay y =0 (1 / 21) = x + 1 21 (1) Similarly, for integers 0 y 5, the marginal PMF of Y is P Y ( y ) = summationdisplay x P X,Y ( x,y ) = 5 summationdisplay x = y (1 / 21) = 6- y 21 (2) The complete expressions for the marginal PMFs are P X ( x ) = braceleftbigg ( x + 1) / 21 x = 0 ,..., 5 otherwise (3) P Y ( y ) = braceleftbigg (6- y ) / 21 y = 0 ,..., 5 otherwise (4) The expected values are E [ X ] = 5 summationdisplay x =0 x x + 1 21 = 70 21 = 10 3 E [ Y ] = 5 summationdisplay y =0 y 6- y 21 = 35 21 = 5 3 (5) To find the covariance, we first find the correlation E [ XY ] = 5 summationdisplay x =0 x summationdisplay y =0 xy 21 = 1 21 5 summationdisplay x =1 x x summationdisplay y =1 y = 1 42 5 summationdisplay x =1 x 2 ( x + 1) = 280 42 = 20 3 (6) The covariance of X and Y is Cov[ X,Y ] = E [ XY ]- E [ X ] E [ Y ] = 20 3- 50 9 = 10 9 (7) 152 Problem 4.7.10 Solution The joint PDF of X and Y and the region of nonzero probability are Y X 1 1-1 f X,Y ( x,y ) = braceleftbigg 5 x 2 / 2- 1 x 1 , y x 2 otherwise (1) (a) The first moment of X is E [ X ] = integraldisplay 1- 1 integraldisplay x 2 x 5 x 2 2 dydx = integraldisplay 1- 1 5 x 5 2 dx = 5 x 6 12 vextendsingle vextendsingle vextendsingle vextendsingle 1- 1 = 0 (2) Since E [ X ] = 0, the variance of X and the second moment are both Var[ X ] = E bracketleftbig X 2 bracketrightbig = integraldisplay 1- 1 integraldisplay x 2 x 2 5 x 2 2 dydx = 5 x 7 14 vextendsingle vextendsingle vextendsingle vextendsingle 1- 1 = 10 14 (3) (b) The first and second moments of Y are E [ Y ] = integraldisplay 1- 1 integraldisplay x 2 y 5 x 2 2 dydx = 5 14 (4) E bracketleftbig Y 2 bracketrightbig = integraldisplay 1- 1 integraldisplay x 2 y 2 5 x 2 2 dydx = 5 26 (5) Therefore, Var[ Y ] = 5 / 26- (5 / 14) 2 = . 0576. (c) Since E [ X ] = 0, Cov[ X,Y ] = E [ XY ]- E [ X ] E [ Y ] = E [ XY ]. Thus, Cov[ X,Y ] = E [ XY ] = integraldisplay 1 1 integraldisplay x 2 xy 5 x 2 2 dydx = integraldisplay 1- 1 5 x 7 4 dx = 0 (6) (d) The expected value of the sum X + Y is E [ X + Y ] = E [ X ] + E [ Y ] = 5 14 (7) (e) By Theorem 4.15, the variance of X + Y is Var[ X + Y ] = Var[ X ] + Var[ Y ] + 2Cov[ X,Y ] = 5 / 7 + 0 . 0576 = 0 . 7719 (8) Problem 4.7.11 Solution Random variables X and Y have joint PDF Y X 1 1 f X,Y ( x,y ) = braceleftbigg 2 0 y x 1 0 otherwise (1)...
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HW 8- solutions - (c) To find the correlation, we evaluate...

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