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Unformatted text preview: PSTAT 120A Spring 2010 HW 9 Solutions 1. (Ross 6.40) (a) p ( x,y ) is the joint PMF p ( x,y ) = p X,Y ( x,y ) = P ( X = x,Y = y ) of discrete random variables X and Y . The conditional PMF of X given Y can be obtained via p X  Y ( x  y ) = p X,Y ( x,y ) p Y ( y ) . So we need the marginal of Y , p Y ( y ) = X x p X,Y ( x,y ) , p Y (1) = p X,Y (1 , 1) + p X,Y (2 , 1) , = 1 4 . p Y (2) = p X,Y (1 , 2) + p X,Y (2 , 2) , = 3 4 . Then the conditional PMF is p X  Y (1  1) = p X,Y (1 , 1) p Y (1) , = 1 / 8 1 / 4 = 1 2 , p X  Y (2  1) = p X,Y (2 , 1) p Y (1) , = 1 / 8 1 / 4 = 1 2 , p X  Y (1  2) = p X,Y (1 , 2) p Y (2) , = 1 / 4 3 / 4 = 1 3 , p X  Y (2  2) = p X,Y (2 , 2) p Y (2) , = 1 / 2 3 / 4 = 2 3 . To be complete, we should also say p X  Y ( x  y ) = 0 , for x / { 1 , 2 } ,y { 1 , 2 } . (b) For X and Y to be independent, it is necessary and sufficient that p X,Y ( x,y ) = p X ( x ) p Y ( y ) , ( x,y ) R 2 . We find the marginal p X ( x ) to check this criterion (we have p Y ( y ) from part (a)). p X ( x ) = X y p X,Y ( x,y ) , p X (1) = p X,Y (1 , 1) + p X,Y (1 , 2) , = 3 8 . p X (2) = p X,Y (2 , 1) + p X,Y (2 , 2) , = 5 8 . 1 Then we note that p X,Y (1 , 1) = 1 8 6 = ( 3 8 )( 1 4 ) = p X (1) p Y (1) , so factorization, and therefore independence, does not hold. Another way to show this, using the condi tional PMF from part (a) is p X  Y (1  1) = 1 2 6 = 1 3 = p X  Y (1  2) whereas if X and Y were independent we would have p X  Y ( x  y 1 ) = p X  Y ( x  y 2 ) , x,y 1 , y 2 , since p X  Y ( x  y ) p X ( x ) = p X,Y ( x,y ) = p X ( x ) p Y ( y ) , ( x,y ) R 2 . which implies that p X  Y ( x  y ) = p X ( x ) , ( x,y ) R 2 . (c) P ( XY 3) = P ( { X = 1 ,Y = 1 } { X = 1 ,Y = 2 } { X = 2 ,Y = 1 } ) = P ( X = 1 ,Y = 1) + P ( X = 1 ,Y = 2) + P ( X = 2 ,Y = 1) = p X,Y (1 , 1) + p X,Y (1 , 2) + p X,Y (2 , 1) = 1 8 + 1 4 + 1 8 = 1 2 . P ( X + Y > 2) = P ( { X = 2 ,Y = 1 } { X = 1 ,Y = 2 } { X = 2 ,Y = 2 } ) = 1 P ( X = 1 ,Y = 1) = 1 p X,Y (1 , 1) = 1 1 8 = 7 8 ....
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 Spring '10
 RohiniKumar

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