HW9_Solutions

# HW9_Solutions - PSTAT 120A Spring 2010 HW 9 Solutions...

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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

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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 . P ( X Y > 1) = P ( X > Y ) = p X,Y (2 , 1) = 1 8 . 2. (Ross 6.41) The joint PDF is f ( x, y ) = xe - x ( y +1) , x > 0 , y > 0 . (a) The conditional densities are defined by f X | Y ( x | y ) = f ( x, y ) f Y ( y ) , 2
and similarly for f Y | X ( y | x ). So we need to compute the marginal of Y . f Y ( y ) = ˆ -∞ f X,Y ( x, y ) dx = ˆ 0 xe - x ( y +1) dx, for y > 0 = ˆ 0 u ( x ) dv ( x ) , for y > 0 where u ( x ) = x and dv ( x ) = e - x ( y +1) . Note that we may (and do) treat y as a constant here since we are integrating only with respect to x . Using integration by parts, we get for y > 0: f Y ( y ) = u ( x ) v ( x ) | x = x =0 - ˆ 0

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