hw12soln

# hw12soln - ECE 310 Spring 2005 HW 12 solutions Problem 13.3...

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ECE 310, Spring 2005, HW 12 solutions Problem 13.3 a) First, notice that P ( Y = k | X = k ) = e - ( x +1) ( x +1) k k ! . By Total Probability Theorem, P ( Y = k ) = P ( X = 0) P ( Y = k | X = 0) + P ( X = 1) P ( Y = k | X = 1) = 1 2 e - 1 1 k k ! + 1 2 e - 2 2 k k ! = e - 2 2( k !) ( e + 2 k ) b) Using the result of part a), we obtain P ( X = 0 | Y = k ) = P ( Y = k | X = 0) P ( X = 0) P ( Y = k ) = e e + 2 k and P ( X = 1 | Y = k ) = 1 - P ( X = 0 | Y = k ) = 2 k e + 2 k . Therefore, the MAP rule is as follows ˆ X ( Y ) = ( 0 , if Y 1; 1 , if Y 2 . Problem 15.1 a) f X ( x ) = Z -∞ f X,Y ( x, y ) dy = e - x U ( x ) So, f Y | X ( y | x ) = f X,Y ( x, y ) f X ( x ) = e - y U ( y ) Then, E ( Y | X ) = Z 0 ye - y dy = 1 b) f X ( x ) = Z -∞ f X,Y ( x, y ) dy = Z 0 ( x + y ) 2 e - ( x + y ) U ( x ) dy Let z = x + y and dz = dy . Then, using this substitution and then integration by parts: f X ( x ) = U ( x ) Z x z 2 e - z dz = U ( x )[ - z 2 e - z | x + Z x 1 2 e - z dz ] = e - x 2 (1+ x ) U ( x ) So, 1

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f Y | X ( y | x ) = f X,Y ( x, y ) f X ( x ) = ( x + y ) (1 + x ) e - y U ( y ) Then, E ( Y | X ) = Z 0 y ( x + y ) (1 + x ) e - y dy = 1 (1 + x ) Z 0 ( xy + y 2 ) e - y dy = ( x + 2) (1 + x ) Problem 15.4 a) Using the subconditioning property: E ( V 1 V 2 ) = E ( E ( V 1 V 2 | V 1 )) = E ( V 1 E ( V 2 | V 1 )) = E ( V 1 · V 3 1 ) = E ( V 4 1 ) = Z 1 - 1 v 4 1 1 2 dv 1 = 1 5 b) No, since E ( V 2 | V 1 ) depends on V 1 .
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