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Unformatted text preview: 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 ) = ( , 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 ∞ ye y dy = 1 b) f X ( x ) = Z ∞∞ f X,Y ( x,y ) dy = Z ∞ ( 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 ] =...
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This homework help was uploaded on 09/26/2007 for the course ECE 3100 taught by Professor Haas during the Spring '05 term at Cornell University (Engineering School).
 Spring '05
 HAAS
 Normal Distribution, Trigraph, Characteristic function, Poisson PMF

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