Midtermsol09 - EE 278 Handout#11 Statistical Signal Processing Monday Midterm Solution 1(a ≥ P A = P A | B P B P A | B c P B c ≥ min P A | B,P

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Unformatted text preview: EE 278 Handout #11 Statistical Signal Processing Monday, July 27, 2009 Midterm Solution 1. (a) ≥ . P ( A ) = P ( A | B ) P ( B )+ P ( A | B c ) P ( B c ) ≥ min { P ( A | B ) ,P ( A | B c ) } ( P ( B )+ P ( B c )) = min { P ( A | B ) ,P ( A | B c ) } . (b) ≤ . F XY ( x,y ) = P ( X ≤ x,Y ≤ y ) 1 ≤ P ( X ≤ x ) = F X ( x ) . The inequality (1) is due to the fact that probability of intersection of two sets is less than the probability of each of those sets. (c) =. First, according to the symmetry of the problem we have, E X 2 1 X 2 1 + X 2 2 = E X 2 2 X 2 1 + X 2 2 . Second, E X 2 1 X 2 1 + X 2 2 + E X 2 2 X 2 1 + X 2 2 = E X 2 1 + X 2 2 X 2 1 + X 2 2 = 1 . Therefore each of the terms should be equal to 1 2 . (d) =. We know that E ( V ar ( X | Y )) is the MSE of the best nonlinear or linear estima- tor of X from Y . Suppose that we call this estimator g ( Y ). Also, E ( V ar ( X | Y 3 )) is the error of the best nonlinear estimator of X from Y 3 and let’s call it h ( Y 3 ). Since h ( Y 3 ) is the best estimator of X given Y 3 we have, E ( X- h ( Y 3 )) 2 ≤ E ( X- g ( 3 √ Y 3 )) 2 = E ( X- g ( Y )) 2 . On the other hand, E ( X- g ( Y )) 2 ≤ E ( X- h ( Y 3 )) 2 . From these two inequalities we see that E ( X- g ( Y )) 2 = E ( X- h ( Y 3 )) 2 . Intu- itively speaking, Y 3 is a one-to-one mapping and it keeps all the information Y has about X ....
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This note was uploaded on 10/26/2011 for the course MATH 180C taught by Professor Eggers during the Spring '09 term at Aarhus Universitet.

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Midtermsol09 - EE 278 Handout#11 Statistical Signal Processing Monday Midterm Solution 1(a ≥ P A = P A | B P B P A | B c P B c ≥ min P A | B,P

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