This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 onetoone mapping and it keeps all the information Y has about X ....
View
Full
Document
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.
 Spring '09
 Eggers

Click to edit the document details