hw5sol - UCSD ECE153 Handout #25 Prof. Young-Han Kim...

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Unformatted text preview: UCSD ECE153 Handout #25 Prof. Young-Han Kim Thursday, May 5, 2011 Solutions to Homework Set #5 (Prepared by TA Lele Wang) 1. Neural net. Let Y = X + Z , where the signal X U[ 1 , 1] and noise Z N (0 , 1) are independent. (a) Find the function g ( y ) that minimizes MSE = E bracketleftbig (sgn( X ) g ( Y )) 2 bracketrightbig , where sgn( x ) = braceleftBigg 1 x +1 x > . (b) Plot g ( y ) vs. y . Solution: The minimum MSE is achieved when g ( Y ) = E (sgn( X ) | Y ). We have g ( y ) = E (sgn( X ) | Y = y ) = integraldisplay - sgn( x ) f X | Y ( x | y )d x. To find the conditional pdf of X given Y , we use f X | Y ( x | y ) = f Y | X ( y | x ) f X ( x ) f Y ( y ) , where f X ( x ) = braceleftBigg 1 2 1 X 1 otherwise . Since X and Z are independent, f Y | X ( y | x ) = f Z ( y x ) Y |{ X = x } N ( x, 1) . To find f Y ( y ) we integrate f Y | X ( y | x ) f X ( x ) over x : f Y ( y ) = integraldisplay - f Y | X ( y | x ) f X ( x )d x = integraldisplay 1- 1 1 2 2 e- ( y- x ) 2 2 d x = 1 2 integraldisplay 1- 1 1 2 e- ( x- y ) 2 2 d x = 1 2 parenleftbiggintegraldisplay - 1 1 2 e- ( x- y ) 2 2 d x integraldisplay 1 1 2 e- ( x- y ) 2 2 d x parenrightbigg = 1 2 ( Q ( y 1) Q ( y + 1)) . Combining the above results, we get 1 g ( y ) = integraldisplay - sgn( x ) f X | Y ( x | y )d x = integraldisplay 1- 1 sgn( x ) 1 2 2 e- ( y- x ) 2 2 f Y ( y ) d x = 1 2 f Y ( y ) parenleftbigg integraldisplay- 1 1 2 e- ( x- y ) 2 2 d x + integraldisplay 1 1 2 e- ( x- y ) 2 2 d x parenrightbigg = Q ( y + 1) 2 Q ( y ) + Q ( y 1) Q ( y 1) Q ( y + 1) . The plot is shown below. Note the sigmoidal shape corresponding to the common neural network activation function.-8-6-4-2 2 4 6 8-1.5-1-0.5 0.5 1 1.5 y g(y) 2. Additive shot noise channel. Consider an additive noise channel Y = X + Z , where the signal X N (0 , 1), and the noise Z |{ X = x } N (0 ,x 2 ), i.e., the noise power of increases linearly with the signal squared. (a) Find E ( Z 2 ). (b) Find the best linear MSE estimate of X given Y . Solution: (a) Since Z |{ X = x } N (0 ,x 2 ), E ( Z | X ) = 0 Var( Z | X = x ) = E ( Z 2 | X = x ) E ( Z | X = x ) 2 = x 2 . Therefore, E ( Z 2 ) = E ( E ( Z 2 | X )) = E ( X 2 ) = 1 . 2 (b) From the the best linear estimate formula, X = Cov( X,Y ) 2 Y ( Y E ( Y )) + E ( X ) . Here we have E ( X ) = 0 , E ( Y ) = E ( X + Z ) = E ( X ) + E ( Z ) = E ( X ) + E ( E ( Z | X )) = 0 + E (0) = 0 , E ( XZ ) = E ( E ( XZ | X )) = E ( X E ( Z | X )) = E ( X 0) = E (0) = 0 , Y 2 = E ( Y 2 ) ( E ( Y )) 2 = E (( X + Z ) 2 ) 2 = E ( X 2 ) + E ( Z 2 ) + 2 E ( XZ ) = 1 + 1 + 2 0 = 2 , Cov( X,Y ) = E (( X E ( X ))( Y E ( Y ))) = E ( XY ) = E ( X ( X + Z )) = E ( X 2 ) + E ( XZ ) = 1 + 0 = 1 ....
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hw5sol - UCSD ECE153 Handout #25 Prof. Young-Han Kim...

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