hw5sol

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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 π e- ( y- x ) 2 2 d x = 1 2 integraldisplay 1- 1 1 √ 2 π e- ( x- y ) 2 2 d x = 1 2 parenleftbiggintegraldisplay ∞- 1 1 √ 2 π e- ( x- y ) 2 2 d x − integraldisplay ∞ 1 1 √ 2 π 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 π e- ( y- x ) 2 2 f Y ( y ) d x = 1 2 f Y ( y ) parenleftbigg − integraldisplay- 1 1 √ 2 π e- ( x- y ) 2 2 d x + integraldisplay 1 1 √ 2 π 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, ˆ X = Cov( X,Y ) σ 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 , σ 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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 14

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online