This preview shows pages 1–3. 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: Utah State University ECE 6010 Stochastic Processes Homework # 4 Solutions 1. Suppose X N ( , ). (a) Show that E [ X ] = and cov( X , X ) = . Using characteristic functions: X ( u ) = exp[ i u T  1 2 u T u ] Taking the gradient with respect to u we have X ( u ) u = exp[ i u T  1 2 T ]( i  u ) Now evaluating at u = and dividing by i we obtain E [ X ] = . Taking the gradient again with respect to u we have exp[ i u T  1 2 T ]( ) + exp[ i u T  1 2 T ]( i  u )( i  u ) T and evaluating at u = we have  Dividing by i 2 we obtain E [ X 2 ] = + T , from which it follows that cov( X , X ) = . (b) Show that A X + b N ( A + b , A A T ). We note that X ( u ) = E [ e i u T X ] = exp[ i u T  1 2 T ]. Then Y ( u ) = E [ e i u T Y ] = E [ e i u T ( A X + b ) ] = e i u T b E [ e i u T A X ] But we can compute the expected value using what we know from X ( u ): E [ e i u T A X ] = X ( v )  v = A T u = exp[ i u T A  1 2 u T A A T u ] 1 So Y ( u ) = e i u T b exp[ i u T A  1 2 u T A A T u ] = exp[ i u ( A + b ) 1 2 u T ( A A T ) u ] . From the form of the characteristic function we identify that Y is Gaussian with E [ Y ] = A + b cov( Y , Y ) = A A T . (c) Suppose > 0 and write = CC T . Show that C 1 ( X ) N ( , I ). If Y = C 1 X C 1 we have, by applying the previous result E [ Y ] = C 1  C 1 = cov( Y , Y ) = C 1 C T = C 1 ( CC T ) C T = I....
View
Full
Document
This note was uploaded on 03/01/2012 for the course ECE 6010 taught by Professor Stites,m during the Spring '08 term at Utah State University.
 Spring '08
 Stites,M

Click to edit the document details