zz-26 - Utah State University ECE 6010 Stochastic Processes...

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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....
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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.

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zz-26 - Utah State University ECE 6010 Stochastic Processes...

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