2 we have lim t t e y t x e zx 2 e y t z x 2 e y t z

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2. We have 0 lim t t 0 E [ Y t X ] - E [ ZX ] 2 = | E [( Y t - Z ) X ] | 2 E [( Y t - Z ) 2 ] E [ X 2 ] = 0 . 2
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ECE 6010: Lecture 7 – Analytical Properties of Random Processes 3 Properties of the derivative Suppose { X t , t T } is mean-square differentiable with mean-square derivative { X 0 t , t T } . Suppose that { X t } is second order. Then: 1. { X 0 t } is also second order. (This follows from the first fact above.) 2. ∂t R X ( t, s ) exists, and equals R X 0 ,X ( t, s ) . Proof R X 0 X ( t, s ) = E [ X 0 t X s ] = E [lim q t X q - X t q - t X s ] = lim q t E [ X q - X t q - t X s ] (from fact 2 above) = lim q t E [ X q X s ] - E [ X t X s ] q - t = lim q t R X ( q, s ) - R X ( t, s ) q - t = ∂t R X ( t, s ) . 2 3. ∂s ∂t R X ( t, s ) exists and is equal to R X 0 ( t, s ) for all t, s T . Proof R X 0 ( t, s ) = E [ X 0 t X 0 s ] = E [ X 0 t lim q s X q - X s q - s = lim q s E [ X 0 t X q ] - E [ X 0 t X s ] q - s (from fact 2 above) = lim q s R X 0 X ( t, s ) - R X 0 X ( t, s ) q - s = lim q s ∂t R X ( t, s ) - ∂t R X ( t, s ) q - s = ∂s ∂t R X ( t, s ) . 2 4. ∂t μ X ( t ) exists and is equal to μ X 0 ( t ) . 5. Suppose T = R and X t is also W.S.S. Then { X 0 t } is also W.S.S. Also, { X t } and { X 0 t } are jointly W.S.S. Also, (a) μ 0 x = 0 . (b) R X 0 X ( τ ) = d R X ( τ ) . (c) R X 0 ( τ ) = - d 2 2 R X ( τ ) . On the existence of the mean-square derivative It can be shown that a sufficient condition for the existence of X 0 t 0 is the existence of 2 ∂t∂s R X ( t, s ) at ( t, s ) = ( t 0 , t 0 ) . A necessary condition is the existence and equality of the mixed partials ∂t ∂s R X ( t, s ) ( t = s = t 0 ) = ∂s ∂t R X ( t, s ) ( t = s = t 0 ) .
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ECE 6010: Lecture 7 – Analytical Properties of Random Processes 4 If X t is W.S.S., then these two conditions are the same. So X 0 t 0 exists if and only if d 2 2 R X ( τ ) τ =0 < .
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  • Fall '08
  • Stites,M
  • Probability theory, Xt

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