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Why E[x^2] = autocorrelation of x evaluated in 0 ?

  1. Nov 16, 2011 #1
    Hi,

    I was studying the derivation of the solution of Wiener filter from the http://en.wikipedia.org/wiki/Wiener_filter#Wiener_filter_problem_setup". There is a step I don't quite understand.
    First, we define the square error between the estimated signal [itex]\hat{s}(t)[/itex] and the original true signal [itex]s(t)[/itex]:

    [tex]e^2(t) = s^2(t) - 2s(t)\hat{s}(t) + s^2(t)[/tex]

    then the authors calculate the mean value of [itex]e^2[/itex], that is [itex]E[e^2][/itex].
    At this point I would note that:

    [tex]E[e^2] = E[s^2] - 2E[s\hat{s}] + E[s^2][/tex]

    Unfortunately we don't know the probability density function of the original signal [itex]s[/itex], so we cannot compute [itex]E[s^2][/itex]. However, the authors of the article seem to suggest that:

    [tex]E[s^2]=R_s(0)[/tex]

    where [itex]R_s(0)[/itex] is the autocorrelation function of s evaluated at 0 (though I might have misunderstood this).
    Could anyone elaborate this point?

    I thought that [itex]E[s^2]=\int f_{s}(x) x^2 dx[/itex], while [itex]R_s(0)=\int s(x)s(x)dx[/itex]
    I don't see either, how they solve this problem without any knowledge on the p.d.f. of s.

    Thanks.
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Nov 17, 2011 #2

    mathman

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    E(s(u)s(v)) = R(|u-v|) by definition. So if u=v we have E(s2(u)) = R(0)
     
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