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

  • Thread starter mnb96
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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.
 
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Answers and Replies

  • #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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