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

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 $\hat{s}(t)$ and the original true signal $s(t)$:

$$e^2(t) = s^2(t) - 2s(t)\hat{s}(t) + s^2(t)$$

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

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

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

$$E[s^2]=R_s(0)$$

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

I thought that $E[s^2]=\int f_{s}(x) x^2 dx$, while $R_s(0)=\int s(x)s(x)dx$
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

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