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

• mnb96
In summary, the conversation discusses the derivation of the solution of Wiener filter from the Wikipedia page. The step of calculating the mean value of e^2 is unclear, but the authors suggest that it equals R_s(0), the autocorrelation function of s evaluated at 0. The conversation also mentions the definitions of E[s^2] and R_s(0) and expresses confusion about how to solve the problem without knowing the probability density function of s.
mnb96
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.

Last edited by a moderator:
E(s(u)s(v)) = R(|u-v|) by definition. So if u=v we have E(s2(u)) = R(0)

## 1. Why is E[x^2] equal to the autocorrelation of x evaluated at 0?

The autocorrelation of a sequence is a measure of the similarity between the sequence and a shifted version of itself. When the shift is 0, the autocorrelation is essentially measuring the similarity between the sequence and itself. This is why the autocorrelation at 0 is equal to the expected value of x^2, as it is measuring the similarity between x and a shifted version of itself by 0.

## 2. How does the expected value of x^2 relate to the autocorrelation of x evaluated at 0?

The expected value of x^2 is one of the components used to calculate the autocorrelation of x at 0. It represents the average squared value of the sequence, which is used in the formula for autocorrelation. Essentially, the expected value of x^2 is a necessary factor in determining the autocorrelation at 0.

## 3. Can you explain the mathematical reasoning behind E[x^2] = autocorrelation of x evaluated at 0?

The autocorrelation of a sequence is calculated using the formula R(t)= E[x(t)x(t-tau)], where x(t) is the original sequence and x(t-tau) is a shifted version of the sequence by tau time units. When tau is equal to 0, the shifted sequence becomes the same as the original sequence. This means that the autocorrelation at 0 is essentially measuring the similarity between the sequence and itself, which is equal to the expected value of x^2.

## 4. Why is the concept of autocorrelation important in statistics and data analysis?

The concept of autocorrelation is important in statistics and data analysis because it helps us understand the patterns and relationships within a sequence of data. It allows us to measure the similarity between a sequence and a shifted version of itself, which can reveal important information about the data. Autocorrelation is commonly used in time series analysis, signal processing, and machine learning to analyze and make predictions based on data.

## 5. Are there any limitations or assumptions when using E[x^2] = autocorrelation of x evaluated at 0?

One limitation is that this equation assumes that the sequence is stationary, meaning that its statistical properties do not change over time. Additionally, this equation only applies to linear autocorrelations and may not be accurate for non-linear relationships. It is important to carefully consider the assumptions and limitations when using this equation in data analysis.

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