If I have two signals $x$ and $y$ and this expression $E\{x[n]y[n-k]\}$, what is the expected value of the product?

I have a basic understanding of EV based on probability, such as $$E[X]=\sum_{i=1}^{\inf}x_ip_i$$ and I understand that EV of flipping coin is 0.5 if $x_1=0$ and $x_2=1$ and both probabilities are 0.5 or for rolling a dice is 3.5 etc.

However, when I see the expression like this the one above, I find it confusing. I don't know for instance where the probability term is.

  • 1
    $\begingroup$ Evaluating a quantity like $E[XY]$ requires knowledge of the joint distribution of $X$ and $Y$. That is, you need to know not just $P(X=x_ii) = p_i$ as in your examples but $p_{i,j}$, the probability that $X$ equals $x_i$ and simultaneously $Y$ equals $y_j$, and no, $p_{i,j}$ doesn't necessarily equal $P(X=x_i)P(Y=y_j)$ or $P(X=i)+P(Y=j)$ or any such thing; you need to find it. That being said, what you need is a generalization of what you have: $$E[X]=\sum_i\sum_j x_iy_j p_{i,j}=\sum_i\sum_j x_iy_jP(\{X=x_i\}\cap\{Y=y_j\}).$$ Note that $$P(X=i)=\sum_jP(\{X=x_i\}\cap\{Y=y_j\}).$$ $\endgroup$ – Dilip Sarwate Jun 22 '15 at 11:05
  • $\begingroup$ @DilipSarwate Thanks for you comment. I still don't understand how all this is related to $E\{x[n]y[n−k]\}$ or $E\{x[n]\}$ if I consider just one signal. Where is probability in $E\{x[n]\}$. Why would one calculate EV of $x$. Probably I am missing something really obvious. $\endgroup$ – Celdor Jun 22 '15 at 11:20
  • $\begingroup$ Ooops, I have a typo, now unfortunately uncorrectable in what I wrote above. That first displayed formula is for $E[XY]$, not $E[X]$ as I miswrote. $\endgroup$ – Dilip Sarwate Jun 22 '15 at 11:34
  • $\begingroup$ @DilipSarwate The second formula you wrote, is it P(X=x_{i}) or P(X=i) as you have written $\endgroup$ – Karan Talasila Jun 22 '15 at 11:46
  • $\begingroup$ More ooops. That second displayed formula should have $P(X=x_i)$ on the left, not $P(X=i)$. Thank God that comments do not get downvotes; else the first one would have garnered a -5 by now. $\endgroup$ – Dilip Sarwate Jun 22 '15 at 12:34

This is the cross-correlation function of the two discrete-time random processes $x[n]$ and $y[n]$. It measures the similarity of the processes for a given time lag between the two. If $x[n]$ and $y[n]$ are jointly wide-sense stationary, then the cross-correlation only depends on the lag $k$, and not on the absolute time index $n$.

In terms of probabilities (or rather probability density functions), the expectation $E\{x[n]y[n-k]\}$ is given by


where $f_{XY}(x,y;n_1,n_2)$ is the joint probability density function of $x[n_1]$ and $y[n_2]$.

In practice, we usually hope that the processes are ergodic, which allows us to estimate such an expectation by taking the appropriate time averages.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.