0
$\begingroup$

Can someone help me understand the difference between a 1-dim autocorrelation function and a 2-dim autocorrelation matrix of a random process aka time series?

My Leon-Garcia textbook defines CX(τ) and RX(τ) as 1-dim functions, yet Haykin's Adaptive Filter Theory defines what looks like the same thing, but as the correlation matrix R instead. I see that the matrix appears to arise when the time series is represented by complex numbers (sinusoids), but is that the only real difference, or is there a fundamental distinction that I'm missing?

And does the correlation matrix R reduce to the 1-dim autocorrelation function when the time series can be represented by only real numbers instead of complex numbers, i.e. in that case the correlation matrix is unnecessary?

thanks, js.

$\endgroup$
  • 1
    $\begingroup$ Does this help? en.wikipedia.org/wiki/… $\endgroup$ – Maxtron Jul 9 at 3:17
  • $\begingroup$ Thanks for responding...but I don't think that helps me... $\endgroup$ – j03y_ Jul 10 at 5:02
  • $\begingroup$ Can you add the equations? I don't have access to these textbooks. $\endgroup$ – Maxtron Jul 10 at 16:22
  • $\begingroup$ Sure. But first, in general, I believe autocorrelation Rx(τ) can be constructed by just multiplying all the time series values (x1, x2, x3, x4, ...) together like this: Rx(0) = x1x1 + x2x2 + x3x3 + ... Rx(1) = x1x2 + x2x3 + x3x4 + ... Rx(2) = x1x3 + x2x4 + x3x5 + ... whereas the first three rows of the corresponding autocovariance matrix would be: Row 1 = x1x1 x1x2 x1x3 ... Row 2 = x2*x1 x2x2 x2x3 ... Row 3 = x3*x1 x3*x2 x3x3 ... In the EE textbooks, assuming RX(τ) = CX(τ) when mean is zero (autocorrelation = autocovariance) CX(τ) = E[ (X(t) X(t - τ) ] $\endgroup$ – j03y_ Jul 11 at 1:17
  • $\begingroup$ Covariance matrix in Haykin is this: R = 𝔼[u(n) uH(n)], where u(n) is the time series and H is the Hermitian operator (conjugate transpose). $\endgroup$ – j03y_ Jul 11 at 1:21

Your Answer

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

Browse other questions tagged or ask your own question.