1
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ Does this help? en.wikipedia.org/wiki/… $\endgroup$
    – Maxtron
    Commented Jul 9, 2020 at 3:17
  • $\begingroup$ Thanks for responding...but I don't think that helps me... $\endgroup$
    – j03y_
    Commented Jul 10, 2020 at 5:02
  • $\begingroup$ Can you add the equations? I don't have access to these textbooks. $\endgroup$
    – Maxtron
    Commented Jul 10, 2020 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_
    Commented Jul 11, 2020 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_
    Commented Jul 11, 2020 at 1:21

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.