0
$\begingroup$

Professor told me that if a random process is weak stationary, and it does not feature any periodic component, then its autocovariance always goes to zero.

I can intuitively understand it, however, where can I find the formal proof of such property?

$\endgroup$
  • 1
    $\begingroup$ What about a process such that every $X(t)$ is the same random variable? In that case the autocovariance function would be a non-zero constant, but the process is WSS. $\endgroup$ – Tendero Jul 20 '19 at 15:27
  • $\begingroup$ Hi Tendero: I don't want to get into a big dicussion but, in statistical time-series/econometrics, one would say that auto-covariance is not defined because process has zero variance. But I'm well aware that, in DSP, things can be quite different. All the best. $\endgroup$ – mark leeds Jul 20 '19 at 17:52
  • $\begingroup$ @Giovanni Castellano: I don't know if what your professor is saying is true but chapter 4 of this document would probably be pretty enlightening if you have the time to check it out. good luck. oh, the answer may also be in brockwell and davis but I can't find that text right now. stat.tamu.edu/~suhasini/teaching673/time_series.pdf $\endgroup$ – mark leeds Jul 20 '19 at 18:25
  • $\begingroup$ @Tendero The autocorrelation function converges to the square of the mean while the autocovariance function does converge to 0 as $\tau$ increases without bound. No problems whatsoever even in the case of your proposed counterexample. $\endgroup$ – Dilip Sarwate Jul 20 '19 at 20:10
  • $\begingroup$ @Dilip Sarwate: : I'm pretty sure the OP's professor assumes a de-meaned process Can you point to somewhere that shows that acf converges to the square of the mean asymptotically. I looked and couldn't find it. I'm not concerned with counter-example bur rather talking about the general case. Thanks. $\endgroup$ – mark leeds Jul 20 '19 at 22:29

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.