2
$\begingroup$

How can I prove that a WSS Gaussian stochastic process with mean 0 is mean-square ergodic in the second moment if and only if:

$$\lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^n r_{xx}^2(k) = 0$$

When $r_{xx}(k)=E[x(n)x(n+k)]$ and $x(n)$ is the process.

$\endgroup$
  • 1
    $\begingroup$ Sorry, I fixed it. $\endgroup$ – nirkov May 31 at 12:54
  • $\begingroup$ Thank-you!!!!!! $\endgroup$ – Peter K. May 31 at 12:55
1
$\begingroup$

A WSS Gaussian process is also a strictly stationary Gaussian process, but beyond that, Gaussianity or the lack thereof has nothing to do with the matter; the question might as well have asked for a proof for a general discrete-time stationary process that the process is second-moment ergodic if and only if the following condition holds: $$\lim_{n\to\infty} \frac{1}{2n+1}\sum_{i=-n}^n |R_X[n]|^2 = 0 \tag{1}$$ (Note that I have changed the condition to be more symmetric). Regardless of all this, I think that this is a question whose answer (the proof of an ergodic theorem) is way too complicated for anyone on dsp.SE to type up for the edification of the OP and any other readers who happen to stray here.

Random processes don't have to be first-moment or second-moment ergodic, and they fail to be ergodic when the autocorrelation function does not decay away fast enough as $n\to\infty$. To see this, note that the sum is one of nonnegative terms, and so if $R_X[n]$ has finite support, there is no problem at all, but if $R_X[n]$ is nonzero for all $n$, then the sum in $(1)$ is increasing with $n$, and if the increase is not slow enough, that $\frac{1}{2n+1}$ will not enough to damp the sum down enough that the limit will be $0$. $R_X[n] = \exp(-|n|)$ or $\frac{1}{1+n^2}$? No problem at all: the process is ergodic. But consider a process in which all the random variables $X[n]$ comprising the process are the same random variable $Y$. The process is stationary, but all the sample paths are (different) constant functions of time. If $Y=a$, the corresponding sample path has value $a$ for all $n$. Thus, while $R_X[n] = E[X[m]X[m+n]] = E[Y^2]$ is a constant, it is not necessarily the same constant as $$\frac{1}{2n+1}\sum_{i=-n}^n y[i]y[i+n] = a^2$$ and thus the process is not second-moment ergodic. It's not mean-ergodic either as noted in this answer of mine. That answer also has an example of a process that is not stationary but is nonetheless second-moment ergodic.

$\endgroup$
  • $\begingroup$ Sorry for bothering but are the terms mean square ergodic and second moment ergodic synonyms ? $\endgroup$ – Fat32 Jun 1 at 19:42
  • $\begingroup$ @Fat32 To me, they are synonymous but I have re-written my answer to conform to the wording in the OP's question. $\endgroup$ – Dilip Sarwate Jun 1 at 22:49
  • $\begingroup$ Ok honestly, when I hear the term Mean-Square Ergodic I compare it with "uniform" vs "mean-square" convergence, hence a little bit confused, but the term Second Moment Ergodic (or even preferrable as Ergodic in the second moment ) is as clear as $E\{ X_t^2 \} = < \bar{ x^2(t)} > $ where the left hand side is the ensemble average of the square of a random variable, and the right hand side is the time average of the square of any sample function of the random process. $\endgroup$ – Fat32 Jun 1 at 23:35

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.