I've already asked this question on math.stack a few days ago and haven't received an answer, so I'm asking here.

In an engineering course, a stationary process was defined to be ergodic if for all $k\in \mathbb{N}$ and for any bounded (measurable) function of $k+1$ variables we have $$\lim_{N\rightarrow \infty}\frac{1}{N}\sum_{n=1}^N g(X_n,\dots,X_{n+k})\overset{\text{a.s}}{=}Eg(X_n,\dots,X_{n+k})$$ From the little I've read about ergodic theory, this does not seem familiar nor does it seem to fit into the ergodic hierarchy I know, i.e ergodic, weak mixing, strong mixing etc. It seems like a different property from ergodicity (in the sense of the Birkhoff ergodic theorem). Here the boundedness of $g$ means a formulation with indicator $g$'s would be equivalent (because of DCT I think). On the other hand, any $k$-tuple of $X_i$'s is allowed.. Is there an insightful bit of intuition for this property as there are for normal ergodicity, and mixing? Where does it fit into the ergodic hierarchy?

Thanks in advance!

  • 4
    $\begingroup$ That’s one extraordinarily formal engineering course! Can I ask in which country this takes place? $\endgroup$ Jun 18, 2014 at 17:30
  • 4
    $\begingroup$ In Israel. I think the course itself is pretty informal, I just thought the material was interesting so I studied probability theory and read a little about the ergodic hierarchy, which led to my question :) $\endgroup$
    – Exterior
    Jun 18, 2014 at 17:36

1 Answer 1


Yes. That is the same thing as ergodicity. To explain it, you have probably seen somewhere that the way to understand random variables formally is functions from a (hidden) underlying space $\Omega$ to $\mathbb R$. That is: knowing the point $\omega$ of $\Omega$, one can recover $X_n(\omega)$ for each $n\in \mathbb Z$.

In fact, a standard $\Omega$ to use is just the space of sequences of real numbers. Then the function $X_n$ is just the $n$th coordinate function. If you do this, then $\Omega$ is equipped with a natural map, namely the shift map, $\sigma$. You can now define a version of your function $g$ on $\Omega$, namely $G(\omega)=g(\omega_0,\ldots,\omega_k)$. The left side of your equality is now $\lim_{N\to\infty}(1/N)\sum_{n=1}^N G(\sigma^n\omega)$. The right side is $\int G(\omega)\,d\mu(\omega)$, where $\mu$ is the distribution on the set of sequences occurring. Notice: invariance of $\mu$ corresponds exactly to stationarity of the sequence of random variables.

  • $\begingroup$ I just looked at the comment on MSE, which seems to be saying the same thing, in a slightly more concise form. $\endgroup$ Jun 19, 2014 at 6:44
  • $\begingroup$ Yep. it was too concise for me to understand, and as it was left as a comment, I waited a few days for an answer. I'll tell the commenter that I will gladly accept his answer if he posts one. $\endgroup$
    – Exterior
    Jun 19, 2014 at 6:54
  • $\begingroup$ I guess a good way to go in future could be to ask (tagging to the poster by name) if they can give a couple more details. $\endgroup$ Jun 20, 2014 at 16:24
  • $\begingroup$ Understood; I'll remember that. $\endgroup$
    – Exterior
    Jun 24, 2014 at 9:02

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