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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!

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That’s one extraordinarily formal engineering course! Can I ask in which country this takes place? –  José Figueroa-O'Farrill Jun 18 at 17:30
3  
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 :) –  Exterior Jun 18 at 17:36

1 Answer 1

up vote 13 down vote accepted

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.

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Beautiful, thanks! –  Exterior Jun 18 at 23:14
    
I just looked at the comment on MSE, which seems to be saying the same thing, in a slightly more concise form. –  Anthony Quas Jun 19 at 6:44
    
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. –  Exterior Jun 19 at 6:54
    
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. –  Anthony Quas Jun 20 at 16:24
    
Understood; I'll remember that. –  Exterior Jun 24 at 9:02

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