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I have a stationary process $\{u_n\}$ and I have a function $f:\mathbb{R}^L\to \mathbb{R}^+$. I want to evaluate the following limit $$\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n g(f(\mathbf{u}_{k}))$$ where $\mathbf{u}_k=\begin{bmatrix} u_k & u_{k-1} & \cdots & u_{k-L} \end{bmatrix}$ and $g$ is a smooth real valued function i.e. $g:\mathbb{R}\to \mathbb{R}$.

My question is

Can I use Birkhoff's Ergodic Theorem here to conclude that $$\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n g(f(\mathbf{u}_{k}))\stackrel{a.s.}{=}\mathbb{E}(g(f(\mathbf{u}_L)))?$$

I know (at least according to my knowledge) that had it been $\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n g(f({u}_{k}))$ the answer would be yes, but I do not have much understanding of ergodic theory to make conclusion about this problem.

Forgive me for my lack of knowledge in this subject which is why I am asking this question, though it maybe trivial to many people here; but I need to understand this. Also it would be great if someone can kindly give some good reference to understand this theorem in the context of this problem (I know basic probability theory and stochastic processes and I am learning measure theory now).

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    $\begingroup$ I apologize @AnthonyQuas if it sounded like I used that as an excuse to post this question here. Actually this problem is related to my research and thought that this place is appropriate for questions related to research level mathematics, though I know that this question may not be at the level of questions asked here but I had no place else to ask this. $\endgroup$
    – Samrat Mukhopadhyay
    Commented Jul 23, 2015 at 18:22

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The answer is no. With stationarity only, the limit is the conditional expectation with respect to the invariant sigma field of the left shift operator applied to path space.

You need ergodicity also (the invariant sigma field is trivial). This is a lot harder to check.

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Unless you have a really weird process going on, the succession of values at different lags: $\mathbf{u}_k=\begin{bmatrix} u_k & u_{k-1} & \cdots & u_{k-L} \end{bmatrix}$ should also be stationary

so you can apply the theorem

However, if you are doing some sort of Monte Carlo scheme, be very careful that different schemes can give you the same marginal distribution for $u_n$ while giving you different joint distributions for $\mathbf{u}_k=\begin{bmatrix} u_k & u_{k-1} & \cdots & u_{k-L} \end{bmatrix}$

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  • $\begingroup$ Thank you very much for the answer. I also had this in mind, but I do not know how to express this idea mathematically so that I can use the theorem in its original form. Should I use some kind of transformation, like shifting which preserves the measure? $\endgroup$
    – Samrat Mukhopadhyay
    Commented Jul 23, 2015 at 18:19
  • $\begingroup$ you should just check that whatever stationary process you want to apply this too gives you a stationary process when you look at multiple values at the same time. just check if the joint is the same (or some other theorem) $\endgroup$
    – Guillaume Dehaene
    Commented Jul 23, 2015 at 18:26

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