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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 don't understand how your statement itself should make sense: the left side no longer depends on $n$ (you have taken the limit already) while the right side depends on $n$. $\endgroup$
    – Ian
    Commented Jul 20, 2015 at 11:01
  • $\begingroup$ Oh, sorry, my mistake, let me correct it. $\endgroup$
    – Samrat Mukhopadhyay
    Commented Jul 20, 2015 at 11:02
  • $\begingroup$ Are the vectors $\mathbf{u}_k$ iid? (In particular, are the correlations $Cov(u_k,u_{k-j})$ dependent only on $j$?) $\endgroup$
    – Ian
    Commented Jul 20, 2015 at 11:09
  • $\begingroup$ Yes, via stationarity of the sequence the correlation $E(u_k u_{k-j})$ is a function of $j$ and you can take ${u_j}$ to be a zero mean process. $\endgroup$
    – Samrat Mukhopadhyay
    Commented Jul 20, 2015 at 11:11

1 Answer 1

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The answer to your question is yes. If you equip $\mathbb{R}^\mathbb{N}$ with the product measure (corresponding to the distribution of $u$) with the shift operator, then you have an ergodic dynamical system, and Birkhoff applies and gives you exactly what you have.

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  • $\begingroup$ I also had this shift operator in my mind, though was not very confident about it due to my lack of knowledge in this subject. Thanks for the answer! $\endgroup$
    – Samrat Mukhopadhyay
    Commented Jul 28, 2015 at 17:54

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