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Let $(X_n)$ be a sequence of independent but not necessarily identically distributed random variables with $\mathbb{E}X_n = 0$ for all $n$. If the $X_n$ are uniformly bounded, Kolmogorov's strong Law of Large Numbers implies that $(1/n)\sum_{i=1}^n X_i \rightarrow 0$ almost surely.

Now let $S$ be a metric space and let $T$ be the shift operator on $S^\mathbb{N}$, that is $T(x_0,x_1,\dots) = (x_1, x_2,\dots)$. Consider a product probability measure $\mu = \prod_{i=0}^\infty \mu_i$ on $S^\mathbb{N}$ (thus $\mu$ is generally not invariant unless $\mu_i = \mu_0$ for all $i$).

For a given bounded, continuous function $f$ on $S^\mathbb{N}$ of the form $f = g \circ \pi$, where $g$ is a continuous bounded function on $S$ and $\pi(x_0,x_1,\dots) = x_0$, let $$X^f_n(\boldsymbol{x}) := f\circ T^n(\boldsymbol{x}) - \int f\circ T^n(\boldsymbol{y})\,\mu(\mathrm{d}\boldsymbol{y}).$$ On the probability space $(\Omega, \mathbb{P}) := (S^\mathbb{N}, \mu)$, we have that $(X_n^f)$ is a sequence of independent random variables with $\mathbb{E}X_n^f = 0$ for all $n$, with $\sup_{\boldsymbol{x},n}X_n^f(\boldsymbol{x}) < \infty$. Thus by the SLLN we have $(1/n)\sum_{i=0}^{n-1}X_n^f \rightarrow 0$ ($\mu$ a.e.).

Therefore, given a product probability measure on $S^\mathbb{N}$, a property analogous to ergodicity (w.r.t. $T$) holds when we restrict ourselves to continuous bounded maps of the form $f = g\circ\pi$ as above.

My question is: does this property still hold for the general $f\in C_b(S^\mathbb{N})$?

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    $\begingroup$ The statement from your first paragraph is incorrect; you need additional assumptions. Consider for example the rv's $X_n=\pm 2^{n^2}$, with prob $1/2$ each. $\endgroup$
    – Christian Remling
    Commented Apr 22, 2017 at 15:49
  • $\begingroup$ I have corrected the statement of the SLLN. $\endgroup$
    – user127022
    Commented Apr 22, 2017 at 23:53

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No. For example, take $S=\mathbb R$ and let $\mu_n$ be Lebesgue measure restricted to $[n,n+1]$. Set $$ f(x_0,x_1,\ldots)=\sin^2(2\pi x_0)\sin(2\pi(x_{2^j-\lfloor x_0\rfloor}) \text{ if $x_0\in [2^{j-1},2^j)$}. $$

Then $f(T^n(x_0,x_1,\ldots))=\sin^2(2\pi x_n)\sin(2\pi x_{2^j})$ for all $n$ in the range $[2^{j-1},2^j)$ (the reason for the $\sin^2$ term is just to make it continuous). The averages don't converge.

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  • $\begingroup$ Should it be $f(T^n(x_0,x_1,\dots)) = \sin^2(2\pi x_n)\sin(2\pi x_{2^j - \lfloor x_n\rfloor})$ for $x_n \in [2^{j-1}, 2^j)$? $\endgroup$
    – user127022
    Commented Apr 27, 2017 at 16:56
  • $\begingroup$ Yes, absolutely. Thanks for the correction. $\endgroup$
    – Anthony Quas
    Commented Apr 28, 2017 at 3:42
  • $\begingroup$ could you perhaps point me to a reference where this (counter-)example appears in more detail? $\endgroup$
    – user127022
    Commented May 8, 2017 at 23:46
  • $\begingroup$ Sorry. I can't because the counter-example was created specifically in response to this question. But: if you agree with the calculation that $f(T^n(x_0,x_1,\ldots))=\sin^2(2\pi x_n)\sin(2\pi x_{2^j})$ for all realizations $(x_0,x_1,\ldots)$ of the random vector $(X_0,X_1,\ldots)$, then the reason for the lack of convergence is that the average over the block $[2^{j-1},2^j)$ is close to $\frac 12\sin(2\pi x_{2^j})$; and as these blocks are doubling in length, brings the overall average towards that value; the next block brings the overall average towards $\frac 12\sin(2\pi x_{2^{j+1}})$ etc. $\endgroup$
    – Anthony Quas
    Commented May 9, 2017 at 1:48

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