# Ergodic theorem and products

If $$f_n (\omega) = \sum_{i=1}^n f_1 (T^i \omega)$$ and $$T$$ is an ergodic action with respect to the measure $$\mu$$ then it is know as Birkhoff's theorem that

$$\lim_{n \rightarrow \infty} \frac{f_n}{n} = \int_{\Omega} f_1(\omega) d\mu.$$

I was wondering what happens if one studies $$h_n (\omega) = \sum_{i=1}^n f (T^i \omega)g(T^{2i} \omega)$$ is it still true that

$$\lim_{n \rightarrow \infty} \frac{h_n}{n} = \int_{\Omega} f(\omega)g(\omega) d\mu?$$

If we assume that $$g(T^{2i}\omega)$$ form a family of i i d random variables.

• I think the integral you have on the right is not the correct limit. The integral of $f\circ T^{i}.g\circ T^{2i}$ is $\int_{\Omega}f(\omega)g\left(T^i\omega\right)d\mu$ by the $T$-invariance of the measure. Assuming that all of the functions are bounded so that we are allowed to swap limit and integral, the integral of the limit $\lim_{n \rightarrow \infty} \frac{h_n}{n}$ then should be $$\lim_{n\to\infty}\int_{\Omega}f(\omega)g_n(\omega)d\mu=\int_{\Omega}f(\omega)d\mu.\int_{\Omega}g(\omega)d\mu$$ since the Birkhoff averages $g_n$ of $g$ tend to $\int_{\Omega}g(\omega)d\mu$ $\mu$-a.e. – KhashF Feb 6 '19 at 20:17
• Having that said, I think it is better to ask whether $\lim_{n \rightarrow \infty} \frac{h_n}{n}=\int_{\Omega}f(\omega)d\mu.\int_{\Omega}g(\omega)d\mu$ $\mu$-a.e.? This is true if you make your assumptions about $T$ stronger, e.g. suppose $T$ is weak mixing. Because then, $T\times T^2$ would be an ergodic transformation of the product space $\left(\Omega\times\Omega,\mu\otimes\mu\right)$ and it suffices to apply the Birkhoff Ergodic Theorem to the function $(\omega,\omega')\mapsto f(\omega).g(\omega')$ on this space. – KhashF Feb 6 '19 at 20:32
• I@KhashF : I had thought of something like what you are suggesting. However, with your outer-product function, do you really get $h_n=\sum_{i=1}^n f (T^i \omega)g(T^{2i} \omega)$? Or do you get $\sum_{i=1}^n f (T^i \omega)g(T^{2i} \omega')$ instead? – Iosif Pinelis Feb 6 '19 at 23:15

Then $$X_0,X_1,\dots$$ are independent Rademacher random variables, with $$P(X_j=\pm1)=1/2$$. Let $$S_n:=\sum_1^n X_i X_{2i}$$. Then the question becomes the following: is it true that $$$$A_n:=\tfrac1n\,S_n\to EX_0^2$$$$ almost surely (or at least in probability), as $$n\to\infty$$?
We have $$$$E S_n^2=\sum_1^n EX_i^2 X_{2i}^2+2\sum_{1\le i so that $$EA_n^2=1/n\to0$$ and hence $$A_n\to0$$ in probability. On the other hand, $$EX_0^2=1\ne0$$.
It is unknown whether averages of the form $$\frac{1}{n} \sum_{k=0}^{n-1} f_1(T^kx)f_2(T^{2k}x)f_3(T^{3k}x)$$ converge almost surely.