# Ergodicity of 2 independent ergodic random processes

I'm wondering if $\{X_i\}$ and $\{Y_i\}$ are 2 independent processes that are ergodic, then would $\{(X_i,Y_i)\}$ be ergodic?

I believe it is the case under the additional assumption that the two processes are either stationary, asymptotically stationary (i.e. $\lim_{n\to\infty}P(T^{-n}E)$ exists) or asymptotically mean stationary (i.e. $\lim_{n\to\infty}n^{-1}\sum_{i=0}^{n-1}P(T^{-i}E)$ exists). But I haven't been able to figure out a way to prove so.

Edit: By ergodic process, I mean it satisfies $P(E)$ is either 0 or 1 for all invariant events $E$ (those that satisfy $T^{-1}E=E$).

Edit #2: $T$ is the left-shift transformation.

• There are various definitions of "ergodic." All those listed in Wikipedia trivially and immediately imply your result without any special assumptions, because all properties (whether ensemble or time averages) of the product process are evaluated component by component. – whuber Feb 14 '15 at 1:21
• Thanks whuber. I'm looking for a formal proof, but the definitions on Wikipedia are vague and insufficient for that. – hikaru Feb 14 '15 at 1:40
• Fine--what is your definition? I see you added something in an edit, but what is $T$? Are you assuming only processes indexed by integers? – whuber Feb 14 '15 at 1:40
• @whuber: Sorry for the confusion. $T$ is the left-shift transformation. And yes, I'm assuming integer-indexed processes. – hikaru Feb 14 '15 at 2:08
• hikaru Your question is not clear because you are talking about possibly non-stationnary processes but you are implicitely assuming stationary processes from your definition of ergodicity and the relation with the shift $T$. Anyway my previous link shows the answer is no (for stationnary processes). – Stéphane Laurent Feb 14 '15 at 10:33

Here invariant sets of non-trivial measure are given by thickened diagonals ${(a,a+b): a\in[0,1), b\in[0,\varepsilon)}$ for $\varepsilon>0$. Those diagonal strips can not be approximated well by rectangles and thus ergodicity of the product fails, even though both irrational rotation factors are ergodic.