If we have two processes and both of them are ergodic. Does this mean that the joint proces is ergodic? Or other way around? If we have the dynamics for both components of the joint process what are the possible ways to show ergodicityof the joint process?


1 Answer 1


Joint behavior cannot always be deduced from individual behavior. For example, if $X$ and $Y$ are (nondegenerate) random variables with finite means, then $P\{X<E[X]\}$ and $P\{Y<E[Y]\}$ both are nonzero. This is because there must be probability mass to the left of the mean (and to the right of the mean too) except in the degenerate case when $X$ has value $E[X]$ with probability $1$. However, $P\{X<E[X],Y<E[Y]\}$ can be zero. Don't believe me? Consider $X\sim$ Bernoulli$(\frac 12)$ and $Y=1−X$ which is also Bernoulli$(\frac 12)$. Both have means $\frac 12$ but $P\{X < \frac 12, Y < \frac 12\} = 0$.

Turning to ergodicity, one characteristic of an ergodic process is that over long periods of time (say duration $T$), a typical sample path has value $\alpha$ or less for approximately $F_X(\alpha)\cdot T$ time units where $F_X(\cdot)$ is the first-order CDF of all the random variables comprising the process. That is, we can get good approximations to the value of the first-order CDF at $\alpha$ by observing what fraction of time a sample path spends below level $\alpha$. So, going back to your question, if $\{X(t)\}$ and $\{Y(t)\}$ are ergodic processes, we cannot deduce that the joint process is also ergodic, for reasons similar to the example I gave in the first paragraph above. Now, if you impose conditions such as $\{X(t)\}$ and $\{Y(t)\}$ are independent processes in addition to being ergodic processes, then Yes, the joint process is also ergodic.

  • $\begingroup$ Thank you for the answer! But do you know what are the mildest assumptions I can ask from the processes so that the joint is again ergodic? I don't want them to be independent $\endgroup$
    – nikolay
    Feb 20, 2016 at 0:57
  • $\begingroup$ @nikolay Ergodic theory is not a good fit for this forum, and most participants here will likely not have an answer to your follow-up query at their fingertips: I certainly don't. You can ask the moderators to migrate this question to math.SE where you might have better luck. $\endgroup$ Feb 22, 2016 at 14:06

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