I have a problem with the entropy rate when two ergodic Markov processes who are independent of each other having a non-ergodic joint. More specifically let us consider two finite-state Markov processes $\mathscr{P}_1$ and $\mathscr{P}_2$ with transition matrices $\Pi_1$ and $\Pi_2$, respectively. Let us assume that $\Pi_i$, $i=1, 2$, is a irreducible row stochastic matrix of period $p_i$, and $\gcd(p_1, p_2) = p > 1$. Because $\Pi_i$, $i=1, 2$, is irreducible, we know that $\mathscr{P}_i$ is ergodic.

Assume that $\mathscr{P}_1$ and $\mathscr{P}_2$ are independent, we have their joint process has transition matrix $\Pi =\Pi_1\otimes\Pi_2$, where $\otimes$ means the Kronecker product. However, since $p>1$, $\Pi$ is not irreducible. In fact, we can reorder the columns and rows of $\Pi$ simultaneously so that it becomes $\textrm{diag}(A_1, \ldots, A_p)$ where $A_i$, $i=1,\dots, p$, is a irreducible row stochastic matrix.

My question is: Let us denote entropy rate of the Markov process given by transition matrix $A$ by $\mathcal{H}(A)$, do we have $$\mathcal{H}(A_i) = \mathcal{H}(\Pi_1) + \mathcal{H}(\Pi_2)$$ for all $i = 1,\dots, p$?

The statement holds for a few examples I tried (some of them are not even Markov process, but were generated by a little bit more complicated models). But I wasn't successful in comping up either a proof or disproof.

Please help! Thank you in advance.

  • $\begingroup$ Yes. That’s a general result of ergodic theory. See for example Walters book. $\endgroup$ – Anthony Quas Jan 6 at 11:24
  • $\begingroup$ Great! Thank you, Anthony! Just to confirm, by Walters' book, you mean "An Introduction to Ergodic Theory (Graduate Texts in Mathematics)" by Peter Walters, right? If so, would you please point to me the chapter/section that I can find relevant information? $\endgroup$ – Yi Huang Jan 6 at 15:07
  • $\begingroup$ Yes. Exactly... $\endgroup$ – Anthony Quas Jan 6 at 18:05

The easiest way to see this is by using the Shannon-McMillan-Breiman equidistribution theorem and noticing that the space of sample paths of the product chain is the product of the sample path spaces of the original chains, so that for a.e. sample path of the product chain the logarithms of the measures of the corresponding cylinder sets are asymptotically $-n(H(\Pi_1)+H(\Pi_2))$.

| cite | improve this answer | |
  • $\begingroup$ Hi R W, thank you for your answer! I think the part I don't quite get is why the statement holds for each individual component. My numerical experiment told me that each individual component is actually NOT the independent product of the two processes. Would you please give me a little more detailed help? Thank you so much! $\endgroup$ – Yi Huang Jan 6 at 17:07
  • $\begingroup$ You apply the Shannon theorem to each of the chains $\mathcal P_1,\mathcal P_2$, and then pass to the product of the path spaces. What do you mean by "each individual component"? $\endgroup$ – R W Jan 6 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.