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. 



Entropy rate problem of ergodic Markov process with non-ergodic joint
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 $-n(H(\Pi\_1)+H(\Pi\_2))$.


