Attitude true for Markov Chains(Maybe duplicate)

Question

Suppose, that we have a Markov-chain with finite domain. For all $i,j$ elements $P_{i,j}>0$, where $P$ is our matrix. Show, that reversible(sorry, I forgot that out) stationary distribution exists only, and only if for all $i,j,k$

$$P_{i,j} P_{j,k} P_{k,i} = P_{i,k} P_{k,j} P{j,i}$$

What I tried to was to make $i$ fix and I made $\pi_j=c\frac{P_{i,j}}{P_{j,i}}$ distribution, but I am stuck how to go forward.

Any help appreciated :)

Answer 1

Not true at all. Any irreducible aperiodic finite Markov chain has a unique stationary distribution. This follows e.g. from the Perron-Frobenius theorem.

Answer 2

For a counterexample, consider $$P=\begin{bmatrix}\frac16& \frac13& \frac12\\ \frac14& \frac 18& \frac58\\\frac15& \frac7{10}&\frac 1{10} \end{bmatrix}. $$ Then $P_{ij}>0$ for all $i,j$ so the process is aperiodic and the (unique) stationary distribution is given by any of the rows of $\lim_{n\to\infty}P^n$. However, $$P_{12}P_{23}P_{31} = \frac13\cdot\frac58\cdot\frac15 = \frac1{24} $$ and $$P_{13}P_{32}P_{21}=\frac12\cdot\frac7{10}\cdot\frac14=\frac7{80} $$ are not equal.