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Let a random walk on a graph $G=(V,E)$ be characterized by the transition matrix $P$.

The usual discrete random walk process is: \begin{equation} p^{t+1}= p^{t} P, \end{equation} where $p^{t}$ is a probability distribution over the graph vertices at time $t$.

Under assumptions that the random walk be ergodic, i.e., reversible and aperiodic, there exists a unique stationary distribution $\pi$ checking: \begin{equation} \pi= \pi P\;,\quad p^{t}\xrightarrow{\ \ t\to\infty\ \ } \pi. \end{equation}

Aperiodicity is achieved provided the graph is not bipartite, or $P$ has all its eigenvalues $>-1$.

Reversibility is the property that the time-reverse random walk has the same transition matrix: \begin{equation} \pi_i P_{ij} = \pi_j P_{ji}. \end{equation}

It is then rather easy to check if the aperiodicity property is achieved.

But how do you check that your random walk is reversible? Is there a simple criteria which does not require to know $\pi$?

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  • $\begingroup$ Reversibility is not required for ergodicity. For the random walk to be reversible, it is necessary and sufficient that it can be coupled with an electrical network, i.e. with conductances attached to the edges where transitions are possible - see e.g. Doyle & Snell book (available online). $\endgroup$ Commented Nov 5 at 7:06
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    $\begingroup$ Sorry, I confunded two properties, ergodicity means the ensemble average equals the time average. A random walk is ergodic if it is irreducible and aperiodic. Reversibility is something else. $\endgroup$ Commented Nov 6 at 12:30

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A test for reversibility, based on $P$ but without having $\pi$ in hand, is Kolmogorov's cycle condition: https://en.wikipedia.org/wiki/Kolmogorov%27s_criterion

(Note: You also need irreducibility to guarantee the uniqueness of solutions to $\pi P = \pi$.)

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