In the text I'm using it says:

Let X = {$X_n : 0 \leq n \leq N$} be an irreducible Markov chain such that $X_n$ has the stationary distribution $\pi$ for all $n$. The chain is called reversible if the transition matricies of X and its time-reversal Y are the same, which is to say

$$\pi_i p_{ij} = \pi_j p_{ji}$$

A transition matrix P and a distribution $\lambda$ are in detailed balance if $\lambda_i p_{ij} = \lambda_jp{ji}$. An irreducible chain X having stationary distribution $\pi$ is called reversible in equilibrium if its transition matrix P is in detailed balance with $\pi$.

Is it because in reversible in equilibrium we are not assuming that X is non-null persisntent? Which is needed to prove the existence of the the X's time reversal Y?

In that case all reversible chains would be reversible in equilibrium, but not all reversible in equilibrium chains would be reversible. Am i correct?

  • 1
    $\begingroup$ What difference? Except perhaps that "reversible in equilibrium" is clumsily worded and not useful... $\endgroup$
    – Did
    Jun 25, 2016 at 6:46

1 Answer 1


The second definition allows a transition matrix to be in detailed balance with respect to a general distribution (here, $\lambda$). If the Markov chain had multiple possible steady state solutions $\{ \pi_i \}_{i\in I}$, each could be in detailed balance with $P$ (or not!).

Since $P$ being in detailed balance with $\lambda$ implies that $\lambda$ is a stationary distribution, in the irreducible case we have only the single candidate $\pi$ to consider. I suppose since $\pi$ is now the equilibrium state, the authors stress the point by saying "reversible in equilibrium".

To your last two points: There should be no question of the existence of the time reversal $Y$: it is the sequence $(X_n, \dots, X_1)$. And "reversible in equilibrium" is exactly the same as the first vanilla definition of "reversible".


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .