1
$\begingroup$

A Markov Chain with state transition matrix $A$ satisfies detailed balance with respect to distribution $\pi$ if for all $i,j$, we have $$\pi_iA_{i,j} = \pi_jA_{j,i}.$$

In such a case, we have $$\sum_i \pi_iA_{i,j} = \sum_j\pi_jA_{j,i} = \pi_j.$$

Isn't this enough to show that if $A$ satisfies the detail balance equations with respect to $\pi$, then $\pi$ is the stationary distribution for $A$?

The text I'm reading states the following theorem:

If a Markov chain $A$ is regular and satisfies detailed balance wrt $\pi$, then $\pi$ is a stationary distribution for $A$.

Why is regularity needed? Above we already showed that $\pi = A\pi$, which is the definition of a stationary distribution for $A$.

Edit:

As a silly example, if $$A = \begin{pmatrix} 1 & 0\\ 0 & 1 \\ \end{pmatrix}$$

then, for example, $\pi = [1, 0]$ satisfies the detailed balance equations and is a stationary distribution for the chain, even though the chain is not irreducible or regular.

$\endgroup$
6
  • $\begingroup$ Detailed balance only guarantees stationarity of $\pi$ when the Markov chain is irreducible. Else $\pi$ cannot be stationary. $\endgroup$ – Xi'an Nov 4 '15 at 21:42
  • $\begingroup$ Then why does the theorem use regularity instead of irreducibility? $\endgroup$ – Fequish Nov 4 '15 at 21:52
  • 2
    $\begingroup$ $\pi=[1/2 \;\; 1/2]$ satisfies the detailed balance equations for the chain with $A=[0 \;\; 1; \;\; 1 \;\; 0]$, but it isn't a stationary distribution. $\endgroup$ – Brian Borchers Nov 4 '15 at 22:05
  • $\begingroup$ Why is it not a stationary distribution? The definition of a stationary distribution in my book is that $\pi = \pi A$, which is satisfied here. The definition on Wikipedia is similar. $\endgroup$ – Fequish Nov 4 '15 at 22:13
  • 2
    $\begingroup$ You have to be careful about whether "stationary distribution" means $\pi A=\pi$ and nothing else or whether "stationary distribution" means that $\pi A = \pi$ and $\pi$ is the limiting distribution for all initial states. If the chain is regular and detailed balance is satisfied than $\pi$ is a limiting distribution for all initial states. It seems likely that either you're misreading your sources or that the authors have been sloppy about this or that they simply included an unnecessary hypothesis in the theorem. $\endgroup$ – Brian Borchers Nov 4 '15 at 23:23

Browse other questions tagged or ask your own question.