# Why does detailed balance not provide a stopping criterion in MCMC?

Like I undestand MCMC sampling, the fulfillment of the detailed balance equation guarantees that our MC has reached its stationary distribution (given we ensure ergodicity).

Detailed Balance is:

$\pi(x)q(x\rightarrow x')=\pi(x')q(x'\rightarrow x)$

with $\pi(x)$ being the probability to be in state $x$ and $q(x\rightarrow x')$ the transition probability from state $x$ to $x'$ at time $T$. (According to Russell, Stuart, and Peter Norvig. "AI a modern approach")

A problem I came across in MCMC is to find the right burn-in time $T$, the amount of samples needed to reach the stationary distribution. Why can we not use DBE as a stopping criterion? Why can we not compute whether DBE is fulfilled after each sample and then stop sampling as soon as it is fulfilled? Naively, it looks like $\pi(x)$ and $q(x\rightarrow x')$ could be computed emperically based on the samples obtained so far.

This is not a correct statement. Detailed balance wrt $\pi(x)$ guarantees that $\pi(x)$ is the stationary distiribution for the Markov chain. The fact that the MC "reaches" this stationary distribution is the definition of ergodicity of the Markov chain. (Note that there is a difference between stationary distribution and limiting distribution).
• No, detailed balance does not guarantee convergence to the stationary distribution. DB is a property of a MC and detailed balance wrt $\pi(x)$ just gives you that $\pi(x)$ is the stationary distribution. The convergence to this stationary distribution depends on other properties being satisfied. There are many Markov chains that do not satisfy a detailed balance condition wrt $\pi$ but still have $\pi$ as the stationary distribution. For e.g., a three variable Gibbs sampler. Jan 14, 2018 at 16:06
• Okay, I phrased that question badly. I rather meant: If I want to sample from a complicated distribution $P(X)$ and I can come up with an ergodic MC and use DBE to show that the stationary distribution of the MC is $P(X)$, then I am lucky in the sense that sampling infinitely long from the MC will certainly yield samples from my target distribution (although I'm still left with my problem to find the burn-in time $T$). Jan 17, 2018 at 10:23