Kolmogorov's criterion

The theorem states that an irreducible, positive recurrent, aperiodic Markov chain with transition matrix P is reversible if and only if its stationary Markov chain satisfies[1]

${\displaystyle p_{j_{1}j_{2}}p_{j_{2}j_{3}}\cdots p_{j_{n-1}j_{n}}p_{j_{n}j_{1}}=p_{j_{1}j_{n}}p_{j_{n}j_{n-1}}\cdots p_{j_{3}j_{2}}p_{j_{2}j_{1}}}$

for all finite sequences of states

${\displaystyle j_{1},j_{2},\ldots ,j_{n}\in S.}$

Here p_ij are components of the transition matrix P, and S is the state space of the chain.

Let ${\displaystyle X}$ be the Markov chain and denote by ${\displaystyle \pi }$ its stationary distribution (such exists since the chain is positive recurrent).

If the chain is reversible, the equality follows from the relation ${\displaystyle p_{ji}={\frac {\pi _{i}p_{ij}}{\pi _{j}}}}$.

Now assume that the equality is fulfilled. Fix states ${\displaystyle s}$ and ${\displaystyle t}$. Then

${\displaystyle {\text{P}}(X_{n+1}=t,X_{n}=i_{n},\ldots ,X_{0}=s|X_{0}=s)}$${\displaystyle =p_{si_{1}}p_{i_{1}i_{2}}\cdots p_{i_{n}t}}$${\displaystyle ={\frac {p_{st}}{p_{ts}}}p_{ti_{n}}p_{i_{n}i_{n-1}}\cdots p_{i_{1}s}}$${\displaystyle ={\frac {p_{st}}{p_{ts}}}{\text{P}}(X_{n+1}}$${\displaystyle =s,X_{n}}$${\displaystyle =i_{1},\ldots ,X_{0}=t|X_{0}=t)}$.

Now sum both sides of the last equality for all possible ordered choices of ${\displaystyle n}$ states ${\displaystyle i_{1},i_{2},\ldots ,i_{n}}$. Thus we obtain ${\displaystyle p_{st}^{(n)}={\frac {p_{st}}{p_{ts}}}p_{ts}^{(n)}}$ so ${\displaystyle {\frac {p_{st}^{(n)}}{p_{ts}^{(n)}}}={\frac {p_{st}}{p_{ts}}}}$. Send ${\displaystyle n}$ to ${\displaystyle \infty }$ on the left side of the last. From the properties of the chain follows that ${\displaystyle \lim _{n\to \infty }p_{ij}^{(n)}=\pi _{j}}$, hence ${\displaystyle {\frac {\pi _{t}}{\pi _{s}}}={\frac {p_{st}}{p_{ts}}}}$ which shows that the chain is reversible.

The theorem states that a continuous-time Markov chain with transition rate matrix Q is reversible if and only if its transition probabilities satisfy[1]

${\displaystyle q_{j_{1}j_{2}}q_{j_{2}j_{3}}\cdots q_{j_{n-1}j_{n}}q_{j_{n}j_{1}}=q_{j_{1}j_{n}}q_{j_{n}j_{n-1}}\cdots q_{j_{3}j_{2}}q_{j_{2}j_{1}}}$

for all finite sequences of states

${\displaystyle j_{1},j_{2},\ldots ,j_{n}\in S.}$

The proof for continuous-time Markov chains follows in the same way as the proof for discrete-time Markov chains.