# Show that an irreducible Markov Chain is always reversible

Suppose we have a Markov chain $$X$$, with transition probabilities $$T_{ij} = P(X_{n+1}=i\mid X_{n}=j )$$.

Let $$(Y_{1}, Y_{2}, .., Y_{N})$$ represent the reversed Markov chain of $$X$$, such that $$Y_{n} = X_{N-n}$$. Then, $$P(Y_{n+1}=j \mid Y_{n}=i) = T_{ij}\frac{P(Y_{n+1}=j)}{P(Y_{n}=i)}.$$ I understand that this result holds if X is homogeneous and is irrelevant whether X is irreducible.

If $$X$$ is an irreducible Markov chain then $$T_{ij}^{(n)} \geq 0$$ for each possible pair $$i, j$$ such that $$i \neq j$$, for some integer $$n$$.

Should one use the idea of detailed balance to show the result that an irreducible (but not necessarily homogeneous) chain must be reversible? I am unsure if the above homogeneous case helps any explanation in the context of the irreducible case.

Thank you in advance for your help and thoughts.

• (Note that Markov was a person, so it is proper to capitalise his name when referring to a Markov chain. Same with Bayes' theorem, Fisher's test, etc.)
– Ben
Commented Apr 12, 2020 at 1:54
• An irreducable Markov chain is not always reversible. E.g., the Markov chain corresponding to a top-to-random shuffle on n cards (at each step taking the top card and inserting it at one of the n positions in the deck chosen uniformly at random) is irreducible, aperiodic but not reversible. Commented Apr 12, 2020 at 9:34

Simplest example of a non-reversible irreducible Markov chain is a deterministic one with three states, that goes $$1,2,3,1,2,3,1,2,3,...$$
I think your definition of an irreducible Markov chain is flawed, also. In fact, it's vacuous; $$T_{ij} \geq 0$$ no matter what. You might be thinking of aperiodicity, one definition of which is that for any fixed $$i,j$$, $$T_{ij}^{(n)}$$ is always $$>0$$ for large $$n$$.