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It is known that if the transition probability matrix of a Markov Chain is regular, the chain has the positive limiting distribution (limiting distribution with all elements positive).

Does the converse hold? i.e., If a Markov Chain has the positive limiting distribution, is its transition probability matrix regular?

If not, what is an additional condition that is required to make this hold?

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If the state space is finite, then regularity is indeed equivalent to the existence of a unique limiting distribution. Let $X=\{X_n:n=0,1,\ldots\}$ be a Markov chain with transition matrix $P_{i,j} = \mathbb P(X_{n+1}=j\mid X_n=i)$ and strictly positive limiting distribution $$\pi_j = \lim_{n\to\infty} \mathbb P(X_n=j\mid X_0=i).$$ This implies that for each pair $(i,j)$ of states, $$ \mathbb P\left(\bigcup_{n=1}^\infty \{X_n=j\mid X_0=i\} \right) = 1, $$ so $X$ is irreducible. If a state $i$ were to have period $d(i)>1$, then $\mathbb P(X_k=i\mid X_0=i)=0$ for $1\leqslant k< d(i)$, and hence $\mathbb P(X_{n\cdot d(i)+k}=i\mid X_0=i)=0$ for $n=1,2,\ldots$ and $1\leqslant k<d(i)$. This contradicts the existence of $\lim_{n\to\infty} \mathbb P(X_n=i\mid X_0=i)$, and so state $i$ is aperiodic. Since periodicity is a class property and $X$ is irreducible, all states are aperiodic.

It follows that if $P^n_{i,j}>0$ then $P^{kn}_{i,j}$ for any positive integer $k$. Let $n_{i,j}=\min\{n: P^n_{i,j}>0\}$ for each pair of states $(i,j)$. Set $$ N = \prod_{i,j=1}^n n_{i,j}, $$ then $P^N_{i,j}>0$ for all $(i,j)$, so that $X$ is a regular Markov chain.

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