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If $\{X_n\}$ is a finite-state Markov chain, which is aperiodic and irreducible, how can I show it is regular?

My attempt:

For all $i,j$ states there exists $n\in\mathbb{N}$ such that $P^n_{ij}>0$. I was trying to proof using the aperiodicity that for some $M\in\mathbb{N}$, $P^n_{ij}>0$ for all $n\geq M$ and taking the maximums would give power of $P$ with only positive entries.

Any suggestions?

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We know that if a (finite state space) Markov Chain is aperiodic, then there is some $n_0$ s.t. for all $n\ge n_0$ and all states $i$, $p_{ii}^n>0$.

Let $i\ne j$ be states of the Markov Chain. Let $m_{ij}$ be the shortest time to go from $i$ to $j$, ie, the mínimum of all $m$ s.t. $p_{ij}^m>0$. (which is finite because the chain is irreducible), and let $M:=\max_{i,j}m_{ij}$.

Now, let $n = M + n_0$ and $i\ne j$. We can write $p_{ij}^{n}=p_{ij}^{(n-m_{ij})+m_{ij}}$. By Chapman-Kolmogorov, we know $p_{ij}^{(n-m_{ij})+m_{ij}}=\sum_{k}p_{ik}^{n-m_{ij}}p_{kj}^{m_{ij}}$. Hence, $$ p_{ij}^{n}=p_{ij}^{(n-m_{ij})+m_{ij}}=\sum_{k}p_{ik}^{n-m_{ij}}p_{kj}^{m_{ij}}\ge p_{ii}^{n-m_{ij}}p_{ij}^{m_{ij}}>0 $$ as $n-m_{ij} = (M - m_{ij})+n_0\ge n_0$.

Hence, $P^{M + n_0}>0$, ie, the chain is regular.

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