# Markov chains: is “aperiodic + irreducible” equivalent to “regular”?

I have two books on stochastic processes.

In one book, it says that the limiting matrix is possible to find if the matrix is regular, that is, if for some $n$ $P^n$ has only positive values. The other book says that the limiting values are possible to find if the Markov chain is recurrent, irreducible and aperiodic; it is then called ergodic.

Does this then hold:

aperiodic + irreducible $\Leftrightarrow$ ergodic $\Leftrightarrow$ regular?

And is there any difference whether it is a finite-state chain or not?

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For a finite MC it holds that

aperiodic + irreducible $\Leftrightarrow$ ergodic $\Leftrightarrow$ regular

as you expected. For an infinite MC it holds that

aperiodic + irreducible + positive recurrent $\Leftrightarrow$ ergodic,

and being "regular" in the infinite setting would require a more precise definition.

................................ explanations following ................................

For every finite or inifinite Markov chain (MC) it holds that

$aperiodic + irreducible + positive~recurrent \Leftrightarrow ergodic$.

See for example here for a proof. For every finite MC, irreducibility already implies positive recurrence, see here for a proof.

Further, for every finite MC we have that

$aperiodic + irreducible \Leftrightarrow regular$.

Proof sketch: the definition of a finite irreducible MC gives that $\forall i, j \in \Omega : \exists k > 0 : P^k[i,j] > 0$. However, there might be no $k$ such that all entries are simultaneously positive - due to periodicities. But if the chain is additionally aperiodic, it follows that $\exists k > 0 : \forall i, j \in \Omega : P^k[i,j] > 0$, which matches your definition of being regular.

Finally, I don't see a canonical way how you would generalize the property "regular" to infinite Markov chains. So, I just ignore the term "regular" for infinite chains here.

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Thank you very much!, it cleared the situation up for me! –  user119615 Jan 21 at 12:01