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# periodicity of a Markov chain

Let $\{X_{n}\}$ be a stationary Markov chain with state space $I$. Let $P_{{ij}}^{{n}}$ be the $n$-step transition probability that the process goes from state $i$ at time $0$ to state $j$ at time $n$:

$P_{{ij}}^{{n}}=P(X_{n}=j\mid X_{0}=i).$ |

Given any state $i\in I$, define the set

$N(i):=\{n\geq 1\mid P_{{ii}}^{n}>0\}.$ |

It is not hard to see that if $n,m\in N(i)$, then $n+m\in N(i)$. The *period* of $i$, denoted by $d(i)$, is defined as

$d(i):=\begin{cases}0&\text{if }N(i)=\varnothing,\\ \gcd(N(i))&\text{otherwise},\end{cases}$ |

where $\gcd(N(i))$ is the greatest common divisor of all positive integers in $N(i)$.

A state $i\in I$ is said to be *aperiodic* if $d(i)=1$. A Markov chain is called *aperiodic* if every state is aperiodic.

Property. If states $i,j\in I$ communicate, then $d(i)=d(j)$.

###### Proof.

We will employ a common inequality involving the $n$-step transition probabilities:

$P_{{ij}}^{{m+n}}\geq P_{{ik}}^{m}P_{{kj}}^{n}$ |

for any $i,j,k\in I$ and non-negative integers $m,n$.

Suppose first that $d(i)=0$. Since $i\leftrightarrow j$, $\displaystyle{P_{{ij}}^{n}>0}$ and $P_{{ji}}^{m}>0$ for some $n,m\geq 0$. This implies that $P_{{ii}}^{{m+n}}>0$, which forces $m+n=0$ or $m=n=0$, and hence $j=i$.

Next, assume $d(i)>0$, this means that $N(i)\neq\varnothing$. Since $i\leftrightarrow j$, there are $r,s\geq 0$ such that $P_{{ji}}^{r}>0$ and $P_{{ij}}^{s}>0$, and so $P_{{jj}}^{{r+s}}>0$, showing $r+s\in N(j)$. If we pick any $n\in N$, we also have $\displaystyle{P_{{jj}}^{{r+n+s}}\geq P_{{ji}}^{r}P_{{ii}}^{n}P_{{ij}}^{s}>0}$, or $r+s+n\in N(j)$. But this means $d(j)$ divides both $r+s$ and $r+s+n$, and so $d(j)$ divides their difference, which is $n$. Since $n$ is arbitrarily picked, $d(j)\mid d(i)$. Similarly, $d(i)\mid d(j)$. Hence $d(i)=d(j)$. ∎

## Mathematics Subject Classification

60J10*no label found*

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