periodicity of a Markov chain

Let $\\{X_{n}\\}$ be a stationary (http://planetmath.org/StationaryProcess) 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 (http://planetmath.org/MarkovChainsClassStructure), 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 $N(i)\eq\\varnothing$ . 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 $d(i)=d(j)$ .∎