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

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

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

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:

$$P_{ij}^{m+n}\ge P_{ik}^m{P}_{kj}^n$$

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

Suppose first that $d(i)=0$. Since $i\leftrightarrow j$,$P_{ij}^n>0$ and $P_{ji}^m>0$ for some $n,m\ge 0$. 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 $N(i)\ne \mathrm{\varnothing }$. Since $i\leftrightarrow j$, there are $r,s\ge 0$ 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 $P_{jj}^{r+n+s}\ge P_{ji}^r{P}_{ii}^n{P}_{ij}^s>0$, 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 $d(i)=d(j)$.∎