Aperiodic chains

Consider the irreducible two-state chain with stochastic matrix

$$P=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$$

it’s clear that the limits $(p^n{)}_{ij}$ don’t exists. This chain has a period of two.

Def A state $i\in I$ is called aperiodic if $(P^n{)}_{ii}>0$ for all $n$ sufficienlty large.

This is the same as $P$ is a regular matrix.

An equivalent definition, more easy to check, is the following:

Proposition If there exists $n_1,\dots ,n_k>0$ such that $p_{ii}^{n_j}>0$ for all $j\in 1,\dots ,k$, and

$$1=\gcd \{n_1,\dots ,n_k\}$$

then the state $i$ is aperiodic.

Proof This follows from the fact that, provided $n$ is big enough, we can always find $a_1,\dots ,a_k\in \mathbb{N}$ such that (Bezou’s lemma)

$$n=a_1{n}_1+\dots a_k{n}_k$$

so that, using Chapman-Kolmogorov

$$(p^n{)}_{ii}\ge (p^{a_1{n}_1}{)}_{ii}\dots (p^{a_k{n}_k}{)}_{ii}>0,\square$$

Also, the period of a state is a class property:

Proposition If $i\leftrightarrow j$ then $i$ and $j$ have the same period. Proof Call the period of state $d_i$ and $d_j$ respectivly. There exists $r,s\ge 0$ such that $(P^r{)}_{ij},(P^s{)}_{ji}>0$. We know that

$$(P^{r+s}{)}_{ii}\ge (P^r{)}_{ij}(P^s{)}_{ji}>0$$

so that $d_i|r+s$. Suppose $d_j|k$, then

$$(P^{k+r+s}{)}_{ii}\ge (P^r{)}_{ij}(P^k{)}_{jj}(P^s{)}_{ji}>0$$

so that $d_i|k+r+s$ but since it divides also $r+s$, this means that

$$d_j|k⟹d_i|k$$

so that $d_i\le d_j$. By swapping $i$ and $j$ we get our result. $\square$

Another useful thing is that if $i$ is an aperiodic in an irreducibile chain, then for evert $j,k\in I$ $(P^n{)}_{ij}>0$ for $n$ large enough (The matrix $P^n$ has all positive entries, i.e. it’s regular). Proof

$$(P^{n+r+s}{)}_{jk}\ge (P^r{)}_{ji}(P^n{)}_{ii}(P^s{)}_{ik}>0\square$$