Proposition 3.2.1. State periodicity is a class property Let $\left(X_n\right)_{n \geq 0}$ be a Markov chain with the state space $S$ and the transition matrix $P$. Then the periodicity of states is a class property, that is all states in the same communicating class have the same period: $ i, j \in S: \quad i \leftrightarrow j \quad \Longrightarrow \quad d_i=d_j $ Proof. Suppose that $i, j \in S$ are such that $i \leftrightarrow j$, that is $i$ and $j$ communicate and thus belong to the same communicating class. We need to show that $i$ and $j$ have the same period, that is $d_i=d_j$. We will show that $d_i \leq d_j$ and that $d_j \leq d_i$ which will then imply that $d_i=d_j$. First we show that $d_i \leq d_j$. Since $i \leftrightarrow j$, then there exist $n, m \in \mathbb{N}$ such that $P_{i j}^n>0$ and $P_{j i}^m>0$, which by Chapman-Kolmogorov equation implies that $ P_{i i}^{n+m}=\sum_{k \in S} P_{i k}^n P_{k j}^m \geq P_{i j}^n P_{j i}^m>0 $ Hence, $d_i$ exists (is well defined) and, by the definition of the period of a state, $d_i$ divides $n+m$ as it is a factor of any $l \in \mathbb{N}$ for which $P_{i i}^l>0$. By following the same argument and swapping $i$ and $j$, it can be shown that $P_{j j}^{n+m}>0$ for some $n, m \in \mathbb{N}$. Hence, there exists at least one $l \in \mathbb{N}$ for which $P_{j j}^l>0$, thus $d_j$ exists. Let $r \in \mathbb{N}$ be such that $P_{j j}^r>0$. Then $ P_{i i}^{n+r+m} \geq P_{i j}^n P_{j j}^r P_{j i}^m>0 $ It follows that $d_i$ divides $n+m+r$. If $d_i$ divides both $n+m$ and $n+m+r$ then $d_i$ divides $r$. Hence, $d_i$ divides any $r \in \mathbb{N}$ for which $P_{j j}^r>0$, that is $d_i$ is a factor of all $r \in \mathbb{N}$ for which $P_{j j}^r>0$. But by the definition of $d_j, d_j$ is the greatest common factor of all such $r \in \mathbb{N}$. Hence, $d_i$ cannot be greater that $d_j$, that is $d_i \leq d_j$. By following the same argument and swapping $i$ and $j$ it can be shown that $d_j \leq d_i$. We deduce from $d_i \leq d_j$ and $d_j \leq d_i$ that $d_i=d_j$