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I've just started studying Markov Chain(MC), and I trying to understand why the following MC is aperiodic with state space $Z =\{0,1\}$

$\begin{bmatrix} 0.5 & 0.5 \\ 1 & 0 \end{bmatrix}$

In my understanding, a state $j$ is aperiodic if it takes one step to come back to its initial position, and a MC is aperiodic if each state is of period 1 (aperiodic). In this case, starting from state $0$, it takes one step to come back in state $0$ with probability 0.5, and so $p_{0,0}>0$.

On the other hand, if I start from state $2$, I'll be back to state 1 with probability $1$, and so it will take 2,3,4,... steps to come back to state $2$. What I'm still missing?

Davide Giraudo
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Maximilian
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1 Answers1

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Assume that the Markov chain has the countable state space $E$. For two states $x,y$ in $E$, let $p^n(x,y)$ denote the $n$-step Markov chain transition probability from $x$ to $y$.

Then the period of a point $x$ is the greatest common divisor of all $n\in\mathbb N_0$ such that $p^n(x,x)>0$.

In your case, $p^n(0,0)>0$ for all $n\ge 0$ and $p^n(1,1)>0$ for all $n\in\mathbb N_0\setminus\{1\}$. But the greatest common divisor of all non-negative integers that are not equal to $1$ is still $1$.

So the period of both points is $1$. So the chain is called aperiodic.

Maximilian Janisch
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    Brilliant answer and super clear. It means that even if it takes more than one step to come back to the initial state, aperiodicity is just about the greatest common divisior of the numer of transitions, which must be equal to 1. Thanks – Maximilian Jan 01 '23 at 17:19