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?
Best Answer
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.