Is the period of an irreducible Markov Chain the same for all states?

Question

I'm a bit confused about this theorem about irreducible Markov Chains:

If a Markov chain is irreducible, then all its states have the same period $d(i) := g.c.d.\{n > 0|P^n(i, i) > 0\}$.[source]

Let's suppose we have the following matrix which represents an irreducible Markov Chain:

\begin{equation} P = \begin{bmatrix} 0 & 0 & 0.8 & 0.2\\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix} \end{equation} if we start from state 1, we can reach it again after two transitions: \begin{equation} 1\rightarrow 3 \rightarrow 1 \end{equation} but if we start from state 2, we can only reach it again after 4 transitions: \begin{equation} 2\rightarrow 3 \rightarrow 1 \rightarrow 4 \rightarrow 2 \end{equation} So what is the period of the system? Is it 2 or 4? We can argue that it is 4 because 4 is a multiple of two and it is acceptable for both cases. But using the prementioned theorem $d(1) = 2$ and all other states should follow. What is it that I am getting wrong here?

Answer 1

The period is $2$ as given in the formula. A Markov chain is periodic if the chain can return to the same state only at multiples of some integer $>1$. See https://www.randomservices.org/random/markov/Periodicity.html

Answer 2

The period $\ d(i)\ $ of a state $\ i\ $ is not the shortest length of tine it takes to return to the state, but the gcd of all the times it can take to return to the state. In your example, if you start from state $2$ you can reach it again after $\ 4,6,8,\dots\ $ steps. That is $\ P^n(2,2)>0\ $ for any $\ n\ $ of the form $\ 4+2r\ $ where $\ r\ $ is any natural number. The gcd of those return times is $2$.