Imagine the following Markov chain:
$$\begin{bmatrix} 0 & 0.5 & 0.5 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix}$$
We always get back to state 1 in two time periods. So, state 1 is periodic and its period is 2.
For states 2 and 3, the time it takes to get back to them can be 2, 4, 6, or any even number of periods. My question is this: are states 2 and 3 periodic or aperiodic?
P.S. I think they are both periodic, since it's always a multiple of 2, but in a lecture it's said that they are aperiodic. So, that's why I'm asking here. Can the reason for those two being aperiodic be the fact that there is a very tiny probability that we never get back to them?
Best Answer
The period of a state is by definition the greatest common divisor of the length of all paths from that state to itself which have positive probability.
So yes, this chain is periodic with period 2, since the paths with positive probability from each state back to itself have length $2,4,6,\dots$. Note that it is actually not important that all multiples of the period are in here. You just need the period to be the GCD. So for example if the possible paths had length $2,6,10,\dots$ then the period would still be $2$.
Incidentally, it is a theorem that period is a class property, i.e. an invariant among all states in a communicating class. So once you know that $1$ has period $2$, you immediately also know that $2$ and $3$ have period $2$.