I am trying to intuitively reconcile the following statement, read from "Probability, Markov Chains, and Queues":
A Markov Chain may possess a stationary distribution but not a limiting distribution.
This is unintuitive to me. I have written down 4 defenitions/facts that I know that I am trying to use:
1) $\pi$ is a limiting distribution of a Markov Chain with transition matrix $P$ if, for some initial distribution $P(0)$, $\pi = P(0)lim_{n \rightarrow \infty}P^{(n)}$. The elements of $\pi$ need not sum to 1.
2) If, for all valid starting distributions $P(0)$, $P(0)lim_{n \rightarrow \infty}P^{(n)} = \pi$, where $\pi$ is a vector of positive reals summing to 1, then $\pi$ is a steady-state distribution.
3) If a Markov Chain has a steady-state distribution, then it is also the unique stationary distribution.
4) A stationary distribution is a vector $\pi$ of positive reals summing to 1 satisfying $\pi = \pi P$.
So the original statement in question is that there is some vector $\pi$ satisfying (4) for some Markov Chains, but not 1. But fact 2 means that steady state distributions are a subset of limiting distributions, and fact 3 means that steady state distributions are stationary distributions, so how can you have a stationary distribution but not a limiting distribution? Where is my logic wrong?
EDIT: after thinking more, if the statement is correct, the chain has some $\pi = \pi P$ (4) but it is NOT true that the same $\pi$ satisfies $\pi = \pi lim_{n \rightarrow \infty}P^{(n)}$, or else $\pi$ would also be a limiting distribution. I guess this means this chain has some kind of fluctuating P matrix when raised to powers. Maybe related to periodicity.
Best Answer
The usual definition of limiting distribution is that a Markov chain has a limiting distribution $\pi$ if for every initial distribution $P(0)$,
$ \lim_{n \rightarrow \infty} P(0)P^{(n)}=\pi $
It's important to get that quantifier right.
A useful example to consider is the Markov chain with
$ P=\left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]. $
Here, $\pi=[1/2 \;\;1/2]$ is a stationary distribution but not a limiting distribution of the Markov chain. In fact this Markov chain does not have a limiting distribution.