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.

  • $\begingroup$ Is the definition in (1) really what you mean? (That the limit exists for some initial distribution?) Take a stationary distribution $\pi$ from (4) and use this as the initial distribution for your chain. Then $\pi = \pi P^n$, i.e. $\pi$ is automagically a "limiting distribution". $\endgroup$ May 13, 2015 at 16:13
  • $\begingroup$ Edited my question by adding a note. $\endgroup$
    – Tommy
    May 14, 2015 at 3:18
  • $\begingroup$ Take the equality $\pi = \pi P$. Right multiply by $P$. You get $\pi P = \pi P^2$. But $\pi P = \pi$. Thus $\pi = \pi P^2$. Repeating gives $\pi = \pi P^n$, for any $n \ge 1$. Take the limit $n\to\infty$ in both sides... $\endgroup$ May 14, 2015 at 9:28
  • $\begingroup$ Are you claiming the statement is false? $\endgroup$
    – Tommy
    May 14, 2015 at 15:12
  • $\begingroup$ Yes, with your definitions. I agree with @Brian Borchers below. $\endgroup$ May 14, 2015 at 17:37

1 Answer 1


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.

  • $\begingroup$ Brian, so then the only difference between a "limiting distribution" and a "steady state distribution" is that steady state must be a valid probability distribution whereas limiting need not sum to 1? $\endgroup$
    – Tommy
    May 15, 2015 at 17:13
  • $\begingroup$ In my experience, "limiting distribution" and "steady state distribution" are synonyms, and both assume that $\pi$ sums to 1. $\endgroup$ May 15, 2015 at 19:11
  • $\begingroup$ The book I mention above gives several examples of limiting distributions that do not sum to 1. $\endgroup$
    – Tommy
    May 15, 2015 at 20:02

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