Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say I have an irreducible Markov chain with state space $\{1, 2, 3 ... m\}$, where $m > 2$ and stationary distribution $s = \{s_1, s_2, ... s_m\}$. The initial state is given by the stationary distribution, so $P(X_0 = i) = s_i$.

Why is it that all of $X_0, X_1 ... X_n$ have the stationary distribution? Apparently, it's because of how $X_0$ does... how does that work?

share|cite|improve this question
This is what "stationary" means. – Nate Eldredge Dec 11 '12 at 4:56
up vote 1 down vote accepted

This is because, by definition, the stationary distribution satisfies $\pi P = \pi$ where $\pi$ is the stationary distribution (row vector) and $P$ is the matrix of transition probabilities.

Remember that at time $t$, $$\pi_{t} P = \pi_{t+1}$$ where $\pi_\tau$ is the probability at time $\tau$. This means if you set $\pi_t=\pi$ you end up with $\pi_{t+1}=\pi$.

share|cite|improve this answer
how can we ensure that the initial distribution $X_0$ will eventually becomes $\pi$ after some steps. – Mathematics Dec 11 '12 at 4:53
@Mathematics That is not the question here. The question is, assume you start with $\pi$, the distribution of $X_0$, why is it then that, at the next step, the distribution of $X_1$ is still $\pi$? Iterating, $X_2,...,X_n$ will all have (marginally) distribution $\pi$. – Learner Dec 11 '12 at 4:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.