# Why do all $X_0 … X_n$ have the stationary distribution?

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?

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This is what "stationary" means. – Nate Eldredge Dec 11 '12 at 4:56

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 $\pi_{t+1}=\pi$.
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