= Stationary distribution
{title2=$\pi P=\pi$}
{wiki}
= Invariant distribution
{synonym}
A stationary distribution is a <probability distribution> $\pi$ preserved by a <Markov chain>'s evolution. For a <Discrete-time Markov chain> in the row convention,
$$
\pi P=\pi,\qquad \pi_i\geq0,\qquad\sum_i\pi_i=1.
$$
Thus $\pi$ is a normalized nonnegative <left eigenvector> of the <transition matrix> for <eigenvalue> one. Starting with $X_0$ distributed as $\pi$ makes every $X_n$ have that same <probability distribution>. For a time-homogeneous <Markov chain>, the path <probabilities> are also unchanged by shifting all observation times together: the process has <strict stationarity>. This does not mean its individual paths are constant. See https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/2edaa8542469ce692226813a8bd1ec88_MIT18_445S15_lecture2.pdf[Hao Wu's MIT lecture 2, stationary distributions].
For the sunny/rainy <transition matrix>, write $\pi=(a,1-a)$. The sunny component of $\pi P=\pi$ gives $a=0.8a+0.4(1-a)$, so $0.6a=0.4$ and
$$
\pi=\left(\frac23,\frac13\right),\qquad
\pi P=\left(\frac{8}{15}+\frac{2}{15},\frac{2}{15}+\frac{3}{15}\right)=\pi.
$$
Moreover,
$$
\pi_SP_{SR}=\frac23\frac15=\frac{2}{15}=\frac13\frac25=\pi_RP_{RS}.
$$
These equal stationary flows verify <detailed balance> and show that the <Markov chain> is <reversible>. The diagonal <detailed balance equations> hold automatically. Only $1/3$ of the stationary days are rainy, but $0.6$ of the days following a rainy day are rainy: <stationarity> permits dependence.
A <finite Markov chain> that is an <irreducible Markov chain> has a unique <stationary distribution>. If it is also an <aperiodic Markov chain>, $\mu_0P^n\to\pi$ for every initial <probability distribution>. The sunny/rainy example satisfies both conditions: both directions are possible and each state has a positive <holding probability of a Markov chain>. See https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/5bd4774f613aa28bd111c36f3a3beeed_MIT18_445S15_lecture4.pdf[Hao Wu's MIT lecture 4, convergence theorem]. For an infinite <state space>, the additional <positive recurrence> hypothesis in the <countable-state Markov chain convergence theorem> matters.
Existence of a <stationary distribution> alone does not give this convergence. Consider the deterministic two-cycle
$$
C=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad\pi=\left(\frac12,\frac12\right).
$$
Here $\pi C=\pi$, but starting in the first state gives alternating <probability distributions> $(1,0),(0,1),(1,0),\ldots$. Its <period of a state in a Markov chain> is two. The running fraction of visits to either state nevertheless tends to $1/2$, illustrating the <Markov chain ergodic theorem>. Starting with $\pi$ instead keeps the one-time <probability distributions> stationary even though every path alternates.
\Image[/markov-periodicity.png]
{title=A deterministic two-cycle has a stationary law, but its one-time probabilities oscillate while its time-averaged visits converge.}
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