= Stationary distribution existence for a finite Markov chain
{title2=$\exists\pi\geq0:\ \pi P=\pi,\quad\pi\mathbf1=1$}
Every <finite Markov chain> has a <stationary distribution>, without requiring <Markov chain irreducibility> or <aperiodicity>. For a row-<stochastic matrix> $P$, put $B=P^T$ and choose a column <probability vector> $q$. Every $B^jq$ lies in the <probability simplex>. Their arithmetic averages $x_N=N^{-1}\sum_{j=0}^{N-1}B^jq$ therefore lie in this <compact set>, while telescoping gives
$$
Bx_N-x_N=\frac{B^Nq-q}{N},\qquad \|Bx_N-x_N\|_1\leq\frac2N.
$$
A <convergent subsequence> has a <limit> $x$ in the <probability simplex>. <Continuity> of <matrix multiplication> gives $Bx=x$, so $\pi=x^T$ is a <stationary distribution>. The individual iterates need not converge: a deterministic two-state swap alternates its point masses forever.
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