Invariant and stationary distribution for a Markov kernel

Let $K$ be a probability kernel on a measurable space $(E,\mathcal{E})$ (Measure kernel and probability kernel), and let $\pi$ be a probability measure on $(E,\mathcal{E})$ (Probability measures and probability spaces). The probability measure $\pi$ is invariant for $K$, and $\pi$ is a stationary distribution of $K$, when

$(\pi K)(A):={\int }_{E}K(x,A)\pi (dx)=\pi (A)\text{for every }A\in \mathcal{E}\mathrm{.}$

The set function $\pi K$ is again a probability measure: for fixed $A$ the map $x\mapsto K(x,A)$ is measurable and lies in $[0,1]$, so the integral exists in $[0,1]$; for pairwise disjoint $A_n\in \mathcal{E}$ the identity $K(x,{\bigcup }_n{A}_n)={\sum }_{n}K(x,A_n)$ of the measure $K(x,\cdot )$ and the monotone convergence theorem for nonnegative functions (Monotone convergence for the integral) give $\sigma$-additivity of $\pi K$; and $K(x,E)=1$ for every $x$ gives $(\pi K)(E)=1$. Thus $(\pi K)(A)=\pi (A)$ is an identity between two probability measures evaluated at $A$.

A $K$-chain whose initial law is $\pi$ is called stationary when $\pi K=\pi$. Stationarity of the process is a statement about all of its finite-dimensional laws and is proved, not assumed, from $\pi K=\pi$; see Invariant initial law makes a Markov chain stationary.

On a countable state space $E$ with sigma-algebra $2^E$ and transition matrix $p(x,y)=K(x,\{y\})$ (Transition matrices and n-step probabilities), a measure on $E$ is its weighted sum of Dirac masses at singletons (Every measure on a countable discrete space is its weighted sum of Dirac measures), so invariance is equivalent to the matrix identity

$\pi (y)={\sum }_{x\in E}\pi (x)p(x,y)\text{for every }y\in E\mathrm{.}$

The definition fixes no irreducibility, aperiodicity or uniqueness hypothesis, and it selects no conditional-expectation versions; the display defines a single measure $\pi K$ and asserts an identity for it. A transient, periodic or reducible chain may have several invariant probability measures or none.