MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Irreducibility and minorization condition

Let $P$ be a stochastic kernel on a measurable space $(E,\mathscr E)$ and $G$ be its potential: $$G = \sum\limits_{n=0}^\infty P^n.$$

• The kernel $P$ is called irreducible if there exists a probability measure $\varphi$ on $(E,\mathscr E)$ such that $$\varphi(A>0)\Rightarrow PG(x,A)>0 \text{ for all }x\in E. \tag{1}$$ Any probability measure satisfying $(1)$ is called an irreucibility measure for $P$.

For each irreducible kernel there is a maximal irreducibility measure $\psi$ such that $P$ is $\psi$-irreducible and if $P$ is $\varphi$-irreducible, $\varphi\ll \psi$. Clearly, it is defined only up to the measure equivalence. One example is $$\psi = \sum\limits_{n=0}^\infty 2^{-n}(\varphi P^n)$$ where $\varphi$ is any irreducibility measure.

• The irreducible kernel $P$ is said to satisfy a minorization condition if there is an integer $m\geq 1$, a probability measure $\nu$ on $(E,\mathscr E)$ and a function $0\leq s(x)\leq 1$ such that $\langle\psi,s\rangle>0$ and $$P^m(x,A)\geq s(x)\nu(A) \text{ for all }x\in E,A\in\mathscr E. \tag{2}$$ In that case the measure $\nu$ is called a small measure and the function $s$ - a small function and $\psi$ was standing for the maximal irreducibility measure of $P$.

It can be proved that for any irreducible kernel satisfies some minorization condition. Moreover, In the book by Numellin it is said that $\nu$ is always an irreducibility measure, so that without loss of generality, we can choose $s$ such that $\langle \nu,s\rangle >0$.

• Let us say that $P$ satisfies a relaxed minorization condition if for some $m\geq 1$, probability measure $\nu$ and a function $0\leq s\leq 1$ such that $\langle \nu,s\rangle >0$ holds $(2)$.

I guess that the relaxed minorization condition is not sufficient for the irreducibility of $P$: for example we can consider a 2-state Markov Chain with states being absorbing and take both $\nu,s$ be equal to $1$ over first state and $0$ over another.

On the other hand, maybe there is another kind of equivalence between minorization conditions and irreducibility?

Update: I've found that the relaxed minoriztion condition implies that $\nu$ is a communicating measure on the set $$B = (x\in E:Gs(x) > 0).$$ Moreover, the set $B^c$ is absorbing. Still I wonder if there is any other connection between the irreducibility and minorization.

Just to clarify things: kernel $P$ means that $P(x,\cdot)$ is a probability measure on $(E,\mathcal E)$ for all $x$ and $P(\cdot, A)$ is a measurable function for all $E$. The product of kernels, say $P$ and $G$ is $$PG(x,A) = \int\limits_E G(y,A)P(x,\mathrm dy)$$ and we use $$\varphi P(A) = \int\limits_E P(y,A)\varphi(\mathrm dy)$$ $$\langle \nu,s\rangle = \int\limits_E s(x)\nu(dx).$$

-
 Isn't $s\nu$ an irreducibility measure ? – Yuri Bakhtin May 4 2012 at 0:46 @Yuri: what did you mean by $s\nu$? In the case you meant $(s\otimes \nu)(x,A)=s(x)\nu(A)$, it is a kernel rather than a measure. – Ilya May 4 2012 at 7:55