# Confusion in the proof of properties for $\psi$-irreducibility

Let $P$ be a stochastic kernel on a measurable space $(\mathsf X,\mathfrak B(\mathsf X))$. The kernel $P$ is called $\varphi$-irreducible if for a positive measure $\varphi$ and for all measurable sets $A$ it holds that: $$\tag{1}\varphi(A)>0 \quad \Rightarrow \quad \sum\limits_{n\geq 1}P^n(x,A) >0 \quad \forall x\in \mathsf X.$$

One of the statements of Proposition 4.2.2 (p. 90 here) is that if $P$ is $\varphi$-irreducible, then it holds that $P$ is $\psi$-irreducible where the measure $\psi$ is given by $$\psi(A) = \int\limits_\mathsf X \sum_{n\geq 0}2^{-(n+1)}P^n(x,A)\varphi(\mathrm dx).$$ The definition of $\psi$ is not that important for my question, though.

In the first part of the proof, also page 90, the following is stated

To see (i), observe that when $\psi(A)>0$, then [...] $$\tag{2} \left\{y:\sum\limits_{n\geq 1}P^n(y,A)>0\right\} = \mathsf X.$$

This fact is further elaborated and used to show the $\psi$-irreducibility of $P$. It seems to me, however, that the cited part explicitly implies irreducibility as it is equivalent to $(1)$. I guess, I am missing something - otherwise it is a cyclic argument. Also, I don't know how to show $(2)$.

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How is Markov chain splitting technique useful for inferring ergodicity of a Markov Chain?Assume that I am working with general state space (uncountable say $R^{N}$ but time is discrete. I want to show that the Markov Process is ergodic. I guess that it suffices to show that it is Harris recurrent. To show Harris recurrence I guess that it suffices to show there exists an atom (obtained via splitting the chain after using minorization criteria) the return time (or hitting time) to which has finite mean. – user24367 Jun 18 '13 at 13:45
@user17523: I think, it is better if you ask a separate question. I'm not an expert in the book by Meyn and Tweedie, and I didn't read much of their chapters on ergodicity. I remember splitting technique, but I haven't reached the part where it's used. – Ilya Jun 19 '13 at 7:47
Hi, I already posted it as separate question, but also posted here since I saw a similar topic being discussed here. – user24367 Jun 19 '13 at 18:32

I agree with you that $(2)$ is equivalent to $\psi$-irreducibility and that it is not clear how the authors derive $(2)$ from the given assumptions.

In the proof, the key point is that $\varphi(\bar{A}(k))>0$ for $k$ sufficiently large where

$$\bar{A}(k) := \left\{y; \sum_{n=1}^k P^n(y,A) >k^{-1} \right\}.$$

Here is possibility to prove this (without using $(2)$):

Set

$$K_{\frac{1}{2}}(y,A) := \sum_{n \geq 1} P^n(y,A) 2^{-(n+1)}.$$

It follows from the very definition of $\psi$ that

\begin{align*} \psi(A) &= \int_{\bar{A}(k)} K_{\frac{1}{2}}(y,A) \, \varphi(dy) + \int_{X \backslash \bar{A}(k)} K_{\frac{1}{2}}(y,A) \, \varphi(dy). \end{align*}

By the definition of $\bar{A}(k)$, we have

\begin{align*} \int_{X \backslash \bar{A}(k)} K_{\frac{1}{2}}(y,A) \, \varphi(dy) &= \int_{X \backslash \bar{A}(k)}\sum_{n=1}^k P^n(y,A) 2^{-(n+1)} \, \varphi(dy) + \int_{X \backslash \bar{A}(k)}\sum_{n=k+1}^{\infty} P^n(y,A) 2^{-(n+1)} \, \varphi(dy) \\ &\leq \int_{X \backslash \bar{A}(k)}\sum_{n=1}^k P^n(y,A) \, \varphi(dy) + \sum_{n=k+1}^{\infty} 2^{-(n+1)} \\ &\leq k^{-1} + \sum_{n=k+1}^{\infty} 2^{-(n+1)} \end{align*}

where we have used that $\varphi(A) \leq 1$ for any measurable set $A$. Combining both computations, we get

$$\int_{\bar{A}(k)} K(y,A) \, \varphi(dy) \geq \psi(A) - k^{-1} - \sum_{n=k+1}^{\infty} 2^{-(n+1)}.$$

Since, by assumption, $\psi(A)>0$, this shows that

$$\int_{\bar{A}(k)} K(y,A) \, \varphi(dy) >0$$

for $k$ sufficiently large. This implies $\varphi(\bar{A}(k))>0$. (Just suppose that $\varphi(\bar{A}(k))=0$, then the left-hand side in the previous equation would equal $0$.) The remaining part of the proof given in the (linked) book goes through.

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I think the "$= \mathsf{X}$" portion is just a typo. If you drop it, everything works fine. I can see that $\bar{A}(k) \uparrow \bar{A}$, but I can't show $\bar{A} = \mathsf{X}$.

To show the "if $\psi(A) > 0$, then $\varphi(\bar{A}(k)) > 0$ for some $k$", use proof by contradiction. If $\varphi(\bar{A}(k)) = 0$ for all $k$ then $\varphi(\bar{A}) = 0$ where $\bar{A} = \{y : \sum_{n=0}^{\infty}P^n(y,A)2^{-(n+1)} > 0\}$. If $\phi(\bar{A}) = 0$, then $\psi(A) = 0$ using its definition, and this contradicts our assumption that $\psi(A) > 0$. Basically when you integrate over $\mathsf{X}$, the positive bits of $\varphi$ and $K_{a_{\frac{1}{2}}}$ are never going to match up anywhere.

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