I am trying to get some intuition on Harris recurrence in Markov chains. Define state space $\mathcal S$ comprising a single communication class, $f_{ii}^{(n)}=P(X_n=i, X_{n-1}\ne i,\ldots X_1\ne i\mid X_0=i)$, $f_{ii}=\sum_n f_{ii}^{(n)}$, $T_{ii}=\inf_n \{X_n=i\mid X_0=i\}$ and $E(T_{ii})=\sum_n nf_{ii}^{(n)}$ and $V_i=\sum_n\mathbb 1_{X_n=i}$, we have the following.

  • Transience: $f_{ii}<1$
  • Null recurrence:$f_{ii}=1$, $E(T_{ii})=\infty$
  • Positive recurrence: $f_{ii}=1$, $E(T_{ii})<\infty$, $E(V_i)=\infty\ \forall i\in \mathcal S$
  • Harris recurrence: $f_{ii}=1$, $P(\omega:V_i(\omega)=\infty)=1\ \forall i\in \mathcal S$

Are the above relations correct? I do not see how the last bullet relates to the definition in Wikipedia. Are there any examples of finite Markov chains that are positive but not Harris recurrent?

  • $\begingroup$ First question: what is the state space of your Markov Chain? I am assuming countable, because you are using summation over i. why don't you check out en.wikipedia.org/wiki/Harris_chain? Also what you have written as defintion of harris recurrence seem to be the same as null recurrence to me. Correct me if I am wrong. If $f_{ii}=\infty$, this implies that $P(T_{ii}<\infty)=1$, right? $\endgroup$
    – Lost1
    Mar 26, 2013 at 13:24
  • $\begingroup$ @Lost1: made some changes... The wiki article was not helpful to my intuition and it has already been referred to in the question. $\endgroup$
    – Bravo
    Mar 26, 2013 at 13:37

3 Answers 3


To the best of my understanding, the Harris recurrence is a strengthened form of recurrence for continuous-state Markov chains. Let $\mathbb{X}$ be the continuous state space and let $A$ be a measurable subset of $\mathbb{X}$. Let $X_k$ be a Markov chain on $\mathbb{X}$ and let $\eta_A = \sum_{k=1}^\infty \mathbb{I}(X_k \in A)$ count the number of times the chain visits the set $A$. If we take $\mathbb{E}_x$ denote the expectation given the chain starts at $X_0 = x$, the set $A$ is $\mu$-irreducible if \begin{equation*} \forall x \in A \hspace{15pt} \mathbb{E}_x(\eta_A) = \infty \end{equation*} Harris recurrence on the other hand requires: \begin{equation*} \forall x \in \mathbb{X} \hspace{15pt} \mathbb{P}_x(\eta_A = \infty) = 1 \end{equation*} The difference between the two notions is (1) the starting position of the chain and (2) the strength of the 'recurrence condition'. Harris recurrence is stronger as it requires that you visit the chain infinitely often with probability 1 no matter where you start from in $\mathbb{X}$, whereas normal recurrence only requires that if you start in $A$ then you will return to $A$ infinitely often on average. Note that you can return infinitely often on average even if you only return infinitely often with positive probability, rather than with the probability 1 Harris recurrence calls for.


This answer may be wrong, but I think it is worth posting and if it is wrong, someone can point it out and I can learn something too.

I think you do not mean a finite Markov Chain, because for a finite state chain, assuming it is irreducible, every state will be visited infinitely often, there is no question.

I think there is only a difference if the state space is uncountable.

This is because the event $V_i=\infty$ is the same as the event "state i is visited infinitely often". This has probability 1 or 0, by Levy's zero-one law.

So, suppose a positive recurrent chain is not Harris recurrent. This means the expected number of visits to $i$ is infinite, but the number visits to $i$ is finite, almost surely, but doesn't this mean it is transient?

  • $\begingroup$ Going from $P(V_i=\infty)=1$ for "Harris recurrent" to $P(V_i<\infty)=1$ for "not Harris recurrent" does not sound like a logical step. $\endgroup$ Nov 17, 2020 at 16:04

A Markov chain $X = (X_0,X_1,\dots)$ is $\pi$-irreducible if for all $B$ such that $\pi(B)>0$, we have

\begin{align} &P_x(\cup_{n=1}^\infty (X_n \in B))> 0 && \mbox{for all}\, x. \end{align}

If the chain $X$ has stationary distribution $\pi$, it is recurrent if for all $B$ such that $\pi(B)>0$:

\begin{align} &P_x(X_n \in B, \mathrm{i.o.}) > 0 && \mbox{for all}\, x.\\ &P_x(X_n \in B, \mathrm{i.o.}) = 1 && \mbox{for $\pi$-almost all}\, x. \end{align}

The chain $X$ is Harris recurrent if

\begin{align} &P_x(X_n \in B, \mathrm{i.o}) = 1 && \mbox{for all}\, x. \end{align}


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