# Intuition on Harris recurrence

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?

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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? – Lost1 Mar 26 '13 at 13:24
@Lost1: made some changes... The wiki article was not helpful to my intuition and it has already been referred to in the question. – Bravo Mar 26 '13 at 13:37

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 definite 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?