Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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 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?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.