Let $\mathbb{S}$ be countable and $X_n$ be the coordinate map on $\mathbb{S}^\mathbb{N}$. Let $p$ be a transitional probability so that $X_n$ is a irreducible transient Markov chain. Let $\tau_A = \inf \{n \ge 0: X_n\in A\}$ be the first time it hits $A$.

Let $A,B \subset \bar{\mathbb{S}}$ be disjoint where $\bar{\mathbb{S}} = \mathbb{S}\cup\partial \mathbb{S}$ and $\partial \mathbb{S}$ is the martin boundary. Suppose that $P^x (\tau_A <\tau_B <\infty) > 0$, does it mean that there is a positive probability that $X_n$ will oscillate between $A$ and $B$, indefinitely? More rigorously, if $\tau_B^0 = -1$ and \begin{align} \tau_A^k &= \tau_B^{k-1} + \tau_A \circ \theta^{\tau_B^{k-1}} \\ \tau_B^k &= \tau_A^k + \tau_B \circ \theta^{\tau_A^k} \end{align} Then is it true that $P^x(\tau_A^k,\tau_B^k < \infty, k \ge 1) > 0$?

EDIT: I'm actually trying to form a contradiction. Since $X_n \rightarrow X_\infty$ where $X_\infty:\mathbb{S}^\mathbb{N} \rightarrow \partial \mathbb{S}$, it cannot oscillate indefinitely and thus $P^x (\tau_A <\tau_B <\infty) = 0$

  • $\begingroup$ Since it is transient and irreducible, it will hit both A and B infinitely many times, yes? $\endgroup$ – jdods May 5 '18 at 17:44
  • $\begingroup$ Why is that? If $A,B \subset \mathbb{S}$ were both finite, then by transience and irreducibility, it must leave $A,B$ after finite steps. $\endgroup$ – Andrew Yuan May 5 '18 at 18:21
  • $\begingroup$ But irreducibility means each state is accessible from every other state, yes? So there is always a positive probability of going from A to B and vice versa, so this will happen infinitely many times both ways (in addition to going to every other state infinitely many times). $\endgroup$ – jdods May 5 '18 at 23:06
  • $\begingroup$ But the probability could go to 0 after infinite many times (the limit is 0 even though each finite term may be positive). How would you show that the probability is uniformly bounded below by some positive number? $\endgroup$ – Andrew Yuan May 6 '18 at 6:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.