Prove that states accessible to an absorbing state are transient

Question

Let $X$ be a Markov chain containing an absorbing state $s$ with which all other states are accessible, in the sense that $p_{is}(n) > 0$ for some $n = n(i)$. I need to prove that that all states other than $s$ are transient.

Intuitively, this is perfectly obvious - there exist paths that will prevent the process from escaping state s. What I'm struggling with is the formal language for expressing this notion. How do I rigorously say "there are situations where there are no paths back to the non-absorbing state"?

Answer 1

If $i$ is recurrent then $P(X_n=i \, \, \text {i.o.}|X_0=i)=1$. But $P(X_n=i \, \, \text {i.o.}|X_0=i)\leq P(X_n \neq s \, \,\forall n|X_0=i)$ since $s$ is absorbing. So we get the contradiction $1<1$.

Answer 2

Consider the chain $(X_n)$ started at a state $i \ne s$ such that $P_{is}(n) > 0$ for some $n = n(i)$. Note that once the chain reaches $s$, it cannot return to $i$. Thus, $f_i = \mathbb{P}_i(X_n \text{ returns to $i$}) < 1$.

Let $V_i$ be the number of times $(X_n)$ hits $i$. By the Markov property, $\mathbb{P}_i(V_i > k) = f_i^k$. Thus, $$ \sum_{k=0}^\infty \mathbb{P}_i(V_i > k) = \sum_{k=0}^\infty f_i^k = \frac{1}{1 - f_i} < \infty . $$ The first Borel-Cantelli lemma thus tells us that $\mathbb{P}(\{ V_i > k \} \text{ infinitely often}) = 0$. (Note that we write infinitely often to mean $\mathbb{P}(V_i > k \text{ i.o.}) = \mathbb{P}\left( \bigcap_{m \geq 0} \bigcup_{k \geq m} \{ V_i > k \} \right)$.) In other words, $V_i < \infty$ with probability one $\iff$ $(X_n)$ hits $i$ finitely many times.

(See Norris' Markov Chains for a nice overview of the different characterisations of recurrence/transience of Markov chains and their relationships.)