The birth-death chain is almost surely never confined in an interval

Question

Consider the birth-death chain on $\mathbb{N} = \{0,1,2,...\}$ \begin{align} p(i,i+1) &= \alpha_i >0 \\ p(i,i-1) &= \beta_i, \qquad \beta_0 = 0\\ p(i,i) &= \gamma_i = 1-\alpha_i-\beta_i >0 \end{align} Let $a<x<b$ and $T_x = \inf\{n >0: X_n = x\}$. If $T = T_a\wedge T_b$, then how would you show that $P^x (T < \infty) = 1$? I saw this come up in Durrett in the proof of Theorem 6.4.6, but it's not clear to me how to prove it.

Answer 1

One way to show this is to dominate $T$ with a geometric random variable. Intuitively: if $m := b-a$, this chain has a positive probability of escaping the interval $[a, b]$ in any collection of $m$ steps, regardless of where inside the interval it starts. So if we consider some disjoint windows of opportunity for it to do this -- such as $[0, m], [m+1, 2m], $ and so forth, then the event of the chain escaping on those intervals is independent of any other intervals (conditionally upon it not having previously occurred). When given infinite independent chances for a rare event to occur, it will almost surely occur eventually.

Now, for some rigor: Let $p = \min_{i \in \{a, \dots, b\}} \{\alpha_i\}$, and note that $p > 0$. If $m := b-a$, then the probability that $T_b \leq m$ is at least $p^m$, corresponding to the case where the first $m$ steps of the chain are all to the right. Hence, $\mathbb P(T > m) \leq \mathbb P(T_b > m) \leq 1 - p^m < 1$. Notice that this bound applied regardless of the choice of $x \in \{a, \dots, b\}$.

On the event $T > m$ we have $X_m \in \{a, \dots, b\}$, so here's where we use the Markov property: \begin{align*} \mathbb P(T > 2m) &= \mathbb P(T > 2m, T > m) \\ &= \mathbb E_x[ \mathbb E_x[1_{T > 2m} 1_{T> m} \mid \mathcal F_m]] \\ &= \mathbb E_x[ \mathbb E_x[1_{T > 2m} \mid \mathcal F_m ]1_{T> m}] \\ &= \mathbb E_x[ \mathbb E_{X_m}[1_{T > m}]1_{T> m}] \\ &< \mathbb E_x[(1-p^m) \cdot 1_{T > m}] \\ &= \mathbb E_x[1_{T > m}] (1-p^m) \\ &< \mathbb (1-p^m)^2. \end{align*} Inductively, one can similarly show that $\mathbb P(T > k \cdot m) < (1 - p^m)^k$, and since $p > 0$ and $m$ are fixed, letting $k \to \infty$ gives the desired result.