Showing that a stopping time is finite for a biased random walk.

Question

If we consider $X_i$ iid with $\mathbb{P}(X_i=1) = p$ and $\mathbb{P}(X_i=-1)=1-p$. Where $p \in (1/2,1)$. The random walk is then given by,

$$S_n=\sum_{i=1}^n X_i. $$

We also define the stopping time $\tau = \inf\{k:S_k \in \{-\alpha,\beta\} \}$ with $\alpha,\beta>0$ in the natural numbers. How to prove that $P(\tau<\infty)=1$? Intuitively it's clear that for sure at some point $S_n$ will hit $\beta$ (because the random walk has a tendincy to move upwards.). I was thinking about showing that $P(S_n=\infty \text{ i.o.})=1$. But I'm not really sure how to make a rigorous argument. Could someone help me with this?

Answer 1

By the law of large numbers, $\frac1nS_n\to2p-1>0$ a.s. In particular, $S_n\to\infty$ a.s., and because the steps are $\pm1$ it will a.s. reach any ordinate $\beta\in\mathbb N$ (when starting from $0$).

Answer 2

Consider $\tau_\beta:=\inf\{n>0:S_n = \beta\}$. We construct the martingale $S_n-n(2p-1)$ which is s.t. $$E[S_n-n(2p-1)|\mathcal{F}_k]=S_k-k(2p-1)$$ Now by optional stopping $$E[S_{\tau_\beta \wedge n}-(\tau_\beta \wedge n)(2p-1)]=0$$ and by monotone convergence $$E[\tau_\beta]=\frac{1}{2p-1}\lim_{n \to \infty}E[S_{\tau_\beta \wedge n}]\leq \frac{\beta}{2p-1}$$ So $$P(\tau_\beta \geq k)\leq \frac{1}{k}E[\tau_\beta]\leq \frac{\beta}{k(2p-1)}\to 0\implies P(\tau_\beta = \infty )=0$$ and so $P(\tau=\infty)=0$.