# Escape speed of random walk on positive integers

Let $$\{X_i\}$$ be the process with transition matrix $$p(X_{i + 1} = \ell | X_{i} = k) = \begin{cases} q & \ell = k + 1 \quad \text{and} \quad k > 0 \\ 1 - q & \ell = k - 1 \quad \text{and} \quad k > 0 \\ 1 & \ell = 1 \quad \text{and} \quad k = 0 \\ 0 & \text{otherwise} \end{cases}$$ where $$X_0 = 0$$ and $$1/2 < q \leq 1$$. This is a 1D biased random walk with a reflecting boundary at $$0$$. My question is

How to show that $$\lim_{n \to \infty} X_n/n = 2q - 1$$ a.s.?

If there was no reflecting boundary, the increments $$\Delta X_{i} = X_{i} - X_{i - 1}$$ would be i.i.d. and, since $$E[\Delta X_{i}] = 2q - 1$$, the limit follows immediately from the strong law. But in this case the increments aren't quite i.i.d. because sometimes the process might be at $$0$$. My thoughts are that, because there will be a last visit to $$0$$ almost surely, there must be some notion of the increments being "effectively" i.i.d. in the limit but I'm not sure how to formalize that.

I think I have it. It essentially follows from \begin{align} \lim_{n \to \infty} P(X_n > 0) &= 1 \\ \lim_{n \to \infty} P(X_n = 0) &= 0 \end{align} By Kingman's theorem, $$\lim_{n \to \infty} X_n/n = \lim_{n \to \infty} E[X_n]/n = c$$ a.s. for a $$c > 0$$. Therefore, letting $$\Delta X_n = X_{n} - X_{n - 1}$$ gives \begin{align} E[\Delta X_n] &= E[\Delta X_n| X_{n - 1} = 0] P(X_{n - 1} = 0) + E[\Delta X_n| X_{n - 1} > 0] P(X_{n - 1} > 0) \\ &= P(X_{n - 1} = 0) + (2q - 1) P(X_{n - 1} > 0) \end{align} Therefore, \begin{align} \lim_{n \to \infty} \frac{X_n}{n} &= \lim_{n \to \infty} \frac{1}{n}\bigg[P(X_{n - 1} = 0) + (2q - 1) P(X_{n - 1} > 0) \bigg] \quad \text{a.s.} \\ &= 2q - 1 \quad \text{a.s.} \end{align}