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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.

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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}

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