# Unbounded stopping time

Suppose we have a sequence of i.i.d. random variables $(X_n)_{n \in \mathbb{N}}$ with $\mathbf{P}(X_n = -1) = \frac{1}{2}, \mathbf{P}(X_n = 0) = \frac{1}{3}, \mathbf{P}(X_n = 1) = \frac{1}{6}$. Denote $\tau = \inf\{n > 0 : X_1 + \ldots + X_n = 1\}$. Show that $\mathbf{P}(\tau = \infty) = \frac{2}{3}$.

As a matter of fact, I have no idea how to approach this question. In the same task I had to prove that $(3^{X_1 + \ldots + X_n}, \sigma(X_1, \ldots, X_n))$ is a martingale but I don't see how it helps in any way to solve the main question, because our stopping time is not bounded.

• $\min(\tau,k)$ is a bounded stopping time and your martingale stopped at $\tau$ is bounded by 3, so use can use bounded/dominated convergence to pass to the limit to see that $1=\mathbb{E}[M_{n\wedge\tau}]=3\times\mathbb{P}[\tau < \infty] + 0\times\mathbb{P}[\tau = \infty]$ and the result follows. – user180850 Jan 17 '15 at 19:10
• When a stopping time $\tau$ is unbounded or even infinite with positive probability, one can always consider the bounded stopping times $\tau_n=\inf\{\tau,n\}$, apply the stopping theorem to each $\tau_n$ and see what happens in the limit $n\to\infty$. – Did Jan 17 '15 at 19:10

1. The stopped process $M_n := 3^{X_1+\ldots+X_{n \wedge \tau}}$ is also a martingale; hence, $$\mathbb{E}M_n = \mathbb{E}M_1.$$ Calculate $\mathbb{E}M_1$.
2. By the strong law of large numbers, $$\frac{X_1+\ldots+X_n}{n} \to \mathbb{E}X_1 = - \frac{1}{3}.$$ Thus, $$X_1+\ldots+X_n \to - \infty \quad \text{almost surely as} \, \, n \to \infty.$$
3. Conclude that \begin{align*} M_n &= 3^{X_1+\ldots+X_n} 1_{\{ \tau=\infty\}} + 3^{X_1+\ldots+X_{n \wedge \tau}} 1_{\{\tau<\infty\}}\\ &\to 0 + 3^1 1_{\{\tau<\infty\}}. \end{align*}
4. Deduce from step 1 and 3 that $$\mathbb{P}(\tau<\infty) = \frac{1}{3}.$$