Exercise: Stopped Martingales — Checking the One-Step Drift

Prerequisites: Stopping Times, Martingales (Discrete Time)

Problem

Let $(M_n)_{n \ge 0}$ be a martingale with respect to $(\mathcal{F}_n)$, and let $\tau$ be a stopping time. Define the stopped process $M_n^\tau := M_{n \wedge \tau}$.

1. Write the increment $M_{n+1}^\tau - M_n^\tau$ in terms of $M_{n+1} - M_n$ and an indicator. Verify that the indicator is $\mathcal{F}_n$-measurable.

2. Prove directly that $(M_n^\tau)$ is a martingale: compute $\mathbb{E}[M_{n+1}^\tau - M_n^\tau \mid \mathcal{F}_n]$ and show it is zero.

3. Concrete computation. Let $S_n$ be a symmetric random walk starting at $0$ and let $\tau = \inf\{n : |S_n| = 10\}$. Verify that the stopped process $(S_{n \wedge \tau})$ has the martingale property at times $n = 1, 2, 5, 100$ by taking expectations of both sides.
4. Where this fails: unbounded stopping times. Still with $S_n$ symmetric random walk at 0 and $\tau = \inf\{n : S_n = 1\}$, we know $\tau < \infty$ a.s. but $\mathbb{E}[\tau] = \infty$. Compute $\mathbb{E}[S_\tau]$ and $\mathbb{E}[S_0]$, and observe they differ. Explain why "the stopped process is a martingale" does not contradict this — that is, why $\mathbb{E}[M_n^\tau] = 0$ for all $n$ is consistent with $\mathbb{E}[M_\tau] \ne 0$.

Hint

For part 1: the key fact is $\mathbf{1}_{n < \tau} = \mathbf{1}_{\tau > n}$, and $\{\tau > n\}$ is the complement of $\{\tau \le n\} \in \mathcal{F}_n$.