Solution: Stopped Martingales — Checking the One-Step Drift

Exercise: Stopped Martingales: Checking the One-Step Drift

Part 1

M_{n+1}^\tau - M_n^\tau = M_{(n+1)\wedge\tau} - M_{n\wedge\tau} = \mathbf{1}_{\tau > n}(M_{n+1} - M_n).

The indicator $\mathbf{1}_{\tau > n} = 1 - \mathbf{1}_{\tau \le n}$ . Since $\tau$ is a stopping time, $\{\tau \le n\} \in \mathcal{F}_n$ , so its complement is also in $\mathcal{F}_n$ . Hence $\mathbf{1}_{\tau > n}$ is $\mathcal{F}_n$ -measurable.

Part 2

\mathbb{E}[M_{n+1}^\tau - M_n^\tau \mid \mathcal{F}_n] = \mathbf{1}_{\tau > n}\cdot \mathbb{E}[M_{n+1} - M_n \mid \mathcal{F}_n] = \mathbf{1}_{\tau > n}\cdot 0 = 0.

First equality: the indicator is $\mathcal{F}_n$ -measurable, pull it out. Second: martingale property of $(M_n)$ . So $(M_n^\tau)$ is a martingale. ✓

Part 3

$S_n$ is a martingale, so $(S_{n \wedge \tau})$ is a martingale by part 2, meaning $\mathbb{E}[S_{n \wedge \tau}] = \mathbb{E}[S_0] = 0$ for every $n$ . Quick Monte Carlo check:

import numpy as np

rng = np.random.default_rng(0)
N = 200_000
max_n = 200

increments = rng.choice([-1, 1], size=(N, max_n))
S = np.concatenate([np.zeros((N, 1)), np.cumsum(increments, axis=1)], axis=1)
abs_S = np.abs(S)

# tau = first n such that |S_n| = 10
tau = np.argmax(abs_S >= 10, axis=1)       # argmax returns first True; 0 if never
hit_mask = (abs_S >= 10).any(axis=1)
tau[~hit_mask] = max_n                     # clip those that didn't hit within window

for n in [1, 2, 5, 100]:
    S_stopped = np.array([S[i, min(n, tau[i])] for i in range(N)])
    print(f"n={n:3d}:  E[S_{{n∧τ}}] = {S_stopped.mean():+.4f}")
# n=  1:  E[S_{n∧τ}] = +0.0016
# n=  2:  E[S_{n∧τ}] = -0.0014
# n=  5:  E[S_{n∧τ}] = -0.0009
# n=100:  E[S_{n∧τ}] = +0.0024

All four close to zero, confirming the martingale property of the stopped process at every fixed finite time.

Part 4

At $\tau = \inf\{n : S_n = 1\}$ (symmetric walk starting at $0$ ), we have $S_\tau = 1$ a.s. (on the event $\tau < \infty$ , which has probability $1$ by recurrence of the walk). So $\mathbb{E}[S_\tau] = 1 \ne 0 = \mathbb{E}[S_0]$ .

Why this is consistent with part 2. Part 2 says

\mathbb{E}[S_{n \wedge \tau}] = 0

for every fixed $n$ . This does not imply

\mathbb{E}[S_\tau] = \lim_{n \to \infty}\mathbb{E}[S_{n \wedge \tau}] = \mathbb{E}[\lim S_{n \wedge \tau}] = \mathbb{E}[S_\tau]

— the interchange of limit and expectation requires an integrability condition (uniform integrability, dominated convergence, or monotone convergence).

For this example the interchange fails:

S_{n \wedge \tau}

converges a.s. to

S_\tau = 1

, but it is not uniformly integrable. At large

n

the conditional distribution of

S_{n \wedge \tau}

\{n < \tau\}

is skewed far to the negative side (waiting for a hit of

+1

after having been driven deep negative), so the positive contribution from

\{\tau \le n\}

(value

+1

) is cancelled by the negative tail — only in the limit does the negative tail escape to

-\infty

with vanishing probability but infinite negative mass, and the mean jumps from

0

1

Moral: the stopped process is always a martingale, but

\mathbb{E}[M_\tau]

need not equal

\mathbb{E}[M_0]

unless you have an integrability condition. The optional stopping theorem supplies precisely the conditions under which the interchange is valid.

Takeaways

Stopped martingales are always martingales — no integrability condition needed. This is the clean structural fact.
$\mathbb{E}[M_\tau] = \mathbb{E}[M_0]$ requires an extra condition — bounded stopping time, bounded martingale, or uniform integrability. Otherwise the interchange of limit and expectation fails.
The symmetric walk's hit-1 stopping time is the canonical failure of naive optional stopping: $\mathbb{E}[\tau] = \infty$ , the martingale hits $+1$ a.s., but the expected value at stopping exceeds the starting value.
Monte-Carlo sanity checks. In a trading simulator, stop-at-strategy profit calculations often implicitly invoke optional stopping. Verify the integrability condition holds (bounded horizon, bounded bankroll) before trusting the expected P&L.