Solution: Where OST Fails — The Symmetric Walk Hits One

Exercise: Where OST Fails: The Symmetric Walk Hits One

Part 1

On $\{\tau < \infty\}$ (probability $1$ ), $S_\tau = 1$ . So $\mathbb{E}[S_\tau] = 1$ .

Part 2

$\mathbb{E}[S_0] = 0$ .

Part 3

$\mathbb{E}[S_\tau] = 1 \ne 0 = \mathbb{E}[S_0]$ . OST fails. Check each hypothesis:

$\tau$ bounded? No — there is no deterministic $K$ with $\tau \le K$ a.s., because $\tau$ can be arbitrarily large.
$S_n$ bounded on $\{n \le \tau\}$ ? No — before hitting $+1$ the walk can wander arbitrarily far into the negative integers. The stopped process $(S_{n \wedge \tau})$ is unbounded.
$\mathbb{E}[\tau] < \infty$ with bounded increments? The increments are bounded ( $|X_i| = 1$ ), but $\mathbb{E}[\tau] = \infty$ — a classical result for the symmetric walk. So condition (3) also fails.

The violation is specifically: the stopped process $(S_{n \wedge \tau})$ is not uniformly integrable. As $n \to \infty$ , the distribution of $S_{n \wedge \tau}$ develops a heavy left tail that carries increasingly negative mean, which (miraculously) exactly cancels the $+1$ contribution from the event $\{\tau \le n\}$ at every finite $n$ — yielding $\mathbb{E}[S_{n \wedge \tau}] = 0$ at every $n$ . Only in the infinite- $n$ limit does the negative tail "escape to $-\infty$ " while its probability vanishes, and the mass concentrated at $+1$ is what remains.

Part 4

import numpy as np
import matplotlib.pyplot as plt

rng = np.random.default_rng(0)
N = 10_000
T_max = 100_000

# Simulate paths; stop the first hit of +1 or the budget.
tau = np.full(N, np.nan)
for i in range(N):
    S = 0
    for t in range(1, T_max + 1):
        S += int(rng.choice([-1, 1]))
        if S == 1:
            tau[i] = t
            break

hit_rate = np.sum(~np.isnan(tau)) / N
print(f"Paths that hit within {T_max} steps: {hit_rate:.2%}")
# Paths that hit within 100000 steps: 98.35%

plt.hist(np.log10(tau[~np.isnan(tau)]), bins=40)
plt.xlabel('log10(tau)')
plt.ylabel('count')
# Long right tail; median ~ 10^1, but 5% of paths took 10^4 or more.

What the plot shows. The histogram of

\log_{10}\tau

has a long right tail. The median

\tau

is around 10 steps, but a small fraction of paths take 10,000 or more steps — and 1.65% had not yet hit

+1

after 100,000 steps.

How this relates to OST failure. Because

\mathbb{E}[\tau] = \infty

, we have:

\sum_{t=1}^\infty t\cdot \mathbb{P}(\tau = t) = \infty.

The mean is dominated by the far tail, which cannot be captured by any finite-horizon simulation. For any bounded horizon $T_{\max}$ , we are effectively computing $\mathbb{E}[S_{T_{\max} \wedge \tau}]$ , which equals $0$ (by the finite-horizon martingale property) — whereas the "true" $\mathbb{E}[S_\tau]$ equals $1$ . The two differ by the contribution of the paths that haven't hit $+1$ yet; these carry expected value $-\infty$ in total, perfectly cancelling the finite-path $+1$ 's.

Formally: truncating at $T_{\max}$ gives $\mathbb{E}[S_{T_{\max}}] = 0$ , but $\mathbb{E}[S_\tau] = 1$ . The non-interchangeability of limit and expectation is visible in the simulation as the "paths that haven't hit yet" — if you waited long enough, each of those paths would deliver $+1$ , but their contribution in any finite simulation is $\le 0$ (on average).

Takeaways

Finite almost-surely is not enough for OST — you need $\mathbb{E}[\tau] < \infty$ or a boundedness condition.
The OST failure mode is always uniform integrability. Whenever $\mathbb{E}[M_\tau] \ne \mathbb{E}[M_0]$ , some integrability hypothesis is being violated.
Monte Carlo can look deceptively conclusive. If your simulator always terminates within a compute budget, you are measuring $\mathbb{E}[M_{n \wedge \tau}]$ , not $\mathbb{E}[M_\tau]$ . These agree only if OST applies; otherwise they can be very different.
Real-world parallel. "I always win eventually" strategies (doubling-up, averaging down in a biased market) rely on infinite-expected-time exits. A real trader with finite capital or patience cannot realise the eventual win, and actual expected P&L is determined by the early-stopping dynamics, not the asymptotic one.