Doob Martingale Convergence

Why Does This Exist?

Bayesian updating, stochastic approximation and sequential gambling all generate fair forecasts that evolve with data. The live question is whether they settle: does the posterior mean lock onto something, or wobble forever? Without a convergence theorem every long-run claim (consistency, stability, final wealth) dangles.

Doob's theorem answers once and for all under one checkable condition. If the martingale stays $L^1$-bounded (expected absolute values capped), it converges almost surely and in $L^1$ to a genuine limit. This page states the theorem and traces it on a stopped walk. Fair-game basics are in Markov chains; the Bayesian payoff is posterior consistency.

The one idea you need first

$L^1$-bounded means $\sup_n \mathbb{E}|M_n| < \infty$: expected magnitudes never blow up. Almost sure convergence means paths settle with probability one; $L^1$ convergence means expected gaps $\mathbb{E}|M_n - M_\infty|$ also vanish, so means converge too.

Think of It Like This

How It Actually Works

Let $M_n$ be a martingale with $\sup_n \mathbb{E}|M_n| < \infty$. Doob's convergence theorem says there is $M_\infty$ with $\mathbb{E}|M_\infty| < \infty$ such that:

In plain words: $M_n$ is the fair forecast at time $n$, the sup condition caps expected size uniformly, and the conclusion gives a final forecast that paths approach and means approach together. The engine is the upcrossing inequality: a bounded martingale can cross any fixed interval only finitely often on average, so oscillation must die out. Nonnegative martingales qualify automatically in $L^1$ reasoning, and uniformly integrable ones keep $\mathbb{E}[M_n] \to \mathbb{E}[M_\infty]$.

Worked example

Symmetric walk from $0$ stopped at $\tau$, the first hit of $+2$ or $-2$. The stopped process $M_n = S_{n \wedge \tau}$ is a martingale with $|M_n| \le 2$, hence $\mathbb{E}|M_n| \le 2$ for all $n$: bounded.

Each path freezes once it hits a barrier, so $M_n \to M_\infty \in \{+2, -2\}$ everywhere: almost sure convergence by inspection. Half the paths end up, half down, so $\mathbb{E}[M_\infty] = 0.5(2) + 0.5(-2) = 0 = \mathbb{E}[M_0]$: $L^1$ convergence holds and the mean survives. Remove the barriers and $|S_n|$ grows like $\sqrt{n}$ with no limit: unbounded, oscillating, exactly what the theorem refuses to cover.

Code

Freeze stopped walks and confirm they sit at the barriers with mean near zero.

1import numpy as np2
3
4def stopped_limits(paths: int, barrier: int = 2, seed: int = 1) -> np.ndarray:5    rng = np.random.default_rng(seed)  # reproducible ensemble6    out = np.empty(paths)7    for i in range(paths):8        pos = 09        while abs(pos) < barrier:  # freeze at first barrier hit10            pos += 1 if rng.random() < 0.5 else -111        out[i] = pos12    return out13
14
15lims = stopped_limits(10000)16print(sorted(set(lims.tolist())))  # -> [-2.0, 2.0]17print(round(float(lims.mean()), 2))  # -> 0.0 approx

Only $\pm 2$ survive, averaging about $0$: the limit the theorem promised.

Watch Out For

The Quick Version

• $L^1$-bounded martingales converge almost surely and in $L^1$ to an integrable limit.
• The upcrossing inequality kills infinite oscillation: bounded fair games cannot cross intervals forever.
• A walk stopped at $\pm 2$ freezes onto one barrier, half up and half down, mean zero.
• Unbounded walks grow like $\sqrt{n}$ and converge nowhere.
• Almost sure alone never moves means inside limits; $L^1$ (via uniform integrability) does.

What to Read Next

• Markov Chains
• Stationary Distributions
• Posterior Consistency
• Optional Stopping Theorem