Doob's Maximal Inequality¶

Martingales come with a remarkable global control: the running maximum \(\sup_{s \le t} |M_s|\) is bounded, in \(L^p\), by the terminal value \(|M_t|\) — with an explicit and sharp constant. This is the canonical tool for turning pointwise bounds into uniform-in-time bounds.

Statement¶

Theorem (Doob's \(L^p\) maximal inequality). Let \(M\) be a right-continuous martingale (or a nonnegative submartingale) and let \(p > 1\). Then

\[ \Bigl\| \sup_{0 \le t \le T} |M_t| \Bigr\|_{L^p} \le \frac{p}{p-1} \|M_T\|_{L^p}. \]

The constant \(p/(p-1)\) is sharp and blows up as \(p \to 1^+\), reflecting that the \(L^1\) case requires a different form.

Weak \(L^1\) form. For a nonnegative submartingale \(X\) and \(\lambda > 0\),

\[ \lambda \cdot \mathbb{P}\Bigl(\sup_{0 \le t \le T} X_t \ge \lambda\Bigr) \le \mathbb{E}\bigl[X_T \mathbf{1}_{\{X_T^* \ge \lambda\}}\bigr] \le \mathbb{E}[X_T]. \]

Intuition¶

The maximum of an arbitrary process can be vastly larger than any single value. For martingales this fails: on the event \(\{X_T^* \ge \lambda\}\), the optional sampling identity forces the terminal value to "pay" for the excursion to \(\lambda\). Integrating this trade-off over all \(\lambda\) via the layer-cake formula upgrades the weak bound to the strong \(L^p\) bound.

Proof Outline¶

Weak \(L^1\) step. Let \(\tau = \inf\{t : X_t \ge \lambda\} \wedge T\). On \(\{X_T^* \ge \lambda\}\), \(X_\tau \ge \lambda\). Optional sampling for submartingales gives \(\mathbb{E}[X_T \mid \mathcal{F}_\tau] \ge X_\tau\), and integrating over \(\{X_T^* \ge \lambda\}\) (which is \(\mathcal{F}_\tau\)-measurable) yields the weak bound.

Layer cake + Hölder. Integrate the weak bound against \(p \lambda^{p-2}\,d\lambda\):

\[ \mathbb{E}[(X_T^*)^p] \le \frac{p}{p-1} \mathbb{E}[X_T (X_T^*)^{p-1}]. \]

Apply Hölder with exponents \(p\) and \(p/(p-1)\) and divide through by \(\|X_T^*\|_{L^p}^{p-1}\). \(\square\)

Brownian Example¶

Brownian motion, \(p = 2\)

\(W_t\) is a martingale, \(|W_t|\) a submartingale. Doob's \(L^2\) inequality gives

\[ \mathbb{E}\Bigl[\sup_{0 \le t \le T} W_t^2\Bigr] \le 4 \mathbb{E}[W_T^2] = 4T, \]

equivalently \(\|\sup_{t \le T} |W_t|\|_{L^2} \le 2\sqrt{T}\). The weak bound gives the tail estimate

\[ \mathbb{P}\Bigl(\sup_{t \le T} |W_t| \ge a\Bigr) \le \frac{\mathbb{E}[W_T^2]}{a^2} = \frac{T}{a^2}. \]

The reflection principle gives the exact value \(2\Phi(-a/\sqrt{T})\) — Doob's bound is a universal upper bound applicable far beyond Brownian motion.

Why This Matters¶

Doob's inequality is the standard route from pointwise to uniform control. It underpins:

convergence of \(L^p\)-bounded martingales (see Martingale Convergence),
existence of solutions to SDEs (controlling Picard iterates uniformly in \(t\)),
\(L^2\) theory of stochastic integrals (via the Burkholder–Davis–Gundy extension, which bounds \((M^*)^p\) by the quadratic variation \([M]^{p/2}\)).

Exercises¶

Exercise 1. Brownian motion.

(a) Use Doob's \(L^2\) inequality to show \(\mathbb{E}[\sup_{t \le T} W_t^2] \le 4T\). (b) Bound \(\mathbb{P}(\sup_{t \le 1} |W_t| \ge 3)\). (c) Compare (b) with the exact value from the reflection principle.

Solution to Exercise 1

(a) Doob with \(p = 2\): \(\mathbb{E}[\sup_{t \le T} W_t^2] \le 4 \mathbb{E}[W_T^2] = 4T\).

(b) The weak \(L^2\) bound (Chebyshev applied to \(W_t^2\), then Doob): \(\mathbb{P}(\sup_{t \le 1} |W_t| \ge 3) \le \mathbb{E}[\sup_{t \le 1} W_t^2]/9 \le 4/9\). A tighter alternative uses Doob's weak \(L^1\) on the submartingale \(W_t^2\): \(\mathbb{P}(\sup_{t \le 1} W_t^2 \ge 9) \le \mathbb{E}[W_1^2]/9 = 1/9\).

(c) Reflection principle: \(\mathbb{P}(\sup_{t \le 1} W_t \ge 3) = 2\Phi(-3) \approx 0.0027\), so \(\mathbb{P}(\sup_{t \le 1} |W_t| \ge 3) \approx 0.0054\). Doob's bound \(1/9 \approx 0.111\) is loose but universally applicable.

Exercise 2. \(L^p\) bounds.

(a) For a martingale \(M\) with \(\mathbb{E}[|M_T|^4] = C\), bound \(\mathbb{E}[\sup_{t \le T} |M_t|^4]\). (b) What happens to the Doob constant as \(p \to 1^+\)?

Solution to Exercise 2

(a) With \(p = 4\): \(\mathbb{E}[\sup_{t \le T} |M_t|^4] \le (4/3)^4 C = 256 C / 81 \approx 3.16 C\).

(b) \((p/(p-1))^p \to \infty\) as \(p \to 1^+\). The strong \(L^p\) bound degenerates; one must use either the weak form or an \(L \log L\) refinement.

Exercise 3. Let \(M_n\) be a discrete martingale with \(\sup_n \mathbb{E}[M_n^2] < \infty\).

(a) Show \(\sup_n |M_n| < \infty\) a.s. (b) Deduce a.s. convergence of \(M_n\).

Solution to Exercise 3

(a) Doob's \(L^2\) inequality applied on \([0, N]\) and letting \(N \to \infty\): \(\mathbb{E}[\sup_n M_n^2] \le 4 \sup_n \mathbb{E}[M_n^2] < \infty\). A finite-expectation random variable is finite a.s.

(b) \(L^2\)-bounded implies \(L^1\)-bounded (Jensen). By the martingale convergence theorem, \(M_n \to M_\infty\) a.s. (and in fact in \(L^2\), since \(L^2\)-boundedness gives uniform integrability).

Exercise 4 (Prove the strong bound). Starting from the weak \(L^1\) inequality for nonnegative submartingales, derive the Doob \(L^p\) inequality for \(p > 1\) in full.

Solution to Exercise 4

Let \(X_T^* = \sup_{t \le T} X_t\). By the layer-cake formula and the weak bound,

\[ \mathbb{E}[(X_T^*)^p] = p \int_0^\infty \lambda^{p-1} \mathbb{P}(X_T^* \ge \lambda)\,d\lambda \le p \int_0^\infty \lambda^{p-2} \mathbb{E}[X_T \mathbf{1}_{\{X_T^* \ge \lambda\}}]\,d\lambda. \]

By Fubini,

\[ p \int_0^\infty \lambda^{p-2} \mathbb{E}[X_T \mathbf{1}_{\{X_T^* \ge \lambda\}}]\,d\lambda = p\, \mathbb{E}\Bigl[X_T \int_0^{X_T^*} \lambda^{p-2}\,d\lambda\Bigr] = \frac{p}{p-1} \mathbb{E}[X_T (X_T^*)^{p-1}]. \]

Hölder with conjugates \(p, p/(p-1)\):

\[ \mathbb{E}[X_T (X_T^*)^{p-1}] \le \|X_T\|_{L^p} \|X_T^*\|_{L^p}^{p-1}. \]

Combine and divide by \(\|X_T^*\|_{L^p}^{p-1}\) (assumed finite and positive; otherwise truncate first):

\[ \|X_T^*\|_{L^p} \le \frac{p}{p-1} \|X_T\|_{L^p}. \quad \square \]

Exercise 5 (Non-UI example). Let \(M_n = \prod_{i=1}^n \xi_i\) where \(\mathbb{P}(\xi_i = 2) = \mathbb{P}(\xi_i = 0) = 1/2\) and \(M_0 = 1\).

(a) Verify \(M_n\) is a martingale with \(\sup_n \mathbb{E}[|M_n|] < \infty\). (b) Show \(M_n \to 0\) a.s. but \(\mathbb{E}[M_n] = 1\) for all \(n\). (c) Why doesn't this contradict Doob's \(L^p\) inequality?

Solution to Exercise 5

(a) \(\mathbb{E}[\xi_i] = 1\), so \(\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] = M_n \cdot \mathbb{E}[\xi_{n+1}] = M_n\) and \(\mathbb{E}[|M_n|] = 1\).

(b) Some \(\xi_i = 0\) a.s., so \(M_n = 0\) eventually, hence \(M_\infty = 0\) a.s. But \(\mathbb{E}[M_n] = 1\) for all \(n\).

(c) Doob's \(L^p\) inequality needs \(p > 1\). Here the mass of \(M_n\) is concentrated on \(\{\xi_1 = \cdots = \xi_n = 2\}\) of probability \(2^{-n}\), where \(M_n = 2^n\). So \(\mathbb{E}[M_n^p] = 2^{n(p-1)} \to \infty\) for \(p > 1\): \(M\) is not \(L^p\)-bounded, and Doob does not apply. The family is \(L^1\)-bounded but not uniformly integrable, which is also why \(\mathbb{E}[M_n] \not\to \mathbb{E}[M_\infty]\).