Doob's Maximal Inequality¶
Why Maximum Control Matters¶
In probability theory, we often need to control not just the value of a stochastic process at a single time, but its extremal behavior over an entire interval. Questions like:
- How large can a martingale get before time \(T\)?
- What is the probability that a random walk ever exceeds level \(a\)?
- How do we bound the fluctuations of a stochastic integral?
All require understanding the running maximum \(M_t^* = \sup_{0 \le s \le t} |M_s|\).
The challenge is clear: while \(\mathbb{E}[M_t]\) might be controlled, the maximum could be much larger. Doob's inequalities provide the remarkable result that for martingales, the maximum is controlled by the terminal value—with explicit constants.
The Maximal Process¶
For a stochastic process \(X = \{X_t\}_{t \ge 0}\), define:
These are increasing processes. The key question: how do we relate \(\mathbb{E}[(X_t^*)^p]\) to \(\mathbb{E}[|X_t|^p]\)?
For general processes, there is no control: the maximum could be arbitrarily larger than any single value. But martingales are special.
Doob's L^p Maximal Inequality¶
Theorem (Doob's \(L^p\) Maximal Inequality): Let \(\{M_t\}_{0 \le t \le T}\) be a right-continuous martingale (or nonnegative submartingale). Then for \(p > 1\):
Equivalently:
The constant: The factor \(\frac{p}{p-1}\) is sharp—it cannot be improved in general. As \(p \to 1^+\), the constant blows up, reflecting that the \(L^1\) case requires different treatment.
| \(p\) | \(\frac{p}{p-1}\) | \(\left(\frac{p}{p-1}\right)^p\) |
|---|---|---|
| 2 | 2 | 4 |
| 3 | 1.5 | 3.375 |
| 4 | 1.33 | 3.16 |
| \(\infty\) | 1 | 1 |
Proof Strategy¶
The proof proceeds in two steps:
Step 1: Doob's L¹ Maximal Inequality (Weak Form)¶
Lemma: For a nonnegative submartingale \(\{X_t\}\) and any \(\lambda > 0\):
Proof: Define the stopping time \(\tau = \inf\{t \ge 0 : X_t \ge \lambda\} \wedge T\). On the event \(\{X_T^* \ge \lambda\}\), we have \(X_\tau \ge \lambda\).
By the optional sampling theorem for submartingales (with bounded stopping times):
Taking expectations and using \(X_\tau \ge \lambda\) on \(\{X_T^* \ge \lambda\}\):
Step 2: From Weak L¹ to Strong L^p¶
Using the layer cake formula and the weak inequality:
Applying Fubini's theorem and integrating:
By Hölder's inequality with exponents \(p\) and \(\frac{p}{p-1}\):
Combining:
Dividing by \(\|X_T^*\|_{L^p}^{p-1}\) (assuming it's positive):
Doob's \(L^1\) Results¶
The case \(p = 1\) requires separate treatment since \(\frac{p}{p-1} \to \infty\) as \(p \to 1\).
Weak \(L^1\) Inequality (already proved above)¶
The first result is the weak \(L^1\) bound already established in the proof:
This yields the tail bound \(\mathbb{P}(M_T^* \ge \lambda) \le \frac{\mathbb{E}[|M_T|]}{\lambda}\) for martingales (by applying to \(|M_t|\), which is a submartingale). Note: this follows from Doob's weak inequality, not directly from Markov's inequality.
Strong \(L^1\) Inequality (Orlicz space bound)¶
For a strong bound on \(\mathbb{E}[M_T^*]\) itself, one needs more than \(L^1\) integrability of \(M_T\). The sharp result involves the \(L \log L\) Orlicz space:
Theorem: For a martingale \(\{M_t\}\) with \(M_T \in L \log L\) (i.e., \(\mathbb{E}[|M_T| \log^+ |M_T|] < \infty\)):
where \(\log^+ x = \max(\log x, 0)\). This bound is sharp. The constant \(e/(e-1)\) arises naturally when optimizing the layer-cake integral using the inequality \(\log^+ x \le x/e\) at the critical point \(x = e\). If \(M_T \in L^1 \setminus L\log L\), then \(M_T^*\) need not be integrable.
Application: Brownian Motion¶
Example: For Brownian motion \(W_t\) on \([0, T]\):
More precisely, using \(p = 2\) in Doob's inequality:
Tail bound: For any \(a > 0\):
Using the reflection principle (which gives the exact distribution), one can show:
where \(\Phi\) is the standard normal CDF. Doob's inequality provides a general upper bound; the reflection principle gives the exact answer for Brownian motion.
The Burkholder-Davis-Gundy Inequality¶
Doob's inequality bounds the maximum in terms of the terminal value. The Burkholder-Davis-Gundy (BDG) inequality bounds it in terms of the quadratic variation.
Theorem (BDG): For any continuous local martingale \(M\) with \(M_0 = 0\) and \(p > 0\), there exist universal constants \(c_p, C_p\) such that:
where \([M]_T\) is the quadratic variation.
For \(p = 2\), the constants are \(c_2 = 1\) and \(C_2 = 4\):
Application to Itô integrals: If \(I_t = \int_0^t H_s \, dW_s\), then \([I]_t = \int_0^t H_s^2 \, ds\), giving:
Applications in Stochastic Analysis¶
1. Existence and Uniqueness of SDEs¶
In proving existence of solutions to stochastic differential equations, one uses Doob's inequality to control the maximum of Picard iterations:
This leads to Gronwall-type arguments establishing convergence.
2. Convergence of Martingales¶
Doob's inequality implies that \(L^p\)-bounded martingales with \(p > 1\) converge almost surely. If \(\sup_n \mathbb{E}[|M_n|^p] < \infty\), then:
By the monotone convergence theorem for the supremum, \(M_n\) converges a.s.
3. Concentration Inequalities¶
Doob's inequality underpins many concentration results. Combined with the exponential martingale:
one obtains exponential tail bounds for martingales.
Discrete-Time Version¶
For discrete-time martingales \((M_n)_{n=0}^N\), Doob's inequality takes the form:
The proof is essentially identical, using the optional sampling theorem for bounded stopping times.
Comparison with Other Inequalities¶
| Inequality | Statement | Application |
|---|---|---|
| Markov | \(\mathbb{P}(X \ge a) \le \frac{\mathbb{E}[X]}{a}\) | Basic tail bound |
| Chebyshev | \(\mathbb{P}(\|X - \mu\| \ge a) \le \frac{\text{Var}(X)}{a^2}\) | Variance-based bound |
| Doob's \(L^p\) | \(\|M^*\|_p \le \frac{p}{p-1}\|M_T\|_p\) | Maximum control for martingales |
| BDG | \(\|M^*\|_p \asymp \|[M]^{1/2}\|_p\) | Two-sided control via quadratic variation |
| Azuma-Hoeffding | Exponential tail for bounded differences | Concentration for bounded martingales |
Summary¶
Doob's maximal inequality is one of the most powerful tools in martingale theory:
Key insights:
- The maximum of a martingale is controlled by its terminal value—with explicit constants.
- The constant \(\frac{p}{p-1}\) is optimal and blows up as \(p \to 1\).
- The inequality extends to submartingales (for the positive part).
- Combined with BDG, it provides comprehensive control of martingale fluctuations.
Applications span:
- SDE theory (existence and uniqueness)
- Martingale convergence
- Concentration inequalities
- Numerical analysis of stochastic algorithms
Exercises¶
Exercise 1: Maximal Inequality Applications¶
(a) Use Doob's \(L^2\) inequality to show \(\mathbb{E}[\sup_{t \le T} W_t^2] \le 4T\).
(b) Find an upper bound for \(\mathbb{P}(\sup_{t \le 1} |W_t| \ge 3)\).
(c) Compare your bound in (b) with the exact value from the reflection principle.
Solution to Exercise 1
(a) Brownian motion \(W_t\) is a martingale, so \(|W_t|\) is a non-negative submartingale. By Doob's \(L^2\) inequality (with \(p = 2\)):
(b) By Doob's weak \(L^1\) inequality (applied to the submartingale \(|W_t|\)) with \(\lambda = 3\):
Alternatively, using the \(L^2\) bound: by Markov's inequality applied to the \(L^2\) result:
The Chebyshev-type bound is tighter: \(1/9\).
(c) The exact value from the reflection principle: \(\mathbb{P}(\sup_{t \le 1} W_t \ge 3) = 2\mathbb{P}(W_1 \ge 3) = 2\Phi(-3) \approx 2 \times 0.00135 = 0.0027\). By symmetry, \(\mathbb{P}(\sup_{t \le 1} |W_t| \ge 3) \le 2 \times 0.0027 = 0.0054\) (this is not exact for the absolute value, but gives the right order). The exact value is approximately \(0.0054\), far smaller than the bounds of \(1/9 \approx 0.111\) or \(4/9 \approx 0.444\). The Doob bounds are correct but conservative.
Exercise 2: L^p Bounds¶
Let \(M_t\) be a martingale with \(\mathbb{E}[|M_T|^4] = C\).
(a) Use Doob's inequality to bound \(\mathbb{E}[\sup_{t \le T} |M_t|^4]\).
(b) What happens to the constant as \(p \to 1\)?
(c) State and prove Doob's \(L^1\) weak inequality.
Solution to Exercise 2
(a) By Doob's \(L^p\) inequality with \(p = 4\):
(b) As \(p \to 1^+\), the constant \(\left(\frac{p}{p-1}\right)^p \to \infty\). Specifically, \(\frac{p}{p-1} \to \infty\) as \(p \to 1^+\), so the bound becomes vacuous. This reflects the fact that for \(p = 1\), the maximal inequality in the strong \(L^p\) form does not hold — one needs the weaker \(L^1\) form or the \(L\log L\) condition.
(c) Doob's \(L^1\) weak inequality: For a non-negative submartingale \(\{X_t\}\) and \(\lambda > 0\):
Proof: Define \(\tau = \inf\{t : X_t \ge \lambda\} \wedge T\). On \(\{X_T^* \ge \lambda\}\), \(X_\tau \ge \lambda\). By optional sampling for submartingales:
Multiplying by \(\mathbf{1}_{\{X_T^* \ge \lambda\}}\) (which is \(\mathcal{F}_\tau\)-measurable) and taking expectations:
Exercise 3: Convergence Application¶
Let \(M_n\) be a discrete martingale with \(\sup_n \mathbb{E}[M_n^2] < \infty\).
(a) Use Doob's inequality to show \(\sup_n |M_n| < \infty\) a.s.
(b) Deduce that \(M_n\) converges a.s.
(c) Give an example where \(\sup_n \mathbb{E}[|M_n|] < \infty\) but \(M_n\) does not converge in \(L^1\).
Solution to Exercise 3
(a) By Doob's \(L^2\) inequality:
Since \(\mathbb{E}[\sup_n |M_n|^2] < \infty\), we have \(\sup_n |M_n|^2 < \infty\) a.s., hence \(\sup_n |M_n| < \infty\) a.s. \(\square\)
(b) Since \(\sup_n \mathbb{E}[M_n^2] < \infty\) implies \(\sup_n \mathbb{E}[|M_n|] \le \sup_n (\mathbb{E}[M_n^2])^{1/2} < \infty\) (by Jensen), the martingale is \(L^1\)-bounded. By Doob's martingale convergence theorem, \(M_n\) converges a.s. to a finite limit \(M_\infty\).
Moreover, \(L^2\)-boundedness implies uniform integrability (since \(p = 2 > 1\)), so \(M_n \to M_\infty\) also in \(L^1\) (and in fact in \(L^2\)). \(\square\)
(c) Let \(M_n = \prod_{i=1}^n \xi_i\) where \(\mathbb{P}(\xi_i = 2) = \mathbb{P}(\xi_i = 0) = 1/2\), with \(M_0 = 1\). Then \(\mathbb{E}[M_n] = 1\) and \(\sup_n \mathbb{E}[|M_n|] = 1 < \infty\).
By the convergence theorem, \(M_n \to M_\infty\) a.s. Since eventually some \(\xi_i = 0\), \(M_\infty = 0\) a.s.
But \(\mathbb{E}[M_n] = 1 \not\to 0 = \mathbb{E}[M_\infty]\), so \(M_n\) does not converge in \(L^1\). The family \(\{M_n\}\) is not uniformly integrable: the expectation is carried by the event \(\{\xi_1 = \cdots = \xi_n = 2\}\) of probability \((1/2)^n\), where \(M_n = 2^n\).
Exercise 4: Maximum of Brownian Motion¶
Let \(M_t = \sup_{s \le t} W_s\).
(a) Prove that \(M_t - W_t \ge 0\) and is increasing in \(t\).
(b) Show that \(M_t\) and \(M_t - W_t\) each have the same distribution as \(|W_t|\).
Hint: Use the reflection principle. The process \(W'_t = W_t - 2(W_t - M_t)^+ = 2M_t - W_t\) (reflecting \(W\) at its running maximum) is also a Brownian motion. Show \(M_t = \sup_{s\le t}W'_s\) and use symmetry.
(c) Use this to find \(\mathbb{P}(M_t \ge a, W_t \le b)\) for \(a > b\).
Solution to Exercise 4
(a) Let \(M_t = \sup_{s \le t} W_s\). Since \(W_t \le M_t\) always and \(M_t\) is the running maximum, \(M_t - W_t \ge 0\).
To show \(M_t - W_t\) is increasing in \(t\): for \(s < t\), \(M_t \ge M_s\) (the maximum can only increase). But \(M_t - W_t\) is not necessarily monotone increasing. Actually, \(M_t - W_t\) is not monotone increasing in general (it can decrease when \(W\) moves up toward a new maximum). The correct statement is that \(M_t\) is increasing in \(t\) (non-decreasing). The process \(M_t - W_t\) is non-negative but oscillates.
What is true: \(M_t - W_t \ge 0\) for all \(t\), and \(M_t - W_t = 0\) precisely when \(W_t\) is at its running maximum (i.e., \(W_t = M_t\)).
(b) By the reflection principle and Lévy's identity:
More precisely, Lévy's identity states:
and \(M_t \stackrel{d}{=} |W_t|\).
To see \(M_t \stackrel{d}{=} |W_t|\): from the reflection principle, \(\mathbb{P}(M_t \ge a) = 2\mathbb{P}(W_t \ge a) = \mathbb{P}(|W_t| \ge a)\) for \(a > 0\) (using \(\mathbb{P}(W_t \ge a) = \mathbb{P}(W_t \le -a)\) by symmetry). So \(M_t\) and \(|W_t|\) have the same distribution.
The identity \(M_t - W_t \stackrel{d}{=} |W_t|\) follows from Lévy's theorem, which states that \((M_t - W_t, M_t) \stackrel{d}{=} (|W_t|, L_t)\) where \(L_t\) is the local time at 0, or more directly that \(M_t - W_t\) is a reflected Brownian motion.
(c) For \(a > b\) (so \(a > 0\) and \(b < a\)):
By the reflection principle: on \(\{M_t \ge a\}\), reflect the path at time \(\tau_a\). The reflected path \(\hat{W}_t = 2a - W_t\) has \(\hat{W}_t \ge 2a - b > a\) when \(W_t \le b\). Therefore:
since the reflection creates a bijection between paths hitting \(a\) and ending at or below \(b\), and paths ending at or above \(2a - b\).
where \(\Phi\) is the standard normal CDF.