From a Tail Bound to an \(L^p\) Estimate
A martingale may have a moderate terminal value while making large excursions before the terminal time. Doob’s \(L^p\) maximal inequality controls those excursions using the \(p\)th moment of the terminal value. The key is to first bound the probability that the running maximum exceeds a level, then integrate that bound over all levels.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{0\le t\le T},\mathbb{P})\). No Brownian motion or stochastic integral is needed for this martingale inequality; the integrals in its proof are ordinary Lebesgue integrals. A martingale \(M=(M_t)_{0\le t\le T}\) is adapted, integrable at each time, and satisfies \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) almost surely for \(s\le t\). Assume its paths are cadlag, meaning right-continuous with left limits, and define
A cadlag function is bounded on a compact interval, so \(M_T^*\) is finite on every cadlag path. The issue is whether its \(p\)th moment is finite.
Here \(\|X\|_p=(\mathbb{E}[|X|^p])^{1/p}\). The theorem assumes a true martingale, not merely a local martingale: integrability and the conditional-expectation identity are part of the martingale hypothesis. The terminal \(L^p\) condition is the additional moment assumption that makes the maximal estimate finite.
The Maximal Tail Estimate
We first establish the tail bound used in the proof. A nonnegative submartingale \(X\) is an adapted, integrable process with \(X_t\ge0\) almost surely and \(\mathbb{E}[X_t\mid\mathcal{F}_s]\ge X_s\) almost surely whenever \(s\le t\).
Proof. First take a finite grid \(0=t_0<\cdots<t_n=T\), and let \(A=\{\max_iX_{t_i}>\lambda\}\). Let \(\tau\) be the first grid time with \(X_\tau>\lambda\) on \(A\); off \(A\), set \(\tau=T\). This is a stopping time taking values in the grid, and \(A\in\mathcal{F}_\tau\). The submartingale property implies \(\mathbb{E}[X_T\mid\mathcal{F}_\tau]\ge X_\tau\): on each event \(\{\tau=t_i\}\), this is the deterministic-time conditional inequality at \(t_i\), and summing over the grid gives the assertion. Since \(X_T\ge0\), we obtain $$ \mathbb{E}[X_T\mathbf{1}_A] =\mathbb{E}[\mathbb{E}[X_T\mid\mathcal{F}_\tau]\mathbf{1}_A] \ge \mathbb{E}[X_\tau\mathbf{1}_A] \ge \lambda\mathbb{P}(A). $$ Now use the nested dyadic grids of \([0,T]\). Right-continuity implies that their union detects every strict crossing: if \(X_t>\lambda\) for some \(t<T\), then \(X_s>\lambda\) for some grid time \(s\) just to the right of \(t\); the endpoint \(T\) belongs to every grid. Thus the grid events increase to \(\{\sup_{t\le T}X_t>\lambda\}\). Taking limits in the finite-grid inequality is valid by monotone convergence on both sides, since \(X_T\ge0\). This proves the lemma. \(\square\)
For a martingale \(M\), the process \(X_t=|M_t|\) is a nonnegative submartingale. Indeed, conditional Jensen’s inequality gives \(\mathbb{E}[|M_t|\mid\mathcal{F}_s]\ge|\mathbb{E}[M_t\mid\mathcal{F}_s]|=|M_s|\). Applying the lemma therefore yields
Proof of the \(L^p\) Bound
For a nonnegative random variable \(Z\) and \(K>0\), the layer-cake identity is \(\mathbb{E}[(Z\wedge K)^p]=\int_0^K p\lambda^{p-1}\mathbb{P}(Z>\lambda)\,d\lambda\). It follows by writing \((Z\wedge K)^p=\int_0^K p\lambda^{p-1}\mathbf{1}_{\{Z>\lambda\}}\,d\lambda\) and applying Tonelli’s theorem to the nonnegative integrand. Truncating at \(K\) avoids assuming in advance that the maximal process has finite \(p\)th moment.
Proof of the theorem’s inequality. Put \(Z=M_T^*\) and \(A_K=\mathbb{E}[(Z\wedge K)^p]\). The layer-cake identity and the tail estimate give
The integral in the first line is finite because \(p>1\), and its value in the second line follows by integrating \(\lambda^{p-2}\) from \(0\) to \(Z\wedge K\). The last line is Hölder’s inequality with exponents \(p\) and \(p/(p-1)\). If \(A_K=0\), the desired truncated bound holds immediately. Otherwise divide by \(A_K^{(p-1)/p}\) and raise both sides to the power \(p\), obtaining \(A_K\le (p/(p-1))^p\mathbb{E}[|M_T|^p]\). As \(K\) increases to infinity, \((Z\wedge K)^p\) increases to \(Z^p\); monotone convergence proves the moment inequality. Taking \(p\)th roots gives the norm inequality. \(\square\)
Worked Applications of the Estimate
Worked Example: A Finite-Time Martingale
Consider a martingale on the times \(0,\tfrac12,1\), with \(M_0=1\). At time \(\tfrac12\), let \(M_{1/2}=0\) or \(2\), each with probability \(\tfrac12\). From the value \(0\), let \(M_1=0\). From the value \(2\), let \(M_1=0\) or \(4\), each with conditional probability \(\tfrac12\). Make the process constant between these times.
The conditional means verify the martingale property: \(\mathbb{E}[M_{1/2}]=\tfrac12(0)+\tfrac12(2)=1=M_0\); conditional on \(M_{1/2}=0\), the terminal mean is \(0\); conditional on \(M_{1/2}=2\), it is \(\tfrac12(0)+\tfrac12(4)=2\). The three possible paths, their probabilities, and their running maxima are: maximum \(1\) with probability \(\tfrac12\), maximum \(2\) with probability \(\tfrac14\), and maximum \(4\) with probability \(\tfrac14\).
For \(p=2\), the terminal second moment is \(\mathbb{E}[|M_1|^2]=\tfrac34(0)^2+\tfrac14(4)^2=4\). The maximal second moment is $$ \mathbb{E}[(M_1^*)^2]=\tfrac12(1)^2+\tfrac14(2)^2+\tfrac14(4)^2 =\tfrac12+1+4=\tfrac{11}{2}. $$ The theorem gives \(\mathbb{E}[(M_1^*)^2]\le 4\mathbb{E}[|M_1|^2]=16\), which this computation satisfies. The direct path calculation also checks that the maximum includes the initial value \(M_0=1\).
Worked Example: A Family Approaching the Sharp Constant
Let \(U\) be uniform on \((0,1)\), with its natural filtration \(\mathcal{F}_t=\sigma(U\wedge t)\) for \(0\le t\le1\). Fix \(0<a<1/p\), set \(Y=(1-U)^{-a}\), and define \(M_t=\mathbb{E}[Y\mid\mathcal{F}_t]\). Conditional on \(U\le t\), the value of \(U\), and hence \(Y\), is known. Conditional on \(U>t\), \(U\) is uniform on \((t,1)\). Therefore $$ M_t= \begin{cases} (1-t)^{-a}/(1-a),& U>t,\\ (1-U)^{-a},& U\le t. \end{cases} $$ This is a martingale by the tower property of conditional expectation. Its paths are right-continuous with left limits: before \(U\) the first expression varies continuously, at \(U\) the process jumps to the revealed value, and after \(U\) it stays there.
For each \(U=u\in(0,1)\), the first expression increases up to its left limit at \(t=u\), and that limit is larger than the value after revelation by the factor \(1/(1-a)>1\). Hence \(M_1^*=(1-U)^{-a}/(1-a)=Y/(1-a)\). Direct integration gives $$ \mathbb{E}[|M_1|^p]=\int_0^1(1-u)^{-ap}\,du=\frac{1}{1-ap}, \qquad \mathbb{E}[(M_1^*)^p]=\frac{1}{(1-a)^p(1-ap)}. $$ Thus \(\|M_1^*\|_p/\|M_1\|_p=1/(1-a)\). As \(a\) increases to \(1/p\), this ratio tends to \(p/(p-1)\). Every \(a<1/p\) gives a finite terminal \(p\)th moment, so any universal norm constant must be at least this limiting value. Together with the proved upper bound, this establishes sharpness.
Check: for \(p=2\) and \(a=\tfrac14\), the terminal second moment is \(1/(1-\tfrac12)=2\), and the maximal second moment is \((\tfrac43)^2\cdot2=\tfrac{32}{9}\). The norm ratio is \(\sqrt{(32/9)/2}=\tfrac43\), matching \(1/(1-a)\). As \(a\) approaches \(\tfrac12\) from below, that ratio approaches \(2\), the sharp constant for \(p=2\).
Worked Example: Why \(p=1\) Has No Such Universal Bound
Use the same construction with \(0<a<1\), now measuring first moments. The terminal variable is integrable, and $$ \mathbb{E}[|M_1|]=\int_0^1(1-u)^{-a}\,du=\frac{1}{1-a}. $$ The pathwise maximum is \(M_1^*=Y/(1-a)\), so $$ \mathbb{E}[M_1^*]=\frac{1}{1-a}\mathbb{E}[Y]=\frac{1}{(1-a)^2}, \qquad \frac{\mathbb{E}[M_1^*]}{\mathbb{E}[|M_1|]}=\frac{1}{1-a}. $$ For example, \(a=\tfrac12\) gives terminal mean \(2\), maximal mean \(4\), and ratio \(2\). As \(a\) increases to \(1\), the ratio becomes arbitrarily large, even though each chosen terminal variable is integrable. Thus an \(L^1\) terminal condition alone cannot yield a finite universal constant for the maximal first moment.
Common Mistake: Confusing the Norm and Moment Constants
The sharp constant \(p/(p-1)\) multiplies the \(L^p\) norm. When the inequality is written for \(p\)th moments, the constant is its \(p\)th power. For a concrete check, take \(p=2\) and \(a=\tfrac13\) in the sharpness family. The terminal second moment is \(1/(1-\tfrac23)=3\), and the maximal second moment is \((\tfrac32)^2\cdot3=\tfrac{27}{4}\). The incorrect moment claim \(\mathbb{E}[(M_1^*)^2]\le2\mathbb{E}[|M_1|^2]\) would give \(\tfrac{27}{4}\le6\), which is false. The correct estimate uses \(2^2=4\), giving \(\tfrac{27}{4}\le12\).
The integrability hypothesis also has a specific role: \(\mathbb{E}[|M_T|^p]<\infty\) is what permits Hölder’s inequality to yield a finite bound. The proof does not assert the same conclusion for a local martingale, nor does it produce a finite \(L^1\) maximal constant.
Check Your Understanding
Use the tail estimate and the theorem’s hypotheses to justify each answer.
- For \(p=3\), what is the sharp constant in the norm inequality, and what constant appears in the third-moment inequality?
- Why is the truncation \(M_T^*\wedge K\) introduced before applying Hölder’s inequality?
- In the finite-time example, verify the conditional mean at the branch where \(M_{1/2}=2\).
- For the sharpness family, explain why \(a<1/p\) is exactly the condition that makes \(\mathbb{E}[|M_1|^p]\) finite.
- Why does the proof apply to \(|M_t|\) even when \(M_t\) can be negative?