Doob's Maximal Inequality in Stochastic Estimates

From a Terminal Bound to a Pathwise Bound

A terminal-time estimate controls a random variable at one fixed time. Many stochastic arguments need more: a bound on how large a process might become at any time before a horizon $T$. Doob’s L² maximal inequality gives this bridge for martingales. In particular, it turns a second-moment calculation at time $T$ into a bound on the expected square of the pathwise supremum.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to the filtration. A process $M_t$ is a martingale on $[0,T]$ if it is adapted, $\mathbb{E}|M_t|<\infty$ for every $t$, and $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ for $0\le s\le t\le T$. It is square-integrable if, in addition, $\mathbb{E}[M_t^2]<\infty$ for all $t$. For a continuous process, write $M_T^*=\sup_{0\le t\le T}|M_t|$.

Theorem: If $M_t$ is a continuous, square-integrable martingale on $[0,T]$, then $$ \mathbb{E}\big[(M_T^*)^2\big]\le 4\mathbb{E}[M_T^2]. $$ The same bound holds for a martingale with right-continuous paths. In particular, $\big(\mathbb{E}[(M_T^*)^2]\big)^{1/2}\le 2\big(\mathbb{E}[M_T^2]\big)^{1/2}$.

The theorem is about the square of the supremum, not the supremum of the second moments. The proof has two parts: first establish the estimate at finitely many times, then pass to a dense set of times using path continuity.

Proof on a Finite Time Grid

Let $0=t_0<t_1<\cdots<t_n=T$ be a finite grid and set $X_k=M_{t_k}$. For $\lambda>0$, define $A=\{\max_{0\le k\le n}|X_k|>\lambda\}$. Let $A_k$ be the event that $k$ is the first index with $|X_k|>\lambda$. These events are disjoint, their union is $A$, and $A_k\in\mathcal{F}_{t_k}$. The martingale property and the conditional triangle inequality give $\mathbb{E}[|X_n|\mid\mathcal{F}_{t_k}]\ge|\mathbb{E}[X_n\mid\mathcal{F}_{t_k}]|=|X_k|$. Thus $\mathbb{E}[\mathbf{1}_{A_k}|X_n|]\ge\lambda\mathbb{P}(A_k)$. Summing over $k$ proves the finite-grid tail bound.

Finite-grid estimate: For $X_k=M_{t_k}$, $$ \lambda\,\mathbb{P}\!\left(\max_{0\le k\le n}|X_k|>\lambda\right) \le \mathbb{E}\!\left[|X_n|\mathbf{1}_{\{\max_{0\le k\le n}|X_k|>\lambda\}}\right]. $$

Proof. For each $k$, the martingale property gives $\mathbb{E}[X_n\mid\mathcal{F}_{t_k}]=X_k$. The conditional triangle inequality therefore gives $\mathbb{E}[|X_n|\mid\mathcal{F}_{t_k}]\ge |X_k|$ almost surely. Because $A_k\in\mathcal{F}_{t_k}$, multiplying by $\mathbf{1}_{A_k}$, taking expectations, and using $|X_k|>\lambda$ on $A_k$ yields $\mathbb{E}[\mathbf{1}_{A_k}|X_n|]\ge\lambda\mathbb{P}(A_k)$. The $A_k$ are disjoint and have union $A$, so summing these inequalities gives the stated estimate. $\square$

To obtain a second-moment estimate, let $X^*=\max_{0\le k\le n}|X_k|$, and fix $K>0$. The tail integral formula for a nonnegative random variable gives $\mathbb{E}[(X^*\wedge K)^2]=\int_0^K2\lambda\mathbb{P}(X^*>\lambda)\,d\lambda$. Apply the finite-grid estimate inside this integral. Tonelli’s theorem applies because the integrand is nonnegative.

\begin{aligned} \mathbb{E}[(X^*\wedge K)^2] &\le 2\int_0^K\mathbb{E}\!\left[|X_n|\mathbf{1}_{\{X^*>\lambda\}}\right]\,d\lambda\\ &=2\mathbb{E}\!\left[|X_n|(X^*\wedge K)\right]\\ &\le 2\big(\mathbb{E}[X_n^2]\big)^{1/2} \big(\mathbb{E}[(X^*\wedge K)^2]\big)^{1/2}. \end{aligned}

The last inequality is Cauchy–Schwarz. If the truncated second moment is zero, the desired bound is immediate. Otherwise, divide by its square root and square both sides to get $\mathbb{E}[(X^*\wedge K)^2]\le4\mathbb{E}[X_n^2]$. Letting $K$ increase to infinity and using monotone convergence proves $\mathbb{E}[(X^*)^2]\le4\mathbb{E}[X_n^2]$.

Passage to Continuous Time

For a continuous martingale, take an increasing sequence of finite grids whose union is dense in $[0,T]$, with $T$ included in every grid. The maximum of $|M_t|$ over the union of the grids equals $\sup_{0\le t\le T}|M_t|$ on every continuous sample path. The finite-grid inequality bounds the expected squared maximum by $4\mathbb{E}[M_T^2]$ at every stage. The maxima increase to $M_T^*$; monotone convergence therefore gives the theorem. This argument proves the stated continuous-path case without assuming in advance that the supremum has finite second moment.

Applying the Inequality to Ito Integrals

Let $H_s$ be progressively measurable and satisfy $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; this is the class $\mathcal{L}^2$. Define the Ito integral $M_t=\int_0^t H_s\,dB_s$. It is a continuous, square-integrable martingale, and the Ito isometry gives $\mathbb{E}[M_T^2]=\mathbb{E}\int_0^T H_s^2\,ds$. For completeness, its martingale property follows first for adapted step integrands from the zero conditional mean of each future Brownian increment. General $\mathcal{L}^2$ integrands are obtained as limits of such integrands in the Ito-isometry norm. Conditional expectation is continuous in $L^2$, so $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ passes to the limit. The isometry also gives finite second moments, and the Ito integral has a continuous version. Doob’s theorem now yields:

Corollary: If $H$ is progressively measurable and $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, then $$ \mathbb{E}\left[\sup_{0\le t\le T}\left|\int_0^t H_s\,dB_s\right|^2\right] \le 4\mathbb{E}\int_0^T H_s^2\,ds. $$

Proof. The Ito integral is a continuous square-integrable martingale by the construction just described. Apply the theorem to this martingale, then substitute the Ito-isometry identity for its terminal second moment. $\square$

Worked Example: A Deterministic Integrand

Let $T=2$ and $H_s=s$ for $0\le s\le2$. Find the Doob bound on $\mathbb{E}[\sup_{0\le t\le2}|M_t|^2]$, where $M_t=\int_0^t s\,dB_s$.

The integrand is deterministic and therefore progressively measurable. Its square is integrable because $\mathbb{E}\int_0^2s^2\,ds=\int_0^2s^2\,ds=[s^3/3]_0^2=8/3$. The Ito isometry gives $\mathbb{E}[M_2^2]=8/3$, and the maximal inequality gives $\mathbb{E}[\sup_{0\le t\le2}|M_t|^2]\le4(8/3)=32/3$.

Check: the terminal second moment computed by the Ito isometry is $8/3$, and the theorem’s constant is $4$, so the resulting upper bound is $32/3$. This estimate controls the whole path, not just $M_2$.

Worked Example: Brownian Motion on a Short Interval

Take $H_s=1$ on $[0,\tfrac14]$, so $M_t=\int_0^t1\,dB_s=B_t$. Compute the bound on the expected squared supremum.

Here $H$ is progressively measurable and $\mathbb{E}\int_0^{1/4}1^2\,ds=1/4$. The Ito isometry gives $\mathbb{E}[B_{1/4}^2]=1/4$. Therefore, $\mathbb{E}[\sup_{0\le t\le1/4}|B_t|^2]\le4(1/4)=1$.

Check: at the endpoint alone, the second moment is $1/4$; the pathwise-supremum bound is $1$, four times that terminal value, in agreement with the theorem. The inequality does not say that the supremum’s second moment equals $1$.

Worked Example: An Integrand Depending on Brownian Motion

On $[0,1]$, set $H_s=B_s$ and $M_t=\int_0^tB_s\,dB_s$. Verify the terminal second moment and find the maximal bound.

Brownian motion is progressively measurable. Since $\mathbb{E}[B_s^2]=s$, the integrability condition is $\mathbb{E}\int_0^1B_s^2\,ds=\int_0^1s\,ds=1/2<\infty$. The Ito isometry consequently gives $\mathbb{E}[M_1^2]=1/2$. The maximal inequality gives $\mathbb{E}[\sup_{0\le t\le1}|M_t|^2]\le4(1/2)=2$.

An independent calculation checks the terminal moment. Apply Ito’s formula to $f(t,x)=\tfrac12(x^2-t)$. Its derivatives are $f_t=-\tfrac12$, $f_x=x$, and $f_{xx}=1$. The Ito correction is $\tfrac12 f_{xx}\,dt=\tfrac12dt$, which cancels $f_t\,dt$. Thus $d f(t,B_t)=B_t\,dB_t$, and, since $f(0,B_0)=0$, $M_1=\tfrac12(B_1^2-1)$. For $B_1\sim N(0,1)$, $\mathbb{E}[B_1^2]=1$ and $\mathbb{E}[B_1^4]=3$, so $\mathbb{E}[M_1^2]=\tfrac14\mathbb{E}[B_1^4-2B_1^2+1]=\tfrac14(3-2+1)=1/2$. This matches the isometry calculation, and the corresponding Doob bound is $2$.

Scope, Use, and a Common Mistake

The inequality is useful when the terminal second moment is easier to estimate than the maximum over time. For example, the corollary reduces a pathwise-supremum estimate for a stochastic integral to estimating the ordinary integral $\mathbb{E}\int_0^T H_s^2\,ds$. It is often a key step in proving that a process remains controlled throughout an interval, rather than only at its endpoint.

The hypothesis matters. A local martingale is a process that becomes a martingale after stopping along a suitable increasing sequence of stopping times. The theorem above is stated for a square-integrable martingale, not an arbitrary local martingale. A local martingale need not be a true martingale, and one cannot invoke the theorem for it without establishing the required martingale and integrability properties. Stopping a process may provide estimates for stopped martingales, but passing from those estimates to the unstopped process requires justification.

Common mistake: replacing the expected squared supremum by the largest fixed-time second moment. For Brownian motion on $[0,1]$, this incorrect step would claim $\mathbb{E}[\sup_{0\le t\le1}|B_t|^2]=\sup_{0\le t\le1}\mathbb{E}[B_t^2]=1$. The equality does not follow: the supremum is taken inside the expectation. In fact, with positive probability $B_{1/2}\in(2,2.1)$ and $B_1-B_{1/2}\in(-2.1,-2)$, since these independent normal increments have positive densities on those intervals. On that event, $\sup_{0\le t\le1}|B_t|>2$ while $|B_1|<0.1$. Hence the squared supremum is strictly larger than $B_1^2$ on an event of positive probability, and its expectation is strictly larger than $\mathbb{E}[B_1^2]=1$. The valid general conclusion is the inequality $\mathbb{E}[\sup_{0\le t\le1}|B_t|^2]\le4$, not equality with the endpoint second moment.

Check Your Understanding

Use the theorem and its hypotheses to answer the following questions.

For a continuous square-integrable martingale $M_t$, what quantity does Doob’s L² inequality bound, and what is its constant?
If $M_t=\int_0^tH_s\,dB_s$, which integrability condition on $H$ allows you to use the Ito isometry and the maximal inequality?
For $H_s=2$ on $[0,3]$, compute $\mathbb{E}\int_0^3H_s^2\,ds$ and the resulting maximal bound.
Why does the finite-grid proof use the event that a grid index is the first to cross the threshold?
Why is a local martingale not automatically covered by the stated theorem?