Doob Bounds for Integral Processes

Why Control the Whole Integral Process?

The Ito isometry controls an integral at one fixed time: its mean square is the expected accumulated squared size of the integrand up to that time. In applications, however, an error may matter at any time in an interval. A bound on the terminal value alone does not control the largest error along the way. Doob’s L2 maximal inequality supplies that missing control. Combined with the Ito isometry, it bounds the expected squared maximum of an Ito integral by four times the integrand’s expected energy.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ adapted to the filtration. Fix $T>0$. Throughout, an Ito integrand $H$ on $[0,T]$ is progressively measurable and satisfies $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; this class is denoted $\mathcal{L}^2$. Its integral process is $M_t=\int_0^t H_s\,dB_s$, $0\le t\le T$. The integral is defined as an L2 limit, and the Ito isometry gives $\mathbb{E}[M_t^2]=\mathbb{E}\int_0^t H_s^2\,ds$.

Definition: A process $M_t$ is a martingale if each $M_t$ is integrable, is adapted, and satisfies $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ almost surely whenever $0\le s\le t$. For $H\in\mathcal{L}^2$, the Ito integral process $M_t=\int_0^t H_u\,dB_u$ is a martingale: it is adapted and square-integrable by the isometry, and its increment from $s$ to $t$ has conditional expectation zero given $\mathcal{F}_s$. This conditional-mean property follows first for simple predictable integrands from the mean-zero Brownian increments and then for $\mathcal{L}^2$ integrands by the L2 construction.

Doob’s L2 Bound for a Finite Martingale

First prove the inequality for finitely many observation times. This avoids assuming any path regularity for the integral process. The argument applies to any square-integrable martingale $M_0,\ldots,M_n$, with its maximum taken over these finitely many times.

Theorem: If $M_0,\ldots,M_n$ is a square-integrable martingale and $M^*=\max_{0\le k\le n}|M_k|$, then $$ \mathbb{E}[(M^*)^2]\le 4\mathbb{E}[|M_n|^2]. $$

Proof. Set $X_k=|M_k|$. Conditional Jensen’s inequality gives $\mathbb{E}[X_{k+1}\mid\mathcal{F}_k]\ge|\mathbb{E}[M_{k+1}\mid\mathcal{F}_k]|=|M_k|=X_k$, so $X_k$ is a nonnegative submartingale. For $\lambda>0$, let $A_k$ be the event that $k$ is the first index with $X_k>\lambda$. The events $A_0,\ldots,A_n$ are disjoint, and $A_k\in\mathcal{F}_k$. The submartingale property iterated from $k$ to $n$ gives $\mathbb{E}[X_n\mid\mathcal{F}_k]\ge X_k$. Therefore $$ \mathbb{E}[X_n\mathbf{1}_{\{M^*>\lambda\}}] =\sum_{k=0}^n\mathbb{E}[X_n\mathbf{1}_{A_k}] \ge\sum_{k=0}^n\mathbb{E}[X_k\mathbf{1}_{A_k}] \ge\lambda\,\mathbb{P}(M^*>\lambda). $$ For any nonnegative random variable $Y$, the tail-integral identity gives $\mathbb{E}[Y^2]=\int_0^\infty 2\lambda\,\mathbb{P}(Y>\lambda)\,d\lambda$. Here $M^*$ is square-integrable because $(M^*)^2\le\sum_{k=0}^n |M_k|^2$. Integrating the displayed tail bound and using Tonelli’s theorem, $$ \mathbb{E}[(M^*)^2]\le 2\mathbb{E}[X_n M^*] \le 2\big(\mathbb{E}[X_n^2]\big)^{1/2}\big(\mathbb{E}[(M^*)^2]\big)^{1/2}. $$ If $\mathbb{E}[(M^*)^2]=0$, the claimed inequality holds. Otherwise divide by its positive square root and square both sides to obtain the result. $\square$

Applying the Bound to an Ito Integral

For a process defined at each time, an unrestricted supremum depends on which version of the process is used. The integral construction at this stage gives an L2 random variable $M_t$ for each fixed $t$; it does not by itself select pathwise values simultaneously at every time. We can state and prove the maximal estimate without making that extra choice by using a countable dense set of times.

Theorem: Let $H\in\mathcal{L}^2$, and let $D$ be the dyadic times in $[0,T]$, together with $T$. For any fixed representatives of $M_t=\int_0^t H_s\,dB_s$ at these countably many times, $$ \mathbb{E}\left[\sup_{t\in D}|M_t|^2\right] \le 4\mathbb{E}\int_0^T H_s^2\,ds. $$ If $M$ is taken in a continuous version, then $D$ is dense and this is exactly $\mathbb{E}[\sup_{0\le t\le T}|M_t|^2]\le4\mathbb{E}\int_0^T H_s^2\,ds$.

Proof. For each integer $n\ge0$, let $D_n=\{kT2^{-n}:0\le k\le2^n\}$. These are nested finite grids, and their union is $D$. At the ordered times in $D_n$, the process $M$ is a square-integrable martingale by the conditional-mean property of the Ito integral. The finite-martingale theorem and the Ito isometry yield $$ \mathbb{E}\left[\max_{t\in D_n}|M_t|^2\right]\le4\mathbb{E}[|M_T|^2] =4\mathbb{E}\int_0^T H_s^2\,ds. $$ The grid maxima increase with $n$, and their limit is $\sup_{t\in D}|M_t|$. Monotone convergence gives the claimed bound. If a continuous version is used, continuity implies that the supremum over the dense set $D$ equals the supremum over all of $[0,T]$ on every continuous sample path. $\square$

The constant $4$ comes from Doob’s inequality; the terminal-time contribution is exactly the Ito isometry. Together they give a bound on the expected squared maximum, not just on the terminal error. The continuous-version formulation is useful once such a version has been selected; the countable-set formulation above does not require us to establish its existence here.

Worked Applications

Worked Example: The Brownian Integral with Constant Integrand

Take $T=1$ and $H_s=1$ for $0\le s\le1$. This deterministic integrand is progressively measurable, and its energy is $\mathbb{E}\int_0^1 1^2\,ds=1$. The Ito integral is $M_t=B_t$. The maximal inequality gives $$ \mathbb{E}\left[\sup_{t\in D}|B_t|^2\right]\le4. $$ For a continuous Brownian version, the same bound holds with the supremum over all $t\in[0,1]$.

Check: the terminal second moment is $\mathbb{E}[B_1^2]=1$, since $B_1$ has mean zero and variance $1$. Thus Doob’s bound is $4\mathbb{E}[B_1^2]=4$, matching the energy calculation. The inequality is an upper bound; it does not assert that the expected squared maximum equals $4$.

Worked Example: An Integrand Switched On by the Past

Let $T=2$, $A=\{B_1>0\}$, and $H_s=2\mathbf{1}_A\mathbf{1}_{(1,2]}(s)$. Since $A\in\mathcal{F}_1$, the random coefficient is known before the interval where the integrand is active, so $H$ is progressively measurable. Also $\mathbb{P}(A)=\tfrac12$, because $B_1$ has the continuous, symmetric distribution $N(0,1)$. Hence $$ \mathbb{E}\int_0^2 H_s^2\,ds =\mathbb{E}[4\mathbf{1}_A]\cdot(2-1) =4\cdot\tfrac12\cdot1=2. $$ The integral process is zero for $t\le1$, and for $1\le t\le2$ it is $M_t=2\mathbf{1}_A(B_t-B_1)$. Therefore Doob’s bound is $$ \mathbb{E}\left[\sup_{t\in D}|M_t|^2\right]\le4\cdot2=8. $$

Check: at the terminal time, independence of $B_2-B_1$ from $\mathcal{F}_1$ gives $$ \mathbb{E}[M_2^2]=4\mathbb{E}\left[\mathbf{1}_A(B_2-B_1)^2\right] =4\mathbb{P}(A)\mathbb{E}[(B_2-B_1)^2] =4\cdot\tfrac12\cdot1=2. $$ This agrees with the Ito isometry. The event has positive probability, so the random switch is nontrivial; importantly, it is measurable at the start of the interval, not chosen using a future Brownian increment.

Worked Example: Controlling the Error from Step Approximation

On $[0,1]$, let $H_s=s$, and approximate it by the left-endpoint step process $H^{(n)}_s=k/n$ for $k/n\le s<(k+1)/n$, $k=0,\ldots,n-1$. Both integrands are deterministic and square-integrable. Apply the maximal inequality to their difference: $$ \mathbb{E}\left[\sup_{t\in D}\left|\int_0^t(H_s-H^{(n)}_s)\,dB_s\right|^2\right] \le4\int_0^1|H_s-H^{(n)}_s|^2\,ds. $$ On each interval write $s=k/n+u$, where $0\le u<1/n$. The difference is $u$, so $$ \int_0^1|H_s-H^{(n)}_s|^2\,ds =n\int_0^{1/n}u^2\,du =n\cdot\frac{(1/n)^3}{3} =\frac{1}{3n^2}. $$ Thus the expected squared supremum of the integral error is at most $4/(3n^2)$, which tends to zero as $n$ increases.

For the specific choice $n=2$, the integral of the squared error is $\int_0^{1/2}s^2\,ds+\int_{1/2}^1(s-\tfrac12)^2\,ds=\tfrac1{24}+\tfrac1{24}=\tfrac1{12}$. The maximal bound is therefore $4\cdot\tfrac1{12}=\tfrac13$. Check: substituting $n=2$ into $4/(3n^2)$ gives $4/(3\cdot4)=\tfrac13$, the same result.

Why the Estimate Matters for Convergence

The maximal inequality makes approximation useful over an entire time interval. If $H^{(n)}$ and $H^{(m)}$ are in $\mathcal{L}^2$, their difference is also in $\mathcal{L}^2$, and linearity of the Ito integral gives $$ \mathbb{E}\left[\sup_{t\in D}\left|\int_0^t(H^{(n)}_s-H^{(m)}_s)\,dB_s\right|^2\right] \le4\mathbb{E}\int_0^T|H^{(n)}_s-H^{(m)}_s|^2\,ds. $$ Consequently, if the integrands are Cauchy in the L2 norm $\big(\mathbb{E}\int_0^T |H_s|^2\,ds\big)^{1/2}$, their integral processes are Cauchy in the expected squared supremum over $D$. This is stronger than knowing that the terminal integrals are Cauchy in mean square: it controls discrepancies at every time in the chosen countable set.

It also yields a probability bound. For any $\varepsilon>0$, Markov’s inequality gives $$ \mathbb{P}\left(\sup_{t\in D}\left|\int_0^t(H^{(n)}_s-H^{(m)}_s)\,dB_s\right|>\varepsilon\right) \le\frac{4}{\varepsilon^2}\mathbb{E}\int_0^T|H^{(n)}_s-H^{(m)}_s|^2\,ds. $$ The right-hand side tends to zero when the integrands approach each other in $\mathcal{L}^2$. This explains the role of the maximal estimate: it transfers an energy estimate on integrands into control of the whole family of integral values in time. No claim of almost-sure uniform convergence follows from this bound alone.

A Common Mistake

A tempting but incorrect argument is to use the Ito isometry at time $T$ and write $\mathbb{E}[\sup_{t\in D}|M_t|^2]\le\mathbb{E}[|M_T|^2]$. The isometry evaluates the terminal second moment, not the maximum. For the constant-integrand example, it gives $\mathbb{E}[B_1^2]=1$, while the maximum includes the terminal value and also all earlier values. The valid conclusion is Doob’s bound $\mathbb{E}[\sup_{t\in D}|B_t|^2]\le4$, not the unjustified bound by $1$.

A second point of care is the meaning of the supremum. Before choosing a continuous version, the integral is constructed at each fixed time up to almost-sure equality. The countable dense-set supremum is well-defined after choosing representatives at those countably many times. Replacing it by a supremum over every time requires a version with suitable path regularity; when a continuous version is available, density gives equality of the two suprema.

Check Your Understanding

Use the finite-martingale inequality and the Ito isometry to justify each answer.

For $H_s=3$ on $[0,2]$, compute the integrand energy and the resulting maximal upper bound.
Why does the proof of Doob’s inequality apply to $|M_k|$, even though $|M_k|$ need not be a martingale?
For $H\in\mathcal{L}^2$, identify the exact quantity that replaces $\mathbb{E}[|M_T|^2]$ in the maximal bound.
If $\mathbb{E}\int_0^T|H^{(n)}_s-H_s|^2\,ds\to0$, what bound follows for the expected squared supremum of the integral difference over $D$?
What additional path property lets the supremum over dyadic times equal the supremum over all $t\in[0,T]$?