Doob's Maximal Inequality for Continuous-Time Submartingales

Controlling a Submartingale’s Maximum

A submartingale tends to increase in conditional expectation, but this does not prevent its paths from making large excursions. Doob’s weak-type maximal inequality bounds the probability of such an excursion using only the expected terminal value. Its proof is a first-crossing argument: on a finite grid, split the event of crossing a threshold according to the first time the process crosses it. Right continuity then allows the grid bound to control the maximum over every time in a finite interval.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. A process $X=(X_t)_{0\le t\le T}$ is a submartingale if it is adapted, each $X_t$ is integrable, and $\mathbb{E}[X_t\mid\mathcal{F}_s]\ge X_s$ almost surely whenever $0\le s\le t\le T$. Assume $T>0$, that $X$ has càdlàg paths (right-continuous paths with left limits), and that it is nonnegative at every time outside one null set. In particular, $\mathbb{E}[X_T]$ is finite and nonnegative.

Theorem: For every $\lambda>0$, $$ \mathbb{P}\!\left(\sup_{0\le t\le T}X_t\ge\lambda\right) \le \frac{\mathbb{E}[X_T]}{\lambda}. $$ The strict-event version also holds: $$ \mathbb{P}\!\left(\sup_{0\le t\le T}X_t>\lambda\right) \le \frac{\mathbb{E}[X_T]}{\lambda}. $$

The quotient is a probability bound, so it is informative when it is below $1$. If it exceeds $1$, the inequality remains true but the trivial bound by $1$ is sharper. The result applies to true submartingales, whose values are integrable and satisfy the conditional-expectation inequality. It does not, without further hypotheses, assert the same bound for a local martingale.

The First-Crossing Bound on a Finite Grid

Let $0=t_0<t_1<\cdots<t_n=T$ be a finite grid, and put $M=\max_{0\le i\le n}X_{t_i}$. The key estimate is $\mathbb{P}(M\ge a)\le \mathbb{E}[X_T]/a$ for $a>0$. The terminal time must be included: the conditional submartingale inequality compares each grid value with $X_T$.

Proof. Define $A_i$ to be the event that $i$ is the first index with $X_{t_i}\ge a$. Then $A_i\in\mathcal{F}_{t_i}$, the events $A_0,\ldots,A_n$ are disjoint, and their union is $\{M\ge a\}$. By the submartingale property and the defining property of conditional expectation,

\mathbb{E}[X_T\mathbf{1}_{A_i}] =\mathbb{E}\!\left[\mathbb{E}[X_T\mid\mathcal{F}_{t_i}]\mathbf{1}_{A_i}\right] \ge \mathbb{E}[X_{t_i}\mathbf{1}_{A_i}] \ge a\,\mathbb{P}(A_i).

Summing over $i$ gives $\mathbb{E}[X_T\mathbf{1}_{\{M\ge a\}}]\ge a\,\mathbb{P}(M\ge a)$. Since $X_T\ge0$, we also have $\mathbb{E}[X_T\mathbf{1}_{\{M\ge a\}}]\le\mathbb{E}[X_T]$. Combining these inequalities proves the finite-grid estimate. The argument also covers $A_0$: the submartingale comparison at $t_0=0$ is valid, and if the crossing occurs at the initial time then $X_{t_0}\ge a$. $\square$

Worked Example: A Martingale That Hits Its Bound Sharply

Let $T=2$, and let $\Omega=\{\omega_1,\omega_2\}$ with probability $1/2$ on each outcome. Set $\mathcal{F}_t$ to be trivial for $t<1$ and to contain all subsets of $\Omega$ for $t\ge1$. Define $X_t=1$ for $t<1$; for $t\ge1$, set $X_t(\omega_1)=2$ and $X_t(\omega_2)=0$. This process is adapted, nonnegative and càdlàg.

For $s<t<1$, both values are $1$. If $s<1\le t$, then $\mathbb{E}[X_t\mid\mathcal{F}_s]=(2+0)/2=1=X_s$. If $1\le s\le t$, the process is constant on each outcome. Thus it is a martingale, and $\mathbb{E}[X_T]=(2+0)/2=1$.

At threshold $\lambda=2$, the event $\{\sup_{0\le t\le2}X_t\ge2\}$ is exactly $\{\omega_1\}$, so its probability is $1/2$. The bound is $1/2$, giving equality. At threshold $\lambda=3/2$, the event is still $\{\omega_1\}$, of probability $1/2$, while the bound is $1/(3/2)=2/3$.

Check: at $\lambda=2$, both sides equal $1/2$. At $\lambda=3/2$, direct enumeration gives $1/2\le2/3$. The expectation used in both bounds is the terminal expectation $1$, not the expected maximum.

Passing from Grids to Continuous Time

For each integer $m\ge0$, use the nested grid $G_m=\{kT2^{-m}:0\le k\le2^m\}$, and let $M_m=\max_{t\in G_m}X_t$. Every grid includes both endpoints, and $G_m\subseteq G_{m+1}$. For each path that is right-continuous, the supremum over these grid values equals the supremum over the whole interval. Indeed, if $t<T$, grid points at or to the right of $t$ can be chosen to converge to $t$; right continuity makes their process values converge to $X_t$. The endpoint $T$ belongs to every grid. Consequently, $\sup_{0\le t\le T}X_t=\sup_{m\ge0}M_m$. This also makes the supremum measurable, since it is a supremum over a countable set of random variables.

Proof of the theorem. First fix $\lambda>0$. By the finite-grid estimate, $\mathbb{P}(M_m\ge\lambda)\le\mathbb{E}[X_T]/\lambda$ for every $m$. The events $\{M_m>\lambda\}$ increase with $m$, and their union is $\{\sup_{0\le t\le T}X_t>\lambda\}$. Countable additivity, in its continuity-from-below form, therefore gives \[ \mathbb{P}\!\left(\sup_{0\le t\le T}X_t>\lambda\right) =\lim_{m\to\infty}\mathbb{P}(M_m>\lambda) \le\frac{\mathbb{E}[X_T]}{\lambda}. \] Here $\{M_m>\lambda\}\subseteq\{M_m\ge\lambda\}$, so the finite-grid estimate applies to each term.

For the inclusive event, take $\lambda_j=\lambda(1-1/j)$ for integers $j\ge2$. These thresholds are positive and increase to $\lambda$. The events $\{\sup_tX_t>\lambda_j\}$ decrease to $\{\sup_tX_t\ge\lambda\}$: a supremum equal to $\lambda$ belongs to every event in the sequence, while a supremum less than $\lambda$ eventually falls below $\lambda_j$. Continuity of probability from above and the strict-event bound give \[ \mathbb{P}\!\left(\sup_{0\le t\le T}X_t\ge\lambda\right) =\lim_{j\to\infty}\mathbb{P}\!\left(\sup_{0\le t\le T}X_t>\lambda_j\right) \le\lim_{j\to\infty}\frac{\mathbb{E}[X_T]}{\lambda_j} =\frac{\mathbb{E}[X_T]}{\lambda}. \] This proves both versions, including the case $\mathbb{E}[X_T]=0$. $\square$

Two Further Applications

Worked Example: A Deterministic Submartingale

On $[0,2]$, let $X_t=1+t/2$, with any filtration. The process is adapted and integrable. For $s\le t$, it is deterministic and $X_t-X_s=(t-s)/2\ge0$, so $\mathbb{E}[X_t\mid\mathcal{F}_s]=X_t\ge X_s$. It is nonnegative and continuous, with $X_T=X_2=1+2/2=2$.

At $\lambda=3/2$, the process first reaches the threshold when $1+t/2=3/2$, that is, at $t=1$. Hence $\mathbb{P}(\sup_{0\le t\le2}X_t\ge3/2)=1$, while the bound is $2/(3/2)=4/3$, which is valid but weaker than $1$. At $\lambda=2$, the threshold is reached at $t=2$, so the inclusive-event probability is $1$, and the bound is $2/2=1$.

Check: the maximum is $X_2=2$, so both event probabilities follow directly from comparing $2$ with the thresholds. The inclusive inequality matters at $\lambda=2$: replacing $\ge$ by $>$ would make that event empty.

Worked Example: A Strict Submartingale on a Finite Space

Let $U,V$ be independent fair bits, so each of the four pairs $(U,V)\in\{0,1\}^2$ has probability $1/4$. Use the filtration that is trivial before time $1$, reveals $U$ from time $1$ to before time $2$, and reveals both bits from time $2$ onward. Define $X_t=1$ for $t<1$; at $t=1$ and for $1\le t<2$, set $X_t=0$ if $U=0$ and $X_t=2$ if $U=1$. At $t=2$, set $X_2=0$ if $U=0$, and set $X_2=2$ when $U=1,V=0$ and $X_2=4$ when $U=1,V=1$. This is an adapted, nonnegative càdlàg process.

Before time $1$, the conditional mean of $X_1$ is $(0+2)/2=1=X_0$. Given $U=0$, the conditional mean of $X_2$ is $0=X_1$; given $U=1$, it is $(2+4)/2=3\ge2=X_1$. Thus the process is a submartingale; the second conditional inequality is strict. Its terminal expectation is $\mathbb{E}[X_2]=(0+0+2+4)/4=3/2$.

For $\lambda=3$, the maximum is at least $3$ only for $(U,V)=(1,1)$, so the probability is $1/4$. The maximal inequality gives $(3/2)/3=1/2$, and direct enumeration confirms $1/4\le1/2$.

Check: the four terminal values are $0,0,2,4$, whose average is $6/4=3/2$. Exactly one of the four paths reaches $3$, giving $1/4$, below the computed bound $1/2$.

Why Nonnegativity and the Threshold Convention Matter

The proof uses nonnegativity at the step $\mathbb{E}[X_T\mathbf{1}_A]\le\mathbb{E}[X_T]$. Without it, the weak-type inequality need not hold. For example, the constant process $X_t=-1$ is an integrable submartingale, but for $\lambda=1$ its maximum is never at least $1$, so this example does not contradict the bound. Instead, applying the stated formula to the event of a large absolute value would give the false claim $\mathbb{P}(\sup_t|X_t|\ge1)\le\mathbb{E}[X_T]= -1$. The theorem concerns the one-sided maximum of a nonnegative process; it is not a bound for the absolute maximum of a signed submartingale.

A second error is to replace the terminal expectation by $\mathbb{E}[\sup_tX_t]$ in the proof. The first-crossing argument compares $X_{t_i}$ with $X_T$ on the crossing event, and nonnegativity then bounds the resulting expectation by $\mathbb{E}[X_T]$. It does not establish a bound involving the expected maximum. The maximal inequality is useful precisely because it controls the probability of a high pathwise maximum from a single terminal-time quantity.

Check Your Understanding

Use the first-crossing proof and keep the distinction between strict and inclusive events explicit.

In the finite-grid proof, why must $A_i$ belong to $\mathcal{F}_{t_i}$?
For a nonnegative submartingale with $\mathbb{E}[X_T]=0$, what does the theorem imply about $\sup_{0\le t\le T}X_t$?
Why does right continuity suffice to recover $X_t$ from grid points approaching $t<T$, and why must $T$ be included separately?
For the finite-space example with $U,V$, compute the probability that the maximum is at least $2$, and compare it with the bound at $\lambda=2$.
Explain why the inclusive-event proof uses thresholds increasing to $\lambda$, rather than applying the strict-event bound directly at $\lambda$.