Doob's L2 Maximal Inequality

Controlling a Martingale’s Running Maximum

A martingale may have modest fluctuations at each fixed time while its largest fluctuation over a whole interval is harder to control. Doob’s L2 maximal inequality gives a uniform bound: the expected square of the largest absolute value up to a fixed time is at most four times the second moment at the terminal time. The proof combines a first-crossing argument with an integral identity for nonnegative random variables.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with a discrete-time filtration $(\mathcal{F}_k)_{0\le k\le N}$, where $N$ is a fixed nonnegative integer. A real-valued process $(M_k)_{0\le k\le N}$, with state space $S=\mathbb{R}$, is a martingale with respect to this filtration if each $M_k$ is $\mathcal{F}_k$-measurable and integrable, and $\mathbb{E}[M_{k+1}\mid\mathcal{F}_k]=M_k$ almost surely for $0\le k<N$. It is square-integrable if $\mathbb{E}[M_k^2]<\infty$ for every $k$. Define the running maximum $M_N^*=\max_{0\le k\le N}|M_k|$.

Theorem (Doob’s L2 maximal inequality): If $(M_k)_{0\le k\le N}$ is a square-integrable martingale, then $$ \mathbb{E}\big[(M_N^*)^2\big]\le 4\,\mathbb{E}[M_N^2]. $$ No assumption that $M_0=0$ is needed. The time horizon $N$ is finite.

The terminal value appears on the right, not a sum of moments over all times. In particular, the inequality controls a maximum taken across the entire finite horizon. We first prove the crossing estimate that makes this possible.

A First-Crossing Bound

The process $Y_k=|M_k|$ is a nonnegative submartingale: it is adapted and integrable, and conditional Jensen’s inequality gives $\mathbb{E}[|M_{k+1}|\mid\mathcal{F}_k]\ge|\mathbb{E}[M_{k+1}\mid\mathcal{F}_k]|=|M_k|$ almost surely. The following argument applies to any nonnegative submartingale $(Y_k)_{0\le k\le N}$ with integrable terms. Write $Y_N^*=\max_{0\le k\le N}Y_k$.

Lemma (First-crossing tail bound): For every $\lambda>0$, $$ \lambda\,\mathbb{P}(Y_N^*>\lambda) \le \mathbb{E}\big[Y_N\mathbf{1}_{\{Y_N^*>\lambda\}}\big]. $$

Proof. For $0\le k\le N$, let $A_k$ be the event that $k$ is the first index with $Y_k>\lambda$. Each $A_k$ belongs to $\mathcal{F}_k$, the events are disjoint, and their union is $\{Y_N^*>\lambda\}$. By the submartingale property and repeated conditioning, $\mathbb{E}[Y_N\mid\mathcal{F}_k]\ge Y_k$ almost surely. Therefore, $\mathbb{E}[Y_N\mathbf{1}_{A_k}]=\mathbb{E}[\mathbb{E}[Y_N\mid\mathcal{F}_k]\mathbf{1}_{A_k}]\ge\mathbb{E}[Y_k\mathbf{1}_{A_k}]\ge\lambda\mathbb{P}(A_k)$, since $Y_k>\lambda$ on $A_k$. Summing over the disjoint events gives $\mathbb{E}[Y_N\mathbf{1}_{\{Y_N^*>\lambda\}}]\ge\lambda\mathbb{P}(Y_N^*>\lambda)$. $\square$

The event in this bound uses a strict crossing, $Y_N^*>\lambda$. This matches the tail-integral identity below; no assumption about whether the maximum can equal a particular threshold is needed.

Proof of the L2 Inequality

Proof. Apply the lemma to $Y_k=|M_k|$. Since the horizon is finite and every $M_k$ is square-integrable, $(M_N^*)^2\le\sum_{k=0}^N M_k^2$, so $\mathbb{E}[(M_N^*)^2]$ is finite. For any nonnegative random variable $Z$, the pointwise identity $Z^2=\int_0^\infty 2\lambda\mathbf{1}_{\{Z>\lambda\}}\,d\lambda$, followed by Tonelli’s theorem, gives $\mathbb{E}[Z^2]=\int_0^\infty 2\lambda\mathbb{P}(Z>\lambda)\,d\lambda$. Using $Z=M_N^*$ and the lemma, \[ \mathbb{E}[(M_N^*)^2] \le 2\int_0^\infty \mathbb{E}\big[|M_N|\mathbf{1}_{\{M_N^*>\lambda\}}\big]\,d\lambda =2\mathbb{E}[|M_N|M_N^*]. \] The equality follows by Tonelli’s theorem, because the integrand is nonnegative and $\int_0^\infty\mathbf{1}_{\{M_N^*>\lambda\}}\,d\lambda=M_N^*$. Cauchy–Schwarz now yields \[ \mathbb{E}[(M_N^*)^2]\le 2\sqrt{\mathbb{E}[M_N^2]}\sqrt{\mathbb{E}[(M_N^*)^2]}. \] If $\mathbb{E}[(M_N^*)^2]=0$, the theorem holds directly. Otherwise, divide by its positive square root and square both sides to obtain $\mathbb{E}[(M_N^*)^2]\le4\mathbb{E}[M_N^2]$. $\square$

Corollary (A probability bound): For any $a>0$, $$ \mathbb{P}(M_N^*\ge a)\le \frac{4\,\mathbb{E}[M_N^2]}{a^2}. $$

Indeed, on $\{M_N^*\ge a\}$, $(M_N^*)^2\ge a^2$, so $a^2\mathbb{P}(M_N^*\ge a)\le\mathbb{E}[(M_N^*)^2]$. The stated bound then follows from the theorem. This consequence is useful when the question concerns the chance of a large excursion rather than its expected squared size.

Worked Applications

Worked Example: A Two-Step Symmetric Walk

Let $\Omega=\{HH,HT,TH,TT\}$, with each outcome having probability $\tfrac14$, and let $\mathcal{F}_k$ reveal the first $k$ coin flips. Set $M_0=0$, and let each head add $1$ while each tail subtracts $1$. The state space is $\mathbb{R}$, and the horizon is $N=2$. Each next increment has conditional mean $\tfrac12(1)+\tfrac12(-1)=0$, so this is a square-integrable martingale.

The paths, their running maxima in absolute value, and their squared maxima are:

Outcome	$(M_0,M_1,M_2)$	$M_2^*$	$(M_2^*)^2$
HH	$(0,1,2)$	2	4
HT	$(0,1,0)$	1	1
TH	$(0,-1,0)$	1	1
TT	$(0,-1,-2)$	2	4

Thus $\mathbb{E}[(M_2^*)^2]=\tfrac14(4+1+1+4)=\tfrac{10}{4}=\tfrac52$. The terminal second moment is $\mathbb{E}[M_2^2]=\tfrac14(4+0+0+4)=2$. The inequality gives $\tfrac52\le4(2)=8$, which holds. Direct enumeration verifies the running-maximum calculation: the two same-flip paths contribute $4$ each, and the two mixed paths contribute $1$ each.

Worked Example: A Conditional-Expectation Martingale

Let $\Omega=\{a,b,c,d\}$, with each outcome having probability $\tfrac14$, and let $\mathcal{F}_0$ be trivial, $\mathcal{F}_1$ distinguish the groups $\{a,b\}$ and $\{c,d\}$, and $\mathcal{F}_2$ reveal the outcome. Take $M_2(a)=2$, $M_2(b)=0$, $M_2(c)=0$, and $M_2(d)=-2$, and set $M_k=\mathbb{E}[M_2\mid\mathcal{F}_k]$ for $k=0,1,2$. These conditional expectations form a square-integrable martingale on the real state space over the horizon $N=2$.

The terminal mean is $\tfrac14(2+0+0-2)=0$, so $M_0=0$. On $\{a,b\}$, the conditional mean is $(2+0)/2=1$; on $\{c,d\}$, it is $(0-2)/2=-1$. The paths are therefore $(0,1,2)$, $(0,1,0)$, $(0,-1,0)$, and $(0,-1,-2)$. Their squared running maxima are $4,1,1,4$, respectively, giving $\mathbb{E}[(M_2^*)^2]=\tfrac{10}{4}=\tfrac52$. Also, $\mathbb{E}[M_2^2]=\tfrac14(4+0+0+4)=2$. Substitution into the inequality again gives $\tfrac52\le8$.

Check: at time $1$, the two possible martingale values $1$ and $-1$ each have probability $\tfrac12$, so $\mathbb{E}[M_1\mid\mathcal{F}_0]=\tfrac12(1)+\tfrac12(-1)=0=M_0$. At time $2$, the conditional means within the two groups are $1$ and $-1$, exactly the corresponding $M_1$ values.

Worked Example: Unequal Increment Sizes

Let $\Omega=\{(++),(+-),(-+),(--)\}$, with each outcome having probability $\tfrac14$, and let $\mathcal{F}_k$ reveal the first $k$ signs. Put $M_0=0$, $M_1=\varepsilon_1$, and $M_2=\varepsilon_1+2\varepsilon_2$, where each sign is $1$ or $-1$. The state space is $\mathbb{R}$, the horizon is $N=2$, and each increment has conditional mean zero: the first sign is symmetric, and given $\mathcal{F}_1$, the second increment has mean $\tfrac12(2)+\tfrac12(-2)=0$. Thus $(M_k)$ is a square-integrable martingale.

For $++$, the path is $(0,1,3)$, so its squared running maximum is $9$. For $+-$, the path is $(0,1,-1)$, giving $1$. For $-+$, the path is $(0,-1,1)$, giving $1$. For $--$, the path is $(0,-1,-3)$, giving $9$. Hence $\mathbb{E}[(M_2^*)^2]=\tfrac14(9+1+1+9)=5$. The terminal values are $3,-1,1,-3$, so $\mathbb{E}[M_2^2]=\tfrac14(9+1+1+9)=5$. The bound is $5\le4(5)=20$, as required. The direct enumeration checks both moments using all four equally likely outcomes.

A Common Error and the Scope of the Bound

A common mistake is to replace the expected squared running maximum by the largest second moment at an individual time. In the four-outcome conditional-expectation example, $\mathbb{E}[M_0^2]=0$, $\mathbb{E}[M_1^2]=\tfrac12(1)+\tfrac12(1)=1$, and $\mathbb{E}[M_2^2]=2$. But the calculation above gives $\mathbb{E}[(M_2^*)^2]=\tfrac52$, so the incorrect claim $\mathbb{E}[(M_2^*)^2]\le\max_{0\le k\le2}\mathbb{E}[M_k^2]$ would assert $\tfrac52\le2$, which is false. The maximum over time is taken separately on each path before expectation; it cannot generally be replaced by the maximum of the separate expectations. Doob’s inequality supplies the valid bound $\tfrac52\le4\mathbb{E}[M_2^2]=8$.

The finite-horizon and square-integrability assumptions matter for the statement proved here: they ensure that the running maximum has finite second moment and that the displayed calculations apply. This result is a bound, not an assertion that the running maximum is itself a martingale or that its second moment equals four times the terminal second moment. Its practical role is to turn control of a terminal L2 quantity into control of the largest absolute value seen along a finite path.

Check Your Understanding

Use the theorem and the first-crossing argument to answer the following questions.

In the first-crossing lemma, why is each event $A_k$ measurable with respect to $\mathcal{F}_k$?
If $\mathbb{E}[M_N^2]=9$, what bound does the theorem give for $\mathbb{E}[(M_N^*)^2]$, and what probability bound follows for $\mathbb{P}(M_N^*\ge6)$?
For the four-outcome conditional-expectation martingale, recompute $\mathbb{E}[M_1^2]$ and compare it with $\mathbb{E}[(M_2^*)^2]$.
Where does the factor $2$ enter when the tail bound is integrated, and where does the second factor $2$ in the final constant come from?
Why does the proof apply to $|M_k|$ even though $M_k$ itself may be negative?

Outcome	\((M_0,M_1,M_2)\)	\(M_2^*\)	\((M_2^*)^2\)
HH	\((0,1,2)\)	2	4
HT	\((0,1,0)\)	1	1
TH	\((0,-1,0)\)	1	1
TT	\((0,-1,-2)\)	2	4

What You'll Learn