From a Maximal Bound to a Threshold
A maximal inequality bounds the probability that a process ever crosses a level during a specified time interval. In applications, the direction of the question is often reversed: a probability tolerance is chosen first, and we need a level that guarantees the crossing probability is below it. That reversal is algebraically simple, but it requires care about which moment is available, which maximal inequality applies, and whether the event uses a strict or non-strict threshold.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with filtration \((\mathcal{F}_n)_{n\ge0}\). Let \((M_n)_{n\ge0}\) be an integrable martingale adapted to this filtration, and fix a finite horizon \(N\ge0\). Define its absolute maximum by \(M_N^*=\max_{0\le k\le N}|M_k|\). For \(a>0\), the event that this maximum reaches or exceeds \(a\) is \(\{M_N^*\ge a\}\). For an event of the form \(\{M_N^*>a\}\), the bounds below also apply, since \(\{M_N^*>a\}\subseteq\{M_N^*\ge a\}\).
The weak bound requires only the first absolute moment of the terminal value. The \(L^p\) bound requires a finite \(p\)th moment for some \(p>1\). In both cases the maximum includes time \(0\), and the terminal moment is that of \(M_N\), not a moment of a randomly selected time. The second probability bound follows from the moment inequality because on \(\{M_N^*\ge a\}\), the nonnegative variable \((M_N^*)^p\) is at least \(a^p\).
Choosing a Level for a Prescribed Risk
Let \(\delta\in(0,1)\) be a desired upper bound on the crossing probability. Solving each inequality for \(a\) gives a threshold sufficient to ensure \(\mathbb{P}(M_N^*\ge a)\le\delta\). The threshold is a guarantee, not necessarily the smallest level at which the actual probability is below \(\delta\).
For the weak bound, multiply \(\mathbb{E}|M_N|/a\le\delta\) by \(a>0\), then divide by \(\delta>0\). For the \(L^p\) bound, write \(C_p=p/(p-1)\). The condition \(C_p^p\mathbb{E}|M_N|^p/a^p\le\delta\) is equivalent, since \(a>0\), to \(a^p\ge C_p^p\mathbb{E}|M_N|^p/\delta\). Taking the positive \(p\)th root gives the stated threshold. This proves both threshold rules.
The zero-moment case causes no problem. If \(\mathbb{E}|M_N|^p=0\), then \(M_N=0\) almost surely. The martingale property gives \(M_k=\mathbb{E}[M_N\mid\mathcal{F}_k]=0\) almost surely for every \(k\le N\); the crossing probability is therefore zero for every \(a>0\). Otherwise, the displayed thresholds are positive. The target \(\delta\) is taken strictly between zero and one; setting \(\delta=0\) cannot generally be handled by a finite threshold from these bounds.
If both the weak bound and one or more \(L^p\) bounds are available, compare their thresholds at the same \(\delta\). Any one of the bounds that is at most \(\delta\) is sufficient, so the smallest of the resulting sufficient thresholds is the strongest guarantee obtained from this collection. Equivalently, at any fixed level \(a>0\), one may take the minimum of the probability upper bounds, also using the universal bound \(1\).
Worked Example: A Two-Step Symmetric Random Walk
Let \(\varepsilon_1,\varepsilon_2\) be independent, each taking values \(1\) and \(-1\) with probability \(1/2\). Use the filtration generated by the first \(n\) increments and set \(M_0=0\), \(M_1=\varepsilon_1\), and \(M_2=\varepsilon_1+\varepsilon_2\). This is a martingale: each increment has conditional mean zero. Find the weak and \(L^2\) bounds at level \(a=2\).
The four increment pairs \( (1,1),(1,-1),(-1,1),(-1,-1) \) each have probability \(1/4\). Their terminal values \(M_2\) are \(2,0,0,-2\), respectively. Thus \(\mathbb{E}|M_2|=(2+0+0+2)/4=1\), and \(\mathbb{E}|M_2|^2=(4+0+0+4)/4=2\). The rows of the calculation are the four equally likely outcomes, whose probabilities sum to \(4(1/4)=1\).
The weak bound at \(a=2\) is \(\mathbb{P}(M_2^*\ge2)\le1/2\). The \(L^2\) bound has \(p/(p-1)=2\), so it gives \(\mathbb{P}(M_2^*\ge2)\le 2^2\cdot2/2^2=2\). Since a probability is at most \(1\), the latter gives only the trivial bound \(1\).
Check: The maximum reaches \(2\) exactly on the two outcomes with equal-sign increments, so its probability is \(2/4=1/2\). The weak bound is exact here, while the \(L^2\) estimate before capping at \(1\) is \(2\), as computed.
Choosing Between Moment Bounds
The \(L^p\) threshold is easiest to compare across exponents when written using the terminal \(L^p\) norm \(\|M_N\|_p=(\mathbb{E}|M_N|^p)^{1/p}\). For a fixed target \(\delta\), the threshold is \(C_p\|M_N\|_p\delta^{-1/p}\), where \(C_p=p/(p-1)\). The constant \(C_p\) decreases as \(p\) increases, but the norm can increase; therefore, having a higher moment does not by itself guarantee a smaller threshold. Calculate the candidate thresholds using the actual moments before choosing one.
Worked Example: Comparing the Weak and \(L^2\) Thresholds
On a two-outcome probability space, let \(\mathcal{F}_0\) be trivial, let \(\mathcal{F}_1=\mathcal{F}\), and define \(M_0=1\). Let \(M_1=2\) on one outcome and \(M_1=0\) on the other, each with probability \(1/2\). Since \(\mathbb{E}[M_1\mid\mathcal{F}_0]=(2+0)/2=1=M_0\), this is a martingale. For target \(\delta=1/4\), calculate both thresholds.
The terminal absolute first moment is \(\mathbb{E}|M_1|=(2+0)/2=1\). The weak threshold is therefore \(1/(1/4)=4\). For the second moment, \(\mathbb{E}|M_1|^2=(4+0)/2=2\). The \(L^2\) threshold is \(2\sqrt{2/(1/4)}=2\sqrt{8}=4\sqrt{2}\). Since \(4<4\sqrt{2}\), the weak bound gives the smaller sufficient threshold.
Check: At level \(4\), the maximum \(\max(|M_0|,|M_1|)\) is either \(1\) or \(2\), so its probability of reaching \(4\) is \(0\), which is at most \(1/4\). Substitution in the weak bound gives \(1/4\). The \(L^2\) bound at the same level gives \(4\cdot2/4^2=1/2\), which is valid but does not meet the target. The thresholds are sufficient bounds, and need not be sharp.
Worked Example: A Fourth Moment Gives a Better Threshold
Let \(M_0=0\), and let \(M_1\) equal \(1\) or \(-1\) with probability \(1/2\) each, under the filtration that is trivial at time \(0\) and reveals \(M_1\) at time \(1\). The conditional mean of \(M_1\) at time \(0\) is zero, so this is a martingale. For \(\delta=1/16\), compare the \(L^2\) and \(L^4\) thresholds.
For each of \(p=2\) and \(p=4\), \(\mathbb{E}|M_1|^p=(1+1)/2=1\), so \(\|M_1\|_p=1\). At \(p=2\), the threshold is \(2(1/(1/16))^{1/2}=2\cdot4=8\). At \(p=4\), it is \((4/3)(1/(1/16))^{1/4}=(4/3)\cdot2=8/3\). Thus the \(L^4\) bound supplies the smaller sufficient threshold for this target.
Check: The maximum is \(1\) on both outcomes, so it never reaches \(8\) or \(8/3\). Both thresholds indeed guarantee a probability no greater than \(1/16\). In this example all terminal absolute moments equal \(1\); the comparison comes from the constants and the factor \(\delta^{-1/p}\). This does not imply that a higher exponent is always preferable, because in other examples the \(L^p\) norm itself changes with \(p\).
Common Errors and Practical Use
A common error is to invert the \(L^p\) probability bound without taking the \(p\)th root. In the preceding example, for \(p=4\), the moment-bound numerator is \((4/3)^4\mathbb{E}|M_1|^4=(256/81)\cdot1=256/81\). The incorrect calculation \(a=(256/81)/(1/16)=4096/81\) treats a fourth-power threshold as a threshold. The correct inequality is \(a^4\ge(256/81)/(1/16)=4096/81\), so \(a\ge(4096/81)^{1/4}=8/3\). Substitution checks it: \((4/3)^4/(8/3)^4=(256/81)/(4096/81)=1/16\).
Another error is to use the \(L^p\) formula at \(p=1\). Its constant \(p/(p-1)\) is undefined there, and Doob's strong \(L^p\) inequality in this form requires \(p>1\). Use the weak \(L^1\) inequality when only the first absolute moment is known. Also distinguish the event \(\{M_N^*\ge a\}\) from \(\{M_N^*>a\}\): the latter is contained in the former at the same \(a\), so an upper bound for the non-strict event also bounds the strict one. A bound for the non-strict event does not automatically describe the exact probability of either event.
In practice, first specify the finite horizon and the desired probability tolerance. Then identify the terminal moments that are actually finite, compute each available threshold at that same tolerance, and choose the smallest sufficient one. Finally, check whether known bounds on the process make a threshold uninformative. Doob's inequalities provide general guarantees from terminal moments; they need not capture the detailed shape of the maximum's distribution.
Check Your Understanding
For each question, state the event being bounded and show the threshold calculation before interpreting it.
- If \(\mathbb{E}|M_N|=3\) and the target crossing probability is \(1/10\), what threshold does the weak inequality provide?
- If \(p=2\), \(\mathbb{E}|M_N|^2=5\), and \(\delta=1/5\), calculate the \(L^2\) threshold.
- Why is it valid to use a bound for \(\mathbb{P}(M_N^*\ge a)\) as an upper bound for \(\mathbb{P}(M_N^*>a)\)?
- In the two-outcome example with terminal values \(1\) and \(-1\), calculate the \(L^2\) threshold for \(\delta=1/4\), and compare it with the \(L^4\) threshold.
- What assumption on \(p\) is required for the stated \(L^p\) threshold rule, and which inequality applies when only the first absolute moment is known?