The Running Maximum of a Submartingale
A submartingale can have modest moments at each fixed time while still making large excursions along a sample path. A maximal inequality controls those excursions: it bounds the moment of the largest value seen up to a fixed time using only the moment at the terminal time. The key is to first control the probability that the process crosses a threshold, then integrate that probability over all thresholds.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a filtration \((\mathcal{F}_n)_{n\ge0}\). A sequence \((X_n)_{n\ge0}\) is a submartingale if each \(X_n\) is integrable and \(\mathcal{F}_n\)-measurable, and \(\mathbb{E}[X_{n+1}\mid\mathcal{F}_n]\ge X_n\) almost surely. Here we consider nonnegative submartingales: \(X_n\ge0\) almost surely for every \(n\). Fix a finite horizon \(N\), and define the running maximum \(X_N^*=\max_{0\le k\le N}X_k\). For \(p\ge1\), write \(\|Y\|_p=(\mathbb{E}|Y|^p)^{1/p}\) when this quantity is finite.
The theorem does not require the process to be a martingale, but it does require nonnegativity and \(p>1\). We first prove a threshold bound that explains how the maximum is controlled by the terminal value.
First Crossing and the Threshold Bound
Proof. Let \(A=\{X_N^*>\lambda\}\), and let \(T\) be the first index \(0\le n\le N\) for which \(X_n>\lambda\), setting \(T=N+1\) if there is no such index. Then \(A=\{T\le N\}\), and \(\{T\le k\}\in\mathcal{F}_k\) for every \(k\le N\). On \(A\), \(X_T>\lambda\). The identity $$ X_N\mathbf{1}_A-X_T\mathbf{1}_A = \sum_{k=0}^{N-1}\mathbf{1}_{\{T\le k\}}(X_{k+1}-X_k) $$ holds: on \(A\), the sum includes exactly the increments from time \(T\) to time \(N\), and off \(A\) both sides are zero. Each summand has nonnegative expectation, since \(\mathbf{1}_{\{T\le k\}}\) is \(\mathcal{F}_k\)-measurable and $$ \mathbb{E}\!\left[\mathbf{1}_{\{T\le k\}}(X_{k+1}-X_k)\right] = \mathbb{E}\!\left[\mathbf{1}_{\{T\le k\}}\mathbb{E}[X_{k+1}-X_k\mid\mathcal{F}_k]\right]\ge0. $$ Therefore \(\mathbb{E}[X_N\mathbf{1}_A]\ge\mathbb{E}[X_T\mathbf{1}_A]\ge\lambda\mathbb{P}(A)\), proving the claim. If \(N=0\), the sum is empty and the same argument reduces to \(X_0>\lambda\) on \(A\). \(\square\)
The event in this estimate uses a strict crossing, \(X_N^*>\lambda\). This convention matches the tail-integral identity below; changing strict to non-strict does not change the integral, but the event used in the first-crossing argument must be specified consistently.
Deriving the \(L^p\) Bound
Proof of Doob’s \(L^p\) maximal inequality. First check that the maximum has a finite \(p\)th moment, so the norm manipulations are justified. Iterating the submartingale property and using the tower property gives \(X_k\le\mathbb{E}[X_N\mid\mathcal{F}_k]\) almost surely for \(0\le k\le N\). Since both sides are nonnegative, conditional Jensen’s inequality yields $$ \mathbb{E}[X_k^p] \le \mathbb{E}\!\left[\big(\mathbb{E}[X_N\mid\mathcal{F}_k]\big)^p\right] \le \mathbb{E}[X_N^p]. $$ Also \((X_N^*)^p\le\sum_{k=0}^N X_k^p\), so \(\mathbb{E}[(X_N^*)^p]<\infty\).
For any nonnegative random variable \(Y\), the layer-cake identity is \(\mathbb{E}[Y^p]=\int_0^\infty p\lambda^{p-1}\mathbb{P}(Y>\lambda)\,d\lambda\). Apply it to \(X_N^*\), then use the first-crossing bound. Tonelli’s theorem allows the nonnegative integral and expectation to be interchanged:
The last integral is finite precisely because \(p>1\): \(\int_0^x\lambda^{p-2}\,d\lambda=x^{p-1}/(p-1)\) for \(x\ge0\). Hölder’s inequality, with conjugate exponents \(p\) and \(p/(p-1)\), now gives $$ \mathbb{E}[(X_N^*)^p] \le \frac{p}{p-1}\|X_N\|_p\|X_N^*\|_p^{p-1}. $$ If \(\|X_N^*\|_p=0\), the desired bound holds immediately. Otherwise divide by \(\|X_N^*\|_p^{p-1}\) to obtain \(\|X_N^*\|_p\le \frac{p}{p-1}\|X_N\|_p\). Raising both sides to the power \(p\) gives the stated moment bound. \(\square\)
The factor \(p/(p-1)\) comes from two separate operations: integrating \(\lambda^{p-2}\) contributes \(1/(p-1)\), and the layer-cake identity contributes \(p\). Hölder’s inequality then converts the mixed moment involving both \(X_N\) and \(X_N^*\) into the two \(L^p\) norms.
Worked Example: A One-Step Martingale
Let \(\mathcal{F}_0\) be trivial and \(\mathcal{F}_1\) reveal a fair coin. Set \(X_0=1\), and let \(X_1=0\) on tails and \(X_1=2\) on heads. This is a nonnegative martingale because \(\mathbb{E}[X_1\mid\mathcal{F}_0]=\tfrac12(0)+\tfrac12(2)=1=X_0\). Take \(N=1\) and \(p=2\).
The maximum is \(1\) on tails and \(2\) on heads. Thus $$ \mathbb{E}[(X_1^*)^2] =\tfrac12(1^2)+\tfrac12(2^2) =\tfrac12+2 =\tfrac52, \qquad \mathbb{E}[X_1^2] =\tfrac12(0^2)+\tfrac12(2^2) =2. $$ Doob’s bound gives \(\tfrac52\le (2/(2-1))^2\cdot2=8\), as required. The terminal moment is \(2\), and the computed maximum moment \(\tfrac52\) is indeed no larger than \(8\).
Worked Example: A Two-Stage Doubling Martingale
Let \(X_0=1\). At each of the next two steps, if the current value is positive, it doubles with probability \(\tfrac12\) and becomes zero with probability \(\tfrac12\); once it is zero, it stays zero. The filtration reveals the outcomes step by step. From a positive value \(x\), the next conditional mean is \(\tfrac12(2x)+\tfrac12(0)=x\); from zero it remains zero. Hence this is a nonnegative martingale. Its possible paths and their probabilities are: first-step failure, probability \(\tfrac12\), with maximum \(1\); first-step success then failure, probability \(\tfrac14\), with maximum \(2\); and two successes, probability \(\tfrac14\), with maximum \(4\).
Consequently, $$ \mathbb{E}[(X_2^*)^2] =\tfrac12(1^2)+\tfrac14(2^2)+\tfrac14(4^2) =\tfrac12+1+4 =\tfrac{11}{2}. $$ The terminal value is \(4\) with probability \(\tfrac14\) and zero with probability \(\tfrac34\), so $$ \mathbb{E}[X_2^2]=\tfrac14(4^2)+\tfrac34(0^2)=4. $$ The bound is \(\tfrac{11}{2}\le4^2=16\). As a direct check of the martingale property at the last step, conditional on \(X_1=2\) the mean of \(X_2\) is \(\tfrac12(4)+\tfrac12(0)=2\); conditional on \(X_1=0\) it is \(0\).
Why the Exponent Must Exceed One
The proof identifies the role of \(p>1\): integrating \(\lambda^{p-2}\) near zero is finite only when \(p>1\). This is not merely a technical limitation. Even for nonnegative martingales, there is no finite constant that bounds the expected running maximum by the expected terminal value uniformly over all finite horizons.
Worked Example: No Uniform \(L^1\) Maximal Bound
For each positive integer \(N\), let \(X_0=1\). At every step, a fair coin is tossed; while the process is positive, a head doubles its value and a tail sets it to zero. After reaching zero, it remains zero. Use the filtration generated by the coin tosses. The conditional mean calculation is \(\tfrac12(2x)+\tfrac12(0)=x\) at every positive value \(x\), so this is a nonnegative martingale.
If the first tail follows exactly \(k\) heads, where \(0\le k<N\), that path has probability \(2^{-(k+1)}\) and its maximum is \(2^k\). If all \(N\) tosses are heads, the probability is \(2^{-N}\) and the maximum is \(2^N\). Therefore $$ \mathbb{E}[X_N^*] =\sum_{k=0}^{N-1}2^{-(k+1)}2^k+2^{-N}2^N =\sum_{k=0}^{N-1}\tfrac12+1 =\tfrac{N}{2}+1. $$ Meanwhile \(X_N=2^N\) with probability \(2^{-N}\), and is zero otherwise, giving \(\mathbb{E}[X_N]=2^{-N}2^N=1\). The ratio \(\mathbb{E}[X_N^*]/\mathbb{E}[X_N]=N/2+1\) grows without bound as \(N\) increases. Thus no horizon-independent \(L^1\) constant can replace the \(p>1\) bound.
Common mistake. Starting with the threshold estimate and replacing \(\mathbb{E}[X_N\mathbf{1}_{\{X_N^*>\lambda\}}]\) by \(\mathbb{E}[X_N]\) loses the event that makes the integral finite. For \(p=2\), that shortcut would give $$ \mathbb{E}[(X_N^*)^2]\le 2\mathbb{E}[X_N]\int_0^\infty 1\,d\lambda, $$ which is infinite and proves nothing. The correction is to retain the indicator, interchange the nonnegative integral and expectation, and obtain \(\mathbb{E}[X_N(X_N^*)^{p-1}]\); Hölder’s inequality then gives the finite \(L^p\) bound.
Supremum Over All Discrete Times
The finite-horizon inequality also controls the supremum over an infinite sequence when the terminal \(p\)th moments are uniformly bounded. This extension follows by applying the finite result at each horizon and taking an increasing limit.
Proof. Set \(X_N^*=\max_{0\le k\le N}X_k\). For every finite \(N\), the theorem gives \(\mathbb{E}[(X_N^*)^p]\le (p/(p-1))^p\mathbb{E}[X_N^p]\), which is at most \((p/(p-1))^p\sup_n\mathbb{E}[X_n^p]\). The sequence \((X_N^*)^p\) increases pointwise to \((\sup_{n\ge0}X_n)^p\). The monotone convergence theorem therefore gives the claimed bound. \(\square\)
This inequality is useful whenever a conclusion concerns paths rather than values at a single time: it turns control of terminal moments into control of the largest observed value. The threshold estimate is the central step; tail integration and Hölder’s inequality turn that probability control into a moment bound.
Check Your Understanding
For each question, identify the assumptions being used and show the relevant calculation.
- For \(p=3\), what factor multiplies \(\mathbb{E}[X_N^3]\) in the moment inequality?
- In the first-crossing bound, why is \(\{T\le k\}\) measurable with respect to \(\mathcal{F}_k\)?
- For the one-step martingale example, calculate \(\mathbb{E}[X_1^*]\) and \(\mathbb{E}[X_1]\). Does this contradict the absence of a uniform \(L^1\) bound?
- Which integral in the proof becomes divergent at \(p=1\), and what is its value for \(p>1\)?
- State the additional moment condition that allows the finite-horizon result to bound \(\sup_{n\ge0}X_n\).