From a Terminal Expectation to a Maximum
A terminal value describes a process at one time, while a maximum records whether it crossed a threshold at any time along the way. For a nonnegative submartingale, the expected terminal value controls the probability of such a crossing. The key is to split the crossing event according to the first time the threshold is exceeded. On each part of that split, the submartingale property compares the value at the crossing time with the later terminal value.
Fix a probability space \((\Omega,\mathcal{F},\mathbb{P})\), a finite horizon \(n\ge0\), and a filtration \((\mathcal{F}_k)_{0\le k\le n}\). A process \((X_k)_{0\le k\le n}\) is a submartingale with respect to this filtration if each \(X_k\) is integrable and \(\mathcal{F}_k\)-measurable, and \(\mathbb{E}[X_{k+1}\mid\mathcal{F}_k]\ge X_k\) almost surely for \(0\le k<n\). Here we require \(X_k\ge0\) almost surely for every \(k\). All threshold events below are defined using a fixed \(\lambda>0\).
The first-crossing event at time \(k\) depends only on \(X_0,\ldots,X_k\), so it belongs to \(\mathcal{F}_k\). This measurability allows us to use the submartingale inequality conditional on \(\mathcal{F}_k\). Notice also the strictness convention: \(X_n^*>\lambda\) means that at least one value is strictly greater than \(\lambda\); a value equal to \(\lambda\) alone does not count as a crossing.
The First-Crossing Argument
Proof. If \(A_k\) occurs, \(k\) is the first index at which the process exceeds \(\lambda\). Therefore one outcome cannot belong to both \(A_k\) and \(A_\ell\) for distinct indices \(k,\ell\), so the events are disjoint. Each condition defining \(A_k\) concerns \(X_j\) with \(j\le k\); adaptedness gives \(A_k\in\mathcal{F}_k\). Finally, an outcome belongs to \(\{X_n^*>\lambda\}\) exactly when there is at least one index \(k\le n\) with \(X_k>\lambda\). The finite, nonempty set of such indices has a first element, and the outcome then belongs to the corresponding \(A_k\). This proves the union identity. \(\square\)
Proof. By iterating the one-step submartingale inequalities and applying the tower property of conditional expectation, \(\mathbb{E}[X_n\mid\mathcal{F}_k]\ge X_k\) almost surely for each \(k\le n\). For the first-crossing events from the proposition, their \(\mathcal{F}_k\)-measurability gives $$ \mathbb{E}[X_n\mathbf{1}_{A_k}] =\mathbb{E}\!\left[\mathbb{E}[X_n\mid\mathcal{F}_k]\mathbf{1}_{A_k}\right] \ge \mathbb{E}[X_k\mathbf{1}_{A_k}] \ge \lambda\,\mathbb{P}(A_k). $$ The last inequality follows because \(X_k>\lambda\) on \(A_k\); replacing a strict lower bound by a weak inequality is valid. Sum over \(k=0,\ldots,n\). Disjointness and the union identity yield $$ \mathbb{E}[X_n\mathbf{1}_{\{X_n^*>\lambda\}}] \ge \lambda\,\mathbb{P}(X_n^*>\lambda). $$ Since \(X_n\ge0\), its expectation on the crossing event is at most its expectation on the whole space: \(\mathbb{E}[X_n\mathbf{1}_{\{X_n^*>\lambda\}}]\le\mathbb{E}[X_n]\). Combining the inequalities proves the strict-threshold bound. For the event \(\{X_n^*\ge\lambda\}\), define the first-crossing events using \(X_k\ge\lambda\) and preceding values \(X_j<\lambda\). The same argument gives \(X_k\ge\lambda\) on each such event and proves the stated version. \(\square\)
The proof has two distinct ingredients. The submartingale property transports information from the crossing time to time \(n\); nonnegativity then lets us bound the terminal expectation restricted to the crossing event by the unrestricted expectation. The argument does not require independent increments or a Markov property. It uses a fixed finite horizon and integrability of the process values.
Worked Example: A Two-Outcome Martingale
Let \(\Omega=\{u,d\}\), with probabilities \(\mathbb{P}(\{u\})=\tfrac14\) and \(\mathbb{P}(\{d\})=\tfrac34\). Take \(\mathcal{F}_0=\{\varnothing,\Omega\}\), and let \(\mathcal{F}_1=\mathcal{F}\). Define \(X_0=1\), and let \(X_1=4\) on \(u\) and \(X_1=0\) on \(d\). This is a nonnegative martingale because \(\mathbb{E}[X_1\mid\mathcal{F}_0]=\tfrac14(4)+\tfrac34(0)=1=X_0\).
Use horizon \(n=1\) and threshold \(\lambda=3\). The maximum is strictly greater than \(3\) exactly on \(u\), since \(X_0=1\) and \(X_1=4\) there, while both values on \(d\) are at most \(3\). Thus \(\mathbb{P}(X_1^*>3)=\tfrac14\), and \(3\mathbb{P}(X_1^*>3)=3\cdot\tfrac14=\tfrac34\). The terminal expectation is \(\mathbb{E}[X_1]=\tfrac14(4)+\tfrac34(0)=1\). Therefore \(\tfrac34\le1\), as the inequality states.
Check: the probability calculation uses the strict event \(X_1^*>3\), which includes \(u\) but not \(d\). The bound compares \(\tfrac34\) with the recomputed terminal expectation \(1\).
Worked Example: Tracking Two First-Crossing Events
Let \(\Omega=\{UU,UD,DU,DD\}\), with each outcome having probability \(\tfrac14\). The filtration \(\mathcal{F}_0\) is trivial, \(\mathcal{F}_1\) records the first letter, and \(\mathcal{F}_2\) records both letters. Set \(X_0=1\). At time \(1\), let \(X_1=2\) after \(U\) and \(X_1=0\) after \(D\). At time \(2\), set \(X_2=2\) after either \(UU\) or \(UD\), \(X_2=2\) after \(DU\), and \(X_2=0\) after \(DD\).
To check the submartingale condition at time \(0\), compute \(\mathbb{E}[X_1\mid\mathcal{F}_0]=\tfrac12(2)+\tfrac12(0)=1=X_0\). After \(U\), \(X_2=2\) on both outcomes, so \(\mathbb{E}[X_2\mid\mathcal{F}_1]=2=X_1\). After \(D\), \(X_2\) is \(2\) or \(0\) with equal conditional probabilities, so \(\mathbb{E}[X_2\mid\mathcal{F}_1]=\tfrac12(2)+\tfrac12(0)=1\ge0=X_1\). All values are nonnegative, so the theorem applies.
For \(\lambda=1\), the first crossing occurs at time \(1\) on \(U\), giving \(A_1=\{UU,UD\}\), with probability \(\tfrac12\). On the branch \(D\), there is a first crossing at time \(2\) only on \(DU\), so \(A_2=\{DU\}\), with probability \(\tfrac14\). There is no crossing on \(DD\), and \(A_0\) is empty because \(X_0=1\) is not strictly greater than \(1\). Hence \(\mathbb{P}(X_2^*>1)=\tfrac12+\tfrac14=\tfrac34\), and \(1\cdot\mathbb{P}(X_2^*>1)=\tfrac34\).
The terminal expectation is \(\mathbb{E}[X_2]=\tfrac14(2)+\tfrac14(2)+\tfrac14(2)+\tfrac14(0)=\tfrac32\). Thus \(\tfrac34\le\tfrac32\). The first-crossing decomposition also shows the proof’s intermediate comparison: \(X_2=2\) on each of the three crossing outcomes, so \(\mathbb{E}[X_2\mathbf{1}_{\{X_2^*>1\}}]=\tfrac14(2)+\tfrac14(2)+\tfrac14(2)=\tfrac32\), which is at least \(\lambda\mathbb{P}(X_2^*>1)=\tfrac34\).
Worked Example: Turning the Bound into a Threshold Choice
Suppose \(X_0=1\), \(X_1=2\) or \(0\) with probabilities \(\tfrac12\) each, and \(X_2=X_1\). Use the filtration that is trivial at time \(0\) and reveals the outcome at time \(1\). The process is nonnegative, and \(\mathbb{E}[X_1\mid\mathcal{F}_0]=\tfrac12(2)+\tfrac12(0)=1=X_0\), while \(\mathbb{E}[X_2\mid\mathcal{F}_1]=X_1\). It is a martingale, hence also a submartingale.
First calculate the terminal expectation: \(\mathbb{E}[X_2]=\tfrac12(2)+\tfrac12(0)=1\). Choose \(c=2\), so the threshold \(c\mathbb{E}[X_2]\) is \(2\). Because the event uses a strict inequality, \(X_2^*>2\) is empty: the only possible maximum values are \(1\) and \(2\). Its probability is \(0\), and the maximal inequality gives the valid but uninformative bound \(0\le\tfrac12\). To obtain a nonzero crossing probability, instead choose \(c=\tfrac12\). The threshold is then \(\tfrac12\), and \(X_2^*>\tfrac12\) occurs on both outcomes, so its probability is \(1\). The bound gives \(1\le 1/c=2\).
This calculation illustrates that the threshold is a choice, not a promise of a sharp estimate. The resulting probability bound may be loose, and a strict threshold matters when the process can equal the threshold exactly.
How to Read and Use the Bound
Proof. When \(a>0\), apply the theorem with \(\lambda=ca>0\) and divide both sides by \(ca\), obtaining \(\mathbb{P}(X_n^*>ca)\le \mathbb{E}[X_n]/(ca)=1/c\). When \(\mathbb{E}[X_n]=0\), the theorem gives \(0\le\lambda\mathbb{P}(X_n^*>\lambda)\le0\). Since \(\lambda>0\), the probability is zero. \(\square\)
The weak \(L^1\) inequality is useful when the quantity of interest is a pathwise maximum but only a terminal expectation is easy to control. It converts an expectation bound into a tail-probability bound at any chosen positive threshold. The estimate can be conservative: it need not give a probability close to the actual crossing probability. Its strength is that it needs no independence assumption and works uniformly over all times up to the specified finite horizon.
Common mistake: dropping nonnegativity. The final proof step uses \(\mathbb{E}[X_n\mathbf{1}_A]\le\mathbb{E}[X_n]\), which can fail if \(X_n\) is sometimes negative. For a concrete counterexample, take a one-step process with \(X_0=0\), and \(X_1=1\) or \(-1\) with probabilities \(\tfrac12\) each. With the trivial initial filtration and the final filtration revealing the outcome, this is an integrable martingale because \(\mathbb{E}[X_1\mid\mathcal{F}_0]=0=X_0\). At \(\lambda=\tfrac12\), \(\mathbb{P}(X_1^*>\tfrac12)=\tfrac12\), so \(\lambda\mathbb{P}(X_1^*>\lambda)=\tfrac12\cdot\tfrac12=\tfrac14\), whereas \(\mathbb{E}[X_1]=\tfrac12(1)+\tfrac12(-1)=0\). The claimed inequality would say \(\tfrac14\le0\), which is false. This does not contradict Doob's weak \(L^1\) maximal inequality: the process is not nonnegative.
Check Your Understanding
For each question, keep the strict crossing convention \(X_n^*>\lambda\) unless the question specifies otherwise.
- In the first-crossing decomposition, why is \(A_k\) measurable with respect to \(\mathcal{F}_k\)?
- Which step in the proof requires nonnegativity of \(X_n\), rather than only the submartingale property?
- If \(\mathbb{E}[X_n]=6\), what bound does the theorem give for \(\mathbb{P}(X_n^*>3)\)?
- For a process whose maximum can equal \(\lambda\) but never exceed it, what are the probabilities of \(X_n^*>\lambda\) and \(X_n^*\ge\lambda\)?
- For the signed one-step martingale in the pitfall, recompute \(\lambda\mathbb{P}(X_1^*>\lambda)\) at \(\lambda=\tfrac12\), and identify the failed proof step.