Why the First Crossing Matters
A process may reach a high level and then fall before the time at which we inspect it. For that reason, a large maximum is not the same event as a large terminal value. The first time the process crosses a threshold gives a way to handle this distinction: separate paths according to when they first cross, then use the submartingale property to compare the value at that time with the conditional expected value at the end of the horizon.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a filtration \((\mathcal{F}_n)_{n=0}^N\), where \(N\) is a fixed nonnegative integer. A real-valued process \((X_n)_{n=0}^N\) is adapted if \(X_n\) is \(\mathcal{F}_n\)-measurable for every \(n\). It is an integrable submartingale if each \(X_n\) is integrable and \(\mathbb{E}[X_{n+1}\mid\mathcal{F}_n]\ge X_n\) almost surely for \(0\le n<N\). In this tutorial assume also \(X_n\ge0\) almost surely for every \(n\). For a threshold \(\lambda>0\), the event of a crossing by time \(N\) is \(A=\{\max_{0\le n\le N}X_n\ge\lambda\}\). The inequality in this event is inclusive: a value exactly equal to \(\lambda\) counts as a crossing.
The Partition and the Expectation Bound
Because the process is adapted, each \(A_k\) belongs to \(\mathcal{F}_k\): the event is determined by \(X_0,\ldots,X_k\), all of which are measurable with respect to \(\mathcal{F}_k\). The events are disjoint, since a path cannot have two different first crossing times. Their union is \(A\), since every path in \(A\) has a first index at which \(X_k\ge\lambda\). If no crossing occurs, the path belongs to none of the \(A_k\).
Proof. Measurability follows because \(A_k\) is an intersection of events of the form \(\{X_j<\lambda\}\) for \(j<k\) and \(\{X_k\ge\lambda\}\), and adaptedness gives \(X_j\) measurable with respect to \(\mathcal{F}_k\) whenever \(j\le k\). If a path belonged to both \(A_k\) and \(A_\ell\) for \(k<\ell\), membership in \(A_k\) would give \(X_k\ge\lambda\), while membership in \(A_\ell\) would require \(X_k<\lambda\); this is impossible. Finally, a path is in the maximum event exactly when the set of indices \(n\in\{0,\ldots,N\}\) with \(X_n\ge\lambda\) is nonempty. A nonempty finite set of indices has a least element \(k\), and the path then belongs to \(A_k\). Every path in any \(A_k\) has \(X_k\ge\lambda\), so belongs to the maximum event. Thus the union is exactly the stated event. \(\square\)
Proof. First, the submartingale inequalities and the tower property of conditional expectation imply \(\mathbb{E}[X_N\mid\mathcal{F}_k]\ge X_k\) almost surely for each \(k\). To verify this, the claim holds at \(k=N\) with equality. If it holds at \(k+1\), then \(\mathbb{E}[X_N\mid\mathcal{F}_k]=\mathbb{E}[\mathbb{E}[X_N\mid\mathcal{F}_{k+1}]\mid\mathcal{F}_k]\ge\mathbb{E}[X_{k+1}\mid\mathcal{F}_k]\ge X_k\). Backward induction proves the claim for every \(k\). Since \(A_k\in\mathcal{F}_k\), the defining property of conditional expectation gives \(\mathbb{E}[X_N\mathbf{1}_{A_k}]=\mathbb{E}[\mathbf{1}_{A_k}\mathbb{E}[X_N\mid\mathcal{F}_k]]\ge\mathbb{E}[X_k\mathbf{1}_{A_k}]\). On \(A_k\), \(X_k\ge\lambda\), so this is at least \(\lambda\mathbb{P}(A_k)\). Sum over \(k=0,\ldots,N\). The first-crossing events are disjoint and have union \(A\), so the left-hand expectations sum to \(\mathbb{E}[X_N\mathbf{1}_A]\), and the probabilities sum to \(\mathbb{P}(A)\). This proves the first inequality. For the second, \(X_N\ge0\) implies \(X_N\mathbf{1}_A\le X_N\) almost surely; taking expectations proves the result. \(\square\)
The first inequality is the part that uses the crossing decomposition. It says that the terminal value, restricted to paths that crossed, has expectation at least the threshold multiplied by the crossing probability. Nonnegativity is then needed for the second inequality, which removes the restriction to \(A\). The argument is for a fixed finite horizon and does not require an unbounded stopping-time theorem.
Worked Examples
Worked Example: A One-Step Bound
Let \(\Omega=\{a,b\}\), with \(\mathbb{P}(\{a\})=\tfrac13\) and \(\mathbb{P}(\{b\})=\tfrac23\). Let \(\mathcal{F}_0=\{\varnothing,\Omega\}\), let \(\mathcal{F}_1\) contain all subsets of \(\Omega\), and set \(X_0=1\), \(X_1(a)=2\), and \(X_1(b)=\tfrac12\). Check the submartingale condition and apply the bound with \(N=1\) and \(\lambda=\tfrac32\).
Since \(\mathbb{E}[X_1\mid\mathcal{F}_0]=\mathbb{E}[X_1]\), compute \(\mathbb{E}[X_1]=\tfrac13(2)+\tfrac23(\tfrac12)=\tfrac23+\tfrac13=1=X_0\). Thus the process is a nonnegative martingale, hence a submartingale. Also \(X_0=1<\tfrac32\), so \(A_0=\varnothing\). At time 1, the crossing occurs exactly on \(\{a\}\), giving \(A_1=A=\{a\}\) and \(\mathbb{P}(A)=\tfrac13\).
On \(A\), \(X_1=2\), so \(\mathbb{E}[X_1\mathbf{1}_A]=\tfrac13(2)=\tfrac23\). The bound reads \(\lambda\mathbb{P}(A)=\tfrac32\cdot\tfrac13=\tfrac12\le\tfrac23\le1=\mathbb{E}[X_1]\). The terminal expectation is \(1\), as calculated above, verifying the upper bound as well.
Worked Example: Crossings at Two Different Times
Let \(N=2\), \(X_0=1\), and construct a finite probability space by choosing \(X_1=2\) with probability \(\tfrac13\) and \(X_1=\tfrac12\) with probability \(\tfrac23\). If \(X_1=2\), let \(X_2=0\) or \(4\), each with conditional probability \(\tfrac12\). If \(X_1=\tfrac12\), let \(X_2=2\) with conditional probability \(\tfrac14\) and \(X_2=0\) with conditional probability \(\tfrac34\). Take \(\mathcal{F}_n\) to record the values observed through time \(n\). Apply the bound at \(\lambda=\tfrac32\).
At time 1, the conditional mean is \(\mathbb{E}[X_1]=\tfrac13(2)+\tfrac23(\tfrac12)=1=X_0\). At time 2, conditional on \(X_1=2\), the mean is \(\tfrac12(0)+\tfrac12(4)=2\). Conditional on \(X_1=\tfrac12\), it is \(\tfrac14(2)+\tfrac34(0)=\tfrac12\). Thus \(\mathbb{E}[X_2\mid\mathcal{F}_1]=X_1\) in both cases. All values are nonnegative, so this is a nonnegative martingale.
Since \(X_0=1<\tfrac32\), there is no crossing at time 0. The event \(A_1\) is \(\{X_1=2\}\), with probability \(\tfrac13\). On the branch \(X_1=\tfrac12\), the process crosses at time 2 exactly when \(X_2=2\), so \(A_2\) has probability \(\tfrac23\cdot\tfrac14=\tfrac16\). Therefore \(A=A_1\cup A_2\) and \(\mathbb{P}(A)=\tfrac13+\tfrac16=\tfrac12\).
For the restricted terminal expectation, the two terminal outcomes on \(A_1\) have probabilities \(\tfrac13\cdot\tfrac12=\tfrac16\) each and values 0 and 4. Their contribution is \(\tfrac16(0)+\tfrac16(4)=\tfrac23\). The crossing path in \(A_2\) has probability \(\tfrac16\) and terminal value 2, contributing \(\tfrac16(2)=\tfrac13\). Hence \(\mathbb{E}[X_2\mathbf{1}_A]=\tfrac23+\tfrac13=1\). Directly, \(\mathbb{E}[X_2]=\tfrac16(0)+\tfrac16(4)+\tfrac16(2)+\tfrac12(0)=1\). The bound checks as \(\tfrac32\cdot\tfrac12=\tfrac34\le1=\mathbb{E}[X_2\mathbf{1}_A]=\mathbb{E}[X_2]\).
Worked Example: Equality in the Bound
Let \(N=1\), \(\Omega=\{a,b\}\), and \(\mathbb{P}(\{a\})=\mathbb{P}(\{b\})=\tfrac12\). Use the trivial \(\mathcal{F}_0\) and the full \(\mathcal{F}_1\), with \(X_0=1\), \(X_1(a)=2\), and \(X_1(b)=0\). For \(\lambda=2\), verify that equality holds in both inequalities.
The conditional mean at time 0 is \(\mathbb{E}[X_1\mid\mathcal{F}_0]=\tfrac12(2)+\tfrac12(0)=1=X_0\), so this is a nonnegative martingale. The crossing event is \(A=\{a\}\), because the threshold is inclusive and \(X_1(a)=2\). Thus \(\mathbb{P}(A)=\tfrac12\), \(\mathbb{E}[X_1\mathbf{1}_A]=\tfrac12(2)=1\), and \(\mathbb{E}[X_1]=1\). Substitution gives \(\lambda\mathbb{P}(A)=2\cdot\tfrac12=1=\mathbb{E}[X_1\mathbf{1}_A]=\mathbb{E}[X_1]\). This example shows that neither inequality can generally be made strict.
A Common Mistake: The Process Can Fall After Crossing
In the two-time example, the path with \(X_1=2\) and \(X_2=0\) has probability \(\tfrac13\cdot\tfrac12=\tfrac16\). It belongs to \(A\), since it crossed \(\tfrac32\) at time 1, even though \(X_2=0<\tfrac32\). Thus the calculation “on \(A\), \(X_N\ge\lambda\)” is false, and it cannot justify the desired inequality.
The correction is to use the value at the first crossing, not to assume that the terminal value remains above the threshold. On each \(A_k\), the submartingale property gives \(\mathbb{E}[X_N\mathbf{1}_{A_k}]\ge\mathbb{E}[X_k\mathbf{1}_{A_k}]\ge\lambda\mathbb{P}(A_k)\). The first comparison is an expectation statement on an event known by time \(k\); it is not a pathwise assertion that \(X_N\ge X_k\). Summing these separate comparisons over the disjoint first-crossing events is what handles paths that later fall.
The nonnegativity condition also has a specific role. The comparison on each \(A_k\) follows from the submartingale property, but concluding \(\mathbb{E}[X_N\mathbf{1}_A]\le\mathbb{E}[X_N]\) uses \(X_N\ge0\). Without that condition, the contribution from outside \(A\) could be negative. The threshold must be positive for the stated probability bound; the first-crossing partition itself is still defined for any real threshold, but a nonpositive threshold does not yield this useful inequality for a nonnegative process.
Check Your Understanding
Use the finite-horizon assumptions stated above. For each question, keep the event convention \(X_n\ge\lambda\).
- Write \(A_0,A_1,A_2\) explicitly for a process observed through time 2, and explain why these events are disjoint.
- In the one-step example, calculate \(\mathbb{E}[X_1\mathbf{1}_A]\) and \(\lambda\mathbb{P}(A)\) when \(\lambda=\tfrac32\).
- In the two-time example, what is the probability of the path \(X_1=2,\ X_2=0\), and why does it refute the claim that \(X_N\ge\lambda\) on \(A\)?
- Identify the step in the theorem’s proof that requires nonnegativity of \(X_N\).
- For the equality example, change the threshold from \(2\) to \(\tfrac32\). Compute the crossing probability and the two sides of the first inequality.