Why Level Sets Describe a Finite Stopping Rule
At a fixed time \(t\), the stopping-time condition asks whether the decision “stop by time \(t\)” can be made using information available by time \(t\). For a rule that can take only finitely many deterministic values, it is enough to check the event that the rule selects each individual value. These events are the rule’s level sets.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). The results below need only the filtration; they do not use Brownian motion or stochastic integrals. Recall that a random time \(\tau:\Omega\to[0,\infty]\) is a stopping time if \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). The filtration is increasing: if \(s\le t\), then \(\mathcal{F}_s\subseteq\mathcal{F}_t\).
The Level-Set Characterization
Proof. First suppose \(\tau\) is a stopping time. For \(k=1\), the range assumption gives \(E_1=\{\tau\le t_1\}\), so \(E_1\in\mathcal{F}_{t_1}\). For \(k\ge2\), because no value of \(\tau\) lies strictly between \(t_{k-1}\) and \(t_k\), we have \(\{\tau=t_k\}=\{\tau\le t_k\}\setminus\{\tau\le t_{k-1}\}\). Both events on the right belong to \(\mathcal{F}_{t_k}\): the first by the stopping-time property, and the second because it belongs to \(\mathcal{F}_{t_{k-1}}\subseteq\mathcal{F}_{t_k}\). A sigma-field is closed under set differences, so \(E_k\in\mathcal{F}_{t_k}\).
Conversely, suppose \(E_k\in\mathcal{F}_{t_k}\) for every \(k\). Fix any \(t\ge0\). If \(t<t_1\), then \(\{\tau\le t\}=\varnothing\in\mathcal{F}_t\). Otherwise let \(j\) be the largest index such that \(t_j\le t\). Since \(\tau\) takes values only among the \(t_k\), we have \(\{\tau\le t\}=\bigcup_{k=1}^j E_k\). Each \(E_k\in\mathcal{F}_{t_k}\subseteq\mathcal{F}_t\), so this finite union is in \(\mathcal{F}_t\). This holds for every \(t\), proving that \(\tau\) is a stopping time. \(\square\)
The theorem converts a condition involving every real time \(t\) into finitely many checks. The level sets also form a partition: they are pairwise disjoint, and their union is \(\Omega\), because \(\tau\) takes exactly one of the listed values on each outcome.
Worked Example: Checking Brownian-Based Level Sets
Let \(t_1=1,t_2=2,t_3=3\). Define \(\tau\) by stopping at time \(1\) if \(B_1\ge0\); if \(B_1<0\), stop at time \(2\) if \(B_2\ge1\); otherwise stop at time \(3\). Find the level sets and verify the stopping-time condition.
The level sets are \(E_1=\{B_1\ge0\}\), \(E_2=\{B_1<0\}\cap\{B_2\ge1\}\), and \(E_3=\{B_1<0\}\cap\{B_2<1\}\). The variable \(B_1\) is \(\mathcal{F}_1\)-measurable, so \(E_1\in\mathcal{F}_1\). Both \(B_1\) and \(B_2\) are \(\mathcal{F}_2\)-measurable, so \(E_2,E_3\in\mathcal{F}_2\subseteq\mathcal{F}_3\).
For a direct check, if \(t<1\), then \(\{\tau\le t\}=\varnothing\). If \(1\le t<2\), then \(\{\tau\le t\}=E_1\). If \(2\le t<3\), then \(\{\tau\le t\}=E_1\cup E_2\). If \(t\ge3\), then \(\{\tau\le t\}=\Omega\). Each event is in the corresponding \(\mathcal{F}_t\), using the increasing-filtration property.
Check: The level sets are disjoint because \(E_2\) and \(E_3\) both require \(B_1<0\), while their conditions \(B_2\ge1\) and \(B_2<1\) cannot both hold. Their union is \(\{B_1\ge0\}\cup\{B_1<0\}=\Omega\), so every outcome receives exactly one stopping time.
Stopping at the First Successful Observation
A common way to use the characterization is to examine observations in order. At time \(t_k\), the rule may test a condition determined by information available then. Write that condition as an event \(A_k\in\mathcal{F}_{t_k}\). Given a finite horizon \(t_1<\cdots<t_n\), stop at the first time \(t_k\) for which \(A_k\) occurs; if none of \(A_1,\ldots,A_{n-1}\) occurs, stop at the terminal time \(t_n\). The terminal fallback ensures the rule always has a value in the finite range.
Proof. For \(k<n\), the event that the rule first stops at \(t_k\) is \(E_k=A_k\cap\bigcap_{j=1}^{k-1}A_j^c\). For \(k=1\), the intersection over no indices is, by convention, \(\Omega\), so \(E_1=A_1\). Since \(A_j\in\mathcal{F}_{t_j}\subseteq\mathcal{F}_{t_k}\) whenever \(j<k\), all the events in the expression for \(E_k\) belong to \(\mathcal{F}_{t_k}\); hence so does \(E_k\). The terminal level set is \(E_n=\bigcap_{j=1}^{n-1}A_j^c\). Every \(A_j\) in this intersection belongs to \(\mathcal{F}_{t_j}\subseteq\mathcal{F}_{t_n}\), so \(E_n\in\mathcal{F}_{t_n}\). These level sets are disjoint and cover \(\Omega\): an outcome either has a first successful index \(k<n\), or has no successful index before \(n\). The level-set characterization now implies that \(\tau\) is a stopping time. \(\square\)
If each observation is an \(\mathcal{F}_{t_k}\)-measurable random variable \(X_k\) and the rule tests whether \(X_k\) lies in a Borel set \(C\), then \(A_k=\{X_k\in C\}\) belongs to \(\mathcal{F}_{t_k}\). Thus the proposition applies. The crucial requirement is that each test use information available at its own time, not a later observation.
Worked Example: A Rule Based on Sequential Coin Observations
Let \(\Omega=\{HH,HT,TH,TT\}\), where the two letters record two successive fair coin tosses. At time \(1\), the first toss is known; at time \(2\), both tosses are known; time \(3\) is a terminal fallback. Define \(\tau=1\) if the first toss is \(H\), \(\tau=2\) if the first toss is \(T\) and the second is \(H\), and \(\tau=3\) if both tosses are \(T\).
The level sets, written as subsets of \(\Omega\), are \(E_1=\{HH,HT\}\), \(E_2=\{TH\}\), and \(E_3=\{TT\}\). The event \(E_1\) is known from the first toss and so lies in \(\mathcal{F}_1\). The event \(E_2\) is determined by both tosses and so lies in \(\mathcal{F}_2\); \(E_3\in\mathcal{F}_2\subseteq\mathcal{F}_3\). The characterization proves that \(\tau\) is a stopping time.
For the fair-coin probability measure, \(\mathbb{P}(E_1)=2/4=1/2\), \(\mathbb{P}(E_2)=1/4\), and \(\mathbb{P}(E_3)=1/4\). These values sum to \(1/2+1/4+1/4=1\), as required by the partition. The cumulative events are \(\{\tau\le1\}=E_1\), \(\{\tau\le2\}=E_1\cup E_2=\{HH,HT,TH\}\), and \(\{\tau\le3\}=\Omega\). Each is available by the time at which it is tested.
Check: For the outcome \(TH\), the first test fails and the second succeeds, so the rule assigns time \(2\), exactly as \(E_2\) specifies. For \(TT\), both tests fail and the fallback assigns time \(3\), exactly as \(E_3\) specifies.
When a Level Set Uses Future Information
Worked Example: A Candidate Rule That Is Not a Stopping Time
Use the completed natural filtration of Brownian motion, for which \(\mathcal{F}_0\) contains only events of probability \(0\) or \(1\), up to null sets. Consider the candidate rule \(\tau=0\) if \(B_1\ge0\), and \(\tau=1\) if \(B_1<0\). Is it a stopping time?
Its level set at time \(0\) is \(E_0=\{B_1\ge0\}\). Since \(B_1\) has the continuous distribution \(N(0,1)\), \(\mathbb{P}(E_0)=1/2\). An event in \(\mathcal{F}_0\) must have probability \(0\) or \(1\), so \(E_0\notin\mathcal{F}_0\). Equivalently, \(\{\tau\le0\}=E_0\notin\mathcal{F}_0\). Therefore \(\tau\) is not a stopping time.
Check: The issue is not that \(\tau\) has finitely many values: it does. The issue is that the rule uses the time-\(1\) observation \(B_1\) to decide whether it would already have stopped at time \(0\). That decision is unavailable at time \(0\).
Common Mistake: Checking the Wrong Time
A tempting but incorrect test is to require \(\{\tau=t_k\}\in\mathcal{F}_{t_{k-1}}\), as though the choice of \(t_k\) must be known at the preceding listed time. The level-set characterization requires measurability at \(t_k\), not at the previous time. For example, in the first Brownian example, \(E_1=\{B_1\ge0\}\) need not be known at time \(0\); it is known at time \(1\), which is exactly what the stopping-time condition requires.
Another incorrect shortcut is to check only that the level sets are measurable in \(\mathcal{F}\). Membership in the full sigma-field says the event is measurable eventually, not that it is known by the time the rule stops. The failed candidate above has \(E_0\in\mathcal{F}\), but \(E_0\notin\mathcal{F}_0\), and this distinction makes the rule invalid.
For a finite-range rule, use this practical check: list every possible stopping value, write its level set, and verify that the level set belongs to the filtration at that value. For a sequential construction, the first-success formula makes the check explicit: every earlier failure and the current success must be measurable by the current time. A terminal fallback is treated separately as the event that all earlier tests failed.
Check Your Understanding
For each question, distinguish measurability at the stopping value from measurability at some later time.
- Suppose \(\tau\) takes values \(2,4,7\). Write \(\{\tau\le t\}\) in terms of the level sets for \(t<2\), \(4\le t<7\), and \(t\ge7\).
- Why does \(\{\tau=t_k\}=\{\tau\le t_k\}\setminus\{\tau\le t_{k-1}\}\) hold for \(k\ge2\) when the range contains no values between \(t_{k-1}\) and \(t_k\)?
- Let \(A_1\in\mathcal{F}_{t_1}\) and \(A_2\in\mathcal{F}_{t_2}\), with \(t_1<t_2<t_3\). Write the three level sets for stopping at the first successful test, with \(t_3\) as fallback.
- If \(X_k\) is \(\mathcal{F}_{t_k}\)-measurable and \(C\) is Borel, why is \(\{X_k\in C\}\) an admissible event for the test at \(t_k\)?
- For the rule that stops at time \(0\) when \(B_1\ge0\), identify the defining stopping-time event at time \(0\) and explain why it fails the condition in the completed natural Brownian filtration.