Continuous-Time Stopping Times and Filtrations

Stopping Times in Continuous Time

In discrete time, information arrives at separate steps. In continuous time, a question such as “has the process crossed the boundary by time \(t\)?” must be answered using only the information available at that exact time. A stopping time formalizes this requirement. The filtration matters: a condition that checks all strict times before \(t\) gives the stopping-time property at \(t\) only when the filtration has an appropriate continuity property.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a filtration \((\mathcal{F}_t)_{t\ge0}\), meaning that each \(\mathcal{F}_t\) is a sigma-field and \(\mathcal{F}_s\subseteq\mathcal{F}_t\) whenever \(s\le t\). A process \((X_t)_{t\ge0}\) is adapted if \(X_t\) is \(\mathcal{F}_t\)-measurable for every \(t\). In this tutorial, processes used for first passage take values in a metric space \(E\), and their path regularity is stated explicitly. A random time is a measurable map \(\tau:\Omega\to[0,\infty]\).

Definition: A random time \(\tau\) is a stopping time with respect to \((\mathcal{F}_t)\) if \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). The value \(\tau=\infty\) is allowed; it means the event being monitored never occurs. A filtration is right-continuous if \(\mathcal{F}_t=\bigcap_{u>t}\mathcal{F}_u\) for every \(t\ge0\).

The definition asks whether, at time \(t\), the information available can determine whether stopping has already occurred. It does not require knowing whether stopping will occur in the future. Right-continuity says that information available just after \(t\), no matter how close to \(t\), is already included in \(\mathcal{F}_t\). This property is useful for converting tests involving strict inequalities in time into the stopping-time definition.

Strict-Time Tests and Right-Continuous Filtrations

Theorem: For a filtration \((\mathcal{F}_t)_{t\ge0}\), a random time \(\tau\) is a stopping time if and only if \(\{\tau<t\}\in\mathcal{F}_t\) for every \(t\ge0\) and the filtration is right-continuous for the reverse implication. More precisely, every stopping time satisfies the strict-time test for any filtration; if the filtration is right-continuous, the strict-time test also implies that \(\tau\) is a stopping time.

Proof. Suppose first that \(\tau\) is a stopping time. For \(t>0\), write \(\{\tau<t\}=\bigcup_{q\in\mathbb{Q},\,0\le q<t}\{\tau\le q\}\). Each event in the union belongs to \(\mathcal{F}_q\subseteq\mathcal{F}_t\), so the union belongs to \(\mathcal{F}_t\). For \(t=0\), the event \(\{\tau<0\}\) is empty and belongs to \(\mathcal{F}_0\). Conversely, suppose the strict-time test holds and the filtration is right-continuous. For each \(n\ge1\), \(\{\tau\le t\}=\bigcap_{n\ge1}\{\tau<t+1/n\}\). The \(n\)-th event belongs to \(\mathcal{F}_{t+1/n}\). Thus their intersection belongs to \(\bigcap_{n\ge1}\mathcal{F}_{t+1/n}\), which equals \(\mathcal{F}_t\): the sequence \(t+1/n\) decreases to \(t\), and the filtration is increasing, so intersecting these sigma-fields gives the same intersection as over all \(u>t\). Therefore \(\{\tau\le t\}\in\mathcal{F}_t\), as required. \(\square\)

The two directions have different hypotheses. A stopping time always passes the strict-time test. The reverse direction uses right-continuity; without it, information that becomes available immediately after \(t\) might not belong to \(\mathcal{F}_t\). The following examples make this distinction concrete.

Worked Example: A Strict Crossing Need Not Be Attained

Take a probability space with one outcome, its trivial sigma-field, and the constant (hence right-continuous) filtration. Let \(X_t=t\), and let \(O=(1,\infty)\). Define the first-entry time \(T_O=\inf\{t\ge0:X_t\in O\}\), with the infimum of the empty set defined as \(\infty\).

The condition \(X_t\in O\) is \(t>1\), so the set of entry times is \((1,\infty)\). Its infimum is \(T_O=1\), even though \(X_1=1\notin O\): the infimum is not attained. For each \(t\ge0\), \(\{T_O\le t\}\) is empty if \(t<1\) and is the whole one-point space if \(t\ge1\). Both events belong to \(\mathcal{F}_t\), so \(T_O\) is a stopping time. The strict-time events are empty for \(t\le1\) and the whole space for \(t>1\), also as expected.

Check: \(T_O=1\), and \(\{T_O\le1\}\) is the whole space although \(X_1\notin O\). The definition uses the infimum of entry times, not a requirement that the process belong to \(O\) at the infimum.

First Entry into an Open Set

For a set \(O\subseteq E\), define \(T_O=\inf\{t\ge0:X_t\in O\}\). When \(O\) is open and the paths are right-continuous, entry can be detected by checking rational times strictly before a given \(t\). Openness is important: if the process is in \(O\) at time \(s\), right-continuity ensures it remains in \(O\) for a short interval immediately after \(s\).

Theorem: Let \((X_t)_{t\ge0}\) be an adapted process with values in a metric space \(E\), and suppose its paths are right-continuous. For every open set \(O\subseteq E\), the first-entry time \(T_O=\inf\{t\ge0:X_t\in O\}\) is a stopping time with respect to any right-continuous filtration to which \(X\) is adapted. This includes \(O=\varnothing\), for which \(T_O=\infty\).

Proof. If \(O=\varnothing\), then \(\{T_O\le t\}=\varnothing\in\mathcal{F}_t\) for every \(t\). Now suppose \(O\ne\varnothing\). For each \(t>0\), right-continuity of the paths and openness of \(O\) give the identity \(\{T_O<t\}=\bigcup_{q\in\mathbb{Q},\,0\le q<t}\{X_q\in O\}\). For the forward inclusion, \(T_O<t\) implies there is an \(s<t\) with \(X_s\in O\), by the definition of an infimum. Since \(O\) is open and \(X\) is right-continuous at \(s\), there is a positive interval immediately to the right of \(s\) on which \(X\) remains in \(O\). That interval contains a rational \(q<t\). For the reverse inclusion, if \(X_q\in O\) for some rational \(q<t\), then \(T_O\le q<t\). Each event \(\{X_q\in O\}\) belongs to \(\mathcal{F}_q\subseteq\mathcal{F}_t\), so the countable union belongs to \(\mathcal{F}_t\). At \(t=0\), \(\{T_O<0\}\) is empty. The strict-time test therefore holds, and the right-continuity of the filtration and the preceding theorem imply that \(T_O\) is a stopping time. \(\square\)

Worked Example: A Random Initial Level

Let \(\Omega=\{a,b\}\), with \(\mathbb{P}(\{a\})=\mathbb{P}(\{b\})=\tfrac12\), and let \(\mathcal{F}_t=2^\Omega\) for all \(t\ge0\). This filtration is right-continuous. Set \(Y(a)=0\), \(Y(b)=2\), and \(X_t=Y+t\). The process is adapted and has continuous paths. Find the first time it enters \(A=[1,\infty)\).

On outcome \(a\), \(X_t=t\), so \(X_t\in A\) exactly when \(t\ge1\), and \(T_A(a)=1\). On outcome \(b\), \(X_t=2+t\in A\) for every \(t\ge0\), so \(T_A(b)=0\). Thus \(\{T_A\le t\}=\{b\}\) when \(0\le t<1\), and \(\{T_A\le t\}=\Omega\) when \(t\ge1\). Both events are in \(\mathcal{F}_t\), confirming the stopping-time property.

Check: At \(t=0\), \(\mathbb{P}(T_A\le0)=\mathbb{P}(\{b\})=\tfrac12\). At \(t=\tfrac12\), the event is still \(\{b\}\), with probability \(\tfrac12\); at \(t=1\), it is \(\Omega\), with probability \(1\). These values agree with the two computed first-entry times.

Hitting a Closed Set with Continuous Paths

For a closed target, right-continuity alone does not provide the short interval inside the target that openness provided. A useful alternative is path continuity. Let \(A\) be a closed subset of a metric space \(E\), and write \(d(x,A)=\inf_{y\in A}d(x,y)\) for the distance from \(x\) to \(A\). If \(A\) is nonempty, then \(d(x,A)=0\) exactly when \(x\in A\). Along a continuous path, the distance to \(A\) is continuous too.

Theorem: Let \((X_t)_{t\ge0}\) be an adapted process with values in a metric space \(E\), and suppose its paths are continuous. For every closed set \(A\subseteq E\), the hitting time \(T_A=\inf\{t\ge0:X_t\in A\}\) is a stopping time with respect to the filtration \((\mathcal{F}_t)\); right-continuity of the filtration is not needed. For \(A=\varnothing\), \(T_A=\infty\).

Proof. If \(A=\varnothing\), the claim follows because \(\{T_A\le t\}\) is empty. Suppose \(A\ne\varnothing\), fix \(t\ge0\), and let \(D_t=(\mathbb{Q}\cap[0,t])\cup\{t\}\), a countable dense subset of \([0,t]\). The function \(s\mapsto d(X_s,A)\) is continuous on \([0,t]\). Its minimum on this compact interval is attained. Consequently, the path hits \(A\) by time \(t\) if and only if \(\inf_{q\in D_t}d(X_q,A)=0\). In one direction, a hit makes the minimum zero. In the other, the minimum is attained at some \(s\in[0,t]\); if it is zero, closedness of \(A\) implies \(X_s\in A\). The event that this countable infimum equals zero can be written as \(\bigcap_{m\ge1}\bigcup_{q\in D_t}\{d(X_q,A)<1/m\}\). Each event in each union belongs to \(\mathcal{F}_q\subseteq\mathcal{F}_t\), including the endpoint \(q=t\). Thus \(\{T_A\le t\}\in\mathcal{F}_t\). This holds for every \(t\), proving the claim. \(\square\)

Worked Example: Hitting a Closed Threshold

Use the two-outcome space and filtration from the preceding example, but now let \(X_t=Y+t\) and take the closed target \(A=\{1\}\). On outcome \(a\), \(X_t=t\), so the target is hit at \(t=1\). On outcome \(b\), \(X_t=2+t\), so the target is never hit. Therefore \(T_A(a)=1\) and \(T_A(b)=\infty\).

For \(0\le t<1\), \(\{T_A\le t\}=\varnothing\); for \(t\ge1\), \(\{T_A\le t\}=\{a\}\). The first event has probability \(0\), and the second has probability \(\tfrac12\). Both are in the filtration. The distance calculation gives the same answer: on outcome \(a\), \(d(X_t,\{1\})=|t-1|\), which is zero at \(t=1\); on outcome \(b\), \(d(X_t,\{1\})=|2+t-1|=1+t\), which is never zero for \(t\ge0\).

Check: At \(t=1\), the event \(\{T_A\le t\}\) is exactly \(\{a\}\), so its probability is \(\tfrac12\). On outcome \(a\), the distance is \(|1-1|=0\); on outcome \(b\), it is \(|3-1|=2\). Thus the distance test identifies precisely the paths that have hit the target by time \(1\).

Why Right-Continuity of the Filtration Matters

The strict-time criterion can fail in the reverse direction when the filtration is not right-continuous. Here is a direct counterexample. Let \(\Omega=\{a,b\}\), with each outcome having probability \(\tfrac12\), and let \(\mathcal{F}\) be the full sigma-field. Define \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for \(0\le t\le1\), and \(\mathcal{F}_t=2^\Omega\) for \(t>1\). This is an increasing filtration, but not a right-continuous one at \(t=1\), since \(\mathcal{F}_1\) is trivial while \(\bigcap_{u>1}\mathcal{F}_u=2^\Omega\).

Set \(\tau(a)=1\) and \(\tau(b)=2\). For \(t\le1\), \(\{\tau<t\}=\varnothing\). For \(1<t\le2\), \(\{\tau<t\}=\{a\}\), which belongs to \(\mathcal{F}_t\). For \(t>2\), \(\{\tau<t\}=\Omega\). Thus the strict-time test holds for every \(t\ge0\). But \(\{\tau\le1\}=\{a\}\notin\mathcal{F}_1\), so \(\tau\) is not a stopping time. Information that distinguishes \(a\) from \(b\) is available immediately after time \(1\), but not at time \(1\) itself.

Common mistake. It is incorrect to replace the event \(\{T_O\le t\}\) by \(\{X_t\in O\}\). For example, let \(X_t=2t\) for \(0\le t\le1\), \(X_t=4-2t\) for \(1<t\le2\), and \(X_t=0\) for \(t>2\), with a one-outcome probability space and the trivial filtration. This is a continuous adapted process. For \(O=(1,\infty)\), it first enters at \(T_O=\tfrac12\). At \(t=2\), \(T_O\le2\) is true, but \(X_2=0\notin O\). The correct test is whether entry has occurred at some time up to \(t\), not whether the process is in the set at the final time \(t\).

For first-passage arguments, check the target set and path regularity together. Right-continuous paths give a rational-time test for entry into an open set. Continuous paths allow closed-set hitting to be tested through the distance to the set over the whole interval up to \(t\). In either case, the stopping-time question is whether \(\{T\le t\}\) is in \(\mathcal{F}_t\); a random time does not become a stopping time merely because it is defined using a process.

Check Your Understanding

For each question, identify the path or filtration assumption needed, and use the event definition of a stopping time.

Why does every stopping time satisfy \(\{\tau<t\}\in\mathcal{F}_t\), even if the filtration is not right-continuous?
For an adapted right-continuous process, why does openness of \(O\) allow first entry into \(O\) to be detected at a rational time strictly before \(t\)?
For an adapted continuous process and a closed set \(A\), express \(\{T_A\le t\}\) using distances evaluated on a countable dense subset of \([0,t]\).
For \(X_t=t\) and \(O=(1,\infty)\), find \(T_O\). Is \(X_{T_O}\) in \(O\)? Explain the role of the infimum.
In the two-outcome filtration example, which event fails to belong to \(\mathcal{F}_1\), and why does this show the need for right-continuity in the reverse strict-time criterion?