What Makes a Hitting Time a Stopping Time?
A first hitting time records when a process reaches a target, but the word “first” alone does not guarantee that the time is a stopping time. The key test is informational: at every fixed time \(t\), can we determine from observations up to \(t\) whether the target has already been reached? For Brownian motion, continuity lets us answer this by looking at the path over the whole interval \([0,t]\), using only information available by time \(t\).
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((B_t)_{t\ge0}\) be standard Brownian motion with state space \(S=\mathbb{R}\). Use its completed natural filtration \((\mathcal{F}_t)_{t\ge0}\), where \(\mathcal{F}_t\) contains the information generated by \(\{B_s:0\le s\le t\}\), together with the relevant null sets. Brownian motion is adapted to this filtration and has continuous paths almost surely. Completion matters if continuity fails on an exceptional null set: events that differ only on that set still belong to each completed \(\mathcal{F}_t\).
For a nonempty closed set \(C\subseteq\mathbb{R}\), define its first hitting time by \(\tau_C=\inf\{s\ge0:B_s\in C\}\), with \(\inf\varnothing=\infty\). A useful quantity is the distance \(d(x,C)=\inf_{y\in C}|x-y|\). It is zero exactly when \(x\in C\), because \(C\) is closed. The next result turns the stopping-time test into a statement about distances along the path.
Proof. Fix \(t\ge0\). On a continuous path, the function \(s\mapsto d(X_s,C)\) is continuous on the compact interval \([0,t]\), so it attains its minimum there. The path hits \(C\) by time \(t\) if and only if this minimum is zero: a hit makes the minimum zero, and a zero minimum means \(X_s\in C\) at a minimizing time because \(C\) is closed. The preimage \(\{s\ge0:X_s\in C\}\) is closed on a continuous path, so if its infimum is at most \(t\), it contains a time in \([0,t]\). Thus, up to the exceptional null set of discontinuous paths, \(\{\tau_C\le t\}=\{\inf_{0\le s\le t}d(X_s,C)=0\}\). Let \(D_t=(\mathbb{Q}\cap[0,t])\cup\{t\}\). This is a countable dense subset of \([0,t]\), and continuity gives \(\inf_{0\le s\le t}d(X_s,C)=\inf_{s\in D_t}d(X_s,C)\). Each \(X_s\) for \(s\in D_t\) is \(\mathcal{F}_t\)-measurable by adaptedness and the filtration property. Therefore the countable infimum is \(\mathcal{F}_t\)-measurable, and its event of being zero belongs to \(\mathcal{F}_t\). Completion handles the exceptional null set. This holds for every \(t\), so \(\tau_C\) is a stopping time. If \(C=\varnothing\), then \(\{\tau_C\le t\}=\varnothing\) for every finite \(t\), which is measurable. \(\square\)
Hitting a Level and Crossing a Threshold
For a fixed level \(a\in\mathbb{R}\), the first hitting time is \(\tau_a=\inf\{s\ge0:B_s=a\}\). The theorem applies because the singleton \(\{a\}\) is closed. In particular, if the process starts at that level, then \(\tau_a=0\); this degenerate case is included in the definition and in the stopping-time argument.
For a positive threshold \(a>0\), Brownian motion starts at \(B_0=0<a\). Continuity and the intermediate value property imply that the path reaches \(a\) by time \(t\) exactly when its running maximum through time \(t\) is at least \(a\). The threshold is inclusive: reaching \(a\) counts as hitting it.
Worked Example: First Passage to a Positive Level
Let \(a=2\), and define \(\tau_2=\inf\{s\ge0:B_s=2\}\). For fixed \(t\ge0\), continuity, \(B_0=0\), and \(2>0\) give
The first equality holds because a path whose maximum is at least \(2\) must take the value \(2\) somewhere between its starting value \(0\) and a time at which it reaches or exceeds \(2\). The second follows from continuity and density of \(D_t\). Every variable \(B_s\) in the countable supremum is observed by time \(t\), so the event belongs to \(\mathcal{F}_t\). At \(t=0\), the supremum is \(B_0=0\), and \(0\ge2\) is false, as required: the process has not hit level \(2\) at time zero.
Check: the event uses \(\ge2\), not \(>2\). A path that reaches \(2\) and never goes above it by time \(t\) has hit the level, so replacing the threshold by a strict inequality would omit valid paths.
There is also a useful threshold version that does not require the target to be a single level. For \(a>0\), put \(H_a=\inf\{s\ge0:B_s\ge a\}\). This is the first time the path enters the closed half-line \([a,\infty)\), so the closed-set theorem applies. By continuity and \(B_0=0<a\), reaching or exceeding \(a\) requires first attaining \(a\), and hence \(H_a=\tau_a\). If \(a\le0\), the threshold set already contains \(B_0=0\), so \(H_a=0\). This distinction checks the initial condition rather than silently assuming the process starts below the threshold.
Worked Example: A Running-Maximum Threshold
Define \(M_t=\sup_{0\le s\le t}B_s\), and let \(H=\inf\{s\ge0:M_s\ge\tfrac32\}\). Since \(M_t\ge\tfrac32>0=B_0\) means that the continuous path has reached level \(\tfrac32\) by time \(t\), the event at a fixed time is
For example, at the test time \(t=0\), the supremum is \(B_0=0\), and \(0\ge\tfrac32\) is false. At any \(t\), the right-hand event is measurable in \(\mathcal{F}_t\), so \(H\) is a stopping time. The calculation does not give a probability that \(H\le t\); it establishes the required measurability without needing the distribution of \(H\).
Check: the threshold is \(\tfrac32\), and the event includes equality. Reaching exactly \(\tfrac32\) is enough to trigger \(H\).
Exit Times from Open Intervals
Let \(l<u\) be finite real numbers, and define the first exit time from the open interval \((l,u)\) by \(\tau_{(l,u)}=\inf\{s\ge0:B_s\notin(l,u)\}\). Its target is the closed complement \((-\infty,l]\cup[u,\infty)\). Therefore it is a special case of hitting a closed set. Equivalently, the path has exited by time \(t\) exactly when its minimum has reached or fallen below \(l\), or its maximum has reached or exceeded \(u\).
Proof. A continuous path is outside \((l,u)\) at some time in \([0,t]\) if and only if at some such time its value is at most \(l\) or at least \(u\). The first alternative holds exactly when the minimum on the compact interval \([0,t]\) is at most \(l\); the second holds exactly when the maximum is at least \(u\). Thus the displayed equality follows, including when the path starts outside the interval, in which case one of the inequalities already holds at time zero. By continuity, each minimum and maximum equals the infimum or supremum over the countable dense set \(D_t=(\mathbb{Q}\cap[0,t])\cup\{t\}\). Each value \(B_s\) for \(s\in D_t\) is \(\mathcal{F}_t\)-measurable, so the two events and their union belong to \(\mathcal{F}_t\). Completion covers the exceptional null set. The event belongs to \(\mathcal{F}_t\) for every \(t\), proving the stopping-time claim. \(\square\)
Worked Example: Exit from a Two-Sided Interval
Take the open interval \((-1,2)\), and let \(\tau=\inf\{s\ge0:B_s\notin(-1,2)\}\). For each fixed \(t\), the event that exit has occurred is
Using \(D_t=(\mathbb{Q}\cap[0,t])\cup\{t\}\), the infimum and supremum can be taken over \(D_t\), so the event depends only on Brownian values observed by time \(t\). At \(t=0\), the minimum and maximum both equal \(B_0=0\). The lower test is \(0\le-1\), which is false, and the upper test is \(0\ge2\), which is also false. Thus the formula correctly says that a Brownian motion starting at zero has not exited \((-1,2)\) at time zero.
The boundary inequalities are inclusive because the interval is open: reaching \(-1\) or \(2\) is already an exit. If the path remains strictly between the boundaries throughout \([0,t]\), neither event occurs and \(\tau>t\).
Why Endpoint Tests and Future Information Fail
A common mistake is to replace the whole path up to time \(t\) by its endpoint. For a level-hitting time, \(\{B_t\ge a\}\) is generally not the same event as \(\{\tau_a\le t\}\): the path could reach \(a\) and then return below it. Here is a continuous path calculation that makes the distinction concrete. Consider \(f(s)=2s\) for \(0\le s\le\tfrac12\), and \(f(s)=2(1-s)\) for \(\tfrac12\le s\le1\). Its maximum on \([0,1]\) is \(f(\tfrac12)=2(\tfrac12)=1\), so its first hit of level \(1\) is at \(s=\tfrac12\). Yet its endpoint is \(f(1)=2(1-1)=0<1\). The path has hit the level by time \(1\), even though the endpoint test says otherwise. The running maximum, not the endpoint, gives the correct event.
A second pitfall is confusing a stopping time with any random time. For example, let \(\sigma=1\) if \(B_1>0\), and \(\sigma=2\) otherwise. Then \(\{\sigma\le1\}=\{B_1>0\}\), which depends on the observation at time \(1\), not information available at time \(0\). In the natural Brownian filtration, this event is not in \(\mathcal{F}_0\): \(B_0=0\) is fixed, whereas \(B_1\) has a nondegenerate centered normal distribution. Thus \(\sigma\) is not a stopping time. By contrast, the hitting and exit events above are expressed using only observations no later than the time being tested.
These arguments establish measurability, not other properties of a hitting time. In particular, the stopping-time definition alone does not show that a hitting time is finite, has finite expectation, or has any specified distribution. Those are separate questions. The practical test is to fix \(t\), express “the target was reached by \(t\)” through the path on \([0,t]\), and verify that the expression uses only \(\mathcal{F}_t\)-measurable observations.
Check Your Understanding
For each question, use the event at a fixed time \(t\) and keep track of whether the boundary is included.
- For \(a>0\), write \(\{\tau_a\le t\}\) in terms of the running maximum of Brownian motion. Why does the formula use \(\ge a\)?
- For exit from \((l,u)\), which two path extrema determine whether exit has occurred by time \(t\)? State the corresponding inequalities.
- If \(B_0=a\), what is the first hitting time of level \(a\)? Does this prevent it from being a stopping time?
- Why is the endpoint event \(\{B_t\ge a\}\) not generally a valid replacement for \(\{\tau_a\le t\}\)?
- For a nonempty closed set \(C\), explain why a zero minimum of \(s\mapsto d(B_s,C)\) on \([0,t]\) implies a hit by time \(t\).