Stopping Is Defined by What Is Known at Each Time
A random time may be measurable on the whole probability space and still fail to be a stopping time. The key question is not simply whether its value is random, but whether, at every time \(t\), the information available by \(t\) suffices to determine whether stopping has already occurred. This event-based definition is central: it lets us distinguish a rule that can be followed as information arrives from one that depends on future observations.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) with respect to this filtration. Thus the filtration is increasing, and Brownian increments after time \(t\) are independent of \(\mathcal{F}_t\). A random time here means an \(\mathcal{F}\)-measurable random variable \(\tau:\Omega\to[0,\infty]\). No stochastic integral is used in this tutorial.
The quantifier “for every \(t\ge0\)” matters. Checking the condition at one time, or at a finite collection of times, does not establish that \(\tau\) is a stopping time. The value \(\infty\) is allowed: \(\tau=\infty\) means that stopping never occurs. For a random time that is finite almost surely, the definition is unchanged.
Since \(\mathcal{F}_t\) is a sigma-field, \(\{\tau\le t\}\in\mathcal{F}_t\) is equivalent to \(\{\tau>t\}\in\mathcal{F}_t\), by taking complements. The first event means “stopped by \(t\),” and the second means “not stopped by \(t\).” Both must be interpreted at the same time \(t\); neither condition asks whether the exact future stopping time is already known.
The Event-Indicator Form
Proof. Fix \(t\ge0\). If \(\{\tau\le t\}\in\mathcal{F}_t\), then the indicator of this event is \(\mathcal{F}_t\)-measurable: it takes value \(1\) on that measurable event and value \(0\) on its measurable complement. Conversely, if \(\mathbf{1}_{\{\tau\le t\}}\) is \(\mathcal{F}_t\)-measurable, then the event \(\{\tau\le t\}\) is the inverse image of the Borel set \(\{1\}\) under that indicator, so it belongs to \(\mathcal{F}_t\). Applying this equivalence separately at every \(t\ge0\) proves both directions of the proposition. \(\square\)
This form makes the information requirement explicit: the binary question “has stopping happened yet?” must have an answer measurable at the current time. It does not require \(\tau\) itself to be \(\mathcal{F}_t\)-measurable for every \(t\). That stronger requirement would generally make the entire value of a future random time known immediately.
Deterministic and Information-Driven Rules
Worked Example: A Deterministic Stopping Time
Let \(c\ge0\) be a fixed constant and define \(\tau(\omega)=c\) for every \(\omega\in\Omega\). For any \(t\ge0\), there are two cases. If \(t<c\), then \(\{\tau\le t\}=\varnothing\), which belongs to every sigma-field. If \(t\ge c\), then \(\{\tau\le t\}=\Omega\), which also belongs to every sigma-field. Thus \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\ge0\), and \(\tau\) is a stopping time.
Check: at the boundary \(t=c\), the defining event is \(\{\tau\le c\}=\Omega\), not the empty set. The weak inequality in the definition includes the stopping time itself.
A genuinely random rule can also be a stopping time if the choice is made using information that has already arrived. The next example makes the event calculation explicit at every range of times.
Worked Example: Choose a Time Using the Brownian Value at Time One
Use the natural filtration \(\mathcal{F}_t=\sigma(B_s:0\le s\le t)\). Let \(A=\{B_1\ge0\}\), and define \(\tau=1\) on \(A\) and \(\tau=2\) on \(A^c\). This is an \(\mathcal{F}\)-measurable random time because \(A\in\mathcal{F}_1\subseteq\mathcal{F}\). For \(0\le t<1\), neither possible value of \(\tau\) is at most \(t\), so \(\{\tau\le t\}=\varnothing\in\mathcal{F}_t\). For \(1\le t<2\), exactly the outcome \(\tau=1\) has stopped, so \(\{\tau\le t\}=A\). Because \(A\in\mathcal{F}_1\) and the filtration is increasing, \(A\in\mathcal{F}_t\) for each such \(t\). For \(t\ge2\), both possible values have stopped, so \(\{\tau\le t\}=\Omega\in\mathcal{F}_t\).
These cases cover all \(t\ge0\), including the boundary times: at \(t=1\) the event is \(A\), and at \(t=2\) it is \(\Omega\). Hence \(\tau\) is a stopping time. Moreover \(B_1\sim N(0,1)\), whose distribution is symmetric about zero and assigns probability zero to the single point \(0\). Therefore \(\mathbb{P}(A)=\mathbb{P}(B_1\ge0)=\tfrac12\), and \(\mathbb{P}(A^c)=\tfrac12\). Its expected value is \(\mathbb{E}[\tau]=1\cdot\tfrac12+2\cdot\tfrac12=\tfrac32\).
Check: the value of \(\tau\) is not known before time \(1\), but the rule does not require it to be known then. At time \(1\), the event that the rule has stopped is precisely \(A\), which is observable at that time. The calculation of the expectation also gives a value between the two possible stopping times, \(1\le\tfrac32\le2\).
More generally, an event used to choose a later stopping time must be available by the time at which the choice can affect whether stopping has occurred. The definition tests the resulting events \(\{\tau\le t\}\) directly; it does not test whether a verbal description sounds like a rule that can be implemented.
Random Times That Use Future Information
Worked Example: A Measurable Random Time That Is Not a Stopping Time
Again use the natural Brownian filtration \(\mathcal{F}_t=\sigma(B_s:0\le s\le t)\), and let \(A=\{B_1>0\}\). Define \(\tau=0\) on \(A\) and \(\tau=1\) on \(A^c\). This is an \(\mathcal{F}\)-measurable random time: its two level events are \(A\) and \(A^c\), both in \(\mathcal{F}\). But at \(t=0\), the event that stopping has already occurred is \(\{\tau\le0\}=A\).
For the natural filtration, \(\mathcal{F}_0=\sigma(B_0)=\{\varnothing,\Omega\}\), since \(B_0=0\) almost surely. The standard normal distribution is symmetric and has no atom at zero, so \(\mathbb{P}(A)=\mathbb{P}(B_1>0)=\tfrac12\). Thus \(A\) is neither \(\varnothing\) nor \(\Omega\), and \(A\notin\mathcal{F}_0\). Consequently \(\tau\) is not a stopping time.
Check: at \(t=1\), \(\{\tau\le1\}=\Omega\), which is measurable. That single successful check does not repair the failure at \(t=0\). The rule declares that stopping occurred at time zero exactly when a future observation, \(B_1\), is positive.
The distinction is between measurability with respect to the whole sigma-field \(\mathcal{F}\) and measurability with respect to the current sigma-field \(\mathcal{F}_t\). In the example, the terminal information reveals which value \(\tau\) takes, but the information at time zero does not reveal whether \(\tau=0\). A random time can therefore be perfectly measurable and still fail the stopping-time condition.
Information Available at Time Zero
Proof. Let \(t\ge0\). Since \(\tau\) is \(\mathcal{F}_0\)-measurable, the inverse image \(\{\tau\le t\}\) of the Borel set \([0,t]\) belongs to \(\mathcal{F}_0\). The filtration is increasing, so \(\mathcal{F}_0\subseteq\mathcal{F}_t\). Therefore \(\{\tau\le t\}\in\mathcal{F}_t\). This holds for every \(t\ge0\), which is exactly the stopping-time definition. \(\square\)
This proposition includes deterministic times, since a constant random variable is \(\mathcal{F}_0\)-measurable. It also shows why a random time chosen using an initial source of information can be a stopping time: the choice is already known at every later time. By contrast, information revealed only in the future cannot be used to declare that stopping occurred earlier.
Common Pitfalls
Checking only the final time. In the non-stopping example, checking \(t=1\) gives \(\{\tau\le1\}=\Omega\), but the definition requires the condition at every time. The correction is to inspect \(t=0\), where the event is \(\{B_1>0\}\notin\mathcal{F}_0\).
Replacing “by time \(t\)” with “at time \(t\).” The defining event is \(\{\tau\le t\}\), not merely \(\{\tau=t\}\). It records all outcomes on which stopping has occurred at or before \(t\). For a rule taking values \(1\) and \(2\), for instance, the event \(\{\tau\le t\}\) for \(1\le t<2\) contains the outcomes with \(\tau=1\), not just outcomes with \(\tau=t\).
Assuming randomness is the obstacle. Randomness itself is not a problem: the Brownian-based rule choosing between times \(1\) and \(2\) is a stopping time. The relevant test is whether each event \(\{\tau\le t\}\) belongs to \(\mathcal{F}_t\). A future-dependent rule may fail that test even though \(\tau\) is an \(\mathcal{F}\)-measurable random variable.
Check Your Understanding
For each proposed rule, identify the event \(\{\tau\le t\}\) at the relevant times and compare it with the information available in \(\mathcal{F}_t\).
- For a constant \(c\ge0\), what are \(\{\tau\le t\}\) when \(\tau=c\) and \(t<c\), \(t=c\), and \(t>c\)?
- In the rule \(\tau=1\) on \(\{B_1\ge0\}\) and \(\tau=2\) otherwise, what is \(\{\tau\le t\}\) for \(1\le t<2\), and why is it in \(\mathcal{F}_t\)?
- For \(\tau=0\) on \(\{B_1>0\}\) and \(\tau=1\) otherwise, which time gives a failure of the stopping-time condition in the natural Brownian filtration?
- Explain why \(\mathcal{F}\)-measurability of \(\tau\) alone does not imply that \(\tau\) is a stopping time.
- Why does \(\mathcal{F}_0\)-measurability of a random time imply that it is a stopping time?