Changing a Stopping Time Without Losing Its Defining Property
A stopping time records when an event has become detectable from the information available so far. Capping that time at a fixed horizon or delaying it by a fixed amount cannot require information from the future: the event that the altered time has occurred by \(t\) is either already determined by time \(t\), or is impossible. Moving a time earlier is different. It can ask at time \(t\) whether the original stopping time will occur by a later time, information that may not yet be available.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), where \(\mathcal{F}_s\subseteq\mathcal{F}_t\) whenever \(s\le t\). No Brownian motion or stochastic integral is needed for these arguments. 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 value \(+\infty\) is allowed. The inclusion of sigma-fields as time increases will be used throughout.
Proof. For the truncation, first suppose \(0\le t<T\). Then \(\tau\wedge T\le t\) holds exactly when \(\tau\le t\), so \(\{\tau\wedge T\le t\}=\{\tau\le t\}\in\mathcal{F}_t\). For \(t\ge T\), the truncated time is at most \(T\) for every outcome, so \(\{\tau\wedge T\le t\}=\Omega\in\mathcal{F}_t\). These cases also cover \(T=0\): only the second case is needed, and the truncated time is identically zero.
For the delayed time, if \(0\le t<c\), then \(\tau+c\ge c>t\), including when \(\tau=+\infty\). Thus \(\{\tau+c\le t\}=\varnothing\in\mathcal{F}_t\). If \(t\ge c\), then \(\{\tau+c\le t\}=\{\tau\le t-c\}\). Since \(\tau\) is a stopping time, this event belongs to \(\mathcal{F}_{t-c}\); because \(t-c\le t\), it also belongs to \(\mathcal{F}_t\). When \(c=0\), this second argument applies for every \(t\ge0\) and gives the original stopping-time condition. Hence both altered times are stopping times. \(\square\)
The proof depends on the direction of the filtration inclusion. At time \(t\), the event for a delayed time is tested using the original stopping time at \(t-c\), an earlier time. The filtration has at least as much information at \(t\) as it had at \(t-c\). Truncation has an even simpler structure: after the horizon, the defining event is all of \(\Omega\).
Worked Examples on a Finite Probability Space
Worked Example: Truncating at a Fixed Horizon
Let \(\Omega=\{a,b,c\}\), with probability \(1/3\) on each outcome. Define the filtration by \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for \(0\le t<1\), \(\mathcal{F}_t=\sigma(\{a\})\) for \(1\le t<2\), and \(\mathcal{F}_t=2^\Omega\) for \(t\ge2\). This is increasing: the trivial sigma-field is contained in \(\sigma(\{a\})\), and \(\sigma(\{a\})\) is contained in \(2^\Omega\).
Set \(\tau(a)=1\), \(\tau(b)=2\), and \(\tau(c)=4\). For \(t<1\), \(\{\tau\le t\}=\varnothing\). For \(1\le t<2\), it is \(\{a\}\), which belongs to \(\mathcal{F}_t\). For \(2\le t<4\), it is \(\{a,b\}\), which belongs to \(2^\Omega=\mathcal{F}_t\). For \(t\ge4\), it is \(\Omega\). Thus \(\tau\) is a stopping time.
Take the horizon \(T=5/2\). The truncated time has values \((\tau\wedge T)(a)=1\), \((\tau\wedge T)(b)=2\), and \((\tau\wedge T)(c)=5/2\). At \(t=3/2\), its defining event is \(\{a\}\in\mathcal{F}_{3/2}\); at \(t=9/4\), it is \(\{a,b\}\in\mathcal{F}_{9/4}=2^\Omega\). At \(t=5/2\) and for every later \(t\), the event is \(\Omega\). For \(t<1\), it is empty. These ranges exhaust all \(t\ge0\), including the horizon itself.
Check: the probability of the event at \(t=3/2\) is \(1/3\), since it contains just \(a\). At \(t=9/4\), its probability is \(2/3\), since it contains \(a\) and \(b\). At \(t=5/2\), all three truncated values are at most \(5/2\), so the event is \(\Omega\) and its probability is \(3(1/3)=1\), as required.
Worked Example: Adding a Deterministic Delay
Use the same filtered space and stopping time \(\tau\), and delay it by \(c=1\). The values of \(\tau+1\) are \(2\) on \(a\), \(3\) on \(b\), and \(5\) on \(c\). If \(t<2\), the event \(\{\tau+1\le t\}\) is empty. If \(2\le t<3\), it is \(\{a\}\); if \(3\le t<5\), it is \(\{a,b\}\); and if \(t\ge5\), it is \(\Omega\).
For \(2\le t<3\), we have \(t\ge2\), so \(\mathcal{F}_t=2^\Omega\) and \(\{a\}\in\mathcal{F}_t\). For \(3\le t<5\), again \(\mathcal{F}_t=2^\Omega\), which contains \(\{a,b\}\). The empty event and \(\Omega\) belong to every sigma-field in the corresponding ranges. Therefore \(\tau+1\) is a stopping time.
Check: at \(t=2\), the event is \(\{a\}\), with probability \(1/3\); at \(t=3\), it is \(\{a,b\}\), with probability \(2/3\); at \(t=5\), it is \(\Omega\), with probability \(1\). These agree with the respective delayed values \(2,3,5\) and verify the boundary cases.
Why Subtracting a Delay Can Fail
For \(c>0\), suppose first that \(\tau\ge c\) on every outcome, so that \(\tau-c\) is a nonnegative random time. Its defining event is \(\{\tau-c\le t\}=\{\tau\le t+c\}\). The original stopping-time property places this event in \(\mathcal{F}_{t+c}\), but the definition of a filtration only gives \(\mathcal{F}_t\subseteq\mathcal{F}_{t+c}\), not the reverse inclusion. In fact, the exact criterion is as follows.
Proof. By the definition of a stopping time, \(\tau-c\) is a stopping time exactly when \(\{\tau-c\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). Subtracting \(c\) from both sides of the inequality \(\tau\le t+c\) shows that \(\tau-c\le t\) holds exactly when \(\tau\le t+c\). Therefore \(\{\tau-c\le t\}=\{\tau\le t+c\}\) for every \(t\), and the stopping-time condition is precisely the stated measurability condition. This proves both directions. \(\square\)
Worked Example: A Backward Shift That Is Not a Stopping Time
Let \(\Omega=\{a,b\}\), with probability \(1/2\) on each outcome. Define \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for \(0\le t<1\) and \(\mathcal{F}_t=2^\Omega\) for \(t\ge1\). This filtration is increasing. Let \(\tau(a)=1\) and \(\tau(b)=2\).
To verify that \(\tau\) is a stopping time, note that for \(t<1\) the event \(\{\tau\le t\}\) is empty; for \(1\le t<2\) it is \(\{a\}\in2^\Omega=\mathcal{F}_t\); and for \(t\ge2\) it is \(\Omega\). Now subtract \(c=1\). The values of \(\tau-1\) are \(0\) on \(a\) and \(1\) on \(b\), so the shifted time is nonnegative. But at \(t=0\), \(\{\tau-1\le0\}=\{a\}\), which is not in \(\mathcal{F}_0=\{\varnothing,\Omega\}\). Therefore \(\tau-1\) is not a stopping time.
Check: \(\mathbb{P}(\{\tau-1\le0\})=\mathbb{P}(\{a\})=1/2\), while the only events in \(\mathcal{F}_0\) have probabilities \(0\) and \(1\). Thus the defining event is indeed absent from \(\mathcal{F}_0\). At \(t=1\), the shifted-time event is \(\Omega\), since both shifted values are at most \(1\); it is measurable in \(\mathcal{F}_1=2^\Omega\). The failure is specifically at the earlier time \(t=0\).
A Common Measurability Pitfall
A tempting but incorrect argument for the backward shift is: “\(\{\tau-c\le t\}=\{\tau\le t+c\}\in\mathcal{F}_{t+c}\), so it belongs to \(\mathcal{F}_t\).” The last implication reverses the filtration inclusion. In the finite example, at \(t=0\) the event is \(\{a\}\in\mathcal{F}_1\), but \(\{a\}\notin\mathcal{F}_0\). Measurability at a later time does not establish measurability at an earlier one.
There is a separate algebraic error to avoid when delaying a time. The correct identity is \(\{\tau+c\le t\}=\{\tau\le t-c\}\) when \(t\ge c\), not \(\{\tau\le t\}\). For the delayed example, at \(t=2\) the correct event is \(\{\tau\le1\}=\{a\}\), whereas the incorrect expression \(\{\tau\le2\}\) gives \(\{a,b\}\). Both happen to be measurable at time \(2\) in that example, but they are different events; the correct identity is essential to the proof.
These transformations are useful whenever a stopping rule is bounded by a fixed horizon or when an action is scheduled a fixed time after the rule triggers. The defining-event test remains the reliable method: write the event that the transformed time is at most \(t\), simplify the inequality, and check membership in \(\mathcal{F}_t\) at every \(t\), with the endpoints included.
Check Your Understanding
For each question, use the defining events and keep the direction of the filtration inclusions explicit.
- If \(\tau\) is a stopping time and \(T=0\), what is \(\{\tau\wedge T\le t\}\) for \(t\ge0\), and why is it measurable?
- For \(c>0\), write \(\{\tau+c\le t\}\) separately for \(t<c\) and \(t\ge c\). Which filtration inclusion is used in the second case?
- In the two-outcome counterexample, list \(\{\tau-1\le t\}\) for \(t=0\), \(0<t<1\), and \(t\ge1\). At which times does the stopping-time condition fail?
- State the exact measurability criterion for \(\tau-c\) to be a stopping time, assuming it is nonnegative.
- Why does \(\{\tau\le t+c\}\in\mathcal{F}_{t+c}\) alone not prove that a backward shift is a stopping time?