Pasting Stopping Times Along Measurable Events

Choosing a Stopping Time Using Information Available in Time

A stopping time can depend on information revealed as time passes, but it cannot use information that becomes available only later. This restriction matters when choosing between two stopping times. If the choice is made on an event that is already measurable by the time the earlier of the two stopping times occurs, the resulting choice is again a stopping time. The key point is to test the event $\{T\le t\}$ using only information in $\mathcal{F}_t$.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$, where the filtration is increasing: $\mathcal{F}_s\subseteq\mathcal{F}_t$ for $s\le t$. A random time $\tau:\Omega\to[0,\infty]$ is a stopping time if $\{\tau\le t\}\in\mathcal{F}_t$ for every $t\ge0$. We use the sigma-field at a stopping time $\rho$ in the following sense.

Definition: For a stopping time $\rho$, the sigma-field $\mathcal{F}_\rho$ consists of the events $A\in\mathcal{F}$ such that $A\cap\{\rho\le t\}\in\mathcal{F}_t$ for every $t\ge0$. Thus, on the event that $\rho$ has occurred by time $t$, whether $A$ holds can be determined using information available at time $t$.

Let $\tau$ and $\sigma$ be stopping times, and write $\rho=\tau\wedge\sigma=\min(\tau,\sigma)$. This minimum is a stopping time because $\{\rho\le t\}=\{\tau\le t\}\cup\{\sigma\le t\}$, a union of two events in $\mathcal{F}_t$. The condition on the choice event is $A\in\mathcal{F}_\rho$, not merely $A\in\mathcal{F}$.

The Pasting Theorem

Theorem: Let $\tau$ and $\sigma$ be stopping times and let $\rho=\tau\wedge\sigma$. If $A\in\mathcal{F}_\rho$, then the random time $$ T=\tau\mathbf{1}_A+\sigma\mathbf{1}_{A^c} $$ is a stopping time. Equivalently, $T=\tau$ on $A$ and $T=\sigma$ on $A^c$; this definition also applies when either stopping time can equal $\infty$.

Proof. Fix $t\ge0$. By the definition of $T$, its sublevel event is $$ \{T\le t\}=(A\cap\{\tau\le t\})\cup(A^c\cap\{\sigma\le t\}). $$ Since $\rho\le\tau$, the event $\{\tau\le t\}$ is contained in $\{\rho\le t\}$. Therefore $$ A\cap\{\tau\le t\}=(A\cap\{\rho\le t\})\cap\{\tau\le t\}. $$ The first factor belongs to $\mathcal{F}_t$ because $A\in\mathcal{F}_\rho$, and the second belongs to $\mathcal{F}_t$ because $\tau$ is a stopping time. Thus their intersection belongs to $\mathcal{F}_t$. Likewise, $A^c\cap\{\rho\le t\}=\{\rho\le t\}\setminus(A\cap\{\rho\le t\})$ belongs to $\mathcal{F}_t$; intersecting it with $\{\sigma\le t\}\in\mathcal{F}_t$ shows $A^c\cap\{\sigma\le t\}\in\mathcal{F}_t$. Both parts of the union are in $\mathcal{F}_t$, so $\{T\le t\}\in\mathcal{F}_t$. This holds for every $t\ge0$, proving that $T$ is a stopping time. $\square$

The proof uses the event $\{\rho\le t\}$ to bring the decision event into $\mathcal{F}_t$. It does not require $A$ itself to be in $\mathcal{F}_t$ at times before $\rho$, or even at times after $t$ when $\rho$ has not yet occurred. That is why $\mathcal{F}_\rho$ is the natural condition.

Worked Example: Pasting in a Four-Outcome Filtration

Let $\Omega=\{a,b,c,d\}$, with each outcome having probability $1/4$. Set $\mathcal{F}_t=\{\varnothing,\Omega\}$ for $0\le t<1$; let $\mathcal{F}_t$ for $1\le t<2$ be generated by the partition $\{\{a,b\},\{c,d\}\}$; and set $\mathcal{F}_t=2^\Omega$ for $t\ge2$. Define $\tau=1$ on $\{a,b\}$ and $\tau=2$ on $\{c,d\}$. Define $\sigma=3$ on $\{a,b,c\}$ and $\sigma=2$ on $\{d\}$.

First check the stopping-time conditions. For $\tau$, $\{\tau\le t\}$ is empty for $t<1$, is $\{a,b\}$ for $1\le t<2$, and is $\Omega$ for $t\ge2$; each set is in the corresponding $\mathcal{F}_t$. For $\sigma$, $\{\sigma\le t\}$ is empty for $t<2$, is $\{d\}$ for $2\le t<3$, and is $\Omega$ for $t\ge3$; these sets are also in the corresponding sigma-fields.

The minimum $\rho=\tau\wedge\sigma$ equals $1$ on $\{a,b\}$ and $2$ on $\{c,d\}$. Choose $A=\{c\}$. To check $A\in\mathcal{F}_\rho$, consider $A\cap\{\rho\le t\}$: it is empty for $t<2$, and equals $\{c\}$ for $t\ge2$. Both are in $\mathcal{F}_t$. Pasting gives $T=\tau$ on $c$ and $T=\sigma$ on the other outcomes, so $$ T(a)=3,\qquad T(b)=3,\qquad T(c)=2,\qquad T(d)=2. $$

Directly, $\{T\le t\}$ is empty for $t<2$, equals $\{c,d\}$ for $2\le t<3$, and equals $\Omega$ for $t\ge3$. For $2\le t<3$, $\mathcal{F}_t=2^\Omega$, so the required event is measurable. The direct check agrees with the theorem.

Finite-Valued Choices and a Multiple-Pasting Result

The same argument works when a finite-valued decision selects among several stopping times. A finite measurable partition means a finite collection of pairwise disjoint events whose union is $\Omega$; empty parts are allowed. The events in the partition specify the choice, and each must be measurable at the minimum of the candidate times.

Theorem: Let $\tau_1,\ldots,\tau_n$ be stopping times, where $n\ge1$, and put $\rho=\min_{1\le i\le n}\tau_i$. Let $A_1,\ldots,A_n$ be a finite measurable partition of $\Omega$ with $A_i\in\mathcal{F}_\rho$ for every $i$. Define $T=\tau_i$ on $A_i$. Then $T$ is a stopping time.

Proof. The minimum $\rho$ is a stopping time because $\{\rho\le t\}=\bigcup_{i=1}^n\{\tau_i\le t\}\in\mathcal{F}_t$. Fix $t\ge0$. Since the $A_i$ form a partition, $$ \{T\le t\}=\bigcup_{i=1}^n(A_i\cap\{\tau_i\le t\}). $$ For each $i$, $\rho\le\tau_i$, so $$ A_i\cap\{\tau_i\le t\}=(A_i\cap\{\rho\le t\})\cap\{\tau_i\le t\}. $$ The first factor is in $\mathcal{F}_t$ by $A_i\in\mathcal{F}_\rho$, and the second is in $\mathcal{F}_t$ because $\tau_i$ is a stopping time. Each intersection, and hence the finite union, belongs to $\mathcal{F}_t$. This is true for every $t$, so $T$ is a stopping time. The argument includes $n=1$ and empty partition parts: an empty part contributes the empty set to the union. $\square$

Worked Example: Three Choices on a Finite Partition

Use the four-outcome filtration from the previous example. Let $\tau_1$ be $1$ on $\{a,b\}$ and $2$ on $\{c,d\}$. Let $\tau_2=2$ on all outcomes and $\tau_3=3$ on all outcomes. These are stopping times: for $\tau_2$, the sublevel event is empty before time $2$ and $\Omega$ from time $2$ onward; for $\tau_3$, it is empty before time $3$ and $\Omega$ from time $3$ onward. The minimum $\rho$ is $1$ on $\{a,b\}$ and $2$ on $\{c,d\}$.

Choose the partition $A_1=\{a,b\}$, $A_2=\{c\}$, $A_3=\{d\}$. For $A_1$, the intersection $A_1\cap\{\rho\le t\}$ is empty for $t<1$ and $\{a,b\}$ for $t\ge1$, which is measurable at those times. For $A_2$ and $A_3$, the intersection is empty for $t<2$, and is the singleton $\{c\}$ or $\{d\}$ for $t\ge2$, when the filtration is full. Thus all three events belong to $\mathcal{F}_\rho$.

The pasted time is $T=\tau_1$ on $\{a,b\}$, $T=\tau_2$ on $\{c\}$, and $T=\tau_3$ on $\{d\}$. Consequently, $$ T(a)=1,\qquad T(b)=1,\qquad T(c)=2,\qquad T(d)=3. $$ Its sublevel events are empty for $t<1$, $\{a,b\}$ for $1\le t<2$, $\{a,b,c\}$ for $2\le t<3$, and $\Omega$ for $t\ge3$. The first nonempty event is in the partition sigma-field $\mathcal{F}_t$ for $1\le t<2$; the next two are in the full sigma-field for $t\ge2$. This independently checks that $T$ is a stopping time.

Why the Measurability Condition Cannot Be Dropped

The requirement $A\in\mathcal{F}_{\tau\wedge\sigma}$ prevents a choice from depending on information unavailable when the earlier candidate time occurs. The following calculation shows what can go wrong when that requirement fails.

Worked Example: A Choice Made Using Unavailable Information

Keep the same four-outcome filtration, and let $\tau=1$ on $\{a,b\}$ and $\tau=2$ on $\{c,d\}$. Set $\sigma=2$ on every outcome. Both are stopping times, and $\rho=\tau\wedge\sigma=\tau$. Take $A=\{a\}$. This event is not in $\mathcal{F}_\rho$, because $$ A\cap\{\rho\le1\}=\{a\}\notin\mathcal{F}_1, $$ where $\mathcal{F}_1$ has atoms $\{a,b\}$ and $\{c,d\}$.

If one nevertheless defines $T=\tau$ on $A$ and $T=\sigma$ on $A^c$, then $T(a)=1$ and $T(b)=T(c)=T(d)=2$. Hence $\{T\le1\}=\{a\}\notin\mathcal{F}_1$, so $T$ is not a stopping time. The tempting but incorrect calculation is to treat $A\cap\{\tau\le1\}$ as measurable at time $1$ merely because $\{\tau\le1\}\in\mathcal{F}_1$. Intersecting a measurable event with a nonmeasurable event need not preserve measurability.

For correction, replace $A$ by $\{a,b\}$, which is in $\mathcal{F}_\rho$: its intersection with $\{\rho\le t\}$ is empty for $t<1$, equals $\{a,b\}$ for $1\le t<2$, and remains $\{a,b\}$ for $t\ge2$. The resulting $T$ equals $1$ on $\{a,b\}$ and $2$ on $\{c,d\}$. Its sublevel event is empty for $t<1$, $\{a,b\}$ for $1\le t<2$, and $\Omega$ for $t\ge2$, all in $\mathcal{F}_t$.

The condition is sufficient, but it need not be necessary for every particular pair of times and event: a choice that fails the condition could still happen to produce a stopping time. The theorem gives a reliable criterion that works uniformly without needing to inspect accidental cancellations between the two candidates. In applications, this distinction is useful when a stopping rule is constructed by branching on observed information. One must verify when that information is available, rather than only verify that the branch event is measurable somewhere in the underlying probability space.

Check Your Understanding

For each question, use the defining sublevel events and state the sigma-field in which each event lies.

Suppose $A\in\mathcal{F}_{\tau\wedge\sigma}$. Why does $\{\tau\le t\}\subseteq\{\tau\wedge\sigma\le t\}$ help prove that $A\cap\{\tau\le t\}\in\mathcal{F}_t$?
In the first worked example, list $\{T\le t\}$ for $t<2$, $2\le t<3$, and $t\ge3$.
For a finite partition $A_1,\ldots,A_n$, explain why $A_i\cap\{\tau_i\le t\}$ is measurable when $A_i\in\mathcal{F}_{\min_j\tau_j}$.
In the unavailable-information example, identify the exact stopping-time condition that fails at $t=1$.
Does the pasting theorem require $A\in\mathcal{F}_t$ for every deterministic $t$? Explain the role of $\mathcal{F}_{\tau\wedge\sigma}$ instead.

Pasting Stopping Times Along Measurable Events

What You'll Learn