The Sigma-Field At A Stopping Time

Information Available at a Random Time

At a deterministic time $t$, the sigma-field $\mathcal{F}_t$ describes the information available by that time. If the observation time itself is random, the corresponding information must account for both what has been observed and when the observation ends. The key test is made at every deterministic time $t$: on the event that the stopping time has already occurred by $t$, the event in question must be measurable using $\mathcal{F}_t$.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. Thus $\mathcal{F}_s\subseteq\mathcal{F}_t\subseteq\mathcal{F}$ whenever $s\le t$. A stopping time $\tau$ is a random variable with values in $[0,\infty]$ such that $\{\tau\le t\}\in\mathcal{F}_t$ for every finite $t\ge0$. No right-continuity or completeness assumption on the filtration is imposed here. The constructions below do not require a Brownian motion.

Definition: The sigma-field at $\tau$ is $$ \mathcal{F}_\tau=\{A\in\mathcal{F}:A\cap\{\tau\le t\}\in\mathcal{F}_t\text{ for every }t\ge0\}. $$ An event $A$ belongs to $\mathcal{F}_\tau$ when, for each deterministic time $t$, knowing the information available by $t$ is enough to decide whether $A$ occurred on the event $\{\tau\le t\}$.

The condition concerns the intersection with $\{\tau\le t\}$, not the event $A$ in isolation. Before $\tau$, the outcome of $A$ need not yet be observable. The definition also includes the requirement $A\in\mathcal{F}$, so every event in $\mathcal{F}_\tau$ is an event in the ambient probability space.

Basic Properties

Theorem: For every stopping time $\tau$, $\mathcal{F}_\tau$ is a sigma-field contained in $\mathcal{F}$, and $\{\tau\le u\}\in\mathcal{F}_\tau$ for every $u\ge0$.

Proof. Containment in $\mathcal{F}$ follows from the definition. To check the sigma-field properties, first note that $\Omega\in\mathcal{F}_\tau$, since $\Omega\cap\{\tau\le t\}=\{\tau\le t\}\in\mathcal{F}_t$. Also $\varnothing\in\mathcal{F}_\tau$. If $A\in\mathcal{F}_\tau$, then for every $t$, \[ A^c\cap\{\tau\le t\}=\{\tau\le t\}\setminus(A\cap\{\tau\le t\})\in\mathcal{F}_t, \] so $A^c\in\mathcal{F}_\tau$. If $A_1,A_2,\ldots\in\mathcal{F}_\tau$, then \[ \left(\bigcup_{n=1}^\infty A_n\right)\cap\{\tau\le t\} =\bigcup_{n=1}^\infty\bigl(A_n\cap\{\tau\le t\}\bigr)\in\mathcal{F}_t \] for every $t$. Thus $\bigcup_n A_n\in\mathcal{F}_\tau$, proving that $\mathcal{F}_\tau$ is a sigma-field. Finally, fix $u\ge0$. For every $t\ge0$, \[ \{\tau\le u\}\cap\{\tau\le t\}=\{\tau\le\min(u,t)\}\in\mathcal{F}_{\min(u,t)}\subseteq\mathcal{F}_t. \] The event $\{\tau\le u\}$ therefore belongs to $\mathcal{F}_\tau$. $\square$

In particular, $\tau$ is measurable with respect to $\mathcal{F}_\tau$: its sublevel events $\{\tau\le u\}$ belong to that sigma-field. This statement uses the usual characterization of measurability of a nonnegative extended-valued random variable by its sublevel events.

Proposition: If $\sigma$ and $\tau$ are stopping times and $\sigma\le\tau$ almost surely, then $\mathcal{F}_\sigma\subseteq\mathcal{F}_\tau$.

Proof. Let $A\in\mathcal{F}_\sigma$. For each $t\ge0$, on $\{\tau\le t\}$ the inequality $\sigma\le\tau$ implies $\sigma\le t$. Hence, up to a $\mathbb{P}$-null set, \[ A\cap\{\tau\le t\} =(A\cap\{\sigma\le t\})\cap\{\tau\le t\}. \] The event $A\cap\{\sigma\le t\}$ belongs to $\mathcal{F}_t$ by $A\in\mathcal{F}_\sigma$, and $\{\tau\le t\}\in\mathcal{F}_t$ because $\tau$ is a stopping time. Their intersection is in $\mathcal{F}_t$. Since the equality is almost sure, the conclusion as written requires that $\mathcal{F}_t$ contain the relevant null sets; without that assumption, the almost-sure ordering alone does not guarantee literal set membership. In particular, if $\sigma\le\tau$ pointwise, the displayed equality holds exactly and proves $A\in\mathcal{F}_\tau$. More generally, the inclusion holds under the usual completeness assumption, or when the ordering holds pointwise. $\square$

This distinction matters when filtrations are not complete: sigma-fields are collections of sets, so changing an event by a null set need not preserve membership. In the examples below, stopping-time inequalities hold pointwise.

Deterministic and Finite-Valued Stopping Times

Proposition: If $\tau=c$ for a deterministic $c\ge0$, then $\mathcal{F}_\tau=\mathcal{F}_c$.

Proof. Suppose first that $A\in\mathcal{F}_c$. If $t<c$, then $A\cap\{\tau\le t\}=\varnothing\in\mathcal{F}_t$. If $t\ge c$, then $A\cap\{\tau\le t\}=A\in\mathcal{F}_c\subseteq\mathcal{F}_t$. Thus $A\in\mathcal{F}_\tau$. Conversely, if $A\in\mathcal{F}_\tau$, use the defining condition at $t=c$: since $\{\tau\le c\}=\Omega$, we have $A=A\cap\{\tau\le c\}\in\mathcal{F}_c$. Both inclusions hold, giving equality. $\square$

Worked Example: A Deterministic Observation Time

Let $\tau=2$ on any filtered probability space. To test an event $A\in\mathcal{F}$, consider its defining intersections. At $t=1$, $A\cap\{\tau\le1\}=\varnothing$, which is measurable in $\mathcal{F}_1$. At $t=2$, the intersection is $A$, so membership in $\mathcal{F}_\tau$ requires $A\in\mathcal{F}_2$. At any $t>2$, the intersection is again $A$; if $A\in\mathcal{F}_2$, then $A\in\mathcal{F}_t$ by the filtration property. Thus the full condition is exactly $A\in\mathcal{F}_2$, and $\mathcal{F}_\tau=\mathcal{F}_2$.

Check: The test at $t=2$ gives the necessary inclusion $\mathcal{F}_\tau\subseteq\mathcal{F}_2$, while the checks before and after $2$ show that every event of $\mathcal{F}_2$ satisfies the definition.

For a stopping time taking finitely many deterministic values, the definition reduces to a level-set test. The filtration must be increasing, and each level set $\{\tau=t_i\}$ must be considered at the time $t_i$ when that value is reached.

Proposition: Suppose $\tau$ takes values in $0\le t_1<t_2<\cdots<t_n<\infty$. Then $$ A\in\mathcal{F}_\tau \quad\Longleftrightarrow\quad A\cap\{\tau=t_i\}\in\mathcal{F}_{t_i}\text{ for every }i=1,\ldots,n. $$

Proof. If $A\in\mathcal{F}_\tau$, then $A\cap\{\tau\le t_i\}\in\mathcal{F}_{t_i}$. For $i=1$, $\{\tau\le t_1\}=\{\tau=t_1\}$. For $i>1$, because $\tau$ takes only the listed values, \[ A\cap\{\tau=t_i\} =(A\cap\{\tau\le t_i\})\setminus(A\cap\{\tau\le t_{i-1}\}). \] Both sets on the right are in $\mathcal{F}_{t_i}$: the second is in $\mathcal{F}_{t_{i-1}}\subseteq\mathcal{F}_{t_i}$. This proves the level-set conditions. Conversely, suppose every $A\cap\{\tau=t_i\}$ belongs to $\mathcal{F}_{t_i}$. For any $t$, $A\cap\{\tau\le t\}$ is the union of those finitely many sets $A\cap\{\tau=t_i\}$ for which $t_i\le t$. Each is in $\mathcal{F}_{t_i}\subseteq\mathcal{F}_t$, so their union is in $\mathcal{F}_t$. If there is no such $i$, the intersection is empty and is still in $\mathcal{F}_t$. Thus $A\in\mathcal{F}_\tau$. $\square$

Worked Example: Computing a Finite-Valued Stopped Sigma-Field

Consider the four-point probability space $\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\}$, with probability $1/4$ assigned to each point and ambient sigma-field $\mathcal{F}=2^\Omega$. Define $E=\{\omega_1,\omega_3\}$. Let $\mathcal{F}_t$ be trivial for $0\le t<1$, let $\mathcal{F}_t=\{\varnothing,E,E^c,\Omega\}$ for $1\le t<2$, and let $\mathcal{F}_t=2^\Omega$ for $t\ge2$. This is an increasing filtration. Set $\tau=1$ on $E$ and $\tau=2$ on $E^c$. It is a stopping time: before $1$ its sublevel event is empty; for $1\le t<2$ it is $E\in\mathcal{F}_t$; and for $t\ge2$ it is $\Omega$.

Apply the finite-valued criterion. On the level $\{\tau=1\}=E$, the set $A\cap E$ must belong to $\mathcal{F}_1=\{\varnothing,E,E^c,\Omega\}$. Since it is a subset of $E$, the only possibilities are $\varnothing$ and $E$. On the level $\{\tau=2\}=E^c$, the set $A\cap E^c$ may be any subset of $E^c$, because $\mathcal{F}_2=2^\Omega$. Therefore $$ \mathcal{F}_\tau=\{A\subseteq\Omega:A\cap E\in\{\varnothing,E\}\}. $$ Its atoms are $E,\{\omega_2\},\{\omega_4\}$, so it has $2^3=8$ events.

For a direct membership check, take $A=\{\omega_1,\omega_3,\omega_2\}$. Its level-set intersections are $A\cap E=E\in\mathcal{F}_1$ and $A\cap E^c=\{\omega_2\}\in\mathcal{F}_2$, so $A\in\mathcal{F}_\tau$. In contrast, for $C=\{\omega_1,\omega_2\}$, the first intersection is $\{\omega_1\}\notin\mathcal{F}_1$, so $C\notin\mathcal{F}_\tau$.

Check: The listed atoms are disjoint and cover all four points, with sizes $2,1,1$. Choosing any subcollection gives $2\cdot2\cdot2=8$ events, agreeing with the count.

Worked Example: Comparing Information at Two Stopping Times

Use the same filtered space and stopping time $\tau$ as in the preceding example, and define $\sigma=1$ everywhere. Both are stopping times, and $\sigma\le\tau$ pointwise. The deterministic-time result gives $\mathcal{F}_\sigma=\mathcal{F}_1=\{\varnothing,E,E^c,\Omega\}$. The computed $\mathcal{F}_\tau$ contains $\{\omega_2\}$, because its intersection with $E$ is empty and its intersection with $E^c$ belongs to $\mathcal{F}_2$. But $\{\omega_2\}\notin\mathcal{F}_1$, since it is neither empty, $E$, $E^c$, nor $\Omega$. Thus in this example $\mathcal{F}_\sigma$ is strictly smaller than $\mathcal{F}_\tau$.

Check: Every event of $\mathcal{F}_1$ intersects $E$ in either $\varnothing$ or $E$, so it satisfies the computed criterion for $\mathcal{F}_\tau$. The event $\{\omega_2\}$ shows that the inclusion is strict.

A Common Membership Pitfall

It is not enough to check the defining condition at the largest possible value of $\tau$. In the finite example, consider $A=\{\omega_1\}$. At $t=2$, $A\cap\{\tau\le2\}=A\in\mathcal{F}_2$, which might tempt one to conclude that $A\in\mathcal{F}_\tau$. But at $t=1$, \[ A\cap\{\tau\le1\}=\{\omega_1\}\notin\mathcal{F}_1. \] The required condition fails, so $A\notin\mathcal{F}_\tau$. Checking only the terminal time misses whether the event is observable on the earlier level set.

A related distinction is that the finite-valued level-set criterion is not a license to check only one level. Each set $A\cap\{\tau=t_i\}$ must be measurable at its corresponding time $t_i$. For a general stopping time, the defining test is imposed for every deterministic $t\ge0$; a finite list of level sets is available only under the finite-valued assumption.

Check Your Understanding

Use the event-based definition in each question, and state which filtration sigma-field is used in every measurability check.

Prove directly from the definition that if $\tau\equiv c$, then $\mathcal{F}_\tau=\mathcal{F}_c$.
If $\tau$ takes values $1$ and $3$, what two level-set intersections must be checked to decide whether $A\in\mathcal{F}_\tau$?
Why does $\{\tau\le u\}\in\mathcal{F}_\tau$ hold for every fixed $u\ge0$?
In the four-point example, determine whether $\{\omega_3,\omega_4\}$ belongs to $\mathcal{F}_\tau$, and justify the answer using both level sets.
What extra issue arises if $\sigma\le\tau$ holds only almost surely and the filtration is not complete?

The Sigma-Field At A Stopping Time

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