The Sigma-Field at a Stopping Time

Information Available at a Random Time

A filtration \((\mathcal F_n)_{n\ge0}\) records what is observable by each discrete time \(n\). If the time at which we stop is itself random, the relevant information is not simply one fixed \(\mathcal F_n\). An event belongs to the sigma-field at the stopping time when, on the outcomes where stopping has occurred by time \(n\), its truth can be determined using \(\mathcal F_n\), for every \(n\). This definition is useful whenever events must be interpreted relative to a random observation horizon.

Work on a probability space \((\Omega,\mathcal F,\mathbb P)\) with a filtration \((\mathcal F_n)_{n\ge0}\), meaning that each \(\mathcal F_n\) is a sub-sigma-field of \(\mathcal F\) and \(\mathcal F_n\subseteq\mathcal F_{n+1}\). A random time \(\tau:\Omega\to\{0,1,2,\ldots\}\cup\{\infty\}\) is a stopping time if \(\{\tau\le n\}\in\mathcal F_n\) for every \(n\ge0\). Here \(\tau\) is allowed to be infinite, and all event equalities in the definitions below are setwise.

Definition: The sigma-field at the stopping time \(\tau\) is \(\mathcal F_\tau=\{A\in\mathcal F:A\cap\{\tau\le n\}\in\mathcal F_n\text{ for every }n\ge0\}\). Thus, for every deterministic time \(n\), the part of \(A\) on which stopping has already happened by \(n\) must be observable at time \(n\).

The condition is imposed at every \(n\), not only at the realized value of \(\tau\) on a particular outcome. For a fixed event \(A\), it asks whether the information available by each possible observation time suffices to decide \(A\) on the outcomes stopped by then.

Why This Collection Is a Sigma-Field

Theorem: If \(\tau\) is a stopping time for \((\mathcal F_n)\), then \(\mathcal F_\tau\) is a sigma-field contained in \(\mathcal F\).

Proof. First, \(\varnothing\in\mathcal F\), and \(\varnothing\cap\{\tau\le n\}=\varnothing\in\mathcal F_n\) for every \(n\), so \(\varnothing\in\mathcal F_\tau\). Also, \(\Omega\in\mathcal F_\tau\), because \(\Omega\cap\{\tau\le n\}=\{\tau\le n\}\in\mathcal F_n\) by the stopping-time property. If \(A\in\mathcal F_\tau\), then \(A^c\cap\{\tau\le n\}=\{\tau\le n\}\setminus(A\cap\{\tau\le n\})\). Both sets on the right are in \(\mathcal F_n\), so their difference is in \(\mathcal F_n\). Hence \(A^c\in\mathcal F_\tau\). Finally, let \(A_1,A_2,\ldots\in\mathcal F_\tau\). For every \(n\), \((\bigcup_{r\ge1}A_r)\cap\{\tau\le n\}=\bigcup_{r\ge1}(A_r\cap\{\tau\le n\})\). Each term in this union belongs to \(\mathcal F_n\), so the union belongs to \(\mathcal F_n\). Also \(\bigcup_{r\ge1}A_r\in\mathcal F\), since \(\mathcal F\) is a sigma-field. Thus \(\bigcup_{r\ge1}A_r\in\mathcal F_\tau\). These properties prove that \(\mathcal F_\tau\) is a sigma-field, and its definition makes it a subset of \(\mathcal F\). \(\square\)

There is an equivalent way to check membership that is especially convenient for finite examples. The event \(\{\tau=n\}\) is the slice of outcomes where stopping occurs exactly at time \(n\). On that slice, an event in \(\mathcal F_\tau\) must be decidable using information available at time \(n\).

Characterization: For a stopping time \(\tau\), an event \(A\in\mathcal F\) belongs to \(\mathcal F_\tau\) if and only if \(A\cap\{\tau=n\}\in\mathcal F_n\) for every finite \(n\ge0\). This characterization also applies when \(\tau\) can equal \(\infty\); the definition imposes no additional restriction on \(A\cap\{\tau=\infty\}\) beyond \(A\in\mathcal F\).

Proof. Suppose first that \(A\in\mathcal F_\tau\). At \(n=0\), \(\{\tau=0\}=\{\tau\le0\}\), so \(A\cap\{\tau=0\}\in\mathcal F_0\). For \(n\ge1\), the stopping-time property and the filtration property give \(\{\tau\le n-1\}\in\mathcal F_{n-1}\subseteq\mathcal F_n\). Therefore \(A\cap\{\tau=n\}=(A\cap\{\tau\le n\})\setminus(A\cap\{\tau\le n-1\})\in\mathcal F_n\). Conversely, suppose \(A\in\mathcal F\) and \(A\cap\{\tau=k\}\in\mathcal F_k\) for every finite \(k\). For each \(n\), the disjoint decomposition of \(\{\tau\le n\}\) gives \(A\cap\{\tau\le n\}=\bigcup_{k=0}^n(A\cap\{\tau=k\})\). Each term belongs to \(\mathcal F_k\subseteq\mathcal F_n\), so the finite union belongs to \(\mathcal F_n\). This is exactly the defining condition for \(A\in\mathcal F_\tau\). \(\square\)

Computing the Stopped Sigma-Field

Worked Example: Stopping After the First Toss or the Second

Let \(\Omega=\{HH,HT,TH,TT\}\), with each outcome having probability \(1/4\), and let \(\mathcal F\) be the set of all subsets of \(\Omega\). The process \((X_n)_{n\ge1}\) records successive coin tosses, with state space \(S=\{H,T\}\), and \(\mathcal F_n=\sigma(X_1,\ldots,X_n)\); take \(\mathcal F_0=\{\varnothing,\Omega\}\) and \(\mathcal F_n=\mathcal F\) for \(n\ge2\). Define \(\tau=1\) when \(X_1=H\) and \(\tau=2\) when \(X_1=T\). This is a stopping time: \(\{\tau\le1\}=\{HH,HT\}\in\mathcal F_1\), and \(\{\tau\le n\}=\Omega\in\mathcal F_n\) for \(n\ge2\).

At time \(1\), the filtration distinguishes the two sets \(\{HH,HT\}\) and \(\{TH,TT\}\), but it cannot distinguish outcomes inside either set. The time-1 slice is \(\{\tau=1\}=\{HH,HT\}\), and \(\mathcal F_1\) distinguishes only that entire slice or the empty set within it. At time \(2\), each outcome is distinguishable. By the slice characterization, the atoms of \(\mathcal F_\tau\) are therefore \(\{HH,HT\}\), \(\{TH\}\), and \(\{TT\}\). An atom here means a nonempty set in the sigma-field that has no smaller nonempty subset in that sigma-field.

For example, \(A=\{HH,HT,TH\}\) belongs to \(\mathcal F_\tau\): its intersection with \(\{\tau=1\}\) is \(\{HH,HT\}\in\mathcal F_1\), and its intersections with \(\{\tau=2\}\) at the individual timeslices are \(\{TH\}\in\mathcal F_2\) and \(\varnothing\in\mathcal F_2\). Equivalently, \(A\) is the union of the first two atoms. Its probability is \(\mathbb P(A)=3/4\). In contrast, \(B=\{HH,TH\}\) is not in \(\mathcal F_\tau\): \(B\cap\{\tau=1\}=\{HH\}\notin\mathcal F_1\), because time \(1\) does not distinguish \(HH\) from \(HT\). There are \(2^3=8\) events in \(\mathcal F_\tau\), one for each union of its three atoms.

Check: The event \(A\) is the union \(\{HH,HT\}\cup\{TH\}\), while \(B\) cuts the time-1 atom \(\{HH,HT\}\) into two pieces. This verifies both the included and excluded cases directly from the definition.

Worked Example: The First Head Within Three Tosses

Let \(\Omega=\{H,T\}^3\) have the uniform probability distribution, let \(X_n\) be toss \(n\) with state space \(S=\{H,T\}\), and let \(\mathcal F_n=\sigma(X_1,\ldots,X_n)\) for \(n=1,2,3\), with \(\mathcal F_0\) trivial and \(\mathcal F_n=\mathcal F_3\) for \(n\ge3\). Stop at the first head, but stop at time \(3\) if the first two tosses are both tails: \[ \tau=\begin{cases} 1,&X_1=H,\\ 2,&X_1=T,\ X_2=H,\\ 3,&X_1=T,\ X_2=T. \end{cases} \]

The stopping-time checks are \(\{\tau\le1\}=\{H**\}\in\mathcal F_1\), \(\{\tau\le2\}=\{H**\}\cup\{TH*\}\in\mathcal F_2\), and \(\{\tau\le3\}=\Omega\in\mathcal F_3\). Here \(H**\) denotes the two outcomes beginning with \(H\), and \(TH*\) denotes the two beginning with \(TH\). On the time-1 slice, the available information cannot distinguish the two outcomes beginning with \(H\), so that slice is one atom of \(\mathcal F_\tau\). On the time-2 slice, the two outcomes beginning with \(TH\) are likewise indistinguishable at time \(2\), so they form another atom. At time \(3\), the two remaining outcomes \(TTH\) and \(TTT\) can be distinguished. Thus the atoms are \(\{HHH,HHT\}\), \(\{THH,THT\}\), \(\{TTH\}\), and \(\{TTT\}\).

For instance, \(C=\{HHH,HHT,TTH\}\) belongs to \(\mathcal F_\tau\): its time-1 slice is the entire \(\{HHH,HHT\}\) atom, its time-2 slice is empty, and its time-3 slice is \(\{TTH\}\in\mathcal F_3\). Since there are eight equally likely outcomes, \(\mathbb P(C)=3/8\). There are four atoms, so \(\mathcal F_\tau\) contains \(2^4=16\) events, all unions of these four sets.

Check: The time-1 and time-2 slices of \(C\) are respectively \(\{HHH,HHT\}\) and \(\varnothing\), both in their required sigma-fields; its time-3 slice is \(\{TTH\}\), which is in \(\mathcal F_3\). The membership test is satisfied at every possible stopping time.

Comparing Information at Different Times

At a deterministic time, the definition returns exactly the usual filtration sigma-field. This also gives a useful check on the definition: stopping at time \(m\) should reveal precisely the information available at time \(m\), neither more nor less.

Proposition: If \(\tau=m\) for every outcome, where \(m\) is a fixed nonnegative integer, then \(\mathcal F_\tau=\mathcal F_m\).

Proof. When \(n<m\), \(\{\tau\le n\}=\varnothing\), so the defining condition at those times is automatic. At \(n=m\), \(\{\tau\le m\}=\Omega\), and the condition requires \(A\in\mathcal F_m\). If \(A\in\mathcal F_m\), then for every \(n\ge m\), \(A\in\mathcal F_m\subseteq\mathcal F_n\), so all remaining conditions hold. Thus \(A\in\mathcal F_\tau\) exactly when \(A\in\mathcal F_m\). \(\square\)

More generally, information available at an earlier stopping time is included in the information available at a later one, provided the times are ordered on every outcome.

Proposition: If \(\sigma\) and \(\tau\) are stopping times and \(\sigma\le\tau\) for every outcome, then \(\mathcal F_\sigma\subseteq\mathcal F_\tau\).

Proof. Take \(A\in\mathcal F_\sigma\). For each \(n\), the pointwise inequality \(\sigma\le\tau\) implies \(\{\tau\le n\}\subseteq\{\sigma\le n\}\). Consequently, \(A\cap\{\tau\le n\}=(A\cap\{\sigma\le n\})\cap\{\tau\le n\}\). Since \(A\in\mathcal F_\sigma\), the first set in the intersection belongs to \(\mathcal F_n\); since \(\tau\) is a stopping time, the second does too. Their intersection is in \(\mathcal F_n\). This holds for every \(n\), so \(A\in\mathcal F_\tau\), proving the inclusion. \(\square\)

A Common Mistake: Using a Time That Is Not a Stopping Time

The condition that \(\tau\) is a stopping time is essential to the sigma-field result. Consider again two fair tosses with \(\mathcal F_1=\sigma(X_1)\) and \(\mathcal F_2\) containing all subsets of the four outcomes. Define a different random time: \(\tau=1\) if \(X_2=H\), and \(\tau=2\) if \(X_2=T\). Then \(\{\tau\le1\}=\{HH,TH\}\), which is not in \(\mathcal F_1\), because time \(1\) reveals only \(X_1\). Therefore this \(\tau\) is not a stopping time.

A tempting but invalid calculation is to apply the stopped-sigma-field definition anyway and claim that \(\Omega\) belongs to the resulting collection. In fact, for \(A=\Omega\), the required time-1 intersection is \(\Omega\cap\{\tau\le1\}=\{HH,TH\}\notin\mathcal F_1\). So even \(\Omega\) fails the defining test. The stopping-time hypothesis ensures that the whole event \(\{\tau\le n\}\) is observable at time \(n\); without it, the collection need not be a sigma-field at all.

When computing \(\mathcal F_\tau\) on a finite sample space, first list the stopping-time slices \(\{\tau=n\}\). On each slice, retain only the distinctions that \(\mathcal F_n\) can make. The atoms of \(\mathcal F_\tau\) are the resulting smallest observable pieces, and every event in \(\mathcal F_\tau\) is a union of them. This method emphasizes the key idea: an event is available at the stopping time only if its truth can be settled using information revealed by the time stopping occurs.

Check Your Understanding

Use the defining condition or the event-slice characterization to answer each question.

If \(\tau=2\) deterministically, what is \(\mathcal F_\tau\), and which condition in the definition forces the answer?
In the first worked example, is \(\{HH,HT,TT\}\) in \(\mathcal F_\tau\)? Check its intersections with the slices \(\{\tau=1\}\) and \(\{\tau=2\}\).
In the three-toss example, how many events belong to \(\mathcal F_\tau\), and what are its atoms?
Why does the proof that \(\mathcal F_\tau\) contains \(\Omega\) require \(\tau\) to be a stopping time?
If \(\sigma\le\tau\) pointwise and both are stopping times, which inclusion holds between \(\mathcal F_\sigma\) and \(\mathcal F_\tau\)?

The Sigma-Field at a Stopping Time

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