What Does It Mean to Decide Using Only Observed Information?
In a sequential experiment, a rule may tell us when to stop collecting observations. The key question is not whether the stopping time can be calculated after the entire experiment is over. It is whether, at each time \(n\), we can determine from the information available by time \(n\) whether stopping has already occurred. That distinction is encoded by the filtration.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A discrete-time filtration is an increasing sequence of sigma-fields \((\mathcal{F}_n)_{n\ge0}\) with \(\mathcal{F}_n\subseteq\mathcal{F}\) and \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\). The events in \(\mathcal{F}_n\) represent information available by time \(n\). A process \((X_n)_{n\ge0}\) is adapted to this filtration if \(X_n\) is \(\mathcal{F}_n\)-measurable for each \(n\). In particular, its observation at time \(n\) is then available at time \(n\).
The event \(\{\tau\le n\}\) is the relevant test: it asks whether stopping has happened by the current time, rather than whether it will happen at some later time. The test must hold at every \(n\), including \(n=0\). If the rule never stops on some outcomes, the value \(\infty\) is allowed; that does not change the definition.
Proof. Suppose first that \(\tau\) is a stopping time. For \(n=0\), \(\{\tau=0\}=\{\tau\le0\}\in\mathcal{F}_0\). For \(n\ge1\), the identity \(\{\tau=n\}=\{\tau\le n\}\setminus\{\tau\le n-1\}\) shows that \(\{\tau=n\}\in\mathcal{F}_n\): both events on the right belong to \(\mathcal{F}_n\), since \(\mathcal{F}_{n-1}\subseteq\mathcal{F}_n\). Conversely, suppose \(\{\tau=k\}\in\mathcal{F}_k\) for every \(k\). For any \(n\ge0\), \(\{\tau\le n\}=\bigcup_{k=0}^{n}\{\tau=k\}\). Each event in this finite union belongs to \(\mathcal{F}_n\), because \(\mathcal{F}_k\subseteq\mathcal{F}_n\) when \(k\le n\). Thus \(\{\tau\le n\}\in\mathcal{F}_n\) for every \(n\), so \(\tau\) is a stopping time. \(\square\)
Worked Examples: Testing the Available Information
Worked Example: The First Observed Success
Let \(\Omega=\{0,1\}^{\mathbb{N}}\) carry the probability measure under which the coordinate observations \(X_1,X_2,\ldots\) are independent, with \(\mathbb{P}(X_k=1)=\mathbb{P}(X_k=0)=\tfrac12\). Set \(X_0=0\), so \((X_n)_{n\ge0}\) has state space \(S=\{0,1\}\). Use \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(X_1,\ldots,X_n)\) for \(n\ge1\). Define \(\tau=\inf\{n\ge1:X_n=1\}\), with \(\inf\varnothing=\infty\). Is \(\tau\) a stopping time?
At time \(0\), \(\{\tau\le0\}=\varnothing\), which belongs to \(\mathcal{F}_0\). For each \(n\ge1\), the event that a success has been observed by time \(n\) is
Each event \(\{X_k=1\}\) belongs to \(\mathcal{F}_k\), and therefore to \(\mathcal{F}_n\) when \(k\le n\). Their finite union belongs to \(\mathcal{F}_n\), so \(\tau\) is a stopping time.
For a numerical check at \(n=3\), the event \(\{\tau\le3\}\) excludes only the outcome \(X_1=X_2=X_3=0\). Independence gives that outcome probability \((\tfrac12)^3=\tfrac18\), so \(\mathbb{P}(\tau\le3)=1-\tfrac18=\tfrac78\). Also, \(\{\tau=3\}=\{X_1=0,X_2=0,X_3=1\}\), which has probability \((\tfrac12)^3=\tfrac18\). These checks agree with the event description: the rule stops at time \(3\) exactly when the first two observations are failures and the third is a success.
Worked Example: A Rule That Looks Ahead
Use the same two-outcome observation space and natural filtration as in the preceding example. Define \(\tau=1\) if \(X_2=1\), and \(\tau=2\) if \(X_2=0\). This is always a finite random time. Is it a stopping time?
At time \(0\), \(\{\tau\le0\}=\varnothing\in\mathcal{F}_0\). At time \(1\), however,
The four possible pairs \((X_1,X_2)\) are \((0,0),(0,1),(1,0),(1,1)\), each with probability \(\tfrac14\). The two atoms of \(\mathcal{F}_1\) are \(\{X_1=0\}\), containing \((0,0),(0,1)\), and \(\{X_1=1\}\), containing \((1,0),(1,1)\). The event \(\{X_2=1\}\) contains \((0,1)\) and \((1,1)\), but not the other outcome in either atom. It is not a union of atoms of \(\mathcal{F}_1\), so it is not in \(\mathcal{F}_1\). Thus \(\tau\) is not a stopping time.
As a probability check, \(\mathbb{P}(\tau\le1)=\mathbb{P}(X_2=1)=\tfrac24=\tfrac12\). Knowing this probability does not make the event observable at time \(1\): the failure is that \(\mathcal{F}_1\) does not reveal \(X_2\).
Worked Example: The Filtration Can Change the Answer
Again let \(X_1\) be a fair binary observation, with state space \(S=\{0,1\}\). Define \(\tau=0\) if \(X_1=1\), and \(\tau=1\) if \(X_1=0\). First use the filtration with \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(X_1,\ldots,X_n)\) for \(n\ge1\). Then compare it with the filtration \(\mathcal{G}_n=\sigma(X_1,\ldots,X_{n+1})\) for \(n\ge0\), which makes \(X_1\) available at time \(0\).
For the first filtration, \(\{\tau\le0\}=\{X_1=1\}\). This event has probability \(\tfrac12\), so it is neither \(\varnothing\) nor \(\Omega\); hence it does not belong to the trivial sigma-field \(\mathcal{F}_0\). The stopping-time condition fails at \(n=0\).
For the second filtration, \(\{\tau\le0\}=\{X_1=1\}\in\mathcal{G}_0\). At every \(n\ge1\), \(\{\tau\le n\}=\Omega\), since \(\tau\) is either \(0\) or \(1\). Thus \(\tau\) is a stopping time relative to \((\mathcal{G}_n)\). Its probabilities are \(\mathbb{P}(\tau=0)=\mathbb{P}(X_1=1)=\tfrac12\) and \(\mathbb{P}(\tau=1)=\mathbb{P}(X_1=0)=\tfrac12\), which sum to \(1\). The rule itself did not change; only the information available at time \(0\) changed.
A Practical Test and a Common Mistake
For a proposed rule, write down \(\{\tau\le n\}\) before deciding whether it is a stopping time. Then ask whether that event can be determined from \(\mathcal{F}_n\). If the rule uses observations \(X_1,\ldots,X_n\), those observations are available when the process is adapted. If the rule depends on \(X_{n+1}\) or on a final outcome not yet revealed, the relevant event may not be in \(\mathcal{F}_n\). A formula for \(\tau\) can use the entire outcome \(\omega\); what matters is whether each decision-by-time-\(n\) event is measurable at time \(n\).
Common mistake. In the look-ahead example, one might argue that \(\{\tau\le1\}=\{X_2=1\}\) is in \(\mathcal{F}_2\), so \(\tau\) is a stopping time. This checks the wrong sigma-field: the defining condition at \(n=1\) requires membership in \(\mathcal{F}_1\), not merely in a later sigma-field. The four-outcome check showed that \(\{X_2=1\}\notin\mathcal{F}_1\), so the stopping-time condition fails. More information eventually becoming available cannot justify a decision that was supposed to be made earlier.
The same caution applies to the words “observed” and “known”: they only have mathematical meaning after the filtration is specified. A rule can be a stopping time under one filtration and not under another, as the third example demonstrated. Conversely, a rule based on future observations does not become a stopping time just because its value is easy to calculate after all observations have been collected.
Check Your Understanding
For each proposed rule, identify the event to test and the sigma-field that must contain it.
- With a natural filtration generated by fair coin tosses, let \(\tau\) be the time of the first tail. Write \(\{\tau\le n\}\) and explain why it belongs to \(\mathcal{F}_n\).
- On a four-outcome space of two fair tosses with the natural filtration, define \(\tau=1\) when \(X_2=0\) and \(\tau=2\) otherwise. Which stopping-time condition should be checked first?
- Suppose \(\tau\) takes only the values \(0\) and \(2\). If \(\{\tau=0\}\in\mathcal{F}_0\), what must be checked at time \(1\), and why?
- Can a random time be a stopping time relative to one filtration but not another? Explain using the role of information at time \(0\).
- Why does knowing that \(\{\tau\le n\}\in\mathcal{F}_{n+1}\) for every \(n\) not establish that \(\tau\) is a stopping time relative to \((\mathcal{F}_n)\)?