Freezing a Process at a Stopping Time
A stopping time lets us specify when to stop by using information available as the process evolves. Once that time arrives, the stopped process keeps the value it had at that moment. Constructing this process carefully is useful in its own right: its paths show exactly what stopping does, while its measurability lets us work with it using the same filtration as the original process. Here we treat discrete time; continuous-time stopped processes require additional considerations.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space with filtration \((\mathcal{F}_n)_{n\ge0}\), an increasing sequence of sub-\(\sigma\)-fields of \(\mathcal{F}\). Let \((X_n)_{n\ge0}\) be a real-valued adapted process, meaning that \(X_n\) is \(\mathcal{F}_n\)-measurable for each \(n\). Let \(\tau\) take values in \(\{0,1,2,\ldots\}\cup\{\infty\}\). It is a stopping time if \(\{\tau\le n\}\in\mathcal{F}_n\) for every \(n\ge0\). Define \(n\wedge\tau=\min(n,\tau)\), with \(n\wedge\infty=n\).
The value at the stopping time is included: if \(\tau=3\), the stopped process makes the move from time 2 to time 3, then makes no further moves. This endpoint convention is reflected in the increment formula below.
Proof. Fix \(n\ge0\). Split the sample space into the disjoint events \(\{\tau=0\},\ldots,\{\tau=n\},\{\tau>n\}\). For \(0\le k\le n\), \(\{\tau=k\}\in\mathcal{F}_k\subseteq\mathcal{F}_n\), because \(\{\tau=k\}=\{\tau\le k\}\setminus\{\tau\le k-1\}\) when \(k\ge1\), and \(\{\tau=0\}=\{\tau\le0\}\). Also, \(\{\tau>n\}\) is the complement of \(\{\tau\le n\}\), so it is in \(\mathcal{F}_n\). On \(\{\tau=k\}\), \(Y_n=X_k\); on \(\{\tau>n\}\), \(Y_n=X_n\). Thus $$ Y_n=\sum_{k=0}^{n}X_k\mathbf{1}_{\{\tau=k\}}+X_n\mathbf{1}_{\{\tau>n\}}. $$ Each \(X_k\) in this finite sum is \(\mathcal{F}_n\)-measurable, and each multiplying indicator is \(\mathcal{F}_n\)-measurable. Therefore \(Y_n\) is \(\mathcal{F}_n\)-measurable, proving adaptedness. Now fix \(k\ge1\). On \(\{\tau\ge k\}\), both \(k\wedge\tau=k\) and \((k-1)\wedge\tau=k-1\), so \(Y_k-Y_{k-1}=X_k-X_{k-1}\). On \(\{\tau<k\}\), integer-valuedness gives \(\tau\le k-1\), so both stopped indices equal \(\tau\), and \(Y_k-Y_{k-1}=0\). These two cases give the stated increment identity. \(\square\)
The indicator in that identity is determined by the information available just before step \(k\): \(\{\tau\ge k\}=\{\tau>k-1\}\in\mathcal{F}_{k-1}\). The formula therefore keeps an increment exactly when the process has not stopped before that increment. This is a pathwise identity; it does not by itself assert that \(Y_n\) is a martingale.
Worked Example: A Random Walk Stopped at a Boundary
Take the probability space of infinite sequences of independent fair signs, with its natural filtration \(\mathcal{F}_n\) generated by the first \(n\) signs. Let \(\xi_k\in\{-1,1\}\) be the \(k\)th sign and \(X_0=0,\ X_n=\sum_{k=1}^n\xi_k\). Stop at \(\tau=\inf\{n\ge0:|X_n|=2\}\), with \(\inf\varnothing=\infty\). This is a stopping time because \(\{\tau\le n\}\) is the event that the observed walk has reached either boundary by time \(n\).
Consider the path whose first five increments are \(+1,+1,-1,+1,+1\). Its probability is \((\tfrac12)^5=\tfrac1{32}\). The successive original values are \(X_0=0,\ X_1=1,\ X_2=2,\ X_3=1,\ X_4=2,\ X_5=3\). The first boundary hit is at \(\tau=2\), so the stopped values are $$ Y_0=0,\quad Y_1=1,\quad Y_2=2,\quad Y_3=2,\quad Y_4=2,\quad Y_5=2. $$
Check by increments: At \(k=1,2\), \(\tau\ge k\), so the stopped increments are \(1\) and \(1\), matching \(X_k-X_{k-1}\). At \(k=3,4,5\), \(\tau<k\), so the stopped increments are all \(0\), even though the original increments are \(-1,+1,+1\). Adding the stopped increments to \(Y_0=0\) gives \(0,1,2,2,2,2\), as required.
Worked Example: A Counting Process Stopped at the First Tail
Use independent fair coin flips on the infinite sequence space and the filtration generated by the flips observed so far. Let \(X_n\) count heads among the first \(n\) flips, with \(X_0=0\). Let \(\tau=\inf\{n\ge1:\text{the }n\text{th flip is a tail}\}\). For every \(n\), whether a tail has appeared by time \(n\) is determined by the first \(n\) flips, so \(\tau\) is a stopping time.
For the path \(H,H,T,H,T\), which has probability \((\tfrac12)^5=\tfrac1{32}\), the original counts are \(X_0=0,\ X_1=1,\ X_2=2,\ X_3=2,\ X_4=3,\ X_5=3\). Here \(\tau=3\), and hence the stopped counts are \(Y_0=0,\ Y_1=1,\ Y_2=2,\ Y_3=2,\ Y_4=2,\ Y_5=2\).
Check by increments: For \(k=1,2,3\), \(\tau\ge k\), and the increments retained are \(1,1,0\). For \(k=4,5\), \(\tau<k\), and the increments retained are \(0,0\). Their cumulative sums are \(0,1,2,2,2,2\), exactly the stopped path. The zero at \(k=3\) is correct: the coin is a tail, so the count does not increase, but the process has not yet frozen before that step.
Why the Stopping-Time Condition Matters
Adaptedness of the stopped process depends on the time being a stopping time. If a proposed time uses information from the future, the decision to have stopped by time \(n\) may not be known at time \(n\). The stopped value can then reveal information unavailable to the filtration at that time.
Worked Example: A Non-Stopping Time Can Break Adaptedness
Let \(\Omega=\{a,b\}\), with \(\mathbb{P}(\{a\})=\mathbb{P}(\{b\})=\tfrac12\). Set \(\mathcal{F}_0=\mathcal{F}_1=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=2^\Omega\) for \(n\ge2\). Define the adapted process \(X_0=0,\ X_1=1,\ X_n=0\) for \(n\ge2\); each of these variables is constant, hence measurable with respect to its filtration. Now define \(\tau(a)=0\) and \(\tau(b)=2\). This is not a stopping time, since \(\{\tau\le0\}=\{a\}\notin\mathcal{F}_0\).
At time 1, the proposed stopped value is \(Y_1(a)=X_0(a)=0\) and \(Y_1(b)=X_1(b)=1\). Thus \(\{Y_1=0\}=\{a\}\notin\mathcal{F}_1\), so \(Y_1\) is not \(\mathcal{F}_1\)-measurable. The failure is not in the adaptedness of \(X\); it is that the rule deciding whether the process has already stopped uses information not available at time 0.
Common indexing mistake. A frequent error is to keep the increment at step \(k\) only when \(\tau>k\). That condition excludes the step that reaches the stopping time. For example, if \(\tau=2\) and the path has \(X_0=0,\ X_1=1,\ X_2=2\), the incorrect rule \(\mathbf{1}_{\{\tau>k\}}(X_k-X_{k-1})\) keeps the first increment but drops the second, producing the false value \(1\) at time 2. The correct condition is \(\tau\ge k\): it retains both increments, giving \(Y_2=2\), and suppresses increments only after the process has stopped.
The stopped process is a pathwise construction, not an optional-stopping conclusion. In particular, the identity \(Y_n=X_{n\wedge\tau}\) and its increment formula do not alone show that a stopped martingale has the same conditional or expected values as the original martingale. Those claims require additional arguments and hypotheses. For constructing paths and checking adaptedness, the essential points are to use a stopping time, include the step at time \(\tau\), and freeze the value afterward.
Check Your Understanding
Use the definition and increment identity to answer each question.
- If \(\tau=0\), what is \(Y_n\) for each \(n\ge0\)?
- For an integer-valued stopping time, express \(\{\tau\ge k\}\) using a strict inequality and identify a filtration field containing it.
- A path has \(X_0=0,\ X_1=2,\ X_2=5,\ X_3=4\) and \(\tau=2\). Write \(Y_0,Y_1,Y_2,Y_3\).
- Why does the increment formula use \(\mathbf{1}_{\{\tau\ge k\}}\), rather than \(\mathbf{1}_{\{\tau>k\}}\)?
- In the two-point example, which event demonstrates that \(\tau\) is not a stopping time, and at what time does the stopped process fail adaptedness?