What Stopping Preserves
Stopping changes a process after a random time, but it does not by itself settle every question about the resulting process. Two separate properties matter here: whether each stopped value is available from the information at that time, and whether the sample paths remain right-continuous. These properties are useful in continuous-time martingale arguments because they let us work with a well-behaved stopped process. They do not, by themselves, show that the stopped process is still a martingale; that requires an additional argument.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with an increasing filtration \((\mathcal{F}_t)_{t\ge0}\). A process \((M_t)_{t\ge0}\) is adapted if \(M_t\) is \(\mathcal{F}_t\)-measurable for every fixed \(t\). A nonnegative, possibly infinite random time \(\tau\) is a stopping time if \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). Define the stopped process by \(M^\tau_t=M_{t\wedge\tau}\), where \(t\wedge\tau=\min(t,\tau)\).
The path statement is a composition fact: for each sample point, the map \(t\mapsto t\wedge\tau\) is continuous and nondecreasing, and never moves backward in time. The adaptedness statement requires more care. Evaluating an adapted process at a random time is not automatically justified just by saying that the random time is at most \(t\); we show it using approximations that use only process values at times no later than \(t\).
Two Facts Behind the Theorem
Proof. For \(0\le x<t\), the event \(\{U_t\le x\}\) equals \(\{\tau\le x\}\), which belongs to \(\mathcal{F}_x\) and hence to \(\mathcal{F}_t\). For \(x\ge t\), \(\{U_t\le x\}=\Omega\); for \(x<0\), the event is empty. These sublevel events show that \(U_t\) is \(\mathcal{F}_t\)-measurable. \(\square\)
Proof of the theorem. First take paths that are right-continuous at every time. For a fixed sample point \(\omega\), write \(c=\tau(\omega)\), with \(c=\infty\) allowed. If \(t_r\downarrow t\), then \(t_r\wedge c\downarrow t\wedge c\). Right-continuity of the path gives \(M_{t_r\wedge c}(\omega)\to M_{t\wedge c}(\omega)\). Thus the stopped path is right-continuous.
It remains to prove adaptedness. Fix \(t>0\), set \(U=t\wedge\tau\), and, for each integer \(r\ge1\), divide \([0,t]\) into \(2^r\) equal intervals. Let \(R_r\) be the smallest grid point greater than or equal to \(U\). By the lemma, \(U\) is \(\mathcal{F}_t\)-measurable, so each event \(\{R_r=s\}\), for a grid point \(s\), is in \(\mathcal{F}_t\). Since \(s\le t\) and \(M\) is adapted, \(M_s\) is also \(\mathcal{F}_t\)-measurable. Therefore the finite-valued sum \(M_{R_r}=\sum_s M_s\mathbf{1}_{\{R_r=s\}}\) is \(\mathcal{F}_t\)-measurable. The grid meshes shrink to zero, and \(U\le R_r\le U+t/2^r\), so \(R_r\downarrow U\). Right-continuity gives \(M_{R_r}\to M_U\). Hence \(M_U=M^\tau_t\) is \(\mathcal{F}_t\)-measurable.
If right-continuity holds only almost surely, the same argument gives an \(\mathcal{F}_t\)-measurable limit that agrees with \(M^\tau_t\) almost surely. Completeness ensures that a random variable equal almost surely to an \(\mathcal{F}_t\)-measurable random variable is itself \(\mathcal{F}_t\)-measurable. This proves adaptedness in that case as well. At \(t=0\), \(M^\tau_0=M_0\), which is adapted by assumption. \(\square\)
There is a useful distinction between the two hypotheses in this proof. The stopping-time property ensures that the clipped time \(t\wedge\tau\) is measurable using information at time \(t\). The right-continuous path condition ensures that approximating the clipped time from above approximates the value of the process itself. Neither role can be replaced by the other.
Worked Applications
Worked Example: Brownian Motion Stopped at a Fixed Time
Let \((B_t)_{t\ge0}\) be Brownian motion with its completed natural filtration, and let \(\tau=1\). Brownian motion is a martingale with continuous paths, and the deterministic time \(1\) is a stopping time because \(\{\tau\le t\}\) is either empty or the whole sample space. The theorem applies.
For every \(t\ge0\), the stopped value is
For \(t<1\), the stopped process follows the original continuous path. At \(t=1\), its value is \(B_1\); for every \(t>1\), it remains \(B_1\). The left and right limits at \(1\) both agree with the value there, so the path is continuous. Adaptedness can also be checked directly: before time \(1\), \(B_t\) is \(\mathcal{F}_t\)-measurable; at and after time \(1\), \(B_1\) is \(\mathcal{F}_1\)-measurable and \(\mathcal{F}_1\subseteq\mathcal{F}_t\).
Check: At \(t=0\), the formula gives \(B^\tau_0=B_0\); at \(t=1\), it gives \(B^\tau_1=B_1\); and at \(t=2\), it gives \(B^\tau_2=B_1\). These values agree with evaluating \(B\) at \(t\wedge1\) in each case.
Worked Example: A Compensated Poisson Process Stopped at Its First Arrival
Let \((N_t)_{t\ge0}\) be a Poisson process of rate \(2\), with its completed natural filtration, and set \(M_t=N_t-2t\). This is an adapted martingale with right-continuous paths. Let \(\tau_1=\inf\{t\ge0:N_t\ge1\}\). This is a stopping time because \(\{\tau_1\le t\}=\{N_t\ge1\}\in\mathcal{F}_t\). The stopped process is therefore adapted and right-continuous.
Before the first arrival, \(N_t=0\), and at and after the arrival, the stopped counting process has value \(1\). Consequently, on paths where \(\tau_1<\infty\),
As \(t\) increases to \(\tau_1\) from below, the first expression tends to \(-2\tau_1\). At \(t=\tau_1\), the value is \(N_{\tau_1}-2\tau_1=1-2\tau_1\), so the path has the right-continuous jump of size \(1\). It then follows the linear function \(1-2t\) only up to the stopping time and remains at \(1-2\tau_1\) afterward. If \(\tau_1=\infty\), the process is never stopped at a finite time and \(M^\tau_t=M_t\) for each finite \(t\).
Check: At a fixed time \(t<\tau_1\), \(N_{t\wedge\tau_1}=N_t=0\), giving \(0-2t=-2t\). At \(t=\tau_1<\infty\), \(N_{\tau_1}=1\), giving \(1-2\tau_1\). For \(t>\tau_1\), the clipped time is still \(\tau_1\), so the value remains \(1-2\tau_1\).
Worked Example: Why Adaptedness of the Original Process Matters
Let \(\Omega=\{-1,1\}\), assign probability \(1/2\) to each point, and let \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for every \(t\ge0\). Define \(Z(\omega)=\omega\) and \(M_t=Z\) for every \(t\). This process has constant, hence right-continuous, paths, but it is not adapted: \(Z\) is not measurable with respect to the trivial \(\mathcal{F}_t\). Stop it at the deterministic time \(\tau=0\), which is a stopping time. Then \(M^\tau_t=Z\) for every \(t\), so the stopped process is still not adapted.
Check: For either state \(z\in\{-1,1\}\), the event \(\{Z=1\}=\{1\}\) is neither empty nor all of \(\Omega\), and therefore is not in the trivial \(\mathcal{F}_t\). Thus \(M^\tau_t\) fails the defining measurability requirement at every fixed time. Right-continuity of paths alone cannot supply missing information.
Version Issues and Common Pitfalls
Right-continuity is a path property, not just a collection of fixed-time distributional or conditional-mean statements. For a concrete illustration, let \(U\) be uniformly distributed on \([0,1]\), put \(U\) in \(\mathcal{F}_0\), and take \(\mathcal{F}_t=\sigma(U)\) for all \(t\). Define \(M_t=\mathbf{1}_{\{t=U\}}\). For each fixed deterministic \(t\), \(\mathbb{P}(U=t)=0\), so \(M_t=0\) almost surely and is integrable. For fixed \(s\le t\), both \(M_s\) and \(M_t\) are zero almost surely, so \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) almost surely. Thus the process satisfies the martingale equalities at every fixed pair of times and is adapted. But for every sample point with \(U<\infty\), its path has value \(1\) at \(t=U\) and value \(0\) at times just to the right of \(U\); it is not right-continuous there. Stopping at \(\tau=\infty\) leaves this path unchanged. Martingale equalities at fixed times do not guarantee that the chosen version has right-continuous paths.
Common mistake. One might write “\(\tau\le t\), so \(M_{t\wedge\tau}\) is \(\mathcal{F}_t\)-measurable.” This is not a valid measurability argument: \(\tau\le t\) does not say which value \(M_{t\wedge\tau}\) takes, and the measurability of the random time’s value needs justification. The correction is to note that \(t\wedge\tau\) is \(\mathcal{F}_t\)-measurable and approximate it from above by grid points no later than \(t\). Adaptedness makes each grid value available at time \(t\), and right-continuity identifies the limit with the stopped value.
A second pitfall is to infer that the stopped process is a martingale merely because it is adapted and right-continuous. The theorem proved here establishes those two properties only. Martingale preservation is a separate assertion with its own hypotheses; it is not a consequence of path regularity and adaptedness alone. In particular, this tutorial does not apply an optional stopping theorem. The practical conclusion is narrower but essential: when a right-continuous adapted process is stopped at a stopping time, its stopped version retains the two regularity properties needed to state later continuous-time stopping results carefully.
Check Your Understanding
For each question, identify which property or hypothesis is being used.
- Why is \(t\wedge\tau\) measurable with respect to \(\mathcal{F}_t\) when \(\tau\) is a stopping time?
- In the grid proof of adaptedness, why are all grid points used to approximate \(t\wedge\tau\) no later than \(t\)?
- What path property ensures that evaluating \(M\) at grid points approaching \(t\wedge\tau\) from above converges to \(M_{t\wedge\tau}\)?
- Why does completeness of the filtration help when right-continuity holds only almost surely?
- Does adaptedness and right-continuity of \(M^\tau\) alone prove that it is a martingale? Explain what the theorem does and does not establish.