From Deterministic Times to Random Times
At a deterministic time \(s\), Brownian motion has a fresh future: the increments \(B_{s+t}-B_s\) are independent of the information available at time \(s\), and have the same laws as Brownian motion started at zero. A stopping time is a random time whose occurrence can be detected from information observed up to that time. The strong Markov property says that the same fresh-future statement holds at such a random time, provided it is bounded here.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to this filtration. In particular, \(B_t\) is adapted, has continuous paths, and for deterministic \(s\), the future increments are independent of \(\mathcal{F}_s\) and have the usual Brownian finite-dimensional laws. No stochastic integral is used in this tutorial.
The process whose behavior matters is the shifted process \(B_{\tau+t}-B_\tau\), \(t\ge0\). The assertion is not that \(B_{\tau+t}\) is independent of the past: its starting value \(B_\tau\) is part of the past. Rather, the increments after subtracting that starting value form a Brownian motion independent of \(\mathcal{F}_\tau\).
Approximate the Stopping Time from Above
Fix a bounded stopping time \(\tau\), and choose \(C\) such that \(\tau\le C\) almost surely. For each integer \(n\ge1\), let \(\delta_n=2^{-n}\) and define the dyadic ceiling approximation \[ \tau_n=\delta_n\left\lceil\frac{\tau}{\delta_n}\right\rceil. \] This takes only finitely many values on the event \(\{\tau\le C\}\), is a stopping time, satisfies \(\tau_n\ge\tau\), and decreases to \(\tau\). The key point is that the deterministic-time independence property can be applied separately on each event \(\{\tau_n=k\delta_n\}\).
Proof. First consider one approximation \(\tau_n\), and write \(\delta=\delta_n\). Its possible values are \(k\delta\) for finitely many nonnegative integers \(k\). If \(A\in\mathcal{F}_\tau\), then \(A\cap\{\tau_n=k\delta\}\in\mathcal{F}_{k\delta}\). For \(k\ge1\), this follows by writing the event as the difference between \(A\cap\{\tau\le k\delta\}\) and \(A\cap\{\tau\le(k-1)\delta\}\); both are in \(\mathcal{F}_{k\delta}\) by the definition of \(\mathcal{F}_\tau\). For \(k=0\), it is \(A\cap\{\tau=0\}\in\mathcal{F}_0\).
Fix the times \(t_1,\ldots,t_m\) and a bounded continuous \(g\). On \(\{\tau_n=k\delta\}\), the vector of increments after \(\tau_n\) is \((B_{k\delta+t_j}-B_{k\delta})_{j=1}^m\). By deterministic-time independent, stationary increments, this vector is independent of \(\mathcal{F}_{k\delta}\) and has the same distribution as \((B_{t_j})_{j=1}^m\). Therefore, summing over the finitely many possible values of \(k\), \[ \mathbb{E}\!\left[\mathbf{1}_A g(B_{\tau_n+t_1}-B_{\tau_n},\ldots,B_{\tau_n+t_m}-B_{\tau_n})\right] =\mathbb{P}(A)\mathbb{E}\!\left[g(B_{t_1},\ldots,B_{t_m})\right]. \]
For each outcome in the event where the Brownian path is continuous and \(\tau_n\to\tau\), we have \(B_{\tau_n+t_j}\to B_{\tau+t_j}\) and \(B_{\tau_n}\to B_\tau\) for every \(j\). Continuity of \(g\) gives convergence of the corresponding \(g\)-values. Since \(g\) is bounded, dominated convergence allows passage to the limit in the expectation. This proves the displayed identity in the theorem.
Bounded continuous test functions determine probability distributions, so the identity gives the Brownian finite-dimensional laws and their independence from \(\mathcal{F}_\tau\). To pass from finite-dimensional laws to the whole process, take cylinder events involving finitely many rational-time coordinates. They form a pi-system; the class of events independent of each fixed \(A\in\mathcal{F}_\tau\) is closed under complements and countable disjoint unions. The pi-lambda theorem therefore extends independence to the sigma-field generated by all rational-time coordinates. For continuous paths, those coordinates determine the path and generate its Borel sigma-field. Hence the shifted process is Brownian and independent of \(\mathcal{F}_\tau\). \(\square\)
Worked Consequences
Worked Example: A Deterministic Stopping Time
Let \(\tau=\tfrac34\), and consider \(Y_t=B_{\tau+t}-B_\tau\) at \(t_1=\tfrac12\) and \(t_2=1\). The time \(\tau\) is bounded and deterministic, so the theorem applies. The vector is \[ (Y_{1/2},Y_1)=(B_{5/4}-B_{3/4},\,B_{7/4}-B_{3/4}). \] Its two coordinates are jointly normal, have mean zero, and have covariance matrix with entries \(\operatorname{Cov}(Y_s,Y_t)=\min(s,t)\). Substitution gives \[ \begin{pmatrix} \operatorname{Var}(Y_{1/2})&\operatorname{Cov}(Y_{1/2},Y_1)\\ \operatorname{Cov}(Y_1,Y_{1/2})&\operatorname{Var}(Y_1) \end{pmatrix} = \begin{pmatrix} \tfrac12&\tfrac12\\ \tfrac12&1 \end{pmatrix}. \] The shifted vector is independent of \(\mathcal{F}_{3/4}\). As a check, its first coordinate is an increment over an interval of length \(5/4-3/4=1/2\), so its variance is \(1/2\); the second spans length \(7/4-3/4=1\), so its variance is \(1\).
Worked Example: A Stopping Time Chosen by an Observed Sign
Define \(\tau=1\) on \(\{B_1\ge0\}\) and \(\tau=2\) on \(\{B_1<0\}\). This is a stopping time: before time \(1\), \(\{\tau\le t\}\) is empty; for \(1\le t<2\), it is \(\{B_1\ge0\}\in\mathcal{F}_1\subseteq\mathcal{F}_t\); and for \(t\ge2\), it is all of \(\Omega\). It is bounded by \(2\).
For any \(h>0\), the theorem gives \[ \mathbb{P}(B_{\tau+h}-B_\tau>0\mid\mathcal{F}_\tau)=\tfrac12 \quad\text{almost surely}. \] Indeed, the shifted increment has distribution \(N(0,h)\), which is symmetric about zero and has no atom at zero. Equivalently, for every \(A\in\mathcal{F}_\tau\), the probability of \(A\cap\{B_{\tau+h}-B_\tau>0\}\) is \(\mathbb{P}(A)/2\). This illustrates that the sign used to choose the stopping time does not bias the subsequent increment.
Check: the sign event has probability \(\mathbb{P}(B_1\ge0)=1/2\), since \(B_1\sim N(0,1)\) is symmetric and has zero probability of being exactly zero. The conditional probability \(1/2\) above is the same on either branch, as required by independence from \(\mathcal{F}_\tau\).
Worked Example: Conditional Mean and Second Moment
Use the same \(\tau\), and compute the mean and second moment of \(B_\tau\), then those of \(B_{\tau+1}\). Write \(Z=B_2-B_1\). The variable \(Z\sim N(0,1)\) is independent of \(\mathcal{F}_1\), and \[ B_\tau=B_1\mathbf{1}_{\{B_1\ge0\}}+(B_1+Z)\mathbf{1}_{\{B_1<0\}}. \] By symmetry, \(\mathbb{E}[B_1\mathbf{1}_{\{B_1\ge0\}}]=1/\sqrt{2\pi}\) and \(\mathbb{E}[B_1\mathbf{1}_{\{B_1<0\}}]=-1/\sqrt{2\pi}\). Also, independence gives \(\mathbb{E}[Z\mathbf{1}_{\{B_1<0\}}]=0\). Adding these terms yields \(\mathbb{E}[B_\tau]=0\).
For the second moment, the nonnegative and negative half-lines each contribute half of \(\mathbb{E}[B_1^2]=1\). On the negative branch, expanding \((B_1+Z)^2\) gives the additional terms \(\mathbb{E}[Z^2\mathbf{1}_{\{B_1<0\}}]=1\cdot\tfrac12=\tfrac12\) and \(2\mathbb{E}[B_1Z\mathbf{1}_{\{B_1<0\}}]=0\). Thus \[ \mathbb{E}[B_\tau^2]=\tfrac12+\tfrac12+\tfrac12=\tfrac32. \] The theorem gives \(\mathbb{E}[B_{\tau+1}-B_\tau\mid\mathcal{F}_\tau]=0\) and \(\mathbb{E}[(B_{\tau+1}-B_\tau)^2\mid\mathcal{F}_\tau]=1\). Therefore \(\mathbb{E}[B_{\tau+1}]=0\) and \(\mathbb{E}[B_{\tau+1}^2]=\tfrac32+1=\tfrac52\); the cross term has expectation zero by the conditional-mean identity. These moments are finite: \(B_\tau\) is square-integrable by the calculation, and the new increment has second moment \(1\). As a check, the conditional second-moment expansion gives \(\mathbb{E}[B_{\tau+1}^2\mid\mathcal{F}_\tau]=B_\tau^2+1\), whose expectation is \(\tfrac32+1=\tfrac52\).
What the Result Does—and Does Not—Say
The result matters whenever a process is observed at a random time and its future behavior is needed. It lets us treat the increments after a bounded stopping time just as we treat Brownian motion after a fixed time, while keeping track of the information that determined the stopping time. It concerns the future increments, not independence of the future levels \(B_{\tau+t}\) from the past.
Common pitfall: substituting a random time into a deterministic-time statement without proof. From the deterministic-time identity \(\mathbb{E}[B_{s+h}-B_s\mid\mathcal{F}_s]=0\), it is not valid simply to replace \(s\) by \(\tau\). The sigma-field \(\mathcal{F}_\tau\) and the random increment both vary with \(\tau\), so the deterministic identity alone does not justify that substitution. The discrete approximation proof supplies the missing argument: condition on each value of \(\tau_n\), use deterministic-time independence there, then pass to the limit using path continuity.
The boundedness assumption ensures that each approximation has only finitely many values, so the discrete sum in the proof is finite. The theorem stated here makes no assertion about unbounded stopping times; extending it requires additional localization or limiting arguments. Also, independence is a statement about the entire shifted process, not merely a zero conditional mean. The finite-dimensional distributions and path continuity are both used to obtain that stronger conclusion.
Check Your Understanding
Use the definition of \(\mathcal{F}_\tau\) and the approximation argument to check each conclusion.
- Why is \(A\cap\{\tau_n=k\delta_n\}\) measurable with respect to \(\mathcal{F}_{k\delta_n}\) whenever \(A\in\mathcal{F}_\tau\)?
- For a fixed \(h>0\), what are the mean and variance of \(B_{\tau+h}-B_\tau\), and what independence statement accompanies them?
- In the sign-based example, verify the stopping-time condition separately for \(t<1\), \(1\le t<2\), and \(t\ge2\).
- Explain why continuity of \(B\) is needed when passing from \(\tau_n\) to \(\tau\).
- Why does independence of \(B_{\tau+t}-B_\tau\) from \(\mathcal{F}_\tau\) not imply independence of \(B_{\tau+t}\) from \(\mathcal{F}_\tau\)?