Brownian Motion After a Stopping Time
At a fixed time, Brownian motion has increments independent of the information available up to that time. A stopping time is different: its value is random, and the decision that it has occurred uses the path observed so far. The strong Markov property says that Brownian motion still restarts at such a time. Precisely, the path of increments after the stopping time has the usual Brownian law and is independent of everything known at the stopping time.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with a standard Brownian motion \((B_t)_{t\ge0}\) taking values in \(S=\mathbb{R}\). Let \((\mathcal{F}_t)_{t\ge0}\) be its completed, right-continuous natural filtration. A random time \(T:\Omega\to[0,\infty]\) is a stopping time for this filtration if \(\{T\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). Its stopped sigma-field is \(\mathcal{F}_T=\{A\in\mathcal{F}:A\cap\{T\le t\}\in\mathcal{F}_t\text{ for every }t\ge0\}\). It represents the information available when \(T\) occurs. The condition \(\mathbb{P}(T<\infty)=1\) ensures the post-\(T\) process is defined almost surely.
The path space \(C([0,\infty))\) here is equipped with the Borel sigma-field for uniform convergence on compact time intervals. The theorem concerns the increment path starting at zero, not the absolute path after \(T\). The latter starts at the random location \(B_T\). Thus, conditionally on \(\mathcal{F}_T\), the future path is distributed as \(B_T\) plus a fresh Brownian path.
Consequences for Future Positions
Proof. Apply the theorem to the bounded Borel function of the path that evaluates it at \(s_1,\ldots,s_m\), or to any bounded Borel function of the successive differences of those evaluations. Under the resulting conditional law, the evaluated path has the same finite-dimensional distribution as standard Brownian motion. Its successive increments are independent, and each increment over an interval of length \(s_r-s_{r-1}\) has the normal distribution with mean \(0\) and variance \(s_r-s_{r-1}\). This conditional distribution does not depend on \(\mathcal{F}_T\), which gives the stated result. \(\square\)
Proof. By the theorem, \(W_t=B_{T+t}-B_T\) has distribution \(\mathcal{N}(0,t)\) and is independent of \(\mathcal{F}_T\). Also, \(B_T\) is \(\mathcal{F}_T\)-measurable. Therefore \(B_{T+t}=B_T+W_t\). For a simple function of \(B_T\) and \(W_t\), independence gives the displayed conditional integral directly; bounded measurable functions follow by bounded approximation. The integral is the conditional expectation because it averages over the law of \(W_t\) while holding the known value \(B_T\) fixed. \(\square\)
Worked Example: The Path After First Hitting 1
Let \(T=\inf\{t\ge0:B_t=1\}\) for Brownian motion started at \(B_0=0\). This is a stopping time, and the reflection principle gives \(\mathbb{P}(T<\infty)=1\). Since \(B_T=1\), the strong Markov property says that \((B_{T+s}-1)_{s\ge0}\) is a standard Brownian motion independent of \(\mathcal{F}_T\).
At one unit of time after \(T\), the increment \(B_{T+1}-B_T\), conditionally on \(\mathcal{F}_T\), has distribution \(\mathcal{N}(0,1)\). Consequently, \(B_{T+1}\) has conditional distribution \(\mathcal{N}(1,1)\). Its conditional mean is \(1\), its conditional variance is \(1\), and \(\mathbb{P}(B_{T+1}>1\mid\mathcal{F}_T)=\mathbb{P}(\mathcal{N}(0,1)>0)=\tfrac12\).
For two successive post-\(T\) increments, the corollary gives \(B_{T+1}-B_T\) and \(B_{T+2}-B_{T+1}\) as independent \(\mathcal{N}(0,1)\) variables, conditionally on \(\mathcal{F}_T\). Hence the conditional probability that both are positive is \(\tfrac12\cdot\tfrac12=\tfrac14\). These calculations concern the future increments; they do not say that \(B_{T+1}\) is independent of the past, since \(B_{T+1}=1+(B_{T+1}-B_T)\).
Check: The conditional mean and variance of \(B_{T+1}\) from the position formula are \(B_T=1\) and \(t=1\), respectively. Subtracting \(B_T\) gives a centered unit-variance increment, as required.
Worked Example: A Stopping Time Chosen at Time 1
Define \(T=1\) when \(B_1\ge0\), and \(T=2\) when \(B_1<0\). This is a stopping time: for \(t<1\), \(\{T\le t\}\) 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 \(\Omega\). It is finite, so the strong Markov property applies even though the time depends on the path.
The post-\(T\) increment path is a standard Brownian motion independent of \(\mathcal{F}_T\). In particular, \(B_{T+1}-B_T\), conditionally on \(\mathcal{F}_T\), has mean \(0\) and variance \(1\). The conditional position is \(B_{T+1}\sim\mathcal{N}(B_T,1)\), in the sense of a conditional distribution, so \(\mathbb{P}(B_{T+1}>B_T\mid\mathcal{F}_T)=\tfrac12\).
The random center \(B_T\) matters. On \(\{B_1\ge0\}\), it is \(B_1\); on \(\{B_1<0\}\), it is \(B_2\). Thus the theorem does not assert that \(B_{T+1}\) itself is a centered normal variable independent of the information at \(T\). It asserts that subtracting the known current position \(B_T\) leaves a fresh Brownian increment.
Check: The conditional mean of \(B_{T+1}-B_T\) is \(0\), and its conditional variance is \(1\). Adding the \(\mathcal{F}_T\)-measurable quantity \(B_T\) gives conditional mean \(B_T\) and leaves the variance equal to \(1\).
Why the Stopping-Time Condition Matters
The requirement that \(T\) be a stopping time cannot be replaced by the weaker condition that \(T\) is merely a finite random time. A stopping time does not use information from the future to decide whether it has occurred. If the time is selected using a future value, the selection can bias the increment that is later examined.
Worked Example: A Future-Dependent Time Is Not a Stopping Time
Set \(U=1\) if \(B_2>0\), and \(U=2\) if \(B_2\le0\). This is finite, but it is not a stopping time: \(\{U\le1\}=\{B_2>0\}\notin\mathcal{F}_1\). To see the effect of this dependence, let \(X=B_2-B_1\) and \(Y=B_2\). On \(\{U=1\}=\{Y>0\}\), the increment \(B_{U+1}-B_U\) equals \(X\).
The pair \((X,Y)\) is jointly normal, with \(\mathbb{E}X=\mathbb{E}Y=0\), \(\operatorname{Var}(Y)=2\), and \(\operatorname{Cov}(X,Y)=\operatorname{Cov}(B_2-B_1,B_2)=2-1=1\). The conditional mean formula for jointly normal variables gives \(\mathbb{E}[X\mid Y]=\tfrac12Y\). Since \(Y\sim\mathcal{N}(0,2)\), \(\mathbb{E}[Y\mid Y>0]=2/\sqrt{\pi}\), so \(\mathbb{E}[B_{U+1}-B_U\mid U=1]=\mathbb{E}[X\mid Y>0]=1/\sqrt{\pi}\ne0\).
For the intermediate value, the positive half-line has probability \(\tfrac12\), and direct integration of the \(\mathcal{N}(0,2)\) density gives \(\mathbb{E}[Y\mathbf{1}_{\{Y>0\}}]=\int_0^\infty y(2\pi\cdot2)^{-1/2}e^{-y^2/4}\,dy=1/\sqrt{\pi}\). Dividing by \(\mathbb{P}(Y>0)=\tfrac12\) yields \(2/\sqrt{\pi}\); multiplying by \(\tfrac12\) yields \(1/\sqrt{\pi}\). The nonzero conditional mean exhibits the selection bias. It does not contradict the strong Markov property, because \(U\) is not a stopping time.
Common pitfall. Applying the strong Markov property at any random time, rather than specifically at an almost surely finite stopping time, can produce an incorrect conditional law. A second error is to claim that the absolute future path is independent of \(\mathcal{F}_T\). The independent object is \((B_{T+s}-B_T)_{s\ge0}\); the absolute path includes \(B_T\), which is already known at time \(T\).
Check Your Understanding
For each question, distinguish the post-stopping increment path from the absolute Brownian path.
- State the conditional law of \(B_{T+3}-B_T\) when \(T\) is an almost surely finite stopping time.
- If \(B_T=2\), what are the conditional mean and variance of \(B_{T+4}\)?
- Why is \(\{U\le1\}=\{B_2>0\}\) enough to show that the time \(U\) in the last example is not a stopping time?
- For \(0<s<t\), identify the conditional distributions of \(B_{T+s}-B_T\) and \(B_{T+t}-B_{T+s}\), and state whether they are conditionally independent.
- Explain why the strong Markov property does not imply that \(B_{T+1}\) is independent of \(\mathcal{F}_T\).