Restarting Brownian Motion at Hitting Times

Brownian Motion After a First Visit

A first hitting time marks a random moment determined by the path already observed. The strong Markov property says that, once Brownian motion reaches a fixed level, its future increments have the law of a fresh Brownian motion and are independent of the information available at that moment. This lets us split calculations at a level visit: the path before the visit determines when the restart occurs, while the path after it can be analyzed from the level onward.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$, and assume throughout that $B_t$ is a standard Brownian motion with respect to this filtration. In particular, for every $0\le s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. This filtration condition is essential: the marginal law of $B$ alone does not guarantee independence from the information recorded in $(\mathcal{F}_t)$.

Definition: For a level $a>0$, the first hitting time is $\tau_a=\inf\{t\ge0:B_t=a\}$, with $\inf\varnothing=\infty$. It is a stopping time because $\{\tau_a\le t\}=\{\sup_{0\le s\le t}B_s\ge a\}\in\mathcal{F}_t$, using path continuity. The stopping-time sigma-field $\mathcal{F}_{\tau_a}$ consists of events $A\in\mathcal{F}$ such that $A\cap\{\tau_a\le t\}\in\mathcal{F}_t$ for every $t\ge0$.

The earlier material on the strong Markov property gives its bounded-stopping-time form: if $\sigma$ is a bounded stopping time, then $(B_{\sigma+s}-B_\sigma)_{s\ge0}$ is Brownian motion independent of $\mathcal{F}_\sigma$. For the first hitting time $\tau_a$, which need not initially be known to be bounded, we will use this result on $\{\tau_a\le n\}$ and then let $n$ increase. We first establish that $\tau_a$ is finite almost surely.

The First-Hitting-Time Transform

Theorem: For $a>0$ and $\lambda>0$, $$ \mathbb{E}\!\left[e^{-\lambda\tau_a}\mathbf{1}_{\{\tau_a<\infty\}}\right] =e^{-a\sqrt{2\lambda}}. $$ In particular, $\mathbb{P}(\tau_a<\infty)=1$.

Proof. Set $\theta=\sqrt{2\lambda}$ and $M_t=\exp(\theta B_t-\lambda t)$. For $0\le s<t$, the Brownian increment $B_t-B_s$ is independent of $\mathcal{F}_s$ and is $N(0,t-s)$. The normal moment-generating function therefore gives $$ \mathbb{E}[M_t\mid\mathcal{F}_s] =M_s\,\mathbb{E}\!\left[e^{\theta(B_t-B_s)-\lambda(t-s)}\right] =M_s e^{(\theta^2/2-\lambda)(t-s)} =M_s. $$ Also $\mathbb{E}[M_t]=1<\infty$, so $M$ is a true martingale. For each integer $n\ge1$, $\tau_a\wedge n$ is bounded, and optional stopping yields $\mathbb{E}[M_{\tau_a\wedge n}]=1$. By continuity, $B_{\tau_a}=a$ on $\{\tau_a\le n\}$, while $B_n<a$ on $\{\tau_a>n\}$; otherwise the path would have reached $a$ by time $n$. Hence $$ 1=e^{\theta a}\mathbb{E}\!\left[e^{-\lambda\tau_a}\mathbf{1}_{\{\tau_a\le n\}}\right] +\mathbb{E}\!\left[e^{\theta B_n-\lambda n}\mathbf{1}_{\{\tau_a>n\}}\right]. $$ The final term is nonnegative and bounded above by $e^{\theta a-\lambda n}$, which tends to zero. As $n\to\infty$, the first expectation increases to $\mathbb{E}[e^{-\lambda\tau_a}\mathbf{1}_{\{\tau_a<\infty\}}]$. Thus this expectation equals $e^{-\theta a}$. Finally, as $\lambda\downarrow0$, the random variables $e^{-\lambda\tau_a}\mathbf{1}_{\{\tau_a<\infty\}}$ increase to $\mathbf{1}_{\{\tau_a<\infty\}}$. Monotone convergence and $e^{-a\sqrt{2\lambda}}\to1$ show that $\mathbb{P}(\tau_a<\infty)=1$. $\square$

This transform gives useful information without requiring an explicit density for $\tau_a$. In particular, its dependence on the level is exponential in $a$, a feature that makes successive level visits especially simple to analyze.

Restarting at the Hitting Time

Theorem: Let $a>0$, and define $Y_s=B_{\tau_a+s}-B_{\tau_a}$ for $s\ge0$, which is well-defined almost surely because $\tau_a<\infty$ almost surely. Then $Y$ is a standard Brownian motion independent of $\mathcal{F}_{\tau_a}$. Consequently, for every $k\ge1$, $0\le s_1<\cdots<s_k$, every bounded Borel function $g:\mathbb{R}^k\to\mathbb{R}$, and every $A\in\mathcal{F}_{\tau_a}$, $$ \mathbb{E}\!\left[\mathbf{1}_A g(Y_{s_1},\ldots,Y_{s_k})\right] =\mathbb{P}(A)\mathbb{E}\!\left[g(B_{s_1},\ldots,B_{s_k})\right]. $$

Proof. Put $\sigma_n=\tau_a\wedge n$. Fix $A\in\mathcal{F}_{\tau_a}$ and let $A_n=A\cap\{\tau_a\le n\}$. Then $A_n\in\mathcal{F}_{\sigma_n}$: for $t<n$, $A_n\cap\{\sigma_n\le t\}=A\cap\{\tau_a\le t\}\in\mathcal{F}_t$, and for $t\ge n$, $A_n\in\mathcal{F}_n\subseteq\mathcal{F}_t$. The bounded-time strong Markov property at $\sigma_n$ applies. On $A_n$, $\sigma_n=\tau_a$, so it gives $$ \mathbb{E}\!\left[\mathbf{1}_{A_n}g(Y_{s_1},\ldots,Y_{s_k})\right] =\mathbb{P}(A_n)\mathbb{E}\!\left[g(B_{s_1},\ldots,B_{s_k})\right]. $$ Since $\tau_a<\infty$ almost surely, $\mathbf{1}_{A_n}\to\mathbf{1}_A$ almost surely. Boundedness of $g$ allows dominated convergence on the left, and $\mathbb{P}(A_n)\to\mathbb{P}(A)$ on the right. This proves the stated identity. The finite-dimensional distributions of $Y$ are therefore those of Brownian motion and are independent of $\mathcal{F}_{\tau_a}$. Since $Y$ has continuous paths, its path law is determined by its values at nonnegative rational times; the finite-dimensional independence thus gives independence of the whole future path. $\square$

For example, the theorem applies to information such as whether the hit occurred before a fixed time: $\{\tau_a\le T\}\in\mathcal{F}_{\tau_a}$. It does not say that the future path is independent of the hitting time as a numerical random variable—the hitting time is itself part of the past information. It says that the increments after the hit are independent of that information.

Worked Example: An Increment After the Hit

For $a>0$, find the conditional probability that $B_{\tau_a+4}$ is at least $a+2$, and the conditional probability that it lies between $a-2$ and $a+2$.

At the hitting time, continuity gives $B_{\tau_a}=a$. The restarted increment $Y_4=B_{\tau_a+4}-a$ has distribution $N(0,4)$. Therefore $Y_4/2$ has distribution $N(0,1)$. If $\Phi$ denotes the standard normal cumulative distribution function, then $$ \mathbb{P}(B_{\tau_a+4}\ge a+2\mid\mathcal{F}_{\tau_a}) =\mathbb{P}(Y_4\ge2\mid\mathcal{F}_{\tau_a}) =1-\Phi(1). $$ Similarly, $$ \mathbb{P}(a-2\le B_{\tau_a+4}\le a+2\mid\mathcal{F}_{\tau_a}) =\mathbb{P}(-2\le Y_4\le2\mid\mathcal{F}_{\tau_a}) =\Phi(1)-\Phi(-1)=2\Phi(1)-1. $$ These conditional probabilities hold almost surely. As a check, symmetry of the centered normal law gives equal probabilities for $Y_4\ge2$ and $Y_4\le-2$; the two tails and the interval $[-2,2]$ partition the real line, with endpoints having probability zero.

Successive Visits to Higher Levels

Let $0<a<b$, and write $\tau_b=\inf\{t\ge0:B_t=b\}$. Since a continuous path starting at $0$ must pass through $a$ before it reaches $b$, define the post-$\tau_a$ waiting time $$ \rho_c=\inf\{s\ge0:Y_s=c\},\qquad c=b-a. $$ The pathwise identity $\tau_b=\tau_a+\rho_c$ holds almost surely. The random variable $\rho_c$ is a function of the restarted path $Y$, so it is independent of $\mathcal{F}_{\tau_a}$. The first-hitting-time transform gives $\mathbb{E}[e^{-\lambda\tau_a}]=e^{-a\sqrt{2\lambda}}$ and $\mathbb{E}[e^{-\lambda\rho_c}]=e^{-c\sqrt{2\lambda}}$. Thus $$ \mathbb{E}[e^{-\lambda\tau_b}] =\mathbb{E}[e^{-\lambda\tau_a}e^{-\lambda\rho_c}] =e^{-a\sqrt{2\lambda}}e^{-(b-a)\sqrt{2\lambda}} =e^{-b\sqrt{2\lambda}}. $$ The argument also explains the restart calculation: the first segment reaches $a$, and an independent Brownian segment, started from zero, then travels the remaining distance $b-a$.

Worked Example: Laplace Transform of a Two-Level Restart

Starting from $0$, Brownian motion first visits level $1$, then level $4$. Compute the Laplace transform at $\lambda=\tfrac12$ of each waiting time and of the total time.

The transform for first reaching a level $d>0$ is $e^{-d\sqrt{2\lambda}}$. Here $\sqrt{2\lambda}=\sqrt{2\cdot\tfrac12}=1$. The wait to reach level $1$ therefore has transform $e^{-1\cdot1}=e^{-1}$. After that visit, the remaining distance is $4-1=3$, so the independent second waiting time has transform $e^{-3\cdot1}=e^{-3}$. Multiplying the transforms gives $e^{-1}e^{-3}=e^{-4}$ for the total time. This agrees with a direct calculation for first reaching level $4$: $e^{-4\cdot1}=e^{-4}$.

Check: the two distances add to $1+3=4$, so their exponential transform factors multiply to the transform for distance $4$. The computation uses $\lambda=\tfrac12>0$, as required for the transform formula.

Worked Example: A Future Event Given an Early Hit

Let $T>0$ and $a>0$. Given that $\tau_a\le T$, find the probability that $B_{\tau_a+1}>a$.

The conditioning event has positive probability: $\{B_T\ge a\}\subseteq\{\tau_a\le T\}$, and $\mathbb{P}(B_T\ge a)=1-\Phi(a/\sqrt{T})>0$. Since $\{\tau_a\le T\}\in\mathcal{F}_{\tau_a}$, the restart theorem gives $$ \mathbb{P}(B_{\tau_a+1}>a\mid\tau_a\le T) =\mathbb{P}(Y_1>0\mid\tau_a\le T) =\mathbb{P}(B_1>0) =\tfrac12. $$ More explicitly, the joint probability is $$ \mathbb{P}(\tau_a\le T,\ B_{\tau_a+1}>a) =\mathbb{P}(\tau_a\le T)\,\mathbb{P}(B_1>0) =\tfrac12\mathbb{P}(\tau_a\le T). $$ Dividing by the positive probability $\mathbb{P}(\tau_a\le T)$ gives the conditional probability. The check is that $Y_1\sim N(0,1)$, whose centered distribution assigns probability $1/2$ to the positive half-line.

Common Pitfall: The Filtration Must Not Reveal the Future

A Brownian path can have standard Brownian distributions and still fail to be Brownian motion with respect to a chosen filtration. For instance, if $\mathcal{F}_t=\sigma(B_s:s\ge0)$ for every $t$, then $B_1$ is already known at time $0$. Although $B$ has its usual marginal law, $B_1$ is not independent of $\mathcal{F}_0$. In that filtration, the strong Markov conclusion at the deterministic stopping time $0$ would be false.

Another error is to replace the random future increment by an increment over a deterministic interval, such as $B_{t+1}-B_t$, without accounting for the random time $t=\tau_a$. The strong Markov property is precisely what justifies the law at a random restart time; deterministic-time independent increments alone do not establish that conclusion.

Quantity	What the restart result says	What it does not say
$B_{\tau_a}$	It equals $a$ almost surely by path continuity.	It is not a random new starting level.
$B_{\tau_a+s}-B_{\tau_a}$	As a process in $s$, it is Brownian and independent of $\mathcal{F}_{\tau_a}$.	It is not independent of the full path when that path includes the future.
$\tau_b-\tau_a$, for $b>a$	It has the hitting-time law for distance $b-a$, independent of the information at $\tau_a$.	It is not the original waiting time to reach $b$.

Check Your Understanding

For each question, use the Brownian-with-respect-to-the-filtration assumption and state any positivity condition needed for conditioning.

For $a>0$, what is the conditional distribution of $B_{\tau_a+3}-B_{\tau_a}$ given $\mathcal{F}_{\tau_a}$?
For $0<a<b$, explain why $\tau_b-\tau_a$ is the first time the restarted process reaches $b-a$.
Using the hitting-time transform, calculate $\mathbb{E}[e^{-2\tau_a}]$.
Why does the bounded-stopping-time strong Markov property need an additional argument before it can be applied directly to $\tau_a$?
For $T>0$ and $a>0$, explain why $\mathbb{P}(\tau_a\le T)>0$, so conditioning on $\{\tau_a\le T\}$ is meaningful.

Quantity	What the restart result says	What it does not say
\(B_{\tau_a}\)	It equals \(a\) almost surely by path continuity.	It is not a random new starting level.
\(B_{\tau_a+s}-B_{\tau_a}\)	As a process in \(s\), it is Brownian and independent of \(\mathcal{F}_{\tau_a}\).	It is not independent of the full path when that path includes the future.
\(\tau_b-\tau_a\), for \(b>a\)	It has the hitting-time law for distance \(b-a\), independent of the information at \(\tau_a\).	It is not the original waiting time to reach \(b\).