Restarting Brownian Paths at First-Passage Times

Restarting at a First-Passage Time

A Brownian path remembers its past, but at a first-passage time its future has a particularly simple description: measured from the level just reached, the path starts afresh as Brownian motion, independently of everything observed up to that time. This lets us break a passage to a distant threshold into a passage to a nearer level and a new passage from there. The same idea describes what happens when the path returns to a level after reaching a different one.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ carrying a standard Brownian motion $(B_t)_{t\ge0}$, with state space $\mathbb{R}$, time index $t\ge0$, and $B_0=0$. Use its usual augmented natural filtration $(\mathcal{F}_t)_{t\ge0}$. For $c>0$, define the first-passage time $T_c=\inf\{t\ge0:B_t=c\}$, with $\inf\varnothing=\infty$. This is a stopping time: by continuity, $\{T_c\le t\}=\{\sup_{0\le s\le t}B_s\ge c\}$, an event determined by the path up to time $t$. The reflection principle gives, for $t>0$,

\mathbb{P}(T_c\le t)=2\left(1-\Phi\left(\frac{c}{\sqrt{t}}\right)\right),

where $\Phi$ is the standard normal distribution function. In particular, as $t\to\infty$, the right side tends to $1$, so $T_c<\infty$ almost surely. We use this first-passage formula, which follows from the reflection principle, as a standard result.

Theorem (Restart at a first-passage time): For $a>0$, define $Y_s=B_{T_a+s}-B_{T_a}$, $s\ge0$. Then $(Y_s)_{s\ge0}$ is a standard Brownian motion independent of $\mathcal{F}_{T_a}$, and $B_{T_a}=a$ almost surely. Consequently, for $b>a$, $$ T_b=T_a+\inf\{s\ge0:Y_s=b-a\}\quad\text{almost surely}. $$

Proof. The strong Markov property for Brownian motion states that if $\tau$ is an almost surely finite stopping time, then $(B_{\tau+s}-B_\tau)_{s\ge0}$ is a standard Brownian motion independent of $\mathcal{F}_\tau$. This is the standard strong Markov theorem established in the earlier material on Brownian motion; here it is applied to the stopping time $T_a$, which is finite almost surely. Continuity of $B$ implies $B_{T_a}=a$: the path first reaches $a$ at $T_a$, and cannot jump across $a$. For the decomposition, the path cannot reach $b>a$ before it reaches $a$, again by continuity. After time $T_a$, it is at $a$, and it reaches $b$ exactly when $Y$ first reaches $b-a$. Thus the first time the original path reaches $b$ is the sum of $T_a$ and that first-passage time of $Y$. $\square$

The important independence is from the entire past $\mathcal{F}_{T_a}$, not merely from the value $B_{T_a}$. In particular, any event determined by the path up to $T_a$ is independent of the restarted path. The displayed decomposition also says that the extra time $T_b-T_a$ has the same distribution as $T_{b-a}$, and is independent of $\mathcal{F}_{T_a}$.

Worked Example: Reaching a Higher Threshold Within Four Time Units

After Brownian motion first reaches level $1$, find the probability that it reaches level $3$ within the next four time units. Also find the probability that it first reaches level $1$ by time $1$ and then reaches level $3$ within four further time units.

Here $a=1$, $b=3$, so the distance still to travel after the first hit is $b-a=3-1=2$. By the restart theorem, the extra time $T_3-T_1$ has the same distribution as $T_2$. Applying the first-passage formula at horizon $4$,

\mathbb{P}(T_3-T_1\le4) =2\left(1-\Phi\left(\frac{2}{\sqrt{4}}\right)\right) =2(1-\Phi(1)).

For the joint event, $\{T_1\le1\}$ is measurable with respect to $\mathcal{F}_{T_1}$, while the extra passage time $T_3-T_1$ is independent of that sigma-field. The first-passage formula also gives

\mathbb{P}(T_1\le1)=2\left(1-\Phi\left(\frac{1}{\sqrt{1}}\right)\right)=2(1-\Phi(1)).

Therefore the requested joint probability is the product

\mathbb{P}(T_1\le1,\ T_3-T_1\le4) =2(1-\Phi(1))\cdot2(1-\Phi(1)) =4(1-\Phi(1))^2.

Check: The factors are respectively the probability of reaching $1$ by time $1$ and the probability of covering the remaining distance $3-1=2$ within time $4$. Their product is justified by independence from $\mathcal{F}_{T_1}$, not by an assumption that passage times at different levels are independent.

Passage-Time Decomposition

The same argument works for any two positive levels $a<b$. It gives both an identity of random variables and an independence statement. The identity is useful because the time to reach $b$ can be studied in two stages; the independence statement allows probabilities involving the first stage and the additional time to factor.

Proposition: If $0<a<b$, then $T_b-T_a$ is independent of $\mathcal{F}_{T_a}$, has the same distribution as $T_{b-a}$, and satisfies $$ T_b=T_a+(T_b-T_a)\quad\text{almost surely}. $$ For $h>0$, its distribution is $$ \mathbb{P}(T_b-T_a\le h) =2\left(1-\Phi\left(\frac{b-a}{\sqrt{h}}\right)\right). $$

Proof. By the restart theorem, $T_b-T_a=\inf\{s\ge0:Y_s=b-a\}$, where $Y$ is a standard Brownian motion independent of $\mathcal{F}_{T_a}$. This proves independence from $\mathcal{F}_{T_a}$. Since $Y$ has the same law as $B$, the displayed infimum has the same distribution as $T_{b-a}$. The first-passage formula, applied with positive level $b-a$ and time $h>0$, gives the stated probability. Finally, the pathwise decomposition in the restart theorem gives the identity. $\square$

Worked Example: A Conditional Threshold Probability

Suppose Brownian motion has just first reached level $2$. Find the conditional probability, given the information up to that time, that it reaches level $5$ within the next $9$ time units.

The remaining distance is $5-2=3$. The restarted path is a standard Brownian motion independent of $\mathcal{F}_{T_2}$, so the conditional probability is the same constant for almost every past history:

\mathbb{P}(T_5-T_2\le9\mid\mathcal{F}_{T_2}) =2\left(1-\Phi\left(\frac{5-2}{\sqrt{9}}\right)\right) =2(1-\Phi(1)).

This conditional statement is meaningful because $T_2$ is finite almost surely. It does not say that $T_5$ itself is independent of the past; rather, the additional time after $T_2$ is independent of the past. Check: substituting the remaining distance $3$ and horizon $9$ gives $3/\sqrt{9}=1$, as used in the final expression.

Returning to the Origin

After a path first reaches a positive level $a$, it may later return to $0$. Define the return duration after the hit by $$ R_a=\inf\{s\ge0:B_{T_a+s}=0\}. $$ This is not the original first-passage time $T_0$, which is zero because $B_0=0$. At the restart time the path is at $a>0$, so $R_a>0$ almost surely. The restarted path shows that this duration has the law of the time for a fresh Brownian motion to reach $-a$.

For standard Brownian motion, symmetry gives $\mathbb{P}(T_{-a}\le t)=\mathbb{P}(T_a\le t)$ for $a>0$: the process $(-B_t)_{t\ge0}$ is also standard Brownian motion. Hence for $t>0$,

\mathbb{P}(R_a\le t) =2\left(1-\Phi\left(\frac{a}{\sqrt{t}}\right)\right), \qquad \mathbb{P}(R_a>t) =2\Phi\left(\frac{a}{\sqrt{t}}\right)-1.

Worked Example: Returning to Zero After Reaching Level Two

After Brownian motion first reaches level $2$, calculate the probability that it returns to zero within $9$ time units. Determine whether the mean duration of this return is finite.

At time $T_2$, the path is at $2$, so returning to zero requires the restarted Brownian path to reach $-2$. The distance to that level is $2$, and the restart formula yields

\mathbb{P}(R_2\le9) =2\left(1-\Phi\left(\frac{2}{\sqrt{9}}\right)\right) =2\left(1-\Phi\left(\frac23\right)\right).

As $t\to\infty$, $\mathbb{P}(R_2>t)=2\Phi(2/\sqrt{t})-1$ tends to zero. Therefore $R_2<\infty$ almost surely. However, its expectation is infinite. To verify this, for $t\ge4$ we have $0\le2/\sqrt{t}\le1$. Write $\varphi(u)=(2\pi)^{-1/2}e^{-u^2/2}$ for the standard normal density. Since $\varphi$ decreases on $[0,1]$,

\mathbb{P}(R_2>t) =2\int_0^{2/\sqrt{t}}\varphi(u)\,du \ge 2\frac{2}{\sqrt{t}}\varphi(1) =\frac{4\varphi(1)}{\sqrt{t}},\qquad t\ge4.

For a nonnegative random variable, its expectation is the integral of its tail probabilities. The lower bound has divergent integral, since $\int_4^\infty t^{-1/2}\,dt=\infty$. Thus $\mathbb{E}[R_2]=\infty$, despite almost-sure finiteness. Check: the bound used the actual level $a=2$, so its endpoint condition is $t\ge a^2=4$, and the coefficient is $2a\varphi(1)=4\varphi(1)>0$.

Why the Restart Matters—and a Common Error

Restarting is useful whenever a question concerns the path after a random first hit. It replaces a conditional future-path calculation with an ordinary Brownian first-passage calculation. The new path is independent of the past, but the original hitting times are not interchangeable with independent copies: the correct decomposition identifies precisely which additional duration has the fresh Brownian law.

Common mistake. For levels $1$ and $3$, the probability of reaching $3$ within four time units after first reaching $1$ is sometimes incorrectly written as $\mathbb{P}(T_3\le4)=2(1-\Phi(3/2))$. That formula uses four time units starting from $0$, so it does not answer the question asked. The elapsed time after reaching $1$ is $T_3-T_1$; the remaining distance is $3-1=2$. The correct probability is therefore $$ \mathbb{P}(T_3-T_1\le4) =2\left(1-\Phi\left(\frac{2}{\sqrt4}\right)\right) =2(1-\Phi(1)). $$ The distinction is between a passage from the original starting point and the extra passage after a stopping time.

The restart statement also has limits that matter. It applies at a stopping time such as $T_a$, not at an arbitrary random time selected using future information. For instance, a time chosen by looking ahead to where the path will be later need not have a future independent of the observed past. Here the stopping-time property and almost-sure finiteness of $T_a$ are both verified, so the strong Markov property applies.

Check Your Understanding

Use the restart theorem and the first-passage formula to answer the questions. In each probability, distinguish the time measured from the original start from the time measured after a hit.

For $0<a<b$, what is the distribution of $T_b-T_a$, and what information is it independent of?
After the first hit of level $3$, find the probability of reaching level $7$ within $16$ time units.
Explain why $\mathbb{P}(T_4-T_1\le h)$ is not calculated by substituting $4$ as the passage distance in the first-passage formula.
For $a>0$, express $\mathbb{P}(R_a>t)$ using $\Phi$, and explain why $R_a$ is finite almost surely.
Why does the fact that $R_a<\infty$ almost surely not imply that $\mathbb{E}[R_a]$ is finite?

Restarting Brownian Paths at First-Passage Times

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