The Reflection Principle for Brownian Paths

Reflecting a Path at Its First Visit

Brownian motion can cross a level and later finish on either side of it. The reflection principle relates paths that have crossed a positive level and finish below it to paths that finish above a reflected endpoint. The key operation is to leave the path unchanged until its first visit to the level, then reverse the signs of all subsequent increments. The result is another Brownian path with the same law.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, adapted to $(\mathcal{F}_t)$. Fix $T>0$ and a level $a>0$. Define the first hitting time $\tau_a=\inf\{t\ge0:B_t=a\}$, with $\inf\varnothing=\infty$. Continuity of Brownian paths implies that $\tau_a$ is a stopping time and, on $\{\tau_a\le T\}$, $B_{\tau_a}=a$.

Definition: On the path space over $[0,T]$, define the reflection map $R_a$ by $(R_aB)_t=B_t$ for $t\le\tau_a$, and $(R_aB)_t=2a-B_t$ for $t>\tau_a$, when $\tau_a\le T$. If $\tau_a>T$, set $(R_aB)_t=B_t$ throughout $[0,T]$. Thus the path is reflected about the horizontal level $a$ after its first visit to $a$.

The map does not change the path up to the first visit. After that time, an increment $B_t-B_{\tau_a}$ becomes its negative. Reflection therefore preserves continuity and fixes the first hitting time. Applying it twice restores every path, so $R_a$ is an involution. The probabilistic fact that makes this operation useful is that Brownian increments after a stopping time are independent Brownian motion.

Why the Reflected Path Has the Same Law

Proposition: For $a>0$ and $T>0$, the reflected path $R_aB$ has the same law on $C([0,T])$ as $B$. Here $C([0,T])$ is the space of continuous real-valued paths with its Borel sigma-field.

Proof. Set $S=\tau_a\wedge T$, a bounded stopping time. The strong Markov property at bounded stopping times says that the process $U_s=B_{S+s}-B_S$, $s\ge0$, is a standard Brownian motion independent of $\mathcal{F}_S$. We use this standard result from the earlier material on restarting Brownian motion at hitting times. The event $H=\{\tau_a\le T\}$ is in $\mathcal{F}_S$: for $t<T$, $H\cap\{S\le t\}=\{\tau_a\le t\}\in\mathcal{F}_t$, and at $t=T$ the required intersection is in $\mathcal{F}_T$. On $H$, $S=\tau_a$ and $B_S=a$, while the remaining time $T-S$ is known given $\mathcal{F}_S$. Conditional on $\mathcal{F}_S$, reflecting the post-$S$ path replaces $U_s$ by $-U_s$ over this remaining time. Brownian sign symmetry gives $U$ and $-U$ the same path law; independence means this replacement does not change the conditional law given the path up to $S$. On $H^c$, the map is the identity. Consequently, for every bounded Borel function $F:C([0,T])\to\mathbb{R}$, $\mathbb{E}[F(R_aB)]=\mathbb{E}[F(B)]$. This equality of expectations establishes equality of path laws. $\square$

The argument uses both the strong Markov property and symmetry. Symmetry at a fixed deterministic time alone would not justify reflecting at a random hitting time: the independence of the restarted Brownian path is what allows the reflection to be made after the path-dependent time $\tau_a$.

Endpoint Identity

Theorem (Reflection Principle): Let $a>0$, $T>0$, and let $A$ be a Borel subset of $(-\infty,a)$. Write $2a-A=\{2a-y:y\in A\}$. Then $$ \mathbb{P}(\tau_a\le T,\ B_T\in A)=\mathbb{P}(B_T\in 2a-A). $$ In particular, for $x<a$, $$ \mathbb{P}(\tau_a\le T,\ B_T\le x)=\mathbb{P}(B_T\ge 2a-x). $$

Proof. On $\{\tau_a\le T,\ B_T\in A\}$, the reflected path has terminal value $2a-B_T\in 2a-A$, and $2a-A\subset(a,\infty)$. Conversely, any continuous path with terminal value in $2a-A$ must have visited $a$ by time $T$, because it starts at $B_0=0<a$ and ends above $a$. Reflecting it gives a path whose terminal value is in $A$. Since $R_a$ preserves path law and is an involution, the probabilities of these two events are equal. Taking $A=(-\infty,x]$, where $x<a$, gives $2a-A=[2a-x,\infty)$, proving the endpoint identity. $\square$

The strict inequality $x<a$ ensures that the endpoint set being reflected lies entirely below the level. Brownian motion has a Gaussian endpoint distribution, so single endpoint values have probability zero; this also lets us interchange a strict or non-strict inequality at a single endpoint when needed. The theorem is an identity for paths that hit $a$ and have a specified endpoint, not a claim that every path ending above $a$ has a reflected counterpart that stays below $a$ throughout.

Worked Example: Hit a Level and Finish Below Zero

For $a=1$, $T=1$, and $x=0$, find $\mathbb{P}(\tau_1\le1,\ B_1\le0)$.

Here $x=0<1=a$, so the endpoint identity applies. The reflected endpoint threshold is $2a-x=2\cdot1-0=2$. Since $B_1$ has the $N(0,1)$ distribution, if $Z$ is standard normal and $\Phi(z)=\mathbb{P}(Z\le z)$,

\mathbb{P}(\tau_1\le1,\ B_1\le0) =\mathbb{P}(B_1\ge2) =1-\Phi(2).

Check: the reflected endpoint set is $[2,\infty)$, which lies above the level $1$, as it must: continuity forces every path ending there to have crossed level $1$. Also $1-\Phi(2)<\tfrac12$, consistent with this being a subset of the event $B_1\le0$, which has probability $\tfrac12$.

Worked Example: Finish in a Specified Interval

For $a=2$ and $T=4$, find the probability that Brownian motion hits level $2$ by time $4$ and ends in $[0,1]$.

The interval $[0,1]$ lies below $a=2$, so reflect each endpoint $y$ to $2a-y=4-y$. The reflected interval is $[3,4]$. Since $B_4\sim N(0,4)$, standardizing by its standard deviation $2$ gives

\begin{aligned} \mathbb{P}(\tau_2\le4,\ 0\le B_4\le1) &=\mathbb{P}(3\le B_4\le4)\\ &=\mathbb{P}\left(\frac32\le\frac{B_4}{2}\le2\right)\\ &=\Phi(2)-\Phi\left(\frac32\right). \end{aligned}

Check: the reflected interval is computed endpoint by endpoint: $4-1=3$ and $4-0=4$. Its standardized endpoints are $3/2$ and $4/2=2$, so the stated normal probability uses the correct variance and interval.

A Crossing Probability as a Consequence

The endpoint identity also yields the probability of hitting $a$ by time $T$, without requiring a separate analysis of the path maximum. A path with $B_T\ge a$ must have hit $a$, by continuity. The remaining paths that hit $a$ end below $a$, and reflection pairs those paths with paths ending above $a$.

Corollary: For $a>0$ and $T>0$, $$ \mathbb{P}(\tau_a\le T)=2\mathbb{P}(B_T\ge a) =2\left[1-\Phi\left(\frac{a}{\sqrt{T}}\right)\right]. $$

Proof. Split the hitting event into the disjoint events $\{\tau_a\le T,\ B_T\ge a\}$ and $\{\tau_a\le T,\ B_T<a\}$. The first event is exactly $\{B_T\ge a\}$, since $B_T\ge a$ implies a visit to $a$, and a path that has hit $a$ may also end at or above it. The endpoint identity applied to $(-\infty,a)$ gives $\mathbb{P}(\tau_a\le T,\ B_T<a)=\mathbb{P}(B_T>a)$. Because $B_T$ has a continuous normal distribution, $\mathbb{P}(B_T=a)=0$, so $\mathbb{P}(B_T>a)=\mathbb{P}(B_T\ge a)$. Adding the two disjoint probabilities gives $2\mathbb{P}(B_T\ge a)$. Finally, $B_T/\sqrt{T}$ is standard normal, giving the displayed expression. $\square$

Worked Example: Probability of Hitting Level One

Find the probability that Brownian motion hits level $1$ by time $1$.

Substituting $a=1$ and $T=1$ into the corollary gives

\mathbb{P}(\tau_1\le1)=2\mathbb{P}(B_1\ge1)=2[1-\Phi(1)].

Check: the endpoint event $\{B_1\ge1\}$ is contained in $\{\tau_1\le1\}$, and the corollary says the full hitting probability is twice its probability. Since $\Phi(1)>\tfrac12$, the result is positive and less than $1$, as required.

Common Pitfall: Reflect the Endpoint Set in the Correct Direction

A frequent error is to keep the endpoint inequality pointing in the same direction after reflection. For example, with $a=1$, $T=1$, and $x=0$, an incorrect calculation might write $\mathbb{P}(\tau_1\le1,\ B_1\le0)=\mathbb{P}(B_1\le2)=\Phi(2)$. This is wrong: reflecting an endpoint $y\le0$ sends it to $2-y\ge2$, not to an endpoint below $2$. The correct calculation is $\mathbb{P}(B_1\ge2)=1-\Phi(2)$. In particular, the incorrect answer $\Phi(2)>\tfrac12$ is impossible, since the event on the left is contained in $\{B_1\le0\}$, whose probability is $\tfrac12$.

The reflection principle is useful because it converts a path event into an endpoint event with a known Gaussian law. Its hypotheses matter: the path starts below the positive level, paths are continuous, and the post-hitting increments have Brownian law independent of the past. The argument here concerns a fixed positive level and the endpoint at a fixed time; it does not require a stochastic integral or an Ito formula.

Check Your Understanding

Use the reflection identity, keeping track of which endpoint set is reflected.

For $a>0$, $T>0$, and $x<a$, what endpoint set is the reflection of $(-\infty,x]$?
Express $\mathbb{P}(\tau_3\le4,\ B_4\le1)$ using $\Phi$.
For $a=2$, $T=1$, and $x=0$, find $\mathbb{P}(\tau_2\le1,\ B_1\le0)$.
Why does the event $\{B_T\ge a\}$ imply $\{\tau_a\le T\}$ when $a>0$?
Which property of Brownian motion justifies reflecting all increments after the random time $\tau_a$?