The Reflection Principle for Brownian Motion

Why Reflection Counts Paths That End Below a Barrier

At a fixed time, Brownian motion can finish below a positive level even if its path has already crossed that level. The reflection principle accounts for precisely these paths: after the first hit of the barrier, reflect the remainder of the path across the barrier. A path that hits the barrier and finishes below it is paired with a path that finishes above it. The pairing preserves Brownian probabilities, so a path-crossing probability can be computed from a Gaussian tail.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ carrying a standard Brownian motion $(B_s)_{s\ge0}$, with $B_0=0$. Its paths are continuous, and its increments over disjoint time intervals are independent; an increment over an interval of length $r$ has distribution $N(0,r)$. Use the natural filtration $\mathcal{F}_s=\sigma(B_u:0\le u\le s)$, completed if desired. For a fixed barrier $a>0$, define the first hitting time $T_a=\inf\{s\ge0:B_s=a\}$, with $\inf\varnothing=\infty$, and the running maximum $M_t=\sup_{0\le s\le t}B_s$.

Definition (Reflection at the first hit): For a continuous path that hits $a$, leave the path unchanged up to and including $T_a$, and replace its value after $T_a$ by its reflection across $a$. Thus the reflected path is $B_s$ for $s\le T_a$ and $2a-B_s$ for $s>T_a$. If the path does not hit $a$, leave it unchanged. Since $B_{T_a}=a$ when $T_a<\infty$, the reflected path is continuous.

The key issue is why this path operation preserves probability. Reflection of increments after a fixed deterministic time preserves Brownian law by symmetry of Gaussian increments. The first hitting time is random, however, so we establish the needed assertion by approximating it from above with grid-valued times. This argument uses only independent increments at deterministic times and continuity of Brownian paths.

Proof That Reflection Preserves Brownian Path Law

Lemma (Reflection principle, path form): Fix $t>0$ and $a>0$. Reflect the Brownian path after its first hit of $a$, leaving it unchanged if there is no hit by time $t$. The resulting path on $[0,t]$ has the same law as $(B_s)_{0\le s\le t}$.

Proof. For an integer $m\ge1$, let $h_m=2^{-m}$ and let $T_a^{(m)}=h_m\lceil T_a/h_m\rceil$ when $T_a<\infty$. This is a grid-valued stopping time: for any grid point $kh_m$, the event $\{T_a^{(m)}\le kh_m\}$ is $\{T_a\le kh_m\}$, which is determined by the path up to $kh_m$. Define a grid reflection by leaving the path unchanged if $T_a^{(m)}>t$; otherwise, leave it unchanged through $T_a^{(m)}$ and reflect every later value across $B_{T_a^{(m)}}$.

For each possible value $kh_m\le t$, the event $\{T_a^{(m)}=kh_m\}$ is determined by the path up to $kh_m$. Conditional on that past, the increments after $kh_m$ are independent of the past and have a distribution unchanged by multiplying them by $-1$. Reflection about $B_{kh_m}$ precisely negates those future increments. It leaves the path up to $kh_m$, and hence the event $\{T_a^{(m)}=kh_m\}$, unchanged. Summing over the finitely many grid points at most $t$, and including the unchanged case $T_a^{(m)}>t$, shows that each grid-reflected path has the same law on $[0,t]$ as Brownian motion. The grid reflection is also an involution: it does not alter the path through $T_a^{(m)}$, so applying it a second time uses the same grid time and restores the original path.

On a path with $T_a<t$, the grid times decrease to $T_a$, with $0\le T_a^{(m)}-T_a\le h_m$. Continuity gives $B_{T_a^{(m)}}\to a$. The grid-reflected paths converge uniformly on $[0,t]$ to the reflection at $T_a$: on the shrinking interval from $T_a$ to $T_a^{(m)}$, continuity makes the discrepancy vanish, and after $T_a^{(m)}$ the reflected values converge uniformly because their reflection centers converge to $a$. If $T_a>t$, all sufficiently fine grid reflections leave the path unchanged on $[0,t]$. If $T_a=t$, reflecting only at the endpoint also gives convergence on $[0,t]$. Therefore the grid-reflected paths converge almost surely, uniformly on $[0,t]$, to the path reflected at $T_a$. For every bounded continuous function of a path on $[0,t]$, dominated convergence shows that its expectation under the limiting reflected path equals its expectation under Brownian motion. Bounded continuous functions determine probability laws on the space of continuous paths, so the reflected path has Brownian law. $\square$

The first-hit reflection is an involution as well: it leaves the path through the first hit unchanged, so the first hit of the reflected path occurs at the same time, and a second reflection restores the original path. Its effect on the endpoint is especially simple. On paths that hit $a$ by time $t$, the endpoint $B_t$ is changed to $2a-B_t$. In particular, an endpoint below $a$ is paired with an endpoint above $a$.

The Gaussian-Tail Relation

Theorem (Reflection principle for a maximum): For standard Brownian motion, $t>0$, and $a>0$, $$ \mathbb{P}(M_t\ge a)=2\mathbb{P}(B_t\ge a) =2\left[1-\Phi\left(\frac{a}{\sqrt{t}}\right)\right], $$ where $\Phi(x)=\mathbb{P}(Z\le x)$ for $Z\sim N(0,1)$.

Proof. By continuity and $B_0=0<a$, the event $\{B_t>a\}$ implies that the path hit $a$ by time $t$. Reflection pairs the event $\{T_a\le t,\ B_t<a\}$ with $\{B_t>a\}$: the first-hit time and the path through that time are unchanged, while the endpoint becomes $2a-B_t>a$. The path-law lemma and the involution therefore give $\mathbb{P}(T_a\le t,\ B_t<a)=\mathbb{P}(B_t>a)$. The Gaussian variable $B_t\sim N(0,t)$ has no atom at $a$, so $\mathbb{P}(B_t=a)=0$. Partitioning the crossing event according to whether the endpoint is above or below $a$ now yields $\mathbb{P}(M_t\ge a)=\mathbb{P}(B_t\ge a)+\mathbb{P}(B_t>a)=2\mathbb{P}(B_t\ge a)$. Finally, $B_t/\sqrt{t}\sim N(0,1)$, so $\mathbb{P}(B_t\ge a)=1-\Phi(a/\sqrt{t})$. $\square$

The restriction $a>0$ matters to the proof because the path starts below the barrier. At $a=0$, the formula still gives $1$: $M_t\ge0$ holds because $B_0=0$, and $2\mathbb{P}(B_t\ge0)=1$. For $a<0$, the displayed formula is not valid: $M_t\ge a$ holds with probability $1$, whereas $2\mathbb{P}(B_t\ge a)$ is generally larger than $1$. The theorem also requires $t>0$; at $t=0$ the endpoint is the constant $B_0=0$, not a nondegenerate Gaussian variable.

Worked Example: Crossing a Barrier of Two by Time Four

For $a=2$ and $t=4$, compute the probability that Brownian motion reaches $2$ by time $4$. The endpoint has distribution $B_4\sim N(0,4)$, so standardization gives

\mathbb{P}(M_4\ge2) =2\mathbb{P}(B_4\ge2) =2\mathbb{P}\left(\frac{B_4}{2}\ge\frac{2}{2}\right) =2[1-\Phi(1)].

Here $\sqrt{4}=2$, and the standardized barrier is $2/2=1$. The endpoint event $\{B_4\ge2\}$ contributes $1-\Phi(1)$; the reflected paths that hit $2$ and end below it contribute another $1-\Phi(1)$. Check: adding the two disjoint endpoint cases gives $2[1-\Phi(1)]$, as required by the theorem.

The same pairing handles endpoint restrictions below the barrier. If $b<a$, reflecting a path that hits $a$ and ends at or below $b$ changes its endpoint to at least $2a-b$. Conversely, an endpoint at least $2a-b$ is above $a$, so continuity forces a hit of $a$, and reflection sends it back to an endpoint at most $b$. Thus

\mathbb{P}(T_a\le t,\ B_t\le b)=\mathbb{P}(B_t\ge 2a-b),\qquad b<a.

This endpoint form follows from the same path-law-preserving involution; equality at the endpoints of intervals does not affect the probabilities because $B_t$ has a continuous Gaussian distribution.

Worked Example: Hit Two and Finish at or Below Zero

Find the probability that Brownian motion hits $a=2$ by $t=4$ and finishes at or below $b=0$. The reflected endpoint threshold is $2a-b=2(2)-0=4$. Since $B_4\sim N(0,4)$, standardizing gives

\mathbb{P}(T_2\le4,\ B_4\le0) =\mathbb{P}(B_4\ge4) =\mathbb{P}\left(\frac{B_4}{2}\ge\frac{4}{2}\right) =1-\Phi(2).

Check: reflecting an endpoint $y\le0$ across $2$ gives $4-y\ge4$. Conversely, an endpoint at least $4$ must have crossed $2$, and reflection gives an endpoint at most $0$. The two endpoint conditions therefore correspond exactly.

Worked Example: A Whole Endpoint Interval

For $t=9$ and barrier $a=3$, find the probability that the path hits $3$ and finishes in the interval $[0,2]$. The reflected endpoint is $2a-B_9=6-B_9$. As the original endpoint ranges from $0$ to $2$, its reflected value ranges from $6$ down to $4$. Hence the event probability equals $\mathbb{P}(4\le B_9\le6)$. Since $B_9\sim N(0,9)$, divide both endpoints by $\sqrt{9}=3$:

\mathbb{P}(T_3\le9,\ 0\le B_9\le2) =\mathbb{P}(4\le B_9\le6) =\Phi(2)-\Phi\left(\frac{4}{3}\right).

Check: reflecting $0$ gives $6-0=6$, and reflecting $2$ gives $6-2=4$; the reflected interval is exactly $[4,6]$. Standardizing $4$ and $6$ by $3$ gives $4/3$ and $2$, respectively.

A Common Mistake and Why the Pairing Matters

A tempting but incorrect calculation is to identify $\{M_t\ge a\}$ with $\{B_t\ge a\}$. The latter event is contained in the former, but the containment is strict: paths can cross $a$ and return below it. For $a=2,t=4$, the incorrect calculation gives $1-\Phi(1)$, because $B_4/2$ is standard normal and the standardized endpoint threshold is $2/2=1$. Reflection shows that paths crossing $2$ and ending below it have the same probability $1-\Phi(1)$. Including both endpoint cases gives $2[1-\Phi(1)]$, not $1-\Phi(1)$.

The principle is useful because it replaces a path-dependent event with a calculation about a single Gaussian endpoint. The path event may involve every time between $0$ and $t$, but reflection reduces its probability to a normal tail with standard deviation $\sqrt{t}$. The essential checks are that the barrier is positive, the time is positive, and the comparison event really counts paths that have hit the barrier—not merely paths whose endpoint lies above it.

Check Your Understanding

For each question, identify the barrier and time conditions before applying reflection.

For $a=1$ and $t=1$, express $\mathbb{P}(M_1\ge1)$ in terms of $\Phi$.
For $a=3,t=4,b=1$, use the endpoint version to express $\mathbb{P}(T_3\le4,\ B_4\le1)$ as a standard normal tail.
Explain why $B_t>a$ implies $T_a\le t$ when $a>0$.
In the grid approximation, why does reflecting future increments preserve the law on the event that the grid hit time equals $kh_m$?
What goes wrong if one computes $\mathbb{P}(M_t\ge a)$ using only $\mathbb{P}(B_t\ge a)$?

The Reflection Principle for Brownian Motion

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