The Reflection Map at a Fixed Barrier

Reflecting a Path at Its First Hit

Brownian increments are symmetric: changing the sign of an increment does not change its distribution. Reflection at a barrier uses this symmetry after the path first reaches a specified level. The path is left unchanged up to that time, then mirrored around the barrier. The definition is simple; the subtlety is that the time at which reflection begins depends on the path itself. We will establish law preservation by approximating that time with deterministic grid times, using only independent increments at fixed times.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space carrying a standard Brownian motion $(B_t)_{t\ge0}$ with state space $\mathbb{R}$, starting at $B_0=0$. Use its natural filtration $\mathcal{F}_t=\sigma(B_s:0\le s\le t)$. Fix a barrier $a>0$, and define the first-hit time pathwise by $\tau_a=\inf\{t\ge0:B_t=a\}$, with $\inf\varnothing=\infty$. The definition makes sense whether or not the path ever reaches $a$.

Definition: The reflection of a continuous path $f:[0,\infty)\to\mathbb{R}$ at its first hit of $a$ is the path $R_af$ given by $$ (R_af)(t)= \begin{cases} f(t),&t\le \tau_a(f),\\ 2a-f(t),&t>\tau_a(f), \end{cases} $$ where $\tau_a(f)=\inf\{t\ge0:f(t)=a\}$. If $\tau_a(f)=\infty$, set $R_af=f$. For Brownian motion, write $\widetilde B_t=(R_aB)(t)$.

The constant $2a$ means that reflection is around the level $a$, not around zero. If $\tau_a<\infty$, continuity gives $B_{\tau_a}=a$. Thus the two formulas agree at the joining time: the reflected path has value $a$ there and remains continuous. For $s,t\ge\tau_a$, its increments satisfy $\widetilde B_t-\widetilde B_s=-(B_t-B_s)$. Increments wholly after the hit have their signs reversed; the path before the hit is not changed.

Worked Example: Reflecting a Continuous Path

Consider a deterministic continuous path that is linear between the listed points $(0,0),(2,2),(3,1),(5,4)$, and take $a=2$. This is an illustration of the map, not a probability calculation. The first hit is at time $\tau_a=2$, since the path rises linearly from $0$ to $2$ on $[0,2]$ and does not equal $2$ earlier.

At times $0$ and $2$, the path is unchanged: $\widetilde f(0)=0$ and $\widetilde f(2)=2$. At time $3$, $f(3)=1$, so $\widetilde f(3)=2(2)-1=3$. At time $5$, $f(5)=4$, so $\widetilde f(5)=2(2)-4=0$. The reflected path therefore passes through $(2,2),(3,3),(5,0)$ after the hit, remaining continuous at time $2$.

Check: The original increments from time $2$ to $3$, and from $3$ to $5$, are $1-2=-1$ and $4-1=3$. The reflected increments are $3-2=1$ and $0-3=-3$, exactly their negatives.

Why the Random Hit Time Needs Care

At a deterministic time $u$, the future Brownian increments are independent of $\mathcal{F}_u$, and their joint distribution is unchanged when all their signs are reversed. It is tempting to apply this directly at $\tau_a$. But $\tau_a$ is selected by the path, not fixed in advance, so deterministic-time independent increments alone do not justify that step. Instead, round the hit time up to a deterministic grid and reflect at the rounded time. At each possible grid time, independent increments apply in their usual deterministic-time form.

Lemma: For $h_n=2^{-n}$, define $\tau_{a,n}=h_n\lceil\tau_a/h_n\rceil$ when $\tau_a<\infty$, and $\tau_{a,n}=\infty$ otherwise. Define $$ B^{(n)}_t= \begin{cases} B_t,&t\le\tau_{a,n},\\ 2B_{\tau_{a,n}}-B_t,&t>\tau_{a,n}, \end{cases} $$ when $\tau_{a,n}<\infty$, and set $B^{(n)}=B$ otherwise. For every finite set of times, the corresponding values of $B^{(n)}$ have the same joint distribution as the values of $B$.

Proof. Fix $n$, let $h=h_n$, and fix observation times $t_1,\ldots,t_m$; put $T=\max_i t_i$. For each positive integer $k$, let $E_k=\{(k-1)h<\tau_a\le kh\}$, so on $E_k$ the rounded time is $kh$. The event $E_k$ belongs to $\mathcal{F}_{kh}$: for a continuous path, whether it hits $a$ by $kh$ is determined by its values up to $kh$. On $E_k$, observations at or before $kh$ are unchanged. For each observation time $t_i>kh$, the original value is $B_{kh}+(B_{t_i}-B_{kh})$, while the reflected value is $B_{kh}-(B_{t_i}-B_{kh})$.

Conditional on $\mathcal{F}_{kh}$, the vector of increments $B_{t_i}-B_{kh}$ for the observation times after $kh$ is independent of $\mathcal{F}_{kh}$. Its distribution is unchanged by multiplying every component by $-1$: it is a centered jointly Gaussian vector, since Brownian increments over disjoint intervals are independent centered normal variables. Consequently, for any bounded measurable function $g$ of the $m$ observations, the conditional expectations of $g$ applied to the reflected and original vectors agree on $E_k$. If $\tau_{a,n}>T$, every observation is unchanged. This remaining event is determined by the path up to the last grid point not exceeding $T$, so it too can be included in the same conditioning argument. Summing over the finitely many $k$ with $kh\le T$, and the event $\{\tau_{a,n}>T\}$, gives equality of the joint distributions at $t_1,\ldots,t_m$. $\square$

There are only finitely many relevant grid times in this argument because $kh\le T$. If the rounded time is later than $T$, reflection has not affected any observation in the chosen vector. This finite-horizon detail avoids making any assumption about whether the path eventually hits the barrier.

Worked Example: A Grid Reflection and Its Increments

Take grid spacing $h=1$ and suppose a path first hits $a=1$ between times $1$ and $2$, so the rounded reflection time is $2$. For illustration, let its values at times $1,2,3,4$ be $\tfrac12,\tfrac65,\tfrac{13}{10},\tfrac9{10}$. These values are consistent with a continuous path that first reaches $1$ between times $1$ and $2$.

The grid-reflected path agrees through time $2$. Its later values are $2B_2-B_t$, so $\,B^{(n)}_3=2\cdot\tfrac65-\tfrac{13}{10}=\tfrac{24}{10}-\tfrac{13}{10}=\tfrac{11}{10}$, and $\,B^{(n)}_4=2\cdot\tfrac65-\tfrac9{10}=\tfrac{24}{10}-\tfrac9{10}=\tfrac{15}{10}=\tfrac32$. The original post-grid increments are $\tfrac{13}{10}-\tfrac65=\tfrac1{10}$ and $\tfrac9{10}-\tfrac{13}{10}=-\tfrac4{10}$. The reflected increments are $\tfrac{11}{10}-\tfrac65=-\tfrac1{10}$ and $\tfrac32-\tfrac{11}{10}=\tfrac4{10}$. Each is the negative of the corresponding original increment.

Check: The endpoint after two reflected increments is $\tfrac65-\tfrac1{10}+\tfrac4{10}=\tfrac{15}{10}=\tfrac32$, agreeing with the direct reflection formula.

The Reflection Map Preserves Brownian Law

Theorem: For standard Brownian motion starting at $0$ and a fixed barrier $a>0$, the process $(\widetilde B_t)_{t\ge0}$ has the same finite-dimensional distributions as $(B_t)_{t\ge0}$. Both have continuous paths, so they also have the same law as random continuous paths.

Proof. Fix finitely many times $t_1,\ldots,t_m$. For each $n$, the lemma shows that $(B^{(n)}_{t_1},\ldots,B^{(n)}_{t_m})$ has the same distribution as $(B_{t_1},\ldots,B_{t_m})$. We next show that the approximating vector converges almost surely to the reflected vector.

If $\tau_a=\infty$, all the approximating paths and the reflected path equal $B$. If $\tau_a<\infty$, then $\tau_{a,n}\ge\tau_a$ and $\tau_{a,n}\to\tau_a$. For a fixed $t<\tau_a$, eventually $t<\tau_{a,n}$, so $B^{(n)}_t=B_t=\widetilde B_t$. At $t=\tau_a$, both values equal $a$. For $t>\tau_a$, eventually $t>\tau_{a,n}$, and continuity gives $B_{\tau_{a,n}}\to B_{\tau_a}=a$. Therefore $B^{(n)}_t=2B_{\tau_{a,n}}-B_t\to2a-B_t=\widetilde B_t$. This proves almost-sure convergence at every one of the finitely many observation times.

For any bounded continuous function $g:\mathbb{R}^m\to\mathbb{R}$, bounded convergence now yields $\mathbb{E}[g(\widetilde B_{t_1},\ldots,\widetilde B_{t_m})] =\lim_{n\to\infty}\mathbb{E}[g(B^{(n)}_{t_1},\ldots,B^{(n)}_{t_m})] =\mathbb{E}[g(B_{t_1},\ldots,B_{t_m})]$. Thus the finite-dimensional distributions agree. The reflected path is continuous: it is unchanged before the hit, and at a finite hit both formulas meet at $a$. For continuous paths, evaluations at nonnegative rational times determine the law on the space of continuous paths with uniform convergence on compact intervals. Hence the two processes have the same path law. $\square$

Worked Example: Sign Symmetry for a Pair of Increments

Suppose two successive unit-length Brownian increments after a deterministic grid time are $U$ and $V$. They are independent $N(0,1)$ variables, so their joint density is $\varphi(u)\varphi(v)$, where $\varphi(x)=(2\pi)^{-1/2}e^{-x^2/2}$. Reversing both signs gives density $\varphi(-u)\varphi(-v)=\varphi(u)\varphi(v)$, because $(-u)^2=u^2$ and $(-v)^2=v^2$.

For example, the probability that both increments are positive is $\tfrac12\cdot\tfrac12=\tfrac14$. After reversing both signs, that event corresponds to both original increments being negative, which also has probability $\tfrac12\cdot\tfrac12=\tfrac14$. The grid proof uses this same symmetry for any finite vector of future increments, including vectors whose coordinates are values at different times after the grid point.

Check: The joint density integrates to $1$, since $\int_{\mathbb{R}}\varphi(u)\,du\int_{\mathbb{R}}\varphi(v)\,dv=1\cdot1=1$; sign reversal therefore preserves the entire joint distribution, not just the probability of this one event.

Common Error and Why the Map Matters

Common mistake: reflecting around zero. Suppose the barrier is $a=2$ and the path hits it at time $\tau_a$. The incorrect formula $-B_t$ after the hit gives value $-2$ at the joining time, while the path still has value $2$ immediately before it. It creates a jump of size $4$. The correct formula is $2a-B_t=4-B_t$, which equals $2$ when $B_t=2$; the reflected path is continuous at the hit.

Another error is to assume that symmetry at a fixed time automatically proves symmetry after a random hit time. The grid argument supplies the missing justification: condition at deterministic grid times, use independence and sign symmetry there, and then pass to the limit using path continuity. This establishes law preservation without calculating any hitting probability or endpoint tail probability. The reflection map is the path transformation needed for such calculations, but its distributional consequences require a separate argument.

Check Your Understanding

Use the definition and the grid approximation to answer the following.

For a path that first hits $a=3$ at time $4$ and has value $5$ at time $6$, what is its reflected value at time $6$?
Why does the reflection formula use $2a-B_t$, rather than $-B_t$?
For a grid approximation with rounded hit time $kh$, which increments are sign-reversed?
What property of Brownian increments justifies the conditional symmetry at a deterministic grid time?
Why does pointwise convergence of the grid-reflected paths at finitely many fixed times suffice to identify the limiting finite-dimensional distributions?

The Reflection Map at a Fixed Barrier

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