Reflection at the first crossing of a level is a measure-preserving transformation of Brownian paths. It converts a pathwise maximum into an endpoint event, producing exact laws for maxima, first-passage times, and barrier exits.
First passage and running maximum
Definition. For \(a\in\mathbb R\), define \[ T_a=\inf\{t\ge0:B_t=a\}, \qquad M_t=\sup_{0\le s\le t}B_s. \] Continuity and adaptedness make \(T_a\) a stopping time, and for \(a>0\), \[ \{T_a\le t\}=\{M_t\ge a\}. \]
The reflection principle
Theorem. Let \(a>0\). For every Borel set \(A\subset(-\infty,a]\), \[ P(T_a\le t,\ B_t\in A) = P(B_t\in2a-A), \] where \(2a-A=\{2a-x:x\in A\}\). In particular, for \(b\le a\), \[ P(M_t\ge a,\ B_t\le b)=P(B_t\ge2a-b). \]
Proof. On \(\{T_a\le t\}\), define \[ \widetilde B_s= \begin{cases} B_s,&s\le T_a,\\ 2a-B_s,&s>T_a. \end{cases} \] By the strong Markov property at \(T_a\), the post-\(T_a\) process \(B_{T_a+u}-a\) is fresh Brownian motion independent of \(\mathcal F_{T_a}\). Its negative has the same law by symmetry, so the reflection map preserves Wiener measure on \(\{T_a\le t\}\). It is an involution. Moreover, \(B_t\in A\) is transformed into \(\widetilde B_t=2a-B_t\in2a-A\). Since any path ending above \(a\) has crossed \(a\), the reflected endpoint event needs no additional crossing condition.
Corollary (maximum law). For \(a>0\), \[ P(M_t\ge a)=2P(B_t\ge a) =2\left\{1-\Phi\!\left(\frac a{\sqrt t}\right)\right\}. \] Hence \[ M_t\stackrel d=|B_t|. \]
Proof. Decompose \[ P(M_t\ge a) =P(B_t\ge a)+P(M_t\ge a,B_t<a). \] Reflection maps the second event to \(\{B_t>a\}\), up to the null endpoint event \(B_t=a\).
Joint law of endpoint and maximum
Proposition. For \(x<m\), \[ P(M_t<m,\ B_t\in\mathrm{d} x) = \{p_t(x)-p_t(2m-x)\}\,\mathrm{d} x, \] where \(p_t(x)=(2\pi t)^{-1/2}e^{-x^2/(2t)}\). Thus \[ P(M_t\in\mathrm{d} m,\ B_t\in\mathrm{d} x) = \frac{2(2m-x)}{t}p_t(2m-x) 1_{\{m>0,x<m\}}\,\mathrm{d} m\,\mathrm{d} x. \]
Proof. Reflection gives \[ P(M_t\ge m,B_t\in\mathrm{d} x) =p_t(2m-x)\,\mathrm{d} x,\qquad x<m. \] Subtract from the unconditional endpoint density to obtain the first formula. Differentiate it with respect to \(m\), using \(p_t'(y)=-(y/t)p_t(y)\), to obtain the joint density.
First-passage distribution
Theorem. For \(a>0\), \[ P(T_a\le t) =2\left\{1-\Phi\!\left(\frac a{\sqrt t}\right)\right\}, \] and \(T_a\) has density \[ f_{T_a}(t) = \frac{a}{\sqrt{2\pi t^3}} \exp\!\left(-\frac{a^2}{2t}\right), \qquad t>0. \] Its Laplace transform is \[ \mathbb Ee^{-\lambda T_a} =e^{-a\sqrt{2\lambda}}, \qquad\lambda\ge0. \]
Proof. The distribution function is the maximum formula. Differentiating \(2\{1-\Phi(a/\sqrt t)\}\) gives the density. For the Laplace transform, use the exponential martingale \[ Z_t=\exp\{\theta B_t-\tfrac12\theta^2t\}. \] Optional stopping at \(T_a\wedge n\), followed by localization and dominated convergence for \(\theta>0\), gives \[ 1=\mathbb EZ_{T_a} =e^{\theta a}\mathbb Ee^{-\theta^2T_a/2}. \] Put \(\lambda=\theta^2/2\).
Corollary. \(P(T_a<\infty)=1\), but \(\mathbb ET_a=\infty\).
Proof. Letting \(t\to\infty\) in the distribution function gives one. Gaussian asymptotics give \[ P(T_a>t) =2\Phi(a/\sqrt t)-1 \sim a\sqrt{\frac2{\pi t}}. \] Therefore \(\mathbb ET_a=\int_0^\infty P(T_a>t)\,\mathrm{d} t=\infty\).
Exit from an interval
Theorem. Let \(a,b>0\) and \[ \tau=\inf\{t\ge0:B_t\notin(-b,a)\}. \] Then \[ P(B_\tau=a)=\frac b{a+b}, \qquad P(B_\tau=-b)=\frac a{a+b}, \qquad \mathbb E\tau=ab. \]
Proof. Continuity gives \(B_\tau\in\{-b,a\}\). The stopped martingale \(B_{t\wedge\tau}\) is bounded, so optional stopping gives \[ 0=\mathbb EB_\tau =aP(B_\tau=a)-bP(B_\tau=-b), \] which, together with total probability one, yields the exit probabilities. Apply optional stopping to \(B_t^2-t\): \[ \mathbb E\tau=\mathbb EB_\tau^2 =a^2\frac b{a+b}+b^2\frac a{a+b}=ab. \]