Maximum Crossing Probabilities for Brownian Motion With Drift

From Endpoint Probabilities to Barrier Crossing

A Gaussian endpoint distribution alone does not determine whether a path crossed a barrier earlier. A path can reach a high level and then return below it before the terminal time. For Brownian motion, reflection relates those returning paths to paths whose endpoints lie on the other side of the barrier. When a constant drift is added, the reflected paths no longer have equal probabilities, but their probability ratio can be calculated from Gaussian densities. Together, these two facts give an exact finite-horizon crossing probability.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \((B_t)_{t\ge0}\), with \(B_0=0\). Fix a time horizon \(T>0\), a positive level \(a>0\), and a constant drift \(\mu\in\mathbb{R}\). Define \(X_t=B_t+\mu t\) and the first hitting time \(\tau_a=\inf\{t\ge0:X_t=a\}\), with the infimum of the empty set taken to be \(+\infty\). By continuity, \(\{\tau_a\le T\}=\{\max_{0\le s\le T}X_s\ge a\}\). The strict inequality \(a>0\) is assumed throughout; the formulas below also have meaningful limits at \(a=0\), but that boundary case is not needed for the derivation.

Why Drift Changes Reflected-Path Weights

We first compute how adding drift changes the probability of a path event. For a fixed \(T\), define a new probability measure on \(\mathcal{F}_T\) by the density \(L_T=\exp(\mu B_T-\tfrac12\mu^2T)\) relative to \(\mathbb{P}\). It is a probability density because, for \(B_T\sim N(0,T)\), the normal moment-generating function gives \(\mathbb{E}[L_T]=1\). Under this measure, the coordinate process has the law of Brownian motion with drift \(\mu\) on \([0,T]\).

Here is a direct verification of that law. Take any finite time grid \(0=t_0<t_1<\cdots<t_n=T\), and write \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\) and \(\Delta t_i=t_i-t_{i-1}\). Under \(\mathbb{P}\), the increments are independent with densities \(p_{\Delta t_i}(x)=(2\pi\Delta t_i)^{-1/2}\exp(-x^2/(2\Delta t_i))\). The density \(L_T\) factors as \(\prod_{i=1}^n\exp(\mu\Delta B_i-\tfrac12\mu^2\Delta t_i)\). Multiplying each increment density by its factor gives \(p_{\Delta t_i}(x-\mu\Delta t_i)\), the density of \(N(\mu\Delta t_i,\Delta t_i)\). Thus under the new measure the increments on every such grid are independent normal variables with the required drifted means and unchanged variances. These finite-dimensional distributions, together with continuity of the paths, identify the law on \([0,T]\).

Consequently, for a path event \(A\) concerning the drifted process, its probability can be computed under the zero-drift Brownian law by weighting each path according to its endpoint: \(\mathbb{P}(B+\mu\cdot\in A)=\mathbb{E}[\mathbf{1}_{\{B\in A\}}\exp(\mu B_T-\tfrac12\mu^2T)]\). In this expression the probability on the left is the law of \(X_t=B_t+\mu t\) under the original measure, while the expectation on the right is under the zero-drift Brownian law. The weight depends on \(B_T\), which is why reflected paths with different endpoints acquire different weights.

For \(y<a\), the zero-drift reflection principle gives the following endpoint density: paths that hit \(a\) by time \(T\) and end in an infinitesimal interval around \(y\) have density \(p_T(2a-y)\,dy\). Indeed, reflecting a path after its first hit of \(a\) sends its endpoint \(y<a\) to \(2a-y>a\), and preserves the zero-drift Brownian path probabilities. Applying the endpoint weight just derived gives the drifted joint density below the barrier.

Proof. Fix a Borel set \(A\subset(-\infty,a)\). The change-of-measure calculation weights each zero-drift endpoint \(y\in A\) by \(\exp(\mu y-\tfrac12\mu^2T)\). By reflection, the zero-drift density of paths that hit \(a\) and end at \(y\) is \(p_T(2a-y)\). Integrating this weighted density over \(A\) proves the first expression. For the second, use \(p_T(2a-y-\mu T)=p_T(2a-y)\exp(\mu(2a-y)-\tfrac12\mu^2T)\). Multiplying by \(e^{2\mu(y-a)}\) gives \(e^{\mu y-\tfrac12\mu^2T}p_T(2a-y)\), as asserted. Since this holds for every such Borel set, it establishes the density. \(\square\)

Deriving the Crossing Formula

Separate crossing paths according to their terminal position. If \(X_T\ge a\), continuity forces the path to have crossed \(a\). Since \(X_T\sim N(\mu T,T)\), this contribution is \(\Phi((\mu T-a)/\sqrt T)\). The event \(X_T=a\) has probability zero, so using \(\ge\) or \(>\) at this endpoint makes no difference. Paths with \(X_T<a\) contribute the integral of the proposition’s density over \(y<a\).

Proof of the Theorem. The crossing event is the disjoint union of \(\{X_T\ge a\}\) and \(\{\tau_a\le T,\ X_T<a\}\). The first event has probability \(\Phi((\mu T-a)/\sqrt T)\). For the second, integrate the density from the proposition and make the substitution \(z=2a-y\). When \(y<a\), this substitution gives \(z>a\), and therefore \(\int_{-\infty}^a e^{\mu y-\frac12\mu^2T}p_T(2a-y)\,dy =e^{2\mu a}\int_a^\infty e^{-\mu z-\frac12\mu^2T}p_T(z)\,dz\). The integrand after the factor \(e^{2\mu a}\) is \(p_T(z+\mu T)\), since completing the square gives \(e^{-\mu z-\frac12\mu^2T}p_T(z)=p_T(z+\mu T)\). The integral is consequently \(e^{2\mu a}\mathbb{P}(N(-\mu T,T)>a) =e^{2\mu a}\Phi((-\mu T-a)/\sqrt T)\). Adding the two disjoint contributions proves the stated formula. \(\square\)

Interpreting the Formula and Avoiding a Sign Error

The two terms in the formula have distinct roles. The endpoint tail \(\Phi((\mu T-a)/\sqrt T)\) is always a subset of the crossing event. The second term is not another endpoint tail for \(X_T\) at the original barrier; it is the reflected-path contribution, corrected for the changed weight of a drifted path. The factor \(e^{2\mu a}\) and the argument \((-\mu T-a)/\sqrt T\) must be used together. Either one in isolation gives the wrong contribution.

A common mistake is to put \(e^{-2\mu a}\) in front of the second normal CDF. For \(\mu=1\), \(a=1\), and \(T=1\), that incorrect calculation gives \(0.5+e^{-2}\Phi(-2)\), or approximately \(0.5+0.1353\cdot0.02275=0.5031\). The correct reflected-path contribution is \(e^2\Phi(-2)\approx0.1681\), not \(e^{-2}\Phi(-2)\approx0.003079\). The error reverses the endpoint-weight ratio: reflection changes an endpoint \(y<a\) into \(2a-y>a\), so the ratio of the drift weights is \(\exp(\mu y-\mu(2a-y))=e^{2\mu(y-a)}\). Integrating this ratio produces the positive factor \(e^{2\mu a}\) in the final expression.

Two further checks help fix the interpretation. First, setting \(\mu=0\) recovers the reflection-principle formula \(2\Phi(-a/\sqrt T)\). Second, for fixed \(a,T>0\), the formula gives a positive crossing probability for either sign of \(\mu\); a negative drift reduces the probability but does not rule out a sufficiently large Brownian fluctuation. The formula concerns a continuous-time maximum, not merely observations on a discrete time grid.