First-Passage Laws for Brownian Motion With Drift

From Brownian Motion to a First-Passage Law

Brownian motion with drift can move toward a target on average while still making random excursions away from it. The first-passage time records when the path reaches the target, not merely where the process is at a fixed time. Its distribution therefore contains information that a single-time Gaussian distribution does not. In particular, a negative drift makes reaching a positive level less likely, but does not make it impossible.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Fix a constant drift \(\mu\in\mathbb{R}\), and define \(X_t=\mu t+B_t\), starting at \(X_0=0\). For a fixed level \(a>0\), define the first-passage time \(\tau_a=\inf\{t\ge0:X_t=a\}\), with \(\inf\varnothing=\infty\). The strict inequality \(a>0\) excludes the degenerate case in which the process starts at the target.

The derivation uses the Brownian transition density killed at the level \(a\): paths that have reached \(a\) are removed from the density of paths still below \(a\). By the reflection principle, for driftless Brownian motion starting at zero, this density at \(y<a\) is \(g_t(y)-g_t(y-2a)\), where \(g_t(z)=(2\pi t)^{-1/2}\exp(-z^2/(2t))\). The reflected term subtracts paths that have crossed the boundary and are reflected back to the endpoint \(y\).

The Killed Kernel with Drift

We first obtain the corresponding density for \(X_t=\mu t+B_t\). Define, for \(t>0\) and \(y<a\),

The exponential factor changes the driftless heat kernel into the constant-drift kernel. To verify the transformation, put \(k(t,y)=g_t(y)-g_t(y-2a)\), so \(q_\mu=e^{\mu y-\mu^2t/2}k\). The reflection principle gives \(k_t=\tfrac12 k_{yy}\) for \(y<a\). Differentiating the transformation gives \(q_{\mu,y}=e^{\mu y-\mu^2t/2}(\mu k+k_y)\), \(q_{\mu,yy}=e^{\mu y-\mu^2t/2}(\mu^2k+2\mu k_y+k_{yy})\), and \(q_{\mu,t}=e^{\mu y-\mu^2t/2}(k_t-\tfrac12\mu^2k)\). Substitution shows \(q_{\mu,t}=-\mu q_{\mu,y}+\tfrac12q_{\mu,yy}\), the forward equation for drift \(\mu\) and diffusion coefficient \(1\). At the boundary \(y=a\), \(k(t,a)=g_t(a)-g_t(-a)=0\). As \(t\) decreases to zero, the kernel has unit point mass at the starting point \(0\), which lies below \(a\). Thus it is the absorbing transition density; uniqueness for the heat equation with this absorbing boundary identifies it as the killed density.

Proof. For each fixed \(t>0\), \(\{\tau_a>t\}\) is the event that the path has not reached \(a\) by time \(t\). Its probability is the integral of the killed density over \(y<a\). Completing the square gives \(e^{\mu y-\mu^2t/2}g_t(y)=g_t(y-\mu t)\). Hence the first term integrates to \(\Phi((a-\mu t)/\sqrt{t})\). For the reflected term, set \(z=y-2a\); then \(y<a\) means \(z<-a\), and \(e^{\mu y-\mu^2t/2}g_t(y-2a)=e^{2\mu a}g_t(z-\mu t)\). Its integral is \(e^{2\mu a}\Phi((-a-\mu t)/\sqrt{t})\). Subtracting this reflected contribution from the first gives \[ \mathbb{P}(\tau_a>t)= \Phi\!\left(\frac{a-\mu t}{\sqrt{t}}\right) -e^{2\mu a}\Phi\!\left(\frac{-a-\mu t}{\sqrt{t}}\right). \] Taking the complement and using \(1-\Phi(z)=\Phi(-z)\) yields the stated \(F(t)\).

To differentiate, put \(u(t)=(\mu t-a)/\sqrt{t}\) and \(v(t)=(-\mu t-a)/\sqrt{t}\). Then \(u'(t)=(\mu t+a)/(2t^{3/2})\) and \(v'(t)=(a-\mu t)/(2t^{3/2})\). Since \(\Phi'(z)=(2\pi)^{-1/2}e^{-z^2/2}\), differentiating the two terms in \(F\) gives \(\phi(u)u'+e^{2\mu a}\phi(v)v'\), where \(\phi(z)=(2\pi)^{-1/2}e^{-z^2/2}\). Directly comparing exponents shows \(e^{2\mu a}\phi(v)=\phi(u)\): the difference between the exponents is \(2\mu a-((\mu t+a)^2-(\mu t-a)^2)/(2t)=0\). Therefore \[ F'(t)=\phi(u)\frac{(\mu t+a)+(a-\mu t)}{2t^{3/2}} =\frac{a}{\sqrt{2\pi t^3}}\exp\!\left(-\frac{(a-\mu t)^2}{2t}\right). \] Also \(F(t)\to0\) as \(t\downarrow0\), since both normal CDF arguments tend to \(-\infty\). Thus the derivative is the density of the finite first-passage times. Finally, as \(t\to\infty\), if \(\mu>0\), then \(u(t)\to\infty\) and \(v(t)\to-\infty\), giving \(F(t)\to1\). If \(\mu<0\), then \(u(t)\to-\infty\) and \(v(t)\to\infty\), giving \(F(t)\to e^{2\mu a}\). If \(\mu=0\), both terms are \(\Phi(-a/\sqrt{t})\), so \(F(t)=2\Phi(-a/\sqrt{t})\to1\). These limits give the claimed total masses. \(\square\)

Using the Distribution and Density

Why the Drift Sign Matters

The density is positive for every \(t>0\), regardless of the drift sign. That does not mean the finite-time density always integrates to one: when \(\mu<0\), some paths never reach the positive level, and the total mass of the density is \(e^{2\mu a}<1\). The remaining probability is the mass at \(\tau_a=\infty\). For \(\mu\ge0\), the finite-time density has total mass one. These are statements under the probability measure \(\mathbb{P}\); no change of measure is being made here.

A common mistake is to use the zero-drift hitting probability \(1\) for every drift, on the grounds that Brownian fluctuations can eventually overcome any fixed adverse trend. For \(\mu=-1\) and \(a=1\), that conclusion would give probability \(1\), but the derived limit is \(e^{-2}\), approximately \(0.1353\). The correction is to retain the exponential multiplier \(e^{2\mu a}\) in the distribution function before taking \(t\to\infty\). It is this term that records the loss of probability to paths that never reach the level.

For a process starting at \(x<a\), replace \(a\) throughout by the distance \(a-x\): the process \(x+\mu t+B_t\) reaches \(a\) precisely when \(\mu t+B_t\) reaches \(a-x\). This reduction relies on constant drift and unit diffusion coefficient. A different diffusion coefficient changes the time and level scaling, while a state-dependent drift does not have this same formula.