Recovering First-Passage Densities from Transforms

From a Transform to a Time-Domain Law

A Laplace transform can encode a first-passage-time distribution in a compact expression, but probabilities over time intervals require the distribution in the time domain. For standard Brownian motion, the transform of the time to reach a positive level has a particularly simple form. Inverting it reveals a density with a long right tail and makes interval probabilities explicit.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) carrying standard Brownian motion \((B_t)_{t\ge0}\), adapted to its Brownian filtration \((\mathcal{F}_t)_{t\ge0}\). Fix a level \(a>0\), and define the first-passage time \(T_a=\inf\{t\ge0:B_t=a\}\), with \(\inf\varnothing=\infty\). This is a stopping time. The earlier result on the one-sided hitting-time transform gives, for \(s>0\), \(\mathbb{E}[e^{-sT_a}]=e^{-a\sqrt{2s}}\), using the convention \(e^{-s\infty}=0\).

The goal is to recover the probability law from this transform, rather than merely to manipulate the transform as an algebraic expression. We will verify a candidate density by direct integration and then use uniqueness of Laplace transforms. The restriction \(a>0\) matters: if \(a=0\), then \(T_0=0\) and its law is a point mass, not a density of the form derived below.

Verifying the Inversion

We first establish the integral needed to check the transform. For \(\beta>0\) and \(\gamma\ge0\), the identity is $$ \int_0^\infty t^{-3/2}e^{-\beta/t-\gamma t}\,dt =\frac{\sqrt{\pi}}{\sqrt{\beta}}e^{-2\sqrt{\beta\gamma}}. $$ Here is a derivation. Substitute \(x=\sqrt{\beta/t}\), so \(t=\beta/x^2\) and \(dt=-2\beta x^{-3}\,dx\). The integral becomes $$ \frac{2}{\sqrt{\beta}}\int_0^\infty e^{-x^2-\beta\gamma/x^2}\,dx. $$ For \(c\ge0\), let \(J(c)=\int_0^\infty e^{-x^2-c^2/x^2}\,dx\). If \(c>0\), differentiation under the integral gives \(J'(c)=-2c\int_0^\infty x^{-2}e^{-x^2-c^2/x^2}\,dx\). In this last integral, the substitution \(y=c/x\) gives \(\int_0^\infty x^{-2}e^{-x^2-c^2/x^2}\,dx=J(c)/c\). Therefore \(J'(c)=-2J(c)\). Also, dominated convergence, with dominating function \(e^{-x^2}\), gives \(J(c)\to J(0)=\sqrt{\pi}/2\) as \(c\downarrow0\). Solving the differential equation yields \(J(c)=(\sqrt{\pi}/2)e^{-2c}\), which proves the identity upon taking \(c=\sqrt{\beta\gamma}\).

Proof. Set \(\beta=a^2/2>0\) and \(\gamma=s>0\) in the identity. The Laplace transform of the proposed density is $$ \begin{aligned} \int_0^\infty e^{-st}f_a(t)\,dt &=\frac{a}{\sqrt{2\pi}}\int_0^\infty t^{-3/2} e^{-a^2/(2t)-st}\,dt\\ &=\frac{a}{\sqrt{2\pi}}\frac{\sqrt{\pi}}{a/\sqrt{2}}\, e^{-2\sqrt{(a^2/2)s}}\\ &=e^{-a\sqrt{2s}}. \end{aligned} $$ The prefactor in the second line equals \(1\), since \(\frac{a}{\sqrt{2\pi}}\frac{\sqrt{\pi}\sqrt{2}}{a}=1\). The same integral identity at \(s=0\) shows that \(\int_0^\infty f_a(t)\,dt=1\), so the proposed function is a probability density. Its transform agrees with that of \(T_a\). By the standard uniqueness theorem for Laplace transforms of finite measures on \([0,\infty)\), two such measures with equal transforms for all \(s>0\) are equal; this standard measure-theoretic uniqueness result is used here without proof. Thus \(T_a\) has density \(f_a\). In particular, its law has no atom at any finite time. \(\square\)

The normalization also shows that \(T_a<\infty\) almost surely. Equivalently, taking \(s\downarrow0\) in the given transform gives \(\mathbb{P}(T_a<\infty)=1\): the random variables \(e^{-sT_a}\) increase to \(\mathbf{1}_{\{T_a<\infty\}}\), and their expectations tend to \(1\). Both checks are consistent with the recovered density.

The Distribution Function and Interval Probabilities

To obtain the distribution function, integrate the density up to \(t>0\). Use \(z=a/\sqrt{u}\), so \(u=a^2/z^2\). As \(u\) increases from \(0\) to \(t\), \(z\) decreases from \(\infty\) to \(a/\sqrt{t}\), and \(du=-2a^2z^{-3}\,dz\). Substitution gives $$ \begin{aligned} \mathbb{P}(T_a\le t) &=\int_0^t \frac{a}{\sqrt{2\pi}\,u^{3/2}} e^{-a^2/(2u)}\,du\\ &=\sqrt{\frac{2}{\pi}}\int_{a/\sqrt{t}}^\infty e^{-z^2/2}\,dz\\ &=2\left[1-\Phi\!\left(\frac{a}{\sqrt{t}}\right)\right]. \end{aligned} $$ For the second line, \(u^{-3/2}=z^3/a^3\), and multiplying by the absolute differential \(2a^2z^{-3}\,dz\) gives the factor \(\sqrt{2/\pi}\). The last line follows from the symmetry of the standard normal density. At \(t=0\), \(T_a>0\) almost surely because Brownian paths are continuous and \(a>0\), so \(F_a(0)=0\).

This endpoint statement follows because a density assigns probability zero to a single point: for any fixed \(r>0\), \(\mathbb{P}(T_a=r)=\int_{\{r\}}f_a(t)\,dt=0\). The interval formula is therefore valid whether the event is written with strict or non-strict inequalities at either positive endpoint. The endpoint \(0\) is also harmless here, since the process starts below \(a\) and \(T_a>0\) almost surely.

Interpretation and a Common Error

The factor \(t^{-3/2}\) is important: first passage is not distributed like the Brownian position at a fixed time. The density is small near zero because the exponential term suppresses very early hits, while for large \(t\) its leading order is proportional to \(t^{-3/2}\). In particular, the hitting time is finite almost surely but has a heavy enough tail that its mean is infinite. This last claim follows directly from the density: for \(t\ge a^2\), \(e^{-a^2/(2t)}\ge e^{-1/2}\), so \(t f_a(t)\ge (a e^{-1/2}/\sqrt{2\pi})t^{-1/2}\), whose integral over \([a^2,\infty)\) diverges. This is consistent with finite hitting probability one and does not contradict it.

Common mistake: omitting the factor \(1/t\). A tempting but incorrect expression is \(g(t)=\frac{a}{\sqrt{2\pi t}}e^{-a^2/(2t)}\). It resembles a Gaussian density in the position variable, but it is not a probability density in time. For \(t\ge a^2\), \(e^{-a^2/(2t)}\ge e^{-1/2}\), so \(g(t)\ge \frac{ae^{-1/2}}{\sqrt{2\pi}}t^{-1/2}\). Consequently \(\int_{a^2}^\infty g(t)\,dt=\infty\), whereas a density must integrate to \(1\). The correct first-passage density contains \(t^{-3/2}\), and its Laplace transform, checked above, is precisely \(e^{-a\sqrt{2s}}\).

For a practical calculation, first specify the level and the interval, then evaluate \(a/\sqrt{u}\) and \(a/\sqrt{v}\) separately before applying the interval rule. Keeping the order straight matters: the lower time endpoint produces the larger normal argument, so the formula takes \(\Phi(a/\sqrt{u})-\Phi(a/\sqrt{v})\), a nonnegative difference. This method gives a probability for a finite time window without needing to simulate paths or solve a new boundary-value problem.