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Distribution of a one-sided Brownian hitting time

Statement

Assume the Axiom of Choice, let B be a standard Brownian motion Brownian motion. Use the everywhere-continuous, zero-start representative of Law of the Brownian maximum: replace paths by zero outside a measurable probability-one event of continuity and zero start, retaining the notation B. Let a>0 and τa:=inf{t0:Bt=a}, with inf=+. This is a measurable [0,]-valued hitting time for this representative; its distribution does not depend on the chosen full-measure event. With Φ the standard normal distribution function Standard normal and normal laws Cumulative distribution function of a real random variable, P(τat)=2(1Φ ⁣(at))(t>0), and on t>0 the law of τa has the density g(t)=a(2πt3)1/2exp ⁣(a22t). Moreover limtP(τat)=1, so τa is finite almost surely and there is no mass at infinity, and P(τa=0)=0.

Facts & Assumptions

Given: AC, a standard Brownian motion B in this everywhere-continuous zero-start representative, a>0 and t>0.

[F1]

For the representative in the statement, Mt=sup[0,t]B is a finite measurable random variable and P(Mtx)=2Φ(x/t)1 for x0 (Law of the Brownian maximum). The closed-set hitting-time lemma applies to an everywhere-continuous Brownian process with its own natural filtration (Brownian closed-set hitting times are stopping times). Brownian motion supplies a measurable full-measure continuity and zero-start event (Brownian motion).

[F2]

Limits of the standard normal distribution function: limx+Φ(x)=1, Φ(0)=1/2 and Φ is continuous, and Φ(v)Φ(u)=uvφ(x)dx for u<v; φ(u)=eu2/2/2π. Standard normal and normal laws Cumulative distribution function of a real random variable The standard normal density has total mass one

[F4]

A probability measure on R is determined by its distribution function; countable choice, which AC supplies, is used there. Probability laws correspond to distribution functions The Axiom of Countable Choice (ACω) The Axiom of Choice

Proof

technique · direct
1.1

Fix the measurable full-measure event A specified in the statement. Replacing the original path by zero on Ac preserves every finite-dimensional law and makes every path continuous with B0=0. Each coordinate remains measurable because A is measurable. Two such choices agree on the intersection of their events, so the resulting measurable hitting times agree there and have the same distribution. By [F1] applied to the closed singleton {a}, τa is a stopping time for the chosen process's own raw natural filtration, hence an extended nonnegative measurable random variable. No stopping-time claim for the original raw filtration is used.

F1given
1.2

Define g(s)=a(2πs3)1/2ea2/(2s) for s>0. For 0<ε<t, the substitution u=a/s on [ε,t], whose derivative a/(2s3/2) is continuous and 2φ is continuous, gives εtg(s)ds=a/ta/ε2φ(u)du=2(Φ(a/ε)Φ(a/t)) by oriented substitution, the compact Riemann/Lebesgue bridge, and the density-integral identity for increments of Φ.

F2F3
2.1

For t>0, if τat, the first hit is attained by continuity (as in the closed-set hitting lemma), so Mta. Conversely Mta gives a time s[0,t] with Bs=Mt by [F5]; since B0=0<aBs, the intermediate value theorem gives a hit by time s. Hence {τat}={Mta} as exact measurable events for this representative. Continuity at zero also gives τa>0 on every path, since B0=0<a.

F1F5step 1.1
3.1

The normal CDF obeys Φ(v)Φ(u)vu/2π, by its density bound, hence is continuous. Symmetry and total mass one give Φ(0)=1/2; monotone convergence of density integrals gives Φ(x)1 as x. Put xn=a(11/n) for n1. The measurable events {Mtxn} increase to {Mt<a}, so [F3] and [F1] give P(Mt<a)=limn(2Φ(xn/t)1)=2Φ(a/t)1=P(Mta). Thus P(Mt=a)=0, and step 2.1 yields P(τat)=2(1Φ(a/t)). Taking integer t and monotone convergence of the events {τat} gives P(τa<)=1.

F1F2F3step 2.1
4.1

Let ε=t/n in step 1.2 and let integers n2 tend to infinity. The nonnegative integrals increase to 0tg(s)ds, and Φ(a/ε)1, so 0tg(s)ds=2(1Φ(a/t))=P(τat). Letting integer t now gives 0g=1.

F2F3step 3.1step 1.2
5.1

Extend g by zero on (,0]. It is nonnegative Borel measurable, and [F5] and step 4.1 make its density measure a Borel probability measure on R. To use [F4] with a real random variable, replace τa= by the value 1 on its measurable null event, obtaining τ~a. This leaves every finite-time distribution probability unchanged, and τ~a>0 by step 2.1. The density measure and τ~a have CDF zero for nonpositive arguments, and the same CDF at every positive argument by step 4.1. Thus [F4] identifies the laws. In particular the original extended hitting time has density g on (0,), no atom there or at zero, and no mass at infinity.

F4F5step 2.1step 3.1step 4.1
6.1

The parameter a>0 and compact substitution bounds 0<ε<t ensure every denominator is positive. At t=0, step 2.1 gives P(τa=0)=0. The infinity limit and total density mass were proved in steps 3.1 and 4.1. AC supplies the Countable Choice hypotheses of both the compact integration bridge and [F4], and the Brownian and hitting-time suppliers. The event equality uses the declared continuous representative throughout.

F1F3F4step 2.1step 3.1step 4.1step 5.1

The proof combines the Brownian maximum law with an exact continuous-path hitting identity. It computes the density integral by compact substitution, the Riemann/Lebesgue bridge and monotone limits, then identifies probability laws through their CDFs on all real arguments.

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