{c}
{title2=$T_a=\inf\{t\geq0:B_t>a\}$}

For standard <Brownian motion> started at zero, the strict first-passage times form a <subordinator> in the level parameter $a$. The gives <independent increments> and <stationary increments>; the strict inverse of the continuous running maximum has <càdlàg> paths. The <Brownian first-passage Laplace transform> gives Laplace exponent $\Phi(\lambda)=\sqrt{2\lambda}$, so the process is strictly stable of index $1/2$. Choosing the non-strict hitting times preserves each fixed-level law but generally loses right continuity at random levels, as in the <fixed-level versus simultaneous Brownian passage-time equality>.

Brownian first-passage subordinator

Let $A=\max_{0\leq t\leq1}B_t$, a random level. The <Brownian reflection principle> gives $A\overset d=|B_1|$, so $A>0$ <almost surely>. Also $A-B_1\overset d=|B_1|$: this follows by applying reflection to the reversed increment process $(B_1-B_{1-s})_{0\leq s\leq1}$, which has the law of a standard <Brownian motion>. Thus $B_1<A$ <almost surely> as well.

By continuity and <compactness>, $A$ is attained at some time strictly between zero and one, so $H_A<1$. It is not exceeded anywhere on $[0,1]$; since $B_1<A$, continuity even excludes an exceedance in a positive interval just after time one. Consequently $T_A>1$. Both conclusions hold on a single probability-one <event>, giving the <fixed-level versus simultaneous Brownian passage-time equality> distinction:
$$
\boxed{\mathbb P\bigl(T_a=H_a\text{ for every }a\geq0\bigr)=0.}
$$
Thus \b[the simultaneous assertion is false]. Equality for each deterministic level, and even simultaneously for all rational levels, cannot be extended to all real levels by an uncountable intersection.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 201 / 6 / c / Solution

For standard <Brownian motion>, set $H_a=\inf\{t:B_t\geq a\}$ and $T_a=\inf\{t:B_t>a\}$. The and immediate positive excursions give $T_a=H_a$ <almost surely> for each fixed $a\geq0$. However, these level-indexed processes are not <indistinguishable stochastic processes>. <Almost surely> the random level $A=\sup_{t\leq1}B_t$ is positive and exceeds $B_1$. It is first attained before time one but is not exceeded until after time one, so $H_A<1<T_A$. The simultaneous-equality <event> therefore has probability zero. The strict passage process $T_a$ is the right-continuous choice needed for a <subordinator>.

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 Incoming links: Fixed-level versus simultaneous Brownian passage-time equality