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For standard Brownian motion started at zero, the strict first-passage times form a subordinator in the level parameter $a$. The Strong Markov property 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.

For each deterministic $a>0$, the Brownian reflection principle givesso $H_a$ is finite almost surely; $H_0=0$. Here $\Phi$ is the standard normal cumulative distribution function. The same reflection formula at levels decreasing to zero gives $\mathbb{P}({\sup }_{s\le \epsilon }{X}_s\le 0)=0$ for every $\epsilon >0$. Taking a countable intersection over $\epsilon =1/n$ shows that Brownian motion exceeds zero arbitrarily soon after its start.

By the Strong Markov property at the finite stopping time $H_a$, the process $X_{H_a+t}-a$ is fresh standard Brownian motion. It therefore exceeds zero arbitrarily soon. Path continuity gives $S_a\ge H_a$, and the preceding fact gives the reverse inequality. ConsequentlyThe exceptional event may depend on $a$; this does not imply simultaneous equality for all levels.

To exhibit the difference, set $A={\max }_{0\le t\le 1}{X}_t$, a random level. The reflection formula gives $A>0$ almost surely. Also $A>X_1$ almost surely: if $X_1$ were the maximum, the reversed increment process $Y_s=X_1-X_{1-s}$ would be nonnegative throughout $[0,1]$. This process has the law of standard Brownian motion, whose probability of staying nonnegative from zero is zero by reflection and symmetry. The level $A$ is attained at some time strictly before one, so $H_A<1$. It is not exceeded by time one, and continuity with $X_1<A$ gives a further positive interval before it can be exceeded. ThusAll these passage times exist simultaneously on the probability-one event where the Brownian path is continuous and unbounded above, obtained by intersecting the finite-hitting events for positive integer levels.

The illustration shows the pathwise mechanism. The Brownian running maximum stays flat during an excursion below an attained record. The non-strict inverse chooses the beginning of that plateau, while the strict generalized inverse of a nondecreasing function chooses its end. The curve is a schematic of this mechanism, not a claim of piecewise linear Brownian paths. The fixed-level equalities give the modification of a stochastic process relation, while the random-level discrepancy rules out indistinguishability of stochastic processes. This is fixed-level versus simultaneous Brownian passage-time equality.

Let $A={\max }_{0\le t\le 1}{B}_t$, a random level. The Brownian reflection principle gives $A\stackrel{d}{=}|B_1|$, so $A>0$ almost surely. Also $A-B_1\stackrel{d}{=}|B_1|$: this follows by applying reflection to the reversed increment process $(B_1-B_{1-s}{)}_{0\le s\le 1}$, 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:Thus 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.