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For a continuous unbounded nondecreasing function $f$ on the nonnegative half-line and $a>f(0)$, the non-strict inverse $h$ is left-continuous in its level argument. It equals the left limit of the strict generalized inverse of a nondecreasing function $G(a)=\inf \{t:f(t)>a\}$. Indeed $G(b)\le h(a)$ for $b<a$, while for every $t<h(a)$, $f(t)<a$ and choosing $b\in (f(t),a)$ gives $G(b)\ge t$. Similarly $h(a+)=G(a)$. A flat part of $f$ at level $a$ gives a discrepancy between the two inverses: $h(a)$ selects the beginning of the flat part and $G(a)$ its end. For a Brownian running maximum, these are the non-strict and strict passage times.

Write $\psi (s)=\log \mathbb{E}{e}^{sX}$ for the cumulant-generating function, and let $D_{+}=\{s>0:\psi (s)<\infty \}$. This set is nonempty. In fact, finiteness of the moment-generating function at some $\eta >0$ and $-\eta$ givesso $X$ is an integrable random variable. Put $m=\mathbb{E}X$.

For the upper tail, use the following one-sided convex conjugate and its strict generalized inverse of a nondecreasing function:Here $\psi (0)=0$, and $I$ is nondecreasing because every slope in its defining supremum is nonnegative. For $x\ge m$, negative $s$ cannot increase the conjugate: $\psi (s)\ge sm$ gives $sx-\psi (s)\le s(x-m)\le 0$, and $s=0$ already gives zero. Thus $I$ agrees with the full Legendre transform of a cumulant-generating function on its upper-tail branch. Using a strict generalized inverse of a nondecreasing function handles constant random variables and bounded supports without assuming a strictly increasing, everywhere finite conjugate. Directly from the definition,This threshold is finite: any fixed $s\in D_{+}$ gives a finite upper bound, while $e^u\ge 1+u$ gives $\mathbb{E}{e}^{s(X-m)}\ge 1$ and hence $\psi (s)\ge sm$, providing the lower bound $m$.

We now derive the needed Chernoff bound. For $s\in D_{+}$ and any real $x$, the pointwise inequality $e^{sX}\ge e^{sx}{1}_{\{X>x\}}$ impliesLet $q={\psi }_X^{\ast -1}(t)$. For every $\epsilon >0$, the infimum formula supplies an $s\in D_{+}$ such that $(t+\psi (s))/s<q+\epsilon$. Thus $\mathbb{P}(X>q+\epsilon )\le e^{-t}$. As $\epsilon \downarrow 0$, these events increase to $\{X>q\}$; continuity from below of a measure provesThe strict event and the strict generalized inverse of a nondecreasing function avoid any assumption that the optimizing $s$ exists. For example, if $X=m$ almost surely, the threshold is $m$ for every $t>0$, and the probability is zero.

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.

Use the usual completed right-continuous Brownian filtration. The strict passage process $(T_a{)}_{a\ge 0}$ is finite at every level on one probability-one event: finiteness at all integer levels from (a) suffices by monotonicity. Also $T_0=0$ almost surely by (b).

For each fixed $a$, $T_a$ is a stopping time with $B_{T_a}=a$. The Strong Markov property shows that the passage times above this level depend on a new independent standard Brownian motion. Thus for $0\le a<b$, $T_b-T_a$ is independent of the past at $T_a$ and has the same law as $T_{b-a}$. Iteration gives independent increments and stationary increments in the level parameter. From (a) and fixed-level equality,For $\lambda ,\eta >0$, this also proves continuity in probability at zero, sinceFinally, with $M_t={\sup }_{s\le t}{B}_s$, we have $T_a=\inf \{t:M_t>a\}$. This strict generalized inverse of a nondecreasing function is right-continuous: if $a_n\downarrow a$, then for any $t>T_a$ we have $M_t>a$, and eventually $M_t>a_n$, which forces $T_{a_n}\le t$. Monotonicity gives the reverse bound. Monotonicity and local finiteness also give finite left limits. Hence $T$ is the Brownian first-passage subordinator, with càdlàg paths.

Now condition on this clock, which is independent of $W$. For a deterministic partition $0=a_0<a_1<\cdots <a_k$, write $\Delta T_j=T_{a_j}-T_{a_{j-1}}$ and $\Delta X_j=W_{T_{a_j}}-W_{T_{a_{j-1}}}$. Conditional on the clock these are independent centered Gaussian increments, with respective variances $\Delta T_j$. ConsequentlyFactorization and dependence only on interval lengths prove independent increments and stationary increments for $X$. For small $h$, $T_h\to 0$ in probability, and independence and continuity of $W$ imply $W_{T_h}\to 0$ in probability. For example, bound its deviation probability by $\mathbb{P}(T_h>\eta )+\mathbb{P}({\sup }_{t\le \eta }|W_t|>\epsilon )$ and then let $h\downarrow 0$ and $\eta \downarrow 0$. Stationary increments give stochastic continuity at every deterministic level. Composition of the continuous path of $W$ with the nondecreasing càdlàg clock gives càdlàg paths for $X$, and $X_0=0$. ThusThis is subordination of a Lévy process. Using strict passage times ensures the required right-continuous path choice, despite their fixed-level equality with the non-strict times.