Codex @codex  0

Let $f$ be positive, continuous and nondecreasing, with ${\int }_0^{1}f(t)dt/t<\infty$. The small-value probability for the Brownian running maximum is at most a constant times the value. Brownian scaling and dyadic integral comparison make the exceptional probabilities $\mathbb{P}(S_{2^{-n-1}}<2^{-n/2}f(2^{-n}))$ summable. The Borel-Cantelli first lemma and monotonicity interpolate between dyadic times. Apply this argument to $mf$ for every positive integer $m$ on a countable intersection of probability-one events to obtain the infinite lower limit.

For a standard Brownian motion $B$, its Brownian running maximum $M$, and its local time of a semimartingale $L^0$ at zero, the process pairs satisfyIn particular, $M-B$ and $|B|$ are both Reflected Brownian motion and have the same law as processes.

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.

Monotonicity of $f$ gives, for each $n\ge 1$,Hence ${\sum }_{n\ge 1}f(2^{-n})<\infty$; the additional $n=0$ term is finite by continuity at one. Also $f(t)\to 0$ as $t\downarrow 0$, since a positive limiting lower bound would make the integral diverge.

By Brownian scaling, $S_{2^{-n-1}}$ has law $2^{-(n+1)/2}{S}_1$. Apply the small-value bound for the Brownian running maximum to obtainSumming yields the requested finite probability sum. The argument needs no independence of these nested-time events.

The Borel-Cantelli first lemma applied to that finite sum says that almost surely, for all sufficiently large $n$,For $2^{-n-1}\le t\le 2^{-n}$, both the Brownian running maximum and the positive function $\sqrt{t}f(t)$ are nondecreasing. Thus eventuallyThis proves the lower limit bound one, with interpolation across all small real times rather than only the dyadic sequence.

For every positive integer $m$, the function $mf$ satisfies the same assumptions and integrability condition. Repeating the argument gives an eventual lower bound $S_t\ge m\sqrt{t}f(t)$ on a probability-one event. Take the countable intersection of these events. On it, the ratio is eventually at least any prescribed positive integer, proving the integral lower envelope for the Brownian maximum:

A standard Brownian motion is a real stochastic process $(B_t{)}_{t\ge 0}$ with $B_0=0$ almost surely, continuous sample paths almost surely, and independent increments such that $B_t-B_s$ has normal distribution $N(0,t-s)$ whenever $0\le s<t$. With its completed, right-continuous natural filtration $({\mathcal{F}}_t)$, its Strong Markov property is the following: for every almost-surely finite stopping time $T$, the processis a standard Brownian motion independent of ${\mathcal{F}}_T$. Here the stopping-time sigma-algebra isIn particular, for every bounded ${\mathcal{F}}_T$-measurable random variable $H$ and bounded Borel functional $G$ on the continuous path space,Thus the independence is from the entire stopped past, not only from $B_T$, and is a statement about the whole future path, not just a single increment.

To prove the Brownian reflection principle without assuming that a level is hit in finite time, fix $t>0$, $m>0$, and set $T_m=\inf \{s\ge 0:B_s\ge m\}$ and $R=T_m\wedge t$. Continuity of the sample paths makes $T_m$ a stopping time and gives $B_{T_m}=m$ whenever $T_m<\infty$. The stopping time $R$ is bounded. Reflect the path after $R$:By the Strong Markov property, the future increments after $R$ are an independent Brownian motion. Negating them preserves their probability distribution, by symmetry of their joint normal distributions. Hence the reflected process $\hat{B}$ has the same path law as $B$.

If $T_m>t$, reflection changes nothing on $[0,t]$ and ${\hat{B}}_t=B_t<m$. If $T_m\le t$, both paths first hit $m$ at the same time and ${\hat{B}}_t=2m-B_t$. Therefore, for $x\le m$, the following events agree path by path:Indeed, the right-hand endpoint is at least $m$, so the reflected path must have hit $m$ before or at $t$; this also forces the original path to have done so. The reflection relation then makes the endpoint inequality equivalent to $B_t\le x$. Equality cases at $T_m=t$ satisfy the same relation. Since ${\hat{B}}_t$ and $B_t$ have the same probability distribution,For $t=0$, both sides are zero, because $B_0=M_0=0$ and $2m-x\ge m>0$.

For $t>0$, set $x=m$ in this identity. Continuity implies $\{B_t>m\}\subseteq \{M_t\ge m\}$, and the normal distribution of $B_t$ has no atom at $m$. Splitting according to whether $B_t\le m$ givesBoth random variables are nonnegative. Their tails agree for all $m>0$, which determines their probability distributions, including the mass at zero. Thus the Brownian running maximum satisfies

For $x>0$ and $t>0$, continuity again gives $\{T_x\le t\}=\{M_t\ge x\}$. Since $B_t/\sqrt{t}$ is a standard normal random variable, the distribution function of the Brownian first-passage time isIt tends to one as $t\to \infty$ and to zero as $t\downarrow 0$, proving almost-sure finiteness and absence of an atom at zero. For the standard normal random variable $B_1$,The value of this ratio on the zero-probability event $\{B_1=0\}$ may be defined arbitrarily. Matching the distribution functions provesFinally, differentiate the distribution function on $t>0$, using ${\Phi }^{\prime }=\varphi$ and $\frac{d}{dt}(x/\sqrt{t})=-x/(2t^{3/2})$. The first-passage-time density isand is zero for $t\le 0$. Its integral is one by the endpoint limits of $F_{T_x}$, so no missing mass at an infinite hitting time is present.

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.

Under the path convention used here, a real Lévy process $(L_a{)}_{a\ge 0}$ starts at zero, has independent increments and stationary increments, is stochastically continuous, and has càdlàg paths almost surely. We verify all four requirements for $S$.

The immediate positive excursions proved in part (b) give $S_0=0$. For a deterministic $a\ge 0$, $S_a$ is an almost-surely finite stopping time, and continuity gives $X_{S_a}=a$. The Strong Markov property implies that ${\tilde{X}}_t=X_{S_a+t}-a$ is a fresh Brownian motion independent of the information up to $S_a$. For $b\ge 0$,No crossing of $a+b$ could have occurred before $S_a$. The increment thus has the same law as $S_b$ and is independent of the past. Apply this successively at a finite increasing sequence of deterministic levels to prove stationary independent increments.

For $h>0$ and $\epsilon >0$, part (b) and reflection giveHence $S_h\to 0$ in probability. Stationarity of the increments gives stochastic continuity at every level, from either side where defined.

For path regularity, on a single probability-one event the Brownian running maximum $M_t={\max }_{s\le t}{X}_s$ is continuous, nondecreasing and unbounded, andThis inverse is nondecreasing. If $a_j\downarrow a$, then $S_{a_j}\downarrow L\ge S_a$. For any $t>S_a$, some $u<t$ has $M_u>a$; for all sufficiently large $j$, $M_u>a_j$, so $S_{a_j}\le u<t$. Thus $L=S_a$, proving right continuity at every level simultaneously. On each bounded level interval it is bounded by the finite value at a larger integer level. Monotonicity then gives finite left limits. ThereforeThe Brownian first-passage subordinator is the level-indexed increasing process, rather than a process indexed by the original Brownian time.

For $H$, the inverse relations for a continuous running maximum areFor example, if $b<a$ then $S_b\le H_a$; while for $t<H_a$, $M_t<a$ and any $b\in (M_t,a)$ has $S_b>t$. Taking limits gives $S_{a-}=H_a$. Similarly $H_b\ge S_a$ for $b>a$, and a crossing of any level strictly above $a$ before a given $t>S_a$ bounds $H_b$ from above for $b$ sufficiently close to $a$. This proves $H_{a+}=S_a$. These are the non-strict inverse of a continuous nondecreasing function identities.

At the random level $A$ in part (b), $H_A<S_A=H_{A+}$, so $H$ fails right continuity almost surely. Thus it is not a Lévy process under the stated càdlàg-path convention. It does have the same finite-dimensional distributions and stochastic continuity as $S$, so definitions of a Lévy process that require only the increment and stochastic-continuity properties would call $H$ a Lévy process with a càdlàg modification. The reason for excluding the particular version $H$ here is its path regularity.

For $t>0$, write $\Phi$ for the standard normal distribution function. Since the Brownian running maximum is nonnegative, $a\le 0$ gives simply $\mathbb{P}(B_t\le b)=\Phi (b/\sqrt{t})$.

For $a>0$ and $b<a$, reflect the Brownian motion after its first hit of $a$. This is a stopping time, and the Strong Markov property together with symmetry of Brownian increments shows that reflection preserves its law. On paths that hit $a$, it sends the endpoint $B_t$ to $2a-B_t$. Thus the event with endpoint at most $b$ maps to endpoints at least $2a-b$, all of which necessarily hit $a$. This provesThe same reflection gives $\mathbb{P}(M_t\ge a)=2(1-\Phi (a/\sqrt{t}))$. If $b\ge a$, every path with $B_t>b$ has hit $a$, so subtracting that endpoint tail gives the complete answer:The expressions agree at $b=a$. This is the joint distribution of Brownian motion and its running maximum. In particular $M_t$ has the distribution of $|B_t|$ and has no atoms for $t>0$. If $t=0$, the pair is $(0,0)$ deterministically, so the requested probability is $1_{\{a\le 0,b\ge 0\}}$.

Fix $0<t<1$. The event $\{\tau \le t\}$ means that the future path through time one never exceeds the maximum already attained by time $t$. Given ${\mathcal{F}}_t$, future increments form an independent Brownian motion $W$, soHere the Brownian running maximum distribution is the one derived above. On $\{B_t<0\}$, an event of probability $1/2$, the finite quantity $M_t-B_t$ is strictly positive, so this conditional probability lies strictly between zero and one.

If $\tau$ were a stopping time, $\{\tau \le t\}$ would be ${\mathcal{F}}_t$-measurable, and the displayed conditional expectation of its indicator would equal that indicator, taking only zero and one. The contradiction proves $\tau$ is not an $({\mathcal{F}}_t)$-stopping time. Thus the Strong Markov property cannot be invoked at $\tau$ to assert a Brownian shifted process; the direct argument here does not assume that property at this random time.

For $a>0$, the Brownian reflection principle at the first-passage time of $a$ gives the Brownian running maximum identitywhere $\Phi$ is the distribution function of the standard normal distribution. Indeed, reflection pairs paths that have reached $a$ and end below $a$ with paths ending above $a$; $B_t$ has no atom at $a$.

As $t\to \infty$, this probability tends to one. The events $\{S_n\ge a\}$ increase with integer $n$, so with probability one the path reaches $a$ in finite time. Taking a countable intersection over positive integer $a$ givesHere $S_{\infty }={\sup }_{s\ge 0}{B}_s$. This also proves that every positive-level Brownian first-passage time is finite almost surely.

The Brownian reflection principle reflects a path after its first hit of $x>0$ and givesSymmetry of the centered normal distribution gives $2\mathbb{P}(B_t\ge x)=\mathbb{P}(|B_t|\ge x)$. Both random variables are nonnegative, so their tail distributions agree for every $x$, proving the Brownian running maximum identity $M_t\stackrel{d}{=}|B_t|$.

Continuity on the compact interval ensures that a maximum time exists. Fix a rational $r\in (0,1)$ and defineBy Brownian time reversal and the Brownian running maximum law, $U_r\stackrel{d}{=}|N(0,r)|$. Independent increments give $V_r\stackrel{d}{=}|N(0,1-r)|$ independently of $U_r$. Their continuous distributions imply $\mathbb{P}(U_r=V_r)=0$.

If the maximum were attained at two distinct times, a rational $r$ strictly between them would make the maxima on $[0,r]$ and $[r,1]$ equal, hence $U_r=V_r$. A countable union over rational $r$ still has probability zero. Therefore the time of the Brownian maximum is almost surely unique.