= Solution
For standard <Brownian motion> $B$ and $h>0$, let $T_h$ be its first hitting time of $h$. Define the reflected path by
$$
\widehat B_t=\begin{cases}B_t,&t\leq T_h,\\2h-B_t,&t>T_h,\end{cases}
$$
leaving a path unchanged if $T_h=\infty$. The \b[<Brownian reflection principle>] says that $\widehat B$ has the same path <probability distribution> as $B$. In its endpoint form, for $t>0$ and every Borel set $A\subset(-\infty,h)$,
$$
\mathbb P(T_h\leq t,\ B_t\in A)=\mathbb P(B_t\in2h-A),
\qquad 2h-A=\{2h-y:y\in A\}.
$$
Here is a direct proof of preservation of the path law. For a fixed integer horizon $L$, let $T^{(m)}=2^{-m}\lceil2^mT_h\rceil\wedge L$, with an infinite first term interpreted as infinity. This is a <stopping time> taking finitely many deterministic grid values. On $\{T^{(m)}=j2^{-m}\}$, the history determining that event is measurable at $j2^{-m}$, and subsequent <Brownian motion> increments are independent of that history. Those increments and their negatives have the same joint law, by symmetry of centered <normal distributions> and <independent increments>. Thus reflecting after $T^{(m)}$ around $B_{T^{(m)}}$ preserves the full path law. For any fixed horizon $t<L$, the reflected paths converge uniformly on $[0,t]$ to the displayed reflected path, by <continuity> as $T^{(m)}$ decreases to $T_h\wedge L$. This includes $T_h=\infty$, since reflection after $L$ has no effect before $t$. Passing to all finite-dimensional <probability distributions> proves the assertion; no prior recurrence claim is needed. Equivalently, the <Strong Markov property> gives the same proof by symmetry of the fresh increments after the hit. Reflection is an involution preserving the hitting event and mapping the indicated endpoint $y$ to $2h-y$. An endpoint above $h$ forces a hit by <continuity>; this proves the endpoint identity, rather than assuming it.
For $S_t=\sup_{0\leq s\leq t}B_s$, split $\{T_h\leq t\}$ according to whether $B_t$ is above or below $h$. Since $B_t$ has no atom at $h$, reflection gives
$$
\mathbb P(S_t\geq h)=\mathbb P(B_t\geq h)+\mathbb P(T_h\leq t,\ B_t<h)
=2\mathbb P(B_t\geq h)=\mathbb P(|B_t|\geq h).
$$
Both variables are nonnegative, so their positive tails identify the law:
$$
\boxed{S_t\stackrel d=|B_t|.}
$$
For $t>0$ this is the <half-normal distribution> of scale $\sqrt t$; at $t=0$ both variables vanish.
For the two separated intervals, put
$$
A_0=\sup_{a\leq t\leq b}B_t-B_b,\qquad Z=B_c-B_b,\qquad H=\sup_{0\leq u\leq d-c}(B_{c+u}-B_c).
$$
Here $A_0$ is measurable for $\mathcal F_b$, $Z$ has the <normal distribution> $N(0,c-b)$, and $H$ is a function of increments after $c$. The <independent increments> make these three objects mutually independent. The later absolute maximum is $B_b+Z+H$, so equality of the two maxima means $Z=A_0-H$. Conditional on $(A_0,H)$, this specifies a single value of the nondegenerate <normal distribution> $Z$, and hence has probability zero. Therefore
$$
\boxed{\mathbb P\left(\sup_{a\leq t\leq b}B_t=\sup_{c\leq t\leq d}B_t\right)=0.}
$$
This proves the <atomless maxima on separated Brownian intervals> without incorrectly treating the two absolute maxima as independent.
Back to article page
