# Supremum of Brownian Motion

We define the supremum process of Brownian Motion by:

$S_t:=\sup_{0\leq s\leq t}B_s.$

Here are two facts about Brownian Motion. Firstly, the Reflection Principle:

$\mathbb{P}(S_t\geq b,B_t\leq a)=\mathbb{P}(B_t\geq 2b-a),$

which we motivate by ‘stopping’ at time $S_t$, and using the SMP for Brownian Motion, even though it isn’t a stopping time. By setting a=b, we get:

$\mathbb{P}(S_t\geq b)=\mathbb{P}(S_t\geq b,B_t\leq b)+\mathbb{P}(B_t\geq b)=2\mathbb{P}(B_t\geq b)=\mathbb{P}(|B|\geq b),$

and conclude that

$S_t\stackrel{d}{=}|B_t|\quad\text{for each }t\geq 0.$

The second fact comes from the decomposition of BM into local times and excursions:

$(S_t,S_t-B_t)_{t\geq 0}\stackrel{d}{=}(L_t,|B_t|)_{t\geq 0},$

where L is the local time process at 0, and this equality in distribution holds for the processes. See the previous post on excursion theory for explanation of what local times mean.

In particular, combining these two facts gives:

$S_t\stackrel{d}{=}S_t-B_t\quad\text{for every }t\geq 0.$

I thought that was rather surprising, and wanted to think of a straightforward reason why this should be true. I think the following works:

Brownian motion is time-reversible. In particular, as processes, we have

$(B_s)_{s\geq 0}\stackrel{d}{=}(B_{t-s}-B_t)_{s\geq 0}$

$\Rightarrow \sup_{0\leq r\leq t}B_r\stackrel{d}{=}\sup_{0\leq r\leq t}(B_{t-r}-B_t)$

$\Rightarrow S_t\stackrel{d}{=}S_t-B_t.$

$S_t, |B_t|, \text{ and } S_t-B_t$