# 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 $(B_s)_{s\geq 0}\stackrel{d}{=}(B_{t-s}-B_t)_{s\geq 0}$ $\Rightarrow \sup_{0\leq r\leq t}B_r\stackrel{d}{=}\sup_{0\leq r\leq t}(B_{t-r}-B_t)$ $\Rightarrow S_t\stackrel{d}{=}S_t-B_t.$

## 6 thoughts on “Supremum of Brownian Motion”

• dominicyeo on said:

Thanks for that. I don’t know enough about financial stochastic models to say under what level of generality it might be appropriate to model this stock data as Brownian motion, but yes, if that is a valid assumption, then you can certainly take an expectation to get the result the OP is asking about.

• Justin on said:

It’s true that Brownian motion with a Gaussian law is not the most accurate model for the movement of stock prices, but it would seem to me that for a continuous Markov martingale process based on any fixed, infinitely divisible, symmetric distribution, it would still be true that $S_t, |B_t|, \text{ and } S_t-B_t$
all have the same distribution for a given t, because your arguments of reflection and time reversal would still apply. By all means do enlighten me if I am wrong.

• dominicyeo on said:

Yes, so I think that’s right. If your model is Markov then it has independent increments, fixed distribution means stationary increments, infinitely divisible is the third property required to be a Levy process. If it is a martingale as well, then it has zero drift and continuity plus the decomposition of the Levy-Khintchine formula (see post a few days ago) confirms that in fact the model must be a Brownian motion up to a scale factor (at least if this is one-dimensional).