# Why has $\sup_{s \in (0,t)} B_s$ the same distribution as $\sup_{s \in (0,t)} B_s-B_t$ for a Brownian motion $(B_t)_{t \geq 0}$?

Here on wikipedia is claimed that the process $X_t:=\sup_{s \in (0,t)} B_s-B_t$ is distributed like $\vert B_t \rvert$ where $B_t$ is standard Brownian motion.

On the other hand, it is claimed here in Corollary $6.21$ that $\sup_{s \in (0,t)} B_s$ is distributed like $\vert B_t \rvert.$

So how is it possible that $\sup_{s \in (0,t)} B_s-B_t$ is distributed like $\sup_{s \in (0,t)} B_s.$ There seems to be something wrong with probability.

If you have any further questions, please let me know.

• What, exactly, is the contradiction? Lots of unequal random variables have the same distribution. I think another thing that would bother you is that $B_t-tB_1$ is distributed as $B_{1-t}-(1-t)B_0$. Commented Jan 14, 2018 at 22:30
• @kimchilover but here we have that $X$ and $X+Y$ have the same distribution where $Y$ is non-trivial.
– user505183
Commented Jan 14, 2018 at 22:31
• In general, what's wrong with that? Take for example $X\sim N(0,1)$ and $Y=-2X$. Here $X$ and $Y$ are far from independent, as are $\sup_{s\in(0,t)}B_s$ and $B_t$. Commented Jan 14, 2018 at 22:54

For each fixed $t>0$ the process
$$W_s := B_{t-s}-B_t, \qquad s \in [0,t],$$
defines a Brownian motion $(W_s)_{s \in [0,t]}$. In particular, $(B_s)_{s \in [0,t]}$ equals in distribution $(W_s)_{s \in [0,t]}$, and so
$$\sup_{s \in [0,t]} B_s \stackrel{d}{=} \sup_{s \in [0,t]} W_s \stackrel{\text{def}}{=} \sup_{s \in [0,t]} B_s-B_t.$$
This shows that the random variables $\sup_{s \in [0,t]} B_s$ and $\sup_{s \in [0,t]} B_s-B_t$ have the same distribution.