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 $(B_t)_{t \geq 0}$?

Question

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.

Answer 1

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.