# Fundamental Contradiction with Brownian motion

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.

• This is indeed the case: for a fixed $t > 0$ both random variables have the same distribution (and are dependent). Furthermore, $X_t$ and $|B_t|$ are equal in law. See any textbook on BM for details. Or ask a question on Math.SE if still needing help. – Mateusz Kwaśnicki Jan 14 '18 at 22:46

To conclude that something here is wrong, you should be able to indicate a contradiction. Of course, you won't be able to do this, since the two random variables, $M_t:=\sup_{s \in [0,t]} B_s$ and $M_t-B_t$, have indeed the same distribution, for each real $t>0$.
To quickly see this, note that $(\tilde B_s)_{s\in[0,t]}:=(B_{t-s}-B_t)_{s\in[0,t]}$ is a standard Brownian motion on the interval $[0,t]$ and hence equals $(B_s)_{s\in[0,t]}$ in distribution. So, $\tilde M_t:=\sup_{s \in [0,t]} \tilde B_s=M_t-B_t$ equals $M_t$ in distribution.
Perhaps it would be a bit easier to comprehend the following discrete version of the same result: Let $X_1,X_2,\dots$ be iid random variables each taking each of the two values $a,-a$ with probability $1/2$, for some real $a>0$. Let $M_n:=\max_{0\le j\le n}S_j$, where $S_j:=X_1+\dots+X_j$ (as usual, let the sum of any empty family equal $0$). Note that $S_{n-j}-S_n=\tilde S_j:=\tilde X_1+\dots+\tilde X_j$ for each $j=0,\dots,n$, where $\tilde X_i:=-X_{n+1-i}$. Since $(\tilde X_i)_{i=1}^n$ equals $(X_i)_{i=1}^n$ in distribution, we see that $(\tilde S_j)_{j=0}^n$ equals $(S_j)_{j=0}^n$ in distribution, whence \begin{equation} M_n-S_n=\max_{0\le j\le n}(S_j-S_n)=\max_{0\le j\le n}(S_{n-j}-S_n)=\max_{0\le j\le n}\tilde S_j \end{equation} equals $M_n$ in distribution.