Q:

Law of Sup of Brownian Motion

Let $(B_t)_{t\ge 0}$ denote a brownian motion. 
If $S_t:=\sup_{[0,t]}B_i$ I want the law of $S_t$. The following gives a formula for the law but it is defined in terms if itself: (all equalities below mean equality in distribution)
$(B_i)_i=(B_{t-i}-B_i)_i$, so $\sup_{[0,t]}B_i=\sup_{[0,t]}(B_{t-i}-B_t)$
which means $S_t=S_t-B_t$.
Is there a way to find a formula for the law of $S_t\,$  in which $S_t$ doesn't appear? 

A:

We calculate $P(S_t\geq s)$ by introducing a stopping time 
$\tau=inf\{t>0|B_t>s\}$.
Obviously $P(S_t\geq s)=P(\tau\leq t)$
$P(\tau\leq t)=P(\tau\leq t,B_t<s)+P(\tau\leq t,B_t\geq s)$
$P(\tau\leq t,B_t\geq s)=P(B_t\geq s)$ as if $B_t\geq s$, therefore ,$\tau\leq t$.
We introduce a process $W$ that is equal to $B_t$ if $\tau>t$, and equal to $2s-B_t$ if $\tau\leq t$. W is also a Brownian motion, therfore 
$P(\tau\leq t,B_t<s)=P(\tau\leq t,W_t<s)=P(\tau\leq t,2s-B_t<s)$
Finally,
$P(\tau\leq t)=2P(B_t>t)$

