Let $B\_t$ be a standard Brownian motion and let $M\_t:=\sup \_{s\le t}B\_s$ be the maximum process. What is the distribution of $2M\_1-B\_1$? is it elementary?



What is the distribution of $2M_1-B_1$ where $M_t$ is the maximum process of the the Brownian motion $B_t$
Yes, the pdf of this distribution is
\begin{equation}
u\mapsto 2u^2 f(u)\,1(u>0) \tag{1}
\end{equation}
where $f$ is the standard normal pdf.


Indeed, by [Proposition 2](https://ocw.mit.edu/courses/sloan-school-of-management/15-070j-advanced-stochastic-processes-fall-2013/lecture-notes/MIT15_070JF13_Lec7.pdf),
\begin{equation}
 G(m,b):=P(M\_1>m,B\_1\le b)=P(B\_1>2m-b)=1-F(2m-b)
\end{equation}
for real $m,b$ such that $m>b\_+:=\max(0,b)$, where $F$ and $f$ are the standard normal cdf and pdf, respectively. So, for the joint pdf $g$ of $(M\_1,B\_1)$ we have
\begin{equation}
 g(m,b)=-\frac{\partial^2 G(m,b)}{\partial m\,\partial b}
 =\frac{\partial^2 F(2m-b)}{\partial m\,\partial b}
 =2(2m-b)f(2m-b)
\end{equation}
if $m>b\_+$, with $g(m,b)=0$ otherwise. So, for
\begin{equation}
 U:=2M\_1-B\_1
\end{equation}
and all real $u>0$ we have
\begin{equation}
\begin{aligned}
 &P(U\le u) \\ 
 &=\iint\_{\mathbb R^2}dm\,db\,g(m,b)\,1(m>b\_+,\,2m-b<u) \\ 
 &=\int\_0^u dm\,\int\_{2m-u}^m db\,g(m,b) \\ 
 &=2 F(u)-2 u f(u)-1,
\end{aligned}
\end{equation}
with $P(U\le u)=0$ for $u\le0$. Differentiating $P(U\le u)$ in $u$, we confirm that the function given by (1) is the pdf of the distribution of $U=2M\_1-B\_1$.


