What is the distribution of $2M_1-B_1$ where $M_t$ is the maximum process of the the Brownian motion $B_t$ 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?
 A: 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,
\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 $U=2M_1-B_1$.
