What is the distribution of $2M_1-B_1$ where $M_t$ is the maximum process of the the Brownian motion $B_t$

Question

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?

Answer 1

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 $U=2M_1-B_1$.