TITLE: Brownian motion and its maximum and its minimum
QUESTION [6 upvotes]: Let $W_u, 0\leq u \leq t$ be Brownian motion.
Let $m_t= min_{0\leq u\leq t} W_u$ and $M_t = max_{0 \leq u \leq t} W_u$.
The fact that $(M_t , W_t)$ is absolutely continuous with respect to Lebesgue measure on $\mathbb{R}^2$ is known in some stochastic calculus book. 
By symmetry of Brownian motion, $(m_t,W_t)$ is absolutely continuous with respect to Lebesgue measure on $\mathbb{R}^2$.
I want to know whether $(m_t, M_t , W_t)$ is absolutely continuous with respect  to Lebesgue measure on $\mathbb{R}^3$. 
Since
$P(x\leq m_t \leq M_t \leq y, W_t \in dx) = \sum_{k=-\infty}^{\infty} \bigg{[} \phi \big( \frac{x-2k(b-a)}{\sqrt{t}}) - \phi \big( \frac{x-2b-2k(b-a)}{\sqrt{t}}) \bigg{]}dx$,
If $(m_t,M_t,W_t)$ is absolutely continuous to Lebesgue measure on $\mathbb{R}^3$, I think I can calculate joint density of $m_t,M_t,W_t$.
Could you help me?

REPLY [4 votes]: It appears that a proof is given in
Choi, ByoungSeon; Roh, JeongHo, On the trivariate joint distribution of Brownian motion and its maximum and minimum, Stat. Probab. Lett. 83, No. 4, 1046-1053 (2013). ZBL1266.60142.