# Brownian motion and its maximum and its minimum

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?

• If it's true, it ought to be possible to prove by similar techniques to $(M_t, W_t)$. Have you tried to work through it? For instance, I would guess that the Malliavin calculus approach would work. On the other hand, if you can calculate formally what the density "should" be, then you can probably turn that into a proof that it really is. – Nate Eldredge Aug 7 '17 at 17:32

• The paper in question does not anywhere use or claim there exists a joint density of $(m_t, M_t, W_t)$. In the cited paper, what the authors' call the 'trivariate joint density' refers to the density $P(a\leq m_t \leq M_t \leq b, W_t \in dx)$ for fixed $a \leq 0 \leq b$. – nullUser Feb 10 at 19:48