# Distribution of $(\sup_{0\leq s\leq t} W_s -W_t)$

I am interest in the law of the $(\sup_{0\leq s\leq t} W_s -W_t)$ where $W$ is a standard brownian motion.

I know that $M_t:=\sup_{0\leq s\leq t} W_s \overset{\mathcal L}{=} |W_t |$ so its density function f is
$$f(x) = \sqrt{\frac{2}{\pi t}} \exp\left(-\frac{x^2}{2t}\right) \mathbb 1_{ \{x \geq 0\}}$$

I am struggling at obtaining the law of $(M_t - W_t)$. Consider a the function $g(m,w) = m-w$. Therefore the question is to find

$$\mathbb P\left(g(M_t, W_t) \leq x\right )$$ knowing that the joint density function of the couple $M_t,W_t$ which is

$$f_{M_t,W_t}(m,w)= \frac{2 (2m-w)} {t\sqrt{2 \pi t}} \exp \left( -\frac{(2m-w)^2} {2 t} \right) \mathbb 1_{ \{m \geq 0\}} \mathbb 1_{ \{w \leq m\}}$$ Any hint is appreciated. Many thanks.

• The joint distribution of a Brownian motion and its running maximum is known. The answer follows. – Did Jan 16 '15 at 11:22
• @Did : Thank you for your answer. I am aware of this result. I am struggling to obtain the law of $g(X_t,W_t)$ where $g(x,w) = x-y$. – Paul Jan 16 '15 at 11:42
• Simply integrate the function $w\mapsto f_{M,W}(w+x,w)$ to get the density of $M-W$ at $x$. – Did Jan 16 '15 at 23:21