Range of a standard brownian motion, using reflection principle

With a standard brownian motion $B_t$, I'm trying to find the distribution of the "range":

$$R_{t} = \sup_{0 \leq s \leq t} B_s - \inf_{0 \leq s \leq t} B_s = \overline{M_t}-\underline{M_t}$$

The reflection principle gives, for $a > 0$, $P(\overline{M_t} \geq a) = 2 P(B_t \geq a)$ (as stated by @A.S. as comment on this question), and clearly $P(\overline{M_t} \geq a) = 1$ for $a \leq 0$. Thus $\overline{M_t} \sim |X|$ with $X \sim \mathcal N(0,\sqrt{t}^2)$, i.e. $\overline{M_t}$ has a folded normal distribution, with well-known density $f_\overline{M_t}$.

Now, in order to find $R_t$'s distribution I would like to do:

$$P(R_t \geq a ) = P(\overline{M_t} - \underline{M_t} \geq a) = \int_{\mathbb R^+} P(s - \underline{M_t}\geq a | \overline{M_t} = s) f_{\overline{M_t}}(s) d s.$$

So now what I need, is to compute:

$$P(-\underline{M_t} \geq y | \overline{M_t} = s)$$

or, for symmetry reason, I need to compute :

$$P(\overline{M_t} \geq y | \underline{M_t} = -s)$$

When $y \geq -s$, it seems that it's possible to compute $$P(\overline{M_t} \geq y | \underline{M_t} = -s)$$ with reflection principle (not 100% sure).

In the other case, how to compute $P(\overline{M_t} \geq y | \underline{M_t} = -s)$ ?

• Just as a remark: The joint density $(\underline{M}_t,\overline{M}_t)$ can be calculated explicitly (this is known as "Lévy's triple law); this might be useful to determine the distribution of $R_t$. – saz Feb 5 '16 at 18:05
• @saz oh I didn't know, do you have a good reference for this? – Basj Feb 5 '16 at 18:53
• E.g. the book by René Schilling & Lothar Partzsch on Brownian motion ("Brownian Motion - An introduction to stochastic processes") – saz Feb 5 '16 at 19:34
• books.google.com/… – A.S. Feb 5 '16 at 21:30
• @A. S. Thanks! Just to be sure, isn't there an integral missing in the expression in this book? What means $B_t \in d x$? In the last formula of the proof there is an integral. – Basj Feb 6 '16 at 23:03