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