TITLE: Running maximum absolute value of Wiener process
QUESTION [6 upvotes]: In Wikipedia a formula is given for the distribution of $$M_t = \max_{0\leq s \leq t} W_s$$
even conditioned on $W_t$.
I wonder if there is also a simple expression for (note the absolute value)
$$\tilde M_t = \max_{0\leq s \leq t} |W_s|$$
maybe conditioned on $|W_t|$?

REPLY [5 votes]: I'd recomment Borodin and Salminen's Handbook of Brownian Motion - Facts and Formulae. There's no simple expression for the distribution, but formula 1.1.8 at page 250 gives for $y > \max(x,z)$:
$$P_x(\max_{0<s<t}|W_s|<y, |W_t| \in dz) = $$
$$\frac{1}{\sqrt{2\pi t}} \sum_{k=-\infty}^\infty(-1)^k \left( e^{-(z-x+2ky)^2/2t} +  e^{-(z+x+2ky)^2/2t} \right)\,dz $$
Isn't that a beautiful work of art... You can use standard techniques to obtain all the formulas you want (but they'll still look like this).
EDIT (as a reply to the a comment): We have the following for the cdf:
$$P_x(\max_{0<s<t}|W_s| \geq y) = $$
$$\sum_{k=-\infty}^\infty (-1)^k \textrm{sign}(x+(2k+1)y) \textrm{Erfc}\left( \frac{|x+(2k+1)y|}{\sqrt{2}t}\right) $$
and to obtain the expectation on does $E_x(\max \ldots) = \int_0^\infty P_x(\max\ldots)\,dy$.