What probability distribution does the Running maximum of brownian motion belong to

Question

I am curious to learn what distribution does the running maximum of Brownian motion $$ f_{M_t}(m) = \sqrt{\frac{2}{\pi t}}\exp\left[-\frac{m^2}{2t}\right], \qquad m \ge 0.$$ with CDF $$\operatorname{erf}\left(\frac{x}{\sqrt{2 t}}\right)$$ with expectation $$E[M_t] = \int_{0}^{\infty} m f_{M_t}(m)\,dm = \int_{0}^{\infty} m \sqrt{\frac{2}{\pi t}}\exp\left[-\frac{m^2}{2t}\right]\,dm = \sqrt{\frac{2t}{\pi}} $$ belong to ?

Thank you

Answer 1

It seem's to me that this is truncated normal, with $a=0$, $b=\infty$, and $\sigma^{2} = t$.

Maybe I'm wrong.

Answer 2

$M_t$ has the same distribution as $|B_t|$, where $(B_t)$ is Brownian motion. Put another way, the distribution of $M_t$ is the same as $\sqrt{t}|Z|$, where $Z$ has the standard normal distribution.