 Solution

The Brownian reflection principle reflects a path after its first hit of $x>0$ and givesSymmetry of the centered normal distribution gives $2\mathbb{P}(B_t\ge x)=\mathbb{P}(|B_t|\ge x)$. Both random variables are nonnegative, so their tail distributions agree for every $x$, proving the Brownian running maximum identity $M_t\stackrel{d}{=}|B_t|$.