Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 / 5

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 $M_t\stackrel{d}{=}|B_t|$.

Continuity on the compact interval ensures that a maximum time exists. Fix a rational $r\in (0,1)$ and defineBy Brownian time reversal and the Brownian running maximum law, $U_r\stackrel{d}{=}|N(0,r)|$. Independent increments give $V_r\stackrel{d}{=}|N(0,1-r)|$ independently of $U_r$. Their continuous distributions imply $\mathbb{P}(U_r=V_r)=0$.

If the maximum were attained at two distinct times, a rational $r$ strictly between them would make the maxima on $[0,r]$ and $[r,1]$ equal, hence $U_r=V_r$. A countable union over rational $r$ still has probability zero. Therefore the time of the Brownian maximum is almost surely unique.

For fixed $s\in (0,1)$, uniqueness givesThe two sides of the comparison are independent and distributed as $\sqrt{s}|Z_1|$ and $\sqrt{1-s}|Z_2|$ for independent standard normal random variables. Rotational invariance of $(Z_1,Z_2)$ makes its angle uniform, soThe endpoint values follow by continuity. Thus the time of the Brownian maximum has the arcsine distribution.