For fixed $s\in(0,1)$, uniqueness gives
$$
\{M^*\leq s\}
=\left\{\max_{t\leq s}B_t-B_s
\geq\max_{s\leq t\leq1}(B_t-B_s)\right\}.
$$
The 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, so
$$
\begin{aligned}
\mathbb P(M^*\leq s)
&=\mathbb P\left(\frac{|Z_2|}{|Z_1|}\leq\sqrt{\frac{s}{1-s}}\right)\\
&=\frac2\pi\arctan\sqrt{\frac{s}{1-s}}
=\frac2\pi\arcsin\sqrt s.
\end{aligned}
$$
The endpoint values follow by continuity. Thus the <time of the Brownian maximum> has the arcsine distribution.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-201/5/c/solution