Brownian Motion : distribution of maximizer time T = argsup B(t) in time interval [0,1]

Question

Consider a standard brownian motion $\{B(t); t \geq 0\}$ on the time interval $[0,1]$, and $$T^* = \mathop{\mathrm{argsup}}\limits_{0\leq t\leq1} B(t)$$ How to find the distribution of $T^*$?

Could someone help me to get a start on the problem?

Answer 1

One way to start: Define $\overline B(t) :=\sup\{B(s): 0\le s\le t\}$. It is known that the process $X(t):=\overline B(t)-B(t)$, $t\ge 0$, has the same distribution as $R(t):=|B(t)|$, $t\ge 0$. The time $T^*$ thus has the same distribution as the time of the last zero of $R(t)$ during the time interval $[0,1]$; namely, the time $G:=\sup\{t\in[0,1]: |B(t)|=0\}=\sup\{t\in[0,1]:B(t)=0\}$. The distribution of $G$ is well known.