Hitting time of the maximum of a Brownian motion

Question

Let B be a standard Brownian motion in 1 dimension. Define $\tau = \inf\{t\in[0,1]:B_t=\underset{s\in[0,1]}{\max}B_s\},$ and $\tau' = \sup\{t\in[0,1]:B_t=\underset{s\in[0,1]}{\max}B_s\}.$ Show that $\tau$ and $\tau'$ are not stopping times. My idea was to first assume that $\tau$ is a stopping time, then by strong Markov property $\{B_{\tau+t}-B_{\tau}:t \geq 0\}$ is a standard Brownian motion, and so the process fluctuates above and below $B_\tau$ at $\tau$, which contradicts the fact that $B_\tau$ is the maximum. I am wondering if I am heading towards the right direction.

Answer 1

As noted, if $\tau$ were a stopping time, then $\beta_t:=B_{t+\tau}-B_\tau$, $t\ge 0$, would be a standard Brownian motion. On the event $\{\tau<1\}$ you would then have $\beta_t\le 0$ for all sufficiently small $t$, which is inconsistent with the Brownian character of $\beta$. Therefore $P[\tau<1]$ would be $0$. But $$ {1\over 2} =P[B_1<0]\le P[\tau<1], $$ because $\max_{0\le s\le 1}B_s\ge 0$. Therefore $\tau$ cannot be a stopping time.

Likewise for $\tau'$. In fact, one can show that the time at which Brownian motion on the time inerval $[0,1]$ attains its maximum is unique; that is $P[\tau=\tau'<1]=1$.