Strong Markov Property Brownian Motion for Non-Stopping Time Let $B$ be a Brownian motion and let $\mathcal{F}^B$ be its natural filtration. Define the random variable
$$ \tau = \inf\{ t \ge 0 \mid B_t = \sup_{0 \le s \le 1} B_s \}.$$
Now, $\tau$ is not an $\mathcal{F}^B$ stopping time (as $\{\tau \le t\}$ is not $\mathcal{F}_t$ measurable). Consider the process
$$\hat{B} = (B_{t+\tau} - B_\tau)_{t \ge 0}.$$
My question is whether or not $\hat{B}$ is a Brownian motion (in its own filtration). Of course, if $\tau$ were a stopping time (in $\mathcal{F}^B$) then it is by the strong Markov property.

Ordinarily with a question like this, it would be fairly clear what the answer is, and then the challenge is proving it. I can't actually convince myself either way in this case! If someone could point me in the right direction (please don't just give me the solutions, as I'd like to be able to work through it myself to get the most out of it!) then I'd be most appreciative. :)

With probability $1$, $\tau<1$, and $\hat B$ is nonpositive on the whole interval $[0,1-\tau]$. As a Brownian motion takes positive and negative values arbitrarily close to $0$, $\hat B$ is not a Brownian motion.
