 Brownian motion shifted at its finite-horizon maximum

Let $\tau$ be the first attainment of the maximum of standard Brownian motion on $[0,T]$, with $T>0$. Time reversal shows $\tau <T$ with probability one. Then $Y_t\le 0$ for $0\le t\le T-\tau$, an initial interval of positive length. Standard Brownian motion has probability zero of this property by the Brownian reflection principle, so $Y$ is not Brownian. Moreover $\tau$ is not a stopping time for the natural Brownian filtration: for $0<t<T$, $\mathbb{P}(\tau \le t\mid {\mathcal{F}}_t)=2\Phi ((M_t-B_t)/\sqrt{T-t})-1$, which is strictly between zero and one on $\{B_t<0\}$. The Strong Markov property is therefore inapplicable at this random time.