Let $B$ be a Brownian motion. I want to show that
$$ \inf\{t\geq0 \mid B(t)=\max_{x\in [0,1]}B(s)\} $$
is not a stopping time w.r.t. the standard filtration.
How can one intuitively see that this is true?
How to handle proofs involving stopping times rigorously, is there a general recipe? What would be a rigorous proof of this specific example?
Best Answer
Intuitively, $T=\inf\{t\geqslant0 \vert B(t)=\max_{x\in [0,1]}B(s)\}$ is not a stopping time of the filtration $(\mathcal F_t)_{t\geqslant0}$ where $\mathcal F_t=\sigma(B(s),s\leqslant t)$ for every $t\geqslant0$, because, for $0\leqslant t\lt1$, the event $[T=t]$ depends on $\mathcal F_t$ but also of the part $(B^t(s))_{s\leqslant1-t}$ of the process $B^t=(B(s+t)-B(t))_{s\geqslant 0}$, which is not $\mathcal F_t$-measurable.
Rigorously, consider $M(t)=\max\{B(s)\mid s\leqslant t\}$, then the process $(M(t))_{t\geqslant0}$ is adapted to the filtration $(\mathcal F_t)_{t\geqslant0}$. Moreover, for $0\leqslant t\leqslant1$, $$ [T\leqslant t]=[M^t(1-t)\leqslant M(t)-B(t)],\qquad M^t(s)=\max\{B^t(u)\mid u\leqslant s\}, $$ where $M(t)-B(t)$ is $\mathcal F_t$-measurable while $M^t(1-t)$ is (non trivial and) independent of $\mathcal F_t$ hence $[T\leqslant t]$ is not in $\mathcal F_t$. For example, $$ \mathbb P(T\leqslant t\mid\mathcal F_t)=u_t(M(t)-B(t)),\quad u_t(x)=\mathbb P(M(1-t)\leqslant x), $$ and the function $u_t$ is not $0-1$ valued when $t\lt1$ hence, except if $t=1$, $$ \mathbb P(T\leqslant t\mid\mathcal F_t)\ne\mathbf 1_{T\leqslant t}. $$ In particular, $[T\leqslant t]$ is not in $\mathcal F_t$ if $0\leqslant t\lt 1$ hence $T$ is not a $(\mathcal F_t)_{t\geqslant0}$-stopping time (but note that $[T\leqslant t]=\Omega$ for every $t\geqslant1$).