Let $B_t$ be a Brownian motion, and let $S_t = \sup_{0\leq s\leq t}B_s$. Show that $T := \inf \{ t \geq 0 : B_t = S_1 \}$ is not a stopping time.

It suffices to show e.g. $\{T \leq 1/2\} \not\in \mathscr{F}_{1/2}$ . This amounts to proving, $$\left\{ \sup_{s\in [0, 1/2]} B_s \geq \sup_{s\in [1/2, 1]} B_s \right\} \not\in \mathscr{F}_{1/2}$$

This seems pretty obvious to me: at time $t = 1/2$, you know $\sup_{s\in [0, 1/2]} B_s$ is some number, but cannot tell whether or not $B_s$ will cross above that number in $s\in [1/2, 1]$. But I am stuck figuring out how to make this intuition rigorous.


1 Answer 1


Let $M = \sup_{0 \le t \le 1/2} B_t$ and $M' = \sup_{1/2 \le t \le 1} (B_t - B_{1/2})$. Then $M$ is in $\mathcal{F}_{1/2}$ and $M'$ is independent of $\mathcal{F}_{1/2}$. Let $F$ be the cdf of $M'$. Note that $F(x) < 1$ for all $x$.

Clearly $T \le 1/2$ iff $M' \le M - B_{1/2}$. Verify using properties of conditional expectation that $$P(T \le 1/2 \mid \mathcal{F}_{1/2}) = P(M' \le M - B_{1/2} \mid \mathcal{F}_{1/2}) = F(M - B_{1/2}) < 1 \qquad \text{a.s.}$$ If $\{T \le 1/2\}$ were $\mathcal{F}_{1/2}$-measurable, then $P(T \le 1/2 \mid \mathcal{F}_{1/2}) = 1_{\{T \le 1/2\}}$ almost surely. The above would then imply $1_{\{T \le 1/2\}} < 1$ a.s., which is to say $P(T \ge 1/2) = 0$. That is absurd.

Alternative, slick solution: clearly $T \le 1$. If $T$ is a stopping time, then it is a bounded stopping time, so by the optional stopping theorem we have $E[B_T] = E[B_0] = 0$. But obviously $B_T = S_1 \ge 0$, so we conclude $B_T = S_1 = 0$ almost surely. That is likewise absurd.


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