Brownian motion a martingale

Question

Let $B(t)$ be a Brownian motion and $\tau$ a finite stopping time. Assuming that I know that $B(\tau)$ is a martingale, does this imply that $E(B(\tau))=0$ and $E((B(\tau))^2)=\tau$ just as one would have for $\tau=t$ and $t \in [0,\infty)$?

Answer 1

Consider the stopping time $\tau:=\inf\{t: B(t)=1\}$. This time satisfies $\tau<\infty$ and $B(\tau)=1$ (both of these holding almost surely), so that $E(B(\tau))=1\not=0$.