Let $B(t)$ be a Brownian motion. Let $a<0<b$ and define corresponding hitting times of those levels $\tau_a$ and $\tau_b$. Put $$\tau := \max \{\tau_a, \tau_b\}.$$ We know $\tau$ is also a stopping time. I want to show that $$0=\mathbb E[ B(\tau)].$$
It is well known that the above equality holds for the stopping time $\tau_a \wedge \tau_b$ and I tried mimicking the proof using martingales, but it does not seem to work since $\mathbb E \tau =\infty$.
Any help is appreciated.
By the (a.s.) continuity of the sample paths of Brownian motion, we have $B_{\tau} \in \{a,b\}$. If we set $\sigma:=\min\{\tau_a,\tau_b\}$, then $$B(\tau,\omega) = a \iff \tau_a(\omega)>\tau_b(\omega) \iff B(\sigma,\omega)=b$$ and, analogously, $$B(\tau,\omega)=b \iff B(\sigma,\omega)=a.$$ Consequently, $$\mathbb{E}(B_{\tau}) = a \mathbb{P}(B_{\sigma}=b) + b \mathbb{P}(B_{\sigma}=a).$$ Since $$\mathbb{P}(B_{\sigma}=a) = \frac{b}{b-a} \qquad \mathbb{P}(B_{\sigma}=b) = \frac{-a}{b-a}$$ (see e.g. this question), we get $$\mathbb{E}(B_{\tau}) =- \frac{a^2}{b-a} +\frac{b^2}{b-a}= \frac{(b-a)(b+a)}{b-a} = b+a.$$
