I have some difficulties to answer a question about stopping time.
We consider $B$ a Brownian motion, $a<0<b$, $\tau_a=\inf\{t : B_t = a\}$ and $\tau_b=\inf\{t : B_t = b\}$. Then I put $\tau=\min(\tau_a,\tau_b)$.
After some computation I found $\mathbb{P}(\tau_a<\tau_b) = \frac{b}{b-a}$.
Now I would like to compute $\mathbb{E}(\tau)$.
Intuitively, I would like to say that $\tau$ take two values such that $\mathbb{E}(\tau)=\tau_a\frac{b}{b-a} +\tau_b(1-\frac{b}{b-a})$
However $\tau_a$ and $\tau_b$ are also random variable that seem not indépendant so I have no idea on how to tackled this problem.
Have you any hints please ?
Here is a new attempt, different from the one I usually seek since it consists to use convergence result on martingale.
First, we can show that $B^{\tau}$ is a uniformly integrable martingale (to show that it is uniformly integrable use the fact that $\lvert B^{\tau}\rvert$ is bounded should be enough).
Then we know that $B_{\min(\tau,t)}$ converges in $L^1$ and almost surely to some $B_{\infty}\in L^1$ and that for all finite stopping time $\tau\leq\sigma$ we have
$$ \mathbb{E}(B_{\sigma} | \mathcal{F}_{\tau}) = B_{\tau} $$
Moreover here $B_{\infty} = B_{\tau}$
Using the fact $B_{\min(\tau,t)}^{2} -\min(\tau,t)$ is still a martingale we have $\mathbb{E}(B_{\min(\tau,t)}^{2}) = \mathbb{E}(\min(\tau,t))$ (using bounded stopping theorem with the constant stopping time $0$).
Now take a sequence $t_n$ of integers that goes to infinity, we have that $\min(\tau,t_n)$ converges to $\tau$ in an increasing way. Then it follows by convergence monotone theorem and the fact that $B_{\infty}$ is the almost sure limit of the sequence $B_{\min(\tau,t_n)}$ that is bounded, we get
$$ \mathbb{E}(B_{\infty}^{2}) = \mathbb{E}(B_{\tau}^{2}) = \mathbb{E}(\tau) $$
We have $B_{\tau}^{2} = a^{2}1_{\tau_{a}<\tau_b} + b^{2} 1_{\tau_{b}<\tau_a}$ so we get
$$ \mathbb{E}(\tau) = a^2\frac{b}{b-a} +b^{2}(1-\frac{b}{b-a}) $$