Expectation of the minimum of two first passage times of a standard Brownian motion Let $W_t$ be a standard Brownian motion. For any real constant $a$, $T_a=\min(t≥0:W_t=a)$. 
I know how to derive the distribution of $T_a$.
Now given $a>0$ and $b<0$, I want to compute $E[\min(T_a,T_b)]$.
Any help? Are $T_a$ and $T_b$ independent?
 A: Hints:


*

*Show that $\min\{T_a,T_b\} = T := \inf\{t \geq 0; W_t \notin (b,a)\}$.

*Show that $W_T \in \{a,b\}$ and $|W_{t \wedge T}| \leq \max\{|a|,|b|\}$ almost surely.

*Use the optional stopping theorem for the martingale $(W_t)_{t \geq 0}$ to prove that $$\mathbb{E}(W_{T \wedge t})=0$$ for any $t \geq 0$. Conclude from step 2 and the dominated convergence theorem that $$0 = \mathbb{E}(W_T) = b \mathbb{P}(W_T=b) + a \mathbb{P}(W_T=a). \tag{1}$$ Combine this with the fact that $$\mathbb{P}(W_T=a)+\mathbb{P}(W_T=b)=1. \tag{2}$$ to compute $\mathbb{P}(W_T=a)$ and $\mathbb{P}(W_T=b)$.

*Apply the optional stopping theorem to the martingale $(W_t^2-t)_{t \geq 0}$ to show that $$\mathbb{E}(W_{T \wedge T}^2) = \mathbb{E}(T \wedge t). \tag{3}$$ By step 2, this implies $$\mathbb{E}(T \wedge t) \leq \max\{a^2,b^2\}< \infty.$$ Conclude that $\mathbb{E}T<\infty$.

*Let $t \to \infty$ in (3) to show that $$\mathbb{E}(W_T^2) = \mathbb{E}(T).$$ Thus, by step 2, $$a^2 \mathbb{P}(W_T=a) + b^2 \mathbb{P}(W_T=b) = \mathbb{E}(T).$$ Now plug in the probabilites you have computed in step 3....

