Expected value of minimum of hitting time of brownian motion. Let $(X_t)_{t\geq0}$ be a Brownian Motion starting from $0$. Define for $a\in\mathbb{R}$, $$T_a = \inf\{t\geq0:X_t=a\}.$$
For $a,b>0$, I know that $\mathbb{P}(T_{-a}<T_b)=\frac{b}{a+b}$, and I am expected to show that $\mathbb{E}(T_{-a}\wedge T_b)=ab$.
One idea I have is to define the process $Q_t = X_t^2-t$. If the stopped process $Q^T$ was uniformly integrable, I could use the optional stopping theorem to easily get the result. However, I'm unsure as to whether $Q^T$ is UI, and if so how to show it.
 A: It is easier to show that $\tau = T_{-a} \land T_b$ has finite expectation than to show that $Q^\tau$ is uniformly integrable.
To that end, define the events
$$E_k = \{|B_{k+1} - B_k| > a+b\}$$
Observe that for each $k$, we have $P(E_k) = r$ for some $r \in (0,1)$. Notice that if $\tau > n+1$, then none of the events $E_1, \ldots , E_n$ could have happened. Since the events $E_1, \ldots , E_n$ are independent, we get:
$$P(\tau > n+1) \leq (1-r)^n$$
It follows that
$$E(\tau) = \sum_{k=0}^\infty P(\tau > k) < \infty$$
To conclude, we consider the martingale $Q_t = X_t^2 - t$. By the triangle inequality, for each $t \geq 0$ we have $$|Q_{t\land \tau}| \leq \max(a,b)^2 + \tau$$ which, after taking expectations, shows that $E|Q_{t\land \tau}| < \infty$. This allows us to use dominated convergence aided by the fact that $E(Q_{t \land \tau}) = 0$  to conclude that $$E(\tau) = \lim_{t \to \infty} E(t \land \tau) = \lim_{t \to \infty} [E(X_{t\land \tau}^2) - E(Q_{t\land \tau})] = \lim_{t \to \infty} E(X_{t \land \tau}^2) = E(X_{\tau}^2) = ab$$
