How to prove that for Brownian motion in $(a, b)$ $\mathbb{E}^x[\min(H_a, H_b)] = (x-a)(b-x)$?

i'm wondering if anyone can help me with proving the fact that for BM in the interval $(a,b)$ and with $$H_y = \inf\{t>0: X_t = y\},$$ the following is true: $$\mathbb{E}^x[\min(H_a, H_b)] = (x-a)(b-x).$$

And the second question here is how to deduce that the speed measure for BM is actually the Lebesgue measure?

Thanks, Anna

Without loss of generality we can assume $x=0$ (otherwise replace $a$ by $a-x$ and $b$ by $b-x$). Moreover, let $$\tau := \inf \{t \geq 0; X_t \notin (a,b)\}$$ $\tau$ is a stopping time and $\min \{H_a, H_b\} = \tau$.

By Wald's identities we have

$$\mathbb{E}(\min\{H_a,H_b\}) = \mathbb{E}(\tau) = \mathbb{E}(X_{\tau}^2) = a^2 \cdot \mathbb{P}[X_{\tau}=a] + b^2 \cdot \mathbb{P}[X_{\tau}=b] \tag{1}$$

where we used

$$X_{\tau} = a \cdot 1_{[X_{\tau}=a]} + b \cdot 1_{[X_{\tau}=b]}$$

As a consequence of Wald's identities

$$\mathbb{P}[X_{\tau}=a] = \frac{b}{b-a} \qquad \qquad \mathbb{P}[X_{\tau}=b] = \frac{-a}{b-a}$$

Hence

$$\mathbb{E}(\min\{H_a,H_b\}) = \frac{a^2 \cdot b - a \cdot b^2}{b-a} = a \cdot b \cdot \frac{b-a}{a-b}=-a \cdot b \tag{2}$$

Concerning the speed measure $m$: Let

$$h_{[a,b]}(x) := \mathbb{E}(\tau) \stackrel{(2)}{=} (a-x) \cdot (x-b)$$

Since $m(dx) = - \frac{1}{2} h^{(2)}_{[a,b]}(dx)$ we have $$m(dx) = - \frac{1}{2} (-2) \, dx=dx$$ i.e. the speed measure is the Lebesgue measure.