I would like to solve exercise 2.8 in Continuous Martingales and Brownian Motion from Daniel Revuz,Marc Yor.
Let $B$ a linear Brownian motion and $a \leq x \leq y \leq b$ prove that :
$E_x[L^y_{T_a \wedge T_b}]=2(x-a)(b-y)/(b-a)$
where $T_a:=\inf\{t \geq 0| B_t=a\},T_b:=\inf\{t \geq 0| B_t=b\}$. and $T_a \wedge T_b:=min(T_a,T_b)$.
My idea is to use Tankas formula:
$|B_t-y|-|B_0-y|-L_t^y= \int_0^t sign(B_t-y) d B_t$ and applying optional stopping theorem because $\int_0^t sign(B_t-y) d B_t$ is a martingal. Then one obtains
$E_x[|B_{T_a \wedge T_b}-y|]-|x-y| -E_x[L^y_{T_a \wedge T_b}]=0$. Now, I'm struggle with the term $E_x[|B_{T_a \wedge T_b}-y|]$. Every idea is welcome.
It is possible to directly compute $\mathbb{E}^x[|B_{T_a \wedge T_b} - y|]$ by noting that $B_{T_a \wedge T_b} \in \{a,b\}$. Indeed, we have $$\mathbb{E}^x[|B_{T_a \wedge T_b} - y|] = |a-y| \mathbb{P}^x (T_a < T_b) + |b-y| \mathbb{P}^x(T_b < T_a)$$ It is well known that $\mathbb{P}^x(T_a < T_b) = \frac{b-x}{b-a}$ and $\mathbb{P}^x(T_b < T_a) = \frac{x-a}{b-a}$. Hence \begin{align*} \mathbb{E}^x[L_{T_a \wedge T_b}^y] &= \mathbb{E}^x[|B_{T_a \wedge T_b} - y|] - (y-x) \\& = (y-a) \frac{b-x}{b-a} + (b-y) \frac{x-a}{b-a} - (y-x) \\& = 2 \frac{(x-a)(b-y)}{b-a} \end{align*} as desired.