Expectation of Brownian hitting time

Question

Let $a,b >0$ and $B_t$ be one dimensional Brownian motion starting in $0$.
I am searching some way to calculate $$E(\tau_{-a}1_{\{\tau_{-a}<\tau_b\}}),$$
where $\tau_x$ denotes the hitting time of some point $x \in \mathbb{R}$.
Of course I know that by symmetry this expectation equals $\frac{a^2}{2}$ if $a=b$. But in this "unsymmetrical" I don't know how to calculate this expectation. Does anybody have an idea?

Answer 1

An alternative approach (motivated by the tediousness of differentiating that ratio of hyperbolic sines) can be made using the martingale $B_t^3 - 3tB_t$. By optional stopping, $E[B_\tau^3-3\tau B_\tau] =0$. Therefore, writing $q$ for $E[\tau_{-a}1_{\{\tau_{-a}<\tau_b\}}]$ (and noting that $E[\tau_b 1_{\{\tau_b<\tau_{-a}\}}]=ab-q$): $$ (-a)^3{b\over a+b}+b^3{a\over a+b} =3(-a)q+3b(ab-q), $$ because $P[\tau_{-a}<\tau_b]=b/(a+b)$, as is well known. Solving for $q$ you get $$ E[\tau_{-a}1_{\{\tau_{-a}<\tau_b\}}]={ab(a+2b)\over 3(a+b)}. $$