Consider two hitting times $T_{a},T_{b}$ for a Brownian motion $B$ such that $a < 0 < b$. Is it possible to recover the fact that $\mathbb P(T_{a} < T_{b} ) = b/(b-a)$ using the fact that I computed:
$$\mathbb E[ e^{-\lambda T_{a}}\chi_{\{T_{a}< T_{b}\}}]=\sinh(\sqrt{2\lambda} b)/\sinh(\sqrt{2\lambda}(b-a))\; ?$$
differentiating both sides of the above the above wrt $\lambda$ and then setting $\lambda = 0$ would leave me with something of the form:
$$ -\mathbb E [T_{a}\chi_{\{T_{a}< T_{b}\}}]=...$$
In other words I cannot isolate $\chi_{\{T_{a}< T_{b}\}}$ under the expectation to get $\mathbb P(T_{a}< T_{b})$. Any other ideas?