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Let $a < 0 < b$ and further for a standard Brownian motion $(B_{t})$ define the stopping times $T_{x}:=\inf\{t \geq 0: B_{t}=x\}$.

Show that:

$$\mathbb E[\exp(-\lambda(T_{a}\land T_{b}))]=\cosh(\frac{a+b}{2}\sqrt{2\lambda})/\cosh(\frac{a-b}{2}\sqrt{2\lambda})$$

My attempt: I have already shown that $\mathbb E[\exp(-\lambda T_{a})]=\exp(-a\sqrt{2\lambda}) \; (*)$ through the Optional sampling theorem as well as the choice of martingale $\left(\exp(\alpha B_{t}-\frac{\alpha^{2}}{2}t)\right)_{t\geq 0}$,

i.e. choosing the martingale $\left(\exp(\sqrt{2\lambda} B_{t}-\lambda t)\right)_{t\geq 0}$, we obtain $(*)$.

Now I am supposing we need to find a suitable exponential martingale again, but I am struggling to find one, any suggestions?

Additional question: How does the knowledge of $$E[\exp(-\lambda(T_{a}\land T_{b}))]=\cosh(\frac{a+b}{2}\sqrt{2\lambda})/\cosh(\frac{a-b}{2}\sqrt{2\lambda})$$

allow us to deduce $\mathbb E[T_{a}\land T_{b}]=b\lvert a \rvert$? I cannot see it.

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    $\begingroup$ why not condition on the event $\{T_a<T_b\}$? Looks like it can be directly bashed that way. $\endgroup$
    – dezdichado
    Commented Nov 5, 2021 at 17:52

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$T:=T_a\wedge T_b$ is still a stopping time.

Apply the optional stopping theorem to the martingale $$M_t:=\exp\Bigl(\sqrt{2\lambda}(B_t-x_0)-\lambda t\Bigr)+\exp\Bigl(-\sqrt{2\lambda}(B_t-x_0)-\lambda t\Bigr),\quad t\ge0,$$ where $x_0:=\frac{a+b}2$.

Regarding the additional question: what's the derivative of $\mathbb E[\exp(-\lambda T)]$ at $\lambda=0$?

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