Let $B$ be a Brownian Motion starting at $0$ and let $\lambda$ be a real number. Define $M_{t}^{\lambda}=e^{\lambda B_{t}-\lambda^{2} t / 2} .$ We recall that $T_{a}=\inf \left\{t \geq 0, B_{t}=a\right\}$
- Prove that the process $\left(M_{t}^{\lambda}\right)_{t>0}$ is a $\left(\mathcal{F}_{t}^{B}\right)_{t \geq 0^{-}}$ -martingale.
I was able to do this question.
- We denote by $\mathbb{P}_{x}$ the law of $\left(x+B_{t}\right)_{t \geq 0},$ i.e. the law of a standard Brownian motion starting at $x \in \mathbb{R} .$ Show that if $0 \leq x \leq a$ and $\lambda>0$ then $$ \mathbb{E}_{x}\left[e^{-\lambda T_{0}} \mathbf{1}_{T_{0}<T_{a}}\right]=\frac{\sinh ((a-x) \sqrt{2 \lambda})}{\sinh (a \sqrt{2 \lambda})} $$
I know we have to somehow apply the optional stopping theorem to both $\left(M_{t}^{\lambda}\right)_{t>0}$ and $\left(M_{t}^{-\lambda}\right)_{t>0}$, but I'm confused by the $\mathbb{E}_x$ notation.
$\mathbb{E}_x$ means that you take the expectation for a BM that starts at $x$ rather than at $0$, which in this context changes the distribution of $T_0$ and $T_a$.