How can I compute $\Bbb{E}\left( e^{-|B_t| \sqrt{2 \lambda}} 1_{B_t\leq -a}\right)$ for a Brownian motion?

Question

Let $a\in \Bbb{R}_+$ and $B$ be a standart Brownian motion. For $\lambda >0$ I want to compute $$\Bbb{E}\left( e^{-|B_t| \sqrt{2 \lambda}} 1_{B_t\leq -a}\right)$$

I am somehow a bit lost where to start. I thought about using the tower property of conditional expectation and get that $$\begin{align}\Bbb{E}\left( e^{-|B_t| \sqrt{2 \lambda}} 1_{B_t\leq -a}\right)&=\Bbb{E}\left(\Bbb{E}\left( e^{-|B_t| \sqrt{2 \lambda}} 1_{B_t\leq -a}\big|\mathcal{F}_t\right)\right)\\&=\Bbb{E}\left(1_{B_t\leq -a} \Bbb{E}\left( e^{-|B_t| \sqrt{2 \lambda}}\big|\mathcal{F}_t\right)\right) \end{align}$$ where I used that $\{B_t\leq -a\}=B_t^{-1}((-\infty, -a])\in \mathcal{F}_t$. But also here I don't see how to continue since I don't have independence of $1_{B_t\leq -a} $ and $\Bbb{E}\left( e^{-|B_t| \sqrt{2 \lambda}}\big|\mathcal{F}_t\right)$.

Can someone give me a hint how do do this or what to use?

Answer 1

$$\begin{align} \Bbb{E}\left( e^{-|B_t| \sqrt{2 \lambda}} \mathbf{1}_{\{B_t\leq -a}\}\right)&=\Bbb{E}\left( e^{B_t \sqrt{2 \lambda}} \mathbf{1}_{\{B_t\leq -a}\}\right) \hspace{1cm}\text{because } B_t\le -a\le0\\ &=\Bbb{E}\left( e^{X \sqrt{2 \lambda t}} \mathbf{1}_{\{X\leq -\frac{a}{\sqrt{t}}}\}\right)\hspace{1cm}\text{with } X:=\frac{B_t}{\sqrt{t}} \text{ follows the }\mathcal{N}(0,1)\\ &=\int_{-\infty}^{-\frac{a}{\sqrt{t}}}\frac{1}{\sqrt{2\pi}}\cdot\exp\left(x\sqrt{2 \lambda t} \right)\cdot \exp\left(-\frac{1}{2}x^2 \right)dx \\ &=\int_{-\infty}^{-\frac{a}{\sqrt{t}}}\frac{1}{\sqrt{2\pi}}\cdot\exp\left(-\frac{1}{2}(x-\sqrt{2\lambda t})^2 \right)\cdot e^{\lambda t}dx\\ &=e^{\lambda t}\int_{-\infty}^{-\frac{a}{\sqrt{t}}-\sqrt{2\lambda t}} \underbrace{\frac{1}{\sqrt{2\pi}}\cdot\exp\left(-\frac{1}{2}x^2 \right)}_{\text{density function of }\mathcal{N}(0,1)}\cdot dx \\ &=e^{\lambda t} \Phi \left(-\frac{a}{\sqrt{t}}-\sqrt{2\lambda t} \right) \end{align}$$

Answer 2

If $a>0$ and $B_t(\omega)\leq -a$ then $|B_t(\omega)|=-B_t(\omega)$. Then $E[e^{-|B_t|\sqrt{2\lambda}}\mathbf{1}_{\{B_t\leq -a\}}]=E[e^{B_t\sqrt{2\lambda}}\mathbf{1}_{\{B_t\leq -a\}}]$ and we get $$\begin{aligned}E[e^{-|B_t|\sqrt{2\lambda}}\mathbf{1}_{\{B_t\leq -a\}}]&=E[e^{Z\sqrt{2t\lambda}}\mathbf{1}_{\{Z\sqrt{t}\leq -a\}}]=\\ &=E[e^{Z\sqrt{2t\lambda}}\mathbf{1}_{\{e^{Z\sqrt{2t\lambda}}\leq e^{-a \sqrt{2\lambda}}\}}]=\\ &=-E[(e^{-a \sqrt{2\lambda}}-e^{Z\sqrt{2t\lambda}})^+]+e^{-a \sqrt{2\lambda}}P(Z\sqrt{t}\leq -a)\end{aligned}$$ We can solve this with Black-Scholes of put options. Set $S_0=1,\sigma =\sqrt{2\lambda},r=\sigma^2/2,\,K=e^{-a \sqrt{2\lambda}},T=t,t_0=0$. We get $$\begin{aligned}E[e^{-|B_t|\sqrt{2\lambda}}\mathbf{1}_{\{B_t\leq -a\}}]&=-e^{rt}\textrm{BS}+K\Phi(-d_2)=\\ &=S_0e^{rt}\Phi(-d_1)=\\ &=e^{\lambda t}\Phi\bigg(\frac{-a\sqrt{2\lambda}-2\lambda t}{\sqrt{2\lambda t}}\bigg) \end{aligned}$$