$E e^{B_s^2 e^{2B_s}}\mathbb{1}_{\{B_s\leq -1\}} < \infty$

Question

I wan't to see that $E e^{2B_s e^{2B_s}}\mathbb{1}_{\{B_s\leq -1\}} < \infty$ when $B_s\sim N(0,s)$ (standard brownian motion). Any help, please?

Answer 1

The expected value you are after is $$\frac{1}{\sqrt{2\pi s}}\int_{-\infty}^{-1}e^{-\frac{x^2}{2 s}}e^{e^{2x}x^2}dx=\frac{1}{\sqrt{2\pi s}}\int_{-\infty}^{-1}e^{-x^2\left(\frac{1}{2 s}-e^{2x}\right)}dx.$$ Now it is easy to show that for some $-\infty<x_0<-1$ we have that $\frac{1}{2 s}-e^{2x}>0$ for all $x<x_0$. And the result you are after follows.