# A Martingale Example

${{\rm exp} (\lambda B_t - \frac{1}{2} \lambda^2t)}$ is a martingale.

We need to show:

$\displaystyle {\mathbb E} (({\rm exp}(\lambda B_t - \frac{1}{2} \lambda^2t)) | {\cal F}_s)) = {\rm exp}(\lambda B_s - \frac{1}{2} \lambda^2s)$

or alternatively that

$\displaystyle {\mathbb E} (({\rm exp}(\lambda (B_t - B_s) - \frac{1}{2} \lambda^2(t-s))) | {\cal F}_s)) = 1$

But ${B_t - B_s \sim {\cal N}(0, t-s)}$ is independent of ${{\cal F}_s}$ and so

$\displaystyle {\mathbb E} (({\rm exp}(\lambda (B_t - B_s) - \frac{1}{2} \lambda^2(t-s))) | {\cal F}_s)) =$

$\displaystyle {\mathbb E} (({\rm exp}(\lambda (B_t - B_s) - \frac{1}{2} \lambda^2(t-s)))) =$

$\displaystyle \int^\infty_{-\infty} {\rm exp}(\lambda u - \frac{1}{2} \lambda^2(t-s)) \frac{{\rm exp} \frac{-u^2}{2(t-s)}}{\sqrt {2\pi(t-s)}} du =$

$\displaystyle \int^\infty_{-\infty} \frac{{\rm exp}(\frac{-(u - \lambda (t-s))^2}{2(t-s)})}{\sqrt {2\pi(t-s)}} du = 1$