How to show that $E[e^{\sigma B(t) - \frac{\sigma ^2 t}{2}} | \mathscr{F}(s)]$ is a martingale? How do you show the following process is a martingale?  My notes say it is a martingale by I can't work it out.
$$
E[e^{\sigma B(t) - \frac{\sigma ^2 t}{2}} | \mathscr{F}(s)]
$$
I tried to multiply and divide by $e^{\sigma B(s) - \frac{\sigma ^2 s}{2}}$ but got stuck.
 A: The $\sigma^2t$ term is deterministic hence one can move it out of the conditional expectation. The $B(t)$ term is $(B(t)-B(s))+B(s)$. The $B(s)$ term is $\mathscr F_s$-measurable hence one can move it out of the conditional expectation. The $B(t)-B(s)$ term is independent of $\mathscr F_s$ hence one can integrate it. To summarize,
$$
\mathbb E(\mathrm e^{\sigma B(t)-\sigma^2t/2}\mid\mathscr F_s)=\mathbb E(\mathrm e^{\sigma\cdot (B(t)-B(s))})\,\mathrm e^{\sigma B(s)}\,\mathrm e^{-\sigma^2t/2}.
$$
The next step is to compute $\mathbb E(\mathrm e^{\sigma\cdot  (B(t)-B(s))})$. To do that, one needs to identify the distribution of $B(t)-B(s)$, and maybe you can proceed from there.
Edit: To be honest, I fail to grasp how one can ask questions about Brownian martingales and be unable to compute basic Gaussian integrals... but here we go: the random variable $\sigma\cdot (B_t-B_s)$ is centered normal with variance $\sigma^2\cdot (t-s)$ and, for any standard normal random variable $X$ and any real number $u$,
$$
\mathbb E(\mathrm e^{uX})=\int_{-\infty}^{+\infty}\frac1{\sqrt{2\pi}}\mathrm e^{ux}\mathrm e^{-x^2/2}\mathrm dx=\int_{-\infty}^{+\infty}\frac1{\sqrt{2\pi}}\mathrm e^{u^2/2}\mathrm e^{-(x-u)^2/2}\mathrm dx=\mathrm e^{u^2/2}.
$$
In particular, for $u=\sigma\cdot \sqrt{t-s}$,
$$
\mathbb E(\mathrm e^{\sigma\cdot  (B(t)-B(s))})=\mathrm e^{\sigma^2(t-s)/2}.
$$
