# How do you show this 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.

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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 $u=\sigma\cdot \sqrt{t-s}$, $$\mathbb E(\mathrm e^{\sigma\cdot (B(t)-B(s))})=\mathrm e^{\sigma^2(t-s)/2}.$$
Ok so its normally distributed with mean 0 and var t-s. How do you solve $E[e^{\sigma (B(t) - B(s))}]$? – riotburn Dec 20 '12 at 7:36
You know the distribution of $B(t)-B(s)$ but you do not know how to compute $E(e^{\sigma(B(t)-B(s)})$? – Did Dec 20 '12 at 7:41
$\int e^{\sigma (x-y)} \frac{1}{\sqrt {2 \pi (t-s)}} e^{-\frac{(x-y)^2}{2(t-s)} }dy$ ? Honestly not sure, the exponential is throwing me off. – riotburn Dec 20 '12 at 7:53