Is $e^{\sigma B_t - \frac{\sigma^2}{2}t}$ a martingale?

Question

I'm trying to prove that $M_t = e^{\sigma B_t - \frac{\sigma^2}{2}t}$ is a martingale (for $(B_t)_{t\geq 0}$ a standard brownian motion), but I cannot show that $E|M_t| <\infty$. Could you help me?

Answer 1

If $X \sim N(\mu, \sigma^{2})$ the $Ee^{aX}=e^{a\mu}e^{a^{2}\sigma^{2}/2}$. [See https://www.statlect.com/probability-distributions/normal-distribution ]. Hence $EM_t=e^{-\sigma^{2}t /2} Ee^{\sigma B_t}=e^{-\sigma^{2}t /2}e^{\sigma^{2}t/2}=1$ for all $t$.

A similar calculation for conditional MGF given $\mathcal F_s$ using independence of $B_t-B_s$ and $\mathcal F_s$ for $s <t$ proves the martingale property.