B$_t$ is a Brownian motion starting from 0. For any fixed constant $\sigma$ $>$ 0, X$_t$ = e $^{\sigma B_t - \sigma^2t/2}$, t$>$0 is a martingale w.r.t. the filtration generated by Brownian motion B$_t$.
From the definition of Martingale, now I stuck in the third property, how can we show
E( X$_s$ | $\mathcal{F}$$_{t}$) = X$_t$ ? ( s $\geq$ t)
My attempt follows:
E( X$_s$ | $\mathcal{F}$$_{t}$) = E( e $^{\sigma B_s - \sigma^2s/2}$ | $\mathcal{F}$$_{t}$} = $\frac{1}{e^{\sigma^2s/2}}$ * E(e$^{\sigma B_s}$ | $\mathcal{F}_t$), then I show
E(e$^{\sigma B_s}$ | $\mathcal{F}_t$) = E(e$^{\sigma B_s}$) by using some properties of Brownian motion. And feel confusing to continue to show the goal. Any help?