# Show a stochastic process is a Martingale

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?

For $$0\le s,
\begin{align} E[X_t|B_r]&=&E[e^{-t\sigma^2/2+\sigma B_t}|B_r, 0\le r\le s]\\ &= & e^{-t\sigma^2/2} E[e^{\sigma (B_t-B_s+B_s}|B_r, 0\le r\le s]\\ &= & e^{-t\sigma^2/2} E[e^{\sigma B_{t-s}}] e^{\sigma B_{s}}\\ &= & e^{-t\sigma^2/2} e^{(t-s)\sigma^2/2} e^{\sigma B_{s}}\\ &=& e^{-s\sigma^2/2+\sigma B_{s}}\\ &=& X_s \end{align} Thus, $$(X_t)_{t>0}$$ is a martingale with respect to the Brownian motion. The fourth inequality is because the moment generating function of a normal random variable with mean 0 and variance $$t − s$$ is $$E[e^{\sigma B_{t−s}} ] = e^{(t−s)\sigma^2/2}$$.