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?
Best Answer
For $0\le s<t$,
\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 $E[e^{\sigma B_{t−s}} ] = e^{(t−s)\sigma^2/2}$.