Why $\mathbb E[e^{i\xi(X_t-X_s)}1_F]=e^{-\frac{1}{2}(t-s)\xi^2}\mathbb P(F)$ for all $F\in \mathcal F_s$ implies that $X_t-X_s\perp F_s$?

Question

Let $(F_t)$ a filtration. In the proof of $(X_t,F_t)$ and $(X_t^2-t,F_t)$ are martingale implies that $(X_t$ is a Brownian motion, they say that it's enough to prove that $$\mathbb E[e^{-i\xi(X_t-X_s)}1_F]=e^{-\frac{1}{2}(t-s)\xi^2}\mathbb P(F)$$ for all $F\in F_s$.

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So I agree that this will implies that $X_t-X_s\sim \mathcal N(0,t-s)$, but why will we get that $X_t-X_s$ is independent of $F_s$ from that ?

Answer 1

For every $F_s$-simple function $Y$ we get $Ee^{-i\xi (T_t-X_s)} Y=Ee^{-\frac 1 2 (t-s)\xi^{2}} EY$. This extends to bounded measurable functions $Y$ so we get $Ee^{-\xi (X_t-X_s)} e^{i\eta Y} =Ee^{-i\xi (X_t-X_s)} Ee^{i\eta Y}$ for any $F_s$- measurable function $Y$ and any real number $\eta$. This implies that $X_t-X_s$ is independent of $Y$. Since $Y$ is arbitrary we see that $X_t-X_s$ is independent of $F_s$.

Answer 2

By Lévy's inversion formula (and the integrability of the normal characteristic function), $$ \eqalign{ \Bbb P[\{a<X_t-X_s\le b\}\cap F] &={1\over 2\pi}\int_{\Bbb R}{e^{-i\xi a}-e^{-i\xi b}\over it} \Bbb E[e^{i\xi(X_t-X_s)}1_F]\,d\xi\cr &={1\over 2\pi}\int_{\Bbb R}{e^{-i\xi a}-e^{-i\xi b}\over it} \Bbb E[e^{i\xi(X_t-X_s)}]\Bbb P[F]\,d\xi\cr &=\Bbb P[a<X_t-X_s\le b]\cdot\Bbb P[F].\cr } $$ for all $a<b$.