If $(B_t)$ is a Brownian motion and adapted to the filtration $(\mathcal F_t)$ does $B_t-B_s$ is independent of $\mathcal F_s$? Let $(B_t)$ a Brownian motion adapted to the filtration $(\mathcal F_t)$.
Q1) If $t> s$, does $B_t-B_s$ is necessarily independent of $\mathcal F_s$ or $(\mathcal F_t)$ must be the natural filtration of $(B_t)$ to have $B_t-B_s$ independent of $\mathcal F_s$ ?
Q2) Does $(B_t)$ is necessarily a martingale w.r.t. $(\mathcal F_t)$ or $(\mathcal F_t)$ must be the natural filtration for $(B_t)$ being a martingale w.r.t. to $(\mathcal F_t)$ ?

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*Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $(B_t)$ a Brownian motion. Take $\mathcal F_t=\mathcal F$ for all $t$. Then $(B_t)$ won't be a martingale. 

*Notice that if $B_t-B_s$ is independent of $\mathcal F_s$ for all $t>s$, then $(B_t)$ will be a martingale. 

*If $(\mathcal F_t)$ is a filtration adapted to $(B_t)$ and such that $B_t-B_s$ is independent of $\mathcal F_s$ for all $t>s$, then $(\mathcal F_t)$ is called an admissible filtration.
