I am reading up on the brownian motion right now and there is sometthing claimed in my textbook that I don't understand.
Let $(B_t)_t$ be a Brownian motion with respect to some filter. Now define the natural filtration of $(B_t)_t$ as $\mathcal{F}_t = \sigma (B_s : 0 \leq s \leq t)$ for $t \geq 0$. Then $(B_t)_t$ is also a Brownian motion with respect to $(\mathcal{F}_t)_t$.
I think the only property that needs checking is that for $0 \leq s \leq t$ the increment $B_t-B_s$ is independent of $\mathcal{F}_s$. But why is this true? Since there is no proof I think this should be trivial, but I can't see how.
Best Answer
The proof depends mainly on properties of multivariate normal distributions.
If $t_1 <t_2<...<t_k \leq s$ then $B_t-B_s,B_{t_1},B_{t_2},...,B_{t_k}$ have a multivariate normal distribution. Since then covariance of $B_t-B_s$ with the other variables are $0$ it follows that $B_t-B_s$ is independent of $\{B_{t_1},B_{t_2},...,B_{t_k}\}$ and this being true for all choices of $t_i$'s it follows that $B_t-B_s$ is independent of $\mathcal F_s$.