filtration generated by function of brownian motion

Question

Consider a filtration $F^{B_1,B_2}$ generated by two independent Brownian motions $B_1,B_2$.

Now I define $W:= a B_1 + \sqrt{1-a^2} B_2$ and consider the filtration $F^W$ generated by W.

Are $F^{B_1,B_2}$ and $F^W$ the same?

Answer 1

No, unless $a=0$ or $a=1$. The inclusion $\mathcal F_t^W\subset\mathcal F_t^B$ always holds as $W_t$ is a function of $(B_{1,t},B_{2,t})$.

However, for $a=1/\sqrt 2$, we get $$\tag{*} \mathbb E\left[B_{1,t}\mid \mathcal F_t^W\right]=\mathbb E\left[B_{2,t}\mid \mathcal F_t^W\right]=W_t/\sqrt 2\neq B_{1,t} $$ hence $B_{1,t}$ is not measurable with respect to $\mathcal F_t^W$.

To see the first equality in (*), observe that $B_{1,t}-B_{2,t}$ is independent of $\mathcal F_t^W$, because for $0\leqslant s_1<\dots<s_N\leqslant t$, $B_{1,t}-B_{2,t}$ is independent of $(W_{s_1},\dots,W_{s_N})$.