Suppose we are given a Brownian motion $\{B_t,\mathcal F_t^B\}_{t\in[0,1]}$. I would like to know if it is possible to construct another Brownian motion $\{W_t,\mathcal F_t^W\}_{t\in[0,1]}$ from $B$, which is different from $B$, such that it is adapted to the filtration $\mathcal F^B$ BUT IS NOT A MARTINGALE UNDER $\mathcal F^B$, and that their joint quadratic variation $[W,B]$ is not zero (the joint quadratic covariation is defined as the limit of approximating sum). (This final condition should exclude the trivial case $W_t=\sqrt 2B_{t/2}$)
If this is the case, is it possible to also have $W$ to be independent of $B$? Is it possible to have $[W,B]_1=a$ for any $a\in(-1,1)$? This question is related to a previous question which has not yet been answered: Brownian motions under different filtrations, quadratic covariation, covergence of approximating sum
For the first part of my question, I have tried a Brownian bridge type construction: Let $\mathcal D_n=\{k/2^n:k=0,1,...,2^n\}=\{t_k^n:k=0,1,...,2^n\}$. For $t\in[t_{k-1}^n,t_k^n]$, define recursively
$W_t^{n+1}=\frac{t-t_{k-1}^n}{t_k^n-t_{k-1}^n}W_{t_k^n}^n+\frac{t_k^n-t}{t_k^n-t_{k-1}^n}W_{t_{k-1}^n}^n+(2B_{t_{2k-1}^{n+1}}-B_{t_k^n}-B_{t_{k-1}^n})F_{k,n+1}(t)$
where $F_{k,n}$ is the Shauder functions. Clearly, this is not adapted to the filtration generated by $B$. So I want to perform a rescale of time (or translation of time) so that it would be adapted. Then, pass $n$ to infinity and hopefully the limit $W$ will be a Brownian motion that satisfies the desired property. Unfortunately, this does not work out very well.
I am now considering Gaussian process if the form $W_t=\int_0^t h(t,s)dB_s$, where $h$ is a Volterra kernal. Are there any condition on $h$ so that $W$ is a Brownian motion in its filtraton? Any such example will helps since I do not have much intuition on this.
Any help would be greatly appreciated!
Let $H=(H_s)$ be an $(\mathcal F^B_t)$-predictable process taking only the values $\pm 1$. The stochastic integral $W_t:=\int_0^t H_s\,dB_s$ is a Brownian motion with respect to the filtration of $B$, and $\langle W,B\rangle = H$.
Conversely, if $(W_t)$ is a Brownian motion with respect to $(\mathcal F^B_t)$ then $W$ admits a stochastic integral representation as above, and because $t=\langle W\rangle_t=\int_0^t H_s^2\,ds$, you have $H_s(\omega)=\pm 1$ for a.e. $(\omega,s)$. In particular, it is not possible to have $W$ independent of $B$.