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: https://math.stackexchange.com/questions/2020801/brownian-motions-under-different-filtrations-quadratic-covariation-covergence

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


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!

  • $\begingroup$ Is $\mathcal{F}_t^B$ required to be the natural filtration of $B$? $\endgroup$ – Nate Eldredge Apr 12 '17 at 14:27
  • $\begingroup$ Yes since $B$ is all I have on the probability space. (In my case, I have Brownian motion $B$ and an independent rv $\Psi$, and the filtration is generated by them. I didn't mention $\Psi$ to make thing simpler. ) $\endgroup$ – Ryan Apr 12 '17 at 15:37
  • $\begingroup$ @NateEldredge In fact, the question I ultimately want to understand is this: math.stackexchange.com/questions/2020801/… For now, I just fix one Brownian motion and construct the other from it. Maybe this is not the right way to go. Please let me know if you have any idea, thanks! $\endgroup$ – Ryan Apr 12 '17 at 15:54

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