Is 1-d Brownian filtration rich enough to admit a 2-d Brownian motion?

Question

Let $B_t$ be the standard 1-d Brownian motion, and let $\mathcal F^B_t$ be the induced filtration. Is it possible to construct a 2-d Brownian motion adapted to $\mathcal F^B_t$?

[EDIT] Come to think of it, I guess it is doable. By picking different segments of 1-d path, shifting and scaling, we can construct 2-d Brownian motion out of the 1-d Brownian motion. So, let's make it more challenging:

Is it possible to construct a 2-d Browning motion $(B_t^{(1)}, B_t^{(2)})$ which is adapted to $(\mathcal F^B_t)$, and $\mathcal F_t^{(B^{(1)}, B^{(2)})}=\mathcal F^B_t$ for all $t$.

Answer 1

Here is the idea for the first part.

In the canonical space, let $B_{[t_1\ t_2]}$ represent a sample path between $[t_1\ t_2]$. We divide $B_{[0\ \infty)}$ into segments: $B_{[2^{-(n+1)}\ 2^{-n}]}$, for $n=...,2,1,0,-1,-2,...$. Now, let's remove all the segments with $n$ being odd, and use the remaining segments to construct another path $W_{[0\ \infty)}$ -- and by similar construction, we can use the ``removed'' segments to construct yet another path (the second component of the 2-d Brownian motion)

First step: we scale (both in time and value) each remaining segment $B_{[2^{-(2k+1)}\ 2^{-2k}]}$ to fill up the hole between $2^{-2k}$ and $2^{-2k+1}$, and define for $t\in (2^{-(2k+1)}, 2^{-2k+1})$ for any integer $k$ ($k$ can be negative).

$W(t) = \sqrt{3} B(2^{-(2k+1)} + \frac{t-{2^{-(2k+1)}}}{3})$

Now, to make the paths continous at the end points, we define for $t\in (2^{-(2k+1)}, 2^{-2k+1}]$,

$\tilde W(t) = W(t) + \sum_{i=k}^\infty \left [ W_+(2^{-(2i+1)}) - W_-(2^{-(2i+1)}) \right ] $

I believe such constructed $\tilde W(t)$ is a Brownian motion, which is adpated to $(\mathcal F^{B}_t)$, and should be independent of the second component (constructed using the ``removed segments'').

There are still some details to be figured out around the convergency of $\sum_{i=k}^\infty \left [ W_+(2^{-(2i+1)}) - W_-(2^{-(2i+1)})\right ]$. I think we can prove the convergency using the iterated logrithm theorem.

Answer 2

The second part is impossible.

Let $B_t$ be a one-dimensional Brownian motion with natural filtration $\mathcal{F}_t$. Suppose that there is a process $(W_t^{(1)}, W_t^{(2)})$ which is a 2-dimensional Brownian motion with respect to the filtration $\mathcal{F}_t$. (This would certainly be true if $(W_t^{(1)}, W_t^{(2)})$ were a Brownian motion whose natural filtration was equal to $\mathcal{F}_t$.)

In particular, $(W_t^{(1)}, W_t^{(2)})$ is a martingale with respect to $\mathcal{F}_t$. By the martingale representation theorem, there are predictable $B_t$-integrable processes $\sigma^{(1)}_t, \sigma^{(2)}_t$ such that $W_t^{(i)} = \int_0^t \sigma^{(i)}_s\,dB_s$ for $i=1,2$.

Now by considering quadratic variation we have $$t = [W^{(i)}]_t = \int_0^t \left|\sigma_t^{(i)}\right|^2\,dt \quad \forall t \ge 0, \text{ a.s.}$$ which by the fundamental theorem of calculus implies $$\left|\sigma_t^{(i)}\right| = 1 \quad \forall t \ge 0, \text{ a.s.} \tag{1}$$ But by considering quadratic covariation we have $$0 = [W^{(1)}, W^{(2)}]_t = \int_0^t \sigma_t^{(1)}\sigma_t^{(2)}\,dt \quad \forall t \ge 0, \text{ a.s.}$$ and thus $$\sigma_t^{(1)}\sigma_t^{(2)} = 0 \quad \forall t \ge 0, \text{ a.s.} \tag{2}$$ Now (1) and (2) are contradictory.

Note: I previously thought this argument contradicted the first part too. The error was that if we just know that $W_t$ is a Brownian motion which is adapted to a filtration $\mathcal{F}_t$, it does not follow that it is a Brownian motion, or even a martingale, with respect to that filtration. The obvious counterexample is $W_t = \sqrt{2} B_{t/2}$.