Given a fixed filtered probability space, can we prove the existence of a Brownian motion on it?

Question

Question 1$\;\;\;$Let $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge 0},\operatorname P)$ be a given filtered probability space. Can we prove that there is a Brownian motion on $(\Omega,\mathcal A,\mathcal F,\operatorname P)$?

All I know is that we can find some $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge 0},\operatorname P)$ such that a Brownian motion on this filtered probability space exists. For example, we can choose $\Omega=\mathbb R^{[0,\infty)}$, $\mathcal A=\mathcal B(\mathbb R)^{\otimes[0,\infty)}$ and $$X_t(\omega):=\omega(t)\;\;\;\text{for }(\omega,t)\in\Omega\times[0,\infty)\;.$$ Then, it's possible to construct a probability measure $\operatorname P$ on $(\Omega,\mathcal A)$ such that $X$ has independent, stationary and normally distributed increments (so, $\mathcal F$ will be the $\sigma$-algebra generated by $X$). The continuity can be concluded by the Kolmogorov-Chentsov theorem.

Question 2$\;\;\;$I suppose the answer to question 1 will be no. In that case, I'm curious about the role of $\mathcal F$. As I said above, we can show the existence of a probability space such that there is a Brownian motion with respect to their generated $\sigma$-algebra. If that's all we can prove, then the role of $\mathcal F$ seems to be very negligible .

Answer 1

It is known that a probability space is atomless if and only if there exists a random variable with continuous distribution. Therefore, in particular, a probability space with atoms cannot have a normal random variable. Thus, the answer is no.

On the other hand, the Levy-Ciesielski construction of the Brownian motion relies on the existence of a sequence of independent standard normal random variables. It is also known that if a probability space is atomless, then for any non-degenate probability distribution there exists a sequence of independent random variables with that distribution (see, for instance the book "Monetary Utility Functions" by Delbaen, where is stated a characterization of atomless probability spaces) and therefore a Brownian motion can be constructed by using the Levy-C. method.

As conclusion, for a given probability space it has a Brownian motion if and only if it is atomless.

Answer 2

You would be hard put to find a Brownian motion with respect to the natural filtration of a standard Poisson process. For example, there is that time interval until the first jump of the Poisson process in which nothing happens, and during that time interval there is not enough randomness to drive a Brownian motion.