# Uniqueness of Brownian motion

May be it is a dumb question, but it vexed me a little bit. I understand the construction of the Brownian motion (first use Kolmogorov extension theorem to construct value at dyadic times and then use (Kolmogorov again?) continuity theorem to fill in the gaps). In short we get a measurable map $f: (\Omega, \mathcal F, P) \to \mathbb R^{[0, +\infty)}$ such that the trajectories are a.s. continuous, of independent increments as Gaussian r.v., etc.

However, I'm considering the following (maybe too pedantic) question: Suppose there is another measurable map $\tilde f: (\tilde \Omega, \mathcal{\tilde F}, \tilde P) \to \mathbb R^{[0, +\infty)}$ whose trajectories are a.s. continuous and has finite dimensional distribution identical to the Brownian motion (constructed above). Then is there a measure preserving isomorphism (maybe modulo the null sets) $\phi: (\Omega, \mathcal F, P) \to (\tilde \Omega, \mathcal{\tilde F}, \tilde P)$ such that f = φf $\tilde f = f\phi$? In other words is there a "universal" (in the sense of category theory) Brownian motion. Maybe some requirements on the space $(\tilde \Omega, \mathcal{\tilde F}, \tilde P)$ is necessary, in which case I'll assume it to be the Standard Borel probability space.

Also a side remark: is such consideration really important in probability theory? Or are we satisfied with equivalences of stochastic processes (having the same finite dimensional distribution), which I suppose is weaker than measure-preserving "isomorphisms"?

Edit: I made a silly mistake in the expression of the (supposed) universal property. It should be $\tilde f = f \phi$ instead of $\tilde f = \phi f$.

• To answer your last sentence, having a measure preserving isomorphism does not imply having the same finite dimensional distributions. Just take some non iid process on $\{0,1\}^\mathbb{N}$ and permute coordinates appropriately. There will be a measure preserving isomorphism, but the finite dimensional distributions will be different. Sep 14, 2013 at 6:38
• Hope I get it right this time... Sep 15, 2013 at 0:41
• Can't you show that all these spaces are isomorphic to $[0,1]$ with the Lebesgue measure?
– Ilya
Jul 28, 2014 at 10:56
• What is a measure preserving isomorphism (sorry, just asking out of curiosity). And: as standard Borel space you mean $\mathbb{R}^I?$ Jan 19, 2016 at 7:56

Consider the measurable space $(C,\mathcal{C})$, where $C$ is the set of continuous functions from $[0,\infty)$ to $\mathbb{R}$ and $\mathcal{C}$ is the $\sigma$-algebra induced by the coordinate projections. If $P$ and $Q$ are two probability measures both satisfying the requirements to be the distribution of a Brownian motion, is $P=Q$?
In this sense, uniqueness is about distributions and not mappings or random variables. Also, the question of uniqueness is rather trivial compared to the question of existence: Since the $P$ and $Q$ measures above have the same finite-dimensional distributions, they are equal on a generating system for $\mathcal{C}$ which is stable under intersections, and therefore standard uniqueness results for probability measures yields $P = Q$.
Also, in probability theory, usually the main result referred to when considering existence of the Brownian motion is the existence of a probability measure on $(C,\mathcal{C})$ satisfying the requirements to be a Brownian motion, and not an actual mapping. If an actual stochastic process is required, one can just use the identity mapping on $(C,\mathcal{C},P)$. These are somewhat abstract considerations, but are useful when making precise what one is constructing. See also the books by Rogers & Williams, in particular Chapter II of Volume I, for more on this.