I am trying to understand how the natural filtration for a Brownian motion might look like.

Definitions: I will start with the definitions for reference. The definition of a natural filtration is $$\mathcal{F}^W(t) = \sigma(\{W_s; 0\le s \le t\}),\quad\text{for all } t\in [0,T].$$ Where for a subset $\mathcal{M}\in \mathcal{P}(\Omega)$ the $\sigma$-Operator is defined as $$\sigma(\mathcal{M})=\bigcap_{\mathcal{A}\in\mathcal{F}(\mathcal{M})}\mathcal{A}.$$ Finally, $\mathcal{F}(M)$ is the set of all $\sigma$-Algebras which contain $\mathcal{M}$: $$\mathcal{F}(M)=\{\mathcal{A} \subseteq \mathcal{P}(\Omega)|\mathcal{M}\subseteq\mathcal{A}, \mathcal{A}\text{ is }\sigma\text{-Algebra}\}$$

Question: How does the natural filtration of a Brownian motion look like?

The concept is still fuzzy to me, so I can't formulate the question as one single statement, but here are the points I find puzzling:

  • Is the intuitive idea correct that the $\sigma$-Operator makes all the sub-sections of the path of the Brownian motion also measurable?
  • Why then do I need to bring in the set of all possible $\sigma$-Algebras into the definition, why isn't it possible to define the measurable sets by intersections of my original $\{W(s); 0\le s \le t\}$?

Here are related questions for further reference:


1 Answer 1


Basically you are confusing two notations here. In fact, $\mathcal F^W(t)$ is the smallest sigma-algebra which contains every event $\{W_s\in B\}$ with $s$ in $[0,t]$ and $B$ in $\mathcal B(\mathbb R)$.

The smallest sigma-algebra which contains every event $\{X\in B\}$ with $X$ in some collection of random variables $\mathcal X$ and $B$ in $\mathcal B(\mathbb R)$ is often denoted $\sigma(\mathcal X)$, in your case $\mathcal X$ is the collection $\mathcal X=\{W_s\,;\,0\leqslant s\leqslant t)$. Finally, recall that $\{X\in B\}$ is a notation for $\{X\in B\}=X^{-1}(B)$.

  • 3
    $\begingroup$ $\mathcal F^W(t)$ contains no path, it contains only events. Roughly speaking, an event is in $\mathcal F^W(t)$ if it depends only on $(W_s)_{s\leqslant t}$. $\endgroup$
    – Did
    Feb 28, 2014 at 14:20

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