I have a problem with understanding independence of a process with respect to say a given r.v $\tau$.

$B$ and $\tau$ are independent by definition iff

$P(B_{t_1} \in A_1, \dots ,B_{t_m} \in A_m, \tau \in B)=P(B_{t_1} \in A_1, \dots ,B_{t_m} \in A_m)P(\tau \in B)$

for any given choice of a finite vector of times.

Can I say that for example, (and is the first set in one of the elements of the natural filtration?):

$P(B([0,t]) \in M, \tau \in B)$

factorizes, where now we think of $B$ as a function $B:\omega\rightarrow \{t\rightarrow B_t(\omega)\}$ (into the space of continuos functions with the Borel sigma algebra $\mathcal{M}$ induced by the topology of local uniform convergence)?

I know that $B$ can be shown to be measurable but is $\sigma(B)=B^{-1}(\mathcal{M})$ equal to the sigma algebra containing the union of all the sigma algebras in the natural filtration of the brownian motion? To me it seems they are because of how the canonical version is built.

That also seems necessary to conclude that $(B, \tau)$ has the same law as $(-B,\tau)$, am I right?


  • $\begingroup$ What is $\tau$? $\endgroup$ – d.k.o. Mar 22 at 20:23
  • $\begingroup$ A given random variable. $\endgroup$ – lucmobz Mar 22 at 20:31
  • $\begingroup$ Also I am using the same name for the process $B$ and for the map $B$ into the space of continuos functions as it's done multiple times in Paolo Baldi's book. $\endgroup$ – lucmobz Mar 22 at 20:39
  • $\begingroup$ Do you mean: $\sigma(B)=\bigvee_{t\ge 0} \sigma(\{B_s: s\le t\})$ instead of the "natural filtration"? $\endgroup$ – d.k.o. Mar 22 at 21:01
  • $\begingroup$ No, when defining the Wiener measure of a continuos process, a function $\psi_B$ is defined and proved to be measurable, this function is what I called B, so I mean $\sigma(\psi_B)$, I wonder if it's equal to the sigma algebra you wrote above. $\endgroup$ – lucmobz Mar 22 at 21:28

To clarify notation let $(B_t)$ be a Brownian motion independent of $\tau$ and $\psi_B:\Omega\to\mathscr{C}$ is defined as $(\psi_B(\omega))_t=B_t(\omega)$. It suffices to show that $\psi^{-1}(\mathcal{M})\subseteq \bigvee_{t\ge 0}\sigma(\{B_s:s\le t\})\equiv \mathcal{G}$. Then for any $M\in \mathscr{M}$, there exists $G\in\mathcal{G}$ and so $$ \mathsf{P}(\psi_B\in M,\tau \in D)=\mathsf{P}(G,\tau\in D)=\ldots=\mathsf{P}(\psi_B\in M)\mathsf{P}(\tau \in D). $$

But $\mathscr{M}$ coinsides with the cylinder $\sigma$-algebra (see e.g. these notes) and the sets of the form $I_n=\{\gamma\in \mathcal{C}:(\pi_{t_1}(\gamma),\ldots,\pi_{t_n}(\gamma))\in A\}$ with $A\in \mathcal{B}(\mathbb{R}^{d\times n})$ generate $\mathscr{M}$. Moreover, $$ \psi_B^{-1}(I_n)=\{\omega\in \Omega:(B_{t_1}(\omega),\ldots,B_{t_n}(\omega))\in D\}\in \mathcal{G}. $$

  • $\begingroup$ It's not clear to me what you mean by $(B_t)$, I am sorry. In Baldi's book, $\psi_B^{-1}$ returns a subset of $\Omega$. $\endgroup$ – lucmobz Mar 22 at 21:47
  • $\begingroup$ @lucmobz You're right. I confused $\psi_B$ with the corresponding canonical process. $\endgroup$ – d.k.o. Mar 22 at 23:56
  • $\begingroup$ Ok, in the end we can say that thanks to the continuity of $B$ and measurability of $\psi_B$ we can factorize (by independence) more complex sets than finite cartesian products at finite times. Right? $\endgroup$ – lucmobz Mar 23 at 8:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.