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?
Thanks.
Best Answer
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, $$ \psi_B^{-1}(I_n)=\{\omega\in \Omega:(B_{t_1}(\omega),\ldots,B_{t_n}(\omega))\in D\}\in \mathcal{G}. $$