What does it mean for a stochastic process to be independent of a sigma algebra?

Question

Does anybody know what it means for a stochastic process $ X = (X_t)_{t \geq 0} $ on a filtered probability space $ (\Omega, \mathcal{F}, \mathbb{F}, P) $ to be independent of a sigma-algebra $ \mathcal{G} \subset \mathcal{F}$ ?

Is this the same as saying that $ \forall 0 \leq t_1 < ... < t_n $ the vector $ (X_{t_1}, ... , X_{t_n}) $ is independent of $ \mathcal{G} $ ?

Thanks for the clarification!

Regards, Si

Answer 1

I think the natural definition would be that the $\sigma$-algebra $\sigma(X_t : t\ge 0)$ generated by the process should be independent of $\mathcal{G}$. That is, for every $A \in \sigma(X_t : t \ge 0)$ and every $B \in \mathcal{G}$, $P(A \cap B) = P(A) P(B)$.

However, this is equivalent to your statement. One can use Dynkin's $\pi$-$\lambda$ lemma to prove it.

Answer 2

We can more generally define the independence between two processes $X=(X_t)_t$ and $Y=(Y_t)_t$ by the following way: if $\mathcal F_X$ and $\mathcal F_Y$ are the smallest $\sigma$-algebra making respectively $X$ and $Y$ measurable, we say that $X$ and $Y$ are independent if and only if so are $\mathcal F_X$ and $\mathcal F_Y$.

We can show the equivalence between

• $X$ and $Y$ are independent;
• The finite-dimensional distributions of $X$ and $Y$ are independent.

If we assume that the finite-dimensional distributions of $X$ and $Y$ are independent, then consider $\pi_1$ and $\pi_2$, the sets which consists of finite intersection of sets of the form $X_t^{-1}(B)$ (respectively $Y_t^{-1}$), where $B$ is a Borel set. These one are $\pi$-systems (i.e. stable by finite intersection), and a monotone class result ensure us that the monotone class generated by $\pi$ and $\pi_2$ are independent.