Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $B$ be a one-dimensional Brownian motion. Let $X, Y:\Omega\to \mathbb R$ be random variables.
Let $\nu: \mathbb R \times \mathcal F \to [0, 1]$ be a regular conditional probability of $\mathcal F$ given $X$, i.e.,
- for every $x \in \mathbb R$, we have $\nu(x, \cdot)$ is a probability measure on $(\Omega, \mathcal F)$,
- for every $A \in \mathcal F$, we have $\nu(\cdot, A)$ is $\mathcal B(\mathbb R)$-measurable.
- for every $A \in \mathcal F$ and $C \in \mathcal B(\mathbb R)$, $$ \mathbb P [A \cap \{X \in C\}] = \int_C \nu(x, A) \ \mathrm d (X_\sharp \mathbb P) (x), $$ where $X_\sharp \mathbb P$ is the push-forward of $\mathbb P$ by $X$.
Then we have a straightforward result, i.e.,
Theorem If $X,Y,B$ are pairwise independent under $\mathbb P$, then $Y, B$ are conditionally independent given $X$.
Proof Let $E := \mathcal C([0, \infty); \mathbb R)$ be the space of paths of $B$. Let $A \in \mathcal B (\mathbb R)$ and $C \in\mathcal B (E)$. By this lemma, $$ \begin{align} \nu(X, \{Y\in A\} \cap \{B \in C\}) &= \mathbb P [\{Y\in A\} \cap \{B \in C\} |X] \quad \text{a.s.} \\ \nu(X, \{Y\in A\}) &= \mathbb P [Y\in A | X] \quad \text{a.s.},\\ \nu(X, \{B\in C\}) &= \mathbb P [B\in C | X] \quad \text{a.s.} \end{align} $$ By pairwise independence of $X,Y,B$, $$ \begin{align} \mathbb P [\{Y\in A\} \cap \{B \in C\} |X] &= \mathbb P [Y\in A ] \cdot \mathbb P [ B \in C] \quad \text{a.s.}, \\ \mathbb P [Y\in A | X] &= \mathbb P [Y\in A] \quad \text{a.s.},\\ \mathbb P [B\in C | X] &= \mathbb P [B\in C] \quad \text{a.s.} \end{align} $$ The claim then follows.
It is mentioned in this answer that
Here independence of $B$ from $X$ guarantees that $B$ is still an independent collection of Brownian motions under each $\nu (x, \cdot)$.
Could you explain how to prove above statement?
