Equivalent condition for a stochastic process to be independent of a $\sigma$-algebra

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra on $\Omega$
• $I$ be a set
• $(E_i,\mathcal E_i)$ be a measurable space for $i\in I$, $$(E,\mathcal E):=\left(⨉_{i\in I}E_i,\bigotimes_{i\in I}\mathcal E_i\right)$$ and $$\pi_J:E\to⨉_{j\in J}\;,\;\;\;x\mapsto\left.x\right|_J$$ for $J\subseteq I$ (if $J=\left\{i\right\}$, we write $\pi_i$ instead of $\pi_J$)
• $X:\Omega\to E$

How can we show that $X$ is independent of $\mathcal F$ if and only if $\pi_J\circ X$ is independent of $\mathcal F$ for all $J\subseteq I$ with $|J|\in\mathbb N$?

Answer 1

Let's start with "$\Rightarrow$": Let $J\subseteq I$. Note that $\pi_J$ is $\left(\mathcal E,\bigotimes_{j\in J}\mathcal E_j\right)$-measurable and hence $\pi_J\circ X$ is $\left(\sigma(X),\bigotimes_{j\in J}\mathcal E_j\right)$-measurable. Thus, $$\sigma\left(\pi_J\circ X\right)\subseteq\sigma(X)\tag1$$ yielding the claim.

For the other direction, we'll need the following fact:

If $\mathcal E_i\subseteq\mathcal A$ and $\mathcal E_i\cup\left\{\emptyset\right\}$ is a $\pi$-System, then $\mathcal E_1$ and $\mathcal E_2$ are independent if and only if $\sigma(\mathcal E_1)$ and $\sigma(\mathcal E_2)$ are Independent.

$\Leftarrow$: By Definition, $$\mathcal E=\sigma(\pi_i:i\in I)\tag2$$ and hence $$\sigma(X)=\sigma\left(X^{-1}\left(\bigcup_{i\in I}\sigma(\pi_i)\right)\right)\tag3.$$ Now, $$X^{-1}\left(\bigcup_{i\in I}\sigma(\pi_i)\right)=\bigcup_{i\in I}\sigma(\pi_i\circ X).\tag4$$ The only problem is that $(4)$ doesn't need to be a $\pi$-system (if I'm not mistaken). However, $$\mathcal R:=\bigcup_{\substack{J\subseteq I\\|J|\in\mathbb N}}\sigma\left(\pi_J\circ X\right)$$ is a $\pi$-System. By assumption, $\mathcal R$ and $\mathcal F$ are independent and hence we obtain the claim by $(3)$ and the mentioned fact.