# If $X_2$ is independent of $\mathcal F$, can we show that $f(X_1,X_2)$ is conditionally independent of $\mathcal F$ given $X_1$?

Let

• $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space
• $$\mathcal F\subseteq\mathcal A$$ be a $$\sigma$$-algebra on $$\Omega$$
• $$(E_i,\mathcal E_i)$$ be a measurable space
• $$X_i$$ be an $$(E_i,\mathcal E_i)$$-valued random variable on $$(\Omega,\mathcal A,\operatorname P)$$
• $$f:E_1\times E_2\to E_3$$ be $$(\mathcal E_1\otimes\mathcal E_2,\mathcal E_3)$$-measurable
• $$X_3:=f(X_1,X_2)$$

Assuming $$X_2$$ is independent of $$\mathcal F$$, are we able to show that $$X_3$$ is conditionally independent of $$\mathcal F$$ given $$X_1$$, i.e. $$\operatorname P\left[X_3\in B_3,F\mid X_1\right]=\operatorname P\left[X_3\in B_3\mid X_1\right]\operatorname P\left[F\mid X_1\right]\;\;\;\text{almost surely}\tag1$$ for all $$B_3\in\mathcal E_3$$ and $$F\in\mathcal F$$?

Let $$B_3\in\mathcal E_3$$ and $$F\in\mathcal F$$. We need to prove that $$\operatorname P\left[X_1\in B_1,X_3\in B_3,F\right]=\operatorname E\left[1_{\{\:X_1\:\in\:A\:\}}\operatorname P\left[X_3\in B_3\mid X_1\right]\operatorname P\left[F\mid X_1\right]\right]\tag2.$$ What's the easiest way to show $$(2)$$? Maybe we are able to reduce the problem to the case $$f^{-1}(B_3)=A_1\times A_2$$ for some $$A_i\in\mathcal E_i$$, but I'm missing the right argument for that.

EDIT: If necessary, feel free to impose a stronger notion of measurability on $$f$$.

• $$\Omega = \{\omega = (\omega_1,\omega_2)\mid \omega_i \in \{0,1\}\}$$,
• $$\mathrm P(\omega) = 1/4, \omega \in \Omega$$,
• $$X_i(\omega) = \omega_i$$, $$i=1,2$$;
• $$X_3 = (X_1+X_2) \mod 2$$, $$\mathcal F = \sigma(X_3)$$.
Then, $$X_1,X_2,X_3$$ are pairwise independent, but not jointly independent. So for $$B_3 = \{0\}$$ and $$F = \{X_3 = 0\}$$ $$\mathrm P \left[X_3=0,F\mid X_1\right] = \mathrm P [F] = 1/2\neq 1/4 = \mathrm P[F]^2 = \mathrm P \left[X_3=0\mid X_1\right] \cdot \mathrm P \left[F\mid X_1\right].$$
Note that in this counterexample we even have that $$X_1$$ and $$X_2$$ are independent and $$X_1$$ and $$\mathcal F$$ are independent. So it is hard to imagine an additional assumption which would make this true (except that the pair $$(X_1,X_2)$$ is independent of $$\mathcal F$$, which makes this trivial).