Let $(X_1,Y_1)$ and $(X_2,Y_2)$ be two independent and identically distributed random pairs. Does it hold that $$E(Y_1Y_2|X_1,X_2)=E(Y_1|X_1)E(Y_2|X_2)?$$ Why?
Comments
By hypothesis, the two sigma-algebras are independent $\sigma(X_1,Y_1)\perp \sigma(X_2,Y_2)$. It implies that $\sigma(Y_1)\perp \sigma(X_2,Y_2)$ and $\sigma(Y_2)\perp \sigma(X_1,Y_1)$. So $E(Y_1Y_2|X_2)=E(Y_1)E(Y_2|X_2)$ and $E(Y_1Y_2|X_1)=E(Y_2)E(Y_1|X_1)$. Is everything correct so far? If so, my intuition says that $E(Y_1Y_2|X_2,X_1)=E(Y_1|X_1)E(Y_2|X_2)$ but I'm not quite sure how to show it.
Let $B \in \sigma(X_1,X_2)$ be a rectangle, namely $B = \{X_1\in B_1,X_2\in B_2\}$. Then $$E(Y_1Y_2\mathbf 1_B) = E(Y_1\mathbf 1(X_1\in B_1))E(Y_2\mathbf 1(X_2\in B_2)) = E(E(Y_1\mid X_1)\mathbf 1(X_1\in B_1))E(E(Y_2\mid X_2)\mathbf 1(X_2\in B_2))$$ The term on the right is $$E[E(Y_1\mid X_1)E(Y_2\mid X_2)\mathbf 1_B ]$$ To get it for general $B\in \sigma(X_1,X_2)$, use a monotone class / $\pi-\lambda$ argument.