Imagine we have $X$ and $Y$ defined on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$ such that $\mathbb{E}|X|, \mathbb{E}|Y|, \mathbb{E}|XY| < \infty$. Let $\mathcal{A}$ and $\mathcal{B}$ be be $\sigma$-fields such that $\mathcal{A} \subset \mathcal{B} \subset \mathcal{F}$.
Assume that $X$ has the property that $\mathbb{E}(X| \: \mathcal{B}) = \mathbb{E}(X| \: \mathcal{A})$ and $Y$ has the property that $\mathbb{E}(Y| \: \mathcal{B}) = \mathbb{E}(X| \: \mathcal{A})$
Must it be the case that $\mathbb{E}(XY| \: \mathcal{B}) = \mathbb{E}(XY| \: \mathcal{A})$?
I cannot find a simple counterexample but I cannot prove it either. I have tried to derive it when $X$ and $Y$ are indicator functions and playing around with the tower property. If it is not true, is there a situation where it could be true (i.e. $X$ and $Y$ are non-negative, $\mathcal{A}$ and $\mathcal{B}$ are countable,...)
Thanks!
Let $(\Omega,\mathcal{F},\mathbb{P}) = ([0,1),\mathcal{B}([0,1]),\mu)$ where $\mu$ is the Lebesgue measure. Let $\mathcal{A}$ be the trivial sigma algebra, and $\mathcal{B} = \sigma([0,1/2),[1/2,1))$.
Define,
\begin{align*} X(\omega) &= \begin{cases} 4\omega &\text{ if } \omega \in [0,1/2)\\ 4\omega - 2 &\text{ if } \omega \in [1/2,1) \end{cases}\\ Y(\omega) &= \begin{cases} 1 &\text{ if } \omega \in [0,1/2)\\ 4\omega-2 &\text{ if } \omega \in [1/2,1) \end{cases} \end{align*}
Then for all $\omega\in \Omega$,
$$\mathbb{E}(X|\mathcal{A})(\omega) = \mathbb{E}(X|\mathcal{B})(\omega) = \mathbb{E}(Y|\mathcal{A})(\omega) = \mathbb{E}(Y|\mathcal{B})(\omega) = 1.$$
For all $\omega\in\Omega$,
$$\mathbb{E}(XY|\mathcal{A})(\omega) = \mathbb{E}(XY) = 1/2 + 2/3 = 7/6.$$
But,
$$\mathbb{E}(XY|\mathcal{B})(\omega) = \begin{cases} 1 &\text{ if } \omega \in [0,1/2)\\ 4/3 &\text{ if } \omega \in [1/2,1) \end{cases}.$$
So $\mathbb{E}(XY|\mathcal{A})\neq\mathbb{E}(XY|\mathcal{B})(\omega).$
Intuition: $\mathcal{A}$ is more coarse than $\mathcal{B}$, so its easiest to look for understanding or counterexamples in simple finite sigma algebras like I presented above. All we need are $X$ and $Y$ that have a constant average on every set in $\mathcal{B}$, but that achieve that average in different ways on different sets of $\mathcal{B}$. Then that average is no longer preserved when $X$ and $Y$ are multiplied together, so $XY$ has different averages on different sets in $B$. This ensures that $\mathbb{E}(XY|\mathcal{B})$ is not $\mathcal{A}$-measurable. I hope that makes sense.