Conditional expectation of the product of two dependent random variables

Question

Let $A$, $B$ and $C$ be random variables, with $A$ and $B$ dependent. Is it true that

$\mathbb{E}(A\cdot B|C)=\mathbb{E}(A|C)\cdot\mathbb{E}(B|A,C)$?

In particular, can I say that \begin{equation*}\mathbb{E}(A\cdot B|C)=\mathbb{E}(\mathbb{E}(A\cdot B|A,C)|C)=\mathbb{E}(A\cdot\mathbb{E}( B|A,C)|C)=\mathbb{E}(A|C)\cdot\mathbb{E}(B|A,C)? \end{equation*}Here I have used the following two properties of the conditional expectation:

1. $\mathbb{E}(A\cdot B|F)=A\cdot\mathbb{E}(B|F)$ if $A$ is measurable with respect to the $\sigma$-algebra F
2. $\mathbb{E}(X|G)=\mathbb{E}(\mathbb{E}(X|F)|G)$ if $G\subset F$ is a $\sigma$-algebra.

If this is not true, under which conditions is it true and what can one say in general about the conditional expectation of the product of two dependent random variables?

Thank you!

Answer 1

Suppose that $\mathsf{E}[A|C]\ne0$ a.s., then $\mathsf{E}[A\cdot B|C]=\mathsf{E}[A|C] \mathsf{E}[B|A,C]$ if and only if $\mathsf{E}[B|A,C]=\mathsf{E}[B|C]$. A sufficient condition of $\mathsf{E}[A\cdot B|C]=\mathsf{E}[A|C] \mathsf{E}[B|A,C]$ is $B\underset{C}{\perp\!\!\perp}A$, i.e. $B$ and $A$ are conditionally independent given $C$(c.f. O. Kallenberg, Foundations of modern Probability, p.109--).