Law of total expectation: $\mathbb{E}( X | A )$ and $p_{A|B}(a|b)$

Question

I've got a question regarding the law of total expectation and conditional probabilities.

Assuming I know the EV: $\mathbb{E}(X | A)$ and the PDF $p_{A|B}(a|b)$ (A,B are not independet but binomially distributed).

What is then the outcome of $\sum_{dom(A)} \mathbb{E}(X | A) ~p_{A|B}(a|b) $ ? Is it $\mathbb{E}(X | B)$?

Answer 1

I would have said something like $$\mathbb{E}(X | B=b) =\sum_{a \in dom(A)} \mathbb{E}(X | A=a, B=b) \;\Pr(A=a|B=b)$$ and this is not necessarily the same as your expression unless $\mathbb{E}(X | A=a, B=b)=\mathbb{E}(X | A=a)$, which would be the case if $X$ and $B$ were conditionally independent given $A$.