Modified Forms of Law of Iterated Expectation

Question

Could you please clarify how we can evaluate the below modified forms of the law of iterated expectation?

The well know law of iterated expectation states,

$$\operatorname{E} (X) = \operatorname{E} ( \operatorname{E} ( X \mid Y))$$

https://en.wikipedia.org/wiki/Law_of_total_expectation

How can this be applied to evaluate the below two modifications?

1)

$$\operatorname{E} ( g(X)\operatorname{E} ( X \mid Y)) = ?? $$

2)

$$\operatorname{E} ( h(Y)\operatorname{E} ( X \mid Y)) = ?? $$

Please let me know if you need any further information or if this is not clear.

Answer 1

Yes and yes. 1) will read $$ \operatorname{E} ( g(X)\operatorname{E} ( X \mid Y)) = \operatorname{E} ( \operatorname{E}(g(X)\mid Y)\operatorname{E} ( X \mid Y)); $$ 2) becomes $$ \operatorname{E} ( h(Y)\operatorname{E} ( X \mid Y)) = \operatorname{E} ( h(Y)X). $$

Both equations can be shown using the fact that $$ \operatorname{E} ( f(Y)Z) = \operatorname{E} ( f(Y)\operatorname{E} ( Z \mid Y)) $$ (this holds for $f(Y) = \mathbf{1}_A(Y)$ by definition and then extends by linearity and some limit procedure to "arbitrary" measurable $f$). Namely, for 1) take $Z = g(X)$, $f(Y) = \operatorname{E} ( X \mid Y)$; for 2), $Z=X$, $f(Y) = h(Y)$.