Suppose $(X_i, Y_i, Z_i)$ are i.i.d. random tuples. What is $E(E(f(X_i)Y_i|X_i, Z_i))$?

Question

Suppose $(X_i, Y_i, Z_i)$ are i.i.d. random tuples. What is $E(E(f(X_i)Y_i|X_i, Z_i))$ where $f(\cdot)$ is some function?

In this case, I think it's clear to me that

$$E(E(f(X_i)Y_i|X_i)) = E(f(X_i)E(Y_i|X_i))$$

However, I'm not sure if I also condition on $Z_i$, would it be $$E(E(f(X_i)Y_i|X_i, Z_i)) = E(f(X_i)E(Y_i|X_i, Z_i))?$$

i.e., that I can pull out the $f(X_i)$ from the inner conditional expectation?

Answer 1

In this case, I think it's clear to me that

$$E(E(f(X_i)Y_i|X_i)) = E(f(X_i)E(Y_i|X_i))$$

Correct, because $f(X_i)$ is $\sigma(X_i)$-measurable.

However, I'm not sure if I also condition on $Z_i$, would it be $$E(E(f(X_i)Y_i|X_i, Z_i)) = E(f(X_i)E(Y_i|X_i, Z_i))?$$

i.e., that I can pull out the $f(X_i)$ from the inner conditional expectation?

Yes, because $f(X_i)$ is $\sigma(X_i, Z_i)$-measurable (which follows from the fact that $\sigma(X_i)$ is a subfield of $\sigma(X_i, Z_i)$).

NOTE: In both cases, the outer expected value is a distraction; the statements $$E(f(X_i)Y_i \mid X_i) = f(X_i) \cdot E(Y_i \mid X_i)$$ and $$E(f(X_i) Y_i \mid X_i, Z_i) = f(X_i) \cdot E(Y_i \mid X_i, Z_i)$$ are true -- specifically, they're almost surely true, since they're equations involving random variables.