How to prove this property of condition expectation?

Question

I'm trying to prove the following property of conditional expectation $$E[f(X)Y+g(X)|X]=f(X)E(Y|X)+g(X)$$ I know that $$E[f(X)Y+g(X)|X]=E[f(X)Y|X]+g(X)$$ But I can't see why $$E[f(X)Y|X]=f(X)E(Y|X)$$ What am I missing out? Thanks very much in advance.

Answer 1

Given $X=x$, $f(X)$ is equal to $f(x)$ and can be treated as a constant while evaluating the conditional expectation:

$$E[f(X)Y|X=x]=E[f(x)Y|X=x]=f(x)E[Y|X=x]$$

or in terms of random variables

$$E[f(X)Y|X]=f(X)E[Y|X]$$