Conditional expectation with multiple conditioning For any r.v.s $X$ and $Y$:
$$E(Y|E(Y|X)) = E(Y|X)$$
But I cannot seem to be able to prove this. I tried using Adam's Law with extra conditioning ($E(Y|X) = E(E(Y|X,Z)|Z)$) but I don't seem to get anywhere with it.
What I tried is the following:
$$g(X) = E(Y|X)$$
$$E(Y|g(X)) = E(E(Y|X,g(X))|g(X))$$
Since the event $X$ happened and $g(X)$ happened are equivalent, conditioning on both $X$ and $g(X)$ is the same as conditioning on only one of them.
Is there any intuitive interpretation of this ?
Does this also mean that conditioning on $X$ or any function $g$ of $X$ is the same ?

Given the level of formality of the book, I think what the exercise is going for is primarily conceptual. I.e., what does the below conditioning mean:
E(Y|E(Y|X), X) | X = x
This represents the expectation of Y, if I know what the expectation of Y given X would be... and now I also know X! So I'm exactly in the situation where I know X, and I know what the expectation of Y given X is, which is what the expectation of Y given that conditional information is. I.e., E(Y|X). And then you Adam's Law it up via taking out what's known given X, which E(Y|X) is as that's a function of X.
(So it's more specific than just any g(X) - its essential to the argument that g(X) = E(Y|X) in particular.)
