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 ?
 A: The argument you already have is a pretty good non-measure theory argument.  I will just formalize that below, it may help to give confidence about some details.
Using your argument structure: Let $g(X)=E[Y|X]$.  Then
\begin{align}
E[Y|g(X)] &\overset{(a)}{=} E[E[Y|g(X),X]|g(X)]\\
&\overset{(b)}{=} E[E[Y|X]|g(X)]\\
&=E[g(X)|g(X)]\\
&\overset{(c)}{=}g(X)
\end{align}
where (a) uses the law of iterated expectations; (b) uses $E[Y|g(X),X]=E[Y|X]$; (c) uses $E[Z|Z]=Z$ for any random variable $Z$. $\Box$

The step (b) more closely examined is:
$$E[Y|g(X),X]=E[Y|X]$$
and this intuitively means that if we already know $X$, then the additional information $g(X)$ adds nothing new.

Notes:

*

*Conditioning on $X$ is generally not the same as conditioning on $g(X)$, but it works in this particular problem.


*A measure-theory derivation could be given along the lines of my first comment on your answer.  You can also justify $E[Y|g(X),X]=E[Y|X]$ more formally by measure theory ("the sigma algebra generated by $(g(X),X)$ is the same as the sigma algebra generated by $X$").


*A formal measure theory definition talks about "versions of" a conditional expectation, and I do not go into such detail in this answer (some people may want to replace my equalities with equalities that hold "with probability 1").
A: This is a special case of the Tower Property of Conditional Expectations, which asserts that if $\mathcal F_1\subset\mathcal F_2$ then
$$
E[E[Y|\mathcal F_1]|\mathcal F_2] = E[E[Y|\mathcal F_2]|\mathcal F_1] = E[Y|\mathcal F_1].
$$
Use the second of these equalities, with $\mathcal F_1=\sigma(E[Y|X])$ and $\mathcal F_2=\sigma(X)$.
A: 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.)
