I have two real-valued random-variables $X$ and $Y$.
I am interested in simplifying the following expression (assuming it even makes sense to write): $$ E\big{[}E[X \vert Y]\ \big{\vert}\ X\big{]} $$
My intuition is that since $E[X|Y]$ is just the first moment of the distribution $p(x|y)$, its value is fixed for a fixed $y$ regardless of any "later conditioning" on $X$. I.e. my instinct was, $$ E\big{[}E[X \vert Y]\ \big{\vert}\ X\big{]} \overset{?}{=} E\big{[}E[X \vert Y]\big{]} = E[X] $$ However, expanding the integrals leads to something that I cannot seem to reduce.
By definition we have, $$ E[X|Y] := \int_\mathbb{R} x p(x|Y) dx $$
The "Law of the Unconscious Statistician" provides, $$ E[f(Y)|X] = \int_\mathbb{R} f(y) p(y|X) dy $$
Letting $f(Y) := E[X|Y]$ we have, \begin{align} E\big{[}E[X \vert Y]\ \big{\vert}\ X\big{]} &= \int_\mathbb{R} \Big{(}\int_\mathbb{R} x p(x|Y) dx \Big{)} p(y|X) dy\\[6pt] &= \int_\mathbb{R}\int_\mathbb{R} x p(x|Y) p(y|X) dxdy\\ \end{align}
There is definitely some weirdness in these expressions coming from the mixing of (uppercase) random-variables and (lowercase) variables-of-integration, which makes me wonder whether $E\big{[}E[X \vert Y]\ \big{\vert}\ X\big{]}$ is even a meaningful expression. If the notation here is misleading and/or the root of the problem, please clarify how I should be writing and thinking of this.
Recap: If the expression $E\big{[}E[X \vert Y]\ \big{\vert}\ X\big{]}$ is meaningful, how can it be simplified? If it's nonsense, why?
Thanks in advance!