Let $F$ be a CDF in $\mathbb R^2$ and $X,Y$ random variables such that $(X,Y)\sim F$. Let $Z=E[X|Y]$. What is the distribution of $Z$ conditional on $X$, i.e $P(Z \in \cdot | X=x)$ ? One could say it is just a Dirac, since "if $X=x$, then $Z=E[x|Y]=E[x]=x$". Now this is not rigorous at all and I definitely feel is wrong. This is the same as saying that the distribution of $Z$ conditional on $X$ is the distribution of $E[X|Y,X]=E[X]$, which is wrong, right? Another way I could convince myself that it is wrong is that $Z$ is a mesurable function of $Y$, not $X$, hence knowing values of $X$ should not influence the outcome of $Z$.
I feel this question should be trivial, but I am not seeing it.
Now what happens if we ask a very similar question: what is the distribution of $Z$ given $X=x$ and $Y=y$? There is nothing random in $Z$ anymore because we know $Y$. So this time it should be a Dirac, i.e $E[X | Y=y]$, correct? Now what role would the conditioning on $X$ play in this scenario?
Let $f_{Y|X=x}(y)$ be the conditional probability density function of $Y$ at $y$ given $X=x$ (assuming the density exists). Let $G_x(z)$ be the CDF of $Z$ conditioned on $X=x$, then $$ G_x(z) =\int_{-\infty}^\infty f_{Y|X=x}(y)\, \mathbf{1}_{\langle X|Y=y \rangle<z}(y)\, dy, $$ where $\mathbf{1}_{\langle X|Y=y \rangle<z}(y)$ is the indicator function of $\langle X|Y=y \rangle<z$.
Only in the simple case where the value $y$ at which $E(X|Y=y)=z$ is a monotonic function of $z$ do we get a simpler expression for $G_x(z)$. In this case, let us call this function $y^*(z)$ (and w.l.o.g., assume $y^*$ is an increasing function of $z$). Then we have $$ G_x(z) =\int_{-\infty}^{y^*(z)} f_{Y|X=x}(y)\, dy = F_{Y|X=x}(y^*(z)), $$ where $F_{Y|X=x}$ is the conditional CDF of $Y$ given $X=x$.