$\mathbb E[\mathbb E[X\mid X=x]]$

Question

By the law of total expectation, $\mathbb E[X]=\mathbb E[\mathbb E[X\mid Y]]$ for any random variables $X,Y$.

What if we replace "$Y$" with "$X=x$", where $x\in\mathbb R$ is arbitrary? Something must go wrong somewhere because this leads to

$\mathbb E[X]=\mathbb E[\mathbb E[X\mid X=x]]=\mathbb E[x]=x, \forall x\in \mathbb R$.

Edit: Actually I remember one course where $E[X\mid Y]$ was defined in terms of $E[X\mid Y=y]$ (and not the other way around). The definition of $E[X\mid Y=y]$ was a function $\varphi=\varphi(y)$ satisfying some constraint (will write it here once I find my notes), and $E[X\mid Y]$ was defined as the function $\varphi$ evaluated at $Y$.

What can be said about $\mathbb E[X\mid Y,Z=z]$?

Answer 1

The mistake is in assuming that $\mathrm{E}[Y]=\mathrm{E}[\mathrm{E}[Y\mid X=x]]$. What holds is the tower property which states that $\mathrm{E}[Y]=\mathrm{E}[\mathrm{E}[Y\mid X]]$.

So the mistake is in assuming that the two objects $\mathrm{E}[Y\mid X]$ and $\mathrm{E}[Y\mid X=x]$ are interchangeable -- which they are not. Instead there's the connection that if we call $\varphi(x)=\mathrm{E}[Y\mid X=x]$, then $\mathrm{E}[Y\mid X]=\varphi(X)$. So indeed $\mathrm{E}[Y\mid X]$ is a random variable (measurable with respect to $\sigma(X)$) whereas $\mathrm{E}[Y\mid X=x]$ is just a constant.