Expectation of the squared conditional expectation

Question

I am considering the expectation of the squared expectation, as asked here but with no answer and so wanted to get the communities thoughts.

Since $E[Y|X]$ is not independent with itself then the first line of the equation below should be true. I then go on to try to derive the answer using the definitions but would be grateful if someone could point out where I am going wrong.

$$\begin{align} E[E(Y|X)^2] & \neq E[E(Y|X)] \cdot E[E(Y|X)] \\[10pt] & = \int_{x \in A_x} E(Y|X)^2 f_{X}(x) dx \\[10pt] & = \int_{x \in A_x} \Big( \int_{y \in A_y} y f_{Y|X}(y|x) dy \Big)^2 f_{X}(x) dx \\[10pt] & = \int_{x \in A_x} \int_{y \in A_y} y \ f_{Y|X}(y|x)f_{X}(x) \ dx \ dy \ \Big( \int_{y \in A_y} y \ f_{Y|X}(y|x) \ dy \Big) \\[10pt] & = \int_{y \in A_y} y \left\{ \int_{x \in A_x} f(y,x)\ dx \right\} \ dy \ \Big( \int_{y \in A_y} y \ f_{Y|X}(y|x) \ dy \Big) \\[10pt] & = \int_{y \in A_y} y f_{Y}(y) \ dy \ \Big( \int_{y \in A_y} y \ f_{Y|X}(y|x) \ dy \Big) \\[10pt] & = E[Y] \cdot E[Y|X] \end{align}$$

In the fourth row I have factored out one of the $E[Y|X]$ expressions and in the fifth row computed the marginal over $x$.

Answer 1

You are going wrong on the fourth line. You cannot distribute the second conditional expectation outside the scope of the outer integral. Both are conditioned on the same variable $(x)$.

$\quad\begin{align}\int_\Bbb R\left(\int_\Bbb R y f_{\small Y\mid X}(y\mid x)\,\mathrm dy\right)^2f_{\small X}(x)\,\mathrm d x &=\int_\Bbb R f_{\small X}(x)\int_\Bbb R yf_{\small Y\mid X}(y\mid x)\int_\Bbb R uf_{\small Y\mid X}(u\mid x)\,\mathrm d u\,\mathrm d y\,\mathrm d x\\[1ex]&= \iint_{\Bbb R^2} y f_{\small X,Y}(x,y)\int_\Bbb R uf_{\small Y\mid X}(u\mid x)\,\mathrm d u\,\mathrm d (x,y)\\[2ex] &= \mathbb E(Y\,\mathbb E(Y\mid X))\end{align}$