Is $\mathbb{E}[\mathbb{E}[Y|X]^{2}] = \mathbb{E}[Y]^{2}$?

Question

I am really sorry to ask such basis questions, but I am unable to find needed information on properties of expectation operator anywhere:

Is $\mathbb{E}[\mathbb{E}[Y|X]^{2}] = \mathbb{E}[Y]^{2}$ ?

Answer 1

Let $Z=E[Y|X]$. Then, $E(Z)=E(Y)$, and we have $$ E[Y]^2=E[Z]^2\leq E[Z^2]=E[E(Y|Z)^2] $$ so what you have is an inequality instead of an equality. The inequality is strict whenever $Z$ has positive variance so an easy example can be obtained by considering $(X,Y)$ bivariate normal in which case $Z$ is an affine transformation of $X$ with generally positive variance.

Answer 2

Begin from $$\mathsf{Var}(Y)=\mathsf{E}Y^2-(\mathsf{E}Y)^2$$ Hence, $$\begin{align} \mathsf{Var}(Y)&=\mathsf{E}(\mathsf{E}Y^2|X)-(\mathsf{E}Y)^2\\ &=\mathsf{E}(\mathsf{Var}(Y|X)+\mathsf{E}(Y|X)^2)-(\mathsf{E}Y)^2\\ &=\mathsf{E}(\mathsf{Var}(Y|X))+\mathsf{E}(\mathsf{E}(Y|X)^2)-(\mathsf{E}Y)^2\\ \end{align} $$ Therefore, $${\mathsf{E}(\mathsf{E}(Y|X)^2)=\mathsf{Var}(Y)-\mathsf{E}(\mathsf{Var}(Y|X))+(\mathsf{E}Y)^2}$$

or using the law of total variance,

$$\boxed{\mathsf{E}(\mathsf{E}(Y|X)^2)=\mathsf{Var}(\mathsf{E}(Y|X))+(\mathsf{E}Y)^2}$$