Proof involving conditional expectation

Question

I am trying to prove or disprove the following: $\mathbb{E}[\mathbb{E}(X\mid Y)^2]-\mathbb{E}[X \mathbb{E}(X\mid Y)]=0$. Using formulas for covariance or something like that just gives back the same problem.

Answer 1

Notice that $E[X \mid Y]$ is $\sigma(Y)$-measurable so that we may denote $E[X \mid Y]$ by $f(Y)$, hence \begin{align*} & E\{[E(X \mid Y)]^2\} \\ = & E\{f(Y)E[X \mid Y]\} \quad \text{ substitute one $E[X \mid Y]$ by $f(Y)$} \\ = & E\{E[f(Y)X \mid Y]\} \quad \text{ move $f(Y)$ inside to the inner conditional expectation as it is $\sigma$-$Y$ measurable}\\ = & E[f(Y)X] \quad \text{law of iterative expectation}\\ = & E[XE[X \mid Y]] \quad \text{ recall the definition of $f(Y)$} \end{align*}

The result then follows.