Prove: $E[\left (Y-E[Y|X] \right )\left ( E[Y|X]-E(Y) \right )]=0$

Question

Context:

My book uses Pythagoras to separate $\left \| Y-E[Y|X] + E[Y|X]-E(Y) \right \|^{2}$ where $\left \|. \right \|$ is the norm induced by $\left \langle X,Y \right \rangle:=E[XY]$.

My attempt:

$ \left \langle (Y-E[Y|X]), ( E[Y|X]-E(Y)) \right \rangle$

$=E[\left (Y-E[Y|X] \right )\left ( E[Y|X]-E(Y) \right )]$

$=E[YE[Y|X]]-(E[Y])^{2}-E[(E[Y|X])^2]+E[Y]E[E[Y|X]]$

$=E[YE[Y|X]]-(E[Y])^{2}-E[(E[Y|X])^2]+(E[Y])^2$

$=E[YE[Y|X]]-E[(E[Y|X])^2]$

Thanks in advance.

Answer 1

Using the tower property again, $$E[YE[Y \mid X]] = E[E[Y E[Y \mid X] \mid X]] = E[(E[Y \mid X])^2].$$