I tried to solve this exercise but got stuck: Assume we have the random variables $X$ and $Y$ where $E(X) = 0$. How can we prove the following inequality $\operatorname{Var} (X-E(X|Y)) \leq\operatorname{Var}(X)$?
I tried to write out the rhs: $\operatorname{Var}(X-E(X|Y)) = \operatorname{Var}(X) + \operatorname{Var}(E(X|Y)) - 2\operatorname{Cov}(X,E(X|Y))$ Since $E(X) = 0$, we have $\operatorname{Var}(E(X|Y)) = E(E(X|Y)^2)$ and $\operatorname{Cov}(X,E(X|Y)) = E(X*E(X|Y))$.
Thus we have $\operatorname{Var}(X) + E(E(X|Y)^2) - 2E(XE(X|Y)) \leq \operatorname{Var}(X)$ giving $E(E(X|Y)^2) \leq 2E(XE(X|Y))$
But that did not get me anywhere as you can see, does anybody have any tips? OR do you think there might be a typo in the exercise?
@Did's hints are spot on and you don't need $E(X)=0$.
Let $Z=E(X|Y)$, consider $$ \text{Var}(X)=\text{Var}(X-Z+Z)=\text{Var}(X-Z)+\text{Var}(Z)+2\text{Cov}(X-Z,Z). $$ Note that \begin{gather} E(X-Z|Y)=E(X-E(X|Y)|Y)=E(X|Y)-E(X|Y)=0\\ \implies E(X-Z)=0\implies \text{Cov}(X-Z,Z)=E[(X-Z)Z]. \end{gather} In going from the first line to the second line, we have used the Law of Iterated Expectations: $$ E(X-Z)=E[E(X-Z|Y)]=E(0)=0. $$ For $E[(X-Z)Z]$, we can condition on $Y$ again: $$ E[(X-Z)Z|Y]=ZE(X|Y)-Z^2=Z^2-Z^2=0\implies E[(X-Z)Z]=0. $$ Thus, we have $$ \text{Var}(X)=\text{Var}(X-Z)+\text{Var}(Z)\geq\text{Var}(X-Z)=\text{Var}[X-E(X|Y)]. $$