How to prove or disprove $ \displaystyle Var(Y-E(Y|X))\le Var (Y)$? Let $X$ and $Y$ be random variables, prove or disprove the following:
$$ Var(Y-E(Y|X))\le Var (Y)$$
After doing many steps, I think it suffices to prove $(EY)^2\le E(Y E(Y|X))$, but I don't know how to prove this one.
 A: Write $U:=E(Y\mid X)$ and $V:=Y-E(Y\mid X)$. We have $E(V)=0$ and $E(V\mid X)=0$, so that
$$E(UV)=E[E(UV\mid X)]=E[UE(V\mid X)]=0,$$
hence $U$ and $V$ are uncorrelated, and so
$$
\operatorname{Var}(Y)=\operatorname{Var}(U+V)=\operatorname{Var}(U) + \operatorname{Var}(V) \ge \operatorname{Var}(V)=\operatorname{Var}(Y-E(Y\mid X)).
$$
A: There's already a nice and short solution by grand_chat, but here's a slightly different proof that uses the known relation
$$
\text{Var}[Z] = \text{E}[\text{Var}[Z|X]] + \text{Var}[\text{E}[Z|X]]
$$
which I immediately thought of when seeing the problem since it splits the variance of $Z$ into one component that is the variance independent of $X$ and one that is the variation caused by its dependence on $X$.
Plugging in $Y$ this becomes
$$
\text{Var}[Y] = \text{E}[\text{Var}[Y|X]] + \text{Var}[\text{E}[Y|X]].
$$
Plugging in $V=Y-\text{E}[Y|X]$ gives
$$
\text{Var}[V] = \text{E}[\text{Var}[V|X]] + \text{Var}[\text{E}[V|X]]
= \text{E}[\text{Var}[Y|X]] + \text{Var}[\text{E}[0|X]]
= \text{E}[\text{Var}[Y|X]]
$$
as $\text{E}[Y|X]$ is constant within the expectancy and variance computed conditional on $X$.
This shows that
$$
\text{Var}[Y] = \text{Var}[Y-\text{E}[Y|X]] + \text{Var}[\text{E}[Y|X]]
\ge  \text{Var}[Y-\text{E}[Y|X]].
$$
