How to prove or disprove $ \displaystyle Var(Y-E(Y|X))\le Var (Y)$?

Question

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.

Answer 1

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 $$ \operatorname{Var}(Y)=\operatorname{Var}(U+V)=\operatorname{Var}(U) + \operatorname{Var}(V) \ge \operatorname{Var}(V)=\operatorname{Var}(Y-E(Y\mid X)). $$

Answer 2

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 $$ \text{Var}[Y] = \text{Var}[Y-\text{E}[Y|X]] + \text{Var}[\text{E}[Y|X]] \ge \text{Var}[Y-\text{E}[Y|X]]. $$