$ E(Var(Y|X)) \leq (1- \rho^2)Var(Y) $

Question

Let $\rho$ be the Pearson correlation coefficient of the random variables X,Y , prove that $$ (*) E[Var(Y|X)] \leq (1- \rho^2)Var(Y) $$

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Since $\rho$ exists we know that $Var(X) \neq 0 \neq Var(Y)$

From the Law of total variance we know that $Var(Y) = E[Var(Y|X)]+Var[E(Y|X)]$

Also, because of Cauchy-Schwarz $ -1\leq\rho \leq1 \Rightarrow 0\leq\rho^2\leq1$

To be more concise let's define $a:= E[Var(Y|X)], b := Var[E(Y|X)] ,c:=Var(Y)$

$a,b,c \geq 0$ ; $a,b \leq c$ ; $a + b = c$ $$ (*) \Leftrightarrow c-b \leq c-c\rho^2 \Leftrightarrow b \geq c\rho^2 \Leftrightarrow ...? $$ or $$ (*) \Leftrightarrow a \geq \rho^2(a+b) \Leftrightarrow a(\rho^2-1)+\rho^2b \leq 0 \Leftrightarrow ...? $$

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Ok, I've been stuck here for a while, there is probably a straightforward approach that I'm missing. Thanks

Answer 1

For any $a \in \mathbb{R}$ we have $$\mathbb{E} Var(Y | X) = \mathbb{E} Var(Y - aX | X) \leq Var(Y - aX).$$ Note $$Var(Y - aX) = Var(Y) - 2aCov(X, Y) + a^2 Var(X).$$ So taking $a = \frac{\rho \sigma_Y}{\sigma_X}$ finishes the proof.