Expectation of a conditional variance

Question

I have to prove this $$E(Var(Y|X))=(1-\rho^2)Var(Y)$$ but I got stuck and don't know how to continue.

This is what I've done so far based on this variance formula $Var(Y)=E(Var(Y|X))+ Var(E(Y|X))$

$$Var(Y|X)= E(Y^2|X)- (E(Y|X))^2$$ $$=Var(Y)- Var(E(Y|X))$$ $$=Cov(Y,Y)- E(E(Y|X))^2- E(Y)^2$$ I get to the part when I relate the covariance in order to get the correlation coefficient, but I don't know what to do from there, or maybe what I've done is wrong, so I'll be grateful if any of you can help me out with this.

Answer 1

Let's work from your decomposition formula (the "law of total variance"): $$\mathrm{Var}(Y) = \mathbb{E}(\mathrm{Var}(Y|X)) + \mathrm{Var}(\mathbb{E}(Y|X))$$ Rearranging this, you get $$\mathrm{Var}(Y)\bigg(1 - \frac{\mathrm{Var}(\mathbb{E}(Y|X))}{\mathrm{Var}(Y)}\bigg) = \mathbb{E}(\mathrm{Var}(Y|X)).$$ So you really just need to show that $\mathrm{Var}(\mathbb{E}(Y|X))\big/\mathrm{Var}(Y) = \rho^2$. Try working this out from here.

Answer 2

We consider the case where $X$ and $Y$ are jointly normal with correlation $\rho$ $(|\rho|<1)$. Let \begin{align*} Z = \frac{\frac{Y-E(Y)}{\sqrt{Var(Y)}}-\rho \frac{X-E(X)}{\sqrt{Var(X)}}}{\sqrt{1-\rho^2}}. \end{align*} Then $Z$ is also normal and independent of $X-E(X)$, as they are jointly normal with zero correlation. Since \begin{align*} Y=\sqrt{Var(Y)}\left(\sqrt{1-\rho^2} Z + \rho \frac{X-E(X)}{\sqrt{Var(X)}}\right) + E(Y), \end{align*} we conclude that \begin{align*} E(Y\mid X) = \rho \sqrt{Var(Y)}\frac{X-E(X)}{\sqrt{Var(X)}} + E(Y). \end{align*} Therefore, \begin{align*} Var\big(E(Y\mid X)\big) = \rho^2\, Var(X)\,Var(Y), \end{align*} and, consequently, \begin{align*} E\big(Var(Y\mid X)\big)=\left(1-\rho^2\right)Var(Y). \end{align*}

Answer 3

As for the useful identity aforementioned by Herschkorn, namely $$ \operatorname{Cov}(X, Y) = \operatorname{Cov}(X, E[Y\mid X]), $$ it follows from the definition of the covariance, by means of the law of total expectation: \begin{align} \operatorname{Cov}(X, Y) &= E[XY] - E[X]\, E[Y]\\ &= E[E[XY\mid X]] - E[X]\cdot E[E[Y\mid X]]\\ &= E[X\cdot E[Y\mid X]] - E[X]\cdot E[E[Y\mid X]]\\ &= \operatorname{Cov}(X, E[Y\mid X]). \end{align}