6

I'm interested in the relationship between $$cov(X,Y)$$ and $$cov\left(E(X|Y),E(Y|X)\right).$$ In particular, can it occur that $cov(X,Y)>0$ but $cov(E(X|Y),E(Y|X))<0$?

In the case of $\begin{bmatrix}X_1\\X_2\end{bmatrix}\sim N\left(\begin{bmatrix}0\\0\end{bmatrix},\begin{bmatrix}1&\rho\\\rho&1\end{bmatrix}\right),$

$$cov(E(X_1|X_2),E(X_2|X_1))=E(\rho X_2\cdot\rho X_1)=\rho^2E(X_1X_2)=\rho^3.$$ So in this case the covariance of the conditional expectations always has the same sign as the original covariance, and smaller magnitude. Is this true in general?

stevo
  • 61

1 Answers1

1

One relation between the two can be arrived at using law of iterated expectation: \begin{align*} cov(E(X|Y), E(Y|X)) & = E[E(X|Y)\, E(Y|X)] - E[E(X|Y)]E[E(Y|X)]\\ & = E[E(X|Y)\, E(Y|X)] - E(X)E(Y)\\ & = E[E(X|Y)\, E(Y|X)] - (E(XY) - cov(X,Y) )\\ cov(E(X|Y), E(Y|X)) - cov(X,Y)& = E[E(X|Y)\, E(Y|X)] - E(XY)\,. \end{align*}

So $cov(E(X|Y), E(Y|X))$ can be larger than $cov(X,Y)$ if $E[E(X|Y)\, E(Y|X)]$ is larger than $E(XY)$.

Greenparker
  • 532