0
$\begingroup$

I am struggling to grasp the intuition behind the E[XY] of 2 Random variables that enters into the Covariance Formula:

Cov[X,Y]=E[XY]-E[X]E[Y]

I’m having a tough time connecting this equation with the information it gives you on how the 2 RV’s move around their means with respect to one another.

Even in its expanded (discrete) form this is making little sense to me as to what the math is saying:

iΣj xiyj pdf(xi,yj)] - [Σixi pdf(xi)] [Σjyj pdf(yj)]

If anyone could provide an intuitive explanation of what E[XY] means by itself, it would be greatly appreciated. Thanks to all. - SDH

$\endgroup$
2
$\begingroup$

If you want intuition about the covariance representing "how the two random variables move around their means with respect to one another," it is better to use the following different (but equivalent) formula.

$$\begin{align}\text{Cov}(X,Y) &= E[(X-E[X])(Y-E[Y])]\\[2ex]&= E[XY-X~E(Y)-Y~E(X)+E(X)~E(Y)]\\[2ex]&=E(XY)-E(X)~E(Y)\end{align}$$

$\endgroup$
  • $\begingroup$ The 1st equation you gave makes perfect sense in terms of the intuition behind the covariance idea... However, still not grasping how that E[XY]-E[X]E[Y] represents the same idea. Thanks for the response sir. $\endgroup$ – SDH May 15 at 23:34
  • $\begingroup$ Do you grasp how the expressions are equal? @SDH $\endgroup$ – Graham Kemp May 16 at 0:07
  • $\begingroup$ I mean I can get from one to the other algebraically. $\endgroup$ – SDH May 16 at 0:08
  • $\begingroup$ Covariance is the expected product of the random variables' displacement from their means. The above shows that this is thus the displacement of the expected product of the variables from the product of the expected values of each. @SDH $\endgroup$ – Graham Kemp May 16 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.