I am trying to figure out how one can deduce from the formula for Covariance:

$ Cov(X,Y) \:=E[(X-\mu_X)(Y-\mu_Y)]= E(XY)-\mu_X\mu_Y $

Correlation measures the degree of linear dependence. Does Covariance too measure the level of linear dependence between $X$ and $Y$? Does it follow from the formula/definition, given $Corr(X,Y)$ is just a standardized version of $Corr(X,Y)$?

Moreover, I am surprised that , given a set of points $(x,y)$ (uniformly-distributed) on the circle, so that we have $y=\pm \sqrt{1-x^2} $ that the correlation between $(Y,X)$ is exactly $0$. Is there someway of seeing this non-analytically? In this sense, a circle or "circular relation" is the opposite ofa linear relation in that If $Y=aX+b $ then $(X,Y)=1$ , while if $Y= \pm \sqrt{ 1- X^2}$ then $Corr(X,Y)=0$?

EDIT: I am trying to understand better just what covariance/correlation measure. So I would appreciate some input on: 1)How can we gather from the formulas for Covariance, Correlation that they measure linear dependence?

2)Is there a way of seeing non-analytically that e.g., points in a circle are uncorrelated? Thanks.

  • 1
    $\begingroup$ I think it's better to stick to one issue per question, but at the least please clearly lay out your specific questions. The functional relation you gave is not a circle, but a half-circle. (please edit to clarify) $\endgroup$ – Glen_b Jun 23 at 23:59
  • $\begingroup$ @Glen_b , thanks, I just edited. $\endgroup$ – MSIS Jun 24 at 0:10
  • 2
    $\begingroup$ Because covariance is proportional to the units of measurement of $X$ and $Y,$ it cannot possibly be any kind of measure of dependence, linear or not. Is it possible that the characterizations I derive at the end of the post at stats.stackexchange.com/a/18200/919 answer your questions? Perhaps the clearest demonstration of zero correlation in the circle example is to observe that because covariance is a bilinear form, the circular symmetry implies the covariance equals its own negative. $\endgroup$ – whuber Jun 24 at 13:13

Your Answer

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

Browse other questions tagged or ask your own question.