# Covariance as Measure of Linear Fit? Circle as “Anti-Line”

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.

• 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) – Glen_b Jun 23 '19 at 23:59
• @Glen_b , thanks, I just edited. – MSIS Jun 24 '19 at 0:10
• 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. – whuber Jun 24 '19 at 13:13