I get that a zero covariance doesn´t imply independence, but everybody says that if there is dependence and the covariance is zero then it is a non linear dependence.

People base their interpretation of Pearson's R in that fact (the closer you are to zero the less linear the relationship is).

Is there a formal proof to that?

I tried to do it by myself but i couldn't. The proposition i think encapsulates the idea is the following:

If $cov(X,Y)\ne0$ then there exists a Z such that $cov(X,Z)=0$ and $E[Y|X]=bX+E[Z|X]$

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    $\begingroup$ It seems to me that this follows directly from the definition of Pearson's correlation coefficient: $$\rho_{X,Y}=\frac{\mathsf{Cov}(X,Y)}{\sigma_X\sigma_Y}$$ If the covariance is zero then the correlation is zero and so they're not linearly related. $\endgroup$ – Remy Oct 19 at 5:41
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    $\begingroup$ Interesting, that means i should turn my attention into what inspired that exact definition. It couldn´t come out of nowhere. $\endgroup$ – Gilbert Ibanez Oct 19 at 18:07
  • $\begingroup$ I think the proposition you put at the end isn't quite right. You want if $cov(X, Y) = 0$ then no linear relationship. But you haveif $cov(X, y) \neq 0$ then linear relationship $\endgroup$ – roundsquare Oct 31 at 20:53

Here is a proof of the mathematical statement at the end of your question: we can find a $Z$ which is uncorrelated to $X$ and satisfies $$ \mathbb{E}(Y|X) = b X + \mathbb{E}(Z|X) $$ by assuming $Z = Y - bX$, and then choosing the $b$ which makes $\mathrm{Cov}(X, Z) = 0$ true. For this $b$ we have $$ 0 = \mathrm{Cov}(X, Z) = \mathrm{Cov}(X, Y - bX) = \mathrm{Cov}(X, Y) - b \mathrm{Var}(X), $$ and thus $$ b = \frac{\mathrm{Cov}(X, Y)}{\mathrm{Var}(X)}. $$ (Note that the same $b$ is found as the slope of the linear regression line.) We have $b = 0$, if and only if $\mathrm{Cov}(X,Y) = 0$.

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    $\begingroup$ wow, it looks "to easy to be true" but i think it is. You also showed that if we choose that specific $Z$ then it is necesarry for $b$ to be $cov(X,Y)/var(X)$. Maybe if someone else constructs a different $Z$, a different requirement for $b$ arises. But i can live with your approach since it shows the best linear approximation, according to linear regression criteria. $\endgroup$ – Gilbert Ibanez Oct 20 at 18:28

If there is a linear relationship between two RVs, i.e. $Y=aX+b$, where $a\neq 0$, then the covariance is $$\operatorname{cov}(X,Y)=a\operatorname{cov}(X,X)=a\operatorname{var}(X)\neq0$$ So, if there a linear relation, covariance is not zero. If the covariance is zero, the linear relation can't exist because we'll contradict.

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    $\begingroup$ This seems as a very specific situation. What if that b was another random variable. Since covariance can be negative it could be the case that cov(B,X)=-aV[X] which makes cov(X,Y)=0 $\endgroup$ – Gilbert Ibanez Oct 19 at 7:56
  • $\begingroup$ If $b$ is random, we can't say anything; Let $Y=X+b$, then what happens when $b=-X$, which is similar to your case actually? $\endgroup$ – gunes Oct 19 at 8:14
  • $\begingroup$ I believe that if we say that if $b$ is random and uncorrelated to $X$ then what you say holds. Thats why I believe we need to find at least one $b$ with that property. Thats why in my question i resume this issue as a proof of existence(with the only difference that i used Z instead of B) $\endgroup$ – Gilbert Ibanez Oct 19 at 18:05
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    $\begingroup$ If $b$ is random, in general we can't say something. And, yes, if $b$ is random and uncorrelated to $X$, and $X$ is lin. related to $Y$, covariance is non-zero. However, there are a lot of situations. If, let's say $X$ is zero mean, and $b=X^2$, covariance of $Y$ and $X$ will be nonzero, where $cov(b,X)=0$. The relation is actually $Y=X+X^2$, we have a linear term, but is this a linear relationship in the end? $\endgroup$ – gunes Oct 20 at 6:36
  • $\begingroup$ Excellent point, $cov(x,y) \ne 0$ and $Y=X+X^2$ is not a perfect linear relationship. However, i feel that there is some linearity in it, since it can be (badly) approximated by a straight line. I got that feeling because the covariance is not zero, but i can´t formally explain that fact. $\endgroup$ – Gilbert Ibanez Oct 20 at 18:08

On a distribution level it should be straightforward to show that a linear correlation implies a non-zero covariance (the other way to prove what you wanted).

But as a word of warning, this may not hold for a sample. If you have a small data set generated with a linear correlation, but by chance a large outlier you can compute a negative correlation or no correlation on the sample.


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