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 $E[Y|X]=bX+E[Z|X]$

**Answer**

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 $\mathrm{Cov}(X,Y) = 0$.

**Attribution***Source : Link , Question Author : Gilbert Ibanez , Answer Author : jochen*