Proof that if covariance is zero then there is no linear relationship 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 $E[Y|X]=bX+E[Z|X]$
 A: 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.
A: 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 $\mathrm{Cov}(X,Y) = 0$.
A: 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.
