It's known that Pearson's correlation is able to measure trends for an observed sample $y$ and a possible linear relationship with a simulated data $y^{(s)}$, being $+1$ if $y^{(s)} = a + by$ and $b >0$.

But, does it matter the shape of the observed data $y$? For instance, if I have $y = f(x) = x^2$, would it make any effect for Pearson's correlation? My intuition says no, but I want to find a more reliable source.

  • $\begingroup$ Are you alluding to the canonical example of zero correlation but a clear relationship: $E[X] = 0$, density of $X$ is symmetric, and $Y = X^2$? Then $Corr(X,Y) = 0$ but there is an entirely deterministic relationship. $\endgroup$ – Matthew Gunn Jun 22 '16 at 6:23
  • $\begingroup$ I think my question is much simpler than that. Does the characteristics of the observed sample matter for the Pearson? Does it have to obey any sort of characteristics? $\endgroup$ – rph Jun 23 '16 at 13:02

Pearson's correlation coefficient is a measure of strength of linear relationship between the variable. So, it may provide false results for non-linear relationship.

Read a more detailed answer on Correlation and dependence

  • $\begingroup$ I would say something like, "It may not pickup any aspect of a non-linear relationship." For example, if $X \sim N(0, 1)$ and $Y = X^2$, then $Corr(X,Y)=0$. That result is true! Saying it provides "false results" in this case isn't quite right. $\endgroup$ – Matthew Gunn Jun 23 '16 at 16:06

Supoose, you have $$y=f(x)=x^2$$ Take log on both sides $$log(y) = 2 \ log(x)$$ Now have a linear relationship. Find Pearson's Correlation Coefficient now. Read this answer for more details.


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