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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.

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  • $\begingroup$ 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
    Commented Jun 22, 2016 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
    Commented Jun 23, 2016 at 13:02
  • $\begingroup$ That has a simple mathematical answer, obtainable from any formula for the correlation: it is defined provided both variables have nonzero variance. $\endgroup$
    – whuber ♦
    Commented Sep 7, 2022 at 12:57
  • $\begingroup$ "Does the characteristics of the observed sample matter for the Pearson?" what does that term 'matter' actually mean in this context?. You can compute a Pearson correlation for any set of paired numbers, whether it is a linear relationship or not, so I would say that it doesn't matter. $\endgroup$
    – Sextus Empiricus
    Commented Sep 7, 2022 at 14:11

2 Answers 2

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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

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    $\begingroup$ 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
    Commented Jun 23, 2016 at 16:06
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Adding on Dr Nisha Arora's answer, the following provides a method to compute meaningful Pearson linear correlation coefficient (PLCC) in such cases.

Suppose, 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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    $\begingroup$ nonlinear transformations e.g. log-transformation will generally affect the correlation between 2 pairs. Reference $\endgroup$
    – Mario
    Commented Sep 7, 2022 at 1:00
  • $\begingroup$ @Mario, of course. You're right. I've updated my answer. To further clarify, we are banking on that. Pearson correlation (PLCC) is defined for linear dependencies. For non-linear dependencies, PLCC does not make much sense. So, we convert our non-linear relationship to a linear one and then compute PLCC. This is also common in the literature. $\endgroup$
    – Nagabhushan S N
    Commented Sep 7, 2022 at 12:24

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