Related1: Pearson's or Spearman's correlation with non-normal data

Related2: How robust is Pearson's correlation coefficient to violations of normality?

Related3: Why is Pearson's ρ only an exhaustive measure of association if the joint distribution is multivariate normal?

I've seen a claim "Pearson's correlation coefficient assumes normality" sometimes. The three answers above explain why the population needs to be normal - the summary is that:

・the variates may have some association other than the correlation, but the Pearson's correration coeff can't tell it anyway, thus it's exhaustive only when bivariate normal is the case.

・the range gets narrower than -1 ~ +1 when bivariate normal is not the case. Especially, the accepted answer in Related2 demonstrates that normal vs lognormal, for example, can get neither -1 nor +1 at its best.

However, uniform vs uniform, for example, apparently can achieve the -1 or +1 coefficient. A trivial demo with Python program:

# This is virtually the same as the demo in Related2
# except here uniform vs uniform is used.
import numpy as np
xs = np.random.uniform(2, 5, size=10**5)
ys = np.random.uniform(1, 2, size=10**5)
np.corrcoef(xs, ys)[0, 1]
# 0.99998421283590666

And now I'm completely lost. Where is the origin of "Pearson's correlation assumes normality" and what is the justification?

  • 1
    $\begingroup$ Attaining $1$ or $-1$ depends on nothing but there being at least two points distinct on both variables and all points lying on a straight line. The bridge here is that sometimes is treacherous is between using Pearson correlation as a purely descriptive measure and trying to attach P-values and test hypotheses. You need some machinery for the latter, which is where normality is often invoked. Personal views: if you have to worry about significance for a correlation, your sample size is too small; bootstrap when you can to get confidence intervals if you must; use regression any way. $\endgroup$
    – Nick Cox
    Apr 12, 2016 at 7:08
  • 1
    $\begingroup$ The "range" point is not exactly about normality, it is about symmetry in the distributions: [-1,1] is theoretically attainable with both distributions identical shape & symmetric. A milder demand, to attain [-1 or 1] one side bound requires identical shapes only. $\endgroup$
    – ttnphns
    Apr 12, 2016 at 7:32
  • 3
    $\begingroup$ The "normality" assumption is for significance testing of $r$ by means of t of F statistics. $\endgroup$
    – ttnphns
    Apr 12, 2016 at 7:34

1 Answer 1


I also think that the assumption of normality concerns only the calculation of a P-value of the correlation coefficient. If the variables are not normally distributed then perhaps the best thing is to calculate the P-value by the means of a permutation procedure.

  • 1
    $\begingroup$ welcome to CV! By our standards, this answer as it currently stands is a bit short. Would you mind elaborating a bit, adding references, etc.? $\endgroup$
    – Antoine
    Jan 21, 2019 at 17:16
  • $\begingroup$ Hey‚ Antoine. Actually‚ this is my current research project so I cannot‚ unfortunately‚ go too much into details. But I promise to update the answer when the publication is ready. $\endgroup$ Jan 22, 2019 at 9:21

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