# Where does the argument “Pearson's correlation coefficient assumes normality” come from?

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)
xs.sort()
ys.sort()
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?

• 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. – Nick Cox Apr 12 '16 at 7:08
• 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. – ttnphns Apr 12 '16 at 7:32
• The "normality" assumption is for significance testing of $r$ by means of t of F statistics. – ttnphns Apr 12 '16 at 7:34