# Why does Pearson's r have a non-normal sampling distribution at high values of ρ?

[W]hen the absolute value of the correlation in the population is low (say less than about 0.4) then the sampling distribution of Pearson's r is approximately normal. However, with high values of correlation, the distribution has a negative skew.

I understand that the point of the Fisher transformation is that the sampling distribution for the transformed value is more normal.

I have two (related) questions.

1. Why does this only happen for high values of r?

2. When we do significance tests we're typically wondering about the probability of obtaining our observed sample (or some extreme sample) if ρ=0. If we're doing that, what's the point of the transformation? The passage above stated that the sampling distribution was approximately normal if the correlation was weak, and there is no weaker correlation than 0.

• The problem with the sampling distribution of $r$ is not just that it can be very asymmetric. This is tied up with an extra fact that the distribution is inevitably bounded by the interval $[-1, 1]$. The Fisher transformation tackles both. Mar 31, 2015 at 9:18

1. Because if your $r$ is high you get a non-symmetric distribution. For example, if the real correlation is $0.9$ you might, by chance, observe a sample correlation of $0.75$. But you will never observe a sample correlation of $1.05$. Generally many normal approximations (e.g. for binomial proportions) don't work well when the real value is close to the bounds of your statistic.
2. If you just test against $r = 0$ you don't need to worry about the sampling distribution being skewed due to high $r$. It becomes relevant when you want to test $r > 0.7$ (for example) or if you want to give a confidence interval for $r$. You always want to give confidence intervals.