I have datasets from two bivariate normal distributions, $\mathcal{N}(\mu_x, \Sigma_x)$, and $\mathcal{N}(\mu_y, \Sigma_y)$ respectively. Now we know the correlation coefficient for these two distributions can be defined as $\rho_x = \frac{\sigma_{12}}{\sigma_1 \sigma_2}$ and $\rho_ y= \frac{\sigma_{34}}{\sigma_3 \sigma_4}$, assuming $\Sigma_x = \left(\matrix{\sigma_{1}^2&\sigma_{12}\\\\\sigma_{12}&\sigma_2^2}\right)$ and $\Sigma_y = \left(\matrix{\sigma_{3}^2&\sigma_{34}\\\\\sigma_{34}&\sigma_4^2}\right)$

Is there a standard test for testing $H_0 : \rho_x = \rho_y$ vs $H_1 : \rho_x \neq \rho_y$, I understand this doable by likelihood ratio tests, but not sure if there is already existing distribution for testing the parameters.


1 Answer 1


If you wish to test if two correlations from two samples for independent pairs of variables are different from each other, the null hypothesis statistical test (NHST) appropriate for this would be the Fisher z-transform test. This test uses the normal distribution, and requires transforming the correlations via $$z = \frac12 \ln\left(\frac{1+r}{1-r}\right) = \text{atanh}(r)$$

More can be found here: https://en.wikipedia.org/wiki/Fisher_transformation

Happy to update this answer further if requested.

  • $\begingroup$ So you mean, after doing the Fisher transformation, we can use t-test to compare atanh(r_1) and atanh(r_2) as they are normal? I am not sure what's the final test is? $\endgroup$ Commented May 25, 2023 at 16:42
  • 1
    $\begingroup$ This would not be a t-test, but a z-test (as the transformations are normally distributed). Once you transform your two correlations, $z_i = \text{atanh}(r_i)$, you find the z-score: $\frac{z_1-z_2}{\sqrt{\frac{1}{n_1-3} + \frac{1}{n_2-3}}}$. $\endgroup$
    – Gregg H
    Commented May 25, 2023 at 16:50
  • $\begingroup$ Posted a similar question stats.stackexchange.com/questions/617002/…, let me know if you have any leads $\endgroup$ Commented May 26, 2023 at 14:33

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