I read that the correlation coefficients have to be converted by using a fisher-transformation when one wants to test their significance (like here When is Fisher's z-transform appropriate? ).

I just don't understand: Why do I have to use this transformation? Here it is given:

Because the correlation is bounded between [-1, 1], the sampling distribution for highly correlated variables is highly skewed. https://blogs.sas.com/content/iml/2017/09/20/fishers-transformation-correlation.html

Ok, so what? What does it mean that it is bounded? When there are no bounds, then "bounds" would be +/- oo. Right? Isn't it something like a normalization finally? The transformation doesn't do anything then something like the opposite, from my pov. I guess it finally does exactly this. But why is it possible to calculate the significance afterwards?


1 Answer 1


The boundedness is not the real problem it just explains the skewness of the sampling distribution. Basically, the transformation approach is so established for historic reasons (just like the prevailing recommendation to log- or whatever-transform your skewed variables to make them more normal for a linear regression - only that Fisher's transformation is formally correct for large enough samples). The reason for such simplifications is mainly to save computing power. Let's have a more detailed look:

First, of all, we need to establish, what the problem actually is. To derive a statistical test for a test statistic such as the correlation, I need to derive the sampling distribution of my statistic. This has been done for the correlation and it turns out that it highly depends on the population value of the correlation. Assuming that your variables follow a bivariate normal distribution, the full formula of the density of the correlation sampling distribution is: enter image description here

Here you see that you need the Gamma function and the Gaussian hypergeometric function to continue calculating probabilities (i.e., areas under this curve). To derive an (asymptotically) exact statistical test, you need to work with this thing to get your p-values or critical values. Well... imagine good old Fisher in the early 20th century long before a calculator (let alone a computer) was a thing. This thing is just intractable for default usage. And basically, this thing just implies that your distribution gets more and more skewed the more you approach the boundaries (see below).

Well, it turns out that there are two lucky properties of this thing. First: if the population correlation (rho) is zero, this simplifies to the well-known central t-distribution. This is something, you can handle (and nearly everyone does in introductory statistics). So no problem for any test against the population value zero.

But what should Fisher do to test against any other value or to compare two sample correlations? Well, Fisher invented one of his famous tricks: By transforming your correlations using Fisher's method, you get scores that approximately follow a normal distribution with mean Fisher-z(r) and variance 1/(n-3), looks like this: enter image description here

Having approximate normality is a great thing from a computational perspective, because you may use a Z-test that is based on the normal distribution which is a well-behaved and easy to handle distribution (and hence, the favorite pet-distribution of nearly all statisticians ;)). Much easier than the exact distribution above which is not tractable without a computer.

To conclude, you are absolutely right that - on a theoretical level - Fisher's transformation is not necessary to conduct the test if you have a computer and a long breath to type this formula into a script that calculates integrals from this density, you can easily conduct your tests on it. If you are a historic statistician or a student in an exam with only a table of the normal distribution, Fisher's transformation is your way to go. Beyond that, I guess, it's as with many beloved traditions, if you have always used the transformation in your tests, you just continue to use it.

Hope this historic anecdote helped to ease your doubts. I hope I got the story right, if I missed something, just leave a comment.

  • $\begingroup$ Reading this in 2023. Thanks for a super clear, fun to read, and informative answer. $\endgroup$
    – Shaz
    Jul 26, 2023 at 9:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.