I have 100 50x50 correlation matrices, which I have all Fisher z-transformed. I understood that this results in the all the entries of one matrix being approximately normally distributed.


  • Now I read somewhere, that this also means that if we take some entry (i,j) of all the matrices (so for instance entry (5, 12) for matrix1, matrix2,...,matrix100), these values are also normally distributed. Is this true, and if so - why?

  • I want to classify these 100 matrices into two groups. The classification assumes that the data from each group is normally distributed. Does the Fisher z-transformation implies that? Alternatively, would the fact that each entry (i,j), $1\le i\le 200$, $1\le j\le 200$, from all matrices being normally distributed, imply that the matrices of each group are normally distributed?


1 Answer 1


The Fisher z-transformation does not guarantee a normal distribution; in particular not within a correlation matrix using different variables.

  • each of your 50 input variables $X_1...X_{50}$ needs to be normally distributed
  • if you repeatedly draw samples from two variables $i$ and $j$ from the same distributions: $Y_i\sim X_i$ and $Y_j \sim X_j$ from these distributions, then the transformed correlation coefficients $\{f_z(\rho_{ij})\}$ will be approximately normally distributed.

So if your 100 correlation matrices stem from the same distribution (and it has not changed in between) then the values of each cell should be approximately normally distributed. However, if you have two classes, this assumption probably does not hold, and the entry will not be normally distributed anymore.

The key point is that you need many sets of samples drawn from the same distribution. The purpose of the Fisher transformation is to estimate confidence intervals of the correlation coefficient. Since the (untransformed) correlation coefficient is bounded by $-1...+1$, it cannot be normally distributed; but using the Fisher transformation you can nevertheless use the known statistics for normal distributions.

So assume you want to estimate the correlation of height and weight (assuming both are normally distributed!). You can take a single sample, and compute the correlation - but how large are your error bounds on the correlation? Instead, you can take 100 independent samples, for each of them compute its correlation, Fisher transform the correlation, estimate the normal distribution errors, and transform these back. Then you can get an average correlation of the two variables and a confidence interval.

  • $\begingroup$ Thanks! So the rows and columns of the matrices (which are the same, obviously) are jointly normally distributed - hence your first point that X1,..,X50 are (marginally) normal follows. Do I understand your second point correctly: If I sample say from X_1 and X_10, even if these are two normal distributions with different parameters, the (repeatedly) sampled data would be approx. normal? If, however, the X_1 and X_10 from different matrices have different normal distributions, this does not hold true (?). Thx! $\endgroup$
    – Pugl
    Sep 14, 2015 at 14:55
  • 1
    $\begingroup$ Indeed. It's literally the same as drawing from different normal distributions. If I have a normal distribution that moves around over time, and I sample at different time points, the resulting overall data won't necessarily be normal distributed. $\endgroup$ Sep 14, 2015 at 16:52
  • $\begingroup$ I have another, related question: If I know that the samples (i,j) from X_i, X_j of all the 100 matrices are approximately normal disributed - does that imply that the X_i, X_j of all these 100 follow the same (normal) distribution? $\endgroup$
    – Pugl
    Sep 14, 2015 at 17:34
  • $\begingroup$ No. All correlations could be normal distributed around 0, i.e. not correlated on average. $\endgroup$ Sep 14, 2015 at 17:49

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