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Suppose $X \in \mathbb{R}^{N \times M}$ with elements sampled i.i.d. from $\mathcal{U}(-\sigma, \sigma)$.

I would like to find the marginal distribution of the unordered singular values of $X$. The case for $X \sim \mathcal{N}(0, \sigma)$ has been extensively studied, but i can't seem to find any work on the uniform distribution case.

Or equivalently the distribution of the eigenvalues of $X^TX$. The distribution of the $X^TX$ seems to follow some gaussian with mean zero and a small gaussian in the tail as observed in some simulations. This leads to the eigenvalues not being distributed as gaussian i.i.d.

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Rescale $Y=(\sqrt{3}/\sigma)X$, so that the matrix elements of $Y$ have uniform distribution with zero mean and unit variance.

In the limit $N,M\rightarrow\infty$ at fixed $M/N=r\in(0,1]$ the $M$ eigenvalues $\lambda_n$ of $N^{-1}Y^{\rm T}Y$ have the Marcenko-Pastur distribution, $$\rho(\lambda)=\frac{1}{2\pi \lambda r}\sqrt{(\lambda_+-\lambda)(\lambda-\lambda_-)},\;\;\lambda_-<\lambda<\lambda_+,\;\;\lambda_\pm=(1\pm\sqrt r)^2.$$ For a proof, see for example these notes.

The plot below is a test for $M=10^3$, $N=10^4$: the histogram shows the eigenvalue distribution for a randomly generated matrix with uniformly i.i.d. matrix elements, the curve is the Marcenko-Pastur distribution, which as you can see agrees very nicely.
(No idea why the histogram in the OP is so different.)


The OP has asked in the comment for non-asymptotic results. I don't think there are exact closed-form expressions for any $N,M$, but the large-$N$,$M$ asymptotics is already reached quite accurately for moderately large values of $N,M$. In the plot below I show the case $M=10,N=20$, which is already quite close to the Marcenko-Pastur limit.

And even $M=5,N=10$ is not too bad...

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  • $\begingroup$ Thanks for the quick reply. I was incomplete in my question. I am aware of the MP distribution, but i am looking for a general pdf or marginal distribution and not the limiting case since the applications requires all sizes of matrices. $\endgroup$ – David Vander Mijnsbrugge Apr 9 at 20:21
  • $\begingroup$ you are looking for an exact result for any $M,N$ ? that will not be forthcoming, I'm afraid; however, I'm pretty confident that the "large-$M,N$" limit is reached with good accuracy already for moderately large values of $M,N$, say upwards of 10 or so. $\endgroup$ – Carlo Beenakker Apr 9 at 20:26
  • $\begingroup$ I have added two plots to illustrate this. $\endgroup$ – Carlo Beenakker Apr 9 at 20:58
  • $\begingroup$ Thank you, it seems that i have some accuracy issues with my code. Thank you for the clarification. $\endgroup$ – David Vander Mijnsbrugge Apr 10 at 7:34

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