I am interested in the determinant of $W = X * X'$, where $X \in \mathbb{R}^{k \times n}$ is a matrix with each row drawn IID from some sub-Gaussian distribution on $\mathbb{R}^{n}$. (I am aware of some universality results, so happy to also consider a "standard" Wishart matrix with parameter $k/n$). Edit: say e.g. that the diagonal of $W$ is all ones, and the off-diagonal are of order $1/\sqrt n$.

Question (short): is there an LDP for the empirical spectral distribution of $W$ to Marchenko-Pastur when $k \approx \sqrt{n}$?

Question (long): There is an argument by Ofer here https://mathoverflow.net/q/372456 that finds the expected determinant of $W$, if $k/n \to c$ for some constant $c \in (0,1)$. The idea is to use an LDP for the convergence of the empirical spectral distribution of Wishart to Marchenko-Pastur.

I am interested in the case of $k/n \to 0$, and in particular, the scaling $k \approx \sqrt{n}$. Letting $\lambda := k/n$, and $\mu(\lambda)$ denote the corresponding M-P law, it is easy to check $$ \mathbb{E}_{X \sim \mu(k,n)}[\log x] \asymp -\lambda/2 + O(\lambda^2) $$ I would hope that this implies something like $$ \mathbb{E}\det(W) = \exp(k \lambda + O(\lambda)) := exp(k^2/n + O(k/n))$$

But if $k/n \to 0$, the referenced LDP gives the trivial answer $\mathbb{E}\det(W) = o(k)$. Is there a more "quantitative" LDP known that allows me to take $k$ and $n$ jointly to $0$?

  • $\begingroup$ LDP = Large Deviation Principle. $\endgroup$ Mar 3 at 20:55
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    $\begingroup$ Actually, the argument I gave in that question does not really need the LDP, only concentration, and the latter is known for in your setup (e.g. if entries satisfy log-sobolev, or are bounded). $\endgroup$ Mar 3 at 22:25
  • $\begingroup$ For a true LDP, unfortunately not much is known for general entries when $k/n\to c$ $\endgroup$ Mar 3 at 22:28
  • $\begingroup$ For $k/n\to 0$, concentration is even better, so see my first comment. $\endgroup$ Mar 3 at 22:28
  • $\begingroup$ You are asking many questions at once. In the regime $k/n\to 0$, the $k\times k$ matrix $W$ is very close to $kI_k$ and then it it not hard to check that $E det(W)\sim n^k$ (unless of course you meant somehow to normalize things differently - seems you did because you talk of convergence to MP. $\endgroup$ Mar 3 at 22:34

1 Answer 1


For standard Gaussians, and with the matrix $W/n$, the proof of the LDP given by Ben Arous-Guionnet adapts to the Wishart setup. However, you will have different scalings and so the non-commutative entropy term (of exponential scaling $k^2$) will disappear, and the proof more or less trivializes. If I did not make a stupid computational mistake, the large deviations will have speed $kn$ and rate function $I(\mu)= \frac12 \int (x-\log x) \mu(dx)$. You see that $I(\mu)=0$ iff $\mu=\delta_1$.


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