LDP for Marchenko Pastur with k/n tending to 0

Question

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 $0$?

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 $\mu=\delta_1$.