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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 en …

I will assume $m=\alpha_d d$ with $\alpha_d\to \alpha \in [1,\infty)$ independent of $d$. The case $\alpha\to\infty$ is actually easier. Define $Z=d^{-1} m^{-2} \sum_{i=1}^d \lambda_i^2$. Then $Z$ con …

This falls within the domain of free probability, at least when the dimension is large. That is, if the matrices are say unitarily invariant and independent, and if we denote by $tr_N$ the trace divid …

The following does not decide on whether the infimum is $0$, but it does give a rough bound. Let $v$ be a normalized eigenvector. Suppose $\|v\|_1<n^{\alpha}$ for some $\alpha<1/2$. Let $I=\{i: |v_i| …

You did not specify whether you have a relation between $n$ and $k$ and whether you care about asymptotics. I will assume that $k/n\to 1$ and that $n\to\infty$, and that your scaling is such that the …

Actually, just to get the semicircle is not hard. Take an $n$-by-$n$ Jacobi matrix whose on diagonal entries are $0$ and $i$th entry on the off diagonal is $\sqrt{i/n}$. The limit ESD will be the semi …

An explicit formula is the hook-product formula, due to Schensted I believe. This formula is used in the classical work of Logan and Shepp, as well as in Vershik-Kerov. See for example equation (1.1) …

In the Gaussian case, you can rewrite $x_i=R^{1/2}y_i$ where $y_i$ now possess iid entries. This leads you to computing the eigenvalues of $Y^*RY$, this is actually the problem solved by Pastur and Ma …

In general you need more than second moment - you need fourth moment finite, otherwise the top eigenvalue can run off to infinity in your scaling. For this and more see the book of Bai and Silverstein …

It is enough to understand the case $p/n\to 0$, as the eigenvalues of $XX^*$ match those of $X^*X$ up to an extra atom at $0$. In that case you can argue as follows, with $\sigma=1$: writing $W=n^{-1} …

I do not understand the role of $k$ in your question, so I will just ignore it. If you take your matrix to consist of independent Gaussian entries (normalized by $\sqrt{n}$), this will give a precise …

All you need is that the empirical measure of the $k_i$s converge. Whenever it does, just apply the original Pastur-Marcenko (1967) results to get that the empirical measure associated with your matr …

The question is not completely clear. Do you fix a random $G(n,p)$ and then replace the $1$ in the adjacency matrix by a uniform random variable? Or is it as you write at the end, namely you just take …

The diagonal elements just shift the spectrum (and the top eigenvalue) by $1$. So we may assume they are $0$. You are essentially dealing with a symmetric matrix whose entries above the diagonal are …

I believe this follows from the standard estimates that show the convergence. For example, look at the proof in Anderson-Guionnet-Zeitouni's book, page 24. Indeed, from the last display there, you get …