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Let $x_1,...,x_n\in\mathbb{R}^p$ be i.i.d. random vectors with mean 0 and covariance $\Sigma_p$. Let $S_{n,p}=\sum_{i=1}^nx_ix_i^T/n$ be the sample covariance. We consider the asymptotics of the empirical distribution of eigenvalues of $S_{n,p}$. We let $p/n\to c\in(0,1)$ where $c$ is a constant.

It is well-known that when $\Sigma_p$ is identity, the limiting distribution of the spectrum of $S_{n,p}$ follows the Marchenko-Pastur distribution

What about for general $\Sigma_p$? What kind of asymptotic regime do we need on $\Sigma_p$ in order to have a meaningful limiting distribution of the spectrum of $S_{n,p}$?

For example, we can assume the empirical distribution of eigenvalues of $\Sigma_p$ converges to a deterministic measure $\mu$. Is this condition enough? I suppose this is not enough and the eigenvectors of $\Sigma_p$ also seems matter. If this is indeed enough, what is the relationship between $\mu$ and the limiting spectrum distribution of $S_{n,p}$?

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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 Marchenko, see Math. USSR. Sbornik vol 1 (1967), with an explicit equation satisfied by the Stieltjes transform of $\mu$. You can find also a discussion in the book of Bai and Silverstein, Chapter 4. All that matters is the asymptotic limit of empirical measure of eigenvalues of $R$.

For a large class of non-Gaussian entries, you can apply https://arxiv.org/pdf/1312.0037.pdf to go back to the Gaussian case.

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  • $\begingroup$ Thanks for pointing out the reference! $\endgroup$
    – neverevernever
    Commented Jan 4, 2020 at 15:58
  • $\begingroup$ Hello and thanks for your answer! I've a very rudilentary follow-up question if it's ok. Since $x_i = \sqrt{R}y_i, R \mathbb{R}^{p \times p} $, then the data matrix is: $X:=[x_1 \dots x_n] \mathbb{R}^{p \times n}$. So if I take the sample cov $1/p XX* = 1/p \sqrt R YY* \sqrt R.$, an dif I take the Gram (or dual covariance) matrix, then $Y*Y = 1/n Y*RY$. Since I'm very new to this area, I'm just checking with you to see if I got this right. I'll read the paper you mentioned. $\endgroup$
    – Learning math
    Commented Mar 11, 2020 at 15:50
  • $\begingroup$ Sorry I wrote $Y*$, but I really meant $Y^{*}$, the adjoint of $Y$. $\endgroup$
    – Learning math
    Commented Mar 11, 2020 at 16:06
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    $\begingroup$ I can't decipher your message, sorry. Y^*Y=1/n Y^*RY?? $\endgroup$
    – ofer zeitouni
    Commented Mar 11, 2020 at 18:42
  • $\begingroup$ @oferzeitouni Sorry I meant: if we take $X:= \sqrt {R} Y,$ of dimension $p \times n,$ then the sample covariance is $1/p XX^{*}= 1/p \sqrt R YY^{*} \sqrt R, $ (dim = $p \times p$) and the Gram matrix (dual covriance) is: $1/n X^{*}X = 1/n Y^{*}R Y$ (dim = $n \times n$). Here, $p=$ no if features, and $n =$ no of samples. None of these two are proportional to $YRY^{*}$, so I'm just checking with you regarding the symbols. I know it's a minor difference, but as a newbie to RMT, it'd still be helpful for me. Thanks in advance! $\endgroup$
    – Learning math
    Commented Mar 11, 2020 at 22:01

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