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Let $X$ be a $N\times P$ matrix with random independent and identically distributed entries $x_{ij}$. I also assume that $\langle x\rangle = 0$ and $\langle x^2\rangle = 1$.

Define the $N\times N$ matrix $C = (1/N)XX^T$.

I am interested in the limit $N\rightarrow\infty$, with $P=\alpha N$ for some finite positive constant $\alpha$. In this limit, the top eigenvalue of $C$ concentrates around a value. If the entries of $X$ are standard normal, then

$$\lambda_\mathrm{max} = (1 + \sqrt\alpha)^2$$

as known from the Marcenko and Pastur result.

I found a paper that claims that the same result is valid for any distribution of the $x_{ij}$, with the same first two moments. But this result is not proved, but only stated in passing. If this is well known, can someone provide a reference or explanation of why this is the case?

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    $\begingroup$ This is done in detail in the book of Bai and Silverstein. It is pretty standard - either by the moment method (after truncating) or computing Stieltjes transforms $\endgroup$ Commented Mar 9, 2020 at 17:35
  • $\begingroup$ @oferzeitouni Thanks! The result is stronger: the full spectrum (not just its extremes) is universal. For future reference, this is theorem 3.7 in the Bail & Silverstein book. $\endgroup$
    – valle
    Commented Mar 9, 2020 at 17:57
  • $\begingroup$ If you post an answer I'll accept it @oferzeitouni $\endgroup$
    – valle
    Commented Mar 9, 2020 at 17:57
  • $\begingroup$ For the extreme you need more. You need 4th moment. Otherwise the top eigenvalue can run off to infinity in the standard scaling (I misread your question - thought you cared only about the convergence of empirical measure) $\endgroup$ Commented Mar 9, 2020 at 18:28

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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 (and the original papers).

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