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?

  • 1
    $\begingroup$ 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 $\endgroup$ – ofer zeitouni Mar 9 '20 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. $\endgroup$ – becko Mar 9 '20 at 17:57
  • $\begingroup$ If you post an answer I'll accept it @oferzeitouni $\endgroup$ – becko Mar 9 '20 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) $\endgroup$ – ofer zeitouni Mar 9 '20 at 18:28

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.