Let $X: \Omega \to \mathbb{R}^{p \times n}$ be a random matrix so that each entry $X_{ij}$ is a random variable with $\mathbb{E}X_{ij}=0, \mathbb{E}X_{ij}^2=\sigma^2$

I was wondering what would happen if we keep every hypothesis in the Marcenko-Pastur theorem, but change only one: assume that $p/n \to 0$ or $p/n \to \infty$. Below I'll refer time and again to the above link and the symbols used therein.

From the numerical experiements that I'm doing and from the expression of $\lambda_{+, -}$ in the link above, it looks like:

(1) When $p/n \to 0$, the empirical spectral ditribution of $Cov := \frac{1}{n}XX^{*}$ approaches the Dirac measure at $\sigma^2$. This is apparent from the numerical experiments, and also from the fact that in this case, when $\lambda:= lim_{p,n\to \infty}p/n < 1$, there's no isolated mass, but the continuous part defined using $\nu$ has support $\lambda_{+, -}$, and as $\lambda \to 0$, this support shrinks to $\sigma^2$.

(2) When $p/n \to \infty$, he empirical spectral ditribution of $Cov := \frac{1}{n}XX^{*}$ approaches the Dirac measure at $0$. This is because when $\lambda:= lim_{p,n\to \infty}p/n > 1$, then the Marcenko Patur distribution has two parts: one isolated mass at $0$ with weight $1-\frac{1}{\lambda}$, and the other is the contonuous distribution with density $\nu$, and with support $[\sigma^2(1-\frac{1}{\lambda})^2, \sigma^2(1+\frac{1}{\lambda})^2]$. Clearly the first isolated pass approach the Dirac measure and the second part approaches $0$ as its support moves infinitely away.

But the above two explanations are heuristics. Is there a mathematically rigorous proof where they actually prove these statements, say perhaps using the Stieltjes' transform method, so proving that for example when $p/n \to \infty$, the Stieltjes' transform ($ST$) of $ESD(cov X)\to -\frac{1}{z}= ST(\delta_0)$

N.B. I tried to look this up on the internet, but was a bit surprised by the lack of resources available. I wonder why there isn't much on these extreme cases?

  • $\begingroup$ I am not sure that I understand why there is a difficulty: first of all, it is helpful to subtract $|n-p|\delta(\lambda)$ from the eigenvalue density and consider only the continuous part $\delta\nu(\lambda)$; we may then without loss of generality assume $p\leq n$, because $XX^\ast$ and $X^\ast X$ have the same $\delta\nu$. Then the usual derivations of the MP law may be applied, which hold in the limit $n\rightarrow\infty$ for any fixed $p/n\in(0,1]$. $\endgroup$ Feb 19 '20 at 19:40
  • $\begingroup$ @CarloBeenakker Thanks for your comment! I'm not sure I understand what exactly you're doing to the random matrix $X$ in question to boil things down to the case where $p/n\to \lambda \in (0, \infty)$, so that usual derivations of the MP law may be applied. From what you wrote it looks like you're transforming the limiting spectral density of the random matrix for which $p/n \to \{0, \infty\}$, but how does that transform the random matrix itself from $X$ to say $Y$ so that the for $Y$, $p/n\to \lambda \in (0, \infty)$? It'd be great if you write a detailed answer. $\endgroup$ Feb 19 '20 at 20:05

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 $W=n^{-1}XX^*$, set $Y=W_I$. Compute $Q:=p^{-1}Etr( YY^*)=p^{-1}\sum_{i,j} E(|Y_{ij}|^2)$. An easy calculation (at least if say $EX_{ij}^4<\infty$) reveals that $Q\to_{n\to\infty} 0$, which implies by standard interlacing of eigenvalues the conclusion you sought.


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