Marcenko Pastur law when the dimensionality/sample size ratio $p/n \to 0, \infty$? Lack of resources?

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?

• 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 $p/n\in(0,1]$. Commented Feb 19, 2020 at 19:40
• @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. Commented Feb 19, 2020 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.