Consider an $N\times K$ random matrix $X$ (defined on a probability space $(Ω,F,μ)$) with i.i.d. entries having zero mean and variance $1/K$.

There are a lot of results regarding the asymptotic behavior of the empirical distribution of eigenvalues of $XX^T$, or more precisely, the asymptotic behavior of the Stieltjes transform of the matrix $XX^T$ (the Marcenko-Pastur law), this result J. W. Silverstein, “Strong convergence of the empirical distribution of eigenvalues of large dimensional random matrices”, J. Multivar. Anal. 54, 175–192 (1995), etc.

Now, all the asymptotic random matrix results that I saw (in this context) are derived under the assumption that both dimensions of the matrix $X$ grows to infinity, but with fixed ratio. Namely, it is assumed that as $K,N→∞$ we fix $N/K→c∈(0,∞)$.

My question is whether there are some results, in case that both dimensions goes to infinity but with vanishing ratio. Concretely, can we say something regard the asymptotic behavior of, e.g. $$ \frac{1}{N}\text{trace}(XX^T+I_N)^{−1} $$ where $N/K→0$. Note that if $N$ is fixed, then we can readily use the SLLN to draw conclusions regard the asymptotic behavior of the above transform.

  • $\begingroup$ this looked initially like a valid question, why has all the text been deleted? $\endgroup$ – Carlo Beenakker Sep 11 '13 at 15:22
  • $\begingroup$ @CarloBeenakker Ops, for some unknown reason, in my last edit, I mistakenly deleted the text. $\endgroup$ – user91011 Sep 11 '13 at 16:51
  • $\begingroup$ Your scaling is not quite right: if $K>>N$ then the matrix $N^{-1} XX^T$ looks (in terms of spectrum) very much like $(K/N) I_N$. So the expression you write will tend to 0 (which is also what you will get if you took $c\to 0$). Did you mean variance $1/K$? In that case, I think you converge to $1/2$. $\endgroup$ – ofer zeitouni Sep 14 '13 at 2:12
  • $\begingroup$ You right, it should be $1/K$, thanks. For fixced $\endgroup$ – user91011 Sep 15 '13 at 9:11
  • $\begingroup$ You should edit the question to reflect that; in any case, my answer covers that case - your statistics converges to 1/2. $\endgroup$ – ofer zeitouni Sep 15 '13 at 12:46

There are many ways to see that, maybe the simplest is the following: let $W=XX^T/K$. Compute $E Tr W^k$ as $K,N\to\infty$. The combinatorics is easy - essentially, the only terms that survive passage to the limit are the terms that involve only diagonal terms of $W$, which converge to $1$. This shows that that the expected empirical measure of $W$ converges to a dirac mass at $1$. On the other hand, by concentration for the empirical measure, this is true for the empirical measure itself (this can be proved either again by moment computations, e.g. like in the proof of lemma 2.1.7 in the book of Anderson-Guionnet-Zeitouni on RM, or directly using the Hoffman-Wielandt inequality to show concentration). This approach will work as soon as your entries possess all moments. If you do not have that, the approach can be modified, by truncating the variables. An alternative would be to use Stieltjes transforms from the beginning.

If $N<<\sqrt{K}$ then an easier approach is to show that the diagonal elements concentrate around $1$, and that the row sum of off diagonal terms converges to $0$. Then use Gershgorin's circle theorem.


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