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Let $A=A(d)$, and $B=B(d)$ be (sequences of) deterministic positive-definite $d \times d$ matrices and let $X$ be an $n \times d$ random matrix with iid rows from $N(0,A)$. Let $R$ be the resolvent of the (recaled) empirical covariance matrix $S := X^\top X$, given by $$ R(z) := (S-z I_d)^{-1}\text{ for any }z \in \mathbb C^+, $$ and define $m(z) := \mathbb E[\operatorname{tr}(BR(z))]$.

Assume that $A$ and $B$ have limiting spectral distributions as $d \to \infty$.

Question. Let $\phi \in (0,\infty)$ be fixed. In the limit $$ \label{1} \tag{1} n,d \to \infty\text{ such that }d/n \to \phi, $$ what is the value of $m(z)$ ?

Notes

  • I'm only interested in computations for small $z$, i.e $z \to 0$.
  • If it helps, it may also be assumed that $BA^{-1}$ has a limiting spectral density.
  • I've looked in Bai and Silverstein's book (Chapter 4) but I don't see anything which applies to my problem.

Solution for the case $\phi \lt 1$

WLOG, let $n \ge d + 2$. Then, it is a standard result that $$ \mathbb E R(0) = \mathbb E (X^\top X)^{-1} = \frac{1}{n-d-1}A^{-1}. $$

We deduce that in the limit \eqref{1}, $$ m(0) = \frac{1}{n-d-1}\operatorname{tr}(B A^{-1}) = \frac{d}{n-d-1}\overline{\operatorname{tr}}(B A^{-1}) \to \frac{\phi}{1-\phi}\cdot\lim_{d \to \infty} \overline{\operatorname{tr}}(BA^{-1}), $$

where $\overline{\operatorname{tr}} := (1/d)\operatorname{tr}$ is the normalized trace operator.

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  • $\begingroup$ Just to clarify: The matrix $B$ is not of special importance, since, by linearity of expectation we have $\mathbb{E}tr(BR(z)) = tr(B\mathbb{E}(R(z)))$, right? $\endgroup$
    – Dirk
    Commented Jun 12, 2023 at 14:47
  • $\begingroup$ Yes, that is true by linearity but doesn't provide any particular insight, and I wrote it that way for good reason. Indeed, it's easier to analyze the limiting spectral density of a resolvent than the expectation of the resolvent. This is already clear in the case where $B=A=I$: It's easier to compute the limit of $tr(BE(R(z)))$ than trying to compute $E(R(z))$ and then computing the limit of its trace. The matrix B does play a big role. I hope I didn't miss anything in what you wrote. $\endgroup$
    – dohmatob
    Commented Jun 12, 2023 at 16:42

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