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 $m(z) := \mathbb E[\operatorname{tr}(BR(z))]$.

Let $\phi \in (0,\infty)$ be fixed.

Question. 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.

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 $R(z) := (S-z I_d)^{-1}$ for any $z \in \mathbb C^+$, and $$ R(z) := (S-z I_d)^{-1}\text{ for any }z \in \mathbb C^+, $$ and define $m(z) := \mathbb E[\operatorname{tr}(BR(z))]$.

Let $\gamma \in (0,\infty)$$\phi \in (0,\infty)$ be fixed.

Question. Assume that $A$ and $B$ have limiting spectral distributions as $d \to \infty$. In the limit $n,d \to \infty$ such that $d/n \to \phi$, what $$ \label{1} \tag{1} n,d \to \infty\text{ such that }d/n \to \phi, $$ 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, assumelet $n \ge d + 2$. 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.

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 $R(z) := (S-z I_d)^{-1}$ for any $z \in \mathbb C^+$, and define $m(z) := \mathbb E[\mbox{tr}(BR(z))]$$m(z) := \mathbb E[\operatorname{tr}(BR(z))]$.

Let $\gamma \in (0,\infty)$ be fixed.

Question. Assume that $A$ and $B$ have limiting spectral distributions as $d \to \infty$. In the limit $n,d \to \infty$ 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.

Solution for the case $\phi \lt 1$

WLOG, assume $n \ge d + 2$. 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 $$ m(0) = \frac{1}{n-d-1}\mbox{tr}(B A^{-1}) = \frac{d}{n-d-1}\overline{\mbox{tr}}(B A^{-1}) \to \frac{\phi}{1-\phi}\cdot\lim_{d \to \infty} \overline{\mbox{tr}}(BA^{-1}), $$$$ m(0) = \frac{1}{n-d-1}\operatorname{tr}(B A^{-1}) = \frac{d}{n-d-1}\overline{\operatorname{tr}}(B A^{-1}) \to \frac{\phi}{1-\phi}\cdot\lim_{d \to \infty} \overline{\operatorname{tr}}(BA^{-1}), $$

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