Limiting value of Stieltjes transform of sum of independent Wishart matrices

Question

Let $n_1$, $n_2$, and $d$ positive integers tending to infinity such that $d/n_k \to \phi_k \in (0,\infty)$ and $n_1/(n_1+n_2) \to p \in (0,1)$. Let $X_k$ be an $n_k \times d$ random matrix with iid rows from $N(0,\Sigma_k)$ and consider the matrix $G_H(z) := (H -z I_d)^{-1}$, where $H := A+B$, $A:=X_1^\top X_1/n$, $B := X_2^\top X_2/n$, and $n := n_1+n_2$ and $z \in \mathbb C \setminus \mathbb R_+$. We make standard assumptions on $\Sigma_{1,2}$, like eigenvalues bounded away from $0$ and $\infty$, etc. It may also be assumed that $\Sigma_{1,2}$ are diagonal.

Question 1. What is the limiting value of $m_H(z) := (1/d)\operatorname{tr} G_H(z)$.

One would expect the above limit to satisfy some fixed-point equation. I stumbled on "Subordination for the sum of two random matrices " https://arxiv.org/abs/1109.5818, but unfortunately this doesn't solve my problem because the "subordination functions" $\omega_A$ and $\omega_B$ therein are themselves defined in terms of $\operatorname{tr}A G_H(z)$ and $\operatorname{tr}B G_H(z)$, which are definitely more complicated than the sought-for $m_H(z)$.

Edit

Take $\Sigma_1 = \Sigma_2 = I_d$. Using the proposal here https://mathoverflow.net/a/157070/78539, the "R-transform" of the LSD of $A$ is $R_1(z) = p/(1-p\phi_1 z)$, and of $B$ is $R_2(z) = q/(1-q\phi_2 z)$ with $q:=1-p$. We deduce that $$ R_H(z) = R_1(z) + R_2(z) = \frac{p}{1-p\phi_1 z} + \frac{q}{1-q\phi_2 z}. $$

Also, it is well-known that $R(-m(z)) -1/m(z) = z$, and so $$ \frac{p}{1+p\phi_1 m_H(z)} + \frac{q}{1+q\phi_2 m_H(z)} - \frac{1}{m_H(z)} = z. $$

Question 2. Experts, please do you think the above is correct ? If so, can it be existended to the anisotropic case ?

It does look about right, because if we eliminate, say the first term on the right (which amounts to setting $p \to 0$), we have $$ \frac{1}{1+\phi_2 m_B(z)} -\frac{1}{m_B(z)} = z, $$ which is the well-known equation defining the Stieltjes transform of the Marchenko-Pastur law.

In the anisotropic case, it seems one should take $$ R_k(z) = \frac{1}{d}\sum_{j=1}^d \frac{p_k\lambda_j^{(k)}}{1-p_k\phi_k \lambda_j^{(k)} z}, $$ where $(\lambda_k^{(j)})_{1 \le j \le d}$ are the eigenvalues of $\Sigma_k$, and continue as in the isotropic case.

Answer 1

Approach 1

As mentioned by user jlewk, Marchenko & Pastur (1967) indeed solved a general problem for which mine is an instance. Indeed, see (1.14) of their paper https://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=4101&option_lang=eng

This, would give $$ m_H(z) = m_A\left(-z+\int_0^\infty \frac{qt}{1+\phi_2qt}\mathrm d\nu_2(t)\right), $$ where $\nu_k$ is the LSD of $\Sigma_k$, and $m_A(w)$ is a appropriate solution to the equation $$ \frac{1}{m_A(w)} = -w +\int_0^\infty \frac{pt}{1+m_A(w)\phi_1 pt}\mathrm d\nu_1(t). $$

Approach 2

Alternatively, Theorem 1 of this RMT monograph https://polaris.imag.fr/romain.couillet/docs/articles/det_eq_MAC.pdf gives

$$ \frac{1}{m_H(z)} = -z + \sum_k \int_0^\infty \frac{p_kt}{1+\phi_k e(z)p_kt}\mathrm d\nu_k(t), $$ where $p_1:=p$, $p_2:=q$, and the function $e$ is defined to be the unique solution nonnegative to the equation $$ \frac{1}{e(z)} = -z + \sum_k \int_0^\infty\frac{p_kt}{1+e(z)\phi_k p_kt}\mathrm d\nu_k(t). $$ Thus, we must have $e(z) = m_H(z)$. We deduce that $m_H(z)$ is the unique nonnegative solution of $$ \frac{1}{m_H(z)} = -z + \sum_k \int_0^\infty \frac{p_kt}{1+m_H(z)\phi_k p_k t}\mathrm d\nu_k(t). $$