Stieltjes Transform of $F^{*}PF$ as a function of the Stieltjes Transform of $P$ where $F$ is drawn from an $n \times n$ Gaussian-like random matrix distribution

Question

I am trying to calculate the Stieltjes Transform of $F^{*}PF$ as a function of the Stieltjes Transform of $P$ where $F$ is drawn from an $n \times n$ Gaussian-like random matrix distribution. I am looking for the solution to look like this:

Let us call the Stieltjes Transform of $F^{*}PF$ to be $S_{F^{*}PF}=t(z)$. I want to show that

\begin{equation} t^2(z)=-\frac{1}{z}S_P\left(-\frac{1}{t(z)}\right) \nonumber \end{equation} where $S_P(z)$ is the Stieltjes Transform of $P.$

I know I have to use Marcenko-Pasture Theorem but couldn't figure out how.

I considered the Marcenko-Pasture Theorem and the iteration they talk about as $B_n=A_n+1/nX_m^{*}T_mX_m$ and compared this to $F^{*}PF$ which means $A_n$ is zero and $X_m=\sqrt{n}F.$ Therefore,

\begin{equation} t(z)=-\frac{1}{z-\int \frac{\tau dH(\tau)}{1+\tau t(z)}} \end{equation}

I cannot go on from here.

Answer 1

Let us call the Stieltjes Transform of $F_i^{*}P_iF_i$ to be $S_{F_i^{*}P_iF_i}=t(z)$. We want to show that

\begin{equation} t^2(z)=-\frac{1}{z}S_{P_i}\left(-\frac{1}{t(z)}\right) \nonumber \end{equation} where $S_{P_i}(z)$ is the Stieltjes Transform of $P_i.$

We consider the Marcenko-Pasture Theorem and see that $A_n$ is zero and $X_m=\sqrt{n}F_i.$ Therefore,

\begin{equation} t(z)=-\frac{1}{z-\int \frac{\tau dH(\tau)}{1+\tau t(z)}} \label{case2_hasibi} \end{equation}

where $H(\tau)$ is the empirical(eigenvalue) distribution of $P_i.$ In general we know that

\begin{eqnarray} \int dH(\tau)&=&1=\int \frac{(\tau-y) dH(\tau)}{(\tau-y)} \nonumber\\ &=&\int \frac{\tau dH(\tau)}{\tau-y}-\int \frac{y dH(\tau)}{\tau-y} \nonumber\\ &=& \int \frac{\tau dH(\tau)}{\tau-y}-y \int \frac{dH(\tau)}{\tau-y} \nonumber\\ &=&\int \frac{\tau dH(\tau)}{\tau-y}-yS_{Z}(y) \nonumber \end{eqnarray} By writing the last equation for $Z=P_i$ and $y=-\frac{1}{t(z)}$, we have \begin{eqnarray} 1&=&\int \frac{\tau dH(\tau)}{\tau+\frac{1}{t(z)}}+\frac{1}{t(z)} \int \frac{dH(\tau)}{\tau+\frac{1}{t(z)}} \nonumber\\ &=& t(z)\int \frac{\tau dH(\tau)}{\tau t(z)+1}+\frac{1}{t(z)} S_{P_i}(z). \nonumber \end{eqnarray} Then, \begin{eqnarray} \frac{1}{t(z)}=\int \frac{\tau dH(\tau)}{\tau t(z)+1}+\frac{1}{t^2(z)} S_{P_i}(z). \nonumber \end{eqnarray} Therefore, $\int \frac{\tau dH(\tau)}{\tau t(z)+1}=\frac{1}{t(z)}-\frac{1}{t^2(z)} S_{P_i}(z).$ By replacing this integration in (\ref{case2_hasibi}) we get \begin{equation} t(z)=-\frac{1}{z-\int \frac{\tau dH(\tau)}{1+\tau t(z)}}=-\frac{1}{z-[\frac{1}{t(z)}-\frac{1}{t^2(z)} S_{P_i}(z)]} \label{case2_hasibi_final} \end{equation} By simplifying both sides of (\ref{case2_hasibi_final}) we have \begin{equation} -t(z)z+1-\frac{1}{t(z)}S_{P_i}\left( -\frac{1}{t(z)} \right) =1. \nonumber \end{equation} And so \begin{equation} t(z)z=-\frac{1}{t(z)}S_{P_i}\left( -\frac{1}{t(z)} \right), \nonumber \end{equation} which means that \begin{equation} t^2(z)=-\frac{1}{zS_{P_i}\left( -\frac{1}{t(z)} \right)}. \label{case2_hasibi_final2} \end{equation}