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 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.
 A: 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}
