2
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.