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 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:31:38Z http://mathoverflow.net/feeds/question/72039 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72039/stieltjes-transform-of-fpf-as-a-function-of-the-stieltjes-transform-of-p 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 Marmar 2011-08-03T21:15:26Z 2011-09-18T22:22:12Z <p>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:</p> <p>Let us call the Stieltjes Transform of $F^{*}PF$ to be $S_{F^{*}PF}=t(z)$. I want to show that</p> <p>$$t^2(z)=-\frac{1}{z}S_P\left(-\frac{1}{t(z)}\right) \nonumber$$ where $S_P(z)$ is the Stieltjes Transform of $P.$</p> <p>I know I have to use Marcenko-Pasture Theorem but couldn't figure out how. </p> <p>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,</p> <p>$$t(z)=-\frac{1}{z-\int \frac{\tau dH(\tau)}{1+\tau t(z)}}$$</p> <p>I cannot go on from here.</p> http://mathoverflow.net/questions/72039/stieltjes-transform-of-fpf-as-a-function-of-the-stieltjes-transform-of-p/72297#72297 Answer by Marmar for 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 Marmar 2011-08-07T19:23:13Z 2011-08-07T19:23:13Z <p>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</p> <p>$$t^2(z)=-\frac{1}{z}S_{P_i}\left(-\frac{1}{t(z)}\right) \nonumber$$ where $S_{P_i}(z)$ is the Stieltjes Transform of $P_i.$</p> <p>We consider the Marcenko-Pasture Theorem and see that $A_n$ is zero and $X_m=\sqrt{n}F_i.$ Therefore,</p> <p>$$t(z)=-\frac{1}{z-\int \frac{\tau dH(\tau)}{1+\tau t(z)}} \label{case2_hasibi}$$</p> <p>where $H(\tau)$ is the empirical(eigenvalue) distribution of $P_i.$ In general we know that</p> <p>\begin{eqnarray} \int dH(\tau)&amp;=&amp;1=\int \frac{(\tau-y) dH(\tau)}{(\tau-y)} \nonumber\ &amp;=&amp;\int \frac{\tau dH(\tau)}{\tau-y}-\int \frac{y dH(\tau)}{\tau-y} \nonumber\ &amp;=&amp; \int \frac{\tau dH(\tau)}{\tau-y}-y \int \frac{dH(\tau)}{\tau-y} \nonumber\ &amp;=&amp;\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&amp;=&amp;\int \frac{\tau dH(\tau)}{\tau+\frac{1}{t(z)}}+\frac{1}{t(z)} \int \frac{dH(\tau)}{\tau+\frac{1}{t(z)}} \nonumber\ &amp;=&amp; 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 $$t(z)=-\frac{1}{z-\int \frac{\tau dH(\tau)}{1+\tau t(z)}}=-\frac{1}{z-[\frac{1}{t(z)}-\frac{1}{t^2(z)} S_{P_i}(z)]} \label{case2_hasibi_final}$$ By simplifying both sides of (\ref{case2_hasibi_final}) we have $$-t(z)z+1-\frac{1}{t(z)}S_{P_i}\left( -\frac{1}{t(z)} \right) =1. \nonumber$$ And so $$t(z)z=-\frac{1}{t(z)}S_{P_i}\left( -\frac{1}{t(z)} \right), \nonumber$$ which means that $$t^2(z)=-\frac{1}{zS_{P_i}\left( -\frac{1}{t(z)} \right)}. \label{case2_hasibi_final2}$$</p>