On the other hand, use $ m_{{\mathsf{MP}}}(z) $ to denote the Stieltjes transform of the Marchenko-Pastur law $$m_{{\mathsf{MP}}}(z) := \int_{{\mathbb{R}}} \dfrac{\rho_{{\mathsf{MP}}}(x)}{x-z}{\mathrm{d}}x=\dfrac{1-\xi-z+\sqrt{(z-\lambda_-)(z-\lambda_+)}}{2\xi z},$$ where $ \sqrt{\quad} $ denotes the square root on the complex plane whose branch cut is the negative real line. With this choice we always have $ {\text{\rm Im}\hspace{0.1cm}}m_{{\mathsf{MP}}}(z)>0 $ when $ {\text{\rm Im}\hspace{0.1cm}}z>0 $. For the singular values, recall the limit distribution $ \rho(x) $ for the empirical measure and use $ m(z) $ to denote its corresponding Stieltjes transform $$m(z) := \int_{{\mathbb{R}}} \dfrac{\rho(x)}{x-z}{\mathrm{d}}x = \int_{{\mathbb{R}}} \dfrac{1}{x-z}\dfrac{1}{2\pi \xi}\sqrt{\dfrac{[(x^2-\lambda_-)(\lambda_+ - x^2)]_+}{x^2}}{\mathrm{d}}x.$$ We have the following relation between $ m(z) $ and $ m_{{\mathsf{MP}}}(z) $ $$\begin{gathered}
\label{e.ContinuousStieltjes}
m(z) = \int_{\sqrt{\lambda_-}}^{\sqrt{\lambda_+}} \left( \dfrac{1}{x-z} - \dfrac{1}{x+z} \right) \dfrac{1}{2\pi \xi}\sqrt{\dfrac{[(x^2-\lambda_-)(\lambda_+ - x^2)]_+}{x^2}}{\mathrm{d}}x\\
= \int_{{\mathbb{R}}}\dfrac{z}{x-z^2}\rho_{{\mathsf{MP}}}(x){\mathrm{d}}x = zm_{{\mathsf{MP}}}(z^2).\end{gathered}$$ It is well known that we have the strong local Marchenko-Pastur law [@bloemendal2014isotropic; @pillai2014universality] for the estimate of $ S_N(z) $, i.e. for any $ D>0 $, there exists $ N_0(D)>0 $ such that for every $ N {\geqslant}N_0 $ we have $${\mathbb{P}}\left( \left| S_N(z) - m_{{\mathsf{MP}}}(z) \right| {\leqslant}\dfrac{\varphi}{N\eta} \right) > 1-N^{-D}.$$ By the relations and , we know that $$|m_N(z) - m(z)| = \left| z \left( S_N(z^2) - m_{{\mathsf{MP}}}(z^2) \right) \right| {\leqslant}|z| \left| S_N(z^2) - 