An isotropic finite-fourth-moment design

The empirical eigenvalue law of $T_n=I_n$ is $H_n={\delta }_1$, and for every $w\mathrm{\notin }[0,\mathrm{\infty })$, $x_n^{\ast }(T_n-wI{)}^{-1}{x}_n=(1-w{)}^{-1}=\int (t-w{)}^{-1}d{\delta }_1(t)\mathrm{.}$ Thus [F1] applies and gives almost-sure convergence to $F_{c,{\delta }_1}$.

Write ${\mu }_c$ for the Marchenko–Pastur probability law in [F3] and $m_c(z)=\int (t-z{)}^{-1}d{\mu }_c(t)$. Its scalar equation $1+z{m}_c=m_c\mathrm{/}(1+c{m}_c)$ rearranges to $m_c=(1-c-z-cz{m}_c{)}^{-1}$, the fixed-point equation for $F_{c,{\delta }_1}$. Moreover $\mathrm{Im}{m}_c(z)=\int \mathrm{Im}z\mathrm{/}\mathrm{\mid }t-z{\mathrm{\mid }}^{2}d{\mu }_c(t)>0$. The signed expression ${\nu }_c=(1-c){\delta }_0+c{\mu }_c$ is in fact a probability measure: for $c\le 1$ both coefficients are nonnegative; for $c>1$ the atom $c(1-1\mathrm{/}c)$ in $c{\mu }_c$ exactly cancels $(1-c){\delta }_0$, leaving a nonnegative measure of total mass one. Consequently $u=-(1-c)\mathrm{/}z+c{m}_c(z)=\int (t-z{)}^{-1}d{\nu }_c(t)$ has strictly positive imaginary part. Both admissibility conditions of [F4] hold, so [F4] gives equality of the transforms and [F5] identifies the two probability laws. Hence the limit is the stated Marchenko–Pastur law.