The Marchenko–Pastur formula defines a probability law with the fixed-point transform

Put $B=2\sqrt{\gamma }$, $A_0=1+\gamma$, and $x(\theta )=A_0+B\cos \theta$ for $0<\theta <\pi$. This substitution maps $(0,\pi )$ onto $(a,b)$ in reverse order, with $\sqrt{(b-x)(x-a)}=B\sin \theta$ and $\mathrm{\mid }dx\mathrm{\mid }=B\sin \theta d\theta$. Thus the continuous part of ${\mu }_{\gamma }$ becomes $(2\mathrm{/}\pi ){\sin }^2\theta d\theta \mathrm{/}x(\theta )$. For real $A>B$, the substitution $t=\tan (\theta \mathrm{/}2)$ gives ${\int }_0^{\pi }\frac{d\theta }{A+B\cos \theta }=2{\int }_0^{\mathrm{\infty }}\frac{dt}{(A+B)+(A-B)t^2}=\frac{\pi }{\sqrt{A^2-B^2}}\mathrm{.}$ Write $I_0(A)={\int }_0^{\pi }(A+B\cos \theta {)}^{-1}d\theta =\pi \mathrm{/}\sqrt{A^2-B^2}$. Since ${\int }_0^{\pi }\cos \theta d\theta =0$ and $\cos \theta \mathrm{/}(A+B\cos \theta )=B^{-1}-A\mathrm{/}[B(A+B\cos \theta )]$, $I_1(A):={\int }_0^{\pi }\frac{\cos \theta }{A+B\cos \theta }d\theta =\frac{\pi }{B}-\frac{A}{B}{I}_0(A),I_2(A):={\int }_0^{\pi }\frac{{\cos }^2\theta }{A+B\cos \theta }d\theta =-\frac{A}{B}{I}_1(A)=-\frac{A\pi }{B^2}+\frac{A^2}{B^2}{I}_0(A)\mathrm{.}$ Therefore $J(A):={\int }_0^{\pi }\frac{{\sin }^2\theta }{A+B\cos \theta }d\theta =I_0(A)-I_2(A)=\frac{\pi (A-\sqrt{A^2-B^2})}{B^2}\mathrm{.}$ For $A=A_0>B$, the root is $\mathrm{\mid }1-\gamma \mathrm{\mid }$; if $\gamma =1$, direct cancellation ${\sin }^2\theta \mathrm{/}(2+2\cos \theta )=(1-\cos \theta )\mathrm{/}2$ gives the same value by continuity. Hence the continuous mass is $2J(A_0)\mathrm{/}\pi =(A_0-\mathrm{\mid }1-\gamma \mathrm{\mid })\mathrm{/}(2\gamma )=\min (1,\gamma )\mathrm{/}\gamma$. Adding $(1-1\mathrm{/}\gamma {)}_{+}$ gives total mass one. Define $f_{\gamma }(x):=\sqrt{(b-x)(x-a)}\mathrm{/}(2\pi \gamma x)$ for $a<x<b$ and $f_{\gamma }(x):=0$ otherwise. This is nonnegative and Borel measurable, and has finite integral by the calculation above. By [F5], $f_{\gamma }$ is Lebesgue measurable. By [F4], $A\mapsto {\int }_A{f}_{\gamma }d\lambda$ is a finite Borel measure; the nonnegative multiple $(1-1\mathrm{/}\gamma {)}_{+}{\delta }_0$ is a finite measure as well. Their sum is the displayed set function, so it is a Borel probability measure supported on $[0,\mathrm{\infty })$.

For $w=r{e}^{i\theta }\in \mathbb{C}\setminus (-\mathrm{\infty },0]$, with $r>0$ and $-\pi <\theta <\pi$, define $q_0(w)=\sqrt{r}{e}^{i\theta \mathrm{/}2}$. This is continuous, satisfies $q_0(w{)}^2=w$, and is holomorphic: for $h\to 0$ with $w+h$ in the slit plane, $\frac{q_0(w+h)-q_0(w)}{h}=\frac{1}{q_0(w+h)+q_0(w)}⟶\frac{1}{2q_0(w)}\mathrm{.}$ Set $\mathrm{\Omega }=\mathbb{C}\setminus (-\mathrm{\infty },B]$ and $q(A)=q_0(A-B)q_0(A+B)$ for $A\in \mathrm{\Omega }$. Both factors are defined and holomorphic there, $q(A{)}^2=A^2-B^2$, and $q(A)>0$ for real $A>B$. The domain is connected: each $A-B=r{e}^{i\theta }$ with $-\pi <\theta <\pi$ can be joined inside it to $B+1$ by continuously changing $r$ to $1$ and $\theta$ to $0$. The integral $J(A)$ is holomorphic on $\mathrm{\Omega }$. Indeed, on any compact subset its denominator is bounded away from zero, and the difference quotient of its integrand tends pointwise to $-{\sin }^2\theta \mathrm{/}(A+B\cos \theta {)}^2$ with a uniform integrable bound; [F3] justifies passing that limit through the integral. The identity theorem [F2] extends the formula from real $A>B$ (which has limit points inside $\mathrm{\Omega }$) to all $A\in \mathrm{\Omega }$. In particular it applies to $A_z=A_0-z$, whose imaginary part is negative.

Let $q=\sqrt{(1+\gamma -z{)}^2-4\gamma }$ on that branch. The substitution in step 1.1 and $1\mathrm{/}[x(x-z)]=z^{-1}((x-z{)}^{-1}-x^{-1})$ give the transform of the continuous part as $[(A_z-q)-(A_0-\mathrm{\mid }1-\gamma \mathrm{\mid })]\mathrm{/}(2\gamma z)$. Adding the atom contributes $-(1-1\mathrm{/}\gamma {)}_{+}\mathrm{/}z$; simplifying in the cases $\gamma \le 1$ and $\gamma >1$ gives $m_{\gamma }(z)=\frac{1-\gamma -z-q}{2\gamma z}\mathrm{.}$ Since $q^2=(1-\gamma -z{)}^2-4\gamma z$, substitution verifies $\gamma z{m}_{\gamma }(z{)}^2+(z+\gamma -1)m_{\gamma }(z)+1=0$, Since step 1.1 gives a probability measure, $\mathrm{Im}{m}_{\gamma }(z)=\int \eta \mathrm{/}\mathrm{\mid }x-z{\mathrm{\mid }}^{2}d{\mu }_{\gamma }(x)>0$. Thus $1+\gamma m_{\gamma }(z)\mathrm{\ne }0$, and rearranging the quadratic gives the fixed-point equation in the Statement.

For $x\ge 0$, $\mathrm{Im}(z\mathrm{/}(x-z))=\eta x\mathrm{/}\mathrm{\mid }x-z{\mathrm{\mid }}^2\ge 0$. Integrating this identity against the probability measure from step 1.1 gives $\mathrm{Im}(z{m}_{\gamma }(z))\ge 0$, so $\mathrm{Im}(1-\gamma -z-\gamma z{m}_{\gamma }(z))=-\eta -\gamma \mathrm{Im}(z{m}_{\gamma }(z))\le -\eta$.