# Distribution of sum of freely independent Marchenko-Pastur measures

Given freely independent random variables $X_i$ with Marchenko-Pastur measures $\mu_i$, $i\in\{1,\dots,n\}$ how can we find the distribution of the scaled sum of these random variables $\sum_ia_iX_i$. (The measures are the empirical distributions of eigenvalues of symmetric random matrices.)

The answer seems to be free convolution of $\mu_i$, $i\in\{1,\dots,n\}$. To do that we can start with the R-transform. The R-transform of M-P measure $\mu_i$ is $R_i(z)=\frac{a_i}{1-\beta a_i z}$, where $\beta$ is the ratio of the dimensions of the random matrices. Now the R-trans of the sum distribution $R(z)$ is the sum of the R-transforms $R_i(z)$, $R(z)=\sum_i\frac{a_i}{1-\beta a_iz}$.

The question is how can one find the inverse of this $R(z)=\sum_i\frac{a_i}{1-\beta a_iz}$? So finally will have the law of $\sum_ia_iX_i$.

P.S.: Related to Random matrix determinant problem

The following method is more or less doable in practice, depending on your coefficients $a_i$'s :
A solution is to use the Cauchy (sometimes called Stieltjes) transform. For a measure $\mu$, the Cauchy transform is defined by $$G(z):= \int \frac{1}{z-t}\mathrm{d}\mu(t)$$
By a combinatorial identity between moments and free cumulants, $$R(G(z)) + \frac{1}{G(z)} = z$$
In the case of general $a_i$'s, finding an analytical expression for $G$ requires finding roots of a polynomial of degree $n$, but the calculation turns out to be very simple when $a_i$'s are equal for example.
Finally, you can use Stieltjes inversion formula to get an expression of the desired density. $$\mathrm{d}\mu(t) = \frac{1}{\pi} \lim_{\epsilon\to 0} \Im{G(t+i\epsilon)} \mathrm{d}t$$