# Combinatorial formula to compute the moments of the product of two free random variables

I found in the PhD thesis Moments method for random matrices with applications to wireless communication the following combinatorial formula to compute the free moments of the product of two random variables.

Given $$a$$ and $$b$$ random variables in free relation in a noncommutative space, denoting $$\kappa_i^a := \kappa_i(a,...,a)$$ and $$\kappa_j^b := \kappa_j(b,...,b)$$, the free moments of the product $$ab$$ are computed from the free cumulants as follows $$$$m_n(a,b)=\sum_{(\pi_1,\pi_2)\in NC(n)} \prod_{i=1}^{|\pi_1|} \prod_{j=1}^{|\pi_2|}\kappa_i^a\kappa_j^b$$$$

where we denote by $$NC(n)$$ the set of non-crossing partitions of $${1,...,n}$$, and where $$|\pi_i|$$ is the cardinality of the block $$\pi_i$$, $$i=1,2$$. For example, it follows that $$m_1(a,b)=\kappa_1^a\kappa_1^b$$.

My question is: does the above formula derive from the relation $$S_{a\boxtimes b}(z)=S_{a}(z)S_{b}(z)$$ on the $$S$$ transform of the multiplicative free convolution of $$a$$ and $$b$$?

Otherwise is there an alternative proof?

• Thanks a lot for your specification! Anyway, for the trivial case n=1, it holds, so $m_1(a,b)=\kappa_1^a\kappa_1^b$, correct? Mar 22, 2021 at 11:21