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 \begin{equation} 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 \end{equation}

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 in advance.


1 Answer 1


The formula as stated in your reference seems to be wrong. The Kreweras complement of the partitions should also show up. You can find the correct version in my original paper https://link.springer.com/content/pdf/10.1023/A:1008643104945.pdf with Andu Nica on this, or also in my book with Andu, which you can find at https://rolandspeicher.com/literature/nica-speicher/ In any case, the formula is equivalent to the description in terms of the S-transform, but in our paper we give an alternative combinatorial proof, and then use this to derive the formula for the S-transform.

  • $\begingroup$ 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? $\endgroup$ Mar 22, 2021 at 11:21
  • 1
    $\begingroup$ yes, for n=1 it is okay, but then you don't need to invoke cumulants, you just have the factorization of the first moment of ab into the product of the first moment of a and the first moment of b - this follows directly by the definition of free independence $\endgroup$ Mar 22, 2021 at 11:50

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