[proofplan] The proof is a direct separation of the one-block term in the [moment-cumulant formula for free cumulants](/theorems/7107). The maximal partition $1_n$ contributes exactly $\kappa_n(x_1,\dots,x_n)$, while every other noncrossing partition has all blocks of size strictly smaller than $n$, so its partition cumulant involves only lower-order cumulants. Rearranging gives the displayed recursion, and induction then gives determination of cumulants from moments. The reverse determination is obtained by evaluating the moment-cumulant formula using the known cumulants. [/proofplan]

[step:Separate the one-block term in the moment-cumulant formula]Fix $n \in \mathbb{N}$ and $x_1,\dots,x_n \in A$. By the moment-cumulant formula for free cumulants (citing a result not yet in the wiki: Moment-cumulant formula for free cumulants), applied to the noncommutative probability space $(A,\varphi)$ and the tuple $(x_1,\dots,x_n) \in A^n$, we have \begin{align*} \varphi(x_1\cdots x_n)=\sum_{\pi\in NC(n)}\kappa_\pi[x_1,\dots,x_n]. \end{align*} The partition $1_n$ is the unique partition of $\{1,\dots,n\}$ with one block, namely $\{1,\dots,n\}$. Therefore its partition cumulant is \begin{align*} \kappa_{1_n}[x_1,\dots,x_n]=\kappa_n(x_1,\dots,x_n). \end{align*} Separating this term from the sum gives \begin{align*} \varphi(x_1\cdots x_n)=\kappa_n(x_1,\dots,x_n)+\sum_{\substack{\pi\in NC(n):\pi\ne 1_n}}\kappa_\pi[x_1,\dots,x_n]. \end{align*}[/step]

[guided]The starting point is the defining moment-cumulant identity for free cumulants. It says that the moment of the ordered product $x_1\cdots x_n$ is obtained by summing one partition cumulant over each noncrossing partition of $\{1,\dots,n\}$. Applied to the present tuple, the identity is \begin{align*} \varphi(x_1\cdots x_n)=\sum_{\pi\in NC(n)}\kappa_\pi[x_1,\dots,x_n]. \end{align*} Here the relevant cited input is the moment-cumulant formula for free cumulants (citing a result not yet in the wiki: Moment-cumulant formula for free cumulants). Its hypotheses are exactly the present ones: $(A,\varphi)$ is a noncommutative probability space and $x_1,\dots,x_n$ are elements of $A$. The useful observation is that the sum contains one term of top order. The maximal partition $1_n$ has a single block, namely $\{1,\dots,n\}$. By the definition of the partition cumulant associated to a one-block partition, this term is \begin{align*} \kappa_{1_n}[x_1,\dots,x_n]=\kappa_n(x_1,\dots,x_n). \end{align*} All other summands are indexed by partitions $\pi \in NC(n)$ with $\pi \ne 1_n$. Therefore the moment-cumulant formula can be rewritten by isolating the $1_n$ term: \begin{align*} \varphi(x_1\cdots x_n)=\kappa_n(x_1,\dots,x_n)+\sum_{\substack{\pi\in NC(n):\pi\ne 1_n}}\kappa_\pi[x_1,\dots,x_n]. \end{align*} This is the whole mechanism behind the recursion: the only unknown top-order cumulant appears once and with coefficient $1$.[/guided]

[step:Show that every remaining term uses only lower-order cumulants] Let $\pi \in NC(n)$ satisfy $\pi \ne 1_n$. Since $1_n$ is the unique one-block partition, $\pi$ has at least two blocks. If $V \in \pi$, then $V$ is a proper subset of $\{1,\dots,n\}$, so \begin{align*} 1 \le |V| \le n-1. \end{align*} Writing each block as $V=\{i_1<\cdots<i_{|V|}\}$, the corresponding factor in $\kappa_\pi[x_1,\dots,x_n]$ is \begin{align*} \kappa_{|V|}(x_{i_1},\dots,x_{i_{|V|}}). \end{align*} Thus every factor appearing in $\kappa_\pi[x_1,\dots,x_n]$ is a cumulant of order strictly smaller than $n$. [/step]

[step:Rearrange the identity to obtain the recursion] From the separated moment-cumulant identity, \begin{align*} \varphi(x_1\cdots x_n)=\kappa_n(x_1,\dots,x_n)+\sum_{\substack{\pi\in NC(n):\pi\ne 1_n}}\kappa_\pi[x_1,\dots,x_n]. \end{align*} Subtracting the finite sum over $\pi \ne 1_n$ from both sides in $\mathbb{C}$ gives \begin{align*} \kappa_n(x_1,\dots,x_n)=\varphi(x_1\cdots x_n)-\sum_{\substack{\pi\in NC(n):\pi\ne 1_n}}\kappa_\pi[x_1,\dots,x_n]. \end{align*} This is the asserted recursive formula. [/step]

[step:Deduce that moments determine cumulants by induction] For $m \in \mathbb{N}$, define the moment functional \begin{align*} M_m: A^m &\to \mathbb{C} \end{align*} \begin{align*} (y_1,\dots,y_m) &\mapsto \varphi(y_1\cdots y_m). \end{align*} We prove by induction on $m$ that $M_1,\dots,M_m$ determine $\kappa_1,\dots,\kappa_m$. For $m=1$, the set $NC(1)$ contains only $1_1$, so the moment-cumulant formula gives \begin{align*} M_1(y_1)=\kappa_1(y_1) \end{align*} for every $y_1 \in A$. Hence $M_1$ determines $\kappa_1$. Assume that $1 < m \le n$ and that $M_1,\dots,M_{m-1}$ determine $\kappa_1,\dots,\kappa_{m-1}$. Applying the recursion with $m$ in place of $n$, for every $y_1,\dots,y_m \in A$ we have \begin{align*} \kappa_m(y_1,\dots,y_m)=M_m(y_1,\dots,y_m)-\sum_{\substack{\pi\in NC(m):\pi\ne 1_m}}\kappa_\pi[y_1,\dots,y_m]. \end{align*} By the previous step, each summand with $\pi \ne 1_m$ uses only cumulants of orders at most $m-1$. These are already determined by the induction hypothesis, and $M_m$ is known by assumption. Therefore $\kappa_m$ is determined. Induction gives that $M_1,\dots,M_n$ determine $\kappa_1,\dots,\kappa_n$. [/step]

[step:Deduce that cumulants determine moments from the moment-cumulant formula] Assume that the cumulant functionals $\kappa_1,\dots,\kappa_n$ are known. For each $m$ with $1 \le m \le n$ and each $y_1,\dots,y_m \in A$, the moment-cumulant formula gives \begin{align*} M_m(y_1,\dots,y_m)=\varphi(y_1\cdots y_m)=\sum_{\pi\in NC(m)}\kappa_\pi[y_1,\dots,y_m]. \end{align*} Every block of every $\pi \in NC(m)$ has cardinality at most $m$, so every factor in every partition cumulant on the right-hand side is one of the known cumulants $\kappa_1,\dots,\kappa_m$. Since $m \le n$, all these cumulants are included among $\kappa_1,\dots,\kappa_n$. Thus each $M_m$ with $1 \le m \le n$ is determined, completing the proof. [/step]