[proofplan] We prove the assertion by induction on the word length. If two adjacent letters lie in the same free component, their product is a single letter in that component, so the desired mixed moment reduces to a shorter word. Once the word is alternating, we split each letter into its scalar part and its centered part. Multilinearity expands the moment into finitely many shorter moments plus the fully centered alternating term, and the latter vanishes by freeness. [/proofplan] [step:Initialize the induction on word length] For each integer $n \geq 1$, let $P(n)$ denote the following assertion: for every index tuple $(i_1,\dots,i_n) \in I^n$ and every choice of elements $a_k \in A_{i_k}$, the scalar $\varphi(a_1\cdots a_n)$ is determined by the marginal functionals $\varphi|_{A_i}$ and the tuple $(i_1,\dots,i_n)$. For $n = 1$, if $a_1 \in A_{i_1}$, then \begin{align*} \varphi(a_1) = \varphi|_{A_{i_1}}(a_1). \end{align*} Thus $P(1)$ holds. Assume now that $n \geq 2$ and that $P(m)$ holds for every integer $m$ with $1 \leq m < n$. We prove $P(n)$. [/step] [step:Merge adjacent letters from the same subalgebra] Let $(i_1,\dots,i_n) \in I^n$ and let $a_k \in A_{i_k}$ for $1 \leq k \leq n$. Suppose first that there exists an index $r \in \{1,\dots,n-1\}$ such that $i_r = i_{r+1}$. Define $b_r \in A_{i_r}$ by \begin{align*} b_r := a_r a_{r+1}. \end{align*} Since $A_{i_r}$ is a subalgebra and $a_r,a_{r+1} \in A_{i_r}$, the product $b_r$ belongs to $A_{i_r}$. Define a shorter word of length $n-1$ by replacing the adjacent pair $a_r,a_{r+1}$ with $b_r$ and leaving all other letters unchanged. Its index tuple is \begin{align*} (i_1,\dots,i_{r-1},i_r,i_{r+2},\dots,i_n). \end{align*} The original product equals this shorter product inside $A$. Therefore $\varphi(a_1\cdots a_n)$ is the moment of a word of length $n-1$, and by the induction hypothesis it is determined by the marginal data and the shorter index tuple. The element $b_r$ is formed using multiplication inside the single algebra $A_{i_r}$, and all marginal moments of expressions involving $b_r$ are values of $\varphi|_{A_{i_r}}$ on products in $A_{i_r}$. Hence the original moment is determined by the allowed data. Consequently, it remains to consider the case \begin{align*} i_k \neq i_{k+1} \quad \text{for every } 1 \leq k \leq n-1. \end{align*} [/step] [step:Decompose an alternating word into scalar and centered parts] Assume now that the word is alternating, meaning $i_k \neq i_{k+1}$ for every $1 \leq k \leq n-1$. For each $k \in \{1,\dots,n\}$, define the scalar $\alpha_k \in \mathbb{C}$ and the centered element $\widetilde{a}_k \in A_{i_k}$ by \begin{align*} \alpha_k := \varphi(a_k) \end{align*} and \begin{align*} \widetilde{a}_k := a_k - \alpha_k 1_A. \end{align*} Because $A_{i_k}$ is unital and contains $1_A$, we have $\widetilde{a}_k \in A_{i_k}$. By linearity and unitality of $\varphi$, \begin{align*} \varphi(\widetilde{a}_k) = \varphi(a_k) - \alpha_k\varphi(1_A) = \alpha_k - \alpha_k = 0. \end{align*} For a subset $S \subseteq \{1,\dots,n\}$, define $c_k(S) \in A_{i_k}$ by \begin{align*} c_k(S) := \widetilde{a}_k \quad \text{if } k \in S, \end{align*} and \begin{align*} c_k(S) := \alpha_k 1_A \quad \text{if } k \notin S. \end{align*} Multilinearity of multiplication and of $\varphi$ gives the finite expansion \begin{align*} \varphi(a_1\cdots a_n) = \sum_{S \subseteq \{1,\dots,n\}} \varphi(c_1(S)c_2(S)\cdots c_n(S)). \end{align*} [guided] The goal is to separate each letter into the part visible to the marginal state and the part to which freeness applies. For every $k$, define \begin{align*} \alpha_k := \varphi(a_k) \end{align*} and \begin{align*} \widetilde{a}_k := a_k - \alpha_k 1_A. \end{align*} The scalar $\alpha_k$ is known from the marginal functional $\varphi|_{A_{i_k}}$, because $a_k \in A_{i_k}$. Also $\widetilde{a}_k$ still lies in $A_{i_k}$, since $A_{i_k}$ is a unital subalgebra containing $1_A$. The reason for this decomposition is that $\widetilde{a}_k$ is centered: \begin{align*} \varphi(\widetilde{a}_k) = \varphi(a_k) - \alpha_k\varphi(1_A) = \alpha_k - \alpha_k = 0. \end{align*} Now expand the product \begin{align*} a_1\cdots a_n = (\alpha_1 1_A + \widetilde{a}_1)\cdots(\alpha_n 1_A + \widetilde{a}_n). \end{align*} To write the expansion without ambiguity, for each subset $S \subseteq \{1,\dots,n\}$ define $c_k(S)$ to be $\widetilde{a}_k$ when $k \in S$ and $\alpha_k 1_A$ when $k \notin S$. Then each subset $S$ records exactly which positions contribute centered parts. Since multiplication in $A$ is bilinear and $\varphi$ is linear, we obtain \begin{align*} \varphi(a_1\cdots a_n) = \sum_{S \subseteq \{1,\dots,n\}} \varphi(c_1(S)c_2(S)\cdots c_n(S)). \end{align*} This expansion is useful because the term with $S = \{1,\dots,n\}$ is precisely the centered alternating word controlled by freeness, while every other term contains at least one scalar factor and therefore reduces to a shorter moment. [/guided] [/step] [step:Use freeness to eliminate the fully centered alternating term] Consider the subset $S_0 := \{1,\dots,n\}$. For this subset, \begin{align*} c_1(S_0)c_2(S_0)\cdots c_n(S_0) = \widetilde{a}_1\widetilde{a}_2 \cdots \widetilde{a}_n. \end{align*} The elements $\widetilde{a}_k \in A_{i_k}$ are centered, and the indices satisfy $i_k \neq i_{k+1}$ for every $1 \leq k \leq n-1$. By the explicit freeness hypothesis in the theorem statement, \begin{align*} \varphi(\widetilde{a}_1\widetilde{a}_2 \cdots \widetilde{a}_n) = 0. \end{align*} Thus the only term in the expansion with all positions centered contributes zero. [/step] [step:Reduce every remaining expansion term to a shorter moment] Let $S \subsetneq \{1,\dots,n\}$ be a proper subset. Define the ordered list $S = \{s_1,\dots,s_m\}$ with \begin{align*} 1 \leq s_1 < \cdots < s_m \leq n, \end{align*} where $m := |S|$. Define the scalar coefficient $\beta_S \in \mathbb{C}$ by \begin{align*} \beta_S := \prod_{k \notin S} \alpha_k. \end{align*} If $m = 0$, then all factors are scalar multiples of $1_A$, and unitality of $\varphi$ gives \begin{align*} \varphi(c_1(S)\cdots c_n(S)) = \beta_S. \end{align*} If $1 \leq m < n$, then scalar factors commute with all elements of $A$ because they are complex multiples of $1_A$, so \begin{align*} \varphi(c_1(S)\cdots c_n(S)) = \beta_S\,\varphi(\widetilde{a}_{s_1}\widetilde{a}_{s_2} \cdots \widetilde{a}_{s_m}). \end{align*} The word $\widetilde{a}_{s_1}\cdots \widetilde{a}_{s_m}$ has length $m<n$, with $\widetilde{a}_{s_j} \in A_{i_{s_j}}$ for each $j$. By the induction hypothesis, its moment is determined by the marginal functionals and the shorter index tuple $(i_{s_1},\dots,i_{s_m})$. The coefficient $\beta_S$ is determined by the marginal values $\alpha_k = \varphi|_{A_{i_k}}(a_k)$. Therefore each proper-subset term in the expansion is determined by the allowed data. [/step] [step:Assemble the recursion for the mixed moment] Combining the expansion with the vanishing of the fully centered term gives \begin{align*} \varphi(a_1\cdots a_n) = \sum_{S \subsetneq \{1,\dots,n\}} \varphi(c_1(S)c_2(S)\cdots c_n(S)). \end{align*} Each summand on the right-hand side is determined by marginal values and shorter mixed moments, and all shorter mixed moments are determined by the induction hypothesis. Hence $\varphi(a_1\cdots a_n)$ is determined by the marginal functionals $\varphi|_{A_i}$ and the index tuple $(i_1,\dots,i_n)$. This proves $P(n)$. By induction, $P(n)$ holds for every $n \geq 1$, which is the desired recursive determination of mixed moments under freeness. [/step]