[proofplan] We construct the example inside the free unital algebra on three noncommuting generators. First we define a reference unital linear functional that extracts the constant term, so the one-generator subalgebras are freely independent in the usual alternating-word sense. Then we modify the functional only on words involving all three generators, forcing the mixed moment $\varphi(x_1x_2x_3)$ to be nonzero while leaving every two-generator subalgebra unchanged. This preserves pairwise freeness but violates the defining centered alternating-word condition for joint freeness. [/proofplan] [step:Build the ambient algebra and define the modified functional] Let $A := \mathbb{C}\langle x_1,x_2,x_3\rangle$ be the free unital associative $\mathbb{C}$-algebra on generators $x_1,x_2,x_3$. Its standard vector-space basis consists of the empty word $1$ and all nonempty words in the alphabet $\{x_1,x_2,x_3\}$. Define the unital linear functional \begin{align*}\psi: A \to \mathbb{C}\end{align*} by setting $\psi(1)=1$ and $\psi(w)=0$ for every nonempty basis word $w$. Now define another unital linear functional \begin{align*}\varphi: A \to \mathbb{C}\end{align*} on the same basis as follows. Set $\varphi(1)=1$. If $w$ is a nonempty word involving at most two of the symbols $x_1,x_2,x_3$, set $\varphi(w):=\psi(w)=0$. Set \begin{align*} \varphi(x_1x_2x_3)=1. \end{align*} For every remaining basis word, namely every basis word involving all three symbols and different from $x_1x_2x_3$, set $\varphi(w):=0$. Extending by linearity gives a well-defined unital linear functional on $A$, so $(A,\varphi)$ is an algebraic noncommutative probability space. [guided] The free algebra $A=\mathbb{C}\langle x_1,x_2,x_3\rangle$ is useful because its elements have unique expansions as finite linear combinations of words in the generators. This means we can define a linear functional by prescribing its value on each basis word. We first define \begin{align*} \psi: A \to \mathbb{C} \end{align*} by $\psi(1)=1$ and $\psi(w)=0$ for every nonempty word $w$. Thus $\psi$ is the constant-term functional: it returns the coefficient of the empty word. The modified functional \begin{align*} \varphi: A \to \mathbb{C} \end{align*} is designed to agree with $\psi$ on every word that uses at most two generators, but to disagree on one genuinely three-generator word. More precisely, $\varphi(1)=1$, $\varphi(w)=0$ for every nonempty word using at most two of the symbols $x_1,x_2,x_3$, and \begin{align*} \varphi(x_1x_2x_3)=1. \end{align*} All other basis words involving all three generators are assigned value $0$. Because every element of $A$ is a finite linear combination of basis words, these assignments determine exactly one linear functional $\varphi:A\to\mathbb{C}$. Since $\varphi(1)=1$, this functional is unital. No positivity condition is required in the algebraic definition of a noncommutative probability space used here. [/guided] [/step] [step:Choose the three one-generator subalgebras] For each $i \in \{1,2,3\}$, define \begin{align*} A_i := \mathbb{C}\langle x_i\rangle \subset A. \end{align*} Thus $A_i$ is the unital subalgebra of $A$ generated by $x_i$. Equivalently, every element of $A_i$ has the form \begin{align*} a = c_0 1 + \sum_{m=1}^{N} c_m x_i^m \end{align*} for some $N \in \mathbb{N}$ and coefficients $c_0,c_1,\dots,c_N \in \mathbb{C}$. Since every nonempty word in $A_i$ involves only the symbol $x_i$, the definition of $\varphi$ gives \begin{align*} \varphi(a)=c_0. \end{align*} In particular, $a \in A_i$ is centered, meaning $\varphi(a)=0$, exactly when its constant coefficient is $0$. [/step] [step:Verify freeness for each pair of subalgebras] Fix distinct indices $i,j \in \{1,2,3\}$. We prove that $A_i$ and $A_j$ are free with respect to $\varphi$. Let $r \in \mathbb{N}$, let $\varepsilon_1,\dots,\varepsilon_r \in \{i,j\}$ satisfy $\varepsilon_k \ne \varepsilon_{k+1}$ for every $k \in \{1,\dots,r-1\}$, and let $a_k \in A_{\varepsilon_k}$ satisfy $\varphi(a_k)=0$ for every $k$. By the previous step, each $a_k$ has zero constant coefficient, so each $a_k$ is a finite linear combination of positive powers of $x_{\varepsilon_k}$. Therefore the product $a_1a_2\cdots a_r$ is a finite linear combination of nonempty words involving only the two symbols $x_i$ and $x_j$. By construction, $\varphi$ agrees with $\psi$ on every word involving at most two symbols, and $\psi$ vanishes on every nonempty word. Hence \begin{align*} \varphi(a_1a_2\cdots a_r)=0. \end{align*} This is exactly the alternating centered-word condition for freeness of $A_i$ and $A_j$. Since $i$ and $j$ were arbitrary distinct indices, the subalgebras $A_1,A_2,A_3$ are pairwise free. [/step] [step:Exhibit a centered alternating word whose mixed moment is nonzero] For each $i \in \{1,2,3\}$, the generator $x_i$ lies in $A_i$ and is centered because $x_i$ is a nonempty one-letter word, so \begin{align*} \varphi(x_i)=0. \end{align*} The word $x_1x_2x_3$ is an alternating product of centered elements from $A_1,A_2,A_3$, since the adjacent subalgebras are distinct. If the family $(A_1,A_2,A_3)$ were free, the definition of freeness would force \begin{align*} \varphi(x_1x_2x_3)=0. \end{align*} But the construction of $\varphi$ gives \begin{align*} \varphi(x_1x_2x_3)=1. \end{align*} This contradiction shows that $A_1,A_2,A_3$ are not jointly free. Therefore pairwise freeness does not imply joint freeness. [/step]