[proofplan]
We construct the ambient algebra as the unital algebraic free product of the algebras $A_i$. Each algebra $A_i$ splits as the direct sum of scalars and the centered subspace $\ker \tau_i$, and the free product has a vector-space normal form consisting of scalars together with finite reduced words in these centered subspaces. We define $\tau$ to be $1$ on the unit and $0$ on every nonempty reduced word. This immediately preserves the original states on each copy and forces alternating products of centered elements from distinct copies to have expectation zero.
[/proofplan]
[step:Separate each algebra into scalar and centered parts]
For each $i \in I$, define the centered subspace
\begin{align*}
A_i^\circ := \ker \tau_i = \{a \in A_i : \tau_i(a)=0\}.
\end{align*}
Since $\tau_i(1_{A_i})=1$, every $a \in A_i$ has the unique decomposition
\begin{align*}
a = \tau_i(a)1_{A_i} + a^\circ,
\end{align*}
where
\begin{align*}
a^\circ := a - \tau_i(a)1_{A_i} \in A_i^\circ.
\end{align*}
Indeed,
\begin{align*}
\tau_i(a^\circ) = \tau_i(a) - \tau_i(a)\tau_i(1_{A_i}) = 0.
\end{align*}
Uniqueness follows because if $\lambda 1_{A_i} \in A_i^\circ$, then applying $\tau_i$ gives $\lambda=0$.
[/step]
[step:Build the free product algebra with its reduced-word decomposition]
If $I=\varnothing$, take $A=\mathbb{C}$ and $\tau: \mathbb{C}\to\mathbb{C}$ to be the identity map; the assertion is vacuous. Assume now that $I$ is nonempty.
Let
\begin{align*}
A := *_{i \in I} A_i
\end{align*}
be the unital algebraic free product of the family $(A_i)_{i \in I}$, amalgamated over the common unit, and let
\begin{align*}
\iota_i: A_i \to A
\end{align*}
be the canonical unital algebra homomorphism for each $i \in I$.
By the standard reduced-word normal form for the unital algebraic free product, applied to the vector-space decompositions $A_i=\mathbb{C}1_{A_i}\oplus A_i^\circ$, the underlying complex [vector space](/page/Vector%20Space) of $A$ is
\begin{align*}
A = \mathbb{C}1_A \oplus \bigoplus_{n=1}^{\infty} \bigoplus_{\substack{i_1, \dots, i_n \in I, i_k \neq i_{k+1}\text{ for }1 \leq k < n}} A_{i_1}^\circ \otimes \cdots \otimes A_{i_n}^\circ.
\end{align*}
Here the inner direct sum is indexed by finite words $(i_1,\dots,i_n)$ with adjacent indices distinct; this is the standard algebraic free product normal form used to define the free product vector space and multiplication.
The summands with $n \geq 1$ are represented in $A$ by finite reduced words
\begin{align*}
\iota_{i_1}(a_1)\cdots \iota_{i_n}(a_n),
\end{align*}
where $a_k \in A_{i_k}^\circ$ and adjacent indices are distinct. Every reduced word is finite, even when the index set $I$ is infinite.
[/step]
[step:Define the free product state by killing nonempty reduced words]
Define the linear functional
\begin{align*}
\tau: A \to \mathbb{C}
\end{align*}
as follows. On the scalar summand set
\begin{align*}
\tau(\lambda 1_A) := \lambda
\end{align*}
for every $\lambda \in \mathbb{C}$. On every nonempty reduced-word summand
\begin{align*}
A_{i_1}^\circ \otimes \cdots \otimes A_{i_n}^\circ
\end{align*}
with $n \geq 1$ and $i_k \neq i_{k+1}$, set $\tau$ equal to $0$. The direct-sum decomposition of $A$ makes this prescription well-defined and linear. Moreover $\tau(1_A)=1$, so $(A,\tau)$ is an algebraic noncommutative probability space.
[guided]
The main point is that the reduced-word decomposition gives a basis-free way to define a linear functional on all of $A$. We are not defining $\tau$ by choosing a presentation of an element and hoping the value is independent of the presentation; instead, the vector-space decomposition says that every element $x \in A$ has a unique form
\begin{align*}
x = \lambda 1_A + r,
\end{align*}
where $\lambda \in \mathbb{C}$ and $r$ belongs to the direct sum of all nonempty reduced-word summands. Therefore the rule
\begin{align*}
\tau(x) := \lambda
\end{align*}
is unambiguous.
Equivalently, $\tau$ keeps only the scalar coefficient of $x$ in the reduced-word normal form. This gives a [linear map](/page/Linear%20Map) $\tau: A \to \mathbb{C}$ because extracting the scalar coefficient from a direct sum decomposition is a linear projection. Since the unit $1_A$ has scalar coefficient $1$, we get
\begin{align*}
\tau(1_A)=1.
\end{align*}
Thus $(A,\tau)$ is a noncommutative probability space in the algebraic sense.
[/guided]
[/step]
[step:Verify that each original state is preserved]
Fix $i \in I$ and $a \in A_i$. Write
\begin{align*}
a = \tau_i(a)1_{A_i} + a^\circ
\end{align*}
with $a^\circ \in A_i^\circ$. Since $\iota_i$ is unital and linear,
\begin{align*}
\iota_i(a)=\tau_i(a)1_A+\iota_i(a^\circ).
\end{align*}
The element $\iota_i(a^\circ)$ is a reduced word of length $1$, so the definition of $\tau$ gives $\tau(\iota_i(a^\circ))=0$. Hence
\begin{align*}
\tau(\iota_i(a))=\tau_i(a)\tau(1_A)+\tau(\iota_i(a^\circ))=\tau_i(a).
\end{align*}
Thus $\tau \circ \iota_i=\tau_i$ for every $i \in I$.
[/step]
[step:Check the centered alternating products vanish]
Let $n \in \mathbb{N}$, let $i_1,\dots,i_n \in I$ satisfy $i_k \neq i_{k+1}$ for $1 \leq k < n$, and let $a_k \in A_{i_k}$ satisfy $\tau_{i_k}(a_k)=0$. Then $a_k \in A_{i_k}^\circ$ for every $k$. Therefore
\begin{align*}
\iota_{i_1}(a_1)\cdots \iota_{i_n}(a_n)
\end{align*}
is a nonempty reduced word in the free product normal form, because adjacent indices are distinct and each letter is centered. By the definition of $\tau$ on nonempty reduced words,
\begin{align*}
\tau(\iota_{i_1}(a_1)\cdots \iota_{i_n}(a_n))=0.
\end{align*}
This is exactly the centered alternating-product criterion for free independence of the subalgebras $\iota_i(A_i)$. Hence the family $(\iota_i(A_i))_{i \in I}$ is freely independent in $(A,\tau)$, and the construction satisfies both required properties.
[/step]
