SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Proposition 2. Assume  $\| v_{n}\|_{\infty}\leq r$  for every  $v_{n}\in \mathcal{G}$ . The estimator based on Eq.(4) requires  $O(\frac{n}{\epsilon^2}\log \frac{n}{\delta})$  utility evaluations to achieve  $P(\|\hat{\phi}(v_n;\delta_w) - \phi(v_n;\delta_w)\|_2\geq \epsilon)\leq \delta$  for every utility function  $v_{n}\in \mathcal{G}$ .
Proof. Note that it only requires  $n + 1$  utility evaluations for exactly yielding 0 and 1-weighted Banzhaf values. Therefore, the MSR estimator is designed for  $0 < w < 1$ . Our proof is adapted from (Wang and Jia 2023, Theorem 4.9).
Let  $v_{n} \in \mathcal{G}$  and  $i \in [n]$  be fixed, and by assumption  $\| v_{n}\|_{\infty} \leq r$ . According to the underlying sampling for  $\{S_j\}$ , there is  $|S_{\ni i}| \sim \mathrm{binomial}(w)$ . Let a bridge estimator defined by
$$
\overline {{\phi}} _ {i} (v _ {n}; \delta_ {w}) = \frac {1}{w m} \sum_ {S \in \mathcal {S} _ {\ni i}} v _ {n} (S) - \frac {1}{(1 - w) m} \sum_ {S \in \mathcal {S} _ {\neq i}} v _ {n} (S).
$$
For simplicity, we write  $\overline{\phi}_i$ ,  $\hat{\phi}_i$  and  $\phi_i$  instead of  $\overline{\phi}_i(v_n; \delta_w)$ ,  $\hat{\phi}_i(v_n; \delta_w)$  and  $\phi_i(v_n; \delta_w)$ . Plus, we also have  $s_{\ni i} = |\mathcal{S}_{\ni i}|$  and  $s_{\not\ni i} = |\mathcal{S}_{\not\ni i}|$ .
First of all, we prove that  $\mathbb{E}[\overline{\phi}_i] = \phi_i$ . Suffice it to prove that  $\mathbb{E}[c_i(S)v_n(S)] = \phi_i$  where  $c_{i}(S) = \frac{1}{w}$  if  $i\in S$  and  $-\frac{1}{(1 - w)}$  otherwise. For every  $S\in [n]\backslash i$
$$
\begin{array}{l} \frac {P (S \cup i)}{w} = \frac {w ^ {s + 1} (1 - w) ^ {n - 1 - s}}{w} = w ^ {s} (1 - w) ^ {n - 1 - s}, \\ - \frac {P (S)}{1 - w} = - \frac {w ^ {s} (1 - w) ^ {n - s}}{1 - w} = - w ^ {s} (1 - w) ^ {n - 1 - s}. \\ \end{array}
$$
On the other hand,  $\phi_{i} = \sum_{S\subseteq [n]\setminus i}w^{s}(1 - w)^{n - 1 - s}(v_{n}(S\cup i) - v_{n}(S))$
Next, we prove the convergence for the MSR estimator. Since  $\mathbb{E}[\bar{\phi}_i] = \phi_i$ , by the Hoeffding's inequality,
$$
P (| \overline {{\phi}} _ {i} - \phi_ {i} | \geq \epsilon) \leq 2 \exp \left(- \frac {m \epsilon^ {2}}{2 r ^ {2} T ^ {2}}\right)
$$
where  $T = \max \left(\frac{1}{w},\frac{1}{1 - w}\right)$ . Assume  $s_{\ni i} > 0$  and  $s_{\neq i} > 0$ , note that  $s_{\ni i} + s_{\neq i} = m$ , there is
$$
\begin{array}{l} | \hat {\phi} _ {i} - \bar {\phi} _ {i} | = | \left(\frac {1}{s _ {\ni i}} - \frac {1}{w m}\right) \sum_ {S \in S _ {\ni i}} v _ {n} (S) + \left(\frac {1}{s _ {\not \ni i}} - \frac {1}{(1 - w) m}\right) \sum_ {S \in S _ {\not \ni i}} v _ {n} (S) \\ \leq \frac {r}{w m} | s _ {\ni i} - w m | + \frac {r}{(1 - w) m} | s _ {\not \ni i} - (1 - w) m | \\ = \frac {r}{w m} | s _ {\ni i} - w m | + \frac {r}{(1 - w) m} | s _ {\ni i} - w m | = \frac {r}{w (1 - w) m} | s _ {\ni i} - w m |. \\ \end{array}
$$
Note that the first inequality in the above still holds if  $s_{\supseteq i} = 0$  or  $s_{\neq i} = 0$ . Since  $s_{\supseteq i} \sim \mathrm{binomial}(w)$ , by the Hoeffding's inequality, there is
$$
P \left(\left| s _ {\ni i} - w m \right| \geq \Delta\right) \leq 2 \exp \left(- \frac {2 \Delta^ {2}}{m}\right).
$$
Therefore,  $|\hat{\phi}_i - \overline{\phi}_i| < \frac{r\Delta}{w(1 - w)m}$  provided that  $|s_{\ni i} - WM| < \Delta$ . Then,
$$
\begin{array}{l} P (| \hat {\phi} _ {i} - \phi_ {i} | \geq \epsilon) = P (| \hat {\phi} _ {i} - \phi_ {i} | \geq \epsilon \cap | s _ {\ni i} - w m | <   \Delta) + P (| \hat {\phi} _ {i} - \phi_ {i} | \geq \epsilon \cap | s _ {\ni i} - w m | \geq \Delta) \\ \leq P \left(\left| \hat {\phi} _ {i} - \phi_ {i} \right| \geq \epsilon \mid | s _ {\ni i} - w m | <   \Delta\right) + 2 \exp \left(- \frac {2 \Delta^ {2}}{m}\right) \\ \leq P \left(\left| \bar {\phi} _ {i} - \phi_ {i} \right| \geq \epsilon - \frac {r \Delta}{w (1 - w) m} \mid \left| s _ {\ni i} - w m \right| <   \Delta\right) + 2 \exp \left(- \frac {2 \Delta^ {2}}{m}\right) \\ \leq \frac {P (| \overline {{\phi}} _ {i} - \phi_ {i} | \geq \epsilon - \frac {r \Delta}{w (1 - w) m})}{1 - 2 \exp (- \frac {2 \Delta^ {2}}{m})} + 2 \exp \left(- \frac {2 \Delta^ {2}}{m}\right) \\ \leq \frac {2 \exp \left(- \frac {m \left(\epsilon - \frac {r \Delta}{w (1 - w) m}\right) ^ {2}}{2 r ^ {2} T ^ {2}}\right)}{1 - 2 \exp \left(- \frac {2 \Delta^ {2}}{m}\right)} + 2 \exp \left(- \frac {2 \Delta^ {2}}{m}\right) \\ \leq 3 \exp \left(- \frac {m \left(\epsilon - \frac {r \Delta}{w (1 - w) m}\right) ^ {2}}{2 r ^ {2} T ^ {2}}\right) + 2 \exp \left(- \frac {2 \Delta^ {2}}{m}\right) \\ \end{array}
$$
where the last inequality is due to that  $1 - 2\exp \left(-\frac{2\Delta^2}{m}\right)\geq \frac{2}{3}$  if  $m$  is sufficiently large. Solving the equation  $\frac{m\left(\epsilon - \frac{r\Delta}{w(1 - w)m}\right)^2}{2r^2T^2} = \frac{2\Delta^2}{m}$  yields  $\Delta = \frac{w(1 - w)m\epsilon}{r(2Tw - 2Tw^2 + 1)}$ . Specifically,  $2Tw - 2Tw^{2} + 1 = 2|w - 0.5| + 2$ , and thus its range is [2, 3) provided that  $w\in (0,1)$ , which indicates  $\epsilon -\frac{r\Delta}{w(1 - w)m}\geq \frac{\epsilon}{2}$ . So, we have  $\frac{2\Delta^2}{m} = Cm\epsilon^2$  where  $C = \frac{2w^2(1 - w)^2}{r^2(2|w - 0.5| + 2)^2}$ . Eventually,
$$
P \left(\left| \hat {\phi} _ {i} - \phi_ {i} \right| \geq \epsilon\right) \leq 5 \exp \left(- C m \epsilon^ {2}\right)
$$
provided that  $1 - 2\exp \left(-\frac{2\Delta^2}{m}\right)\geq \frac{2}{3}$ , or equivalently  $m\geq \frac{\log 6}{C\epsilon^2}$ . Then,
$$
P (\| \hat {\phi} - \phi \| _ {2} \geq \epsilon) \leq P (\bigcup_ {i \in [ n ]} | \hat {\phi} _ {i} - \phi_ {i} | \geq \frac {\epsilon}{\sqrt {n}}) \leq 5 n \exp \left(- \frac {C m \epsilon^ {2}}{n}\right).
$$
Solving  $5n\exp \left(-\frac{Cm\epsilon^2}{n}\right)\leq \delta$  leads to  $m\geq \frac{n}{C\epsilon^2}\log \frac{5n}{\delta}$ . Eventually, we conclude that the MSR estimator requires
$$
\max  \left(\frac {\log 6}{C \epsilon^ {2}}, \frac {n}{C \epsilon^ {2}} \log \frac {5 n}{\delta}\right) = O \left(\frac {n}{\epsilon^ {2}} \log \frac {n}{\delta}\right)
$$
samples (equivalently, utility evaluations) to achieve  $P(\| \hat{\phi} -\phi \| _2\geq \epsilon)\leq \delta$
Proof. By Assumption 1,  $v_{n} - \mathbb{E}[v_{n}] = \pmb{\epsilon}_{n} \otimes \pmb{\epsilon}_{n-1} \otimes \dots \otimes \pmb{\epsilon}_{1}$  such that  $(\pmb{\epsilon}_{i})_{1 \leq i \leq n}$  are independent continuous random vectors in  $\mathbb{R}^2$  that have the same covariance matrix  $\pmb{\Sigma}$ . For simplicity, we write  $\pmb{\epsilon}^{[n]} = \pmb{\epsilon}_{n} \otimes \pmb{\epsilon}_{n-1} \otimes \dots \otimes \pmb{\epsilon}_{1}$ .
Since  $(\mathbf{A}\otimes \mathbf{B})(\mathbf{C}\otimes \mathbf{D}) = (\mathbf{AC})\otimes (\mathbf{BD})$  , there is
$$
\boldsymbol {\epsilon} ^ {[ n ]} \big (\boldsymbol {\epsilon} ^ {[ n ]} \big) ^ {\top} = \big (\boldsymbol {\epsilon} _ {n} \boldsymbol {\epsilon} _ {n} ^ {\top} \big) \otimes \big (\boldsymbol {\epsilon} _ {n - 1} \boldsymbol {\epsilon} _ {n - 1} ^ {\top} \big) \otimes \dots \otimes \big (\boldsymbol {\epsilon} _ {1} \boldsymbol {\epsilon} _ {1} ^ {\top} \big).
$$
By the independence, we have
$$
\operatorname {C o v} \left(v _ {n}\right) = \operatorname {C o v} \left(\boldsymbol {\epsilon} ^ {[ n ]}\right) = \overbrace {\boldsymbol {\Sigma} \otimes \boldsymbol {\Sigma} \otimes \cdots \otimes \boldsymbol {\Sigma}} ^ {n \text {t i m e s}}. \tag {11}
$$
Next, we prove the rest by induction. It is clear that it holds for  $n = 1$ . Suppose it holds for  $n = k > 0$ , i.e.,
$$
v _ {k} (S) = \mathbb {E} \left[ v _ {k} (S) \right] + \prod_ {i \in S} \epsilon_ {i, 2} \prod_ {i \notin S} \epsilon_ {i, 1} \text {f o r e v e r y} S \subseteq [ k ],
$$
$$
\operatorname {C o v} \left(v _ {k} (S), v _ {k} (T)\right) = \sigma_ {1 1} ^ {| [ k ] \backslash (S \cup T) |} \cdot \sigma_ {1 2} ^ {| (S \backslash T) \cup (T \backslash S) |} \cdot \sigma_ {2 2} ^ {| S \cap T |} \text {f o r e v e r y} S, T \subseteq [ k ] \tag {12}
$$
and we proceed for  $n = k + 1$ . Suppose the binary ordering for  $\{S\}_{S \subseteq [k]}$  is
$$
S _ {1}, S _ {2}, \dots , S _ {2 ^ {k}},
$$
then, the binary ordering for  $\{S\}_{S\subseteq [k + 1]}$  is
$$
S _ {1}, S _ {2}, \dots , S _ {2 ^ {k}}, S _ {1} \cup \{k + 1 \}, S _ {2} \cup \{k + 1 \}, \dots , S _ {2 ^ {k}} \cup \{k + 1 \}, \tag {13}
$$
which aligns well with  $\pmb{\epsilon}_{k + 1}\otimes (\pmb{\epsilon}_k\otimes \pmb{\epsilon}_{k - 1}\otimes \dots \otimes \pmb{\epsilon}_1)$ . Thus, we have  $v_{k + 1}(S) = \mathbb{E}[v_{k + 1}(S)] + \prod_{i\in S}\epsilon_{i,2}\prod_{i\notin S}\epsilon_{i,1}$  for every  $S\subseteq [k + 1]$ . On the other hand, write  $\mathbf{C}^{[k]} = \mathrm{Cov}(v_k)$ , by Eq. (11), we have
$$
\mathbf {C} ^ {[ k + 1 ]} = \left( \begin{array}{l l} \sigma_ {1 1} \mathbf {C} ^ {[ k ]} & \sigma_ {1 2} \mathbf {C} ^ {[ k ]} \\ \sigma_ {1 2} \mathbf {C} ^ {[ k ]} & \sigma_ {2 2} \mathbf {C} ^ {[ k ]}. \end{array} \right) \tag {14}
$$
Let  $S, T \subseteq [k + 1]$  be given, there are four cases in total: i)  $S, T \subseteq [k]$ ; ii)  $S, T \nsubseteq [k]$ ; iii)  $S \subseteq [k]$ ,  $T \nsubseteq [k]$ ; and iv)  $S \nsubseteq [k]$ ,  $T \subseteq [k]$ . For the first case, combing Eqs. (13), (14) and (12), there is
$$
\mathbf {C} ^ {[ k + 1 ]} (S, T) = \sigma_ {1 1} \mathbf {C} ^ {[ k ]} (S, T) = \sigma_ {1 1} ^ {| [ k + 1 ] \backslash (S \cup T) |} \cdot \sigma_ {1 2} ^ {| (S \backslash T) \cup (T \backslash S) |} \cdot \sigma_ {2 2} ^ {| S \cap T |}.
$$
The rest of the cases can be done similarly.
C Prerequisites and Proof for Theorem 1
To prove Theorem 1, our first step is to lift each  $\phi(v_n; P)$  into  $\mathbb{R}^{2^n}$  w.r.t. the binary ordering. Precisely, we extend each semi-value  $\phi_P$ , a shorthand for  $\phi(\cdot; P)$ , to be a semi-index  $\mathcal{I}_P$  that maps each  $v \in \mathcal{G}_n$  (for every  $n \geq 1$ ) into another utility function in  $\mathcal{G}_n$ , whose matrix representations on all  $\mathcal{G}_n$  are inductive. Then, we prove a theorem of robustness for semi-indices, which is then used to prove Theorem 1.
C.1 Extend Semi-values to Be Semi-indices
Index is another concept developed in cooperative game theory that serves as an extension for value (Fujimoto et al. 2006; Grabisch and Roubens 1999), which has also been employed in the community of machine learning (Sundararajan et al. 2020; Zhang et al. 2021). Specifically, an index  $\mathcal{I}$  is a function that maps each utility function  $v\in \mathcal{G}_n$  (for every  $n\geq 1$ ) into another utility function in  $\mathcal{G}_n$ . Note that each utility function  $v\in \mathcal{G}_n$  can be viewed as a vector indexed by all subsets of  $[n]$ . In other words,  $\mathcal{I}$  is to impute contributions to all possible subsets of  $[n]$ , not just to each datum in  $[n]$ , so as to account for the contributions of interactions among all data  $[n]$ . For convenience, we write  $\mathcal{I}(v_n,S)$  for the value of  $\mathcal{I}(v_n)$  indexed at  $S$ .
Notably, the axiom of recursion was proposed by Grabisch and Roubens (1999) to extend the Banzhaf and Shapley values into the Banzhaf and Shapley indices, respectively. To state the axiom of recursion, we need further definitions. Suppose  $v \in \mathcal{G}_n$  and  $S \subseteq [n]$  are given. The restriction of  $v$  on  $S$ , denoted by  $v_S$ , is defined by
$$
v _ {S} (T) = v (T), \forall T \subseteq S. \tag {15}
$$
In other words, the domain  $2^{[n]}$  of  $v$  is restricted to be  $2^S$ .
Plus, let  $j \in [n] \backslash S$  be given, the restriction of  $v$  on  $S$  in the presence of  $j$ , denoted by  $v_{S}^{\cup j}$ , is defined by
$$
v _ {S} ^ {\cup j} (T) = v (T \cup j), \forall T \subseteq S. \tag {16}
$$
Remark 3. The definition of  $v_{S}^{\cup j}$  is not the same as the one defined by Grabisch and Roubens (1999), which is instead  $v_{S}^{\cup j}(T) = v(T \cup j) - v(j), \forall T \subseteq S$ . The term  $-v(j)$  is only to make  $v_{S}^{\cup j}(\emptyset) = 0$  and we comment that it is not necessary for all the results therein. Moreover, this extra term will make Eq. (17) false.
1. Recursion: For every  $v \in \mathcal{G}_n$ ,  $S \subseteq [n]$  with  $s > 1$ , and  $j \in S$ , there is
$$
\mathcal {I} (v, S) = \mathcal {I} (v _ {[ n ] \backslash j} ^ {\cup j}, S \backslash j) - \mathcal {I} (v _ {[ n ] \backslash j}, S \backslash j).
$$
The axiom of recursion dictates that the contribution imputed to the subset  $S$  should be the difference of the contributions imputed to  $S \setminus j$  in the presence and absence of any player  $j \in S$ . So, starting from  $\mathcal{I}_P(v_n, i) = \phi_i(v_n; P)$  for every  $v_n \in \mathcal{G}$  and  $i \in [n]$ , the axiom of recursion can be used to define  $\mathcal{I}_P(v_n, S)$  recursively for every non-empty subsets  $S \subseteq [n]$ . Particularly, the defined value of  $\mathcal{I}(v_n, S)$  is independent of the choice of  $j \in S$ .
One thing that remains is to determine the contribution for the empty subset. For each semi-value  $\phi_P$ , we argue that it should be what is stated as the axiom of empty.
2. Empty: To extend a semi-value  $\phi_P$  to be a semi-index  $\mathcal{I}_P$ , there should be  $\mathcal{I}_P(v_n, \emptyset) = \sum_{S \subseteq [n]} p_{s+1}^{n+1} \cdot v(S)$  for every  $v_n \in \mathcal{G}$ . Recall that  $p_{s+1}^{n+1} = \int_0^1 t^s (1 - t)^{n-s} \mathrm{d}P(t)$  for every  $0 \leq s \leq n$ .
Observe that  $\sum_{S\subseteq [n]}p_{s + 1}^{n + 1} = \sum_{s = 0}^{n}\binom {n}{s}p_{s + 1}^{n + 1} = 1$ , the imposed value for  $\mathcal{I}_P(v_n,\emptyset)$  is just the expected value of  $v_{n}$  w.r.t. the probability distribution  $(p_{s + 1}^{n + 1})_{S\subseteq [n]}$  induced by the underlying  $P\in \mathcal{P}$ . One can verify the axiom of empty consistently extends the axiom of recursion, i.e.,
$$
\mathcal {I} _ {P} (v, i) = \mathcal {I} _ {P} \left(v _ {[ n ] \backslash i} ^ {\cup i}, \emptyset\right) - \mathcal {I} _ {P} \left(v _ {[ n ] \backslash i}, \emptyset\right) \text {f o r e v e r y} v \in \mathcal {G} _ {n} \text {a n d} i \in [ n ]. \tag {17}
$$
Definition 1 (Semi-indices). A semi-index  $\mathcal{I}_P$  is extended from a semi-value  $\phi_P$  using the axioms of recursion and empty.
Remark 4. We can therefore extend each  $w$ -weighted Banzhaf value to be  $w$ -weighted Banzhaf index. Particularly, 0-weighted Banzhaf index is exactly the Möbius transform Eq. (18), while 1-weighted Banzhaf index is instead the co-Möbius transform (Grabisch et al. 2000, Eq. (12)). In addition, extending the Shapley value and the Banzhaf value in this way will yield the Shapley interaction index and the Banzhaf interaction index presented in (Grabisch and Roubens 1999, Theorem 3).
Proposition 5 (Formulas for Semi-indices). For each semi-index  $\mathcal{I}_P$ , there is
$$
\mathcal {I} _ {P} (v _ {n}, S) = \sum_ {R \subseteq [ n ] \backslash S} p _ {r + 1} ^ {n + 1 - s} \cdot \Delta_ {S} v _ {n} (R) f o r e v e r y v _ {n} \in \mathcal {G} a n d S \subseteq [ n ]
$$
$$
w h e r e p _ {j} ^ {m} = \int_ {0} ^ {1} t ^ {j - 1} (1 - t) ^ {m - j} \mathrm {d} P (t) f o r 1 \leq j \leq m
$$
$$
a n d \Delta_ {S} v _ {n} (R) = \sum_ {T \subseteq S} (- 1) ^ {s - t} v _ {n} (R \cup T).
$$
Proof. For simplicity, we write  $v$  instead of  $v_{n}$ . It is clear that it holds when  $s = 0$  and  $s = 1$ . Assume it holds for all  $S$  such that  $|S| = k$ . Let  $S \subseteq [n]$  that satisfies  $|S| = k + 1$ , and some  $i \in S$  be given, there is
$$
\begin{array}{l} \mathcal {I} _ {P} (v _ {n}, S) \\ = \mathcal {I} _ {P} \left(v _ {[ n ] \backslash i} ^ {\cup i}, S \backslash i\right) - \mathcal {I} _ {P} \left(v _ {[ n ] \backslash i}, S \backslash i\right) \quad \text {t h e a x i o m o f r e c u r s i o n} \\ = \sum_ {R \subseteq [ n ] \backslash S} p _ {r + 1} ^ {n - k} \cdot \Delta_ {S \backslash i} v _ {[ n ] \backslash i} ^ {\cup i} (R) - \sum_ {R \subseteq [ n ] \backslash S} p _ {r + 1} ^ {n - k} \cdot \Delta_ {S \backslash i} v _ {[ n ] \backslash i} (R) \quad \text {b y} \\ = \sum_ {R \subseteq [ n ] \backslash S} p _ {r + 1} ^ {n - k} \cdot \left(\Delta_ {S \backslash i} v (R \cup i) - \Delta_ {S \backslash i} v (R)\right) \\ = \sum_ {R \subseteq [ n ] \backslash S} p _ {r + 1} ^ {n + 1 - s} \cdot \Delta_ {S} v (R). \\ \end{array}
$$
Note the above reasoning is independent of the choice of  $i \in S$ .
The reader can also verify that each semi-index  $\mathcal{I}_P$  satisfies the following axioms, the first three of which are often laid out for how indices should be in the fields of cooperative game theory and machine learning (Fujimoto et al. 2006; Grabisch and Roubens 1999; Sundararajan et al. 2020).
3. Linearity:  $\mathcal{I}_P(a\cdot u_n + v_n,S) = a\cdot \mathcal{I}_P(u_n,S) + \mathcal{I}_P(v_n,S)$  for every  $v_{n},u_{n}\in \mathcal{G},a\in \mathbb{R}$  and  $S\subseteq [n]$
4. Dummy Interaction: If  $S \subseteq [n]$  with  $s > 1$  contains a datum  $i$  such that for some constant  $C$ ,  $v_{n}(T \cup i) = v_{n}(T) + C$  for all  $T \subseteq [n] \backslash i$ , then  $\mathcal{I}_{P}(v_{n}, S) = 0$ .
5. Symmetry:  $\mathcal{I}_P(v_n\circ \pi ,S) = \mathcal{I}_P(v_n,\pi (S))$  for every utility function  $v_{n}\in \mathcal{G}$ , every permutation  $\pi$  of  $[n]$  and every  $S\subseteq [n]$ .
6. Consistency: For each utility function  $v_{n} \in \mathcal{G}$ , its extension  $\overline{v}_{n+1}$  is defined by  $\overline{v}_{n+1}(S) = v_{n}(S \cap [n])$  for every  $S \subseteq [n+1]$ . Then,  $\mathcal{I}_P(\overline{v}_{n+1}, S) = \mathcal{I}_P(v_n, S)$  for every  $S \subseteq [n]$ .
C.2 Interaction Transform
The concept of interaction transform is crucial to establish well-structured matrix representations for each semi-index  $\mathcal{I}_P$ . Specifically, Definitions 3 and 4 are inspired by (Denneberg and Grabisch 1999).
Definition 2 (Mobius Transform). The Möbius transform  $\mu$  is a function that maps each utility function  $v \in \mathcal{G}_n$  (for every  $n \geq 1$ ) into another utility function in  $\mathcal{G}_n$ . For each  $v_n \in \mathcal{G}$ , we write  $\mu(v_n, S)$  for the value of  $\mu(v_n)$  indexed at  $S \subseteq [n]$ . The definition is
$$
\mu \left(v _ {n}, S\right) = \sum_ {R \subseteq S} (- 1) ^ {s - r} v _ {n} (R) f o r e v e r y S \subseteq [ n ]. \tag {18}
$$
Definition 3 (Convolution). The convolution  $\iota \star u$  between each  $\iota \in \mathbb{R}^{\mathbb{N}}$  and each  $u \in \mathcal{G}_n$  is defined by
$$
(\iota \star u) (S) = \sum_ {[ n ] \supseteq T \supseteq S} \iota (t - s) u (T) f o r e v e r y S \subseteq [ n ]. \tag {19}
$$
In other words, each  $\iota \in \mathbb{R}^{\mathbb{N}}$  may be seen as a function that maps each utility function  $v \in \mathcal{G}_n$  (for every  $n \geq 1$ ) into another utility function in  $\mathcal{G}_n$ , and thus we may write  $\iota(u)$  instead of  $\iota \star u$ .
Definition 4 (Interaction Transform). For each semi-index  $\mathcal{I}_P$ , its interaction transform is defined to be a vector  $\iota_P \in \mathbb{R}^{\mathbb{N}}$  that satisfies
$$
\mathcal {I} _ {P} = \iota_ {P} \circ \mu .
$$
Proposition 6 (Interaction Transforms for Semi-indices). For each semi-index  $\mathcal{I}_P$ , its interaction transform  $\iota_P \in \mathbb{R}^{\mathbb{N}}$  is
$$
\iota_ {P} (i) = \int_ {0} ^ {1} t ^ {i} \mathrm {d} P (t) \text {f o r e v e r y} i \in \mathbb {N}. \tag {20}
$$
Here,  $\iota_P(i)$  refers to the  $i$ -th entry of  $\iota_P$ .
Proof. Let  $v \in \mathcal{G}_n$  be given, and for simplicity, we write  $\mu$  instead for  $\mu(v)$ . For every  $S \subseteq [n]$ , there is
$$
\begin{array}{l} \mathcal {I} _ {P} (v, S) = \sum_ {T \subseteq [ n ] \backslash S} p _ {t + 1} ^ {n + 1 - s} \Delta_ {S} v (T) = \sum_ {T \subseteq [ n ] \backslash S} p _ {t + 1} ^ {n + 1 - s} \sum_ {R \subseteq T} \mu (S \cup R) \\ = \sum_ {R \subseteq [ n ] \backslash S} \sum_ {R \subseteq T \subseteq [ n ] \backslash S} p _ {t + 1} ^ {n + 1 - s} \mu (S \cup R) = \sum_ {[ n ] \supseteq W \supseteq S} \sum_ {W \backslash S \subseteq T \subseteq [ n ] \backslash S} p _ {t + 1} ^ {n + 1 - s} \mu (W) \\ = \sum_ {[ n ] \supseteq W \supseteq S} \left(\sum_ {i = 0} ^ {n - w} \binom {n - w} {i} p _ {w - s + i + 1} ^ {n + 1 - s}\right) \mu (W). \\ \end{array}
$$
We refer the reader to (Grabisch et al. 2000, Eq. (10)) for the equality  $\Delta_S v(T) = \sum_{R\subseteq T}\mu (S\cup R)$  used in the second equality.
Since  $p_k^m = \int_0^1 t^{k - 1}(1 - t)^{m - k}$  for  $1 \leq k \leq m$ , there is
$$
\sum_ {i = 0} ^ {n - w} \binom {n - w} {i} p _ {w - s + i + 1} ^ {n + 1 - s} = \int_ {0} ^ {1} t ^ {w - s} \mathrm {d} P (t).
$$
$$
\mathcal {I} _ {P} (v, S) = \sum_ {[ n ] \supseteq W \supseteq S} \left(\int_ {0} ^ {1} t ^ {w - s} \mathrm {d} P (t)\right) \mu (W) = \sum_ {[ n ] \supseteq W \supseteq S} \iota_ {P} (w - s) \mu (W) = [ \iota_ {P} \star \mu (v) ] (S).
$$
C.3 Matrix Representation for Semi-indices
From now on, let  $2^{[n]}$  be ordered w.r.t. the binary ordering. Then,  $v_{n} \in \mathcal{G}$  is a vector in  $\mathbb{R}^{2^n}$ , and so is  $\mathcal{I}_P(v_n)$ . Since  $\mathcal{I}_P$  is linear in  $v_{n} \in \mathcal{G}_{n}$ ,  $\mathcal{I}_P$  has a matrix representation on each  $\mathcal{G}_n$ .
Proposition 7 (Matrix Representations for Semi-indices). Let  $2^{[n]}$  be ordered using the binary ordering, the matrix representation of semi-index  $\mathcal{I}_P$  on  $\mathcal{G}_n$  w.r.t. the standard basis is
$$
\mathbf {M} _ {P} ^ {[ n ]} = \int_ {0} ^ {1} \mathbf {B} ^ {[ n ]} (t) \mathrm {d} P (t)
$$
$$
\mathbf {B} ^ {[ n ]} (t) = \overbrace {\mathbf {B} ^ {[ 1 ]} (t) \otimes \mathbf {B} ^ {[ 1 ]} (t) \otimes \cdots \otimes \mathbf {B} ^ {[ 1 ]} (t)} ^ {n r e p e t i t i o n s} a n d \mathbf {B} ^ {[ 1 ]} (t) = \left( \begin{array}{c c} 1 - t & t \\ - 1 & 1 \end{array} \right).
$$
Here,  $\otimes$  is the Kronecker product.
Proof. Since  $\mathcal{I}_P = \iota_P \circ \mu$ , suffice it to find the matrix representations for both  $\mu$  and  $\iota_P$  w.r.t. the standard basis in  $\mathcal{G}_n$ .
As provided by Grabisch et al. (2000, Eq. (45)), which can be verified by induction using Eq. (13), the matrix representation for the Möbius transform  $\mu$  on  $\mathcal{G}_n$  w.r.t. the standard basis is
$$
\mathbf {W} ^ {[ n ]} = \overbrace {\mathbf {W} ^ {[ 1 ]} \otimes \mathbf {W} ^ {[ 1 ]} \otimes \cdots \otimes \mathbf {W} ^ {[ 1 ]}} ^ {n \text {r e p e t i t i o n s}} \text {w h e r e} \mathbf {W} ^ {[ 1 ]} = \left( \begin{array}{c c} 1 & 0 \\ - 1 & 1 \end{array} \right)
$$
We first prove by induction that the matrix representation of  $\iota_P$  on  $\mathcal{G}_n$  w.r.t. the standard basis is
$$
\mathbf {T} _ {P} ^ {[ n ]} = \int_ {0} ^ {1} \mathbf {A} ^ {[ n ]} (t) \mathrm {d} P (t) \tag {21}
$$
$$
\mathbf {A} ^ {[ n ]} (t) = \overbrace {\mathbf {A} ^ {[ 1 ]} (t) \otimes \mathbf {A} ^ {[ 1 ]} (t) \otimes \cdots \otimes \mathbf {A} ^ {[ 1 ]} (t)} ^ {n \text {r e p e t i t i o n s}} \text {a n d} \mathbf {A} ^ {[ 1 ]} (t) = \left( \begin{array}{c c} 1 & t \\ 0 & 1 \end{array} \right)
$$
It is clear that it holds for  $n = 1$ . Assume it holds for some  $k > 0$ , i.e., for every  $u^{[k]} \in \mathcal{G}_k$
$$
\iota_ {P} \star u ^ {[ k ]} = \mathbf {T} _ {P} ^ {[ k ]} u ^ {[ k ]}, \tag {22}
$$
and we proceed with  $k + 1$ . Let  $u^{[k + 1]} \in \mathcal{G}_{k + 1}$  and  $S \subseteq [k + 1]$  be given, there are two cases: i)  $S \subseteq [k]$  and ii)  $S \nsubseteq [k]$ . If  $S \subseteq [k]$ , by Eq. (19),