SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\sum_{\substack{S\subseteq D\\ |S| = s_{0}}}I^{SII}(S) = \sum_{\substack{T\subseteq D\\ t <   s_{0}}}(-1)^{t}r(t)\left[(-1)^{s_{0}}\nu (T) + \nu (D\setminus T)\right],
$$
where  $r(t) \coloneqq \frac{1}{s_0} \binom{d - t}{s_0 - t - 1}$  and  $m_s^{\mathrm{II}}(t) \coloneqq \frac{(d - t - s_0)!t!}{(d - s_0 + 1)!}$ .
Proof. For SII, we let  $m(t) \coloneqq m_s^{\mathrm{SII}}(t) = \frac{1}{d - s + 1} \binom{d - s}{t}^{-1}$  and have by Theorem 4.7 and the definition of s-efficiency
$$
\sum_{\substack{S\subseteq D\\ |S| = s_{0}}}I^{m}(S) = \sum_{\substack{S\subseteq D\\ |S| = s_{0}}}c_{s_{0}}(S) = \sum_{\substack{T\subseteq D\\ t <   s_{0}}}\nu (T)\sum_{\substack{S\subseteq D\\ |S| = s_{0}}}\gamma_{s}^{m}(t,|T\cap S|) + \sum_{\substack{T\subseteq D\\ t > \bar{d} -s_{0}}}\nu (T)\sum_{\substack{S\subseteq D\\ |S| = s_{0}}}\gamma_{s}^{m}(t,|T\cap S|)
$$
For  $t < s_0$  we have
$$
\begin{array}{l} \rho (t):= \sum_{\substack{S\subseteq D\\ |S| = s_{0}}}\gamma_{s}^{m}(t,|T\cap S|) = \sum_{k = 0}^{t}\binom {t}{k}\binom {t - k}{s_{0} - k}\gamma_{s}^{m}(t,|T\cap S|) \\ = \frac {1}{d - s _ {0} + 1} \sum_ {k = 0} ^ {t} (- 1) ^ {s _ {0} - k} \left( \begin{array}{c} t \\ k \end{array} \right) \left( \begin{array}{c} d - t \\ s _ {0} - k \end{array} \right) \left( \begin{array}{c} d - s _ {0} \\ t - k \end{array} \right) ^ {- 1}. \\ \end{array}
$$
For  $t > d - s$  there are at least  $k_{\min} = t - (d - s)$  elements in the intersection of  $|T \cap S|$  and thus with  $\bar{t} := d - t < s_0$  and  $k_{\min} = s_0 - \bar{t}$
$$
\begin{array}{l} \sum_{\substack{S\subseteq D\\ |S| = s_{0}}}\gamma_{s}^{m}(t,|T\cap S|) = \sum_{k = k_{\min}}^{s_{0}}\binom {t}{k}\binom {t - k}{s_{0} - k}\gamma_{s}^{m}(t,|T\cap S|) \\ = \frac {1}{d - s _ {0} + 1} \sum_ {k = k _ {\min}} ^ {s _ {0}} (- 1) ^ {s _ {0} - k} \binom {t} {k} \binom {d - t} {s _ {0} - k} \binom {d - s _ {0}} {t - k} ^ {- 1} \\ = \frac {1}{d - s _ {0} + 1} \sum_ {k = s _ {0} - \bar {t}} ^ {s _ {0}} (- 1) ^ {s _ {0} - k} \binom {d - \bar {t}} {k} \binom {\bar {t}} {s _ {0} - k} \binom {d - s _ {0}} {d - \bar {t} - k} ^ {- 1} \\ = \frac {1}{d - s _ {0} + 1} \sum_ {k = 0} ^ {\bar {t}} (- 1) ^ {k} \left( \begin{array}{c} d - \bar {t} \\ s _ {0} - k \end{array} \right) \left( \begin{array}{c} \bar {t} \\ k \end{array} \right) \left( \begin{array}{c} d - s _ {0} \\ d - \bar {t} - s _ {0} + k \end{array} \right) ^ {- 1} \\ = \frac {1}{d - s _ {0} + 1} \sum_ {k = 0} ^ {\bar {t}} (- 1) ^ {k} \left( \begin{array}{c} d - \bar {t} \\ s _ {0} - k \end{array} \right) \left( \begin{array}{c} \bar {t} \\ k \end{array} \right) \left( \begin{array}{c} d - s _ {0} \\ \bar {t} - k \end{array} \right) ^ {- 1} \\ = (- 1) ^ {s _ {0}} \rho (\bar {t}). \\ \end{array}
$$
We can explicitly compute  $\rho(t)$  as
$$
\begin{array}{l} \rho (t) = \frac{1}{d - s_{0} + 1}\sum_{k = 0}^{t}(-1)^{s_{0} - k}\binom {t}{k}\binom {d - t}{s_{0} - k}\binom {d - s_{0}}{t - k}^{-1} \\ = (- 1) ^ {s _ {0}} \frac {t ! (d - t) !}{(d - s _ {0} + 1) !} \sum_ {k = 0} ^ {t} (- 1) ^ {k} \frac {1}{(s _ {0} - k) ! k !} \\ = (- 1) ^ {s _ {0}} \frac {t ! (d - t) !}{s _ {0} ! (d - s _ {0} + 1) !} \sum_ {k = 0} ^ {t} (- 1) ^ {k} \binom {s _ {0}} {k} \\ = (- 1) ^ {s _ {0}} \frac {t ! (d - t) !}{s _ {0} ! (d - s _ {0} + 1) !} (- 1) ^ {t} \binom {s _ {0} - 1} {t} = \frac {(- 1) ^ {s _ {0} + t}}{s _ {0}} \binom {d - t} {s _ {0} - t - 1}, \\ \end{array}
$$
where we have used that  $\sum_{k=0}^{t}(-1)^{k}\binom{s_{0}}{k} = (-1)^{t}\binom{s_{0}-1}{t}$  for  $t < s_0$ . Hence,
$$
\sum_{\substack{S\subseteq D\\ |S| = s_{0}}}I^{m}(S) = \frac{1}{s_{0}}\sum_{\substack{T\subseteq D\\ t <   s_{0}}}\left[(-1)^{s_{0}}\nu (T) + \nu (D\setminus T)\right](-1)^{t}\binom {d - t}{s_{0} - t - 1}.
$$
This section contains the proofs of the claims made in the main paper.
Proof. By definition, the sum  $I^{m}(S) \coloneqq \sum_{T \subseteq D \setminus S} m(t) \delta_{S}^{\nu}(T)$  ranges over all subsets  $T \subseteq D$ , where every subset is exactly once evaluated. On the one hand, it is easy to see that every evaluated subset in  $I^{m}(S)$  is different, as  $T \cup L$  is unique. Furthermore, given any subset  $T \subseteq D$ , we decompose  $T = \tilde{T} \cup L$ , where  $\tilde{T} \subseteq D \setminus S$  and  $L \coloneqq T \cap S \subseteq S$ . The corresponding weight is  $m(\tilde{t}) = m(t - l) = m(t - |T \cap S|)$  and the sign from  $\delta_{S}^{\nu}(\tilde{T})$  is  $(-1)^{s - l} = (-1)^{s - |T \cap S|}$ . This yields with  $\nu_{0}(T) \coloneqq \nu(T) - \nu(\emptyset)$
$$
\begin{array}{l} I ^ {m} (S) = \sum_ {T \subseteq D} \nu (T) \gamma_ {s} ^ {m} (t, | T \cap S |) = \sum_ {T \subseteq D} \nu_ {0} (T) \gamma_ {s} ^ {m} (t, | T \cap S |) + \nu (\emptyset) \sum_ {T \subseteq D} \gamma_ {s} ^ {m} (t, | T \cap S |) \\ = \sum_ {T \subseteq D} \nu_ {0} (T) \gamma_ {s} ^ {m} (t, | T \cap S |), \\ \end{array}
$$
as the sum over all  $\gamma^m$  is zero, if the dummy axiom is fulfilled.
Remark B.1. It is important to note, that the sum  $\sum_{T\subseteq D}\gamma_s^m (t,|T\cap S|)$  is not zero, if not all subsets are considered, which makes it crucial to use  $\nu_0$  instead of  $\nu$ . In fact, the estimates of  $I^{m}$  would be heavily skewed by  $\nu (\emptyset)$ . While the estimator would still be unbiased, its variance would scale with  $\nu (\emptyset)^2$ .
Proof. We aim to show that  $\hat{I}^m(S)$  is unbiased and consistent, i.e.  $\mathbb{E}\left[\hat{I}_{k_0}^m(S)\right] = I^m(S)$  and  $\lim_{K\to \infty}\hat{I}_{k_0}^m(S) = I^m(S)$ . Given
$$
\hat {I} _ {k _ {0}} ^ {m} (S) := c _ {k _ {0}} (S) + \frac {1}{K} \cdot \sum_ {k = 1} ^ {K} \nu_ {0} (T _ {k}) \frac {\gamma_ {s} ^ {m} (t _ {k} , | T _ {k} \cap S |)}{p _ {k _ {0}} (T _ {k})},
$$
it is clear that due to the linearity of the expectation
$$
\mathbb {E} _ {T \sim p _ {k _ {0}} (T)} \left[ \hat {I} _ {k _ {0}} ^ {m} (S) \right] = c _ {k _ {0}} (S) + \frac {1}{K} \sum_ {k = 1} ^ {K} \mathbb {E} _ {T \sim p _ {k _ {0}} (T)} \left[ \nu_ {0} (T) \frac {\gamma_ {s} ^ {m} (| T | , | T \cap S |)}{p _ {k _ {0}} (T)} \right] = I ^ {m} (S).
$$
Furthermore, let  $\sigma^2 (S)\coloneqq \mathbb{V}_{T\sim p_{k_0}(T)}\left[\nu_0(T)\frac{\gamma_s^m(|T|,|T\cap S|)}{p_{k_0}(T)}\right]$  be the variance of each estimate, then, by the law of large numbers
$$
\frac {1}{K} \cdot \sum_ {k = 1} ^ {K} \nu_ {0} (T _ {k}) \frac {\gamma_ {s} ^ {m} (t _ {k} , | T _ {k} \cap S |)}{p _ {k _ {0}} (T _ {k})} \stackrel {K \to \infty} {\longrightarrow} \mathbb {E} _ {T \sim p _ {k _ {0}} (T)} \left[ \nu_ {0} (T) \frac {\gamma_ {s} ^ {m} (| T | , | T \cap S |)}{p _ {k _ {0}} (T)} \right],
$$
and thus  $\lim_{K\to \infty}\hat{I}^m (S) = I^m (S)$ . Lastly, as  $\hat{I}^m (S)$  is unbiased, we have for  $\epsilon >0$  by Chebyshev's inequality
$$
\mathbb {P} (| \hat {I} _ {k _ {0}} ^ {m} (S) - I ^ {m} (S) | > \epsilon) \leq \frac {\mathbb {V} [ \hat {I} _ {k _ {0}} ^ {m} (S) ]}{\epsilon^ {2}} = \frac {1}{K ^ {2}} \frac {K \sigma^ {2} (S)}{\epsilon^ {2}} = \frac {1}{K} \frac {\sigma^ {2} (S)}{\epsilon^ {2}}.
$$
B.3 Proof of Theorem 4.4
Proof. We let  $m(t) \coloneqq \frac{(d - t - 1)!t!}{d!}$  and apply Theorem 4.1. With  $\gamma_s^m (0,0) = -m(0),\gamma_s^m (d,1) = m(d - 1)$  and  $m(0) = m(d - 1) = \frac{1}{d}$ , we have
$$
\begin{array}{l} I ^ {\mathrm {S V}} (i) \stackrel {\text {T h e o r e m 4 . 1}} {=} \sum_ {T \subseteq D} \nu (T) \gamma_ {s} ^ {m} (t, \mathbf {1} (i \in T)) = \frac {\nu (D) - \nu (\emptyset)}{d} + \sum_ {T \in \mathcal {T} _ {1}} \nu (T) \gamma_ {s} ^ {m} (t, \mathbf {1} (i \in T)) \\ = c _ {1} (i) + \sum_ {T \in \mathcal {T} _ {1}} \nu (T) \left[ \mathbf {1} (i \in T) \gamma_ {s} ^ {m} (t, 1) + \mathbf {1} (i \notin T) \gamma_ {s} ^ {m} (t, 0) \right] \\ = c _ {1} (i) + \sum_ {T \in \mathcal {T} _ {1}} \nu (T) [ \mathbf {1} (i \in T) (\gamma_ {s} ^ {m} (t, 1) - \gamma_ {s} ^ {m} (t, 0)) + \gamma_ {s} ^ {m} (t, 0) ] \\ = c _ {1} (i) + \sum_ {T \in \mathcal {T} _ {1}} \nu (T) [ \mathbf {1} (i \in T) (m (t - 1) + m (t)) - m (t) ] \\ = c _ {1} (i) + \sum_ {T \in \mathcal {T} _ {1}} \nu (T) \left[ \mathbf {1} (i \in T) \left(\frac {(d - t) ! (t - 1) !}{d !} + \frac {(d - t - 1) ! t !}{d !}\right) - \frac {(d - t - 1) ! t !}{d !} \right] \\ = c _ {1} (i) + \sum_ {T \in \mathcal {T} _ {1}} \nu (T) \frac {(d - t - 1) ! (t - 1) !}{(d - 1) !} \left[ \mathbf {1} (i \in T) - \frac {t}{d} \right] \\ = c _ {1} (i) + \sum_ {T \in \mathcal {T} _ {1}} \nu (T) \mu (t) \left[ \mathbf {1} (i \in T) - \frac {t}{d} \right]. \\ \end{array}
$$
B.4 Proof of Theorem 4.5
Proof. According to Proposition E.3, our goal is to show that
$$
\hat {I} _ {U} ^ {\mathrm {S V}} (i) = c _ {1} (i) + \frac {2 h _ {d - 1}}{K} \sum_ {k = 1} ^ {K} \nu_ {0} (T _ {k}) \left[ \mathbf {1} (i \in T _ {k}) - \frac {t _ {k}}{d} \right]
$$
with  $T_{k} \stackrel{\mathrm{iid}}{\sim} p(T) \coloneqq \mu(t) / (2h_{d-1})$ , where  $p$  is a probability distribution over  $\mathcal{T}_1$  and  $h_n \coloneqq \sum_{t=1}^{n} t^{-1}$ . The proof is structured in the following steps:
1. Exact computation of  $A^{-1}$  using the exact structure of  $A$  with diagonal entries  $\mu_{1}$  and off-diagonal entries  $\mu_{2}$ , cf. [8, Appendix A].
2. Exact computation of  $\hat{I}_U^{\mathrm{SV}}$ , which yields with Proposition E.3  $\hat{I}_U^{\mathrm{SV}}(i) = \hat{I}_1^m (i)$ , if  $(\mu_{1} - \mu_{2})2h_{d - 1} = 1$
3. We show that  $(\mu_{1} - \mu_{2})2h_{d - 1} = 1$
Calculation of  $A^{-1}$ . It has been shown in [8, Appendix A] that all off-diagonal entries are equal and all diagonal entries are equal, i.e.  $A$  may be written as  $A = \mu_2\mathbf{J} + (\mu_1 - \mu_2)\mathbf{I}$  with off-diagonal entries  $\mu_2 \coloneqq p(Z_i = Z_j = 1)$  and diagonal entries  $\mu_1 \coloneqq p(Z_i = 1)$ , where  $Z_i$  refers to the i-th component of the binary vector  $Z$  and  $\mathbf{J}$  is a matrix of ones and  $\mathbf{I}$  is the identity matrix. The simple structure of  $A$  allows to compute the inverse exactly by using the following Lemma.
Lemma B.2. Let  $\mu_1, \mu_2 > 0$  with  $\mu_1 \neq \mu_2$ , then
$$
\left(\mu_ {2} \mathbf {J} + \left(\mu_ {1} - \mu_ {2}\right) \mathbf {I}\right) ^ {- 1} = \tilde {\mu} _ {2} \mathbf {J} + \left(\tilde {\mu} _ {1} - \tilde {\mu} _ {2}\right) \mathbf {I}
$$
$$
\tilde {\mu} _ {2} = \frac {- \mu_ {2}}{(\mu_ {1} - \mu_ {2}) (\mu_ {1} + (d - 1) \mu_ {2})}
$$
$$
\tilde {\mu} _ {1} = \frac {\mu_ {1} + (d - 2) \mu_ {2}}{(\mu_ {1} - \mu_ {2}) (\mu_ {1} + (d - 1) \mu_ {2})}.
$$
$$
\begin{array}{l} \mathbf {I} = \left(\mu_ {2} \mathbf {J} + \left(\mu_ {1} - \mu_ {2}\right) \mathbf {I}\right) \cdot \left(\tilde {\mu} _ {2} \mathbf {J} + \left(\tilde {\mu} _ {1} - \tilde {\mu} _ {2}\right) \mathbf {I}\right) \\ = \left(\left(\mu_ {1} + (d - 1) \mu_ {2}\right) \tilde {\mu} _ {2} + \left(\tilde {\mu} _ {1} - \tilde {\mu} _ {2}\right) \mu_ {2}\right) \mathbf {J} \\ + (\mu_ {1} - \mu_ {2}) (\tilde {\mu} _ {1} - \tilde {\mu} _ {2}) \mathbf {I}, \\ \end{array}
$$
which yields  $(\mu_{1} - \mu_{2})(\tilde{\mu}_{1} - \tilde{\mu}_{2}) = 1$  and  $\mu_2\tilde{\mu}_2d + (\mu_1 - \mu_2)\tilde{\mu}_1 + (\tilde{\mu}_1 - \tilde{\mu}_2)\mu_2 = 0$ . From the first equation we have  $\tilde{\mu}_1 - \tilde{\mu}_2 = 1 / (\mu_1 - \mu_2)$  and thus by the second equation
$$
\tilde {\mu} _ {1} = \frac {\mu_ {1} + (d - 2) \mu_ {2}}{(\mu_ {1} - \mu_ {2}) (\mu_ {1} + (d - 1) \mu_ {2})}
$$
Calculation of  $\hat{I}_U^{\mathbf{SV}}$ . By Lemma B.2, we proceed to compute the different components of
$$
\hat {I} _ {U} ^ {\mathrm {S V}} := A ^ {- 1} \left(\hat {b} _ {L} - \mathbf {1} \frac {\mathbf {1} ^ {T} A ^ {- 1} \hat {b} _ {L} - \nu_ {0} (\mathbf {1})}{\mathbf {1} ^ {T} A ^ {- 1} \mathbf {1}}\right).
$$
$$
\mathbf {1} ^ {T} A ^ {- 1} = \left((d - 1) \tilde {\mu} _ {2} + \tilde {\mu} _ {1}\right) \mathbf {1} ^ {T} = \frac {1}{\mu_ {1} + (d - 1) \mu_ {2}} \mathbf {1} ^ {T}.
$$
Then the denominator yields  $\mathbf{1}^T A^{-1}\mathbf{1} = \frac{d}{\mu_1 + (d - 1)\mu_2}$ . We then obtain
$$
\mathbf {1} \frac {\mathbf {1} ^ {T} A ^ {- 1} \hat {b} _ {L} - \nu_ {0} (\mathbf {1})}{\mathbf {1} ^ {T} A ^ {- 1} \mathbf {1}} = \frac {1}{d} \mathbf {1} \mathbf {1} ^ {T} \hat {b} _ {L} - \frac {\mu_ {1} + (d - 1) \mu_ {2}}{d} \nu_ {0} (\mathbf {1}) \mathbf {1} = \frac {1}{d} \mathbf {J} \cdot \hat {b} _ {L} + \frac {\mu_ {1} + (d - 1) \mu_ {2}}{d} \nu_ {0} (\mathbf {1}) \cdot \mathbf {1},
$$
which, with  $A^{-1}\mathbf{1} = (\tilde{\mu}_1 + (d - 1)\tilde{\mu}_2)\mathbf{1} = \frac{1}{\mu_1 + (d - 1)\mu_2}\mathbf{1}$ , yields
$$
\hat {I} _ {U} ^ {\mathrm {S V}} = A ^ {- 1} \left(\hat {b} _ {L} - \frac {1}{d} \mathbf {J} \cdot \hat {b} _ {L}\right) + \frac {\nu_ {0} (\mathbf {1})}{d} \cdot \mathbf {1} = c _ {1} + A ^ {- 1} \left(\hat {b} _ {L} - \frac {1}{d} \mathbf {J} \cdot \hat {b} _ {L}\right).
$$
It remains to show that
$$
\left(A ^ {- 1} \left(\hat {b} _ {L} - \frac {1}{d} \mathbf {J} \cdot \hat {b} _ {L}\right)\right) _ {i} = \frac {1}{K} \sum_ {k = 1} ^ {K} \nu_ {0} (T _ {k}) \frac {\gamma_ {s} ^ {m} \left(t _ {k} , \mathbf {1} (i \in T _ {k})\right)}{p \left(T _ {k}\right)}.
$$
With  $\hat{b}_L = \frac{1}{K}\sum_{k = 1}^{K}z_k\nu_0(z_k)$  it follows
$$
A ^ {- 1} (\hat {b} _ {L} - \frac {1}{d} \mathbf {J} \cdot \hat {b} _ {L}) = \frac {1}{K} \sum_ {k = 1} ^ {K} \left(A ^ {- 1} z _ {k} - \frac {1}{d} A ^ {- 1} \mathbf {J} z _ {k}\right) \nu_ {0} (z _ {k}).
$$
Then  $A^{-1}z_{k} = t_{k}\tilde{\mu}_{2}\mathbf{1} + (\tilde{\mu}_{1} - \tilde{\mu}_{2})z_{k}$  where  $t_k$  is the subset size, i.e.  $t_k$  is the sum of all entries in  $z_{k}$ . It follows with  $A^{-1}J = \frac{1}{\mu_1 + (d - 1)\mu_2} J$
$$
A ^ {- 1} z _ {k} - \frac {1}{d} A ^ {- 1} {\bf J} z _ {k} = t _ {k} \left(\tilde {\mu} _ {2} - \frac {1}{d (\mu_ {1} + (d - 1) \mu_ {2})}\right) {\bf 1} + (\tilde {\mu} _ {1} - \tilde {\mu} _ {2}) z _ {k} = \frac {1}{\mu_ {1} - \mu_ {2}} \left(z _ {k} - \frac {t _ {k}}{d} {\bf 1}\right).
$$
For the i-th component, we have with set notation  $T_{k}$  and for the SV weights  $m(t) \coloneqq \frac{(d - t - 1)!t!}{d!}$  then
$$
\begin{array}{l} \hat {I} _ {U} ^ {\mathrm {S V}} (i) - c _ {1} (i) = \left(A ^ {- 1} \left(\hat {b} _ {L} - \frac {1}{d} \mathbf {J} \cdot \hat {b} _ {L}\right)\right) _ {i} = \frac {1}{K} \sum_ {k = 1} ^ {K} \nu_ {0} \left(T _ {k}\right) \frac {1}{\mu_ {1} - \mu_ {2}} \left(\mathbf {1} (i \in T _ {k}) - \frac {t _ {k}}{d}\right) \\ = \frac {1}{(\mu_ {1} - \mu_ {2}) 2 h _ {d - 1}} \left(\hat {I} _ {1} ^ {m} (i) - c _ {1} (i)\right), \\ \end{array}
$$
where we have used Proposition E.3 for  $\tilde{I}_1^m (i)$ . It remains to show that  $(\mu_{1} - \mu_{2})2h_{d - 1} = 1$
Show that  $(\mu_1 - \mu_2)2h_{d-1} = 1$ . We let  $p(T_k) \coloneqq \frac{\mu(t)}{R}$  be a probability distribution over  $\mathcal{T}_1$ . By definition, and as subsets of size  $t$  have equal probability, we have
$$
\begin{array}{l} \mu_ {1} = p (Z _ {i} = 1) = \sum_ {t = 1} ^ {d - 1} p (Z _ {i} = 1 | \mathbf {1} ^ {T} Z = t) p (\mathbf {1} ^ {T} Z = t) \\ = \sum_ {t = 1} ^ {d - 1} \frac {\binom {d - 1} {t - 1}}{\binom {d} {t}} \frac {\mu (t)}{R} \binom {d} {t} = \sum_ {t = 1} ^ {d - 1} \binom {d - 1} {t - 1} \frac {\mu (t)}{R} \\ \end{array}
$$
$$
\begin{array}{l} \mu_ {2} = p (Z _ {i} = Z _ {j} = 1) = \sum_ {t = 1} ^ {d - 1} p (Z _ {i} = Z _ {j} = 1 | \mathbf {1} ^ {T} Z = t) p (\mathbf {1} ^ {T} Z = t) \\ = \sum_ {t = 2} ^ {d - 1} \frac {\binom {d - 2} {t - 2}}{\binom {d} {t}} \frac {\mu (t)}{R} \binom {d} {t} = \sum_ {t = 2} ^ {d - 1} \binom {d - 2} {t - 2} \frac {\mu (t)}{R}. \\ \end{array}
$$
$$
\mu_ {1} - \mu_ {2} = \frac {\mu (1)}{R} + \sum_ {t = 2} ^ {d - 1} \frac {\mu (t)}{R} \left(\binom {d - 1} {t - 1} - \binom {d - 2} {t - 2}\right) = \sum_ {t = 1} ^ {d - 1} \frac {\mu (t)}{R} \binom {d - 2} {t - 1} = \frac {1}{R},
$$
where we have used the recursion for the binomial coefficient and  $\mu(t) = \frac{1}{d-1} \binom{d-2}{t-1}^{-1}$ . Lastly, we have seen in the proof of Proposition E.3 that  $R = 2h_{d-1}$ , which finishes the proof.
B.5 Proof of Theorem 4.7
Proof. We prove the statements in separate subsections. We consider interactions of maximum order  $s_0 \geq 1$ , summarized in  $\mathcal{I} \coloneqq \{S \subset D \mid s = s_0\}$ . To show that SII and STI are s-efficient, by Theorem 4.1 it suffices to show that  $\sum_{S \in \mathcal{I}} \gamma_s^m (t, |T \cap S|) = 0$  for all  $T \in \mathcal{T}_{s_0}$ . Given a subset  $T \in \mathcal{T}_{s_0}$  with  $|T \cap S| = k$  and  $k \in \{0, \dots, s_0\}$ , we have
$$
\sum_ {S \in \mathcal {I}} \gamma_ {s} ^ {m} (t, | T \cap S |) = \sum_ {k = 0} ^ {s _ {0}} {\binom {t} {k}} {\binom {t - k} {s _ {0} - k}} \gamma_ {s} ^ {m} (t, k) = \sum_ {k = 0} ^ {s _ {0}} (- 1) ^ {s _ {0} - k} {\binom {t} {k}} {\binom {d - t} {s _ {0} - k}} m (t - k).
$$
B.5.1 Proof of s-efficiency for SII.
We let  $m(t) \coloneqq m_{s_0}^{\mathrm{SII}}(t) = \frac{(d - t - s_0)!t!}{(d - s_0 + 1)!} = \frac{1}{d - s_0 + 1}\binom{d - s_0}{t}^{-1}$  and obtain
$$
\sum_{S\in \mathcal{I}}\gamma_{s}^{m}(t,|T\cap S|) = \frac{1}{d - s_{0} + 1}\sum_{k = 0}^{s_{0}}(-1)^{s_{0} - k}\binom {t}{k}\binom {d - t}{s_{0} - k}\binom {d - s_{0}}{t - k}^{-1}.
$$
We now consider
$$
\begin{array}{l} \sum_ {k = 0} ^ {s _ {0}} (- 1) ^ {k} \binom {t} {k} \binom {d - t} {s _ {0} - k} \binom {d - s _ {0}} {t - k} ^ {- 1} = \sum_ {k = 0} ^ {s _ {0}} (- 1) ^ {k} \frac {t !}{k ! (t - k) !} \frac {(d - t) !}{(s _ {0} - k) ! (d - t - s _ {0} + k) !} \frac {(t - k) ! (d - s _ {0} - t + k) !}{(d - s _ {0}) !} \\ = \frac {t ! (d - t) !}{(d - s _ {0}) !} \sum_ {k = 0} ^ {s _ {0}} (- 1) ^ {k} \frac {1}{(s _ {0} - k) ! k !} \\ = \frac {t ! (d - t) !}{s _ {0} ! (d - s _ {0}) !} \sum_ {k = 0} ^ {s _ {0}} (- 1) ^ {k} \binom {s _ {0}} {k} \\ = \frac {t ! (d - t) !}{s _ {0} ! (d - s _ {0}) !} (1 - 1) ^ {s _ {0}} \\ = 0, \\ \end{array}
$$
where we have used the binomial expansion for  $(1 - 1)^{s_0}$ . Hence,
$$
\sum_ {S \in \mathcal {I}} \gamma_ {s} ^ {m} (t, | T \cap S |) = 0,
$$
which finishes the proof of SII s-efficiency.
B.5.2 Proof of s-efficiency of STI.
For STI, i.e.  $m_{s_0}^{\mathrm{STI}} \coloneqq s_0 \frac{(d - t - 1)!t!}{d!}$ , we have
$$
\begin{array}{l} \sum_{S\in \mathcal{I}}\gamma_{s}^{m}(t,|T\cap S|) = \sum_{k = 0}^{s_{0}}(-1)^{s_{0} - k}\binom {t}{k}\binom {d - t}{s_{0} - k} s_{0}\frac{(d - t + k - 1)!(t - k)!}{d!} \\ = \frac {s _ {0} (- 1) ^ {s _ {0}} t ! (d - t) !}{s _ {0} ! d !} \sum_ {k = 0} ^ {s _ {0}} (- 1) ^ {k} \binom {s _ {0}} {k} \frac {(d - t + k - 1) !}{(d - t - s _ {0} + k) !} \\ \end{array}
$$
As  $\frac{(d - t + k - 1)!}{(d - t - s_0 + k)!}$  is a polynomial with orders less than  $s_0$ , we can use  $\sum_{k=0}^{s_0} \binom{s_0}{k} k^m = 0$  for  $m < s_0$  [38] to obtain
$$
\frac{s_{0}(-1)^{s_{0}}t!(d - t)!}{s_{0}!d!}\sum_{k = 0}^{s_{0}}(-1)^{k}\binom {s_{0}}{k}\frac{(d - t + k - 1)!}{(d - t - s_{0} + k)!} = 0,
$$
which finishes the proof of STI s-efficiency.
B.5.3 Proof of efficiency for STI.
It is clear that s-efficiency of a CII implies by Theorem 4.1 with sampling order  $k \geq s_0$  that the sum of top-order interaction estimates is
$$
\sum_ {S \in \mathcal {I}} \hat {I} _ {s _ {0}} ^ {\mathrm {S T I}} (t) = \sum_ {S \in \mathcal {I}} c _ {s _ {0}} ^ {\mathrm {S T I}} (S).
$$
On the other hand, it also implies that the sum of STI scores for the top-order interactions are
$$
\sum_ {S \in \mathcal {I}} I _ {s _ {0}} ^ {\mathrm {S T I}} (t) = \sum_ {S \in \mathcal {I}} c _ {s _ {0}} ^ {\mathrm {S T I}} (S).
$$
While lower-order estimates are computed exactly for STI, it follows that the sum of STI scores and sum of SHAP-IQ estimates over all interaction sets  $\mathcal{S}_{s_0}$  are equal. Furthermore, due to the definition of STI, they must fulfill the efficiency axiom, which finishes the proof.
B.5.4 Proof of efficiency for n-SII.
This result follows from the aggregation suggested by n-SII, which is independent of the index. The efficiency condition follows directly from the SV efficiency and the Bernoulli numbers, independent of higher-order interaction values, cf. proof in [3, Proposition 12]. The proof is based on induction, starting from the SV, where the efficiency condition holds. The specific aggregation based on the Bernoulli numbers then ensures that this efficiency condition is maintained. To apply this observation to SHAP-IQ estimates of n-SII, we observe that due to Proposition E.3, SHAP-IQ maintains the efficiency condition for order  $s = 1$ , i.e. the SV estimates. The aggregation of n-SII for the SHAP-IQ estimates of higher orders then immediately implies that this efficiency condition is always preserved, as the arguments presented therein hold independent of the interaction index, cf. [3, Proof of Proposition 12].
C Algorithmic Details of SHAP-IQ
In this section, we describe further algorithmic details that are used in the SHAP-IQ implementation.
C.1 Algorithm of SHAP-IQ
The pseudo-code for SHAP-IQ is outlined in Algorithm 1.
Algorithm 1 SHAP-IQ for any-order interactions  $S_{s_0}$  up to order  $s_0$    
Require: Budget  $K > 0$  weights  $q(t)\geq 0$  with  $t\in [d]$  ,precomputed weights  $\gamma_s^m (t,\ell)$  for  $s = 1,\dots ,s_0,t = 0\dots ,d$  and  $\ell = 0,\ldots ,s$    
1:  $k_{0}\gets \mathrm{GETSAMPLINGORDER}(q,K)$    
2: for  $T\notin T_{k_0}$  do Deterministic   
3:  $\eta \leftarrow \nu_0(T)$    
4: for  $S\in S_{s_0}$  do   
5:  $c_{k_0}(S)\gets c_{k_0}(S) + \eta \cdot \gamma_s^m (t,|T\cap S|)$  Update deterministic part for all  $S$    
6: end for   
7:  $K\gets K - 1$    
8: end for   
9: for  $t = k_0,\dots ,d - k_0$  do   
10:  $p(t)\gets q(t) / \left(\sum_{k = k_0}^{d - k_0}q(k)\binom {d}{k}\right)$  compute probabilities  $\mathbb{P}_{k_0}(|T| = t)$    
11: end for   
12: for  $k = 1,\dots ,K$  do ▷ Sampling   
13:  $T\gets \mathrm{SAMPLE}(p,k_0)$    
14:  $\eta \leftarrow \nu_0(T)$    
15: for  $S\in S_{s_0}$  do   
16:  $\Delta (S)\gets \eta \cdot \gamma_s^m (t,|T\cap S|)(\stackrel {d}{t}) / p[t]$  Use probabilities   
 $\mathbb{P}_{k_0}(T) = \mathbb{P}_{k_0}(|T| = t) / (\stackrel {d}{t})\propto q(t)$    
17: end for   
18:  $\hat{\mu},\hat{s}^2\gets \text{WELFORDUPDATE} (\hat{\mu},s2,k,\Delta)$    
19: end for   
20: mean  $\hat{I}_{k_0}^m\gets c_{k_0} + \hat{\mu}^m$  and variance  $\hat{\sigma}^2\gets s2 / (n - 1)$    
21: return  $\hat{I}_{k_0}^m$  and  $\hat{\sigma}^2$
C.2 Sampling Weights  $q(t)$
In our implementation, we rely on  $q(t) \propto \mu(t)$  for  $1 \leq t \leq d - 1$ , where the weights  $q(0) = q(d) = q_0 \gg 0$  are set to a high positive constant, which favors these subsets before weighting the remaining subsets in  $\mathcal{T}_1$ . These weights ensure that SHAP-IQ is equal to U-KSH for the SV. Another choice of weights is  $q(t) = \frac{(d - t - s_0)!(t - s_0)!}{(d - s_0 + 1)!}$  for  $s_0 \leq t \leq d - s_0$  and  $q(t) = q_0$  otherwise, which prefers all orders up to  $s_0$  and from  $d$  to  $d - s_0$ . This choice of subset weights may be beneficial for very low budgets, as it is important to ensure that  $k_0 \geq s_0$  for SHAP-IQ to maintain the efficiency condition. The algorithm to find  $k_0$  given weight  $q$  and budget  $K$  is outlined in Algorithm 2. The sampling procedure to generate a subset according to  $p(T)$  is outlined in Algorithm 3.
C.3 Welford's Algorithm
Welford's algorithm [45] allows to iteratively update the mean and variance using a single pass. The algorithm is outlined in Algorithm 4.
Algorithm 2 Determine the sampling order  $k_{0}$  for the deterministic part
Require: weights  $q$  over  $0,\ldots ,d$  budget  $K > 0$    
1: initialize  $k_{0} = 0$    
2: for  $t = 0,\dots ,FLOO R(d / 2)$  do   
3:  $R = \left(\sum_{k = k_0}^{d - k_0}q[k]\binom {d}{k}\right)$  ▷ Normalization   
4:  $p[t]\gets q[t]\binom {d}{t} /R$    
5:  $p[d - t]\gets q[d - t]\binom {d}{t} /R$    
6: if  $K\cdot q[t] > R$  and  $K\cdot q[d - t] > R$  then   
7:  $k_{0}\gets k_{0} + 1$    
8:  $K\gets K - 2\binom{d}{t}$    
9: end if   
10: end for   
11: return  $k_{0}$
Algorithm 3 Sample a subset  $T \sim p(T)$
Require:  $p$  with  $\sum_{k = k_0}^{d - k_0}p[k] = 1$  sampling order  $k_{0}$    
1: for  $t = k_0,\dots ,d - k_0$  do   
2:  $p(|T| = t)\gets p[t]\binom {d}{t}$    
3: end for   
4: choose subset size  $t_0$  with probability  $p(|T| = t)$    
5: choose subset  $T$  of size  $t_0$  with probability  $\left( \begin{array}{l}d\\ t_0 \end{array} \right)^{-1}$    
6: return  $T$