SlowGuess/NeurIPS_2023_OCR huggingface dataset:
The KL divergence between Gaussian distributions has a closed-form solution, and in the case with equal covariance matrices reduces to the following:
$$
D _ {\mathrm {K L}} \left(p \left(\boldsymbol {x} _ {1} \mid x _ {2}\right) \mid p \left(\boldsymbol {x} _ {1} \mid x _ {2} ^ {\prime}\right)\right) = \frac {1}{2} \left(x _ {2} - x _ {2} ^ {\prime}\right) ^ {\top} \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1} \left(\Sigma_ {1 1} - \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1}\right) ^ {- 1} \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \left(x _ {2} - x _ {2} ^ {\prime}\right)
$$
To provide an upper bound on this term based on  $\| x_2 - x_2' \|_2$ , it suffices to find the maximum eigenvalue of the above matrix. To characterize the matrix, we observe that it is present in the formula for the block inversion of the joint covariance matrix:
$$
\Sigma^ {- 1} = \left( \begin{array}{c c} \left(\Sigma_ {1 1} - \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1}\right) ^ {- 1} & - \left(\Sigma_ {1 1} - \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1}\right) ^ {- 1} \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \\ - \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1} \left(\Sigma_ {1 1} - \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1}\right) ^ {- 1} & \Sigma_ {2 2} ^ {- 1} + \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1} \left(\Sigma_ {1 1} - \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1}\right) ^ {- 1} \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \end{array} \right)
$$
We therefore have the following, because the matrix we're interested in is a principal minor:
$$
\lambda_ {\max } \left(\Sigma_ {2 2} ^ {- 1} + \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1} \left(\Sigma_ {1 1} - \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1}\right) ^ {- 1} \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1}\right) \leq \lambda_ {\max } (\Sigma^ {- 1}).
$$
Weyl's inequality also gives us:
$$
\begin{array}{l} \lambda_ {\min } \left(\Sigma_ {2 2} ^ {- 1}\right) + \lambda_ {\max } \left(\Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1} \left(\Sigma_ {1 1} - \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1}\right) ^ {- 1} \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1}\right) \\ \leq \lambda_ {\max } \left(\Sigma_ {2 2} ^ {- 1} + \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1} \left(\Sigma_ {1 1} - \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1}\right) ^ {- 1} \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1}\right) \\ \end{array}
$$
Combining this with the previous inequality, we arrive at the following:
$$
\lambda_ {\max } \left(\Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1} \left(\Sigma_ {1 1} - \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1} \Sigma_ {2 1}\right) ^ {- 1} \Sigma_ {1 2} \Sigma_ {2 2} ^ {- 1}\right) \leq \lambda_ {\max } (\Sigma^ {- 1}) - \lambda_ {\min } (\Sigma_ {2 2} ^ {- 1})
$$
For the subtractive term, we can write:
$$
\lambda_ {\min} (\Sigma_ {2 2} ^ {- 1}) = \frac {1}{\lambda_ {\max} (\Sigma_ {2 2})} \geq \frac {1}{\lambda_ {\max} (\Sigma)} = \lambda_ {\min} (\Sigma^ {- 1}).
$$
We therefore have the following upper bound on the KL divergence:
$$
D _ {\mathrm {K L}} \left(p (\boldsymbol {x} _ {1} \mid x _ {2}) \mid p (\boldsymbol {x} _ {1} \mid x _ {2} ^ {\prime})\right) \leq \frac {1}{2} \cdot \| x _ {2} - x _ {2} ^ {\prime} \| _ {2} ^ {2} \cdot \left(\lambda_ {\max } (\Sigma^ {- 1}) - \lambda_ {\min } (\Sigma^ {- 1})\right).
$$
Combining this with Pinsker's inequality, we arrive at the final result:
$$
d _ {T V} \left(p \left(\boldsymbol {x} _ {1} \mid x _ {2}\right), p \left(\boldsymbol {x} _ {1} \mid x _ {2} ^ {\prime}\right)\right) \leq \frac {1}{2} \cdot \| x _ {2} - x _ {2} ^ {\prime} \| _ {2} \cdot \sqrt {\lambda_ {\max } \left(\Sigma^ {- 1}\right) - \lambda_ {\min } \left(\Sigma^ {- 1}\right)}.
$$
This means that we have  $M = \frac{1}{2}\sqrt{\lambda_{\max}(\Sigma^{-1}) - \lambda_{\min}(\Sigma^{-1})}$ .
Notice that this does not imply  $M = 0$  when the features  $\pmb{x}$  are independent: it only captures this for the case of isotropic variance. However, if we assume that  $\pmb{x}$  is standardized, so that each dimension has variance equal to 1,  $M = 0$  is implied for independent features because we have  $\Sigma = I$ .
Proof. The result follows directly from Lemma 2 and Assumption 3:
$$
\begin{array}{l} \left| f \left(x _ {S}\right) - f \left(x _ {S} ^ {\prime}\right) \right| \leq L \cdot \left\| x _ {S} - x _ {S} ^ {\prime} \right\| _ {2} + 2 B \cdot d _ {T V} \left(p \left(x _ {\bar {S}} \mid x _ {S}\right), p \left(x _ {\bar {S}} \mid x _ {S} ^ {\prime}\right)\right) \\ \leq L \cdot \| x _ {S} - x _ {S} ^ {\prime} \| _ {2} + 2 B M \cdot \| x _ {S} - x _ {S} ^ {\prime} \| _ {2} \\ = (L + 2 B M) \cdot \| x _ {S} - x _ {S} ^ {\prime} \| _ {2}. \\ \end{array}
$$
C Proofs: prediction robustness to model perturbations
We now re-state and prove our results from Section 3.2.
$$
| f (x _ {S}) - f ^ {\prime} (x _ {S}) | \leq \| f - f ^ {\prime} \| _ {\infty , \mathcal {X} ^ {\prime}} \cdot Q _ {x _ {S}} (\mathcal {X} ^ {\prime}) + 2 B \cdot \left(1 - Q _ {x _ {S}} (\mathcal {X} ^ {\prime})\right),
$$
where  $Q_{x_S}(\mathcal{X}')$  is the probability of imputed samples lying in  $\mathcal{X}'$  based on the distribution  $q(\pmb{x}_{\bar{S}})$ :
Proof. The main idea here is that when integrating over the difference in function outputs, we can separate the integral over  $(x_S, x_{\bar{S}})$  values falling in the domains  $\mathcal{X}'$  and  $\mathbb{R}^d \setminus \mathcal{X}'$ . Because we have a single value for the observed variables  $x_S$ , the distribution  $q(x_{\bar{S}})$  is identical for both terms regardless of the removal technique.
$$
\begin{array}{l} | f (x _ {S}) - f ^ {\prime} (x _ {S}) | = \left| \int f (x _ {S}, x _ {\bar {S}}) q (x _ {\bar {S}}) d x _ {\bar {S}} - \int f ^ {\prime} (x _ {S}, x _ {\bar {S}}) q (x _ {\bar {S}}) d x _ {\bar {S}} \right| \\ \leq \int | f (x _ {S}, x _ {\bar {S}}) - f ^ {\prime} (x _ {S}, x _ {\bar {S}}) | \cdot q (x _ {\bar {S}}) d x _ {\bar {S}} \\ = \int \left| f \left(x _ {S}, x _ {\bar {S}}\right) - f ^ {\prime} \left(x _ {S}, x _ {\bar {S}}\right) \right| \cdot \mathbb {I} \left\{\left(x _ {S}, \boldsymbol {x} _ {\bar {S}}\right) \in \mathcal {X} ^ {\prime} \right\} \cdot q \left(x _ {\bar {S}}\right) d x _ {\bar {S}} \\ + \int | f (x _ {S}, x _ {\bar {S}}) - f ^ {\prime} (x _ {S}, x _ {\bar {S}}) | \cdot \mathbb {I} \left\{\left(x _ {S}, \boldsymbol {x} _ {\bar {S}}\right) \notin \mathcal {X} ^ {\prime} \right\} \cdot q (x _ {\bar {S}}) d x _ {\bar {S}} \\ \leq \| f - f ^ {\prime} \| _ {\infty , \mathcal {X} ^ {\prime}} \int \mathbb {I} \left\{\left(x _ {S}, \boldsymbol {x} _ {\bar {S}}\right) \in \mathcal {X} ^ {\prime} \right\} \cdot q \left(x _ {\bar {S}}\right) d x _ {\bar {S}} \\ + 2 B \cdot \int \mathbb {I} \left\{\left(x _ {S}, \boldsymbol {x} _ {\bar {S}}\right) \notin \mathcal {X} ^ {\prime} \right\} \cdot q \left(x _ {\bar {S}}\right) d x _ {\bar {S}} \\ = \| f - f ^ {\prime} \| _ {\infty , \mathcal {X} ^ {\prime}} \cdot Q _ {x _ {S}} \left(\mathcal {X} ^ {\prime}\right) + 2 B \cdot \left(1 - Q _ {x _ {S}} \left(\mathcal {X} ^ {\prime}\right)\right) \\ \end{array}
$$
Proof. The result follows directly from Lemma 4 when we set  $\mathcal{X}' = \mathcal{X}$ . Note that when we remove features using the conditional distribution, we have  $Q_{x_S}(\mathcal{X}) = 1$  for all  $x_S$  lying on the data manifold, or where there exists  $x_{\bar{S}}$  such that  $p(x_S,x_{\bar{S}}) > 0$ .
$$
| f (x _ {S}) - f ^ {\prime} (x _ {S}) | \leq \| f - f ^ {\prime} \| _ {\infty}.
$$
Proof. The result follows directly from Lemma 4 when we set  $\mathcal{X}' = \mathbb{R}^d$ , because we have  $Q_{x_S}(\mathbb{R}^d) = 1$  for all  $x_S \in \mathbb{R}^{|S|}$ .
D Proofs: summary technique robustness
We now re-state and prove our results from Section 3.3.
Proof. This result follows directly from the definition of each method. Following the approach of Covert et al. [17], we disentangle between each method's feature removal implementation and how it generates attribution scores. Assuming a fixed input  $x$  where  $f(x_{S})$  denotes the prediction with a feature set, each method's attributions are defined as follows.
- Occlusion [67]: this method simply compares the prediction given all features and the prediction with a single missing feature. The attribution score  $\phi_i(f, x)$  is defined as
$$
\phi_ {i} (f, x) = f (x) - f \left(x _ {[ d ] \backslash \{i \}}\right).
$$
We refer to this summary technique as "leave-one-out" in Table 3.
- Shapley value [45, 54]: this method calculates the impact of removing a single feature, and then averages this over all possible preceding subsets. The attribution score  $\phi_i(f,x)$  is defined as
$$
\phi_ {i} (f, x) = \frac {1}{d} \sum_ {S \subseteq [ d ] \setminus \{i \}} \binom {d - 1} {| S |} ^ {- 1} \Big (f (x _ {S \cup \{i \}}) - f (x _ {S}) \Big).
$$
- Banzhaf value [9, 13]: this method is similar to the Shapley value, but it applies a uniform weighting when averaging over all preceding subsets. The attribution score  $\phi_i(f,x)$  is defined as
$$
\phi_ {i} (f, x) = \frac {1}{2 ^ {d - 1}} \sum_ {S \subseteq [ d ] \backslash \{i \}} \left(f (x _ {S \cup \{i \}}) - f (x _ {S})\right).
$$
- RISE [51]: this method is similar to the Banzhaf value, but it does not subtract the value for subsets where a feature is not included. The attribution score  $\phi_i(f,x)$  is defined as
$$
\phi_ {i} (f, x) = \frac {1}{2 ^ {d - 1}} \sum_ {S \subseteq [ d ] \backslash \{i \}} f (x _ {S \cup \{i \}}).
$$
We refer to this summary technique as "mean when included" in Table 3
- LIME [53]: this method is more flexible than the others and generally involves solving a regression problem that treats the features as inputs and the predictions as labels. LIME can be viewed as a removal-based explanation when the features are represented in a binary fashion, indicating whether they are provided to the model or not [17]. There is also a choice of weighting kernel  $w(S) \geq 0$ , of a regularization term, and of whether to fit an intercept term.
We assume that the regularization is omitted and that we fit an intercept term. Denoting the model's coefficients as  $(\beta_0, \dots, \beta_d)$ , the problem is the following:
$$
\min  _ {\beta_ {0}, \dots , \beta_ {d}} \sum_ {S \subseteq [ d ]} w (S) \left(\beta_ {0} + \sum_ {i \in S} \beta_ {i} - f (x _ {S})\right) ^ {2}.
$$
We can rewrite this problem in a simpler fashion to derive its solution. First, we let  $W \in [0,1]^{2^d \times 2^d}$  be a diagonal matrix containing the weights  $w(S)$  for each subset. Next, we let
the predictions be represented by a vector  $v \in \mathbb{R}^{2^d}$  where  $v_{S} = f(x_{S})$ , and we let the matrix  $Z \in \{0,1\}^{2^d \times d}$  enumerate all subsets in a binary fashion. Finally, to accommodate the intercept term, we let  $\beta = (\beta_0, \dots, \beta_d) \in \mathbb{R}^{d+1}$  represent the parameters, and we prepend a columns of ones to  $Z$ , resulting in  $Z' \in \{0,1\}^{2^d \times (d+1)}$ . The problem then becomes:
$$
\min  _ {\beta} \left(Z ^ {\prime} \beta - v\right) ^ {\top} W \left(Z ^ {\prime} \beta - v ^ {\prime}\right) = \left\| Z ^ {\prime} \beta - v \right\| _ {W} ^ {2}.
$$
Assuming that  $Z'^\top WZ'$  is invertible, which is guaranteed when  $w(S) > 0$  for all  $S \subseteq [d]$ , this problem has a closed-form solution:
$$
\beta^ {*} = (Z ^ {\prime \top} W Z ^ {\prime}) ^ {- 1} Z ^ {\prime \top} W v.
$$
LIME's attribution scores are given by  $\phi_i(f,x) = \beta_i^*$ , for  $i = 1,\dots,d$ , where we discard the intercept term. We can therefore conclude that each attribution is a linear function of  $v$ , with coefficients given by all but the first row of  $(Z^{\prime \top}WZ^{\prime})^{-1}Z^{\prime \top}W$ . If we denote all but the first row of  $(Z^{\prime \top}WZ^{\prime})^{-1}$  as  $(Z^{\prime \top}WZ^{\prime})_{[1:]}^{-1}$ , then we have  $A = (Z^{\prime \top}WZ^{\prime})_{[1:]}^{-1}Z^{\prime \top}W$ .
The formula for the attributions remains roughly the same if we do not fit an intercept term: if the intercept is zero, we have  $A = (Z^{\top}WZ)^{-1}Z^{\top}W$ . We can similarly allow for ridge regularization: given a penalty parameter  $\lambda > 0$ , we then have  $A = (Z^{\top}WZ + \lambda I_{2^d})^{-1}Z^{\top}W$ .
These coefficients depend on the specific choice of weighting kernel  $w(S)$ , and previous work has shown that leave-one-out, Shapley and Banzhaf values all correspond to specific weighting kernels [17]. Our analysis of those methods therefore pertains to LIME, at least in certain special cases. However, LIME often uses a heuristic exponential kernel in practice, which is more difficult to characterize analytically. We therefore omit the specific entries and norms for LIME in Table 3, but show norms for the exponential kernel in our computed results.
We refer to this summary technique as "weighted least squares" in Table 3.
We can therefore see that in each method,  $\phi_i(f,x)$  is a linear function of the predictions for each  $S\subseteq [d]$ . Across all methods, we can let the predictions be represented by a vector  $v\in \mathbb{R}^{2^d}$  and the coefficients by a matrix  $A\in \mathbb{R}^{d\times 2^d}$ .
$$
\| A v - A v ^ {\prime} \| _ {2} \leq \| A \| _ {2} \cdot \| v - v ^ {\prime} \| _ {2} \quad o r \quad \| A v - A v ^ {\prime} \| _ {2} \leq \| A \| _ {1, \infty} \cdot \| v - v ^ {\prime} \| _ {\infty},
$$
Proof. We first derive the two matrix inequalities. The first inequality follows from the definition of the spectral norm, which is defined as the maximum singular value:
$$
\| A \| _ {2} \equiv \sup  _ {u \neq 0} \frac {\| A u \| _ {2}}{\| u \| _ {2}} = \sigma_ {\max } (A) = \sqrt {\lambda_ {\max } (A A ^ {\top})}.
$$
We prove the second bound using Holder's inequality as follows, where  $A_{i}$  denotes the ith row:
$$
\| A u \| _ {2} = \sqrt {\sum_ {i = 1} ^ {d} \left(A _ {i} ^ {\top} u\right) ^ {2}} \leq \sqrt {\sum_ {i = 1} ^ {d} \| A _ {i} \| _ {1} ^ {2} \cdot \| u \| _ {\infty} ^ {2}} = \| u \| _ {\infty} \cdot \sqrt {\sum_ {i = 1} ^ {d} \| A _ {i} \| _ {1} ^ {2}} = \| u \| _ {\infty} \cdot \| A \| _ {1, \infty}.
$$
In the special case where all of  $A$ 's row 1-norms are equal, or where we have  $\| A_i \|_1 = \| A_j \|_1$  for all  $i, j \in [d]$ , we can use the simplified expression  $\| A \|_{1,\infty} = \| A_1 \|_1 \cdot \sqrt{d}$ .
Next, we derive the specific norm values for each linear operator in Table 3.
Beginning with the operator norm  $\| A \|_{1,\infty}$ , we remark that several methods have equal 1-norms across all rows: following their interpretation as semivalues [46], leave-one-out, Shapley and Banzhaf all have  $\| A_i \|_1 = 2$  for all  $i \in [d]$ . Similarly, RISE has  $\| A_i \|_1 = 1$  for all  $i \in [d]$ . This yields the norms  $\| A \|_{1,\infty} = 2\sqrt{d}$  for leave-one-out, Banzhaf and Shapley, and  $\| A \|_{1,\infty} = \sqrt{d}$  for RISE.
For the spectral norm, we consider each method separately. We find that for all methods, it is simplest to calculate  $AA^{\top}$  and then  $\lambda_{\max}(AA^{\top})$  to find the maximum singular value  $\sigma_{\max}(A)$ . Note that the entries of the matrix are determined by the inner product  $(AA^{\top})_{ij} = A_i^{\top}A_j$ .
For leave-one-out, we derive  $AA^{\top}$  as follows:
$$
(A A ^ {\top}) _ {i j} = \left\{ \begin{array}{l l} 2 & \text {i f} i = j \\ 1 & \text {o t h e r w i s e .} \end{array} \right.
$$
The matrix  $AA^{\top}$  is therefore the sum of a rank-one component and an identity term, and we have  $AA^{\top} = \mathbf{1}_d\mathbf{1}_d + I_d$ . Recognizing that  $\mathbf{1}_d$  is the leading eigenvector, we get the following result:
$$
\sigma_ {\max } (A) = \sqrt {\lambda_ {\max } \left(A A ^ {\top}\right)} = \sqrt {\frac {(\mathbf {1} _ {d} \mathbf {1} _ {d} + I _ {d}) \mathbf {1} _ {d}}{\| \mathbf {1} _ {d} \| _ {2}}} = \sqrt {\frac {\mathbf {1} _ {d} d + \mathbf {1} _ {d}}{\| \mathbf {1} _ {d} \| _ {2}}} = \sqrt {d + 1}.
$$
For the Banzhaf value, we derive  $AA^{\top}$  as follows:
$$
(A A ^ {\top}) _ {i j} = \left\{ \begin{array}{l l} 1 / 2 ^ {2 d - 2} & \text {i f} i = j \\ 0 & \text {o t h e r w i s e .} \end{array} \right.
$$
The max eigenvalue is therefore  $\lambda_{\max}(AA^{\top}) = 1 / 2^{d - 2}$ , which yields  $\| A\| _2 = 1 / 2^{d / 2 - 1}$ .
For the mean when included technique used by RISE, we derive  $AA^{\top}$  as follows:
$$
(A A ^ {\top}) _ {i j} = \left\{ \begin{array}{l l} 1 / 2 ^ {d - 1} & \text {i f} i = j \\ 1 / 2 ^ {d} & \text {o t h e r w i s e .} \end{array} \right.
$$
The matrix  $AA^{\top}$  is therefore the sum of a rank-one component and an identity term, or we have  $AA^{\top} = \mathbf{1}_d\mathbf{1}_d / 2^d +I_d / 2^d$ . Recognizing that  $\mathbf{1}_d$  is the leading eigenvector, we get the following result:
$$
\sigma_ {\max} (A) = \sqrt {\lambda_ {\max} (A A ^ {\top})} = \sqrt {\frac {(\mathbf {1} _ {d} \mathbf {1} _ {d} / 2 ^ {d} + I _ {d} / 2 ^ {d}) \mathbf {1} _ {d}}{\| \mathbf {1} _ {d} \| _ {2}}} = \sqrt {\frac {\mathbf {1} _ {d} d / 2 ^ {d} + \mathbf {1} _ {d} / 2 ^ {d}}{\| \mathbf {1} _ {d} \| _ {2}}} = \sqrt {\frac {d + 1}{2 ^ {d}}}.
$$
For the Shapley value, we derive  $AA^{\top}$  by first considering the diagonal entries:
$$
\begin{array}{l} (A A ^ {\top}) _ {i i} = A _ {i} ^ {\top} A _ {i} = \sum_ {S: i \in S} \left(\frac {(| S | - 1) ! (d - | S |) !}{d !}\right) ^ {2} + \sum_ {S: i \notin S} \left(\frac {| S | ! (d - | S | - 1) !}{d !}\right) ^ {2} \\ = \sum_ {k = 0} ^ {d} \left[ \binom {d - 1} {k - 1} \left(\frac {(k - 1) ! (d - k) !}{d !}\right) ^ {2} + \binom {d - 1} {k} \left(\frac {k ! (d - k - 1) !}{d !}\right) ^ {2} \right] \\ = \frac {1}{d} \sum_ {k = 1} ^ {d} \frac {(k - 1) ! (d - k) !}{d !} + \frac {1}{d} \sum_ {k = 0} ^ {d - 1} \frac {k ! (d - k - 1) !}{d !} \\ = \sum_ {k = 1} ^ {d - 1} \frac {(k - 1) ! (d - k - 1) !}{d !} + \underbrace {\frac {2}{d ^ {2}}} _ {k = 0, k = d} \\ = \frac {1}{d (d - 1)} \sum_ {k = 1} ^ {d - 1} \binom {d - 2} {k - 1} ^ {- 1} + \frac {2}{d ^ {2}} \\ \end{array}
$$
We next consider the off-diagonal entries:
$$
\begin{array}{l} (AA^{\top})_{ij} = A_{i}^{\top}A_{j} = \sum_{S:i\in S,j\in S}\frac{(|S| - 1)!(d - |S|)!}{d!}\frac{(|S| - 1)!(d - |S|)!}{d!} \\ - \sum_ {S: i \in S, j \notin S} \frac {(| S | - 1) ! (d - | S |) !}{d !} \frac {| S | ! (d - | S | - 1) !}{d !} \\ - \sum_ {S: i \notin S, j \in S} \frac {| S | ! (d - | S | - 1) !}{d !} \frac {(| S | - 1) ! (d - | S |) !}{d !} \\ + \sum_ {S: i \notin S, j \notin S} \frac {| S | ! (d - | S | - 1) !}{d !} \frac {| S | ! (d - | S | - 1) !}{d !} \\ = \sum_ {k = 0} ^ {d} \binom {d - 2} {k - 2} \frac {(k - 1) ! (d - k) !}{d !} \frac {(k - 1) ! (d - k) !}{d !} \\ - 2 \sum_ {k = 0} ^ {d} \binom {d - 2} {k - 1} \frac {(k - 1) ! (d - k) !}{d !} \frac {k ! (d - k - 1) !}{d !} \\ + \sum_ {k = 0} ^ {d} \binom {d - 2} {k} \frac {k ! (d - k - 1) !}{d !} \frac {k ! (d - k - 1) !}{d !} \\ = \sum_ {k = 2} ^ {d} \frac {(k - 1)}{d (d - 1)} \frac {(k - 1) ! (d - k) !}{d !} \\ - 2 \sum_ {k = 1} ^ {d - 1} \frac {1}{d (d - 1)} \frac {k ! (d - k) !}{d !} \\ + \sum_ {k = 0} ^ {d - 2} \frac {(d - k - 1)}{d (d - 1)} \frac {k ! (d - k - 1) !}{d !} \\ \end{array}
$$
We first focus on the case where  $d > 2$ , and we later return to the case where  $d = 2$ . We can group terms that share the same  $k$  values to simplify the above:
$$
\begin{array}{l} A _ {i} ^ {\top} A _ {j} = \frac {1}{d (d - 1)} \left[ - \sum_ {k = 2} ^ {d - 2} \frac {(k - 1) ! (d - k - 1) !}{(d - 1) !} + \underbrace {\frac {2 (d - 1)}{d}} _ {k = 0, k = d} - \underbrace {\frac {2}{d - 1}} _ {k = 1, k = d - 1} \right] \\ = - \frac {1}{d (d - 1)} \sum_ {k = 2} ^ {d - 2} \frac {(k - 1) ! (d - k - 1) !}{(d - 1) !} + \frac {2}{d ^ {2}} - \frac {2}{d (d - 1) ^ {2}} \\ = - \frac {1}{d (d - 1) ^ {2}} \sum_ {k = 2} ^ {d - 2} \binom {d - 2} {k - 1} ^ {- 1} + \frac {2}{d ^ {2}} - \frac {2}{d (d - 1) ^ {2}}. \\ \end{array}
$$
By convention, we let the summation above evaluate to zero for  $d = 3$ . We therefore have the following result for  $AA^{\top}$ :
$$
(A A ^ {\top}) _ {i j} = \left\{ \begin{array}{l l} \frac {1}{d (d - 1)} \sum_ {k = 1} ^ {d - 1} \binom {d - 2} {k - 1} ^ {- 1} + \frac {2}{d ^ {2}} & \text {i f} i = j \\ - \frac {1}{d (d - 1) ^ {2}} \sum_ {k = 2} ^ {d - 2} \binom {d - 2} {k - 1} ^ {- 1} + \frac {2}{d ^ {2}} - \frac {2}{d (d - 1) ^ {2}} & \text {o t h e r w i s e}. \end{array} \right.
$$
From this, we can see that  $AA^{\top}$  is the sum of a rank-one component and an identity term, or that we have  $AA^{\top} = \mathbf{1}_{d}\mathbf{1}_{d}b + I_{d}(c - b)$  with values for  $b = (AA^{\top})_{ij}$  and  $c = (AA^{\top})_{ii}$  given above. This matrix has two eigenvalues,  $c - b$  and  $db + (c - b)$ , and we must consider which of the two is larger. Under our assumption of  $d > 2$ , we can observe that  $b > 0$ :
$$
\begin{array}{l} b = (A A ^ {\top}) _ {i j} = \frac {2}{d ^ {2}} - \frac {1}{d (d - 1) ^ {2}} \sum_ {k = 2} ^ {d - 2} \binom {d - 2} {k - 1} ^ {- 1} - \frac {2}{d (d - 1) ^ {2}} \\ \geq \frac {2}{d ^ {2}} - \frac {1}{d (d - 1) ^ {2}} \sum_ {k = 2} ^ {d - 2} \binom {d - 2} {1} ^ {- 1} - \frac {2}{d (d - 1) ^ {2}} \\ = \frac {2}{d ^ {2}} - \frac {d - 3}{d (d - 1) ^ {2} (d - 2)} - \frac {2}{d (d - 1) ^ {2}} \\ = \frac {2 (d - 1) ^ {2} (d - 2) - d (d - 3) - 2 d (d - 2)}{d ^ {2} (d - 1) ^ {2} (d - 2)} \\ \end{array}
$$
It can be verified that the numerator of the above fraction is positive for  $d > 2$ . This implies that  $db + (c - b)$  is the larger eigenvalue. We can therefore calculate  $\lambda_{\max}(AA^{\top})$  as follows:
$$
\begin{array}{l} \lambda_ {\max } \left(A A ^ {\top}\right) = d b + (c - b) = c + (d - 1) b \\ = \frac {2}{d ^ {2}} + \underbrace {\frac {2}{d (d - 1)}} _ {k = 1, k = d - 1} + \frac {2 (d - 1)}{d ^ {2}} - \frac {2}{d (d - 1)} \\ = \frac {2 (d - 1) + 2 (d - 1) ^ {2}}{d ^ {2} (d - 1)} \\ = \frac {2}{d} \\ \end{array}
$$
On the other hand, when we have  $d = 2$  we have the following entries for  $AA^{\top}$ :
$$
(A A ^ {\top}) _ {i j} = \left\{ \begin{array}{l l} 1 & \text {i f} i = j \\ 0 & \text {o t h e r w i s e .} \end{array} \right.
$$
This yields  $\lambda_{\max}(AA^{\top}) = 1$ , which coincides with the formula  $\lambda_{\max}(AA^{\top}) = 2 / d$  derived above. Finally, this let us conclude that the spectral norm for the Shapley value is the following:
$$
\left\| A \right\| _ {2} = \sqrt {\lambda_ {\operatorname* {m a x}} \left(A A ^ {\top}\right)} = \sqrt {\frac {2}{d}}.
$$
Proof. Following Theorem 1 in Monderer and Samet [46], solution concepts satisfying linearity and the boundedness property (also known as the Milnor axiom) are probabilistic values. This means that each attribution is the average of a feature's marginal contributions, or that for each feature  $i \in [d]$  there exists a distribution  $p_i(S)$  with support on the power set of  $[d] \setminus \{i\}$ , such that the attributions are given by
$$
\phi_ {i} (f, x) = \mathbb {E} _ {p _ {i} (S)} \left[ f \left(x _ {S \cup \{i \}}\right) - f \left(x _ {S}\right) \right] = \sum_ {S \subseteq [ d ] \backslash \{i \}} p _ {i} (S) \left(f \left(x _ {S \cup \{i \}}\right) - f \left(x _ {S}\right)\right). \tag {D.0.1}
$$
The corresponding matrix  $A \in \mathbb{R}^{d \times 2^d}$  therefore has rows  $A_i$  with entries  $A_{iS} \geq 0$  when  $i \in S$  and  $A_{iS} \leq 0$  when  $i \notin S$ , where the positive and negative entries sum to 1 and -1 respectively. We can therefore conclude that each row satisfies  $\| A_i \|_1 = 2$ , and that  $\| A \|_{1,\infty} = 2\sqrt{d}$ .
Proof. Following Monderer and Samet [46], solution concepts satisfying linearity, boundedness and symmetry are semvalues. This means that they are probabilistic values for which the probability distribution  $p_i(S)$  depends only on the cardinality  $|S|$  (see the definition of probabilistic values above). As a result, we can parameterize the summary technique by how much weight it places on each cardinality.
Let  $\alpha \in \mathbb{R}^d$  denote a probability distribution over the cardinalities  $k = 1, \ldots, d$ . We can write the attributions generated by any semvalue as follows:
$$
\phi_{i}(f,x) = \sum_{k = 1}^{d}\alpha_{k}\sum_{\substack{S\subseteq [d]\setminus \{i\} \\ |S| = k - 1}}\binom {d - 1}{k - 1}^{-1}\Big(f(x_{S\cup \{i\}}) - f(x_{S})\Big).
$$