SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \left(\iota_ {P} \star u ^ {[ k + 1 ]}\right) (S) \\ = \sum_ {[ k + 1 ] \supseteq T \supseteq S} \iota_ {P} (t - s) u ^ {[ k + 1 ]} (T) \\ = \sum_ {[ k ] \supseteq T \supseteq S} \iota_ {P} (t - s) u ^ {[ k + 1 ]} (T) + \sum_ {[ k ] \supseteq T \supseteq S} \iota_ {P} (t - s) u ^ {[ k + 1 ]} (T) \\ = \sum_ {[ k ] \supseteq T \supseteq S} \iota_ {P} (t - s) u _ {[ k ]} ^ {[ k + 1 ]} (T) + \sum_ {[ k ] \supseteq T \supseteq S} \iota_ {P} (t - s) u ^ {[ k + 1 ]} (T) (byEq.15) \\ = \left[ \mathbf {T} _ {P} ^ {[ k ]} u _ {[ k ]} ^ {[ k + 1 ]} (S) + \sum_ {[ k ] \nsubseteq T \supseteq S} \iota_ {P} (t - s) u ^ {[ k + 1 ]} (T) \right. (22) \\ = \left[ \mathbf {T} _ {P} ^ {[ k ]} u _ {[ k ]} ^ {[ k + 1 ]} \right] (S) + \sum_ {[ k ] \supseteq R \supseteq S} \iota_ {P} (r + 1 - s) u _ {[ k ]} ^ {[ k + 1 ], \cup \{k + 1 \}} (R) \quad \text {b y} R = T \backslash \{k + 1 \} \text {a n d E q .} (16) \\ = \left[ \mathbf {T} _ {P} ^ {[ k ]} u _ {[ k ]} ^ {[ k + 1 ]} \right] (S) + \left[ \left(\int_ {0} ^ {1} t \cdot \mathbf {A} ^ {[ k ]} (t) \mathrm {d} P (t)\right) u _ {[ k ]} ^ {[ k + 1 ], \cup \{k + 1 \}} \right] (S) \quad \text {b y E q s .} (2 0), (2 1) \\ = \left[ \mathbf {T} _ {P} ^ {[ k + 1 ]} u ^ {[ k + 1 ]} \right] (S) \quad \text {b y E q .} (13) \\ \end{array}
$$
The remaining case can be done in a similar fashion. After all, the matrix representation of  $\iota_{P}$  on  $\mathcal{G}_n$  w.r.t. the standard basis is
$$
\begin{array}{l} \mathbf {M} _ {P} ^ {[ n ]} = \mathbf {T} _ {P} ^ {[ n ]} \times \mathbf {W} ^ {[ n ]} \\ = \int_ {0} ^ {1} \overbrace {\mathbf {A} ^ {[ 1 ]} (t) \otimes \mathbf {A} ^ {[ 1 ]} (t) \otimes \dots \otimes \mathbf {A} ^ {[ 1 ]} (t)} ^ {n \text {r e p e t i t i o n s}} \mathrm {d} P (t) \times (\overbrace {\mathbf {W} ^ {[ 1 ]} \otimes \mathbf {W} ^ {[ 1 ]} \otimes \dots \otimes \mathbf {W} ^ {[ 1 ]}} ^ {n \text {r e p e t i t i o n s}}) \\ = \int_ {0} ^ {1} \left(\mathbf {A} ^ {[ 1 ]} (t) \otimes \mathbf {A} ^ {[ 1 ]} (t) \otimes \dots \otimes \mathbf {A} ^ {[ 1 ]} (t)\right) \times \left(\mathbf {W} ^ {[ 1 ]} \otimes \mathbf {W} ^ {[ 1 ]} \otimes \dots \otimes \mathbf {W} ^ {[ 1 ]}\right) d P (t) \\ = \int_ {0} ^ {1} \mathbf {B} ^ {[ 1 ]} (t) \otimes \mathbf {B} ^ {[ 1 ]} (t) \otimes \dots \otimes \mathbf {B} ^ {[ 1 ]} (t) d P (t) \\ \end{array}
$$
$$
\mathbf {B} ^ {[ 1 ]} (t) = \mathbf {A} ^ {[ 1 ]} (t) \times \mathbf {W} ^ {[ 1 ]}.
$$
C.4 A Theorem of Robustness for Semi-indices
Definition 5 (Truncated Semi-indices). Given a semi-index  $\mathcal{I}_P$  and a non-empty collection of subsets  $S\subseteq 2^{[n]}$ , the truncated semi-index  $\mathcal{I}_{P,S}$  is obtained by only retaining the values of  $\mathcal{I}_P$  indexed by  $S$ .
Theorem 2. Under Assumption 1, and further assume  $\sigma_{12} < \min(\sigma_{11}, \sigma_{22})$ , for every non-empty collection of subsets  $S \subseteq 2^{[n]}$ , the Dirac delta distribution  $\delta_{w^*}$  with
$$
w ^ {*} = \frac {\sigma_ {1 1} - \sigma_ {1 2}}{\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}}
$$
is optimal to the optimization problem
$$
\underset {P \in \mathcal {P}} {\operatorname {a r g m i n}} \det  \left(\operatorname {C o v} \left(\mathcal {I} _ {P, \mathcal {S}} (v _ {n})\right)\right)
$$
$$
s. t. \operatorname {C o v} \left(v _ {n}\right) = \boldsymbol {\Sigma} \otimes \boldsymbol {\Sigma} \otimes \dots \otimes \boldsymbol {\Sigma} (n \text {r e p e t i t i o n s}).
$$
Proof. The matrix representation of  $\mathcal{I}_P$  on  $\mathcal{G}_n$  w.r.t. the standard basis is  $\mathbf{M}_P^{[n]} = \mathbf{T}_P^{[n]} \times \mathbf{W}^{[n]}$ , as presented in Proposition 7. For simplicity, we write  $\mathbf{V}_{[n],P} = \mathrm{Cov}(\mathcal{I}_P(v_n))$ . Since  $\mathcal{I}_P$  is linear in  $v_n$ , there is
$$
\mathbf {V} _ {[ n ], P} = \mathbf {M} _ {P} ^ {[ n ]} \mathbf {C} ^ {[ n ]} \left(\mathbf {M} _ {P} ^ {[ n ]}\right) ^ {\top}. \tag {23}
$$
Note that  $\operatorname*{det}(\mathbf{T}_P^{[n]}) = \operatorname*{det}(\mathbf{W}^{[n]}) = 1$  as they are triangular matrices with all diagonal entries being one, and thus we have
$$
\det  \left(\mathbf {M} _ {P} ^ {[ n ]}\right) = 1, \tag {24}
$$
$$
\det (\mathbf {V} _ {[ n ], P}) = \det (\mathbf {C} ^ {[ n ]}). \tag {25}
$$
Observe that  $\operatorname{Cov}(\mathcal{I}_{P,S}(v_n))$  is just a sub-matrix of  $\mathbf{V}_{[n],P}$ . To represent such sub-matrices, for every non-empty  $\alpha, \beta \subseteq [2^n]$ ,  $\mathbf{V}_{[n],P}^{\alpha,\beta}$  is defined to be the sub-matrix of  $\mathbf{V}_{[n],P}$  whose rows and columns are indexed by  $\alpha$  and  $\beta$ , respectively. Plus, we write  $\mathbf{V}_{[n],P}^{\alpha}$  and  $\mathbf{V}_{[n],P}^{;\alpha}$  instead for  $\mathbf{V}_{[n],P}^{\alpha,\alpha}$  and  $\mathbf{V}_{[n],P}^{[2^n],\alpha}$ , respectively. Therefore, for each  $\mathcal{I}_{P,S}(v_n)$ , there exists a non-empty  $\alpha \subseteq [2^n]$  such that  $\mathbf{V}_{[n],P}^{\alpha} = \operatorname{Cov}(\mathcal{I}_{P,S}(v_n))$ .
We prove by induction that for every  $n > 0$ , it holds that for every non-empty  $\alpha \subseteq [2^n]$  and for every  $P \in \mathcal{P}$ ,  $\operatorname*{det}(\mathbf{V}_{[n], P}^{\alpha}) \geq \operatorname*{det}(\mathbf{V}_{[n], \delta_{w^*}}^{\alpha})$ .
For  $n = 1$ , let  $s = \int_0^1 t\mathrm{d}P(t)$ , there is
$$
\mathbf {V} _ {[ 1 ], P} = \left( \begin{array}{c c} \sigma_ {1 1} (1 - s) ^ {2} + \sigma_ {2 2} s ^ {2} - 2 \sigma_ {1 2} s (1 - s) & (\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}) s + \sigma_ {1 2} - \sigma_ {1 1} \\ (\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}) s + \sigma_ {1 2} - \sigma_ {1 1} & \sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2} \end{array} \right),
$$
by which one can verify the statement by noticing that the function  $s \mapsto \sigma_{11}(1 - s)^2 + \sigma_{22}s^2 - 2\sigma_{12}s(1 - s)$  achieves its minimum at point  $s = w^*$ . Suppose the statement holds for some  $k > 0$ , and we proceed for  $k + 1$ .
Before proceeding with the induction, we have to explore the Kronecker structure between  $\mathbf{V}_{[k + 1],P}$  and  $\mathbf{V}_{[k],P}$ . By Eq. (23), and Proposition 7,
$$
\begin{array}{l} \mathbf {V} _ {[ k + 1 ], P} = \left(\int_ {0} ^ {1} \binom {(1 - t) \mathbf {B} ^ {[ k ]} (t)} {- \mathbf {B} ^ {[ k ]} (t)} \quad t \mathbf {B} ^ {[ k ]} (t) \quad \mathrm {d} P (t)\right) \times \left( \begin{array}{l l} \sigma_ {1 1} \mathbf {C} ^ {[ k ]} & \sigma_ {1 2} \mathbf {C} ^ {[ k ]} \\ \sigma_ {1 2} \mathbf {C} ^ {[ k ]} & \sigma_ {2 2} \mathbf {C} ^ {[ k ]} \end{array} \right) \\ \times \left(\int_ {0} ^ {1} \left( \begin{array}{c c} (1 - t) \mathbf {B} ^ {[ k ]} (t) & t \mathbf {B} ^ {[ k ]} (t) \\ - \mathbf {B} ^ {[ k ]} (t) & \mathbf {B} ^ {[ k ]} (t) \end{array} \right) \mathrm {d} P (t)\right) ^ {\top}. \\ \end{array}
$$
Denote  $\mathbf{L}_P^{[k]} = \int_0^1 t\mathbf{B}^{[k]}(t)\mathrm{d}P(t)$ , Eq. (26) can be rewritten as
$$
\begin{array}{l} \mathbf {V} _ {[ k + 1 ], P} = \left( \begin{array}{c c} \mathbf {M} _ {P} ^ {[ k ]} - \mathbf {L} _ {P} ^ {[ k ]} & \mathbf {L} _ {P} ^ {[ k ]} \\ - \mathbf {M} _ {P} ^ {[ k ]} & \mathbf {M} _ {P} ^ {[ k ]} \end{array} \right) \left( \begin{array}{c c} \sigma_ {1 1} \mathbf {C} ^ {[ k ]} & \sigma_ {1 2} \mathbf {C} ^ {[ k ]} \\ \sigma_ {1 2} \mathbf {C} ^ {[ k ]} & \sigma_ {2 2} \mathbf {C} ^ {[ k ]} \end{array} \right) \left( \begin{array}{c c} \mathbf {M} _ {P} ^ {[ k ]} - \mathbf {L} _ {P} ^ {[ k ]} & \mathbf {L} _ {P} ^ {[ k ]} \\ - \mathbf {M} _ {P} ^ {[ k ]} & \mathbf {M} _ {P} ^ {[ k ]} \end{array} \right) ^ {\top} \tag {27} \\ = \left( \begin{array}{c c} \mathbf {U} & \mathbf {F} ^ {\top} \\ \mathbf {F} & \mathbf {E} \end{array} \right) \\ \end{array}
$$
$$
\begin{array}{l} \mathbf {U} = \sigma_ {1 1} \left(\mathbf {M} _ {P} ^ {[ k ]} - \mathbf {L} _ {P} ^ {[ k ]}\right) \mathbf {C} ^ {[ k ]} \left(\mathbf {M} _ {P} ^ {[ k ]} - \mathbf {L} _ {P} ^ {[ k ]}\right) ^ {\top} + \sigma_ {2 2} \mathbf {L} _ {P} ^ {[ k ]} \mathbf {C} ^ {[ k ]} \left(\mathbf {L} _ {P} ^ {[ k ]}\right) ^ {\top} (28) \\ + \sigma_ {1 2} (\mathbf {M} _ {P} ^ {[ k ]} - \mathbf {L} _ {P} ^ {[ k ]}) \mathbf {C} ^ {[ k ]} (\mathbf {L} _ {P} ^ {[ k ]}) ^ {\top} + \sigma_ {1 2} \mathbf {L} _ {P} ^ {[ k ]} \mathbf {C} ^ {[ k ]} (\mathbf {M} _ {P} ^ {[ k ]} - \mathbf {L} _ {P} ^ {[ k ]}) ^ {\top}, \\ \mathbf {F} = \mathbf {M} _ {P} ^ {[ k ]} \mathbf {C} ^ {[ k ]} \mathbf {K}, (29) \\ \end{array}
$$
$$
\mathbf {K} = \left(\sigma_ {1 2} - \sigma_ {1 1}\right) \left(\mathbf {M} _ {P} ^ {[ k ]} - \mathbf {L} _ {P} ^ {[ k ]}\right) ^ {\top} + \left(\sigma_ {2 2} - \sigma_ {1 2}\right) \left(\mathbf {L} _ {P} ^ {[ k ]}\right) ^ {\top}, \tag {30}
$$
$$
\mathbf {E} = \left(\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}\right) \mathbf {V} _ {[ k ], P}. \tag {31}
$$
Our goal is to simplify the expression
$$
\mathbf {S} = \mathbf {U} - \mathbf {F} ^ {\top} \mathbf {E} ^ {- 1} \mathbf {F}.
$$
Note that  $\mathbf{E}$  and  $\mathbf{M}_P^{[k]}$  are invertible by Eqs. (24) and (25) and that  $\pmb{\Sigma}$  is assumed to be positive definite (and thus  $\operatorname*{det}(\mathbf{C}^{[n]}) > 0$ ). Observe that  $\sigma_{11} + \sigma_{22} - 2\sigma_{12} > 0$  by assumption, using Eqs. (23) and (31), Eq. (29) is equal to
$$
\mathbf {F} = \mathbf {E} \left(\frac {1}{\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}} ((\mathbf {M} _ {P} ^ {[ k ]}) ^ {\top}) ^ {- 1} \mathbf {K}\right),
$$
$$
\begin{array}{l} \mathbf {F} ^ {\top} \mathbf {E} ^ {- 1} \mathbf {F} \\ = \left(\frac {1}{\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}} \mathbf {K} ^ {\top} \left(\mathbf {M} _ {P} ^ {[ k ]}\right) ^ {- 1}\right) \mathbf {E E} ^ {- 1} \mathbf {E} \left(\frac {1}{\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}} \left(\left(\mathbf {M} _ {P} ^ {[ k ]}\right) ^ {\top}\right) ^ {- 1} \mathbf {K}\right) \\ = \frac {1}{\left(\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}\right) ^ {2}} \mathbf {K} ^ {\top} \left(\mathbf {M} _ {P} ^ {[ k ]}\right) ^ {- 1} \mathbf {E} \left(\left(\mathbf {M} _ {P} ^ {[ k ]}\right) ^ {\top}\right) ^ {- 1} \mathbf {K} \\ = \frac {1}{\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}} \mathbf {K} ^ {\top} \mathbf {C} ^ {[ k ]} \mathbf {K}. \\ \end{array}
$$
The last equality is obtained by using Eqs. (23) and (31) again. Write  $\mathbf{R} = \mathbf{M}_P^{[k]} - \mathbf{L}_P^{[k]}$ . From Eqs. (28) and (30), there is
$$
\begin{array}{l} \mathbf {U} = \sigma_ {1 1} \mathbf {R C} ^ {[ k ]} \mathbf {R} ^ {\top} + \sigma_ {2 2} \mathbf {L} _ {P} ^ {[ k ]} \mathbf {C} ^ {[ k ]} (\mathbf {L} _ {P} ^ {[ k ]}) ^ {\top} + \sigma_ {1 2} (\mathbf {R C} ^ {[ k ]} (\mathbf {L} _ {P} ^ {[ k ]}) ^ {\top} + \mathbf {L} _ {P} ^ {[ k ]} \mathbf {C} ^ {[ k ]} \mathbf {R} ^ {\top}), \\ \mathbf {K} ^ {\top} \mathbf {C} ^ {[ k ]} \mathbf {K} = (\sigma_ {1 2} - \sigma_ {1 1}) ^ {2} \mathbf {R C} ^ {[ k ]} \mathbf {R} ^ {\top} + (\sigma_ {2 2} - \sigma_ {1 2}) ^ {2} \mathbf {L} _ {P} ^ {[ k ]} \mathbf {C} ^ {[ k ]} (\mathbf {L} _ {P} ^ {[ k ]}) ^ {\top} \\ + (\sigma_ {1 2} - \sigma_ {1 1}) (\sigma_ {2 2} - \sigma_ {1 2}) (\mathbf {R C} ^ {[ k ]} (\mathbf {L} _ {P} ^ {[ k ]}) ^ {\top} + \mathbf {L} _ {P} ^ {[ k ]} \mathbf {C} ^ {[ k ]} \mathbf {R} ^ {\top}). \\ \end{array}
$$
$$
\begin{array}{l} \left(\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}\right) \mathbf {U} - \mathbf {K} ^ {\top} \mathbf {C} ^ {[ k ]} \mathbf {K} \\ = \left(\sigma_ {1 1} \sigma_ {2 2} - \sigma_ {1 2} ^ {2}\right) \left(\mathbf {R C} ^ {[ k ]} \mathbf {R} ^ {\top} + \mathbf {L} _ {P} ^ {[ k ]} \mathbf {C} ^ {[ k ]} \left(\mathbf {L} _ {P} ^ {[ k ]}\right) ^ {\top} + \mathbf {R C} ^ {[ k ]} \left(\mathbf {L} _ {P} ^ {[ k ]}\right) ^ {\top} + \mathbf {L} _ {P} ^ {[ k ]} \mathbf {C} ^ {[ k ]} \mathbf {R} ^ {\top}\right) \\ = \left(\sigma_ {1 1} \sigma_ {2 2} - \sigma_ {1 2} ^ {2}\right) \left(\left(\mathbf {R} + \mathbf {L} _ {P} ^ {[ k ]}\right) \mathbf {C} ^ {[ k ]} \left(\mathbf {R} ^ {\top} + \left(\mathbf {L} _ {P} ^ {[ k ]}\right) ^ {\top}\right)\right) \tag {32} \\ = (\sigma_ {1 1} \sigma_ {2 2} - \sigma_ {1 2} ^ {2}) \mathbf {M} _ {P} ^ {[ k ]} \mathbf {C} ^ {[ k ]} (\mathbf {M} _ {P} ^ {[ k ]}) ^ {\top}. \\ \end{array}
$$
Combining Eqs. (32) and (23), we eventually get the simplified formula for  $\mathbf{S}$
$$
\mathbf {S} = \frac {\sigma_ {1 1} \sigma_ {2 2} - \sigma_ {1 2} ^ {2}}{\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}} \mathbf {V} _ {[ k ], P}. \tag {33}
$$
Particularly,  $\sigma_{11}\sigma_{22} - \sigma_{12}^2 > 0$  as  $\pmb{\Sigma}$  is assumed to be positive definite. So far so good, we are ready to complete the induction by cases while looking into Eq. (27).
Case 1: If  $[2^{k + 1}]\backslash [2^k ]\subseteq \alpha \subseteq [2^{k + 1}]$ , let  $\beta = \alpha \backslash ([2^{k + 1}]\backslash [2^k])$ , which could be empty. Then, there is
$$
\mathbf {V} _ {[ k + 1 ], P} ^ {\alpha} = \left( \begin{array}{c c} \mathbf {U} ^ {\beta} & (\mathbf {F} ^ {:, \beta}) ^ {\top} \\ \mathbf {F} ^ {:, \beta} & (\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}) \mathbf {V} _ {[ k ], P} \end{array} \right).
$$
Note that we can write
$$
\mathbf {F} ^ {:, \beta} = \mathbf {F D} _ {\beta} \tag {34}
$$
where  $\mathbf{D}_{\beta}$  is the corresponding column-selecting matrix. It also gives that
$$
\mathbf {D} _ {\beta} ^ {\top} \mathbf {U} \mathbf {D} _ {\beta} = \mathbf {U} ^ {\beta}. \tag {35}
$$
Using Eqs. (33), (34) and (35), according to Lemma 2, there is
$$
\begin{array}{l} \det (\mathbf {V} _ {[ k + 1 ], P} ^ {\alpha}) \\ = \det  \left(\left(\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}\right) \mathbf {V} _ {[ k ], P}\right) \det  \left(\mathbf {U} ^ {\beta} - \left(\mathbf {F} ^ {:, \beta}\right) ^ {\top} \left(\left(\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}\right) \mathbf {V} _ {[ k ], P}\right) ^ {- 1} \mathbf {F} ^ {:, \beta}\right) \\ = \det  \left((\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}) \mathbf {V} _ {[ k ], P}\right) \det  (\mathbf {S} ^ {\beta}) = L \det  (\mathbf {S} ^ {\beta}) \\ = L \det  \left(\frac {\sigma_ {1 1} \sigma_ {2 2} - \sigma_ {1 2} ^ {2}}{\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}} \mathbf {V} _ {[ k ], P} ^ {\beta}\right) \\ \end{array}
$$
where  $L = \operatorname*{det}\left((\sigma_{11} + \sigma_{22} - 2\sigma_{12})\mathbf{V}_{[k],P}\right) = (\sigma_{11} + \sigma_{22} - 2\sigma_{12})^{2^k}\operatorname*{det}(\mathbf{C}^{[k]})$  by Eq. (25).
Therefore, by inductive assumption, for every  $P\in \mathcal{P}$
$$
\begin{array}{l} \det  \left(\mathbf {V} _ {[ k + 1 ], P} ^ {\alpha}\right) = L \det  \left(\frac {\sigma_ {1 1} \sigma_ {2 2} - \sigma_ {1 2} ^ {2}}{\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}} \mathbf {V} _ {[ k ], P} ^ {\beta}\right) \tag {36} \\ \geq L \det  \left(\frac {\sigma_ {1 1} \sigma_ {2 2} - \sigma_ {1 2} ^ {2}}{\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}} \mathbf {V} _ {[ k ], \delta_ {w ^ {*}}} ^ {\beta}\right) = \det  (\mathbf {V} _ {[ k + 1 ], \delta_ {w ^ {*}}} ^ {\alpha}). \\ \end{array}
$$
Case 2: If  $[2^{k + 1}]\backslash [2^k ]\nsubseteq \alpha \subseteq [2^{k + 1}]$  and  $\alpha \neq \emptyset$ , let  $\beta = [2^{k + 1}]\backslash [2^k ]$  and  $\gamma = \alpha \cap \beta$ . Note that  $\gamma$  could be empty. By Lemma 5, there is
$$
\det ((\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}) \mathbf {V} _ {[ k ], P} ^ {\gamma - 2 ^ {k}}) \det (\mathbf {V} _ {[ k + 1 ], P} ^ {\beta \cup \alpha}) \leq \det (\mathbf {V} _ {[ k + 1 ], P} ^ {\alpha}) \det ((\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}) \mathbf {V} _ {[ k ], P}),
$$
which implies, for every  $P\in \mathcal{P}$
$$
\begin{array}{l} \det  (\mathbf {V} _ {[ k + 1 ], P} ^ {\alpha}) \geq \frac {\det  ((\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}) \mathbf {V} _ {[ k ] , P} ^ {\gamma - 2 ^ {k}}) \det  (\mathbf {V} _ {[ k + 1 ], P} ^ {\beta \cup \alpha})}{(\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}) ^ {2 ^ {k}} \det  (\mathbf {C} ^ {[ k ]})} \\ \geq \frac {\operatorname* {d e t} \left(\left(\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}\right) \mathbf {V} _ {[ k ] , \delta_ {w ^ {*}}} ^ {\gamma - 2 ^ {k}}\right) \operatorname* {d e t} \left(\mathbf {V} _ {[ k + 1 ] , \delta_ {w ^ {*}}} ^ {\beta \cup \alpha}\right)}{\left(\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}\right) ^ {2 ^ {k}} \operatorname* {d e t} \left(\mathbf {C} ^ {[ k ]}\right)} \tag {37} \\ = \det  (\mathbf {V} _ {[ k + 1 ], \delta_ {w} ^ {*}} ^ {\alpha}). \\ \end{array}
$$
The second inequality comes from the inductive assumption as well as case 1, while the last equality is due to that  $\mathbf{V}_{[k + 1],\delta_{w^*}}$  is diagonal according to Lemma 7
C.5 Proof for Theorem 1
Proof. Let  $\mathcal{S} = \{\{1\}, \{2\}, \ldots, \{n\}\}$ , then
$$
\mathcal {I} _ {P, \mathcal {S}} (v _ {n}) = \phi (v _ {n}; P).
$$
Therefore, by Theorem 2,  $\delta_{w^*}$  is an optimal solution. For the uniqueness, it does not hold if  $n \leq 2$ . For example, let
$$
\mathcal {O} = \left\{P \in \mathcal {P} \mid \int_ {0} ^ {1} t d P (t) = w ^ {*} \right\},
$$
we point out that every  $P \in \mathcal{O}$  is optimal if  $n = 1$  or  $n = 2$ . To see  $|\mathcal{O}| > 1$ , for example, if  $\sigma_{11} = \sigma_{22}$  and  $\sigma_{12} = 0$ , the probability distribution for the Shapley value lies in  $\mathcal{O}$ . Nevertheless, we prove the uniqueness as follows: 1) first show that if  $n = k$  possesses such a uniqueness, so does  $n = k + 1$ , and 2) then demonstrate that it holds for  $n = 4$ .
For the inductive part, it can be completed by reusing the induction in the proof for Theorem 2. Specifically, for case 1, let  $\alpha = ([2^{k + 1}]\backslash [2^k])\cup \beta$  where  $\beta$  contains all indices pointing to subsets  $\{1\} ,\{2\} ,\ldots$  , and  $\{k\}$  , then the inequality in Eq. (36) is strict if  $P\neq \delta_{w^{*}}$  . On the other hand, for case 2, let  $\alpha$  contain all indices pointing to subsets  $\{1\} ,\{2\} ,\ldots$  , and  $\{k + 1\}$  , the second inequality in Eq. (37) is also strict if  $P\neq \delta_{w^{*}}$  by using the aforementioned strict inequality in case 1.
To show the uniqueness for  $n = 4$ , it can also be done by looking into the proof for Theorem 2. Specifically, in the induction, we focus on case 2 for  $k = 3$  and  $\alpha = \{2,3,5,9\}$  that points to subsets  $\{1\}, \{2\}, \{3\}$  and  $\{4\}$ . Note that  $\beta = [2^4] \backslash [2^3] = \{9,10,\ldots,16\}$  in case 2. By Lemma 5, a necessary condition for the first inequality in Eq. (37) to be equal is that each column of  $\mathbf{V}_{[4],P}^{\{2,3,5\},\{10,11,\ldots,16\}}$  lies in  $\operatorname{span}(\mathbf{V}_{[4],P}^{\{2,3,5\},\{9\}})$ . We prove that this condition can only be met by  $\delta_{w^*}$ , and thus the first inequality in Eq. (36) is strict if  $P \neq \delta_{w^*}$ .
Regarding Eq. (23), using the fact that
$$
\begin{array}{l} \mathbf {V} _ {[ n ], P} = \left(\int_ {0} ^ {1} \mathbf {B} ^ {[ n ]} (t) \mathrm {d} P (t)\right) \times \mathbf {C} ^ {[ n ]} \times \left(\int_ {0} ^ {1} \mathbf {B} ^ {[ n ]} (t) \mathrm {d} P (t)\right) ^ {\top} \\ = \int_ {0} ^ {1} \mathrm {d} P (t) \int_ {0} ^ {1} \mathbf {B} ^ {[ n ]} (t) \mathbf {C} ^ {[ n ]} (\mathbf {B} ^ {[ n ]} (s)) ^ {\top} \mathrm {d} P (s), \\ \end{array}
$$
one can verify that  $\mathbf{V}_{[4],P}^{\{2,3,5\},\{9\}} = c_P\mathbf{1}$  for some  $c_{P}\in \mathbb{R}$ . Therefore, to meet the necessary condition, all entries in each column of  $\mathbf{V}_{[4],P}^{\{2,3,5\},\{10,11,\dots,16\}}$  should be the same. Particularly, there is
$$
\begin{array}{l} \mathbf {V} _ {[ 4 ], P} ^ {\{3, 5 \}, \{1 2 \}} \\ = \left( \begin{array}{c} \int_ {0} ^ {1} \mathrm {d} P (t) \int_ {0} ^ {1} (\sigma_ {1 1} + \sigma_ {2 2} - 2 \sigma_ {1 2}) p (t) ^ {2} (p (s \cdot t) + (\sigma_ {1 2} - \sigma_ {1 1}) (s + t - 1) + \sigma_ {1 1}) \mathrm {d} P (s) \\ \int_ {0} ^ {1} \mathrm {d} P (t) \int_ {0} ^ {1} p (s) p (t) ^ {3} \mathrm {d} P (s) \end{array} \right) \\ \end{array}
$$
where  $p(w) = (\sigma_{11} + \sigma_{22} - 2\sigma_{12})w + \sigma_{12} - \sigma_{11}$ .
The equation  $\mathbf{V}_{[4],P}^{\{3\},\{12\}} - \mathbf{V}_{[4],P}^{\{5\},\{12\}} = 0$  leads to
$$
\int_ {0} ^ {1} \left(\sigma_ {1 1} \sigma_ {2 2} - \sigma_ {1 2} ^ {2}\right) p (t) ^ {2} \mathrm {d} P (t) = 0.
$$
Set  $g(w) = (\sigma_{11}\sigma_{22} - \sigma_{12}^2)p(w)^2$ , since  $g(w)$  attains its minimum 0 only at the point  $w^{*}$ , by Lemma 6, the necessary condition is met if and only if  $P = \delta_{w^{*}}$ .
For the ease of the reader to verify the calculation in Theorem 1, we include below a Matlab code to help compute.
syms sigma11 sigma12 sigma22 s t real;   
Sigma  $=$  [sigma11,sigma12；sigma12，sigma22];   
B1  $\equiv$  [1-t,t；-1,1]；   
Bt  $\equiv$  B1;   
Sigma  $\equiv$  Sigma1;   
for i  $\equiv$  1:3 Sigma  $\equiv$  kron(Sigma1,Sigma); Bt  $\equiv$  kron(B1,Bt);   
end   
Bs  $\equiv$  subs(Bt,t,s);   
V  $\equiv$  Bt\*Sigma\*Bs';   
disp(simplify(V([2,3,5],9)))；%V^{\{2,3,5\},\{9\}}   
Vp  $\equiv$  simplify(V([3,5],12));   
disp(Vp)；%V^{\{3,5\},\{12\}}   
disp(simplify(Vp(1)-Vp(2))）；%V^{\{3\},\{12\}}-V^{\{5\},\{12\}}
C.6 Lemmas
In this section, we provide all lemmas necessary to prove Theorems 2 and 1.
Lemma 1. Suppose  $\mathbf{A}$  is positive definite and  $\mathbf{B}$  is positive semi-definite, there is
$$
\det  (\mathbf {A} + \mathbf {B}) \geq \det  (\mathbf {A})
$$
where the equality holds if and only if  $\mathbf{B} = \mathbf{0}$
Proof. Since  $\mathbf{A} + \mathbf{B} = \mathbf{A}^{\frac{1}{2}}(\mathbf{I} + \mathbf{A}^{-\frac{1}{2}}\mathbf{B}\mathbf{A}^{-\frac{1}{2}})\mathbf{A}^{\frac{1}{2}}$ , there is  $\operatorname*{det}(\mathbf{A} + \mathbf{B}) = \operatorname*{det}(\mathbf{A})\operatorname*{det}(\mathbf{I} + \mathbf{A}^{-\frac{1}{2}}\mathbf{B}\mathbf{A}^{-\frac{1}{2}})$ . By spectral decomposition,  $\mathbf{A}^{-\frac{1}{2}}\mathbf{B}\mathbf{A}^{-\frac{1}{2}} = \mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}$  where  $\mathbf{Q}\mathbf{Q}^{\top} = \mathbf{I}$  and  $\mathbf{D}$  is diagonal with all entries being non-negative. Then,  $\operatorname*{det}(\mathbf{I} + \mathbf{A}^{-\frac{1}{2}}\mathbf{B}\mathbf{A}^{-\frac{1}{2}}) = \operatorname*{det}(\mathbf{I} + \mathbf{D}) \geq 1$ , and thus  $\operatorname*{det}(\mathbf{A} + \mathbf{B}) \geq \operatorname*{det}(\mathbf{A})$ .
For the claim on the equality, suppose  $\operatorname*{det}(\mathbf{I} + \mathbf{A}^{-\frac{1}{2}}\mathbf{B}\mathbf{A}^{-\frac{1}{2}}) = 1$ , it leads to  $\mathbf{D} = \mathbf{0}$ , and thus  $\mathbf{B} = \mathbf{0}$ .
Lemma 2. Given a positive definite matrix  $\mathbf{M} \in S_{++}$  that is partitioned as
$$
\mathbf {M} = \left( \begin{array}{c c} \mathbf {A} _ {m \times m} & \mathbf {B} _ {m \times k} \\ \mathbf {B} _ {k \times m} ^ {\top} & \mathbf {C} _ {k \times k} \end{array} \right),
$$
there is  $\operatorname{det}(\mathbf{M}) = \operatorname{det}(\mathbf{A} - \mathbf{BC}^{-1}\mathbf{B}^{\top})\operatorname{det}(\mathbf{C})$
$$
\mathbf {Q} \left( \begin{array}{c c} \mathbf {A} _ {m \times m} & \mathbf {B} _ {m \times k} \\ \mathbf {B} _ {k \times m} ^ {\top} & \mathbf {C} _ {k \times k} \end{array} \right) \mathbf {Q} ^ {\top} = \left( \begin{array}{c c} \mathbf {A} - \mathbf {B C} ^ {- 1} \mathbf {B} ^ {\top} & \mathbf {0} \\ \mathbf {0} & \mathbf {C} \end{array} \right) \text {w h e r e} \mathbf {Q} = \left( \begin{array}{c c} \mathbf {I} _ {m \times m} & - \mathbf {B C} ^ {- 1} \\ \mathbf {0} _ {k \times m} & \mathbf {I} _ {k \times k} \end{array} \right).
$$
Lemma 3. Let  $\mathbf{M}$  be the matrix given in Lemma 2, then
$$
\det (\mathbf {M}) \leq \det (\mathbf {A}) \det (\mathbf {C}).
$$
Moreover, the equality holds if and only if  $\mathbf{B} = \mathbf{0}$
Proof. By Lemmas 2 and 1,  $\operatorname*{det}(\mathbf{M}) = \operatorname*{det}(\mathbf{A} - \mathbf{BC}^{-1}\mathbf{B}^{\top})\operatorname*{det}(\mathbf{C})\leq \operatorname*{det}(\mathbf{A})\operatorname*{det}(\mathbf{C})$ . Notice that the proof for Lemma 2 ensures that  $\mathbf{A} - \mathbf{BC}^{-1}\mathbf{B}^{\top}$  is positive definite.
By Lemma 1, the equality holds if and only if  $\mathbf{BC}^{-1}\mathbf{B}^{\top} = \mathbf{0}$ . Note that for a positive definite matrix  $\mathbf{C}^{-1}$ ,  $\mathbf{x}^{\top}\mathbf{C}^{-1}\mathbf{x} = 0$  if and only if  $\mathbf{x} = \mathbf{0}$ . Therefore,  $\mathbf{BC}^{-1}\mathbf{B}^{\top} = \mathbf{0}$  if and only if  $\mathbf{B} = \mathbf{0}$ .