Tournament solutions based on cooperative game theory

Abstract

A tournament can be represented as a set of candidates and the results from pairwise comparisons of the candidates. In our setting, candidates may form coalitions. The candidates can choose to fix who wins the pairwise comparisons within their coalition. A coalition is winning if it can guarantee that a candidate from this coalition will win each pairwise comparison. This approach divides all coalitions into two groups and is, hence, a simple game. We show that each minimal winning coalition consists of a certain uncovered candidate and its dominators. We then apply solution concepts developed for simple games and consider the desirability relation and the power indices which preserve this relation. The tournament solution, defined as the maximal elements of the desirability relation, is a good way to select the strongest candidates. The Shapley–Shubik index, the Penrose–Banzhaf index, and the nucleolus are used to measure the power of the candidates. We also extend this approach to the case of weak tournaments.

Appendix

Proof of Lemma 1

We demonstrate that if candidate $a\in UC$ then $a\cup {\overline{D}}(a)$ is a minimal winning coalition. Indeed, $a\cup {\overline{D}}(a)$ is winning because a dominates D(a). Using the principle of contradiction, assume that there exists a coalition $B \subset a \cup {\overline{D}}(a)$ such that $B \ne a \cup {\overline{D}}(a)$ and B is winning. Then a certain candidate $b\in B$ dominates the countercoalition $B'$ such that $D(a)\subset B'$ and $D(a) \ne B'$. Therefore, $b \ne a$, $b\in {\overline{D}}(a)$ and b dominates $a\cup D(a)$. Hence, b covers a, which contradicts $a\in UC$.

On the other hand, consider a candidate $a\notin B$, a minimal winning coalition $a \cup B$ and its countercoalition $B'{\setminus } a$ such that a dominates $B'{\setminus } a$. Suppose that there exists a candidate $b\in B$ dominated by a. Then, the coalition $a \cup B {\setminus } b$ is also winning, which contradicts the minimality of $a \cup B$. Therefore, B dominates a, and $B={\overline{D}}(a)$. If candidate a were covered by a certain candidate $b\in B={\overline{D}}(a)$, then b would dominate $B'$, and B would be winning, which contradicts the minimality of $a \cup B$. Hence, a belongs to UC. $\square$

Lemma 3

For a certain pair of candidates $\{a,b\}$ and each coalition $K\subset A{\setminus } \{a,b\}$, let characteristic functions v and ${\widehat{v}}$ satisfy the conditions

$$\begin{aligned} v(K)= & {} {\widehat{v}}(K), \quad v(K\cup b)\ge {\widehat{v}}(K\cup b), \\ v(K\cup a)\le & {} {\widehat{v}}(K\cup a), \quad v(K\cup a \cup b) = {\widehat{v}}(K\cup a \cup b). \end{aligned}$$

Then, the transition from v to ${\widehat{v}}$ does not decrease the Shapley value $\varphi _a$ for candidate a and also does not increase the Shapley value $\varphi _b$ for candidate b.

For every other candidate $c\in A{\setminus } \{a,b\}$, the Shapley value increment $\varphi _c({\widehat{v}})-\varphi _c(v)$ is not greater than the increment $\varphi _a({\widehat{v}})-\varphi _a(v)$ and not smaller than the increment $\varphi _b({\widehat{v}})-\varphi _b(v)$.

Hence, the transition from v to ${\widehat{v}}$ does not reduce the ordinal rank of candidate a and also does not raise the ordinal rank of candidate b.

Proof

By the hypotheses of this lemma and formula (3), the Shapley value for candidate a does not decrease while for candidate b it does not increase.

Now, demonstrate that for every other candidate $c\in A{\setminus } \{a,b\}$, the Shapley value increment does not exceed that for a. According to the hypotheses of Lemma 3, we have the inequalities

$$\begin{aligned} v(K \cup a) \le {\widehat{v}}(K\cup a), \quad v(K \cup c) \ge {\widehat{v}}(K\cup c) \quad \text{ for } \text{ each } \quad K\subset A{\setminus } \{a,c\}. \end{aligned}$$

Therefore,

$$\begin{aligned} \varphi _a({\widehat{v}})-\varphi _c({\widehat{v}})= & {} \sum _{K\subset A{\setminus } \{a,c\}} \frac{k!(m-k-2)!}{(m-1)!} \left( {\widehat{v}}(K\cup a)-{\widehat{v}}(K\cup c)\right) \ge \\\ge & {} \sum _{K\subset A{\setminus } \{a,c\}} \frac{k!(m-k-2)!}{(m-1)!} \left( v(K\cup a)-v(K\cup c)\right) = \varphi _a(v)-\varphi _c(v), \end{aligned}$$

whence it follows that $\varphi _a({\widehat{v}})-\varphi _a(v)\ge \varphi _c({\widehat{v}})-\varphi _c(v).$

In a similar fashion, we receive the inequalities

$$\begin{aligned} v(K \cup c) \le {\widehat{v}}(K\cup c), \quad v(K \cup b) \ge {\widehat{v}}(K\cup b) \quad \text{ for } \text{ each } \quad K\subset A{\setminus } \{b,c\}. \end{aligned}$$

As a result,

$$\begin{aligned} \varphi _c({\widehat{v}})-\varphi _b({\widehat{v}})= & {} \sum _{K\subset A{\setminus } \{b,c\}} \frac{k!(m-k-2)!}{(m-1)!} \left( {\widehat{v}}(K\cup c)-{\widehat{v}}(K\cup b)\right) \ge \\\ge & {} \sum _{K\subset A{\setminus } \{b,c\}} \frac{k!(m-k-2)!}{(m-1)!} \left( v(K\cup c)-v(K\cup b)\right) = \varphi _c(v)-\varphi _b(v), \end{aligned}$$

which directly gives $\varphi _c({\widehat{v}})-\varphi _c(v) \ge \varphi _b({\widehat{v}})-\varphi _b(v)$. $\square$

Remark 3

Lemma 3 remains in force for the Banzhaf value (5). The proof is analogous. Really, for each candidate $c\in A{\setminus } \{a,b\}$,

$$\begin{aligned} {\widetilde{\beta }}_a({\widehat{v}})-{\widetilde{\beta }}_c({\widehat{v}})= & {} 2 \sum _{K\subset A{\setminus } \{a,c\}} \left( {\widehat{v}}(K\cup a)-{\widehat{v}}(K\cup c)\right) \ge \\\ge & {} 2 \sum _{K\subset A{\setminus } \{a,c\}} \left( v(K\cup a)-v(K\cup c)\right) = {\widetilde{\beta }}_a(v)-{\widetilde{\beta }}_c(v), \end{aligned}$$

By analogy, ${\widetilde{\beta }}_a({\widehat{v}})-{\widetilde{\beta }}_b({\widehat{v}})\ge {\widetilde{\beta }}_a(v)-{\widetilde{\beta }}_b(v)$.

Lemma 4

For each weak tournament $T=(A,H)$, the Shapley value $\varphi (A,H)$ satisfies the following properties:

• (1) For any candidate a, the Shapley value is $\varphi _a=1$ if and only if $a\in CW(A,H)$ is the Condorcet winner in A;

• (2) For each candidate $b\in A{\setminus } TC$, the Shapley value is $\varphi _b=0$ and hence $\varphi _{+}\subset TC$;

• (3) The Shapley value remains the same after removing any candidate that receives zero Shapley value, that is, $\varphi _a(A,H)=\varphi _a(A{\setminus } b, H)$ for each pair $\{a, b\}$, whenever $\varphi _b (A,H) =0$ and $a\in A{\setminus } b$;

• (4) The support of the Shapley value is $\varphi _{+} = MNL$.

Proof

(1) Let candidate $a\in CW(A,H)$ be the Condorcet winner in A. As is easily seen, the optimal strategy of the coalition that contains a is to nominate him. Therefore, $v(B)=1$ if $a\in B$, and $v(B)=0$ if $a\notin B$. Subsequently, the contribution of candidate a to each coalition is 1, and $\varphi _a=1$.

We prove the converse result by contradiction. Suppose that, in a certain weak tournament, $a\notin CW(A,H)$. Then, $v(A{\setminus } a)\ge 1/2$ and $v(A)-v(A{\setminus } a) = 1 -v(A{\setminus } a) \le 1/2$. This inequality implies $\varphi _a < 1$.

(2) Consider the game of an arbitrary coalition $B \subset A$ against the countercoalition $B'=A{\setminus } B$; in the payoff matrix of this game, the rows and columns for players $a\in TC$ dominate the rows and columns for players $b\in A{\setminus } TC$. Take arbitrary $b\in A{\setminus } TC, B\subset A{\setminus } b, K=B'{\setminus } b$, the payoff matrix in the game of the coalition $B\cup b$ against the countercoalition K that is defined by

and also the payoff matrix in the game of the coalition B against the countercoalition $K\cup b$ that is defined by

Introduce the notation $B_1=B\cap TC=\{b^1_1,\ldots ,b^1_l\}$, $B_2=B\cap (A{\setminus } TC)=\{b^2_1,\ldots ,b^2_p\}$, $K_1=K\cap TC=\{k^1_1,\ldots ,k^1_r\}$, and $K_2=K\cap (A{\setminus }{\setminus } TC)=\{k^2_1,\ldots ,k^2_t\}$. If $B_1=\emptyset$, then $v(B)=v(B\cup b)=0$. If $B_1 \ne \emptyset$ and $K_1=\emptyset$, then $v(B)=v(B\cup b)=1$. If $B_1 \ne \emptyset$ and $K_1 \ne \emptyset$, then in both games the dominating strategies are from the sets $B_1$ and $K_1$, which means that $v(B)=v(B\cup b)$.

(3) Consider any candidate b with zero Shapley value, $\varphi _b =0$. Thus, for each coalition $K\subset A{\setminus } b$ we have $v(K\cup b) - v(K) =0$. Let ${\widehat{v}}$ be the characteristic function of the weak tournament $(A{\setminus } b, H)$. Then, we have the following inequalities

$$\begin{aligned} v(K\cup b)= & {} \max _{c\in K \cup b}{\min _{d\in A{\setminus } b \setminus K}{H(c,d)}} \ge {\widehat{v}}(K) = \max _{c\in K}{\min _{d\in A{\setminus } b \setminus K}{H(c,d)}} \ge \\\ge & {} v(K) = \max _{c\in K}{\min _{d\in A {\setminus } K}{H(c,d)}}, \end{aligned}$$

and so ${\widehat{v}}(K) = v(K) = v(K\cup b)$.

For each candidate $a\in A{\setminus } b$, the Shapley value takes the form

$$\begin{aligned} \varphi _a= & {} \sum _{K\subset A{\setminus } a} \frac{k!(m-k-1)!}{m!} \left( v(K\cup a)-v(K)\right) \\= & {} \sum _{K\subset A{\setminus } \{a,b\}} \frac{k!(m-k-1)!}{m!} \left( v(K\cup a)-v(K)\right) \\&+ \sum _{K\subset A{\setminus } \{a,b\}} \frac{(k+1)!(m-k-2)!}{m!} \left( v(K\cup a\cup b)-v(K\cup b)\right) \\= & {} \sum _{K\subset A{\setminus } \{a,b\}} \left( \frac{k!(m-k-1)!}{m!}+\frac{(k+1)!(m-k-2)!}{m!}\right) \left( {\widehat{v}}(K\cup a)-{\widehat{v}}(K)\right) \\= & {} \sum _{K\subset A{\setminus } \{a,b\}} \frac{k!(m-k-2)!}{(m-1)!} \left( {\widehat{v}}(K\cup a)-{\widehat{v}}(K)\right) =\widehat{\varphi _a}. \end{aligned}$$

(4) We have to show that $\varphi _{+} \supset MNL$. Each candidate $a\in MNL(A,H)$ belongs at least to one minimal nonlosing coalition $a\in B\in M_1 \cup M_{1/2}$; hence, $v(B)-v(B{\setminus } a)>0$ and $\varphi _a>0$.

To prove $\varphi _{+} \subset MNL$, we use the principle of contradiction. Assume that $a\in \varphi _{+}(T)$, $a\notin MNL(T)$, in a certain tournament $T=(A,H)$. Since $a\in \varphi _{+}(T)$, then there exists a set $K\cup a$ such that $v(K\cup a)>v(K)$. On the strength of $a\notin MNL(T)$, for each $B\in M_1(T)\cup M_{1/2}(T)$ we have $a\notin B$. If $v(K\cup a)=i\in \{1, 1/2\}$, then there exists $B\subset K\cup a$, where $B\in M_i$ and $a\notin B$. As a result, $B\subset K$ and the conclusion follows from the contradiction $v(B)=i=v(K\cup a)\le v(K)$. $\square$

Remark 4

Lemma 4 remains in force for the Banzhaf value (5). The only difference in the proof concerns Item 3. To argue this result, it suffices to observe that

$$\begin{aligned} {\widetilde{\beta }}_a(A,H)= & {} \sum _{K\subset A{\setminus } a} \left( v(K\cup a)-v(K)\right) \\= & {} 2 \sum _{K\subset A{\setminus } \{a,b\}} \left( v(K\cup a)-v(K)\right) =2 {\widetilde{\beta }}_a(A{{\setminus }} b,H) \end{aligned}$$

for each b, whenever $\varphi _b(A,H)=0$ and $a\in A{\setminus } b$.

Lemma 5

For each weak tournament $T=(A,H)$ and each pair of candidates $\{a,b\}$ such that a covers b, we have the following properties:

• (1) For each coalition $B\subset A{\setminus } \{a,b\}$, the payoffs satisfy the inequalities $v(B\cup a)$$\ge$$v(B\cup b)=v(B)$;

• (2) Candidate a is at least as influential as candidate b: $a\succeq b$;

• (3) The Shapley value satisfies the inequality $\varphi _a\ge \varphi _b$.

  Moreover, if the tournament T allows no ties, then also:

• (4) If $a\in UC$ is uncovered, then a is more influential than b: $a\succ b$;

• (5) If $a\in UC$, then $\varphi _a>\varphi _b$.

Proof

(1) Using the principle of contradiction, assume that $v(B\cup b)>v(B)$ for a certain coalition $B\subset A{\setminus } \{a,b\}$. Consider the game of the coalition $B\cup b$ against the coalition $K\cup a$, where $K=A{\setminus } B\setminus \{a,b\}$. Candidate b cannot be a nominated candidate from the coalition $B\cup b$, as long as $H(b,a)=0$. Denote by $c\in B$ a nominated candidate from the coalition $B\cup b$ that guarantees the payoff $v(B\cup b)$. Therefore, $H(c,a)\ge v(B\cup b)$ and $H(c,d)\ge v(B\cup b)$ for each $d\in K$. In the case of $H(c,b)\ge v(B\cup b)$, candidate $c\in B$ would guarantee the payoff $v(B)=v(B\cup b)$ for the coalition B. Hence, $H(c,b) < v(B\cup b)\le H(c,a)$, which obviously contradicts the covering relation a over b.

Thus, we have demonstrated that $v(B\cup b)=v(B)$. The inequality $v(B\cup a)$$\ge$v(B) holds by monotonicity of v.

(2) The desired relation follows from Item 1 and the influence relation definition.

(3) The desired inequality follows from Item 1 and formula (6).

(4,5) Under items 2 and 3 we have $a\succeq b$ and $\varphi _a\ge \varphi _b$. Under Lemma 1 coalition $a\cup {\overline{D}}(a) \in M$ is minimal winning; thus, $v(a\cup {\overline{D}}(a)) = 1$ and $v({\overline{D}}(a))=0$. According to Item 1 we have $v(b\cup {\overline{D}}(a)) = v({\overline{D}}(a)) =0$. The inequality $v(a\cup {\overline{D}}(a)) > v(b\cup {\overline{D}}(a))$ implies $a\succ b$ and $\varphi _a>\varphi _b$.

$\square$

Remark 5

Lemma 5 remains in force for the Banzhaf value (5). The only difference in the proof is that it uses formula (7).

Proof of Theorem 1

(1) Suppose that the adjacency matrices ${\widehat{H}}$ and H satisfy ${\widehat{H}}(a,b) > H(a,b)$ and ${\widehat{H}}(b,a) < H(b,a)$ for a certain pair of candidates $\{a, b\}$ and the other elements of the matrices coincide. Let ${\widehat{v}}$ and v be the characteristic functions for the tournaments $(A,{\widehat{H}})$ and (A, H), respectively. The definition of the characteristic function under formula (1) directly implies that

$$\begin{aligned} {\widehat{v}}(B\cup a) \ge v(B\cup a), \quad {\widehat{v}}(B) \le v(B) \quad \text{ for } \text{ each } \quad B\subset A{\setminus } a. \end{aligned}$$

(10)

To prove that $a\in IN(A,{\widehat{H}})$, whenever $a\in IN(A,H)$, we use the principle of contradiction. Assume that a certain candidate c is more influential than candidate a in the tournament $(A,{\widehat{H}})$, that is,

$$\begin{aligned} {\widehat{v}}(B\cup c) \ge {\widehat{v}}(B\cup a) \quad \text{ for } \text{ each } \quad B\subset A{\setminus } \{a,c\}, \end{aligned}$$

and the inequality is strict for some B. These inequalities and the inequalities (10) imply that c is more influential than a in (A, H), which contradicts $a\in IN(A,H)$.

(2) The desired inclusion follows from Item 4 of Lemma 5.

(3) Under Lemma 1 the set of minimal winning coalitions can be computed in polynomial time. Hence, for each pair of candidates, under Lemma 2 we can check the influence relation in polynomial time. $\square$

Proof of Theorem 4

Items 1 and 2 have been proved in Lemmas 5 and 4, respectively.

(3) If we change an adjacency matrix H for ${\widehat{H}}$ so that ${\widehat{H}}(a,b)>H(a,b)$ and ${\widehat{H}}(b,a)<H(b,a)$ for a certain pair of candidates $\{a,b\}$ (the other values remain the same), then the hypotheses of Lemma 3 hold and the rank of candidate a is not decreased.

(4) The desired inclusion $\varphi _{+} = MNL \subset TC$ has been proved in Lemma 4.

(5) Suppose candidate b covered by candidate $a\in A$ is added to a tournament (A, H), that is, ${\widehat{H}}(a,b)=1$, ${\widehat{H}}(a,c)\ge {\widehat{H}}(b,c)$ for each candidate $c\in A{\setminus } a$, and ${\widehat{H}}=H$ in the other cases.

Let ${\widehat{v}}$ be the characteristic function for the tournament $(A\cup b, {\widehat{H}})$. As is easily observed, for each coalition $B\subset A{\setminus } a$ not containing candidate a, we have

$$\begin{aligned} v(B)={\widehat{v}}(B), \quad v(B\cup a)={\widehat{v}}(B\cup a \cup b),\quad B\subset A{\setminus } a. \end{aligned}$$

(11)

Under Item 1 of Lemma 5, for each coalition $B\subset A{\setminus } a$ the contribution is ${\widehat{\varDelta }}_b(B) = {\widehat{v}}(B\cup b) -{\widehat{v}}(B)=0$.

Consider an arbitrary permutation of length m in which a certain candidate $c\in A{\setminus } \{a,b\}$ stands after candidate a, and $\sigma =b_1$, $\ldots ,b_l,a,k_1$, $\ldots$, $k_r,c,t_1$, $\ldots ,t_p$. Construct the sets $B=\{b_1,\ldots ,b_l\}$ and $K=\{k_1,\ldots ,k_r\}$. There exist $m+1$ ways to place candidate b, among which $l+1$ ways to place it before a and $m-l$ ways after a. Taking into account (11), averaging over all the $m+1$ ways yields

$$\begin{aligned} {\widehat{\varDelta }}_{a}(\sigma +b)= & {} \frac{(l+1){\widehat{\varDelta }}_a(B\cup b)+(m-l){\widehat{\varDelta }}_a(B)}{m+1}\\= & {} \frac{(l+1)(\varDelta _a(B) - {\widehat{\varDelta }}_{b}(B))+(m-l)(\varDelta _a(B)- {\widehat{\varDelta }}_{b}(B\cup a)) }{m+1}\\= & {} \varDelta _a(B)-\frac{(l+1) {\widehat{\varDelta }}_{b}(B)+(m-l){\widehat{\varDelta }}_{b}(B\cup a) }{m+1} \\= & {} \varDelta _a(B)-\frac{(m-l){\widehat{\varDelta }}_{b}(B\cup a) }{m+1}. \end{aligned}$$

Let us associate each permutation $\sigma$ considered with a permutation $\sigma '=b_1,\ldots ,b_l, c, k_1,\ldots ,k_r, a, t_1,\ldots ,t_p$ of length m, where candidate a stands after candidate c. The averaged contribution of candidate c over all the $m+1$ ways to permute candidate b makes up

$$\begin{aligned} {\widehat{\varDelta }}_{c}(\sigma '+b)= & {} \frac{(l+1){\widehat{\varDelta }}_c(B\cup b)+(m-l){\widehat{\varDelta }}_c(B)}{m+1}\\= & {} \frac{(l+1)(\varDelta _c(B) - {\widehat{\varDelta }}_{b}(B)+{\widehat{\varDelta }}_{b}(B\cup c))+(m-l)\varDelta _c(B)}{m+1}\\= & {} \varDelta _c(B)+\frac{(l+1)({\widehat{\varDelta }}_{b}(B\cup c) -{\widehat{\varDelta }}_{b}(B)) }{m+1} = \varDelta _c(B). \end{aligned}$$

Hence, candidate c loses in $\sigma '$ no more than candidate a in $\sigma$.

By analogy, one may obtain the formulas

$$\begin{aligned} {\widehat{\varDelta }}_{a}(\sigma '+b)= & {} \varDelta _a(B\cup c\cup K)- \frac{(m-l-r-1){\widehat{\varDelta }}_{b}(B\cup c \cup K \cup a) }{m+1}, \\ {\widehat{\varDelta }}_{c}(\sigma +b)= & {} \varDelta _c(B\cup a\cup K)\\&+\frac{(m-l-r-1)( {\widehat{\varDelta }}_{b}(B\cup a \cup K)- {\widehat{\varDelta }}_{b}(B\cup a \cup K \cup c)) }{m+1}. \end{aligned}$$

Thus, candidate c loses in $\sigma$ no more than candidate a in $\sigma '$. Since the Shapley value for an arbitrary candidate is calculated as its average score over all permutations, adding candidate b does not increase the ordinal rank and the score of candidate a. $\square$

Proof of Theorem 2

Items 1 and 2 have been proved in Lemmas 5 and 4, respectively. Items 3 and 5 have been proved in Theorem 4.

(4) The inclusion $\varphi _{max} \subset IN$ follows from formula (6) and the definition of the influence relation. The inclusion $IN \subset UC$ has been proved in Theorem 1.

The formula for $\varphi _{+}$ follows from Lemmas 1 and  4. The inclusion $\varphi _{+} \subset TC$ has been proved in Lemma 4. $\square$

Proof of Theorem 3

(1) Consider a certain candidate $a\in UC(T)$ that covers candidate b. For each coalition $B\subset A{\setminus } \{a,b\}$, we have $v(B\cup b)=v(B)$ according to Item 1 of Lemma 5 in Appendix. Therefore, each minimal winning coalition $K\in M(A,H)$ containing candidate b always includes candidate a that covers the former. However, candidate b is not a member of the coalition $a\cup {\overline{D}}(a)\in M(A,H)$ by Lemma 1. The largest excesses belong to the coalitions $K\in M(A,H)$,

$$\begin{aligned} e(x,K)=1-\sum _{a_i\in K}{x_i}\ge 0. \end{aligned}$$

The excesses of losing coalitions do not exceed zero. Hence, to minimize the excess vector, the whole imputation $x_b$ is transferred from candidate b to candidate a, and $N_b=0$.

Now show that removing candidate $b\notin UC$ does not affect the nucleolus. Let the imputation be $x=N(A,H)$, $x_b=0$. Denote by ${\widehat{v}}$ the corresponding characteristic function and by ${\widehat{e}}$ the excesses for ${\widehat{A}}=A{\setminus } b$.

Consider an arbitrary coalition $B\subset A{\setminus } \{a,b\}$. As is easily seen,

$$\begin{aligned} v(B)=v(B\cup b)={\widehat{v}}(B), \quad v(B\cup a)\le v(B\cup a \cup b)={\widehat{v}}(B\cup a). \end{aligned}$$

For each imputation ${\widehat{x}}$, the excesses are

$$\begin{aligned} {\widehat{e}}({\widehat{x}},B)= & {} {\widehat{v}}(B)-\sum \limits _{k\in B} {{\widehat{x}}_k}=e({\widehat{x}},B)=e({\widehat{x}},B\cup b), \\ {\widehat{e}}({\widehat{x}},B\cup a)= & {} {\widehat{v}}(B\cup a)-\sum \limits _{k\in B\cup a} {{\widehat{x}}_k} =e({\widehat{x}},B\cup a \cup b) \ge e({\widehat{x}}, B\cup a). \end{aligned}$$

Consequently, the lexicographic minimum of the excesses ${\widehat{e}}({\widehat{x}})$ is achieved at the imputation ${\widehat{x}} : {\widehat{x}}_c=x_c, c\ne b$.

(2) Item 1 directly implies that $NU\subset UC^{\infty }$. $\square$