3.1 Axioms
Simply speaking, CON implies that the social ranking at
\(\succsim\) must be consistent with the social rankings at its partitioned weak orders
\(\succsim _{1}\) and
\(\succsim _{2}\) (where
\(\succsim \in \succsim _{1} \oplus \succsim _{2}\)). CON is an analog of standard consistency in social choice theory,
^{3} which demands that the social outcome at a society
N must be consistent with the social outcomes at its partitioned sub-societies
\(N_{1}\) and
\(N_{2}\) (where
\(N = N_{1} + N_{2}\)). Nevertheless, the difference between CON and the standard consistency is often more than expected.
The standard consistency is often considered in anonymous contexts. Voters (the elements of N) are not distinguished. Therefore, the ballot profile of society N is usually represented as the list of the numbers of each type of ballot. In such a case, partitioning the ballot profile of society N into partitioned sub-societies, \(N_{1}\) and \(N_{2}\), does not lose any information: the number of a ballot, B, in society, N, is found as the sum of such number for sub-societies, \(N_{1}\) and \(N_{2}\).
In our model, however, the partitions of weak orders is considered. As a result, two weak orders
\(\succsim _{1}\) and
\(\succsim _{2}\) usually have multiple sums (recall (e) in Example
1). Since CON requires that the social ranking
\(R_{\succsim _{1}}\) and
\(R_{\succsim _{2}}\) are consistent with
\(R_{\succsim }\) for all
\(\succsim \in \succsim _{1} \oplus \succsim _{2}\), its requirement can be stronger than what the standard consistency means in the anonymous contexts. This can lead to the fact that the Borda count fails to satisfy CON (Proposition
2). Nevertheless, the multiplicity of the sum of two weak orders do not make the idea completely ruined. We will prove that the lexicographic SRSs surely satisfy CON (Proposition
1). This is attractive especially when the whole coalitions are too large or too mixed, as only the small domain of coalitions are considered (e.g., interdisciplinary research teams or young research teams in Example
1 (d) and (e)). Even in such a case, our results show that the lexicographic SRSs can judge individuals consistently in the above sense.
First of all, Lemmas
1 and
2 identify straightforward but useful facts about the partition of weak orders. When
\(\succsim \in \succsim _{1} \oplus \succsim _{2}\), Lemma
1 states the relationship between the quotient orders of
\(\succsim , \succsim _{1}\), and
\(\succsim _{2}\), and Lemma
2 states the relationship between the appearance vectors of
\(\succsim , \succsim _{1}\), and
\(\succsim _{2}\).
The intuition of CD is straightforward. If every coalition containing
x but not
y is ranked above those coalitions containing
y but not
x, then
x is socially superior than
y. In the previous literature, similar dominance properties have been proposed. Moretti and Öztürk (
2017) introduce dominance (DOM), which demands that (1) if
x is at least as good as
y in every ceteris paribus comparison (
x dominates
y),
x is socially as good as
y; and (2) if
x dominates
y and
x wins
y in at least one ceteris paribus comparison, then
x is socially better than
y. Later, Suzuki and Horita (
2021) introduce a weaker notion called weak dominance (WDOM); this demands that if
x wins
y in every ceteris paribus comparison,
x is socially better than
y. Let us demonstrate the difference between WDOM (DOM) and CD. Suppose
\(X = \{ 1,2,3 \}\), and
\(\succsim :\{ 1,3 \} \succ \{ 2,3 \} \succ \{ 1 \} \succ \{ 2 \}\). Since 1 wins 2 in every ceteris paribus comparison (i.e.,
\(\{ 1,3 \} \succ \{ 2,3 \}\) and
\(\{ 1 \} \succ \{ 2 \}\)), WDOM implies
\(1P_{\succsim }2\). However, 1 does not completely dominate 2; this is because
\(\{ 1 \} \in {{{\,\textrm{und}\,}}}_{1}( \succsim ) \backslash {{{\,\textrm{und}\,}}}_{2} ( \succsim )\) is ranked lower than
\(\{ 2,3 \} \in {{{\,\textrm{und}\,}}}_{2}( \succsim ) \backslash {{{\,\textrm{und}\,}}}_{1}( \succsim )\). Therefore, CD says nothing about this situation.
Let \(x \in X\). A bijection \(\pi\) on \(\mathcal {X}\) is called x-invariant if for all \(S \in \mathcal {X}\), \([ x \in S \Rightarrow x \in \pi (S) ]\). For \(\succsim \in \mathfrak {D}\) and a bijection \(\pi\) on \(\mathcal {X}\), we define \(\succsim _{\pi }\) as \(\pi (C) \succsim _{\pi }\pi (D) \Leftrightarrow C \succsim D\) for all \(C,D \in \mathcal {X}\). Two notes are in order. First, if a bijection \(\pi\) on \(\mathcal {X}\) is both x-invariant and y-invariant, we say that \(\pi\) is x, y-invariant. Second, suppose that a bijection \(\pi\) on \(\mathcal {X}\) is x-invariant. Then, \(\pi\) maps every element of \(\mathcal {Y}:= \{ S \in \mathcal {X:}x \in S \}\) to an element of \(\mathcal {Y}\). Since \(\pi\) is a bijection and \(\mathcal {Y}\) is finite, we have that for all \(S \in \mathcal {X}\), \([ x \in S \Leftrightarrow x \in \pi (S) ]\).
WCA
^{6} was introduced in Algaba et al. (
2021) as a weaker version of CA by Bernardi et al. (
2019). Let us briefly explain WCA. Suppose that
\(\pi\) is
x,
y-invariant bijection on
\(\mathcal {X}\). Then, by the note just above Definition
4, we have that for all
\(S \in \mathcal {X}\),
\(S \cap \left\{ x,y \right\} = \pi (S) \cap \left\{ x,y \right\}\). Thus, the only difference between
\(\succsim\) and
\(\succsim _{\pi }\) is the teammates of
x and
y. WCA demands that such change does not affect the ranking between
x and
y.
For a bijection \(\sigma\) on X and \(\succsim \in \mathfrak {D}\), let \(\sigma ( \succsim )\in \mathfrak {D}\) be such that \(C \succsim D \Leftrightarrow \sigma (C)\sigma (\succsim )\) \(\sigma (D)\) for all \(C,D \in {{\,\textrm{und}\,}}( \succsim )\) (i.e., \(\sigma ( \succsim )\) is a ranking obtained from \(\succsim\) by changing the names of candidates according to \(\sigma\)). The last axiom, neutrality, demands that the name of candidates do not matter.
3.2 Characterization
Our primary result dictates that the four axioms characterize the LES and DLES (Theorem
1). Three lemmas (Lemmas
3,
4, and
5) are provided for the proof of Theorem 1. Lemmas
3 and
4 state that if an SRS satisfies the four axioms (while CD is not necessary in Lemma
3), then the social ranking is determined by the appearance vectors
\(\theta _{\succsim }(x)\) and
\(\theta _{\succsim }(y)\). Lemma
3 is on the indifference case, and Lemma
4 is on the strict case. Lemma
5 is a technical lemma guaranteeing that the two cases considered in Lemma
4 are exhaustive.
Let us begin with two preliminary claims.
3.2.1 Induction step
Assume that for any
\(\succsim \in \mathfrak {D}\) with
\(\Vert \theta _{\succsim }(x) \Vert ,\Vert \theta _{\succsim }(y) \Vert \ge 1\) and
\(\Vert \theta _{\succsim }(x) \Vert + \Vert \theta _{\succsim }(y) \Vert \le s\),
\([ \theta _{\succsim }(x) >_{E}\theta _{\succsim }(y) \Rightarrow xP_{\succsim }y ]\). Claims
1 and
2 cover the whole case of
\(s \le 3\) (i.e.,
\(( \Vert \theta _{\succsim }(x) \Vert ,\Vert \theta _{\succsim }(y) \Vert ) = (1,1),(1,2),(2,1)\)). Therefore, we can assume
\(s \ge 3\) to prove the induction step.
Let
\(\succsim \in \mathfrak {D}\) with
\(\Vert \theta _{\succsim }(x) \Vert ,\Vert \theta _{\succsim }(y) \Vert \ge 1\),
\(\Vert \theta _{\succsim }(x) \Vert + \Vert \theta _{\succsim }(y) \Vert = s + 1( \ge 4)\), and
\(\theta _{\succsim }(x) >_{E}\theta _{\succsim }(y)\). We prove
\(xP_{\succsim }y\). By Claims
1 and
2, we have only to prove the case of
\(\Vert \theta _{\succsim }(x) \Vert ,\Vert \theta _{\succsim }(y) \Vert \ge 2\). Let
\(\Sigma _{1} \succ \Sigma _{2} \succ \cdots \succ \Sigma _{K}\) be the quotient order of
\(\succsim\).
(i)
Suppose that there exists
k with
\(x_{k},y_{k} > 0\). If there exists
\(D_{1} \in \Sigma _{k}\) such that
\(x,y \in D_{1}\), let
\(\succsim _{1}:D_{1}\). Otherwise,
\(x_{k},y_{k} > 0\) implies that
\(D_{2},D_{3} \in \Sigma _{k}\) exist such that
\(y \notin D_{2} \ni x\) and
\(x \notin D_{3} \ni y\). In this case, let
\(\succsim _{1}:D_{2}\sim D_{3}\). In either case, we have
\(\theta _{\succsim _{1}}(x) = \theta _{\succsim _{1}}(y)\). Therefore, Lemma
3 says that
\(xI_{\succsim _{1}}y\). Let
\(\succsim _{2} := {\succsim |}_{{{\,\textrm{und}\,}}( \succsim ) \backslash {{\,\textrm{und}\,}}( \succsim _{1} )}\). Because
\(\theta _{\succsim }(x) >_{E}\theta _{\succsim }(y)\), (b) in Lemma 2 says that
\(\theta _{\succsim _{2}}(x) >_{E}\theta _{\succsim _{2}}(y)\). By the assumption of induction, we have
\(xP_{\succsim _{2}}y\). Because
\(\succsim \in \succsim _{1} \oplus \succsim _{2}\), CON implies that
\(xP_{\succsim }y\).
(ii)
Suppose that for any
\(k \in [ K]\), either
\(x_{k} = 0\) or
\(y_{k} = 0\). In this case,
\(\{ x,y \} \notin {{\,\textrm{und}\,}}( \succsim )\). Let
\(\succsim _{1}:\{ x,y \}\). Lemma
3 states that
\(xI_{\succsim _{1}}y\). Let
\(\succsim _{2} \in \succsim \oplus \succsim _{1}\). CON says that
\(R_{\succsim } |_{\{ x,y \}} = R_{\succsim _{2}} |_{\{ x,y \}}\).
Because of
\(\theta _{\succsim }(x) >_{E}\theta _{\succsim }(y)\), there exists
\(k \in [ K]\), such that
\(x_{i} = 0\) for all
\(i < k\) and
\(x_{k} > 0\). Let
\(D \in {{{\,\textrm{und}\,}}}_{x}( \succsim ) \cap \Sigma _{k}\). Because of the assumption of (ii), we have
\(y_{k} = 0\). Furthermore, because
\(x_{i} = 0\) for all
\(i < k\) and
\(\theta _{\succsim }(x) >_{E}\theta _{\succsim }(y)\), it follows that any element of
\({{{\,\textrm{und}\,}}}_{y}( \succsim )\) is ranked below
D at
\(\succsim\). Let
\(\succsim _{3} := {\succsim _{2} |}_{\{ D \} \cup {{{\,\textrm{und}\,}}}_{y}( \succsim _{2} )}\) and
\(\succsim _{4} := {\succsim _{2} |}_{{{\,\textrm{und}\,}}( \succsim _{2} ) \backslash {{\,\textrm{und}\,}}( \succsim _{3} )}\). By the above discussion, it follows that
\(\theta _{\succsim _{3}}(x) >_{E}\theta _{\succsim _{3}}(y)\) and
\(\Vert \theta _{\succsim _{3}}(x) \Vert + \Vert \theta _{\succsim _{3}}(y) \Vert < \Vert \theta _{\succsim }(x) \Vert + \Vert \theta _{\succsim }(y) \Vert\). By the assumption of induction, we have
\(xP_{\succsim _{3}}y\). It also follows that
\(\theta _{\succsim _{4}}(x) \ge _{E}\theta _{\succsim _{4}}(y)\) and
\(\Vert \theta _{\succsim _{4}}(x) \Vert + \Vert \theta _{\succsim _{4}}(y) \Vert < \Vert \theta _{\succsim }(x) \Vert + \Vert \theta _{\succsim }(y) \Vert\). By the assumption of induction (and Lemma
3), we have
\(xR_{\succsim _{4}}y\). By CON, we have
\(xP_{\succsim _{2}}y\). Because of
\(R_{\succsim }|_{\{ x,y \}} = R_{\succsim _{2}} |_{\{ x,y \}}\), this means that
\(xP_{\succsim }y\).
\(\blacksquare\)
Now, we are ready to show the main result.