Measuring the cohesiveness of preferences: an axiomatic analysis

Abstract

In this paper, we axiomatically study how to measure the similarity of preferences in a group of individuals. For simplicity, we refer to this as the cohesiveness. First, we provide axioms that characterize a family of linear and additive measures whose intersection is a partial ordinal criterion similar to first order stochastic dominance. The introduction of some additional properties isolates a one-parameter subfamily. This parameter evaluates the effect on the cohesiveness if one individual changes his ranking on a single pair of objects, as a function of how many of the remaining individuals in the group rank the first object over the second and vice versa. Finally, we characterize the focal measures of this subfamily separately showing that they coincide with measures constructed using two, at first sight, totally different approaches suggested in the literature.

Appendix

1.1 Proof of Theorem 1

It is straightforward to show that the measures of the family of cohesiveness measures defined in the theorem satisfies the properties. Therefore, we consider from now on the other implication. Take any cohesiveness measure $M$ that satisfies AN, IND, NEU, and MON. First, we show that for all groups $N \subset \mathbb{N }$, $M^N$ is a function of $d(P_N)$; that is, we are going to prove that for any two ranking profiles $P_N,P^{\prime }_N \in \mathcal{P }^N$ such that $d(P_N)=d(P^{\prime }_N)$, $M(P_N)=M(P^{\prime }_N)$.

Take any group $N \subset \mathbb{N }$ and any two ranking profiles $P_N,P^{\prime }_N \in \mathcal{P }^N$ with the property that $d(P_N)=d(P^{\prime }_N)$. Now, consider any unanimous ranking profile $\bar{P}_N \in \mathcal{P }^N$ (i.e., for all $i,j \in N, \bar{P}_i=\bar{P}_j$) and observe that it is possible to arrive from $P_N$ (and $P^{\prime }_N$, respectively) to $\bar{P}_N$ by means of successive changes of contiguous pairs of objects. Then, given any pair of distinct objects $\{x,y\} \in \bar{X}$, suppose without loss of generality that $\#(x P_N y) \ge \#(y P_N x)$. Now, there are two possibilities: $x \bar{P}_i y$ for all $i \in N$ or $y \bar{P}_i x$ for all $i \in N$.

• 1. If $x \bar{P}_i y$ for all $i \in N$, in the process of arriving from $P_N$ to $\bar{P}_N$, the contiguous pair of objects $(y,x)$ has to be changed exactly $\#(y P_N x)$-times to $(x,y)$. Consider first the particular case when $P_N$ is such that all individuals agree on how to order all pairs $\{w,z\}$ different from $\{x,y\}$. It follows from MON that each of these changes raises the cohesiveness. Then, by IND, each $xP_iy$-change raises the cohesiveness also if there is no unanimity in the rankings for pairs distinct from $\{x,y\}$. By IND and AN, each incremental only depends on the number of individuals that rank $x$ over $y$ in the profile at which this change is applied. Formally, denote by $t_j(x,y)$ the incremental in the cohesiveness when a contiguous pair $(y,x)$ is changed to $(x,y)$ and there were $j$ individuals that rank $x$ over $y$. Also by NEU, $t_j(x,y) = t_j(z,w) \equiv t_j$ for all ordered pairs of objects $(x,y),(z,w)$. Observe that the total incremental in the cohesiveness from the changes of the contiguous pair $(y,x)$ to $(x,y)$ is $T_{x,y} = \sum _{j= \#(x P_N y)}^{n-1} t_j$.

• 2. On the other hand, if $y \bar{P}_i x$ for all $i \in N$, in the process of arriving from $P_N$ to $\bar{P}_N$, the total incremental in the cohesiveness from the changes of the contiguous pair $(x,y)$ to $(y,x)$ equals $T_{y,x}=\sum _{j= \#(yP_Nx)}^{n-1} t_j$. By IND and NEU, $t_j+t_{n-1-j}=0$ for all $j \ne \frac{n-1}{2}$ and, by AN and IND, $t_{\frac{n-1}{2}}=0$ when $n$ is odd. Consequently, $T_{y,x}=T_{x,y}$, which states that the total incremental in the cohesiveness for the changes in the distinct pair of objects $\{x,y\} \in \bar{X}$ in the process of arriving from $P_N$ to $\bar{P}_N$ is independent on which of the two objects is preferred at any $\bar{P}_i$.

Since $d(P_N)=d(P^{\prime }_N)$ by assumption, there exists a bijection $\phi : \bar{X} \rightarrow \bar{X}$ such that $n_{\phi (\{x,y\})}(P^{\prime }_N)=n_{x,y}(P_N)$. We can thus conclude that there exists a pair of objects $\phi (\{x,y\})$ such that all changes of this pair in the process of arriving at $\bar{P}_N$ from $P^{\prime }_N$ raises the cohesiveness in the same quantity as changes of the pair of objects $\{x,y\}$ in the process of arriving at $\bar{P}_N$ from $P_N$. Repeating this process for all elements of $\bar{X}$, one can conclude that $M(\bar{P}_N) - M(P^{\prime }_N) = M(\bar{P}_N) - M(P_N)$. Hence, $M(P^{\prime }_N)=M(P_N)$.

It remains to be shown that $M^N$ is a linear and additive function of $d(P_N)$. To do that, observe that MON and IND imply some restrictions on $M^N$ depending on the values of $d(P_N)$. In particular, MON implies that for all ranking profiles $P_N,P^{\prime }_N \in \mathcal{P }^N$ such that $d_n(P^{\prime }_N) \ge d_n(P_N) = 1 - \frac{2}{k(k-1)}$, and for some $i \le n-2$ $d_i(P^{\prime }_N)=d_i(P_N) - \frac{2}{k(k-1)}$ and $d_{i+2}(P^{\prime }_N)=d_{i+2}(P_N) + \frac{2}{k(k-1)}$, $M(P^{\prime }_N)>M(P_N)$. This implication is called Condition 1 from now on. Furthermore, IND implies that for all ranking profiles $P_N,P^{\prime }_N,\bar{P}_N,\bar{P}^{\prime }_N \in \mathcal{P }^N$ such that for some $i \le n-2$, $d_i(P^{\prime }_N)=d_i(P_N)-\frac{2}{k(k-1)}$, $d_{i+2}(P^{\prime }_N)=d_{i+2}(P_N)+\frac{2}{k(k-1)}$, $d_i(\bar{P}^{\prime }_N)=d_i(\bar{P}_N)-\frac{2}{k(k-1)}$, $d_{i+2}(\bar{P}^{\prime }_N)=d_{i+2}(\bar{P}_N) +\frac{2}{k(k-1)}$, $d_j(P^{\prime }_N)=d_j(P_N)$ and $d_j(\bar{P}^{\prime }_N)=d_j(\bar{P}_N)$ for all $j \ne \{i, i+2\}$, $M(P^{\prime }_N) -M(P_N)= M(\bar{P}^{\prime }_N)-M(\bar{P}_N)$. We refer to this implication as Condition 2. It is easy to see that, in the presence of Condition 2, Condition 1 can be strengthened in the following way: for all ranking profiles $P_N,P^{\prime }_N \in \mathcal{P }^N$ such that for some $i \le n-2$, $d_i(P^{\prime }_N)=d_i(P_N)-\frac{2}{k(k-1)}$, $d_{i+2}(P^{\prime }_N)=d_{i+2}(P_N)+\frac{2}{k(k-1)}$ and $d_j(P^{\prime }_N)=d_j(P_N)$ for all $j \not \in \{i,i+2\}$, $M(P^{\prime }_N)>M(P_N)$. This implication is denoted Condition 3. Next, we proceed by induction separating the proof depending on whether the size of the group is even or odd.

• 1. Suppose that $n$ is even. It follows directly from Condition 3 that $M^N$ would attain its minimum at any ranking profile $P_N \in \mathcal{P }^N$ such that $d_0(P_N) = 1$ and $d_j(P_N) = 0$ for all $j \ne 0$. It is easy to see that this type of ranking profile exists. An example is a ranking profile such that $\frac{n}{2}$ individuals have any ranking $P_i \in \mathcal{P }$ and $\frac{n}{2}$ individuals have the ranking $P_i^{OP}$. Without loss of generality, let the cohesiveness at these profiles be $a^N_1$. Consider now any profile $P^{\prime }_N$ such that $d_i(P^{\prime }_N) = 0$ for all $i > 2$. Then, by Condition 2, $M(P^{\prime }_N) = a^N_1 + d_2(P^{\prime }_N) \cdot \frac{k(k-1)}{2} \cdot t_{\frac{n}{2}}$. Fix $a^N_2 = a^N_1 + t_{\frac{n}{2}} \cdot \frac{k(k-1)}{2}$.

  Suppose now that for any ranking profile $P^{\prime \prime }_N$ such that $d_i(P^{\prime \prime }_N) = 0$ for all $i>q$, with $q$ being an even number from $\{4,\ldots ,n-2\}$, $M(P^{\prime \prime }_N) = a^N_1 + d_2(P^{\prime \prime }_N) \cdot t_{\frac{n}{2}} \cdot \frac{k (k-1)}{2} + \ldots + d_q(P^{\prime \prime }_N) \cdot t_{\frac{n+q-2}{2}} \cdot \frac{k (k-1)}{2}$ and that $a^N_{\frac{q}{2}+1} = a^N_{\frac{q}{2}}+t_{\frac{n+q-2}{2}} \cdot \frac{k (k-1)}{2}$. Next, consider any ranking profile $\tilde{P}_N$ such that $d_i(\tilde{P}_N) = 0$ for all $i > q+2$. Then, by Condition 2, we have that $M(\tilde{P}_N) = a^N_1 + d_2(\tilde{P}_N) \cdot t_{\frac{n}{2}} \cdot \frac{k (k-1)}{2} + \ldots + d_q(\tilde{P}_N) \cdot t_{\frac{n+q-2}{2}} \frac{k (k-1)}{2} + d_{q+2}(\tilde{P}_N) \cdot t_{\frac{n+q}{2}} \cdot \frac{k (k-1)}{2}$. Define $a^N_{\frac{q}{2}+2} = a^N_{\frac{q}{2}+1}+ t_{\frac{n+q}{2}} \cdot \frac{k (k-1)}{2}$. Hence, the vector $a^N$ is completely defined and it is easy to see that for any arbitrary ranking profile $P_N \in \mathcal{P }^N$, $M(P_N) = a^N \cdot d(P_N)$.

• 2. Suppose that $n$ is odd. It follows directly from Condition 3 that $M^N$ would attain its minimum at any ranking profile $P_N \in \mathcal{P }^N$ such that $d_1(P_N) = 1$ and $d_i(P_N) = 0$ for all $i \ne 1$. It is easy to see that this type of ranking profile exists. An example is a ranking profile such that $\frac{n-1}{2}$ individuals have any ranking $P_i \in \mathcal{P }$ and $\frac{n+1}{2}$ individuals have the ranking $P_i^{OP}$. Without loss of generality, let the cohesiveness at these profiles be $a^N_1$. Consider now any profile $P^{\prime }_N$ such that $d_i(P^{\prime }_N) = 0$ for all $i > 3$. Then, by Condition 2, $M(P^{\prime }_N) = a^N_1 + d_3(P^{\prime }_N) \cdot t_{\frac{n+1}{2}} \cdot \frac{k (k-1)}{2}$. Fix $a^N_2 = a^N_1 + t_{\frac{n+1}{2}} \cdot \frac{k (k-1)}{2}$. Suppose now that for any ranking profile $P^{\prime \prime }_N$ such that $d_i(P^{\prime \prime }_N) = 0$ for all $i>q$, with $q$ being an odd number from $\{5,\ldots ,n-2\}$, $M(P^{\prime \prime }_N) = a^N_1 + d_3(P^{\prime \prime }_N) \cdot t_{\frac{n+1}{2}} \cdot \frac{k (k-1)}{2} + \ldots + d_q(P^{\prime \prime }_N) \cdot t_{\frac{n+q-2}{2}} \cdot \frac{k (k-1)}{2}$ and that $a^N_{\frac{q+1}{2}}= a^N_{\frac{q-1}{2}} + t_{\frac{n+q-2}{2}} \cdot \frac{k (k-1)}{2}$. Next, consider any ranking profile $\tilde{P}_N$ such that $d_i(\tilde{P}_N) = 0$ for all $i > q+2$. Then, by Condition 2, we have that $M(\tilde{P}_N) = a^N_1 + d_3(\tilde{P}_N) \cdot t_{\frac{n+1}{2}} \cdot \frac{k (k-1)}{2} + \ldots + d_q(\tilde{P}_N) \cdot t_{\frac{n+q-2}{2}} \cdot \frac{k (k-1)}{2} + d_{q+2}(\tilde{P}_N) \cdot t_{\frac{n+q}{2}} \cdot \frac{k (k-1)}{2}$. Let $a^N_{\frac{q+3}{2}} = a^N_{\frac{q+1}{2}} + t_{\frac{n+q}{2}} \cdot \frac{k (k-1)}{2}$. Hence, the vector $a^N$ is completely defined and it is easy to see that for any arbitrary ranking profile $P_N \in \mathcal{P }^N$, $M(P_N) = a^N \cdot d(P_N)$.

It follows from AN that $a^N = a^{\bar{N}}$ whenever the groups $N$ and $\bar{N}$ are equally sized. Therefore, it can be concluded that for any arbitrary ranking profile $P_N \in \mathcal{P }^N$, $M(P_N) = a^n \cdot d(P_N)$. By Condition 3, $t_i > 0$ for all $i \ge \frac{n}{2}$. This implies that for all $n$ and all $j \in \{1,2,\ldots ,\lfloor \frac{n}{2}\rfloor \}, a^n_j < a^n_{j+1}$. Given also that the minimal value of $M^N$ is attained at all ranking profiles $P_N \in \mathcal{P }^N$ such that $d_0(P_N) = 1$ if $n$ is even or $d_1(P_N)= 1$ if $n$ is odd, and by definition this minimal value has to be non-negative, we get that $a_1^n \ge 0$ for all $n$. Finally, by Condition 3, the maximal value of $M^N$ would be attained at all ranking profiles $P_N \in \mathcal{P }^N$ such that $d_n(P_N) = 1$. Since these ranking profiles exist (i.e., the unanimous ranking profiles $P_N$ such that $P_i = P_j$ for all $i, j \in N$) and, by definition, the maximal value of $M^N$ cannot be greater than 1, the restriction $a^n_{\lfloor \frac{n}{2}\rfloor +1} \le 1$ has to be satisfied.

1.2 Proof of Theorem 2

It is straightforward to show that the measures of the family of cohesiveness measures defined in the theorem satisfies the properties. Therefore, we consider from now on the other implication. Take any cohesiveness measure $M$ that satisfies IND, NEU, MON, REP, FR, and CON. By Theorem 1, $M \in \Omega$. In the following, we investigate the additional restrictions REP, FR, and CON impose on the set of vectors $a \equiv \{a^n\}_{n \ge 2}$. First, we establish six preliminary claims that will help us to develop the proof later on.

Claim 1

If $n$ is odd, $a_i^n = a_{2i}^{2n}$ for all $i \in \{1,\ldots ,\lfloor \frac{n}{2}\rfloor +1\}$.

Proof

Take any odd $n$, any $i \in \{1,\ldots ,\lfloor \frac{n}{2}\rfloor + 1\}$, and any ranking profile $P_N \in \mathcal{P }^N$ such that $d_{2i-1}(P_N) = 1$ and $d_j(P_N)=0$ for all $j\ne 2i-1$. Such a ranking profile always exists. For example, it is a ranking profile in which $\frac{n-1}{2}+i$ individuals have a ranking $P_i$ and $\frac{n+1}{2}-i$ have the ranking $P_i^{OP}$. Then, by Theorem 1, we have that $M(P_N) = a_i^n$. Consider now a group $N^{\prime }$ of size $2n$ and a ranking profile $P^{\prime }_{N^{\prime }}$ such that it is the union of $2$ isomorphic and disjoint copies of $P_N$. Then, by definition, $d_{4i-2}(P^{\prime }_{N^{\prime }}) = 1$ and $d_j(P^{\prime }_{N^{\prime }})=0$ for all $j \ne 4i-2$. Then, by Theorem 1, we have that $M(P^{\prime }_{N^{\prime }}) = a_{2i}^{2n}$. By REP, $M(P_{N}) = M(P^{\prime }_{N^{\prime }})$. Therefore, $a^n_i = a_{2i}^{2n}$. $\square$

Claim 2

If $n$ is even, $a_i^n = a_{2i-1}^{2n}$ for all $i \in \{1,\ldots ,\lfloor \frac{n}{2}\rfloor +1\}$.

Proof

Consider any even $n$ and any $i \in \{1,\ldots ,\frac{n}{2} + 1\}$. Take any ranking profile $P_N \in \mathcal{P }^N$ such that $d_{2i-2}(P_N) = 1$ and $d_j(P_N)=0$ for all $j\ne 2i-2$. Such a ranking profile always exists. For example, it is a ranking profile in which $\frac{n}{2}+i-1$ individuals have a ranking $P_i$ and $\frac{n}{2}-i+1$ have the ranking $P_i^{OP}$. Then, by Theorem 1, we have that $M(P_N) = a_i^n$. Consider now a group $N^{\prime }$ of size $2n$ and a ranking profile $P^{\prime }_{N^{\prime }}$ such that it is the union of $2$ isomorphic and disjoint copies of $P_N$. Then, by definition, $d_{4i-4} (P^{\prime }_{N^{\prime }}) = 1$ and $d_j(P^{\prime }_{N^{\prime }})=0$ for all $j \ne 4i-4$. Then, by Theorem 1, we have that $M(P^{\prime }_{N^{\prime }}) = a_{2i-1}^{2n}$. By REP, $M(P_{N}) = M(P^{\prime }_{N^{\prime }})$. Therefore, $a^n_i = a_{2i-1}^{2n}$. $\square$

Claim 3

If $n$ is odd, $a_1^n > 0$.

Proof

Suppose otherwise. Then, there is some odd $n$ such that $a_1^n =0$. Then, by Claim 1, $a^{2n}_{2} = 0$. Since $a^t_j < a^t_{j+1}$ for all $t$ and all $j \in \{1, \ldots , \lfloor \frac{t}{2}\rfloor \}$ by Theorem 1, it must be the case that $a^{2n}_1 < 0$. But this is not possible given that $a^t_1 \ge 0$ for all $t$ by Theorem 1. $\square$

Claim 4

If $n$ is even, $a_1^n = 0$.

Proof

Consider any even $n$. By FR, there is some group $C$ of size $c$ and some ranking profile $P^{\prime }_C \in \mathcal{P }^C$ such that $M(P^{\prime }_C)=0$. It follows from Theorem 1 that the minimal value of $M^C$ is $a_1^c$. Then, it follows from Claim 3 that $c$ is an even number. By definition, $P^{\prime }_C$ should be one ranking profile with the minimal value of $M^C$. By Theorem 1, this minimal value is attained in ranking profiles $P^{\prime \prime }_C$ satisfying that $d_0(P^{\prime \prime }_C)=1$, which always exist. An example is $P^{\prime \prime }_C$ such that $\frac{c}{2}$ individuals have the ranking $P^{\prime \prime }_i$ and the other $\frac{c}{2}$ individuals have the ranking $(P^{\prime \prime }_i)^{OP}$. By Theorem 1, $M(P^{\prime \prime }_C)=M(P^{\prime }_C)=0$.

Take any pair of individuals $\{i,j\} \subseteq C$ such that $P^{\prime \prime }_i \ne P^{\prime \prime }_j$. By REP, $M(P^{\prime \prime }_{\{i,j\}})=M(P^{\prime \prime }_C)=0$. Since $d_0(P^{\prime \prime }_{\{i,j\}}) = 1$ and $d_2(P^{\prime \prime }_{\{i,j\}})=0$, we know that $M(P^{\prime \prime }_{\{i,j\}})= a^2_1$ by Theorem 1. Hence, $a^2_1=0$. Now, consider the ranking profile $P_N$ that consists of the union of $\frac{n}{2}$ isomorphic and disjoint copies of $P^{\prime \prime }_{\{i,j\}}$. Then, $d_0(P_N) = 1$ and $d_j(P_N) = 0$ for all $j \ne 0$. By Theorem 1, we have that $M(P_N) = a^n_1$. By REP, we have that $M(P_N) = M(P^{\prime \prime }_{\{i,j\}}) = 0$. Therefore, $a_1^n = 0$. $\square$

Claim 5

For all $n$, $a^n_{\lfloor \frac{n}{2}\rfloor + 1} = 1$.

Proof

Consider any $n$. Observe first that, by FR, there exists a group $C$ of size $c$ and a ranking profile $P^{\prime }_C \in \mathcal{P }^C$ for which the cohesiveness is 1. By definition, $P^{\prime }_C$ should be one ranking profile with the maximal value of $M^C$. By Theorem 1, this maximal value is attained in ranking profiles $P^{\prime \prime }_C$ satisfying that $d_c(P^{\prime \prime }_C) = 1$ and $d_i(P^{\prime \prime }_C) = 0$ for all $i \ne c$, which always exist. An example is $P^{\prime \prime }_C$ such that $P^{\prime \prime }_i = P^{\prime \prime }_j$ for all $i, j \in C$. By Theorem 1, $M(P^{\prime }_C)=M(P^{\prime \prime }_C)= a_{\lfloor \frac{c}{2}\rfloor +1}^c = 1$. Consider now the profile $\bar{P}_B$ consisting in the union of $n$ isomorphic and disjoint copies of $P^{\prime }_C$. Then, $d_{nc}(\bar{P}_B) = 1$ and $d_i(\bar{P}_B) = 0$ for all $i \ne nc$. By REP, $M(\bar{P}_B) = M(P^{\prime }_C) = 1$. Therefore, $a^{nc}_{\lfloor \frac{nc}{2}\rfloor + 1} = 1$. Consider the profile $P_N$ of size $n$ such that all individuals $i \in N$ have the ranking $P^{\prime \prime }_i$. Note that $d_n(P_N) = 1$ and $d_i(P_N)=0$ for all $i \ne n$. Then, $M(P_N) = a^n_{\lfloor \frac{n}{2}\rfloor + 1}$ and $\bar{P}_B$ consists in the union of $c$ isomorphic and disjoint copies of $P_N$. Then, by REP, $M(P_N)=M(\bar{P}_B)=1$. Therefore, $a^n_{\lfloor \frac{n}{2}\rfloor + 1} = 1$. $\square$

Claim 6

If $n$ is a multiple of 4, $a_2^4 = a_{\frac{n}{4} + 1}^n$.

Proof

Consider any $n$ multiple of 4, a group $C$ of size $4$ and a ranking profile $P^{\prime }_C \in \mathcal{P }^C$ such that $d_2(P^{\prime }_C) = 1$ and $d_j (P^{\prime }_C)=0$ for all $j \ne 2$. Then, $M(P^{\prime }_C) = a_2^4$. Consider now the ranking profile $P_N$ of size $n$ consisting in the union of $\frac{n}{4}$ isomorphic and disjoint copies of $P^{\prime }_C$. Note that $d_{\frac{n}{2}} (P_N) = 1$ and $d_j (P_N)=0$ for all $j \ne \frac{n}{2}$. Then, $M(P_N) = a_{\frac{n}{4} + 1}^n$. By REP, $M(P_N) = M(P^{\prime }_C)$. Therefore, $a_2^4 = a_{\frac{n}{4} + 1}^n$. $\square$

To find the exact description of the set of vectors $a$, we divide the proof into two parts, depending on whether the size of the group $N$ is even or odd.

• 1. Suppose that $n$ is even. We know from Claims 4 and 5 that $a_1^n = 0$ and $a^n_{\frac{n}{2}+1} = 1$. This concludes the proof for the case $n=2$. Therefore, suppose next that $n\ge 4$. If $n = 4$, the unique value not determined by the axioms is $a_2^4$. Define $\gamma \equiv a^4_2$ and observe that, for the moment, by Theorem 1, $\gamma \in (0,1)$. If $n \ge 6$, it follows from CON and IND that for all $i \in \{1, \ldots , \frac{n}{2} - 2\}$, $\left((a_{i+2}^n - a_{i+1}^n) - (a_{i+1}^n - a_{i}^n)\right) = \left((a_{i+3}^n - a_{i+2}^n) - (a_{i+2}^n - a_{i+1}^n)\right)$. Consequently, if we define $\bar{p}(n,\gamma )\equiv a_2^n$ and $\bar{q}(n,\gamma ) \equiv \left((a_{i+2}^n - a_{i+1}^n) - (a_{i+1}^n - a_{i}^n)\right)$ for any $i \in \{1, \ldots , \frac{n}{2} - 2\}$, we have that $a_i^n = (i-1) \, \bar{p}(n,\gamma ) + \frac{(i-2)(i-1)}{2}\, \bar{q}(n,\gamma )$ for all $i \in \{1, \ldots , \frac{n}{2} + 1\}$. In case $n$ is a multiple of 4, $a^n_{\frac{n}{4}+1} = a^4_2 = \gamma$ by Claim 6. Using this and Claim 5 we obtain the following set of linear equations:

  $$\begin{aligned} \frac{n}{2} \, \bar{p}(n,\gamma ) + \frac{\frac{n}{2}(\frac{n}{2}-1)}{2} \, \bar{q}(n,\gamma )&= 1\\ \frac{n}{4} \, \bar{p}(n,\gamma ) + \frac{\frac{n}{4}(\frac{n}{4}-1)}{2} \, \bar{q}(n,\gamma )&= \gamma . \end{aligned}$$

  Solving for $\bar{p}(n,\gamma )$ and $\bar{q}(n,\gamma )$ yields

  $$\begin{aligned} \bar{p}(n,\gamma )&= \frac{1}{n^2} \, (16 \gamma + (8\gamma -2)(n-4))\\ \bar{q}(n,\gamma )&= \frac{1}{n^2} \, (16-32\gamma ). \end{aligned}$$

  Letting $p(n,\gamma )=n^2 \, \bar{p}(n,\gamma )$ and $q(n,\gamma )=n^2 \, \bar{q}(n,\gamma )$, one can easily check that if $n$ is a multiple of 4, then for all $i \in \{1, \ldots , \frac{n}{2}+1\}.$

  $$\begin{aligned} a^n_i=\frac{i-1}{2n^2} \, \left(2 \, p(n,\gamma )+(i-2)\, q(n,\gamma )\right). \end{aligned}$$

  Then, we have that

  $$\begin{aligned} a^n_i = \frac{i-1}{2n^2} \, \left[ 2 (16 \gamma + (8 \gamma - 2) (n-4)) + (i-2) (16-32 \gamma )\right]. \end{aligned}$$

  Slightly tedious but straightforward manipulations of this equation imply that

  $$\begin{aligned} a^n_i = \frac{2(i-1)}{n^2}\, [ n (4 \gamma -1) + 2 (i-1) (2-4\gamma ) ], \end{aligned}$$

  which can be easily rewritten in the desired form:

  $$\begin{aligned} a^n_i = \frac{2(i-1)}{n^2}\, [ 4(\gamma -0.25) \cdot n + (1-4(\gamma -0.25)) \cdot 2 (i-1)]. \end{aligned}$$

  For all other even values of $n$, observe that $a_i^n =a_{2i-1}^{2n}$ by Claim 2. Then, given that $2n$ is a multiple of $4$, we have that for all $i \in \{1, \ldots , \frac{n}{2}+1\}$,

  $$\begin{aligned} a_i^n = \frac{2(2i-2)}{4n^2} \, [ 4(\gamma -0.25) \cdot 2n + (1-4(\gamma -0.25)) \cdot 2 (2i-2)]. \end{aligned}$$

  It can be easily seen that this is the desired result.

• 2. Suppose now that $n$ is odd. We know from Claim 1 that $a_i^n = a_{2i}^{2n}$. Consequently, we have that for all $i \in \{1, \ldots , \lfloor \frac{n}{2}\rfloor +1\}$,

  $$\begin{aligned} a_i^n = \frac{2(2i-1)}{4n^2} \, [ 4(\gamma -0.25) \cdot 2n + (1-4(\gamma -0.25)) \cdot 2 (2i-1)]. \end{aligned}$$

  It can be easily seen that this is the desired result.

Finally, it remains to be shown that $\gamma \in [\frac{1}{4}, \frac{3}{4}]$. We know from Theorem 1 that for all $n$ and all $i \in \{2, \ldots , \lfloor \frac{n}{2}\rfloor +1\}$, it is necessary that $a_i^n - a_{i-1}^n > 0$. To guarantee it, we can focus on the cases in which $n$ is even and strictly greater than 2.⁵ We have then that for all $i \in \{2, \ldots , \frac{n}{2}+1\}$, $a_i^n - a_{i-1}^n = \bar{p}(n,\gamma )+ (i-2) \, \bar{q}(n,\gamma )$. Since this equation reduces to $a_2^n - a_1^n = \bar{p}(n,\gamma )$ for $i=2$, it is necessary that $p(n,\gamma ) > 0$ for all $n$ even and strictly greater than 2. Consequently, it is required that $16 \gamma + (8\gamma -2)(n-4) > 0$ for all $n$ even and strictly greater than 2. It is easy to see that if this inequality holds for $n \rightarrow \infty$, it also holds for all $n$ even and strictly greater than 2. For $n \rightarrow \infty$ this equation reduces to $8\gamma -2\ge 0$, which is equivalent to $\gamma \ge \frac{1}{4}$.

To establish that $a_i^n - a_{i-1}^n > 0$ for all $i \in \{3, \ldots , \frac{n}{2}+1\}$, we only have to show that $a_{\frac{n}{2}+1}^n - a_{\frac{n}{2}}^n = \bar{p}(n,\gamma ) + (\frac{n}{2}-1) \, \bar{q}(n,\gamma ) > 0$. This is because $\bar{p}(n,\gamma )>0$ for $\gamma \ge \frac{1}{4}$ and for any negative $\bar{q}(n,\gamma )$, $a_i^n - a_{i-1}^n$ is minimized for $i=\frac{n}{2}+1$. Using some algebra, it can be verified that the necessary condition is $n(6-8\gamma )+8(2\gamma -1)>0$. Again, it is sufficient to check the condition for $n \rightarrow \infty$. For $n \rightarrow \infty$ this equation reduces to $6-8\gamma \ge 0$, which is equivalent to $\gamma \le \frac{3}{4}$.

1.3 Proof of Theorem 3

It is straightforward to show that $\hat{\tau }$ satisfies the set of axioms. To show the other implication, consider any cohesiveness measure $M$ that satisfies REP, NEU, PROP, MON, and FR. Since IND and CON are implied by the other five properties, Theorem 2 applies and we only have to determine $\gamma = a^4_2$. Since it follows from Theorem 2 that $a^4_1=0$ and $a^4_3=1$, we can apply PROP to obtain that $\frac{\gamma }{1 - \gamma }=\frac{1}{3}$. This equation solves for $\gamma =0.25$.

1.4 Proof of Theorem 4

It is straightforward to show that $\bar{\sigma }$ satisfies the set of axioms. Similarly, some basic calculus can show that $\bar{\sigma }$ can be formulated as stated in points 2 and 3. Then, we have to show only that the set of axioms of point 1 imply that the cohesiveness measure is exactly the measure of $\Gamma$ with $\gamma = \frac{1}{2}$. To show it, consider any cohesiveness measure $M$ that satisfies REP, SIND, NEU, MON, and FR. Since IND and CON are implied by the other five properties, Theorem 2 applies and we only have to determine $\gamma = a^4_2$. Since it follows from Theorem 2 that $a_1^4=0$ and $a_3^4 = 1$, we can apply SIND to obtain that $\gamma - 0 = 1 - \gamma$. Hence, $\gamma =0.5$.

1.5 Independence in Theorem 1

We show by means of four examples that the properties in Theorem 1 are independent.

Anonymity: For every group $N \subset \mathbb{N }$, assign to each group of individuals $A \subseteq N$ a non-negative number $p_N(A)$ such that $p_N(A) = 0$ if $\# A \le \frac{n}{2}$, $p_N(A) > p_N(B)$ if $\# A > \# B$ and $\#A > \frac{n}{2}$, $p_N(N) = \frac{2}{k(k-1)}$, and $p_{N^{\prime }}(A) \ne p_{N^{\prime }}(B)$ for some group $N^{\prime } \subset \mathbb{N }$ and some $A, B \subset N^{\prime }$ such that $A \ne B$ and $\# A = \# B$. Given a ranking profile $P_N$ and a pair of objects $\{x,y\} \in \bar{X}$, denote the value assigned to the maximal set of individuals who prefer the object that is favored by the majority as $s_{x,y}(P_N)=p_N(\{A \subseteq N: \#(xP_Ay) \ge \#(yP_Ax) \text{ and} \text{ there} \text{ is} \text{ no} B \subseteq N \text{ s.t.} \#(xP_By) > \#(xP_Ay)\})$. Now, let cohesiveness measure $M_1$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_1(P_N)=\sum _{\{x,y\} \in \bar{X}} s_{x,y}(P_N)$. This cohesiveness measure satisfies NEU, IND, and MON. The following example shows that it is not anonymous. Let $N=\{1,2,3\}$ and $X=\{x,y\}$. Suppose that the ranking profiles $P_N$ and $P^{\prime }_N$ are such that $xP_1y$, $xP_2y$, $yP_3x$, $yP^{\prime }_1x$, $xP^{\prime }_2y$, and $xP^{\prime }_3y$. Moreover, let $p_N(\{1, 2\}) = p_N(\{1, 3\}) = \frac{1}{2}$ and $p_N(\{2, 3\}) = \frac{3}{4}$. Then, $M_1(P_N) = \frac{1}{2}$ and $M_1(P^{\prime }_N) = \frac{3}{4}$. AN would imply that $M_1(P_N)=M_1(P^{\prime }_N)$.

Neutrality: Let $q: \bar{X} \rightarrow \mathbb{R }_{++}$ be a function that assigns to each pair of objects $\{x,y\} \in \bar{X}$ a strictly positive weight $q_{x,y} > 0$ in such a way that $q_{x,y} \ne q_{w,z}$ for some $\{w, z\} \in (\bar{X} \setminus \{\{x,y\}\})$. Now, let cohesiveness measure $M_2$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_2(P_N)= \sum _{\{x,y\} \in \bar{X}} \frac{q_{x,y}}{\sum _{\{w,z\} \in \bar{X}} q_{w,z}} \sigma _{x,y}(P_N)$. This cohesiveness measure satisfies AN, IND, and MON. The following example shows that it is not neutral. Let $N=\{1,2\}$ and $X=\{x,y,z\}$. Suppose that the ranking profiles $P_N$ and $P^{\prime }_N$ are such that $xP_1yP_1z$, $yP_2xP_2z$, $zP^{\prime }_1yP^{\prime }_1x$, and $yP^{\prime }_2zP^{\prime }_2x$. Moreover, let $q_{x,z}=q_{y,z}=1$ and $q_{x,y}=2$. Then, $M_2(P_N)=\frac{2}{4}$ and $M_2(P^{\prime }_N)=\frac{3}{4}$. NEU would imply that $M_2(P_N)=M_2(P^{\prime }_N)$.

Independence: Let $f : \{1, \ldots , k\} \rightarrow \mathbb{R }_{++}$ be a function that assigns to each position in a ranking a strictly positive weight such that $f(v) \ne f(w)$ for some $v,w \in \{1, \ldots , k\}$. Then, for each group $N \subset \mathbb{N }$, define the function $F_N : X \times \mathcal{P }^N \rightarrow \mathbb{R }_{++}$ as $F_N(x, P_N) = \sum _{i \in N} f( \# \{y \in X : \lnot y P_i x\})$. Let the cohesiveness measure $M_3$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_3(P_N)=\sum _{\{x,y\} \in \bar{X}} \frac{F_N(x, P_N) + F_N(y, P_N)}{\sum _{\{w,z\} \in \bar{X}} F_N(w, P_N) + F_N(z, P_N)} \, \sigma _{x,y}(P_N)$. This cohesiveness measure satisfies AN, NEU, and MON. The following example shows that it is not independent. Let $N=\{1,2,3\}$ and $X=\{x,y,z\}$. Suppose that the ranking profiles $P_N$ and $\bar{P}_N$ are such that $xP_1yP_1z$, $P_2=P_1$, $yP_3xP_3z$, $z\bar{P}_1x\bar{P}_1y$, $\bar{P}_2=\bar{P}_1$, and $z\bar{P}_3y\bar{P}_3x$. Let the ranking profiles $P^{\prime }_N$ and $\bar{P}^{\prime }_N$ be obtained by performing a $yP_3x$-change and a $y\bar{P}_3x$-change, respectively. Moreover, let $f(1) = 3$, $f(2) = 2$ and $f(3) = 1$. Then $M_3(P^{\prime }_N)-M_3(P_N)=1-\frac{2}{3}=\frac{1}{3}$ and $M_3(\bar{P}^{\prime }_N)-M_3(\bar{P}_N)=1-\frac{8}{9}=\frac{1}{9}$. IND would imply that $M_3(P^{\prime }_N)-M_3(P_N)=M_3(\bar{P}^{\prime }_N)-M_3(\bar{P}_N)$.

Monotonicity: Let the cohesiveness measure $M_4$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_4(P_N)= 1-\bar{\sigma }(P_N)$. This cohesiveness measure satisfies AN, NEU, and IND. The following example shows that it is not monotonic. Let $N=\{1,2\}$ and $X=\{x,y\}$. Suppose that the ranking profiles $P_N$ and $P^{\prime }_N$ are such that $xP_1y$, $yP_2x$, $xP^{\prime }_1y$, and $xP^{\prime }_2y$. Then, $M_4(P_N)=1$ and $M_4(P^{\prime }_N)=0$. MON would imply that $M_4(P^{\prime }_N)>M_4(P_N)$.

1.6 Independence in Theorem 2

We show by means of six examples that the properties in Theorem 2 are independent.

Replication Invariance: As we showed in Sect. 5.1, the measure $\bar{\tau }$ satisfies NEU, IND, MON, FR and CON, but it does not satisfy REP.

Neutrality: The cohesiveness measure $M_2$ satisfies REP, IND, MON, FR, and CON. However, it is not neutral.

Independence: The cohesiveness measure $M_3$ satisfies REP, NEU, MON, FR, and CON. However, it is not independent.

Monotonicity: The cohesiveness measure $M_4$ satisfies REP, NEU, IND, FR, and CON. However, it is not monotonic.

Full range: Let the cohesiveness measure $M_5$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_6(P_N)= \frac{1}{2} \, \bar{\sigma }(P_N)$. This cohesiveness measure satisfies REP, NEU, IND, MON, and CON. Since its maximum is $\frac{1}{2}$, it does not satisfy Full Range.

Concordance: For every group $N \subset \mathbb{N }$, denote the vector of weights associated with the measure $\bar{\sigma }$ by $a^n(\bar{\sigma })$. Now, let the cohesiveness measure $M_6 \in \Omega$ be such that for all $n$, $a_{\lfloor \frac{n}{2}\rfloor +1}^n = 1$ and $a_i^n= \frac{a_i^n \, (\bar{\sigma })}{2}$ for all $i \in \{1,\ldots ,\lfloor \frac{n}{2}\rfloor \}$. This cohesiveness measure satisfies REP, NEU, IND, MON, and FR. The following example shows that it is not concordant. Suppose that $N=\{1, 2, 3, 4, 5, 6\}$ and $X=\{x,y\}$. Let $P_N$ be such that $x P_1 y$, $y P_4 x$, $P_1 = P_2 = P_3$, and $P_4 = P_5 = P_6$. Then, $M_6(P_N) = 0$. If the ranking profile $P^{\prime }_N$ is $(x,y)$-different from $P_N$ for $4$, then $M_6(P^{\prime }_N)=\frac{1}{6}$. If the ranking profile $P^{\prime \prime }_N$ is $(x,y)$-different from $P^{\prime }_N$ for individual $5$, then $M_6(P^{\prime \prime }_N)=\frac{1}{3}$. Finally, if the ranking profile $P^{\prime \prime \prime }_N$ is $(x,y)$-different from $P^{\prime \prime }_N$ for individual $6$, then $M_6(P^{\prime \prime \prime }_N)=1$. Consequently, $(M_6(P^{\prime \prime }_N)-M_6(P^{\prime }_N)) - (M_6(P^{\prime }_N)-M_6(P_N))= \frac{1}{6}-\frac{1}{6}=0$, whereas $(M_6(P^{\prime \prime \prime }_N)-M_6(P^{\prime \prime }_N)) - (M_6(P^{\prime \prime }_N)-M_6(P^{\prime }_N))= \frac{2}{3}-\frac{1}{6}=\frac{1}{2}$. CON would imply that $(M_6(P^{\prime \prime }_N)-M_6(P^{\prime }_N))-(M_6(P^{\prime }_N)-M_6(P_N))=(M_6(P^{\prime \prime \prime }_N)-M_6(P^{\prime \prime }_N)) - (M_6(P^{\prime \prime }_N)-M_6(P^{\prime }_N))$.

1.7 Independence in Theorem 3

We show by means of five examples that the properties in Theorem 3 are independent.

Replication Invariance: The cohesiveness measure $\bar{\tau }$ satisfies NEU, PROP, MON, and FR. However, it does not satisfy REP.

Neutrality: Let $q: \bar{X} \rightarrow \mathbb{R }_{++}$ be a function that assigns to each pair of objects $\{x,y\} \in \bar{X}$ a strictly positive weight $q_{x,y} > 0$ in such a way that $q_{x,y} \ne q_{w,z}$ for some $\{w, z\} \in (\bar{X} \setminus \{\{x,y\}\})$. Given a group $N \in \mathbb{N }$, a ranking profile $P_N$, and two distinct individuals $i,j \in N$, let $w_{i,j}(P_N)= \sum _{\{x,y\} \in \bar{X}} \frac{q_{x,y}}{\sum _{\{w,z\} \in \bar{X}} q_{w,z}} \cdot \tau _{i,j}^{x,y}(P_N)$ be the weighted percentage of pairwise comparisons individual $i$ and $j$ agree upon. Define $\bar{w}(P_N)=\frac{2}{n(n-1)}\sum _{i,j \in N:j \ne i} w_{i,j}(P_N)$ as the average weighted tau. Now, let cohesiveness measure $M_7$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_7(P_N) = \frac{\bar{w}(P_N) - \frac{1}{n}}{\frac{n-1}{n}}$ if $n$ is even and $M_7(P_N) = \frac{1}{n^2} + \frac{\bar{w}(P_N) - \frac{n-1}{2n}}{\frac{n+1}{2n}} \, \frac{n^2 - 1}{n^2}$ if $n$ is odd. This cohesiveness measure satisfies REP, PROP, MON, and FR. The following example shows that it is not neutral. Let $N=\{1,2\}$ and $X=\{x,y,z\}$. Suppose that the ranking profiles $P_N$ and $P^{\prime }_N$ are such that $xP_1yP_1z$, $yP_2xP_2z$, $zP^{\prime }_1yP^{\prime }_1x$, and $yP^{\prime }_2zP^{\prime }_2x$. Moreover, let $q_{x,z}=q_{y,z}=1$ and $q_{x,y}=2$. Then, $M_7(P_N)=\frac{2}{4}$ and $M_7(P^{\prime }_N)=\frac{3}{4}$. NEU would imply that $M_7(P_N)=M_7(P^{\prime }_N)$.

Proportionality: Let the cohesiveness measure $M_8$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_8(P_N)=\bar{\sigma }(P_N)$. This cohesiveness measure satisfies REP, NEU, MON, and FR. The following example shows that it is not proportional. Let $N=\{1,2,3,4\}$ and $X=\{x,y\}$. Suppose that the ranking profiles $P_N$ and $\bar{P}_N$ are such that $xP_1y$, $yP_2x$, $P_3=P_1$, $P_4=P_2$, $\bar{P}_1=\bar{P}_2=\bar{P}_3=P_1$, and $\bar{P}_4=P_2$. Let the ranking profiles $P^{\prime }_N$ and $\bar{P}^{\prime }_N$ be obtained by performing a $yP_4x$-change and a $y\bar{P}_4x$-change, respectively. Then, $M_8(P^{\prime }_N)-M_8(P_N)=\frac{1}{2}-0=\frac{1}{2}$ and $M_8(\bar{P}^{\prime }_N)-M_8(\bar{P}_N)=1-\frac{1}{2}=\frac{1}{2}$. PROP would imply that $\frac{M_8(P^{\prime }_N)-M_8(P_N)}{M_8(\bar{P}^{\prime }_N)-M_8(\bar{P}_N)}=\frac{1}{3}$.

Monotonicity: Let the cohesiveness measure $M_{9}$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_{9}(P_N)= 1-\hat{\tau }(P_N)$. This cohesiveness measure satisfies REP, NEU, PROP, and FR. The following example shows that it is not monotonic. Let $N=\{1,2\}$ and $X=\{x,y\}$. Suppose that the ranking profiles $P_N$ and $P^{\prime }_N$ are such that $xP_1y$, $yP_2x$, $xP^{\prime }_1y$, and $xP^{\prime }_2y$. Then, $M_{9}(P_N)=1$ and $M_{9}(P^{\prime }_N)=0$. MON would imply that $M_{9}(P^{\prime }_N)>M_{9}(P_N)$.

Full range: Let the cohesiveness measure $M_{10}$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_{10}(P_N)= \frac{1}{2} \, \hat{\tau }(P_N)$. This cohesiveness measure satisfies REP, NEU, PROP, and MON. Since its maximum is $\frac{1}{2}$, it does not satisfy FR.

1.8 Independence in Theorem 4

We show by means of five examples that the properties in Theorem 4 are independent.

Replication invariance: Let the cohesiveness measure $M_{11}$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_{11}(P_N)=\bar{\sigma }(P_N)$ whenever $n$ is even and $M_{11}(P_N)=\frac{\bar{\sigma }(P_N)-\frac{1}{n}}{1-\frac{1}{n}}$ if $n$ is odd. This cohesiveness measure satisfies NEU, SIND, MON, and FR. The following example shows that it does not satisfy REP. Let $N=\{1,2,3\}$, $\bar{N}=\{1,2,3,4,5,6\}$, and $X=\{x,y\}$. Let $xP_1y$, $yP_2x, P_3=P_4=P_6=P_1$, and $P_5=P_2$. Then, $M_{11}(P_N)=0$ and $M_{11}(P_{\bar{N}})=\frac{1}{3}$. REP would imply that $M_{11}(P_{\bar{N}})=M_{11}(P_N)$.

Neutrality: The cohesiveness measure $M_2$ satisfies REP, SIND, MON, and FR. However, it is not neutral.

Strong independence: Let the cohesiveness measure $M_{12}$ be such that for all groups $N \in \mathbb{N }$ and all ranking profiles $P_N$, $M_{12}(P_N)=\hat{\tau }(P_N)$. This cohesiveness measure satisfies REP, NEU, MON, and FR. The following example shows that it is not strongly independent. Let $N=\{1,2,3,4\}$ and $X=\{x,y\}$. Suppose that the ranking profiles $P_N$ and $\bar{P}_N$ are such that $xP_1y$, $yP_2x$, $P_3=P_1$, $P_4=P_2$, $\bar{P}_1=\bar{P}_2=\bar{P}_3=P_1$, and $\bar{P}_4=P_2$. Let the ranking profiles $P^{\prime }_N$ and $\bar{P}^{\prime }_N$ be obtained by performing a $yP_4x$-change and a $y\bar{P}_4x$-change, respectively. Then, $M_{12}(P^{\prime }_N)-M_{12}(P_N)=\frac{1}{4}-0=\frac{1}{4}$ and $M_{12}(\bar{P}^{\prime }_N)-M_{12}(\bar{P}_N)=1-\frac{1}{4}=\frac{3}{4}$. SIND would imply that $M_{12}(P^{\prime }_N)-M_{12}(P_N)=M_{12}(\bar{P}^{\prime }_N)-M_{12}(\bar{P}_N)$.

Monotonicity: The cohesiveness measure $M_4$ satisfies REP, NEU, SIND, and FR. However, it is not monotonic.

Full range: The cohesiveness measure $M_5$ satisfies REP, NEU, SIND, and MON. However, it does not satisfy FR.