Modeling Value Disagreement

Abstract

In this article, monist values are expressed as preferences like in economics and decision making. On the basis of this formalization, various ways of defining value disagreement of agents within a group are investigated. Twelve notions of categorical value disagreement are laid out. Since these are too coarse-grained for many purposes, known distance-based approaches like Kendall’s Tau and Spearman’s footrule are generalized from linear orders to preorders and position-sensitive variants are developed. The account is further generalized to allow for agents with incomplete information. The article ends with a discussion of known limitations of preference-based accounts of values and how these might be overcome by accounting for parity and essential incompleteness. It is also shown that one intuitively compelling notion of disagreement does not give rise to a proper distance measure.

Appendix

1.1 Appendix 1: Distance Measures

A distance measure is a function D(x, y) of two arguments with domains X, Y into the real numbers that satisfies the following properties (Kemeny 1959, p. 587), (Deza and Deza 2009, p. 16):

$$\begin{aligned} D(x, y)&=0\Leftrightarrow x=y\quad \text {coincidence} \end{aligned}$$

(24)

$$\begin{aligned} D(x, y)&=D(y,x)\quad \text {symmetry}\end{aligned}$$

(25)

$$\begin{aligned} D(x, z)&\leqslant D(x, y)+D(y, z)\quad \text {triangle}\, \text{inequality} \end{aligned}$$

(26)

It defines a metric over the space $X\times Y$. Throughout the article the domain is the same for both arguments, namely the set of all linear ordering relations over the set of alternatives A.

1.2 Appendix 2: Proofs

In the following proofs it is assumed that the respective functions are based on a non-empty permutation $\pi (p, q)$. Since this permutation is not given explicitly as an argument, the following proofs concern in fact families of functions, but for simplicity we will speak as if they were single functions in what follows. As a shortcut, $A^\alpha _{x, y}{:=}\{(a, b)\mid \Delta _\alpha (x, y, a, b)>0\}$ is written for the set of disagreement pairs based on $\alpha$.

Proposition 1

(Table-based measures) $D_W$ and $D_T$ are distance measures.

Proof

• (a) Coincidence: we prove both directions of the biconditional separately. For proving the direction from left to right, assume (1) $D_W(p_x, q_y)=D_T(p_x, q_y)=0$ but (2) $p_x\ne q_y$. From (2) it follows that there is at least one pair $a, b\in A$ s.t. $a\succ _x b$ but not $a\succ _y b$. But for this pair Table 2b yields 1 and Table 3b either 1 or $\frac{1}{2}$, and so it follows from (13), according to which the result is half of the total sum, that $D_W(p_x, q_x)\ge \frac{1}{2}$ and $D_T(p_x, q_x)\ge \frac{1}{4}$, contradicting the assumption. For the other direction, assume (3) $p=q$ but (4) $D_W(p_x, q_x)\ne 0$ and $D_T(p_x, q_x)\ne 0$. From (13) and the tables it follows that the measures cannot be negative, hence because of (4) that $D_W(p_x, q_x)>0$ and $D_T(p_x, q_x)>0$. It follows from the tables that this can only be true if there is at least one pair $a, b\in A$ for which p and q differ, contradicting the assumption.

• (b) Symmetry: observe that $a\sim b$ is symmetric and that, moreover, Tables 2b and 3b are symmetric for $\succ$, i.e. $a\succ _x b$ and $b\succ _y a$ yields one and $b\succ _x a$ and $a\succ _y b$ yields one, and correspondingly for combinations of $\succ$ with $\sim$. From this it follows directly that instances of (13) by $\Delta _W$ and $\Delta _T$ are symmetric, too.

• (c) Triangle inequality: the proof is direct by induction on the size of the set of alternatives A. Notice first that positivity, i.e. $D(x, y)\ge 0$ for any x, y, holds trivially because the tables contain no negative values. Case 1: Wide disagreement. Let A be the domain and let $A'=A\cup \{b\}$ be the domain extended by one element b and $D'$ be a new measure obtained from D and the extended domain. If $A=\emptyset$, then $A'=\{b\}$. In this case, $b\sim _x b, b\sim _y b$ and $b\sim _z b$ hold by reflexivity and totality of ‘$\succeq$’, hence $D'(x, y)=D'(y, z)=D'(x, z)$ and so the inequality holds. Suppose now that A contains alternatives $a_1, a_2, \dots , a_n$ and that the inequality holds for D. The revised measure between x and z is $D'(x, z)=D(x, z)+k$ and we need to show that $D'(x, z)\leqslant D'(x, y)+D'(y, z)$. To do this, we assume a scenario in which k is maximal, try to minimize the other distances and show that the inequality still holds.

  The parameter k is maximal if for all $a_i\in A$ either (1) $a_i\succ _x b$ and $b\succ _z a_i$, or (2) $b\succ _x a$ and $a_i\succ _z b$, or (1) is the case for some $a_i$ in a subset B of A and (2) for all remaining $a_{j}\in (A\backslash B)$. The proof of case (2) is parallel to that of case (2) and therefore omitted. Case (3) is a mixture of case (1) and (2) and can be omitted without loss of generality as well (the only new element in $A'$ is b, so the new measure can be constructed as the sum of the measures for B and $A\backslash B$ like in cases (1) and (2) and their proofs carry over). Continuing with case (1), we assume $a_i\succ _x b$ and $b\succ _z a_i$ for all $a_i\in A$ and first try to make $D'(x, y)$ minimal. This is the case when $a_i\succ _y b$ for all $a_i\in A$, since then $D'(x, y)=D(x, y)$. But then $D'(y, z)-D(y, z)=k$ and so the triangle inequality is fulfilled for the new measure $D'$. Likewise, if $D'(y, z)$ is made minimal by assuming that $b\succ _y a_i$ for all $a_i\in A$, then $D'(x, y)-D(x, y)=k$ because by assumption $a_i\succ _x b$ holds, and the inequality is fulfilled as well. It is easy to see that this holds in general: Any equality of the agent’s preferences at a point $a_i, b$ between x, y (y, z) will not increase the respective measure, but then there will be a corresponding increase between y, z (x, y) that will suffice to ensure the triangle inequality.

  Case 2: Trivalent disagreement. The proof is analogous to the previous case. We start by induction and consider an extended measure with one additional alternative b. Let $\delta (x, y)$ be a shortcut for $\Delta _T (x, y, a, b)$. This time we list all possibilities for $a\succ _x b$:

  The five columns to the left can be read as a horizontally drawn ternary tree with the respective distances as labels on the edges. The two rightmost columns list the sum of the distances and D(x, y) respectively; clearly, $\Delta _T(x, y, a, b)+\Delta _T(y, z, a, b)\ge \Delta _T(x, z, a, b)$ holds for every possibility for arbitrary $a\in A$. The tables for $a\sim _x b$ and $b\succ _x a$ are analogous. Induction over A completes the proof. $\square$

x	$\delta (x, y)$	y	$\delta (y, z)$	z	$\delta (x, y)+\delta (y, z)$	$\delta (x, z)$
			0	$a\succ _z b$	0	0
	0	$a\succ _y b$	$\frac{1}{2}$	$a\sim _z b$	$\frac{1}{2}$	$\frac{1}{2}$
			1	$b\succ _z a$	1	1
			$\frac{1}{2}$	$a\succ _z b$	1	0
$a\succ _x b$	$\frac{1}{2}$	$a\sim _y b$	0	$a\sim _z b$	$\frac{1}{2}$	$\frac{1}{2}$
			$\frac{1}{2}$	$b\succ _z a$	1	1
			1	$a\succ _z b$	2	0
	1	$b\succ _y a$	$\frac{1}{2}$	$a\sim _z b$	$\frac{3}{2}$	$\frac{1}{2}\frac{1}{2}$
			0	$b\succ _z a$	1	1

The underlying reason for the following negative result is that loose agreement is not transitive:

Proposition 2

$D_N$ does not satisfy triangle inequality.

Proof

Let $A=\{a, b\}$ and $a\succ _x b, b\succ _z a$, and $a\sim _y b$. Clearly, $(a, b)\in A^N_{x, z}$ and in this case $\Delta _N (x, z, a, b)=1, \Delta _N(x, y, a, b)=0$ and $\Delta _N(y, z, a, b)=0$. Thus, $|A^N_{x, z}|=1$ but $D_N(x, y)+D_N(y, z)=0$, providing a counter-example. $\square$

Proposition 3

(Position-sensitive measures) ${\mathcal {D}}_\alpha$, ${\mathcal {I}}, {\mathcal {S}}$, and ${\mathcal {S}}^*$ are not symmetric.

Proof

By counterexample. Suppose preference p and q are such that $2\,3\,4\,1$ is the permutation that takes p to q. Then $4\,1\,2\,3$ is the inverse permutation taking q to p. Table 7 shows that $D(p, q)\ne D(q, p)$ when D is ${\mathcal {D}}_W, {\mathcal {I}}$, ${\mathcal {S}}$ or ${\mathcal {S}}^*$. Since trivalent disagreement collapses to wide disagreement whenever the orderings are strict, this example also works for ${\mathcal {D}}_T$. $\square$