Plural identities and preference formation

Abstract

This article explores under what conditions an agent can derive a transitive all-things-considered preference from a plurality of non-comparable objectives, values or judgements mirroring her plural identity. In contrast to existing contributions, the multiple values are reflected by partial (viz. incomplete) orderings. It is shown that a slight modification of the conditions employed by Arrow implies a spread of the dictate of one identity to its ‘incomplete parts’. The second result reveals that if one requires the decision making power to be spread a bit more equally across the various parts of a person’s identity, then even the derivation of an acyclic all-things-considered preference is rendered impossible. A third result shows that requiring some minimal consistency among a person’s plural identities, introduced via a domain restriction, allows the avoidance of the highlighted impossibilities.

Appendix

Definition 1

Identity Aggregation Procedure An Identity Aggregation Procedure (IAP) is a function $f: \fancyscript{D} \rightarrow \fancyscript{R}_{a},\; \fancyscript{D}\subseteq \fancyscript{R}_{s}^{M}$.

Condition,

Unrestricted Domain $f(.)$ satisfies Unrestricted Domain (UD) iff $\fancyscript{D}=\fancyscript{R}_{s}^{M}$.

Definition 2

Minimal Decisiveness For all $i \in M$, for all $x\in X,\; i$ is minimally decisive for $\{x,y\}$ (or for $x$ against $y$) iff for all $d \in \fancyscript{D}$: [$xP_iy$ and $xN_jy$ for all $j \not = i$], implies $xPy$, and [$yP_ix$ and $xN_jy$ for all $j \not = i$], implies $yPx$.

Condition,

Minimal Identity Decisiveness $f(.)$ satisfies Minimal Identity Decisiveness (MID) iff for all $x,y \in X$ there exists $i \in M$ such that $i$ is minimally decisive for $x$ against $y$.

Condition,

Monotonicity $f(.)$ satisfies Monotonicity (M) iff for all $d, d' \in \fancyscript{D},$ for all $x,y \in X$, if $[$ for all $i \in M (xP_{i}y$ implies $xP'_{i}y)$ and there does not exist $j \in M$ such that $(xN_{j}y$ and $yP'_{j}x)]$ then $[xPy$ implies $xP'y]$.

Condition,

Strong Imposition $f(.)$ is strongly imposed iff there exists $i \in M$ such that for all $x,y \in X, xP_{i}y$ implies $xPy$ and $xN_{i}y$ implies $xNy$.

Condition,

All-things-considered Transitivity $f(.)$ satisfies All-things-considered Transitivity (TP) iff its range is restricted to $\fancyscript{R}_{s}$.

Lemma 1

Suppose $f(.)$ satisfies conditions TP, UD, MID, and M. For $m \ge 2$ and $\# X \ge 3$, for all $x,y \in X$, if for all $i \in M, xN_{i}y$, then $xNy$.

Proof

Suppose to the contrary, namely that TP, UD, MID, and M are satisfied, but there exists $x,y \in X$ and $d^{1} \in \fancyscript{D}$ such that for all $i \in M, xN_{i}y$ and $xP^{1}y$. Let $z \in X-\{x,y\}$.

By MID there is some element of $M$ that is minimally decisive for $\{y,z\}$. Say $j$ is so, and consider a profile $d^2$ in which

• (a) for all   $i \not = j$: $xN^2_iy$, $yN^2_iz$ and $xN^2_iz$

• (b) $xN^2_jy$, $yP^2_jz$ and $xN^2_jz$.

By M we obtain $xP^2y$, by MID $yP^2z$, and by TP $xP^2z$.

By MID some $k \in M$ is minimally decisive for $\{x,z\}$. Suppose $j \not = k$ and consider a profile $d'^2$ in which the following holds:

• (a) for all   $i \not = j, k$: $xN'^2_iy$, $yN'^2_iz$ and $xN'^2_iz$

• (b) $xN'^2_jy$, $yP'^2_jz$ and $xN'^2_jz$

• (c) $xN'^2_ky$, $yN'^2_kz$ and $zP'^2_k x$.

By M we obtain $xP'^2y$, by MID $yP'^2z$, and $zP'^2x$, violating TP. Hence $j = k$ and $j$ is also minimally decisive for $\{x,z\}$.

Consider a profile $d^3$ in which

• (a) for all   $i \not = j$: $xN^3_iy, yN^3_iz, xN^3_iz$

• (b) $xN^3_jy, yN^3_jz, zP^3_jx$.

By M we obtain $xP^3y$, by MID $zP^3x$, and by TP $zP^3y$.

By MID some $l \in M$ is minimally decisive for $\{x,y\}$. Take a profile $d^4$:

• (a) for all   $i \not = l$: $xN^4_iy, yN^4_iz, xN^4_iz$

• (b) $yP^4_lx, yN^4_lz, xN^4_lz$.

By MID we obtain $yP^4x$, and by M, $zP^3y$ and $xP^2z$, we derive $zP^4y$, and $xP^4z$, violating TP. $\square$

Theorem 1

For $m \ge 2$ and $\# X \ge 3$, every $f(.)$ that satisfies TP, UD, MID, and M is strongly imposed.

Proof

(1) In the first step we prove that for any triple $\{x,y,z\}$ there is some $i \in M$ that is minimally decisive for all distinct $a, b \in \{x,y,z\}$.

By MID for each pair of alternatives $a, b \in \{x,y,z\}$, there is some part of the person’s identity that is minimally decisive for it. Assume there is no identity part that is minimally decisive for at least two pairs in $\{x,y,z\}$. There must then be three different identity parts $i$, $j$ and $k$, such that $i$ is minimally decisive for $x,y$, $j$ for $y, z$ and $k$ for $x,z$. Now consider a profile $d$ satisfying:

• (a) $xP_{i}y$ and $aN_{i}b$   for all   $\{a,b\} \ne \{x,y\}$

• (b) $yP_{j}z$ and $bN_{j}c$   for all   $\{a,b\} \ne \{y,z\}$

• (c) $zP_{k}x$ and $aN_{k}c$   for all   $\{a,b\} \ne \{x,z\}$

• (d) $aN_lb$   for all   $a, b \in X$   and all   $l \not \in \{i,j,k\}$.

By UD, any such $d$ is in the domain. By MID we get $xPy, yPz$ and $zPx$, which violates TP. Hence, there is some $i$ that is minimally decisive for at least two pairs in $\{x,y,z\}$.

We now show that $i$ is minimally decisive for all pairs in $\{x,y,z\}$. Without loss of generality, assume $i$ is minimally decisive for the pairs $\{x,y \}$ and $\{y,z\}$. Consider any profile $d$, such that:

• (a) $xP_{i}y$, $yP_{i}z$ and $xP_{i}z$

• (b) $xN_{k}y$, $yN_{k}z$ and $zN_{k}x$   for all $k\not = i$.

By MID we get $xPy$ and $yPz$ and thus by TP, $xPz$.

In a similar way, we can construct a profile $d''$ in which $zP''_i x$ and $xN''_jz$ for all $j \not =i$ and in which $zP''x$. Hence, $i$ is minimally decisive for $\{x,z\}$.

(2) We next show that for any triple $\{x,y,z\}$ if some $i \in N$ is minimally decisive for all distinct $a, b \in \{x,y,z\}$, then for all profiles $d$, and all $a,b \in \{x,y,z\}$, $aP_ib$ implies $aPb$ and $aN_ib$ implies $aNb$, that is, $i$ is decisive for $a$ against $b$.

Let $i$ be as defined and assume $xP_iy$. We want to show that $xPy$. Take an arbitrary profile $d'$ satisfying:

• (a) $xP'_{i}y$, $zP'_{i}y$, $xP'_{i}z$

• (b) $yP'_{j}x$, $yN'_jz$ $xN'_{j}z$   for all $j \not = i$.

By MID we get $xP'z$ and $zP'y$ and, by TP, $xP'y$.

Since $xP'y$, by M also $xPy$, which had to be shown.

Next let $d$ be an arbitrary profile in which $xN_{i}y$. We now have to show that $xNy$. Assume to the contrary that it is not the case that $xNy$, say we have $xPy$. Let $D$ be the set of all profiles $d'$ satisfying:

• (a) $xN'_{i}y$, $zN'_{i}y$, $zP'_{i}x$

• (b) $zN'_{j}x$ and $zN'_jy$,   for all $j \not = i$.

There must be some profile $d' \in D$ such that the restriction of $d'$ to $\{x,y\}$ is exactly the same as it is in $d$. Since $xPy$, by M also $xP'y$. By MID we have $zP'x$, and thus by TP $zP'y$, contradicting Lemma 1.

(3) We show that if there is some $i$ such that $i$ is decisive for any pair over the triple $\{x,y,z\}$, then $i$ is decisive for any pair of alternatives in $X$.

Let $i$ be as defined. We have to show that $i$ is decisive for any $v,w \in X$. If $\{v,w\} \subseteq \{x,y,z\}$, we already have the desired result. Say $v \not \in \{x,y,z\}$. Consider the triple $\{v,x,y\}$. By step 1 some $j$ is minimally decisive for each pair in $\{v,x,y\}$, and thus also for $\{x,y\}$. By step 2, at most one individual can be decisive for $x$ and $y$, and therefore $i = j$ and $i$ is decisive for $\{v,y\}$. If $w \in \{v,x,y\}$, the result follows immediately. Let $w \not \in \{v,x,y\}$. By step 1 some $j$ is minimally decisive for each pair in $\{v,w,y\}$. Again, by step 2, there is at most one individual decisive for $v$ and $y$ and we have just shown this to be $i$. Hence, $i$ is decisive for each pair in $\{v,w,y\}$, and thus in particular for $\{v, w\}$. $\square$

Condition,

Identity Decisiveness $f(.)$ satisfies Identity Decisiveness (ID) iff for all $d \in \fancyscript{D}$, all $i \in M$, and all $x,y \in X$: if $[xP_{i}y$ and $xN_{j}y$ for all $j \not =i$], then $xPy$.

Theorem 2

For $m \ge 2$ and $\# X > 3$, there is no $f(.)$ that satisfies UD and ID.

Proof

Consider the set of profiles $D$ satisfying for some distinct elements $x, y, z, w$ of $X$:

• (a) $xP_{i}y$, $yN_{i}z$, $zP_{i}w$, $wN_{i}x$

• (b) $xN_{j}y$, $yP_{j}z$, $zN_{j}w$, $wP_{j}x$

• (c) for all   $k \not = i, j$ and all $x,y \in X$: $xN_ky$.

By UD, the class of profiles $D$ is within our domain. By ID, one obtains $xPy$, $yPz$, $zPw$ and $wPx$, violating acyclicity. $\square$

Condition,

Imposition $f(.)$ is imposed iff there exists $i \in M$ such that for all $x,y \in X, xP_{i}y$ implies $xPy$.

Condition,

Value Consistency $f(.)$ satisfies Value Consistency (VC) iff $\fancyscript{D}=\{d \in \fancyscript{R}_{s}^{M}\mid$ for all $x,y,z \in X$, for all $i,j \in M, xP_i y$ and $yP_j z$ implies that there exists $k \in M$ such that $xP_{k}z\}$.

Theorem 3

If VC is satisfied there is an IAP that satisfies ID, M and TP and is not imposed.

Proof

An Identity Aggregation Procedure which satisfies the conditions is the Modified Intersection Rule $\hat{f}$. It is defined as follows (where $\hat{R}, \hat{P}$ and $\hat{N}$ denote the relations $R, P$ and $N$ assigned by $\hat{f}$):

For all $x, y \in X$, $x \hat{R}y$ if, and only if, (a) $x =y$, or (b) for all $i \in M$, not $yP_i x$ and for some $j \in M$, $xP_jy$.

The proof that the modified intersection rule satisfies ID, M and is not imposed is straightforward. Therefore we only prove TP.

Transitivity: We prove by contradiction. Suppose there exists some $x,y,z \in X$ such that $x\hat{P}y, y\hat{P}z$ and not $(x\hat{P}z)$. By definition of the modified intersection rule, we have

• 1. For all $i \in M,$ not $(yP_{i}x)$, not $(zP_{i}y)$;

• 2. there exists $i,j\in M$, such that $xP_{i}y$ and $yP_{j}z$.

By VC, $xP_{i}y$ and $yP_{j}z$ implies that there exists $k \in M$ such that $xP_{k}z$. Since, by assumption, not $(x\hat{P}z)$ there exists some $l \in M$ such that $zP_{l}x$. By VC, $zP_{l}x$ and $xP_{i}y$ imply that there exists $m \in M$ such that $zP_{m}y$, contradicting 1. $\square$