Convergent Menus of Social Choice Rules

Abstract

Suzuki and Horita [11] proposed the notion of convergence as a new solution for the procedural choice problem. Given a menu of feasible social choice rules (SCRs) $F$ and a set of options $X$, a preference profile $L^{0}$ is said to (weakly) converge to $C \,\subseteq\, X$ if every rule to choose the rule (or every rule to choose the rule to choose the rule, and so on) ultimately designates C under a consequential sequence of meta-preference profiles. Although its frequency is shown, for example, under a large society with F = {plurality, Borda, anti-plurality}, a certain failure (trivial deadlock) occurs with small probability. The objective of this article is to find a convergent menu (a menu that can “always” derive the convergence). The results show that (1) several menus of well-known SCRs, such as {Borda, Hare, Black}, are convergent and that (2) the menu {plurality, Borda, anti-plurality} and a certain class of scoring menus can be expanded so that they become convergent.

Appendix A

Proof of Lemma 1.

Given a menu $F$ and a sequence of CI profiles $L^{0} , \ldots ,L^{k - 1}$ which satisfy the stated conditions, let

$$ F_{x} : = \left\{ {f \in F\left| {C_{f} } \right.\left[ {L^{0} ,L^{1} , \ldots ,L^{k - 1} } \right] = \left\{ x \right\}} \right\}, $$

$$ F_{y} \text{ := }\left\{ {f \in F\left| C \right.\left[ {f{:}L^{0} ,L^{1} , \ldots ,L^{k - 1} } \right] = \left\{ y \right\}} \right\}, $$

and let $\alpha \text{ := }\left| {F_{x} } \right|$ and $\beta \text{ := }\left| {F_{y} } \right|$. We label the elements as $F_{x} = \left\{ {g_{1} ,g_{2} , \ldots ,g_{\alpha } } \right\}$ and $F_{y} = \left\{ {h_{1} ,h_{2} , \ldots ,h_{\beta } } \right\}$. Also $N_{x} = \left\{ {\left. {i\,{ \in }\,N} \right|xL_{i}^{0} y} \right\}$, $N_{y} = \left\{ {\left. {i\,{ \in }\,N} \right|yL_{i}^{0} x} \right\}$, $n_{x} = \left| {N_{x} } \right|$, and $n_{y} = \left| {N_{y} } \right|$. Since $\alpha + \beta = 3$, we have two possible cases: (a) $\left( {\alpha ,\beta } \right) = \left( {2,1} \right)$ and (b) $\left( {\alpha ,\beta } \right) = \left( {1,2} \right)$.

(a) The case of $\left( {\varvec{\alpha},\varvec{\beta}} \right) = \left( {\bf 2,1} \right)$

Define $L^{k} \,\in\, {\mathcal{L}}^{k} \left[ {L^{0} , \ldots ,L^{k - 1} } \right]$ as follows.

$$ L_{i}^{k} :\left\{ {\begin{array}{*{20}c} {g_{1} ,g_{2} ,h_{1} } & {if\,i \,\in\, N_{x} } \\ {h_{1} ,g_{1} ,g_{2} } & {if\,i \,\in\, N_{y} .} \\ \end{array} } \right. $$

It is easy to see that every $f \,\in\, {\mathcal{F}}$ chooses a subset of $\left\{ {g_{1} ,g_{2} } \right\}$. So, $L^{0}$ weakly converges to $\left\{ x \right\}$.

(b) The case of $\left( {\varvec{\alpha},\varvec{\beta}} \right) = \left( {\bf 1,2} \right)$

Define $L^{k} \,\in\, {\mathcal{L}}^{k} \left[ {L^{0} ,L^{1} , \ldots ,L^{k - 1} } \right]$ as follows.

$$ L_{i}^{k} :\left\{ {\begin{array}{*{20}l} {g_{1} ,h_{1} ,h_{2} }& {for} &~ {\left( {n_{y} + 1} \right) \,{\text{individuals}}\,{\text{in}} \,N_{x} } \\ {g_{1} ,h_{2} ,h_{1} } & { for} &~{n_{x} - \left( {n_{y} + 1} \right) \,{\text{individuals }}{\text{in}}\,N_{x} } \\ {h_{1} ,h_{2} ,g_{1} } & {for} &~{\left\lfloor\frac{n}{2}\right\rfloor - \left( {n_{y} + 1} \right)\,{\text{individuals }} {\text{in}}\, N_{y} } \\ {h_{2} ,h_{1},g_{1} } & {for} &~ {the\, other\, individuals.} \\\end{array} } \right. $$

Intuitively, $L^{k}$ is a profile such that the score of $g_{1}$ is the largest and the scores of $h_{1}$ and $h_{2}$ are the smallest. First, it is easy to see that each level-$\left( {k + 1} \right)$ $f_{P} ,f_{H} ,f_{\text{C}} ,f_{Bl} ,f_{M}$ chooses $\left\{ {g_{1} } \right\}$.

Consider $f_{B}$. For simplicity, we denote by $s\left( f \right)$ the score of $f \,\in\, F^{k}$ evaluated by $f_{B}$. Since each $i \,\in\, N_{x}$ ranks $g_{1}$ at the first position and each $i \,\in\, N_{y}$ ranks it at the third, we have $s\left( {g_{1} } \right) = n_{x}$. Since $\left\lfloor\frac{n}{2}\right\rfloor$ individuals rank $h_{1}$ above $h_{2}$ and $n - \left\lfloor\frac{n}{2}\right\rfloor \, {=}\,$ individuals rank $h_{2}$ above $h_{1}$, we have $s\left( {h_{2} } \right) \ge s\left( {h_{1} } \right)$. Furthermore,

$$ \begin{aligned} s\left( {h_{2} } \right) & = \frac{1}{2}\left[ {\left\{ {n_{x} - \left( {n_{y} + 1} \right)} \right\} + \left\{ {\left\lfloor\frac{n}{2}\right\rfloor - \left( {n_{y} + 1} \right)} \right\}} \right] + \left( {2n_{y} + 1 -\left\lfloor\frac{n}{2}\right\rfloor} \right) \\ & = \frac{1}{2}n_{x} + n_{y} - \frac{1}{2} \cdot \left\lfloor\frac{n}{2}\right\rfloor \\ & \le \frac{1}{2}n_{x} + n_{y} - \frac{1}{2} \cdot n_{y} \left( {\because \left\lfloor\frac{n}{2}\right\rfloor \ge n_{y} } \right) \\ & < n_{x} = s\left( {g_{1} } \right) \left( {\because n_{x} > n_{y} } \right). \\ \end{aligned} $$

Therefore, $f_{B}^{k + 1} \left( {L^{k} } \right) = \left\{ {g_{1} } \right\}$. It is straightforward to check that $f_{N}^{k + 1} \left( {L^{k} } \right) = \left\{ {g_{1} } \right\}$.

Finally, consider $f_{A}$. Since $n_{y}$ individuals rank $g_{1}$ at the third position and $\left( {n_{y} + 1} \right)$ individuals rank $h_{2}$ at the third, it follows that $f_{A}^{k + 1} \left( {L^{k} } \right) \,\subseteq\, \left\{ {g_{1} ,h_{1} } \right\}$. Recall that each $f \,\in\, {\mathcal{F}\backslash }\left\{ {f_{A} } \right\}$ chooses $\left\{ {g_{1} } \right\}$ at $L^{k}$. If $f_{A}^{k + 1} \left( {L^{k} } \right) = \left\{ {g_{1} } \right\}$, this implies that $L^{0}$ weakly converges to $\left\{ x \right\}$. If $f_{A}^{k + 1} \left( {L^{k} } \right) = \left\{ {h_{1} } \right\}$, we can apply the case (a) to the CI sequence $L^{0} ,L^{1} , \ldots ,L^{k}$ (instead of the sequence $L^{0} ,L^{1} , \ldots ,L^{k - 1}$) to find the convergence. Suppose $f_{A}^{k + 1} \left( {L^{k} } \right) = \left\{ {g_{1} ,h_{1} } \right\}$. Then, it follows that

$$ n_{x} - \left( {n_{y} + 1} \right) = n_{x} . $$

This implies that $n_{x} = 2n_{y} + 1$. Then, let $M^{k} \,\in\, {\mathcal{L}}^{k} \left[ {L^{0} ,L^{1} , \ldots ,L^{k - 1} } \right]$ as

$$ M_{i}^{k} :\left\{ {\begin{array}{*{20}l} { {g_{1} ,h_{1} ,h_{2} } \,\,\,for\,\,\, \left( {n_{y} + 2} \right) \rm{individuals} \,\rm{in} \,\,{\it N}_{\it x} } \hfill \\ { {g_{1} ,h_{2} ,h_{1} } \,\,\,for \,\,\,\left( {n_{y} - 1} \right) \rm{individuals} \, \rm{in} \,{\it N}_{\it x} } \hfill \\ { {h_{1} ,h_{2} ,g_{1} } \,\,\,for \,\,\,\left\lfloor\frac{n}{2}\right\rfloor - \left( {n_{y} + 2} \right) \rm{individuals}\, \rm{in}\, {\it N}_{\it y} } \hfill \\ { {h_{2} ,h_{1} ,g_{1} } \,\,\,for \,\,\,the \,other\, individuals.} \hfill \\ \end{array} } \right. $$

In a similar way, we can check that $f_{P} ,f_{H} ,f_{\text{C}} ,f_{Bl} ,f_{M} ,f_{B} ,f_{N}$ chooses $\left\{ {g_{1} } \right\}$ at $M^{k}$. Also, we have $f_{A}^{k + 1} \left( {M^{k} } \right) = \left\{ {h_{1} } \right\}$. So, we can apply the case (a) to the CI sequence $L^{0} ,L^{1} , \ldots ,M^{k}$ to find the convergence. ■

Proof of Theorem 1.

Let $f_{1} ,f_{2} ,f_{3}$ be distinct SCRs among $f_{P} ,f_{B} ,f_{A} ,f_{H} ,f_{N} ,f_{C} ,f_{M} ,f_{Bl}$. When $n \to \infty$ under IAC, it is easy to see that the probability of tied outcomes by some of $f_{1}^{1} ,f_{2}^{2} ,f_{3}^{1}$ is negligible. So, we can discuss only $L^{0} \,\in\, {\mathcal{L}}\left( X \right)^{n}$ such that each $f_{1}^{1} \left( {L^{0} } \right),f_{2}^{1} \left( {L^{0} } \right),f_{3}^{1} \left( {L^{0} } \right)$ is a singleton.

• (1) The Case of $\left| {\left\{ {\varvec{f}_{1} \left( {\varvec{L}^{0} } \right),\varvec{f}_{2} \left( {\varvec{L}^{0} } \right),\varvec{f}_{3} \left( {\varvec{L}^{0} } \right)} \right\}} \right| = 2$

Let $\left\{ {f_{1} \left( {L^{0} } \right),f_{2} \left( {L^{0} } \right),f_{3} \left( {L^{0} } \right)} \right\} = \left\{ {x,y} \right\}$. When $n \to \infty$, the probability of the event

$$ {\# }\left\{ {i \,\in\, N\left| {xL_{i}^{0} y} \right.} \right\} = {\# }\left\{ {i \,\in\, N\left| {yL_{i}^{0} x} \right.} \right\} $$

is negligible. Hence, we can apply Lemma 1 to derive weak convergence.

• (2) The Case of $\left| {\left\{ {\varvec{f}_{1} \left( {\varvec{L}^{0} } \right),\varvec{f}_{2} \left( {\varvec{L}^{0} } \right),\varvec{f}_{3} \left( {\varvec{L}^{0} } \right)} \right\}} \right| = 3$

In this case, the level-$1$ CI profile $L^{1}$ is uniquely determined. It is also straightforward to show that the probability of tied outcomes among some of the level-$2$ SCRs is negligible. If $\left| {\left\{ {f_{1}^{2} \left( {L^{1} } \right),f_{2}^{2} \left( {L^{1} } \right),f_{3}^{2} \left( {L^{1} } \right)} \right\}} \right| = 2$, we can apply Lemma 1 again to derive weak convergence. We next show that $\left| {\left\{ {f_{1}^{2} \left( {L^{1} } \right),f_{2}^{2} \left( {L^{1} } \right),f_{3}^{2} \left( {L^{1} } \right)} \right\}} \right|$ cannot be 3 if the menu $F$ is one of the $56$ menus. Suppose to the contrary that it is $3$. Note that

$$ {\mathcal{L}}\left( {F^{1} } \right) = \left\{ {f_{1} f_{2} f_{3} ,f_{1} f_{3} f_{2} ,f_{2} f_{1} f_{3} ,f_{2} f_{3} f_{1} ,f_{3} f_{1} f_{2} ,f_{3} f_{2} f_{1} } \right\}. $$

Let $n_{j}$ be the number of individuals who have the $j^{\text{th}}$ preference. For example, $n_{1}$ and $n_{4}$ are the numbers of individuals whose level-$1$ CI preferences are $f_{1} f_{2} f_{3}$ and $f_{2} f_{3} f_{1}$, respectively.

From now on, the proof is similar for all 10 menus. Let us prove the case of $F = \left\{ {f_{B} ,f_{H} ,f_{Bl} } \right\}$. Without loss of generality, we can assume $f_{B}^{2} \left( {L^{1} } \right) = f_{1}$, $f_{H}^{2} \left( {L^{1} } \right) = f_{2}$, and $f_{Bl}^{2} \left( {L^{1} } \right) = f_{3}$. With $n_{1} , \ldots ,n_{6}$, we can rephrase these conditions as follows:

$$ f_{Bo}^{2} \left( {L^{1} } \right) = f_{1}^{1} :\left\{ {\begin{array}{*{20}c} {n_{1} + 2n_{2} + n_{5} > n_{3} + 2n_{4} + n_{6} } \\ {2n_{1} + n_{2} + n_{3} > n_{4} + n_{5} + 2n_{6} } \\ \end{array} } \right. $$

$$ f_{H}^{2} \left( {L^{1} } \right) = f_{2}^{1} :\left\{ {\begin{array}{*{20}c} {\left\{ {\begin{array}{*{20}c} {n_{3} + n_{4} > n_{1} + n_{2} } \\ {n_{5} + n_{6} > n_{1} + n_{2} } \\ {n_{1} + n_{3} + n_{4} > n_{2} + n_{5} + n_{6} } \\ \end{array} } \right.} \\ {\rm or} \\ {\left\{ {\begin{array}{*{20}c} {n_{1} + n_{2} > n_{5} + n_{6} } \\ {n_{3} + n_{4} > n_{5} + n_{6} } \\ {n_{1} + n_{2} + n_{5} < n_{3} + n_{4} + n_{6} } \\ \end{array} } \right.} \\ \end{array} } \right. $$

$$f_{Bl}^{2} \left( {L^{1} } \right) = f_{3}^{1} : \left\{ {\begin{array}{*{20}l} {\left( {n_{4} + n_{5} + n_{6} > n_{1} + n_{2} + n_{3} \,{\text{and }}\,n_{2} + n_{5} + n_{6} > n_{1} + n_{3} + n_{4} } \right)} \hfill \\ {\left[ {\begin{array}{*{20}c} {\text{or}} \\ {\left( {n_{3} + n_{4} + n_{6} > n_{1} + n_{2} + n_{5} \,{\text{or }}\,n_{4} + n_{5} + n_{6} > n_{1} + n_{2} + n_{3} } \right)} \\ {\text{and}} \\ {\left( {n_{1} + n_{2} + n_{5} > n_{3} + n_{4} + n_{6} \,{\text{or }}\,n_{2} + n_{5} + n_{6} > n_{1} + n_{3} + n_{4} } \right)} \\ {\text{and}} \\ {\left( {n_{1} + n_{2} + n_{3} > n_{4} + n_{5} + n_{6} \,{\text{or}}\, n_{1} + n_{3} + n_{4} > n_{2} + n_{5} + n_{6} } \right)} \\ {\text{and}} \\ {n_{4} + n_{5} + 2n_{6} > 2n_{1} + n_{2} + n_{3} } \\ {\text{and}} \\ {n_{2} + 2n_{5} + n_{6} > n_{1} + 2n_{3} + n_{4} } \\ \end{array} } \right]} \hfill \\ \end{array} } \right. $$

With elementary verification⁶, we can see that there is no non-negative integer solution $\left( {n_{1} ,n_{2} , \ldots ,n_{6} } \right)$ for this system of inequalities. ■

Proof of Theorem 2.

The probability of tied outcomes among level-$1$ SCRs can be assumed to be negligible, and so we can expect that each $f_{P}^{1} \left( {L^{0} } \right),f_{B}^{1} \left( {L^{0} } \right),f_{A}^{1} \left( {L^{0} } \right),\varphi^{1} \left( {L^{0} } \right)$ is a singleton. If $\left| {F^{1} \left( {L^{0} } \right)} \right| \le 2$, we can apply Lemma 2 of Suzuki and Horita [11] to be assured of weak convergence. Because of the definition of $\varphi$, we know that $\varphi \left( {L^{0} } \right) \,\subseteq \,f_{P}^{1} \left( {L^{0} } \right) \cup f_{B}^{1} \left( {L^{0} } \right) \cup f_{A}^{1} \left( {L^{0} } \right)$. So, we can assume $\left| {F^{1} \left( {L^{0} } \right)} \right| = 3$. Without loss of generality, we can assume $g_{1}^{1} \left( {L^{0} } \right) = g_{2}^{1} \left( {L^{0} } \right) = \left\{ {x_{1} } \right\}$, $g_{3}^{1} \left( {L^{0} } \right) = \left\{ {x_{2} } \right\}$, and $g_{4}^{1} \left( {L^{0} } \right) = \left\{ {x_{3} } \right\}$, where $F^{1} = \left\{ {g_{1}^{1} ,g_{2}^{1} ,g_{3}^{1} ,g_{4}^{1} } \right\}$.

Let $L^{1} \,\in\, {\mathcal{L}}^{1} \left[ {L^{0} } \right]$ such that all individuals rank $g_{1}^{1}$ above $g_{2}^{1}$. Note that the probability of tied outcomes among some of $f_{P}^{2} ,f_{B}^{2} ,f_{A}^{2} ,\varphi^{2}$ can also be assumed to be negligible. So, we can expect that $f_{P}^{2} \left( {L^{1} } \right),f_{B}^{2} \left( {L^{1} } \right),f_{A}^{2} \left( {L^{1} } \right),\varphi^{2} \left( {L^{1} } \right)$ are also singletons. Then, $L^{1}$ is characterized with be the number $n_{1} , \ldots ,n_{6}$ that expresses the number of individuals who have each specific type of preference, i.e., $g_{1}^{1} ,g_{2}^{1} ,g_{3}^{1} ,g_{4}^{1}$ ($n_{1}$ individuals), $g_{1}^{1} ,g_{2}^{1} ,g_{3}^{1} ,g_{4}^{1}$ ($n_{2}$ individuals), $g_{3}^{1} ,g_{1}^{1} ,g_{2}^{1} ,g_{4}^{1}$ ($n_{3}$ individuals), $g_{3}^{1} ,g_{4}^{1} ,g_{1}^{1} ,g_{2}^{1}$ ($n_{4}$ individuals), $g_{4}^{1} ,g_{1}^{1} ,g_{2}^{1} ,g_{3}^{1}$ ($n_{5}$ individuals), and $g_{4}^{1} ,g_{3}^{1} ,g_{1}^{1} ,g_{2}^{1}$ ($n_{6}$ individuals), where $n = n_{1} + n_{2} + n_{3} + n_{4} + n_{5} + n_{6}$. Note also that if $\left| {F^{2} \left( {L^{1} } \right)} \right| \le 2$, then weak convergence is also guaranteed. So, we assume that $\left| {F^{2} \left( {L^{1} } \right)} \right| = 3$. At this time, $\varphi^{2} \left( {L^{1} } \right)$ is either $f_{P}^{2} \left( {L^{1} } \right)$ or $f_{B}^{2} \left( {L^{1} } \right)$. We can also expect $n_{i} > 0$ for $i = 1,2,3,4,5,6$ when $n \to \infty$, and so we have that $f_{A}^{2} \left( {L^{1} } \right) = \left\{ {g_{1}^{1} } \right\}$. Now, we have only two possibilities: (1) $f_{B}^{2} \left( {L^{1} } \right) = \varphi^{2} \left( {L^{1} } \right) = \left\{ {g_{3}^{1} } \right\}$, $f_{P}^{2} \left( {L^{1} } \right) = \left\{ {g_{4}^{1} } \right\}$, and $f_{A}^{2} \left( {L^{1} } \right) = \left\{ {g_{1}^{1} } \right\}$, or (2) $f_{B}^{2} \left( {L^{1} } \right) = \left\{ {g_{3}^{1} } \right\}$, $f_{B}^{2} \left( {L^{1} } \right) = \varphi^{2} \left( {L^{1} } \right) = \left\{ {g_{4}^{1} } \right\}$, and $f_{A}^{2} \left( {L^{1} } \right) = \left\{ {g_{1}^{1} } \right\}$.

• (1) The case of $f_{B}^{2} \left( {L^{1} } \right) = \varphi^{2} \left( {L^{1} } \right) = \left\{ {g_{3}^{1} } \right\}$, $f_{P}^{2} \left( {L^{1} } \right) = \left\{ {g_{4}^{1} } \right\}$, and $f_{A}^{2} \left( {L^{1} } \right) = \left\{ {g_{1}^{1} } \right\}$.

Let $L^{2} \,\in\, \left( {{\mathcal{L}}\left( {F^{2} } \right)} \right)^{n}$ be as follows: $f_{A}^{2} ,f_{B}^{2} ,\varphi^{2} ,f_{P}^{2}$ ($n_{1}$ individuals), $f_{A}^{2} ,f_{P}^{2} ,f_{B}^{2} ,\varphi^{2}$ ($n_{2}$ individuals), $f_{B}^{2} ,\varphi^{2} ,f_{A}^{2} ,f_{P}^{2}$ ($n_{3}$ individuals), $f_{B}^{2} ,\varphi^{2} ,f_{P}^{2} ,f_{A}^{2}$ ($n_{4}$ individuals), $f_{P}^{2} ,f_{A}^{2} ,f_{B}^{2} ,\varphi^{2}$ ($n_{5}$ individuals), and $f_{P}^{2} ,f_{B}^{2} ,\varphi^{2} ,f_{A}^{2}$ ($n_{6}$ individuals). Clearly, we have $L^{2} \,\in\, {\mathcal{L}}^{2} \left[ {L^{0} ,L^{1} } \right]$.

Because $n_{1} , \ldots ,n_{6}$ are positive, we obtain $f_{A}^{3} \left( {L^{2} } \right) = \left\{ {f_{B}^{2} } \right\}$. Furthermore, $f_{P}^{2} \left( {L^{1} } \right) = \left\{ {g_{4}^{1} } \right\}$ and so the plurality score of $g_{4}^{1}$ is greater than those of $g_{1}^{1}$ and $g_{3}^{1}$:

$$ \left\{ {\begin{array}{*{20}c} {n_{5} + n_{6} > n_{1} + n_{2.} } \\ {n_{5} + n_{6} > n_{3} + n_{4} .} \\ \end{array} } \right. $$

This also shows that the plurality score of $f_{P}^{2}$ is greater than those of $f_{A}^{2}$ and $f_{B}^{2}$ at $L^{2}$. Hence, we have that $f_{P}^{3} \left( {L^{2} } \right) = \left\{ {f_{P}^{2} } \right\}$. Next, we show that $f_{B}^{3} \left( {L^{2} } \right) = \left\{ {f_{B}^{2} } \right\}$.

Because $f_{B}^{2} \left( {L^{1} } \right) = \left\{ {g_{3}^{1} } \right\}$, the Borda scores at $L^{1}$ are as follows:

$$ \begin{aligned} s_{B} \left( {g_{3}^{1} } \right) & > s_{B} \left( {g_{1}^{1} } \right) \Leftrightarrow n_{3} + n_{4} + \frac{2}{3}n_{6} + \frac{1}{3}n_{1} > n_{1} + n_{2} + \frac{2}{3}\left( {n_{3} + n_{5} } \right) + \frac{1}{3}\left( {n_{4} + n_{6} } \right). \\ s_{B} \left( {g_{3}^{1} } \right) & > s_{B} \left( {g_{4}^{1} } \right) \Leftrightarrow n_{3} + n_{4} + \frac{2}{3}n_{6} + \frac{1}{3}n_{1} > n_{5} + n_{6} + \frac{2}{3}n_{4} + \frac{1}{3}n_{2} . \\ \end{aligned} $$

At $L^{2}$, we have:

$$ \begin{aligned} s_{B} \left( {f_{P}^{2} } \right) & = n_{5} + n_{6} + \frac{2}{3}n_{2} + \frac{1}{3}n_{4} . \\ s_{B} \left( {f_{B}^{2} } \right) & = n_{3} + n_{4} + \frac{2}{3}\left( {n_{1} + n_{6} } \right) + \frac{1}{3}\left( {n_{2} + n_{5} } \right). \\ s_{B} \left( {f_{A}^{2} } \right) & = n_{1} + n_{2} + \frac{2}{3}n_{5} + \frac{1}{3}n_{3} . \\ s_{B} \left( {\varphi^{2} } \right) & < s_{B} \left( {f_{B}^{2} } \right). \\ \end{aligned} $$

These equations show that $s_{B} \left( {f_{B}^{2} } \right) > \hbox{max} \left\{ {s_{B} \left( {f_{P}^{2} } \right),s_{B} \left( {f_{A}^{2} } \right),s_{B} \left( {\varphi^{2} } \right)} \right\}$.

• (2) The case of $f_{B}^{2} \left( {L^{1} } \right) = \left\{ {g_{3}^{1} } \right\}$, $f_{P}^{2} \left( {L^{1} } \right) = \varphi \left( {L^{1} } \right) = \left\{ {g_{4}^{1} } \right\}$, and $f_{A}^{2} \left( {L^{1} } \right) = \left\{ {g_{1}^{1} } \right\}$. $f_{B}^{2} \left( {L^{1} } \right) = \left\{ {g_{3}^{1} } \right\}$, and so the score of $g_{3}^{1}$ at $L^{1}$ is strictly greater than those of $g_{1}^{1}$ and $g_{4}^{1}$. Formally, we have that:

  $$ \begin{aligned} n_{3} + n_{4} + \frac{2}{3}n_{6} + \frac{1}{3}n_{1} & > n_{1} + n_{2} + \frac{2}{3}\left( {n_{3} + n_{5} } \right) + \frac{1}{3}\left( {n_{4} + n_{6} } \right). \\ n_{3} + n_{4} + \frac{2}{3}n_{6} + \frac{1}{3}n_{1} & > n_{5} + n_{6} + \frac{2}{3}n_{4} + \frac{1}{3}n_{2} . \\ \end{aligned} $$

  (1)

Let $L^{2} \,\in\, {\mathcal{L}}^{2} \left[ {L^{0} ,L^{1} } \right]$ be such that:

$$ \begin{aligned} & n_{1}\, {\text{individuals}}{:}\, f_{A}^{2} ,f_{B}^{2} ,f_{P}^{2} ,\varphi^{2} . \\ & n_{2}\, {\text{individuals}}{:}\, f_{A}^{2} ,f_{P}^{2} ,\varphi^{2} ,f_{B}^{2} . \\ & n_{3} \,{\text{individuals}}{:}\, f_{B}^{2} ,f_{A}^{2} ,f_{P}^{2} ,\varphi^{2} . \\ & n_{4} \,{\text{individuals}}{:}\, f_{B}^{2} ,f_{P}^{2} ,\varphi^{2} ,f_{A}^{2} . \\ & n_{5}\, {\text{individuals}}{:}\, f_{P}^{2} ,\varphi^{2} ,f_{A}^{2} ,f_{B}^{2} . \\ & n_{6}\, {\text{individuals}}{:}\, f_{P}^{2} ,\varphi^{2} ,f_{B}^{2} ,f_{A}^{2} . \\ \end{aligned} $$

In words, this is a consequentially induced preference where everyone ranks $f_{P}^{2}$ above $\varphi^{2}$. Similar to (1), we can check that $f_{P}^{3} \left( {L^{2} } \right) = f_{A}^{3} \left( {L^{2} } \right) = \left\{ {f_{P}^{2} } \right\}$. Furthermore, the scores of $f_{A}^{2} ,f_{P}^{2} ,f_{B}^{2} ,\varphi^{2}$ evaluated by $f_{B}^{3}$ are as follows:

$$ \begin{aligned} s_{B} \left( {f_{A}^{2} } \right) & = n_{1} + n_{2} + \frac{2}{3}n_{3} + \frac{1}{3}n_{5} . \\ s_{B} \left( {f_{B}^{2} } \right) & = n_{3} + n_{4} + \frac{2}{3}n_{1} + \frac{1}{3}n_{6} . \\ s_{B} \left( {f_{P}^{2} } \right) & = n_{5} + n_{6} + \frac{2}{3}\left( {n_{2} + n_{4} } \right) + \frac{1}{3}\left( {n_{1} + n_{3} } \right). \\ \end{aligned} $$

With the inequalities (1) we have that $s_{B} \left( {f_{B}^{2} } \right) > s_{B} \left( {f_{A}^{2} } \right)$. Note that ties between $f_{B}^{2} ,f_{P}^{2}$ occur only if

$$ n_{3} + n_{4} + \frac{2}{3}n_{1} + \frac{1}{3}n_{6} = n_{5} + n_{6} + \frac{2}{3}\left( {n_{2} + n_{4} } \right) + \frac{1}{3}\left( {n_{1} + n_{3} } \right). $$

This event is negligible as $n \to \infty$.■