Compromise in combinatorial vote

Abstract

We consider collective choice problems where the set of social outcomes is a Cartesian product of finitely many finite sets. Each individual is assigned a two-level preference, defined as a pair involving a vector of strict rankings of elements in each of the sets and a strict ranking of social outcomes. A voting rule is called (resp. weakly) product stable at some two-level preference profile if every (resp. at least one) outcome formed by separate coordinate-wise choices is also an outcome of the rule applied to preferences over social outcomes. We investigate the (weak) product stability for the specific class of compromise solutions involving q-approval rules, where q lies between 1 and the number I of voters. Given a finite set $\mathcal {X}$ and a profile of I linear orders over $\mathcal {X}$, a q-approval rule selects elements of $\mathcal {X}$ that gathers the largest support above q at the highest rank in the profile. Well-known q-approval rules are the Fallback Bargaining solution ($q=I$) and the Majoritarian Compromise ($q=\left\lceil \frac{I}{2}\right\rceil$). We assume that coordinate-wise rankings and rankings of social outcomes are related in a neutral way, and we investigate the existence of neutral two-level preference domains that ensure the weak product stability of q-approval rules. We show that no such domain exists unless either $q=I$ or very special cases prevail. Moreover, we characterize the neutral two-level preference domains over which the Fallback Bargaining solution is weakly product stable.

Appendix A Appendix: Proof of theorem 1

1.1 A.1 Proof of sufficiency part

We want to show that for any $\widetilde{q}\in \mathcal {L}_{\mathcal {M}}$, $F_{I}$ is weakly product stable at $\delta ^{\widetilde{q}}$.

Take any $\widetilde{q}\in \mathcal {L}_{\mathcal {M}}$. Up to a reshuffling of coordinates $m\in \mathcal {M}$, one can assume w.l.o.g. that $\widetilde{q }=[12\ldots M]$. Suppose that there exists a coordinate profile $\mathbf {p}$ such that $[\prod _{m\in \mathcal {M}}F_{I}(\mathbf {p}_{m})]\cap F_{I}[{ \delta }(\mathbf {p})]=\emptyset$, where ${\delta }=(\delta ^{ \widetilde{q}},\ldots ,\delta ^{\widetilde{q}}).$ Take any $\overrightarrow{b }=(b_{1},\ldots ,b_{M})\in \prod _{m\in \mathcal {M}}F_{I}(\mathbf {p}_{m})$ and $\overrightarrow{a}=(a_{1},\ldots ,a_{M})\in F_{I}[{\delta }( \mathbf {p})]$. Define $m_{1}\in \mathcal {M}$ as the coordinate with the lowest index such that $a_{m_{1}}\ne b_{m_{1}}$. Hence, we can write $\overrightarrow{a}=(\overrightarrow{b}_{<m_{1}},\overrightarrow{a}_{\ge m_{1}})$. Moreover, define $r^{*}$ as the I-value of $\overrightarrow{b }$ at ${\delta }(\mathbf {p})$. By definition of the I-value, $r^{*}$ is the highest rank given to $\overrightarrow{b}$ by some linear order in ${\delta }(\mathbf {p})$. Pick $i^{*}\in \mathcal {I}$ for whom $\delta ^{\widetilde{q}}(p^{i^{*}})(\overrightarrow{b})=r^{*}$. Since $p_{m_{1}}^{i^{*}}$ $\in \mathcal {L}_{\mathcal {A}_{m_{1}}}$ and $a_{m_{1}}\ne b_{m_{1}}$, we have $p_{m_{1}}^{i^{*}}(b_{m_{1}})\ne p_{m_{1}}^{i^{*}}(a_{m_{1}})$. Moreover, since $\overrightarrow{a}\in F_{I}[{\delta }(\mathbf {p})]$ and whereas $\overrightarrow{b}\mathbf { \notin }F_{I}[{\delta }(\mathbf {p})]$, $V_{I}[\overrightarrow{a}, {\delta }(\mathbf {p})]$, that is the I-value of $\overrightarrow{a}$ in ${\delta }(\mathbf {p})$ is strictly less than $r^{*}$. It follows from the definition of $\delta ^{\widetilde{q}}$ that $p_{m_{1}}^{i^{*}}(a_{m_{1}})<p_{m_{1}}^{i^{*}}(b_{m_{1}})$. Now, pick any $i\ne i^{*}$ and suppose that $p_{m_{1}}^{i}(a_{m_{1}})>p_{m_{1}}^{i^{*}}(b_{m_{1}})$. By the definition of $\delta ^{\widetilde{q}}$, we get $\delta ^{\widetilde{q} }(p^{i})(\overrightarrow{a})>\delta ^{\widetilde{q}}(p^{i^{*}})( \overrightarrow{b})=r^{*}$. As this would imply $V_{I}[\overrightarrow{a},{\delta }(\mathbf {p})]>$ $r^{*}$, we contradict $\overrightarrow{ a}$ $\mathbf {\in }$ $F_{I}[{\delta }(\mathbf {p})]$. Hence, we have shown that (1) $p_{m_{1}}^{i^{*}}(a_{m_{1}})<p_{m_{1}}^{i^{*}}(b_{m_{1}})$ and (2) $p_{m_{1}}^{i}(a_{m_{1}})\le p_{m_{1}}^{i^{*}}(b_{m_{1}})$ for all $i\in \mathcal {I}\setminus \{i^{*}\}$. Observe that $p_{m_{1}}^{i}(a_{m_{1}})<p_{m_{1}}^{i^{*}}(b_{m_{1}})$ for all $i\in \mathcal {I}$ contradicts with $b_{m_{1}}\in F_{I}(\mathbf {p}_{m_{1}})$. Thus, $p_{m_{1}}^{i}(a_{m_{1}})=p_{m_{1}}^{i^{*}}(b_{m_{1}})$ for some $i\in \mathcal {I}$, which in turn implies $a_{m_{1}}\in F_{I}(\mathbf {p} _{m_{1}})$.

Next consider $\overrightarrow{b}^{(1)}=(b_{1}^{(1)},\ldots ,b_{M}^{(1)})=(a_{m_{1}},\overrightarrow{b}_{-m_{1}})$. Since $\overrightarrow{b}\in \prod _{m\in \mathcal {M}}F_{I}(\mathbf {p}_{m})$ and $a_{m_{1}}\in F_{I}(\mathbf {p}_{m_{1}})$, then $\overrightarrow{b}^{(1)}\in \prod _{m\in \mathcal {M}}F_{I}(\mathbf {p}_{m})$. All is done if $\overrightarrow{b}^{(1)}\in F_{I}[{\delta }(\mathbf {p})]$. Suppose that $\overrightarrow{b}^{(1)}\notin F_{I}[{\delta }(\mathbf {p})]$. Consider again $\overrightarrow{a}$ and define $m_{2}\in \mathcal {M}$ as the coordinate with the lowest index $\overrightarrow{a}$ and $\overrightarrow{b} ^{(1)}$ disagree upon. By construction, we have $m_{2}>m_{1}$. Hence, we can write $\overrightarrow{a}=(\overrightarrow{b}_{<m_{2}}^{(1)},\overrightarrow{ a}_{\ge m_{2}})\in F_{I}[{\delta }(\mathbf {p})]$. By applying to $m_{2}$ the same argument as the one for $m_{1}$, we get $a_{m_{2}}\in F_{I}( \mathbf {p}_{m_{2}})$. This shows that $\overrightarrow{b}^{(2)}=(a_{m_{2}}, \overrightarrow{b}_{-m_{2}}^{(1)})\in \prod _{m\in \mathcal {M}}F_{I}(\mathbf {p }_{m})$. By replicating the same argument, and because of the finiteness of $\mathcal {M}$, there must exist $T\le M$ such that $\overrightarrow{b}^{(T)}= \overrightarrow{a}$. By construction, we have $\overrightarrow{b} ^{(T)}=(a_{m_{T}},\overrightarrow{b}_{-m_{T}}^{(T-1)})\in \prod _{m\in \mathcal {M}}F_{I}(\mathbf {p}_{m})$. Since $\overrightarrow{a}\in F_{I}[ {\delta }(\mathbf {p})]$, then $[\prod _{m\in \mathcal {M}}F_{I}(\mathbf { p}_{m})]\cap F_{I}[{\delta }(\mathbf {p})]\ne \emptyset$, which shows that $F_{I}$ is weakly product stable at $\delta ^{\widetilde{q}}$.

1.2 A.2 Proof of necessity part

The proof of the necessity part is organized in two lemmas. The first one states that the weak product stability of $F_{I}$ requires all voters to use a lexicographic preference.

Lemma 6

Let $F_{I}$ be product stable over $\mathcal {D}$. If $I\ge I^{*}+3$, then $\delta \in \mathcal {D}$ only if $\delta =\delta ^{ \widetilde{q}}$ for some $\widetilde{q}\in \mathcal {L}_{\mathcal {M}}$.

Proof

Let $\mathcal {R}^{*}=\{R\in \mathcal {R}:R=(2,\overrightarrow{1}_{-m})$ for some $m\in \mathcal {M}\}$. Take any $\delta \in \mathcal {D}$. By Proposition 1, $\delta$ is responsive. Write $\delta |_{ \mathcal {R}^{*}}$ as $(2,\vec {1}_{-m_{M}})$ $\delta |_{\mathcal {R}^{*}}$ $(2,\vec {1}_{-m_{M-1}})$ $\delta |_{\mathcal {R}^{*}}$ $\ldots$ $\delta |_{\mathcal {R}^{*}}(2,\vec {1}_{-m_{2}})$ $\delta |_{\mathcal {R} ^{*}}$ $(2,\vec {1}_{-m_{1}})$. Observe that responsiveness implies that $\delta$ ranks $(2,\vec {1}_{-m_{M}})$ second in $\mathcal {R}$. Moreover, define $\widetilde{q}\in \mathcal {L}_{\mathcal {M}}$ by $\widetilde{q} ^{-1}(n)=m_{n}$ for all $n\in \mathcal {M}$. For notational simplicity, assume that $\widetilde{q}=[1,2,\ldots ,M]$. Hence, $\delta$ is responsive, ranks $\vec {1}$ at top, $(2,\vec {1}_{-M})$ second and $\delta |_{\mathcal {R} ^{*}}=[(2,\vec {1}_{-M}),(2,\vec {1}_{-(M-1)}),\ldots ,(2,\vec {1}_{-1})]$.

Suppose, towards a contradiction, that $\delta \ne \delta ^{\widetilde{q}}$. Thus, there exist $m^{*}\in \mathcal {M}$ and $R,R^{\prime }\in \mathcal {R}$ with $R=(R_{<m^{*}},r_{m^{*}},r_{m^{*}+1},\ldots ,r_{M})$, where $R_{<m^{*}}=(\overset{m^{*}-1}{\overbrace{ r_{1},\ldots ,r_{m^{*}-1}})}$, $R^{\prime }=$ $(R_{<m^{*}},r_{m^{*}}^{\prime },r_{m^{*}+1}^{\prime },\ldots ,r_{M}^{\prime })$ with $r_{m^{*}}>r_{m^{*}}^{\prime }$, and R $\delta$ $R^{\prime }$.

Observe that, by the responsiveness of $\delta$, we must have $m^{*}<M$. Again by the responsiveness of $\delta$, one can assume w.l.o.g. that $r_{m}=1$ and $r_{m}^{\prime }=A_{m}$ for all $m\in \{m^{*}+1,\ldots ,M\}$. Thus,

– $R=(R_{<m^{*}},r_{m^{*}},1,\ldots ,1)$, where $R_{<m^{*}}=( \overset{m^{*}-1}{\overbrace{r_{1},\ldots ,r_{m^{*}-1}})}$,

– $R^{\prime }=$ $(R_{<m^{*}},r_{m^{*}}^{\prime },A_{m^{*}+1},\ldots ,A_{M})$ with $r_{m^{*}}>r_{m^{*}}^{\prime }$,

– R $\delta$ $R^{\prime }$.

We distinguish between two cases.

Case 1: $r_{m^{*}}<A_{m^{*}}$.

Define

– $\tilde{R}=(R_{<m^{*}},r_{m^{*}}+1,1,\ldots ,1)$,

– $\tilde{R}^{\prime }=(R_{<m^{*}},r_{m^{*}}^{\prime }+1,A_{m^{*}+1},\ldots ,A_{M})$.

Case 1a: $\tilde{R}$ $\delta$ $\tilde{R}^{\prime }$.

Consider a coordinate 4-voter profile $\mathbf {p}$ having the general form below:

– $\forall m<m^{*}$, $\mathbf {p}_{m}=\left( \begin{array}{ccccc} Rank &{} p_{m}^{1} &{} p_{m}^{2} &{} p_{m}^{3} &{} p_{m}^{4} \\ \hline 1 &{} \cdot &{} \cdot &{} \cdot &{} a_{m} \\ r_{m}=r_{m}^{\prime } &{} a_{m} &{} a_{m} &{} a_{m} &{} \cdot \end{array} \right)$,

– $\mathbf {p}_{m^{*}}=\left( \begin{array}{ccccc} Rank &{} p_{m^{*}}^{1} &{} p_{m^{*}}^{2} &{} p_{m^{*}}^{3} &{} p_{m^{*}}^{4} \\ \hline 1 &{} \cdot &{} \cdot &{} \cdot &{} a_{m^{*}} \\ r_{m^{*}}^{\prime } &{} a_{m^{*}} &{} a_{m^{*}} &{} b_{m^{*}} &{} \cdot \\ r_{m^{*}}^{\prime }+1 &{} b_{m^{*}} &{} b_{m^{*}} &{} \cdot &{} b_{m^{*}} \\ r_{m^{*}}+1 &{} \cdot &{} \cdot &{} a_{m^{*}} &{} \cdot \end{array} \right)$,

– $\forall m>m^{*}$, $\mathbf {p}_{m}=\left( \begin{array}{ccccc} Rank &{} p_{m}^{1} &{} p_{m}^{2} &{} p_{m}^{3} &{} p_{m}^{4} \\ \hline 1 &{} \cdot &{} \cdot &{} a_{m} &{} a_{m} \\ 2 &{} a_{m} &{} \cdot &{} \cdot &{} \cdot \\ A_{m} &{} \cdot &{} a_{m} &{} \cdot &{} \cdot \end{array} \right)$¹⁹.

Let $\overrightarrow{a}=(a_{1},\ldots ,a_{M}),\overrightarrow{b} =(b_{1},\ldots ,b_{M}),\overrightarrow{c}=(b_{m^{*}},\overrightarrow{a} _{-m^{*}})\in \mathcal {A}$, $\mathcal {A}^{\prime }=\{(b_{m^{*}}, \overrightarrow{\alpha }_{-m^{*}}):\overrightarrow{\alpha }\in \mathcal {A }\}{\setminus }\{\overrightarrow{c}\}$, and consider a coordinate profile $\overline{\mathbf {p}}$ involving $I\ge \prod _{m\in \mathcal {M}\backslash \{m^{*}\}}A_{m}$ $+3$ voters and such that:

(1) $\forall i\in \{1,2,3\}$, $\overline{p}^{i}=p^{i}$,

(2) $\forall i\in \{4,\ldots ,I\}$, $\overline{p}^{i}$ agrees with $p^{4}$ on the ranks given to $a_{m}$ for all m, $\overline{p}^{i}$ agrees with $p^{4}$ on the rank given to $b_{m^{*}}$, and $\forall c_{m^{*}}\in \mathcal {A}_{m^{*}}{\setminus }\{a_{m^{*}},b_{m^{*}}\}$ there exists $i(c_{m^{*}})\in \{4,\ldots ,I\}$ such that $r[c_{m^{*}}, \overline{p}^{i(c_{m^{*}})}]=A_{m^{*}}$,

(3) $\forall \overrightarrow{d}\in \mathcal {A}^{\prime }$, there exists $i( \overrightarrow{d})\in \{4,\ldots ,I\}$ such that $R[\overrightarrow{d}, \overline{p}^{i(\overrightarrow{d})}]=\tilde{R}^{\prime }$.

As $I^{*}\ge \prod _{m\in \mathcal {M}{\setminus }\{m^{*}\}}A_{m}$, condition (3) can be satisfied. Moreover, we have $\prod _{m\in \mathcal {M} }F_{I}(\overline{\mathbf {p}})\subseteq \mathcal {A}^{\prime }\cup \{ \overrightarrow{c}\}$ and $\overrightarrow{c}\in \prod _{m\in \mathcal {M} }F_{I}(\overline{\mathbf {p}})$. Table 1 gives the rank vectors of $\overrightarrow{a}$ and $\overrightarrow{c}$ for each coordinate preference $\overline{p}^{i}$:

Table 1 The rank vectors of $\overrightarrow{a}$ and $\overrightarrow{c}$ in Case 1a

Pick ${\delta }=(\delta ,\ldots ,\delta )$. Observe that the responsiveness of $\delta$ implies $R(\overrightarrow{c}{\small ,}\overline{ p}^{i})$ $\delta$ $R(\overrightarrow{c},\overline{p}^{2})$ for all $i\ne 2$. Hence $V_{I}[\overrightarrow{c},{\delta }(\overline{\mathbf {p}} \mathbf {)}]$, the I-value of $\overrightarrow{c}$ in ${\delta }( \overline{\mathbf {p}}\mathbf {)}$, is equal to $r[\overrightarrow{c},\delta ( \overline{p}^{2})]$. Similarly, the responsiveness of $\delta$ implies $R( \overrightarrow{a},\overline{p}^{i})$ $\delta$ $R(\overrightarrow{a}, \overline{p}^{j})$ for all $i\ge 4$ and all $j<4$. Moreover, as $A_{m}\ge 3$ for all $m\in \mathcal {M}$, $R(\overrightarrow{a},\overline{p}^{1})$ $\delta$ $R(\overrightarrow{a},\overline{p}^{2})$. Thus, $V_{I}[ \overrightarrow{a},{\delta }(\overline{\mathbf {p}}\mathbf {)}]\in \{r[ \overrightarrow{a},\delta (\overline{p}^{2})],$ $r[\overrightarrow{a},\delta (\overline{p}^{3})]\}$. By definition of case 1a, we have $\tilde{R}$ $\delta$ $\tilde{R}^{\prime }$, and by responsiveness $R^{\prime }$ $\delta$ $\tilde{R}^{\prime }$, which implies $V_{I}[\overrightarrow{a},{ \delta }(\overline{\mathbf {p}}\mathbf {)}]<V_{I}[\overrightarrow{c},{ \delta }(\overline{\mathbf {p}}\mathbf {)}]$. Therefore, $\overrightarrow{c} \notin F_{I}[{\delta }(\overline{\mathbf {p}}\mathbf {)]}$. Finally, pick any $\overrightarrow{d}\in \mathcal {A}^{\prime }$. By definition of $\overline{\mathbf {p}}$, we have $V_{I}[\overrightarrow{d},{\delta }( \overline{\mathbf {p}}\mathbf {)}]\ge r[\overrightarrow{d},\overline{p}^{i( \overrightarrow{d})}]=r[\overrightarrow{c},\delta (\overline{p}^{2})]=V_{I}[ \overrightarrow{c},{\delta }(\overline{\mathbf {p}}\mathbf {)}]$. As $V_{I}[\overrightarrow{a},{\delta }(\overline{\mathbf {p}}\mathbf {)} ]<V_{I}[\overrightarrow{c},{\delta }(\overline{\mathbf {p}}\mathbf {)}]$, we get $[\prod _{m\in \mathcal {M}}F_{I}(\mathbf {p})]{\cap }F_{I}[ {\delta }(\overline{\mathbf {p}}\mathbf {)]=\emptyset }$, in contradiction with the assumption that $F_{I}$ is weakly stable over $\mathcal {D}$.

Case 1b: $\tilde{R}^{\prime }$ $\delta$ $\tilde{R}$.

We use an argument similar to the one for case 1a. Consider the coordinate 4-voter profile $\mathbf {p}$ with the form below:

– $\forall m<m^{*}$, $\mathbf {p}_{m}=\left( \begin{array}{c|cccc} Rank &{} p_{m}^{1} &{} p_{m}^{2} &{} p_{m}^{3} &{} p_{m}^{4} \\ \hline 1 &{} \cdot &{} \cdot &{} \cdot &{} a_{m} \\ r_{m}=r_{m}^{\prime } &{} a_{m} &{} a_{m} &{} a_{m} &{} \cdot \end{array} \right)$,

– $\mathbf {p}_{m^{*}}=\left( \begin{array}{c|cccc} Rank &{} p_{m}^{1} &{} p_{m}^{2} &{} p_{m}^{3} &{} p_{m}^{4} \\ \hline 1 &{} \cdot &{} \cdot &{} \cdot &{} a_{m^{*}} \\ r_{m^{*}}^{\prime } &{} \cdot &{} \cdot &{} a_{m^{*}} &{} \cdot \\ r_{m^{*}}^{\prime }+1 &{} a_{m^{*}} &{} \cdot &{} \cdot &{} \cdot \\ r_{m^{*}}+1 &{} \cdot &{} a_{m^{*}} &{} \cdot &{} \cdot \end{array} \right)$,

– $\forall m>m^{*}$, $\mathbf {p}_{m}=\left( \begin{array}{c|cccc} Rank &{} p_{m}^{1} &{} p_{m}^{2} &{} p_{m}^{3} &{} p_{m}^{4} \\ \hline 1 &{} b_{m} &{} a_{m} &{} a_{m} &{} a_{m} \\ 2 &{} \cdot &{} b_{m} &{} b_{m} &{} b_{m} \\ A_{m} &{} a_{m} &{} \cdot &{} \cdot &{} \cdot \end{array} \right)$

Let $\overrightarrow{a}=(a_{1},\ldots ,a_{M}),\overrightarrow{b} =(b_{1},\ldots ,b_{M}),\overrightarrow{c}=(a_{\le m^{*}}, \overrightarrow{b}_{>m^{*}})\in \mathcal {A}$, $\mathcal {A}^{\prime }=\{( \overrightarrow{\alpha }_{\le m^{*}},b_{>m^{*}}):\overrightarrow{ \alpha }\in \mathcal {A}\}{\setminus }\{\overrightarrow{c}\}$. Now, extend $\mathbf {p}$ to coordinate profile $\overline{\mathbf {p}}$ involving $I\ge \prod _{m\le m^{*}}A_{m}$ $+3$ voters and such that:

(1) $\forall i\in \{1,2,3\}$, $\overline{p}^{i}=p^{i}$,

(2) $\forall i\in \{4,\ldots ,I\}$, $\overline{p}^{i}$ agrees with $p^{4}$ on the ranks given to $a_{m}$ for all $m\in \mathcal {M}$, $\overline{p}^{i}$ agrees with $p^{4}$ on the ranks given to $b_{m}$ for all $m>m^{*}$, and for all $m>m^{*}$, $\forall c_{m}\in \mathcal {A}_{m}{\setminus }\{a_{m},b_{m}\}$ there exists $i(c_{m})\in \{4,\ldots ,I\}$ such that $r[c_{m},\overline{p}^{i(c_{m})}]=A_{m}$,

(3) $\forall \overrightarrow{d}\in \mathcal {A}^{\prime }$, there exists $i( \overrightarrow{d})\in \{4,\ldots ,I\}$ such that $R[\overrightarrow{d}, \overline{p}^{i(\overrightarrow{d})}]=(R_{<m^{*}},r_{m^{*}}+1,2,\ldots ,2)$.

As $I^{*}\ge$ $\prod _{m\le m^{*}}A_{m}$, condition (3) can be satisfied. We have $\prod _{m\in \mathcal {M}}F_{I}(\overline{\mathbf {p}})$ $\subseteq \mathcal {A}^{\prime }\cup \{\overrightarrow{c}\}$ and $\overrightarrow{c}\in \prod _{m\in \mathcal {M}}F_{I}(\overline{\mathbf {p}})$. Table 2 gives the rank vectors of $\overrightarrow{a}$ and $\overrightarrow{c}$ for each coordinate preference $\overline{p}^{i}$:

Table 2 The rank vectors of $\overrightarrow{a}$ and $\overrightarrow{c}$ in Case 1b

By the responsiveness of $\delta$, $R(\overrightarrow{c},\overline{p}^{i})$ $\delta$ $R(\overrightarrow{c},\overline{p}^{2})$ for all $i\ne 2$. Thus, $V_{I}[\overrightarrow{c},{\delta }(\overline{\mathbf {p}}\mathbf {)}]=r[ \overrightarrow{c},\delta (\overline{p}^{2})]$. By assumption, $R( \overrightarrow{a},\overline{p}^{1})$ $\delta$ $R(\overrightarrow{a}, \overline{p}^{2})$, and by the responsiveness of $\delta$, $R( \overrightarrow{a},\overline{p}^{i})$ $\delta$ $R(\overrightarrow{a}, \overline{p}^{2})$ for all $i\ge 3$. Thus, $V_{I}[\overrightarrow{a}, {\delta }(\overline{\mathbf {p}}\mathbf {)}]=r[\overrightarrow{a},\delta (\overline{p}^{2})]$. Moreover, again by the responsiveness of $\delta$, $R(\overrightarrow{a},\overline{p}^{2})$ $\delta$ $R( \overrightarrow{c},\overline{p}^{2})$, which implies $V_{I}[\overrightarrow{a },{\delta }(\overline{\mathbf {p}}\mathbf {)}]<V_{I}[\overrightarrow{c}, {\delta }(\overline{\mathbf {p}}\mathbf {)}]$. It follows that $\overrightarrow{c}\notin F_{I}[{\delta }(\overline{\mathbf {p}}\mathbf { )]}$.

Finally, pick any $\overrightarrow{d}\in \mathcal {A}^{\prime }$. By definition of $\overline{\mathbf {p}}$, $V_{I}[\overrightarrow{d},{ \delta }(\overline{\mathbf {p}}\mathbf {)}]\ge r[\overrightarrow{d},\delta ( \overline{p}^{i(\overrightarrow{d})})]=r[\overrightarrow{c},\delta ( \overline{p}^{2})]=V_{I}[\overrightarrow{c},{\delta }(\overline{ \mathbf {p}}\mathbf {)}]$. As $V_{I}[\overrightarrow{a},{\delta }( \overline{\mathbf {p}}\mathbf {)}]<V_{I}[\overrightarrow{c},{\delta }( \overline{\mathbf {p}}\mathbf {)}]$, we get $[\prod _{m\in \mathcal {M}}F_{I}( \mathbf {p})]$ ${\cap }$ $F_{I}[{\delta }(\overline{\mathbf {p}} \mathbf {)]=\emptyset }$, in contradiction the assumption that $F_{I}$ is weakly stable over $\mathcal {D}$.

This shows that case 1 is impossible.

Case 2: $r_{m^{*}}=A_{m^{*}}$.

First, observe that one must have $r_{m^{*}}^{\prime }=A_{m^{*}}-1$. To see why, suppose that $r_{m^{*}}^{\prime }<A_{m^{*}}-1$. Define $R^{\prime \prime }=(R_{<m^{*}},A_{m^{*}}-1,1,\ldots ,1)$. By the responsiveness of $\delta$, $R^{\prime \prime }$ $\delta$ R, and by the transitivity of $\delta$, $R^{\prime \prime }$ $\delta$ $R^{\prime }$. Since $r_{m^{*}}^{\prime \prime }=A_{m^{*}}-1<A_{m^{*}}$, this contradicts with case 1 being impossible. We complete the proof by using an argument similar to the one for case 1. Consider the coordinate 3-voter profile $\mathbf {p}$ with the form below:

– $\forall m<m^{*}$, $\mathbf {p}_{m}=\left( \begin{array}{c|ccc} Rank &{} p_{m}^{1} &{} p_{m}^{2} &{} p_{m}^{3} \\ 1 &{} \cdot &{} \cdot &{} a_{m} \\ r_{m} &{} a_{m} &{} a_{m} &{} \cdot \end{array} \right)$,

– $\mathbf {p}_{m^{*}}=\left( \begin{array}{c|ccc} Rank &{} p_{m^{*}}^{1} &{} p_{m^{*}}^{2} &{} p_{m^{*}}^{3} \\ \hline 1 &{} \cdot &{} a_{m^{*}} &{} a_{m^{*}} \\ A_{m^{*}}-1 &{} b_{m^{*}} &{} b_{m^{*}} &{} b_{m^{*}} \\ A_{m^{*}} &{} a_{m^{*}} &{} \cdot &{} \cdot \end{array} \right)$,

– $\forall m>m^{*}$, $\mathbf {p}_{m}=\left( \begin{array}{c|ccc} Rank &{} 1 &{} 2 &{} 3 \\ \hline 1 &{} a_{m} &{} \cdot &{} a_{m} \\ A_{m} &{} \cdot &{} a_{m} &{} \cdot \end{array} \right)$.

Let $\overrightarrow{a}=(a_{1},\ldots ,a_{M}),\overrightarrow{b} =(b_{1},\ldots ,b_{M}),\overrightarrow{c}=(b_{m^{*}},\overrightarrow{a} _{-m^{*}})\in \mathcal {A}$, $\mathcal {A}^{\prime }=\{(b_{m^{*}}, \overrightarrow{\alpha }_{-m^{*}}):\overrightarrow{\alpha }\in \mathcal {A }\}{\setminus }\{\overrightarrow{c}\}$. Now, extend $\mathbf {p}$ to a coordinate profile $\overline{\mathbf {p}}$ involving $I\ge \prod _{m\le m^{*}}A_{m}$ $+3$ and such that

(1) $\forall i\in \{1,2\}$, $\overline{p}^{i}=p^{i}$,

(2) $\forall i\in \{3,...,I\}$, $\overline{p}^{i}$ agrees with $p^{3}$ on the ranks given to $a_{m}$ for all $m\in \mathcal {M}$, and $\overline{p}^{i}$ agrees with $p^{3}$ on the rank given to $b_{m^{*}}$,and $\forall c_{m^{*}}\in \mathcal {A}_{m^{*}}{\setminus }\{a_{m^{*}},b_{m^{*}}\}$ there exists $i(c_{m^{*}})\in \{4,\ldots ,I\}$ such that $r[c_{m^{*}},\overline{p}^{i(c_{m^{*}})}]=A_{m^{*}}$,

(3) $\forall \overrightarrow{d}\in \mathcal {A}^{\prime }$, there exists $i( \overrightarrow{d})\in \{3,\ldots ,I\}$ with $r[(\overrightarrow{d}, \overline{p}^{i(\overrightarrow{d})}]=R^{\prime }=(R_{<m^{*}},A_{m^{*}}-1,A_{m^{*}+1},\ldots ,A_{M})$.

As $I^{*}\ge \prod _{m\le m^{*}}A_{m}$, condition (3) can be satisfied. Check that $\prod _{m\in \mathcal {M}}F_{I}(\overline{\mathbf {p}} _{m})\subseteq \mathcal {A}^{\prime }\cup \{\overrightarrow{c}\}$, and $\overrightarrow{c}\in \mathbf {\prod }_{m\in \mathcal {M}}F_{I}(\overline{ \mathbf {p}}_{m})$. Table 3 gives the rank vectors of $\overrightarrow{a}$ and $\overrightarrow{c}$ for each $\overline{p}^{i}$:

Table 3 The rank vectors of $\overrightarrow{a}$ and $\overrightarrow{c}$ in Case 2

The responsiveness of $\delta$ implies $V_{I}[\overrightarrow{c},{ \delta }(\overline{\mathbf {p}}\mathbf {)}]=r[\overrightarrow{c},\delta ( \overline{p}^{2})]$. Moreover, again by the responsiveness of $\delta$, we have $V_{I}[\overrightarrow{a},{\delta }(\overline{\mathbf {p}}\mathbf { )}]\in \{r[\overrightarrow{a},\delta (\overline{p}^{1})],r[\overrightarrow{a},\delta (\overline{p}^{2})]\}$. Since R $\delta$ $R^{\prime }$, then $r[ \overrightarrow{a},\delta (\overline{p}^{1})]<V_{I}[\overrightarrow{c}, {\delta }(\overline{\mathbf {p}}\mathbf {)}]$. Finally, the responsiveness of $\delta$ implies $r[\overrightarrow{a},\delta (\overline{p }^{2})]<V_{I}[\overrightarrow{c},{\delta }(\overline{\mathbf {p}} \mathbf {)}]$. Hence, $V_{I}[\overrightarrow{a},{\delta }(\overline{ \mathbf {p}}\mathbf {)}]<V_{I}[\overrightarrow{c},{\delta }(\overline{ \mathbf {p}}\mathbf {)}]$. Thus, $\overrightarrow{c}\notin F_{I}[{ \delta }(\overline{\mathbf {p}}\mathbf {)]}$.

Finally, pick any $\overrightarrow{d}\in \mathcal {A}^{\prime }$. By definition of $\overline{\mathbf {p}}$, $V_{I}[\overrightarrow{d},{ \delta }(\overline{\mathbf {p}}\mathbf {)}]\ge r[\overrightarrow{d},\overline{ p}^{i(\overrightarrow{d})}]=r[\overrightarrow{c},\delta (\overline{p} ^{2})]=V_{I}[\overrightarrow{c},{\delta }(\overline{\mathbf {p}} \mathbf {)}]$. As $V_{I}[\overrightarrow{a},{\delta }(\overline{ \mathbf {p}}\mathbf {)}]<V_{I}[\overrightarrow{c},{\delta }(\overline{ \mathbf {p}}\mathbf {)}]$, we get $[\prod _{m\in \mathcal {M}}F_{I}(\mathbf {p} )]$ ${\cap }$ $F_{I}[{\delta }(\overline{\mathbf {p}}\mathbf { )]=\emptyset }$, in contradiction the assumption that $F_{I}$ is weakly stable over $\mathcal {D}$. $\square$

The second step in the proof of the necessity part consists in showing that $F_{I}$ is weak product stable only if all voters are assigned to the same lexicographic preference.

Lemma 7

Take $\widetilde{q},\widetilde{r}\in \mathcal {L}_{\mathcal { M}}$ with $\widetilde{q}\ne \widetilde{r}$. If $I\ge 3$, and if $\{\delta ^{\widetilde{q}},\delta ^{\widetilde{r}}\}\subseteq \mathcal {D}$, then $F_{I}$ is not weakly product stable over $\mathcal {D}$.

Proof

Take $\widetilde{q},\widetilde{r}\in \mathcal {L}_{\mathcal {M}}$ with $\widetilde{q}\ne \widetilde{r}$ and $\{\delta ^{\widetilde{q}},\delta ^{ \widetilde{r}}\}\subseteq \mathcal {D}$. Up to a relabelling of coordinates $m\in \mathcal {M}$, one can suppose w.l.o.g. that $\widetilde{q}=[12\ldots M]$ and $m^{*}$ $\widetilde{r}$ $\overline{m}$ for some $\overline{m},m^{*}\in \mathcal {M}$ with $\overline{m}<m^{*}$. Take a 3-voter coordinate profile $\mathbf {p}$ such as below:

– $\forall m\in \mathcal {M}{\setminus }\{m^{*},\overline{m}\}$, $\mathbf {p}_{m}=\left( \begin{array}{c|ccc} Rank &{} p_{m}^{1} &{} p_{m}^{2} &{} p_{m}^{3} \\ \hline 1 &{} b_{m} &{} b_{m} &{} b_{m} \\ \ldots &{} \ldots &{} \ldots &{} \ldots \end{array} \right)$,

– $\mathbf {p}_{m^{*}}=\left( \begin{array}{c|ccc} Rank &{} p_{m^{*}}^{1} &{} p_{m^{*}}^{2} &{} p_{m^{*}}^{3} \\ \hline 1 &{} b_{m^{*}} &{} a_{m^{*}} &{} b_{m^{*}} \\ 2 &{} a_{m^{*}} &{} b_{m^{*}} &{} c_{m^{*}} \\ \ldots &{} \ldots &{} \ldots &{} \ldots \end{array} \right)$, $\mathbf {p}_{\overline{m}}=\left( \begin{array}{c|ccc} Rank &{} p_{\overline{m}}^{1} &{} p_{\overline{m}}^{2} &{} p_{\overline{m} }^{3} \\ \hline 1 &{} a_{\overline{m}} &{} b_{\overline{m}} &{} a_{\overline{m}} \\ 2 &{} b_{\overline{m}} &{} c_{\overline{m}} &{} b_{\overline{m}} \\ \ldots &{} \ldots &{} \ldots &{} \ldots \end{array} \right)$.

Consider $\overrightarrow{b}=(b_{1},\ldots ,b_{M})\in \mathcal {A}$. Clearly, $\prod _{m\in \mathcal {M}}F_{I}(\mathbf {p})=\{\overrightarrow{b}\}$. Now, define $\overrightarrow{c}=(c_{1},\ldots ,c_{M})\in \mathcal {A}$ by $\forall m\in \mathcal {M}$, $c_{m}=\left\{ \begin{array}{l} a_{m} \ \ \hbox {if }m\in \{m^{*},\overline{m}\} \\ b_{m} \ \ otherwise \end{array} \right.$. Take ${\delta }=(\delta ^{1},\delta ^{2},\delta ^{3})=(\delta ^{\widetilde{q}},\delta ^{\widetilde{r}},\delta ^{\widetilde{q} })$. Since $a_{\overline{m}}$ $p_{\overline{m}}^{i}$ $b_{\overline{m}}$ for $i=1,3$, the definition of $\delta ^{\widetilde{q}}$ ensures that $\overrightarrow{c}$ $\delta (p^{i})$ $\overrightarrow{b}$ for $i=1,3$. Similarly, since $a_{m^{*}}$ $p_{m^{*}}^{2}$ $b_{m^{*}}$, the definition of $\delta ^{\widetilde{r}}$ ensures that $\overrightarrow{c}$ $\delta (p^{2})$ $\overrightarrow{b}$. Thus, $V_{I}[\overrightarrow{c}, {\delta }(\mathbf {p)}]<V_{I}[\overrightarrow{b},{\delta }( \mathbf {p)}]$, which in turn implies $[\prod _{m\in \mathcal {M}}F_{I}( \mathbf {p})]$ ${\cap }$ $F_{I}[{\delta }(\overline{\mathbf {p}} \mathbf {)]=\emptyset }$. This shows that $F_{I}$ is not weakly product stable over $\mathcal {D}$. $\square$

The necessity part of Theorem 1 follows from combining Lemmas 6 and 7.