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Characterizing referenda with quorums via strategy-proofness

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Abstract

The paper works with a formal model of referenda, where a finite number of voters can choose between two options and abstention. A referendum will be invalid if too many voters abstain, otherwise the referendum will return one of the two options. We consider quorum rules where an option is chosen if it is preferred by the majority of voters and if at least a certain number of voters (the quorum) votes for the alternative. The paper characterizes these rules as the only referenda which are strategy-proof over certain preferences.

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Notes

  1. While local referenda in the Netherlands are always non-binding to the city council, the city council will usually self-bind itself to the referendum outcome provided the quorum is met.

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Correspondence to Marc Pauly.

Appendix

Appendix

Proof of Lemma 1

By axiom SP, we know that individual 1 has a dominant strategy \(v_{*}\) under preference relation \(\ge _1\). First note that by axiom A, \(v_{*}\) must be a dominant strategy for all individuals under \(\ge _1\). If \(v_{*} = 1\), we are done. So consider the other two cases. Suppose first that \(v_{*}=0\). Then using axiom WP and the assumption that \(0\) is a dominant strategy, we have \(C(0,1,\ldots ,1)=1\). Now we can apply for every individual one by one that \(0\) is a dominant strategy and obtain \(C(0,0,\ldots ,0)=1\). Using neutrality, we also have \(C(0,0,\ldots ,0)=-1\), a contradiction. Now suppose second that \(v_{*}=-1\). Then using axiom WP and the assumption that \(-1\) is a dominant strategy, we have \(C(-1,1,\ldots ,1)=1\), and repeated application for every individual gives us \(C(-1,-1,\ldots ,-1)=1\). By axiom N, \(C(1,1,\ldots ,1)=-1\), contradicting WP.

Finally, we also know that individual 1 has a dominant strategy \(v_{*}\) under preference relation \(\ge _{-1}\). By axioms N and WP, \(C(-1,-1,\ldots ,-1)=-1\). The rest of the argument is analogous to the argument for \(\ge _1\). \(\square \)

Proof of Lemma 2

(1.) Using axiom SP*, we have \(C(Y{+}1,N) \ge C(Y,N) = 1\), so \(C(Y{+}1,N)=1\). Repeatedly applying this argument yields the desired conclusion. (2.) Assuming that \(N>0\), using SP*, we have \(1=C(Y,N) \le C(Y,N{-}1)\). Hence, \(C(Y,N{-}1)=1\). Repeatedly applying this argument yields the desired conclusion. \(\square \)

Proof of Lemma 3

Since \(C\) satisfies A, for any \(V \in \mathcal{V }\) we can write \(C(V)\) as \(C(Y,N)\) with \(Y=n^+(V)\) and \(N=n^-(V)\). Suppose that \(C(Y,N)=1\). For a proof by contradiction, suppose that \(Y \le N\). We distinguish two cases. First, suppose that \(Y=N\). Then by neutrality, \(C(N,Y)=-1\). But since \(C(N,Y)=C(Y,N)\), we have obtained a contradiction. Second, suppose that \(Y<N\). By Lemma 2, \(C(Y,Y)=1\). But then by neutrality again, \(C(Y,Y) = -1\) as well, a contradiction. \(\square \)

Proof of Theorem 1

It is easy to check that any \(M_q\) rule satisfies the axioms, so we will only prove the converse. As a visual aid to the proof, it may be useful to construct the diagram of \(C\) step by step. Using the axioms we will see that the diagram must look like Fig. 2.

Suppose some decision rule \(C\) satisfies the axioms. Given anonymity, we can write \(C\) as \(C(n^+(V),n^-(V))\). Now let \(q\) be the smallest number such that \(C(q,r)=1\) for some \(r\). By WP we know that such a q exists. By axiom M (implied by Lemmas 1 and 3), we know that \(q>r\). We will show below that \(C=M_q\).

By Lemma 2, we know that for all \(Y \ge q\), we have \(C(Y,r)=1\). Again, by Lemma 2, we know that for all \(Y \ge q\) and \(N \le r\), we have \(C(Y,N)=1\).

Now suppose we have \(Y \ge q\) and \(N \ge r\) with \(Y > N\). By axioms M and N, we know that \(C(Y,N) \ne -1\). Also, by Q, \(C(Y,N) \ne 0\), otherwise we would have \(C(q,r)=0\). Hence, we know that as long as \(Y > N, C(Y,N) = 1\) for all \(Y \ge q\). Note that we have just proved the first clause of the definition of \(M_q\).

To show the second clause of the definition of \(M_q\), suppose that \(N>Y\) and that \(N \ge q\). Then by neutrality, \(C(Y,N) = - C(N,Y) = -1\).

Finally, as for the third clause in the definition of \(M_q\), we distinguish three cases. First, suppose that \(Y=N\). Then by axioms M and N, we know that \(C(Y,N)=0\). Second, consider that \(Y > N\) but that \(Y < q\). We know that \(C(Y,N) \ne 1\) since \(q\) was taken to be the lowest point for which \(C\) yields 1. Also, we know that \(C(Y,N) \ne -1\) for that would contradict the majority axiom M using neutrality. Hence, \(C(Y,N)=0\). The third case where \(Y < N\) with \(N < q\) follows by neutrality from this second case: \(C(Y,N) = -C(N,Y) =0\).

Since we have shown that in all cases \(C\) yields the same result as \(M_q\), we have completed the proof. \(\square \)

Proof of the independence of axioms A, N, WP, Q, and SP:

NOT SP We have already seen earlier that the participation quorum rule \(P_q\) is not strategy-proof. It can be checked that it satisfies all the other axioms.

NOT WP Consider the referendum rule that constantly returns 0 on all inputs, i.e., \(C(V)=0\) for all \(V \in \mathcal{V }\). It satisfies all the axioms except WP.

NOT N Consider the acceptance quorum rule \(A_q\), defined in Sect. 6 and depicted in Fig. 3. It satisfies all axioms except neutrality (see also Theorem 2).

NOT Q Let \(C\) be defined just like \(M_q\), except that for all \(X>0\) we have \(C(X,0)=1\) and \(C(0,X)=-1\). Intuitively, this referendum rule yields the majority opinion if the quorum is met, but if nobody opposes the majority opinion, the majority opinion results even if the quorum is not met. \(C\) does not satisfy Q since for \(q=3\), we have \(C(1,2)=0\) while \(C(1,0)=1\). However, \(C\) satisfies the other axioms. To see that \(C\) satisfies SP* (and hence also SP), we only show part (1.) of the axiom since part (2.) can be shown analogously. For part (1.), it suffices to verify that \(C(V[i:1]) \ge C(V)\) for the cases where one of the two sides of the inequality is \(C(X,0)\) or \(C(0,X)\). If \(C(X,0)\) occurs on the left side of the inequality or \(C(0,X)\) occurs on the right side, the inequality is automatically satisfied. Furthermore, \(C(0,X)\) cannot occur on the left side. Hence, we are left with the case that \(C(X,0)\) occurs on the right side. In this case, we need to check that \(C(X+1,0) =C(X,0)=1\) which holds by definition of \(C\).

NOT A Consider the referendum rule \(C^n\) which uses a majority quorum but counts voter 1 twice. Formally, we have \(C^n(v_1,\ldots ,v_n) = M^{n+1}_q(v_1,\ldots ,v_n,v_1)\). This rule is clearly not anonymous. Neutrality and WP are easily seen to be satisfied by \(C\). To show that \(C\) satisfies SP*, we use twice the fact that \(M^{n+1}_q\) satisfies SP*:

$$\begin{aligned} \begin{array}{rl} C(1,v_2,\ldots ,v_n) &{} = M^{n+1}_q(1,v_2,\ldots ,v_n,1) \\ &{} M^{n+1}_q(k,v_2,\ldots ,v_n,1) \\ &{} M^{n+1}_q(k,v_2,\ldots ,v_n,k) = C(k,v_2,\ldots ,v_n) \end{array} \end{aligned}$$

Finally, we note that \(C\) also satisfies Q: \(M^{n+1}_q\) satisfies Q over the set of all vote vectors, so in particular it satisfies Q over the restricted set of vote vectors where the first and the last voter submit identical votes. \(\square \)

Proof of Theorem 2

It is easy to check that any \(A_q\) rule satisfies the axioms, so we will only prove the converse.

Suppose some decision rule \(C\) satisfies the axioms. Given anonymity, we can write \(C\) as \(C(n^+(V),n^-(V))\). Now let \(q\) be the smallest number such that \(C(q,r)=1\) for some \(r\). By WP we know that such a q exists. By axiom M, we know that \(q>r\). We will show below that \(C=A_q\).

By Lemma 2, we know that for all \(Y \ge q\), we have \(C(Y,r)=1\). Again, by Lemma 2, we know that for all \(Y \ge q\) and \(N \le r\), we have \(C(Y,N)=1\).

Now suppose we have \(Y \ge q\) and \(N \ge r\) with \(Y > N\). By RQ, we know that \(C(Y,N) \ne -1\) (otherwise we would have \(C(q,r)=-1\) as well), and so by axiom D, \(C(Y,N)=1\). Hence, we know that as long as \(Y > N, C(Y,N) = 1\) for all \(Y \ge q\), proving the first clause of the definition of \(A_q\).

For the second clause of the definition, we have two cases. First, suppose that \(Y \le N\). Then by axioms M and D, \(C(Y,N) = -1\). Second, suppose that \(Y < q\). Then we know that \(C(Y,N)=-1\) by axiom D and by our assumption that \(q\) was the smallest value for \(C\) to yield \(1\). \(\square \)

Proof of the independence of axioms A, M, WP, RQ, D, and SP*:

NOT M As was already mentioned, we can consider the constant referendum rule \(C(V)=1\) for all \(V\).

NOT D The majority quorum rule \(M_q\) does not satisfy decisiveness but does satisfy RQ (and all the other axioms).

NOT WP Consider the referendum rule that constantly returns \(-1\) on all inputs. It violates WP but satisfies all the other axioms.

NOT SP* The participation quorum rule \(P^{\prime }\) satisfies all the axioms except strategy-proofness. Consider the case of 5 voters with \(q=3\), where voter 3 has preference \(\ge _{-1}\), and the following ballot profile \(V=(1,1,-1,0,0)\). When submitted, we have \(P^{\prime }_q(V)=1\), the opposite of what voter 3 wants. However, voter 3 can behave strategically by abstaining. We then obtain \(P^{\prime }_q(V[3:0])=-1\). Hence, \(P^{\prime }_q(V) \not \le P^{\prime }_q(V[3:0])\), a violation of axiom SP*.

NOT A Define \(C(k,1,\ldots ,1)=1\) for all \(k \in \{-1,0,1\}\) and \((V)=-1\) otherwise. This rule is not anonymous, but all the other axioms are satisfied by \(C\).

NOT RQ Let \(C(V) = 1\) iff \(n^+(V) > n^-(V), n^+(V) \ge q\) and \(n^-(V) \le q\). This referendum rule adds to the acceptance quorum rule an additional requirement for acceptance, namely that there are at most \(q\) no-votes. It does not satisfy RQ, since \(C(q+2,q+1)=-1\) while \(C(q+2,q)=1\). Axioms A, M, D, and WP are clearly satisfied. For SP*, first we must show that \(C(V[i:1]) \ge C(V)\) for all \(V\). If \(C(V)=1\), all three conjuncts in the definition of \(C\) are true, and they will remain true when increasing \(i\)’s ballot. Second, we need to show that \(C(V) \ge C(V[i:-1])\). If \(C(V)=-1\), one of the three conjuncts in the defining clause of \(C\) must be false, and lowering \(i\)’s ballot will not change this fact.\(\square \)

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Pauly, M. Characterizing referenda with quorums via strategy-proofness. Theory Decis 75, 581–597 (2013). https://doi.org/10.1007/s11238-013-9358-3

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