Formation of Parties and Coalitions in Multiple Referendums

Abstract

We consider a thought experiment in which voters could submit binary preferences regarding each of a pre-determined list of independent relevant issues, so that majorities could be tallied per issue. It might be thought that if such voting became technically feasible and widespread, parties and coalitions could be circumvented altogether and would become irrelevant. In this paper, we show, however, why and how voters would spontaneously self-organize into parties, and parties would self-organize into coalitions, prior to elections. We will see that such coordination is possible, even assuming very limited capabilities of communication and coordination. Using both analytical and empirical methods, we show that the average voter in a majority coalition would gain more than if no parties were formed, but the average voter overall (in or out of the coalition) would be worse off. Furthermore, the extent of these gains and losses is inversely proportional to the degree to which voters line along a unidimensional left–right axis.

Appendix: Proofs

Proof of Theorem 1

To prove the theorem, we first define intra-party regret, a measure of voters’ dissatisfaction with their respective parties.

Definition 5 [Intra-party regret]

Given a partition P, for each party $p \in P$, let $\langle p\rangle$ be the (not necessarily natural) profile of p.The intra-party regret for a party p is $ir(p,\langle p\rangle) = \sum\nolimits_{v \in p} {d(v,\langle p\rangle)}$. Let $\langle p\rangle$ represent the set of profiles of parties in P. The total intra-party regret of the partition P is $ir(P,\langle P\rangle) = \sum\nolimits_{p \in P} {d(p,\langle p\rangle)}$.

It is important to distinguish between intra-party regret, which is the dissatisfaction of voters with their respective parties, and total regret, which is the dissatisfaction of voters with the outcome of an election.

We will show that Procedure 1 converges to a stable partition. The key idea is that each step of the procedure decreases the value of $ir(P,\langle P\rangle)$. Since $ir(P,\langle P\rangle)$ is bounded from below, the process must converge. Furthermore, since at the last step, we require that the party be natural and the voters be stable, the final result is a stable partition.

It remains to show that $ir(P,\langle P\rangle)$ diminishes at each step.

First, we show that each application of Step 2 diminishes $ir(P,\langle P\rangle)$. Specifically, we must show that intra-party regret is diminished when each voter joins the nearest party. For a given voter v, let p^j(v) be the party of v prior to Step 2 and let p*^(v) be the party of v subsequent to Step 2. Then, p*^(v) is the nearest party to v. Let P be the partition prior to Step 2 and let P′ be the partition that results from Step 2. Note that in Step 2, voters switch parties but the party profiles remain fixed. Then we have:

$$ ir(P^{{\prime}},\langle P\rangle) = \mathop \sum \limits_{v \in V} d(v,\langle P^{*(v)} \rangle) \le \mathop \sum \limits_{v \in V} d(v,\langle P^{j(v)} \rangle) = ir(P,\langle P\rangle) $$

Now we show that each application of Step 3 diminishes $ir(P,\langle P\rangle)$. In Step 3, we change the profile of each party to its natural profile, but we do not change the membership of any party. For a party p, let its profile prior to Step 3 be $\langle p\rangle$ and let its profile subsequent to Step 3 be $\langle p\rangle^{*} = sign\left({\sum\nolimits_{v \in p} {\langle v\rangle}} \right)$. We need to show that:

$$ ir(P,\langle P\rangle^{*}) = \mathop \sum \limits_{p \in P} \mathop \sum \limits_{v \in p} d(v,\langle p\rangle^{*}) \le \mathop \sum \limits_{p \in P} \mathop \sum \limits_{v \in p} d(v,\langle p\rangle) = ir(P,\langle P\rangle) $$

To see this, consider each issue i separately. Let $v_{i} \in \{1, - 1\}$ be the ith element in the profile of v. That is, it represents the preference of voter v regarding issue i. Likewise, for a party p, let $p_{i} \in \{1, - 1\}$ and $p_{i}^{*} \in \{1, - 1\}$ be the ith element in $\langle P\rangle$ and $\langle P\rangle^{*}$, respectively. Regarding issue i, the preference of the majority of voters in p is $sign\left({\sum\nolimits_{v \in p} {v_{i}}} \right)$ and the difference between support for the majority and minority preference is $m_{i} = \left| {\sum\nolimits_{v \in p} {v_{i}}} \right|$. The intra-party regret for p is thus:

$$ \mathop \sum \limits_{v \in p} d(v_{i},p_{i}) = \left\{{\begin{array}{*{20}l} {\frac{{\left| p \right| - m_{i}}}{2},} \hfill & {p_{i} = sign\left({\mathop \sum \limits_{v \in p} v_{i}} \right) = p_{i}^{*}} \hfill \\ {\frac{{\left| p \right| + m_{i}}}{2},} \hfill & {otherwise} \hfill \\ \end{array}} \right. $$

We now have the following inequality:

$$ ir(P,\langle P\rangle) = \mathop \sum \limits_{p \in P} \mathop \sum \limits_{v \in p} d(v,\langle p\rangle) = \mathop \sum \limits_{p \in P} \mathop \sum \limits_{v \in p} \mathop \sum \limits_{i = 1}^{q} d(v_{i},p_{i}) = \mathop \sum \limits_{p \in P} \mathop \sum \limits_{i = 1}^{q} \mathop \sum \limits_{v \in p} d(v_{i},p_{i}) = \mathop \sum \limits_{p \in P} \mathop \sum \limits_{i = 1}^{q} \frac{{\left| p \right| \pm m_{i}}}{2} \ge \mathop \sum \limits_{p \in P} \mathop \sum \limits_{i = 1}^{q} \frac{{\left| p \right| - m_{i}}}{2} = \mathop \sum \limits_{p \in P} \mathop \sum \limits_{i = 1}^{q} \mathop \sum \limits_{v \in p} d(v_{i},p_{i}^{*}) = \mathop \sum \limits_{p \in P} \mathop \sum \limits_{v \in p} \mathop \sum \limits_{i = 1}^{q} d(v_{i},p_{i}^{*}) = \mathop \sum \limits_{p \in P} \mathop \sum \limits_{v \in p} d(v,\langle p^{*} \rangle) = ir(P,\langle P^{*} \rangle) $$

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Proof of Theorem 2

Consider each issue i separately. Let $v_{i} \in \{1, - 1\}$ be the ith element in the profile of v. That is, it represents the preference of voter v regarding issue i. Regarding issue i, the natural outcome of the individualist partition is $sign\left({\sum\nolimits_{v \in V} {v_{i}}} \right)$ and the difference between support for the majority and minority preference in V is $m_{i} = \left| {\sum\nolimits_{v \in p} {v_{i}}} \right|$. Since for the individualist partition each issue is decided in favor of the majority of voters, the overall regret of voters for the individualist partition regarding issue i is $\frac{{\left| V \right| - m_{i}}}{2}$. For any other partition, each issue is decided either in favor of the majority or the minority, so the overall regret of voters regarding issue i is either $\frac{{\left| V \right| - m_{i}}}{2}$ or $\frac{{\left| V \right| + m_{i}}}{2}$. Thus we have for any partition $P:\;r(P) = \mathop \sum \limits_{i = 1}^{q} \frac{{\left| V \right| \pm m_{i}}}{2} \ge \mathop \sum \limits_{i = 1}^{q} \frac{{\left| V \right| - m_{i}}}{2} = r(I)$ □

Proof of Theorem 3

As in the proof of Theorem 2, consider each issue i separately and let $v_{i} \in \{1, - 1\}$ be the ith element in the profile of v. Let $n_{i} = \sum\nolimits_{{p_{j} \ne p}} {p_{i}^{j}}$ and let $t_{i} = \sum\nolimits_{v \in p} {v_{i}}$. Then the outcome of the vote for issue i when the members of party p vote as individuals is sign(n_i + t_i). The outcome of the vote for issue i when the members of party p vote as a party is sign(n_i+ sign($\sum\nolimits_{v \in p} {v_{i}}$)|p|). Denote the regret of members of p for issue i by r_i(p, P) when they vote as a party and by r_i(p, P/p) when they vote as individuals. If sign(n_i)= sign(t_i) or if |t_i| > |n_i|, the outcome accords with the majority preference in p, whether the members of p vote as a party or as individuals; r_i(p, P) = r_i(p, P/p)= $\frac{{\left| p \right| - m_{i}}}{2}$, where $m_{i} = \left| {\sum\nolimits_{v \in p} {v_{i}}} \right|$. Otherwise, if |p| < |n_i|, then the outcome accords with the minority preference in p, whether the members of p vote as a party or as individuals; r_i(p, P) = r_i(p, P/p)= $\frac{{\left| p \right| + m_{i}}}{2}$. Finally, if |p| > |n_i|, the outcome accords with the majority preference of p if and only if they vote as a party. In this case r_i(p, P/p) = $\frac{{\left| p \right| + m_{i}}}{2}$ and r_i(p, P) = $\frac{{\left| p \right| - m_{i}}}{2}$ . Thus, for every i, r_i(p, P/p) ≥ r_i(p, P) so that r(p, P/p) = ∑_ir_i(p, P/p) ≥ ∑_ir_i(p, P) = r(p, P). □

Proof of Theorem 4

Consider each issue i separately. Regarding any issue i, the outcome given a majority coalition C is according to the majority preference within C. That is, $o(P/C) = sign\sum\nolimits_{v \in C} {v_{i}}$. The difference between support for the majority and minority preference within the coalition is $m_{i} = \left| {\sum\nolimits_{v \in C} {v_{i}}} \right|$. Thus, the overall regret of voters in the coalition regarding issue i is $\frac{{\left| C \right| - m_{i}}}{2}$. For the individualist partition, each issue is decided in favor of the majority of all voters, which might be the same as the majority of coalition members or not. Thus, the overall regret of coalition members regarding issue i is either $\frac{{\left| C \right| - m_{i}}}{2}$ (when the majority of voters share the preference of the majority of coalition members) or $\frac{{\left| C \right| + m_{i}}}{2}$ (otherwise). Thus, we have,

$$ \mathop \sum \limits_{v \in C} d(o(I),v) = \mathop \sum \limits_{i = 1}^{q} \frac{{\left| C \right| \pm m_{i}}}{2} \ge \mathop \sum \limits_{i = 1}^{q} \frac{{\left| C \right| - m_{i}}}{2} = \mathop \sum \limits_{v \in C} d(o(P/C),v). $$

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