Electoral competition under best-worst voting rules

Abstract

We characterise multi-candidate pure-strategy equilibria in the Hotelling–Downs spatial election model for the class of best-worst voting rules, in which each voter is endowed with both a positive and a negative vote, i.e., each voter votes in favour of their most preferred candidate and against their least preferred. The importance of positive and negative votes in calculating a candidate’s net score may be different, so that a negative vote and a positive vote need not cancel out exactly. These rules combine the first-place seeking incentives of plurality with the incentives to avoid being ranked last of antiplurality. We show that, in our simple model, arbitrary best-worst rules admit equilibria, which (except for three candidates) are nonconvergent if and only if the importance of a positive vote exceeds that of a negative vote. The set of equilibria in the latter case is very similar to that of plurality, except the platforms are less extreme due to the moderating effect of negative votes. Moreover: (i) any degree of dispersion between plurality, at one extreme, and full convergence, at the other, can be attained for the correct choice of the weights; and, (ii) when they exist (and there are at least five candidates), there always exist nonconvergent equilibria in which none of the most extreme candidates receive the most electoral support.

Appendix A

1.1 Preliminary results and lemmata

We include here a number of lemmata that are needed for our main results. Several of these minor results are adapted from results in Cahan and Slinko (2017), though similar conditions have appeared in various form in the previous literature since at least Eaton and Lipsey (1975).

The first lemma tells us that the most extreme occupied positions cannot be occupied by single candidates, and they cannot be located at the most extreme points on the issue space.

Lemma A.1

(Cahan and Slinko 2017) In an NCNE, we must have $n_1,n_q\ge 2$. Moreover, no candidate may adopt the most extreme positions on the issue space. That is, $0<x^1$ and $x^q<1$.

Proof

Evidently, an unpaired candidate at $x^1$ could move to the right and capture a larger share of positive votes and, at the same time, reduce the number of negative votes.

To see the second part, suppose $x^1=0$. Then the at least two candidates at $x^1$ are ranked last equal by a positive measure of voters in the interval $(1-\frac{1}{2}x^q,1]$. By moving to $x^{1+}$, however, a candidate originally at $x^1$ is no longer ranked last by any voters, but still receives the same number of first-place rankings. $\square$

The next lemma puts a condition on the continuity of the function $v_i(t,x_{-i})$ when, in equilibrium, i is at a position occupied by one other candidate and makes a small deviation.

Lemma A.2

Suppose at profile x candidate i is at $x^l$ and $n_l=2$. Then $v_i(x^{l-},x_{-i})+v_i(x^{l+},x_{-i})=2v_i(x)$. In particular, when x is in NCNE, we have $v_i(x^{l-},x_{-i})=v_i(x^{l+},x_{-i}) =v_i(x)$.

Proof

The issue space can be divided into subintervals of voters who all rank i in the same position. The immediate interval around $x^l$, $I_l=I_l^L\cup I_l^R$, is the set of voters from which the candidate receives positive votes. Let J be the interval of voters from which i receives negative votes. In particular, J is nonempty only if $l=1$ or $l=q$, and it is located at the opposite side of the issue space. Thus, if $l\notin \{1,q\}$, we have $v_i(x)=\frac{1}{2}(\ell (I_l^L)+\ell (I_l^R))$. Then $v_i(x^{l-},x_{-i})=\ell (I_l^L)$ and $v_i(x^{l+},x_{-i})=\ell (I_l^R)$. For NCNE, we need $v_i(x^{l-},x_{-i})\le v_i(x)$ and $v_i(x^{l+},x_{-i})\le v_i(x)$. Summing these inequalities, we need $v_i(x^{l-},x_{-i})+v_i(x^{l+},x_{-i})\le 2v_i(x)$. This, in fact, turns out to be an equality, so that we must have $v_i(x^{l-},x_{-i})=v_i(x^{l+},x_{-i})=v_i(x)$.

If $l=1$ (symmetrically for $l=q$), we have $v_i(x)=\frac{1}{2}(\ell (I_l^L)+\ell (I_l^R))-\frac{c}{2}\ell (J).$ Also, $v_i(x^{l-},x_{-i})=\ell (I_l^L)-c\ell (J)$ and $v_i(x^{l+},x_{-i})=\ell (I_l^R)$. As in the previous case, summing the requirements that these two moves not be beneficial, we find that $v_i(x^{l-},x_{-i})=v_i(x^{l+},x_{-i})=v_i(x)$.$\square$

Lemma A.3

If $n_i=n_j=2$, then $v_i(x)=v_j(x)$ in NCNE.

Proof

Let k be a candidate at $x^i$ and l be a candidate at $x^j$. Note that if k moves to $x^{j+}$ or $x^{j-}$, due to the nature of the voting rule, k receives exactly the same score as l would recieve on moving to $x^{j+}$ or $x^{j-}$. So $v_k(x^{j+},x_{-k})=v_l(x^{j+},x_{-l})$. Then, if x is in NCNE, using Lemma A.2 gives that $v_l(x)=v_l(x^{j+},x_{-l})=v_k(x^{j+},x_{-k})\le v_k(x)$. Similarly, $v_l(x^{i+},x_{-l})=v_k(x^{i+},x_{-k})$, from which it follows that $v_k(x)=v_k(x^{i+},x_{-k})=v_l(x^{i+},x_{-l})\le v_l(x)$. So $v_k(x)=v_l(x)$. $\square$

Next, we note that there cannot be more than two candidates at any position. In particular, this implies that there cannot exist NCNE for $m=3$, a well known result.

Lemma A.4

In any NCNE, at any given position there are no more than two candidates. Moreover, $n_1=n_q=2$.

Proof

By Corollary 4.2 we have to consider only the case when $c<1$.

First, we show that, in NCNE, $n_i\le 2$ for all $2\le i \le q-1$. If $n_i>2$, where $2\le i\le q-1$, then candidate k, located at $x^i$ is not ranked last by any voter. Moreover, she is not ranked last by any voter even on deviating to $x^{i+}$ or $x^{i-}$. So the only change in her score on making these moves is from voters in the immediate subintervals $I_1=[(x^{i-1}+x^i)/2,x^{i}]$ and $I_2=[x^{i},(x^{i}+x^{i+1})/2]$, where voters change candidate k from first equal to first, and from first equal to $n_i$th, respectively.

In NCNE we must have

$$\begin{aligned}v_k(x^{i-},x_{-k})-v_k(x)= \ell (I_1)- \frac{1}{n_i}(\ell (I_1)+\ell (I_2))\le 0 \end{aligned}$$

and

$$\begin{aligned} v_k(x^{i+},x_{-k})-v_k(x)= \ell (I_2) - \frac{1}{n_i}(\ell (I_1)+\ell (I_2))\le 0. \end{aligned}$$

Adding together these two inequalities we get the requirement that $n_i\le 2$.

To show that $n_1=n_q=2$, let us introduce the following notation: $I_1=[0,x^1]$, the voters to the left of candidate 1 (note that by Lemma A.1, this set has positive measure); $I_2=[x^1,(x^1+x^2)/2]$, the voters in half the interval between candidates 1 and 2; $I_3= [(x^1+x^q)/2,1]$, the voters for whom 1 is ranked last equal.

Note that $v_1(x) =\frac{1}{n_1}(\ell (I_1)+\ell (I_2))-\frac{c}{n_1}\ell (I_3).$ Consider if 1 moves to $x^{1-}$. Then $v_1(x^{1-},x_{-1})= \ell (I_1)-c\ell (I_3)) .$ If 1 moves to $x^{1+}$ then $v_1(x^{1+},x_{-1})= \ell (I_2).$ For NCNE we require that these moves not be beneficial to candidate 1. That is, $v_1(x^{1-},x_{-1})\le v_1(x)$ which implies we need

$$\begin{aligned} \ell (I_1)-c\ell (I_3))\le \frac{1}{n_1}(\ell (I_1)+\ell (I_2))-\frac{c}{n_1}\ell (I_3), \end{aligned}$$

or

$$\begin{aligned} \left( 1-\frac{1}{n_1}\right) c \ell (I_3)\ge \ell (I_1)-\frac{1}{n_1}(\ell (I_1)+\ell (I_2)). \end{aligned}$$

(1)

Similarly, for the other move we have $v_1(x^{1+},x_{-1})\le v_1(x)$ which gives us

$$\begin{aligned} \ell (I_2)\le \frac{1}{n_1}(\ell (I_1)+\ell (I_2))-\frac{c}{n_1}\ell (I_3), \end{aligned}$$

implying

$$\begin{aligned} \left( 1-\frac{1}{n_1}\right) c\ell (I_3)\le \left( 1-\frac{1}{n_1}\right) (\ell (I_1)+\ell (I_2))-(n_1-1)\ell (I_2). \end{aligned}$$

Combining this last equation with (1) yields $(2-n_1)\ell (I_2)\ge 0$, which means $n_1\le 2$. Hence, $n_1=2$, since we cannot have a lone candidate at $x^1$. A similar argument gives $n_q=2$. $\square$

1.2 Win maximisation and generalised best-worst voting rules

The following result and example relate to the discussion in Sect. 5.

Proposition A.5

When candidates only care about winning, the profile $x=(x_1,x_2,x_3)$, $x_1\le x_2 <x_3$, is an NCNE in which candidate 3 wins if and only if the following are satisfied (by symmetry, a corresponding set of NCNE exists where candidate 1 wins):

• (i) $2(1+c)>(2c+3)x_2+(2c+1)x_3$;

• (ii) $2(1+c)<(2c+3)x_3+(2c+1)x_1$;

• (iii) $x_3-x_1<\frac{2}{3}(1-c)$.

Proof

First, there cannot be an equilibrium with a tie for first, since candidates 1 or 3 could move inwards and break the tie. Second, 2 could never win in an NCNE since we would then require that both $v_1(x_2,x_{-1})<v_3(x_2,x_{-1})$ and $v_3(x_2,x_{-3})<v_1(x_2,x_{-3})$, that is, neither 1 nor 3 should want to deviate to $x_2$. Summing together these inequalities leads to the requirement that $v_2(x)=\frac{1}{2}(x_3-x_1)<\frac{1-c}{3(1+c)}$. If $c\ge 1$, this is impossible. If $c<1$, note that $v_2(x)<\frac{1-c}{3(1+c)}<\frac{1}{3}(1-c)$, which contradicts that 2 is winning since to be winning 2 must receive more than 1/3 of the total votes. So there must be a unique winner—suppose without loss of generality it is candidate 3.

Assume $x_1<x_2<x_3$. For NCNE, 1 should not want to move to $x_2^-$, $x_2$ or $x_2^+$. For the first move, we need $v_1(x_2^-,x_{-1})<v_3(x_2^-,x_{-1})$, which yields (i). If this move is not beneficial, neither will be the other two. For the second move, 1 and 2 are now tied, and since 3’s score does not change (that is, $v_3(x_2,x_{-1})=v_3(x_2^-,x_{-1})$), 3 must still be winning. For the third move, 3 is still winning, since this is the same as the first move but swapping the labels on candidates 1 and 2. Since none of these three moves are beneficial for 1, it is clear that 2 would not want to move to $x_1^-$, $x_1$ or $x_1^+$ (all three moves would lead to 3 winning by an even bigger margin than if 1 deviated).

We also require that 2 not want to move to $x_3^+$, $x_3$ or $x_3^-$. For the first move, we need $v_1(x_3^+,x_{-2})>v_2(x_3^+,x_{-2})$, which yields (ii). If this is true then, for the second move, 2 and 3 are now tied, and since 1’s score does not change (that is, $v_1(x_3,x_{-2})=v_1(x_3^+,x_{-2})$), 1 would win here too. For the third move, 1 would again win, since this is the same as the first move but swapping the labels on candidates 2 and 3. Since 2 does not want to move to any of these three positions, it is clear that 1 would not want to move to either (candidate 2 would certainly win).

Next, suppose (iii) is not satisfied, so $x_3-x_1\ge \frac{2}{3}(1-c)$. Consider if candidate 2 deviates to any point t between $x_1$ and $x_3$, in which case 2’s score remains constant at $(x_3-x_1)/2$. Note also that $v_1(t ,x_{-2})$ and $v_3(t ,x_{-2})$ are increasing and decreasing, respectively, in t. By the above, when $t=x_1^+$, 3 is the sole winner, while 2 is the sole winner when $t=x_3^-$. Therefore, at some point $t'$ it must be the case that $v_1(t',x_{-2})=v_3(t',x_{-2})$. For NCNE, it must be that $v_1(t',x_{-2})=v_3(t',x_{-2})>v_2(t',x_{-2})$. However, the sum of the scores is fixed at $1-c$, which contradicts the previous statement and the assumption that $x_3-x_1\ge \frac{2}{3}(1-c)$. So (iii) is necessary. Since 2 does not benefit from any move between $x_1$ and $x_3$, it is also the case that 1 would not benefit from moving between $x_2$ and $x_3$.

Sufficiency of (i)–(iii) follows by construction, since we have checked all potentially profitable deviations. The above arguments also apply when $x_1=x_2$, and are simpler because some of the deviations become redundant.$\square$

Example A.6

We investigate equilibria of the form $x=((x^1,3),(1-x^1,3))$. That is, NCNE with three candidates apiece at two symmetric occupied positions. Recall that for a standard best-worst rule, it is never possible to have three candidates at a single position in an NCNE. For generalised best-worst rules, such NCNE may indeed exist.

We may apply Theorem 5 of Cahan and Slinko (2017), which characterises “bipositional symmetric” equilibria for arbitrary scoring rules. First, we find that NCNE cannot exist for $d_1\ge 3$, so there can be at most two positive votes. If $d_1=2$, we find that bipositional symmetric NCNE exist for any $d_2<5$ and they exist if and only if one of the following is true:

• (i) If $d_2=0$ (equivalently, $d_2=4$) and $1/6\le x^1\le 1/3$.

• (ii) If $d_2=1$, $c<2$, and $(1+c)/6\le x^1\le (1+c)/3$.

• (iii) If $d_2=2$, $c<1$, and $(1+2c)/6\le x^1\le (2+c)/6$.

• (iv) If $d_2=3$, $c\le 1$, and $(1+3c)/(6(1+c))\le x^1\le 1/3$.

In cases (ii)–(iv), we note some interesting similarities to case of standard best-worst rules described in Theorem 4.3. First, the weight placed on the negative votes should not be too high for NCNE to exist, although the bound need not always be 1. Second, the presence of negative votes again induces moderation in the set of possible equilibrium profiles. As c approaches its upper bound, the extreme most positions possible in NCNE move closer to the median voter (though need not converge).¹⁶

If $d_1=1$, bipositional symmetric NCNE only exist if $d_2=4$ and $c\ge 1$. Any $x^1$ satisfying $(2+c)/(6(1+c)) \le x^1\le (1+2c)/(6(1+c))$ is an NCNE. This is quite different to the NCNE described above as well as NCNE for standard best-worst rules—here, there is a lower bound on the value of c and, as c increases, the range of possible NCNE positions becomes wider and includes NCNE that are more dispersed. This shows that while negative votes may have similar moderating effects for generalized best-worst rules as for standard best-worst rules in some cases, in other cases these patterns may break down.