On stable outcomes of approval, plurality, and negative plurality games

Abstract

We prove two results on the generic determinacy of Nash equilibrium in voting games. The first one is for negative plurality games. The second one is for approval games under the condition that the number of candidates is equal to three. These results are combined with the analogous one obtained in De Sinopoli (Games Econ Behav 34:270–286, 2001) for plurality rule to show that, for generic utilities, three of the most well-known scoring rules, plurality, negative plurality and approval, induce finite sets of equilibrium outcomes in their corresponding derived games—at least when the number of candidates is equal to three. This is a necessary requirement for the development of a systematic comparison amongst these three voting rules and a useful aid to compute the stable sets of equilibria Mertens (Math Oper Res 14:575–625, 1989) of the induced voting games. To conclude, we provide some examples of voting environments with three candidates where we carry out this comparison.

Appendices

Appendix 1: Generic determinacy of Nash equilibrium in negative plurality games

Henceforth, for every $i\in N$ we fix the set of pure strategies to be equal to

$$\begin{aligned} {\fancyscript{V}}_ np \equiv \left\{ (v^1,\ldots ,v^k)\in \{0,-1\}^k:\sum _{c\in K}{v^c}\in \{0,-1\}\right\} . \end{aligned}$$

(4.1)

With slight abuse of notation we denote by $c$ both candidate $c\in K$ and the ballot that gives a negative vote to candidate $c$. When that is the case, we say that a voter casts a negative vote against $c$ or, simply, that she votes against $c$. The symbol $0$ represents abstention. Therefore, we may write ${\fancyscript{V}}_{ np }=\{0\}\cup K$.

Before proving Proposition 1 we point out one complication of the analysis of negative plurality. There are examples robust to slight utility perturbations such that, even if more than one candidate wins with positive probability, abstention can be a best response for some voters. Note that the same is not true with plurality or approval because voting for the most preferred candidate among those who win with positive probability always yields a strictly larger payoff than abstention (as long as the voter is not completely indifferent among all the winning candidates).

Example 5

Consider a negative plurality voting game with set of voters $N=\{1,2,3,4\}$ and set of candidates $K=\{a,b,c,d\}$. Writing $u_i=(u_i(a),u_i(b),u_i(c),u_i(d))$ for voter $i$’s utility vector, voters’ preferences are given by:

$$\begin{aligned} u_1 = (0,0,-1,0),\quad u_2 = (4,3,0,6),\quad u_3 = (6,3,0,4),\quad u_4 = (0,-\varepsilon ,-1,0). \end{aligned}$$

where $\varepsilon >0$ is a suitable small number. A Nash equilibrium of this voting game is $\sigma =(c,\frac{1}{2}a+\frac{1}{2}b,\frac{1}{2}b+\frac{1}{2}d,0)$. Under this Nash equilibrium, candidates $a$ and $d$ win with probability $3/8$, and candidate $b$ wins with probability $1/4$. Note that for voter 4, voting against $b$, her least preferred candidate among those who win with positive probability, is not a best response. The reason is that if voter 4 votes against $b$ then candidate $c$, her least preferred candidate overall, wins with positive probability. Moreover, every game in a neighborhood has a close by Nash equilibrium with the same characteristics.

Nevertheless, we must point out that both abstention and voting against candidate $c$ are best responses for voter 4 and that abstaining is always a dominated strategy (by voting against the least preferred candidate overall).

Thus, in a Nash equilibrium of a negative plurality game, a voter may find it optimal to abstain even in close races. It should also be clear that voting against a candidate that wins with zero probability is “similar” to abstention in the sense that, once we fix the behavior of the rest of the voters, it does not affect the probability distribution over winning candidates.

We focus on Nash equilibria of negative plurality where more than one candidate wins with positive probability. We call such Nash equilibria nondegenerate equilibria.

Definition 2

The Nash equilibrium $\sigma$ is nondegenerate if the probability distribution $p(\sigma )\equiv {(p(c\mid \sigma ))_{c\in K}}$ that it induces on candidates satisfies $p(c\mid \sigma )<1$ for every $c\in K$.

Given that there are exactly $k$ probability distributions where only one candidate wins (the point masses on the elements of $K$), it is enough to prove that the set of equilibrium distributions induced by nondegenerate equilibria is finite.

Recall that $W(v)$ is the set of candidates that receive the least negative votes under the ballot profile $v$. Given some collection $C$ of ballot profiles, let us write ${W(C)=\bigcup _{v\in C}{W(v)}}$. For any strategy profile $\sigma$ we let $\fancyscript{C}(\sigma )\equiv \{v:\sigma (v)>0\}$ denote the carrier of $\sigma$. Note that $\fancyscript{C}(\sigma )$ has a product structure, i.e. $\fancyscript{C}(\sigma )=\prod _{i\in N}{\fancyscript{C}_i(\sigma _i)}$, where $\fancyscript{C}_i(\sigma _i)=\{v_i\in {\fancyscript{V}}:\sigma _i(v_i)>0\}$. Any strategy profile $\sigma$ with carrier $C$ satisfies $p(c\mid \sigma )<1$ for every $c\in K$ if and only if $\#W(C)>1$. Therefore, if $\#W(C)>1$ we say that the carrier $C$ is nondegenerate. Given a nondegenerate carrier $C$, we can construct the set of candidates that cannot win if voter $i$ abstains. Every ballot in $C_i$ that consists of a negative vote against one of such candidates is equivalent to abstention. That set of ballots, together with abstention, is denoted $\fancyscript{A}_i(C)$. In symbols, $\fancyscript{A}_i(C)\equiv C_i{\setminus } W(C_{-i})$. Note that for any $v_i$, $v'_i\in \fancyscript{A}_i(C)$ and any $v_{-i}\in C_{-i}$ we have $W(v_{-i},v_i)=W(v_{-i},v'_i)$ and, consequently, if $\fancyscript{C}(\sigma )=C$ we also have $p(c\mid \sigma _{-i},v_i)=p(c\mid \sigma _{-i},v'_i)$ for every $c\in K$.

Insofar as we aim to establish a result that holds for generic utilities, we can restrict the analysis to utility vectors where no player is indifferent between two candidates. The set of all such utility vectors is denoted $\tilde{\fancyscript{U}}$. The set $\tilde{\fancyscript{U}}$ is obtained removing a finite number of lower-dimensional hyperplanes from $\fancyscript{U}$ and its closure coincides with $\fancyscript{U}$.

Assumption 1

For every voter $i\in N$ and every pair of candidates $c$, $d\in K$ we have $u_i(c)\ne u_i(d)$.

We fix a point $u\in \tilde{\fancyscript{U}}$, a nondegenerate carrier $C$ and a Nash equilibrium $\sigma$ such that $\fancyscript{C}(\sigma )=C$. Take an arbitrary ballot profile $v^*\in C$ that satisfies $v^*_i\in \fancyscript{A}_i(C)$ whenever $\fancyscript{A}_i(C)\ne \varnothing$ (otherwise $v^*_i$ is an arbitrary element of $C_i$). For each $i\in N$, let $\hat{K}_i\equiv C_i{\setminus }\left( \fancyscript{A}_i(C)\cup \{v^*_i\}\right)$.

Since $\sigma$ is a Nash equilibrium, for each voter $i\in N$ and each pure strategy $c\in C_i$, the following equality holds:

$$\begin{aligned} \sum _{d\in K}{p(d\mid \sigma _{-i}, c)u_i(d)}=\sum _{d\in K}{p(d\mid \sigma _{-i}, v^*_i)u_i(d)}. \end{aligned}$$

Subtracting from both sides voter $i$’s expected utility if she abstains and letting ${\pi (d\mid \sigma _{-i},c)}\equiv p(d\mid \sigma _{-i},c)-p(d\mid \sigma _{-i},0)$, we can rewrite the previous equality as:

$$\begin{aligned} \sum _{d\in K}{\pi (d\mid \sigma _{-i},c)u_i(d)}=\sum _{d\in K}{\pi (d\mid \sigma _{-i}, v^*_i)u_i(d)}. \end{aligned}$$

(4.2)

For each voter $i$, let us select from (4.2) the equalities corresponding to ballots $c\in \hat{K}_i$. Rearranging those equalities, for all $c\in \hat{K}_i$, we obtain:

$$\begin{aligned}&-\sum _{d\in \hat{K}_i}{\left[ \pi (d\mid \sigma _{-i}, c)-\pi (d\mid \sigma _{-i}, v^*_i)\right] u_i(d)} \nonumber \\&\qquad \qquad =\sum _{d\notin \hat{K}_i}{\left[ \pi (d\mid \sigma _{-i}, c)-\pi (d\mid \sigma _{-i}, v^*_i)\right] u_i(d)}. \end{aligned}$$

(4.3)

Therefore, for each voter $i\in N$ we have $\hat{k}_i\equiv \#\hat{K}_i$ equalities. Suppose that we know the values assumed by $u_i$ over candidates in $K{\setminus }\hat{K}_i$. We call this vector $u_i^*$. We can interpret the $\hat{k}_i$ equalities as a system of $\hat{k}_i$ equations in $\hat{k}_i$ unknowns; the set of unknowns being the values assumed by $u$ over candidates in $\hat{K}_i$. Let us call this vector of unknowns $u^\circ _i$ so that $u_i=(u^\circ _i,u^*_i)$. We let $X^{C}_i$ denote the $\hat{k}_i\times \hat{k}_i$ matrix of coefficients of this system of equations. Hence, the $(c,d)$-th entry of $X^{C}_i$ is

$$\begin{aligned} X^{C}_i(c,d)= -\pi (d\mid \sigma _{-i}, c)+\pi (d\mid \sigma _{-i}, v^*_i). \end{aligned}$$

(4.4)

We need to show that the matrix $X_i^C$ is always invertible. To this end, we need the next Lemma. Recall that a square matrix is an M-matrix (Ostrowski 1955, p. 95) if all diagonal elements are strictly positive, all nondiagonal elements weakly negative, and all principal minors of all orders strictly positive.

Lemma 1

(Ostrowski 1955, p. 97). Let $\Pi$ be an $n\times n$ M-matrix and let $\pi = (\pi _1,\ldots ,\pi _j,\ldots ,\pi _n)$ be a weakly positive vector. The determinant of the $n\times n$ matrix $X$ whose $(i,j)$-th element is given by $X_{ ij } = \Pi _{ ij } + \pi _{j}$ is strictly positive.

Note that the matrix $\Pi ^{C}_i$ whose $(c,d)$-th element is $-\pi (d\mid \sigma _{-i}, c)$ and the vector ${(\pi (d\!\!|\sigma _{-i}, v^*_i))_{d\in \hat{K}_i}}$ decompose the matrix $X_i^C$ in the way described in Lemma 1. Therefore, our task is to prove now that $\Pi ^{C}_i$ is an M-matrix and that $(\pi (d\!\mid \sigma _{-i}, v^*_i))_{d\in \hat{K}_i}$ is a weakly positive vector. We start with the latter result.

Lemma 2

Every element of the vector $(\pi (d\mid \sigma _{-i}, v^*_i))_{d\in \hat{K}_i}$ is weakly positive.

Proof

Recall that we chose $v^*_i$ so that $v_i^*\in \fancyscript{A}_i(C)$ if $\fancyscript{A}_i(C)\ne \emptyset$. If $v_i^*\in \fancyscript{A}_i(C)$ then $\pi (d\mid \sigma _{-i},v^*_i)=0$ for every candidate $d$ because $v_i^*$ is equivalent to abstention. If $v_i^*\notin \fancyscript{A}_i(C)$ then a negative vote against candidate $v^*_i$ can never decrease the probability that some other candidate $d\ne v_i^*$ gets elected, therefore, ${\pi (d\mid \sigma _{-i},v^*_i)\ge 0}$ for every $d\in \hat{K}_i$ (note that $v^*_i\notin \hat{K}_i$ by definition). $\square$

To prove that $\Pi ^{C}_i$ is an M-matrix we proceed in several stages. The first one is to show some properties about its entries.

Lemma 3

The following assertions hold:

• (i) Every nondiagonal element of $\Pi ^{C}_i$ is weakly negative.

• (ii) Every diagonal element of $\Pi ^{C}_i$ is strictly positive.

• (iii) Every row in $\Pi ^{C}_i$ adds up to some weakly positive number.

Proof

Part (i) follows from the proof of Lemma 2.

To prove part (ii) we need to show that a negative vote against a candidate always decreases the probability that she wins the election. For any $c\in \hat{K}_i$ we have $c\notin \fancyscript{A}_i(C)$, so there exists a ballot profile $v_{-i}\in C_{-i}$ such that $c$ is the candidate that collects the least number of negative votes under $(v_{-i},0)$. If some other candidate receives as many negative votes, or just one negative vote more than $c$, then candidate $c$ wins with less probability under $(v_{-i},c)$ than under $(v_{-i},0)$. Given that every ballot profile in $C_{-i}$ receives positive probability under $\sigma _{-i}$ we obtain $p(c\mid \sigma _{-i},c)<p(c\mid \sigma _{-i},0)$. Hence, suppose that candidate $c$ receives at least two negative votes fewer than any other candidate under every ballot profile such that candidate $c$ wins with positive probability. The carrier $C$ is nondegenerate, so there must be another ballot profile $v'_i$ such that $c$ is not the only candidate that receives the least negative votes under $(v'_{-i},0)$. Since $C_{-i}$ has a product structure, we can obtain the ballot profile $v'_{-i}$ from $v_{-i}$ by changing one coordinate at a time. During this process we must find some ballot profile $v''_{-i}\in C_{-i}$ such that, under $(v''_{-i},0)$, candidate $c$ either obtains as many negative votes as some other winning candidate or wins the election outright but collecting just one negative vote fewer than another candidate. Given that every ballot profile in $C_{-i}$ receives positive probability under $\sigma _{-i}$, we conclude again that $p(c\mid \sigma _{-i},c)<p(c\mid \sigma _{-i},0)$.

Part (iii) follows because the decrease $\pi (c\mid \sigma _{-i},c)$ in the probability that candidate $c$ gets elected when player $i$ votes negatively for $c$ is necessarily equal to the increase in the probability that some candidate $d\in K{\setminus }\{c\}$ is elected. That is,

$$\begin{aligned} -\pi (c\mid \sigma _{-i},c)=\sum _{d\in K{\setminus }\{c\}}{\pi (d\mid \sigma _{-i},c)}\ge \sum _{d\in \hat{K}_i{\setminus }\{c\}}{\pi (d\mid \sigma _{-i},c)}, \end{aligned}$$

with strict inequality whenever $\pi (d'\mid \sigma _{-i},c)>0$ for some candidate $d'\notin \hat{K}_i$. $\square$

In view of Lemma 3(i)–(ii), to prove that $\Pi ^{C}_i$ is an M-matrix we now just need to show that all principal minors of all orders are strictly positive. To establish this, we use result $\mathrm {C_9}$ in Plemmons (1977) which says that it is enough to prove that all real eigenvalues of $\Pi ^{C}_i$ are strictly positive. The next Lemma is the first step in this direction.

Lemma 4

The real part of every eigenvalue of $\Pi ^{C}_i$ is weakly positive.

Proof

The Gershgorin Circle Theorem (Gershgorin 1931) tells us that every eigenvalue of a square matrix $A=(a_{cd})$ can be found in one of the closed disks $D(a_{cc},R_c)$ with center $a_{cc}$ and radius $R_c=\sum _{d\ne c}{\vert a_{cd}\vert }$. Therefore, every eigenvalue of $\Pi _i^{C}$ lies in some closed disk with center the strictly positive (by Lemma 3(ii)) real number $-\pi (c\mid \sigma _{-i},c)$ and with radius $\sum _{d\in \hat{K}_i{\setminus }\{c\}}{\pi (d\mid \sigma _{-i},c)}$. Lemma 3(iii) implies that the real part of every eigenvalue of $\Pi _i^{C}$ is weakly positive. $\square$

In order to prove that every real eigenvalue is indeed strictly positive we now show that $\Pi _i^{C}$ is nonsingular. Lemma 3(i)–(iii) show that $\Pi _i^{C}$ is a dominant diagonal matrix. Recall that a matrix $A=(a_{cd})$ is dominant diagonal if $\vert a_{cc}\vert \ge \sum _{d\ne c}{\vert a_{cd}\vert }$ for every row $c$.

Price (1951) gives the following bound on the determinant $\vert A\vert$ of a dominant diagonal matrix:

$$\begin{aligned} \prod _{c}{\left( \vert a_{cc}\vert -\sum _{d>c}\vert a_{cd}\vert \right) }\le \vert A\vert . \end{aligned}$$

(4.5)

Now we can prove:

Lemma 5

The matrix $\Pi ^{C}_i$ in nonsingular and, therefore, the matrix $X^{C}_i$ is also nonsingular.

Proof

Reorder the rows and columns of $\Pi ^{C}_i$ so that columns (rows) corresponding to voter $i$’s more preferred candidates appear before columns (rows) corresponding to voter $i$’s less preferred candidates. With this reordering of the matrix, if $-\pi (c\mid \sigma _{-i},c)=\sum _{d>c}{\pi (d\mid \sigma _{-i},c)}$ then the decrease in the probability that candidate $c$ is elected is equal to the increase in the probability that candidates worse than $c$ (according to voter $i$’s preferences) win the election. This provides a contradiction because, using Assumption 1, voter $i$’s utility from voting against $c$ would be strictly lower than under abstention, which contradicts the fact that $\sigma$ is a Nash equilibrium. Consequently, $-\pi (c\mid \sigma _{-i},c)>\sum _{d>c}{\pi (d\mid \sigma _{-i},c)}$ for every candidate $c$.

In light of Lemma 3(i)–(ii) we can apply Eq. (4.5) to $\Pi ^{C}_i$ knowing that every term on the left-hand side of (4.5) is strictly positive. Therefore, $\Pi ^{C}_i$ is nonsingular and, given that we already established that every real eigenvalue of this matrix is weakly positive, $\Pi ^{C}_i$ is also an M-matrix. We can now apply Lemma 1 to conclude that $X^{C}_i$ is nonsingular. $\square$

Therefore, if for each voter $i$ we know $u^*_i$ then we can reconstruct the entire vector of utilities $u$ using the strategy profile $\sigma$ and the system of Eq. (4.3). This allows us to construct a continuous function $(u^*,\sigma )\mapsto (u^*,u^\circ )$ from the set of Nash equilibria with carrier $C$ to the set of utility vectors $\tilde{\fancyscript{U}}$. The next step of the proof is to apply the following result to such a function. It follows from the Generic Local Triviality Theorem (Bochnak et al. 1998).

Lemma 6

(Govindan and Wilson 2001) Let $X$ and $Y$ be semialgebraic subsets of $\mathbb {R}^m$ and let $f:X\rightarrow Y$ be a continuous semi-algebraic function. If $\mathrm {dim}(X)\le \mathrm {dim}(Y)$ then, for generic $y\in Y,\, f^{-1}(y)$ is a finite or empty set.

We now have all the necessary ingredients to prove Proposition 1:

Proposition 1

For generic negative plurality games, the set of probability distributions on candidates induced by Nash equilibria is finite.

Proof

If only one candidate can win under the carrier $C$ (i.e. if $\#\left( \bigcup _{v\in C}{W(v)}\right) =1$) then the set of equilibrium distributions induced by Nash equilibria with carrier $C$ is a singleton and, therefore, necessarily finite. Hence, let the carrier $C$ be nondegenerate.

Furthermore, let us first consider those carriers $C$ such that $\fancyscript{A}_i(C)$, i.e. the set of votes equivalent to abstention, satisfies $\#\fancyscript{A}_i(C)\le 1$ for every player $i$.

Given a utility vector $u$, the set of Nash equilibria of the corresponding negative plurality game is denoted by $\mathrm {NE}_{\mathrm {np}}(u)$. The graph of the Nash equilibrium sub-correspondence that contains only Nash equilibria with carrier $C$ is

$$\begin{aligned} GNE _{\mathrm {np}}^C\equiv \ \left\{ (\sigma ,u)\in \Sigma \times \tilde{\fancyscript{U}}:\sigma \in \mathrm {NE}_{\mathrm {np}}(u)\text { and }\fancyscript{C}(\sigma )=C\right\} . \end{aligned}$$

Write $E _{\mathrm {np}}^{C}$ for the projection of $GNE _{\mathrm {np}}^{C}$ on $\Sigma$ and on those coordinates of $\tilde{\fancyscript{U}}$ that contain, for each voter $i$, her utility to candidates in $K{\setminus }\hat{K}_i$ (that is, those coordinates of $\tilde{\fancyscript{U}}$ where we can find the entries of the subvector $u^*=(u^*_1,\ldots ,u^*_n)$).

Lemma 5 implies that there is a continuous function $f_{\mathrm {np}}^C: E _{\mathrm {np}}^C\rightarrow \tilde{\fancyscript{U}}$ mapping $(u^*,\sigma )$ into $u=(u^\circ ,u^*)$. The function $f_{\mathrm {np}}^C$ is also semi-algebraic. Since $\mathrm {dim}(\tilde{\fancyscript{U}})=nk$, in view of Lemma 4, the only thing remaining to show is $\mathrm {dim}( E _{\mathrm {np}}^C)\le nk$.

For each voter $i$, the dimension of her subset of strategies that have carrier $C_i$ is $\mathrm {dim}\left( \Delta (C_i)\right) =\#C_i-1=\hat{k}_i$ (recall that we assumed that $\#\fancyscript{A}_i(C)\le 1$ for every player $i$). Therefore, the dimension of $GNE _{\mathrm {np}}^C$ is at most $\sum _{i\in N}{\hat{k}_i}+nk$. Consequently,

$$\begin{aligned} \mathrm {dim}\left( E_{\mathrm {np}}^C\right) \le \sum _{i\in N}{\hat{k}_i}+nk-\sum _{i\in N}{\hat{k}_i}=nk. \end{aligned}$$

Applying Lemma 6 to the function $f_{\mathrm {np}}^C: E _{\mathrm {np}}^C\rightarrow \tilde{\fancyscript{U}}$ shows that for generic games $u\in \tilde{\fancyscript{U}}$ the set of Nash equilibria with carrier $C$ is finite.

We now consider those carriers $C$ such that $\#\fancyscript{A}_i(C)> 1$ for some player $i$. As argued before, Example 5 shows that there is an open set of utilities for which the negative plurality game has a continuum of Nash equilibria so, clearly, we cannot hope to prove that for generic games $u\in \tilde{\fancyscript{U}}$ the set of Nash equilibria with carrier $C$ is finite. However, in that example, different Nash equilibria contained in the same continuum only differ on how the strategy of each player $i$ assigns probabilities over the different elements of $\fancyscript{A}_i(C)$. Moreover, different such assignments do not affect the resulting probability distribution on candidates because elements of $\fancyscript{A}_i(C)$ are all equivalent to abstention.

Recall that, for each player $i$, we defined $\hat{K}_i\equiv C_i{\setminus }(\fancyscript{A}_i(C)\cup \{v^*_i\})$. (Also recall that our choice of $v^*_i$ was such that $v^*_i\in \fancyscript{A}_i(C)$ whenever such a set was nonempty.) With abuse of notation, let $\Sigma ^{\hat{K}}\equiv \prod _{i=1}^n{\mathbb {R}^{\hat{K}_i}}$. We define as $GNEO _{\mathrm {np}}^C$ the projection of $GNE _{\mathrm {np}}^{C}$ onto $\Sigma ^{\hat{K}}\times \tilde{\fancyscript{U}}$. That is, in this projection, we only eliminate those components of the strategy profile that we do not need to compute the equilibrium outcome $p(\sigma )$.¹⁰ Note that $\mathrm {dim}\left( GNEO _{\mathrm {np}}^C\right) \le \sum _{i\in N}{\hat{k}_i}+nk$.

We write $EO _{\mathrm {np}}^{C}$ for the projection of $GNEO _{\mathrm {np}}^{C}$ on $\Sigma$ and on those coordinates of $\tilde{\fancyscript{U}}$ where we can find the entries of the subvector $u^*=(u^*_1,\ldots ,u^*_n)$. For each player $i$, the values that her mixed strategy assumes on $\fancyscript{A}_i(C)$ are neither needed to construct the system (4.3) nor used to show that it always has a solution. Hence, Lemma 5 implies that there is a continuous function $g_{\mathrm {np}}^C: EO _{\mathrm {np}}^C\rightarrow \tilde{\fancyscript{U}}$ mapping $(u^*,\sigma )$ into $u=(u^\circ ,u^*)$. The proof now follows the same lines as before. The only difference is that when applying Lemma 6 to the function $g_{\mathrm {np}}^C: EO _{\mathrm {np}}^C\rightarrow \tilde{\fancyscript{U}}$ we conclude that for generic games $u\in \tilde{\fancyscript{U}}$ the set $(g_{\mathrm {np}}^C)^{-1}(u)$ is finite so that $\Big \{p(\sigma ):\sigma \in \mathrm {NE}_{\mathrm {np}}(u)\text { and }\fancyscript{C}(\sigma )=C\Big \}$ is also finite. $\square$

Appendix 2: Generic determinacy of Nash equilibrium in approval games with three candidates

In this section we fix the set of candidates $K\equiv \{a,b,c\}$, Hence, we write each player’s set of pure strategies as ${\fancyscript{V}}_a\equiv \{ abc , ab , ac , bc , a , b , c ,0\}$, where each ballot represents the set of candidates approved by the voter.

As in the case of negative plurality, we point out the main complication that we face when trying to prove that for generic approval voting games the set of equilibrium outcomes is finite. The next example is a generic approval game with three candidates and a continuum of Nash equilibria.

Example 6

Consider an approval game with set of candidates $K=\{a,b,c\}$ and five voters with preferences:

$$\begin{aligned} u_1&= u_2 =(4,2,1),\\ u_3&= (1,4,3),\\ u_4&= u_5\,=\,(2,1,3) \end{aligned}$$

Consider the following continuum of Nash equilibria of this game:

$$\begin{aligned} \left\{ \frac{3}{5}a+\frac{2}{5}\gamma ab +\frac{2}{5}(1-\gamma ) b , ab ,\frac{2}{5} bc +\frac{3}{5}b,c,c):0\le \gamma \le 1\right\} . \end{aligned}$$

At every point in this continuum, voter 1 faces with probability $\frac{2}{5}$ a ballot profile where candidates $a$, $b$, and $c$ receive, respectively, 1, 2, and 3 approval votes. With the remaining probability $\frac{3}{5}$, voter 1 faces a ballot profile where candidates $a,\, b$, and $c$ receive, respectively, 1, 2, and 2 approval votes. Thus, voter 1 can make candidate $a$ win with positive probability by approving that candidate as long as she does not approve candidate $b$ as well. In more general terms, we can think of ballots $b$ and $ab$ as being equivalent from the viewpoint of voter 1 in the sense that, given the opponents’ behavior, they generate the same probability distribution over winning candidates.

Note that every Nash equilibrium in the continuum induces the same probability distribution $(p_a,p_b,p_c)=(\frac{3}{25},\frac{11}{25},\frac{11}{25})$. Voters 2, 4 and 5 are playing a strict best reply while voters 1 and 3 equilibrate each other’s expected utilities by playing mixed strategies. Hence, every game in a neighbourhood of this one has a continuum of Nash equilibria with the same characteristics.

Besides $K\equiv \{a,b,c\}$ and the set of possible ballots ${\fancyscript{V}}_a$, we also fix the set of voters $N$, and a utility vector $u$ that satisfies Assumption 1. Take a Nash equilibrium $\sigma$ such that every candidate wins with positive probability, i.e. letting $C\equiv \fancyscript{C}(\sigma )$, we have $W(C)=K$.

Given the carrier $C$, we partition $C_i$ into equivalence classes. Two ballots $v_i,\, v'_i\in C_i$ are equivalent if $W(v_{-i},v_i)=W(v_{-i},v'_i)$ for every $v_{-i}\in C_{-i}$. For instance, in Example 6, strategies $b$ and $ab$ are equivalent for voter 1 because $W(v_{-1},b)=W(v_{-1}, ab )$ for every $v_{-1}\in \{ ab \}\times \{b, bc \}\times \{c\}\times \{c\}$.

For the time being we focus on a voter $i$ with preferences $a\succ _i b\succ _i c$. Furthermore, with slight abuse of notation, we let $a$ denote the ballot that approves candidate $a$, $b$ denote the ballot that approves candidate $b$, and $ab$ denote the ballot that approves both candidate $a$ and candidate $b$.

Lemma 7

Let $C$ be the carrier of a Nash equilibrium $\sigma$ of an approval game with set of candidates $K\equiv \{a,b,c\}$. Suppose $W(C)=K$ and let player $i$ have preferences $a\succ _i b\succ _i c$. Then,

• (i) abstention is not in $C_i$,

• (ii) no ballot in $C_i$ approves candidate $c$, and

• (iii) if $b\in C_i$ and $ab \in C_i$ then $ab$ and $b$ belong to the same equivalence class in $C_i$.

Proof

We start with part (i). Looking for a contradiction let $v=(0,v_{-i})\in C$ be such that $a\in W(v)$. If, under $v$, candidate $a$ is tied in the first place with other candidate then we are done because $a$ would be a better response to $\sigma$ than abstention. If $a$ wins unopposed under $v$ then substitute, for one voter at a time, the ballot where they approve $a$ to some other ballot in their carrier where they do not. Given $W(C)=K$ and the fact that $C$ has a product structure, proceeding in this way we must find a ballot profile $v'=(0,v'_{-i})$ such that, either $a$ is tied in the first place with some other candidate or candidate $a$ receives just one vote fewer than the winning candidates. In either case, this shows that approving candidate $a$ is a better response to $\sigma$ than abstention, thus, $v\notin C$.

To prove (ii) by contradiction let $v_i\in C_i$ be a ballot that approves candidate $c$ and let $v'_i$ be the ballot that we obtain from $v_i$ by removing the approval vote for $c$. Ballot $v'_i$ is also best response against $\sigma _{-i}$ and it is not a better response than $v_i$ only if the additional approval vote for $c$ in $v_i$ does not increase the probability that $c$ wins. That means that for every $v_{-i}$ such that $c\in W(v_{-i},v'_i)$, under $(v_{-i},v'_i)$, candidate $c$ must receive at least one approval vote more than candidates $a$ and $b$. At least one such a $v_{-i}$ exists because $c\in W(C)$. In turn, for every $v'_{-i}\in C_{-i}$ such that $c\notin W(v'_{-i},v'_i)$ candidate $c$ must receive at least two approval votes fewer than $a$ and $b$. There is at least one such a $v'_{-i}$ because $a,\, b\in W(C)$. Since $C_{-i}$ has a product structure, by changing the ballot of one voter at a time we must find a $v''_{-i}$ such that, under $(v''_{-i},v'_i)$ candidate $c$ is either tied at the first place or just one approval vote behind the winning candidates. Such a $v''_{-i}$ receives positive probability under $\sigma _{-i}$ and, therefore, $v'_i$ is a better response than $v_i$ against $\sigma _{-i}$. We conclude that no ballot in $C_i$ approves candidate $c$.

Let us move to (iii). If $b\in C_i$ then $ab$ is also a best response against $\sigma _{-i}$ because $C$ is the carrier of a Nash equilibrium and voter $i$’s utility can never decrease when her most preferred candidate receives an additional approval vote. Furthermore, $ab$ is not a better response only if that additional approval vote does not increase the probability that $a$ wins. That is, if $W(v_{-i},b)=W(v_{-i}, ab )$ for every $v_{-i}\in C_{-i}$. $\square$

A consequence of this Lemma is that $C_i$ is partitioned into at most two equivalence classes. When this is indeed the case and voter $i$ has preferences $a\succ _i b\succ _i c$, for any Nash equilibrium $\sigma$ with carrier $C$, we can write

$$\begin{aligned} \sum _{d\in K}{p(d\mid \sigma _{-i},a)u_i(d)}=\sum _{d\in K}{p(d\mid \sigma _{-i}, ab )u_i(d)}. \end{aligned}$$

(We show below that $a$ and $ab$ cannot belong to the same equivalence class, so we are in fact taking $a$ and $ab$ to be the representatives of the two equivalence classes in $C_i$.) Rearranging,

$$\begin{aligned}&\left[ p(b\mid \sigma _{-i}, ab )-p(b\mid \sigma _{-i},a)\right] u_i(b)=\left[ p(a\mid \sigma _{-i},a)-p(a\mid \sigma _{-i}, ab )\right] u_i(a) \\&\qquad \qquad +\left[ p(c\mid \sigma _{-i},a)-p(c\mid \sigma _{-i}, ab )\right] u_i(c). \end{aligned}$$

We want to show that if $u_i(a)$ and $u_i(c)$ are known then we can use the equilibrium strategy to find out $u_i(b)$. That is true as long as $p(b\mid \sigma _{-i}, ab )\ne p(b\mid \sigma _{-i},a)$. The next Lemma establishes that this is indeed the case.

Lemma 8

Let $C$ be the carrier of a Nash equilibrium of an approval game with set of candidates $K\equiv \{a,b,c\}$. Suppose $W(C)=K$ and take a player $i$ with preferences $a\succ _i b\succ _i c$. If $ab ,\, a\in C_i$ then $ab$ and $a$ do not belong to the same equivalence class.

Proof

The proof follows the same lines as the proof of Lemma 7(ii). Assume that $W(v_{-i},a)=W(v_{-i}, ab )$ for every $v_{-i}\in C_{-i}$. This implies that for every $v_{-i}\in C_{-i}$ such that $b\in W(v_{-i},a)$, under $(v_{-i},a)$, candidate $b$ receives at least one approval vote more than $a$ and $c$. There is at least one such a $v_{-i}$ because $b\in W(C)$. In turn, in every $v'_{-i}\in C_{-i}$ such that $b\notin W(v'_{-i},a)$, under $(v'_{-i},a)$, candidate $b$ receives at least two approval votes fewer than the winner. There is at least one such a $v'_{-i}$ because $a$, $c\in W(C)$. Since $C_{-i}$ has a product structure, by changing the ballot of one voter at a time we must find a $v''_{-i}$ such that, under $(v''_{-i},a)$, candidate $b$ is either tied at the first place or just one approval vote behind the winning candidates. For such $v''_{-i}$ we have $W(v''_{-i},a)\ne W(v''_{-i}, ab )$. $\square$

Thus, in more general terms, if we know the utility derived by each voter from her top- and bottom-ranked candidates then there is a semi-algebraic continuous function that, knowing $\sigma$, gives us the whole vector of utilities.¹¹ Paralleling the proof of Proposition 1, in the proof of the next proposition we apply Lemma 6 to such a function.

Lemma 9

For generic approval voting games with three candidates, the set of Nash equilibrium outcomes is finite.

Proof

Clearly, there are three probability distributions such that just one candidate wins with positive probability. If only two candidates win with positive probability then the strategic interaction reduces to the one in a plurality voting game. In such a case, a similar argument to the one applied in De Sinopoli (2001) proves that for generic utilities the set of Nash equilibria where two candidates win with positive probability is finite.

Thus, take a nondegenerate carrier $C$ such that all three candidates win with positive probability. Given a utility vector $u$, let $\mathrm {NE}_{\mathrm {a}}(u)$ be the set of Nash equilibria of the corresponding approval voting game. The graph of the Nash equilibrium sub-correspondence that contains only Nash equilibria with carrier $C$ is

$$\begin{aligned} GNE _{\mathrm {a}}^{C}\equiv&\ \left\{ (\sigma ,u)\in \Sigma \times \tilde{\fancyscript{U}}:\sigma \in \mathrm {NE}_{\mathrm {a}}(u)\text { and }\fancyscript{C}(\sigma )=C\right\} . \end{aligned}$$

We have $\dim ( GNE _{\mathrm {a}}^C)\le \sum _{i=1}^n{(\#C_i-1)}+nk$. Let $\hat{N}$ be the set of voters $i$ with two equivalence classes in $C_i$. We decompose $u=(u^*,u^\circ )$ so that ${u}^{\circ }$ contains the utility to each voter $i\in \hat{N}$ if her second-ranked candidate wins the election. Write $\tilde{\fancyscript{U}}^*$ for the projection of $\tilde{\fancyscript{U}}$ on the corresponding coordinates so that ${u}^*\in \tilde{\fancyscript{U}}^*$. Letting $\hat{n}=\#\hat{N}$, we obtain $\dim (\tilde{\fancyscript{U}}^*)=nk-\hat{n}$. Furthermore, let $EO _{\mathrm {a}}^C$ be the projection of $GNE _{\mathrm {a}}^C$ on $\tilde{\fancyscript{U}}^*$ and on those coordinates of the strategy space that capture the probability with which each voter only approves her corresponding top-ranked candidate. That is, $EO _{\mathrm {a}}^C$ contains just the part of the strategy profile that we actually need to compute the set of Nash equilibrium outcomes for a given Nash equilibrium with carrier $C$.¹²

We argued above that there is a semi-algebraic continuous function $f_{\mathrm {a}}^C: EO _{\mathrm {a}}^C\rightarrow \tilde{\fancyscript{U}}$ mapping $(u^*,\sigma )$ into $u=(u^\circ ,u^*)$. We have $\mathrm {dim}(\tilde{\fancyscript{U}})=nk$. Hence, in order to apply Lemma 6, we now prove that $\mathrm {dim}( EO _{\mathrm {a}}^C)\le nk$. However, for every voter $i$ such that $C_i$ has two equivalence classes the set of possible probabilities that she can attach to the ballot that just approves her top-preferred candidate is one dimensional. Thus, we obtain:

$$\begin{aligned} \mathrm {dim}\left( EO _{\mathrm {a}}^C\right) \le \hat{n}+\dim (\tilde{\fancyscript{U}}^*)=\hat{n}+nk-\hat{n}=nk. \end{aligned}$$

Applying Lemma 6 to the function $f_{\mathrm {a}}^C: EO _{\mathrm {a}}^C\rightarrow \tilde{\fancyscript{U}}$ shows that for generic games $u\in \tilde{\fancyscript{U}}$ the set of outcomes induced by Nash equilibria with carrier $C$ is finite. Since there are only finitely many carriers, the desired result follows. $\square$

Remark 1

Extending the result to any number of candidates is challenging. Vaguely speaking, to apply Lemma 6, for each player, we need to recover as many utility values as the dimensionality of the set of probability distributions that that player can induce by changing her strategy. However, note that if the number of candidates is $x$ then the number of strategies is $2^x$. Hence, it seems that we need a better understanding about how the set of best replies looks like in an approval voting game. Note, for instance, that Nash equilibrium strategies are not necessarily sincere (De Sinopoli et al. 2006), that is, if a voter approves a candidate $c$ she does not necessarily also approve every candidate that she prefers to $c$.

We now finish the proof of Proposition 2.

Proposition 2

For generic approval voting games, the set of probability distributions on three or fewer candidates induced by Nash equilibria is finite.

Proof

Take an arbitrary set of candidates $K$ and a Nash equilibrium $\sigma$ that induces a probability distribution that gives positive probability to exactly three candidates, say, $c_1$, $c_2$ and $c_3$. Construct a three-candidate approval game by choosing those three candidates. Interpreting ballots under approval as subsets of candidates, we define the strategy profile $\sigma '$ of the three-candidate game by $\sigma '_i(v'_i)=\sum _{\{v_i:v'_i\subset v_i\}}{\sigma _i(v_i)}$ for every $i\in N$ and every $v'_i\subset \{c_1,c_2,c_3\}$.¹³ It is not difficult to see that $\sigma '$ is a Nash equilibrium of the three-candidate approval game. A similar thing can be done if two candidates win with positive probabilities. Finally, we note that the set of degenerate distributions on candidates is necessarily finite. $\square$