Mixed equilibriums in a three-candidate spatial model with candidate valence

Abstract

We study a spatial model of electoral competition among three office-motivated candidates of unequal valence (one advantaged and two equally disadvantaged candidates) under majority rule assuming that candidates are uncertain about the voters’ policy preferences and that the policy space consists of three alternatives (one at each extreme of the linear policy spectrum and one in the center) and we characterize mixed strategy Nash equilibriums of the game. Counterintuitively, we show that (a) when uncertainty about voters’ preferences is high, the advantaged candidate might choose in equilibrium a more extremist strategy than the disadvantaged candidates and that (b) when uncertainty about voters’ preferences is low, there exist equilibriums in which one of the disadvantaged candidates has a larger probability of election than the disadvantaged candidate of the equivalent two-candidate (one advantaged and one disadvantaged candidate) case.

Appendix: Proofs

Proof (Proposition 2)

We will prove the proposition by contradiction. Assume that $\hat{\sigma} _{1}^{D_{1}}=\hat{\sigma}_{1}^{D_{2}}>\hat{\sigma}_{3}^{D_{1}}=\hat{\sigma} _{3}^{D_{2}}$. Then, $\varPi_{A}(x_{3},\hat{\sigma}^{D_{1}},\hat{\sigma} ^{D_{2}})=r+(1-2r)(1-\hat{\sigma}_{1}^{D_{1}}-\hat{\sigma} _{3}^{D_{1}})^{2}+r(\hat{\sigma}_{3}^{D_{1}})^{2}$ and $\varPi_{A}(x_{1},\hat{ \sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=r+(1-2r)(1-\hat{\sigma}_{1}^{D_{1}}- \hat{\sigma}_{3}^{D_{1}})^{2}+r(\hat{\sigma}_{1}^{D_{1}})^{2}$. It is obvious that $\varPi_{A}(x_{3},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})<\varPi_{A}(x_{1},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$ and therefore that $\hat{\sigma}_{3}^{A}$ must be equal to zero. Notice that $\hat{\sigma} _{1}^{A}$ must be positive, because from the arguments in support of Proposition 1 it is made clear that no equilibrium exists in which A uses a pure strategy. Then we have that $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{1},\hat{ \sigma}^{D_{2}})=r(1-\hat{\sigma}_{1}^{A})(1-\hat{\sigma}_{1}^{D_{2}}+\frac{ \hat{\sigma}_{1}^{D_{2}}}{2})=r(1-\hat{\sigma}_{1}^{A})(1-\frac{\hat{\sigma} _{1}^{D_{2}}}{2})$ and $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{3},\hat{\sigma} ^{D_{2}})=r(1-\hat{\sigma}_{3}^{D_{2}}+\frac{\hat{\sigma}_{3}^{D_{2}}}{2} )=r(1-\frac{\hat{\sigma}_{3}^{D_{2}}}{2})$, which implies that $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{1},\hat{\sigma}^{D_{2}})<\varPi_{D_{1}}(\hat{\sigma}^{A},x_{3},\hat{\sigma}^{D_{2}})$; a disadvantaged player cannot place a positive probability on x ₁. The equivalent occurs if we assume that $\hat{\sigma}_{1}^{D_{1}}=\hat{\sigma}_{1}^{D_{2}}<\hat{\sigma}_{3}^{D_{1}}=\hat{\sigma}_{3}^{D_{2}}$. Therefore, in a type-symmetric equilibrium the disadvantaged candidates must mix symmetrically about the center of the policy space. To conclude the proof we must investigate the case in which the advantaged player mixes asymmetrically about the center of the policy space ($\hat{\sigma}_{1}^{A}\neq \hat{\sigma}_{3}^{A}$). Since we have proved that in a type-symmetric equilibrium we must have $\hat{\sigma}_{1}^{D_{1}}=\hat{\sigma}_{1}^{D_{2}}=\hat{\sigma}_{3}^{D_{1}}=\hat{\sigma}_{3}^{D_{2}}$, then it follows that if $\hat{\sigma}_{1}^{A}>\hat{\sigma}_{3}^{A}$ (the equivalent occurs if we assume $\hat{\sigma}_{1}^{A}<\hat{\sigma}_{3}^{A}$) we have $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{1},\hat{\sigma}^{D_{2}})=r(1-\hat{\sigma}_{1}^{A})(1-\frac{\hat{\sigma}_{1}^{D_{2}}}{2})<\varPi_{D_{1}}(\hat{\sigma}^{A},x_{3},\hat{\sigma}^{D_{2}})=r(1-\hat{\sigma}_{3}^{A})(1-\frac{\hat{\sigma}_{3}^{D_{2}}}{2})$ and, therefore, a disadvantaged player cannot place positive probability on x ₁. This contradicts the fact that in a type-symmetric equilibrium we must have $\hat{\sigma}_{1}^{D_{1}}=\hat{\sigma}_{1}^{D_{2}}=\hat{\sigma}_{3}^{D_{1}}=\hat{\sigma}_{3}^{D_{2}}$ and therefore it should also be the case that in a type-symmetric equilibrium $\hat{\sigma}_{1}^{A}$ must be equal to $\hat{\sigma}_{3}^{A}$. □

Proof (Proposition 3)

By Proposition 2 we have that $\hat{\sigma}_{1}^{A}=\hat{\sigma}_{3}^{A}$ and $\hat{\sigma}_{1}^{D_{1}}=\hat{\sigma}_{1}^{D_{2}}=\hat{\sigma}_{3}^{D_{1}}=\hat{\sigma}_{3}^{D_{2}}$ in every type-symmetric equilibrium. Therefore, since we are looking for an equilibrium in which candidates mix symmetrically about the center of the policy space we can impose: $\hat{\sigma}_{1}^{D_{1}}=\hat{\sigma}_{3}^{D_{1}}=\hat{\sigma}_{1}^{D_{2}}=\hat{\sigma}_{3}^{D_{2}}=\hat{s}$. We need to check for the values of $\hat{\sigma }_{1}^{A}$ and $\hat{s}$ at which we get $\varPi_{A}(x_{1},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=\varPi_{A}(x_{2},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$ and $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{1},\hat{\sigma}^{D_{2}})=\varPi_{D_{1}}(\hat{\sigma}^{A},x_{2},\hat{\sigma}^{D_{2}})$ (due to the symmetry of the problem if these equalities hold then $\varPi_{A}(x_{3},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=\varPi_{A}(x_{2},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$, $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{3},\hat{\sigma}^{D_{2}})= \varPi_{D_{1}}(\hat{\sigma}^{A},x_{2},\hat{\sigma}^{D_{2}})$, $\varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{1})= \varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{2})$ and $\varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{3})=\varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{2})$ should hold as well). We know that $\varPi_{A}(x_{1},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=r+(1-2r)(4\hat{s}^{2})+r\hat{s}^{2}$ and that $\varPi_{A}(x_{2},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=(1-2r)+2r(1-\hat{s})^{2}$ and therefore $\varPi_{A}(x_{1},\hat{\sigma }^{D_{1}},\hat{\sigma}^{D_{2}})=\varPi_{A}(x_{2},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$ holds if and only if

Equivalently, $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{1},\hat{\sigma} ^{D_{2}})=r((1-\hat{\sigma}_{1}^{A})(1-\hat{s}+\frac{\hat{s}}{2}))$ and $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{1},\hat{\sigma}^{D_{2}})=(1-2r)(2\hat{\sigma} _{1}^{A}(2\hat{s}+\frac{1-2\hat{s}}{2}))+2r(\hat{\sigma}_{1}^{A}(\hat{s}+ \frac{1-2\hat{s}}{2}))$, and therefore $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{1}, \hat{\sigma}^{D_{2}})=\varPi_{D_{1}}(\hat{\sigma}^{A},x_{2}, \hat{\sigma}^{D_{2}})$ holds if and only if

$$\hat{\sigma}_{1}^{A}= \frac{r\hat{s}-2r}{9r\hat{s}-4\hat{s}-2}=\frac{rf(r)-2r}{9rf(r)-4f(r)-2}. $$

Therefore, the presented equilibrium is the unique type-symmetric equilibrium of our game. □

Proof (Proposition 4)

We will provide a proof for (i). It is effortless to see that the proofs for (ii), (iii) and (iv) are equivalent. We have to demonstrate that (a) $\varPi_{A}(x_{1},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})\leq \varPi_{A}(x_{2}, \hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=\varPi_{A}(x_{3},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$, that (b) $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{1},\hat{\sigma}^{D_{2}})\leq \varPi_{D_{1}}(\hat{\sigma}^{A},x_{2},\hat{\sigma}^{D_{2}})=\varPi_{D_{1}}(\hat{\sigma}^{A},x_{3},\hat{\sigma}^{D_{2}})$, that (c) $\varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{1})\geq \max \{\varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{2}), \varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{3})\}$ and that (d) (a), (b) and (c) all hold simultaneously only if $r\in (0,\frac{1}{3}]$. Starting with (a) we compute $\varPi_{A}(x_{1},\hat{ \sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=r+(1-2r)\frac{1-2r}{1-r}$, $\varPi_{A}(x_{2},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=(1-2r)+r\frac{r}{1-r}$ and $\varPi_{A}(x_{3},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=r+(1-2r)\frac{1-2r}{1-r}$ and we observe that $\varPi_{A}(x_{1},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=\varPi_{A}(x_{2},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})=\varPi_{A}(x_{3},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$ holds for any r. That is, $\hat{\sigma}^{A}$ is a best response to $\{\hat{\sigma }^{D_{1}},\hat{\sigma}^{D_{2}}\}$. We proceed to point (b). We compute $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{1},\hat{\sigma}^{D_{2}})=\frac{r}{2}$, $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{2},\hat{\sigma}^{D_{2}}) =(1-2r)\frac{r}{1-r}$ and $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{3},\hat{\sigma}^{D_{2}})=r\frac{1-2r}{1-r}$ and we observe that $\varPi_{D_{1}}(\hat{\sigma}^{A},x_{1},\hat{\sigma}^{D_{2}})\leq \varPi_{D_{1}}(\hat{\sigma}^{A},x_{2},\hat{\sigma}^{D_{2}})=\varPi_{D_{1}}(\hat{\sigma}^{A},x_{3},\hat{\sigma}^{D_{2}})$ holds for any $r\in (0,\frac{1}{3}]$; that is, $\hat{\sigma}^{D_{1}}$ is a best response to $\{\hat{\sigma}^{A},\hat{\sigma}^{D_{2}}\}$ for any $r\in (0,\frac{1}{3}]$ (this proves point (d)). Finally we go to point (c) and we compute $\varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{1})=r$, $\varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{2})=\frac{r}{1-r}(1-r)(\frac{1-2r}{1-r}+\frac{1}{2}\frac{r}{1-r})$ and $\varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{3})=\frac{1-2r}{1-r}r(\frac{r}{1-r}+\frac{1}{2}\frac{1-2r}{1-r})$ and we observe that $\varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{1})\geq \max \{\varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{2}), \varPi_{D_{2}}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},x_{3})\}$ holds for any r; that is, $\hat{\sigma}^{D_{2}}$ is a best response to $\{\hat{\sigma}^{A},\hat{\sigma}^{D_{1}}\}$ for any r. □

Proof (Proposition 5)

We first prove part (a) of this proposition. Consider a type-symmetric equilibrium $\{\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}}\}$ (that is, $\hat{\sigma}^{D_{1}}=\hat{\sigma}^{D_{2}}=\hat{\sigma}$). Now let us create a parallel game in which instead of A we have a player B (we also rank her/him in first place) who is identical to the other two disadvantaged candidates. If all players had the same valence and they were using the same mixed strategy $\hat{\sigma}$ then they would obviously enjoy the same election probability ($\frac{1}{3}$). Observe that $\varPi_{A}(\hat{\sigma},\hat{\sigma},\hat{\sigma})>\varPi_{B}(\hat{\sigma},\hat{\sigma}, \hat{\sigma})=\frac{1}{3}$, as for any possible platform realization in our game, we have that A should receive a payoff at least as large as B in the parallel game, and for some platform realizations (for the case in which all candidates end up offering the same policy to the voters, for example) A should receive a payoff strictly larger than the payoff of B in the parallel game. That is, $\varPi_{A}(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})\geq \varPi_{A}(\hat{\sigma},\hat{\sigma},\hat{\sigma})>\frac{1}{3}$.

As far as part (b) is concerned notice that for $[\frac{1}{3}+\frac{1}{n-1}, \frac{2}{3}-\frac{1}{n-1}]$ to be well defined and non-degenerate we must have $\frac{1}{3}+\frac{1}{n-1}<\frac{2}{3}-\frac{1}{n-1}\Longrightarrow n>7$. Assume that for some equilibrium we have $[\frac{1}{3}+\frac{1}{n-1}, \frac{2}{3}-\frac{1}{n-1}]\varsubsetneq S(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$. Since both sets are convex, this implies that either $\frac{1}{3}+\frac{1}{n-1}<z$ for any $z\in S(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$ or $\frac{2}{3}-\frac{1}{n-1}>z$ for any $z\in S(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$, or both. Without loss of generality assume that $\frac{1}{3}+\frac{1}{n-1}<z$ for any $z\in S(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$ holds. This implies that the policy x _k such that $x_{k}\notin S(\hat{ \sigma}^{A},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$ and $x_{k+1}\in S( \hat{\sigma}^{A},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$ satisfies $x_{k}>\frac{1}{3}$. In any equilibrium there exists at least one candidate who receives a payoff that does not exceed $\frac{1}{3}$. If this candidate deviates to x _k and since the probability that some other candidate announces a policy identical or to the left of x _k is zero then she/he will receive a payoff larger than $\frac{1}{3}$. Therefore, in any equilibrium of the game $[\frac{1}{3}+\frac{1}{n-1},\frac{2}{3}-\frac{1}{n-1}] \subseteq S(\hat{\sigma}^{A},\hat{\sigma}^{D_{1}},\hat{\sigma}^{D_{2}})$. □