Individual preferences and democratic processes: two theorems with implications for electoral politics

Abstract

The paper provides a complete characterization of Nash equilibria for games in which n candidates choose a strategy in the form of a platform, each from a circle of feasible platforms, with the aim of maximizing the stretch of the circle from where the candidate’s platform will receive support from the voters. Using this characterization, it is shown that if the sum of all players’ payoffs is 1,  the Nash equilibrium payoff of each player in an arbitrary Nash equilibrium must be restricted to the interval $[1/2(n-1),2/(n+1)].$ This implies that in an election with four candidates, a candidate who is attracting less than one-sixth of the voters can do better by changing his or her strategy.

Appendix

1.1 Proof of claim (5) in Sect. 3

Since $k\ne j,$ the definition of $s^{\prime }$ implies that $s_{k}^{\prime }=s_{k}\in Y_{k}(s).$ If there is some $s_{r}^{\prime }$ in $Y_{k}(s)$ with $s_{r}^{\prime }\ne s_{k}^{\prime },$ then $r\ne k,j,$ and so by definition of $s^{\prime },$$s_{r}^{\prime }=s_{r}$ must hold. In this case, we have $s_{r}$ in $Y_{k}(s)$ with $s_{r}\ne s_{k},$ a contradiction to the definition of $Y_{k}(s).$ This establishes that there is no $s_{r}^{\prime }$ in $Y_{k}(s)$ with $s_{r}^{\prime }\ne s_{k}^{\prime }.$ Since $Y_{k}(s)$ is a connected set containing $s_{k}^{\prime },$ and there is no $s_{r}^{\prime }$ in $Y_{k}(s)$ with $s_{r}^{\prime }\ne s_{k}^{\prime },$ we must have $Y_{k}(s)\subset Y_{k}(s^{\prime }),$ and consequently ${\bar{Y}} _{k}(s)\subset {\bar{Y}}_{k}(s^{\prime })$ as well.

Using (NBD), let us write $Y_{k}(s)=arc(s_{L(k)},s_{R(k)})\equiv arc(s_{p},s_{q}).$ Note that $s_{L(k)}\ne s_{k},$ and $s_{R(k)}\ne s_{k},$ and $s_{L(k)}\ne s_{R(k)}.$ Now, if $s_{L(k)}=s_{j},$ then since $s_{L(k)}\in {\bar{Y}}_{k}(s)$ and $s_{j}\in Y_{j}(s),$ we would have ${\bar{Y}} _{k}(s)\cap Y_{j}(s)\ne \phi ,$ a contradiction to the definition of case (b)(I). Thus, we must have $s_{L(k)}\ne s_{j},$ and similarly $s_{R(k)}\ne s_{j}.$ By definition of $s^{\prime }$ it follows that $s_{p}^{\prime }=s_{p}$ and $s_{q}^{\prime }=s_{q}.$

Both $s_{p}$ and $s_{q}$ belong to ${\bar{Y}}_{k}(s),$ and since ${\bar{Y}} _{k}(s)\subset {\bar{Y}}_{k}(s^{\prime }),$ they belong to ${\bar{Y}} _{k}(s^{\prime }).$ However, they cannot be interior points of ${\bar{Y}} _{k}(s^{\prime });$ that is, they cannot belong to $Y_{k}(s^{\prime }).$ To see this, note that $s_{p}\ne s_{k},$$s_{p}=s_{p}^{\prime }$ and $s_{k}=s_{k}^{\prime },$ and so $s_{p}^{\prime }\ne s_{k}^{\prime }.$ Then, by definition of $Y_{k}(s^{\prime }),$ we infer that $s_{p}^{\prime }$ cannot be in $Y_{k}(s^{\prime });$ that is, $s_{p}$ cannot be in $Y_{k}(s^{\prime }).$ A similar reasoning establishes that $s_{q}$ cannot be an interior point of ${\bar{Y}}_{k}(s^{\prime }).$ Thus, both are boundary points of ${\bar{Y}}_{k}(s^{\prime }).$ This means that ${\bar{Y}}_{k}(s^{\prime })$ is either the set $arc[s_{p},s_{q}]$ or the set $arc[s_{q},s_{p}].$ Since $Y_{k}(s)=arc(s_{p},s_{q}),$ and ${\bar{Y}}_{k}(s^{\prime })$ has to contain the set ${\bar{Y}}_{k}(s)=arc[s_{p},s_{q}],$ we conclude that ${\bar{Y}} _{k}(s^{\prime })$ must be the set $arc[s_{p},s_{q}].$ Thus, $Y_{k}(s^{\prime })$ must be the set $arc(s_{p},s_{q}),$ and so $Y_{k}(s)=Y_{k}(s^{\prime }).$

1.2 Proof of claim (8) in Sect. 3

We break up our analysis into two cases: (A1) $s_{L(j)}=s_{R(k)}$ and (A2) $s_{L(j)}\ne s_{R(k)}.$

Case (A1): In this case, $s_{j}$ belongs to $arc(s_{R(k)},s_{L(k)}].$ If $s_{j}$ actually belongs to $arc(s_{R(k)},s_{L(k)}),$ then $Y_{j}(s)=arc(s_{R(k)},s_{L(k)}),$ which is disjoint from ${\bar{Y}}_{k}(s),$ contradicting the fact that $Y_{j}(s)\cap {\bar{Y}}_{k}(s)\ne \phi$ in subcase (b)(II). Thus, $s_{j}=s_{L(k)},$ and so $s_{k}=s_{R(j)},$ so that (8)(i) in claim (8) holds.

Case (A2): We subdivide our analysis into the following two possibilities (A2a) $s_{L(j)}\in {\bar{Y}}_{k}(s),$ (A2b) $s_{L(j)}\notin \bar{ Y}_{k}(s).$

In subcase (A2a), $s_{L(j)}$ must be equal to $s_{L(k)}$ or $s_{k}$ or $s_{R(k)}.$ The last possibility is ruled out since in (A2) we have $s_{L(j)}\ne s_{R(k)}.$ If $s_{L(j)}=s_{L(k)},$ then $s_{j}$ must be equal to $s_{k},$ which is ruled out since s is a scattered placement (recall that $k\ne j$ in case (b)). Thus, we must have $s_{L(j)}=s_{k}$ and consequently $s_{j}=s_{R(k)}.$ That is, (8)(ii) in claim (8) holds.

In subcase (A2b), $s_{L(j)}\in$$arc(s_{R(k)},s_{L(k)}),$ and consequently $s_{R(j)}\notin arc(s_{R(k)},$$s_{L(k)}).$ This is because if $s_{R(j)}\in$$arc(s_{R(k)},s_{L(k)}),$ then $Y_{j}(s)$ is entirely contained in $arc(s_{R(k)},s_{L(k)}),$ which is disjoint from ${\bar{Y}}_{k}(s),$ contradicting the fact that $Y_{j}(s)\cap {\bar{Y}}_{k}(s)\ne \phi$ in subcase (b)(II).

Since $s_{R(j)}\notin$$arc(s_{R(k)},s_{L(k)}),$ we must have $s_{R(j)}\in {\bar{Y}}_{k}(s)\equiv$$arc[s_{L(k)},s_{R(k)}].$ Thus, $s_{R(j)}$ must be equal to $s_{L(k)}$ or $s_{k}$ or $s_{R(k)}.$ The first possibility is ruled out since $s_{R(j)}=s_{L(k)}$ would imply that $Y_{j}(s)\cap {\bar{Y}} _{k}(s)=\phi ,$ a contradiction. The third possibility is ruled out because $s_{R(j)}=s_{R(k)}$ would imply that $s_{j}$ must be equal to $s_{k},$ which would contradict the fact that s is a scattered placement (recall that $k\ne j$ in case (b)). Thus, we must have $s_{R(j)}=s_{k}$ and consequently $s_{j}=s_{L(k)},$ so that (8)(i) in claim (8) holds.