# Electoral Competition: Policy-Motivated Candidates ## The Second Model: What If Candidates Have Policy Preferences? In Lecture 7, we found that office-motivated candidates both converge to the median voter's bliss point $v^M$. The model is elegant, but its prediction of complete convergence sits uncomfortably with the observable fact of substantial policy divergence between parties. One natural explanation is that candidates are not purely office-motivated — they also care about which policies get implemented. A politician who spent her career fighting for a particular vision of healthcare, immigration, or economic policy does not simply abandon those views to chase the median voter. She may accept some electoral risk in order to remain closer to her true policy preferences. This lecture develops the second spatial model, in which candidates are **policy-motivated**: they care only about the policy that gets implemented, not about winning per se. ## Setup The setup is nearly identical to Model 1. There are two candidates, $D$ (Democrat) and $R$ (Republican), each choosing a platform $x_i \in \mathbb{R}$. The electorate consists of $N$ voters (odd) with symmetric single-peaked preferences, and the winner is determined by simple majority rule. The median voter's bliss point is $v^M$. The only change from Model 1 is the **payoff function**. ## Policy-Motivated Payoffs Each candidate $i$ has a **bliss point** $z_i$ — the policy she would most like to see implemented. We assume $z_D < v^M < z_R$: the Democrat prefers a left-leaning policy and the Republican prefers a right-leaning one, and neither has the median voter's ideal point as their own. Candidate $i$'s payoff depends on which policy gets implemented: $$u_i(x_D, x_R) = \begin{cases} -|x_D - z_i| & \text{if D wins} \\ -|x_R - z_i| & \text{if R wins} \\ -\dfrac{1}{2}|x_D - z_i| - \dfrac{1}{2}|x_R - z_i| & \text{if tie} \end{cases}$$ The candidate receives the negative absolute deviation between the implemented policy and her own bliss point. A tie is treated as each policy being implemented with probability $\frac{1}{2}$. Note that winning is not intrinsically valuable in this model. Winning is valuable only insofar as it allows the candidate to implement her preferred policy. If the opponent's platform is very close to the candidate's bliss point, the candidate may actually prefer to lose — letting the opponent implement a policy close to her ideal — rather than winning with a platform far from her ideal. ## The Electoral Outcome Formula The electoral outcome formula is unchanged from Model 1. With symmetric SP preferences and majority rule, $D$ wins iff $|x_D - v^M| < |x_R - v^M|$, there is a tie iff $|x_D - v^M| = |x_R - v^M|$, and $R$ wins iff $|x_D - v^M| > |x_R - v^M|$. ### A Worked Example Suppose $v^M = 0$, $z_D = -1$, $z_R = 1$, and consider the platform profile $(x_D, x_R) = (-0.6, 0.2)$. We have $|x_D - v^M| = 0.6 > 0.2 = |x_R - v^M|$, so $R$ wins. The implemented policy is $x_R = 0.2$. $D$'s payoff: $-|x_R - z_D| = -|0.2 - (-1)| = -1.2$. $R$'s payoff: $-|x_R - z_R| = -|0.2 - 1| = -0.8$. Despite the fact that $R$'s own platform is $x_R = 0.2$ rather than $z_R = 1$, $R$ wins the election and implements $x_R = 0.2$, which is only 0.8 away from $z_R$. ## Best Responses for Policy-Motivated Candidates The best-response analysis is more subtle than in Model 1. The key insight is that **a policy-motivated candidate may prefer to lose**. Consider Candidate $D$'s problem, taking $x_R$ as given. **The key tension**: $D$ wants to implement $z_D$, but to win she must be closer to $v^M$ than $R$. This requires $x_D$ to be close to $v^M$. But if $v^M$ is far from $z_D$, winning means implementing a policy far from $D$'s ideal. $D$ might prefer to let $R$ win with $x_R$ closer to $z_D$ than $v^M$ would be. **Case: $x_R = v^M$.** If $D$ plays $x_D = v^M$, there is a tie, and the implemented policy is $v^M$ with probability $\frac{1}{2}$ and $v^M$ with probability $\frac{1}{2}$ — effectively $v^M$ for certain. $D$'s payoff is $-|v^M - z_D|$. If $D$ deviates to any other $x_D \neq v^M$, she loses, and $R$ implements $v^M$. $D$'s payoff is still $-|v^M - z_D|$. So any $x_D$ is a best response when $x_R = v^M$ — $D$ is indifferent. **Case: $x_R \neq v^M$.** Here $D$ must weigh the payoff from winning (implementing $x_D$, which must be close to $v^M$ to win) against the payoff from losing (letting $R$ implement $x_R$). Whether $D$ prefers to win or lose depends on the relative positions of $x_D$, $x_R$, and $z_D$. ## Nash Equilibria Let us check candidate equilibrium profiles. **Case: $x_D < v^M < x_R$, both away from $v^M$.** Consider $D$'s incentive. $D$ wants to win: deviating to a platform closer to $v^M$ wins the election and implements a policy closer to $v^M$. But $v^M > z_D$, so a policy closer to $v^M$ is farther from $z_D$. Whether $D$ prefers to deviate depends on specifics. Similarly for $R$. In general, this is not a Nash equilibrium because either candidate can typically improve by moving toward $v^M$. **Case: $x_D = v^M$, $x_R > v^M$.** Here $D$ wins (or ties if $x_D = x_R$, but $x_D = v^M \neq x_R$ so $D$ wins outright). $D$ implements $v^M$. But $v^M > z_D$, so $D$ is implementing a policy she does not like. $D$ could deviate to some $x_D' < v^M$, moving toward $z_D$. If this still wins ($|x_D' - v^M| < |x_R - v^M|$), $D$ is better off. So this is generally not a Nash equilibrium. **Case: $x_D = v^M = x_R$.** Both candidates play $v^M$. There is a tie, and the implemented policy is $v^M$ with probability $\frac{1}{2}$ from each. $D$'s payoff is $-\frac{1}{2}|v^M - z_D| - \frac{1}{2}|v^M - z_D| = -|v^M - z_D|$. Now check if $D$ can profitably deviate. If $D$ deviates to any $x_D \neq v^M$, $R$ wins (since $|x_D - v^M| > 0 = |x_R - v^M|$), and $R$ implements $v^M$. $D$'s payoff is $-|v^M - z_D|$. Exactly the same as before. $D$ is indifferent between deviating and not — the payoff is the same regardless of what $x_D$ $D$ chooses, because in every outcome $v^M$ ends up being implemented. But wait: if $D$ deviates to some $x_D$ and wins — impossible, since $x_R = v^M$ means $D$ can only tie (by also playing $v^M$) or lose. In a tie, both platforms are implemented with probability $\frac{1}{2}$. So if $D$ deviates to $x_D \neq v^M$, she loses and $R$ implements $v^M$: payoff $-|v^M - z_D|$. If $D$ plays $v^M$, tie: each implements $v^M$: payoff $-|v^M - z_D|$. No improvement possible. $D$ cannot strictly gain by deviating. The same logic applies symmetrically to $R$. Neither player has a strict incentive to deviate. This **is** a Nash equilibrium. The **unique Nash equilibrium** is $(x_D^*, x_R^*) = (v^M, v^M)$. ## Intuition The result is the same as in Model 1 — full convergence to $v^M$ — but the logic is different. Policy-motivated candidates converge to $v^M$ not because they love $v^M$, but because it is the only way to influence the implemented policy. A candidate who loses has no influence whatsoever — the opponent implements whatever she likes. The only way to have any impact is to win, and winning requires being at least as close to $v^M$ as the opponent. In equilibrium, both are exactly at $v^M$. Put differently: a policy-motivated candidate would love to implement $z_i$, but she knows that if she campaigns on $z_i$ and the opponent campaigns on $v^M$, she will lose and $v^M$ will be implemented anyway. Given that the implemented policy will be $v^M$ regardless (in equilibrium), both candidates are best-responding by also playing $v^M$. ## Observed Polarization Both models predict full convergence to $v^M$, yet in practice major parties are often substantially polarized. In the United States during the 2010s, the Democratic Party moved left on healthcare (Medicare for All, the Affordable Care Act), climate (Green New Deal), and redistribution (wealth taxes), while the Republican Party moved right on immigration, trade, and social policy. The distance between the parties' platforms grew, not shrank, over this period. In the United Kingdom between 2015 and 2019, the Conservatives under Theresa May and Boris Johnson ran on hard Brexit, deep immigration cuts, and nationalist rhetoric. Labour under Jeremy Corbyn ran on nationalizing railways, mail, water, and energy, reversing austerity, and taxing the wealthy. Both parties were at or near the extremes of the one-dimensional policy space. In France since 2022, La France Insoumise campaigns on repealing pension reform and a large minimum wage increase; the Rassemblement National campaigns on immigration restriction and "national preference" in social policy. Both are far from the center. Both Models 1 and 2 predict convergence to $v^M$, yet convergence is nowhere to be seen. This motivates Model 3, which introduces uncertainty about the location of $v^M$. 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