Lecture 7

Electoral Competition: Office-Motivated Candidates

The Template for Electoral Competition Models

We now bring together the social choice theory from Lectures 2–4 and the game theory from Lectures 5–6 to build models of electoral competition. The general template for these models has four components:

Players: The players in the game are the candidates — not the voters. Voters are not strategic agents; their behavior is fixed by their preferences and the voting rule, and it enters the model as the mechanism that translates candidates' platforms into electoral outcomes and thus payoffs.

Actions: Each candidate $i$ chooses a policy platform $x_i \in X \subseteq \mathbb{R}$ , typically $X = [-1, 1]$ .

Electoral outcome: Voters have single-peaked preferences over $X$ . The electoral outcome — who wins — is determined by majority rule. By the Median Voter Theorem, the candidate whose platform is closer to the median voter's bliss point $v^M$ wins.

Payoffs: What candidates care about. This is the key variable across our three models. Office-motivated candidates care only about winning. Policy-motivated candidates care only about the implemented policy. Uncertain candidates have the same policy motivations but face uncertainty about $v^M$ .

In each model, we find the Nash equilibrium in platforms by applying the best-response algorithm to candidates' payoff functions.

Model 1: Office-Motivated Candidates

Setup

There are two candidates, $A$ and $B$ , each choosing a platform $x_i \in \mathbb{R}$ . (We allow the full real line here for generality; the equilibrium platforms will turn out to be in any bounded policy space.) There are $N$ voters with odd $N$ , each with symmetric single-peaked preferences. Majority rule determines the winner. The median voter's bliss point is $v^M$ .

Payoffs

The candidates care only about winning office — they have no intrinsic policy preferences. The payoffs are:

Candidate $A$ receives $1$ if $A$ wins, $\frac{1}{2}$ if there is a tie, and $0$ if $B$ wins.
Candidate $B$ receives $1$ if $B$ wins, $\frac{1}{2}$ if there is a tie, and $0$ if $A$ wins.

A tie can occur even with an odd number of voters if the median voter is exactly indifferent between the two platforms — she splits her "vote" equally, effectively contributing a half-vote to each candidate.

Electoral Outcome with Symmetric Preferences

With symmetric single-peaked preferences and majority rule, voters to the left of the midpoint $m = \frac{x_A + x_B}{2}$ prefer $A$ (if $x_A < x_B$ ) and voters to the right prefer $B$ . More precisely, the electoral outcome depends entirely on whether the median voter prefers $A$ or $B$ .

$\text{A wins} \iff |x_A - v^M| < |x_B - v^M|$ $\text{Tie} \iff |x_A - v^M| = |x_B - v^M|$ $\text{B wins} \iff |x_A - v^M| > |x_B - v^M|$

Warning: this formula relies on symmetric single-peaked preferences and majority rule. In models without symmetry, the electoral outcome requires more careful analysis.

Best Responses

Now we find each candidate's best response. Consider Candidate $A$ 's problem, taking $x_B$ as given.

Case 1: $x_B = v^M$ . If $A$ chooses $x_A = v^M$ , there is a tie and $A$ gets $\frac{1}{2}$ . If $A$ deviates to any $x_A \neq v^M$ , then $|x_A - v^M| > 0 = |x_B - v^M|$ , so $B$ wins and $A$ gets 0. So $A$ 's best response is $x_A = v^M$ when $x_B = v^M$ — the tie at $v^M$ is the best $A$ can do.

Case 2: $x_B \neq v^M$ . The opponent is not at the median. $A$ can choose any $x_A$ with $|x_A - v^M| < |x_B - v^M|$ — that is, any platform strictly closer to $v^M$ than $x_B$ — and win outright, getting payoff 1. This is better than a tie ( $\frac{1}{2}$ ) or a loss (0). Note that $A$ has a whole interval of best responses: any platform in the open ball of radius $|x_B - v^M|$ around $v^M$ will do. There is no unique best response, but the best response correspondence is the interior of this interval (plus possibly $v^M$ itself, which gives a tie).

Finding the Nash Equilibria

We now check all possible cases for $(x_A^*, x_B^*)$ exhaustively.

Case (i): $x_A^* = v^M$ and $x_B^* = v^M$ . Is $A$ best-responding to $v^M$ ? As shown above, yes — $v^M$ is $A$ 's best response when $x_B = v^M$ . By symmetry, $B$ is also best-responding. This is a Nash equilibrium. Both candidates choose the median voter's bliss point, the race is a tie, and each receives $\frac{1}{2}$ .

Case (ii): $x_A^* = v^M$ and $x_B^* \neq v^M$ . In this profile, $B$ is losing: $|x_B^* - v^M| > 0 = |x_A^* - v^M|$ , so $A$ wins and $B$ gets 0. But $B$ can deviate to $x_B = v^M$ and achieve a tie, getting $\frac{1}{2} > 0$ . $B$ has a profitable deviation, so this is not a Nash equilibrium.

Case (iii): $x_A^* \neq v^M$ and $x_B^* = v^M$ . Symmetric to Case (ii). $A$ is losing and can profitably deviate to $v^M$ . Not a Nash equilibrium.

Case (iv): $x_A^* \neq v^M$ and $x_B^* \neq v^M$ . In this profile, one of three things holds: $A$ wins, $B$ wins, or there is a tie (with both platforms equidistant from $v^M$ ). In any sub-case, the losing candidate (or either tied candidate) can deviate to $v^M$ and improve. If $A$ is losing, $A$ deviates to $v^M$ and wins outright. If both are tied at some $x \neq v^M$ , either can deviate to a platform closer to $v^M$ and win outright. In all sub-cases, some candidate has a profitable deviation. Not a Nash equilibrium.

The unique Nash equilibrium is $(x_A^*, x_B^*) = (v^M, v^M)$ . Both candidates converge to the median voter's bliss point.

Downsian Convergence

This result is known as Downsian convergence or the Downs Median Voter Result, after Anthony Downs's 1957 book An Economic Theory of Democracy. It says that two office-motivated candidates, competing for votes among an electorate with symmetric single-peaked preferences, will both adopt the median voter's ideal policy as their platform.

The intuition is straightforward: winning is all that matters, winning requires being closer to $v^M$ than your opponent, and the only position that cannot be outflanked is $v^M$ itself. Any other platform leaves an opening for the opponent to move between you and the median, stealing the election.

Real-World Parallels

Downsian convergence has been cited to explain several historical episodes of centrist drift.

In the United Kingdom during the 1990s, both major parties moved toward the center. Labour under Tony Blair adopted many Thatcherite policies it had previously opposed — keeping anti-union legislation, granting independence to the Bank of England, accepting means-testing in welfare. The Conservatives under John Major and later William Hague accepted the broad architecture of the welfare state and increased health and education spending. Both parties, reading the electorate's center of gravity, moved toward it.

In the United States, Bill Clinton's "triangulation" strategy in the 1990s explicitly aimed to occupy the center: welfare reform, balanced budgets, free trade agreements, and crime bills were designed to position Clinton at or near the median voter against Republican opponents.

In Germany, Angela Merkel's CDU drifted left over her chancellorship — adopting a legal minimum wage, abandoning nuclear power after Fukushima, and embracing a more welcoming refugee policy — all moves that can be interpreted as attempts to track a shifting median voter.

Limitations and What Comes Next

Downsian convergence is a clean and elegant result, but it predicts complete policy convergence — both candidates adopting identical platforms. In practice, political parties often diverge considerably. The two key assumptions driving the result are (1) candidates care only about winning (office motivation) and (2) the location of $v^M$ is known with certainty. Lectures 8 and 9 relax these assumptions in turn, and we will see that each relaxation can produce equilibrium divergence.