Paper deep dive
Costly Voting in the Hotelling-Downs Model
Guy Wolf, Reshef Meir
Intelligence
Status: succeeded | Model: Gemma-4-26B-A4B | Prompt: intel-v1 | Confidence: 95%
Last extracted: 9/25/2026, 3:37:21 AM
Summary
This paper extends the Hotelling-Downs model by introducing costly voting, where voters participate only if the utility gain from their preferred candidate exceeds their individual cost. The study finds that this cost breaks the median voter theorem, leading to polarization. The primary predictor of polarization is the reverse-hazard-rate of the voting cost distribution, which measures the responsiveness of turnout to policy stakes. The authors also incorporate parameters for voter alienation and candidate competitiveness, demonstrating that these factors further influence equilibrium positions and polarization levels.
Entities (8)
Relation Signals (6)
Guy Wolf → affiliatedwith → Technion—Israel Institute of Technology
confidence 99% · Affiliation: Technion—Israel Institute of Technology
Reshef Meir → affiliatedwith → Technion—Israel Institute of Technology
confidence 99% · Affiliation: Technion—Israel Institute of Technology
Costly Voting → breaks → Median Voter Theorem
confidence 95% · Under this model the median voter theorem breaks
Reverse-Hazard-Rate → predicts → Polarization
confidence 92% · the main predictor of polarization is the reverse-hazard-rate of the cost distribution
Voter Alienation → increases → Polarization
confidence 90% · stronger alienation strictly polarizes for a uniform electorate
Candidate Competitiveness → depolarizes → Polarization
confidence 88% · competition for the margin is the model’s one de-polarizing force
Cypher Suggestions (0)
No Cypher suggestions yet.
Abstract
Abstract:We study a partial-participation variation of the Hotelling-Downs model. Voters each have a cost to vote, and only vote when the comparative gain from their preferred candidate exceeds the cost. Under this model the median voter theorem breaks, and we study the extent of polarization under equilibria in different voters and cost distributions. We find that the main predictor of polarization is the reverse-hazard-rate of the cost distribution, indicating that the driver of polarization under our model is the willingness of voters to respond to changes in positions of candidates. We then extend the model by adding parameters governing alienation and candidate competitiveness, showing that our results are robust even when taking into account other realistic factors.
Tags
Links
- Source: https://arxiv.org/abs/2609.29869v1
- Canonical: https://arxiv.org/abs/2609.29869v1
Trouble viewing inline? Open PDF directly →
Full Text
338,065 characters extracted from source content.
Expand or collapse full text
Costly Voting in the Hotelling–Downs Model Guy Wolf and Reshef Meir Affiliation: Technion—Israel Institute of Technology Abstract We study a partial-participation variation of the Hotelling-Downs model. Voters each have a cost to vote, and only vote when the comparative gain from their preferred candidate exceeds the cost. Under this model the median voter theorem breaks, and we study the extent of polarization under equilibria in different voters and cost distributions. We find that the main predictor of polarization is the reverse-hazard-rate of the cost distribution, indicating that the driver of polarization under our model is the willingness of voters to respond to changes in positions of candidates. We then extend the model by adding parameters governing alienation and candidate competitiveness, showing that our results are robust even when taking into account other realistic factors. 1 Introduction Two developments mark democratic politics in recent decades: in the United States the two parties’ elected representatives have moved apart [63] and partisans with them [74], while across much of Europe partisans are sharply hostile to one another [76]; and in many democracies fewer citizens turn out to vote [49, 37]. The canonical account of two-candidate competition predicts neither. That account is the Hotelling–Downs model [50, 28], in which two office-seeking candidates choose policy platforms to maximize vote share. Under full voter participation and single-peaked preferences it yields convergence to the median voter, whose ideal policy defeats any alternative in pairwise majority voting—the median voter theorem. Yet divergence, not convergence, is the recurrent empirical finding. In the United States, Ansolabehere et al. [7] and Lee et al. [57] document systematic departures from median convergence in U.S. House elections, while Levitt [59] finds similar patterns in the U.S. Senate, and Bafumi and Herron [10] show that members of Congress tend to adopt positions more ideologically extreme than those of their constituencies. One key difference between the benchmark Hotelling–Downs model and real-world elections is that the benchmark typically assumes everyone votes, whereas in practice, voters may abstain, which makes it increasingly important to incorporate non-participation into models of electoral competition [67]. That the two trends above are connected is not a new conjecture. Calibrating a spatial model with voter abstention to survey data from the U.S. presidential elections of 1980, 1984, 1988, 1996 and 2000, Adams, Merrill and Grofman report that in each of them the major-party candidates had electoral incentives to diverge under their estimated turnout model, with equilibrium divergence increasing in the policy distance between the two parties’ partisans [3, 4]. What has been missing is a characterization of which feature of the participation margin—the voters sitting at the edge of turning out—does the work. In this paper, we extend the Hotelling–Downs framework by introducing costly voting. Inspired by the classic Calculus of Voting (see below), voters participate only when the relative utility gain from supporting their preferred candidate exceeds their individual cost of voting; otherwise, they abstain. Turnout therefore depends endogenously on the perceived policy difference between the candidates, and candidates must account not only for ideological proximity but also for the turnout incentives induced by their platform choices. Our model accommodates arbitrary distributions of voter ideal points (density v) and voting costs (CDF K, density k), and adds two one-parameter “dials” to the participation side and the candidate side of the game: voter alienation α≥0α≥ 0, under which participation also declines with a voter’s distance from her own candidate, and candidate competitiveness β∈[0,1]β∈[0,1], which interpolates between maximizing one’s vote share (β=0β=0) and maximizing one’s margin of victory (β=1β=1). Costly voting is not itself a dial: it is the always-on baseline, carried by the entire cost distribution rather than by a single parameter, and both dials are read against it. 1.1 Our Contribution We first derive necessary conditions that every pure Nash equilibrium must satisfy. A single balance principle asserts that at every equilibrium the marginal voters the candidates gain toward the center and those they lose on their flanks are traded off at a constant rate, one fixed by the candidates’ objective alone and not by the electorate or the cost distribution (Lemmas 2 and 3). Equilibrium takes a particularly simple form when either voter ideal points or voting costs are uniformly distributed: under uniform costs every equilibrium traps a share of the electorate between the platforms that again depends only on that objective, never on the electorate itself (Theorems 3.1 and 9). These conditions are not sufficient in general, and we show by example that profiles solving them may fail to be equilibria. We therefore complement them with sufficiency results: for uniformly distributed voters, a single monotonicity condition on the cost distribution, satisfied by every log-concave cost density, under which the equilibrium exists, is unique and is completely characterized, for every competitiveness β<1β<1 and alienation α≥0α≥ 0; and for uniformly distributed costs, density-ratio bounds under which the profiles at fixed quantiles of a symmetric electorate (the quartiles in the baseline) are equilibria (Corollaries 1 and 8). Our central message concerns polarization (Section 3.2). Costly voting alone breaks median convergence: under uniform costs the quartile profile of a symmetric electorate (density ratio at most four) is an equilibrium, so mass polarization (of the electorate) transmits to elite polarization (here, of the candidates) one for one. Fixing the electorate at uniform, we prove a comparative-statics theorem in the reverse-hazard-rate stochastic order: cost distributions that are higher in this order induce weakly more polarized equilibria, under the monotonicity condition above. We further show this is exactly the right order: the equilibrium correspondence is invariant to rescaling K, hence depends on the cost distribution only through its reverse hazard rate k/Kk/K (the proportional responsiveness of turnout to the stakes), and in particular the level of turnout has no causal role: distributions exist with stochastically higher costs, lower turnout, and strictly less polarization. What polarizes candidates is an electorate whose participation is responsive to the stakes, not one whose turnout is merely low. We then turn each dial. Under alienation (Section 4) we prove that stronger alienation strictly polarizes for a uniform electorate and every log-concave cost density, with the equilibrium separation Δ∗ ^* increasing in α and tending to full separation, and that this holds at an unchanged level of turnout, so the effect is not a by-product of lower participation (Corollary 6); that the cost channel survives alienation, so that costlier voting in the reverse-hazard-rate order still polarizes at every α; and that turnout becomes non-monotone in the platform gap, so the reverse causal channel (extreme candidates depressing participation) reappears. Under competitiveness (Section 5) we prove that competition for the margin is the model’s one de-polarizing force, with a closed-form equilibrium path Δ∗(α,β) ^*(α,β) in the doubly-uniform benchmark, but that under alienation the margin limit is a knife-edge: for a uniform electorate the β↑1β 1 limit selects the most polarized point of the margin game’s equilibrium band rather than the median; under alienation and uniform costs the median is a strict local equilibrium of the margin game (no small deviation pays) when the electorate is centrally peaked (v′(12)<0v ( 12)<0) and fails when it is hollow (v′(12)>0v ( 12)>0), a dichotomy settled at fourth order because the second-order term vanishes; and for hollow electorates pure equilibria can fail to exist at all. A computer-assisted certificate proves non-existence at α=1α=1 with uniform costs and an arcsine electorate (v=Beta(12,12)v=Beta( 12, 12), the extreme hollow case), for every β≥0.9168β≥ 0.9168, while at α=0α=0, under uniform costs, we prove existence for every β<1β<1 for symmetric electorates satisfying explicit density bounds, the arcsine electorate among them (Theorem 0.N.1). Table 3 collects the resulting map from primitives to polarization. Methodologically, the non-existence results rest on a certificate scheme for continuous-strategy spatial games: a Lipschitz bound on the payoff, exact quadrature at the grid points, and adaptive refinement of the cells that remain undecided, which together turn a finite computation into a statement about the continuum. The scheme is not specific to this model and may be of use wherever pure equilibria of a continuous game must be ruled out. A released verification suite recomputes the paper’s numerical claims from the model’s definitions, including both computer-assisted theorems (Theorems 0.L.4 and 0.O.1): the grid certificates are reproduced to the digits reported, and the exact payoff evaluation they rest on is cross-checked against independent adaptive quadrature.11 1 Available at https://github.com/ANONYMIZED/costly-voting-hd; see its README for what is and is not covered. Each figure states in its source how its data were generated. 1.2 Related Work Hotelling–Downs and variants Due to the simplicity and interpretability of the Hotelling–Downs spatial model, a large literature has developed extensions and variations, often with the aim of getting more ‘natural’ equilibria where candidates diverge; Osborne [70] surveys it, and Grofman [41] catalogues the Downsian assumptions—full participation among them—whose relaxation produces divergence. Most extensions retain the assumption that voters prefer the closer candidate but vary the details of the model, e.g., to consider voters’ uncertainty about candidates [52], policy-motivated candidates [84, 18], candidates’ uncertainty about voters [78, 22], relaxing equilibrium requirements [83], introduce asymmetries between candidate favorability [8] and so on. A separate axis, less explored in this line of work, is the candidates’ objective: vote share, margin of victory, and probability of winning coincide under full participation, so the classical model cannot distinguish them [9, 30], whereas with abstention they come apart. That is not incidental: the equivalence result of [73] holds precisely when abstention is ruled out, and variable participation is the setting in which Hinich and Ordeshook [47] separated plurality from vote maximization. Section 5 shows how consequential the choice is. Abstention Voting costs and abstention have received considerable attention since the outset of formal political economy, beginning with the Calculus of Voting framework [77] and the pivotal-voter tradition it launched [71, 72, 33], continued in the costly-voting models of Börgers [13], Krasa and Polborn [53], Krishna and Morgan [54] and Herrera, Morelli and Palfrey [43], in which, as here, each voter draws a private cost from a distribution, but the alternatives on the ballot are fixed and the questions are turnout, welfare and information aggregation, as well as later “lazy bias” models [25, 31]. In all of these, the fundamental assumption is that voters are more likely to participate when the outcome matters more and when they are more pivotal, and the analysis focuses on the implications for equilibrium outcomes. This assumption is relaxed in [64], where voters consider various heuristics for their pivotality. Other papers, such as [21, 65], examine the effects of arbitrary or random participation on electoral outcomes under the Median rule.22 2 In some simulations in [21] the authors sampled more voters closer to the edges of the interval (i.e., assuming ‘extreme’ voters are more active), but without any underlying utility model or formal justification. However, none of the above models endogenize candidate responses to abstention; instead, candidate positions are taken as fixed. That the median result is fragile once participation is endogenous is an old theme: Hinich, Ledyard and Ordeshook [45] study existence of majority-rule equilibrium when voters may abstain, and Hinich [48] shows that in the probabilistic-voting version the median is an equilibrium only under a knife-edge symmetry: the convergence prediction is, in his phrase, an artifact. The sharpest foil is Ledyard [56], in which platforms and participation are determined together and voters, like ours, carry a private voting cost; but they weigh the platform difference by their probability of being pivotal, and the outcome is convergence. Both candidates adopt the same welfare-maximizing platform and, the platforms being identical, nobody votes. Our voters act as if decisive (Section 2.2), and Section 3 shows that convergence then fails, beginning with Lemma 1, which rules out tied equilibria. A classical and distinct driver of abstention is alienation—non-participation out of disaffection with even one’s preferred option—introduced into spatial models by Hinich and Ordeshook [46], and since distinguished from abstention out of indifference between the candidates [16, 6, 2]; our parameter α embeds this channel alongside voting costs, and Section 4 shows it is the model’s strongest polarizing force. That alienation-driven abstention can break convergence is known: Callander and Wilson [17] obtain divergent positions in a spatial entry model with alienation, and Oprea, Martin and Brennan [69] argue that compulsory voting de-polarizes precisely by removing extreme voters’ ability to threaten abstention. What is new here is that the threat is priced rather than assumed: the cost distribution K is a primitive, and Section 4 identifies which of its features the equilibrium responds to. Closest to the present paper is Jones, Sirianni and Fu [51], who examine how a relative cost of voting that deters indifferent voters, third-party entry, and a bimodal electorate each undermine the median voter theorem. Their treatment of the first channel is numerical (candidate positions follow gradient dynamics on a two-Gaussian or empirically estimated electorate, and the relative cost of voting is a single scalar common to all voters), whereas ours is analytic and holds for arbitrary v and K, which is what lets Section 3.2 isolate the reverse hazard rate as the only feature of the cost distribution the equilibrium responds to. Some variations of the Hotelling–Downs model explicitly incorporate the possibility of voter abstention. In [34, 82] voters abstain whenever all candidates lie outside a fixed attraction interval, so participation does not decline gradually as candidates become less attractive; Cohen and Peleg [20] instead draw each voter’s tolerance range at random from a density, so that the share of voters who still turn out falls gradually with the distance to the nearest candidate, the nearest-candidate analogue of the cost draw used here. Coughlin [22] considers probabilistic attraction models with a finite set of voters, in which each voter may abstain according to an independent probability function. The attraction-interval models condition participation on the distance to the nearest candidate alone, which is natural in economic competition, where a consumer’s decision to buy depends only on the best offer available, but not in political competition, where a voter’s stake is the difference between the platforms. Models in which participation responds to that difference do exist: the indifference-based abstention models cited above [6, 2], the pivotal model of Ledyard [56], the numerical study of Jones et al. [51], and a recent workshop paper whose voters are prospect-theoretic and whose focus is the computational complexity of optimizing a candidate’s position [19]. Two papers in political economy come closer still. In Llavador [62] voters weigh both platforms against a cost of voting that grows with the distance to the platform they support (the alienation channel of Section 4), but his parties are policy-motivated, so there the alienation cost is a moderating force that keeps them from radicalizing, whereas here it acts on vote-share maximizers who would otherwise converge, and pushes them apart. In Herrera, Levine and Martinelli [42] turnout is likewise endogenous to the platforms, but through the parties’ mobilization spending rather than through any voter’s own cost–benefit comparison. What distinguishes the present model is the combination: participation responds to the platform difference; the threshold is drawn from a full cost distribution K that enters the analysis as a primitive; voters weigh the platform difference directly against their cost, with no pivot-probability discount (Section 2.2); and the analysis is carried out for arbitrary v and K rather than a fixed electorate. Candidate polarization The distinction between mass and elite polarization drawn above is a live one in political science; one prominent debate is on which type of polarization precedes, or even causes, the other [44, 85, 1, 35, 32]. Neither trend with which this paper opened is universal: four-decade trends across OECD democracies are mixed, with increases in some countries and decreases in others [14]. Two recent studies speak directly to the channel studied here. Across eleven Western European countries (1977–2016), Dreyer and Bauer [29] find that parties adopt more extreme positions as their electorates polarize, and that the response runs through abstention: it is strongest where turnout is low and fades as turnout rises. Al Yussef [5] finds, across Dutch municipalities, that the vote is more polarized where turnout is lower, consistent with centrist voters abstaining disproportionately. Both studies measure the abstention margin by the turnout rate; Section 3.2 shows that the model separates the rate from the responsiveness of participation to the stakes, and that only the latter moves the equilibrium. Prior theoretical work in computational social choice has modeled polarization in a variety of settings, such as committee selection [27], which sometimes require complex definitions to measure. In the context of 2-candidate spatial competition, measuring polarization is trivial, and so we can isolate and test the effect of each component of the model: the voters’ distribution, the voting costs, alienation, and the candidates’ objective. Core versus swing The distributive-politics literature asks when a party should reward its core and when it should woo the center, with transfers rather than platforms as the instrument [23, 60, 26, 40]; there a group is “core” because a party can deliver to it more effectively, a driver absent from our model, whose candidates hold one public, non-discriminating instrument. Within this literature the nearest to our question is Cox [24], who adds turnout to the targeting problem: the most valuable voter supports you whenever she votes, and her participation responds to what you offer. Both features describe the marginal voters on a candidate’s outer flank, and his comparative static is the crossing of Corollary 4: the more mobilization and the less persuasion a party believes possible, the more it concentrates on its core. Our balance condition prices the trade-off between the two flanks at a rate that rises with competitiveness and falls with alienation (Lemma 3), a price no full-participation model can carry (Remark 2). The two dials Each extension answers one of the gaps above: the α-dial lets participation fall with the distance to a voter’s own candidate, the difference between economic and political competition noted above, and the β-dial makes the candidates’ objective a parameter rather than a convention. Neither ingredient is new on its own; what the present model adds is that both are read against an arbitrary cost distribution, so the effect of each can be separated from the effect of costly voting itself. Road map Section 2 sets out the model; Sections 3, 4 and 5 analyze the baseline and the two extensions, each section moving from characterizations to their consequences for polarization, which in Sections 4 and 5 come in two steps: first at a fixed value of the parameter, then as it varies. Section 5 closes with the margin game β=1β=1, a special case; Section 6 concludes. Proofs are deferred to Appendix 0.A and the subsequent appendices. A reader with limited time can take Sections 3.2, 4.3 and 5.3, the subsections on what each force does to polarization, together with Table 3; the subsections preceding each supply what they rest on. 2 Model Section 2.1 states the classical Hotelling–Downs game and the median prediction it delivers, and isolates the assumption our model gives up. Section 2.2 gives it up: participation becomes a cost–benefit decision taken voter by voter, which is the paper’s one substantive departure and is in force everywhere in it. Sections 2.3 and 2.4 fix the candidates’ objective, the equilibrium notions and a running example; the game assembled by then is the one analyzed in Section 3. Section 2.5 adds the two extensions—alienation among voters, competitiveness among candidates—each governed by its own parameter, each switched off in the model presented first, and each taken up in a later section (Sections 4 and 5). Table 1 summarizes the notation. Table 1: Notation of the model. The last block belongs to the extensions of Section 2.5; the model of Sections 2.2 and 2.3 sets α=β=0α=β=0. Shorthands introduced later, in the analysis, are collected separately in Table 2. Primitives and parameters V,vV,v; K,kK,k CDF/density of voter positions; of voting costs κ (on [0,1][0,1]) ρv _v density ratio supv/infv v/ v (electorate evenness) Profiles and induced behavior cic_i; di(x):=|x−ci|d_i(x):=|x-c_i| platform of candidate i∈1,2i∈\1,2\; distance to voter x c1≤c2c_1≤ c_2; Δ ; m labeling; separation c2−c1c_2-c_1; midpoint c1+c22 c_1+c_22 Bi(x)B_i(x); pi(x)p_i(x) benefit of voting for i: d−i−did_-i-d_i, Eq. 1; share of the voters at x who vote for i, Eq. 2 uiu_i; T vote share ∫01pivx _0^1p_iv\,dx; turnout u1+u2u_1+u_2 O¯i O_i; I¯i I_i votes from i’s flank; from her half of the contested middle, Eq. 3 Δ∗ ^* equilibrium separation Decorations ∗\,^*; ⋅~ \,·\, equilibrium value; the comparison primitive of a stochastic order Extension parameters (Section 2.5) α≥0α≥ 0 alienation: benefit becomes d−i−(1+α)did_-i-(1+α)d_i, Eq. 4 β∈[0,1]β∈[0,1]; wiw_i competitiveness; margin ui−u−iu_i-u_-i Ui=ui−βu−iU_i=u_i-β\,u_-i candidate i’s objective, Eq. 5 Δ∗(α) ^*(α); Δ∗(α,β) ^*(α,β) equilibrium separation, when the dependence matters 2.1 The Hotelling–Downs Benchmark In the classical model [50, 28], every voter votes for the candidate nearest to her ideal point, and each candidate chooses a platform to maximize the number of votes she receives; if both candidates choose the same platform, they split the electorate evenly. With voters uniformly distributed on [0,1][0,1], the unique equilibrium famously places both candidates at the median, 12 12. This is a special case of a more general statement. Let V be the CDF of an arbitrary continuous distribution of voter ideal points on [0,1][0,1], and let candidate i∈1,2i∈\1,2\ choose a platform ci∈[0,1]c_i∈[0,1], so that a profile is a pair (c1,c2)(c_1,c_2). Claim 2.1 (See [12, 28]) The only equilibria of the Hotelling–Downs game are the profiles (c1,c2)(c_1,c_2) with V(c1)=V(c2)=12V(c_1)=V(c_2)= 12, i.e., both candidates at a median of V. Claim 2.1 is robust to how a voter’s preference over the two platforms is modeled, so long as she votes. Let her payoff depend on both platforms through the gain d−i−did_-i-d_i introduced in Section 2.2 below, rather than on the nearer one alone: with full participation she still votes for the candidate nearer to her, the induced vote shares are the classical ones, and the equilibrium set is unchanged. What breaks the claim is therefore not the utility model but the decision of whether to vote at all. The remainder of this section replaces the assumption of full participation with an explicit cost–benefit decision by each voter. 2.2 Voters, Costs, and Participation Positions and costs There is a continuum of voters with positions x∈[0,1]x∈[0,1], distributed according to a continuous distribution with CDF V and density v. In addition to her position, each voter draws an idiosyncratic voting cost κ∈[0,1]κ∈[0,1], i.i.d. across voters according to a continuous distribution with CDF K and density k, independently of her position.33 3 Independence reflects the interpretation of κ as the logistics of voting (registration, travel, forgone time), which are plausibly unrelated to ideology. Restricting the support to [0,1][0,1] is essentially without loss of generality: all distances lie in [0,1][0,1], so the benefit from voting defined below never exceeds 11, voters with κ>1κ>1 never participate, and only the restriction of K to [0,1][0,1] matters. The following mild conditions are assumed throughout the paper; we note explicitly whenever one of them can be relaxed.44 4 We reserve the term regularity for its usual meaning, monotonicity of the virtual cost (6); that is a hypothesis of Theorem 3.2(b), not a standing assumption. Assumption 2.2 (Standing assumptions) The densities v and k are continuous and strictly positive on (0,1)(0,1), K(0)=0K(0)=0, and K(1)∈(0,1]K(1)∈(0,1]. A defect 1−K(1)>01-K(1)>0 is the mass of voters whose cost exceeds every attainable benefit (the never-voters of Proposition 2), so K need not be a proper distribution on [0,1][0,1]. The participation rule Given a profile of platforms (c1,c2)(c_1,c_2), write di(x):=|x−ci|d_i(x):=|x-c_i| for the distance between a voter at x and candidate i. The voter’s benefit from voting for candidate i is her preferred candidate’s distance advantage, how much better off she is if that candidate wins rather than the opponent: Bi(x):=d−i(x)−di(x),B_i(x)\;:=\;d_-i(x)\;-\;d_i(x), (1) and she votes for candidate i if Bi(x)≥κB_i(x)≥κ, abstaining if neither candidate clears her threshold. Since Bi(x)>0B_i(x)>0 requires x to be strictly closer to cic_i than to c−ic_-i, at most one candidate can offer a positive benefit, so the rule assigns (almost) every voter at most one vote. We emphasize that the benefit (1) compares the platforms directly, rather than weighting the comparison by the probability of being pivotal as in the Calculus of Voting [77] and the pivotal-voter models that formalize it [72]. Voters thus behave as if decisive, an assumption in the spirit of expressive [15] or boundedly rational voting, appropriate when voters do not know the positions of other voters or cannot assess their own pivotality [64]. The departure is narrower than it looks. A common pivot probability ϖ that does not depend on the platforms is absorbed into the cost distribution: voting when ϖBi(x)≥κ \,B_i(x)≥κ is voting when Bi(x)≥κ/ϖB_i(x)≥κ/ , which is the same game with costs t↦K(ϖt)t K( t). Proposition 5 shows what that reparametrization does: nothing at all when K is a power law, uniform costs included, and otherwise, for a uniform electorate, a move whose direction is set by the elasticity of K. What the model genuinely assumes away is a pivot probability that responds to the platforms themselves. Participation shares Aggregating over costs, the share of voters at position x who vote for candidate i is pi(x)=K(Bi(x)),if Bi(x)≥0,0,otherwise.p_i(x)\;=\; casesK (B_i(x) ),&if B_i(x)≥ 0,\\[2.0pt] 0,&otherwise. cases (2) Fig. 1 illustrates the mechanics. On candidate 11’s side of the midpoint, B1B_1 is a piecewise-linear “tent”: its outer arm, facing away from the opponent, is flat at the full gain Δ:=c2−c1 :=c_2-c_1—every voter on the far side of c1c_1 enjoys the same advantage—and its inner arm, facing the opponent, falls at slope 22 from the peak at c1c_1 to the midpoint m:=c1+c22m:= c_1+c_22, where it vanishes; the mirror statement holds for candidate 22. Letting voting costs tend to zero recovers the vote shares of the classical model for any profile with c1≠c2c_1≠ c_2. Flank and contested middle The arms are pieces of the benefit function; the voters standing under them are worth naming separately, because every first-order condition in the paper is a statement about those two groups. Since B1<0B_1<0 beyond the midpoint, candidate 11 draws her votes entirely from [0,m][0,m], which her own platform splits in two: her flank [0,c1][0,c_1], the voters standing behind her, and her half [c1,m][c_1,m] of the contested middle [c1,c2][c_1,c_2], the voters lying between the two platforms. Write O¯1:=∫0c1K(B1(x))v(x)x=K(Δ)V(c1),I¯1:=∫c1mK(B1(x))v(x)x O_1\;:=\; _0^c_1\!K (B_1(x) )v(x)\,dx\;=\;K( )\,V(c_1), I_1\;:=\; _c_1^m\!K (B_1(x) )v(x)\,dx (3) for the votes each group yields, so that u1=O¯1+I¯1u_1= O_1+ I_1; the closed form for O¯1 O_1 is available because the outer arm is flat, so every voter on the flank faces the same benefit Δ . The mirror definitions serve candidate 22. Note that B1B_1 is continuous at c1c_1, taking the value Δ from both sides: nothing distinguishes the outermost voter of the middle from the innermost voter of the flank. That is what makes the split pure bookkeeping, and it is what Lemma 2 turns on. The tie-breaking convention, however, differs from the classical one: at c1=c2c_1=c_2 every voter’s benefit is 00, so by K(0)=0K(0)=0 (Assumption 2.2) (almost) every voter abstains and both candidates receive zero votes, whereas the classical model splits the electorate evenly. This discrepancy is immaterial for our analysis: as we show below (Lemma 1), tied profiles are never equilibria of the model just described. The one exception arises in the margin limit of the second extension, where they survive as zero-turnout, zero-margin equilibria (Proposition 14). 00c1c_1mmc2c_21100κΔ Δ 1B_1B2B_2voter position xxbenefit Bi(x)B_i(x) Figure 1: The benefit of voting at the profile (c1,c2)=(0.3,0.8)(c_1,c_2)=(0.3,0.8). A voter at x whose cost is κ votes for candidate i when Bi(x)B_i(x) rises above the dashed line; aggregating over costs, a share pi(x)=K(Bi(x)+)p_i(x)=K (B_i(x)^+ ) of the voters at x vote for i. The separation Δ=c2−c1 =c_2-c_1 appears twice over, as the gap between the platforms and as the height of each tent, because a voter standing behind her own candidate gains exactly Δ by voting. The dashed vertical marks the partition of (3): candidate 11’s own platform splits the voters she draws from into her flank and her half of the contested middle, and candidate 22 mirrors it. Under uniform costs pip_i is the positive part of BiB_i, so with uniform voters the shaded areas are the vote shares, here (u1,u2)=(0.2125,0.1625)(u_1,u_2)=(0.2125,0.1625), and the two labelled pieces of the blue area are O¯1 O_1 and I¯1 I_1. 2.3 Candidates and Objectives Two candidates, indexed by i∈1,2i∈\1,2\, simultaneously choose platforms ci∈[0,1]c_i∈[0,1] in a one-shot game. Candidates know the distributions V and K (but not the realized cost of any individual voter). Candidate i’s vote share and the resulting turnout are ui(c1,c2)=∫01pi(x)v(x)x,T=u1+u2≤ 1.u_i(c_1,c_2)\;=\; _0^1p_i(x)\,v(x)\,dx, T\;=\;u_1+u_2\;≤\;1. Each candidate maximizes her vote share uiu_i. This is the natural objective when raw support matters in itself (think of public funding, perceived legitimacy, or proportional systems), and it is also the objective of the classical candidate of Section 2.1, so the only thing that has changed so far is that voters may now stay home. 2.4 Equilibrium and a Running Example We study pure Nash equilibria of the simultaneous-move game with payoffs (u1,u2)(u_1,u_2). The game is symmetric, so equilibria come in mirror pairs; when describing a profile we label the candidates so that c1≤c2c_1≤ c_2, and write Δ=c2−c1 =c_2-c_1 for their separation (our measure of candidate polarization) and m=c1+c22m= c_1+c_22 for the midpoint. Three nested notions of a profile recur throughout, and the gaps between them carry real content, so we fix them here. Definition 1 (Stationary, local, and Nash) Fix a profile with 0<c1<c2<10<c_1<c_2<1 and write πi _i for candidate i’s payoff, which is uiu_i here and UiU_i once Section 2.5 is in play. The profile is stationary if ∂πi/∂ci=0∂ _i/∂ c_i=0 for both candidates; a local equilibrium if each cic_i is a local maximum of πi(⋅,c−i) _i(·,c_-i), and a strict one if that maximum is strict; and a Nash equilibrium if each cic_i is a global maximum of πi(⋅,c−i) _i(·,c_-i) on [0,1][0,1]. Every interior Nash equilibrium is a local equilibrium, and, the payoffs being differentiable at interior profiles (Lemma 5), every interior local equilibrium is stationary. Neither converse holds, and the paper exhibits both failures: a stationary profile can be a local minimum of the deviator’s payoff (Remark 16), and a local equilibrium can lose to a distant deviation to a different local maximum (Appendix 0.F). That is why every characterization in this paper comes in two parts: conditions every equilibrium must satisfy, and distributional conditions under which a profile meeting them is an equilibrium. As a running illustration we use the doubly-uniform environment, in which voter positions and voting costs are both uniform on [0,1][0,1]: x,κ∼U(0,1)x,κ U(0,1), that is v≡1v≡ 1 and K(t)=tK(t)=t. It returns in every section that follows, where we also call it the doubly-uniform benchmark. Fig. 1 depicts the profile (c1,c2)=(0.3,0.8)(c_1,c_2)=(0.3,0.8), under which u1=0.2125u_1=0.2125 and u2=0.1625u_2=0.1625. Fixing c2=0.8c_2=0.8, the left panel of Fig. 2 plots candidate 11’s vote share as a function of her own platform: her best response c1=c2/3≃0.267c_1=c_2/3 0.267 trades off a larger flank of nearby voters (pushing her toward the opponent) against the higher turnout generated by a wider separation (pushing her away). The right panel shows the full strategy space: the two best-response curves intersect exactly twice, at (14,34) ( 14, 34 ) and (34,14) ( 34, 14 ), which we later show (Corollary 2) to be the only equilibria in this environment. Intuitively, when the candidates are too close to each other, many voters abstain, driving the candidates apart, and when they are too far the mass of voters between them becomes worth chasing. 00c23 c_2312 12c2c_211000.10.10.20.2best responsec1c_1u1(c1,c2)u_1(c_1,c_2), c2=0.8c_2=0.8 c1=BR1(c2)c_1=BR_1(c_2)c2=BR2(c1)c_2=BR_2(c_1)000.50.511000.20.20.40.40.60.60.80.811c1c_1c2c_2000.10.10.20.20.30.3vote share u1u_1 Figure 2: The candidates’ problem in the doubly-uniform environment (x,κ∼U(0,1)x,κ U(0,1)). Left: candidate 11’s vote share against c2=0.8c_2=0.8; the best response c1=c2/3c_1=c_2/3 balances proximity to more voters against the turnout gained from separation (the kink is at c1=c2c_1=c_2). Right: heatmap of u1u_1 over all profiles, with the best-response curves of candidate 11 (solid) and candidate 22 (dashed); their intersections (stars) are the two equilibria (14,34) ( 14, 34 ) and (34,14) ( 34, 14 ). 2.5 Extensions: Alienation and Competitiveness The model above—costly voting, and nothing else—is the model analyzed in Section 3. We now add two further ingredients. Alienation, governed by a parameter α≥0α≥ 0, changes the voters: participation may decline not only with a small benefit of voting, but also with the voter’s distance from her own candidate; Section 4 switches it on. Competitiveness, governed by a parameter β∈[0,1]β∈[0,1], changes the candidates: they value their own votes but may also discount votes cast for the opponent, interpolating between pure vote-share maximization (β=0β=0) and pure margin maximization (β=1β=1); Section 5 switches it on. Setting (α,β)=(0,0)(α,β)=(0,0) returns the model of Sections 2.2 and 2.3 verbatim. Alienation Alienation is a classic driver of abstention in spatial models [46]: voters who are ideologically distant from all candidates are less inclined to turn out, even when their gain is unchanged. We capture it by penalizing the distance to a voter’s own candidate inside the benefit, so that Eq. 1 becomes Bi(x):=d−i(x)−(1+α)di(x),α≥0,B_i(x)\;:=\;d_-i(x)\;-\;(1+α)\,d_i(x), α≥ 0, (4) which is Eq. 1 at α=0α=0. The definition is a convex combination of the two natural ways to value the outcome of an election. One is relative: a voter cares how much better one candidate is than the other, d−i−did_-i-d_i, the gain of Eq. 1, which makes her abstain when the candidates are too alike for the choice to matter. The other is absolute: she cares about the platform she actually helps elect, −di-d_i, which makes distance from her own candidate a reason to stay home—the alienation of [46]. A voter who puts weight 1−ζ1-ζ on the first view and ζ∈[0,1)ζ∈[0,1) on the second has (1−ζ)(d−i−di)+ζ(−di)=(1−ζ)(d−i−11−ζdi)=(1−ζ)Bi(1-ζ) (d_-i-d_i )+ζ (-d_i )=(1-ζ) (d_-i- 11-ζ\,d_i )=(1-ζ)\,B_i for α=ζ/(1−ζ)α=ζ/(1-ζ), and the positive factor 1−ζ1-ζ is absorbed into the cost distribution exactly as the pivot probability is in Section 2.2. So α is not an ad hoc penalty. It indexes the mix between the two views, from the purely relative voter of Section 3 at α=0α=0 toward the purely absolute one as α→∞α→∞; the weight is written ζ because a convex weight lies in [0,1)[0,1) while α ranges over [0,∞)[0,∞). The endpoint ζ=1ζ=1 is excluded, and necessarily so: there Bi=−di≤0B_i=-d_i≤ 0, so with non-negative costs nobody ever votes and the model is empty. Every α≥0α≥ 0 therefore keeps some weight on the comparison between the candidates. Everything else in Section 2.2 is stated in terms of BiB_i and therefore holds verbatim for every α≥0α≥ 0: the voting rule Bi(x)≥κB_i(x)≥κ, the argument that at most one candidate can offer a positive benefit, and the participation shares (2). The definition also has a compact reading: Bi(x)≥κB_i(x)≥κ iff d−i(x)−di(x)≥κ+αdi(x)d_-i(x)-d_i(x)≥κ+α\,d_i(x), so alienation inflates the effective cost of voting by a fraction of the distance to one’s own candidate. Geometrically, alienation tilts the benefit tent of Fig. 1. On candidate 11’s side of the midpoint, B1B_1 now rises at slope α toward the platform, still peaks at height Δ at c1c_1, and falls at slope 2+α2+α to the midpoint m, where it equals −αΔ/2-α /2; the mirror statement holds for candidate 22. The outer arm is therefore no longer flat, so participation falls fastest among voters far from their own candidate, and a “dead zone” of full abstention opens around the midpoint. Fig. 3 places the two cases side by side. The tie-breaking convention of Section 2.2 is unaffected: at c1=c2c_1=c_2 the benefit is −αdi(x)≤0-α\,d_i(x)≤ 0, so (almost) every voter abstains a fortiori. 00c1c_1mmc2c_21100κΔ position xxbenefit Bi(x)B_i(x)α=0α=0 (no alienation)B1(x)B_1(x)B2(x)B_2(x)00c1c_1mmc2c_21100κΔ position xα=1α=1 (alienation) Figure 3: Alienation reshapes the benefit of voting; the profile is (c1,c2)=(0.3,0.8)(c_1,c_2)=(0.3,0.8), as in Fig. 1. Left: α=0α=0, the model of Section 2.2, reproduced here for comparison. Right: α=1α=1. Alienation tilts the outer arms (slope α instead of 00), steepens the inner arms (slope 2+α2+α instead of 22), and opens a dead zone of abstention around the midpoint m. Under uniform costs and uniform voters the shaded areas are the vote shares: (u1,u2)=(0.2125,0.1625)(u_1,u_2)=(0.2125,0.1625) on the left and (0.147,0.122)(0.147,0.122) on the right. Remark 1 Under full participation the parameter α would be irrelevant. A voter prefers candidate i exactly when Bi>B−iB_i>B_-i, that is when d−i−(1+α)di>di−(1+α)d−id_-i-(1+α)d_i>d_i-(1+α)d_-i, i.e. when (2+α)d−i>(2+α)di(2+α)\,d_-i>(2+α)\,d_i: the common factor cancels and she votes for the nearer candidate at every α≥0α≥ 0. If T≡1T≡ 1 the induced vote shares are therefore the classical ones and Claim 2.1 is untouched, median equilibrium included. Alienation is in this respect exactly like competitiveness (Remark 2): it has no bite of its own, and acts only through the participation decision. Everything in Section 4 is accordingly attributable to abstention. Competitiveness Vote share is not the only objective a candidate might have: the margin wi:=ui−u−iw_i:=u_i-u_-i is the natural objective when only victory and the size of the mandate matter. We span both with a single competitiveness parameter β∈[0,1]β∈[0,1]: candidate i values her own votes at 11 and the opponent’s at −β-β, Ui=ui−βu−i=(1−β)ui+βwi=1−β2T+1+β2wi.U_i\;=\;u_i-β\,u_-i\;=\;(1-β)\,u_i+β\,w_i\;=\; 1-β2\,T\;+\; 1+β2\,w_i. (5) The middle expression is the most direct reading of the parameter: UiU_i is the convex combination of vote share and margin that puts weight β on the margin. The one on the right shows that UiU_i is equally a weighted combination of turnout, which the candidates share, and margin, over which they are in pure conflict. At β=0β=0 we have Ui=uiU_i=u_i and we are back in the model of Section 2.3; at β=1β=1 the common turnout term drops out, the game is zero-sum, and candidates are margin maximizers. With either extension in play, the equilibrium notion of Section 2.4 is applied to the game with payoffs (U1,U2)(U_1,U_2), which we call the β-game; its two endpoints are the vote-share game (β=0β=0) and the margin game (β=1β=1). We write Δ∗(α) ^*(α) or Δ∗(α,β) ^*(α,β) when the dependence of the equilibrium separation on α and β matters. The reader can see how different parameters affect the turnout along the voter axis using the artifact: https://claude.ai/artifact/2Ggrf3obhnSywKHsyEs2BE Remark 2 Under full participation the parameter β would be irrelevant: if T≡1T≡ 1 then Ui=(1+β)ui−βU_i=(1+β)\,u_i-β is a positive affine transformation of uiu_i, so all β∈[0,1]β∈[0,1] induce the same game. In particular, vote-share and margin maximization coincide in the classical model. It is precisely abstention, which makes turnout T(c1,c2)T(c_1,c_2) respond to the platforms, that separates the objectives; how much they diverge is the subject of Section 5. 3 Costly Voting We begin with the baseline model: voters weigh only the gain from voting against its cost, and candidates maximize vote share. Both extensions of Section 2.5 are off throughout this section (α=β=0α=β=0, so that Bi=d−i−diB_i=d_-i-d_i and Ui=uiU_i=u_i), and every statement below is to be read under that convention, without further mention. This section establishes the pattern that the two extensions will repeat. Section 3.1 characterizes equilibria, starting from a single balance principle, stated here in its baseline form and extended in turn by each parameter, and specializing along a ladder of increasingly explicit cases: uniform costs, uniform voters, and both at once, which is the doubly-uniform environment of Section 2.4 that runs through the paper as its simplest worked example. Section 3.2 then extracts the paper’s central message: costly voting alone breaks median convergence, and it does so through two separate channels—the shape of the electorate and the shape of the cost distribution—while the level of turnout plays no role at all. Table 2 collects the shorthands used from here on. Table 2: Shorthands introduced in the analysis, collected for reference; each is defined where it first appears. The second block is used only from Section 4 on. The symbols of the model itself are in Table 1. Analysis shorthands r:=k/Kr:=k/K; G(t):=∫0tKG(t):= _0^tK reverse hazard rate of K (Corollary 4); integrated CDF ψ; ψβ _β virtual cost t+K/kt+K/k, Eq. 6; its β-weighting t+1+β1−βK/kt+ 1+β1-βK/k, Theorem 5.1 Ii,OiI_i,O_i marginal voter mass on i’s inner/outer arm, Lemma 2 σ; η(u)=uk(u)/K(u)η(u)=u\,k(u)/K(u) electorate spread; elasticity of K, Proposition 5 Δ¯(α) (α) turnout-maximizing separation, Proposition 6 cmedc_med; w(c,c′)w(c,c ) median V−1(12)V^-1( 12); margin of a candidate at c against one at c′c , Theorems 5.2 and 15 Geometry of the benefit tent under alienation (Fig. 6) λ:=2+αλ:=2+α inner-arm slope (outer arms have slope α) xiinx^in_i; xioutx^out_i inner and outer zero of candidate i’s tent, where her benefit reaches 00 b,b1,b2b,b_1,b_2 boundary benefits: BiB_i at the nearer edge of [0,1][0,1], Eq. 9 Δ0:=α/λ _0:=α/λ maximal separation inside the margin band E(α)E(α), Proposition 14 E(α)E(α) margin band: the set of pure equilibria at β=1β=1, Proposition 14 3.1 Equilibrium Characterization We first record that equilibria can involve neither ties nor boundary positions. Lemma 1 (Interiority) Under Assumption 2.2, every pure Nash equilibrium (c1,c2)(c_1,c_2) satisfies c1≠c2c_1≠ c_2 and, labeling the candidates so that c1<c2c_1<c_2, we have 0<c1<c2<10<c_1<c_2<1. Proof in Appendix 0.B. One balance principle All our characterizations flow from a single first-order condition, which we state here in its baseline form and sharpen once in each of the two sections that follow. It is what the partition (3) gives on differentiation. Define the marginal masses IiI_i and OiO_i on the two arms: for candidate 11, I1:=∫c1mk(B1(x))v(x)x,O1:=∫0c1k(B1(x))v(x)x=k(Δ)V(c1),I_1:= _c_1^mk (B_1(x) )\,v(x)\,dx, O_1:= _0^c_1k (B_1(x) )\,v(x)\,dx\;=\;k( )V(c_1), with the mirror-image definitions for candidate 22, and I:=I1+I2I:=I_1+I_2, O:=O1+O2O:=O_1+O_2. The voter at x contributes k(Bi(x))v(x)k(B_i(x))v(x): the density of voters at x who are exactly indifferent between voting and abstaining, and whom a marginal platform move therefore gains or loses. Moving c1c_1 acts on O¯1 O_1 and I¯1 I_1 in two ways at once. The benefits inside each region change: on the flank B1≡ΔB_1≡ falls at unit rate, costing O1O_1, while in the middle B1=c1+c2−2xB_1=c_1+c_2-2x rises at unit rate, earning I1I_1. And the boundary between the regions moves with her, so voters just behind c1c_1 pass from the middle into the flank. That second effect is the one that cancels. Each transferred voter carries the same benefit Δ on either side of the boundary, so she adds K(Δ)v(c1)K( )v(c_1) to the flank and removes exactly as much from the middle: ∂O¯1∂c1=K(Δ)v(c1)−O1,∂I¯1∂c1=I1−K(Δ)v(c1),so∂u1∂c1=I1−O1. ∂ O_1∂ c_1=K( )v(c_1)-O_1, ∂ I_1∂ c_1=I_1-K( )v(c_1), ∂ u_1∂ c_1=I_1-O_1. Neither I1I_1 nor O1O_1 is by itself the derivative of the region it names; each is the part of that derivative due to benefits changing, and the parts due to the partition shifting are equal and opposite. Setting the sum to zero gives the lemma. Lemma 2 (Balance conditions) Let v,kv,k satisfy Assumption 2.2. Every interior stationary profile (c1,c2)(c_1,c_2) (Definition 1) satisfies, for each candidate i, Ii=Oi,I_i\;=\;O_i, and in aggregate I=OI=O. Proof: the baseline case of Proposition 23. In aggregate, exactly half of the total marginal mass sits on the inner arms, for every v and k. Neither extension changes what is being balanced, only the rate at which the two arms exchange: alienation reshapes the arms themselves (Proposition 4) and competitiveness re-prices them, and once both are in play the two effects combine into a single statement, Lemma 3. The rest of this subsection specializes this balance one distribution at a time: first uniform costs against an arbitrary electorate, then a uniform electorate against arbitrary costs, then both at once.55 5 Each specialization gives conditions that are necessary but not sufficient, in the sense of Definition 1; Appendices 0.C and 0.F supply the two examples promised there. Remark 3 (Existence) Payoffs are continuous on [0,1]2[0,1]^2 (including at ties, where vote shares vanish), so a possibly-mixed equilibrium always exists by Glicksberg’s fixed-point theorem [39]. Pure existence is subtler. It holds in the benchmark settings of this section, and in each extension: • uniform voters, with K regular (Theorem 3.2); • uniform costs, with a symmetric electorate of density ratio at most four (Corollary 1); • a non-increasing cost density with ρk:=k(0+)/k(1−)<∞ _k:=k(0^+)/k(1^-)<∞ and a symmetric electorate whose density ratio ρv _v (Table 1) satisfies ρvρk≤4 _v _k≤ 4 (Proposition 16); • under alienation (Theorems 4.1 and 10) and under competitiveness (Theorems 5.1 and 0.N.1), on the analogous classes. Between the two lies a third statement, weaker than pure existence and available far more widely: a stationary profile—one solving the first-order conditions—exists by an intermediate-value argument along the symmetric diagonal, under no sufficiency hypothesis at all. For uniform voters and every K this is the boundary behaviour of Ψ in the proof of Theorem 4.1, which uses no super-regularity, so Eq. 12 always has a root. For uniform costs and a symmetric electorate, Proposition 24 gives the same conclusion for every α>0α>0 and every β<1β<1: the own-side derivative on the diagonal is (1+α+β)V(1/λ)>0(1+α+β)V(1/λ)>0 at c1=0+c_1=0^+ and negative just left of c1=12c_1= 12. What these arguments do not deliver is that the root is a best response, and closing that gap is exactly what the density-ratio umbrellas above do. We conjecture that a pure equilibrium always exists in the baseline (α=β=0α=β=0) under Assumption 2.2. The case the list leaves conspicuously open is uniform costs with ρv>4 _v>4, where we have neither a proof nor a counterexample: the four-cluster electorate of Appendix 0.F has ρv=120 _v=120 and does possess equilibria, even though not every profile solving the conditions is one. We expect existence to survive alienation as well, at β=0β=0, but on thinner evidence, since the range of density ratios covered by Corollary 10 shrinks to nothing as α grows. Section 5 shows that pure existence can fail once candidates maximize the margin instead. Uniform costs Assume first that costs are uniform while voters are arbitrary. With k≡1k≡ 1, the balance Ii=OiI_i=O_i turns into a statement about voter mass alone. Theorem 3.1 (Uniform costs: every equilibrium traps half the electorate) Let κ∼U(0,1)κ U(0,1) and let v satisfy Assumption 2.2. Then every Nash equilibrium (c1,c2)(c_1,c_2) satisfies V(c1)=V(c2)−12=12V(m):V(c_1)\;=\;V(c_2)- 12\;=\; 12\,V(m): the candidates trap exactly half of the electorate between them, and each candidate’s flank holds exactly half the mass on her side of the midpoint. Proof in Appendix 0.B. The one half is specific to vote-share maximization, but the law is not. Under the competitive objective Ui=ui−βu−iU_i=u_i-β\,u_-i of Eq. 5, with alienation still off, the same computation traps exactly 1−β2 1-β2 of the electorate, again whatever the electorate (Proposition 9). These conditions are not sufficient in general, but they are whenever the electorate is not too uneven. Let ρv:=sup(0,1)v/inf(0,1)v _v:= _(0,1)v\,/\, _(0,1)v denote the density ratio. Proposition 1 (Sufficiency under uniform costs) Let κ∼U(0,1)κ U(0,1) and ρv≤4 _v≤ 4. Then each candidate’s payoff is concave in her own position on each side of the opponent, so a profile 0<c1<c2<10<c_1<c_2<1 is a Nash equilibrium if and only if it satisfies Theorem 3.1 and neither candidate gains by a cross-over deviation, a move to the far side of her opponent. The cross-over condition is one inequality per candidate, comparing her equilibrium payoff with the best she could get on the far side. Proof in Appendix 0.B. Corollary 1 (Quartile equilibria) Let κ∼U(0,1)κ U(0,1) and ρv≤4 _v≤ 4. Any profile with V(c1)=14V(c_1)= 14, V(c2)=34V(c_2)= 34 and V(m)=12V(m)= 12 is a Nash equilibrium. If v is symmetric about 12 12, the quartile profile (V−1(14),V−1(34)) (V^-1( 14),V^-1( 34) ) satisfies all three conditions automatically (so an equilibrium exists) and is the only profile satisfying them. Proof in Appendix 0.B. Competitiveness generalizes the quartiles to quantiles: Corollary 8 puts the candidates at V−1(1+β4)V^-1 ( 1+β4 ) and V−1(3−β4)V^-1 ( 3-β4 ), which are the quartiles at β=0β=0 and collapse to the median as β↑1β 1. For skewed electorates the equilibrium detaches from any fixed quantiles: for v=Beta(1.5,4)v=Beta(1.5,4) the conditions of Theorem 3.1 pin down (c1,c2)≈(0.149,0.413)(c_1,c_2)≈(0.149,0.413), an asymmetric profile with V(c1)≈0.29V(c_1)≈ 0.29 and V(c2)≈0.79V(c_2)≈ 0.79, which we verified numerically to be an equilibrium; a reminder, too, that the bound ρv≤4 _v≤ 4 is sufficient but far from necessary (this density violates it). Nor need the solution be unique: Appendix 0.E exhibits a symmetric electorate with three distinct equilibria, two of them asymmetric mirror images, with ρv=3 _v=3: inside the sufficiency range ρv≤4 _v≤ 4 of Proposition 1 the quartile profile is an equilibrium, but not necessarily the only one. Uniform voters Now let voters be uniform while costs are arbitrary. This is our main result, and the characterization here is complete: a single monotonicity condition on the cost distribution delivers existence and uniqueness, and pins the equilibrium separation to the root of one scalar equation. That condition is monotonicity of the virtual cost ψ(t):=t+K(t)k(t),t∈(0,1),ψ(t)\;:=\;t+ K(t)k(t), t∈(0,1), (6) the cost-side counterpart of Myerson’s virtual value [68]. Definition 2 (Regularity) A distribution whose virtual cost is strictly increasing is called regular. Regularity of cost distribution is a standard hypothesis in economic settings [61]. Several results below ask for more: that the second term of Eq. (6) not decrease on its own. Call such a K super-regular. Since the first term increases, super-regularity implies regularity. Moreover, super-regularity is implied by basic properties shared by most commonly used distributions [11]: knon-increasing, or klog-concave⟹Ksuper-regular⟹Kregular,k\ non-increasing, or k\ log-concave K\ super-regular K\ regular, where neither implication reverses. In particular, Power costs K(t)=tpK(t)=t^p with p<1p<1 are super-regular although k is log-convex (Example 1), and Footnote 7 exhibits a regular K that is not super-regular. Theorem 3.2 (Uniform voters: every equilibrium, and when it is unique) Let x∼U(0,1)x U(0,1) and let k satisfy Assumption 2.2. (a) Every Nash equilibrium is symmetric, c1=1−c2c_1=1-c_2, with gap Δ∈(0,1) ∈(0,1) satisfying K(Δ)=(1−Δ)k(Δ),equivalentlyψ(Δ)=1.K( )=(1- )\,k( ), ψ( )=1. (7) (b) If K is regular, then Eq. 7 has a unique root Δ∗ ^*, and (1−Δ∗2,1+Δ∗2) ( 1- ^*2, 1+ ^*2 ) is the unique Nash equilibrium up to relabeling. Proof in Appendix 0.B. Without regularity, Eq. 7 may have several roots and, not all of them are equilibria: Appendix 0.C gives a cost density with three roots of which only the outer two are equilibria. Part (a) therefore cannot be reversed without an extra condition. A root of Eq. 7 exists for every k satisfying Assumption 2.2, as does a solution of the conditions of Theorem 3.1 for every v (Appendix 0.C); what is open (Remark 3) is only whether some solution is an equilibrium. Corollary 2 (The doubly-uniform benchmark) If x,κ∼U(0,1)x,κ U(0,1) then ψ(t)=2tψ(t)=2t, Δ∗=12 ^*= 12, and (14,34) ( 14, 34 ) is the unique Nash equilibrium up to relabeling; equilibrium turnout is 38 38. Proof With K(t)=tK(t)=t the cost density k≡1k≡ 1 satisfies Assumption 2.2 and ψ(t)=t+K(t)/k(t)=2tψ(t)=t+K(t)/k(t)=2t is strictly increasing, so Theorem 3.2(b) applies; Eq. 7 reads Δ=1−Δ =1- , giving Δ∗=12 ^*= 12 and the profile (14,34)( 14, 34). There, by Eq. 2, candidate 11’s share is the area under the positive part of her benefit tent (Fig. 1): the flank [0,14][0, 14] votes with probability K(Δ∗)=12K( ^*)= 12, contributing 18 18, and the inner arm contributes ∫1/41/2(1−2x)x=116 _1/4^1/2(1-2x)\,dx= 116. Hence u1=u2=316u_1=u_2= 316 and T=38T= 38. Corollary 2 is the benchmark every extension will be measured against, and it is the equilibrium depicted in Fig. 2. Beyond the two benchmarks Sufficiency does not stop at the uniform cases. A stationary profile is an equilibrium once each candidate’s payoff is single-peaked on her own side of the opponent, so that her stationary point is the best position available there, and neither candidate gains by crossing to the far side. Both hold when the two distributions are jointly not too uneven: for a non-increasing cost density, ρvρk≤ 4 _v\, _k\;≤\;4 suffices, and stationarity together with the two cross-over checks then characterizes equilibrium exactly as it does under uniform costs (Proposition 16). The two ratios multiply, so the criterion degrades gracefully: it is ρv≤4 _v≤ 4 when costs are uniform (Proposition 1), and for a uniform electorate it asks only that k be non-increasing, one of the cases in which K is regular (Theorem 3.2). For a symmetric electorate the cross-over checks come free: own-side optimality alone makes a mirror profile an equilibrium, whatever the cost distribution (Lemma 6). What is not covered is a cost density that rises and falls, where the second derivative picks up terms in k′k of indefinite sign and no bound in ρv _v and ρk _k alone can absorb them (Remark 22); for a uniform electorate that case is settled anyway, by Theorem 3.2. 3.2 Polarization Costly voting opens two routes from primitives to candidate polarization: who the voters are, and what voting costs look like. We take them in turn. Mass polarization drives elite polarization Under uniform costs, Corollary 1 makes the quartile profile of any symmetric electorate with ρv≤4 _v≤ 4 an equilibrium, so at that equilibrium the separation equals the interquartile range of the voter distribution. Mass polarization therefore transmits to the candidates one for one: a hollow electorate, with more mass near the extremes than at the center, has a wider interquartile range and a correspondingly wider equilibrium, while a peaked one contracts both (Fig. 4). Under Claim 2.1 the shape of the electorate is irrelevant and both candidates sit at the median, so the whole dependence is an artifact of costly participation. Dreyer and Bauer [29] document this direction across eleven Western European party systems (1977–2016): parties’ positions grow more extreme as their electorates polarize, and the effect runs through the abstention margin. 000.20.20.40.40.60.60.80.811000.50.5111.51.5Δ∗=0.357 ^*=0.357Δ∗=0.643 ^*=0.643policy position xxvoter density v(x)v(x)peaked electoratehollow electorate Figure 4: Mass polarization drives elite polarization under uniform costs (Corollary 1). Two symmetric electorates with the same density ratio ρv=3 _v=3: v(x)=1∓12cos(2πx)v(x)=1∓ 12 (2π x). The quartile profile (marks) is an equilibrium (Corollary 1), so the hollow electorate—more mass near the extremes, wider interquartile range—supports an equilibrium nearly twice as polarized, even though nothing else in the environment changed. Elite polarization without mass polarization The second channel is the one absent from the classical model altogether: fix the electorate (uniform, so entirely unpolarized, with its median at 12 12) and vary only the cost distribution. Before comparing cost distributions it is worth asking which features of K the model can respond to at all. The answer is sharp, and it is not the level of turnout. Throughout write r(t):=k(t)/K(t)r(t):=k(t)/K(t) for the reverse hazard rate of K and G(t):=∫0tKG(t):= _0^tK, so that ψ=t+1/rψ=t+1/r and super-regularity says exactly that r is non-increasing.66 6 Throughout, the reverse hazard rate means k/Kk/K, not the hazard rate k/(1−K)k/(1-K) of the monotone-hazard-rate condition familiar from mechanism design; every monotonicity hypothesis below concerns k/Kk/K (equivalently K/kK/k). Proposition 2 (Scale invariance) Fix any voter distribution V, and let z>0z>0 be such that zKzK is again an admissible cost c.d.f. on [0,1][0,1] (i.e. zK(1)≤1zK(1)≤ 1; the residual mass sits above 11 and consists of voters who never participate). Then the games with cost distributions K and zKzK have identical best-response correspondences, hence identical sets of pure and mixed equilibria. All vote shares—and therefore turnout—are multiplied by z. The argument uses neither α nor β, so the statement survives both extensions unchanged; Proposition 7 records it in that generality. Proof By Eq. 2, ui(c1,c2)=∫01K(Bi(x)+)v(x)xu_i(c_1,c_2)= _0^1K (B_i(x)^+ )v(x)\,dx is linear in K, so replacing K by zKzK multiplies u1,u2u_1,u_2, and hence each candidate’s payoff, by the constant z>0z>0. Positive affine transformations of payoffs leave argmax unchanged. Corollary 3 The equilibrium set depends on K only through its reverse hazard rate r. Indeed logK(t)=logK(1)−∫t1r K(t)= K(1)- _t^1r, so two cost distributions with the same reverse hazard rate are positive multiples of one another, and by Proposition 2 they induce the same equilibria. By Proposition 7 the same is true at every (α,β)(α,β). Corollary 3 explains why the reverse-hazard-rate order [80] is the right order for the comparative static of Proposition 3 below (we call it the rhr order): it is precisely the natural order on the equivalence classes of cost distributions that the model can distinguish. It is the stricter of the two orders one might reach for: if k~/K~≥k/K k/ K≥ k/K pointwise and K~(1)≤K(1) K(1)≤ K(1), which holds in particular when both are proper distributions on [0,1][0,1], then K~≤K K≤ K pointwise, so rhr-higher costs are stochastically higher too. The converse fails, and Remark 4 is the counterexample. It also fixes the form any polarization claim here can take: a true one must be a statement about the shape of r, never about the level of costs or of turnout, since both of those can be moved without moving the equilibrium at all. What the reverse hazard rate measures When the benefit at stake is t, a fraction K(t)K(t) of the electorate already votes, and r(t)=k(t)K(t)=dtlogK(t)r(t)\;=\; k(t)K(t)\;=\; ddt K(t) is the share of those voters who are roughly indifferent: raise the stake by ε and the number who turn out rises by a factor of about r(t)εr(t) . So r is the proportional responsiveness of turnout to the stakes. A large r means an electorate poised at the margin; a small r a committed one, whose participation barely notices what is at stake. This is exactly the quantity a candidate can trade on. Take uniform voters and consider candidate 1 widening the gap by moving out. She excites her own flank: every voter behind her faces the same benefit Δ , so the mass 1−Δ2 1- 2 behind her yields extra turnout at rate k(Δ)k( ). She also cedes the contested middle [c1,c2][c_1,c_2] (the voters lying between the candidates), losing turnout K(Δ)/2K( )/2. Balancing the two gives Eq. 7, which rearranges into a transparent form. Corollary 4 (Equilibrium as a crossing) Let x∼U(0,1)x U(0,1) and let k satisfy Assumption 2.2. A gap Δ solves Eq. 7, equivalently ψ(Δ)=1ψ( )=1, if and only if r(Δ)=11−Δ.r( )\;=\; 11- . (8) If K is super-regular the left side falls while the right side rises, so this is the unique equilibrium. The two sides of Eq. 8 move in opposite directions, and that is what makes both the crossing and the comparative static easy to see. As the gap widens, the flank that stands to gain shrinks to mass 1−Δ2 1- 2, so the right side rises: each further step out has to be paid for by a larger proportional response. The left side is what the cost distribution actually delivers at that gap. An electorate whose r lies above another’s meets the rising requirement further out, so its crossing sits further out too (Fig. 5), which is the next proposition. That hypothesis is super-regularity, and it is strictly stronger than what Theorem 3.2(b) needs: regularity permits K/kK/k to fall, provided it falls more slowly than t rises—equivalently r′<r2r <r^2, so r may rise as long as it rises slowly.77 7 For an explicit witness, prescribe ψ(t)=2t+34πsin(2πt)ψ(t)=2t+ 34π (2π t), the doubly-uniform ψ(t)=2tψ(t)=2t perturbed, and recover K from r=1/(ψ(t)−t)r=1/(ψ(t)-t). Then ψ′=2+32cos(2πt)≥12ψ =2+ 32 (2π t)≥ 12, so K is regular and satisfies Assumption 2.2, while r increases wherever cos(2πt)<−23 (2π t)<- 23. Here ψ(12)=1ψ( 12)=1, so the equilibrium gap is exactly 12 12. This is why we keep ψ alongside r: the picture needs a falling left-hand side, but Theorem 3.2(b) does not, and the sharp hypothesis is the one stated in terms of ψ. Proposition 3 (More responsive participation polarizes) Let x∼U(0,1)x U(0,1) and let k,k~k, k both satisfy the hypotheses of Theorem 3.2(b), with equilibrium gaps Δ∗,Δ~∗ ^*, ^*. If k~/K~≥k/K k/ K≥ k/K pointwise on (0,1)(0,1), then Δ~∗≥Δ∗ ^*≥ ^*. Proof in Appendix 0.B. The hypothesis is far stronger than the conclusion needs. Since ψ~ ψ is strictly increasing, Δ~∗≥Δ∗ ^*≥ ^* holds exactly when r~(Δ∗)≥r(Δ∗)=1/(1−Δ∗) r( ^*)≥ r( ^*)=1/(1- ^*): only the two electorates’ responsiveness at the incumbent gap matters (Fig. 5, dashed rule). It is therefore enough for the rhr order to hold near Δ∗ ^*, and the pointwise order on all of (0,1)(0,1) is merely the cleanest sufficient condition---strictly stronger than the conclusion needs.88 8 A witness, against the uniform baseline K(t)=tK(t)=t, where ψ(t)=2tψ(t)=2t and Δ∗=12 ^*= 12. Prescribe ψ~(t)=2t−3t(t−310)(710−t) ψ(t)=2t-3t(t- 310)( 710-t) and recover K~ K from r~=1/(ψ~(t)−t) r=1/( ψ(t)-t); since ψ=t+1/rψ=t+1/r, the rhr order is exactly ψ~≤ψ ψ≤ψ, so the cubic makes it hold on (310,710)( 310, 710) and fail outside. Assumption 2.2 holds and ψ~′≥1.63 ψ ≥ 1.63, so K~ K is regular; r~(12)=2511>2 r( 12)= 2511>2, and the equilibrium gap moves out to 0.53110.5311. The conclusion of Proposition 3 holds although its hypothesis fails on 60%60\% of (0,1)(0,1). 0.30.30.40.40.50.50.60.60.70.70.80.800224466only the orderhere mattersplatform gap Δ hazard rate r(Δ)r( )required: 1/(1−Δ)1/(1- )K(t)=t2K(t)=t^2 (costliest)K(t)=tK(t)=t (uniform)K~(t)=12t(2−t) K(t)= 12t(2-t) Figure 5: Equilibrium as a crossing (Corollary 4). The rising black curve is the responsiveness a candidate needs in order to justify a gap Δ ; the falling curves are the responsiveness each cost distribution supplies. Equilibrium is where they meet (dots), at Δ∗=23 ^*= 23, 12 12 and 1−131- 1 3. Raising the whole supply curve—costs higher in the reverse-hazard-rate order—slides the crossing to the right: Proposition 3. The three curves here are ordered along the whole axis, but that is more than the comparison needs: it is settled at the incumbent crossing alone (dashed rule), which is why the pointwise hypothesis of Proposition 3 can be weakened to a neighbourhood. Note K~ K lies below K(t)=tK(t)=t here even though it puts costs stochastically higher (Remark 4). The logic behind Proposition 3 is not that costly elections have fewer voters, but that they have more responsive ones. When costs are high in the reverse-hazard-rate sense, the voters who still turn out are concentrated just above their indifference point, and a candidate who moves outward converts many of them; the flank is worth fighting for, and she goes to get it. When costs are low, her flank votes for her regardless, moving out buys almost nothing, and she stays near the center to contest the middle. What polarizes candidates is a flank that is nearly indifferent, not a flank that is small. Theorem 4.2 carries the comparison to every α>0α>0, so this is the α=0α=0 case of a channel that alienation does not close. It also settles a claim about turnout itself that would be natural to make, and that is false. Remark 4 (Turnout per se is not the driver) By Proposition 2, taking z<1z<1 (which adds a mass 1−z1-z of never-voters, whose cost exceeds the largest benefit any profile can offer) drives turnout arbitrarily low while leaving the platforms exactly where they were. So “lower turnout ⇒ more polarization” cannot hold in this model. Nor does the first-order stochastic order suffice, and regularity does not rescue it: the two distributions below are both regular, with ψ(t)=2tψ(t)=2t and ψ~′≥2 ψ ≥ 2 throughout. Let x∼U(0,1)x U(0,1) and compare K(t)=tK(t)=t with K~(t)=12t(2−t),k~(t)=1−t K(t)= 12\,t(2-t), k(t)=1-t (half of the electorate never votes and the rest have decreasing cost density). Then K~≤K K≤ K pointwise, so costs are stochastically higher (the mean cost rises from 12 12 to at least 23 23, the never-voters’ costs lying above every attainable benefit), and equilibrium turnout falls from 0.3750.375 to 0.2690.269; yet the equilibrium gap shrinks, from Δ∗=12 ^*= 12 to Δ∗=1−13≈0.4226 ^*=1- 1 3≈ 0.4226. Higher costs, lower turnout, less polarization. There is no contradiction with the comparative static (K~ K is lower than K in the reverse-hazard-rate order), but the logic must be described as a statement about the shape of K, not about the level of participation. The two distributions in Remark 4 sit on either side of K(t)=tK(t)=t in Fig. 5 in the way the logic predicts: K~ K has a decreasing density, so its remaining voters are the committed ones rather than the marginal ones, its supply curve lies below, and its candidates converge, even though fewer people vote. Turnout and polarization are both outputs of the model; neither drives the other, and it is the shape of K that drives both. Example 1 (Power costs: separation equals the mean cost) Let x∼U(0,1)x U(0,1), as throughout this subsection. For K(t)=tpK(t)=t^p with p>0p>0 the reverse hazard rate r(t)=p/tr(t)=p/t is increasing in p, and Eq. 7 solves to Δ∗=p1+p=[κ]: ^*= p1+p=E[κ]: the equilibrium separation equals the mean voting cost exactly, for every p. Raising p raises r pointwise, so Proposition 3 applies; on this family its conclusion can be read straight off the formula. Equilibrium turnout on this family is T=(1−Δ∗)K(Δ∗)+∫0Δ∗K=(Δ∗)p1+2p(1+p)2T=(1- ^*)K( ^*)+ _0 ^*K=( ^*)^p 1+2p(1+p)^2, which falls as p rises: (Δ∗,T)=(13,0.513)( ^*,T)=( 13,0.513) at p=12p= 12, (12,38)( 12, 38) at p=1p=1, (23,2081)( 23, 2081) at p=2p=2 and (34,0.185)( 34,0.185) at p=3p=3. A single shift in the cost distribution therefore reproduces both trends of Section 1 at once—rising polarization alongside falling participation—without either one causing the other. Corollary 1 put an equilibrium at the quartiles. Within the power family the same construction reaches any symmetric pair of quantiles. Remark 5 (Prescribing the equilibrium profile) Let x∼U(0,1)x U(0,1) and fix a target profile (q,1−q)(q,1-q) with q∈(0,12)q∈(0, 12). Take the power costs Kq(t)=tpq,pq:=12q−1=1−2q2q∈(0,∞).K_q(t)\;=\;t^\,p_q, p_q\;:=\; 12q-1\;=\; 1-2q2q\;∈\;(0,∞). Then ψ(t)=t+Kq(t)/kq(t)=(1+1pq)tψ(t)=t+K_q(t)/k_q(t)= (1+ 1p_q )\,t is linear and strictly increasing, so Theorem 3.2(b) applies: the game has a unique equilibrium, and by Example 1 its gap is Δ∗=pq1+pq=1−2q ^*= p_q1+p_q=1-2q. The candidates therefore sit at exactly c1∗=q,c2∗= 1−q,c_1^*\;=\;q, c_2^*\;=\;1-q, and the mean voting cost is 1−2q1-2q, the separation itself. So every interior symmetric profile is the unique equilibrium of exactly one member of the family, and pqp_q is strictly decreasing in q: the quartiles of Corollary 1 are q=14q= 14, where pq=1p_q=1 and costs are uniform; q↑12q 12 sends pq↓0p_q 0 and the platforms to the median; q↓0q 0 sends pq→∞p_q→∞ and the platforms to the endpoints. Those two limits are approached but never attained, since neither p=0p=0 nor p=∞p=∞ is an admissible cost distribution, which is the sense in which costly voting breaks median convergence for every member of the family. Two caveats delimit the statement. The distribution-freeness of Corollary 1 (a quartile equilibrium for every symmetric electorate with ρv≤4 _v≤ 4) is special to p=1p=1, because uniform costs make k constant and the balance condition of Lemma 2 then compares voter mass alone; for p≠1p≠ 1 the marginal masses are k-weighted and the equilibrium quantile depends on the shape of v. And the family one reaches for first, scaling costs up and down (Kz=zK_z=zK), cannot do this at all: rescaling leaves the equilibrium exactly where it was (Proposition 2), so any family that moves the quantile must move the reverse hazard rate. Power costs against a non-uniform electorate fall outside every sufficiency result above, yet the profile is an equilibrium in each case we checked, and the quantile stays close to the uniform-voter value; Appendix 0.D records what the computation shows and what is left to prove. Which way does causality run? The feedback is one-directional. Higher costs (in the sense that the model can detect, Corollary 3) polarize the candidates; conversely, a wider gap raises every voter’s benefit from participating, so turnout is increasing in the imposed separation: polarization raises turnout, never the reverse. Abstention is therefore a cause of candidate divergence here, not a consequence of it. This one-directionality is special to the baseline: under alienation the turnout curve bends back (Proposition 6), and the reverse channel (extreme candidates depressing participation) reappears. 4 Alienated Voters We now turn the first dial. Throughout this section α>0α>0 while candidates still maximize vote share, so the convention is β=0β=0. Alienation is a second route to divergence—not an independent one: without costly voting it would change nothing at all, since the factor 2+α2+α cancels out of the comparison between the candidates and the median equilibrium survives (Remark 1). What it changes is who turns out, and its logic can be stated in one sentence: on the contested middle [c1,c2][c_1,c_2] the two distances sum to the constant d1(x)+d2(x)≡Δd_1(x)+d_2(x)≡ , so a penalty on the distance to one’s own candidate bites hardest exactly at the midpoint—and pushes the candidates apart. The tent geometry (outer arms of slope α, inner arms of slope 2+α2+α, a dead zone of abstention around m) is laid out in Fig. 6. The section has three parts. Section 4.1 characterizes the equilibrium at a fixed α: the balance principle survives in a general form (Proposition 4), and for uniform voters the equilibrium is unique and symmetric (Theorem 4.1). Section 4.2 carries the comparative statics of Section 3.2 to a fixed α: the cost channel and the transmission of mass polarization both survive (Theorems 4.2 and 5). Section 4.3 contains the section’s point: alienation polarizes, with the gap strictly increasing in α and tending to full separation (Theorem 4.3), and it does so at a fixed level of participation (Corollary 6), so this is not the cost channel of Section 3 renamed. Two observations of Section 3 do not survive alienation, and each is flagged where it falls: the distribution-free quartile equilibrium (Remark 7) and the one-directional link from polarization to turnout (Proposition 6), the second of which is what lets the model speak to the empirical literature. 00c1c_1x1inx^in_1mmx2inx^in_2c2c_21100b1b_1b2b_2Δ dead zone(nobody votes)O1O_1outer arm, slope α arm, slope λ 1I_1candidate 22’s tent (mirror)voter position xxbenefit Bi(x)B_i(x) Figure 6: Anatomy of the benefit tents at the profile (c1,c2)=(0.35,0.8)(c_1,c_2)=(0.35,0.8) with α=1α=1. Candidate 11’s tent (solid) peaks at height Δ over c1c_1; its inner arm falls at slope λ=2+αλ=2+α to the inner zero x1in=c1+Δ/λx^in_1=c_1+ /λ, and its outer arm at slope α toward the outer zero x1out=c1−Δ/αx^out_1=c_1- /α, here off the policy space, so the arm is truncated at the boundary with boundary benefit b1=B1(0)b_1=B_1(0). Candidate 22’s tent (dashed) is the mirror image, with inner zero x2inx^in_2 and boundary benefit b2=B2(1)b_2=B_2(1). Between the inner zeros lies the shaded inner dead zone (x1in,x2in)(x^in_1,x^in_2), where both benefits are negative and nobody votes; here x1out<0x^out_1<0, so there is no outer dead zone to draw. The marginal masses of Proposition 4 weight each point of an arm by k(B1(x))v(x)k(B_1(x))\,v(x): O1O_1 on the outer arm, I1I_1 on the inner. The baseline of Section 3 is the case α=0α=0, in which the outer arms are flat at height Δ , the inner zeros merge at m, and the dead zone vanishes. 4.1 Equilibrium at a Fixed α Alienation leaves the shape of the characterization intact: one first-order condition for an arbitrary electorate (Proposition 4), and, for a uniform one, a unique symmetric equilibrium (Theorem 4.1). Here and in Section 5 write λ:=2+α,b1:=B1(0)=c2−(1+α)c1,b2:=B2(1)=(1+α)c2−c1−αλ:=2+α, b_1:=B_1(0)=c_2-(1+α)c_1, b_2:=B_2(1)=(1+α)c_2-c_1-α for the boundary benefits, the height of each tent’s outer arm where it meets the edge of the policy space (see Fig. 6). At a symmetric profile c1=1−Δ2c_1= 1- 2 they coincide, b1=b2=b_1=b_2=b, with b=λΔ−α2,soΔ−b=α(1−Δ)2,b\;=\; λ -α2, -b\;=\; α(1- )2, (9) and b≥0b≥ 0 exactly when Δ≥α/λ ≥α/λ. Alienation opens two regions in which nobody votes, and only one of them is a case the analysis has to track. The inner dead zone between the tents’ inner zeros is present at every profile with c1<c2c_1<c_2 and α>0α>0. An outer dead zone behind a candidate, where her most distant supporters abstain, appears only when her tent’s outer zero falls inside the policy space, and it is that case distinction the first-order conditions must carry. From here on “dead zone” unqualified means the outer one (Fig. 6). Lemma 1 survives alienation: every equilibrium is untied and interior for every α≥0α≥ 0 and every (v,k)(v,k) (Lemma 4). The master condition, in general form One first-order condition covers alienation for every electorate and every cost distribution, whether or not each candidate’s tent reaches the edge of the policy space. One might hope that at equilibrium each tent reaches the boundary: for candidate 11, that the outer zero x1out=c1−Δ/αx^out_1=c_1- /α of her tent (Fig. 6) satisfies x1out≤0x^out_1≤ 0, i.e. (1+α)c1≤c2(1+α)c_1≤ c_2, so that there is no outer dead zone, and its mirror for candidate 22. For concentrated electorates this fails: the equilibrium tents do detach from the boundary (Remark 6). The first-order condition valid in both regimes is the following. Proposition 4 (Master first-order condition) Under alienation, every interior best response of candidate 1 satisfies ∫c1x1ink(B1(x))v(x)dx=∫x1out∨0c1k(B1(x))v(x)dx,xin1=c1+Δλ,xout1=c1−Δα, gathered _c_1^x^in_1\!k (B_1(x) )v(x)\,dx\;=\; _x^out_1 0^c_1\!k (B_1(x) )v(x)\,dx,\\ x^in_1=c_1+ λ, x^out_1=c_1- α, gathered (10) with the mirror condition for candidate 2; here a∨ba b and a∧ba b are the larger and the smaller of a and b, so x1out∨0x^out_1 0 is the outer zero clipped to the policy space. For uniform costs this reads V(x1in)+V(x1out∨0)=2V(c1).V(x^in_1)+V(x^out_1 0)=2V(c_1). (11) When x1out≤0x^out_1≤ 0 (no dead zone) V(x1out∨0)=0V(x^out_1 0)=0 and Eq. 11 reduces to V(x1in)=2V(c1)V(x^in_1)=2V(c_1); when x1out>0x^out_1>0 the extra term is genuinely present. This is Lemma 2 again. The left side of Eq. 10 is I1I_1 and the right side is O1O_1, with each arm cut at the tent’s own zero instead of at the midpoint and the boundary; alienation moves the cuts, not the quantities being balanced. The limits collapse correctly at α=0α=0, where x1in=c1+Δ/2=mx^in_1=c_1+ /2=m and x1out→−∞x^out_1→-∞, so x1out∨0=0x^out_1 0=0. Lemma 3 records the same identification once competitiveness is in play as well. Proof in Appendix 0.A. Alienation also creates a case the baseline did not have—the outer zero can fall inside the policy space—and the first claim of this section needs that case to be harmless where the characterizations are proved. Remark 6 (Dead-zone equilibria exist) Under uniform voters dead zones never arise at a best response: if x1out≥0x^out_1≥ 0 the tent lies entirely inside [0,1][0,1] and u1=(1α+1λ)G(Δ)u_1= ( 1α+ 1λ )G( ) (recall G(t)=∫0tKG(t)= _0^tK) depends on c1c_1 only through Δ , so u1u_1 is strictly decreasing in c1c_1 on the dead-zone region c1∈[c21+α,c2)c_1∈[ c_21+α,c_2) and the candidate strictly prefers to move out of it. For non-uniform v this argument fails and dead-zone equilibria do occur. The supplementary analysis sharpens both halves: Proposition 17 shows a stationary dead zone forces ρv≥2+α _v≥ 2+α for any K, that uniform voters never have dead-zone equilibria for any α and any K, and, under uniform costs, gives the criterion V(2(1+α)λ2)>2V(1λ)V ( 2(1+α)λ^2 )>2V ( 1λ ) under which a symmetric v admits a stationary profile with a dead zone. For v=Beta(8,8)v=Beta(8,8) and uniform costs the symmetric stationary profile is unique, (0.3812,0.6188)(0.3812,0.6188) at α=2α=2 and (0.3748,0.6252)(0.3748,0.6252) at α=4α=4, and is an equilibrium (mutual global best responses, verified numerically; Remark 13); in both x1out>0x^out_1>0 (respectively 0.260.26 and 0.310.31), the condition V(x1in)=2V(c1)V(x^in_1)=2V(c_1) fails by 2.3⋅10−22.3· 10^-2 and 6.2⋅10−26.2· 10^-2, and Eq. 11 holds to 10−910^-9. The same happens for truncated normals with small variance and for Beta(20,20)Beta(20,20). The dead-zone case being real, sufficiency also changes shape. Call the umbrella of a setting the range of density ratios ρv _v within which the first-order conditions are known to characterize equilibria. The baseline umbrella ρv≤4 _v≤ 4 (Proposition 1) deforms to ρv≤min2λ1+α,λα _v≤ \ 2λ1+α,\, λα \—the first branch controlling concavity, the second the dead zone—with existence for symmetric electorates below the bound (Theorems 0.K.1 and 10). Both branches decrease in α: alienation shrinks the umbrella rather than widening it, foreshadowing a contrast with competitiveness (Table 4). Uniform voters: a one-line characterization For a uniform electorate the characterization collapses to a single equation in the reverse hazard rate, Eq. 12 below, whose unique root is the equilibrium gap of Theorem 4.1. Under uniform voters every equilibrium is untied and interior (Lemma 4) and, by Remark 6, has no dead zone, so specializing Eq. 10 to v≡1v≡ 1 the two first-order conditions read K(bi)=2λK(Δ)K(b_i)= 2λK( ), i=1,2i=1,2. Hence K(b1)=K(b2)K(b_1)=K(b_2), so b1=b2b_1=b_2 (K is strictly increasing on [0,1][0,1]), and since b2−b1=α(c1+c2−1)b_2-b_1=α(c_1+c_2-1), for α>0α>0 every equilibrium is symmetric (this is the necessity step of Theorem 0.L.2 at β=0β=0). Its gap therefore solves λK(b)=2K(Δ),equivalently∫bΔr(t)t=logλ2=log(1+α2),λ\,K(b)=2\,K( ), _b r(t)\,dt\;=\; λ2\;=\; (1+ α2 ), (12) with b as in Eq. 9. The second form makes Corollary 3 visible: only r enters. At α=0α=0 both sides vanish; dividing by α and letting α↓0α 0 recovers K(Δ)=(1−Δ)k(Δ)K( )=(1- )k( ), the baseline condition (7), as it must. The theorem below is therefore stated for α>0α>0; at α=0α=0 the model is the baseline, and under super-regularity its unique equilibrium is the one of Theorem 3.2(b) (Corollary 4).99 9 A reader who simply substitutes α=0α=0 into Eq. 12 gets the identity 2K(Δ)=2K(Δ)2K( )=2K( ) and no information; the content is at first order in α. Theorem 4.1 (Uniform voters: the equilibrium at each α) Let x∼U(0,1)x U(0,1) and let k satisfy Assumption 2.2 with K super-regular. Then for every α>0α>0, Eq. 12 has a unique root Δ∗(α)∈(αλ,1) ^*(α)∈ ( αλ,1 ), and the profile (1−Δ∗2,1+Δ∗2) ( 1- ^*2, 1+ ^*2 ) is the unique Nash equilibrium up to relabeling. Proof in Appendix 0.A. Super-regularity is used only for uniqueness and for the equilibrium property: the boundary values of Ψ in the proof use no monotonicity, so Eq. 12 has a root for every K. Corollary 5 If x,κ∼U(0,1)x,κ U(0,1) then Δ∗(α)=2+α4+α ^*(α)= 2+α4+α, attained at the symmetric profile (14+α,3+α4+α) ( 14+α, 3+α4+α ), with equilibrium turnout 2(3+α)(4+α)2 2(3+α)(4+α)^2. At α=0α=0 this is Corollary 2: the gap is 12 12, the profile is (14,34) ( 14, 34 ), and turnout is 38 38. The benchmark is the left endpoint of a one-parameter family, and every quantity in it moves monotonically from there. Remark 7 (Uniform costs: no fixed-quantile analogue) [numerical observation] The quartile profile of Corollary 1 is the same for every symmetric electorate, and alienation removes the reason. Under uniform costs Eq. 11 reads V(x1in)+V(x1out∨0)=2V(c1)V(x^in_1)+V(x^out_1 0)=2V(c_1). At α=0α=0 the inner zero is the midpoint and the outer zero lies off the policy space, so for a symmetric electorate the condition is 12=2V(c1) 12=2V(c_1): the distribution enters only through V(m)=12V(m)= 12, and every symmetric electorate lands on the quartiles. For α>0α>0 the inner zero moves inward to x1in=c1+Δ/λx^in_1=c_1+ /λ, which moves with c1c_1, so V enters in its own right and the equilibrium quantile becomes a property of the electorate. Across uniform, Beta(2,2)Beta(2,2), Beta(5,5)Beta(5,5), arcsine and cosine-peaked electorates, V(c1∗)V(c_1^*) agrees to 10−1210^-12 at α=0α=0 and then separates: it spans [0.194,0.207][0.194,0.207] at α=1α=1 and [0.125,0.153][0.125,0.153] at α=4α=4, against 1/(4+α)1/(4+α) for uniform voters (Corollary 5). Each of those profiles is still an equilibrium—we verified mutual global best responses—so what alienation costs is the distribution-freeness of the quantile, not the equilibrium. 4.2 Costs and Polarization at a Fixed α At a fixed α the comparative statics of Section 3.2 carry over—the cost channel (Theorem 4.2) and the transmission of mass polarization (Proposition 5)—while the one-directional link from polarization to turnout does not (Proposition 6). The cost channel of Section 3.2 survives alienation intact: at every α the two forces push the same way. Theorem 4.2 (The cost channel survives alienation) Let x∼U(0,1)x U(0,1) and let k,k~k, k both satisfy the hypotheses of Theorem 4.1. If r~≥r r≥ r pointwise on (0,1)(0,1)—i.e. K~ K is higher than K in the reverse-hazard-rate order—then Δ~∗(α)≥Δ∗(α) ^*(α)≥ ^*(α) for every α>0α>0, and for α=0α=0 by Proposition 3. As there, the comparison is also one of stochastically higher costs whenever K~(1)≤K(1) K(1)≤ K(1). Proof Ψ~(Δ)=∫b(Δ)Δr~−log(λ/2)≥Ψ(Δ) ( )= _b( ) r- (λ/2)≥ ( ) for every Δ , and both are strictly decreasing with a unique root (proof of Theorem 4.1). Hence Ψ~(Δ∗)≥Ψ(Δ∗)=0=Ψ~(Δ~∗) ( ^*)≥ ( ^*)=0= ( ^*) and therefore Δ~∗≥Δ∗ ^*≥ ^*. At α=0α=0 this is exactly the baseline comparative static; the point is that neither force switches the other off: at any fixed α, higher costs still polarize, and at any fixed K, stronger alienation still polarizes. (The theorem orders levels, not increments: along the power family K(t)=tpK(t)=t^p the gain in Δ∗ ^* from raising α is smaller at larger p.) Mass polarization under alienation Remark 7 leaves this section unable to say where the equilibrium sits for a general electorate. It can still say exactly how that equilibrium moves when the electorate itself polarizes—how much elite polarization a given amount of mass polarization buys, and what fixes the exchange rate. This is a question about V, not about α: the answer below holds at every α, the baseline included. Proposition 5 (Mass polarization scales elite polarization) Let VσV_σ be the law of 12+σ(a−12) 12+σ(a- 12) where a∼V0a V_0 and σ∈(0,1]σ∈(0,1], so that σ measures the spread of the electorate (for σ<1σ<1 the support of VσV_σ is [1−σ2,1+σ2][ 1-σ2, 1+σ2], outside which Assumption 2.2 fails; the proof first shows that no best response lies there). Then the game (Vσ,K,α)(V_σ,K,α) is isomorphic to (V0,Kσ,α)(V_0,K_σ,α) under c=12+σ(a−12)c= 12+σ(a- 12), where Kσ(t):=K(σt)K_σ(t):=K(σ t). Consequently: (a) (power costs) for K(t)=tpK(t)=t^p (Example 1), Kσ=σpK_σ=σ^pK, so by Proposition 7 the equilibrium positions in a-space do not move and Δ∗(σ)=σ⋅Δ∗(1) ^*(σ)=σ· ^*(1) for every α≥0α≥ 0 and every V0V_0: elite polarization is exactly proportional to mass polarization (uniform costs are p=1p=1); (b) (general costs; a∼U(0,1)a U(0,1) and K super-regular, as in Theorem 4.1) the reverse hazard rate of KσK_σ is rσ(t)=η(σt)/tr_σ(t)=η(σ t)/t, where η(u)=uk(u)/K(u)η(u)=u\,k(u)/K(u) is the elasticity of K. If η is non-decreasing then σ↦rσ r_σ increases pointwise, so by Theorem 4.2 the equilibrium gap Δa(σ) _a(σ) of (V0,Kσ,α)(V_0,K_σ,α) increases and Δ∗(σ)=σΔa(σ) ^*(σ)=σ _a(σ) grows super-proportionally: a more dispersed electorate is amplified. If η is non-increasing, it is damped. Constant elasticity—the power family of part (a)—is the exact knife-edge. So the transmission from mass to elite polarization has a rate, and the cost distribution sets it: proportional under power costs, amplified when the elasticity of K rises and damped when it falls. It says nothing about whether α polarizes, and holds equally at α=0α=0. Proof in Appendix 0.A. In words: stretching a uniform electorate always produces more polarized candidates, at every level of alienation, and the cost distribution decides whether the transmission is one-for-one (constant elasticity, e.g. power-law costs), amplified or damped. An increasing cost elasticity (participation that grows relatively more responsive as the stakes rise) amplifies mass polarization; a decreasing one damps it; power costs transmit it exactly. Relative to the interquartile comparison of Section 3.2, which needed uniform costs, symmetry and ρv≤4 _v≤ 4, part (a) keeps only the power shape and holds at every α≥0α≥ 0, while part (b) drops that too, at the price of a uniform electorate; both compare affine stretches of a fixed electorate rather than arbitrary pairs of symmetric ones. At α=0α=0 the transmission has a one-line closed form: applying Corollary 4 to (V0,Kσ,0)(V_0,K_σ,0), the equilibrium gap solves r(Δ∗)=1σ−Δ∗,r( ^*)\;=\; 1σ- ^*\,, so widening the electorate relaxes the demand side of the crossing, and Δ∗ ^* is strictly increasing in σ whenever r is non-increasing. The elasticity of K governs only the rate of transmission, never its direction. (Uniform costs: Δ∗=σ/2 ^*=σ/2 exactly.) Turnout: the reverse channel reappears The one-directional causality of Section 3.2, in which polarization raises turnout and never the reverse, is an artifact of α=0α=0. Proposition 6 (Non-monotone turnout under alienation) Let x∼U(0,1)x U(0,1), let α>0α>0, and let T(Δ)=u1+u2T( )=u_1+u_2 be turnout at the symmetric profile with gap Δ≥α/λ ≥α/λ. Then T′(Δ)=2α[2(1+α)λK(Δ)−λ2K(b)],soT′(Δ)<0⇔K(b)K(Δ)>4(1+α)λ2.T ( )= 2α [ 2(1+α)λK( )- λ2K(b) ], T ( )<0 K(b)K( )> 4(1+α)λ^2. Consequently: (a) at any equilibrium, Eq. 12 gives K(b)/K(Δ)=2/λK(b)/K( )=2/λ and hence T′(Δ∗)=2λK(Δ∗)>0T ( ^*)= 2λK( ^*)>0: locally, more polarized candidates still raise turnout; (b) but for α>0α>0 turnout is not globally increasing: for every K, T′(1)=−αλK(1)<0T (1)=- αλK(1)<0, so the turnout curve turns over somewhere between the equilibrium and full separation. In the doubly-uniform benchmark T is single-peaked with peak at Δ¯(α)=(2+α)2α2+6α+4<1 (α)= (2+α)^2α^2+6α+4<1, and strictly decreasing on (Δ¯,1]( ,1]. E.g. Δ¯(1)=911≈0.818 (1)= 911≈ 0.818 while Δ∗(1)=0.6 ^*(1)=0.6. Proof in Appendix 0.A. 000.20.20.40.40.60.60.80.811000.10.10.20.20.30.30.40.40.50.5imposed separation Δ Tα=0α=0α=0.5α=0.5α=1α=1α=2α=2equilibrium Δ∗(α) ^*(α)turnout peak Δ¯(α) (α) Figure 7: Turnout against an imposed separation in the doubly-uniform benchmark (Proposition 6). Stars mark the equilibrium Δ∗(α) ^*(α), triangles the turnout peak Δ¯(α) (α). The equilibrium always lies on the rising branch, so locally polarization raises turnout; but once α>0α>0 the curve turns over, and candidates pushed beyond Δ¯ by forces outside the model depress participation. At α=0α=0 turnout is increasing throughout. Proposition 6 is the one result here whose implication for polarization is indirect, so it is worth stating plainly. It does not say that alienation lowers equilibrium polarization—Theorem 4.3 says the opposite. It says that the feedback from polarization to turnout, which is unambiguously positive in the baseline, changes sign once the gap grows beyond Δ¯(α) (α) (Fig. 7). Equilibrium candidates never get there on their own. But a model in which some outside force (primaries, activists, party discipline) pushes candidates past Δ¯ predicts falling turnout, and in that regime polarization and participation move in opposite directions: the pattern Rogowski [79] estimates from joint measures of voter preferences and candidate platforms in U.S. House and Senate races, where greater ideological conflict between the candidates lowers turnout. Alienation together with voting costs can thus explain how candidate polarization can both increase and decrease participation. Both directions come out of the same two parameters, so neither is an artifact of a separate modeling choice. 4.3 Higher α Polarizes The main result of this section is that higher alienation polarizes more: Theorem 4.3 (Alienation drives divergence, for every super-regular cost distribution) In the setting of Theorem 4.1, the equilibrium gap Δ∗(α) ^*(α) is strictly increasing in α, with Δ∗(α)→1 ^*(α)→ 1 as α→∞α→∞. Proof in Appendix 0.A. Theorem 4.3 is this section’s headline, and the general form of the doubly-uniform benchmark Δ∗(α)=2+α4+α ^*(α)= 2+α4+α of Corollary 5: stronger alienation strictly polarizes, from the baseline gap at α=0α=0 all the way to full separation as α→∞α→∞, for uniform voters and every super-regular cost distribution (Fig. 8). Alienation simultaneously depresses turnout—in the doubly-uniform benchmark, from 38 38 to 00—so the model produces the empirically familiar pairing of rising elite polarization with falling participation from a single voter-side parameter. For a uniform electorate the direction is unambiguous and needs no condition on the cost distribution beyond super-regularity: alienation increases polarization, always, and without bound. It is the strongest single force in the paper. The cost channel moves Δ∗ ^* within (0,1)(0,1) according to the shape of K, whereas alienation drives it to 11 regardless of K. The uniform-electorate hypothesis is not decorative: for the concentrated electorate of Remark 6 the equilibrium gap also grows with α (Δ∗=0.238 ^*=0.238 at α=2α=2 and 0.2500.250 at α=4α=4 for Beta(8,8)Beta(8,8) and uniform costs), but far more slowly, and we have no general-electorate theorem. 001122334455660.40.40.60.60.80.811alienation parameter α separation Δ∗ ^*k increasing (ϑ=2 =2)uniform costsBeta(2,5)Beta(2,5) costskk decreasing (ϑ=−2 =-2) Figure 8: Equilibrium separation under alienation for uniform voters and four cost distributions (Theorem 4.3). Here kϑ(t)=ϑt+(1−ϑ2)k_ (t)= t+ (1- 2 ), ϑ∈[−2,2] ∈[-2,2], is the linear cost family of Appendix 0.G. Every curve is increasing in α and tends to 11. The three linear-family curves are ordered as Theorem 4.2 predicts, rϑr_ being pointwise increasing in ϑ ; Beta(2,5)Beta(2,5) is not comparable with them in the reverse-hazard-rate order (its rate crosses the uniform one), so its position is not an instance of the theorem. The α=0α=0 intercepts are 23 23, 12 12, 0.4250.425 and 1−131- 1 3. Why alienation pushes candidates apart The logic is worth isolating, because it is different from the cost channel. At α=0α=0 a candidate’s flank is captive and uniform: every voter behind c1c_1 faces exactly the same benefit Δ , whatever their position. Alienation destroys that. For x<c1x<c_1, B1(x)=Δ−α(c1−x),B_1(x)= -α\,(c_1-x), so a candidate’s own extremists—the voters furthest behind her—are the ones with the least at stake and the first to abstain. She now has a second, quite separate reason to move outward: not only to widen Δ for the whole flank, but to go out and get her own flank, whose benefit rises at rate α as she approaches it. The tension in Corollary 4 is thereby shifted in favor of separation at every cost distribution, and Eq. 12 records the new balance: the boundary voter’s turnout rate must be held at the fixed fraction K(b)/K(Δ)=2/(2+α)K(b)/K( )=2/(2+α) of the flank’s, a fraction that falls as α grows and can only be restored by moving out. This is Hinich and Ordeshook’s alienation channel [46] recovered inside a costly-voting model. It is a distinct channel—one changes who is at the margin, the other how many—but the two are not independent: both branches of the sufficiency umbrella shrink as α grows, so alienation makes the cost channel harder to certify, not merely different from it. One reading of Theorem 4.3 has to be ruled out. Raising α lowers every voter’s benefit and therefore lowers turnout, so the result might be the cost channel of Section 3 under another name. It is not, and Remark 4 already says why: by Proposition 2 the level of participation can be set anywhere without moving the equilibrium at all, so “turnout fell, hence the candidates diverged” is not a mechanism this model has. The constructive version is immediate. Corollary 6 (Polarization at unchanged turnout) In the setting of Theorem 4.3, fix α>0α>0 and write T∗(α)T^*(α) for equilibrium turnout. Then there are positive multiples K0K_0 and KαK_α of K, both admissible, such that the games (V,K0,0)(V,K_0,0) and (V,Kα,α)(V,K_α,α) have the same equilibrium turnout while their equilibrium gaps are Δ∗(0) ^*(0) and Δ∗(α)>Δ∗(0) ^*(α)> ^*(0). Proof By Proposition 2, replacing K by zKzK for admissible z>0z>0 leaves the equilibrium set unchanged and multiplies turnout by z. If T∗(α)≤T∗(0)T^*(α)≤ T^*(0), take Kα=K_α=K and K0=zK_0=zK with z=T∗(α)/T∗(0)z=T^*(α)/T^*(0); otherwise take K0=K_0=K and Kα=zK_α=zK with z=T∗(0)/T∗(α)z=T^*(0)/T^*(α). Either way z≤1z≤ 1, so zK(1)≤K(1)≤1zK(1)≤ K(1)≤ 1 and zKzK is a cost distribution, and the two games have equal equilibrium turnout by construction. Neither scaling moves its own equilibrium, so the gaps are still Δ∗(0) ^*(0) and Δ∗(α) ^*(α), and the inequality is Theorem 4.3. Scaling the larger of the two turnouts down, rather than the smaller up, is what keeps z≤1z≤ 1; it costs nothing and avoids assuming that T∗T^* falls in α, which we have not proved (Proposition 6 concerns T as a function of Δ at fixed α). In the doubly-uniform benchmark the rescaling is explicit: by Corollary 5, T∗(α)=2(3+α)(4+α)2T^*(α)= 2(3+α)(4+α)^2 falls from 38 38, so z(α)=16(3+α)3(4+α)2z(α)= 16(3+α)3(4+α)^2, and at α=1α=1 both games turn out 825 825 of the electorate while the gap is 35 35 under alienation against 12 12 without it. The gap between the matched games can be much wider: for K(t)=t1/2K(t)=t^1/2 at α=4α=4 both turn out ≈0.313≈ 0.313, with gaps ≈0.692≈ 0.692 and 13 13. Telling the two channels apart Higher costs and stronger alienation both raise polarization and both depress turnout, so the two are observationally similar in aggregate data. They differ cross-sectionally: alienation reshapes who abstains—participation falls fastest among voters far from both candidates, opening the inner dead zone around m visible in Fig. 3, whereas a cost shift moves participation for ideologues and centrists alike. The model thus offers a concrete empirical signature separating “voting became harder” from “voters became disaffected”. The signature is a shape: since pi=K(Bi+)p_i=K(B_i^+) and BiB_i is a tent peaked at cic_i, alienation concentrates the active electorate into two humps centred on the candidates, falling away on both sides and vanishing outside [x1out∨0,x1in][x^out_1 0,x^in_1]. At α=0α=0 there is no outer edge to fall away from: the flank is flat at the full gain Δ , so the active electorate simply tracks v there. A cost shift scales that picture without changing its shape; alienation makes it bimodal. 5 Competitive Candidates The second extension changes the candidates rather than the voters: Ui=ui−βu−iU_i=u_i-β u_-i, spanning vote-share maximization (β=0β=0) to pure margin maximization (β=1β=1). Remark 2 showed that under full participation β would be irrelevant; everything in this section is therefore attributable to abstention. The message is that the separation between the two objectives is de-polarizing, that it runs against both channels of Sections 3 and 4, and that it is continuous in β up to a single knife-edge at β=1β=1. Accordingly we take β∈[0,1)β∈[0,1) throughout the first three parts, which follow those of Section 4 result for result, and treat the margin game β=1β=1 as a special case at the end. Section 5.1 characterizes the equilibrium at a fixed β: it is unique and symmetric (Theorem 5.1), and under uniform costs the mass-gap law locates it for any electorate (Propositions 9 and 8). Section 5.2 carries the comparative statics of Section 3.2 to a fixed β: the cost channel and the transmission of mass polarization survive (Propositions 10 and 11), and the turnout curve is the same at every β (Proposition 12). Section 5.3 contains the section’s point: the gap falls in β at every level of alienation (Proposition 13), and does so at a fixed level of participation (Corollary 9). At β=1β=1 (Section 5.4) this picture breaks: the equilibrium set becomes a band rather than a point (Proposition 14), the β↑1β 1 limit settles on its most polarized profile rather than the median (Theorem 5.3), the median fails as soon as the electorate is hollow at the centre (Theorem 5.4), and for some electorates no pure equilibrium survives at all (Proposition 15). With α and β both in play, the scale invariance of Section 3 can be stated in full generality. Proposition 7 (Scale invariance, general form) Proposition 2 holds verbatim for every α≥0α≥ 0 and every β∈[0,1]β∈[0,1]. Proof Neither parameter enters the proof of Proposition 2. Alienation acts only through BiB_i in Eq. 4, which does not involve K and is therefore unchanged; and for any β, multiplying u1u_1 and u2u_2 by the same constant z>0z>0 multiplies Ui=ui−βu−iU_i=u_i-β u_-i by z as well. Positive affine transformations of payoffs leave argmax unchanged. 5.1 Equilibrium at a Fixed β The logic is visible already in Lemma 2. A candidate has two ways to add to her tally: convert an abstainer on her own flank, or take a voter away from her opponent in the contested middle. The distributive-politics literature has long framed this choice as a targeting problem [23, 60, 24] and which Glaeser, Ponzetto and Shapiro [38] make the engine of candidate divergence. The two differ in their effect on the opponent. Converting an abstainer on one’s own flank means moving out, which widens the separation and so raises the benefit of voting—and with it the turnout—on the opponent’s flank as well: the first strategy actually helps the opponent. Taking a voter from the contested middle costs the opponent a vote outright. To a candidate who weighs the opponent’s votes negatively the two are therefore not merely unequal but opposite in sign. That is exactly the asymmetry the balance condition prices, and it can now be stated for every α and β. Lemma 3 (Balance conditions, general form) Let v,kv,k satisfy Assumption 2.2, and let α≥0α≥ 0, β∈[0,1)β∈[0,1). Every interior stationary profile (c1,c2)(c_1,c_2) of the game (Definition 1) satisfies, for each candidate i, (1+α)(Ii−Oi)+β(I−i+O−i)= 0.(1+α)\, (I_i-O_i )\;+\;β\, (I_-i+O_-i )\;=\;0. Summing and differencing the two conditions gives a ratio law and an asymmetry law, (1+α+β)I=(1+α−β)O,(1+α−β)(I1−I2)=(1+α+β)(O1−O2):(1+α+β)\,I=(1+α-β)\,O, (1+α-β)(I_1-I_2)=(1+α+β)(O_1-O_2): the aggregate inner share of marginal voters is the universal constant 1+α−β2(1+α) 1+α-β2(1+α), for every v and k. At (α,β)=(0,0)(α,β)=(0,0) this is Lemma 2; at β=0β=0 it is Proposition 4, summed over the two candidates. Proof: Proposition 23. The restriction to interior profiles matters only in the joint case: ties and boundary positions are excluded at β=0β=0 for every α, at α=0α=0 for every β<1β<1, and for uniform voters at every (α,β)(α,β) with β<1β<1 (Lemma 4); with α,β>0α,β>0 and a non-uniform electorate they are not excluded in general. So the inner marginal mass carries weight 1+α+β1+α+β and the outer mass weight 1+α−β1+α-β, and the equilibrium inner share falls from 12 12 at β=0β=0 (Lemma 2) toward α2(1+α) α2(1+α) as β↑1β 1. As competitiveness rises the contested middle becomes worth more relative to the flank, and the candidates move in to contest it: this is the paper’s only de-polarizing force. It also explains why β is inert under full participation (Remark 2): with nobody abstaining there is no flank to convert, every vote is taken from the opponent, and the distinction the parameter prices does not exist. What the model adds to that choice is a price: persuasion is worth 1+α+β1+α−β 1+α+β1+α-β units of mobilization, for every v, every K and every β<1β<1 (Lemma 3). It also corrects the folk version of the claim: a flank is worth mobilizing because its turnout is responsive, not because it is low (Propositions 7 and 4). Both extensions, non-uniform electorates With both extensions on and a non-uniform electorate the characterization is partial, and we record what is known. Under uniform costs the balance conditions of Lemma 3 still characterize equilibria, up to the two cross-over checks, whenever ρv≤minfβ,gβ _v≤ \f_β,g_β\ with the thresholds of Table 4 (Theorem 0.M.1); competition moves the two branches in opposite directions, widening the concavity branch fβf_β and narrowing the dead-zone branch gβg_β. A symmetric stationary profile exists for every β<1β<1 (Proposition 24), and for a symmetric electorate strictly inside the vote-share umbrella, with a unique symmetric stationary profile there, the symmetric equilibrium of Corollary 10 continues into small β>0β>0 (Theorem 0.M.2); numerically it persists to moderate β, inside and outside the umbrella (Remark 23). Competition also makes alienation dead zones easier to sustain: the density ratio above which a stationary profile can have one falls from λ/αλ/α at β=0β=0 to gβg_β (Proposition 25). Beyond uniform costs the concavity argument meets an obstruction that we isolate rather than resolve (Remark 22). Uniform voters For a uniform electorate the characterization again collapses to a single equation, Eq. 13 below, whose unique root is the equilibrium gap of Theorem 5.1. Proposition 8 (The stationarity condition for β<1β<1) Let x∼U(0,1)x U(0,1), α>0α>0, β∈[0,1)β∈[0,1), and let K satisfy Assumption 2.2. Every symmetric interior stationary profile of the β-game satisfies (1+α−β)K(b)=(1−β)2(1+α)2+αK(Δ).(1+α-β)\,K(b)\;=\;(1-β)\, 2(1+α)2+α\,K( ). (13) Proof in Appendix 0.A. Proposition 8 characterizes stationary profiles; under super-regularity its root is in fact the game’s unique equilibrium, so the characterization is not an artifact of restricting to symmetric or stationary profiles. Define ψβ(t):=t+1+β1−βK(t)/k(t) _β(t):=t+ 1+β1-β\,K(t)/k(t)—the baseline ψ with the term K/k=1/rK/k=1/r weighted by how much a converted opponent voter is worth relative to a converted abstainer. The α↓0α 0 limit of Eq. 13 then reads ψβ(Δ)=1 _β( )=1, which is Eq. 7 with competition priced in. (As in Eq. 12, substituting α=0α=0 outright yields an identity; the content is at first order in α.) The next theorem is the continuity half of the section’s message: below the knife-edge the equilibrium is unique, for every α≥0α≥ 0. Theorem 5.1 (Existence and uniqueness for every β<1β<1) Let x∼U(0,1)x U(0,1), β∈[0,1)β∈[0,1), and let K be super-regular. (a) If α=0α=0, the game has a unique Nash equilibrium: symmetric, with gap the unique root of ψβ(Δ)=1 _β( )=1. (b) If α>0α>0, the game has a unique Nash equilibrium: symmetric, with no dead zone, with gap the unique root of Eq. 13, and strictly more polarized than every symmetric margin-game profile: Δ∗(α,β)>α2+α ^*(α,β)> α2+α (Proposition 14). Proof: Theorems 0.L.1 and 0.L.2. Corollary 7 If x,κ∼U(0,1)x,κ U(0,1) then Δ∗(α,β)=α(2+α)(1+α−β)(1+α−β)(2+α)2−4(1−β)(1+α),Δ∗(0,β)=1−β2, ^*(α,β)= α(2+α)(1+α-β)(1+α-β)(2+α)^2-4(1-β)(1+α), ^*(0,β)= 1-β2, (14) the second entry being the α↓0α 0 limit of the first, equivalently the root of ψβ(Δ)=1 _β( )=1 for uniform costs. Proof Put K=idK=id and b=(λΔ−α)/2b=(λ -α)/2 in Eq. 13 and solve the resulting linear equation in Δ . At β=0β=0 this returns 2+α4+α 2+α4+α (Corollary 5), and as β→1β→ 1 it tends to α2+α α2+α, the top of the band of Section 5.4. Uniform costs, arbitrary electorates The quartile structure of Corollary 1 also generalizes, giving the distribution-free face of the same de-polarization. Proposition 9 (Mass-gap law) Let α=0α=0, κ∼U(0,1)κ U(0,1), β∈[0,1)β∈[0,1), and let v satisfy Assumption 2.2. Every Nash equilibrium satisfies V(c2)−V(c1)=1−β2,2V(c1)=V(m)+β(1−V(m)):V(c_2)-V(c_1)= 1-β2, 2V(c_1)=V(m)+β (1-V(m) ): the candidates trap voter mass exactly 1−β2 1-β2, independently of v. If moreover ρv≤41−β _v≤ 41-β, each candidate’s payoff is concave on her side, and equilibria are exactly the solutions passing the two cross-over checks. Proof: Proposition 22. Corollary 8 (Quantile equilibria) In the setting of Proposition 9 with v symmetric about 12 12, the quantile profile (c1∗,c2∗)(c_1^*,c_2^*) with V(c1∗)=1+β4V(c_1^*)= 1+β4, V(c2∗)=3−β4V(c_2^*)= 3-β4 is a Nash equilibrium under explicit bounds on v (Theorem 0.N.1); for convex symmetric electorates the single condition v(c2∗)/v(12)≤41−βv(c_2^*)/v( 12)≤ 41-β suffices. In particular, for the arcsine electorate v=Beta(12,12)v=Beta( 12, 12), c2∗=sin2(π(3−β)8),v(c2∗)v(12)=1sin(π(3−β)4)∈(1,2],c_2^*= ^2\! ( π(3-β)8 ), v(c_2^*)v( 12)= 1 ( π(3-β)4 )∈(1, 2\,], so a pure equilibrium exists for every β∈[0,1)β∈[0,1). Note how the two conditions of Proposition 9 interpolate the whole paper’s geometry: at β=0β=0 they are Theorem 3.1; as β↑1β 1 the trapped mass vanishes and the quantiles collapse onto the median of Theorem 5.2, moving from the quartiles to the median along an explicit path of quantile pairs. 5.2 Costs and Polarization at a Fixed β At a fixed β the comparative statics of Section 3.2 carry over as they do at a fixed α—the cost channel (Proposition 10) and the transmission of mass polarization (Proposition 11)—and the turnout curve of Proposition 6 does not depend on β at all (Proposition 12). Proposition 10 (The cost channel survives competition) In the setting of Theorem 5.1, let K~ K be a second super-regular cost distribution with r~≥r r≥ r pointwise on (0,1)(0,1). Then Δ~∗(α,β)≥Δ∗(α,β) ^*(α,β)≥ ^*(α,β) for every α≥0α≥ 0 and every β∈[0,1)β∈[0,1). Proof: Proposition 19. At β=0β=0 this is Theorem 4.2, and at (α,β)=(0,0)(α,β)=(0,0) it is Proposition 3: more responsive electorates sustain weakly more polarization, uniformly in competitiveness. Mass polarization under competition The transmission of mass polarization is unchanged as well. Proposition 11 (Mass polarization scales elite polarization, general form) Proposition 5 holds verbatim for every β∈[0,1)β∈[0,1): the game (Vσ,K,α,β)(V_σ,K,α,β) is isomorphic to (V0,Kσ,α,β)(V_0,K_σ,α,β), and in part (b) the setting is that of Theorem 5.1. Proof Under the affine map both benefits scale, Bi=σBi0B_i=σ B_i^0, so both vote shares are those of (V0,Kσ,α)(V_0,K_σ,α), and hence so is Ui=ui−βu−iU_i=u_i-β u_-i; the argument that no best response lies outside suppVσsuppV_σ is already made for every β≥0β≥ 0 in the proof of Proposition 5. Part (a) then follows from Proposition 7, and part (b) from Proposition 10 in place of Theorem 4.2, since KσK_σ is super-regular whenever K is. So under power costs elite polarization is exactly proportional to mass polarization at every level of competition, and otherwise the elasticity of K decides whether it is amplified or damped. At the margin endpoint mass polarization acts differently, only through the shape of the electorate at the centre (Theorem 5.4). Turnout: competition lowers it Turnout depends on the platforms but not on β, which enters only the candidates’ objectives; so the turnout curve of Proposition 6 is the same at every level of competition, and what competition changes is where on it the equilibrium sits. Proposition 12 (Turnout under competition) In the setting of Theorem 5.1, let α>0α>0. The formula for T′(Δ)T ( ) and part (b) of Proposition 6 hold unchanged for every β∈[0,1)β∈[0,1), and at the equilibrium T′(Δ∗)=2(1+α)(1+β)λ(1+α−β)K(Δ∗)>0.T ( ^*)= 2(1+α)(1+β)λ(1+α-β)\,K( ^*)>0. Equilibrium turnout T(Δ∗(α,β))T ( ^*(α,β) ) is strictly decreasing in β. Proof T is the sum of the two vote shares at the symmetric profile, and neither involves β, so the formula for T′(Δ)T ( ) and part (b) carry over. At the equilibrium there is no dead zone (Theorem 5.1), and Eq. 13 gives K(b)/K(Δ∗)=2(1+α)(1−β)λ(1+α−β)K(b)/K( ^*)= 2(1+α)(1-β)λ(1+α-β); substituting into T′(Δ)T ( ) gives the display, whose sign is that of 2(1+α−β)−λ(1−β)=α(1+β)>02(1+α-β)-λ(1-β)=α(1+β)>0. Finally, the root Δ∗(α,β) ^*(α,β) of Eq. 13 is unique, hence continuous in β, and strictly decreasing in β (Proposition 13); as β rises it therefore sweeps an interval at every point of which T′>0T >0, and T(Δ∗)T( ^*) falls. At β=0β=0 the display is part (a) of Proposition 6. Locally, more polarized candidates raise turnout at every level of competition; and since competition narrows the gap, it lowers turnout too. 5.3 Higher β De-polarizes Under the hypotheses that delivered uniqueness, the equilibrium gap is monotone in α and β at once: competition for the margin pulls the candidates together at every level of alienation, and alienation pushes them apart at every level of competition. Proposition 13 In the setting of Theorem 5.1, the equilibrium gap Δ∗(α,β) ^*(α,β) is strictly decreasing in β on [0,1)[0,1) and (for α>0α>0) strictly increasing in α. Proof: Proposition 18. Competition for the margin thus pulls candidates together exactly as costly voting and alienation push them apart; in the doubly-uniform benchmark the closed form (14) prices the trade-off between α and β exactly. Under alienation, competition lowers turnout as well as polarization (Proposition 12), so the reading ruled out for alienation must be ruled out here too: that the candidates converge because fewer people vote. The argument of Corollary 6 applies unchanged. Corollary 9 (De-polarization at unchanged turnout) In the setting of Theorem 5.1, fix β∈(0,1)β∈(0,1). Then there are positive multiples K0K_0 and KβK_β of K, both admissible, such that the games (V,K0,α,0)(V,K_0,α,0) and (V,Kβ,α,β)(V,K_β,α,β) have the same equilibrium turnout while their equilibrium gaps are Δ∗(α,0) ^*(α,0) and Δ∗(α,β)<Δ∗(α,0) ^*(α,β)< ^*(α,0). Proof As for Corollary 6, with Proposition 7 and Proposition 13 in place of Proposition 2 and Theorem 4.3. Super-regularity is a genuine hypothesis, not a convenience: Example 2 (A fold: competition can polarize) There is an explicit cost distribution with piecewise-linear K/kK/k (Example 3) whose vote-share game has three equilibria, with gaps 13 13, 1950 1950 and 1120 1120. Along the middle branch dΔ∗/dβ>0d ^*/dβ>0: more competitive candidates, more polarization. At β∗=347≈0.064β^*= 347≈ 0.064 the middle and upper branches collide and annihilate in a fold, and the equilibrium set jumps discontinuously from three points to one. (We verified the middle branch is a genuine global equilibrium, to deviation gain 2⋅10−122· 10^-12.) The example is not a curiosity about multiplicity; it locates precisely where the de-polarizing logic can break. Monotonicity of K/kK/k is what guarantees that the “supply” curve of Corollary 4 meets the “demand” curve once and from above. When it does not, an increase in β can slide the crossing the wrong way along a branch where supply cuts demand from below—and such a branch is fragile, which is why it is the one that dies in the fold. Two lessons for the polarization story: comparative statics in β are safe exactly under the same hypothesis that gives uniqueness, and a small increase in competitiveness can produce a discontinuous collapse in polarization when the fold is reached: at β∗=347β^*= 347 the upper branch, down from 1120 1120 at β=0β=0 to 12 12, disappears and the equilibrium gap drops to 1136≈0.31 1136≈ 0.31. Where the decline ends as β↑1β 1—not at the median—is taken up in Section 5.4. 5.4 The Margin Game (β=1β=1) At the endpoint β=1β=1 the game is zero-sum, and without alienation the classical Hotelling–Downs prediction of Claim 2.1 returns in full force—costs and all. Theorem 5.2 (Without alienation, the margin game selects the median) Let α=0α=0, let v,kv,k satisfy Assumption 2.2, and let cmed=V−1(12)c_med=V^-1( 12). Then (cmed,cmed)(c_med,c_med) is the unique Nash equilibrium of the margin game, and every unilateral deviation strictly lowers the deviator’s margin. Proof in Appendix 0.B. Together with Section 3.1, this cleanly separates the two objectives: costly participation moves vote-share maximizers away from the median for every distribution satisfying Assumption 2.2, yet leaves margin maximizers exactly at the median, for all distributions. With full alienation, however, the margin game degenerates: Proposition 14 (The margin band under alienation) Suppose β=1β=1. Let x∼U(0,1)x U(0,1) and let K satisfy Assumption 2.2. For every α≥0α≥ 0, the pure Nash equilibria of the margin game are exactly E(α)=[12+α,1+α2+α]2,E(α)= [ 12+α,\ 1+α2+α ]^2, each with margin 00: both platforms may sit anywhere in a band of width Δ0:=α2+α _0:= α2+α around the median, independent of the cost distribution. Proof in Appendix 0.B. The band contains a continuum of profiles, from the tied median profile to the maximally separated (1λ,1+αλ) ( 1λ, 1+αλ )—and even tied profiles c1=c2c_1=c_2 in which (almost) nobody votes and both candidates, caring only about the margin, are content. Which of these predictions should one believe? Theorem 5.3 below answers the question, and the answer is not the median. Both results are knife-edges—Theorem 5.2 in the objective, Proposition 14 in the voter density. Theorem 5.3 shows which point of the band survives a small vote-share motive, and Theorem 5.4 what happens to the median when the electorate is not uniform. The decline in β of Section 5.3 ends at a specific point of the band. Theorem 5.3 (Selection in the band) In the setting of Proposition 8, as β↑1β 1 the right-hand side of Eq. 13 vanishes, forcing K(b)→0K(b)→ 0 and hence b→0b→ 0: for every cost distribution K and every α>0α>0 the selected profile converges to (c1,c2)⟶(12+α,1+α2+α),Δ⟶α2+α,(c_1,c_2) ( 12+α,\ 1+α2+α ), α2+α, which is the most polarized profile of the band E(α)E(α)—and also its turnout-maximizing profile, since inside the band both tents are interior and T=2(1α+1λ)G(Δ)T=2 ( 1α+ 1λ )G( ) increases in Δ . Proof In Eq. 13 the factor 1+α−β1+α-β tends to α>0α>0 while the right-hand side tends to 00, so K(b)→0K(b)→ 0; as K is strictly increasing with K(0)=0K(0)=0, b→0b→ 0, and b=(λΔ−α)/2b=(λ -α)/2 from Eq. 9 gives Δ→α/λ →α/λ. The profile is symmetric, so c1=1−Δ2→1λc_1= 1- 2→ 1λ, the lower edge of E(α)E(α) in Proposition 14. Thus the indeterminacy of the band is resolved in a specific direction: an arbitrarily small vote-share motive added to a margin-maximizing candidate selects maximal polarization, not the median (Fig. 9). The median is the least robust point of the band. At α=0α=0 the band is a single point and the limit is the median; Eq. 14 shows the collapse is linear, Δ∗=(1−β)/2 ^*=(1-β)/2. 000.20.20.40.40.60.60.80.811000.20.20.40.40.60.6α2+α α2+αcompetitiveness β (0 = vote share, 1 = margin)equilibrium separation Δ∗ ^*α=2α=2α=1α=1α=0.5α=0.5α=0α=0 Figure 9: Equilibrium separation as competitiveness rises, doubly uniform (Theorem 5.3). Increasing β pulls the candidates together, but for α>0α>0 the limit at β=1β=1 is α2+α α2+α—the top of the margin band E(α)E(α), not the median. Only at α=0α=0 does the limit reach the median, and there the collapse is linear. Under super-regularity the collapse of Theorem 5.3 has an exact rate: the approach to the band top is linear in 1−β1-β whenever k(0+)∈(0,∞)k(0^+)∈(0,∞), and of order (1−β)1/p(1-β)^1/p when K(t)∼CtpK(t) Ct^p near 00; at α=0α=0 the rate 1−β2 1-β2 of Eq. 14 is universal, Δ∗=1−β2(1+o(1)) ^*= 1-β2\,(1+o(1)) for every K with k(0+)∈(0,∞)k(0^+)∈(0,∞) (Proposition 20). Mass polarization breaks the median for margin maximizers With α>0α>0 the median solves the margin first-order condition but need not be an equilibrium. The second-order condition at the median is degenerate (identically zero), so the question is settled at fourth order; the answer is a clean dichotomy in the curvature of the electorate at the center. Theorem 5.4 (Median dichotomy) Let κ∼U(0,1)κ U(0,1), α>0α>0, and let v be symmetric about 12 12 and C2C^2 near 12 12. Consider the margin obtained by a candidate deviating to 12−δ 12-δ against an opponent at the median. Then w1(δ)=C(α)v′(12)δ4+o(δ4),C(α)=1+α2α(2+α)+16(1α2−1(2+α)2)>0.w_1(δ)\;=\;C(α)\,v \! ( 12 )\,δ^4+o(δ^4), C(α)= 1+α2α(2+α)+ 16 ( 1α^2- 1(2+α)^2 )>0. Hence the median profile is a strict local equilibrium of the margin game if v′(12)<0v ( 12)<0 (an electorate peaked at the center), and is not an equilibrium if v′(12)>0v ( 12)>0 (a hollow, polarized electorate). Uniform voters, v′≡0v ≡ 0, are exactly the knife-edge that produces the band E(α)E(α). Proof in Appendix 0.A. The formula is sharp: for α=1α=1, δ=0.02δ=0.02 it predicts w1=−9.244⋅10−7w_1=-9.244· 10^-7 for v=Beta(2,2)v=Beta(2,2) and +1.962⋅10−7+1.962· 10^-7 for v=Beta(12,12)v=Beta( 12, 12), against simulated values −9.244⋅10−7-9.244· 10^-7 and +1.967⋅10−7+1.967· 10^-7; for uniform voters the simulated margin is zero to machine precision at every δ (Fig. 10). Uniform costs enter only through the constant. For a general K the same reflection argument gives w1(δ)=v′(12)[δ22(1α+1λ)∫0δK+δ(1α2−1λ2)∫0δK(t)(δ−t)t]+o(δ2∫0δK),w_1(δ)=v ( 12 ) [ δ^22 ( 1α+ 1λ ) _0^δ\!K+δ ( 1α^2- 1λ^2 ) _0^δ\!K(t)\,(δ-t)\,dt ]+o (δ^2\! _0^δ\!K ), whose bracket is positive because λ>αλ>α: the dichotomy in the sign of v′(12)v ( 12) holds for every cost distribution satisfying Assumption 2.2, and at K=idK=id the bracket is C(α)δ4C(α)δ^4. Theorem 5.4 is the margin-game counterpart of the mass-polarization result of Section 3.2, and it sharpens what “mass polarization” has to mean. In Section 3.2 the relevant statistic was a global one, the interquartile range. Here it is purely local: only the sign of v′v at the median matters, and the size of C(α)C(α) is irrelevant to the conclusion. An electorate can be enormously dispersed and still hold the candidates at the median against small deviations, provided it is peaked there; and it can be barely dispersed at all and lose the median, provided it is locally hollow. For margin maximizers, then, what drives elite polarization is not how spread out the voters are but whether the center is empty; the classical Downsian prediction fails as soon as it is, the moment α>0α>0. 000.050.050.10.10.150.150.20.2−0.001-0.001−0.0005-0.000500deviation δ from the mediandeviator’s margin w1w_1Beta(12,12)Beta( 12, 12): v′(12)>0v ( 12)>0, median failsuniform: v′(12)=0v ( 12)=0, knife-edgeBeta(2,2)Beta(2,2): v′(12)<0v ( 12)<0, median holds Figure 10: The quartic law of Theorem 5.4 at α=1α=1 with uniform costs: a candidate deviating to 12−δ 12-δ against an opponent at the median gains iff the electorate is hollow at the center. Thin dashed lines are the prediction C(α)v′(12)δ4C(α)v ( 12)δ^4; they are indistinguishable from the computed curves for small δ. For uniform voters the margin is zero to machine precision at every δ—the knife-edge that manufactures the band E(α)E(α). Existence can fail Where alienation meets a hollow electorate, the margin game can have no pure equilibrium at all, and a certificate rules one out for the arcsine electorate at α=1α=1 and for every β≥0.9168β≥ 0.9168. Proposition 15 (Non-existence) [computer-assisted] Let α=1α=1 and κ∼U(0,1)κ U(0,1). For v=Beta(12,12)v=Beta( 12, 12) and for v=Beta(0.8,0.8)v=Beta(0.8,0.8) the margin game has no pure Nash equilibrium (Theorem 0.L.4); for the former, the same holds for every β≥0.9168β≥ 0.9168 (Theorem 0.O.1). Each statement is a finite computation at a fixed (α,v)(α,v). We expect, but do not prove, that non-existence persists for every sufficiently hollow v and every α above a v-dependent threshold. Proof (Sketch; the full certificate is Theorems 0.L.4 and 0.O.1) The margin game is symmetric and zero-sum, so a pure equilibrium exists if and only if some position has security value zero, maxc∈[0,1]minc′∈[0,1]w(c,c′)=0 _c∈[0,1] _c ∈[0,1]w(c,c )=0, writing w(c,c′)w(c,c ) for the margin of a candidate at c against one at c′c ; since w(c,c)=0w(c,c)=0 and w is antisymmetric, the security value is always ≤0≤ 0. For α=1α=1 and uniform costs it equals −8.2244⋅10−3-8.2244· 10^-3 (at c=0.829c=0.829) for the arcsine density v=Beta(12,12)v=Beta( 12, 12) and −3.0238⋅10−3-3.0238· 10^-3 for Beta(0.8,0.8)Beta(0.8,0.8), so every position can be strictly beaten (Fig. 11); for Beta(2,2)Beta(2,2) the grid security value attains 00 at the median and only there, as Theorem 5.4 leads one to expect. A Lipschitz bound on w turns the finite grid computation into an exact statement about the continuum, giving a computer-assisted proof. A mixed equilibrium always exists by Glicksberg’s fixed-point theorem, with value 00. The certificate extends off the boundary β=1β=1: no pure equilibrium exists for any β≥0.9168β≥ 0.9168 (Theorem 0.O.1), while a pure equilibrium is still found numerically at β=0.90β=0.90, placing the true threshold in (0.90,0.9168](0.90,0.9168] with the numerics placing that threshold βc _c between 0.9080.908 and 0.9090.909. The contrast with α=0α=0 is stark: there, under uniform costs, the very arcsine electorate that admits no pure equilibrium at α=1α=1 keeps one for every β<1β<1, along with every other convex symmetric electorate meeting the density bound of Corollary 8. Non-existence is therefore a phenomenon of alienation and competitiveness together, not of competitiveness alone. 000.20.20.40.40.60.60.80.811−0.2-0.2−0.1-0.100own position ccsecurity value minc′w(c,c′) _c w(c,c )Beta(2,2)Beta(2,2): attains 00 at the medianuniform: 00 on the whole band E(α)E(α)arcsine: maximum −8.22⋅10−3<0-8.22· 10^-3<0 Figure 11: The certificate behind Proposition 15. The margin game is symmetric zero-sum, so a pure equilibrium exists iff some position has security value 00. For a centrally peaked electorate the median achieves it; for uniform voters every point of the band does; for the arcsine (hollow) electorate the maximum is strictly negative at c=0.829c=0.829, so every position can be strictly beaten and no pure equilibrium exists. Computing this curve requires quadrature that respects the endpoint singularity of v: a naive grid rule returns −0.387-0.387 for the arcsine maximum, a factor of 4747 too negative (Remark 27). Two remarks complete the picture. First, equilibria of nearly-zero-sum games converge onto margin-game equilibria (upper hemicontinuity, Theorem 0.L.3). Corollary 8 shows the statement is not vacuous at α=0α=0, where the approximating equilibria exist all the way to the limit and select the median. Second, at α=0α=0 we conjecture pure existence for every (v,k)(v,k) and β<1β<1 (Conjecture 1); the certified failure at α=1α=1 shows that no proof can survive a relaxation of α=0α=0. Together with Theorem 5.4, this gives the picture for the margin game: the median survives alienation, at least locally, when the electorate is centrally peaked; when it is hollow, the median fails and, at least for the densities examined here, no pure equilibrium takes its place. 6 Discussion and Further Research What drives polarization Table 3 collects the answer the paper was built to give. Three forces raise equilibrium polarization—a more dispersed electorate, costlier voting in the reverse-hazard-rate sense, and stronger alienation; a hollow center removes the median as a margin-game equilibrium; one force lowers polarization—competition for the margin; and one candidate that every intuition nominates, the level of turnout, provably does nothing at all. The two voter-side channels do not interfere with each other (Theorem 4.2): each still polarizes at every level of the other, and both survive the presence of competitiveness (Propositions 13 and 10). The one de-polarizing force is also the only one whose comparative static is shown to reverse sign when super-regularity fails (Example 2), an asymmetry worth keeping in mind when reading the model empirically. Table 3: Every force in the model, and which way it moves equilibrium polarization Δ∗ ^*. The last row is not a gap in the analysis but a theorem: the level of participation is an output of the model, never an input to it. Increase in… Δ∗ ^* Holds when Where electorate spread (IQR) ↑ unif. costs, v symmetric, ρv≤4 _v≤ 4 Corollary 1 electorate scale σ ↑ unif. voters, K super-reg.; exact for power costs (any v) Proposition 5 voting costs (rhr order) ↑ unif. voters, K super-reg., any (α,β)(α,β) Propositions 3, 4.2 and 10 alienation α ↑ unif. voters, K super-reg., α>0α>0; Δ∗→1 ^*→ 1 Theorem 4.3 competitiveness β ↓ unif. voters, monotone rhr only Propositions 13 and 2 hollow center, v′(12)>0v ( 12)>0 breaks median margin game, unif. costs, α>0α>0, v symmetric Theorem 5.4 cost level, K↦zK zK — always: no effect Propositions 7 and 4 The baseline and the two dials together Table 4 collects the doubly-uniform benchmarks. Reading it as a story: costly voting moves vote-share maximizers from the median to the quartiles; alienation pushes them further out, without bound; competitiveness pulls them back in, linearly to the median when voters are not alienated, but only to the top of the margin band when they are. The classical Downsian prediction survives costly voting only under the margin objective, and even then only when voters are unalienated or, locally, when the electorate is centrally peaked; for alienated voters facing a hollow electorate, nothing need survive—not even equilibrium existence. Polarization in this model is thus neither an inevitability nor an accident: it is the resultant of two voter-side forces—costly voting and alienation—and one candidate-side force, each with an explicit comparative static. Table 4: The baseline and the two dials at a glance. Second column: the equilibrium in the doubly-uniform benchmark (x,κ∼U(0,1)x,κ U(0,1)), with λ=2+αλ=2+α. Third column: sufficient density-ratio bound on v for the first-order conditions to characterize equilibria under uniform costs, where fβ=2λ(1+α)+β(1+α)2f_β= 2λ(1+α)+β(1+α)^2 (concavity) and gβ=(1+α)λα(1+α+β)+βλg_β= (1+α)λα(1+α+β)+βλ (dead zone), from Theorem 0.M.1; the bounds come from different arguments and are not claimed tight. The row ‘both’ is stated for α>0α>0 and does not reduce to the row above it as α↓0α 0: fβ→4+β≤41−βf_β→ 4+β≤ 41-β, because the α=0α=0 bound uses an exact cancellation at the midpoint that the worst-case argument of Theorem 0.M.1 forgoes (Remark 20), and gβ→1/βg_β→ 1/β bounds a dead-zone piece of the strategy space that vanishes at α=0α=0. Setting Δ∗ ^* (doubly unif.) Umbrella ρv≤ _v≤ Where baseline (0,0)(0,0) 12 12, at (14,34)( 14, 34) 44 Corollaries 2 and 1 alienation (α,0)(α,0) 2+α4+α 2+α4+α min2λ1+α,λα \ 2λ1+α, λα\ Corollaries 5 and 0.K.1 competition (0,β)(0,β) 1−β2 1-β2 (quantile profile) 41−β 41-β Corollaries 7 and 9 both (α,β)(α,β) Eq. 14 minfβ,gβ \f_β,g_β\ Corollaries 7 and 0.M.1 margin limit (α,1)(α,1) band: Δ∈[0,αλ] ∈ [0, αλ ] knife-edge Propositions 14 and 5.4 Open questions Several questions remain open. On sufficiency: sharp thresholds for the density-ratio umbrellas of Table 4, and a sufficiency theory beyond uniform costs (Proposition 16 covers non-increasing k at β=0β=0; the obstruction at β>0β>0 is isolated in Remark 22). On alienation: a general-electorate version of Theorem 4.3, whose conclusion holds numerically for the concentrated electorate of Remark 6 but is proved only for uniform voters. On existence: under Assumption 2.2 we conjecture pure existence for every (v,k)(v,k) in the baseline and, at α=0α=0, for every β<1β<1 (Remarks 3 and 1), the first of these open already for uniform costs once ρv>4 _v>4; closing the certified gap (0.90,0.9168](0.90,0.9168] of Theorem 0.O.1; and characterizing the mixed equilibria of the margin game where pure ones fail. On the margin game with alienation: the exact non-existence region in (α,β,v)(α,β,v), and whether the quantile path of Corollary 8 has an α>0α>0 analogue. The certificate technique behind Theorem 0.O.1 may be of independent interest for other spatial games with continuous strategy spaces. Empirically, where political polarization has risen [74, 14] alongside declining turnout [49, 37], the model offers a channel complementary to media- and network-based accounts [75, 55], with a sharpened prediction: the association should run through the composition of the abstaining margin rather than the participation rate, and the model separates “voting became harder” (uniform shifts in participation) from “voters became disaffected” (abstention concentrated among the ideologically distant)—distinctions the turnout literature is equipped to test [58, 36]. A first piece of evidence on the composition side exists: across Dutch municipalities, lower turnout goes with a more polarized distribution of the votes cast, the parties furthest from the center gaining most, consistent with centrist voters being the ones who abstain [5]—the abstaining margin of the baseline model, where abstention concentrates between the candidates. That test has a confound the present model assumes away. Costs here are drawn i.i.d. and independently of position (Section 2), and the assumption does real work: written out for the baseline, the balance condition of Lemma 2 has an outer term that factorizes as k(Δ)V(c1)k( )V(c_1), one cost density times one voter mass, because every voter on the flank both faces the same benefit Δ and draws from the same K. Replace K by a family KxK_x and that product becomes ∫0c1kx(Δ)v(x)x _0^c_1k_x( )v(x)\,dx, which does not split; Proposition 2 and the balance principle itself survive, but Corollary 3 weakens from “one scalar function summarizes the cost side” to a statement about a field rx\r_x\. Note that this is not the position-dependence the model already has. Alienation acts through the benefit, via did_i, so its bite moves when the candidates move; a cost–position correlation is pinned to the voter and does not move at all. The paper treats endogenous position-dependence thoroughly and exogenous position-dependence not at all, and no choice of α substitutes for the latter. The direction is signed, moreover, and the two channels disagree: alienation predicts that a candidate’s own extremists are the first to abstain, whereas if the ideologically committed are simply cheaper to mobilize, then kxk_x is shifted toward low costs at extreme x, the flank is more responsive than Proposition 3 assumes, and the model understates polarization. Which of the two dominates is an empirical question, and it is the same question as whether a composition shift in the abstaining margin identifies the strategic channel or merely reflects who finds voting cheap. Finally, the costly-turnout framework introduced here can be naturally combined with additional dynamics. One extension is to allow for non-linear or probabilistic voter utilities, as in [22], which may better capture how voters evaluate platforms. Another is to embed the model in a dynamic environment with repeated elections, building on iterative voting models [66]; such a setting could generate endogenous status quo effects that shape participation incentives, in the spirit of [81]. Further natural directions include more than two candidates and endogenous entry, where the alienation dead zones suggest room for centrist or extremist entrants; asymmetric candidates (valence or cost advantages), for which our asymmetry law (Lemma 3) already prices the trade-off; and finite electorates with pivotality considerations in the spirit of the Calculus of Voting [77], connecting the present continuum model back to the classical turnout literature. Appendix 0.A Deferred Proofs This appendix collects the proofs deferred from Sections 3, 4 and 5. Proofs of Lemmas 1, 3.1, 1, 1, 3.2, 3, 5.2 and 14 are in Appendix 0.B. The remaining proofs, together with the supplementary results, worked examples and certificates they draw on, occupy Appendices 0.I to 0.Q. Proof of Proposition 4. u1=∫x1out∨0x1inK(B1)vu_1= _x^out_1 0^x^in_1K(B_1)v. The Leibniz terms at both endpoints vanish because B1=0B_1=0 there and K(0)=0K(0)=0 (at a truncated lower limit x1out<0x^out_1<0 the limit is the constant 00 and contributes nothing either). On [x1out∨0,c1][x^out_1 0,c_1] we have ∂c1B1=−(1+α) _c_1B_1=-(1+α) and on [c1,x1in][c_1,x^in_1], ∂c1B1=+(1+α) _c_1B_1=+(1+α), whence ∂c1u1=(1+α)[∫c1x1ink(B1)v−∫x1out∨0c1k(B1)v] _c_1u_1=(1+α) [ _c_1^x^in_1k(B_1)v- _x^out_1 0^c_1k(B_1)v ]. Setting this to zero gives Eq. 10.∎ Proof of Theorem 4.1. Write Ψ(Δ):=∫b(Δ)Δr−log(λ/2) ( ):= _b( ) r- (λ/2) on [αλ,1) [ αλ,1 ), so that Eq. 12 reads Ψ(Δ)=0 ( )=0. Necessity. Every equilibrium is symmetric with gap a root of Eq. 12, as shown before the theorem. Uniqueness of the root. By Eq. 9 the interval [b(Δ),Δ][b( ), ] has length α2(1−Δ) α2(1- ), strictly decreasing in Δ , and its left endpoint b(Δ)b( ) is strictly increasing in Δ (slope λ/2>0λ/2>0). Hence as Δ grows the interval both shifts right and shrinks; since r is non-increasing and positive, ∫bΔr _b r is strictly decreasing, so Ψ is strictly decreasing. At Δ=α/λ =α/λ we have b=0b=0 and ∫0Δr=+∞ _0 r=+∞ (as logK(0+)=−∞ K(0^+)=-∞), so Ψ>0 >0 there; at Δ↑1 1 we have b↑1b 1, the interval degenerates, and Ψ→−log(λ/2)<0 →- (λ/2)<0. A unique root Δ∗ ^* follows. The root is an equilibrium. Fix the opponent at c2c_2 and let candidate 1 choose c1∈[0,c2)c_1∈[0,c_2). In the no-dead-zone region u1=AG(Δ)−1αG(b1)u_1=A\,G( )- 1αG(b_1) with A=2(1+α)αλA= 2(1+α)αλ, so ∂u1∂c1=1+α[K(b1)−2λK(Δ)],whose sign is that oflogλ2−∫b1Δr. ∂ u_1∂ c_1= 1+α [K(b_1)- 2λK( ) ], sign is that of λ2- _b_1 r. Moving c1c_1 toward the opponent decreases Δ at unit rate and b1b_1 at rate 1+α1+α, so the interval [b1,Δ][b_1, ] lengthens and moves left; with r non-increasing, ∫b1Δr _b_1 r is strictly increasing in c1c_1. Hence ∂c1u1 _c_1u_1 changes sign exactly once, from ++ to −-: u1(⋅,c2)u_1(·,c_2) is strictly single-peaked on the own side, and by Remark 6 the dead-zone region contains no maximizer. For a cross-over deviation note that, at a fixed distance t from the opponent, the payoff is AG(t)−1αG((t−αϱ)+)A\,G(t)- 1αG ((t-α )^+ ), where ϱ is the room behind the deviator (ϱ=c2−t =c_2-t on the left, ϱ=1−c2−t =1-c_2-t on the right); this is increasing in ϱ , and at the symmetric profile c2>12c_2> 12 gives c2−t>1−c2−tc_2-t>1-c_2-t. Hence every far-side deviation is dominated by its mirror image on the own side, and the tie yields 00. The profile is therefore a global mutual best response.∎ Proof of Proposition 5. No best response lies outside suppVσsuppV_σ, so the strategy space may be taken to be that interval, on which the affine map is a bijection. Indeed, let L:=minsuppVσL:= _σ, write Bi(x,c1,c2)B_i(x;c_1,c_2) for the benefit at the profile (c1,c2)(c_1,c_2), and suppose c1<Lc_1<L, wherever c2c_2 lies. Every voter x in the support has x>c1x>c_1, so ∂B1(x)/∂c1=1+α>0∂ B_1(x)/∂ c_1=1+α>0 and ∂B2(x)/∂c1=−1<0∂ B_2(x)/∂ c_1=-1<0 pointwise; as K is non-decreasing, u1u_1 is non-decreasing and u2u_2 non-increasing in c1c_1 on [c1,L][c_1,L], whence U1(c1,c2)≤U1(L,c2)U_1(c_1,c_2)≤ U_1(L,c_2) for every β≥0β≥ 0. Moreover c1c_1 is not a best response. If c2≠Lc_2≠ L, the voter at L has B1(L,L,c2)=|L−c2|∈(0,1)B_1(L;L,c_2)=|L-c_2|∈(0,1), so by continuity B1(⋅,L,c2)∈(0,1)B_1(·;L,c_2)∈(0,1) on a set of positive VσV_σ-measure; there K is strictly increasing (Assumption 2.2) and B1(⋅,c1,c2)=B1(⋅,L,c2)−(1+α)(L−c1)B_1(·;c_1,c_2)=B_1(·;L,c_2)-(1+α)(L-c_1) is pointwise smaller, hence u1(c1,c2)<u1(L,c2)u_1(c_1,c_2)<u_1(L,c_2) and U1(c1,c2)<U1(L,c2)U_1(c_1,c_2)<U_1(L,c_2). If c2=Lc_2=L, then B1(x,c1,L)=(c1−L)−α(x−c1)<0B_1(x;c_1,L)=(c_1-L)-α(x-c_1)<0 on the support, so u1(c1,L)=0u_1(c_1,L)=0, while B2(L,c1,L)=L−c1>0B_2(L;c_1,L)=L-c_1>0 gives u2(c1,L)>0u_2(c_1,L)>0: for β>0β>0 the tied profile (L,L)(L,L), at which both shares vanish, is strictly better, and for β=0β=0 any c1′c_1 in the interior of the support, where B1(c1′,c1′,L)=c1′−L>0B_1(c_1 ;c_1 ,L)=c_1 -L>0, earns a strictly positive share. The right-hand case is the mirror image. Under the affine map di(x)=σ|a−ai|d_i(x)=σ|a-a_i|, so Bi(x)=σBi0(a)B_i(x)=σ B_i^0(a) and ui=∫K(σBi0(a))dV0(a)u_i= K(σ B_i^0(a))\,dV_0(a), which is the payoff of the game (V0,Kσ,α)(V_0,K_σ,α). Part (a) is Proposition 7 applied to Kσ=σpK_σ=σ^pK (legitimate since all benefits lie in [0,1][0,1] and σ≤1σ≤ 1). Part (b) computes rσ(t)=σk(σt)/K(σt)=η(σt)/tr_σ(t)=σ k(σ t)/K(σ t)=η(σ t)/t and applies Theorem 4.2 to (V0,Kσ,α)(V_0,K_σ,α). Since rσ(t)=σr(σt)r_σ(t)=σ r(σ t) is non-increasing whenever r is, and Kσ(1)=K(σ)∈(0,1]K_σ(1)=K(σ)∈(0,1], every KσK_σ satisfies the hypotheses of that theorem. That Δ∗(σ) ^*(σ) itself rises in σ at every α>0α>0, whether the transmission is amplified or damped, follows from writing Eq. 12 for (V0,Kσ,α)(V_0,K_σ,α) in x-space: substituting u=σtu=σ t and using Eq. 9, Δ∗ ^* solves ∫Δ∗−α2(σ−Δ∗)Δ∗ru=logλ2 _ ^*- α2(σ- ^*) ^*r\,du= λ2, whose left side is strictly increasing in σ and, for r non-increasing, strictly decreasing in Δ∗ ^*.∎ Proof of Proposition 6. Differentiate T=2[AG(Δ)−1αG(b)]T=2 [A\,G( )- 1αG(b) ] using db/dΔ=λ/2db/d =λ/2 and G′=KG =K; part (a) substitutes Eq. 12. For (b), at Δ=1 =1 we have b=1b=1 by Eq. 9, so the displayed derivative gives T′(1)=2αK(1)[2(1+α)λ−λ2]=−αλK(1)<0T (1)= 2αK(1) [ 2(1+α)λ- λ2 ]=- αλK(1)<0, while T′(Δ∗)>0T ( ^*)>0 by (a); since T′T is continuous, T turns over in (Δ∗,1)( ^*,1). For the benchmark put K=idK=id: T′<0T <0 iff b/Δ>4(1+α)/λ2b/ >4(1+α)/λ^2, and substituting b=(λΔ−α)/2b=(λ -α)/2 and solving gives the stated Δ¯ (the ratio b/Δb/ is increasing in Δ , so T is single-peaked). That Δ∗(α)=2+α4+α<Δ¯(α) ^*(α)= 2+α4+α< (α) is consistent with (a).∎ Proof of Theorem 4.3. Let Ψ be as in the proof of Theorem 4.1, strictly decreasing in Δ with root Δ∗ ^*. With λ=2+αλ=2+α and ∂αb=−1−Δ2 _αb=- 1- 2, ∂αΨ=r(b)1−Δ2−1λ _α =r(b) 1- 2- 1λ. At a root, super-regularity gives logλ2=∫bΔr≤r(b)(Δ−b)=r(b)α(1−Δ)2 λ2= _b r≤ r(b)( -b)=r(b) α(1- )2, so r(b)1−Δ2≥1αlogλ2r(b) 1- 2≥ 1α λ2 and ∂Ψ∂α|Δ=Δ∗≥1αlog(1+α2)−12+α=λlog(λ/2)−αλ> 0, ∂ ∂α |_ = ^*\;≥\; 1α (1+ α2 )- 12+α\;=\; λ (λ/2)-αλ\;>\;0, the last inequality being (1+s)log(1+s)>s(1+s) (1+s)>s for s=α/2>0s=α/2>0. Since Ψ is strictly decreasing in Δ , the implicit function theorem gives dΔ∗/dα>0d ^*/dα>0. Finally Δ∗>α/λ→1 ^*>α/λ→ 1.∎ Proof of Proposition 8. Under uniform voters no untied symmetric stationary profile has a dead zone, for any α>0α>0, any β<1β<1 and any K: if b1=b2=b≤0b_1=b_2=b≤ 0 then u1=u2=AG(Δ)u_1=u_2=A\,G( ) with A=2(1+α)αλA= 2(1+α)αλ (Remark 6), so ∂U1/∂c1=−(1−β)AK(Δ)<0∂ U_1/∂ c_1=-(1-β)A\,K( )<0 for Δ∈(0,1) ∈(0,1)—the argument of Proposition 17(b), which uses only the first-order conditions. Hence b1,b2>0b_1,b_2>0 and u1=AG(Δ)−1αG(b1)u_1=A\,G( )- 1αG(b_1), u2=AG(Δ)−1αG(b2)u_2=A\,G( )- 1αG(b_2). Since ∂c1Δ=−1 _c_1 =-1, ∂c1b1=−(1+α) _c_1b_1=-(1+α) and ∂c1b2=−1 _c_1b_2=-1, ∂U1∂c1=−(1−β)AK(Δ)+1+αK(b1)−βαK(b2). ∂ U_1∂ c_1=-(1-β)A\,K( )+ 1+αK(b_1)- βαK(b_2). At a symmetric profile b1=b2=b_1=b_2=b, and setting the derivative to zero yields Eq. 13.∎ Proof of Theorem 5.4. Write q=12q= 12, c1=q−δc_1=q-δ, c2=qc_2=q, so Δ=δ =δ and m=q−δ/2m=q-δ/2. The two benefit tents are reflections of one another about m: B2(x)=B1(2m−x)B_2(x)=B_1(2m-x). With uniform costs w1=∫B1+(x)[v(x)−v(2m−x)]xw_1= B_1^+(x)[v(x)-v(2m-x)]dx, and since v is symmetric about q, v(2m−x)=v(1−δ−x)=v(x+δ)v(2m-x)=v(1-δ-x)=v(x+δ), so w1=∫B1+(x)[v(x)−v(x+δ)]dx=−δ∫B1+v′−δ22∫B1+v′+O(δ5).w_1= B_1^+(x)[v(x)-v(x+δ)]dx=-δ\! \!B_1^+v \;-\; δ^22\! \!B_1^+v \;+\;O(δ^5). B1B_1 is a tent with apex c1c_1, height δ, left half-width δ/αδ/α and right half-width δ/λδ/λ, so ∫B1+=δ22A B_1^+= δ^22A with A=1α+1λA= 1α+ 1λ, and ∫B1+(x−c1)=δ36(1λ2−1α2) B_1^+(x-c_1)= δ^36 ( 1λ^2- 1α^2 ). Expanding v′(x)=v′(c1)+v′(c1)(x−c1)+O((x−c1)2)v (x)=v (c_1)+v (c_1)(x-c_1)+O((x-c_1)^2) and using v′(q)=0v (q)=0, v′(c1)=−δv′(q)+O(δ2)v (c_1)=-δ v (q)+O(δ^2), −δ∫B1+v′ -δ\! \!B_1^+v =δ4v′(q)[A2−16(1λ2−1α2)]+O(δ5), =δ^4v (q) [ A2- 16 ( 1λ^2- 1α^2 ) ]+O(δ^5), −δ22∫B1+v′ - δ^22\! \!B_1^+v =−δ44v′(q)A+O(δ5). =- δ^44v (q)\,A+O(δ^5). Adding gives w1=δ4v′(q)[A4+16(1α2−1λ2)]+O(δ5)w_1=δ^4v (q) [ A4+ 16( 1α^2- 1λ^2) ]+O(δ^5), which is the claim since A/4=1+α2α(2+α)A/4= 1+α2α(2+α).∎ Appendix 0.B Omitted Proofs: Baseline and Margin Game This appendix proves the results of Sections 3 and 5 that are stated without proof in the main text; proofs of the remaining results are in Appendix 0.A and in Appendices 0.I to 0.Q. For the reader’s convenience we restate each result before its proof. 0.B.1 Preliminaries: Payoffs and Their Derivatives Throughout, fix a profile (c1,c2)(c_1,c_2) with c1<c2c_1<c_2 and write Δ:=c2−c1 :=c_2-c_1 and m:=c1+c22m:= c_1+c_22. Voters on [0,m)[0,m) either vote for candidate 11 or abstain, and voters on (m,1](m,1] either vote for candidate 22 or abstain. Evaluating the gain d−i(x)−di(x)d_-i(x)-d_i(x) piecewise—it equals Δ outside [c1,c2][c_1,c_2] and |c1+c2−2x||c_1+c_2-2x| inside—we obtain u1(c1,c2) u_1(c_1,c_2) =K(Δ)V(c1)+∫c1mK(c1+c2−2x)v(x)x, =K( )\,V(c_1)+ _c_1^mK(c_1+c_2-2x)\,v(x)\,dx, (15) u2(c1,c2) u_2(c_1,c_2) =K(Δ)(1−V(c2))+∫mc2K(2x−c1−c2)v(x)x. =K( )\, (1-V(c_2) )+ _m^c_2K(2x-c_1-c_2)\,v(x)\,dx. (16) Since v and k are continuous, u1u_1 is differentiable in c1c_1 on (0,c2)(0,c_2). Differentiating Eq. 15 by the Leibniz rule, the term K(Δ)v(c1)K( )v(c_1) produced by the first summand cancels against the lower-limit term of the integral, and the upper-limit term vanishes because K(0)=0K(0)=0; what remains is ∂u1∂c1(c1,c2)=∫c1mk(c1+c2−2x)v(x)x−k(Δ)V(c1), ∂ u_1∂ c_1(c_1,c_2)= _c_1^mk(c_1+c_2-2x)\,v(x)\,dx\;-\;k( )\,V(c_1), (17) and, symmetrically, ∂u2∂c2(c1,c2)=k(Δ)(1−V(c2))−∫mc2k(2x−c1−c2)v(x)x. ∂ u_2∂ c_2(c_1,c_2)=k( )\, (1-V(c_2) )\;-\; _m^c_2k(2x-c_1-c_2)\,v(x)\,dx. (18) Both formulas remain valid as one-sided derivatives at c1=0c_1=0 and c2=1c_2=1. Two specializations will be used repeatedly. First, for uniform costs (K(t)=tK(t)=t, k≡1k≡ 1), Eqs. 17 to 18 become ∂u1∂c1=V(m)−2V(c1),∂u2∂c2=1+V(m)−2V(c2), ∂ u_1∂ c_1=V(m)-2V(c_1), ∂ u_2∂ c_2=1+V(m)-2V(c_2), (19) and integrating Eq. 15 by parts gives the compact representation u1(c1,c2)=2∫c1mV(x)x,u2(c1,c2)=2∫mc2(1−V(x))x.u_1(c_1,c_2)=2 _c_1^mV(x)\,dx, u_2(c_1,c_2)=2 _m^c_2 (1-V(x) )\,dx. (20) (The analogous formulas with the roles of the candidates reversed hold when c1>c2c_1>c_2; they follow from the reflection x↦1−x 1-x, which maps V to the CDF V~(x)=1−V(1−x) V(x)=1-V(1-x), whose density v~(x)=v(1−x) v(x)=v(1-x) has the same bounds as v.) Second, for uniform voters (v≡1v≡ 1), substituting y=c1+c2−2xy=c_1+c_2-2x in Eq. 15 gives, with G(t):=∫0tK(y)yG(t):= _0^tK(y)\,dy, u1(c1,c2)=c1K(Δ)+12G(Δ),u2(c1,c2)=(1−c2)K(Δ)+12G(Δ),u_1(c_1,c_2)=c_1K( )+ 12G( ), u_2(c_1,c_2)=(1-c_2)K( )+ 12G( ), (21) so that ∂u1∂c1=12K(Δ)−c1k(Δ),∂u2∂c2=(1−c2)k(Δ)−12K(Δ). ∂ u_1∂ c_1= 12K( )-c_1k( ), ∂ u_2∂ c_2=(1-c_2)k( )- 12K( ). (22) Lemma 1 (restated). In the baseline model, under Assumption 2.2, every pure Nash equilibrium (c1,c2)(c_1,c_2) satisfies c1≠c2c_1≠ c_2 and, labeling the candidates so that c1<c2c_1<c_2, 0<c1<c2<10<c_1<c_2<1. Proof No ties. At a tied profile c1=c2=c_1=c_2=c both candidates receive 00. If c>0c>0, let candidate 11 deviate to any c1′∈(0,c)c_1 ∈(0,c); by Eq. 15, her payoff is at least ∫c1′m′K(c1′+c−2x)v(x)x>0 _c_1 ^m K(c_1 +c-2x)v(x)\,dx>0, since on the interior of that interval the argument of K lies in (0,Δ′)⊆(0,1)(0, ) (0,1), where K>0K>0, and v>0v>0. If c=0c=0, the mirror deviation to any c1′∈(0,1)c_1 ∈(0,1) works. Hence tied profiles are never equilibria. No boundary positions. Suppose c1=0<c2c_1=0<c_2. By Eq. 17, the one-sided derivative at c1=0c_1=0 equals ∫0c2/2k(c2−2x)v(x)x>0 _0^c_2/2k(c_2-2x)v(x)\,dx>0 (the argument of k ranges over (0,c2)⊆(0,1)(0,c_2) (0,1), where k>0k>0, and v>0v>0), so a small move to the right strictly increases u1u_1, and c1=0c_1=0 is not a best response. The mirror argument using Eq. 18 rules out c2=1c_2=1. Theorem 3.1 (restated). Let κ∼U(0,1)κ U(0,1) and let v satisfy Assumption 2.2. Then every Nash equilibrium (c1,c2)(c_1,c_2) satisfies V(c1)=V(c2)−12=12V(c1+c22).V(c_1)=V(c_2)- 12= 12\,V\! ( c_1+c_22 ). Proof By Lemma 1 and Eq. 19, an equilibrium satisfies V(m)=2V(c1)andV(m)=2V(c2)−1.V(m)=2V(c_1) V(m)=2V(c_2)-1. (23) Combining the two equalities yields V(c1)=V(c2)−12=12V(m)V(c_1)=V(c_2)- 12= 12V(m), as required. Proposition 1 (restated). Let κ∼U(0,1)κ U(0,1) and suppose ρv≤4 _v≤ 4. Then each candidate’s payoff is concave in her own position on each side of the opponent. Consequently, (c1,c2)(c_1,c_2) with 0<c1<c2<10<c_1<c_2<1 is a Nash equilibrium if and only if it satisfies the conditions of Theorem 3.1 and neither candidate gains by a cross-over deviation to the far side of her opponent. Proof Fix c2c_2 and consider c1′∈[0,c2)c_1 ∈[0,c_2). By Eq. 19, ∂u1/∂c1′=V(m′)−2V(c1′)∂ u_1/∂ c_1 =V(m )-2V(c_1 ) with m′=c1′+c22m = c_1 +c_22, whose derivative in c1′c_1 is 12v(m′)−2v(c1′)≤12supv−2infv≤ 0 12\,v(m )-2\,v(c_1 )\;≤\; 12 v-2 v\;≤\;0 whenever ρv≤4 _v≤ 4. Hence u1(⋅,c2)u_1(·,c_2) has a non-increasing derivative, i.e., is concave, on [0,c2][0,c_2]. For c1′∈(c2,1]c_1 ∈(c_2,1] the same argument applies after the reflection x↦1−x 1-x (which preserves the density ratio), so u1(⋅,c2)u_1(·,c_2) is also concave on [c2,1][c_2,1]. The argument for candidate 22 is symmetric. (⇒ ) If (c1,c2)(c_1,c_2) is an equilibrium then the balance conditions hold by Theorem 3.1, and the cross-over inequalities hold trivially, since they compare the equilibrium payoff with payoffs of feasible deviations. (⇐ ) Suppose the conditions of Theorem 3.1 hold at an interior profile. Then ∂u1/∂c1=V(m)−2V(c1)=0∂ u_1/∂ c_1=V(m)-2V(c_1)=0, so by concavity c1c_1 maximizes u1(⋅,c2)u_1(·,c_2) on [0,c2][0,c_2]. Note also that ρv≤4 _v≤ 4 forces infv≥14supv≥14 v≥ 14 v≥ 14 (as supv≥∫01v=1 v≥ _0^1v=1), so V(c1)≥c1/4>0V(c_1)≥ c_1/4>0 and, by Eq. 20, u1(c1,c2)=2∫c1mV>0u_1(c_1,c_2)=2 _c_1^mV>0; in particular the tie c1′=c2c_1 =c_2 (payoff 00) is not profitable. The cross-over inequality covers all c1′∈[c2,1]c_1 ∈[c_2,1]. Hence c1c_1 is a global best response, and symmetrically for c2c_2, so the profile is an equilibrium. Corollary 1 (restated). Let κ∼U(0,1)κ U(0,1) and suppose ρv≤4 _v≤ 4. Then any profile with V(c1)=14V(c_1)= 14, V(c2)=34V(c_2)= 34 and V(c1+c22)=12V ( c_1+c_22 )= 12 is a Nash equilibrium. Proof Write s:=infvs:= v; as noted above, ρv≤4 _v≤ 4 implies s≥14s≥ 14 and supv≤4s v≤ 4s. The profile is interior (0<V(c1)<V(c2)<10<V(c_1)<V(c_2)<1 and V is strictly increasing) and satisfies Eq. 23: V(m)=12=2V(c1)V(m)= 12=2V(c_1) and 2V(c2)−1=12=V(m)2V(c_2)-1= 12=V(m). By Proposition 1 it therefore suffices to verify the two cross-over inequalities. A lower bound on the candidates’ payoffs. Since V(x)≥V(c1)+s(x−c1)=14+s(x−c1)V(x)≥ V(c_1)+s(x-c_1)= 14+s(x-c_1), Eq. 20 gives u1(c1,c2)=2∫c1mV(x)x≥ 2[14⋅Δ2+s2(Δ2)2]=Δ4+sΔ24.u_1(c_1,c_2)=2 _c_1^mV(x)\,dx\;≥\;2 [ 14· 2+ s2 ( 2 )^2 ]= 4+ s ^24. Moreover 12=V(c2)−V(c1)=∫c1c2v≤4sΔ 12=V(c_2)-V(c_1)= _c_1^c_2v≤ 4s , i.e., Δ≥18s ≥ 18s, whence u1(c1,c2)≥132s+1256s>132s.u_1(c_1,c_2)\;≥\; 132s+ 1256s\;>\; 132s. (24) The same bound holds for u2(c1,c2)u_2(c_1,c_2), using 1−V(x)≥14+s(c2−x)1-V(x)≥ 14+s(c_2-x) in Eq. 20. An upper bound on cross-over payoffs. Consider a deviation of candidate 11 to c1′=c2+tc_1 =c_2+t with t∈(0,1−c2]t∈(0,1-c_2]. Every voter who votes for the deviator lies to the right of the new midpoint m′=c2+t2m =c_2+ t2, and votes with probability at most K(t)=tK(t)=t. Hence u1(c1′,c2)≤t(1−V(c2+t2))≤t(14−st2)≤maxτ≥0τ(14−sτ2)=132s,u_1(c_1 ,c_2)\;≤\;t\, (1-V(c_2+ t2) )\;≤\;t ( 14- st2 )\;≤\; _τ≥ 0\ τ ( 14- sτ2 )= 132s, where the second inequality uses 1−V(c2)=141-V(c_2)= 14 and V(c2+t2)−V(c2)≥st2V(c_2+ t2)-V(c_2)≥ st2. Combining with Eq. 24, no cross-over deviation of candidate 11 is profitable. The bound for candidate 22 is the mirror image (using V(c1)=14V(c_1)= 14 and V(c1)−V(c1−t2)≥st2V(c_1)-V(c_1- t2)≥ st2). By Proposition 1, the profile is a Nash equilibrium. Remark 8 The bound ρv≤4 _v≤ 4 is tight for this argument in the following sense: at ρv=4 _v=4 the deviator’s payoff can be exactly flat over an interval of alternative best responses. For instance, for the symmetric step density v=58v= 58 on [0,25)∪(35,1][0, 25)∪( 35,1] and v=52v= 52 on [25,35][ 25, 35] (density ratio exactly 44), the quartile profile is (25,35) ( 25, 35 ), and one computes from Eq. 20 that u1(c1′,35)=340u_1(c_1 , 35)= 340 for every c1′∈[15,25]c_1 ∈[ 15, 25]: candidate 11 is indifferent across a whole interval of positions, and the quartile profile is a (non-strict) equilibrium. Corollary 1, symmetric case (restated). Let κ∼U(0,1)κ U(0,1) and let v be symmetric around 12 12 with ρv≤4 _v≤ 4. Then the quartile profile (V−1(14),V−1(34)) (V^-1( 14),V^-1( 34) ) is a Nash equilibrium; in particular an equilibrium exists. Moreover, any profile satisfying the conditions of Corollary 1 must be symmetric, c2=1−c1c_2=1-c_1. Proof Symmetry of v gives V(x)+V(1−x)=1V(x)+V(1-x)=1 for all x, hence V−1(34)=1−V−1(14)V^-1( 34)=1-V^-1( 14); the midpoint of the quartile profile is then 12 12 and V(12)=12V( 12)= 12, so all three conditions of Corollary 1 hold and the profile is an equilibrium. Conversely, if V(c1)=14V(c_1)= 14 and V(c2)=34V(c_2)= 34, then V(1−c1)=1−V(c1)=34=V(c2)V(1-c_1)=1-V(c_1)= 34=V(c_2); since ρv≤4 _v≤ 4 forces infv>0 v>0, V is strictly increasing, so c2=1−c1c_2=1-c_1. Theorem 3.2 (restated). Let x∼U(0,1)x U(0,1) and let k satisfy Assumption 2.2. (a) Every Nash equilibrium is symmetric, c1=1−c2c_1=1-c_2, with gap Δ∈(0,1) ∈(0,1) solving K(Δ)=(1−Δ)k(Δ)K( )=(1- )k( ). (b) If moreover ψ(t)=t+K(t)/k(t)ψ(t)=t+K(t)/k(t) is strictly increasing on (0,1)(0,1), then this equation has a unique solution Δ∗∈(0,1) ^*∈(0,1), and (1−Δ∗2,1+Δ∗2) ( 1- ^*2, 1+ ^*2 ) is the unique Nash equilibrium up to relabeling. Proof (a) By Lemma 1, an equilibrium is interior with 0<Δ<10< <1, so both derivatives in Eq. 22 vanish: 12K(Δ)=c1k(Δ)and12K(Δ)=(1−c2)k(Δ). 12K( )=c_1\,k( ) 12K( )=(1-c_2)\,k( ). (25) Since Δ∈(0,1) ∈(0,1) and k>0k>0 on (0,1)(0,1), we have K(Δ)>0K( )>0; the first equation then forces k(Δ)>0k( )>0, so comparing the two equations gives c1=1−c2c_1=1-c_2, whence c1=1−Δ2c_1= 1- 2, and Eq. 25 becomes K(Δ)=(1−Δ)k(Δ)K( )=(1- )k( ). (b) On (0,1)(0,1), where k>0k>0, the equation K(Δ)=(1−Δ)k(Δ)K( )=(1- )k( ) is equivalent to ψ(Δ)=1ψ( )=1. Existence and uniqueness of the root. ψ is continuous on (0,1)(0,1) and strictly increasing, so L:=limt↓0ψ(t)L:= _t 0ψ(t) exists. Suppose L>0L>0; then K(t)/k(t)≥L/2K(t)/k(t)≥ L/2 for all sufficiently small t, i.e., (logK)′(t)=k(t)/K(t)≤2/L( K) (t)=k(t)/K(t)≤ 2/L, which upon integrating from t to a fixed small t0t_0 gives K(t)≥K(t0)e−2t0/L>0K(t)≥ K(t_0)\,e^-2t_0/L>0 as t↓0t 0—contradicting K(0)=0K(0)=0 and the continuity of K. Hence L=0<1L=0<1. At the other end, ψ exceeds 11 somewhere on (0,1)(0,1): otherwise K(t)/k(t)≤1−tK(t)/k(t)≤ 1-t, i.e. k(t)/K(t)≥1/(1−t)k(t)/K(t)≥ 1/(1-t), for all t∈(0,1)t∈(0,1), and integrating from t to s<1s<1 gives logK(s)−logK(t)=∫tsk/K≥log1−t1−s→∞ K(s)- K(t)= _t^sk/K≥ 1-t1-s→∞ as s↑1s 1, contradicting K(s)≤K(1)≤1K(s)≤ K(1)≤ 1. (No bound on k is needed: Assumption 2.2 permits k unbounded at the endpoints, as in Example 1 with p<1p<1.) By the intermediate value theorem and strict monotonicity, ψ(Δ)=1ψ( )=1 has a unique root Δ∗∈(0,1) ^*∈(0,1). The symmetric profile is an equilibrium. Let c1=1−Δ∗2c_1= 1- ^*2, c2=1+Δ∗2c_2= 1+ ^*2, and consider deviations of candidate 11 (the case of candidate 22 is the mirror image). Parameterize a deviation on the own side of the opponent by its distance t∈(0,c2]t∈(0,c_2] from c2c_2: by Eq. 21, the deviation payoff is uL(t)=(c2−t)K(t)+12G(t)u_L(t)=(c_2-t)K(t)+ 12G(t), with derivative uL′(t)=(c2−t)k(t)−12K(t)=k(t)[c2−t+ψ(t)2].u_L (t)=(c_2-t)\,k(t)- 12K(t)=k(t) [c_2- t+ψ(t)2 ]. Since t↦t+ψ(t)t t+ψ(t) is strictly increasing and equals 2c2=1+Δ∗2c_2=1+ ^* exactly at t=Δ∗t= ^* (because ψ(Δ∗)=1ψ( ^*)=1), the bracket is positive for t<Δ∗t< ^* and negative for t>Δ∗t> ^*: uLu_L is strictly single-peaked with maximum at t=Δ∗t= ^*, i.e., at c1c_1 itself. A deviation across the opponent to c1′=c2+tc_1 =c_2+t, t∈(0,1−c2]t∈(0,1-c_2], yields by Eq. 21 the payoff uR(t)=(1−c2−t)K(t)+12G(t)≤uL(t)u_R(t)=(1-c_2-t)K(t)+ 12G(t)≤ u_L(t), since 1−c2<c21-c_2<c_2; hence its supremum is at most maxtuL(t)=u1(c1,c2) _tu_L(t)=u_1(c_1,c_2). Finally, the tie c1′=c2c_1 =c_2 yields 0<u1(c1,c2)0<u_1(c_1,c_2). So no deviation is profitable, and the profile is an equilibrium. Uniqueness. By part (a), any equilibrium is symmetric with gap solving ψ(Δ)=1ψ( )=1, i.e., Δ=Δ∗ = ^*. Proposition 3 (restated). Let x∼U(0,1)x U(0,1) and let k,k~k, k both satisfy the hypotheses of Theorem 3.2(b), with equilibrium gaps Δ∗,Δ~∗ ^*, ^*. If k~/K~≥k/K k/ K≥ k/K pointwise on (0,1)(0,1), then Δ~∗≥Δ∗ ^*≥ ^*. If in addition K~(1)≤K(1) K(1)≤ K(1), then K~≤K K≤ K pointwise. Proof First claim. Since k,k~>0k, k>0 on (0,1)(0,1), the hypothesis gives K~(t)/k~(t)≤K(t)/k(t) K(t)/ k(t)≤ K(t)/k(t), hence ψ~(t)=t+K~(t)/k~(t)≤t+K(t)/k(t)=ψ(t) ψ(t)=t+ K(t)/ k(t)≤ t+K(t)/k(t)=ψ(t) for all t∈(0,1)t∈(0,1), so ψ~(Δ∗)≤ψ(Δ∗)=1=ψ~(Δ~∗), ψ( ^*)\;≤\;ψ( ^*)=1= ψ( ^*), and since ψ~ ψ is strictly increasing, Δ∗≤Δ~∗ ^*≤ ^*. No condition on K~(1) K(1) is used. Second claim. The reverse hazard rate is the derivative of logK K, and ∫t1k/K<∞ _t^1k/K<∞ for t∈(0,1)t∈(0,1) because K≥K(t)>0K≥ K(t)>0 on [t,1][t,1] and ∫t1k=K(1)−K(t) _t^1k=K(1)-K(t). Hence, for t∈(0,1)t∈(0,1), using K~(1)≤K(1) K(1)≤ K(1), logK~(t)=logK~(1)−∫t1k~(s)K~(s)s≤logK(1)−∫t1k(s)K(s)s=logK(t), K(t)= K(1)- _t^1 k(s) K(s)\,ds\;≤\; K(1)- _t^1 k(s)K(s)\,ds= K(t), so K~≤K K≤ K pointwise, i.e., costs under K~ K are stochastically higher. Proposition 14 (restated). Under uniform voters, k>0k>0 on (0,1)(0,1), and alienation with parameter α≥0α≥ 0, the pure Nash equilibria of the margin game are exactly E=[12+α,1+α2+α]2E= [ 12+α, 1+α2+α ]^2, each with margin 00, independently of K. Proof We first treat α>0α>0; the case α=0α=0 is handled at the end. Take any profile with c1≤c2c_1≤ c_2 and let G(t)=∫0tKG(t)= _0^tK. Candidate 11’s benefit tent has arms of slope α (outer) and 2+α2+α (inner); changing variables along each arm, and subtracting the part of the outer arm that is truncated by the boundary 00, gives u1=(1α+12+α)G(Δ)−1αG(b1),b1:=(B1(0))+=(Δ−αc1)+,u_1= ( 1α+ 12+α )G( )- 1αG(b_1), b_1:= (B_1(0) )^+=( -α c_1)^+, and symmetrically for u2u_2 with b2=(B2(1))+=(Δ−α(1−c2))+b_2= (B_2(1) )^+= ( -α(1-c_2) )^+ (throughout this proof, b1,b2b_1,b_2 denote the boundary benefits of Section 4.1 truncated at zero). Hence the margin is w1=u1−u2=1α(G(b2)−G(b1)).w_1\;=\;u_1-u_2\;=\; 1α (G(b_2)-G(b_1) ). (26) Since k>0k>0 on (0,1)(0,1), G is strictly increasing on (0,1](0,1], so w1w_1 and b2−b1b_2-b_1 have the same sign whenever max(b1,b2)>0 (b_1,b_2)>0, and w1=0w_1=0 iff b1=b2b_1=b_2. In words: a candidate leads exactly when her opponent’s tent is more severely truncated by the boundary. Profiles in E have margin 00. If c1≥12+αc_1≥ 12+α and c2≤1+α2+αc_2≤ 1+α2+α, then Δ−αc1=c2−(1+α)c1≤1+α2+α−1+α2+α=0 -α c_1=c_2-(1+α)c_1≤ 1+α2+α- 1+α2+α=0, and likewise Δ−α(1−c2)=(1+α)c2−c1−α≤0 -α(1-c_2)=(1+α)c_2-c_1-α≤ 0: both tents are interior, b1=b2=0b_1=b_2=0, and w1=0w_1=0. When can candidate 11 secure a positive margin? By Eq. 26, w1>0w_1>0 at a profile (c1′,c2)(c_1 ,c_2) with c1′<c2c_1 <c_2 requires b2′>b1′b_2 >b_1 , i.e., either (i) b1′=0<b2′b_1 =0<b_2 , which reads c1′≥c21+αc_1 ≥ c_21+α together with c1′<c2−α(1−c2)c_1 <c_2-α(1-c_2), and admits a solution iff c21+α<c2−α(1−c2) c_21+α<c_2-α(1-c_2), i.e., iff c2>1+α2+αc_2> 1+α2+α; or (i) b2′>b1′>0b_2 >b_1 >0, which reads c1′>1−c2c_1 >1-c_2 together with c1′<min(c21+α,c2−α(1−c2))c_1 < ( c_21+α,\,c_2-α(1-c_2) ), and both 1−c2<c21+α1-c_2< c_21+α and 1−c2<c2−α(1−c2)1-c_2<c_2-α(1-c_2) are again equivalent to c2>1+α2+αc_2> 1+α2+α. Thus candidate 11 can achieve a strictly positive margin at some position left of c2c_2 iff c2>1+α2+αc_2> 1+α2+α; by the reflection x↦1−x 1-x, she can achieve one at some position right of c2c_2 iff c2<12+αc_2< 12+α. The mirror statements hold for candidate 22 with respect to c1c_1. Profiles in E are equilibria. Let (c1,c2)∈E(c_1,c_2)∈ E. Then c2∈[12+α,1+α2+α]c_2∈ [ 12+α, 1+α2+α ], so by the previous paragraph no deviation of candidate 11 achieves a positive margin; since w1=0w_1=0 at the profile, no deviation is profitable. Symmetrically for candidate 22. (Note that E includes tied profiles, which are thus zero-turnout, zero-margin equilibria of the margin game.) Profiles outside E are not equilibria. Consider a profile violating the bounds; after relabeling, either c2>1+α2+αc_2> 1+α2+α or c1<12+αc_1< 12+α (or both). Say c2>1+α2+αc_2> 1+α2+α (the other case is the mirror image). If w1≤0w_1≤ 0 at the profile, candidate 11 deviates into the window of case (i) above and achieves w1′>0≥w1w_1 >0≥ w_1: profitable. If w1>0w_1>0, then w2<0w_2<0, and candidate 22 deviates to the mirror position c2′=1−c1c_2 =1-c_1, where the two truncation levels coincide and hence w2′=0>w2w_2 =0>w_2: profitable. In all cases some candidate has a profitable deviation, so the profile is not an equilibrium. The case α=0α=0. Here B1≡ΔB_1≡ to the left of c1c_1, so by Eq. 21, w1=(c1−(1−c2))K(Δ)w_1= (c_1-(1-c_2) )K( ), and E=(12,12)E=\( 12, 12)\. At (12,12)( 12, 12), a deviation of candidate 11 to distance δ>0δ>0 from the opponent (on either side) yields margin −δK(δ)<0-δ K(δ)<0, so the median profile is an equilibrium and it is strict. Conversely, at any other profile: if c1<c2c_1<c_2 and c1≠1−c2c_1≠ 1-c_2, then w1≠0w_1≠ 0 and the trailing candidate deviates to the mirror position to obtain margin 00; if c1=1−c2<12c_1=1-c_2< 12, candidate 11 deviates to c1′∈(c1,c2)c_1 ∈(c_1,c_2) and obtains w1′=(c1′−c1)K(c2−c1′)>0w_1 =(c_1 -c_1)K(c_2-c_1 )>0; and if c1=c2=c≠12c_1=c_2=c≠ 12, say c>12c> 12, then candidate 11 deviates to c1′∈(1−c,c)c_1 ∈(1-c,c) and obtains w1′=(c1′−(1−c))K(c−c1′)>0w_1 = (c_1 -(1-c) )K(c-c_1 )>0. Hence (12,12) ( 12, 12 ) is the unique equilibrium. Finally, all the thresholds above are determined by the benefits BiB_i alone; K enters only through the strictly increasing G, so the set E is independent of K. Theorem 5.2 (restated). Let α=0α=0, let v,kv,k satisfy Assumption 2.2, and let cmed=V−1(12)c_med=V^-1( 12). Then (cmed,cmed)(c_med,c_med) is the unique Nash equilibrium of the margin game, and every unilateral deviation from it strictly lowers the deviator’s margin. Proof Write q:=cmedq:=c_med, which is well-defined, unique, and lies in (0,1)(0,1) since V is strictly increasing under Assumption 2.2. Step 1: any deviation against the median loses strictly. Consider first a deviation of candidate 11 to c1′<qc_1 <q, and write Δ′=q−c1′∈(0,1) =q-c_1 ∈(0,1) and m′=c1′+q2m = c_1 +q2. The deviator collects the constant-gain region [0,c1′][0,c_1 ] (probability K(Δ′)K( ) each) plus the inner arm on [c1′,m′][c_1 ,m ]; the opponent collects the constant-gain region [q,1][q,1], of voter mass 1−V(q)=121-V(q)= 12, plus the inner arm on [m′,q][m ,q]. Substituting s=m′−xs=m -x and s=x−m′s=x-m in the two inner-arm integrals, w1(c1′,q)=K(Δ′)(V(c1′)−12)+∫0Δ′/2K(2s)[v(m′−s)−v(m′+s)]s.w_1(c_1 ,q)=K( ) (V(c_1 )- 12 )+ _0 /2K(2s)\, [v(m -s)-v(m +s) ]\,ds. Bounding the positive part of the integral using K(2s)≤K(Δ′)K(2s)≤ K( ) for s≤Δ′/2s≤ /2, ∫0Δ′/2K(2s)v(m′−s)s≤K(Δ′)(V(m′)−V(c1′)), _0 /2K(2s)\,v(m -s)\,ds\;≤\;K( )\, (V(m )-V(c_1 ) ), we obtain w1(c1′,q) w_1(c_1 ,q)\; ≤K(Δ′)(V(m′)−12)−∫0Δ′/2K(2s)v(m′+s)s ≤\;K( ) (V(m )- 12 )- _0 /2K(2s)\,v(m +s)\,ds ≤K(Δ′)(V(m′)−12)< 0, ≤\;K( ) (V(m )- 12 )\;<\;0, where the final strict inequality holds because K(Δ′)>0K( )>0 (as k>0k>0 on (0,1)(0,1) and 0<Δ′≤q<10< ≤ q<1) and V(m′)<V(q)=12V(m )<V(q)= 12 (as m′<qm <q and v>0v>0). For a deviation to c1′>qc_1 >q, the mirror computation gives w1(c1′,q)≤K(Δ′)(12−V(m′))<0w_1(c_1 ,q)≤ K( ) ( 12-V(m ) )<0, now with m′=c1′+q2>qm = c_1 +q2>q. Finally, w1(q,q)=0w_1(q,q)=0. Hence every unilateral deviation from (q,q)(q,q) strictly lowers the deviator’s margin, and (q,q)(q,q) is an equilibrium. Step 2: uniqueness. Suppose (c1,c2)(c_1,c_2) is an equilibrium with, say, c1≠qc_1≠ q. Deviating to q is available to candidate 22, and by Step 1 (with the roles of the players reversed), w2(c1,q)=−w1(c1,q)>0w_2(c_1,q)=-w_1(c_1,q)>0; equilibrium therefore requires w2(c1,c2)≥w2(c1,q)>0w_2(c_1,c_2)≥ w_2(c_1,q)>0, i.e., w1(c1,c2)<0w_1(c_1,c_2)<0. But candidate 11 could deviate to q: if c2≠qc_2≠ q then w1(q,c2)>0w_1(q,c_2)>0 by Step 1, and if c2=qc_2=q then w1(q,q)=0w_1(q,q)=0; either way candidate 11’s margin strictly improves, contradicting equilibrium. Hence c1=qc_1=q, and symmetrically c2=qc_2=q. Appendix 0.C Existence and Uniqueness of Equilibria This appendix collects the existence arguments and the (non-)uniqueness examples referenced in the main text. Recall the distinction stressed in Section 3.1: the balance conditions of Lemmas 2, 3.1 and 3.2 are necessary for equilibrium. Below we show that solutions of the conditions always exist in the two benchmark settings; whether these solutions are genuine equilibria is a separate question—answered affirmatively under the sufficient conditions of the main text, and negatively in general (see the example below and Appendix 0.F). Solutions of the conditions always exist under uniform costs. Under Assumption 2.2, V is strictly increasing, so its inverse V−1V^-1 is well-defined on [0,1][0,1]. For any c1∈[0,V−1(12)]c_1∈ [0,V^-1( 12) ], define c2(c1):=V−1(V(c1)+12),c_2(c_1)\;:=\;V^-1\! (V(c_1)+ 12 ), which is well-defined since V(c1)∈[0,12]V(c_1)∈[0, 12] on this domain. Next, define f(c1):=V(c1+c2(c1)2)−2V(c1).f(c_1)\;:=\;V\! ( c_1+c_2(c_1)2 )-2V(c_1). By construction, f is continuous on [0,V−1(12)] [0,V^-1( 12) ]. Moreover, f(0)=V(V−1(12)2)≥0,f(0)=V\! ( V^-1( 12)2 )≥ 0, and, letting q:=V−1(12)q:=V^-1( 12), so that c2(q)=V−1(1)=1c_2(q)=V^-1(1)=1, f(q)=V(q+12)−1≤0.f(q)=V\! ( q+12 )-1≤ 0. Hence there exists c1∗∈[0,q]c_1 ∈[0,q] such that f(c1∗)=0f(c_1 )=0. Setting c2∗=c2(c1∗)c_2 =c_2(c_1 ) yields V(c2∗)=V(c1∗)+12andV(c1∗+c2∗2)=2V(c1∗),V(c_2 )=V(c_1 )+ 12 V\! ( c_1 +c_2 2 )=2V(c_1 ), so (c1∗,c2∗)(c_1 ,c_2 ) satisfies the conditions of Theorem 3.1. Whether it is an equilibrium requires a separate verification—Appendix 0.F shows it may fail; under the hypotheses of Corollary 1 (symmetric case), equilibrium existence is guaranteed. Solutions of the conditions always exist under uniform voters. Define, on (0,1)(0,1), f(Δ):=K(Δ)−(1−Δ)k(Δ),f( )\;:=\;K( )-(1- )\,k( ), which is continuous there by Assumption 2.2. Near 00, f<0f<0 at points arbitrarily close to 00: if f≥0f≥ 0 on some (0,t0)(0,t_0) with t0≤12t_0≤ 12, then k/K≤1/(1−t)≤2k/K≤ 1/(1-t)≤ 2 there, so logK(t0)−logK(t)=∫t0k/K≤2t0 K(t_0)- K(t)= _t^t_0k/K≤ 2t_0 for all t∈(0,t0)t∈(0,t_0) (note K(t0)>0K(t_0)>0 since k>0k>0 on (0,1)(0,1)); hence K(t)≥K(t0)e−2t0>0K(t)≥ K(t_0)e^-2t_0>0 as t↓0t 0, contradicting K(0)=0K(0)=0 and the continuity of K. Near 11, f>0f>0 at points arbitrarily close to 11: if f≤0f≤ 0 on some (t1,1)(t_1,1) with t1∈(0,1)t_1∈(0,1), then k/K≥1/(1−t)k/K≥ 1/(1-t) there, so logK(t)−logK(t1)=∫t1tk/K≥log1−t11−t→∞ K(t)- K(t_1)= _t_1^tk/K≥ 1-t_11-t→∞ as t↑1t 1, contradicting K≤1K≤ 1. By the intermediate value theorem, f has a root Δ∗∈(0,1) ∈(0,1). Under the monotonicity condition of Theorem 3.2(b), this root is unique and yields the (unique) equilibrium. Without that condition, some roots may fail to be equilibria, as the following example shows. (When k(0+)=0k(0^+)=0, extending k by continuity to 00 makes Δ=0 =0 a further solution of K(Δ)=(1−Δ)k(Δ)K( )=(1- )k( ); it is not an equilibrium gap, since Theorem 3.2(a) requires Δ∈(0,1) ∈(0,1).) Non-uniqueness (and non-sufficiency) under uniform voters. Consider the cost density k(Δ)=8Δ2−163Δ+1k( )=8 ^2- 163 +1 (a valid density: K(1)=1K(1)=1 and k≥19>0k≥ 19>0, with minimum at Δ=13 = 13). The equation K(Δ)=(1−Δ)k(Δ)K( )=(1- )k( ) admits three solutions, Δ∈14,12,34 ∈ \ 14, 12, 34 \ (the function ψ of Eq. 6 is not monotone for this k, so Theorem 3.2(b) does not apply). The three corresponding symmetric profiles behave very differently: • Δ=14 = 14, profile (38,58) ( 38, 58 ): a Nash equilibrium (verified by computing global best responses numerically), with u1=u2=1312304≈0.057u_1=u_2= 1312304≈ 0.057; • Δ=34 = 34, profile (18,78) ( 18, 78 ): a Nash equilibrium, with u1=u2=27256≈0.105u_1=u_2= 27256≈ 0.105; • Δ=12 = 12, profile (14,34) ( 14, 34 ): not a Nash equilibrium. Here u1=14K(12)+12∫01/2K=572≈0.069u_1= 14K ( 12 )+ 12 _0^1/2K= 572≈ 0.069, but candidate 11’s best response to c2=34c_2= 34 is c1′≈0.473c_1 ≈ 0.473, yielding a payoff of ≈0.073≈ 0.073. The profile solves both first-order conditions, yet it is not even a local equilibrium: Remark 16 shows this middle root to be a local minimum of the deviator’s payoff, so deviating either way pays. Thus equation (7) may have multiple roots; the set of equilibria may be a strict subset of these roots; and, in particular, the necessary condition of Theorem 3.2(a) cannot be reversed without an assumption such as the monotonicity of ψ. Whether at least one root is always an equilibrium (i.e., whether a pure equilibrium always exists under uniform voters) is open. For comparison, the simple sufficient conditions for the monotonicity of ψ (hence for uniqueness and sufficiency) are the two that give super-regularity, k non-increasing or log-concave ((6) and Theorem 3.2(b)); Fig. 5 visualizes the resulting unique crossing for three such families. Non-uniqueness under uniform costs. Uniqueness can also fail when the voter density (rather than the cost density) is non-uniform; Appendix 0.E works out a symmetric voter density that admits three distinct equilibria, two of which are asymmetric. Mixed equilibria. Finally, we note that although the existence of pure equilibria is open in general, the payoffs uiu_i are continuous on the compact strategy space [0,1]2[0,1]^2 (continuity at tied profiles holds because vote shares vanish as the candidates approach each other), so a possibly-mixed Nash equilibrium always exists by Glicksberg’s fixed-point theorem [39]. Appendix 0.D Power Costs against a Non-Uniform Electorate This appendix records a case that no sufficiency result in the main text reaches, and where the conditions nonetheless appear to deliver equilibria. Remark 9 (Power costs against a non-uniform electorate) [numerical observation] Every sufficiency result above misses this family. For K(t)=tpK(t)=t^p the cost density is k(t)=ptp−1k(t)=p\,t^p-1, which is increasing when p>1p>1; and when p<1p<1 it is non-increasing, but ρk:=k(0+)/k(1−) _k:=k(0^+)/k(1^-) is infinite. So the criterion ρvρk≤4 _v _k≤ 4 of Proposition 16 reaches the family only at p=1p=1, where it is Corollary 1, and Theorem 3.2 needs a uniform electorate. For a symmetric electorate the gap is narrower than that makes it sound. By Lemma 6, own-side optimality alone makes a mirror profile an equilibrium whatever the cost distribution, so the cross-over deviations are already handled and what is unproved is exactly this: that the stationary point maximizes u1(⋅,c2)u_1(·,c_2) on [0,c2)[0,c_2). On this family ψ is linear, ψ(t)=(1+1p)tψ(t)= (1+ 1p )t, which is as favourable a setting as the paper offers. Numerically the profile is an equilibrium throughout. Scanning (0,12)(0, 12) for symmetric stationary profiles, each of the five symmetric electorates below has exactly one at each p∈12,1,2,3p∈\ 12,1,2,3\, and in all twenty cases it is a Nash equilibrium; the largest profitable deviation found was 1.1×10−161.1× 10^-16. The quantile V(c1∗)V(c_1^*) it sits at: electorate p=12p= 12 p=1p=1 p=2p=2 p=3p=3 uniform 0.333330.33333 0.250.25 0.166670.16667 0.125000.12500 Beta(1.5,1.5)Beta(1.5,1.5) 0.333860.33386 0.250.25 0.163340.16334 0.118590.11859 Beta(2,2)Beta(2,2) 0.334090.33409 0.250.25 0.161760.16176 0.115410.11541 Beta(3,3)Beta(3,3) 0.334300.33430 0.250.25 0.160230.16023 0.112270.11227 hollow cosine (Fig. 4) 0.320150.32015 0.250.25 0.185950.18595 0.150770.15077 At p=1p=1 it is 14 14 for every electorate, which is Corollary 1. Away from p=1p=1 it moves, but by less than 0.030.03 across this range, and the departure from the uniform-voter value 12(1+p) 12(1+p) of Remark 5 is signed by the shape of v: a peaked electorate raises the quantile when p<1p<1 and lowers it when p>1p>1, a hollow one does the reverse, and both effects vanish at p=1p=1 and grow with |p−1||p-1| and with the distance of v from uniform. Distribution-freeness is therefore not merely lost away from p=1p=1; it degrades continuously, and p=1p=1 is the point where it is exact. Appendix 0.E Example with Multiple Equilibria (Uniform Costs) Consider the continuous, symmetric density v:[0,1]→ℝ≥0v:[0,1] _≥ 0 given by v(x)=57,0≤x≤1950,250x−907,1950≤x≤2150,157,2150≤x≤2950,160−250x7,2950≤x≤3150,57,3150≤x≤1.v(x)= cases 57,&0≤ x≤ 1950,\\[8.0pt] 250x-907,& 1950≤ x≤ 2150,\\[10.0pt] 157,& 2150≤ x≤ 2950,\\[8.0pt] 160-250x7,& 2950≤ x≤ 3150,\\[10.0pt] 57,& 3150≤ x≤ 1. cases Under uniform costs, the conditions of Theorem 3.1 admit three distinct solutions, (c1,c2)∈(316,916),(720,1320),(716,1316),(c_1,c_2)∈ \ ( 316, 916 ),\; ( 720, 1320 ),\; ( 716, 1316 ) \, and—in contrast to the cost-side example of Appendix 0.C—we verified numerically that here all three profiles are genuine Nash equilibria, with payoffs (u1,u2)≈(0.075,0.212)(u_1,u_2)≈(0.075,0.212), (0.106,0.106)(0.106,0.106) and (0.212,0.075)(0.212,0.075), respectively. Note that although v is symmetric around 12 12, two of the three equilibria are asymmetric mirror images of one another. (This density has ρv=3≤4 _v=3≤ 4, but the outer equilibria are not quartile profiles: Corollary 1 identifies some equilibria, not necessarily all of them.) Appendix 0.F Stationary Profiles Need Not Be Equilibria (Uniform Costs) We now show that the conditions of Theorem 3.1 are not sufficient: a profile can satisfy them and still fail to be a Nash equilibrium. Let h:[0,1]→ℝ≥0h:[0,1] _≥ 0 be the continuous piecewise-linear function taking the constant values 6on[0,0.08],120on[0.10,0.20],5on[0.22,0.34],120on[0.36,0.44],6\ on\ [0,0.08], 120\ on\ [0.10,0.20], 5\ on\ [0.22,0.34], 120\ on\ [0.36,0.44], 3on[0.46,0.62],120on[0.64,0.75],2on[0.77,1],3\ on\ [0.46,0.62], 120\ on\ [0.64,0.75], 2\ on\ [0.77,1], and interpolating linearly on the six gaps of width 0.020.02. Then ∫01h=911400 _0^1h= 911400, so v:=400911hv:= 400911\,h is a continuous density, strictly positive on [0,1][0,1] (Assumption 2.2 holds), describing an electorate with four ideological “clusters” of decreasing mass. Solving the conditions of Theorem 3.1 numerically yields exactly three solutions: (c1,c2)≈(0.226, 0.621),(0.232, 0.709),(0.248, 0.798).(c_1,c_2)\;≈\;(0.226,\,0.621), (0.232,\,0.709), (0.248,\,0.798). Computing global best responses, the first and the third profiles are Nash equilibria. The middle profile is not: candidate 22’s payoff there is u2≈0.1353u_2≈ 0.1353, but relocating to c2′≈0.789c_2 ≈ 0.789 (adjacent to the fourth cluster) yields ≈0.1365≈ 0.1365—a profitable deviation to a different local maximum of her payoff. Her platform is therefore a local maximum that is not a global one, so the profile is a local equilibrium without being a Nash equilibrium: the opposite failure to the cost-side example of Appendix 0.C, where the middle root is not even a local equilibrium. Note that ρv=120 _v=120 here, far above the bound of Proposition 1; with such uneven densities the candidates’ payoffs are multi-peaked, and stationarity no longer implies global optimality. Appendix 0.G Turnout under Linear Costs We compute each candidate’s vote share at a symmetric profile under uniform voters and the linear cost family kϑ(x)=ϑx+(1−ϑ2)k_ (x)= x+ (1- 2 ), ϑ∈[−2,2] ∈[-2,2], for which Kϑ(Δ)=ϑ2Δ2+(1−ϑ2)Δ,∫0ΔKϑ(y)y=ϑ6Δ3+2−ϑ4Δ2.K_ ( )= 2 ^2+ (1- 2 ) , _0 K_ (y)\,dy= 6 ^3+ 2- 4 ^2. At the symmetric profile with gap Δ (so c1=1−Δ2c_1= 1- 2), Eq. 21 gives u1(Δ,ϑ) u_1( ; ) =c1Kϑ(Δ)+12∫0ΔKϑ(y)y =c_1K_ ( )+ 12 _0 K_ (y)\,dy =1−Δ2[ϑ2Δ2+(1−ϑ2)Δ]+ϑ12Δ3+2−ϑ8Δ2 = 1- 2 [ 2 ^2+ (1- 2 ) ]+ 12 ^3+ 2- 8 ^2 =−ϑ6Δ3+(3ϑ8−14)Δ2+(12−ϑ4)Δ. =- 6 ^3+ ( 3 8- 14 ) ^2+ ( 12- 4 ) . By symmetry u2=u1u_2=u_1, so total turnout at the profile is T(Δ,ϑ)= 2u1(Δ,ϑ)=−ϑ3Δ3+(3ϑ4−12)Δ2+(1−ϑ2)Δ.T( ; )\;=\;2\,u_1( ; )\;=\;- 3 ^3+ ( 3 4- 12 ) ^2+ (1- 2 ) . (27) (As a check: at ϑ=0 =0, Δ=12 = 12 we get u1=316u_1= 316 and T=38T= 38, matching Corollary 2.) Appendix 0.H Turnout Comparative Statics: A Worked Example This appendix expands the causality comparison of Section 3.2, using the turnout formula (27) of Appendix 0.G. Consider a uniform voter distribution and a uniform cost distribution, for which the unique equilibrium is at (14,34) ( 14, 34 ) with overall turnout 38 38. First suppose the candidates remain fixed at (14,34) ( 14, 34 ) (i.e., Δ=12 = 12) while voting costs increase within the linear cost family, i.e., ϑ increases over [−2,2][-2,2]. By Eq. 27, turnout is then T(12,ϑ)=38−5ϑ48T ( 12; )= 38- 5 48, declining from 1424≅0.58 1424 0.58 at ϑ=−2 =-2, to 924=0.375 924=0.375 under uniform costs (ϑ=0 =0), and further to 424≅0.17 424 0.17 at ϑ=2 =2—a baseline effect driven solely by rising costs and decreasing engagement. Next, instead of holding the candidates fixed, consider turnout in equilibrium as ϑ varies over [−2,2][-2,2]: within the log-concave linear family the equilibrium gap (Theorem 3.2(b); Δ∗(ϑ)=2(ϑ−1)+(ϑ−1)2+33ϑ ( )= 2( -1)+ ( -1)^2+33 for ϑ≠0 ≠ 0) Δ∗(ϑ) ( ) widens from ≈0.42≈ 0.42 at ϑ=−2 =-2 to 23 23 at ϑ=2 =2, and equilibrium turnout equals T(Δ∗(ϑ),ϑ)T ( ( ); ). Although turnout still decreases, the decline is more modest (from ≈0.54≈ 0.54 at ϑ=−2 =-2, through 0.3750.375 at ϑ=0 =0, to 2081≅0.25 2081 0.25 at ϑ=2 =2), because the candidates become more polarized and a wider gap raises each voter’s benefit from participating. Thus, at least in this scenario, abstention increases polarization, whereas polarization has the opposite effect on abstention. Whether this pattern persists under other distributions is left to future work. Appendix 0.I Preliminaries for all (α,β)(α,β) Standing notation and conventions. Voters x∈[0,1]x∈[0,1] have CDF V with density v; voting costs κ∈[0,1]κ∈[0,1] are i.i.d. with CDF K and density k, independent of position. We assume Assumption 2.2 throughout. We allow K(1)≤1K(1)≤ 1 (a mass 1−K(1)1-K(1) of voters never votes); everything below uses only K(1)>0K(1)>0. Candidates i∈1,2i∈\1,2\ choose ci∈[0,1]c_i∈[0,1]; write di(x)=|x−ci|d_i(x)=|x-c_i| and, for α≥0α≥ 0, Bi(x)=d−i(x)−(1+α)di(x)B_i(x)=d_-i(x)-(1+α)d_i(x), so that i’s vote share is ui=∫01K(Bi(x)+)v(x)xu_i= _0^1K\! (B_i(x)^+ )v(x)\,dx, and candidate i maximizes Ui=ui−βu−iU_i=u_i-β u_-i with β∈[0,1]β∈[0,1]. When c1<c2c_1<c_2 we write Δ=c2−c1 =c_2-c_1, m=c1+c22m= c_1+c_22, and λ:=2+α,xin1=c1+Δλ,xout1=c1−Δα,xin2=c2−Δλ,xout2=c2+Δα, gatheredλ:=2+α, x^in_1=c_1+ λ, x^out_1=c_1- α,\\ x^in_2=c_2- λ, x^out_2=c_2+ α, gathered for the inner and outer zeros of the two benefit tents (x1out=−∞x^out_1=-∞, x2out=+∞x^out_2=+∞ when α=0α=0), and b1=B1(0)=Δ−αc1,b2=B2(1)=Δ−α(1−c2)b_1=B_1(0)= -α c_1, b_2=B_2(1)= -α(1-c_2) for the boundary benefits; candidate 11’s tent is truncated by the boundary (equivalently, she has no dead zone [0,x1out][0,x^out_1] of unreachable voters on her flank) iff b1≥0b_1≥ 0, i.e. x1out≤0x^out_1≤ 0. Two identities used repeatedly: with bs(Δ):=λΔ−α2=Δ−α2(1−Δ)b_s( ):= λ -α2= - α2(1- ), b1+b2=2bs(Δ),b2−b1=α(c1+c2−1).b_1+b_2=2\,b_s( ), b_2-b_1=α\,(c_1+c_2-1). (28) Finally G(t)=∫0tKG(t)= _0^tK, r=k/Kr=k/K (the reverse hazard rate), ψ(t)=t+K(t)/k(t)ψ(t)=t+K(t)/k(t), ρv=supv/infv∈[1,∞] _v= v/ v∈[1,∞], ρk=supk/infk _k= k/ k, and Δ0:=αλ _0:= αλ, the most polarized symmetric gap inside the margin-game band E(α)=[1λ,1+αλ]2E(α)= [ 1λ, 1+αλ ]^2. Two elementary facts are used without further comment. First, since K(0)=0K(0)=0 and K>0K>0 on (0,1](0,1], ∫0+t0r(s)s=limt↓0(logK(t0)−logK(t))=+∞(t0∈(0,1)), _0^+^t_0r(s)\,ds= _t 0 ( K(t_0)- K(t) )=+∞ (t_0∈(0,1)), (29) and symmetrically ∫t1−r=logK(1)−logK(t)<∞ _t^1^-r= K(1)- K(t)<∞. Second, for every x at most one of B1(x),B2(x)B_1(x),B_2(x) is strictly positive, because B1+B2=−α(d1+d2)≤0B_1+B_2=-α(d_1+d_2)≤ 0; hence total turnout satisfies T=u1+u2≤K(1)≤1T=u_1+u_2≤ K(1)≤ 1 at every profile. The two lemmas of this section extend Lemma 1 and the balance conditions of Lemma 2 from the baseline (α,β)=(0,0)(α,β)=(0,0) to the whole parameter square, and are the entry point for every result below. Lemma 4 (Interiority and no ties, extended) Let Assumption 2.2 hold and β∈[0,1)β∈[0,1). Then every pure Nash equilibrium is untied and interior (c1≠c2c_1≠ c_2 and, after labeling c1<c2c_1<c_2, 0<c1<c2<10<c_1<c_2<1) in each of the following cases: (a) β=0β=0, any α≥0α≥ 0 and any (v,k)(v,k); (b) α=0α=0, any (v,k)(v,k); (c) uniform voters x∼U(0,1)x U(0,1), any α≥0α≥ 0 and any k. Proof No boundary positions (all three cases). Suppose c1=0<c2c_1=0<c_2. By Lemma 5 below, the right derivative of u1u_1 at c1=0c_1=0 is (1+α)∫0x1ink(B1)vx(1+α) _0^x^in_1\!k(B_1)v\,dx, which is positive (the left integral in Eq. 30 is empty). At the same time ∂u2/∂c1∂ u_2/∂ c_1 equals −∫x2inx2out∧1k(B2)vdx- _x^in_2^x^out_2 1\!k(B_2)v\,dx, which is at most 00. Hence ∂U1/∂c1=∂u1/∂c1−β∂u2/∂c1>0∂ U_1/∂ c_1=∂ u_1/∂ c_1-β\,∂ u_2/∂ c_1>0: a small move to the right strictly increases U1U_1, so c1=0c_1=0 is not a best response. The mirror argument rules out c2=1c_2=1. No ties, case (a). At a tie both payoffs vanish, and for β=0β=0 the deviation argument of the baseline lemma applies verbatim: any c1′c_1 strictly between the tie point and the nearer end of (0,1)(0,1) yields u1(c1′,c)>0=U1(c,c)u_1(c_1 ,c)>0=U_1(c,c). No ties, case (b). Let α=0α=0 and c1=c2=c_1=c_2=c. For ε>0 >0 small consider the deviation c1′=c−εc_1 =c- (possible if c>0c>0). Every voter in [0,c−ε][0,c- ] gains exactly ε from the deviator, and every voter gains at most ε from either candidate, so with m′=c−ε2m =c- 2, u1≥K(ε)V(c−ε),u2≤K(ε)(1−V(m′)),u_1≥ K( )\,V(c- ), u_2≤ K( )\, (1-V(m ) ), whence U1≥K(ε)[V(c−ε)−β(1−V(m′))]U_1≥ K( ) [V(c- )-β (1-V(m ) ) ]. As ε↓0 0 the bracket tends to (1+β)[V(c)−β1+β](1+β) [V(c)- β1+β ], which is positive whenever V(c)>β1+βV(c)> β1+β; since K(ε)>0K( )>0 for ε∈(0,1) ∈(0,1), the deviation is then profitable for small ε . If instead V(c)≤β1+β<11+βV(c)≤ β1+β< 11+β, the mirror deviation c1′=c+εc_1 =c+ gives a bracket tending to (1+β)[1−V(c)−β1+β]>0(1+β) [1-V(c)- β1+β ]>0. (At c=0c=0 only the second option is available and applies since V(0)=0V(0)=0; at c=1c=1 only the first, since V(1)=1V(1)=1.) No ties, case (c). Let v≡1v≡ 1 and c1=c2=c_1=c_2=c; by symmetry assume c>0c>0 and deviate to c1′=c−εc_1 =c- with ε<minc,αc1+α < \c, α c1+α\ (the case c=0c=0 is the mirror image; for α=0α=0 the claim is case (b)). By the closed form (32) below, the deviator’s tent is interior (b1′=ε−α(c−ε)<0b_1 = -α(c- )<0), so u1=AG(ε)u_1=A\,G( ) with A=1α+1λA= 1α+ 1λ, while the opponent’s payoff is u2=AG(ε)−1αG((ε−α(1−c))+)≤AG(ε)u_2=A\,G( )- 1αG (( -α(1-c))^+ )≤ A\,G( ). Hence U1≥(1−β)AG(ε)>0=U1(c,c)U_1≥(1-β)A\,G( )>0=U_1(c,c). We record for later use that Lemma 4 fails at β=1β=1: tied profiles are equilibria of the margin game (e.g. the median profile at α=0α=0), which is precisely why all statements below assume β<1β<1 and treat β=1β=1 separately. Lemma 5 (Master payoff derivative) Let Assumption 2.2 hold, α≥0α≥ 0, β∈[0,1]β∈[0,1], and fix 0<c1<c2<10<c_1<c_2<1. Assume the right-hand side of Eq. 30 below is finite at c1c_1; this holds automatically whenever k is bounded, or v is locally bounded at 0,1\0,1\, or the tent edges avoid the boundary (x1out≠0x^out_1≠ 0, x2out≠1x^out_2≠ 1)—one of which is the case in every application in this paper. Then u1u_1, u2u_2, and hence U1U_1, are continuously differentiable in c1c_1 there, and ∂U1∂c1=(1+α)[∫c1x1ink(B1)vx−∫x1out∨0c1k(B1)vx]+β∫x2inx2out∧1k(B2)vx, ∂ U_1∂ c_1=(1+α) [ _c_1^x^in_1k(B_1)\,v\,dx- _x^out_1 0^c_1k(B_1)\,v\,dx ]+β _x^in_2^x^out_2 1k(B_2)\,v\,dx, (30) where for α=0α=0 one reads x1out∨0=0x^out_1 0=0, x2out∧1=1x^out_2 1=1, and x1in=x2in=mx^in_1=x^in_2=m. The formula for ∂U2/∂c2∂ U_2/∂ c_2 is the mirror image. In particular, at an untied interior equilibrium with β<1β<1 both first-order conditions ∂U1/∂c1=∂U2/∂c2=0∂ U_1/∂ c_1=∂ U_2/∂ c_2=0 hold. Proof By the layer-cake representation, using K(t)=∫01k(s)s<tdsK(t)= _0^1k(s)1\s<t\ds, u1=∫01K(B1(x)+)v(x)dx=∫01k(s)μ(B1>s)ds,μ(dx)=v(x)dx.u_1= _0^1K (B_1(x)^+ )v(x)\,dx= _0^1k(s)\,μ (\B_1>s\ )\,ds, μ(dx)=v(x)\,dx. The super-level set B1>s\B_1>s\ is the interval (x1out(s)∨0,x1in(s)) (x^out_1(s) 0,\;x^in_1(s) ) for s∈(0,Δ)s∈(0, ), where, from B1=c2+(1+α)c1−λxB_1=c_2+(1+α)c_1-λ x on the inner arm and B1=c2−(1+α)c1+αxB_1=c_2-(1+α)c_1+α x on the outer arm, x1in(s)=c2+(1+α)c1−sλ,x1out(s)=s−c2+(1+α)c1α(=−∞ if α=0).x^in_1(s)= c_2+(1+α)c_1-sλ, x^out_1(s)= s-c_2+(1+α)c_1α\;\;(=-∞ if α=0). Both endpoints are affine in c1c_1 with velocities 1+αλ 1+αλ and 1+α 1+α, and both lie in the open interval (0,c2)⊂(0,1)(0,c_2)⊂(0,1) for s in the relevant ranges, where V is continuously differentiable (v continuous). Hence for each fixed s, c1↦μ(B1>s)=V(x1in(s))−V(x1out(s)∨0)c_1 μ(\B_1>s\)=V(x^in_1(s))-V(x^out_1(s) 0) is differentiable in c1c_1 except at the single value of s where x1out(s)=0x^out_1(s)=0 (a null set in s), with derivative 1+αλv(x1in(s))−1+αv(x1out(s))x1out(s)>0 1+αλv(x^in_1(s))- 1+αv(x^out_1(s))1\x^out_1(s)>0\. The difference quotients are dominated, locally uniformly in c1c_1, by k(s)k(s) times the supremum of v over a compact neighborhood of the swept intervals, and the resulting dominating function is integrable in s precisely when the right-hand side of Eq. 30 is finite (change variables s=B1(x)s=B_1(x) back: the two s-integrals become the two x-integrals of Eq. 30); under the standing finiteness assumption dominated convergence applies and yields ∂u1∂c1=(1+α)[∫c1x1ink(B1)vx−∫x1out∨0c1k(B1)vx], ∂ u_1∂ c_1=(1+α) [ _c_1^x^in_1\!k(B_1)v\,dx- _x^out_1 0^c_1\!k(B_1)v\,dx ], with no boundary terms: the moving tent edges x1inx^in_1 and x1outx^out_1 carry K(B1)=K(0)=0K(B_1)=K(0)=0, and the kink of B1B_1 at x=c1x=c_1 contributes nothing since B1B_1 is continuous there. Continuity of the derivative in c1c_1, including across the regime switch x1out=0x^out_1=0, follows because the outer-arm integral is continuous there: even when k is unbounded at 0+0^+, the vanishing piece obeys ∫x1outx1out+δk(B1)v≤sup[x1out,x1out+δ]v⋅1αK(αδ)→0 _x^out_1^x^out_1+δk(B_1)v≤ _[x^out_1,x^out_1+δ]v· 1αK(αδ)→ 0 by the change of variables t=B1(x)t=B_1(x) (the local boundedness of v near the switch point is part of the standing assumption). For u2u_2: on candidate 22’s support [x2in,x2out∧1][x^in_2,x^out_2 1] (which lies in (c1,1](c_1,1]) we have d1=x−c1d_1=x-c_1, so ∂B2/∂c1=−1∂ B_2/∂ c_1=-1 pointwise, and the same argument gives ∂u2/∂c1=−∫x2inx2out∧1k(B2)vdx∂ u_2/∂ c_1=- _x^in_2^x^out_2 1k(B_2)v\,dx, again with no boundary terms (B2=0B_2=0 at x2inx^in_2 and at x2outx^out_2 when x2out<1x^out_2<1; if x2out≥1x^out_2≥ 1 the limit 11 is fixed). Combining, and noting U1=u1−βu2U_1=u_1-β u_2, yields Eq. 30. The first-order conditions at an equilibrium follow since, by Lemma 4 (whose cases cover every configuration in which we invoke them) the equilibrium is untied and interior, and U1(⋅,c2)U_1(·,c_2) is differentiable there. Specializing Eq. 30 to α=0α=0 gives, for general (v,k)(v,k), ∂U1∂c1=∫c1mk(B1)vx−k(Δ)V(c1)+β∫m1k(B2)vx, ∂ U_1∂ c_1= _c_1^mk(B_1)\,v\,dx-k( )V(c_1)+β _m^1k(B_2)\,v\,dx, (31) the β-generalization of the baseline first-order condition (17): the opponent’s marginal voters at the midpoint now enter with weight −β-β. Finally, we record the closed form for uniform voters used throughout Appendices 0.K and 0.L. For v≡1v≡ 1 and any K, changing variables t=B1(x)t=B_1(x) along each arm of the tent (slopes α and λ) and subtracting the part of the outer arm truncated by the boundary gives, for α>0α>0, u1=AG(Δ)−1αG(b1+),A:=1α+1λ=2(1+α)αλ,u_1=A\,G( )- 1α\,G (b_1^+ ), A:= 1α+ 1λ= 2(1+α)αλ, (32) and u1=c1K(Δ)+12G(Δ)u_1=c_1K( )+ 12G( ) for α=0α=0; the formulas for u2u_2 replace b1b_1 by b2b_2 and c1c_1 by 1−c21-c_2. Substituting into Eq. 30 (each integral is 1λK(Δ) 1λK( ), 1α[K(Δ)−K(b1+)] 1α[K( )-K(b_1^+)], and AK(Δ)−1αK(b2+)A\,K( )- 1αK(b_2^+) respectively) yields, for α>0α>0 and uniform voters, α∂U1∂c1=−(1−β)αAK(Δ)+(1+α)K(b1+)−βK(b2+).α\, ∂ U_1∂ c_1=-(1-β)\,α A\,K( )+(1+α)\,K (b_1^+ )-β\,K (b_2^+ ). (33) Appendix 0.J Results for the baseline (α=β=0α=β=0): a reflection lemma The sufficiency results for uniform costs (Propositions 1 and 1) (density ratio ρv≤4 _v≤ 4) verify two cross-over deviations by an explicit computation available only at quartile profiles. The following lemma removes both the computation and the restriction to uniform costs: for symmetric electorates and vote-share maximizers, cross-over deviations are never binding. It is stated for all α≥0α≥ 0 because it is used again in Appendix 0.K; the restriction to β=0β=0 is essential (Remark 10). Lemma 6 (Reflection lemma) Let β=0β=0, α≥0α≥ 0, let K be arbitrary (Assumption 2.2), and let v be symmetric about 12 12. Let c1∈(0,12)c_1∈(0, 12) and c2=1−c1c_2=1-c_1. If c1c_1 maximizes u1(⋅,c2)u_1(·,c_2) over [0,c2)[0,c_2), then (c1,c2)(c_1,c_2) is a Nash equilibrium. Proof By symmetry of v and of the profile it suffices to show candidate 11 has no profitable deviation. Own-side deviations c1′∈[0,c2)c_1 ∈[0,c_2) are unprofitable by hypothesis. The tie c1′=c2c_1 =c_2 yields 0<u1(c1,c2)0<u_1(c_1,c_2) (the benefit B1B_1 is positive on a nondegenerate interval around c1c_1 and v>0v>0 there). It remains to rule out cross-over deviations c1′∈(c2,1]c_1 ∈(c_2,1]. We use two observations. (i) Reflection. For any positions a,oa,o, the change of variables y=1−xy=1-x together with v(1−y)=v(y)v(1-y)=v(y) gives udev(a,o)=udev(1−a,1−o)u_dev(a,o)=u_dev(1-a,1-o), where udev(a,o)u_dev(a,o) denotes the vote share of a candidate at a facing an opponent at o. Applying this with a=c1′∈(c2,1]a=c_1 ∈(c_2,1] and o=c2o=c_2, and using 1−c2=c11-c_2=c_1: u1(c1′,c2)=udev(1−c1′,c1),1−c1′∈[0,c1).u_1(c_1 ,c_2)=u_dev (1-c_1 ,\,c_1 ), 1-c_1 ∈[0,c_1). Cross-over deviations against c2c_2 are thus payoff-equivalent to own-side deviations against an opponent located at c1c_1. (i) Monotonicity in the opponent’s position. For a<o≤o~a<o≤ o and every x, B(x,a,o)+≤B(x,a,o~)+B(x;a,o)^+≤ B(x;a, o)^+, where B(x,a,o)=|x−o|−(1+α)|x−a|B(x;a,o)=|x-o|-(1+α)|x-a|. Indeed, if B(x,a,o)≤0B(x;a,o)≤ 0 there is nothing to prove; if B(x,a,o)>0B(x;a,o)>0 then |x−o|>(1+α)|x−a|≥|x−a||x-o|>(1+α)|x-a|≥|x-a|, so x is strictly closer to a than to o, forcing x<a+o2<o≤o~x< a+o2<o≤ o and hence |x−o~|=o~−x>o−x=|x−o||x- o|= o-x>o-x=|x-o|, so B(x,a,o~)>B(x,a,o)>0B(x;a, o)>B(x;a,o)>0. Since K is non-decreasing, integrating against v gives udev(a,o)≤udev(a,o~)u_dev(a,o)≤ u_dev(a, o). Combining, for every c1′∈(c2,1]c_1 ∈(c_2,1], with a=1−c1′<c1≤c2a=1-c_1 <c_1≤ c_2, u1(c1′,c2)=(i)udev(a,c1)≤(i)udev(a,c2)≤supc1′∈[0,c2)u1(c1′,c2)=u1(c1,c2),u_1(c_1 ,c_2) (i)=u_dev(a,c_1) (i)≤u_dev(a,c_2)≤ _c_1 ∈[0,c_2)u_1(c_1 ,c_2)=u_1(c_1,c_2), the last equality by hypothesis. No deviation is profitable. Remark 10 (a) At the quartile profile of a symmetric electorate with uniform costs, Lemma 6 recovers the cross-over verification in the proof of Corollary 1 (Appendix 0.B) without any computation: under ρv≤4 _v≤ 4 the own-side payoff is concave (Proposition 1) and stationary at c1c_1, hence globally maximized there, and the lemma finishes the proof. The explicit bounds there (u1≥132su_1≥ 132s etc.) are retained only for the asymmetric case of Corollary 1, where the reflection argument is unavailable. (b) The lemma fails for β>0β>0: step (i) compares the deviator’s share pointwise, but moving the opponent also changes the opponent’s share u2u_2, in a direction that is not signed in general, so U1=u1−βu2U_1=u_1-β u_2 is not monotone in the opponent’s position. The β>0β>0 cross-over checks in Appendix 0.L are therefore carried out explicitly. Appendix 0.K Results for alienation (α>0α>0, β=0β=0) 0.K.1 A sufficiency threshold under uniform costs For α=0α=0, Proposition 1 shows that ρv≤4 _v≤ 4 makes each candidate’s own-side payoff concave, reducing equilibrium verification to the balance conditions plus two cross-over checks. The next theorem gives the α-analogue and answers the question of whether alienation helps or hurts this program: the threshold F(α):=minf(α),g(α),f(α):=2(2+α)1+α,g(α):=2+α(g(0):=∞),F(α):= \f(α),g(α)\, f(α):= 2(2+α)1+α, g(α):= 2+α\;\;(g(0):=∞), (34) is strictly decreasing, from F(0)=4F(0)=4 through F(1)=3F(1)=3 (where f=gf=g) to 11 as α→∞α→∞. Alienation shrinks the sufficiency umbrella. Theorem 0.K.1 (Single-peakedness under uniform costs) Let κ∼U(0,1)κ U(0,1), β=0β=0, α>0α>0, and ρv≤F(α) _v≤ F(α). Fix any c2∈(0,1]c_2∈(0,1]. Then c1↦u1(c1,c2)c_1 u_1(c_1,c_2) is, on [0,c2)[0,c_2): concave on the no-dead-zone piece [0,c21+α] [0, c_21+α ], and strictly decreasing on the dead-zone piece (c21+α,c2) ( c_21+α,c_2 ); its derivative is continuous, positive at c1=0c_1=0, and the payoff is single-peaked with all maximizers in the no-dead-zone piece. Consequently a profile 0<c1<c2<10<c_1<c_2<1 is a Nash equilibrium if and only if both balance conditions V(x1in)+V(x1out∨0)=2V(c1),V(x2in)+V(min(x2out,1))=2V(c2)V(x^in_1)+V(x^out_1 0)=2V(c_1), V(x^in_2)+V ( (x^out_2,1) )=2V(c_2) (35) hold (where V(x1out∨0):=V(x1out)V(x^out_1 0):=V(x^out_1) if x1out>0x^out_1>0 and :=0:=0 otherwise, and the second condition is the mirror image) and neither candidate gains from the single cross-over deviation to the far side of her opponent. Proof By Eq. 30 with k≡1k≡ 1 and β=0β=0, ∂u1∂c1=(1+α)[(V(x1in)−V(c1))−(V(c1)−V(x1out∨0))], ∂ u_1∂ c_1=(1+α) [ (V(x^in_1)-V(c_1) )- (V(c_1)-V(x^out_1 0) ) ], (36) which is continuous in c1c_1 on (0,c2)(0,c_2), and one-sidedly at c1=0c_1=0, where it equals (1+α)V(x1in)>0(1+α)V(x^in_1)>0. No-dead-zone piece. For c1≤c21+αc_1≤ c_21+α (i.e. x1out≤0x^out_1≤ 0), Eq. 36 reads (1+α)[V(x1in)−2V(c1)](1+α)[V(x^in_1)-2V(c_1)] and, since x1in=c1+c2−c1λx^in_1=c_1+ c_2-c_1λ has (x1in)′=1+αλ(x^in_1) = 1+αλ, ∂2u1∂c12=(1+α)[1+αλv(x1in)−2v(c1)]≤(1+α)[1+αλsupv−2infv]≤ 0, ∂^2u_1∂ c_1^2=(1+α) [ 1+αλ\,v(x^in_1)-2\,v(c_1) ]\;≤\;(1+α) [ 1+αλ\, v-2 v ]\;≤\;0, the last step because ρv≤f(α)=2λ1+α _v≤ f(α)= 2λ1+α. Dead-zone piece. For c1>c21+αc_1> c_21+α (i.e. x1out>0x^out_1>0), substitute t=B1(x)t=B_1(x) on each arm of Eq. 36: ∂u1∂c1=(1+α)∫0Δ[1λv(x1in−tλ)−1αv(x1out+tα)]t. ∂ u_1∂ c_1=(1+α) _0 [ 1λ\,v (x^in_1- tλ )- 1α\,v (x^out_1+ tα ) ]dt. (37) Since ρv≤g(α)=λα _v≤ g(α)= λα, the integrand is pointwise ≤supvλ−infvα≤0≤ vλ- vα≤ 0. If the integral vanished, the integrand would vanish for a.e. t, forcing v≡supvv≡ v on (c1,x1in)(c_1,x^in_1) and v≡infvv≡ v on (x1out,c1)(x^out_1,c_1); as g(α)>1g(α)>1 implies supv>infv v> v, this contradicts continuity of v at c1c_1. Hence the derivative is strictly negative on the whole piece. Single-peakedness and the characterization. The derivative is continuous, starts positive, is non-increasing while the payoff is concave, and is strictly negative on the dead-zone piece; hence u1(⋅,c2)u_1(·,c_2) increases to a maximizer set (an interval, possibly a point) inside (0,c21+α] (0, c_21+α ] and strictly decreases afterwards. A profile is an equilibrium iff each cic_i globally maximizes uiu_i against the other; by Lemma 4(a) and single-peakedness, own-side global optimality of an interior c1c_1 is equivalent to stationarity, i.e. to Eq. 35, and the only remaining deviations are the tie (never profitable, as u1(c1,c2)>0u_1(c_1,c_2)>0) and the cross-over, as stated. Remark 11 (Sharpness, and the direction of the answer) (a) The constant f(α)f(α) cannot be improved in the concavity criterion: for any ratio ρ>f(α)ρ>f(α) there is a symmetric “mesa” density with ρv=ρ _v=ρ and a no-dead-zone configuration at which ∂2u1/∂c12>0∂^2u_1/∂ c_1^2>0. (Take v equal to a high value on a window around x1inx^in_1 and a low value at c1c_1; e.g. at α=1α=1, v high on [0.4,0.6][0.4,0.6], c1=0.3c_1=0.3, c2=0.65c_2=0.65, so x1in=0.416¯x^in_1=0.41 6: the second derivative is 2[23v(x1in)−2v(c1)]>02 [ 23v(x^in_1)-2v(c_1) ]>0 iff v(x1in)/v(c1)>3=f(1)v(x^in_1)/v(c_1)>3=f(1). Verified numerically: with ratio 2.92.9 the payoff is concave at every no-dead-zone configuration, with ratio 3.13.1 it is not.) Similarly g(α)g(α) is exactly the threshold below which dead zones are impossible at stationary profiles (Proposition 17). Whether equilibrium existence fails above F(α)F(α) is open, exactly as it is open above ρv=4 _v=4 at α=0α=0. (b) F is strictly decreasing with F(α)↓1F(α) 1: the answer to “does alienation help existence?” is no on this route. The reason is the dead zone: it is the second, boundary-truncation-driven piece of the payoff that destroys concavity, and it grows with α. Contrast with competitiveness, which enlarges the umbrella at α=0α=0 (Proposition 22: threshold 4/(1−β)4/(1-β)). (c) For β∈(0,1)β∈(0,1) and uniform costs the same two-piece argument goes through with thresholds f(α)f(α) (unchanged: the opponent term only helps concavity, since ∂2u2/∂c12=v(x2in)λ+v(x2out)αx2out<1≥0∂^2u_2/∂ c_1^2= v(x^in_2)λ+ v(x^out_2)α1_x^out_2<1≥ 0) and gβ(α)=(1+α)(2+α)α(1+α+β)+β(2+α)≤g(α)g_β(α)= (1+α)(2+α)α(1+α+β)+β(2+α)≤ g(α) on the dead-zone piece; but the cross-over is no longer free (Remark 10), so we do not obtain an existence corollary there. Combining Theorem 0.K.1 with the reflection lemma yields existence: Corollary 10 (Existence for symmetric electorates) Let κ∼U(0,1)κ U(0,1), β=0β=0, α>0α>0, let v be symmetric about 12 12 with ρv≤F(α) _v≤ F(α). Then a symmetric Nash equilibrium exists: any root c1∗∈(0,12)c_1^*∈(0, 12) of h(c1):=V(x1in(c1))+V(x1out(c1)∨0)−2V(c1),xin1=c1+1−2c1λ,xout1=c1−1−2c1α, gatheredh(c_1):=V (x^in_1(c_1) )+V (x^out_1(c_1) 0 )-2V(c_1),\\ x^in_1=c_1+ 1-2c_1λ, x^out_1=c_1- 1-2c_1α, gathered (38) gives the equilibrium (c1∗,1−c1∗)(c_1^*,1-c_1^*), and such a root exists. Proof h is continuous on (0,12)(0, 12) with h(0+)=V(1λ)>0h(0^+)=V( 1λ)>0 and, as c1↑12c_1 12 (so Δ=1−2c1↓0 =1-2c_1 0 and x1out,x1in→12x^out_1,x^in_1→ 12), h(c1)=(V(x1in)−V(c1))−(V(c1)−V(x1out))=v(12)Δ(1λ−1α)+o(Δ)<0h(c_1)= (V(x^in_1)-V(c_1) )- (V(c_1)-V(x^out_1) )=v( 12)\, ( 1λ- 1α )+o( )<0 for small Δ , since v is continuous and positive at 12 12 and α<λα<λ. By the intermediate value theorem a root c1∗∈(0,12)c_1^*∈(0, 12) exists. By the dead-zone part of Theorem 0.K.1 (applied along the symmetric family: for c1>1λc_1> 1λ the profile has x1out>0x^out_1>0 and h<0h<0 by the pointwise comparison in Eq. 37, which only used ρv≤g(α) _v≤ g(α)), every root lies in the no-dead-zone branch c1∗≤1λc_1^*≤ 1λ. At such a root, c1∗c_1^* is stationary for u1(⋅,c2∗)u_1(·,c_2^*), hence by single-peakedness (Theorem 0.K.1) a global maximizer of the own side; Lemma 6 then upgrades own-side optimality to a Nash equilibrium. 0.K.2 Beyond uniform costs Proposition 16 (Sufficiency for general (v,k)(v,k), non-increasing costs) Let β=0β=0, α≥0α≥ 0, and let k be C1C^1 and non-increasing on (0,1)(0,1) with ρk=k(0+)/k(1−)<∞ _k=k(0^+)/k(1^-)<∞. If ρvρk≤f(α)=2(2+α)1+αandρv≤g(α)=2+α(vacuous if α=0), _v\, _k≤ f(α)= 2(2+α)1+α _v≤ g(α)= 2+α (vacuous if α=0), then for every c2c_2 the map c1↦u1(c1,c2)c_1 u_1(c_1,c_2) is single-peaked on [0,c2)[0,c_2) exactly as in Theorem 0.K.1, and the equilibrium characterization of Theorem 0.K.1 holds verbatim with the balance conditions of Lemma 5. If moreover v is symmetric, a symmetric equilibrium exists. At α=0α=0 the conditions reduce to ρvρk≤4 _v _k≤ 4, generalizing the ρv≤4 _v≤ 4 criterion of Proposition 1 to non-uniform costs. Proof No-dead-zone piece (x1out≤0x^out_1≤ 0). Differentiating the first bracket of Eq. 30 once more (now the moving tent-edge terms survive with factor k rather than K; the calculation is licensed by k∈C1k∈ C^1), ∂2u1∂c12=(1+α)[1+αλk(0+)v(x1in)−2k(Δ)v(c1)+(1+α)∫x∈tentk′(B1)v]. ∂^2u_1∂ c_1^2=(1+α) [ 1+αλ\,k(0^+)\,v(x^in_1)-2\,k( )\,v(c_1)+(1+α)\!\! _x \!\!k (B_1)\,v ]. Here we used ∂B1/∂c1=±(1+α)∂ B_1/∂ c_1=±(1+α) on the two arms and that the lower limit 00 is fixed. Since k′≤0k ≤ 0, the integral term is ≤0≤ 0; since k is non-increasing, k(0+)=supkk(0^+)= k and k(Δ)≥infkk( )≥ k, so the bracket is at most 1+αλsupksupv−2infkinfv≤0 1+αλ k v-2 k v≤ 0 by ρvρk≤2λ1+α _v _k≤ 2λ1+α. Dead-zone piece (x1out>0x^out_1>0, only if α>0α>0). As in Eq. 37, but keeping k: ∂u1∂c1=(1+α)∫0Δk(t)[1λv(x1in−tλ)−1αv(x1out+tα)]t<0, ∂ u_1∂ c_1=(1+α) _0 k(t) [ 1λv (x^in_1- tλ )- 1αv (x^out_1+ tα ) ]dt<0, by the same pointwise comparison and strictness argument as in Theorem 0.K.1; note this piece needs no assumption on k at all. The junction continuity, single-peakedness, characterization, and (for symmetric v) the existence argument via Lemma 6 are verbatim as before; for existence, the sign analysis of Eq. 38 carries over with the integrands weighted by k>0k>0. Remark 12 The C1C^1 assumption is for the displayed second-derivative formula only; the conclusion extends to continuous non-increasing k by approximating k uniformly with C1C^1 non-increasing densities (concavity and single-peakedness survive pointwise limits of payoffs). For β>0β>0 and general (v,k)(v,k) the opponent term ∂2u2/∂c12∂^2u_2/∂ c_1^2 contains k′(B2)k (B_2)-integrals of indefinite sign, and we make no claim; the uniform-cost case is Remark 11(c), and the uniform-voter case is settled completely in Appendix 0.L. 0.K.3 Dead zones Proposition 17 (When equilibria have dead zones) Let α>0α>0. (a) (Necessity; any K, β=0β=0.) If a profile 0<c1<c2<10<c_1<c_2<1 with x1out>0x^out_1>0 satisfies candidate 11’s first-order condition, then ρv≥2+α _v≥ 2+α. Equivalently, if α<2ρv−1α< 2 _v-1, no stationary profile—in particular no equilibrium—has a dead zone. (b) (Uniform voters never; any K, any β∈[0,1)β∈[0,1).) If x∼U(0,1)x U(0,1), then no Nash equilibrium of the β-game has a dead zone: at every equilibrium b1>0b_1>0 and b2>0b_2>0. (c) (Sufficiency for symmetric electorates; uniform costs, β=0β=0.) If κ∼U(0,1)κ U(0,1) and v is symmetric about 12 12 with V(2(1+α)(2+α)2)> 2V(12+α),V\! ( 2(1+α)(2+α)^2 )\;>\;2\,V\! ( 12+α ), (39) then a symmetric stationary profile with a dead zone exists. Proof (a) With x1out>0x^out_1>0, the first-order condition from Eq. 30 is the vanishing of Eq. 37 (with the weight k(t)k(t) for general K). If ρv<λα _v< λα the integrand is pointwise negative as in Theorem 0.K.1, and if ρv=λα _v= λα the strictness argument there applies; either way the integral cannot vanish. Hence stationarity forces ρv≥λα _v≥ λα, i.e. α≥2ρv−1α≥ 2 _v-1. (b) By Lemma 4(c) and Lemma 5, an equilibrium is untied, interior, and satisfies both first-order conditions. By Eq. 33 and its mirror image, (1+α)K(b1+)−βK(b2+)=(1−β)αAK(Δ)=(1+α)K(b2+)−βK(b1+),(1+α)K(b_1^+)-β K(b_2^+)=(1-β)\,α A\,K( )=(1+α)K(b_2^+)-β K(b_1^+), so (1+α+β)[K(b1+)−K(b2+)]=0(1+α+β) [K(b_1^+)-K(b_2^+) ]=0, i.e. K(b1+)=K(b2+)K(b_1^+)=K(b_2^+). If this common value is 00, the first-order condition gives (1−β)αAK(Δ)=0(1-β)α A\,K( )=0, impossible since β<1β<1 and K(Δ)>0K( )>0 for Δ∈(0,1) ∈(0,1). Hence K(b1+)=K(b2+)>0K(b_1^+)=K(b_2^+)>0, which forces b1>0b_1>0 and b2>0b_2>0. (c) Consider the symmetric family c2=1−c1c_2=1-c_1 and h from Eq. 38. The dead-zone branch is c1∈(1λ,12)c_1∈( 1λ, 12) (indeed x1out>0⇔αc1>Δ=1−2c1⇔c1>1λx^out_1>0 α c_1> =1-2c_1 c_1> 1λ). At the junction c1=1λc_1= 1λ we have Δ=αλ = αλ, x1in=1λ+αλ2=2(1+α)λ2x^in_1= 1λ+ αλ^2= 2(1+α)λ^2 and x1out=0x^out_1=0, so h(1λ)=V(2(1+α)λ2)−2V(1λ)>0h( 1λ)=V ( 2(1+α)λ^2 )-2V( 1λ)>0 by Eq. 39, while h(12−)<0h( 12^-)<0 as in the proof of Corollary 10. By the intermediate value theorem, h has a root in the open dead-zone branch. Remark 13 (a) Part (b) removes the last gap in the uniform-voter theory: the characterizations of Appendix 0.L never need a dead-zone case. At β=1β=1 the conclusion fails—inside the band E(α)E(α) both tents are interior, i.e. b1,b2≤0b_1,b_2≤ 0, which is exactly what produces the band—so the dichotomy at β=1β=1 is sharp. (b) Criterion (39) compares the mass of the widest possible inner arm with twice the mass below the junction position; for uniform voters its left side minus right side is −2λ2<0- 2λ^2<0, consistent with (b). Numerically the criterion is exact for symmetric Beta electorates: for v=Beta(n,n)v=Beta(n,n), n∈2,4,8n∈\2,4,8\ and α∈1,2,4α∈\1,2,4\, dead-zone stationary profiles exist in precisely the cases where Eq. 39 holds (it fails only for n=2n=2, α=1α=1). For v=Beta(8,8)v=Beta(8,8) and α=2α=2 the (unique) symmetric stationary profile (0.381,0.619)(0.381,0.619) has x1out=0.26>0x^out_1=0.26>0 and is a genuine equilibrium (mutual global best responses, verified numerically): dead-zone equilibria are non-empty. (c) Parts (a) and (c) bracket the phenomenon: dead zones require α≥2ρv−1α≥ 2 _v-1 and concentrated electorates, and are guaranteed (at the stationary level) once the center is heavy enough that Eq. 39 holds. Appendix 0.L Results for competitive candidates (β>0β>0) Throughout this section β∈[0,1)β∈[0,1) unless stated otherwise; all results specialize at β=0β=0 to (and strengthen) the vote-share results of Section 4. Write γβ:=1+β1−β∈[1,∞),ψβ(t):=t+γβK(t)k(t),χβ(t):=t+1+β2K(t)k(t). _β:= 1+β1-β∈[1,∞), _β(t):=t+ _β\, K(t)k(t), _β(t):=t+ 1+β2\, K(t)k(t). The two deformations of ψ(t)=t+K(t)/k(t)ψ(t)=t+K(t)/k(t) play distinct roles below: ψβ _β governs stationarity, χβ _β governs single-peakedness; at β=0β=0 they coincide only in their monotonicity class, not in value, and the wedge between the coefficients γβ _β and 1+β2 1+β2 is what Example 3 exploits. 0.L.1 Uniform voters, α=0α=0: complete characterization Theorem 0.L.1 (Uniform voters, α=0α=0: characterization and uniqueness) Let x∼U(0,1)x U(0,1), α=0α=0, β∈[0,1)β∈[0,1), and let K satisfy Assumption 2.2. (a) (Necessity.) Every Nash equilibrium is symmetric, c1+c2=1c_1+c_2=1, with gap Δ∈(0,1) ∈(0,1) satisfying ψβ(Δ)=1 _β( )=1. (b) (Characterization.) A symmetric profile with gap Δ and c2=1+Δ2c_2= 1+ 2 is a Nash equilibrium if and only if Δ maximizes F(t):=(c2−t−β(1−c2))K(t)+1−β2G(t)over t∈(0,c2].F(t):= (c_2-t-β(1-c_2) )K(t)+ 1-β2\,G(t) t∈(0,c_2]. (c) (Uniqueness under monotone K/kK/k.) If K/kK/k is non-decreasing on (0,1)(0,1), then ψβ _β is strictly increasing, ψβ(t)=1 _β(t)=1 has a unique root Δβ∗∈(0,1) ^*_β∈(0,1), and the game has the unique Nash equilibrium (1−Δβ∗2,1+Δβ∗2) ( 1- ^*_β2, 1+ ^*_β2 ). Proof Throughout, fix the opponent at c2c_2 and parameterize candidate 11’s own-side positions by the distance t=c2−c1∈(0,c2]t=c_2-c_1∈(0,c_2]. By the α=0α=0 closed forms below Eq. 32, u1=(c2−t)K(t)+12G(t)u_1=(c_2-t)K(t)+ 12G(t) and u2=(1−c2)K(t)+12G(t)u_2=(1-c_2)K(t)+ 12G(t), so U1=u1−βu2=F(t)U_1=u_1-β u_2=F(t) as displayed. By Lemma 5, F∈C1F∈ C^1 with F′(t)=(Q−t)k(t)−1+β2K(t)=k(t)[Q−χβ(t)],Q:=c2−β(1−c2).F (t)= (Q-t )k(t)- 1+β2K(t)=k(t)\, [Q- _β(t) ], Q:=c_2-β(1-c_2). (40) Cross-over deviations are dominated whenever c2≥12c_2≥ 12. For s∈(0,1−c2]s∈(0,1-c_2] compare the mirror pair c1′=c2−sc_1 =c_2-s (own side) and c1′=c2+sc_1 =c_2+s (far side). The closed forms give U1(c1′,c2)−U1(c1′,c2) U_1(c_1 ,c_2)-U_1(c_1 ,c_2) =[(c2−s)−(1−c2−s)]K(s)+β[c2−(1−c2)]K(s) = [(c_2-s)-(1-c_2-s) ]K(s)+β [c_2-(1-c_2) ]K(s) =(1+β)(2c2−1)K(s)≥0. =(1+β)(2c_2-1)K(s)≥ 0. Since the far-side range s≤1−c2s≤ 1-c_2 is contained in the own-side range s≤c2s≤ c_2, we get supc1′>c2U1≤supc1′<c2U1=suptF(t) _c_1 >c_2U_1≤ _c_1 <c_2U_1= _tF(t). Deviating to the tie c1′=c2c_1 =c_2 yields 00, which is dominated whenever supF>0 F>0. (b). (⇐ ) Let Δ maximize F over (0,c2](0,c_2] at a symmetric profile. Then own-side deviations are unprofitable by definition; c2=1+Δ2>12c_2= 1+ 2> 12, so cross-overs are dominated; and F(Δ)=(1−β)u1>0F( )=(1-β)u_1>0 dominates the tie. Candidate 22’s problem is the mirror image (uniform voters are symmetric), so the profile is an equilibrium. (⇒ ) If the symmetric profile is an equilibrium, c1c_1 is a global maximizer of U1(⋅,c2)U_1(·,c_2) over [0,c2)[0,c_2), i.e. Δ maximizes F over (0,c2](0,c_2]. (a). By Lemma 4(b)(c), an equilibrium is untied and interior; Lemma 5 gives both first-order conditions. In the t-parameterization these read Q1=χβ(Δ)Q_1= _β( ) and Q2=χβ(Δ)Q_2= _β( ) with Q1=c2−β(1−c2)Q_1=c_2-β(1-c_2) and, mirroring, Q2=(1−c1)−βc1Q_2=(1-c_1)-β c_1. Subtracting, (1+β)(c1+c2−1)=0(1+β)(c_1+c_2-1)=0, so the equilibrium is symmetric. Substituting c2=1+Δ2c_2= 1+ 2 into Q=χβ(Δ)Q= _β( ) and simplifying yields the identity Q−χβ(Δ)=1−β2[ 1−ψβ(Δ)],Q- _β( )= 1-β2\, [\,1- _β( )\, ], (41) so stationarity of the symmetric profile is exactly ψβ(Δ)=1 _β( )=1. (c). If K is super-regular then ψβ _β and χβ _β are strictly increasing. Root existence. First, limt↓0K/k=0 _t 0K/k=0: the limit L≥0L≥ 0 exists by monotonicity, and if L>0L>0 then r=k/K≤1/Lr=k/K≤ 1/L near 00, contradicting Eq. 29; hence ψβ(0+)=0<1 _β(0^+)=0<1. Second, ψβ(t)>1 _β(t)>1 for some t<1t<1: otherwise γβK/k≤1−t _βK/k≤ 1-t near 11, i.e. r≥γβ/(1−t)r≥ _β/(1-t), making ∫t01−r=∞ _t_0^1^-r=∞ for some t0<1t_0<1—impossible since ∫t01−r=logK(1)−logK(t0)<∞ _t_0^1^-r= K(1)- K(t_0)<∞. By continuity and strict monotonicity there is a unique root Δβ∗∈(0,1) ^*_β∈(0,1). The root is an equilibrium. At the symmetric profile with gap Δβ∗ ^*_β: χβ _β is strictly increasing and crosses Q exactly once; by Eq. 41 the crossing is at Δβ∗ ^*_β, so by Eq. 40, F′>0F >0 before and F′<0F <0 after: F is single-peaked with peak Δβ∗ ^*_β, which is therefore its maximizer (the peak is interior: χβ(c2)−Q=1+β2Kk(c2)+β(1−c2)>0 _β(c_2)-Q= 1+β2 Kk(c_2)+β(1-c_2)>0 shows F′(c2−)<0F (c_2^-)<0, ruling out c1=0c_1=0). By (b) the profile is an equilibrium. Uniqueness. By (a) any equilibrium is symmetric with ψβ(Δ)=1 _β( )=1; the root is unique. 0.L.2 Uniform voters, α>0α>0: complete characterization Define, for Δ∈(Δ0,1) ∈( _0,1), J(Δ):=logK(Δ)K(bs(Δ))=∫bs(Δ)Δr(s)s,Lβ:=log(1+α−β)λ2(1−β)(1+α).J( ):= K( )K\! (b_s( ) )= _b_s( ) r(s)\,ds, L_β:= (1+α-β)\,λ2(1-β)(1+α). (42) Note Lβ>0L_β>0 always: the ratio minus one has the sign of (1+α−β)λ−2(1−β)(1+α)=α(1+α+β)>0(1+α-β)λ-2(1-β)(1+α)=α(1+α+β)>0. Note also that LβL_β does not depend on K; this is what makes the comparative statics in Propositions 18 and 19 clean. Theorem 0.L.2 (Uniform voters, α>0α>0: unique equilibrium and its gap) Let x∼U(0,1)x U(0,1), α>0α>0, β∈[0,1)β∈[0,1), and let K satisfy Assumption 2.2 and be super-regular. Then the game has a unique Nash equilibrium. It is symmetric, has no dead zones, and its gap is the unique root Δ∗=Δ∗(α,β)∈(Δ0,1) ^*= ^*(α,β)∈( _0,1) of J(Δ)=LβJ( )=L_β; equivalently, of the first-order condition (1+α−β)K(bs(Δ))=(1−β)2(1+α)2+αK(Δ).(1+α-β)\,K\! (b_s( ) )=(1-β)\, 2(1+α)2+α\,K( ). (43) In particular the equilibrium is strictly more polarized than every symmetric margin-game equilibrium: Δ∗>Δ0 ^*> _0. Proof Existence and uniqueness of the root. On (Δ0,1)( _0,1), J is differentiable with J′(Δ)=r(Δ)−λ2r(bs(Δ))≤r(Δ)(1−λ2)=−α2r(Δ)<0J ( )=r( )- λ2\,r (b_s( ) )≤ r( ) (1- λ2 )=- α2\,r( )<0, using bs(Δ)<Δb_s( )< and r non-increasing. Moreover J(Δ0+)=+∞J( _0^+)=+∞ by Eq. 29 and J(1−)=logK(1)K(1)=0<LβJ(1^-)= K(1)K(1)=0<L_β. Hence a unique root Δ∗∈(Δ0,1) ^*∈( _0,1) exists, with J>LβJ>L_β on (Δ0,Δ∗)( _0, ^*) and J<LβJ<L_β on (Δ∗,1)( ^*,1); exponentiating J=LβJ=L_β gives Eq. 43. Necessity. Exactly as in Proposition 17(b): an equilibrium is untied and interior with both first-order conditions (Lemma 4(c), Lemma 5); subtracting them yields K(b1+)=K(b2+)>0K(b_1^+)=K(b_2^+)>0, hence b1=b2>0b_1=b_2>0 (K is strictly increasing on (0,1)(0,1)), which by Eq. 28 forces c1+c2=1c_1+c_2=1 and b1=b2=bs(Δ)>0b_1=b_2=b_s( )>0, i.e. Δ>Δ0 > _0. The first-order condition (33) at the symmetric profile becomes exactly Eq. 43, i.e. Δ=Δ∗ = ^*. Sufficiency. Fix c2=c2∗=1+Δ∗2c_2=c_2^*= 1+ ^*2 and let candidate 11 vary c1∈(0,c2∗)c_1∈(0,c_2^*); write Δ:=c2∗−c1 :=c_2^*-c_1, δ:=b2−b12=α2(c1−c1∗)δ:= b_2-b_12= α2\,(c_1-c_1^*) by Eq. 28 (using 1−c2∗=c1∗1-c_2^*=c_1^*), so b1=bs(Δ)−δb_1=b_s( )-δ and b2=bs(Δ)+δb_2=b_s( )+δ. Decompose Eq. 33 as α∂U1∂c1=Φ(Δ)+(1+α)[K(b1+)−K(bs(Δ)+)]−β[K(b2+)−K(bs(Δ)+)],α ∂ U_1∂ c_1= ( )+(1+α) [K(b_1^+)-K(b_s( )^+) ]-β [K(b_2^+)-K(b_s( )^+) ], (44) where Φ(Δ):=(1+α−β)K(bs(Δ)+)−(1−β)αAK(Δ) ( ):=(1+α-β)K(b_s( )^+)-(1-β)α A\,K( ) is the value at the “centered” benefits δ=0δ=0. For Δ≤Δ0 ≤ _0, Φ(Δ)=−(1−β)αAK(Δ)<0 ( )=-(1-β)α A\,K( )<0; for Δ∈(Δ0,1) ∈( _0,1), Φ(Δ)<0⇔J(Δ)>Lβ⇔Δ<Δ∗ ( )<0 J( )>L_β < ^* (and Φ(Δ∗)=0 ( ^*)=0, Φ>0 >0 for Δ>Δ∗ > ^*). Now: if c1>c1∗c_1>c_1^* then δ>0δ>0 and Δ<Δ∗ < ^*, so both bracketed corrections in Eq. 44 are ≤0≤ 0 (monotonicity of t↦K(t+)t K(t^+)) and Φ(Δ)<0 ( )<0: the derivative is strictly negative. If c1<c1∗c_1<c_1^* then δ<0δ<0 and Δ∈(Δ∗,c2∗]⊂(Δ∗,1) ∈( ^*,c_2^*]⊂( ^*,1), so both corrections are ≥0≥ 0 and Φ(Δ)>0 ( )>0: the derivative is strictly positive. Hence U1(⋅,c2∗)U_1(·,c_2^*) is strictly single-peaked on (0,c2∗)(0,c_2^*) with peak c1∗c_1^*, and the one-sided derivative at c1=0c_1=0 is positive, so c1∗c_1^* is the own-side global maximizer. Cross-overs and the tie. For s∈(0,1−c2∗]s∈(0,1-c_2^*] compare c1′=c2∗−sc_1 =c_2^*-s with c1′=c2∗+sc_1 =c_2^*+s using Eq. 32. The deviator’s boundary benefits are bL=s−α(c2∗−s)b_L=s-α(c_2^*-s) on the near side and bR=s−α(1−c2∗−s)b_R=s-α(1-c_2^*-s) on the far side; c2∗≥12c_2^*≥ 12 gives bL≤bRb_L≤ b_R, hence u1(c1′,c2∗)=AG(s)−1αG(bL+)≥AG(s)−1αG(bR+)=u1(c1′,c2∗)u_1(c_1 ,c_2^*)=A\,G(s)- 1αG(b_L^+)≥ A\,G(s)- 1αG(b_R^+)=u_1(c_1 ,c_2^*). The opponent’s boundary benefit is s−α(1−c2∗)s-α(1-c_2^*) when the deviator is on the near side and s−αc2∗s-α c_2^* on the far side; c2∗≥12c_2^*≥ 12 gives u2(c2∗,c1′)≤u2(c2∗,c1′)u_2(c_2^*,c_1 )≤ u_2(c_2^*,c_1 ). Hence U1(c1′,c2∗)≥U1(c1′,c2∗)U_1(c_1 ,c_2^*)≥ U_1(c_1 ,c_2^*), and since the far-side range is contained in the own-side range, cross-overs are dominated. The tie yields 0<(1−β)u1(c1∗,c2∗)0<(1-β)u_1(c_1^*,c_2^*) (the equilibrium turnout is positive because Δ∗>bs(Δ∗) ^*>b_s( ^*) makes Eq. 32 strictly positive). By symmetry candidate 22 has no profitable deviation either, so the profile is the unique Nash equilibrium. Remark 14 (Consistency as α→0α→ 0) Applying the mean value theorem to K(Δ)−K(bs(Δ))K( )-K(b_s( )) in Eq. 43 produces the exact identity 1+α+β2+αK(Δ∗)=1+α−β2(1−Δ∗)k(ξα),ξα∈(bs(Δ∗),Δ∗). 1+α+β2+α\,K( ^*)= 1+α-β2\,(1- ^*)\,k( _α), _α∈ (b_s( ^*), ^* ). As α↓0α 0, bs(Δ∗)→Δ∗b_s( ^*)→ ^*, so ξα→Δ∗ _α→ ^* and, by continuity of k, the identity converges to (1+β)K(Δ)=(1−β)(1−Δ)k(Δ)(1+β)K( )=(1-β)(1- )k( ), i.e. ψβ(Δ)=1 _β( )=1: the α>0α>0 theory deforms continuously onto Theorem 0.L.1. (Verified numerically: the closed forms for doubly-uniform and polynomial K match the roots of Eq. 43 and of ψβ=1 _β=1 to 10−610^-6 across α∈0,12,1,2,β∈0,…,0.9α∈\0, 12,1,2\,β∈\0,…,0.9\.) 0.L.3 Comparative statics and their limits Proposition 18 (Monotone comparative statics) In the setting of Theorem 0.L.1(c) (resp. Theorem 0.L.2), the equilibrium gap Δ∗(α,β) ^*(α,β) is strictly decreasing in β on [0,1)[0,1), and (for α>0α>0) strictly increasing in α. Proof Decreasing in β, α=0α=0. γβ _β is strictly increasing in β and K/k(Δ∗)>0K/k( ^*)>0, so ψβ(Δ∗) _β( ^*) is strictly increasing in β; since ψβ _β is strictly increasing in t, the root of ψβ=1 _β=1 strictly decreases. (No differentiability in β is needed.) Decreasing in β, α>0α>0. Write Ψ(Δ,α,β)=J(Δ)−Lβ ( ,α,β)=J( )-L_β, so Ψ(Δ∗,⋅)=0 ( ^*,·)=0, ∂Ψ/∂Δ=J′<0∂ /∂ =J <0, and ∂Ψ∂β=−dLβdβ=−(11−β−11+α−β)=−α(1−β)(1+α−β)<0, ∂ ∂β=- dL_βdβ=- ( 11-β- 11+α-β )=- α(1-β)(1+α-β)<0, whence dΔ∗/dβ=−Ψβ/ΨΔ<0d ^*/dβ=- _β/ _ <0. Increasing in α. Here ∂Ψ/∂α=∂J/∂α−dLβ/dα∂ /∂α=∂ J/∂α-dL_β/dα with ∂J/∂α=−r(bs)∂bs∂α=r(bs(Δ))1−Δ2>0∂ J/∂α=-r (b_s ) ∂ b_s∂α=r (b_s( ) ) 1- 2>0 and dLβ/dα=1λ+11+α−β−11+α=:RdL_β/dα= 1λ+ 11+α-β- 11+α=:R. At Δ=Δ∗ = ^*, since r is non-increasing, Lβ=J(Δ∗)=∫bsΔ∗r≤r(bs)(Δ∗−bs)=r(bs)α2(1−Δ∗),L_β=J( ^*)= _b_s ^*r≤ r (b_s ) ( ^*-b_s )=r (b_s )\, α2\,(1- ^*), so r(bs)1−Δ∗2≥Lβαr(b_s) 1- ^*2≥ L_βα and it suffices to show g(β):=Lβ−αR>0g(β):=L_β-α R>0. Compute g′(β)=α(1−β)(1+α−β)−α(1+α−β)2=α2(1−β)(1+α−β)2>0,g (β)= α(1-β)(1+α-β)- α(1+α-β)^2= α^2(1-β)(1+α-β)^2>0, and, with s=α2s= α2, g(0)=logλ2−αλ=log(1+s)−s1+s>0g(0)= λ2- αλ= (1+s)- s1+s>0. Hence g(β)>0g(β)>0 for all β∈[0,1)β∈[0,1), ∂Ψ/∂α≥g(β)/α>0∂ /∂α≥ g(β)/α>0, and dΔ∗/dα=−Ψα/ΨΔ>0d ^*/dα=- _α/ _ >0. Proposition 19 (Reverse-hazard-rate comparison at every β) Let K,K~K, K satisfy the hypotheses of Theorem 0.L.1(c) (resp. Theorem 0.L.2) and suppose r~(t)≥r(t) r(t)≥ r(t) for all t∈(0,1)t∈(0,1). Then for every β∈[0,1)β∈[0,1) (and every α), Δ~∗(α,β)≥Δ∗(α,β) ^*(α,β)≥ ^*(α,β): more eager electorates sustain weakly more polarization, uniformly in competitiveness. Proof α=0α=0: r~≥r r≥ r means K~/k~≤K/k K/ k≤ K/k pointwise, so ψ~β≤ψβ ψ_β≤ _β pointwise; hence ψ~β(Δ∗)≤ψβ(Δ∗)=1=ψ~β(Δ~∗) ψ_β( ^*)≤ _β( ^*)=1= ψ_β( ^*) and, since ψ~β ψ_β is increasing, Δ~∗≥Δ∗ ^*≥ ^*. α>0α>0: J~(Δ)=∫bsΔr~≥J(Δ) J( )= _b_s r≥ J( ) pointwise, while LβL_β is the same for both games (K-free). Thus J~(Δ~∗)=Lβ=J(Δ∗)≤J~(Δ∗) J( ^*)=L_β=J( ^*)≤ J( ^*), and J~ J decreasing gives Δ~∗≥Δ∗ ^*≥ ^*. Proposition 20 (Collapse as β↑1β 1, with exact rate) Let x∼U(0,1)x U(0,1) and α>0α>0. (a) (Universal collapse; arbitrary K.) Let βn↑1 _n 1 and let (c1n,c2n)(c_1^n,c_2^n) be symmetric stationary profiles of the βn _n-games; by the proof of Proposition 17(b) (which uses no super-regularity) this includes every Nash equilibrium. Then Δn→Δ0=α2+α _n→ _0= α2+α: polarization collapses onto the most polarized symmetric margin-game profile. (b) (Exact finite-β identity.) Under the hypotheses of Theorem 0.L.2, for every β, Δ∗−Δ0=2λK−1((1−β)cβK(Δ∗)),cβ:=2(1+α)λ(1+α−β). ^*- _0= 2λ\,K^-1\! ((1-β)\,c_β\,K( ^*) ), c_β:= 2(1+α)λ\,(1+α-β). (c) (Rates.) If k(0+)∈(0,∞)k(0^+)∈(0,∞) then Δ∗−Δ0=(1−β)4(1+α)K(Δ0)λ2αk(0+)(1+o(1)) ^*- _0=(1-β)\, 4(1+α)\,K( _0)λ^2α\,k(0^+)\,(1+o(1)): linear in 1−β1-β, with K entering only through K(Δ0)/k(0+)K( _0)/k(0^+). If instead K(t)∼CtpK(t) Ct^p as t↓0t 0 (p>1p>1), the rate slows to order (1−β)1/p(1-β)^1/p. At α=0α=0 (under Theorem 0.L.1(c)): if K(t)/k(t)∼ctK(t)/k(t) ct as t↓0t 0 then Δβ∗=1−β2c(1+o(1)) ^*_β= 1-β2c(1+o(1)); in particular c=1c=1, and the rate is the universal 1−β2 1-β2, whenever k(0+)∈(0,∞)k(0^+)∈(0,∞). Proof (a) A symmetric stationary profile satisfies Eq. 43 with β=βnβ= _n (the derivation of Eq. 43 from Eq. 33 used no super-regularity), so K(bs(Δn)+)=(1−βn)cβnK(Δn)≤(1−βn)2(1+α)λαK(1)⟶ 0.K (b_s( _n)^+ )=(1- _n)\,c_ _nK( _n)≤(1- _n)\, 2(1+α)λα\,K(1)\; \;0. Since K≥K(t0)>0K≥ K(t_0)>0 on [t0,1][t_0,1] for each t0>0t_0>0, this forces bs(Δn)+→0b_s( _n)^+→ 0, i.e. Δn→Δ0 _n→ _0 (using bs(Δ)=λ2(Δ−Δ0)b_s( )= λ2( - _0) and, for the lower side, that Δn≤Δ0 _n≤ _0 would make the right-hand side of Eq. 43 positive and the left zero). (b) Immediate from Eq. 43: K(bs(Δ∗))=(1−β)cβK(Δ∗)K(b_s( ^*))=(1-β)c_βK( ^*) and bs(Δ∗)=λ2(Δ∗−Δ0)b_s( ^*)= λ2( ^*- _0). (c) For k(0+)∈(0,∞)k(0^+)∈(0,∞), K−1(ε)=ε/k(0+)(1+o(1))K^-1( )= /k(0^+)\,(1+o(1)); insert ε=(1−β)cβK(Δ∗) =(1-β)c_βK( ^*), cβ→2(1+α)λαc_β→ 2(1+α)λα and K(Δ∗)→K(Δ0)K( ^*)→ K( _0) (by (a) and continuity). For K∼CtpK Ct^p, K−1(ε)≍ε1/pK^-1( ) ^1/p. At α=0α=0: ψβ(Δ∗)=1 _β( ^*)=1 reads γβ(K/k)(Δ∗)=1−Δ∗ _β\,(K/k)( ^*)=1- ^*; the right side tends to 11 (as Δ∗→0 ^*→ 0 by the same argument as (a)), so (K/k)(Δ∗)∼γβ−1(K/k)( ^*) _β^-1 and K/k∼ctK/k ct gives Δ∗∼1cγβ∼1−β2c ^* 1c _β 1-β2c. Remark 15 (Numerical confirmation) For K=t2K=t^2 (p=2p=2, k(0+)=0k(0^+)=0) at α=0α=0 the root is exactly Δβ∗=2(1−β)3−β ^*_β= 2(1-β)3-β: rate (1−β)(1-β), constant 1/c=21/c=2, not the universal 1−β2 1-β2. At α=1α=1, β=0.99β=0.99, K=t2K=t^2: the computed gap is Δ∗=0.360984 ^*=0.360984 and 2λK−1((1−β)cβK(Δ∗))=0.027651=Δ∗−13 2λK^-1 ((1-β)c_βK( ^*) )=0.027651= ^*- 13 to six digits, matching (b) exactly. The reverse-hazard-rate comparison of Proposition 19 was confirmed at every β∈0,0.5,0.9,0.99β∈\0,0.5,0.9,0.99\ for K=tK=t versus K=t2K=t^2 at α∈0,1α∈\0,1\. The monotone-reverse-hazard-rate hypothesis in Proposition 18 is not removable: without it, more competition can increase polarization. Example 3 (An explicit cost distribution with three equilibria and an increasing branch) Let α=0α=0, x∼U(0,1)x U(0,1), and define K through φ:=K/k :=K/k by K(t)=exp(−∫t1φ(s)−1ds)K(t)= (- _t^1 (s)^-1ds ), where φ is piecewise linear: φ(t)=2t,t∈[0,1750],1725−32(t−1750),t∈[1750,12],1125+15(t−12),t∈[12,1]. (t)= cases2t,&t∈ [0, 1750 ],\\[2.0pt] 1725- 32 (t- 1750 ),&t∈ [ 1750, 12 ],\\[2.0pt] 1125+ 15 (t- 12 ),&t∈ [ 12,1 ]. cases Then φ>0 >0 on (0,1](0,1], K is a valid cost CDF satisfying Assumption 2.2 with K(1)=1K(1)=1 (k(t)=K(t)/φ(t)∼const⋅t−1/2k(t)=K(t)/ (t) · t^-1/2 as t↓0t 0, continuous and positive on (0,1)(0,1)), and: (i) For every β∈[0,13)β∈[0, 13), all symmetric stationary profiles are Nash equilibria; the equilibrium gaps are exactly the roots of ψβ(t)=t+γβφ(t)=1 _β(t)=t+ _β (t)=1. (i) For β∈[0,347)β∈ [0, 347 ) there are exactly three equilibria, with gaps (writing γ=γβγ= _β) Δ1=11+2γ,Δ2=100−119γ100−150γ,Δ3=50−17γ50+10γ, _1= 11+2γ, _2= 100-119γ100-150γ, _3= 50-17γ50+10γ, and dΔ2dγ=3100(100−150γ)2>0 d _2dγ= 3100(100-150γ)^2>0: along the middle branch, polarization strictly increases in β, from Δ2=1950 _2= 1950 at β=0β=0 to 12 12 at the fold. The outer branches are strictly decreasing. (i) At β∗=347β^*= 347 (where γβ∗=2522 _β^*= 2522) the middle and upper branches collide at Δ=12 = 12 and annihilate: for β∈(β∗,13)β∈(β^*, 13) the branch Δ1 _1 is the unique equilibrium. The equilibrium count drops from three to one through a fold—a non-smooth one, since the two gaps meet at the kink of ψβ _β at t=12t= 12 and so approach 12 12 at finite opposite rates rather than at the square-root rate of a smooth saddle-node: increasing competitiveness discontinuously de-polarizes the equilibrium set. Proof K is a CDF with K(0)=0K(0)=0 by divergence of ∫0+φ−1=∫0+ds2s _0^+ ^-1= _0^+ ds2s, and Assumption 2.2 is immediate. (i) By Theorem 0.L.1(b) and Eqs. 40 to 41, a symmetric stationary profile is an equilibrium as soon as χβ _β is strictly increasing (then F is single-peaked with peak at the stationary gap). Here χβ=t+1+β2φ(t) _β=t+ 1+β2 (t) is piecewise linear with slopes 1+1+β2⋅2=2+β,1−34(1+β),1+1+β10,1+ 1+β2· 2=2+β, 1- 34(1+β), 1+ 1+β10, all strictly positive iff β<13β< 13. (i) ψβ=t+γφ(t) _β=t+γ (t) is piecewise linear with slopes 1+2γ>01+2γ>0, 1−32γ<01- 32γ<0, 1+γ5>01+ γ5>0; solving ψβ=1 _β=1 on each piece gives the displayed formulas, and the membership conditions Δ1≤1750 _1≤ 1750, Δ2∈(1750,12] _2∈( 1750, 12], Δ3∈(12,1) _3∈( 12,1) hold precisely for γ<2522γ< 2522, i.e. β<347β< 347 (at γ=1γ=1: Δ1=13≤0.34 _1= 13≤ 0.34, Δ2=1950 _2= 1950, Δ3=1120 _3= 1120). The derivative of Δ2 _2 in γ is as displayed, and γβ _β is strictly increasing in β; monotonicity of the outer branches is a one-line computation. (i) At γ=2522γ= 2522, ψβ(12)=12+γ⋅1125=1 _β( 12)= 12+γ· 1125=1, and for γ>2522γ> 2522 the minimum of ψβ _β over [1750,1][ 1750,1] exceeds 11 while the first piece still crosses: exactly one root remains; for β∈(347,13)β∈( 347, 13) it is, by (i), the unique equilibrium. Remark 16 (A local criterion, and the three-root example of Appendix 0.C) The proof of Example 3 isolates a criterion of independent use: at a symmetric stationary profile with gap Δ at which φ=K/k =K/k is differentiable, χβ′(Δ)=1+1+β2φ′(Δ)<0⟹Δ is a strict local minimum of F, _β ( )=1+ 1+β2\, ( )<0\; \;$ $ is a strict local minimum of $F$, hence not an equilibrium, since F′=k[Q−χβ]F =k\,[Q- _β] then crosses 00 from below. This explains, conceptually rather than numerically, why the three-root cost distribution has non-equilibrium middle root: there χ0′=1+φ′/2 _0 =1+ /2 evaluates (numerically) to +4.5+4.5, −0.5-0.5, +0.94+0.94 at the three roots Δ∈14,12,34 ∈\ 14, 12, 34\—the middle root is a local minimum of the deviator’s payoff, while in Example 3 the middle root has χβ′>0 _β >0 and is a genuine equilibrium. Which of the two behaviors a middle root exhibits is thus read off the slope of K/kK/k at the root. 0.L.4 The β↑1β 1 limit for general electorates Theorem 0.L.3 (Upper hemicontinuity onto the margin game) Let Assumption 2.2 hold and α≥0α≥ 0. The payoffs (u1,u2)(u_1,u_2) are jointly continuous on [0,1]2[0,1]^2, and supc|Uiβ(c)−Ui1(c)|≤(1−β) _c\,|U_i^β(c)-U_i^1(c)|≤(1-β) for all i and β. Consequently, if βn↑1 _n 1 and (cn)(c^n) are Nash equilibria of the βn _n-games with cn→c∞c^n→ c^∞, then c∞c^∞ is a Nash equilibrium of the margin game (β=1)(β=1). In particular: (a) If α=0α=0, then cn→(q,q)c^n→(q,q), the median profile: equilibria of the nearly-zero-sum games select the margin game’s unique equilibrium. (No monotonicity of r=k/Kr=k/K and no bound on ρv _v is assumed.) (b) If α>0α>0 and x∼U(0,1)x U(0,1), every limit point lies in the margin-game equilibrium set (c1,c2):1λ≤c1≤c2≤1+αλ\(c_1,c_2): 1λ≤ c_1≤ c_2≤ 1+αλ\; under the hypotheses of Theorem 0.L.2 the full limit is the maximally polarized symmetric profile, Δ∗(α,β)↓Δ0 ^*(α,β) _0. Proof Continuity. For positions (a,o)(a,o) and (a′,o′)(a ,o ) and every voter x, |B(x,a,o)−B(x,a′,o′)|≤(1+α)|a−a′|+|o−o′| |B(x;a,o)-B(x;a ,o ) |≤(1+α)|a-a |+|o-o |, and t↦K(t+)t K(t^+) is uniformly continuous on [−1,1][-1,1] (continuous on a compact); since ∫v=1 v=1, |u1(a,o)−u1(a′,o′)|≤ωK((1+α)|a−a′|+|o−o′|),|u_1(a,o)-u_1(a ,o )|≤ _K ((1+α)|a-a |+|o-o | ), with ωK _K the modulus of continuity of K(⋅+)K(·^+)—valid for arbitrary v, including unbounded densities. The same bound holds for u2u_2. Continuity holds across ties: as a→oa→ o all benefits tend to 00 uniformly, so ui→0u_i→ 0, the tie payoff. Next, |Uiβ−Ui1|=(1−β)u−i≤(1−β)K(1)≤1−β|U_i^β-U_i^1|=(1-β)\,u_-i≤(1-β)K(1)≤ 1-β. Limit. Fix any c1′c_1 . For each n, U1βn(c1n,c2n)≥U1βn(c1′,c2n)U_1 _n(c_1^n,c_2^n)≥ U_1 _n(c_1 ,c_2^n). Passing to the limit using joint continuity of (u1,u2)(u_1,u_2) and the uniform bound, U11(c∞)≥U11(c1′,c2∞)U_1^1(c^∞)≥ U_1^1(c_1 ,c_2^∞); similarly for candidate 22. Hence c∞c^∞ is a margin-game equilibrium. (a) then follows from Theorem 5.2 (at α=0α=0 the margin game has the unique equilibrium (q,q)(q,q), q the median), plus compactness of [0,1]2[0,1]^2: every subsequence has a sub-subsequence converging to (q,q)(q,q), so the whole sequence does. (b) follows from the uniform-voter margin-game characterization and Propositions 18 and 20. Upper hemicontinuity presumes the βn _n-equilibria exist. The next results show that for concentrated-at-the-extremes electorates they may fail to exist for all β close to 11—so the median selection of Theorem 0.L.3(a) is genuinely a statement about limits, not a guarantee. The engine is a quantitative margin-game certificate. Lemma 7 (Lipschitz transfer) Let V be arbitrary, α≥0α≥ 0, and let K be LKL_K-Lipschitz on [0,1][0,1] (e.g. LK=1L_K=1 for κ∼U(0,1)κ U(0,1)). Then the margin w(c,c′):=udev(c,c′)−udev(c′,c)w(c,c ):=u_dev(c,c )-u_dev(c ,c) satisfies, for all c,c~,c′c, c,c , |w(c,c′)−w(c~,c′)|≤(2+α)LK|c−c~|,|w(c,c′)−w(c,c~′)|≤(2+α)LK|c′−c~′|.|w(c,c )-w( c,c )|≤(2+α)L_K\,|c- c|, |w(c,c )-w(c, c )|≤(2+α)L_K\,|c - c |. Proof Moving one candidate by h changes every voter’s benefit toward her by at most (1+α)|h|(1+α)|h| and toward her opponent by at most |h||h|; since t↦K(t+)t K(t^+) is LKL_K-Lipschitz and ∫v=1 v=1, the two vote shares move by at most (1+α)LK|h|(1+α)L_K|h| and LK|h|L_K|h|. Sum. (No boundedness of v is used.) Proposition 21 (Grid certificate) In the setting of Lemma 7, let g1,…,gNg_1,…,g_N be the uniform grid on [0,1][0,1] with step h, and suppose that maximinjw(gi,gj)≤−ε^< 0. _i _jw(g_i,g_j)\;≤\;- \;<\;0. Set ε′:=ε^−(2+α)LKh/2 := -(2+α)L_K\,h/2 and assume ε′>0 >0. Then: (a) minc′∈[0,1]w(c,c′)≤−ε′ _c ∈[0,1]w(c,c )≤- for every c∈[0,1]c∈[0,1]; in particular the margin game (β=1β=1) has no pure Nash equilibrium. (b) For every c′∈[0,1]c ∈[0,1] there exists c′c with w(c′,c′)≥ε′w(c ,c )≥ . Proof (a) Given c, take the nearest grid point gig_i (|c−gi|≤h/2|c-g_i|≤ h/2) and the certified gjg_j with w(gi,gj)≤−ε^w(g_i,g_j)≤- ; then w(c,gj)≤w(gi,gj)+(2+α)LKh/2≤−ε′w(c,g_j)≤ w(g_i,g_j)+(2+α)L_Kh/2≤- . A pure equilibrium (c1,c2)(c_1,c_2) of the symmetric zero-sum margin game would need w(c1,⋅)≥0w(c_1,·)≥ 0 everywhere (its value is 00 by antisymmetry), which (a) excludes. (b) Apply (a) at c=c′c=c to get gjg_j with w(c′,gj)≤−ε′w(c ,g_j)≤- , hence by antisymmetry w(gj,c′)≥ε′w(g_j,c )≥ . Theorem 0.L.4 (Non-existence for hollow electorates) [computer-assisted] Let α=1α=1, κ∼U(0,1)κ U(0,1). (a) For v=Beta(12,12)v=Beta( 12, 12) (the arcsine electorate): on the uniform grid with N=1201N=1201 (h=11200h= 11200), every w(gi,gj)w(g_i,g_j) evaluates exactly (in terms of the regularized incomplete beta function) and maximinjw(gi,gj)=−8.2244×10−3 _i _jw(g_i,g_j)=-8.2244× 10^-3; the Lipschitz slack is (2+α)h/2=1.25×10−3(2+α)h/2=1.25× 10^-3, so Proposition 21 applies with ε′=6.97×10−3 =6.97× 10^-3. The margin game has no pure equilibrium. (b) The same holds for v=Beta(0.8,0.8)v=Beta(0.8,0.8) with N=2401N=2401, ε^=3.0238×10−3 =3.0238× 10^-3, slack 6.25×10−46.25× 10^-4, ε′=2.40×10−3 =2.40× 10^-3: non-existence is not an artifact of the unbounded arcsine density. Proof Lemmas 7 and 21 reduce the claim to the finite computation, which is exact modulo floating-point evaluation of the incomplete beta function: each w(gi,gj)w(g_i,g_j) is a finite signed combination of terms V(⋅)V(·) and ∫⋅xv(x)x _·^·x\,v(x)dx at algebraic break points (the closed-form uniform-cost payoff), evaluated in double precision; we deduct a numerical guard of 10−1010^-10—more than seven orders of magnitude below the certified margins—inside ε′ . The computation is in the accompanying verification suite (check_competition.py, claim C1.arcsine). A sanity check on v=Beta(2,2)v=Beta(2,2) (centrally peaked), at the same α=1α=1, returns grid security exactly 00, attained at the median and only there on the grid, consistent with Theorem 5.4 (v′(12)=−12<0v ( 12)=-12<0, so the median is a strict local equilibrium of the margin game). Corollary 11 (Non-existence for an interval of β) In the setting of Theorem 0.L.4, the β-game has no pure Nash equilibrium for any β>1−ε′1+ε′=0.98615…v=Beta(12,12),0.99521…v=Beta(0.8,0.8).β> 1- 1+ = cases0.98615…&v=Beta( 12, 12),\\ 0.99521…&v=Beta(0.8,0.8). cases Proof Suppose (c1,c2)(c_1,c_2) is a pure equilibrium of the β-game. Decompose U1=1−β2T+1+β2wU_1= 1-β2T+ 1+β2w with T=u1+u2T=u_1+u_2. By Proposition 21(b) there is c′c with w(c′,c2)≥ε′w(c ,c_2)≥ , so candidate 11’s equilibrium payoff satisfies U1≥U1(c′,c2)≥1+β2ε′U_1≥ U_1(c ,c_2)≥ 1+β2 (both terms of the decomposition are then nonnegative). Likewise U2≥1+β2ε′U_2≥ 1+β2 . Summing, (1−β)T(c1,c2)=U1+U2≥(1+β)ε′(1-β)\,T(c_1,c_2)=U_1+U_2≥(1+β) , while T≤1T≤ 1 always; hence β≤1−ε′1+ε′β≤ 1- 1+ . Remark 17 (How conservative is the bound?) [numerical observation] The corollary certifies non-existence on a left neighborhood of 11; it is not tight. Numerically (best-response iteration with global verification), for v=Beta(12,12)v=Beta( 12, 12), α=1α=1: pure equilibria exist for β≤0.90β≤ 0.90 (at β=0.90β=0.90: (0.234,0.766)(0.234,0.766)) and none are found for β≥0.91β≥ 0.91. We label the location of the true threshold a numerical observation; the theorem-grade statements are Corollary 11 and, sharper, Theorem 0.O.1 (β≥0.9168β≥ 0.9168). Together with Theorems 0.L.1 and 0.L.2 (existence for all β<1β<1 under uniform voters and super-regularity) and Theorems 0.N.1 and 14 (existence for all β<1β<1 at α=0α=0 under uniform costs, the arcsine electorate included), this places the certified failure of pure existence at the joint corner α=1α=1, β near 11, hollow v: non-existence is a phenomenon of alienation and competitiveness together, not of competitiveness alone. The two existence conjectures stay well clear of that corner: Remark 3 is the baseline, with α=β=0α=β=0, and Conjecture 1 switches off alienation alone. 0.L.5 Quantile structure at α=0α=0 under uniform costs Proposition 22 (Mass-gap law and concavity under uniform costs) Let α=0α=0, κ∼U(0,1)κ U(0,1), β∈[0,1)β∈[0,1), and let v satisfy Assumption 2.2. (a) Every Nash equilibrium satisfies V(c2)−V(c1)=1−β2,2V(c1)=V(m)+β(1−V(m)):V(c_2)-V(c_1)= 1-β2, 2V(c_1)=V(m)+β (1-V(m) ): (45) the platforms trap voter mass exactly 1−β2 1-β2, decreasing in competitiveness, independently of v. (b) If ρv≤41−β _v≤ 41-β then U1(⋅,c2)U_1(·,c_2) is concave on [0,c2)[0,c_2) for every c2c_2 (and mirror), so equilibria are exactly the profiles satisfying Eq. 45 at which neither candidate gains from her cross-over deviation. Proof (a) By Lemma 4(b) and Eq. 31 with k≡1k≡ 1: ∂U1/∂c1=V(m)−2V(c1)+β(1−V(m))=0∂ U_1/∂ c_1=V(m)-2V(c_1)+β(1-V(m))=0 and, mirroring, ∂U2/∂c2=1−V(m)−2(1−V(c2))+βV(m)=0∂ U_2/∂ c_2=1-V(m)-2(1-V(c_2))+β V(m)=0. The first display is the second condition of Eq. 45; the second rearranges to 2(1−V(c2))=(1−V(m))+βV(m)2 (1-V(c_2) )= (1-V(m) )+β V(m). Adding this to 2V(c1)=V(m)+β(1−V(m))2V(c_1)=V(m)+β(1-V(m)) gives 2+2V(c1)−2V(c2)=1+β2+2V(c_1)-2V(c_2)=1+β, i.e. the mass-gap law. (b) Differentiating once more, ∂2U1/∂c12=1−β2v(m)−2v(c1)≤1−β2supv−2infv≤0∂^2U_1/∂ c_1^2= 1-β2\,v(m)-2v(c_1)≤ 1-β2 v-2 v≤ 0. Concavity makes stationarity equivalent to own-side global optimality; ties are dominated as before, and the only remaining deviations are cross-overs. Corollary 12 (Quantile equilibrium for symmetric electorates) In the setting of Proposition 22, let v be symmetric about 12 12 with ρv≤41−β _v≤ 41-β. Then the profile placing the candidates at the 1+β4 1+β4 and 3−β4 3-β4 quantiles, V(c1)=1+β4,V(c2)=3−β4,V(c_1)= 1+β4, V(c_2)= 3-β4, is a Nash equilibrium. At β=0β=0 this is the quartile rule of Corollary 1; as β↑1β 1 the platforms converge to the median from the 14 14/34 34 quantiles, sweeping the entire inner quartile range. Proof By symmetry of v, c2=1−c1c_2=1-c_1, m=12m= 12, V(m)=12V(m)= 12, and Eq. 45 holds: 2⋅1+β4=12+β⋅122· 1+β4= 12+β· 12. By Proposition 22(b), c1c_1 is a global maximizer of U1(⋅,c2)U_1(·,c_2) on [0,c2)[0,c_2). It remains to compare the equilibrium payoff with the tie and with cross-over deviations. Write s:=infv>0s:= v>0 and ρ:=ρvρ:= _v; the mass-gap law gives 1−β2=∫c1c2v≤(supv)Δ≤ρsΔ 1-β2= _c_1^c_2v≤( v)\, ≤ρ s\, , so Δ≥1−β2ρs. \;≥\; 1-β2ρ s. (46) Equilibrium payoff. By symmetry u2=u1u_2=u_1 at the profile, so U1=(1−β)u1U_1=(1-β)u_1. With K(t)=tK(t)=t, u1 u_1 =ΔV(c1)+∫c1m(c1+c2−2x)vx = \,V(c_1)+ _c_1^m(c_1+c_2-2x)\,v\,dx ≥1+β4Δ+s∫c1m(c1+c2−2x)x=1+β4Δ+sΔ24, ≥ 1+β4\, +s\! _c_1^m(c_1+c_2-2x)\,dx= 1+β4\, + s ^24, hence U1(c1,c2)≥(1−β2)Δ4> 0,U_1(c_1,c_2)\;≥\; (1-β^2)\, 4\;>\;0, (47) which in particular dominates the tie. Cross-over. Let c1′=c2+tc_1 =c_2+t, t∈(0,1−c2]t∈(0,1-c_2], and m′=c2+t2m =c_2+ t2. The deviator’s supporters lie in (m′,1](m ,1] with benefit at most t, so, using V(m′)≥V(c2)+st2V(m )≥ V(c_2)+s t2, u1′≤t(1−V(m′))≤t1+β4−st22.u_1 ≤ t (1-V(m ) )≤ t\, 1+β4- s\,t^22. The opponent’s supporters include all of [0,c2][0,c_2], each with benefit exactly t, so her share satisfies u2′≥tV(c2)=t3−β4u_2 ≥ t\,V(c_2)=t\, 3-β4. Hence U1(c1′,c2)≤t1+β4−βt3−β4−st22=(1−β)24t−s2t2≤(1−β)432s,U_1(c_1 ,c_2)≤ t\, 1+β4-β t\, 3-β4- st^22= (1-β)^24\,t- s2\,t^2\;≤\; (1-β)^432\,s, the last step by maximizing the concave quadratic over t≥0t≥ 0. Finally, by Eqs. 46 to 47, U1(c1,c2)≥(1−β2)4⋅1−β2ρs=(1+β)(1−β)28ρs≥(1−β)432s⇔ρ≤4(1+β)(1−β)2,U_1(c_1,c_2)≥ (1-β^2)4· 1-β2ρ s= (1+β)(1-β)^28ρ s\;≥\; (1-β)^432s ρ≤ 4(1+β)(1-β)^2, and ρ≤41−β≤4(1+β)(1−β)2ρ≤ 41-β≤ 4(1+β)(1-β)^2 always. Candidate 22 is the mirror image. Remark 18 (a) The mass-gap law in Eq. 45 is distribution-free and gives a clean empirical signature of competitiveness: the fraction of the electorate strictly between the platforms is 1−β2 1-β2, whatever v. It was verified numerically to 10−1110^-11 at asymmetric densities, and the quantile equilibrium of Corollary 12 to 10−410^-4 against best-response search for v=1+12cos2πxv=1+ 12 2π x at β∈0.2,0.5,0.8β∈\0.2,0.5,0.8\ (unique equilibrium found in each case). (b) The concavity threshold 41−β 41-β increases in β: competitiveness enlarges the sufficiency umbrella at α=0α=0, in exact antithesis to alienation, which shrinks it (Theorem 0.K.1: F(α)↓F(α) ). The two extensions are not symmetric levers on existence. (c) Multiplicity is not excluded and is real: non-uniqueness examples at β=0β=0 deform continuously to small β>0β>0 (Example 3 is the cost-side analogue). Appendix 0.M The joint corner: α>0α>0, β>0β>0, non-uniform voters Standing notation. As in Appendix 0.I: voters x∼Vx V on [0,1][0,1] with density v, costs κ∼Kκ K with density k, both under Assumption 2.2 (with K(1)∈(0,1]K(1)∈(0,1] permitted). Candidate positions c1≤c2c_1≤ c_2, Δ=c2−c1 =c_2-c_1, λ=2+αλ=2+α, benefit Bi(x)=d−i(x)−(1+α)di(x)B_i(x)=d_-i(x)-(1+α)d_i(x), payoffs ui=∫K(Bi+)Vu_i= K(B_i^+)\,dV and Ui=ui−βu−iU_i=u_i-β u_-i with β∈[0,1)β∈[0,1). Tent edges for c1<c2c_1<c_2: xin1=c1+Δλ,xout1=c1−Δα,xin2=c2−Δλ,xout2=c2+Δα(α>0), gatheredx^in_1=c_1+ λ, x^out_1=c_1- α,\\ x^in_2=c_2- λ, x^out_2=c_2+ α (α>0), gathered with the conventions x1out=−∞x^out_1=-∞, x2out=+∞x^out_2=+∞ at α=0α=0. We write ρv=supv/infv _v= v/ v (possibly ∞), S=supvS= v, s=infvs= v. The master formula (Lemma 5) reads, for c1<c2c_1<c_2 interior, ∂U1∂c1=(1+α)(I1−O1)+β(I2+O2), ∂ U_1∂ c_1=(1+α)\,(I_1-O_1)+β\,(I_2+O_2), (48) where I1=∫c1x1ink(B1)v,O1=∫x1out∨0c1k(B1)v,I2=∫x2inc2k(B2)v,O2=∫c2x2out∧1k(B2)v, gatheredI_1= _c_1^x^in_1\!k(B_1)v, O_1= _x^out_1 0^c_1\!k(B_1)v,\\ I_2= _x^in_2^c_2\!k(B_2)v, O_2= _c_2^x^out_2 1\!k(B_2)v, gathered the k-weighted voter masses on the inner and outer arms of the two benefit tents (at α=0α=0 the outer arms carry the constant benefit Δ , so O1=k(Δ)V(c1)O_1=k( )V(c_1) and O2=k(Δ)(1−V(c2))O_2=k( )(1-V(c_2))). 0.M.1 Two distribution-free laws of stationarity The first result generalizes the mass-gap law of Proposition 22: that law is the α=0α=0, uniform-cost shadow of a ratio law that is free of both distributions and holds at every interior stationary profile of the joint game. Lemma 8 (Mirror first-order condition) For c1<c2c_1<c_2 interior, ∂U2∂c2=(1+α)(O2−I2)−β(I1+O1). ∂ U_2∂ c_2=(1+α)\,(O_2-I_2)-β\,(I_1+O_1). (49) Proof Apply the reflection σ(x)=1−xσ(x)=1-x. In the reflected game the voter density is v^(x)=v(1−x) v(x)=v(1-x), the cost distribution is unchanged, and the profile (c1,c2)(c_1,c_2) becomes (c^1,c^2)=(1−c2,1−c1)( c_1, c_2)=(1-c_2,1-c_1) with c^1<c^2 c_1< c_2; the change of variables x↦1−x 1-x in the payoff integrals gives u^1(c^1,c^2)=u2(c1,c2) u_1( c_1, c_2)=u_2(c_1,c_2) and u^2=u1 u_2=u_1, hence U^1=U2 U_1=U_2. Since c2=1−c^1c_2=1- c_1, ∂U2/∂c2=−∂U^1/∂c^1∂ U_2/∂ c_2=-\,∂ U_1/∂ c_1, and Eq. 48 applied in the reflected game gives ∂U^1/∂c^1=(1+α)(I^1−O^1)+β(I^2+O^2)∂ U_1/∂ c_1=(1+α)( I_1- O_1)+β( I_2+ O_2). Changing variables back, I^1=I2 I_1=I_2, O^1=O2 O_1=O_2, I^2=I1 I_2=I_1, O^2=O1 O_2=O_1, which is Eq. 49. Proposition 23 (Inner–outer ratio law and asymmetry law) Let (c1,c2)(c_1,c_2) with 0<c1<c2<10<c_1<c_2<1 be a stationary profile of the joint game, i.e. a profile at which both first-order conditions hold — in particular, any interior untied Nash equilibrium. Write I:=I1+I2I:=I_1+I_2 and O:=O1+O2O:=O_1+O_2 for the total k-weighted inner- and outer-arm masses. Then, for every (v,k,α,β)(v,k,α,β) under Assumption 2.2: (1+α+β)I (1+α+β)\,I =(1+α−β)O, =(1+α-β)\,O, (50) (1+α−β)(I1−I2) (1+α-β)\,(I_1-I_2) =(1+α+β)(O1−O2). =(1+α+β)\,(O_1-O_2). (51) Proof Subtracting Eq. 49 from Eq. 48, both being zero, 0=(1+α)[(I1−O1)−(O2−I2)]+β[(I2+O2)+(I1+O1)]=(1+α)(I−O)+β(I+O),0=(1+α) [(I_1-O_1)-(O_2-I_2) ]+β [(I_2+O_2)+(I_1+O_1) ]=(1+α)(I-O)+β(I+O), which rearranges to Eq. 50. Adding the two conditions instead, 0=(1+α)[(I1−I2)−(O1−O2)]−β[(I1−I2)+(O1−O2)],0=(1+α) [(I_1-I_2)-(O_1-O_2) ]-β [(I_1-I_2)+(O_1-O_2) ], which rearranges to Eq. 51. Corollary 13 (Special cases) At every interior stationary profile: (a) β=0β=0: I=OI=O — each candidate pair balances inner against outer k-mass in aggregate (the balance conditions of Lemma 5, summed). (b) α=0α=0, k≡1k≡ 1: I=V(c2)−V(c1)I=V(c_2)-V(c_1) and O=1−IO=1-I, so Eq. 50 reads V(c2)−V(c1)=1−β2V(c_2)-V(c_1)= 1-β2: the mass-gap law of Proposition 22 is the α=0α=0 shadow of Eq. 50. (c) For any α>0α>0, the inner share of the total marginal k-mass is the universal constant I/(I+O)=1+α−β2(1+α)I/(I+O)= 1+α-β2(1+α); it decreases from 12 12 at β=0β=0 to α2(1+α)>0 α2(1+α)>0 as β↑1β 1. Competition drains the contested middle, but under alienation it can never empty it. (d) By Eq. 51, since 1+α+β1+α−β>1 1+α+β1+α-β>1, any asymmetry between the candidates’ outer masses is amplified in their inner masses, with the universal gain 1+α+β1+α−β 1+α+β1+α-β; at symmetric profiles both sides vanish. Remark 19 (Verification) [numerical observation] The suite verifies Eq. 50 and the inner-share constant of Corollary 13(c) at polished stationary profiles of thirteen games. The games span α∈0,0.5,1,2α∈\0,0.5,1,2\ and β∈0,…,0.5β∈\0,…,0.5\, with uniform, cosine, Beta(2,2)Beta(2,2), Beta(8,8)Beta(8,8), and asymmetric Beta(2,5)Beta(2,5) electorates, and with uniform and quadratic (K(t)=t2K(t)=t^2) costs. The cases include a dead-zone equilibrium — Beta(8,8)Beta(8,8), α=2α=2, β=0.2β=0.2, where I/O=7/8I/O=7/8 exactly — and asymmetric equilibria. Relative errors are 10−810^-8–10−1210^-12. 0.M.2 The joint single-peakedness threshold For uniform costs, Theorem 0.K.1 gave the β=0β=0 umbrella ρv≤F(α)=minf(α),g(α) _v≤ F(α)= \f(α),g(α)\ with f(α)=2λ1+αf(α)= 2λ1+α (concavity off the dead zone) and g(α)=λαg(α)= λα (monotone decrease inside it). The joint threshold deforms the two branches in opposite directions. Theorem 0.M.1 (Joint single-peakedness, uniform costs) Let κ∼U(0,1)κ U(0,1), α>0α>0, β∈[0,1)β∈[0,1), and let v satisfy Assumption 2.2 with ρv≤Fβ(α):=minfβ(α),gβ(α),fβ(α)=2λ(1+α)+β(1+α)2,gβ(α)=(1+α)λα(1+α+β)+βλ. gathered _v\;≤\;F_β(α):= \f_β(α),\,g_β(α)\,\\ f_β(α)= 2λ(1+α)+β(1+α)^2, g_β(α)= (1+α)λα(1+α+β)+βλ. gathered (52) Then for every opponent position c2∈(0,1]c_2∈(0,1] the own-side map c1↦U1(c1,c2)c_1 U_1(c_1,c_2) on [0,c2)[0,c_2) is concave on the no-dead-zone piece c1:x1out≤0=[0,c21+α]\c_1:x^out_1≤ 0\=[0, c_21+α] and strictly decreasing on the dead-zone piece (c21+α,c2)( c_21+α,c_2); hence it is single-peaked, with its peak in the no-dead-zone piece. The mirror statement holds for candidate 22 on (c1,1](c_1,1]. Moreover fβf_β is strictly increasing and gβg_β strictly decreasing in β, with f0=f_0=f and g0=g_0=g. Proof Fix c2c_2 and write derivatives in c1c_1. With uniform costs Eq. 48 reads U1′=(1+α)[(V(x1in)−V(c1))−(V(c1)−V(x1out∨0))]+β[V(x2out∧1)−V(x2in)].U_1 =(1+α) [ (V(x^in_1)-V(c_1) )- (V(c_1)-V(x^out_1 0) ) ]+β [V(x^out_2 1)-V(x^in_2) ]. (53) No-dead-zone piece (x1out≤0x^out_1≤ 0). Here V(x1out∨0)=0V(x^out_1 0)=0 is constant. Differentiating Eq. 53, using (x1in)′=1+αλ(x^in_1) = 1+αλ, (x2in)′=1λ(x^in_2) = 1λ, (x2out)′=−1α(x^out_2) =- 1α, and treating the kink of V(x2out∧1)V(x^out_2 1) at x2out=1x^out_2=1 one-sidedly, U1′=(1+α)[1+αλv(x1in)−2v(c1)]−β[1λv(xin2)+1αv(xout2)xout2<1]a.e.U_1 =(1+α) [ 1+αλv(x^in_1)-2v(c_1) ]\\ -β [ 1λv(x^in_2)+ 1αv(x^out_2)1_\x^out_2<1\ ] .e. Dropping the nonnegative x2outx^out_2-term and pairing the two 1λ 1λ-terms, U1′ U_1 ≤(1+α)2v(x1in)−βv(x2in)λ−2(1+α)v(c1) \;≤\; (1+α)^2v(x^in_1)-β v(x^in_2)λ-2(1+α)v(c_1) ≤(1+α)2S−βsλ−2(1+α)s≤ 0 \;≤\; (1+α)^2S-β sλ-2(1+α)s\;≤\;0 whenever (1+α)2ρv≤2(1+α)λ+β(1+α)^2 _v≤ 2(1+α)λ+β, i.e. ρv≤fβ(α) _v≤ f_β(α). Since U1′U_1 is continuous in c1c_1 (the integrands and limits in Eq. 53 are continuous), a continuous function with a.e. nonpositive derivative of its derivative is concave. Dead-zone piece (x1out>0x^out_1>0). From Eq. 53, using V(x1in)−V(c1)≤SΔλV(x^in_1)-V(c_1)≤ S λ, V(c1)−V(x1out)≥sΔαV(c_1)-V(x^out_1)≥ s α, and V(x2out∧1)−V(x2in)≤SΔ(1λ+1α)V(x^out_2 1)-V(x^in_2)≤ S ( 1λ+ 1α), U1′≤(1+α)Δ(Sλ−sα)+βSΔ(1λ+1α),U_1 \;≤\;(1+α) ( Sλ- sα )+β S ( 1λ+ 1α ), (54) which is negative iff (1+α)λ>ρv[α(1+α+β)+βλ](1+α)λ> _v [α(1+α+β)+βλ ], i.e. ρv<gβ(α) _v<g_β(α). At ρv=gβ(α) _v=g_β(α) equality in Eq. 54 would force v≡Sv≡ S on (c1,x1in)(c_1,x^in_1) and v≡sv≡ s on (x1out,c1)(x^out_1,c_1) simultaneously, contradicting continuity of v at c1c_1 unless S=sS=s, in which case ρv=1<gβ _v=1<g_β. Hence U1′<0U_1 <0 strictly on the dead-zone piece whenever ρv≤gβ(α) _v≤ g_β(α). A function that is concave on [0,c21+α][0, c_21+α], strictly decreasing on (c21+α,c2)( c_21+α,c_2), and continuous, is single-peaked with its maximum in the first piece. The mirror statement follows from Lemma 8 (equivalently, reflection). Monotonicity of fβf_β in β is immediate; for gβg_β, the denominator is strictly increasing in β since α+λ>0α+λ>0. Finally f0=2λ1+αf_0= 2λ1+α and g0=λαg_0= λα. Remark 20 (Competition rescues concavity but taxes the dead zone) The joint threshold Fβ(α)F_β(α) is neither a β-dependent rescaling of the β=0β=0 threshold F(α)F(α) nor moved by β in a single direction. On the no-dead-zone piece the competitiveness term −βu2-β u_2 contributes the stabilizing second derivative −β[v(x2in)λ+v(x2out)α]-β[ v(x^in_2)λ+ v(x^out_2)α1] (hurting the opponent is easiest exactly where the deviator’s own tent is thinnest), so fβf_β grows with β: competition partly rescues the umbrella that alienation shrinks. Inside the dead zone the same term is destabilizing — moving toward the opponent hurts her even when it no longer helps oneself — so gβg_β shrinks. Consequently for small α (where f binds) the umbrella widens with β, while for large α (where g binds) it narrows further; the crossover alienation level solves fβ=gβf_β=g_β and moves left as β grows. At α=0α=0 the exact computation of Proposition 22 gives the threshold 41−β 41-β, strictly better than limα↓0fβ=4+β _α 0f_β=4+β: there the inner tent edges x1inx^in_1 and x2inx^in_2 coincide at the midpoint, so the stabilizing term cancels against the destabilizing one pointwise rather than through the worst-case pair (S,s)(S,s). For α>0α>0 the two edges are distinct and the worst case is achievable; see the sharpness evidence below. Remark 21 (Sharpness) [numerical observation] A numerical search over continuous piecewise-linear “mesa” densities of ratio ρ with the high plateau covering x1inx^in_1 and verifies, at (α,β)∈(1,.5),(.5,.8),(2,.3)(α,β)∈\(1,.5),(.5,.8),(2,.3)\, that U1′U_1 changes sign as ρ crosses fβ(α)f_β(α) (±7%± 7\%). Section 0.M.4 shows gβg_β is essentially attained as well (Remark 24). The constant fβf_β is therefore the exact worst-case concavity threshold within this proof scheme, though densities realizing the worst case are non-generic. Remark 22 (General k: the obstruction) [open] For non-increasing k∈C1k∈ C^1 the threshold ρvρk≤2λ/(1+α) _v _k≤ 2λ/(1+α) of Proposition 16 does not deform as cleanly. Differentiating the β-term of Eq. 48 in c1c_1 produces, besides stabilizing boundary terms proportional to −βk(0+)-β k(0^+), the bulk term −β∫x2inx2out∧1k′(B2)v-β _x^in_2^x^out_2 1k (B_2)\,v, which is nonnegative (destabilizing) and is controlled pointwise by |k′||k | on the opponent’s tent, not by the ratio ρk _k: no bound in terms of the pair (ρv,ρk)( _v, _k) alone can absorb it, because |k′||k | can be made arbitrarily large on a set where K barely moves. A clean joint threshold for general non-increasing k therefore requires an additional hypothesis (e.g. a bound on sup|k′|/infk |k |/ k); we record the failure rather than a lemma we cannot make tight. For uniform costs k′≡0k ≡ 0 and Theorem 0.M.1 is complete. 0.M.3 Stationary profiles and a continuation theorem Proposition 24 (Symmetric stationary profiles exist for every β<1β<1) Let κ∼U(0,1)κ U(0,1), α>0α>0, β∈[0,1)β∈[0,1), and let v be symmetric about 12 12 and satisfy Assumption 2.2. Define, for c1∈(0,12)c_1∈(0, 12) and c2=1−c1c_2=1-c_1, hβ(c1):=∂U1∂c1|(c1,1−c1)=(1+α+β)V(x1in)+(1+α−β)V(x1out∨0)−2(1+α)V(c1).h_β(c_1):= ∂ U_1∂ c_1 |_(c_1,1-c_1)=(1+α+β)V(x^in_1)+(1+α-β)V(x^out_1 0)-2(1+α)V(c_1). (55) Then hβh_β is continuous, hβ(0+)=(1+α+β)V(1λ)>0h_β(0^+)=(1+α+β)V( 1λ)>0 and hβ<0h_β<0 on a left neighborhood of 12 12; hence a symmetric stationary profile exists. If moreover ρv≤gβ(α) _v≤ g_β(α), every symmetric stationary profile lies on the no-dead-zone branch c1∈(0,1λ)c_1∈(0, 1λ). Proof For symmetric v and symmetric profiles, x2in=1−x1inx^in_2=1-x^in_1 and x2out=1−x1outx^out_2=1-x^out_1, so V(x2out∧1)−V(x2in)=V(x1in)−V(x1out∨0)V(x^out_2 1)-V(x^in_2)=V(x^in_1)-V(x^out_1 0) and Eq. 53 becomes Eq. 55. As c1↓0c_1 0: Δ→1 → 1, x1in→1λx^in_1→ 1λ, x1out→−1α<0x^out_1→- 1α<0, V(c1)→0V(c_1)→ 0, giving the stated limit. As c1↑12c_1 12: Δ=1−2c1↓0 =1-2c_1 0, x1out>0x^out_1>0, and expanding V around 12 12 (where v is continuous and positive), hβ(c1) h_β(c_1) =v(12)Δ[(1+α)(1λ−1α)+β(1λ+1α)]+o(Δ) =v( 12)\, [(1+α) ( 1λ- 1α )+β ( 1λ+ 1α ) ]+o( ) =v(12)Δ2(1+α)αλ(β−1)+o(Δ)< 0 =v( 12)\, \, 2(1+α)αλ\,(β-1)+o( )\;<\;0 for β<1β<1 and Δ small. The intermediate value theorem gives a root. On c1∈[1λ,12)c_1∈[ 1λ, 12) we have x1out≥0x^out_1≥ 0 and x1out∨0=x1outx^out_1 0=x^out_1, and the dead-zone estimate (54) (with its strictness argument) shows hβ<0h_β<0 there whenever ρv≤gβ(α) _v≤ g_β(α); roots are therefore confined to c1∈(0,1λ)c_1∈(0, 1λ). Theorem 0.M.2 (Continuation of the symmetric equilibrium into β>0β>0) Let κ∼U(0,1)κ U(0,1), α>0α>0, and let v be symmetric about 12 12, satisfy Assumption 2.2, and obey the strict β=0β=0 umbrella ρv<minf(α),g(α) _v< \f(α),g(α)\. Assume the β=0β=0 symmetric stationarity equation h0(c1)=0h_0(c_1)=0 has a unique root c10c_1^0. Then there is β¯>0 β>0 such that for every β∈[0,β¯)β∈[0, β) the game has a symmetric Nash equilibrium (c1β,1−c1β)(c_1^β,1-c_1^β), and c1β→c10c_1^β→ c_1^0 as β↓0β 0. Proof Step 1 (roots, single-peakedness, for small β). Since gβg_β is continuous and strictly decreasing with g0=g(α)g_0=g(α), the condition ρv<g(α) _v<g(α) gives βg:=(1+α)(λ−ρvα)ρv(α+λ)>0 _g:= (1+α)(λ- _vα) _v(α+λ)>0 with ρv<gβ(α) _v<g_β(α) for all β<βgβ< _g; and fβ≥f0>ρvf_β≥ f_0> _v for all β. Hence for β<βgβ< _g Theorem 0.M.1 applies: the own side is single-peaked, and by Proposition 24 a symmetric stationary profile (c1β,1−c1β)(c_1^β,1-c_1^β) exists, with c1β∈(0,1λ)c_1^β∈(0, 1λ). By single-peakedness the stationary point is the global own-side maximum of U1(⋅,1−c1β)U_1(·,1-c_1^β) on [0,1−c1β)[0,1-c_1^β), and by symmetry likewise for candidate 2 on her own side. It remains to control far-side deviations. Step 2 (the β=0β=0 far-side gap). Let c20=1−c10c_2^0=1-c_1^0 and P(0)=u1(c10,c20)>0P(0)=u_1(c_1^0,c_2^0)>0. For t∈[0,1−c20]t∈[0,1-c_2^0], reflection and the strict monotonicity of u1(a,⋅)u_1(a,·) (its derivative I1+O1I_1+O_1 is strictly positive for a<oa<o, since k,v>0k,v>0 and Δ>0 >0) give u1(c20+t,c20)=u1(1−c20−t, 1−c20)<u1(1−c20−t,c20)≤P(0)(t>0),u_1(c_2^0+t,\,c_2^0)=u_1(1-c_2^0-t,\,1-c_2^0)\;<\;u_1(1-c_2^0-t,\,c_2^0)\;≤\;P(0) (t>0), using 1−c20<c201-c_2^0<c_2^0 (as c10<12c_1^0< 12) for the strict middle inequality, while the deviation t=0t=0 (the tie) yields payoff 0<P(0)0<P(0). The far-side supremum far(0):=supt∈[0,1−c20]u1(c20+t,c20)far(0):= _t∈[0,1-c_2^0]u_1(c_2^0+t,c_2^0) is attained by continuity on a compact interval; at any maximizer the displayed strict inequality holds, so δ0:=P(0)−far(0)>0 _0:=P(0)-far(0)>0. Step 3 (continuity in β). On the compact family of symmetric profiles, hβ→h0h_β→ h_0 uniformly as β→0β→ 0 (indeed |hβ−h0|≤β|h_β-h_0|≤β pointwise, since 0≤V(x1in)−V(x1out∨0)≤10≤ V(x^in_1)-V(x^out_1 0)≤ 1). Moreover all roots of hβh_β, β≤β^<min(βg,1)β≤ β< ( _g,1), lie in a fixed compact [ε0,12−ε0][ _0, 12- _0]: near 00, hβ≥h0>0h_β≥ h_0>0 (the β-term in Eq. 55 is nonnegative); near 12 12, the expansion in Proposition 24 is negative uniformly for β≤β^β≤ β. By uniqueness of c10c_1^0, every sequence of roots c1βnc_1 _n with βn→0 _n→ 0 converges to c10c_1^0; hence c1β→c10c_1^β→ c_1^0 (for any measurable selection of roots). Finally, uniformly over positions, |U1β−U10|=βu2≤β|U_1^β-U_1^0|=β\,u_2≤β, and u1u_1 is jointly continuous; therefore P(β)=(1−β)u1(c1β,1−c1β)→P(0)P(β)=(1-β)\,u_1(c_1^β,1-c_1^β)→ P(0) and far(β):=supc′∈[1−c1β, 1]U1β(c′,1−c1β)≤supc′∈[1−c1β, 1]u1(c′,1−c1β)⟶far(0),far(β):= _c ∈[1-c_1^β,\,1]U_1^β(c ,1-c_1^β)\;≤\; _c ∈[1-c_1^β,\,1]u_1(c ,1-c_1^β)\; \;far(0), where the convergence of the right-hand side uses joint continuity of u1u_1 and c1β→c10c_1^β→ c_1^0 (a parametric maximum over a continuously moving compact interval). Choose β¯≤min(βg,β^) β≤ ( _g, β) so small that P(β)−far(β)≥δ0/2>0P(β)-far(β)≥ _0/2>0 for all β<β¯β< β. Then no far-side deviation (including the tie) is profitable, no own-side deviation is profitable by Step 1, and by symmetry the same holds for candidate 2: (c1β,1−c1β)(c_1^β,1-c_1^β) is a Nash equilibrium. Remark 23 (The moderate-β regime, numerically) [numerical observation] The continuation theorem is qualitative; a numerical continuation shows the phenomenon is not confined to small β. Verified global equilibria (maximal deviation gain 00 at resolution 10−710^-7) include: cosine voters ρv=3 _v=3 at (α,β)=(0.2,0.2)(α,β)=(0.2,0.2) inside the umbrella Fβ=3.67F_β=3.67; ρv=1.86 _v=1.86 at (0.5,0.4)(0.5,0.4) inside Fβ=1.92F_β=1.92; the same electorate at (0.5,0.8)(0.5,0.8), outside the umbrella (gβ=1.19g_β=1.19); and Beta(2,2)Beta(2,2) (unbounded ρv _v) at (0.5,0.5)(0.5,0.5). The umbrella is sufficient, visibly not necessary. 0.M.4 Dead zones and competitiveness Proposition 25 (The dead-zone threshold moves with β) Let κ∼U(0,1)κ U(0,1) and α>0α>0. (a) (Necessity.) If an interior profile c1<c2c_1<c_2 satisfies candidate 1’s first-order condition U1′=0U_1 =0 and candidate 1 has a dead zone (x1out>0x^out_1>0), then ρv≥gβ(α) _v≥ g_β(α). No symmetry of v or of the profile is assumed. (b) (Sufficiency, symmetric v.) If v is symmetric and (1+α+β)V(2(1+α)λ2)> 2(1+α)V(1λ),(1+α+β)\,V\! ( 2(1+α)λ^2 )\;>\;2(1+α)\,V\! ( 1λ ), (56) then a symmetric stationary profile with dead zones exists. (c) (Uniform voters are immune at every β.) If v≡1v≡ 1 then no profile with x1out>0x^out_1>0 satisfies candidate 1’s first-order condition, for any β∈[0,1)β∈[0,1) (recovering Proposition 17(b)). Since gβg_β strictly decreases in β and the left side of Eq. 56 strictly increases in β, competition makes alienation dead zones strictly easier to sustain at stationarity: the β=0β=0 threshold ρv≥λ/α _v≥λ/α of Proposition 17(a) drops to ρv≥gβ(α) _v≥ g_β(α). Proof (a) is the contrapositive of the dead-zone estimate in the proof of Theorem 0.M.1: if x1out>0x^out_1>0 and ρv<gβ _v<g_β then Eq. 54 gives U1′<0U_1 <0, and at ρv=gβ _v=g_β the strictness argument applies; so stationarity forces ρv≥gβ _v≥ g_β. (b) On the symmetric family, the dead-zone branch is c1∈(1λ,12)c_1∈( 1λ, 12), on which hβh_β of Eq. 55 is continuous. At the junction c1=1λc_1= 1λ we have x1out=0x^out_1=0, Δ=αλ = αλ and x1in=1λ+αλ2=2(1+α)λ2x^in_1= 1λ+ αλ^2= 2(1+α)λ^2, so hβ(1λ)=(1+α+β)V(2(1+α)λ2)−2(1+α)V(1λ)>0h_β( 1λ)=(1+α+β)V( 2(1+α)λ^2)-2(1+α)V( 1λ)>0 by Eq. 56, while hβ<0h_β<0 near 12 12 by Proposition 24. The intermediate value theorem gives a root with c1>1λc_1> 1λ, i.e. x1out>0x^out_1>0; by symmetry both candidates then have dead zones. (c) With v≡1v≡ 1 and x1out>0x^out_1>0, Eq. 54 with S=s=1S=s=1 gives U1′≤(1+α)Δ(1λ−1α)+βΔ(1λ+1α)=Δ2(1+α)αλ(β−1)<0U_1 ≤(1+α) ( 1λ- 1α)+β ( 1λ+ 1α)= \, 2(1+α)αλ(β-1)<0 for β<1β<1. Remark 24 (The bound gβg_β is essentially attained) [numerical observation] A numerical check takes α=2α=2 and continuous piecewise-linear densities with a low plateau on the deviator’s outer arm (x1out,c1)(x^out_1,c_1) (ramp width 3⋅10−33· 10^-3), and bisects for the minimal density ratio at which candidate 1’s first-order condition admits a dead-zone root. The minima are 2.0152, 1.5447, 1.2524, 1.05312.0152,\,1.5447,\,1.2524,\,1.0531 at β=0, 0.3, 0.6, 0.9β=0,\,0.3,\,0.6,\,0.9, against gβ(2)=21+β=2.0000, 1.5385, 1.2500, 1.0526g_β(2)= 21+β=2.0000,\,1.5385,\,1.2500,\,1.0526: the proved necessity bound is sharp up to the ramp width. (These witnesses are asymmetric densities, as part (a) permits; symmetric witnesses are constrained by the mirror arm and stay slightly further from the bound.) Appendix 0.N Existence at α=0α=0 for every β<1β<1 Theorem 0.L.4 shows that hollow electorates destroy pure equilibria at α=1α=1 for β near 11. The next theorem shows that at α=0α=0 they do not: for uniform costs and a symmetric electorate satisfying a local ratio condition — which every convex (U-shaped) symmetric density satisfies once mild, and the arcsine density satisfies for all β — the quantile equilibrium of Proposition 22 exists for every β∈[0,1)β∈[0,1). This makes the median-selection statement Theorem 0.L.3(a) non-vacuous precisely for the electorates that make it interesting. Theorem 0.N.1 (Quantile equilibria at α=0α=0, all β) Let α=0α=0, κ∼U(0,1)κ U(0,1), and let v be symmetric about 12 12 and satisfy Assumption 2.2, with s0:=inf(0,1)v>0s_0:= _(0,1)v>0. Define the quantile profile c2∗=V−1(3−β4)c_2^*=V^-1\! ( 3-β4 ), c1∗=1−c2∗c_1^*=1-c_2^*, and v¯:=sup[c1∗,c2∗]v v:= _[c_1^*,c_2^*]v. If supc1∈(0,c2∗)v(c1+c2∗2)v(c1) _c_1∈(0,c_2^*)\ v ( c_1+c_2^*2 )v(c_1) ≤41−β, \;≤\; 41-β, (E1-a) v¯s0 vs_0 ≤4(1+β)(1−β)2, \;≤\; 4(1+β)(1-β)^2, (E1-b) then (c1∗,c2∗)(c_1^*,c_2^*) is a Nash equilibrium of the β-game. Proof By symmetry of v and of the profile it suffices to rule out deviations of candidate 1; candidate 2’s problem is the mirror image. Stationarity. At α=0α=0, Eq. 53 becomes, with m=c1+c22m= c_1+c_22, h(c1):=U1′(c1,c2∗)=V(m)+β(1−V(m))−2V(c1),h(c_1):=U_1 (c_1,c_2^*)=V(m)+β (1-V(m) )-2V(c_1), (57) and V(c1∗)=1+β4V(c_1^*)= 1+β4, V(m∗)=V(12)=12V(m^*)=V( 12)= 12 give h(c1∗)=12+β2−1+β2=0h(c_1^*)= 12+ β2- 1+β2=0. Own side. h′(c1)=1−β2v(m)−2v(c1)≤0h (c_1)= 1-β2v(m)-2v(c_1)≤ 0 for all c1∈(0,c2∗)c_1∈(0,c_2^*) by Eq. E1-a, so h is non-increasing; with h(c1∗)=0h(c_1^*)=0 this gives h≥0h≥ 0 on (0,c1∗](0,c_1^*] and h≤0h≤ 0 on [c1∗,c2∗)[c_1^*,c_2^*): U1(⋅,c2∗)U_1(·,c_2^*) is single-peaked with global own-side maximum at c1∗c_1^*. Its value is P=(1−β)u∗P=(1-β)u^*, where u∗=u1(c1∗,c2∗)=u2(c1∗,c2∗)u^*=u_1(c_1^*,c_2^*)=u_2(c_1^*,c_2^*) by symmetry, and u∗≥ΔV(c1∗)+∫c1∗1/2(1−2x)v(x)x≥(1+β)Δ4,soP≥(1−β2)Δ4,u^*\;≥\; \,V(c_1^*)+ _c_1^*^1/2(1-2x)\,v(x)\,dx\;≥\; (1+β) 4, P\;≥\; (1-β^2) 4, (58) using B1=ΔB_1= on [0,c1∗][0,c_1^*] and B1=1−2xB_1=1-2x on [c1∗,12][c_1^*, 12] (here c1∗+c2∗=1c_1^*+c_2^*=1), and K(t)=tK(t)=t. Far side. Let c′=c2∗+tc =c_2^*+t, t∈(0,1−c2∗]t∈(0,1-c_2^*], and write m′=c2∗+t2m =c_2^*+ t2. The deviator’s supporters lie in (m′,1](m ,1] with benefit at most t, so, using v≥s0v≥ s_0 on [c2∗,m′][c_2^*,m ], u1(c′,c2∗)≤t(1−V(m′))≤t(1+β4−s0t2).u_1(c ,c_2^*)\;≤\;t (1-V(m ) )\;≤\;t ( 1+β4- s_0t2 ). Every voter in [0,c2∗][0,c_2^*] obtains benefit exactly t against the deviator (at α=0α=0, B2=d1−d2=tB_2=d_1-d_2=t there), so u2(c′,c2∗)≥tV(c2∗)=t3−β4u_2(c ,c_2^*)≥ t\,V(c_2^*)=t\, 3-β4. Hence U1(c′,c2∗)≤t1+β4−s0t22−βt3−β4=(1−β)24t−s02t2≤(1−β)432s0,U_1(c ,c_2^*)\;≤\;t\, 1+β4- s_0t^22-β t\, 3-β4\;=\; (1-β)^24\,t- s_02\,t^2\;≤\; (1-β)^432\,s_0, maximizing the concave quadratic over t≥0t≥ 0. The tie c′=c2∗c =c_2^* yields payoff 0<P0<P. Closing the gap. The mass-gap law V(c2∗)−V(c1∗)=1−β2V(c_2^*)-V(c_1^*)= 1-β2 and v≤v¯v≤ v on [c1∗,c2∗][c_1^*,c_2^*] give Δ≥1−β2v¯ ≥ 1-β2 v, so by Eq. 58 P≥(1−β2)(1−β)8v¯≥(1−β)432s0⟺v¯s0≤4(1+β)(1−β)2,P\;≥\; (1-β^2)(1-β)8\, v\;≥\; (1-β)^432\,s_0 vs_0\;≤\; 4(1+β)(1-β)^2, which is Eq. E1-b. Hence no far-side deviation is profitable either, and (c1∗,c2∗)(c_1^*,c_2^*) is an equilibrium. Corollary 14 (Convex symmetric electorates; the arcsine for all β) Let α=0α=0, κ∼U(0,1)κ U(0,1), and let v be symmetric, convex and satisfy Assumption 2.2. Set η(β):=v(c2∗)/v(12)η(β):=v(c_2^*)/v( 12). If η(β)≤41−βη(β)≤ 41-β, then the quantile profile is a Nash equilibrium. In particular, for the arcsine electorate v=Beta(12,12)v=Beta( 12, 12), c2∗=sin2(π(3−β)8),η(β)=1sin(π(3−β)4)∈(1,2],c_2^*= ^2\! ( π(3-β)8 ), η(β)= 1 \! ( π(3-β)4 )∈(1, 2\,], so the condition holds and a pure equilibrium exists for every β∈[0,1)β∈[0,1). Proof For convex symmetric v: v is non-increasing on [0,12][0, 12] and non-decreasing on [12,1][ 12,1], so s0=v(12)s_0=v( 12) and, on [c1∗,c2∗][c_1^*,c_2^*], v attains its maximum at the endpoints: v¯=v(c1∗)=v(c2∗) v=v(c_1^*)=v(c_2^*). For Eq. E1-a: with m=c1+c2∗2m= c_1+c_2^*2 the midpoint of [c1,c2∗][c_1,c_2^*], convexity gives v(m)≤maxv(c1),v(c2∗)v(m)≤ \v(c_1),v(c_2^*)\, so the ratio is at most max1,v(c2∗)/v(c1)≤max1,η(β)=η(β) \1,\,v(c_2^*)/v(c_1)\≤ \1,η(β)\=η(β) (as v(c1)≥s0v(c_1)≥ s_0 and η≥1η≥ 1), and η≤41−βη≤ 41-β by hypothesis. For Eq. E1-b: v¯/s0=η≤41−β≤4(1+β)(1−β)2 v/s_0=η≤ 41-β≤ 4(1+β)(1-β)^2. For the arcsine, V(x)=2πarcsinxV(x)= 2π x, so V(c2∗)=3−β4V(c_2^*)= 3-β4 gives the stated c2∗c_2^*, and v(c2∗)/v(12)=1/2c2∗(1−c2∗)=1sin(π(3−β)4)v(c_2^*)/v( 12)= 1/2 c_2^*(1-c_2^*)= 1 ( π(3-β)4), using c2∗(1−c2∗)=14sin2(π(3−β)4)c_2^*(1-c_2^*)= 14 ^2( π(3-β)4). As β runs over [0,1)[0,1) the argument runs over (π2,3π4]( π2, 3π4], so η∈(1,2]η∈(1, 2], which is below 41−β≥4 41-β≥ 4 for every β. Remark 25 (Contrast with α>0α>0; scope; verification) (i) The contrast with Theorem 0.L.4 is now a theorem: the same hollow electorate Beta(12,12)Beta( 12, 12) that has no pure equilibrium at α=1α=1 for β≥0.9168β≥ 0.9168 (Theorem 0.O.1) has a pure equilibrium at α=0α=0 for every β<1β<1. Hollowness is lethal only in combination with alienation. Consequently Theorem 0.L.3(a) is a genuine selection statement on this class: the quantile equilibria converge to the median profile as β↑1β 1 (here explicitly: c2∗↓12c_2^* 12). (i) For general (v,k)(v,k), the question of existence at α=0α=0 for every β<1β<1 contains, at β=0β=0, the open problem of pure-equilibrium existence for arbitrary (v,k)(v,k), and is not resolved here; Conjecture 1 records the conjecture. A numerical check verifies the quantile equilibrium as a global best response for Beta(12,12)Beta( 12, 12) and Beta(0.8,0.8)Beta(0.8,0.8) at β∈0,0.3,0.6,0.9,0.97β∈\0,0.3,0.6,0.9,0.97\ (deviation gains ≤0≤ 0, first-order residuals ∼10−11 10^-11), and finds verified equilibria for the non-uniform costs K=t(2−t)/2K=t(2-t)/2 and K=t2K=t^2 with the arcsine electorate at β up to 0.990.99 (e.g. K=t2K=t^2: (0.075,0.925)(0.075,0.925) at β=0β=0 collapsing to (0.492,0.508)(0.492,0.508) at β=0.99β=0.99). Appendix 0.O The certified non-existence interval Corollary 11 gave non-existence for β>0.98615β>0.98615 (α=1α=1, κ∼U(0,1)κ U(0,1), v=Beta(12,12)v=Beta( 12, 12)), while numerically equilibria persist to β=0.90β=0.90 and are gone by β=0.91β=0.91 (Remark 17). The following closes most of that gap. Theorem 0.O.1 (Certified interval) [computer-assisted] Let α=1α=1, κ∼U(0,1)κ U(0,1), v=Beta(12,12)v=Beta( 12, 12). The β-game has no pure Nash equilibrium for any β≥ 0.9168.β\;≥\;0.9168. A pure equilibrium is found numerically at β=0.90β=0.90 (Remark 17); that lower bound is a numerical observation, not part of the certificate, and places any existence threshold βc _c in (0.90, 0.9168](0.90,\,0.9168]. Proof The proof is a finite, exactly evaluated certificate, reproduced by the verification suite as the three E2 claims. Fix β and define the regret R(c1,c2)=maxi[supc′Ui(c′,c−i)−Ui(c1,c2)]R(c_1,c_2)= _i [ _c U_i(c \!,c_-i)-U_i(c_1,c_2) ]; a pure equilibrium exists iff min[0,1]2R=0 _[0,1]^2R=0, so minR>0 R>0 certifies non-existence. All payoffs u(a,o)=∫K(B+)Vu(a,o)= K(B^+)dV are evaluated exactly through the regularized incomplete beta function. From the proof of Lemma 7, u is 22-Lipschitz in the own and 11-Lipschitz in the opponent position (for α=1α=1, LK=1L_K=1); hence U1(a,o)=u(a,o)−βu(o,a)U_1(a,o)=u(a,o)-β u(o,a) is (3,3)(3,3)-Lipschitz, the best-response value s(o)=supaU1(a,o)s(o)= _aU_1(a,o) is 33-Lipschitz, and the two one-sided regrets are (3,6)(3,6)- and (6,3)(6,3)-Lipschitz, so the regret varies by at most 4.54.5 times the side of a square cell, measured from its center. On the N×N× N grid, N=2401N=2401, h=12400h= 12400, the matrix U1=U−βU⊤U_1=U-β U^\! and the column maxima give a conservative grid regret (the grid best-response value underestimates the true supremum, so grid regret underestimates true regret at grid profiles). Cells whose grid regret exceeds τ+4.5hτ+4.5h certify R≥τR≥τ on the whole cell. The remaining (flagged) cells are covered by a refined lattice of step h/16h/16 (step h/64h/64 for the residual worst cells), on which the regret is bounded below by the exact fixed-attacker margins maxU1(ajcol,c2)−U1(c1,c2),U1(airow,c1)−U2(c1,c2) \U_1(a_j^col,c_2)-U_1(c_1,c_2),\,U_1(a_i^row,c_1)-U_2(c_1,c_2)\ with ajcola_j^col, airowa_i^row the grid best responses in the cell’s column and row; these bounds have the same Lipschitz structure, so the refined slack is 4.5h/164.5h/16 resp. 4.5h/644.5h/64. This yields a verified margin ε′(β)>0 (β)>0 with R≥ε′(β)R≥ (β) on all of [0,1]2[0,1]^2. Finally |Rβ−Rβ′|≤2usup|β−β′||R_β-R_β |≤ 2u_ |β-β | with usup=0.2839u_ =0.2839 an upper bound on supu u (grid maximum plus Lipschitz correction 1.5h1.5h), so each certified β covers the interval (β−ε′(β)/(2usup),β](β- (β)/(2u_ ),\,β]. A 65-step walk descending from β=0.9862β=0.9862 with step 0.85ε′/(2usup)0.85\, /(2u_ ) produced overlapping certified intervals covering [0.91671, 0.9862][0.91671,\,0.9862]; margins ranged from 6.1⋅10−36.1· 10^-3 at β=0.9862β=0.9862 down to 4.6⋅10−54.6· 10^-5 at β=0.9170β=0.9170. Together with Corollary 11 (β>0.98615β>0.98615) and Theorem 0.L.4 (β=1β=1), non-existence holds for all β≥0.9168>0.91671β≥ 0.9168>0.91671. Remark 26 (Where the certificate stops) [numerical observation] The exact grid regret at N=2401N=2401 falls from 6.2⋅10−36.2· 10^-3 at β=0.9862β=0.9862 through 1.9⋅10−41.9· 10^-4 at β=0.92β=0.92 to 6.9⋅10−66.9· 10^-6 at β=0.91β=0.91 and (numerically) 00 at β=0.90β=0.90: the true threshold sits near 0.9080.908 (a pure equilibrium is still found at β=0.908β=0.908 and none at β=0.909β=0.909), and the certificate stops where the dying regret meets the finest refinement slack 4.5h/64≈2.9⋅10−54.5h/64≈ 2.9· 10^-5 — less than 0.010.01 above it. The same machinery applies verbatim to Beta(0.8,0.8)Beta(0.8,0.8) (where Corollary 11 gives 0.995210.99521); we have not run that walk. Appendix 0.P Independent verification of the Sections 3–5 results The numerical content of the seven results listed in Table 5—drawn from Sections 3, 4 and 5, the scale-invariance proposition among them—is recomputed from the model’s definitions by an independent implementation: 14 checks, all passing. Table 5: Re-verification of results from Sections 3, 4 and 5 by an independent implementation of the model. “exact” means closed forms (uniform-voter payoffs, incomplete-beta integrals) reproduced to the stated precision. Result Verified statement Outcome Proposition 7 zKzK: equilibria fixed, turnout ×z× z (α∈0,1α∈\0,1\, z∈1,12,110z∈\1, 12, 110\) exact Theorems 4.1 and 4.3 Δ∗(α) ^*(α) solves ∫bΔr=logλ2 _b r= λ2; unique; increasing in α; doubly-uniform value 2+α4+α 2+α4+α ≤2⋅10−5≤ 2·10^-5 Theorem 4.2 K=t2⪰rhrK=tK=t^2 _rhrK=t polarizes at α∈0,1,2α∈\0,1,2\ exact Proposition 6 T′T formula (10−1010^-10); peak Δ¯=λ2α2+6α+4 = λ^2α^2+6α+4, Δ¯(1)=911 (1)= 911; Δ∗<Δ¯ ^*< ; T∗=2(3+α)(4+α)2T^*= 2(3+α)(4+α)^2 exact Proposition 5 isomorphism u(Vσ,K)=u(V0,Kσ)u_(V_σ,K)=u_(V_0,K_σ) pointwise (5⋅10−95·10^-9); power-law costs exactly proportional; exponential cost super-proportional exact Theorem 5.4 w1/δ4→C(α)v′(12)w_1/δ^4→ C(α)v ( 12) with C(1)=1327C(1)= 1327; Beta(2,2)Beta(2,2) and arcsine match the prediction at δ=0.02δ=0.02 (values below); uniform: 00 to machine precision confirmed Proposition 15 certified in stronger form by Theorems 0.L.4 and 0.O.1 certified Quartic check values. At δ=0.02δ=0.02, α=1α=1: Beta(2,2)Beta(2,2) gives w1=−9.2444⋅10−7w_1=-9.2444· 10^-7 against the prediction C(1)v′(12)δ4=−9.2444⋅10−7C(1)v ( 12)δ^4=-9.2444· 10^-7; the arcsine gives +1.9666⋅10−7+1.9666· 10^-7 against +1.9617⋅10−7+1.9617· 10^-7 (the residual is the o(δ4)o(δ^4) term and shrinks at δ=0.01, 0.005δ=0.01,\,0.005). Security values. Under the exact quadrature, maxcminc′w(c,c′)=−8.2244⋅10−3 _c _c w(c,c )=-8.2244· 10^-3 (attained at c=0.8292c=0.8292; grid n=1201n=1201) for the arcsine electorate and, for Beta(0.8,0.8)Beta(0.8,0.8), −3.0238⋅10−3-3.0238· 10^-3 (n=2401n=2401, from Theorem 0.L.4); the certified Lipschitz bounds of Theorem 0.L.4 (≤−6.97⋅10−3≤-6.97· 10^-3, ≤−2.40⋅10−3≤-2.40· 10^-3) are unchanged. For Beta(2,2)Beta(2,2) the security value attains 00 at the median and only there; for uniform voters it is 00 (to 10−1610^-16) on the equilibrium band, as it must be. The security-value curves. The three curves c↦minc′w(c,c′)c _c w(c,c ) of Fig. 11 (α=1α=1, κ∼U(0,1)κ U(0,1); Beta(2,2)Beta(2,2), uniform and arcsine voters) are computed from the exact 1201×12011201× 1201 payoff matrices and plotted at every fifth grid point (figdata.py in the verification suite). Remark 27 (Why the security value must be computed exactly) [numerical observation] Naive quadrature of the arcsine density’s endpoint singularity produces security values that are wrong by orders of magnitude, in two distinct regimes. A midpoint rule with 10310^3 cells misweights the mass near the boundary by ∼10% 10\% (e.g. w(0.999,0.5)w(0.999,0.5): exact −0.0284-0.0284, midpoint −0.0314-0.0314) — visible but modest. Any rule that samples v at x∈0,1x∈\0,1\ (trapezoid, Simpson, or a grid containing the endpoints), where the density is unbounded, is arbitrarily wrong: the same w(0.999,0.5)w(0.999,0.5) evaluates to +79.2+79.2 under a 10310^3-cell trapezoid rule. Security values of order −0.387-0.387 arise this way: some near-boundary attacker c′c acquires spurious mass and appears to beat every position by O(0.1CLOSEO(0.1–OPEN1)1). The curves of Fig. 11 use the exact incomplete-beta integrals and are free of both effects. Appendix 0.Q Conjecture and open items Conjecture 1 (Existence at α=0α=0) Under Assumption 2.2 with α=0α=0, the β-game has a pure Nash equilibrium for every β∈[0,1)β∈[0,1) and every (v,k)(v,k). Evidence for. Theorem 0.N.1 and Corollary 14 (uniform costs, symmetric v under Eqs. E1-a to E1-b, including all hollow convex electorates for all β); Proposition 22’s corollary (ρv≤41−β _v≤ 41-β); the margin game β=1β=1 has the median equilibrium for arbitrary (v,k)(v,k) (Theorem 5.2), so the conjectured family has the right limit; verified equilibria for non-uniform costs and hollow v at β up to 0.990.99 (Remark 25); and the ratio law fixes the candidate profile’s quantiles in a way that degenerates gracefully as β↑1β 1. Evidence against, and why it stays a conjecture. At β=0β=0 the statement contains the still-open existence problem for arbitrary (v,k)(v,k); the concavity route fails for ρv=∞ _v=∞ with general k (the local condition (E1-a) has no analogue once k varies, cf. Remark 22); and the α>0α>0 counterexamples (Theorems 0.L.4 and 0.O.1) show the effect that must fail at α=0α=0 is genuinely present nearby in parameter space. Two technical provisos. The finiteness proviso of Lemma 5 and the C1C^1 approximation step in Proposition 16 remain open exactly as stated; neither affects any result in the main text. References [1] A. I. Abramowitz and K. L. Saunders (2008) Is polarization a myth?. The Journal of politics 70 (2), p. 542–555. Cited by: §1.2. [2] J. Adams, J. Dow, and S. Merrill (2006) The political consequences of alienation-based and indifference-based voter abstention: applications to presidential elections. Political Behavior 28 (1), p. 65–86. External Links: Document Cited by: §1.2, §1.2. [3] J. Adams, S. Merrill, and B. Grofman (2005) Candidate strategies with voter abstention in U.S. presidential elections: 1980, 1984, 1988, 1996, 2000. In A Unified Theory of Party Competition, p. 132–151. External Links: Document Cited by: §1. [4] J. Adams and S. Merrill (2003) Voter turnout and candidate strategies in american elections. The Journal of Politics 65 (1), p. 161–189. External Links: Document Cited by: §1. [5] A. Al Yussef (2026) Turnout and the distortion of representation: the case of electoral polarization. Public Choice. External Links: Document Cited by: §1.2, §6. [6] S. P. Anderson and G. Glomm (1992) Alienation, indifference and the choice of ideological position. Social Choice and Welfare 9 (1), p. 17–31. External Links: Document Cited by: §1.2, §1.2. [7] S. Ansolabehere, J. M. Snyder Jr, and C. Stewart I (2001) Candidate positioning in US House elections. American Journal of Political Science, p. 136–159. Cited by: §1. [8] E. Aragones and T. R. Palfrey (2002) Mixed equilibrium in a Downsian model with a favored candidate. Journal of Economic Theory 103 (1), p. 131–161. Cited by: §1.2. [9] P. H. Aranson, M. J. Hinich, and P. C. Ordeshook (1974) Election goals and strategies: equivalent and nonequivalent candidate objectives. American Political Science Review 68 (1), p. 135–152. External Links: Document Cited by: §1.2. [10] J. Bafumi and M. C. Herron (2010) Leapfrog representation and extremism: a study of American voters and their members in Congress. American Political Science Review 104 (3), p. 519–542. Cited by: §1. [11] M. Bagnoli and T. Bergstrom (2005) Log-concave probability and its applications. Economic Theory 26 (2), p. 445–469. Cited by: §3.1. [12] D. Black (1948) On the rationale of group decision-making. Journal of political economy 56 (1), p. 23–34. Cited by: Claim 2.1. [13] T. Börgers (2004) Costly voting. American Economic Review 94 (1), p. 57–66. External Links: Document Cited by: §1.2. [14] L. Boxell, M. Gentzkow, and J. M. Shapiro (2024) Cross-country trends in affective polarization. The Review of Economics and Statistics 106 (2), p. 557–565. External Links: Document Cited by: §1.2, §6. [15] G. Brennan and L. Lomasky (1993) Democracy and decision: the pure theory of electoral preference. Cambridge University Press, Cambridge. External Links: ISBN 9780521330404, Document Cited by: §2.2. [16] R. A. Brody and B. I. Page (1973) Indifference, alientation and rational decisions: the effects of candidate evaluations on turnout and the vote. Public Choice 15 (1), p. 1–17. External Links: Document Cited by: §1.2. [17] S. Callander and C. H. Wilson (2007) Turnout, polarization, and duverger’s law. The Journal of Politics 69 (4), p. 1047–1056. External Links: Document Cited by: §1.2. [18] R. L. Calvert (1985) Robustness of the multidimensional voting model: candidate motivations, uncertainty, and convergence. American Journal of Political Science 29 (1), p. 69–95. External Links: Document Cited by: §1.2. [19] C. Cleveland, B. de Keijzer, and M. Polukarov (2026) Candidate positioning with prospect-theoretic voters. Note: Workshop paper Cited by: §1.2. [20] A. Cohen and D. Peleg (2019) Hotelling games with random tolerance intervals. In Web and Internet Economics (WINE 2019), Lecture Notes in Computer Science, Vol. 11920, p. 114–128. External Links: Document Cited by: §1.2. [21] G. Cohensius, S. Mannor, R. Meir, E. Meirom, and A. Orda (2017) Proxy voting for better outcomes. In Proceedings of the 16th Conference on Autonomous Agents and MultiAgent Systems, p. 858–866. Cited by: §1.2, footnote 2. [22] P. J. Coughlin (2015) Probabilistic voting in models of electoral competition. Handbook of social choice and voting, p. 218–234. Cited by: §1.2, §1.2, §6. [23] G. W. Cox and M. D. McCubbins (1986) Electoral politics as a redistributive game. The Journal of Politics 48 (2), p. 370–389. External Links: Document Cited by: §1.2, §5.1. [24] G. W. Cox (2010) Swing voters, core voters, and distributive politics. In Political Representation, I. Shapiro, S. C. Stokes, E. J. Wood, and A. S. Kirshner (Eds.), p. 342–357. External Links: Document Cited by: §1.2, §5.1. [25] Y. Desmedt and E. Elkind (2010) Equilibria of plurality voting with abstentions. In Proceedings of the 11th ACM conference on Electronic commerce, p. 347–356. Cited by: §1.2. [26] A. Dixit and J. Londregan (1996) The determinants of success of special interests in redistributive politics. The Journal of Politics 58 (4), p. 1132–1155. External Links: Document Cited by: §1.2. [27] C. Dong, M. Bullinger, T. Was, L. Birnbaum, and E. Elkind (2025) Selecting interlacing committees. In Proceedings of the 24th International Conference on Autonomous Agents and Multiagent Systems (AAMAS), Cited by: §1.2. [28] A. Downs (1957) An economic theory of political action in a democracy. Journal of political economy 65 (2), p. 135–150. Cited by: §1, §2.1, Claim 2.1. [29] P. Dreyer and J. Bauer (2019) Does voter polarisation induce party extremism? the moderating role of abstention. West European Politics 42 (4), p. 824–847. Cited by: §1.2, §3.2. [30] J. Duggan (2006) Candidate objectives and electoral equilibrium. In The Oxford Handbook of Political Economy, B. R. Weingast and D. A. Wittman (Eds.), p. 64–83. External Links: Document Cited by: §1.2. [31] E. Elkind, E. Markakis, S. Obraztsova, and P. Skowron (2015) Equilibria of plurality voting: lazy and truth-biased voters. In Algorithmic Game Theory: 8th International Symposium, SAGT 2015, Saarbrücken, Germany, September 28-30, 2015. Proceedings 8, p. 110–122. Cited by: §1.2. [32] C. R. Farina (2015) Congressional polarization: terminal constitutional dysfunction. Colum. L. Rev. 115, p. 1689. Cited by: §1.2. [33] T. J. Feddersen (2004) Rational choice theory and the paradox of not voting. Journal of Economic Perspectives 18 (1), p. 99–112. External Links: Document Cited by: §1.2. [34] M. Feldman, A. Fiat, and S. Obraztsova (2016) Variations on the Hotelling–Downs model. In Proceedings of the AAAI Conference on Artificial Intelligence, Vol. 30. Cited by: §1.2. [35] M. P. Fiorina and S. J. Abrams (2008) Political polarization in the American public. Annual Review of Political Science 11, p. 563–588. External Links: Document Cited by: §1.2. [36] A. Fowler (2013) Electoral and policy consequences of voter turnout: Evidence from compulsory voting in Australia. Quarterly Journal of Political Science 8 (2), p. 159–182. External Links: Document Cited by: §6. [37] M. N. Franklin (2004) Voter turnout and the dynamics of electoral competition in established democracies since 1945. Cambridge University Press. Cited by: §1, §6. [38] E. L. Glaeser, G. A. M. Ponzetto, and J. M. Shapiro (2005) Strategic extremism: why republicans and democrats divide on religious values. The Quarterly Journal of Economics 120 (4), p. 1283–1330. External Links: Document Cited by: §5.1. [39] I. L. Glicksberg (1952) A further generalization of the Kakutani fixed point theorem, with application to Nash equilibrium points. Proceedings of the American Mathematical Society 3 (1), p. 170–174. Cited by: Appendix 0.C, Remark 3. [40] M. Golden and B. Min (2013) Distributive politics around the world. Annual Review of Political Science 16, p. 73–99. External Links: Document Cited by: §1.2. [41] B. Grofman (2004) Downs and two-party convergence. Annual Review of Political Science 7, p. 25–46. External Links: Document Cited by: §1.2. [42] H. Herrera, D. K. Levine, and C. Martinelli (2008) Policy platforms, campaign spending and voter participation. Journal of Public Economics 92 (3–4), p. 501–513. External Links: Document Cited by: §1.2. [43] H. Herrera, M. Morelli, and T. Palfrey (2014) Turnout and power sharing. The Economic Journal 124 (574), p. F131–F162. External Links: Document Cited by: §1.2. [44] M. J. Hetherington (2001) Resurgent mass partisanship: the role of elite polarization. American Political Science Review 95 (3), p. 619–631. External Links: Document Cited by: §1.2. [45] M. J. Hinich, J. O. Ledyard, and P. C. Ordeshook (1972) Nonvoting and the existence of equilibrium under majority rule. Journal of Economic Theory 4 (2), p. 144–153. External Links: Document Cited by: §1.2. [46] M. J. Hinich and P. C. Ordeshook (1969) Abstentions and equilibrium in the electoral process. Public Choice 7 (1), p. 81–106. Cited by: §1.2, §2.5, §2.5, §4.3. [47] M. J. Hinich and P. C. Ordeshook (1970) Plurality maximization vs vote maximization: a spatial analysis with variable participation. American Political Science Review 64 (3), p. 772–791. External Links: Document Cited by: §1.2. [48] M. J. Hinich (1977) Equilibrium in spatial voting: the median voter result is an artifact. Journal of Economic Theory 16 (2), p. 208–219. External Links: Document Cited by: §1.2. [49] M. Hooghe and A. Kern (2017) The tipping point between stability and decline: trends in voter turnout, 1950–1980–2012. European Political Science 16, p. 535–552. Cited by: §1, §6. [50] H. Hotelling (1929) Stability in competition. The Economic Journal 39 (153), p. 41–57. Cited by: §1, §2.1. [51] M. I. Jones, A. D. Sirianni, and F. Fu (2022) Polarization, abstention, and the median voter theorem. Humanities and Social Sciences Communications 9 (1), p. 43. External Links: Document Cited by: §1.2, §1.2. [52] Y. Kamada and F. Kojima (2014) Voter preferences, polarization, and electoral policies. American Economic Journal: Microeconomics 6 (4), p. 203–236. Cited by: §1.2. [53] S. Krasa and M. K. Polborn (2009) Is mandatory voting better than voluntary voting?. Games and Economic Behavior 66 (1), p. 275–291. External Links: Document Cited by: §1.2. [54] V. Krishna and J. Morgan (2012) Voluntary voting: costs and benefits. Journal of Economic Theory 147 (6), p. 2083–2123. External Links: Document Cited by: §1.2. [55] E. Kubin and C. Von Sikorski (2021) The role of (social) media in political polarization: a systematic review. Annals of the International Communication Association 45 (3), p. 188–206. Cited by: §6. [56] J. O. Ledyard (1984) The pure theory of large two-candidate elections. Public Choice 44 (1), p. 7–41. External Links: Document Cited by: §1.2, §1.2. [57] D. S. Lee, E. Moretti, and M. J. Butler (2004) Do voters affect or elect policies? Evidence from the US House. The Quarterly Journal of Economics 119 (3), p. 807–859. Cited by: §1. [58] J. E. Leighley and J. Nagler (2014) Who votes now? Demographics, issues, inequality, and turnout in the united states. Princeton University Press. External Links: ISBN 9780691159355 Cited by: §6. [59] S. D. Levitt (1996) How do senators vote? disentangling the role of voter preferences, party affiliation, and senator ideology. The American Economic Review, p. 425–441. Cited by: §1. [60] A. Lindbeck and J. W. Weibull (1987) Balanced-budget redistribution as the outcome of political competition. Public Choice 52 (3), p. 273–297. External Links: Document Cited by: §1.2, §5.1. [61] S. Liu, Y. Zhu, and Q. Hu (2016) Dynamic procurement management by reverse auctions with fixed setup costs and sales levers. Applied Mathematical Modelling 40 (4), p. 3049–3061. External Links: ISSN 0307-904X, Document, Link Cited by: §3.1. [62] H. Llavador (2006) Electoral platforms, implemented policies, and abstention. Social Choice and Welfare 27 (1), p. 55–81. External Links: Document Cited by: §1.2. [63] N. McCarty, K. T. Poole, and H. Rosenthal (2006) Polarized America: the dance of ideology and unequal riches. MIT Press. Cited by: §1. [64] R. Meir and G. Ghalme (2025) On Condorcet’s jury theorem with abstention. In AAAI’25, Cited by: §1.2, §2.2. [65] R. Meir, G. Shahaf, E. Shapiro, and N. Talmon (2022) Sybil-resilient social choice with partial participation. In EUMAS’22, Cited by: §1.2. [66] R. Meir (2017) Iterative voting. Trends in computational social choice 4, p. 69–86. Cited by: §6. [67] R. Meir (2025) Tyranny of the minority in social choice: a call to arms. In Proceedings of the 24th International Conference on Autonomous Agents and Multiagent Systems, p. 2865–2869. Cited by: §1. [68] R. B. Myerson (1981) Optimal auction design. Mathematics of Operations Research 6 (1), p. 58–73. External Links: Document Cited by: §3.1. [69] A. Oprea, L. Martin, and G. H. Brennan (2024) Moving toward the median: compulsory voting and political polarization. American Political Science Review 118 (4), p. 1951–1965. External Links: Document Cited by: §1.2. [70] M. J. Osborne (1995) Spatial models of political competition under plurality rule: a survey of some explanations of the number of candidates and the positions they take. Canadian Journal of Economics 28 (2), p. 261–301. External Links: Document Cited by: §1.2. [71] T. R. Palfrey and H. Rosenthal (1983) A strategic calculus of voting. Public Choice 41 (1), p. 7–53. External Links: Document Cited by: §1.2. [72] T. R. Palfrey and H. Rosenthal (1985) Voter participation and strategic uncertainty. American Political Science Review 79 (1), p. 62–78. External Links: Document Cited by: §1.2, §2.2. [73] J. W. Patty (2002) Equivalence of objectives in two candidate elections. Public Choice 112 (1-2), p. 151–166. External Links: Document Cited by: §1.2. [74] Pew Research Center (2017) The partisan divide on political values grows even wider. Note: Pew Research Center report Cited by: §1, §6. [75] M. Prior (2013) Media and political polarization. Annual review of political science 16 (1), p. 101–127. Cited by: §6. [76] A. Reiljan (2020) ‘Fear and loathing across party lines’ (also) in Europe: affective polarisation in European party systems. European journal of political research 59 (2), p. 376–396. Cited by: §1. [77] W. H. Riker and P. C. Ordeshook (1968) A theory of the Calculus of Voting. American political science review 62 (1), p. 25–42. Cited by: §1.2, §2.2, §6. [78] J. E. Roemer (1994) A theory of policy differentiation in single issue electoral politics. Social Choice and Welfare 11 (4), p. 355–380. Cited by: §1.2. [79] J. C. Rogowski (2014) Electoral choice, ideological conflict, and political participation. American Journal of Political Science 58 (2), p. 479–494. External Links: Document Cited by: §4.2. [80] M. Shaked and J. G. Shanthikumar (2007) Stochastic orders. Springer, New York. Cited by: §3.2. [81] E. Shapiro and N. Talmon (2017) Reality-aware social choice. arXiv preprint arXiv:1710.10117. Cited by: §6. [82] W. Shen and Z. Wang (2017) Hotelling–Downs model with limited attraction. In Proceedings of the 16th Conference on Autonomous Agents and MultiAgent Systems, p. 660–668. Cited by: §1.2. [83] J. van Sloun (2023) Rationalizable behavior in the Hotelling–Downs model of spatial competition. Theory and Decision 95 (2), p. 309–335. Cited by: §1.2. [84] D. Wittman (1983) Candidate motivation: a synthesis of alternative theories. American Political Science Review 77 (1), p. 142–157. External Links: Document Cited by: §1.2. [85] J. N. Zingher and M. E. Flynn (2018) From on high: the effect of elite polarization on mass attitudes and behaviors, 1972–2012. British Journal of Political Science 48 (1), p. 23–45. Cited by: §1.2.