Single-Issue Campaigns and MultidimensionalPolitics∗

Georgy EgorovNorthwestern University

February 2014

Abstract

In most elections, voters care about several issues, but candidates may have to choose onlya few to build their campaign on. The information that voters will get about the politiciandepends on this choice, and it is therefore a strategic one. In this paper, I study a model ofelections where voters care about the candidates’competences (or positions) over two issues,e.g., economy and foreign policy, but each candidate may only credibly signal his competenceor announce his position over at most one issue. Voters are assumed to get (weakly) betterinformation if the candidates campaign on the same issue rather than on different ones. Ishow that the first mover will in equilibrium set the agenda for both himself and the opponentif campaigning on a different issue is uninformative, but otherwise the other candidate mayactually be more likely to choose the other issue. The social (voters’) welfare is a non-monotonefunction of the informativeness of different-issue campaigns, but in any case the voters are betteroff if candidates are free to pick an issue rather than if an issue is set by exogenous events or byvoters. If the first mover is able to reconsider his choice if the follower picked a different issue,then politicians who are very competent on both issues will do so. If the decision to move firstor second is endogenized, choosing to move first signals incompetence in one of the issues andthus politicians may wait until one has to make a move; however, the one moving first is morelikely to be overall more competent and more likely to be elected. The model and these resultsmay help understand endogenous selection of issues in political campaigns and the dynamics ofthese decisions.

Keywords: Elections, campaigns, issues, salience, competence, probabilistic voting.JEL Classification: D72, D82 .

∗I am grateful to David Austen-Smith, V. Bhaskar, César Martinelli, Mattias Polborn, Maria Socorro Puy, Kon-stantin Sonin, and participants of the PECA 2012 conference, NES 20th Anniversary conference, the EconometricSociety meeting in Los Angeles in 2013, Wallis 2013 Conference in Rochester, and seminars at the Universityof Pennsylvania, New York University, MIT, University of British Columbia, University College London, andQueen’s University at Belfast for valuable comments.

1 Introduction

Political competition in one-dimensional policy space with two candidates is one of the most

well-studied areas in political economy, and formal analysis goes at least as far as Downs (1957).

The assumption of two candidates is not unrealistic; moreover, Duverger’s Law (Riker 1982,

see also Lizzeri and Persico 2005) predicts emergence of two major candidates or parties in

majoritarian (winner-take-all) elections, as is evident from the USA. Thus, we can expect, under

certain conditions, that political systems with multiple political actors endogenously evolve into

ones with only two. This observation prompts a natural question: Is it true, in some sense,

that political preferences over multiple dimensions nevertheless endogenously result in political

competition over a single issue, and if so, how is it chosen?

In this paper, I study endogenous selection of issues that make political campaigns. I as-

sume that voter preferences are two-dimensional (and separable), and two politicians decide the

issue they want to run on. The following friction is key: I assume that a politician can only

credibly announce his position on (at most) one issue, but not on both.1 The motivation is that

persuading voters about his/her position or about his/her talents in a particular sphere is hard,

and there is only a limited amount of time. Many voters make their decision based on a single

debate performance, or a single rally they attend, so losing focus may be costly for candidates.

Even though occasionally candidates talk about several issues, the broad idea of the campaign

is typically clear: e.g., Bill Clinton ran on economy in 1992 while George H.W. Bush on foreign

policy; in 1972, George McGovern ran against the Vietnam war; in 2012, the Romney campaign

attacked the economic record of Barack Obama, and the Obama campaign eventually switched

from focusing on social issues to defending his economic record of his first term. Of course, in

the course of long campaigns, the focuses of the campaign, and even issue salience itself might

change, but it is probably reasonably to assume that at each stage (e.g., in a given debate) the

candidate has to adopt a particular line of attack or defense, which may or may not coincide

with the one chosen by his opponent.

The second key assumption I make is that voters get better information about a candidate’s

position or competence in a given issue if both of them choose to talk about it. In other words,

there is a certain complementarity: the more likely a politician’s opponent to talk about, say,

economy, the more credibly the politician himself will sound if he chooses to talk about economy

as well. Indeed, the other politician may criticize his positions, check whether the facts are

1Polborn and Yi (2006) consider politicians who can either run positive (revealing information about them-selves) or negative (revealing information about their opponents) informative campaigns, but not both. Alterna-tively, one can assume that politicians have a limited resource (time) that they need to spend on different issues.Whenever this last problem has corner solutions, the two approaches yield identical results.

1

correct, and thus indirectly add credibility to the politician’s statements.2 On the other hand,

if the other politician talks on another topic, then one’s own credibility is undermined. I do

not model the process of making statements and having the other party check their veracity

explicitly; but the assumption that voters learn more if politicians have a sensible debate on the

same issue seems reasonable.3

These two assumptions lead to a simple and tractable model of issue selection. I first study a

basic case where one politician (dubbed “Incumbent”) commits to build his campaign on one of

the two dimensions. The other politician (“Challenger”) observes this, and decides whether to

reciprocate or to talk about the other dimension instead. Voters are Bayesian and they update on

the politician’s decisions and the information that they observe, and vote, in a probabilistic way,

for one of the two candidates. The model gives the following predictions. First, if campaigning

on different issues gives politicians very little credibility, there is a strong unraveling effect,

which forces Challenger to respond to Incumbent and talk about the same issue. However, if a

politician is quite likely to be able to credibly announce his position or establish his competence

even when talking on a different issue, then divergence is possible, and in fact it is slightly more

likely that Challenger will choose a different issue. Second, the social welfare (as measured by

the expected competence of the elected politician) does not necessarily increase in the ability

of politicians to make credible announcements when the opponent campaigns on a different

issue. The reason is that when this ability is low, politicians will endogenously choose the same

issue, and this will help voters rather than hurt them. Third, if the first-mover (Incumbent)

is allowed to reconsider his initial choice of issue (e.g., because the opponent picked a different

issue, or because the campaign is long enough), then switching focus is not a signal of weakness;

rather, it signals competence in both issues. Fourth, if voters are more informed on Incumbent’s

competence in one of the issues, say, economy, then he is more likely to campaign on the other

issue that voters are more uncertain about. This implies, in particular, that Incumbent is likely

to run for reelection on an issue different than the one where he had a chance to demonstrate his

competence (or incompetence) during the first term; doing the opposite would be interpreted as

lack of competence in the other dimension. Finally, if politicians have some discretion whether

2Bhattacharya (2011) models the process of information revelation through positive and negative ads.3One way to microfound this assumption is by introducing a third-party fact-checker which is active only some

of the time; however, if both candidates campaign on the same issue, their opponents serve this role automatically.An alternative way is to assume that political positions of politicians are correlated, and when one candidate stateshis position, voters update on the other candidate’s position as well. For example, Barack Obama’s open supportof gay marriage might be interpreted as a belief shared by all politicians of his generation and thus have littleimpact for voters’decision to support or oppose him – unless his opponent makes a clear stance, too. In this case,again, voters will get more precise signals about the difference in the candidate’s positions when they campaignon the same issue. I do not model either mechanism explicitly in order to focus on consequences of single-issuepolitical competition.

2

to be the first or second to start a campaign, they would prefer to postpone a campaign as

this would signal their relative indiffence between the two issues; ultimately, those who choose

to move first are likely to be more competent (and also disproportionately competent in one

issue relative to the other). As such, there is a first-mover advantage, even though ex-ante, the

expected competences of the two politicians are equal.

Strategic choice of campaign issues by politicians has been the topic of a number of descriptive

and formal studies. Riker (1996, see also 1993), in an extensive study of the U.S. Constitution

ratification, observes that politicians were likely to abandon an issue where they could not

beat their opponents; this means that debating the same issue should be rarely observed, and

attributed to overconfidence or lack of information. This observation predicts “issue ownership”

(as in Petrocik 1996, see also Petrocik et al., 2003), but it relies on the assumption of non-

Bayesian and somewhat “automatic” voters. An influential body of the literature focuses on

priming, i.e., that campaigning on an issue attracts voters’ attention, and in the end of the

day, the voters begin to put a higher weight on this issue, i.e., their preferences change. In this

spirit, Aragonès, Castanheira, and Giani (2012) study a model where parties invest in generating

alternative proposals to the status quo, and then advertise this proposal to increase the salience

of this issue; they show that parties are most likely to invest in issues where they hold ex-ante

advantage (own), although sometimes “issue stealing”is also possible. Amorós and Puy (2007)

consider a model where campaigns allocate an advertisement budget to increase the salience of

an issue (see also Colomer and Llavador, 2011, for a related model). Other papers emphasize

signaling considerations; e.g., Morelli and van Weelden (2011) predict that a politician will spend

a disproportionate amount of effort on a divisive issue for the purpose of credible signaling.

The paper most closely related to this one is Polborn and Yi (2006). There, politicians

also campaign on their qualities, and there may be positive and negative information available

about each of them (four dimensions totally). Politicians choose the campaign issue among two

options: each politician can reveal positive information about himself or negative information

about his opponent. The authors characterize a unique equilibrium, in which running a negative

campaign reveals lack of positive information about oneself. This setting corresponds to the case

where politicians have perfect ability to reveal their competence even if they talk about different

issues. This paper generalizes Polborn and Yi (2006) for the case of generic campaign issues

where campaigning on different issues may undermine voters’ability to learn the truth.4

The model in this paper assumes that the relevant characteristic of candidates is their com-

petence in each of the two issues, and all voters have the same preferences (more competence

4Duggan and Martinelli (2011) study a model of media slant, where media are assumed to collapse a multidi-mensional policy position to a one-dimensional one.

3

is better), but similar forces would be in effect if candidates were competing on more divisive

issues; in this case, the counterpart of competence would be proximity of candidate’s ideal point

(in a given dimension) to the median voter’s position. The model would predict that politicians

would have an incentive to campaign on an issue where their position is close to the median

voter’s one, and if politician turns out to be far from the median voter on the issue that he

chose, voters would suspect that he is even more radical on the other issue. The results would

be valid under the following assumptions: that the candidates cannot commit to any policy

position other than their ideal one in the course of the campaign, and also that they cannot lie

about their position (or, more precisely, that they cannot lie if the other party is campaigning

on the same issue and can reveal this lie to voters).5 In the current model, voters’preferences

are aligned and competence is unambiguously good, so pandering must take the form of exag-

gerating one’s competence. The results are driven by the assumption that doing so is easier if

the opponent talks about a different issue; the assumption that exaggeration is either infinitely

costly (or at least that competence is fully revealed) or totally uncontrolled to the point that

the candidate has zero credibility makes the model tractable, but hardly drives the results.

The rest of the paper is organized as follows. Section 2 introduces the basic model. Section

3 studies the equilibria of the basic model, with sequential or simultaneous moves, and obtains

implications for social welfare and for election probabilities. In Section 4, I study a dynamic

version, where both politicians get a chance to respond to each other’s choice of issue. Section 5

discusses several extensions of the basic model. Section 6 concludes. Appendix A contains the

main proofs; Appendix B (not-for-publication) contains auxiliary proofs and out-of-equilibrium

cases.

2 Model

Consider a two-dimensional policy space, one dimension being economy (E) and the other being

foreign policy (F ). There are two politicians, which we will refer to as Incumbent, indexed by

i, and Challenger, indexed by c; this is mainly done for brevity, and otherwise they will be

symmetric until we relax this assumption in Section 5. Each politician has a two-dimensional

type a = (e, f), which corresponds to his ability in economic and foreign policy questions,

respectively (in what follows, a|s will denote the projection of a on s ∈ E,F). Consider an5There is a large literature on pandering to voters by partisan politicians as well as obscuring one’s positions,

both on campaign trail and in offi ce, starting with Shepsle (1972). Alesina and Cukierman (1990) suggest thatincumbents have an incentive to be ambiguous (see also Heidhues and Lagerlof, 2003). Callander and Wilkie (2007)talk about lying on the campaign trail, as does Bhattacharya (2011). Kartik and McAfee (2007) consider signalingmotive in policy choices; Acemoglu, Egorov, and Sonin (2012) suggest that signaling may make politicians choosepolicies further from the median voter rather than closer to his position.

4

electorate with perfectly aligned preferences: there is a continuum of voters, the utilities of each

of which if politician of type (e, f) is elected is

U (e, f) = e+ f ; (1)

in doing so, I assume that voters weigh both issues equally (this is done for tractability). The

politician’s type is his private information and is not known to the other politician and voters.

At the time of voting, all voters have the same information on both Incumbent and Challenger:

they know the history of the candidates’moves as well as the moves of the Nature; we capture

this information (or, more precisely, the distribution of (ei, fi, ec, fc) conditional on the history

by I for brevity). Voting is probabilistic: voter j votes for Incumbent if and only if

E (U (ei, fi)− U (ec, fc) | I) > θ + θj , (2)

where θ is a common shock and θj is voter j’s individual shock. As is standard, we assume

that θ is distributed uniformly on[− 1

2A ,1

2A

]and θj distributed uniformly on

[− 1

2B ,1

2B

], where

A < 12 and B < A

2A+1 .

During the campaign, each politician can only talk about one issue, economy or foreign policy.

If both talk about the same issue, they have a reasonable conversation or debates, during which

the voters perfectly learn their competences on this dimension (ei and ec or fi and fc). However,

if they end up talking about different issues, it is much harder for them to do it credibly (e.g.,

the other side is not actively engaged in responding or fact-checking). In this case, the chance

that the voters will actually find out his competence on this issue is µ ∈ [0, 1], where µ = 0

corresponds to zero credibility, and µ = 1 is the other extreme where a politician’s ability to

make credible statements of his competence does not depend on the issue his opponent chose.

In a sense, a lower µ corresponds to a higher noise in communication between politician and

voters if there is nobody around to limit exaggerations or bluffi ng.6

More precisely, the types of Incumbent and Challenger are given, respectively, by ai =

(ei, fi) ∈ Ωi = [0, 1]2 and ac = (ec, fc) ∈ Ωc = [0, 1]2. Incumbent moves first and chooses the

issue of his campaign, di ∈ E,F; Challenger observes this and chooses his issue dc ∈ E,F.In other words, the set of Incumbent’s strategies is Si : Ωi → E,F and the set of Challenger’sstrategies is Sc : Ωc × E,F → E,F.7 Nature then decides whether each of the candidates

6One may assume that a politician may announce any competence, but is heavily penalized if he is found tohave exaggerated. When politicians talk about the same issue, there is only chance µ that some third party iswilling to check the politician’s announcements. However, when they talk about the same issue, there is alwayssomeone to do this, and in this case the politicians have to be credible, so voters learn the true competences onthis issue. On the other hand, if nobody puts a check on politicians, then all announce that they are the mostcompetent, and Bayesian voters learn nothing. I do not model this explicitly to simplify the exposition.

7 It should be emphasized that the two candidates do not have private information about the types of their

5

is successful in announcing his competence, and voters get signals κi, κc ∈ [0, 1]∪∅ about theissues that each politician is talking about. Thus, for Incumbent, κi = ai|di if di = dc; whereas

if dc 6= di, then κi = ai|di with probability µ and κc = ∅ with probability 1 − µ. Similarly, forChallenger, κc = ac|dc if dc = di, and if dc 6= di, then κc = ac|dc with probability µ and κc = ∅otherwise. Each voter observes the history of moves (di, dc) and signals (κi, κc), and updates

accordingly, thus getting conditional distribution I, which allows him to compute the difference

in competence between Incumbent and Challenger,

V (di, dc, κi, κc) = E (U (ei, fi)− U (ec, fc) | di, dc, κi, κc) = E (U (ei, fi)− U (ec, fc) | I) .

After that, the common shock θ and idiosyncratic shocks θj are realized for each voter j, and

voter j votes for Incumbent if and only if (2) holds.

Both politicians are expected utility maximizers. The utility of each is normalized to 0 if he

is not elected and to 1 if he is elected, and therefore each maximizes the chance of being elected.

Incumbent, therefore, solves the problem

maxdiE (V (di, dc, κi, κc) | ei, fi) , (3)

and Challenger maximizes

maxdcE (V (di, dc, κi, κc) | ec, fc; di) (4)

where both in (3) and (4) the expectations are taken over the opponent’s type (ac and ai,

respectively), as well as moves by Nature.

The equilibrium concept is the following refinement of the standard Perfect Bayesian equilib-

rium (PBE) in pure strategies: monotone strategies.8 This means that if Incumbent’s strategy

satisfies di (ei, fi) = E, then di (e′i, fi) = E for e′i > ei and di (ei, f′i) = E for f ′i < fi, and similar

requirements are satisfied for dc (ec, fc; di = E) and dc (ec, fc; di = F ).9 This refinement means

that if some type chooses to discuss economy, then increasing his competence in this dimension

without changing the other does not make him switch to foreign policy, and vice versa. In

addition, I do not distinguish between equilibria which differ on a subset of types of measure

zero; for example, if all incumbents with ei = fi are indifferent between economy and foreign

policy and can choose either way, this is treated as a single equilibrium.

opponent, and in particular Incumbent makes his decision knowing the issue that Challenger chose, but notChallenger’s competence in that issue. This is a simplification of reality, but from a technical standpoint, itprevents politicians from strategically jamming the opponent’s signal if they know it to be very high.

8Polborn and Yi (2006) introduce a similar refinement in a game where politicians choose to run a positivecampaign or a negative campaign.

9Suppose that e′i > ei and the candidate with (ei, fi) strictly prefers E while (e′i, fi) strictly prefers F . In thatcase, the latter would be strictly better off if he could fake a lower competence, i.e., choose E and pretend to be(ei, fi). We do not allow candidates to pretend to have lower competence to simplify the exposition.

6

3 Analysis

In this section, I start by analyzing the game introduced in Section 2. I then consider an

alternative story where both candidates choose issues simultaneously. I conclude this Section by

studying social welfare and compare the results for different values of the credibility parameter

µ and for both timings.

3.1 Sequential game

Let us first compute the probability of each politician to be elected for any possible posterior

distribution I. For a given θ, citizen j votes for Incumbent with probability

Pr (θj < E (U (ei, fi)− U (ec, fc) | I)− θ) =1

2+B (E (U (ei, fi)− U (ec, fc) | I)− θ) , (5)

which is also the share of votes Incumbent gets. Incumbent wins if and only if (5) exceeds 12 ,

which happens with probability

Pr (θ < E (U (ei, fi)− U (ec, fc) | I)) =1

2+AE (U (ei, fi)− U (ec, fc) | I) . (6)

Since A is a constant, Incumbent chooses di to maximize

E (E (U (ei, fi)− U (ec, fc) | I))

and Challenger chooses dc to minimize it. Notice that at the time either politician makes a

decision, he knows that he can affect the voters’posterior about his opponent (e.g., Challenger

can make a very competent Incumbent appear worse if he chooses the opposite field, because

the latter could then fail to make a credible announcement). However, if he takes expecta-

tion conditional only on the information he knows at the time of decision-making, the expected

voters’posterior is given and he cannot change it. This simplifies the problem a lot by effec-

tively separating the problems of Incumbent and Challenger (Appendix A contains the details).

From now on, Incumbent and Challenger maximize the expectations of E (U (ei, fi) | I) and

E (U (ec, fc) | I), respectively; more precisely, Incumbent maximizes

E (E (U (ei, fi) | di, κi) | ei, fi) ,

and Challenger maximizes

E (E (U (ec, fc) | dc, κc; di) | ec, fc; di) .

The properties of equilibrium critically depend on whether µ exceeds 12 or not. The next

Proposition gives characterization of equilibrium for a relatively high µ.

7

Proposition 1 If µ > 12 , there is a unique equilibrium. In this equilibrium, Incumbent chooses

the issue that he is more competent in: economy if ei > fi and foreign policy if fi > ei (and he

is indifferent otherwise). If di = E, then Challenger chooses E if and only if

ec >2µ− 1

2− µ fc + C, where C =4− 5µ+

√µ (8− 7µ)

4 (2− µ), (7)

and if di = F ; then a symmetric formula applies.

Not surprisingly, Incumbent always chooses the issue that he is most competent in. The

Challenger’s response depends on the chance that he will be heard if he chooses a different

issue. If µ = 1, then his credibility is the same regardless of the issue, and then his strategy is

independent from the choice of Incumbent: he will always choose the issue his is most competent

in. If 12 < µ < 1, then he may not be as credible when choosing a different issue. For a very

competent (in both dimensions) Challenger, this gives a reason to choose the same issue as

Incumbent, in order to signal his competence in either issue and avoid being pooled with the

less competent mass of potential challengers. Conversely, if Challenger lacks competence in both

dimensions, he is better off choosing an issue different from Incumbent’s choice, because he will

have a better chance to be viewed as an average rather than a very bad type. In equilibrium,

if Incumbent chose Economy, then the set of Challengers who are indifferent between the two

alternatives forms a straight line which is steeper than the diagonal (see Figure 1). Intuitively, in

this case, Challenger will have a higher chance to reveal ec if he chooses E than he has to reveal

fc if he chooses F ; thus, it is not surprising that the choice of Challenger is more “sensitive”to ec

than to fc: for example, if µ is close to 12 , then Challenger’s decision depends almost exclusively

on ec.

We therefore have the following equilibrium strategies of the politicians. Incumbent always

chooses the issue he is best at. Challengers who are good at one issue and bad at the other

also choose their preferred issue, regardless of the choice of Incumbent. However, Challengers

which have roughly equal abilities in both dimensions and who would otherwise be relatively

indifferent respond to Incumbent’s pick of issue in a non-trivial way: those who excel in both

dimensions pick the same issue, whereas very incompetent ones choose a different dimension.

The equilibrium strategies are depicted on Figure 2. Notice that for µ close to 1, the lines

separating the four regions converge to the diagonal, and Challenger’s decision becomes largely

independent from that of Incumbent. Conversely, if µ is close to 12 , the four regions have (almost)

equal size, and half of Challengers will condition their choice of issue on that of Incumbent (the

northeastern corner will always pick the same issue and the southwestern one will always pick

the other issue).

8

Figure 1: Challenger’s equilibrium response if Incumbent picked Economy; 12 < µ ≤ 1.

Figure 2: Equilibrium strategies of Incumbent and Challenger if 12 < µ < 1.

9

The formal proof of Proposition 1 is in the Appendix, but the idea is relatively straight-

forward. The Incumbent’s problem is symmetric, and thus it is natural to expect symmetric

equilibrium strategies. At the same time, picking the issue where one has a disadvantage cannot

happen in equilibrium: then deviating and choosing the other issue would send the voters a

better signal on both issues. The Challenger’s problem is more complicated. For simplicity,

suppose that both issues E and F are picked with a positive probability, and suppose that

Challenger with type (ec, fc) = (x, y) is indifferent, whereas those with a higher ec or lower

fc choose E and those with lower ec or higher fc choose F (Appendix fills in the details on

why such equilibrium exists and why there are no other equilibria). Suppose that Incumbent

chose E; then, if Challenger chooses E, the voters will perceive him as having total competence

x+E (fc | di = E, d (ec, fc) = E, ec = x) = x+ y2 . If he chooses F , then with probability µ, sim-

ilarly, voters will perceive his total competence as y + E (ec | di = E, dc (ec, fc) = F, fc = y) =

y + x2 . With complementary probability 1 − µ, however, they will only know that he chose F ,

and will think of him as E (ec + fc | di = E, dc (ec, fc) = F ), which we denote by ZF . Then the

indifference condition for Challenger of this type may be written as

x+y

2= µ

(y +

x

2

)+ (1− µ)ZF , (8)

and thus we get an upward-sloping boundary between E and F with a slope 2−µ2µ−1 (this is positive

only if µ > 12). Following the intuition above, conjecture that Challenger of type (1, 1) chooses

E and one of type (0, 0) chooses F ; then (8) intersects the boundary of Ωc at some points (a, 0)

and (b, 1). If so, one can compute ZF directly: ZF = a2+ab+a+b2+2b3(a+b) . Substituting this into (8),

we have an equation on (x, y) which should hold for (a, 0) and (b, 1). It is then straightforward

to find that a =4−5µ+

√µ(8−7µ)

4(2−µ) , b =3µ+√µ(8−7µ)

4(2−µ) , which establishes (7).

Are the two politicians ex ante more likely to campaign on the same issues or on different

issues? Surprisingly, the answer is that campainging on different issues is more likely (and this

is captured on Figure 2): Challenger if more likely to choose economy if Incumbent chose foreign

policy, and vice versa. More precisely, we have the following result.

Proposition 2 If µ > 12 , then the probability of having politicians talk about different issues is

higher than 12 : it increases in µ on

(12 ,

45

)and decreases on

(45 , 1), thus reaching its maximum

at µ = 45 . Nevertheless, the probability that either politician successfully communicates his

competence on the chosen dimension is strictly increasing in µ.

It is true that politicians who choose the opposite issue have worse expected quality, and thus

candidates have an incentive to pool with those who choose the same one. However, the result

will be more intuitive if we recall that a campaign where one can reveal one’s competence on one

10

dimension but not the other punishes politicians who are equally competent (or incompetent) on

both issues disproportionately harshly, and these types are key to determine whether Challenger

is more likely to choose the same or a different issue. In particular, suppose that type(

12 ,

12

)is indifferent, which would be true if both cases are equally likely. If Challenger of this type

reveals his competence in either issue, the voters’posterior belief about his competence will be12 + 1

4 = 34 . However, it is easy to see that the average competence of politicians choosing the

opposite issue is (weakly) larger than this, and therefore, as long as µ < 1, the type(

12 ,

12

)would

choose the opposite issue, too. Intuitively, politicians who excel in only one dimension are more

likely to be interested in revealing their competence in this dimension, whereas more symmetric

politicians are relatively less interested in communicating credibly and thus are more likely to

campaign on a different issue. The difference reaches its maximum at µ = 45 , where campaigns

on different issues are almost eight percentage points (precisely,√

33 −

12 ≈ 0.077) more likely

than campaigns on the same issue.

If µ ≤ 12 , the equation (8) no longer defines an upward-sloping boundary, and thus equilibrium

must be different. As it turns out, there are multiple equilibria.

Proposition 3 If µ ≤ 12 , then there are multiple equilibria, and Incumbent’s strategy is the

same (di = E iff ei > fi) in every symmetric equilibrium. The strategy of Challenger may fall

into one of the two classes:

(i) Challenger always conforms to Incumbent’s choice of issue: dc = di;

(ii) If Incumbent chose di = E, then Challenger chooses E if ec > a and chooses F if ec < a;

similarly, if di = F , then Challenger chooses F if fc > a and E if fc < a. There is an equilibrium

with such equilibrium strategies if and only if a ∈ [µ, 1− µ].

The equilibrium strategy of Challenger may, therefore, be either to reciprocate Incumbent or

to play a strategy which only depends on his competence in the issue chosen by Incumbent. If

Challenger is expected to play a similar strategy (mutatis mutandis) for the two possible choices

of Incumbent, the latter expects that he would be successful in communicating his competence

with the same probability regardless of the issue he chooses. In this case, Incumbent will

use a symmetric strategy. In principle, there exist equilibria where Challenger plays different

strategies if Incumbent chose E and F , and then Incumbent must play asymmetrically as well:

for example, if Challenger always picks the same issue as Incumbent if di = E but not if

di = F , then Incumbent of type (1, 1) should strictly prefer E and Incumbent of type (0, 0)

should strictly prefer F . To avoid these complications, from now on I focus on symmetric

equilibria, i.e., equilibria where the equilibrium play if Incumbent and Challenger have types

11

(ei, fi, ec, fc) = (a, b, c, d) and (ei, fi, ec, fc) = (b, a, d, c) will be the opposite.10

To illustrate the equilibria of type (i), suppose Incumbent chose E. If for all (ec, fc),

Challenger chooses E, then E (fc | di = E, d (ec, fc) = E, ec = x) = 12 for all x; at the same

time, E (ec | di = E, dc (ec, fc) = F, fc = y) = zy may be chosen arbitrarily, as may ZF =

E (ec + fc | di = E, dc (ec, fc) = F ). If Challenger of type (x, y) chooses E, voters believe his

competence is x+ 12 , and if he chooses F , they believe it is µ (y + zy) + (1− µ)ZF . Clearly, the

type most likely to deviate is (x, y) = (0, 1); Challenger of this type does not deviate if and only

if1

2≥ µ (1 + z1) + (1− µ)ZF .

Since z1, ZF ≥ 0, such equilibrium is possible only if µ ≤ 12 . At the same time, for such values

of µ, it is indeed an equilibrium; it suffi ces to set zy = 0 for all y and ZF = 0.

Equilibria of type (ii) are also possible only if µ ≤ 12 . Indeed, fix a, and suppose that

Incumbent chose E and Challenger chooses E if ec > a and chooses F if ec < a. Then

E (fc | di = E, d (ec, fc) = E, ec = x) equals 12 if x > a and may be chosen arbitrarily if x < a

(denote it zx). At the same time, E (ec | di = E, dc (ec, fc) = F, fc = y) = a2 for all y; we also

have E (ec + fc | di = E, dc (ec, fc) = F ) = a2 + 1

2 . To verify whether threshold a constitutes an

equilibrium, it suffi ces to concentrate on the types most likely to deviate. In equilibrium, type

(a+ ε, 1) must prefer E and type (a− ε, 0) must prefer F ; this yields two conditions:

a+ ε+1

2≥ µ

(1 +

a

2

)+ (1− µ)

(a

2+

1

2

),

a− ε+ za−ε ≤ µ(

0 +a

2

)+ (1− µ)

(a

2+

1

2

).

Since these conditions need to hold for ε arbitrarily close to 0, the first condition implies a ≥ µ,whereas the second one (setting za−ε = 0) implies a ≤ 1 − µ. It is now straightforward to

show that Challengers with ec 6= a do not have incentives to deviate. Technically, it remains to

define strategies for ec = a (this may not be done arbitrarily) and show that there are still no

deviations, but this is simple and is done in the Appendix.

In Subsection 3.3, I compute welfare of voters under different equilibria and parameter values,

and it turns out that if µ ≤ 12 , the equilibrium of type (i) dominates every equilibrium of type

(ii) in terms of welfare. The reason for this is that the two politicians always campaign on the

same issue, and a low µ does not result to loss of information due to their campaigns lacking

credibility. The equilibrium of type (i) is unique for any µ, and thus this equibrium refinement

10There are many alternative ways to define the same restriction. E.g., it would be suffi cient to assume thatthe chances of Incumbents (Challengers) with type (x, y) and with type (y, x) to be elected are the same. It isremarkable that for µ > 1

2, such symmetry need not be assumed, but may rather be proved (see Proposition 1).

12

extends the uniqueness result from Proposition 1 to the case µ ≤ 12 . This is the equilibrium we

focus from now on.

3.2 Simultaneous game

In this subsection, consider an alternative game, where the two politicians must choose the issue

simultaneously (or, which is equivalent, Challenger starts his campaign before he gets a chance

to observe the choice of Incumbent). Since there is no way to know the exact sequence of moves,

this makes it a priori an equally interesting case to consider. Thus, we consider exactly the same

game as the one introduced in Section 2, except that Challenger, when deciding on dc, does not

observe the choice of Incumbent, di.

Given the symmetry of the game, it is not surprising that there is a symmetric equilibrium,

where each politician j picks dj (ej , fj) = E if ej > fj and dj (ej , fj) = F if ej < fj . Indeed, for

either politician, the probability of ending up campaigning on the same issue is exactly 12 and

this does not depend on the issue; consequently, each politician is able to send a credible signal

with probability 12 + µ

2 ≥12 , regardless of the issue he chooses. Consequently, if one politician

follows this symmetric strategy, then the other politician also must do so in equilibrium (see

Appendix B for a comprehensive analysis of all alternatives). Hence, symmetric strategies by

the politicians are “best responses”to one another, and thus such an equilibrium exists for all

µ.

This argument does not preclude existence of equilibria which are not symmetric in the two

issues: for example if µ ≤ 12 , then there is an equilibrium where di (ei, fi) = dc (ec, fc) = E for

both politicians and all types; indeed, in Subsection 3.1, it was shown that if Incumbent plays E,

then there is an equilibrium where Challenger picks E regardless of his type as well.11 However,

for µ > 12 , only symmetric equilibria exist. Formally, we have the following result.

Proposition 4 For any µ there exists a symmetric equilibrium where each politician j ∈ i, cchooses dj (ej , fj) = E if ej > fj and dj (ej , fj) = F if ej < fj. Moreover, if µ > 1

2 , then this is

the only equilibrium.

3.3 Social welfare

In this Subsection, we study the consequences of the issue selection game on social welfare and

provide comparative statics results. For the society, the relevant variable is the expected com-

petence of the elected politician. The following lemma shows that in the probabilistic voting

11There are other, more exotic equilibria. For example, the following is an equilibrium for µ = 15: campaign on

E whenever ej ≥ 14, campaign on F otherwise.

13

model as above, there is a simple formula for this expected competence. This formula ap-

plies to sequential and simultaneous games and as well as some other situations, and we use it

throughout.

Lemma 1 Let I (ei, fi, ec, fc) be the distribution of the skills of the two politicians,

(ei, fi, ec, fc) ∈ Ωi × Ωc, that voters will get if these politicians follow the equilibrium play.

The expected quality of the elected politician equals

1 +A

∫(ei,fi,ec,fc)∈Ωi×Ωc

E(

(U (ei, fi))2 + (U (ec, fc))

2 | I (ei, fi, ec, fc))dλ− 2

, (9)

where λ is the uniform measure on Ωi × Ωc.

Proof. The expected competence of the elected politician equals (dropping the argument at Ifor brevity): ∫

Ω

( (12 +AE (U (ei, fi)− U (ec, fc) | I)

)E (U (ei, fi) | I)

+(

12 −AE (U (ei, fi)− U (ec, fc) | I)

)E (U (ec, fc) | I)

)dλ

= 1 +A

∫Ω

E(

(U (ec, fc)− U (ei, fi))2 | I

)dλ.

The result would follow immediately if the expected utilities inside the integral were independent.

They are not; for example, a high realization of E (U (ei, fi) | I) means that most likely the

politicians are talking about the same issue, and therefore Challenger is likely to have high

average competence. However, conditional on the choices of issues by both politicians, these

variables are independent. In addition, the choice of issue by one politician does not depend on

the competence of the other one, except through the choice of issue, even if the case of sequential

moves. This implies that social welfare may be computed using (9). Appendix A fills in the

details.

Lemma 1 simplifies the computations considerably. In particular, it shows that the expected

competence of the elected politician only depends on the sum of variances of posterior beliefs

that politicians’equilibrium play generates. This allows to do the computations for Incumbent

and Challenger separately, only taking into account the equilibrium strategies.

In evaluating the welfare consequences, the following benchmarks are useful. First, if the

winner were picked at random, the average competence would be the unconditional expectation,

i.e., 1. At the opposite extreme is the full information case: if both candidates revealed their

competences to voters on both dimensions, then the expected quality of the winner would be

1 +A

(2

∫ 1

0

∫ 1

0(x+ y)2 dxdy − 2

)= 1 +

1

3A. (10)

14

Thus, 13A is the extra benefit of having elections as opposed to picking the candidate randomly;

this is the maximum one can achieve with probabilistic voting where a less competent candidate

has a chance due to a shock to preferences θ. As the variance of common shock decreases (A

becomes higher), the expected competence of the elected politician would increase.

In the case of sequential voting, we have the following result.

Proposition 5 The expected competence of the elected politician is increasing in A. It is non-

monotone in µ; more precisely, it equals 1+ 524A if µ ≤

12 (if type (i) equilibrium from Proposition

3, which yields the highest social welfare, is played), and for µ > 12 , it monotonically increases

from 1 + 316A < 1 + 5

24A to 1 + 14A > 1 + 5

24A.

The nonmonotonicity result is not surprising if one takes into account that for µ < 12 , the two

politicians are guaranteed to discuss the same issue (in the equilibrium that maximizes social

welfare at least), whereas for µ slightly exceeding 12 they will talk about different issues half of

the time, and thus there is a chance of one-fourth that they will fail to announce their respective

competences. In fact, the expected competence exceeds 1 + 524A only if µ > 0.7. In all cases,

this falls short of the maximal possible gain of 13A, although if µ is close to 1, then 75% of this

gain is realized, and even in the worst-case scenario this chance exceeds 56% ( 916).

Consider now the welfare implications of a game where strategies are chosen simultaneously.

Proposition 4 established that a symmetric equilibrium exists for all µ, and arguably it is the

most plausible one (and unique if µ > 12). We get the following comparison in this case.

Proposition 6 In the symmetric equilibrium of the game with simultaneous moves, the expected

quality of the elected politician equals 1 + 1+µ8 A; this is lower than the expected quality in the

game of sequential moves if µ ≤ 12 , but it is higher in the case

12 < µ < 1.

This intuition for this result is simple: all things equal, voters make a more informed choice,

and therefore get a higher utility, if politicians reveal more information about their competences.

This is more likely to happen (again, all things equal) if µ is high, and given that strategies are

the same, the expected quality is increasing in µ. The probability of campaigning on the same

issue is in the game with simultaneous moves is 12 ; however, Proposition 2 states that in the

game with sequential moves, it is less than 12 for µ ∈

(12 , 1). This explains higher voter welfare

for such µ. If µ ≤ 12 , then sequential moves allow the politicians to converge on the same issue

and have an informative discussion at least on that issue. With simultaneous moves, they lack

the ability to converge, and thus the welfare of voters is lower in this case.

If there is a concern that social welfare is lower if politicians fail to coordinate on the same

issue, then one alternative would be to choose an issue (say, at random), and then require that

15

both politicians campaign on that issue. The formula (9) applies to this case as well, and the

expected quality of the winner equals

1 +A

(2

∫ 1

0

∫ 1

0

(x+

1

2

)2

dxdy − 2

)= 1 +

1

6A. (11)

Comparing this result to Proposition 5 reveals that for all µ, the expected competence in the

case where politicians are free to choose their issue is higher than if they are not. This seems

surprising, because in this case, there is no loss due to possible failure to announce their compe-

tences credibly in any dimension. However, there is a different force at play: with endogenous

choice of issues, the announcement of a politician of his competence over one dimension carries

quite a bit of information about his competence on the other issue, which is not the case if they

were forced to talk on a given issue (this is not true for Challenger, who is the second-mover, if

µ < 12 , but for Incumbent this is true for any µ). It turns out that the latter effect dominates.

Proposition 7 If politicians were forced to campaign on the same dimension, then social welfare

would be strictly lower than in the case of sequential moves for all values of µ. It would be lower

than in the game with simultaneous moves as well, provided that µ > 13 ; if, however, µ < 1

3 ,

then forcing politicians to campaign on an exogenously given dimension results in a higher social

welfare than a simultaneous-move game.

Figure 3 illustrates the expected qualities of elected politicians under different scenarios.

Proposition 4, in particular, suggests that campaigning on different issues may decrease social

welfare, if this prevents voters from getting precise information about candidates’ policies.12

This will not happen if politicians choose issues in a well-established sequence, or at least if

the one who moves second has a chance to observe the choice of the first politician and to

respond. This leads to the following nontrivial implication. When candidates choose an agenda

for an entire campaign, the decisions are unlikely to be made at once, and hence constraining

the candidates with an exogenously given agenda is not a good idea. However, when it comes

to some particular event, such as debates, where candidates are likely to prepare their strategy

without observing the opponent’s plan, fixing a particular issue or set of related issues may

make sense. Notice that for µ ∈(

12 , 1), it is better for voters if politicians made their campaign

decisions independently; intuitively, it is due to the fact that with sequential decisions, it is

more likely that they will campaign on different issues, while with independent decisions this

probability is fixed at 12 .

12Caselli and Morelli (2004) and Mattozzi and Merlo (2007) consider very different models that lead to selectionof incompetent politicians. In this paper, incompetent politicians may get elected because voters do not necessarilymake a strong inference about a politician’s incompetence in the issue he is not campaigning on.

16

Figure 3: Social welfare under different scenarios and values of µ.

3.4 Probability of being elected

Another natural question is whether the timing of the game gives an advantage to Incumbent or

Challenger. At first glance, the timing allows Incumbent to pick the issue that he prefers, and

thus can guarantee that he will talk about the issue that he excels at. However, the voters are

aware of this incentive and behavior, and discount Incumbent’s competence on the other issue

accordingly. Challenger, on the other hand, does not have such flexibility, and if µ < 12 he is

forced to choose the same issue as the challenger, which might not be his strong side. However,

voters understand this, and do not infer Challenger’s competence in the other dimension from his

announcement, and thus if Challenger turns out to be incompetent in the issue that Incumbent

picked, he is not penalized further. It turns out that in the probabilistic voting model, these

effects cancel each other out; the expected probability of winning only depends on the expected

competence, which is equal in our case for the two politicians. This implies, for example, that

before knowing his type, a politician is indifferent between being a first-mover and a second-

mover in the game.

Proposition 8 The probability of Incumbent winning and Challenger winning are equal.

At the same time, a given type of politician need not be indifferent. For example, if µ < 1,

then all politicians who are equally competent on both dimensions, with ej = fj < 1, would

prefer to be second-movers rather than first-movers. In Section 5, Incumbent will be given an

option to postpone his announcement and allow Challenger to move first. It turns out that while

17

Figure 4: Timing of proposal-making by Incumbent and Challenger in the dynamic game.

not all types of Incumbent will use this option, those who do are likely to be competent, and

this in the end of the day will result in observable first-mover (incumbent’s) advantage.13

4 Dynamics of campaign

In this Section, I extend the baseline model by allowing the Incumbent (who moves first) to

reconsider his initial choice if the Challenger made a different suggestion. The idea here is to

capture the process of finding a common theme for the campaign or debates and to make sure

that both players have an opportunity to react to each other’s suggestion. Specifically, consider

the following timing. First, Incumbent picks a (tentative) issue for his campaign, di. Challenger

observes this choice and responds with his own (final) decision dc. If the two issues coincide,

di = dc, then the parties proceed to campaigning on this issue, and so Incumbent’s final decision

is di = di. If, however, di 6= dc, then with some probability ν, Incumbent has an opportunity

to revise his initial decision, and is free to pick any di ∈di, dc

= E,F.14 This timing (as

well as a possible conversation) is shown on Figure 4.15 I assume that voters observe the entire

sequence of moves(di, dc, di

); in particular, they know whether Incumbent proposed the issue

of his campaign di himself, or he started with a different issue and then switched.

13 In Ashworth and Bueno de Mesquita (2008), incumbency advantage arises in a probabilistic voting modeldue to his ex-ante higher competence (which in turn is present because he had won elections before). Since thispaper assumes, for simplicity, that the candidates are ex-ante symmetric, this effect is not present.14An alternative interpretation for ν is the share of voters who will listen to Incumbent after he switches, and

thus 1− ν is the share fo voters who paid attention in the beginning of the campaign.15 If Incumbent were allowed to revise his initial pick even if Challenger chose the same issue, then the first-stage

announcement by Incumbent would have to be “babbling”. The reason is that in a candidate equilibrium wherefirst announcement is informative, switching to another issue should be interpreted as competence in both issues.Thus, it would make sense to announce the weaker issue for at least some types, and this cannot be true inequilibrium. This is discussed later in more detail. I am indebted to V. Bhaskar for the suggestion to explore thepossibility that Incumbent must stick to his original choice if Challenger approves the choice of issue.

18

We are interested in symmetric equilibria, so the first proposal by Incumbent is di = E if

ei > fi and di = F otherwise. If Challenger responds with dc = di, then Incumbent does not

have another move; thus, consider the case where di = E and dc = F . In this case, Incumbent

may either agree to discuss foreign policy or insist on talking about economy. Suppose first

that Incumbent is very competent in economy but not in foreign policy, for example, his type is

(1, 0). For such Incumbent, switching to foreign policy is unlikely to do him any good: indeed, he

would have a (weakly) higher chance to signal his competence credibly, but in the foreign policy

dimension, this is precisely what he prefers to avoid. In contrast, suppose that Incumbent is very

competent, say, type (1, 1− ε). In this case, it makes perfect sense to comply with Challenger’sproposal and switch to foreign policy. By doing so, not only he would be able to signal that

he has a very high competence in foreign policy, in addition, voters will take into account that

foreign policy is his weaker issue, and therefore he must be even more competent in economy

(which he truly is). For such Incumbent, therefore, switching allows to signal competence in both

dimensions, which is otherwise diffi cult to achieve. If he insisted in talking about economy, he

would, with some probability, reveal his highest competence in this issue, but voters would view

his foreign policy credentials to be average at best (and, in fact, worse than average, because they

expect those who excel in foreign policy to switch). Therefore, one can expect an equilibrium

where, having chosen an issue in the beginning, Incumbent will insist on campaigning on this

issue if his competence in the other issue is low, but will be open to switching if he is competent

on the other issue as well.

What is the choice of Challenger who received a proposal to talk about, say, economy, and

anticipates that if he proposes foreign policy instead, then Incumbent will follow the strategy

above? Such Challenger knows that if he agrees on E, then both will campaign on economy, and

his signal will be credible with probability 1. At the same time, if he chooses F instead, he will

talk about F , but will only be able to send a credible signal with probability µ′ = p+ (1− p)µ,where p is the probability of Incumbent switching to F . Assuming µ > 1

2 (which we do here to

avoid dealing with multiple equilibria), we are guaranteed to have µ′ ≥ µ > 12 , and therefore the

characterization of Challenger’s strategies from Proposition 1 applies. It remains to verify that

Incumbent’s strategy to start with the issue he is most competent in is indeed an equilibrium;

one can verify that this is true whenever ν < 2(2√

3− 3)≈ 0.93.16 We therefore have the

following result, which is illustrated by Figure 5.16 If Incumbent always had the possilibity to switch, then the types who are very competent in both dimensions

would have incentives to pick the weaker issue. Indeed, they plan to switch if Challenger proposes a differentissue. However, we saw, in Proposition 2, that this will happen with a probability exceeding 1

2, therefore, an

incumbent who, on the margin, would prefer to talk about economy, has incentives to start with proposing foreignpolicy. The assumption that with some small (less than 10% ) probability Incumbent will not have a chance toswitch ensures that this effect is not strong enough.

19

Figure 5: Equilibrium strategies by Incumbent and Challenger in the dynamic game if µ > 12 .

Proposition 9 Suppose that ν < 2(2√

3− 3)≈ 0.93. If µ > 1

2 , there is a unique symmetric

equilibrium. In this equilibrium, Incumbent initially chooses the issue that he is more competent

at, and if Challenger picks a different issue, then a positive share of incumbents switch. The

boundary between those willing to switch and those that are not is linear; for µ = 1, Incumbent

is flexible if min (ei, fi) >12 max (ei, fi), i.e., if Incumbent is relatively symmetric; at µ close to

12 , Incumbent is flexible if min (ei, fi) >

4−√

103 ≈ 0.28. Challenger follows the strategy described

in Proposition 1.

5 Extensions

In this section, I consider several extensions of the baseline model of Section 2.

5.1 Asymmetric uncertainty

The baseline model assumed that the ex-ante distributions of Incumbent and Challenger’s abil-

ities are the same. This was obviously a simplification; first, the voters are likely to be better

informed about Incumbent’s competence, and also Incumbent is more likely to be more compe-

tent on the grounds that he was selected into the offi ce earlier. To study this possibility, suppose

that, from voters’perpective, Incumbent’s two-dimensional type is taken not from a uniform

distribution on [0, 1] × [0, 1], but instead from a uniform distribution on Ω′i = [e1, e2] × [f1, f2];

in particular, e1 = e2 would imply that the voters are perfectly informed about Incumbent’s

20

Figure 6: Information structure and equilibrium strategies.

ability on economy, and f1 = f2 would imply the same on foreign policy. The new distribution

of Incumbent’s type is shown on Figure 6.

The Challenger’s strategies, for either choice made by Incumbent, are the same as specified

in Proposition 1 for µ > 12 and Proposition 3 for µ ≤

12 . The strategy of Incumbent, as it turns

out, critically depends on the shape of the set Ω′i. In particular, if it is a square, albeit smaller

than Ωi = [0, 1] × [0, 1], which means that the residual uncertainty of Incumbent’s competence

in the two issues is the same, then Incumbent will have equal probabilities of campaigning

on both issues. Interestingly, this does not depend on whether he is known to be competent or

incompetent in either dimension; all that matters is residual uncertainty. If, however, uncertainty

about Incumbent’s competence in one of the dimensions is less, then Incumbent is more likely to

campaign in the other dimension (and in particular, the most competent and the least competent

of Incumbent’s types will choose the other dimension, provided that µ is not equal to exactly 1.

More formally, we have the following result.

Proposition 10 Suppose that Incumbent’s type is distributed on Ω′i = [e1, e2]× [f1, f2]. Then:

(i) If e2− e1 = f2− f1, then Incumbent is equally likely to choose either dimension, and will

choose E if ei − e1 > fi − f1 and will choose F if ei − e1 < fi − f1.

(ii) If e2 − e1 < f2 − f1, then in any equilibrium, Incumbent is more likely to choose F than

E. More precisely, let ξ = p + µ (1− p) be the probability of being credible in communication;then if e2−e1

f2−f1 ≤ 1 − ξ, then all types of Incumbent will choose F . A similar characterization

applies if e2 − e1 > f2 − f1.

In other words, Incumbent is always more likely to campaign on the issue where voters are

more uncertain about his competence. At the extreme, if voters perfectly know his competence

21

on either of the dimensions, Incumbent must campaign on a different issue. Interestingly, this

perfect knowledge is not required for this result. If µ 6= 1, so there is a slightest chance of

miscommunication, then Incumbent will never choose economy if e2−e1f2−f1 is small enough, and

will never choose foreign policy if f2−f1e2−e1 is small enough. Intuitively, if e2 − e1 is small and

f2 − f1 is not, then even Incumbent with ei = e2 will not want to waste his campaign on the

issue of economy where the campaign cannot make a big difference, and will talk about foreign

policy instead. Thus, one can expect that an Incumbent who had a chance to demonstrate

his true competence (or true incompetence) in one of the issues will build his campaign on a

different issue.

5.2 Asymmetric issues

We have so far assumed that voters care about both issues equally. However, more generally,

the weights voters put on economy and foreign policy may be different, i.e., it is possible that

voter’s utility is wee+ wff with we 6= wf .

The case of different weights is similar to the previous one in Subsection 5.1, with issue

that has a higher weight corresponding to an issue where there is more uncertainty about both

politicians types. (The difference is that both the incumbent and the challenger would have the

same ex-ante distribution, with higher variance on one dimension.) The easiest way to see this is

to renormalize the policitians’types to the scale that voters use to evaluate them. Namely, one

can define ej = weej , fj = wffj , for j ∈ i, c (so that both ai and ac are distributed uniformlyon [0, we]× [0, wf ]), and from then on assume that the weights are equal: we = wf = 1.

It will follow immediately that both the challenger and the incumbent are more likely to

campaign on the issue that voters care about more. Indeed, the effective uncertainty about the

politicians’types would be greater over that dimension rather than over the issue that voters

care little about. This is realistic: there is little reason for politicians to campaign on the issue

that does not interest voters, and even more, doing so would be interpreted negatively as lack

of competence (or occupying an extreme position) on the issue that concerns voters more. If

the issues in the model were ones where voters had disagreement about, then campaigning on

more divisive issues would sway fewer voters; and so the model predicts that campaigns would

focus on candidates’qualities on issues where preferences are the same, rather than on the issues

where they are different.17

One less trivial insight is the impact of different weight that voters put on issues on social

17The difference with Morelli and van Weelden (2011) is that there, taking positions on divisive issues servesas signaling, while the issues where voters’preferences are similar does not have the competence component thatvoters care about a lot.

22

welfare. In addition to the trade-offbetween losing information on the other important dimension

and the chance of learning nothing at all, a new problem is that politicians will end up talking

about the less important issue. It turns out that this is not a concern: in equilibrium, politicians

are more likely to choose the issue that voters care about more, and this still makes it optimal

to give politicians the freedom to choose the issue for their campaign rather than force them to

focus on a single (even ex-ante more important) issue. This result, of course, need not hold if

politicians made choices simultaneously.

5.3 Delaying campaign

So far, Incumbent was assumed to be the first mover and Challenger was assumed to follow.

Let us augment this game by assuming that Incumbent may instead postpone starting a cam-

paign and wait for Challenger’s choice of issue. More precisely, suppose that with probability

p1, Incumbent must move first, with probability p2 he must wait and move second, and with

probability p0 = 1 − p1 − p2 he has discretion. The question we study here is under which

conditions he will decide to move first and when he will prefer to wait.

The voters observe whether Incumbent moves first or second, but are not aware about the

reason (whether he had the discretion or had to for exogenous reasons). The following two

parameters are key: κ = p0p0+p1

and λ = p0p0+p2

; here, κ is the probability that moving first was a

strategic decision rather than imposed by Nature, and similarly, λ captures the probability that

Incumbent who waited had an option to move. As we will see, voters will also update on the

competence that the politician announces; in particular, a first move, coupled with a declaration

of a very low competence, is likely to be attributed to the politician being forced to move first,

rather than to a conscious decision.

The equilibrium is easy to characterize in the extreme situations, with κ and λ equal (or

close) to 0 or 1.18 We therefore study the two limit cases which are tractable analytically. The

first is κ = λ = 0 (equivalently, p0 = 0), where voters attribute decision to move first or second

to exogenous factors. In this case, if Incumbent gets a chance to decide, he will effectively

compare the costs and benefits of being the first-mover and the second-mover. The following

result shows that, the types of Incumbent who are very asymmetric will decide to act first, and

those who are equally competent in both issues will pass.

18For Bayesian voters who observe that Incumbent did not make the first move, information on whether In-cumbent did not have a chance to move first or had this chance but decided not to is payoff-relevant information,because this tells them something about his competence on the other issue. However, the probability that hehad this chance will depend on the competence over the issue he campaigns on. This complicates the problemconsiderably; in particular, the boundaries of incumbent types choosing whether to move first or to pass, andlater whether to campaign on economy or foreign policy need not be linear.

23

Proposition 11 Suppose that κ = λ = 0, and also µ < 12 , so politicians are guaranteed to

campaign on the same issue in equilibrium. In this case:

(i) Incumbent will move first if|1− ei||1− fi|

/∈[

1

2, 2

](12)

and will pass otherwise. If he moves first, he will choose the issue he is more competent in;

(ii) conditional on moving first, the chance that Incumbent wins is 12 + 1

6A > 12 ;

(iii) the expected competence of the elected politician is higher if Incumbent chose to move

first than if he chose to move second or did not have a chance to choose strategically.

Proposition 11 is intuitive, and the only role of the assumption µ < 12 is to make the condition

(12) so simple to analyze. The most competent Incumbent, (ei, fi) = (1, 1), will be indifferent

between acting first and passing. However, if he is competent in one dimension but incompetent

in the other, he will find waiting too risky, and thus candidates close to (1, 0) and (0, 1) will

move first. On the other hand, candidates with ei = fi < 1 will strictly prefer to wait: they

know that if they move first, their competence in the dimension they choose not to campaign on

will be heavily discounted, and this is less of a problem if they are moving second. Indeed, in the

latter case, the voters will think that they were forced to campaign on a dimension chosen by the

challenger, and as such the penalty would not be so high. It is also intuitive that least competent

politicians (in particular, those with ei, fi < 12) prefer to pass and become second-movers: for

them, moving first and campaigning on either issue is going to release a (justified) negative

signal about their competence in both dimensions; at the same time, if they move second, this

would only be true for one dimension. These strategies are illustrated on Figure 7; notice that

in the case µ < 12 , the candidates will eventually campaign on the same issue regardless of who

moves first.

According to Proposition 11, if Incumbent gets a chance to move first, it is equally likely

that he will use this chance and that he will not. However, there is another difference in the

types of Incumbent who choose either strategy. Not only the ones who act first are more likely

to be competent in one dimension and challenged in the other, but they are more likely to be

more competent overall. Indeed, suppose, for example, that such Incumbent has ei > fi; then

simple calculation shows that his expected competence in economy E is 56 , and his expected

competence in foreign policy F is 13 , which makes his overall expected competence

76 . Conse-

quently, incumbents who move first are also more competent than challengers, in expectation.

This insight leads to the other two results in the proposition; first, there is first-mover advantage,

in the sense that an incumbent who gets a chance to move first is more likely to win, and second,

when campaigns choose their issues earlier, the expected competence of the elected politician is

24

Figure 7: Incumbent’s strategy when he can postpone his campaign.

higher. The latter effect arises in the equilibrium because more competent politicians are more

likely to move first, and incompetent ones prefer to wait.19

Consider now the opposite case, κ = λ = 1, corresponding to the case where the politician

almost always has discretion (p0 = 1). The trade-off that the incumbent faces now looks some-

what differently. Moving first allows them to reveal their competence on the issue they are good

at, but the cost is that they will be thought of as very incompetent on the other issue. On

the other hand, if they wait, voters will think that their skills on the two issues are relatively

close, and therefore announcing their competence on one issue will not hurt them on the other

dimension, or will hurt very little. Hence the types of Incumbent close to the boundary have an

incentive to pass, and this only increases the penalty for moving early. When κ = λ = 1, this

leads to full unraveling: no Incumbent will ever move first.

Proposition 12 If κ, λ = 1, then the unique equilibrium requires Incumbent to postpone choos-

ing the campaign issue until Challenger emerges and makes the choice.

This results suggests that when voters know that a politician is free to decide between

moving first and moving second, the best response for the politician is to wait. In this model,

this means that strategies on choosing issues do not change, and neither does social welfare.

More broadly, this result implies that while making a first announcement is an attractive option

for at least some politicians, not using this option is even more attractive, as it serves as a

positive signal of possessing balanced competence in the two issues, which is valuable in the

19Another effect, which is not modeled explicitly, is that deciding on an issue earlier gives politicians more time,and allows voters to get more precise signals about politicians’ competences, which again raises the expectedcompetence of the winner.

25

environment where campaigns may be run on one issue only. As a result, the opportunity to

make an early commitment to the issue of the campaign will not be used. This insight may be

extended further, to a game where both politicians get, e.g., alternating opportunities to move

first; backward induction will immediately suggest that both politicians will wait until the very

last opportunity, and on the equilibrium path, waiting will be something expected rather than

a positive signal. (Interestingly, off equilibrium path, once one politician makes the choice, the

other one has no incentive to wait further.)

The insights obtained in these extreme cases suggest the following implications. First, politi-

cians are not likely to seize the very first opportunity to pick an issue, and the reason is not the

aggregate uncertainty (i.e., they might want to learn, which issues voters find most important),

but rather signaling considerations. Second, politicians will use opportunities which present a

good reason not to wait. For example, if some event or story makes it impossible or very hard

for a politician not to react, the politician might well make the first move (e.g., a stock market

crash creates a good reason to start campaigning on economy). A politician may also want to

use an opportunity which would otherwise go unnoticed (and therefore he would not get enough

credit for waiting). If none of these event types get realized, politicians are likely to wait until

the last moment, when the remaining time becomes a binding constraint.

6 Conclusion

The paper studies the incentives of politicians to choose issues to run their campaigns on. Voters

are Bayesian, and by campaigning, the candidates cannot change voters’preferences, but can

affect the information they possess. I assume that both candidates are more credible when they

run on the same issue, and this creates a non-trivial interplay between their incentives. The

first mover (Incumbent) has a disproportionate influence on the course of the campaign, but

this does not necessarily help him win. Whenever a politician gets a chance to postpone the

announcement of his issue, he will, because this will signal his competence on the issues he will

not focus his campaign on. The politicians who nevertheless choose to act first are more likely to

be competent, and more likely to win. The model predicts that allowing politicians a free choice

of campaign issues reveals more information in the course of campaign, and ultimately raises the

chance of electing the most competent candidate. The model offers non-trivial insights on the

nature of issue selection in campaigns, and is tractable enough to allow for different extensions.

In particular, building on the dynamic model of Section 4 and further studying dynamics of

political campaigns as sequential issue selection seems to be an interesting avenue for future

research.

26

Appendix A – Main Proofs

Proof of Proposition 1. Suppose that di = E. For Challenger, we have, in notation of Lemma

5 from Appendix B, ξE = 1 and ξF = µ, thus r = 1µ ∈ [1, 2). By Lemma 5, there is unique

upward-sloping boundary with types who are indifferent, and Lemma 6 and Lemma 7 imply

that there is no other equilibrium response by Challenger in this case. We therefore need to find

the required boundary.

Since the slope of the boundary is given by 2−µ1−2µ , there are several cases to consider. If types

(a, 0) and (1, b) are indifferent (and a > 0, b < 1), then we would get a contradiction with Lemma

8. Similarly, if types (0, a) and (b, 1) are indifferent, then we again get a contradiction with the

same Lemma 8, because ξE = 1 > 23 . Therefore, we must have that types (a, 0) and (b, 1) are

indifferent, and a < b. Then the expected quality of types choosing F is a2+ab+b2

3(a+b) + a+2b3(a+b) . The

conditions that types (a, 0) and (b, 1) are indifferent are indifferent are then

a = µa

2+ (1− µ)

a2 + ab+ a+ b2 + 2b

3 (a+ b),

b+1

2= µ

(1 +

b

2

)+ (1− µ)

a2 + ab+ a+ b2 + 2b

3 (a+ b).

For µ > 12 , this has solution a =

4−5µ+√µ(8−7µ)

8−4µ , b =3µ+√µ(8−7µ)

8−4µ .

The probability that Challenger chooses the same issue is found in the proof of Proposition

2; it equals p =3(2−µ)−

√µ(8−7µ)

4(2−µ) . Given the symmetry of Challenger’s strategies, Incumbent

has ξE = ξF = p + (1− p)µ = µ+34 −

(1−µ)√

8µ−7µ2

4(2−µ) > 23 for all µ. Lemma 5 implies that for

Incumbent r = 1, and there is a unique equilibrium.

Proof of Proposition 2. The probability that the incumbent choosing issue F when the

challenger chose E (or vice versa) equals 0 if µ < 12 and

1

2

(4− 5µ+

√µ (8− 7µ)

8− 4µ+

3µ+√µ (8− 7µ)

8− 4µ

)=

2− µ+√µ (8− 7µ)

4 (2− µ)

if µ > 12Differentiating this with respect to µ yields

d

dµ

(2− µ+

√µ (8− 7µ)

4 (2− µ)

)=

4− 5µ

2 (2− µ)2√µ (8− 7µ)

.

This is increasing for µ < 45 and decreasing for µ >

45 , and the maximal value equals

14 +

√3

6 ≈0.539. The probability that a politician communicates his competence on the chosen dimension

successfully equals µ+34 −

(1−µ)√

8µ−7µ2

4(2−µ) , which is increasing on(

12 , 1). This completes the proof.

27

Proof of Proposition 3. This result immediately follows from Lemma 6 and Lemma 7 in

the case of Challenger; the case of Incumbent is considered similarly to Proposition 1

Proof of Proposition 4. The proof of this result is trivial and is omitted.

Proof of Lemma 1. The expected competence of the elected politician is

EE (U (ei, fi) | I) = E (U (ei, fi)) = 1 (and similarly for Challenger).

Denote Ω = Ωi × Ωc, Let us show that∫

ΩE ((U (ei, fi)U (ec, fc)) | I) dλ = 1. We split the

entire space of types into four regions, (di, dc) ∈ (E,E) , (E,F ) , (F,E) , (F, F ) according tothe dimension chosen in equilibrium (note that these need not be independent, as the second

player’s choice depends on the first player’s one). We have∫Ω

E ((U (ei, fi)U (ec, fc)) | I) dµ

=∑

(di,dc)∈Sµ(di,dc)

∫Ω(di,dc)

E((U (ei, fi)U (ec, fc)) | Ω(di,dc), ai|di , ac|dci

)dλ

=∑

(di,dc)∈Sµ(di,dc)

∫Ω(di,dc)

E(U (ei, fi) | Ω(di,dc), ai|di

)E(U (ec, fc) | Ω(di,dc), ac|dc

)dλ

=∑

di∈E,F

∫Ωdi

∑dc∈E,F

µ(di,dc)

∫Ωdc (di)

E (U (ei, fi) | Ωdi , ai|di)E (U (ec, fc) | Ωdi ,Ωdc (di) , ac|dc) dλdcdλdi

=∑

di∈E,F

∫Ωdi

E (U (ei, fi) | Ωdi , ai|di)∑

dc∈E,Fλ(di,dc)

∫Ωdc (di)

E (U (ec, fc) | Ωdi ,Ωdc (di) , ac|dc) dλdcdλdi

=∑

di∈E,F

∫Ωdi

E (U (ei, fi) | Ωdi , ai|di)λdiE (U (ec, fc) | Ωdi) dλdi

= 1 ∗∑

di∈E,F

∫Ωdi

µdiE (U (ei, fi) | Ωdi , ai|di) dλdi = 1 ∗ 1 = 1.

(where ai|di is the projection of type ai on the dimension di). We used that E (U (ec, fc) | Ωdi)

does not in fact depend on di and equals 1. Consequently,

1+A

∫Ω

E(

(U (ei, fi)− U (ec, fc))2 | I

)dµ = 1+A

∫Ω

E(

(U (ei, fi))2 + (U (ec, fc))

2 | I)dµ− 2

.Within each region, E (U (ei, fi) | I) and E (U (ec, fc) | I) are independent, and thus

E ((U (ei, fi)U (ec, fc)) | I) = E (U (ei, fi) | I)E (U (ec, fc) | I). This completes the proof.

28

Proof of Proposition 5. First, suppose µ ≤ 12 . Without loss of generality, assume that

the challenger has ec > fc and thus chooses E, and the incumbent does the same. Computing

the relevant integrals in the right-hand side of (9), we obtain∫ 1

0

(3

2x

)2

2xdx+

∫ 1

0

(y +

1

2

)2

dy =53

24= 2 +

5

24.

Now suppose that µ ≥ 12 . We first calculate the integral for the challenger (again, as-

suming that ec > fc as the opposite case may be considered similarly). With probability

1− 2−µ+√µ(8−7µ)

4(2−µ) , the incumbent chooses E, too. This gives∫ 1

0

(32x)2

2xdx = 98 . With probabil-

ity2−µ+

√µ(8−7µ)

4(2−µ) , the incumbent chooses F , and the integral equals µ ∗ 98 + (1− µ) ∗ 1. Totally,

the contribution of the challenger is

9

8

(1− 2− µ+

√µ (8− 7µ)

4 (2− µ)

)+

2− µ+√µ (8− 7µ)

4 (2− µ)

(µ ∗ 9

8+ (1− µ) ∗ 1

)=

(9

8− 1− µ

32 (2− µ)

(2− µ+

√µ (8− 7µ)

)).

Now, consider the contribution of the incumbent. The part of the integral coming from the

set where he chooses E is (as before, a =4−5µ+

√µ(8−7µ)

8−4µ , b =3µ+√µ(8−7µ)

8−4µ ):∫ b

a

(x+

1

2

x− ab− a

)2 x− ab− a dx+

∫ 1

b

(x+

1

2

)2

dx =13

12−3a+ 9b+ 4ab2 + 4a2b+ 8ab+ 4a2 + 4a3 + 12b2 + 4b3

48.

The part of the integral coming from the set where he chooses F may be found as follows: with

probability µ, it is∫ 1

0

(y +

1

2(a+ (b− a) y)

)2

(a+ (b− a) y) dy =1

48

(4a+ 12b+ 3ab2 + 3a2b+ 8ab+ 4a2 + 3a3 + 12b2 + 3b3

),

and with probability 1− µ, it is(a2 + ab+ b2

3 (a+ b)+

a+ 2b

3 (a+ b)

)2a+ b

2=

1

18 (a+ b)

(a2 + ab+ a+ b2 + 2b

)2.

This means that the total integral for the challenger and the incumbent, after substituting for

the values of a and b, is

2 +

(19µ+ 6µ2 − 68

)(2− µ)2 +

(29µ2 − 6µ3 − 52µ+ 24

)√µ (8− 7µ)

192 (2− µ)3 .

It may be shown directly that this is an increasing function of µ, and that its value for µ = 12 is

2 + 316 and for µ = 1 is 2 + 1

4 . This completes the proof.

Proof of Proposition 6. It was proved earlier that an equilibrium where both politicians

always choose E or always choose F is only possible if µ > 12 . On the other hand, it is obvious

29

that picking one’s better dimension is an equilibrium for all values of µ. For this strategy, each

politician is able to announce his competence credibly with probability 12 + 1

2µ, and he fails to

do so with probability 12 (1− µ). Consequently, the integral in the right-hand side of (9) equals

2

((1

2+

1

2µ

)∫ 1

0

(3

2x

)2

2xdx+1

2(1− µ) ∗ 1

)= 2 +

1 + µ

8.

Therefore, the expected competence of the elected politician is monotonically increasing, and it

exceeds the competence in the case of an exogenously given issue (2 + 16) if and only if µ >

13 .

This completes the proof.

Proof of Proposition 7. Proved in the text.

Proof of Proposition 8. The expected probability of challenger winning is obtained by

taking expectation of (6) over all possible realizations of (ec, fc), (ei, fi), as well as κc and κi.

The law of iterated expectations implies that this equals 12 , as thus the expected probability of

incumbent winning also equals 12 . This completes the proof.

Proof of Proposition 9. In the symmetric equilibrium, Incumbent chooises di = E if

ei > fi and di = F if ei < fi. Consider the case where di = E and dc = F . Consider an

agent with competence (x, y); suppose that he is indifferent. If so, then if he chooses E, he

gets µ(x+ y

2

)+ (1− µ)ZE , where ZE = E

(ei + fi | di (ei, fi) = di (ei, fi) = E, dj = F

). If he

chooses F instead, he gets y + x+y2 . Consequently, he is indifferent if and only if

y =2µ− 1

3− µ x+2 (1− µ)

3− µ ZE . (A1)

This defines an upward-sloping line for 12 < µ ≤ 1. Moreover, for such µ, y > 0, provided that

x > 0. Consequently, if there are positive measures of Incuments who stay and who switch,

the line should connect some point (a, a) with some point (1, b); moreover, it must be that

b = a+ 2µ−13−µ (1− a) and 0 ≤ a < b < 1.

Let us now find ZE as a function of a. The area E consists of two triangles, ob-

tained from the bottom-right triangle by connecting (a, a) with (1, 0). One triangle has ver-

tices (0, 0) , (1, 0) , (a, a); its area is a2 and the sum of the coordinates of its mass center is

13 (1 + a+ a) = 2a+1

3 . The other triangle has vertices (1, b) , (1, 0) , (a, a); its area is (1−a)b2 =

(1−a)(a+ 2µ−1

3−µ (1−a))

2 and the sum of coordinates of its mass center is 13 (1 + 1 + a+ b+ a) =

2a+b+23 =

2a+(1−a)(a+ 2µ−1

3−µ (1−a))

+2

3 . Therefore,

ZE =1

3

−2a2b+ 2a2 − ab2 + a+ b2 + 2b

a+ b− ab .

30

Thus, (A1), when evaluated at point (x, y) = (a, a), simplifies to

H (a, µ) =(3µ3 + 23µ2 − 84µ+ 64

)a3 +

(288µ− 108µ2 − 192

)a2 (A2)

+(81µ2 − 171µ+ 84

)a+

(30µ− 20µ2 − 10

)= 0.

If µ = 1, the equation (A2) has a unique root on [0, 1), a = 0, and in this case the indifference

line connects (0, 0) and(1, 1

2

). In the other extreme, µ = 1

2 , we must have a = b, and thus a > 0;

in this case, the only root is a = 4−√

103 ≈ 0.28.

Let us show that for all µ ∈ (0, 1), there is a unique root a ∈ (0, 1). The left-hand side of

(A2) equals 10 (2µ− 1) (1− µ) > 0 if a = 0 and it equals −3 (2− µ) (3− µ)2 < 0 if a = 1. Thus,

there exists a root a ∈ (0, 1) for any such µ. Moreover, if µ = 1, the three roots of the cubic

equation are

1−√

2, 0, 1 +√

2, and so a = 0 is a simple root. Therefore, for µ close to 1,

the equation (A2) has a unique simple root on (0, 1). If for some µ ∈(

12 , 1)there are two or

three roots on (0, 1), then at least one of the following three alternatives must be true: either

for some (perhaps other) µ ∈(

12 , 1), a = 0 is a root, or a = 1 is a root, or there is a double root

a′ ∈ (0, 1). The first two possibilities are ruled out, because we know that a = 0 may be a root

only for µ ∈

1.2 , 1

, and a = 1 may be a root only for µ ∈ 2, 3. Suppose that some a ∈ (0, 1)

is a double root, then the second derivative of (A2) must vanish at some a between the single

root and the new double root:

6(3µ3 + 23µ2 − 84µ+ 64

)a+ 2

(288µ− 108µ2 − 192

)= 0. (A3)

Since µ 6= 43 , this is equivalent to

a =12µ− 16

9µ+ µ2 − 16, (A4)

which is decreasing from 89 to

23 as µ increases from

12 to 1. For a given by (A4), (A2) simplifies

to−2 (3− µ)2

(16− 9µ− µ2)2

(10µ4 − 261µ3 + 1053µ2 − 1536µ+ 768

).

It is straightforward to check that the last factor has no roots on(

12 , 1); this proves that the

root is unique.

We can show that this root is decreasing in µ. Indeed, the root 0 for µ = 1 satisfied∂H(a,µ)∂a < 0; since we proved that there is no double root, we must have that ∂H(a,µ)

∂a < 0 for all

µ. It remains to show that ∂H(a,µ)∂µ < 0 at any root. We have

∂H (a, µ)

∂µ=(9µ2 + 46µ− 84

)a3 + (288− 216µ) a2 + (162µ− 171) ,

31

Let us show that it is positive for all µ ∈[

12 , 1], a ∈

[0, 4−

√10

3

]. Indeed, for such values, ∂H(a,µ)

∂µ

is increasing in µ:

∂2H (a, µ)

∂µ2= 9a3µ− 108a2 + 23a3 + 81

≥ 9

2a3 − 108a2 + 23a3 + 81

≥ 27a3 − 108a2 + 81 = 27 (1− a)(3 + 3a− a2

)> 0.

Consequently, it remains to prove that

∂H (a, µ)

∂µ

∣∣∣∣µ=1

= 72a2 − 29a3 − 9 < 0

for a ∈[0, 4−

√10

3

], which is straightforward.

We can also prove that b is increasing in µ. Indeed, given a = 3b−2µ−bµ+14−3µ , we can rewrite

(A2) as

H (b, µ) =(µ3 + 6µ2 − 43µ+ 48

)b3 +

(6µ3 − 57µ2 + 165µ− 144

)b2 (A5)

+(12µ3 − 57µ2 + 63µ

)b+

(8µ3 − 18µ2 + 7µ

)= 0.

This function has no double root: indeed, otherwise there would be a point with ∂2H(a,µ)∂b2

= 0,

in which case b = 16−13µ+2µ2

16−9µ−µ2 ; plugging this into (A5) yields

−2 (4− 3µ)2

(16− 9µ− µ2)2

(10µ4 − 261µ3 + 1053µ2 − 1536µ+ 768

),

which, as we know, has no roots on µ ∈(

12 , 1). Now, this means that ∂H(b,µ)

∂b < 0 for all µ,

because this is true for µ = 1. It remains to prove that ∂H(b,µ)∂µ > 0. We have

∂H (b, µ)

∂µ=(3µ2 + 12µ− 43

)b3+

(18µ2 − 114µ+ 165

)b2+

(36µ2 − 114µ+ 63

)b+(24µ2 − 36µ+ 7

).

One can show that this is positive at the root b.

Proof of Proposition 10. Without loss of generality, we can assume that e2−e1 ≤ f2−f1.

If so, renormalize the rectangle [e1, e2] × [f1, f2] by mapping (x, y) 7→(x−e1f2−f1 ,

y−f1f2−f1

); we then

have a uniform distribution on [0,m]× [0, 1], were m is the ratio e2−e1f2−f1 .

If Incumbent chooses E (F ), he will talk about E (F ) with probability p =3(2−µ)−

√µ(8−7µ)

4(2−µ) ,

and will thus be able to reveal his competence credibly with probability ξE = ξF = p+(1− p)µ >23 ; then the ratio r = ξE/ξF = 1. Then, as in Lemma 5, we can prove that if both E and F

32

are picked by Incumbent with a positive probability, then the line separating them must be

upward-sloping, with slope 1, and the points on this line must satisfy

∆ (x, y) = ξE

(x+

y

2

)+ (1− ξE)ZE − ξF

(y +

x

2

)− (1− ξF )ZF = 0.

Suppose, to obtain a contradiction, that ∆ (0, 0) > 0, so that the line ∆ (x, y) = 0 lies

above the “diagonal” y = x. Then, evidently, ZF > ZE , but if so, we would have ∆ (0, 0) =

(1− ξE) (ZE − ZF ) < 0, a contradiction. Thus, ∆ (0, 0) < 0, and the intersection of the line

∆ (x, y) = 0 with ∂Ωi are some points (a, 0) and (m,m− a), with 0 ≤ a < m. The equilibrium

condition is (where we set ξ = ξE = ξF and (x, y) = (a, 0)):

ξ (a) + (1− ξ)ZE = ξ(a

2

)+ (1− ξ)ZF .

Then1− ξξ

(ZF − ZE) =a

2.

We have ZE = a+ m−a3 + 2

3 (m− a) = m, and ZF may be found from

ZE(m− a)2

2m+ ZF

(1− (m− a)2

2m

)=m+ 1

2.

This implies ZF = m−2m−2am+a2+m2−1−2m−2am+a2+m2 , and thus ZF − ZE = m(1−m)

2m+2am−a2−m2 = m(1−m)

2m−(m−a)2.

Consequently, a is found from the equation

1− ξξ

m (1−m)

2m− (m− a)2 =a

2;

since the left-hand side is decreasing in a, it has at most one solution. It has a solution a < m

if and only if ξ +m > 1, i.e., m > 1− ξ.Consider the cases where all or almost all types choose the same issue. If all types choose

F , then ZF = m+12 , and the condition that type (m, 0) does not deviate is

ξ (m+ zm) + (1− ξ)ZE ≤ ξ(m

2

)+ (1− ξ) m+ 1

2.

This is satisfied (for zm = ZE = 0) if m ≤ 1−ξ2ξ−1 . If all types choose E, then ZE = m+1

2 , and the

condition that type (0, 1) does not deviate is

ξ

(1

2

)+ (1− ξ) m+ 1

2≥ ξ (1 + z1) + (1− ξ)ZF .

This is satisfied (for z1 = ZF = 0) if m ≥ 2ξ−11−ξ > 1, since ξ > 2

3 . Therefore, there is only one

equilibrium with both E and F chosen with a positive probability if m > 1−ξ2ξ−1 , two equilibria if

1− ξ < m ≤ 1−ξ2ξ−1 , and only one equilibrium where only F is chosen when m < 1− ξ.

33

Proof of Proposition 11. The set of types that decide to wait should be symmetric around

the line ei = fi. In addition, if the incumbent decides to act first, he should choose E if and only

if ei > fi, and choose F otherwise. The remainder of the proof involves considering all possible

cases as in the proof of Proposition 1 and applying (9).

Proof of Proposition 12. Suppose not, i.e., that some types choose to act first. Notice

that it cannot be that all types act first: then waiting would make voters believe that the voter

is taken from the same uniform distribution on [0, 1]2, which means that the types with low ei

and fi would wait.

Now suppose that the border between the regions where the agent acts first by choosing E

and the region where he waits is given by fi = g (ei), and the border between F and waiting

is symmetric and given by ei = g (ei). Consider an individual with (ei, fi) = (x, y) such that

y = g (x). By choosing to act first and announce E, he expects to get x + g(x)2 . By waiting,

he has an equal chance of having to campaign on E and F ; in the first case he expects to get

x+ g−1(x)+g(x)2 and in the second case y+ g−1(y)+g(y)

2 = g (x)+ x+g2(x)2 (where g2 (x) ≡ g (g (x))).

Consequently, we must have

x+g (x)

2=

1

2

(x+

g−1 (x) + g (x)

2

)+

1

2

(g (x) +

x+ g2 (x)

2

).

Simplification yields x = g−1 (x) + g (x) + g2 (x). But this is impossible, since x < g−1 (x). This

contradiction completes the proof.

34

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36

Appendix B – For Online Publication

The following Lemmas study the strategies of a player j in the following situation. Suppose he

knows that if he picks issue E, then his opponent will choose E with probability λE , and if he

picks F , then his opponent will choose F with probability λF . Denote ξE = λE +(1− λE)µ and

ξF = λF + (1− λF )µ and suppose that at least one of these values is non-zero: ξE + ξF > 0.

In what follows, we let ZE and ZF be the beliefs that voters have of a player j if he

chooses E and F , respectively (in equilibrium, ZE = E (ej + fj | dj (ej , fj) = E) and ZF =

E (ej + fj | dj (ej , fj) = F ) whenever these expectations are well defined). Slightly abusing no-

tation, let E be the set of (ej , fj) such that dj (ej , fj) = E and F be the set of (ej , fj) such

that dj (ej , fj) = F . In other words, (re)define E,F ⊂ Ωj ; E = x, y ∈ Ωj : dj (x, y) = E, andF = Ωj \ E.

In the following Lemmas, suppose that both areas, E and F , have a positive mass, unlike

otherwise noted.

Lemma 2 Let G = E ∩ F = ∂E ∩ ∂F be the set of points that have points from both E and

F in any ε-neighborhood. Then G is a simple curve which may be parametrized by a (closed)

interval T ⊂ [0, 2] such that t ∈ T is mapped into (x, y) such that x+ y = t.

Proof. Since area E has a positive mass, there is a point (x, y) ∈ E with x < 1 and y > 0,

and thus point (1, 0) and all points (x, y) within some ε-radius of (1, 0) belong to E; similarly,

all points within ε-radius of (0, 1) belong to F . Since G is intersecion of two closed sets, it

is closed; define t1 = min(x,y)∈G (x+ y) and t2 = max(x,y)∈G (x+ y) and let φ : G → [t1, t2]

be defined by φ (x, y) = x + y. It is injective: indeed, if for two points (x1, y1) , (x2, y2) ∈ G,we had x1 < x2 and y1 > y2, then there would be a points (x3, y3) ∈ E and (x4, y4) ∈ F

with x3 < x4 and y3 > y4 (because we could take these points arbitrarily close to (x1, y1) and

(x2, y2), respectively). But this would contradict monotonicity: indeed, monotonicity would

imply that point (x4, y3) ∈ E because x4 > x3, but also that (x4, y3) ∈ F because y3 > y4,

which is impossible. At the same time, it is also surjective. Indeed, take some t ∈ (t1, t2) which

is not part of Imφ. Since Imφ is a closed set (as φ is a continuous function with a compact as

its range), take t3 = max [Imφ ∩ [t1,t]] and t4 = min [Imφ ∩ [t, t2]]. Let (x5, y5) = φ−1 (t3) and

(x6, y6) = φ−1 (t4) and let ε = min (x6 − x5, y6 − y5) > 0. Within ε2 -neighborhood of (x5, y5),

pick (x7, y7) ∈ E and within ε2 -neighborhood of (x6, y6), pick (x8, y8) ∈ F . Then by monotonicity,

(x8, y7) ∈ E, and (x7, y8) ∈ F (if not, then (x8, y8) ∈ E by monotonicity, a contradiction). Now,

B-1

observe that

x8 + y7 > x6 −ε

2+ y5 −

ε

2= x6 − ε+ y5 ≥ x5 + y5 = t1,

x8 + y7 < x6 +ε

2+ y5 +

ε

2= x6 + (y5 + ε) < x6 + y6 = t2,

and thus φ (x8, y7) ∈ (t1, t2); similarly, φ (x7, y8) ∈ (t1, t2). Thus, for any λ ∈ [0, 1], we have

φ (λx8 + (1− λ7)x7, λy7 + (1− λ7) y8) ∈ (t1, t2). However, there is λ for which this point is in

E ∩ F , a contradiction.Consider the following expression:

∆ (x, y) = ξE

(x+

y

2

)+ (1− ξE)ZE − ξF

(y +

x

2

)− (1− ξF )ZF . (B1)

Lemma 3 Let (x, y) ∈ G\∂Ωj and suppose that (ej , y) ∈ G⇒ ej = x and (x, fj) ∈ G⇒ fj = y,

i.e., there are no other points in G with ej or fj. Then ∆ (x, y) = 0.

Proof. Suppose, to obtain a contradiction, that ∆ (x, y) 6= 0. Suppose that ∆ (x, y) > 0

(the opposite case is considered similarly). Take ε = ∆(x,y)3 ; then, since curve G is continuous,

there exists δ ∈ (0, ε) for which there exists a unique point (x′, y′) ∈ G with x′ = x − δ andy′ ∈ (y − ε, y) (let ε′ = y − y′ < ε), and also ∆ (x′, y) > ∆ (x, y) − ε. Now consider the point(x′, y); we know, by definition of G, that (x′, y) ∈ F , and by continuity of ∆, ∆ (x′, y) > 0.

By construction, we have E (ej | (ej , fj) ∈ F, fj = y) = x2 , and E (fj | (ej , fj) ∈ E, ej = x) =

y′

2 . Consequently, if the player chooses E, he gets uE = ξE

(x′ + y′

2

)+ (1− ξE)ZE , and if he

chooses F , he gets uF = ξF(y + x

2

)+ (1− ξF )ZF . But (x′, y) ∈ F , thus uF ≥ uE . Note,

however, that

uE − uF = ∆ (x, y) + ξE

(x′ − x+

y′ − y2

)= ∆ (x, y)− ξE

(δ +

ε′

2

)> ∆ (x, y)− ε−

(ε+

ε

2

)> ∆ (x, y)− 3ε = 0,

We get a contradiction which completes the proof.

The next lemma considers the case where (x, y) lies on a horizontal or vertical segment of G.

Lemma 4 Suppose that G contains a vertical segment not entirely on the border: for some

a ∈ (0, 1), y : (a, y) ∈ G = [b, c]. Then ∆ (a, b) ≤ 0 and ∆ (a, c) ≥ 0, in particular, for some

y ∈ [b, c], ∆ (a, y) = 0. Similarly, suppose that G contains a horizontal segment not entirely on

the border: for some a ∈ (0, 1), y : (x, a) ∈ G = [b, c]. Then ∆ (b, a) ≥ 0 and ∆ (c, a) ≤ 0, in

particular, for some x ∈ [b, c], ∆ (x, a) = 0.

B-2

Proof. Suppose that G has a vertical segment with (x, y) where x = a, y ∈ [b, c] and suppose

a ∈ (0, 1). If there is y ∈ (b, c) such that (a, y) ∈ E for y < y and (a, y) ∈ F for y > y,

then it must be that ∆ (a, y) = 0. Indeed, a person (a, y) with y ∈ (b, c) expects to get

uE = ξE

(a+ y

2

)+ (1− ξE)ZE from choosing E and to get ξF

(y + a

2

)+ (1− ξF )ZF from

choosing F . Now, we have uE ≥ uF for y close to y but less than y and uE ≤ uF for y close toy but greater than y. Taking limits, we get ∆ (a, y) = 0.

Consider the case b > 0. Take a small ε and consider the point (x′, y′) = (a− ε, b+ ε). If ε is

small enough, then there is some b′ such that (a− ε, b′) ∈ G; for almost all ε, b′ is unique (takesuch ε) and, moreover, 0 < b′ < b; if we take ε small enough, then b′ will be arbitrarily close to b.

Now, E (fj | (ej , fj) ∈ E, fj = x′) = b′

2 and E (ej | (ej , fj) ∈ F, fj = y′) = a2 This player chooses

F in equilibrium. At the same time, he expects to get uE = ξE

(a− ε+ b′

2

)+ (1− ξE)ZE if he

chooses E and to get ξF(b+ ε+ a

2

)+ (1− ξF )ZF . Taking the limit ε→ 0, we get ∆ (a, b) ≤ 0.

Now consider the case b = 0. In this case, again take a small ε and consider the point

(x′, y′) = (a− ε, ε). Now, E (fj | (ej , fj) ∈ E, fj = x′) need not be well-defined, but in any case

it is some zx′ ≥ 0 and E (ej | (ej , fj) ∈ F, fj = y′) = a2 This player chooses F in equilibrium. At

the same time, he expects to get uE = ξE(a− ε+

zx′2

)+ (1− ξE)ZE if he chooses E and to

get ξF(ε+ a

2

)+ (1− ξF )ZF . Consequently, we have ξE (a− ε) + (1− ξE)ZE ≤ ξF

(ε+ a

2

)+

(1− ξF )ZF ; taking the limit ε→ 0, we again get ∆ (a, b) ≤ 0.

We can similarly prove that ∆ (a, c) ≥ 0. Thus, in any case, ∆ (a, b) ≤ 0 and ∆ (a, c) ≥ 0

and therefore there is (a, y) ∈ G with ∆ (a, c) = 0.

In the case of a horizontal segment y = a, x ∈ [b, c], we can similarly prove that ∆ (a, b) ≥ 0

and ∆ (a, c) ≤ 0, and thus there is (x, a) with ∆ (x, a) = 0.

In what follows, let r = ξE/ξF ∈ [0,+∞] (since we assumed that ξE and ξF are not equal

to zero at the same time). The next results shows that for 12 < r < 2, G is precisely the set of

points with ∆ (x, y) = 0.

Lemma 5 If r ∈(

12 , 2), then (x, y) ∈ Ωj \ ∂Ωj : ∆ (x, y) = 0 = G \ ∂Ωj. If r ≤ 1

2 or r ≥ 2,

then G is either a horizontal line or a vertical line.

Proof. Suppose r ∈(

12 , 2). Then the set of points with ∆ (x, y) = 0 is an upward-sloping

straight line with slope 2r−12−r ; moreover, ∆ (x, y) is strictly increasing in x and strictly decreasing

in y. Consequently, G may contain no horizontal or vertical segments, as this would contradict

Lemma 4. If so, Lemma 3 implies that all points (x, y) ∈ G \ ∂Ωj satisfy ∆ (x, y) = 0.

If r = 2, then the set of points with ∆ (x, y) = 0 defines a vertical line. In this case, G

cannot have a non-vertical upper-sloping part (by Lemma 3), and every vertical segment must

lie on the set ∆ (x, y) = 0. A horizontal segment could have one end on the line ∆ (x, y) = 0. If

B-3

this is the left end, then Lemma 4 implies that ∆ (x, y) ≤ 0 on the right end. However, this is

impossible, since ∆ is strictly increasing in x in this case. We would get a similar contradiction

if the right end of the horizontal segment satisfied ∆ (x, y) = 0. Thus, G is a vertical line.

If r > 2, then the set of points with ∆ (x, y) = 0 defines a downward sloping line. This means

that G may have only one point of intersection with this set, and then Lemma 3 implies that G

must be either a horizontal or a vertical line. If it is horizontal, then its left end should satisfy

∆ (x, y) ≥ 0 and its right end should satisfy ∆ (x, y) ≤ 0 by Lemma 4. Again, this contradicts

the fact that ∆ is strictly increasing in x. Thus, G is a vertical line in this case, too.

The cases r = 12 and r <

12 are considered similarly.

The next Lemma characterizes the conditions under which the curve G separating regions

E and F may be a horizontal or a vertical line.

Lemma 6 There exists an equilibrium for which the line G separating regions E and F is

the vertical line connecting points [a, 0] and [a, 1], where 0 < a < 1 if and only if a ∈[1 + ξF

ξE− 1

ξE, 2− ξF

ξE− 1

ξE

]. Similarly, there exists an equilibrium for which the line G sepa-

rating regions E and F is the horizontal line connecting points [0, b] and [1, b], where 0 < b < 1

if and only if b ∈[1 + ξE

ξF− 1

ξF, 2− ξE

ξF− 1

ξF

].

Proof. It suffi ces to prove the first part of the result, as the proof of the second part is completely

symmetric.

Under the conditions of the Lemma, we have ZE = a+12 + 1

2 = a+22 and ZF = a

2 + 12 = a+1

2 .

For (x, y) such that x > a, we have E (fj | (ej , fj) ∈ E, ej = x) = 12 ; for x < a, we can set this

value arbitrarily; denote it by E (fj | (ej , fj) ∈ E, ej = x) = zx. At the same time, for any (x, y),

E (ej | (ej , fj) ∈ F, fj = y) = a2 . For any (x, y) with x < a, we must have

ξE (x+ zx) + (1− ξE)a+ 2

2≤ ξF

(y +

a

2

)+ (1− ξF )

a+ 1

2; (B2)

in particular, this should hold for y = 0 and x arbitrarily close to a; since zx ≥ 0, we have

ξE (a) + (1− ξE)a+ 2

2≤ ξF

(a2

)+ (1− ξF )

a+ 1

2. (B3)

For any (x, y) with x > a, we must have

ξE

(x+

1

2

)+ (1− ξE)

a+ 2

2≥ ξF

(y +

a

2

)+ (1− ξF )

a+ 1

2; (B4)

in particular, this should be true for y = 1 and x arbitrarily close to a, therefore,

ξE

(a+

1

2

)+ (1− ξE)

a+ 2

2≥ ξF

(1 +

a

2

)+ (1− ξF )

a+ 1

2. (B5)

B-4

Now, (B3) is equivalent to ξEa ≤ 2ξE − ξF − 1, and (B5) is equivalent to ξEa ≥ ξE + ξF − 1.

This implies 2ξE − ξF − 1 ≥ ξE + ξF − 1, so ξE ≥ 2ξF , and in particular, since ξE and ξF are

not equal to 0 together, ξE > 0. Thus, a ∈[1 + ξF

ξE− 1

ξE, 2− ξF

ξE− 1

ξE

]. One can easily verify

that for any such a (provided that a ∈ (0, 1)) there is an equilibrium, where we define zx = 0

and for x = a, we let (x, y) ∈ E if and only if y < c, where c satisfies

ξE

(a+

c

2

)+ (1− ξE)

a+ 2

2= ξF

(c+

a

2

)+ (1− ξF )

a+ 1

2(B6)

(existence of such value follows from inequalities (B3) and (B5); it is unique if ξE > 2ξF .

The next Lemma characterizes conditions under which there is an equilibrium where (almost)

all types choose E.

Lemma 7 There exists an equilibrium where all types choose E if and only if 12ξE + ξF ≤ 1,

and there exists an equilibrium where all types choose F if and only if ξE + 12ξF ≤ 1.

Proof. It suffi ces to consider the case where everyone chooses E. We have ZE =

12 + 1

2 = 1. For any (x, y), we have E (fj | (ej , fj) ∈ E, ej = x) = 12 , and we can pick

E (fj | (ej , fj) ∈ E, ej = x) = zy arbitrarily. We can also pick ZF arbitrarily.

Player (x, y) can choose E in equilibrium if and only if

ξE

(x+

1

2

)+ (1− ξE) ≥ ξF

(y +

zy2

)+ (1− ξF )ZF .

This must be satisfied for x = 0, y = 1; we can find suitable zy and ZF (for example, zeroes) if

and only if

ξE

(1

2

)+ (1− ξE) ≥ ξF .

Thus, a necessary condition for such equilibrium is 1− 12ξE − ξF ≥ 0. At the same time, since

the type (0, 1) was most prone to deviation, this condition is also suffi cient.

The next Lemma restricts the possible separtion lines G if r ∈(

12 , 2).

Lemma 8 Suppose 1 ≤ r < 2. Then there exists (x, y) ∈ G such that y ≥ x, and if all

(x, y) ∈ G \ ∂Ωj satisfy y > x, then ξE < 23 . Similarly, suppose

12 < r ≤ 1. Then there exists

(x, y) ∈ G such that x ≥ y, and if all (x, y) ∈ G \ ∂Ωj satisfy x ≥ y, then ξF < 23 .

Proof. Consider the case 1 ≤ r < 2 (the case 12 < r ≤ 1 is considered similarly). By Lemma 3,

all points (x, y) ∈ G satisfy ∆ (x, y) = 0. Solving the equation ∆ (x, y) = 0 for y, we get

y =2r − 1

2− r x+ 2(1− ξE)ZE − (1− ξF )ZF

2ξF − ξE. (B7)

B-5

This defines an upward-sloping line for 2r−12−r ≥ 0, i.e., for r < 2. Moreover, since we assumed

r ≥ 1, we have 2r−12−r ≥ 1.

Suppose that all (x, y) ∈ G satisfy x ≥ y. Then we can denote the points of intersection of

G with ∂Ωj by (a, 0) and (1, b) (with a < 1, b > 0 because of the assumption of positive mass).

If so, we have b = 2r−12−r (1− a). We have

ZE =2

3+a

3+b

3=

2

3+

1− b 2−r2r−1

3+b

3= 1 +

r − 1

2r − 1b.

Since r ≥ 1, ZE ≥ 1. On the other hand, in this case, Pr (dj = E) = b(1−a)2 ≤ 1

2 ≤ Pr (dj = F ),

and thus ZF ≤ 0 We can rewrite ∆ (x, y) = 0 as

ξE

(x+

y

2− ZE

)− ξF

(y +

x

2− ZF

)+ ZE − ZF = 0, (B8)

which must be true for any (x, y), for which the indifference condition is satisfied, in particular,

(a, 0). A suffi cient condition for the left-hand side to be positive is

ξE (a− ZE)− ξF(a

2− 1)

+ ZE − 1 > 0

Substituting a = 1− 2−r2r−1b and ZE = 1 + r−1

2r−1b, we have

ξE

(− b

2r − 1

)+

1

2ξF

(1 +

2− r2r − 1

b

)+

r − 1

2r − 1b.

This is positive whenever

ξF

(−r +

1

2

(2r − 1

b+ 2− r

))+ r − 1 > 0.

Since b ≤ 1, this is positive if

−1

2ξF (r − 1) + r − 1 > 0.

Since ξF ≤ 0, this is true, and therefore such equilibrium is impossible.

Now, suppose that all (x, y) ∈ G \ ∂Ωj satisfy x < y. Then we can denote the points of

intersection of G with ∂Ωj by (a, 1) and (0, b) (with a > 0, b < 1 because of the assumption of

positive mass). If so, we have 1− b = 2r−12−r a. We also have

ZF =a

3+

2

3+b

3=a

3+

2

3+

1− 2r−12−r a

3= 1− r − 1

2− ra.

Since 1 ≤ r < 2, it must be that ZF ≤ 1. As before, since Pr (dj = F ) = a(1−b)2 ≤ 1

2 ≤Pr (dj = E), we must have ZE ≥ 1 and 1 − ZF ≥ ZE − 1, so ZE ≤ 2 − ZF . In equilibrium, we

B-6

must have (B8) satisfied for all (x, y) for which the indifference condition holds, in particular,

(0, b). The left-hand side is negative whenever

ξE

(b

2− 2 + ZF

)− ξF (b− ZF ) + 2 (1− ZF ) < 0

Substituting b = 1− 2r−12−r a and ZF = 1− r−1

2−ra, we have

1

2ξE

(−4r − 3

2− r a− 1

)+ ξF

(r

2− ra)

+ 2r − 1

2− ra.

This is negative whenever

1

2ξF r

(− (4r − 3)− 2− r

a+ 2

)+ 2 (r − 1) < 0. (B9)

Now consider two subcases. If r > 1, then, since a ≤ 2−r2r−1 (because b = 1 − 2r−1

2−r a ≥ 0), (B8)

holds if

−3r (r − 1) ξF + 2 (r − 1) < 0, (B10)

which is negative, provided that ξF >23r and r > 1. On the other hand, if r = 1, then b = 1−a,

and x < y for all points on G implies a < 2−r2r−1 strictly. Hence (B8) holds if (B10) is satisfied

weakly, which is true since r = 1 in this case. Therefore, if ξE = ξF r ≥ 23 , this case is impossible

in equilibrium.

B-7