Investing in the Peace: Economic Interdependence and ...egartzke/publications/gartzkeetal_io_01.pdf · Work by Robert O. Keohane and Joseph S. Nye, James A. Caporaso, Karl W. Deutsch,

Investing in the Peace EconomicInterdependence andInternational Con ictErik Gartzke Quan Li and Charles Boehmer

Research appears to substantiate the liberal conviction that trade fosters globalpeace Still existing understanding of linkages between con ict and internationaleconomics is limited in at least two ways First cross-border economic relationshipsare far broader than just trade Global capital markets dwarf the exchange of goodsand services and states engage in varying degrees of monetary policy coordinationSecond the manner in which economics is said to inhibit con ict behavior isimplausible in light of new analytical insights about the causes of war We discussand then demonstrate formally how interdependence can in uence statesrsquo recourseto military violence The risk of disrupting economic linkagesmdashparticularly accessto capitalmdashmay occasionally deter minor contests between interdependent statesbut such opportunity costs will typically fail to preclude militarized disputesInstead interdependence offers nonmilitarized avenues for communicating resolvethrough costly signaling Our quantitative results show that capital interdependencecontributes to peace independent of the effects of trade democracy interest andother variables

Students of world politics have long argued that peace is a positive externality ofglobal commerce Theorists like Montesquieu and Kant and practitioners likeWoodrow Wilson asserted that economic relations between states pacify politicalinteraction Mounting evidence in recent years appears to substantiate these claimsMultiple studies many identi ed with the democratic peace link interstate tradewith reductions in militarized disputes or wars1 While we concur with the evolvingconsensus we see existing analyses of economics and peace as incomplete On the

We thank Erick Duchesne James McCormick Neil J Mitchell Robert Pahre Jon Pevehouse WilliamReed Gerald Schneider two anonymous reviewers and the editors of IO for helpful commentsLawrence Broz and John R Oneal provided data We also thank Dong-Joon Jo for valuable researchassistance in data collection An earlier version of this article was presented at the 95th Annual Meetingof the American Political Science Association

1 See McMillan 1997 for a review of the literature See also the Journal of Peace Research 36 (4)July 1999

International Organization 55 2 Spring 2001 pp 391ndash438copy 2001 by The IO Foundation and the Massachusetts Institute of Technology

one hand a rich history of theorizing offers speculation addressing virtually everyaspect of the relationship between economics and con ict On the other handempirical studies of interdependence often adopt excessively narrow indicators ofeconomic activity It may be rewarding to take an intellectual step backmdashto brie yassess the broader theoretical question of how interdependence is likely to affectcon ict behaviormdashand then to examine promising aspects of the relationship usingmore appropriate indictors

We begin with a theory of disputes A valid explanation for the effect ofeconomics on peace must be placed in the context of an account of why most statesoccasionally resort to military violence Using a theory of dispute onset based onwork by James Fearon and others we deduce conditions under which interdepen-dence likely contributes to peace2 In contrast to conventional interpretations weshow that opportunity costs associated with economic bene ts generally cannotdeter disputes Instead interdependence creates the means for states to demonstrateresolve without resorting to military violence Liberal states more ably address theinformational problems that give rise to costly contests credibly communicatingthrough costly signals using nonviolent methods of con ict3

Our analysis calls for a notion of interdependence involving aspects of economicactivity besides trade Most studies of interdependence and con ict focus solely onbilateral or aggregate trade ows but interdependence through international capitalis substantially larger than exchanges of goods and services Capital markets linkaspects of domestic economies that otherwise have little global exposure Apreoccupation with risk leads capital to react to political violence in ways that arearguably both more sensitive and more unwavering States can trade with theenemy but political shocks to capital market equilibria invariably imply capital ight andor higher rents in the shadow of costly contests Peace may be a positivepolitical externality of commerce but risk is clearly a negative economic externalityof political contests Other macropolitical aspects of international economicsmdashsuchas the need for monetary policy coordinationmdashare also omitted in previous studiesof interdependence

Thus while accepting as valid the correlation between interdependence andpeace we seek to alter both the logic underpinning the observation and the scope ofindicators used in assessing the relationship Through a series of formal illustrationsand models we show that the opportunity cost conception cannot account for theimpact of interdependence on peace We also show that costly signaling offers asatisfactory alternative We broaden empirical assessment of interdependence byintroducing measures of other aspects of economy We test our ideas by replicatingthe work of a prominent research program on liberalism and peace4 Results support

2 See Blainey 1973 Fearon 1995 Gartzke 1999 and Wagner 20003 Morrow outlines an argument about trade interdependence that in important respects parallels our

own Morrow 1999 We develop our ideas independently derive them formally apply them morebroadly and offer empirical tests

4 See Oneal and Russett 1997 and Russett Oneal and Davis 1998

392 International Organization

our broader interpretation of interdependencemdashmonetary and nancial indicatorsare typically signi cant while standard measures of trade and especially jointdemocracy are marginal in their impact or insigni cant

Together these extensions form a comprehensive and theoretically satisfyingaccount of the relationship between interdependence and peace We review therelevant literature discuss the theory and derive predictions We then outline ourtests and present the results We conclude with speculation about implications ofsignaling and interdependence for globalization and peace

Existing Arguments About Economics and Peace

The literature on interdependence international con ict and the nexus of thesetopics is vast We survey work in four areas the democratic peace and tradecapital and monetary interdependence

The Democratic Peace

Scholarly attention has focused in recent years on the ldquodemocratic peacerdquo theobservation that liberal polities rarely ght each other5 though appearing about aslikely to engage in disputes generally6 Democracies behave differently toward eachother than toward nondemocracies7

Researchers initially sought to verify the statistical observation but work increas-ingly focuses on augmenting theoretical bases for the democratic peace A strongstrain in the literature argues that domestic political factors explain the relativeabsence of military violence among liberal states States sharing republican normsmay be more willing to bargain compromise and ful ll contracts than stateswithout these norms8 Alternately democratic institutions may constrain leadersfrom using force against leaders who are likewise constrained9 Still others contendthat in democracies domestic audiences or opposition groups force the revelation ofprivate information responsible for costly contests averting war10

Trade Interdependence

Democratic peace research was inspired by the Kantian prophesy of a ldquoperpetualpeacerdquo but Kantrsquos recipe (often called the ldquoliberal peacerdquo) consists of much broader

5 See Bremer 1992 and 1993 Bueno de Mesquita and Lalman 1992 Maoz and Russett 1993 Onealand Russett 1997 Rousseau et al 1996 Rummel 1997 Russett 1993 and Russett Oneal and Davis1998

6 See Benoit 1996 Ray 1997 Chan 1984 Rousseau et al 1996 and Weede 19847 The consensus is that the monadic effect of democracy on peace is at most much weaker than the

dyadic effect8 Dixon 1993 and 19949 See Babst 1972 Bueno de Mesquita and Lalman 1992 and Morgan and Campbell 1991

10 On domestic audiences see Fearon 1995 on opposition groups see Schultz 1998 and 1999

Economic Interdependence and Con ict 393

conditions including republican government a league of nations and commonmarkets11 Beginning in the 1970s students of political economy began to evaluateevidence that interdependence inhibits con ict behavior Debate continues butconsensus appears to be that interdependence is associated with peace

Work by Robert O Keohane and Joseph S Nye James A Caporaso Karl WDeutsch James M Rosenau and John A Kroll made conceptual contributions byclarifying de nitions of interdependence12 However these studies lack theoreticalprecision and fail to delineate key processes One is left to ponder the origins ofinterdependence Exactly how do multiple channels alter incentives to competeAlso complex interdependence appears to imply dyadic consequences but theargument as posed is almost exclusively systemic

There are many ways to conceive of interdependence The central logic of moststudies of con ict and interdependence is that states are less likely to ght if thereexist additional opportunity costs associated with military force ldquoInternationalcommerce being a transaction between nations could conceivably also have adirect impact on the likelihood of peace and war once again the [economic]interests might overcome the passions speci cally the passion for conquestrdquo13

Evidence has mounted that trade interdependence reduces interstate disputes14

John R Oneal and Bruce M Russett argue that Kant was rightmdashliberalism leads topeace15 In addition to interdependence law civil liberties executive constraintsand a bargaining culture all reduce disputes Interdependence has a greater effectthan democracy growth or alliances in reducing con ict in contiguous states

However ldquotheoretically liberalism does not specify what types of con ict aremost likely to decrease in the presence of high levels of interdependencerdquo16 Gartzkeand Dong-Joon Jo nd that while liberal dyads are less likely to engage inmilitarized con ict they have more nonmilitarized con icts17 Mark J Gasiorowski nds that short-term capital ows increase con ict while trade reduces con ict18

Gasiorowski and Mary Ann Tetreault emphasize that the quantitative literaturemeasures not interdependence but interconnectedness19 Trade ows alone may notbe an optimal measure of interdependence

Other recent work directly challenges the validity of research on the trade-con ictnexus Using a measure of interdependencebased on the salience of trade Katherine

11 Kant [1795] 195712 See Caporaso 1978 Deutsch 1978 Keohane and Nye 1989 Kroll 1993 and Rosenau 198413 Hirschman 197714 See Oneal et al 1996 Oneal and Russett 1997 and 1999a Polachek 1980 and Polachek Robst

and Chang 199915 See Oneal and Russett 1997 and 1999a Though not directly related to our research other studies

assess the effect of con ict and alliances on trade See Gowa and Mans eld 1993 Mans eld and Bronson1997 and Pollins 1989

16 McMillan 1997 5417 Gartzke and Jo 200018 Gasiorowski 198619 See Gasiorowski 1986 and Tetreault 1980


Barbieri nds that trade increases con ict20 Like others we suspect that Barbierirsquosmeasure of interdependence is responsible for her ndings21 Oneal and Russettassess differences between the two programs They nd that positive trade-con ictrelationships are isolated to nonrelevant dyads

Barbieri and Gerald Schneider are concerned about discrepant ndings and warnthat bias may be a product of tainted trade data They question the reliability ofexisting empirical ndings22 Barbieri and Jack S Levy provide evidence that statesoften trade with the enemy while at war and suggest that liberalism and realismreconsider expectations regarding interdependence and con ict23 Han Dorussendemonstrates that trade has a pacifying effect on interstate con ict mainly whenthere are minimal barriers to trade and few states in the system24 Numerouspotential trading partners combined with barriers increase the incentives to engagein military contests

In a project that anticipates aspects of this study James D Morrow offers acoherent basis for questioning the statistical association between trade and con ictHe begins by outlining an explanation for the causes of international crises anddisputes and provides two reasons why trade and con ict may not interact the wayresearchers typically expect25 First because rms anticipate con ict between stateswith volatile relations trade will be reduced ex ante where the risk of con ict isgreatest26 Thus trade and con ict are both endogenous states will not be deterredfrom con ict if the threat of con ict deters trade Second the deterrent effect oftrade should be modest Any factor that discourages aggression by one partyencourages aggression in others States can use trade to signal informing others bydemonstrating a willingness to pursue costly acts (harming trade)

Finally interdependence may affect con ict indirectly by transforming statepreferences in such a way that states no longer desire to compete Etel Solingenargues that domestic coalitions with internationalist preferences may forge cross-national bonds at the regional level facilitating greater economic interdependenceand prosperity The efforts of domestic internationalist coalitions to act in concertmay in turn improve their stability and in uence in domestic politics27 Statepreferences will converge producing regional zones of peace Still peace may notfollow from interdependence between status quo and revisionist states Paul APapayoanou contends economic linkages act as signals of resolve and credibility28

Because domestic economic actors in status quo states only support con icts that

20 Barbieri 199621 See Oneal et al 1996 and Oneal and Russett 1999a22 Barbieri and Schneider 199923 Barbieri and Levy 199924 Dorussen 199925 Morrow 199926 Bueno de Mesquita found that during the Seven Weeksrsquo War money markets signaled expectations

about the risk of a contest as well as the probability that Prussia would win (and thus that Denmark wouldlose) Bueno de Mesquita 1990

27 Solingen 199828 Papayoanou 1999


protect their interests these states are more easily constrained from balancingagainst revisionist states with which they share economic relations If confrontationsarise revisionist states may threaten to disrupt economic relations increasingopportunity costs for status quo states

Monetary Interdependence

Monetary interactions may also be a source of interdependence States may chooseto subordinate monetary sovereignty to a foreign power through a xed exchange-rate regime pool sovereignty in a monetary union or assert their own sovereigntyunder a oating exchange-rate regime29 Interstate monetary relations can becharacterized by intermittent cooperation competition and coercion30 Attempts byone state to increase its monetary authority (a relative gain) may produce ldquopublicbadsrdquo that diminish absolute gains Hence monetary interaction may be consideredas part of the general notion of economic interdependence31 Although they reducestate autonomy in monetary policymaking higher levels of monetary dependenceraise the incentives to cooperate

Capital Interdependence

In Spirit of the Laws Montesquieu argues that ldquomovable wealthrdquo encourages peacebetween and within states Mobile capital constrains the sovereign domesticallyldquoThe richest trader had only invisible wealth which could be sent everywherewithout leaving any trace [so that] rulers have been compelled to govern withgreater wisdom than they themselves would have thoughtrdquo32

Trade is only one manifestation of the global spread of capitalism33 Since capitalmarkets dwarf the exchange of goods and services rms should weigh the risks ofinvestment much more heavily than trade Foreign production facilities are vulner-able to nationalization in a way that trade is not Further even the threat of lostrevenues makes investors skittish Globalization has increased capital mobility andmonetary cooperation even as it rede nes the terms on which states compete

State policies aim to preserve political autonomy but states are faced with adilemma when seeking to in uence interstate nance Vittorio Grilli and Gian MariaMilesi-Ferretti suggest that states impose capital controls for four reasons limitingvolatile short-term capital ows retaining domestic savings sustaining structuralreform and stabilization programs and maintaining the tax base34 States engage in

29 Cohen 199830 Kirshner 199531 See Hamada 1976 and Suzuki 199432 Montesquieu [1748] 1989 389 Montesquieu also discusses the political effects of currency

market speculation33 Andrews 199434 Grilli and Milesi-Ferretti 1995


nancial repression for similar reasons35 However states may nd capital controlsless useful when facing integrated capital markets and pressures to liberalize frompowerful interest groups such as multinational corporations and nancial institu-tions36 Edward L Morse points out that when states fail to reduce their vulnera-bility and solve the crises that arise through interdependence they may seek toexternalize problems37 Interdependence may even transmit economic crises38

Thus the literature suggests that interdependence could increase con ict betweenstates while decreasing the chances of violent militarized behavior

Theory Economic Interdependence and Peace

The literature generally reports that states that trade are less likely to ght Stillthere is substantial room to expand and to better account for the processes linkinginterdependence to disputes Our argument follows in three stages We rst offer atheory of contests based on recent developments in the literature on war39 Discus-sions of this logic appear elsewhere so we focus here on intuition We use simpleformal examples in the text and provide rigorous formal proofs in the appendix Wethen use the theory of contests to demonstrate how interdependence affects milita-rized behavior We show that conventional opportunity cost accounts of the effectof interdependence on disputes are inconsistent with a logic of costly contests butthat an alternative explanation based on signaling does anticipate paci c effectsDiscussion is again informal with a proof in the appendix Finally we coversupporting topics needed to link signaling to capital and monetary processes

Why States Fight A Theory of Costly Contests

Explanations for war are legion However work by James Fearon and others showsthat most purposive theories of war are internally inconsistent in that they do notaccount for the behavior of interest40 Fearon points out that theories of warcommonly con ate the motives for con ict with the choice of method for con ictresolution Costly contests involve at least two elements First there is zero-sum

35 Leblang 199736 See Frieden 1991 and Goodman and Pauly 199337 Morse 197638 Ibid 12939 See for example Fearon 1995 Gartzke 1999 and Wagner 2000 Morrow offers a parallel

account Morrow 199940 Fearon identi es three explanations why purposive unitary states in con ict fail to achieve

bargains ex ante that they accept ex post We address only the rst explanation here For a more detaileddiscussion see Gartzke 1999


competition for an excludable good41 States differ over issues or territory that eachcannot possess simultaneously Second states choose a settlement method Thechoice of method is non-zero-sum Transaction costs deprive ldquowinnersrdquo of bene tsand increase the burden for ldquolosersrdquo so that all are better off selecting methods thatminimize costs Since war is expensive ghting makes sense only if equivalentsettlements cannot be obtained using cheaper methods A theory of war thenexplains why ef cient settlements are at times unobtainable ex ante

Fearon follows Geoffrey Blainey in arguing that wars result from uncertaintyabout conditions likely to in uence eventual settlements as well as incentives stateshave to misrepresent these conditions42 States possess private information aboutstrategic variables (capabilities resolve and so on) If states could credibly shareprivate information ef cient ex ante bargains could be identi ed Instead uncer-tainty provides weak or unresolved states an opportunity to conceal weakness evenas competition creates incentives to bluff States ldquopoolrdquo claiming to be resolved andcapable regardless of their true nature Such ldquocheap talkrdquo claims do not allowobservers to differentiate resolved or capable opponents from the weak or unre-solved Only by imposing costly contestsmdashby ghting or similar actsmdashcan statesdistinguish resolute opponents from those seeking to bluff States ght largelybecause they cannot agree on bargains that each prefers to what each expects toobtain from ghting If states can agree about the nature of eventual settlementsthen there is always some mutually preferable bargain Therefore uncertainty aboutthe allocation of spoils from the contest accounts for the contest itself

Imagine that two states compete over a sum of money (say $100)43 States keepany division of the stakes but pay a fee for ghting (say $20) Suppose that thewinner gets all the money (minus costs) and the loser gets nothing (again minuscosts) For simplicity assume that states have equal chances of victory Thus ghting has an expected value of [05($100) + 05($0) ndash $20] or $30 If states arerisk neutral then any offer yielding each side at least $30 is at least as valuable as ghting Negotiated settlements from ($31 $69) to ($69 $31) are available andpreferred to war44

Suppose instead that war costs remain $20 for one state (A) but that the other state(B) has private information about its costs (c) Suppose that war costs for states likeB range from $0 to $40 Obviously if Brsquos costs are high then A can make ademanding proposal B thus has an incentive to bluff A in turn recognizes that B will

41 Excludability means that property rights are enforceable To the degree that a good is nonexclud-able states stand to internalize bene ts regardless of whether they ght There are few incentives toengage in costly contests to obtain what cannot be denied even the loser Realism implies that rivalry isnot necessary for con ict States that care about relative gains still have incentives to ght even if oneactorrsquos consumption does not diminish the consumption of others

42 Blainey 1973 For a brief and highly intuitive account of the argument see also Morrow 199943 Proofs in the appendix are more general Examples in the text are meant only to illustrate the

arguments44 Risk propensity affects the size of the Pareto space (the range over which bargains are preferable

to the lottery of war) Risk aversion increases the range of acceptable bargains while risk acceptancereduces the Pareto space


claim low costs whether or not Brsquos costs are modest A thus cannot infer anythingmeaningful from Brsquos ldquocheap talkrdquo claims45

For simplicity imagine that A makes a demand (d) that B either accepts (givingA [$100 ndash d]) or rejects leading to war If A knows Brsquos cost then Arsquos optimal offeris one that B just prefers to ghting If A is uncertain what B prefers then Arsquosoptimal offer is the best demand of each type of B weighted by the odds that a giventype is the real opponent Technically Arsquos demand equals the integral over thedistribution of types of Brsquos marginal product for ghting Arsquos offer is thus linked tothe central tendency of opponent types46 If Brsquos costs are distributed uniformly thisimplies that A offers $30 Types B with lower than average costs ght whereashigh-cost types B accept Arsquos offer

Fearonrsquos reassessment renders a nite set of necessary conditions for costlycontests States must have incentives to compete This rst set of conditions iswidely recognized and is referred to by several labels (opportunity and willingnesscapability and resolve probability and utility)47 Yet while the rst set of conditionsimplies con ict these conditions do not determine the method by which con ict isresolved Costly contests only follow when states are also uncertain about likelyconsequences of contests The second set of conditions requires that states possessprivate information about some aspect of the rst set of conditions and that states beunable to credibly dissipate uncertainty

Now that we have identi ed conditions leading to costly contests explaining theabsence of such contests in certain dyads is tantamount to removing a necessarycondition In the model wars are unlikely if ghting is prohibitively expensive (warsare anticipated to cost more than the payoffs) or if information about strategicvariables is revealed Interdependence can contribute to peace by making costlycontests more costly than the stakes or by revealing private information for at leastone of the actors We show later that revelation of private information is the likelymechanism

Why Some States Do Not Fight Contribution(s) ofInterdependence to Peace

For interdependence to promote peace economic processes must either removeincentives for states to engage in con ict or reduce the uncertainty states face when

45 Cheap-talk signals can be informative if costs are applied conditionally See Fearon 1994 Sartori1996 and Smith 1998 Sartori points out that states do appear to communicate credibly through cheaptalk (diplomacy and so on) Yet diplomats are notorious for bluf ng (someone once referred to adiplomat as a gentleman who lies for his country) While reputation or audience costs add to thecredibility of cheap talk credible communication is clearly problematic Here we adopt the useful ctionthat the effects of cheap-talk signaling can be normalized to zero while noting that our results are general

46 Gartzke 199947 Rationalist theories can motivate costly contests with full information by imposing restrictive

assumptions about actorsrsquo ability to bargain Examples include Bueno de Mesquita 1981 and Bueno deMesquita and Lalman 1992


bargaining in the shadow of costly contests Since removing incentives to actaggressively only increases incentives for opponents the former explanation musttypically occur in special ldquoboundaryrdquo conditions (discussed later) We argue thatinterdependence makes it easier to substitute nonviolent contests for militarizeddisputes in signaling resolve States that possess a range of methods of con ictresolution have less need to resort to the most destructive (and costly) techniquesLiberal dyads can damage mutually valuable linkages to communicate crediblyStates without linkages must choose between a very limited set of optionsincludingmdashmore oftenmdashwar The con ict model with uncertainty shows why thisis so

Recall that Arsquos best response is an offer that an opponent weakly prefers to ghting If the opponent (B) has private information about its war costs (c) then Arsquosoptimal offer derives from a rational guess (the distribution of reservation prices fordifferent types of player B) A calculates its offer as the best demand it can make toeach opponent weighted by the odds that a given opponent ldquotyperdquo is the actualadversary Players B whose war costs are high accept whereas those with low costs ght

Conventionaldescriptions of interdependence see war as less likely because statesface additional opportunity costs for ghting The problem with such an account isthat it ignores incentives to capitalize on an opponentrsquos reticence to ght If anopponent (B) is reluctant then state A can make larger demands without risking warAssume that interdependent dyads are those that derive some bene t from economiclinkages (h say h = $10) If A and B avoid a ght then each receives the settlementplus the bene t ($100 ndash d + $10 and d + $10 respectively) Brsquos war costs are againbetween $0 and $40 Conventional explanations for interdependence identify thefact that B receives (d + $10) instead of (d) for accepting Arsquos demand as leading topeace If demands are the same then not ghting is more bene cial in interdepen-dent dyads and B should more often prefer Arsquos demand to ghting Yet unless weassume that A is ignorant of its own interdependencewith B (not very plausible) Arsquosdemand must be different Arsquos best offer is one that B just prefers to a ght Sincebene ts increase under interdependence A simply demands commensurately moreIn the previous example A offers $30 (A receives $100 ndash d = $70) If interdepen-dent A proposes that B accept $20 plus the bene t ($10) The same range of statesB that accepted $30 previously (since $30 ugrave $50 ndash c if c ugrave $20) now accepts $20(since $20 + $10 [the bene t] ugrave $50 ndash c if c ugrave $20) State A again makes an offerthat a given opponent just prefers to ghting weighted by the odds that B is thegiven opponent Interdependence is simply subsumed in bargaining Since they failto reduce uncertainty opportunity costs generally do not alter the prospects ofengaging in costly contests48

Economic interdependence can motivate peace in two ways First con ict mayoccasionally be so expensive relative to the expected value of ghting that states

48 The result is general to symmetric bargains (see proposition 4 in the appendix)


prefer any offer rather than enduring a contest Suppose Brsquos war costs range from$50 to $90 Brsquos expected value for war thus ranges from $0 to ndash$40 Because Bstands to lose more from ghting than its value for the stakes B prefers to concedeWe refer to this as a boundary solution because it is possible only by assuming thatstakes in the contest are bounded Bounded stakes are reasonable especially whenissues are of tertiary importance or when costs are extreme (as in nuclear war)Interdependent dyads may avoid costly contests if economic linkages decrease theexpected value of competition to the point where one party prefers conceding tocompeting Yet economic bene ts seldom equate in consequence to nuclear warIssues over which states may consider major contests are unlikely to meet boundaryconditions for interdependence Instead boundary solutions are relevant whenliberal states experience relatively minor con ict Finally competition can continueeven given boundary conditions Liberal dyads deterred from war can still competeby manipulating the risk of contests49

Second instead of deterring con ict interdependence can convey credible sig-nals obviating the need for costly military contests Actorsrsquo behaviors potentiallyinform observers about the value of strategic variables dissipating private informa-tion Interdependent states that endure opportunity costs in pursuit of politicalobjectives differentiate themselves from other less resolved competitors To thedegree that nonviolent con ict allows observers to identify opponents costlysignaling also allows ef cient ex ante bargaining States seek to obtain settlementswhile competing for preferable terms War is less often necessary when statespossess nonviolent methods that credibly inform

Suppose that in the earlier model state B can choose to preempt interdependence(h) with a sanction (s) B rst chooses whether to eliminate interdependence A thenmakes a demand which B accepts or rejects The solution to this signaling game istechnical and is detailed in the appendix The intuition however is simple StatesB that are unresolved (high war costs) retain the bene ts of interdependenceUnresolved states prefer interdependence and a demanding settlement to ghtingResolved states (low-war-cost states) prefer to ght rather than to accept a poorsettlement Since states willing to ght expect to forfeit the bene t these states aremore willing to signal destroying interdependence at the outset Costly signalscredibly inform A about opponent resolve A in turn makes a more generous offerto resolved opponents so that they prefer the settlement to ghting50

49 Schelling notes that any contest can be converted into a lottery over additional outcomes byplaying brinkmanship games so that even contests that are unacceptably costly (for example generalnuclear war) are imaginable as risky probabilistic consequences of competition Schelling 1960 and 1966

50 The result is general (see proposition 4 in the appendix) Banks offers a proof implying that theresults are generalizable to the entire class of bargaining games in which offers are endogenous andunrestricted Banks 1990


How Interdependence Functions in the Context of Politics

The task remaining is to link con ict behavior (or its absence) to economic activityEconomic ties between states may be portrayed as lying along a continuum frompure autarky (closed economies) to perfectly integrated (one economy) It makes nosense to discuss the role of interstate economics under autarchy Instead foreconomies that are not closed we must delineate the channels through which statesare interrelated We distinguish between market interactions and policy interactions

Economic exchanges between market agents occur in a particular market Ifmarket agents happen to belong to different national political entities then theireconomic exchanges relate states in that market For the processes of markets toin uence con ict behavior two conditions must hold First states must be able tointercede to obstruct economic bene ts This may seem a trivial point but manyeconomic exchangesmdashsuch as trade between an opponent and third partiesmdashmaynot be readily obstructed or may only be interfered with under extreme conditions(violations of the law of the sea and so on) Second bene ts must generally bejointly contingent States can technically signal through almost any costly act butdoing so makes little sense given incentives to compete A man may seek todemonstrate his resolve by severing a limb but subsequent efforts to bargain will behindered by the fact that his opponent retains all his initial appendages51 The manis better off proving his resolve by severing the opponentrsquos limb To the degree thatone is able an actor should attempt to demonstrate resolve in a manner that does notdamage the actorrsquos own bargaining position

While trade may or may not be affected by political shocks we argue that capitalmarkets are particularly vulnerable to interventionAssume an equilibrium in capitalmarkets at time t Next imagine that a political shock occurs that increases risk(investments yield a lower return or principal is in greater danger) Re-equilibratingsuch a system requires capital out ows andor larger returns If political shocksimpose more costs on users of capital then to the degree that states rely on capitalmarkets for prosperity they also rely on political stability Political shocks thatendanger capital also threaten economic prosperity The more interdependent statesbecome the greater the effect on capital markets of small changes in political riskWe have shown that opportunity costs cannot alter the probability of costly contestsInstead a willingness to endure costs in pursuit of political goals demonstratesresolve and informs opponents Capital interdependence promotes peace by allow-ing states to engage in costly signaling and reducing the need to resort to violenceto obtain settlements

51 A reviewer notes that the example is reminiscent of the Black Knight scene in Monty Pythonrsquos TheHoly Grail The Black Knight guards a small bridge ldquoNone shall passrdquo he exclaims as King Arthur andthe Knights of the Round Table approach The problem is that previous contests have left the BlackKnight short several limbs While no one doubts the Black Knightrsquos resolve his costly signals have alsoeffectively undermined his bargaining position Hence the humor


Events surrounding the Agadir crisis may serve to illustrate the argument Thegold standard prior to World War I was based on the commitment by three majorpowers (Britain France and Germany) to maintain their currency convertibility intogold The three states attached top priority to defense of the gold standard to ensurebene ts from a stable international monetary system a commitment that wascredible among investors52 In June 1911 France sent troops to the Moroccancapital Fez ostensibly to protect European residents Germany saw the move as aneffort to further French claims in Morocco and sought compensation Negotiationsbroke down in July and war seemed imminent At this point France and Britainbegan withdrawing funds from German banks leading to a nancial crisis inSeptember that threatened the solvency of the German currency and risked suspen-sion of the gold standard In November the Germans decided to concede signing atreaty recognizing Morocco as a French protectorate53 Britain and France were ableto demonstrate their resolve short of military violence through a costly signal thatthreatened a common economic asset

At the macroeconomic policy level states construct scal and monetary policiesto manage aggregate demand and supply For nonautarchic economies statesrsquomonetary and scal policies produce externalities that are transmitted to foreigneconomies through trade and exchange rates To manage such externalities statescoordinate bilateral and international monetary arrangements When statesrsquo marketsare integrated monetary policy autonomy in one state depends on the type ofexchange rate regime If state A pegs its currency to state B A must then pursue thepolicy of state B Yet the two states have asymmetric in uence As anchor currencyB enjoys policy autonomy but abuse of policy independence leads to externalimbalances that threaten the credibility of a xed-rate regime B has a stake over thelong run in maintaining the currency peg54 Manipulation of the relationship is thusrisky and can demonstrate resolve to the degree that coordination is valuable55

By increasing the economic interdependence of members monetary policycoordination creates a mechanism that allows credible signals of political resolvethrough economic acts For example the postndashWorld War II Bretton Woods systemwas essentially a zone in which members pegged currencies to the dollar expectingconvertibility of the dollar into gold On 26 July 1956 Egyptian president GamalAbdel Nasser nationalized the Suez Canal On 31 October British and French forcesattacked Egypt after negotiations to resolve the crisis failed Despite vocal opposi-tion from the United States Britain and France decided to continue efforts to seizethe canal and overthrow the Nasser regime On 5 November the US governmentstarted to sell pounds British reserves fell by 15 percent within a month US

52 Eichengreen 199653 Kirshner 199554 Simmons offers examples of governments torn between international commitment to the gold

standard and domestic needs for more policy autonomy during the interwar years Simmons 199455 Gilpin argues that the Bretton Woods system failed due to irresponsible US economic policies

Gilpin 1987


Treasury secretary George Humphrey informed Britain that unless it obeyed the UNresolution and withdrew from Suez the United States would continue to sell poundsand block British access to International Monetary Fund (IMF) reserves On6 November Britain ordered a cease- re in effect forcing the French to end militaryoperations as well56 It may be questioned whether the United States would haveintervened militarily to block British and French efforts in Egypt fortunately suchan effort was not necessary The Bretton Woods system made it possible for theUnited States to demonstrate resolve short of military force jeopardizing valuableeconomic linkages but averting the need for costlier actions

In summary traditional studies of economics and international con ict only payattention to a particular type of market the goods market and one channel ofeconomic linkages international trade The in uence of economics is underesti-mated and the causal mechanisms overly simpli ed We argue that linkages betweeneconomics and peace are more complex than previously postulated Interdepen-dence through capital and trade acts as a costly signal reducing uncertainty aboutrelative resolve and lessening the need for militarized disputes Instead of beingdeterred by opportunity costs interdependent states can use opportunity costs ascostly signals demonstrating resolve

Research Design

A large body of literature shows that economic interdependence is related tointerstate con ict even after controlling for a variety of factors We build on thisbody of work speci cally we adopt most aspects of the research design representedby Oneal and Russett and Russett Oneal and Davis57 By using largely the samemethods data and research design we offer a clear look at the implications of ouranalysis Our sample and unit of analysis is thus politically relevant dyad years1951 to 198558

In adopting many aspects of the research design in Oneal and Russett andRussett Oneal and Davis we also inherit limitations The data are limited tocontiguous and major powers in a period largely coterminous with the Cold War aperiod of rising trade and nancial interdependence and increased monetary policycoordination Although we believe our theory is general readers are nonethelesscautioned to keep these data limitations in mind when interpreting our resultsFuture research may collect data for earlier periods or test our argument in broaderempirical contexts59

56 Kirshner 199557 See Oneal and Russett 1997 and Russett Oneal and Davis 199858 Oneal and Russett describe the temporal domain as 1950ndash85 but the data contain no observations

for 1950 Oneal and Russett 199759 Integrated capital markets are arguably a feature relatively unique to recent decades Data

collection in the prendashWorld War II era are unlikely to alter the results reported here because so manydyads would report ows and stocks near zero


The Dependent Variable

The dependent variable is the onset of a militarized interstate dispute (MID) coded1 for any dyad year in which a threat display use of force or war begins and zerootherwise (including the subsequent years of a multiyear dispute)60 We use data byZeev Maoz (DYMID10) which offers an authoritative dyadic coding and correctsa number of coding errors in previous versions of the MID data61

Construction for our dependent variable is novel in other ways Russett Onealand Davis look at MID involvement (the presence or absence of a MID in a givendyad year)62 Since statistical models assume cases are independent we preferassessing MID onset rather than involvement As a result our sample contains 622dyad years of MID onset whereas Russett Oneal and Davis report 947 dyad yearsof MID involvement a difference of 325 dyad years Results using the RussettOneal and Davis dependent variable are included for comparison but are discussedonly brie y63

Signaling suggests that interdependent states avoid escalating crises to militarizedthreats or violence Ideally analysis would involve assessing the presence orabsence of signals followed by the presence or absence of escalated behaviorUnfortunately existing data do not allow for direct assessment of escalation MIDsare coded based on the most intense dispute behavior for each actor in a crisisAssessing escalation with MIDs is thus highly problematic Instead we anticipatethat much of the behavior indicative of signaling occurs below the threshold ofMIDs Signaling is assessed indirectly by associating the presence or absence of theability to signal with changes in the dispute propensity of dyads in the form of theprobabilistic presence or absence of MIDs

Measuring Monetary Interdependence

A state may peg its currency to a precious metal or a foreign currency engage in acooperative arrangement with a group of other countries to maintain its exchangerate within a certain ldquobandrdquo or oat the currency to allow the market to determineits equilibrium exchange rate Whatever exchange-rate regime a state chooses theregime links national economies so that local economic shocks and policies haveforeign externalities The choice of exchange-rate regime implies different degrees

60 For discussions of the MID data see Gochman and Maoz 1984 and Jones Bremer and Singer1996

61 Maoz 199962 See Russett Oneal and Davis 1998 and Oneal and Russett 199763 Russett Oneal and Davis appear to code disputes using initial and terminal years of the dispute

as a whole rather than from individual participants Russett Oneal and Davis 1998 Coding errors resultfrom the entry and exit of actors in multiparticipant disputes of greater than one-year duration Forexample Thailand and the Soviet Union are considered a dispute dyad from 1964 to 1975 (dispute 611Vietnam War) even though the two statesrsquo participation in the contest never overlaps (Soviet partici-pation begins on 25 February 1964 and ends on 18 April 1965 Thailand 21 September 1967 to 28January 1973) Russett Oneal and Davis report 160 ldquodisputesrdquo between actors not coded as disputes inthe MID data (they also code 165 subsequent-year disputes) Russett Oneal and Davis 1998


of monetary interdependence A peg demands greater interstate commitment and anassociated loss of autonomy Pegging makes it easier to exchange currencies and forthe country to maintain price stability Yet a state that pegs its currency to a foreigncurrency relies heavily on the economic management of the foreign economyTherefore states maintaining xed exchange rates face a double-edged sword Theregime may facilitate exchange and provide incentives to avoid con ict butasymmetry may also increase uncertainty about policy acts and ultimately fail todeter disputes Similarly states embarked on a cooperative exchange-rate arrange-ment such as the European Monetary System (EMS) have greater commitment toeach other compared with an independently oating system

We see three relevant aspects of currency areas and monetary pegs for signalingFirst such arrangements can demonstrate resolve States that possess bene cialregimes can reveal information about the relative value of competitive politicalobjectives by threats or acts that jeopardize the status quo Second integrating oneeconomy with others restricts a statersquos ability to shelter itself from negativeeconomic consequences of political shocks A state must not only stand to losewealth in signaling but also be unable to avert this loss Finally it is important tonote that other factors that historically lead to monetary integration confound thesignaling effect of regimes Old colonial or imperialist ties proximity dependencyand other factors may cause developing states to participate in regimes but resentparticipation The underlying factors may be correlated with strong motives forcon ict so that signaling appears to fail when in fact it mitigates disputes Monetarypolicy coordination is theoretically consistent with and supportive of the signalingargument Further we see this role as expanding in the future as states likeArgentina peg their currencies for purely economic purposes

We construct two variables to measure the impact of monetary interdependenceon interstate con ict The rst variable PEGGING measures the existence coded 1 orabsence coded zero of pegging between the anchor currency and another currencySince the 1960s eleven currencies have served as anchor currencies Our sampleinvolves the four largest anchor currencies (the American dollar the British poundthe French franc and the South African rand) PEGGINGrsquos expected sign depends onwhich effect is stronger signaling (pegging is negatively correlated with disputes)or the selection effects associated with pegging which are positively correlated tointerstate disputes

JOINT CURRENCY AREA measures whether states in a dyad peg their currencies to thesame anchor currency coded 1 or not coded zero The measure involves membersof the four currency areas and one cooperative arrangement system (the EMS)during the sample period This variable captures both currency area linkage and theeffect of the cooperative arrangements64

JOINT CURRENCY AREA is expected to reducethe onset of interstate con ict (a negative sign)

64 Ideally currency area membership and cooperative arrangements could be tested separately Wecannot estimate the effect of the EMS alone There is no variance in dispute behavior because no MIDsoccur among EMS members JOINT CURRENCY AREA when purged of EMS dyads is still signi cant


Data for these two variables are obtained from the IMFrsquos Annual Reports onExchange Arrangements and Exchange Controls and the International FinancialStatistics Yearbooks Data for the exchange-rate regime variables start from 1966Inclusion of PEGGING and JOINT CURRENCY AREA leaves 10399 usable cases (out of theoriginal 20990)

Measuring the Impact of Capital Investment

Capital seeks higher risk-adjusted returns Risk is contingent on governmentrestrictions the degree of domestic capital market integration into world marketsand the overall exposure of the economy to direct investments This has threeimplications for international con ict First states in con ict may place morestringent government restrictions on foreign exchange payments settlement capitalrepatriation or even nationalization Since con ict threatens investments amongdisputing states it makes such investments less desirable and capital becomesrelatively scarce Second political shocks produce negative externalities affectinginvestments Military con ict increases uncertainty and risk to any capital invest-ment all else being equal and reduces risk-adjusted rates of return The moreglobally integrated a statersquos capital market the more likely that capital will eeThird states that are heavily exposed to capital ows are more vulnerable todisruptions Policy actions that increase risk for capital are costlier for politicalleaders and thus demonstrate stronger resolve

States that are heavily dependent on international capital markets for nationaleconomic well being are much more vulnerable to the will of these markets Statescan disengage their economies from the global system They can also seek to restrictthe movement of capital across their borders However attempts to limit thein uence of international markets on domestic economies also limit growth Statescannot restrict the free movement of capital without raising the cost of production

We construct two measures to capture dyadic levels of governmental opennessand exposure to capital CAPOPENL measures eight types of government restrictionson foreign exchange current and capital account transactions Restrictions includelimits on payments for capital transactions limits on payments for current transac-tions prescription of currency import surcharges advance import deposits surren-der requirements for export proceeds bilateral payments arrangements with IMFmembers and bilateral payments arrangements with nonmembers65 We computethe difference between eight (the maximum number of restrictions) and the sum ofrestrictions for each country in a dyad We then follow the weak link assumptionused by Russett Oneal and Davis The lower of the two monadic values in a dyadmeasures the openness of a dyad to capital investments We expect the variable tohave a negative effect on disputes

65 These data are collected from summary tables of the IMFrsquos Annual Reports on ExchangeArrangements and Exchange Controls since 1966


CAP FLOWSL indicates a statersquos exposure to foreign direct investments (FDI)measured as the ratio of gross FDI over gross domestic product (GDP) in purchasingpower parity (PPP)66 Direct investments include equity capital reinvestment ofearnings and other long- and short-term capital The measure is lagged one periodto control for endogeneity (disputes may repress current and future investments)We use the lower of the two monadic values in a dyad to measure the exposure ofa dyad to capital investments CAP FLOWSL is expected to negatively affect the onsetof disputes Data are from the World Bankrsquos World Development Indicatorsdatabase67 The variable contains many missing cases Including the variable dropsthe sample size to about 2200 dyad years so results should be interpreted with somecaution

Control Variables

We include control variables from Russett Oneal and Davis We also add temporalspline variables for duration dependence and later include a control for endogenouspreferences among states

DEPEND measures trade interdependence in a dyad It equals the lower of the twomonadic trade dependence scores (bilateral tradeGDP) in the dyad for a given yearlagged one year to avoid reciprocal effects of con ict on trade

68

DEPEND shoulddecrease the risk of disputes

Two variables are included to measure regime type Monadic democracy is rstcalculated as the difference between reported values for democracy (DEMOC) andautocracy (AUTOC) in Polity III (values range from 10 [democratic] to ndash10 [auto-cratic] for each variable)69 Democracy low (DEML) is then calculated using thelower of the two monadic democracy values whereas democracy high (DEMH)equals the higher value in a dyad year DEML thus measures the threshold of dyadicdemocracy (weak-link assumption) whereas DEMH measures asymmetry in regimetype Research on the democratic peace leads us to expect DEML to be negative whileDEMH is positive

States that are growing economically may be disinclined to engage in militarizedcon icts Alternately economic growth may encourage expansionist tendenciesThe rates of change in GDP per capita for states in the dyad are rst obtained usingmoving averages over a three-year period (one- or two-year periods when missingvalues are present) Based on the weak-link assumption low economic growth(GROWTHL) equals the lower of the two growth rates in a dyad

CONTIGUITY is a dummy variable for geographic contiguity coded 1 when bothstates in a dyad are contiguous (or within 150 miles by sea) and zero otherwise

66 Ideally we would also include portfolio investments but the data are extremely limited67 World Bank 199868 Trade data are from IMF 1993 GDP data (purchasing power parity) are from Summers and Heston

199169 Jaggers and Gurr 1995


Either because of opportunity willingness or both bordering states have a higherrisk of experiencing disputes than distant states

ALLIES is also a dummy coded 1 when both states of a dyad share a militaryalliance or if each is separately allied with the United States Allies may bedisinclined to ght each other

CAPRAT controls for the balance of power in a dyad by measuring the ratio ofcomposite national capabilities scores (CINC) CINC scores from the Correlates ofWar (COW) project measure a statersquos share of world capabilities in three dimen-sions demographic (total and urban population) economic (energy consumptionand ironsteel production) and military (expenditures and total personnel)70 Dis-putes may be less (or more) likely in dyads with an imbalance of power

We also follow Nathaniel Beck Jonathan N Katz and Richard Tucker andconstruct temporal splines (_SPLINE1 _SPLINE2 and _SPLINE3) to control for durationdependence71 Because they are simply statistical corrective measures we excludesplines that are not signi cant The statistical ndings remain substantially the same

Several analysts suggest either that statesrsquo interests confound the effect ofinterdependence or that interdependence itself leads states to form similar inter-ests72 (In Table 4 we add AFFINITY a variable based on an index of the similarityof statesrsquo voting behavior in the UN General Assembly We add the variable toassess and control for the effect of preference similarity73)

Finally our analysis omits two variables used by Russett Oneal and Davis Firstwe found that once temporal dependence is controlled for joint membership inintergovernmental organizations (IGOs) becomes positive and highly statisticallysigni cant implying that IGO membership increases interstate con ict74 We doubtthat IGO membership induces states to ght one another Rather we suspect thatIGOs differ as to purpose with some IGOs serving as proxies for rivalries and othersperforming the constraining role imagined by Kant and others We suspect a aw in

70 Singer Bremer and Stuckey 197271 See Beck Katz and Tucker 1998 Other work in the research program incorporates this technique

See Oneal and Russett 1999ab We also examined a xed-effects model Green Kim and Yoon offera rationale for controlling dyad-speci c effects in binary cross-sectional time-series analysis by estimat-ing intercepts for each dyad Results of xed-effects models are discouraging for quantitative IR researchGreen Kim and Yoon 2001 441 We see the technique as problematic Coef cients lack theoreticaljusti cation tend to absorb or encumber much of the limited variance in statistical models of con ict andyield nonintuitive results (for example contiguity becomes insigni cant as a predictor of disputebehavior)

72 Morrow argues that measures of interdependence may really be capturing interest similarityMorrow 1999 Papayoanou and Solingen each suggest (though in different ways) that interdependencecan lead states to form similar interests See Papayoanou 1999 and Solingen 1998

73 Oneal and Russett 1999b and others have applied this technique to assess political closeness SeeOneal and Russett 1999b and Gartzke 1998 and 2000

74 The authors of the study acknowledge the problem Personal communication with John OnealIncluding the IGO variable weakens results for most of the capital and monetary variables (though mostare still signi cant) due to collinearity


variable construction rather than in theory Thus given the anomalous nding andthe lack of need for the variable in this study we remove the indicator75

We also omit Russett Oneal and Davisrsquo variable for trend in trade dependenceOneal and Russett themselves often omit the variable which is rarely statisticallysigni cant Including the trend variable does not alter our results though it makesthe model less parsimonious

Plan for Model Estimation and Presentation

We begin with a baseline speci cation from Russett Oneal and Davis excludingthe variables for joint-IGO membership and trend in trade dependence but addingtemporal splines

MIDij t 5 b 0 1 b 1DEMLij 1 b 2DEMHij 1 b 3DEPLijt 2 1 1 b 4GROWTHLij

1 b 5ALLIES ij 1 b 6CONTIGij 1 b 7CAPRATij 1 b 8_SPLINE1ij

1 b 9_SPLINE2ij 1 b 10SPLINE3ij

(1)

We extend the basic model by rst adding the monetary variables and then thecapital variables76 One might argue that our research design errs in light of previousstudies that appear to support the opportunity cost model How do we show thatpeace is caused by signaling The problem is really with previous studies thatassume (but fail to demonstrate) that a negative relationship results from opportunitycosts Rigorous theory shows that a negative relationship is the wrong hypothesis forthe opportunity cost model that opportunity costs predict no relationship We alsoshow that the negative relationship is the correct hypothesis for the signaling modelThus although our empirical results appear to substantiate previous claims ourtheory directly contradicts previous purported causes

Our statistical ndings are presented in four tables Table 1 presents the resultsfrom ve speci cations Model 1 reports the baseline (listed earlier) model 2introduces PEGGING and JOINT CURRENCY AREA model 3 introduces CAPOPENL andCAP FLOWSL model 4 includes PEGGING JOINT CURRENCY AREA and CAPOPENL andmodel 5 reports the full model including all variables Table 2 provides a substantiveinterpretation of some of the results from Table 1 Table 3 presents model resultsusing Russett Oneal and Davisrsquo dependent variable MID involvement Table 4 thenreestimates the models in Table 1 while controlling for preference similarity byadding AFFINITY

Ideally we would only present the results of the ldquotruerdquo causal model (or at leastthe one that we believe captures the true causal process) However the variables of

75 We explore the anomalous resultmdashand seek to formulate a better indicator of IGO member-shipmdashin other research

76 Analysis is conducted using probit in Stata 60 with robust standard errors and adjusting forclustering on dyads


interest limit the sample size For example model 5 in Table 1 including allvariables contains only 2133 observations By comparison model 1 has 20990observations Multiple speci cations show whether variables of interest are robustacross samples Insigni cant results may still be due to inadequacies in the data

TABLE 1 Probit estimates of the effects of economic interdependence on MIDonset 1951ndash85 (using MID data from Maoz)

Independent variables Model 1 Model 2 Model 3 Model 4 Model 5

DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)

ECONOMIC GROWTHL 2 00178 2 00308 2 00207 2 00309 2 00184(00059) (00078) (00196) (00074) (00193)

ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)

CAPABILITY RATIO 2 00010 2 00011 2 00004 2 00007 2 00008(00003) (00004) (00003) (00003) (000049)

JOINT CURRENCY AREA mdash 2 01618 mdash 2 01503 2 02368(00758) (00781) (03979)

PEGGING mdash 02990 mdash 02346 05285(01495) (01520) (02215)

CAPOPENLt2 1 mdash mdash 2 00962 2 00777 2 00866(00509) (00241) (00523)

CAP FLOWSLt2 1 mdash mdash 2 03988 mdash 2 04473(01973) (02001)

SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)

SPLINE3 00007 00007 mdash 00006 mdash(00002) (00002) mdash (00002) mdash

Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)

N 10399 20990 2202 9303 2133Wald test 28025 27024 12631 29549 14349p-value 00000 00000 00000 00000 00000Log likelihood 2 934795 2 227358 2 202087 2 869763 2 198943Pseudo R2 02237 01884 03103 02322 03161

Source Maoz 1999 (DYMID10 data)Note Numbers in parentheses are robust standard errors adjusted for clustering over dyadsp lt 01 two-tailed testp lt 05 two-tailed testp lt 10 two-tailed test


However an incremental approach shows whether some variables are sensitive toother variables In the next section we discuss our ndings and explore theoreticalimplications of the results

Results

Table 1 provides broad support for our argument Results for democracy aresurprising and perhaps noteworthy Adding economic variables generally makesDEML the low democracy score statistically insigni cant77


In Model 2 PEGGING is positive and signi cant at the 5 percent level Asymmetrybetween the pegging state and the pegged appears to correlate with dyads in which

77 We use the following method to assess the effect of missing values First we set missing valuesfor four key economic variables to high values within each dyad and estimate the models We then setmissing values to low values in dyads and reestimate the models The two sets of results identify upperand lower bounds of possible results Results for the key economic variables remain largely the same (Weomit reporting results since they are equivalent to those in Table 1)

TABLE 2 The relative risk of MID onset in ve different scenarios(based on coefcient estimates in model 5 of Table 1)

Scenarios Model 5Relative risk of MIDcompared to baseline

1 Baseline scenario democracy low democracy hightrade dependence low economic growth low capabilityratio CAP FLOWSLt2 1 and CAPOPENLt2 1 at samplemean values allies JOINT CURRENCY AREA andPEGGING equal 0 and contiguity equals 1

00105 100

2 JOINT CURRENCY AREA equals 1 other variables atbaseline values

00054 51( 2 00051)

3 CAP FLOWSLt2 1 and CAPOPENLt2 1 increase by onestandard deviation other variables at baseline values

00028 27( 2 00077)

4 JOINT CURRENCY AREA equals 1 CAP FLOWSLt2 1 andCAPOPENLt2 1 and trade dependence low increase byone standard deviation other variables at baselinevalues

00013 12( 2 00092)

Note Numbers in parentheses are changes in probability


TABLE 3 Probit estimates of the effects of economic interdependence on MIDinvolvement 1951ndash85 (using data from Russett Oneal and Davis)


DEMOCRACY SCOREL 2 00309 2 00027 00157 00035 00199(00072) (00072) (00125) (00072) (00128)

DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)

DEPENDENCE SCORELt2 1 2 77653 2 51846 2 42870 2 17752 2 41857(67218) (43416) (17588) (42965) (17412)

THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)


JOINT CURRENCY AREA mdash 2 01419 mdash 2 01144 02242(00831) (00878) (03487)




SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)

SPLINE3 00015 00014 mdash 00014 mdash(00002) (00002) (00002)

Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)


Source Russett Oneal and Davis 1998Note Numbers in parentheses are robust standard errorsp lt 01 two-tailed testp lt 05 two-tailed testp lt 10 two-tailed test


TABLE 4 Probit estimates of the effects of economic interdependence andpreference similarity on MID onset 1951ndash85 (using data from Maoz)


DEMOCRACY SCOREL 2 00098 00034 00058 00082 00079(0006) (00068) (00123) (00069) (00124)

DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)

CAPABILITY RATIO 2 00006 2 00009 2 00004 07591 2 00009(00002) (00003) (00004) (01201) (00005)

PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)


Source Maoz 1999Note Numbers in parentheses are robust standard errors adjusted for clustering in dyadsp lt 01 two-tailed testp lt 05 two-tailed testp lt 10 two-tailed test


there is greater friction swamping any signaling effect78JOINT CURRENCY AREA is

signi cant (and negative) at the 5 percent level Membership in a common currencyarea or in the EMS reduces disputes Note that in model 2 both DEML and DEPL havethe expected negative sign but neither is signi cant

PEGGING is positive but statistically insigni cant once we control for joint nancialopenness (CAPOPENL) in model 4 However the variable becomes positive andstatistically signi cant again in model 5 (the full model with a limited sample N =2133) JOINT CURRENCY AREA also has the expected negative sign in both models 4and 5 but is signi cant only in model 4

Impact of Capital Investments

Model 3 includes the baseline plus CAPOPENL and CAP FLOWSL CAPOPENL measuring nancial openness in a dyad over current and capital account transactions isstatistically signi cant and has the expected negative sign Governmental opennessto interstate nancial activity is negatively related to con ict behavior CAP FLOWSLmeasuring joint state exposure to direct investments is also statistically signi cantand in the expected negative direction Even though they are correlated (r = 053)the signi cance of both variables suggests that they tap into different dimensions ofthe same economic process and that both dimensions help to dampen the onset ofmilitarized disputes

In model 4 CAPOPENL is also negative and statistically signi cant at the 1 percentlevel based on a much larger sample compared to model 2 and controlling for themonetary variables In model 5 (the full model) both CAPOPENL and CAP FLOWSL

remain statistically signi cant in the expected directions The robustness of theseresults favors the validity of our theoretical arguments

Model Comparison and Substantive Implications

Comparing log likelihoods and pseudo R2 it is clear that model 5 has the bestoverall t with the caveat that the ve models are based on different samplesIncluding trade dependence openness to capital ows and monetary coordinationseems to best re ect underlying processes

Note that Russett Oneal and Davisrsquo measure of trade dependence (DEPL) isstatistically signi cant only in models 3 and 5 based on a limited sample of over2000 observations Much of the in uence of trade appears to be absorbed by capitaland currency area variables along with the splines Correlation is 025 between CAP

FLOWSL and DEPL 028 between CAPOPENL and DEPL and 011 between DEPL and JOINT

CURRENCY AREA The results highlight the importance of examining cross-nationaleconomic linkages other than trade It is misleading to draw inferences about theimpact of economic exchanges on interstate con ict just from trade interdepen-dence

The results also suggest that it is premature to conclude that all economic linkagescontribute to peace PEGGING shows that asymmetric interstate monetary relations


correlate positively with con ict a result consistent with other studies Asymmetri-cally interdependent dyads (where one state is vulnerable) are somewhat more likelyto experience a militarized dispute The result may be a consequence of selectioneffects that associate historical pegs with asymmetric dependent linkages Wespeculate that contemporary efforts at pegging may be more benign (a topic offuture research)

Table 2 reports the probability of MID onset in four scenarios based on parameterestimates in model 5 The rst row in Table 2 reports baseline probabilitiesof a MIDin absolute and relative terms Holding DEML DEMH DEPL GROWTHL CAPRAT thesplines CAP FLOWSL and CAPOPENL at their sample mean values setting ALLIESPEGGING and JOINT CURRENCY AREA to zero and setting CONTIG to 1 the likelihood ofa MID is 00105 (the relative-risk column reports 100 percent since this scenario isthe baseline) In row 2 when JOINT CURRENCY AREA equals 1 (currency area linkage)the probability of a MID declines to 00054 or only 51 percent of baselineconditions Row 3 shows that increasing exposure to direct investments (CAP

FLOWSL) and openness to capital (CAPOPENL) by 1 standard deviation reduces theprobability of dispute onset to 00028 or only 27 percent of baseline conditions

Finally we argue that these economic processes all represent elements ofinterdependence and that they tend to be mutually reinforcing Changes in tradeinterdependence are likely to involve similar shifts in investment (CAP FLOWSL) andgovernment liberalization of capital ows restrictions (CAPOPENL) States sharingcurrency area membership tend to trade and invest among themselves Thusassessing one aspect of interdependence while holding others constant likelyunderestimates the actual contribution of economics to peace Row 4 Table 2reports the effect of an increase of one standard deviation in DEPL CAPOPENL and CAP

FLOWSL together with currency area membership (JOINT CURRENCY AREA = 1) Theprobability of MID onset decreases from 00105 in the baseline to 00013 so thatthese dyads are only about 12 percent as likely to experience dispute onset

Control Variables

Table 1 reveals surprising results for some of the control variables Addingmonetary and capital variables in models 2ndash5 makes joint democracy (DEML)insigni cant These variables CAP FLOWSL in particular shrink the sample sizedramatically so the results may be due to changes in the sample Variance in regimetype may be reduced by nonrandom missing values for the capital variablesDemocracy may also lose signi cance because of covariance with capital variablesYet monetary and capital variables are signi cant even while democracy is notindicating that variation in regime type is subsumed by greater variation in theseeconomic variables While adding observations may increase variance in regimetype it will certainly increase variance in capital and monetary variables79

79 We expect to continue exploring this topic in subsequent research


Of the other control variables economic growth has a robust pacifying effect onmilitarized disputes in models 1 2 and 4 Geographic contiguity has the expectedpositive effect on the chance of a military dispute in all models The spline variablescontrolling duration dependence are also all signi cant80 Power preponderance (animbalance of power in a dyad) reduces the odds of a dispute but not signi cantlyin models 3 and 5 Except in model 1 alliance ties are not signi cant

Results Based on Dispute Involvement

Table 3 reports results using Russett Oneal and Davisrsquo dependent variable (MIDinvolvement plus discrepant cases) Results for variables of interest depart in somecases from those in Table 1 but results also show the consistency of our ndingsModel 2 in Table 3 shows that JOINT CURRENCY AREA is statistically signi cant at the10 percent level and has the expected negative sign whereas PEGGING is insigni cantIn model 3 CAPOPENL and CAP FLOWSL are both signi cant in the expected negativedirection In model 4 CAPOPENL is statistically signi cant and negative but JOINT

CURRENCY AREA and PEGGING are not signi cant In model 5 CAP FLOWSL remainssigni cant in the expected direction and CAP OPENL is close to being signi cant at the10 percent level PEGGING is signi cant and positive JOINT CURRENCY AREA howeveris not signi cant In none of the last four models in Table 3 is democracy signi cantTrade dependence (DEPL) is signi cant in models 3 and 5

Results After Controlling for Preference Similarity

Morrow argues that trade correlates with con ict in part because trade patternsre ect political closeness81 If so then the effect of economic variables may actuallybe attributable to a measure of interest similarity Some analysts go further arguingthat interdependenceactually creates interest similarity82 Omitting a measure of thecloseness of political interests might produce spurious ndings To examine thesensitivity of our results we replicate models in Table 1 but include a measure ofinterstate preference similarity AFFINITY The statistical results appear in Table 4

In model 2 of Table 4 JOINT CURRENCY AREA is signi cant at the 5 percent level(negative sign) and PEGGING is insigni cant In model 3 CAPOPENL is marginallysigni cant at the 10 percent level CAP FLOWSL is signi cant at the 5 percent level andboth variables have the expected negative sign In model 4 CAPOPENL and JOINT

CURRENCY AREA are negative and signi cant and PEGGING is positive and insigni cantFinally model 5 shows that CAPOPENL CAP FLOWSL and JOINT CURRENCY AREA arenegative with only CAP FLOWSL being statistically signi cant (though CAPOPENL issigni cant at the 10 percent level for a one-tailed test) PEGGING is positive and

80 Models 3 and 5 omit a spline variable that is not signi cant Including the third spline does notchange the results

81 Morrow 199982 See Papayoanou 1999 and Solingen 1998


statistically signi cant Note that the measure of similar interests is signi cant onlyin models 1 and 4 This may result from sample differences and from the correlationof AFFINITY with DEPL CAPOPENL CAP FLOWSL and JOINT CURRENCY AREA States witheconomic ties tend to have similar political preferences or vice versa Still evenafter controlling for preference similarity monetary and capital ties reduce disputesStatesrsquo interests may converge with interdependence but their contribution to peacedoes not appear to eclipse the direct effects of our broader set of economic variablesThe results are consistent with our argument that economic ties increase the rangeand ef ciency of signaling between states thus reducing the need for militaryviolence

Conclusion

We have reviewed arguments for the effect of economic interdependence on peaceWe show that existing accounts do not adequately explain why liberal economiesare less likely to ght but that a signaling argument is consistent with theobservation of a liberal peace We also expand interdependence to include nancialand monetary integration offering a set of variables that measure these processesOur results corroborate our hypotheses This study is limited by data and by atheoretical framework that necessarily simpli es reality Still despite weaknessesthe combination of theory and analysis offers a compelling and not-inconsiderablere nement of the relationship between economics and peace

Trade and direct investment increase cross-border economic contact and raisea statersquos stake in maintaining linkages Monetary coordination and interdepen-dence demand that states strike deals Through such interactions states create abroad set of mutually bene cial economic linkages While these linkages maydeter very modest clashes their main impact is as a substitute method forresolving con ict Political shocks that threaten to damage or destroy economiclinkages generate information reducing uncertainty when leaders bargainThreats from interdependent states carry more weight than threats from autar-chic states precisely because markets inform observers as to the veracity ofpolitical ldquocheap talkrdquo Multiple channels of economic interactions help states tocredibly communicate increasing the ldquovocabularyrdquo available to states in at-tempting to assess relative resolve

A signaling interpretation of interdependence offers some promise both analyti-cally and in terms of international events If costly signaling through economicinterdependence reduces statesrsquo recourse to military violence then increasingeconomic interdependence (globalization) implies the prospect of a more paci cglobal system The magnitude of the paci c effect of interdependence is dif cult toassess however since other factors such as increasing polarization may add to themotives for con ict At the same time the signaling argument implies that much of


the variance in the propensity for con ict is unknowable83 Before we can havegreater con dence in our results we need to examine a larger data sample includingall dyads and longer time spans Precise measures may be obtained by limiting thesample to US dyads Finally the effects of democracy on con ict appear to requireadditional assessment However our ndings provide evidence (and a rationale)suggesting that liberal economics may be at least as salient to peace as liberalpolitics

Appendix

Here we provide formal proofs of key propositionsin the text in the form of simple bargaininggames In each game two states (A and B) compete over the disposition of spoils in a unitspace (representingissues territory and so on) where Arsquos ideal point is arbitrarily coded zeroand Brsquos ideal point is coded 1 Utilities for outcomes decline linearly (generalizable to anymonotonic function) in the distance from playersrsquo ideal points (disposition of spoils iszero-sum) Games in propositions 1ndash3 offer parsimonious conditions suf cient to motivatethe propositions In these games Arsquos utility for its ideal point (zero) is coded 1 Brsquos utility foroutcomes varies with ldquotyperdquo and is selected randomly from a uniform distribution of unitinterval (t Un[01]) Types weight outcomes so that types B near 1 are ldquoresolvedrdquo andtypes B near zero care relatively little about the stakes Jeffrey S Banks offers a proofshowing that proofs like those in propositions 1ndash3 are generalizable for the entire class ofbargaining games with endogenous offers and one-sided asymmetric information84 Propo-sition 4 shows that propositions 1ndash3 are robust to two-sided asymmetric information (A hastype space tA Un[01])

The sequence of play for games in propositions1ndash3 is for A to propose a take-it-or-leave-itoffer (d) to B (A receives 1 ndash d) Player B then either accepts the demand or rejects it leadingto a costly contest with a lottery over the stakes We assume that the ex ante probability thateither player wins a contest equals the ratio of military capabilities in the dyad (a for playerB and 1 ndash a for player A) Brsquos decision to ght is dichotomous (f where f = 1 is ldquo ghtrdquo andzero is ldquoacceptrdquo Arsquos beliefs about types (b) may be probabilistic)Fighting imposes a positivecost on both players (the non-zero-sum component of the game) Proofs for proposition 4involve an additional stage of bargainingand two-sided asymmetric informationModels withasymmetric information involve Perfect Bayesian equilibria Results are generalizable toother distributions

PROPOSITION 1 UNCERTAINTY IS NECESSARY FOR COSTLY CONTESTS (WARS AND SO ON)The proposition requires demonstrating that costly contests occur only with strategicuncertainty (asymmetric information) States A and B never ght if A knows Brsquos type If A isuncertain about Brsquos type then war can occur We rst model the full information game andthen introduce asymmetric information (for B)

83 Gartzke shows that an informational theory of war implies that researchers cannot anticipatecontests from crises unless researchers are better informed about participantsrsquo private information thanare the participants themselves Gartzke 1999

84 Banks 1990


Variables

a = Brsquos odds of victory (ratio of military capabilities in the dyad [0 oslash a oslash 1])bA = Arsquos beliefs about Brsquos type (t)c = Each statersquos cost for ghting c [0 lt c oslash 12]d = Arsquos proposal (0 oslash d oslash 1) A receives utility 1 ndash d (UA (d = 0) = 1)f = Brsquos decision to ght (and Arsquos estimate) [0 oslash f oslash 1]t = Brsquos type (0 oslash t oslash 1) drawn at random from type space T Un[01]

State Brsquos Problem

max UB wrt f st f [ [01]

UB 5 [(1 2 f ) z (t z d)] 1 [ f z ((t z a) 2 c)]shy UB

shy f5 t z (a 2 d) 2 c

If

d a 2c

t

then Brsquos utility is increasing in f B accepts Arsquos proposal ( f = 0) if

d ugrave a 2c

t

else f = 1

State Arsquos Problem

max UA wrt d st 0 oslash d oslash 1

UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)

Arsquos utility is decreasing in the offer (d) up to the point where B is willing to ght ( f ) Arsquosoptimal demand is one B is just willing to accept If A knows Brsquos type then A makes thisdemand If A does not know Brsquos type then A will sometimes make a demand that B prefersto reject leading to war

Full Information Solution (A knows Brsquos type)

Arsquos calculation of d Since UA is decreasing in d A prefers

d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c

t(Arsquos payoff for not fighting)

is preferred to 1 2 a 2 c (Arsquos payoff for ghting) for all types B ( F t) Since A knows Brsquostype (and thus Brsquos reservation price) and since d is endogenous capabilities resolve costsand other factors are subsumed in Arsquos offer and do not motivate a contest

Equilibrium In all cases A proposes

d 5 a 2c

t

and B accepts ( f = 0)

Asymmetric Information Solution (A does not know Brsquos type)

Arsquos calculation of d Given Brsquos private information about its type A calculates its bestresponse to each type of B weighted by the odds of encountering any given type Arsquos optimalproposal (d) is a function of the de nite integral of Brsquos reservation price over the domain ofBrsquos type space (T)

E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)

Setting the integral equal to zero and solving for d yields d = a ndash 2c This implies that f = 1 if

(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0

Changing only the informational conditions leads to a motive for costly contests

Equilibrium A proposes d = a ndash 2c B accepts ( f = 0) if

t oslash1

2


and rejects (war) otherwise Arsquos beliefs equal (bA u tt [ T Un[01]) but there is noopportunity to update before the offer QED

PROPOSITION 2 OPPORTUNITY COST ARGUMENTS FOR INTERDEPENDENCE HAVE LITTLE OR

NO EFFECT ON THE PROBABILITY OF COSTLY CONTESTS The second proposition requiresshowing that the addition of a mutual benet for actors (where the bene t is conditional oncooperation)does not alter the probabilityof costly contests We add a mutual bene t (h) Weshow that interdependence through opportunity costs is subsumed in bargaining and has noeffect on conict behavior

Variables The variables are the same as in the previous model with the addition ofopportunity costs

h = Each actorrsquos bene t from economic exchange [h ugrave 0]



UB 5 [(1 2 f ) z (t z d 1 h)] 1 [ f z ((t z a) 2 c)]shy UB

shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t

then Brsquos utility is increasing in f B accepts ( f = 0) if

d ugrave a 2c 1 h

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d 1 h)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)

Arsquos utility again decreases in d until B prefers ghting ( f = 1) to Arsquos demand Arsquos optimaldemand is again one B will just accept The addition of a mutual bene t threatened by acontest (h) shifts Brsquos reservation price If bene ts can be anticipated A simply subsumes Brsquosopportunity costs into a more extractive demand Unanticipated bene ts (states probably


know if they are interdependent) are a potential motivemdashnot a palliativemdashfor costly contestssince they contribute to asymmetric information

SolutionArsquos calculation of d The solution is similar to the asymmetric information game in

proposition 1 A calculates its best response to each type B weighted by the odds ofencountering any given type

E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)

Solving for d yields d = a ndash 2c ndash 2h A expects f = 1 if

(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0

Opportunity costs fail to motivate a decline in costly contests Given endogenous bargainsaltering payoffs without changing informational conditions in the game cannot change theodds of a ght Player A simply subsumes changes in Brsquos reservation price in a more exactingdemand Uncertainty about interdependencecan lead A to make a demand that B rejects butthis cannot account for peace

Equilibrium A proposes d = a ndash 2c ndash 2h B accepts ( f = 0) if

t oslash1

2

and rejects otherwise Arsquos beliefs equal

(bA u t t [ T Un[0 1])

Again updating occurs after the offer QED

PROPOSITION 3 COSTLY SIGNALING CAN EXPLAIN THE APPARENT EFFECT OF INTERDEPEN-DENCE ON THE PROPENSITY FOR COSTLY CONTESTS Proposition 3 involves a more complexframework and solution We need to show that a mutual bene t can allow actors to signalleading to peace We again begin with asymmetric information and a mutual bene t (h)Instead of directly inhibiting contests however opportunity costs offer a cheaper method ofdemonstrating resolve In the signaling game B has the option to preemptively eliminate


interdependence (h) B might signal using a portion of h (v oslash h) but results are monotonicand this complicates presentationAlso to simplify the game (and with no loss of generality)we assume B weakly prefers signaling if indifferent

Variables The variables are the same as in the previous model though B now has asecond decision variable

s = Brsquos signaling decision (whether to delete the mutual bene t h) [0 oslash s oslash 1]


max UB wrt s f sts f [ [01]

UB 5 (1 2 s)[t z d0 1 h] 1 s z [(1 2 f )(t z d1)] 1 [ f((t z a) 2 c)]

shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB

shy s5 t z (d 1(1 2 f ) 2 d0 1 a z f ) 2 h 2 c z f

De ne t [ T st

shy UB

shy s5 0 if t 5 t

B signals (s = 1) if

d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t

then B also ghts ( f = 1)


max UA wrt d i st 0 oslash d i oslash 1

where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)

Arsquos best response remains an offer each B will just accept discountedby the odds of any giventype but A must also gauge how offers affect Brsquos choice of subgame The less generous thebaseline (d0 ) the more often B ends the bene t (h) The greater the signaling demand (d1)the more often B prefers ghting Arsquos optimal demands make A indifferent betweensubgames Since B can play Arsquos game the calculation is the basis for common conjecture IfB expects to reject the offer B always signals

Solution

Arsquos calculation of di A rst calculates a best response to each type B willing to signal (d1 t to 1) Both players then use d1 t Arsquos estimate of f and Arsquos utility to solve for a baselinedemand (d0 )

Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]

Solving for d1 yields

d 1 5 a 22c

1 1 t

Substituting d1 into

shy UB

shy s

and solving for t produces

t 52 a 1 h 1 d0 1 c(2 2 f ) 1 Icirc ( 2 a 1 h 1 d0 1 c(2 2 f ))2 1 4(a 2 d0)(b 1 f z c)

2(a 2 d0)

Using Bayesrsquos theorem the conditional probability of a ght P( f= 1 u s= 1) is

P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2


Substituting for t d1 s and f in Arsquos utility function and solving the optimization problem weobtain

d 0 54h2 1 c2 1 4ah 2 2ac 1 4hc 2 2 Icirc 2 z c32 z Icirc 2h 1 c

2(2h 2 c)

Substituting d0 and solving for d1 we now obtain both of Arsquos demands in terms of gameparameters

d1 52c(4h2 1 c2 1 4hc 2 2 Icirc 2 z c32 z Icirc 2h 1 c)

2c2 2 Icirc 2 z c32 z Icirc 2h 1 c 1 2(c 2 2h) z Icirc c(4h3 1 c3 1 2h2c 2 2h Icirc 2 z c32 z Icirc 2h 1 c

(c 2 2h)2

B compares payoffs in each subgame to decide whether to signal (s) Substituting and solving for (t )

S 2(c 2 2h)(h 1 cf ) S 2 Icirc 2 z c32 z Icirc 2h 1 c 2 2c2 1 (4h 2 2) S Icirc c(4h3 1 c3 1 2h2c 2 2h Icirc 2 z c32 z Icirc 2h 1 c(c 2 2h)2 D D D

S (4h2 1 c3 1 4hc 2 Icirc 2 z c32 z Icirc 2h 1 c)(c(c(2 2 4f ) 1 8h( f 2 1) 1 Icirc 2 z c32 z Icirc 2h 1 c))

1 (4h 2 2c) z S Icirc c(4h3 1 c3 1 2h2c 2 2h Icirc 2 z c32 z Icirc 2h 1 c(c 2 2h )2 D

Arsquos demands nullify relative power (a) A expects f = 1 if

d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0

the probability of ghting in both the opportunity cost and signaling games is the same onlyif h = 0

Equilibrium A offers di i [ [01] B accepts d0 (s = 0 f = 0) if t oslash t d1 (s = 1

f = 0) if

1

2(1 1 t ) ugrave t t

and ghts (s = 1 f = 1) otherwiseArsquos beliefs equal

(bAs= 1u t t [ T Un[ t 1] bA

s= 0 u t t [ T Un[0 t ])


The graphs in Figure 1 show the relationship between interdependence (h horizontal axis)and types B (t vertical axis c = 14) The horizontal line at 05 is for opportunity costsTypes B larger than 05 ght at all levels of interdependence The line sloping upward

FIGURE 1 The range of types B that ght (probability of war)


represents signaling The greater the mutual bene t sacri ced as a signal the fewer types Bthat ght The nding is general to situations where B sacri ces only part of the bene t ofinterdependence Signaling reduces dispute frequency QED

PROPOSITION 4 PROPOSITIONS 1ndash3 HOLD FOR TWO-SIDED ASYMMETRIC INFORMATIONSYMMETRIC BARGAINING AND TEMPORAL DISCOUNTS Proposition 4 claims that previouspropositions are robust to relaxing model assumptions Science favors the simplest accountconsistent with behavior of interest Still it is worth conrming the generality of results fromparsimonious models We show that our conclusions do not depend on bargaining symmetryor the treatment of uncertainty or time Other aspects of the games such as structure areunlikely to alter results Banks shows that for all two-player bargaining games where oneactor is privately informed both the odds of a contest and the equilibrium settlement increasemonotonically in the informed actorrsquos payoffs85 Given Banksrsquo proof and equilibriumexpectations it follows that propositions 1ndash3 hold for at least the class of two-playerbargaining games in which offers are endogenous and unrestricted The following gamesfeature a two-stage sequence of play B can now make a counteroffer A then accepts Brsquoscounter or ghts

Proposition 1 Proposition 1 is trivially satis ed Since one-sided asymmetric informa-tion is suf cient to motivate costly contests (and full information is not) two-sidedasymmetric information is also suf cient

Proposition 2 Utilities appear with the interdependence parameter (hi) but the value ofbene ts can be zero

Variables Variables are the same though symmetric play requires the use of subscripts

d i = Playerrsquos discount for payoffs in the second period (after Brsquos counter i [ [AB])p = Brsquos proposal decision (whether to propose dB or accept d) [0 oslash dB oslash 1]


max UA wrt d st 0 oslash d oslash 1 and fA st fA [ [0 1]

UA 5 [p d A(1 2 fA)(tA(1 2 dB) 1 hA)] 1 [ fA(tA(1 2 a) 2 c)]

1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1

shy fA5 d A[tA(dB 2 a) 2 c 2 hA]


max UB wrt dB st 0 oslash dB oslash 1 and p st p [ [01]

85 Ibid


UB 5 p d B[(1 2 fA)(tdB 1 hB)] 1 [fA(ta 2 c)] 1 (1 2 p)(td 1 hB(1 1 d B))

shy UB

shy p5 t[ d B(dB(1 2 fA) 1 fAa) 2 d] 2 b(1 1 d B fA) 2 d B fAc

shy UBp= 1

shy dB5 d B[t(1 2 fA)]

Arsquos offer is again one each B just accepts but this time B can make a counteroffer B in turnmakes an offer A just accepts The less generous Arsquos initial offer (d) the more often B prefersto counter (dB) A then chooses between ghting ( fA ) and accepting Brsquos counterproposal

Solution

Brsquos calculation of dB i (p = 1) B makes a best counteroffer weighted by the odds of eachtype A

E0

1 shy UAp= 1

shy fAz f (tA)dtA 5 E

0

1

d A[tA(dB 2 a) 2 c 2 hA] z f (tA)dtA 51

2d A(dB 2 a 2 2(c 1 hA))

Solving

dB 5 a 1 2(c 1 hA) fA 5 1 if a 1 2(c 1 hA) a 1c 1 hA

tA3 tA

1

2

else if tA oslash 12 fA = 0

Brsquos calculation of p Brsquos optimal counter

(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0

Arsquos calculation of d A can play Brsquos game anticipating Brsquos offer Brsquos estimate of fA and t tocalculate P(p = 1) = (1 ndash t ) Substituting values then recalculating

shy UB

shy p

and solving the integral

E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)

Solving d = d B (a + hA ) 2 h(2 + d B ) Substituting d dB t = 12 B accepts Arsquos offer if t oslash12 else p = 1 If p = 1 A accepts Brsquos offer if tA oslash 12 This is so regardless of values assignedto h and hA

Equilibrium A proposes d = d B (a + hA ) 2 h(2 + d B ) and B accepts (p = 0) if t oslash 12else if p = 1 B offers dB = a + 2(c + hA ) and A accepts ( fA = 0) if tA oslash 12 else ( fA = 1)Arsquos beliefs equal (bA u tt [ T Un[01]) Brsquos beliefs are (bB u tA tA [ TA Un[01])

The probability of a contest is independent of the bene t of interdependence (h or hA )because such bene ts (and all other parameters that can be anticipated) are subsumed inplayersrsquo offers

Proposition 3 Some terms are presented in implicit form to save space and because theequations are quite bulky

Variables Variables are the same as in the previous model subscripts are again used toidentify player offers and so on



where i [ [01] and sA fA st sA fA [ [01]

UA 5 (1 2 s)[tA(1 2 d0) 1 hA(1 1 d A)] 1 s(1 2 p)(tA(1 2 d 1) 1 hA d A)

1 p[(1 2 sA)( d A(tA(1 2 dB0) 1 hA)) 1 sA( d A((1 2 fA)tA(1 2 dB1))

1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1

shy fA5 d A[tA(dB1 2 a) 2 c]

shy UAs= 1p= 1

shy sA5 d A[tA(dB0 2 dB1(1 2 fA) 2 fAa) 2 fAc 2 hA]

De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0

if tA = t A A signals (sA = 1) if

d 0 dB1(1 2 fA) 1 afA 1hA 1 fAc

tA

and if

dB1 a 1c

tA A fights ( fA 5 1)


max UB wrt dBi st 0 oslash dBi oslash 1

where i [ [01] and p s st p s [ [01]

UB 5 (1 2 s)[td0 1 h(1 1 d B)] 1 s(1 2 p)(td1 1 h d B)

1 p[(1 2 sA)( d B(tdB0 1 h)) 1 sA( d B((1 2 fA)tdB1 1 fA(ta 2 c)))]

shy UB

shy dB05 d B pst(1 2 sA)

shy UB

shy dB15 d B pssAt(1 2 fA)

shy UB

shy s5 t[d1((1 2 p) 2 d0 1 d Bp(dB0(1 2 sA) 1 sA(dB1(1 2 fA) 1 afA))] 2 d BpsA(h 1 fAc) 2 h

shy UBs= 1

shy p5 t[ d B(sA(dB1(1 2 fA) 1 afA) 1 dB0(1 2 sA)) 2 d1] 2 d BsA(h 1 fAc)

t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if

d0 d1(1 2 p) 1 d B p(dB0(1 2 sA) 1 sA(dB1(1 2 fA) 1 afA)) 2(d B psA(h 1 fAc) 1 h)

t

and B also proposes (p = 1) if

d 1 d B(dB0(1 2 sA) 1 sA(dB1(1 2 fA) 1 afA)) 2d BsA(h 1 fAc)

t

Solution

Brsquos calculation of dB i B calculates a best response to each type A that signals (dB 1 t A to1) States then use dB 1 t A Arsquos estimate of fA and Brsquos utility to solve for (dB 0 )

Et A

1 shy UAs= 1p= 1sA= 1

shy fAf(tA)dtA 5 E

t A

1

[ d A(tA(dB1 2 a) 2 c)] f(tA)dtA

51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A

Substituting dB 1 into

shy UAs= 1p= 1

shy sA

and solving for t A

t A 5dB0 2 a 2 hA 2 c(2 2 fA) 2 Icirc (a 1 hA 2 dB0 1 c(2 2 fA))

2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)

Using Bayesrsquos theorem the conditional probability of a ght P( fA = 1 sA = 1) is

P( fA 5 1) z P(sA 5 1 u fA 5 1)

P( fA 5 1) z P(sA 5 1 u fA 5 1) 1 P( fA 5 0) z P(sA 5 1 u fA 5 0)

5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2


Substituting for t A dB 1 sA and fA in Brsquos optimization problem

S Et p

1 shy UBs= 1p= 1

shy dB0 Dand solving for dB 0

dB0 51

2(1 1 t p)(2h 2 t pc)[2a(1 1 t p)(2h 2 t pc) 2 2b(c(1 1 t p)

+ 2 Icirc c Icirc 2h 1 c Icirc 1 1 t p 1 2hA(1 1 t p))] 1 Icirc c[(2h(1 1 t p)

3 ( 2 Icirc c Icirc 2h 1 c Icirc 1 1 t p) 1 c( 2 Icirc c(1 1 t p) 1 Icirc 2h 1 c Icirc 1 1 t p(1 1 3 t p))]

Brsquos calculation of p t p is the type B just indifferent between accepting Arsquos proposal (d1) andcountering Setting

shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)

d B(sA(dB1(1 2 fA) 1 afA) 1 dB0(1 2 sA)) 2 d1

If A offers


t

p 5 1 else p 5 0

Arsquos calculation of di A calculates best response offers (di) to each type B given Brsquos signal(s) common conjectures about t A and t p and anticipating Brsquos offers (dB 0 and dB 1 ) and Brsquosestimate of P( fA ) If B signals (s = 1) A offers d1 equal to the solution to the optimizationproblem

S Et

1 shy UBs= 1

shy p D

d 1 5 d B(dB0(1 2 sA) 1 sA(dB1(1 2 fA) 1 afA)) 22( d BsA(h 1 fAc))

1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)

Using Bayesrsquos theorem Arsquos estimate P(p = 1 u s = 1) is

P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB

d sD 5 0 and solving for t

t 5d BsA(h 1 fAc) 1 h

d B p(sA(dB1(1 2 fA) 1 afA) 1 dB0(1 2 sA)) 1 d1(1 2 p) 2 d 0

If A offers

d 0 d B p(dB0(1 2 sA) 1 sA(dB1(1 2 fA) 1 afA)) 1 d 1(1 2 p)

2d B psA(h 1 fAc) 1 h

t p 5 1 else p 5 0

Substituting known or estimated values of t t A d1 dB 0 dB 1 sA and fA in Arsquos optimizationproblem

S E0

1 shy UB

shy d0 D


and solving

d 0 5 d B p(dB0(1 2 sA) 1 sA(dB1(1 2 fA) 1 afA)) 1 d1(1 2 p) 2d B psA(h 1 fAc) 1 h

t

where t is as de ned earlier (Solutions for terms in explicit form were obtained using thesoftware package Mathematica The explicit form equations are extremely cumbersome andare omitted here)

Equilibrium A offers di i [ [01] B accepts d0 (s = 0) if t oslash t else if t gt t B signals(s = 1) accepting d1 if 12 (1 + t ) z t ugrave t and proposes (p = 1) otherwise B offers dB i i [[01] A accepts dB 0 (sA = 0) if tA oslash t A else if tA gt t A A signals (sA = 1) accepting dB 1 (sA

= 1 fA = 0) if 12(1 + t A ) z tA ugrave t A and ghts (sA = 1 fA = 1) otherwise Arsquos beliefsequal


s= 0 u t t [ T Un[0 t ])

Brsquos beliefs equal

(bBsA= 1u tA tA [ T Un[ t A 1] bB

s= 0 u tA tA [ T Un[0 t A])

Contests occur in the symmetric signaling game if 1 ugrave ti gt 12(1 + t i) whereas contests occurin the opportunity cost game if 1 ugrave ti gt 12 (where i is for t and tA ) Since

1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2

contests in the signaling game are always weakly less likely to occur than in the opportunitycost game More important since hi gt 0 3 t i gt 0 and since

1

2(1 1 t i)

1

2if t A 0

it follows that interdependence reduces the frequency of costly contests but only in thesignaling game These results corroborate the ndings in proposition 3 The results alsotestify to the parsimony of the previous model QED

References

Andrews David M 1994 Capital Mobility and State Autonomy Toward a Structural Theory ofInternational Monetary Relations International Studies Quarterly 38 (2)193ndash218

Babst Dean V 1972 A Force for Peace Industrial Research 14 (April)55ndash58


Banks Jeffrey S 1990 Equilibrium Behavior in Crisis Bargaining Games American Journal of PoliticalScience 34 (3)599ndash614

Barbieri Katherine 1996 Economic Interdependence A Path to Peace or a Source of InternationalCon ict Journal of Peace Research 33 (1)29ndash49

Barbieri Katherine and Jack S Levy 1999 Sleeping with the Enemy The Impact of War on TradeJournal of Peace Research 36 (4)463ndash79

Barbieri Katherine and Gerald Schneider 1999 Globalization and Peace Assessing New Directions inthe Study of Trade and Con ict Journal of Peace Research 36 (4)387ndash404

Beck Nathaniel Jonathan N Katz and Richard Tucker 1998 Taking Time Seriously Time-SeriesCross-Section Analysis with a Binary Dependent Variable American Journal of Political Science 42(4)1260ndash88

Benoit Kenneth 1996 Democracies Really Are More Paci c (in General) Reexamining Regime Typeand War Involvement Journal of Con ict Resolution 40 (4)636ndash57

Blainey Geoffrey 1973 The Causes of War New York Free PressBremer Stuart A 1992 Dangerous Dyads Conditions Affecting the Likelihood of Interstate War

1816ndash1965 Journal of Con ict Resolution 36 (2)309ndash41mdashmdashmdash 1993 Democracy and Militarized Interstate Con ict 1816ndash1965 International Interactions 18

(3)231ndash49Bueno de Mesquita Bruce 1981 The War Trap New Haven Conn Yale University Pressmdashmdashmdash 1990 Pride of Place The Origins of German Hegemony World Politics 43 (1)28ndash52Bueno de Mesquita Bruce and David Lalman 1992 War and Reason Domestic and International

Imperatives New Haven Conn Yale University PressCaporaso James A 1978 Dependence Dependency and Power in the Global System A Structural and

Behavioral Analysis International Organization 32 (1)13ndash43Chan Steve 1984 Mirror Mirror on the Wall Are the Freer Countries More Paci c Journal of

Conict Resolution 28 (4)617ndash48Cohen Benjamin J 1998 The Geography of Money Ithaca NY Cornell University PressDeutsch Karl W 1978 The Analysis of International Relations 2d ed Englewood Cliffs NJ

Prentice-HallDixon William J 1993 Democracy and the Management of International Con ict Journal of Con ict

Resolution 37 (1)42ndash68mdashmdashmdash 1994 Democracy and the Peaceful Settlement of International Con ict American Political

Science Review 88 (1)14ndash32Dorussen Han 1999 Balance of Power Revisited A Multi-Country Model of Trade and Con ict

Journal of Peace Research 36 (4)443ndash62Eichengreen Barry 1996 Globalizing Capital Princeton NJ Princeton University PressFearon James D 1994 Domestic Political Audiences and the Escalation of International Disputes

American Political Science Review 88 (3)577ndash92mdashmdashmdash 1995 Rationalist Explanations for War International Organization 49 (3)379ndash414Frieden Jeffry 1991 Invested Interests The Politics of National Economic Policies in a World of Global

Finance International Organization 45 (4)425ndash51Gartzke Erik 1998 Kant We All Just Get Along Opportunity Willingness and the Origins of the

Democratic Peace American Journal of Political Science 42 (1)1ndash27mdashmdashmdash 1999 War Is in the Error Term International Organization 53 (3)567ndash87mdashmdashmdash 2000 Preferences and the Democratic Peace International Studies Quarterly 44 (2)191ndash210Gartzke Erik and Dong-Joon Jo 2000 Tipping the Scale Signaling and the Democratic Peace

Unpublished manuscript Pennsylvania State UniversityGasiorowski Mark J 1986 Economic Interdependence and International Con ict Some Cross-National

Evidence International Studies Quarterly 30 (1)23ndash38Gilpin Robert 1987 The Political Economy of International Relations Princeton NJ Princeton

University Press


Gochman Charles S and Zeev Maoz 1984 Militarized Interstate Disputes 1816ndash1976 ProcedurePatterns and Insights Journal of Conict Resolution 28 (4)585ndash616

Goodman John B and Louis W Pauly 1993 The Obsolescence of Capital Controls EconomicManagement in an Age of Global Markets World Politics 46 (1)50ndash82

Gowa Joanne and Edward D Mans eld 1993 Power Politics and International Trade AmericanPolitical Science Review 87 (2)408ndash21

Green Donald P Soo Yeon Kim and David Yoon 2001 Dirty Pool International Organization 55(2)441ndash68

Grilli Vittorio and Gian Maria Milesi-Ferretti 1995 Economic Effects and Structural Determinants ofCapital Controls IMF Staff Papers 42 (3)517ndash51

Hamada Koichi 1976 A Strategic Analysis of Monetary Interdependence Journal of Political Economy84 (4)677ndash700

Hirschman Albert O 1977 The Passions and the Interests Political Arguments for Capitalism BeforeIts Triumph Princeton NJ Princeton University Press

International Monetary Fund Various years Annual Report on Exchange Arrangements and ExchangeRestrictions Washington DC IMF

mdashmdashmdash 1993 Direction of Trade ICPSR 7628 Washington DC IMF Distributed by InteruniversityConsortium for Political and Social Research Ann Arbor Mich

mdashmdashmdash Various years International Financial Statistics Yearbook Washington DC IMFJaggers Keith and Ted Robert Gurr 1995 Tracking Democracyrsquos Third Wave with the Polity III Data

Journal of Peace Research 32 (4)469ndash82Jones Daniel M Stuart A Bremer and J David Singer 1996 Militarized Interstate Disputes

1816ndash1992 Rationale Coding Rules and Empirical Patterns Conict Management and PeaceScience 15 (2)163ndash213

Kant Immanuel [1795] 1957 Perpetual Peace Reprint New York The Liberal Arts PressKeohane Robert and Joseph S Nye Jr 1989 Power and Interdependence 2d ed Glenview Ill Scott

ForesmanKirshner Jonathan 1995 Currency and Coercion The Political Economy of International Monetary

Power Princeton NJ Princeton University PressKroll John A 1993 The Complexity of Interdependence InternationalStudies Quarterly 37 (3)321ndash48Leblang David A 1997 Domestic and Systemic Determinants of Capital Controls in the Developed and

Developing World International Studies Quarterly 41 (3)455ndash74Mans eld Edward D and Rachel Bronson 1997 Alliances Preferential Trading Arrangements and

International Trade American Political Science Review 91 (1)94ndash106Maoz Zeev 1999 Dyadic Militarized Interstate Disputes (DYMID10) Data Set Available at lt http

spirittauacil zeevmaozgt Maoz Zeev and Bruce Russett 1993 Normative and Structural Causes of Democratic Peace 1946ndash

1986 American Political Science Review 87 (3)624ndash38McMillan Susan M 1997 Interdependence and Con ict Mershon International Studies Review 41

(1)33ndash58Montesquieu Baron de [1748] 1989 Spirit of the Laws Reprint Cambridge Cambridge University

PressMorgan T Clifton and Sally Howard Campbell 1991 Domestic Structure Decisional Constraints and

War So Why Kant Democracies Fight Journal of Conict Resolution 35 (2)187ndash211Morrow James D 1999 How Could Trade Affect Con ict Journal of Peace Research 36 (4)481ndash89Morse Edward L 1976 Modernization and the Transformation of International Relations New York

Free PressOneal John R and Bruce M Russett 1997 The Classical Liberals Were Right Democracy Interde-

pendence and Con ict 1950ndash1985 International Studies Quarterly 41 (2)267ndash93mdashmdashmdash 1999a Assessing the Liberal Peace with Alternative Speci cations Trade Still Reduces

Con ict Journal of Peace Research 36 (4)423ndash42


mdashmdashmdash 1999b Is the Liberal Peace Just an Artifact of Cold War Interests Assessing Recent CritiquesInternational Interactions 25 (3)213ndash41

Oneal John R Frances H Oneal Zeev Maoz and Bruce Russett 1996 The Liberal Peace Interde-pendence Democracy and International Con ict 1950ndash1985 Journal of Peace Research 33 (1)11ndash28

Papayoanou Paul A 1999 Power Ties Economic Interdependence Balancing and War Ann ArborUniversity of Michigan Press

Polachek Solomon W 1980 Con ict and Trade Journal of Con ict Resolution 24 (1)55ndash78Polachek Solomon W John Robst and Yuan-Ching Chang 1999 Liberalism and Interdependence

Extending the Trade-Con ict Model Journal of Peace Research 36 (4)405ndash22Pollins Brian M 1989 Does Trade Still Follow the Flag American Political Science Review 83

(2)465ndash80Ray James Lee 1997 On the Level(s) Does Democracy Correlate with Peace Unpublished manuscript

Vanderbilt UniversityRosenau James M 1984 A Pre-Theory Revisited World Politics in an Era of Cascading Interdepen-

dence International Studies Quarterly 28 (3)245ndash305Rousseau David L Christopher Gelpi Dan Reiter and Paul K Huth 1996 Assessing the Dyadic

Nature of the Democratic Peace 1918ndash88 American Political Science Review 90 (3)512ndash33Rummel Rudolph J 1997 Power Kills Democracy as a Method of Nonviolence New Brunswick NJ

Transaction PublishersRussett Bruce 1993 Grasping the Democratic Peace Principles for a PostndashCold War World Princeton

NJ Princeton University PressRussett Bruce John R Oneal and David R Davis 1998 The Third Leg of the Kantian Tripod for Peace

International Organizations and Militarized Disputes 1950ndash1985 International Organization 52(3)441ndash67

Sartori Anne 1996 The Use of Effective Cheap-Talk Signals in International Disputes Unpublishedmanuscript University of Michigan

Schelling Thomas C 1960 The Strategy of Con ict Cambridge Mass Harvard University Pressmdashmdashmdash 1966 Arms and Inuence New Haven Conn Yale University PressSchultz Kenneth A 1998 Domestic Opposition and Signaling in International Crises American

Political Science Review 92 (4)829ndash44mdashmdashmdash 1999 Do Domestic Institutions Constrain or Inform Contrasting Two Institutional Perspectives

on Democracy and War International Organization 53 (2)233ndash66Simmons Beth A 1994 Who Adjusts Domestic Sources of Foreign Economic Policy During the

Interwar Years Princeton NJ Princeton University PressSinger J David Stuart Bremer and John Stuckey 1972 Capability Distribution Uncertainty and Major

Power War 1820ndash1965 In Peace War and Numbers edited by Bruce M Russett 19ndash48 BeverlyHills Calif Sage

Smith Alastair 1998 International Crises and Domestic Politics American Political Science Review 92(3)623ndash38

Solingen Etel 1998 Regional Orders at Centuryrsquos Dawn Global and Domestic In uences on GrandStrategy Princeton NJ Princeton University Press

Summers Robert and Alan Heston 1991 The Penn World Table (Mark 5) An Expanded Set ofInternational Comparisons 1950ndash1988 Quarterly Journal of Economics 106 (2)327ndash69

Suzuki Motoshi 1994 Economic Interdependence Relative Gains and International Cooperation TheCase of Monetary Policy Coordination International Studies Quarterly 38 (3)475ndash98

Tetreault Mary Ann 1980 Measuring Interdependence International Organization 34 (3)429ndash43Wagner R Harrison 2000 Bargaining and War American Journal of Political Science 44 (3)469ndash84Weede Erich 1984 Democracy and War Involvement Journal of Con ict Resolution 28 (4)649ndash64World Bank 1998 The 1998 World Development Indicators CD-ROM Washington DC World Bank


one hand a rich history of theorizing offers speculation addressing virtually everyaspect of the relationship between economics and con ict On the other handempirical studies of interdependence often adopt excessively narrow indicators ofeconomic activity It may be rewarding to take an intellectual step backmdashto brie yassess the broader theoretical question of how interdependence is likely to affectcon ict behaviormdashand then to examine promising aspects of the relationship usingmore appropriate indictors

We begin with a theory of disputes A valid explanation for the effect ofeconomics on peace must be placed in the context of an account of why most statesoccasionally resort to military violence Using a theory of dispute onset based onwork by James Fearon and others we deduce conditions under which interdepen-dence likely contributes to peace2 In contrast to conventional interpretations weshow that opportunity costs associated with economic bene ts generally cannotdeter disputes Instead interdependence creates the means for states to demonstrateresolve without resorting to military violence Liberal states more ably address theinformational problems that give rise to costly contests credibly communicatingthrough costly signals using nonviolent methods of con ict3

Our analysis calls for a notion of interdependence involving aspects of economicactivity besides trade Most studies of interdependence and con ict focus solely onbilateral or aggregate trade ows but interdependence through international capitalis substantially larger than exchanges of goods and services Capital markets linkaspects of domestic economies that otherwise have little global exposure Apreoccupation with risk leads capital to react to political violence in ways that arearguably both more sensitive and more unwavering States can trade with theenemy but political shocks to capital market equilibria invariably imply capital ight andor higher rents in the shadow of costly contests Peace may be a positivepolitical externality of commerce but risk is clearly a negative economic externalityof political contests Other macropolitical aspects of international economicsmdashsuchas the need for monetary policy coordinationmdashare also omitted in previous studiesof interdependence

Thus while accepting as valid the correlation between interdependence andpeace we seek to alter both the logic underpinning the observation and the scope ofindicators used in assessing the relationship Through a series of formal illustrationsand models we show that the opportunity cost conception cannot account for theimpact of interdependence on peace We also show that costly signaling offers asatisfactory alternative We broaden empirical assessment of interdependence byintroducing measures of other aspects of economy We test our ideas by replicatingthe work of a prominent research program on liberalism and peace4 Results support

2 See Blainey 1973 Fearon 1995 Gartzke 1999 and Wagner 20003 Morrow outlines an argument about trade interdependence that in important respects parallels our

own Morrow 1999 We develop our ideas independently derive them formally apply them morebroadly and offer empirical tests

4 See Oneal and Russett 1997 and Russett Oneal and Davis 1998




































































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press






















































































































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press








































































































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press



























































































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press


















































































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press








































































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press






























































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press





















































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press











































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press





































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press































































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press

























































Gilpin 1987




Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press






















































Research Design
































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press













































































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press



































































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press




























































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press





















































Control Variables



68



























(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press







































































(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press



























































(1)










DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press























































DEMOCRACYL 2 00244 2 00001 00072 00058 00092(00065) (00064) (00122) (00063) (00123)

DEMOCRACYH 00195 00168 00202 00128 00138(0006) (00062) (00112) (00064) (00114)

DEPENDENCELt2 1 2 62620 2 46895 2 321126 2 00442 2 30536(56863) (38974) (15895) (3900) (15825)


ALLIES 2 01539 2 17665 2 10442 2 14858 2 10665(00065) (00064) (00122) (00063) (00123)

CONTIGUITY 06491 07659 07001 07283 07552(00996) (01050) (01751) (01125) (01796)






SPLINE1 00029 00029 00021 00029 00021(00003) (00005) (00004) (00005) (00004)

SPLINE2 2 00022 2 00021 2 00011 2 00021 2 00010(00003) (00005) (00002) (00005) (00002)


Constant 2 18176 2 17665 2 10442 2 14858 2 10665(01047) (01141) (02436) (01395) (02610)





Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press





















































Results








00105 100


00054 51( 2 00051)


00028 27( 2 00077)


00013 12( 2 00092)






DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press























































DEMOCRACYH 00143 00130 00219 00066 00151(00066) (00074) (00121) (00079) (00128)


THREE-YEAR ECONOMIC

GROWTHL

2 00168 2 00287 2 00182 2 00297 2 00157(00069) (00080) (00197) (00081) (00198)

ALLIES 2 02121 2 17665 2 10442 2 14858 2 10665(00897) (00980) (001926) (00971) (01916)

CONTIGUITY 04582 05086 05769 04670 06329(01039) (01252) (01918) (01320) (01993)






SPLINE1 00049 00048 00028 00048 00029(00004) (00005) (00004) (00006) (00004)

SPLINE2 2 00040 2 00037 2 00014 2 00038 2 00014(00004) (00005) (00002) (00005) (00002)


Constant 2 13502 2 12308 2 06073 2 09737 2 06058(01088) (01369) (02352) (01548) (02520)







DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press























































DEMOCRACYH 00141 00147 00214 00127 00151(00058) (00069) (00112) (00068) (00114)

DEPENDENCELt2 1 2 60440 2 38462 2 35156 2 00611 2 36023(50769) (39350) (16474) (01674) (16481)


ALLIES 2 00281 01164 2 01005 2 00320 2 01336(00968) (00964) (001945) (00076) (01874)

CONTIGUITY 07604 08014 06865 00804 07672(01089) (01163) (01770) (00937) (01932)


PREFERENCE

SIMILARITYt2 1

2 03314 2 01003 02240 2 00007 02732(01066) (01560) (02889) (00003) (02833)





SPLINE1 00027 00029 00023 00029 00023(00004) (00005) (00004) (00005) (00004)

SPLINE2 2 00020 2 00021 2 00012 2 00021 2 00012(00004) (00005) (00002) (00005) (00002)


Constant 2 18080 2 17651 2 10991 2 14842 2 11377(00877) (01125) (02306) (01314) (02483)























Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press







































































Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press

























































Control Variables
















Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press
































































Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press





















































Conclusion






Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press





















































Appendix





84 Banks 1990


Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press




















































Variables






If

d a 2c

t


d ugrave a 2c

t

else f = 1



UA 5 [(1 2 f ) z (1 2 d)] 1 [ f z ((1 2 a) 2 c)]shy UA

shy d5 ( f 2 1)




d 5 a 2c

t


to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press




















































to any

d a 2c

t

and

(1 2 d) 5 1 2 a 1c




d 5 a 2c

t




E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c] z f(t)dt 51

2(a 2 2c 2 d)


(a 2 2c) S a 2c

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2










shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press




























































shy f5 t z (a 2 d) 2 c 2 h

If

d a 2c 1 h

t


d ugrave a 2c 1 h

t

else f = 1




shy d5 ( f 2 1)






E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press























































E0

1 shy UB

shy fz f(t)dt 5 E

0

1

[t(a 2 d) 2 c 2 h] z f(t)dt 51

2(a 2 2c 2 d 2 2h)


(a 2 2c 2 2h) S a 2c 1 h

t D 3 t 1

2

else if

t oslash1

2 f 5 0



t oslash1

2


(bA u t t [ T Un[0 1])










shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press


























































shy UBs= 1

shy f5 t z (a 2 d1) 2 c

shy UB


De ne t [ T st

shy UB

shy s5 0 if t 5 t


d 0 af 1 d1(1 2 f ) 2h 1 c z f

t

and if

d 1 a 2c

t




where i [ [01]

UA 5 (1 2 s)[1 2 d0 1 h] 1 s z [(1 2 f )(1 2 d1)] 1 [ f((1 2 a) 2 c)]


shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press




















































shy UA

shy 05 (s 2 1)

shy UB

shy 15 s( f 2 1)


Solution


Et

1 shy UBs= 1

shy fz f(t)dt 5 E

t

1

[t(a 2 d1) 2 c] z f (t)dt 51

2[(t 2 1)(a 2 2c 2 d1 1 t (a 2 d1))]


d 1 5 a 22c

1 1 t


shy UB

shy s



2(a 2 d0)


P( f 5 1) z P(s 5 1 u f 5 1)

P( f 5 1) z P(s 5 1 u f 5 1) 1 P( f 5 0) z P(s 5 1 u f 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2




2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press






















































2(2h 2 c)




(c 2 2h)2






d 1t at 2 c 3 2ct c(1 1 t ) 3 t c(1 1 t )

2c5

1

2(1 1 t )

else f = 0 Since

1

2(1 1 t )

1

2 t 0



f = 0) if

1




s= 0 u t t [ T Un[0 t ])














1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press
































































1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press





























































1 (1 2 p)(tA(1 2 d) 1 hA(1 1 d )

shy UA

shy d5 tA(p 2 1)

shy UAp= 1




85 Ibid



shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press





















































shy UB


shy UBp= 1



Solution


E0

1 shy UAp= 1


0

1


2d A(dB 2 a 2 2(c 1 hA))

Solving


tA3 tA

1

2



(dB) 3 P( fA 5 1) 51

2

Simplifying

shy UB

shy p

and solving for

t 5d B(h 1 c) 1 2h

2( d B(a 1 hA 1 c) 2 d)


If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press




















































If A offers

d d B(a 1 hA 1 c) 2d B(h 1 c) 2 2h

2 t p 5 1 else p 5 0


shy UB

shy p


E0

1 shy UB

shy pz f(t)dt 5 E

0

1

t[ d B(a 1 c 1 hA) 2 d] 21

2( d B(c 1 h) 1 1) z f(t)dt

51

2( d B(a 1 hA) 2 h(2 1 d B) 2 d)











1 fA(tA(1 2 a) 2 c))]


shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press




















































shy UA

shy d05 tA(s 2 1)

shy UA

shy d15 tAs(p 2 1)

shy UAs= 1p= 1sA= 1


shy UAs= 1p= 1


De ne

t A [ T stshy UA

s= 1p= 1

shy sA5 0



tA

and if

dB1 a 1c







shy UB


shy UB


shy UB


shy UBs= 1


t [ T stshy UB

s= 1

shy s5 0


if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press




















































if t = t (s = 1) if


t



t

Solution


Et A


shy fAf(tA)dtA 5 E

t A

1


51

2d A( t A 2 1)(2c 1 (a 2 dB1)(1 1 t A))

Solving for dB 1

dB1 5 a 22c

1 1 t A


shy UAs= 1p= 1

shy sA

and solving for t A


2 1 4(a 2 dB0)(hA 1 fAc)

2(a 2 dB0)


P( fA 5 1) z P(sA 5 1 u fA 5 1)


5

S 1 21

2(1 1 t A) D 1

S 1 21

2(1 1 t A) D 1 1

1

2(1 1 t A) 1

1

2(1 1 t A) 2 t A

1

2(1 1 t A) 2

51

2



S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press





















































S Et p

1 shy UBs= 1p= 1


dB0 51





shy UBs= 1

shy p5 0

and solving for t

t p 5d BsA(h 1 fAc)


If A offers


t

p 5 1 else p 5 0


S Et

1 shy UBs= 1

shy p D


1 1 t


Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press




















































Substituting d1 in

t p 3 t p 51 1 t

2

B accepts d1 if

t 1 1 t

2 else ( p 5 1)


P( p 5 1) z P(s 5 1 u p 5 1)

P( p 5 1) z P(s 5 1 u p 5 1) 1 P( p 5 0) z P(s 5 1 u p 5 0)

5

S 1 21

2(1 1 t ) D 1

S 1 21

2(1 1 t ) D 1 1

1

2(1 1 t ) 1

1

2(1 1 t ) 2 t

1

2(1 1 t ) 2

51

2

Setting S d UB




If A offers



t p 5 1 else p 5 0


S E0

1 shy UB

shy d0 D


and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press




















































and solving


t





s= 0 u t t [ T Un[0 t ])





1 21

2(1 1 t i) 5

1

2(1 2 t i) oslash

1

2


1

2(1 1 t i)

1

2if t A 0


References

























University Press









































































University Press






























































































































Documents

Investing in the Peace: Economic Interdependence and ...egartzke/publications/gartzkeetal_io_01.pdf · Work by Robert O. Keohane and Joseph S. Nye, James A. Caporaso, Karl W. Deutsch,