Online Appendix to The Paradox of Civilization

Ernesto Dal Bó

UC Berkeley

Pablo Hernández-Lagos

NYU Abu Dhabi

Sebastián Mazzuca

Johns Hopkins

July 9, 2020

1

Part I

Theory: Modeling choices and proofs for

the main text

A Modeling choices, interpretation, and robustness

Internal vs external conflict and social structure of incumbent polity The civilization process

precedes nations and hence distinctions between domestic and external conflict. The classification of

challengers as internal vs. external (akin to classifying conflict as about internal order vs. sovereignty)

partly depends on whether the eventual civilization incorporated or excluded formerly hostile popula-

tions. Both cases occurred in history. Also, we do not distinguish between ruler and subjects within

each actor. The incumbent and challenger in our model can be taken to be representative agents

of their respective groups, perfectly benevolent rulers acting on their behalf, or perfectly extractive

rulers who are residual claimants.1 A long tradition of work on the political sources of prosperity has

focused on internal order considerations (Fried 1960, Olson 1993, 2000, North, Wallis and Weingast

2009, Boix 2015). We abstract from those to keep our focus on the incumbent-challenger dynamic.

Asymmetries We have kept as many aspects as possible symmetric between the incumbent

and the challenger, and only introduced asymmetries that we deemed helpful to analyze the type of

interaction of interest. The incumbent acting as a Stackelberg leader helps make deterrence possible.

Also, while defense effort costs the incumbent resources, it does not deplete a budget for the challenger.

This is for tractability. It would be possible to include a budget constraint for the challenger and the

advantage of a wealthier incumbent at attaining deterrence would operate in similar fashion. However,

the reaction function of the challenger would have a kink where the budget constraint becomes binding,

making the analysis less elegant. A third asymmetry is that the incumbent can accumulate and the

challenger does not. This maintains tractability, and matches historical situations where one settled,

food-producing, group has more room to accumulate than rival nomadic groups.

A related aspect is whether results are robust to individual actors choosing which roles to play -

something akin to an entry stage. An extension available upon request shows that such an extension

1The latter interpretation is more suitable if growth in our model is interpreted in per working capita terms, becauseextraction and concentration of surplus in a few hands would be a way in which an ancient society could escapeMalthusian population adjustment. Alternatively we can interpret growth in our model at the level of groups, wherewealth becomes population. For models of conflict over land in prehistoric settings that put Malthusian dynamics frontand center, see Dow and Reed (2003) and Dow, Mitchell and Reed (2017).

2

is possible and a no-arbitrage condition holds under which both incumbent and challenger groups get

populated, previous to the action proceeding to the game modeled in the paper.

Trade and other interactions We do not consider trade between the incumbent and the chal-

lenger. Exchange between populations has been documented around the time of the birth of the first

cities (see Mann 1986: 79). In this paper we seek to understand the interaction between prosperity–as

driven only by investment, also the focus of traditional growth models–and security. The way this in-

teraction is further affected by the possibility of exchange is an important next step for future research

(Martin et al. (2008), for instance, address trade and war in modern states). Similarly, we do not

consider potential interactions between multiple incumbents (food producers who can accumulate),

which are also part of the complex civilizational process. These are certainly limitations in the current

analysis. Our model has empirical traction to the extent that (i) conflict among incumbents involves

asymmetries similar to those in our model or (ii) however such conflict is resolved, civilization still

requires prevailing against non-accumulating challengers.

Dynamic considerations Our basic model has two periods. Would things change qualitatively if

more periods were considered? With two periods, it does not matter whether the challenger is different

across periods or not, nor whether the challenger aims to steal the flow income or permanently replace

the incumbent. An extension to three periods is direct if a different challenger arrives each period

who only aims to steal the income flow. If the challenger aims to replace the incumbent, an extension

to three periods is not trivial for at least two reasons. First, the challenger now cares directly about ρ

as he may want to invest upon becoming the incumbent, and he values the asset for its future, as well

as current, yield. In addition, with a longer horizon the incumbent may want to invest in order to be

able to finance a larger army in the future and attain more security. We show in Part II below that

an extension to more periods preserves our qualitative results. Therefore, an extension to many or

even infinite periods, while extremely difficult, would add more in terms of elegance than substance.

The use of contests Like many authors before us, we use a Tullock contest and abstract from

transfers.2 As is well known, even when transfers are possible inefficient conflict may occur, for ex-

ample due to inconsistent priors across players, agency issues, commitment problems, or asymmetric

information (see Jackson and Morelli (2011) for a survey of reasons for war). Every paper on conflict

must take a stand on whether or not to microfound conflict onset by reference to one of those phe-

nomena. The cost of the added structure is justified when a specific distortion responsible for conflict

is particularly likely or germane given the problem under investigation. When the researcher remains

2Generalized versions of the ratio-based contest success function exist but are less tractable. Hirshleifer (2001)explores some of the difficulties. The typical generalization is to consider functions of the type a

α

aα+bα . The mostimportant feature of our model, which is the possibility of generating deterrence, obtains for any function satisfyingα ∈ [0, n/(n− 1)] in a symmetric contest with n players. For α > n/(n− 1) pure strategy equilibria cease to exist.

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agnostic about such connections, the more parsimonious approach that we take seems justifiable.

Still, we explore two simple extensions that include the possibility of transfers paid by the incum-

bent to the challenger in Part III below and show that the equilibrium characterized in the main text

remains unchanged.

Destruction from conflict Our results do not change qualitatively if conflict generates some

destruction.3 The model incorporates dissipation via contest efforts.

Do we really need a model? The takeaways of the model are to some extent intuitive: in order

to attain civilization security threats must be kept in check by natural or artificial defenses. But the

simplicity of this takeaway covers up a number of non-obvious effects. First, the static productive

advantage captured by initial income v plays a very different role than the dynamic advantage captured

by growth capability ρ. Second, we did not expect that the partition of the parameter space in

Proposition 1 would be invariant in initial income, or that investments would not go to zero as growth

capability increases and the challenger becomes more aggressive. Likewise, it is not obvious that

higher income will make the erection of artificial defense incentive compatible when both the higher

income and the investments in defense make the challenger more aggressive in the present period. We

wrote this model strictly thinking about what the right ingredients and strategic tensions were, and

before knowing what the solution would be. In other words, the model was not backward engineered

to produce a set of results.

References

Boix, C. (2015). Political Order and Inequality: Their Foundations and their Consequences for

Human Welfare. Cambridge University Press.

Dow, G. and C. Reed (2003). “The Origins of Inequality: Insiders, Outsiders, Elites, and Common-

ers.” Journal of Political Economy 121(3), 609-641.

Dow, G., Mitchell, L., and C. Reed (2017). “The Economics of Early Warfare Over Land,” Journal

of Development Economics 127, 297-305.

Fried, M. (1960), On the Evolution of Social Stratification and the State. Indianapolis: Bobbs-

Merrill.

Hirshleifer, J. (2001). The Dark Side of the Force: Economic Foundations of Conflict Theory.

Cambridge: Cambridge University Press.

3The model presented here represents the limit case of a more general model where a fraction σ ∈ [0, 1] of the assetsurvives the war. The solution of that model is continuous in σ, so the solution remains qualitatively similar when σdips below 1 (proof available upon request).

4

Jackson, M. and M. Morelli (2011). “The Reasons for Wars: An Updated Survey.” In: C. Coyne and

R. Mathers (eds.), The Handbook on the Political Economy of War, Edward Elgar Publishing,

chapter 3.

Mann, M. (1986), The Sources of Social Power Vol. I: A History of Power from the Beginning to

A.D. 1760. Cambridge University Press.

Martin, P., Mayer, T., and M. Thoenig (2008). “Make trade not war?.” The Review of Economic

Studies, 75(3), 865-900.

North, D., Wallis, J., and B. Weingast (2009). “Violence and Social Orders: A Framework for

Interpreting Recorded Human History.” Cambridge University Press.

Olson, M. (1993). “Dictatorship, Democracy, and Development.” American Political Science Review,

87(3), 567-576.

Olson, M. (2000). “Power And Prosperity: Outgrowing Communist And Capitalist Dictatorships.”

Basic Books.

B Proof of Proposition 1

This is a particular case of the model with general values of κc, studied in Proposition 3.�

C Proof of Proposition 2

We use the convention that in period 0 the incumbent starts by selecting m0, and then picks a0 and

i0.4 Backward inducting within period 0, we first characterize solutions to (a0, i0) given m0 and then

prove the statements in the proposition concerning m∗0.

From the expressions for V1 in Proposition 1, as the incumbent lands in each of the regions

PS− SS, the value V1 is,4The assumption that m0 is decided before a0 and i0 is just to simplify exposition. The results do not vary if

(m0, a0, i0) are selected simultaneously since the challenger moves after the incumbent.

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V1(i0,m0) = (v0 + ρi0)×

(κ0+m0)(1+ρ)κ0+m0+ρ

(κ0 +m0, ρ) ∈ PS√(κ0+m0)

ρ(1+ρ)

2(κ0 +m0, ρ) ∈ PDI(

1 + κ0+m04

)(κ0 +m0, ρ) ∈ SI(

2− 1κ0+m0

)(κ0 +m0, ρ) ∈ SS

≡ (v0 + ρi0)S(m0).

Once m0 is set, the problem of choosing (a0, i0) to solve the program in equation (7) in the paper

is similar to the choice of (a1, i1) in the baseline model, except now the continuation value depends

explicitly on m0 through S(m0). The objective function is differentiable and concave in a0 and i0,

and the constraints are linear, so the first order and complementary slackness conditions below are

necessary and sufficient for a unique maximum:

a0 :1

2

√(v0 + ρi0)S(m0)

a0− 1κ0− λBC

κ0− λDC + λa = 0 (1)

i0 :ρS(m0)

2

√a0

(v0 + ρi0)S(m0)− 1− λBC + λDCρS(m0) + λi = 0 (2)

λBC(v0 −m0 −a0κ0− i0) = 0, λDC ((v0 + ρi0)S(m0)− a0) = 0, λaa0 = 0, λii0 = 0 (3)

As before, λBC , λDC are the Lagrange multipliers for the budget constraint and deterrence constraints,

and λa, λi are the multipliers for the non-negativity constraints for the control variables. The infinite

marginal utility of a0 at zero implies a0 > 0 and λa = 0, so there are eight possible cases depending on

whether the remaining three Lagrange multipliers are positive or zero. The following Lemma shows

that given our Assumption 1 there are only two feasible cases in period 0.

Lemma 1 If Assumption 1 holds, then in period 0 the incumbent chooses:

i) i0 = 0 and a0 =κ204v0S(m0) when

v0S(m0)(v0−m0) <

4κ0

; or

ii) i0 = 0 and a0 = κ0(v0 −m0) when v0S(m0)(v0−m0) >4κ0.

C.1 Proof of Lemma 1

There are two cases that constitute a solution out of eight possible ones. We will show that the

assumption 1 implies that the first case holds when v0S(m0)v0−m0 <

4κ0

and the second case holds whenv0S(m0)v0−m0 >

4κ0

. The remaining six cases can be shown to be either inconsistent with any set of parameter

values or consistent only with a non-generic set.

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1. Case λBC = 0 (BC does not bind),λDC = 0 (DC does not bind, conflict), and λi > 0

(i0 = 0)

The FOCs are,

a0 :1

2

√v0S(m0)

a0− 1κ0

= 0 (4)

i0 :ρS(m0)

2

√a0

v0S(m0)− 1 + λi = 0 (5)

This implies,

a0 =κ204v0S(m0)

λi = 1−κ0ρS(m0)

4

The necessary and sufficient conditions for this case to hold are,

λBC = 0⇔v0S(m0)

v0 −m0<

4

κ0(from the BC not binding)

λDC = 0⇔ 1 >κ204⇔ κ0 < 2 (from the DC not binding)

λi > 0⇔ 1−κ0ρS(m0)

4> 0⇔ κ0ρS(m0) < 4 (from the FOC for i0)

The first inequality holds for values of m0. The second inequality holds given κ0 < 2. The

third holds by assumption 1. To see this, note that assumption 1 implies that κ0ρS(m0) < 4 for all

m0 ∈ [0,∞+). This follows from noting that for a given ρ, limm0→∞+ S(m0) = 1 + ρ, because for m0large enough, precisely m0 > ρ/(ρ− 1), S(m0) = (κ0 +m0)(1 + ρ)/((κ0 +m0) + ρ). Given that S(m0)in strictly increasing in m0, S(m0) < 1 + ρ for all m0. As a result, κ0ρS(m0) 0 (BC binds),λDC = 0 (DC does not bind, conflict), and λi > 0 (i0 = 0)

The FOCs are,

a0 :1

2

√v0S(m0)

a0− 1κ0− λBC

κ0= 0 (6)

i0 :ρS(m0)

2

√a0

v0S(m0)− 1− λBC + λi = 0 (7)

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λBC > 0 implies κ0 (v0 −m0) = a0.From the FOCs, we obtain,

λBC =κ02

√v0S(m0)

a0− 1

λi =κ02

√v0S(m0)

a0− ρS(m0)

2

√a0

v0S(m0).

The parameter conditions for this to be a solution are,

λBC =κ02

√v0S(m0)

a0− 1 > 0⇔ v0S(m0)

v0 −m0>

4

κ0

λDC = 0⇔v0S(m0)

v0 −m0> κ0

λi =κ02

√v0S(m0)

κ0(v0 −m0)− ρS(m0)

2

√κ0(v0 −m0)v0S(m0)

> 0⇔ v0S(m0)v0 −m0

> ρS(m0).

The inequalities v0S(m0)(v0−m0) > κ0 and

v0S(m0)(v0−m0) > ρS(m0) are both implied by the condition

v0S(m0)v0−m0 >

4κ0

,

by assumption 1. To see this, note that assumption 1 implies that κ0ρS(m0) < 4 for all m0 ∈ [0,∞+).This follows from noting that for a given ρ, limκ0→∞+ S(m0) = 1 + ρ, because for κ0 large enough,

κ0 > ρ/(ρ−1), S(m0) = κ0(1+ρ)/(κ0+ρ). Given that S(m0) in strictly increasing inm0, S(m0) < 1+ρ.As a result, κ0ρS(m0) κ0 (because κ0 < 2) and 4/κ0 > ρS(m0)

(because κ0ρS(m0) 0 (BC binds),λDC > 0 (DC binds), and λi = 0

If λDC > 0 then

(v0 + ρi0)S(m0) = a0,

FOC

κ02− 1− λBC − λDCκ0 = 0 (8)

ρS(m0)

2− 1− λBC + λDCρS(m0) = 0 (9)

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If λBC > 0 then

v0 −m0 −a0κ0− i0 = 0

We have four equations and four unknowns. Therefore

λDC =1

2

(κ0 − ρS(m0)κ0 + ρS(m0)

)

λBC =

[ρS(m0)

2− 1 + 1

κ0

(κ02− 1)ρS(m0)

](1 +

ρS(m0)

κ0

)−1

i0 =v0κ0 −m0κ0 − v0S(m0)

κ0 + ρS(m0)

a0 =

(v0κ0 (1 + ρ)− ρm0κ0

κ0 + ρS(m0)

)S(m0)

We need to establish the conditions on the parameters and m0 such that those parameters support

this equilibrium.

λDC =1

2

(κ0 − ρS(m0)κ0 + ρS(m0)

)> 0 ⇐⇒ κ0 > ρS(m0)

λBC =

[ρS(m0)

2− 1 + 1

κ0

(κ02− 1)ρS(m0)

](1 +

ρS(m0)

κ0

)−1> 0 ⇐⇒ κ0 >

ρS(m0)

ρS(m0)− 1

i0 =v0κ0 −m0κ0 − v0S(m0)

κ0 + ρS(m0)> 0 ⇐⇒ v0κ0 −m0κ0 − v0S(m0) > 0

The first and second conditions are enough to prove this case is infeasible in equilibrium. To see

this, note that the minimum value of κ0 is ρS(m0) from the first inequality and ρS(m0)/(ρS(m0)− 1)from the second. If either one is larger than the other, then there are values of κ0 such that violate

the most stringent (largest) constraint. For both to hold, the minimum has to coincide: ρS(m0) =

ρS(m0)/(ρS(m0)− 1), or ρS(m0) = 2. In this case, however, κ0 = 2 which is strictly greater than κ0under assumption 1.

4. Case λBC > 0 (BC binds),λDC = 0 (DC does not bind, conflict), and λi = 0 (i0 > 0)

FOC

9

κ02

√(v0 + ρi0)S(m0)

a0− 1− λBC = 0

ρS(m0)

2

√a0

(v0 + ρi0)S(m0)− 1− λBC = 0

If λBC > 0 then

v0 −m0 −a0κ0

= i0

Equating the two FOCs after solving for λBC yieldsκ0

ρS(m0)(v0 + ρi0)S(m0) = a0, and using this into

the investment equation above we get i0 =v0(1− 1ρ)−m0

2which then yields a0 =

κ0ρ

(v0 + ρ

v0(1− 1ρ)−m02

).

Using these expressions for a0 and i0 we can write,

λBC =κ02

√√√√√√√(v0 + ρ

v0(1− 1ρ)−m02

)S(m0)

κ0ρ

(v0 + ρ

v0(1− 1ρ)−m02

) − 1 = 12

√κ0ρS(m0)− 1.

For λBC > 0, it must be the case that√κ0ρS(m0) > 2, or κ0ρS(m0) > 4, which violates assumption

1, because κ0ρS(m0) < κ0ρ(1 + ρ) < 4, for all m0 ∈ [0,∞+).5. Case λBC = 0 (BC does not bind),λDC = 0 (DC does not bind), and λi = 0 (i0 > 0)

The FOCs are,

a0 :κ02

√(v0 + ρi0)S(m0)

a0− 1 = 0 (10)

i0 :ρS(m0)

2

√a0

(v0 + ρi0)S(m0)− 1 = 0. (11)

The first FOC yields√

(v0+ρi0)S(m0)a0

= 2κ0

, and substituting into the second FOC we get

κ0ρS(m0) = 4

which violates the assumption κ0ρS(m0) < 4, making this an infeasible case.

6. Case λBC = 0 (BC does not bind,λDC > 0 (DC binds, deterrence), and λi = 0 (i0 > 0)

10

The FOCs are,

a0 :1

2− 1κ0− λDC = 0

i0 :ρS

2− 1 + λDCρS(m0) = 0.

From the first FOC, λDC =12− 1

κ0, which is negative because κ0 < 2 under assumption 1.

7. Case λBC > 0 (BC binds),λDC > 0 (DC binds, deterrence), and λi > 0 (i0 = 0)

The FOCs are,

a0 :1

2− 1κ0− λBC

κ0− λDC = 0 (12)

i0 :ρS

2− 1− λBC + λDCρS(m0) + λi = 0 (13)

v0 −m0 −a0κ0

= 0, v0S(m0)− a0 = 0. (14)

From the first FOC, however, λDC =12− 1

κ0− λBC

κ0is less than zero because κ0 < 2 under assumption

1, making this case unfeasible.

8. Case λBC = 0 (BC does not bind),λDC > 0 (DC binds, deterrence), and λi > 0

(i0 = 0)

a0 :1

2− 1κ0− λDC = 0 (15)

i0 :ρ

2− 1 + λDCρS(m0) + λi = 0 (16)

v0 −m0 −a0κ0

> 0, v0S(m0) = a0 (17)

From the first FOC,

λDC =1

2− 1κ0

is negative because κ0 < 2, under assumption 1.�

We now proceed with the statements in the proposition. We analyze separately the cases where

ρ ≤ 2 and ρ > 2.

11

C.2 Case ρ ≤ 2

There are three steps. Step 1 is to compute the expected utility in period t = 0 as a function of m0

fixing all the other parameters for the two cases highlighted in Lemma 1. Step 2 is to note that there

are levels of m0 that shift the regime the polity is in, both in period 0 (depending on which case of

Lemma 1 obtains) and also in period 1 (depending on which region PS,PDI,SI,SS, the polity lands

in period 1). Each of these regime changes is reflected in the expected utility function. Step 3, having

correctly characterized expected utility, is to show that a low enough v0 yields m∗0 = 0 for part 1, and

for part 2(i) that high enough v0 yields a positive m∗0 that is as high as desired.

C.2.1 Proposition 2, ρ ≤ 2, part 1

We prove this by showing that for v0 small enough, m∗0 = 0.

Step 1 - We write the incumbent’s expected utility in each of the two cases in Lemma 1:

Case λBC = 0 (BC not binding),λDC = 0 (DC not binding, conflict), and λi > 0 (i0 = 0)

The proof to Lemma 1 shows that the Lagrange multiplier conditions defining the case implyv0S(m0)v0−m0 <

4κ0

, κ0 < 2, and ρS(m0) <4κ0

, and expected utility is, V0 = v0 −m0 + κ04 v0S(m0).Case λBC > 0 (BC binds),λDC = 0 (DC not binding, conflict) and λi > 0 (i0 = 0)

The Lagrange multiplier conditions imply, v0S(m0)v0−m0 >

4κ0

, v0S(m0)v0−m0 > κ0,

v0S(m0)v0−m0 > ρS(m0), and

expected utility is,

V0 =√κ0 (v0 −m0) v0S(m0).

Step 2 - We denote with m̄ the value of m0 that satisfiesv0S(m̄)v0−m̄ =

4κ0

and which makes the polity

switch from case 1 to case 2 in Lemma 1 in period 0. We denote with mRX|RY the value of m0 such that

regimes change in period 1 from RX to RY. In the case of SI and SS that value is mSI|SS = 2− κ0.Because m̄ is an implicit function of S(.), we compute conditions on the parameters when m̄ lies

respectively below and above mSI|SS. The reason that it is important to know where m̄ lies relative

to mSI|SS is that it will indicate which expected utility expression to use to evaluate choices of m0. It

turns out that if v0 is low enough, more precisely, if v0 <8(2−κ0)8−3κ0 , then m̄ < mSI|SS. To see this, notice

that v0S(m0)v0−m0 is increasing in m0. Using the definition for m̄, note m̄ < mSI|SS ⇔

v0S(mSI|SS)

v0−mSI|SS> 4

κ0⇐⇒

v0 <8(2−κ0)8−3κ0 .

Step 3 - Now we know how to write expected utility depending on the value of v0, given all other

parameters. We only need to find a cutoff in the space of possible values for v0 such that when v0 is

under the cutoff the incumbent prefers to stay in SI. We propose v≡ 8(2−κ0)8−3κ0 . In this case, v0 <

8(2−κ0)8−3κ0

12

is equivalent to a regime described by m̄ < mSI|SS, which means we have to use two different expected

utility expressions in the interval[0,mSI|SS

]depending on whether m0 < m̄, or m0 > m̄. A useful

fact is that 8(2−κ0)8−3κ0 is strictly decreasing in κ0 so its maximum value is at κ0 = 0 (since κ0 ≥ 0). In

this case 8(2−1)8−3×1 = 2.

Segment [0, m̄] Expected utility in period t = 0 is given by case (i) in Lemma 1 and by considering

S(m0) to be given by the expectation of remaining in SI:

V0 = v0 −m0 + κ04 v0S(m0) = v0 −m0 +κ04v0(1 + κ0+m0

4

)= v0(1 +

κ04

+(κ04

)2) +m0

(κ0v016− 1)

Since m̄ < mSI|SS ⇐⇒ v0 < 8(2−κ0)8−3κ0 thenκ0v016− 1 < 0. To see why, replace v0 = 8(2−κ0)8−3κ0 in

κ0v016

so

κ0v016

=κ0

8(2−κ0)8−3κ016

≤ κ0×216

< 1. This implies V0 is decreasing in m0 and the optimal choice is m0 = 0.

Segment [m̄,mSI|SS] Expected utility in period t = 0 is given by case (ii) in Lemma 1 and by

considering S(m0) to be given by the expectation of remaining in SI:

V0 =√κ0 (v0 −m0) v0S(m0) =

√κ0 (v0 −m0) v0

(1 + κ0+m0

4

), and we now show this to decrease

in m0. Note,

dV0dm

= 12

(κ0 (v0 −m0) v0

(1 + κ0+m0

4

))− 12(−κ0v0

(1 + κ0+m0

4

)+ 1

4κ0 (v0 −m0) v0

)and,

dV0dm

< 0 ⇐⇒ 14κ0 (v0 −m0) v0 < κ0v0

(1 + κ0+m0

4

), or, iff v0 < 4 + κ0 + 2m0.

If 4 + κ0 + 2m0 is higher than8(2−κ0)8−3κ0 , the condition m̄ < mSI|SS

(⇐⇒ v0 < 8(2−κ0)8−3κ0

)is also a

suficient condition for V0 in the segment [m̄,mSI|SS] to be decreasing. So, it is sufficient to show that

4 + κ0 + 2m0 >8(2−κ0)8−3κ0 . Because the right hand side is decreasing in κ0, it attains a maximum at

κ0 = 0 and it is equal to 2 which is smaller than any feasible value of the expression in the left hand

side, which is at least 4. Therefore, in this segment utility is maximized at m0 = m̄, and equals

V0 =√κ0 (v0 −m0) v0S(m0) =

√4(v0 − m̄)2 = 2 (v0 − m̄).

Segment [mSI|SS,mSS|PS] Since now m0 can only be larger than m̄, we know expected utility

will be given by case (ii) in Lemma 1 and by the expectation of landing in SS, implying V0 =√κ0 (v0 −m0) v0S (m0) =

√κ0 (v0 −m0) v0

(2− 1

κ0+m0

). Computing the first derivative with respect

to m0, we get,

dV0dm0

=

(κ0 (v0 −m0) v0

(2− 1

κ0 +m0

))− 12[−κ0v0

(2− 1

κ0 +m0

)+κ0 (v0 −m0) v0

(κ0 +m0)2

]

which is negative whenever 2 (κ0 +m0)−1 > (v0−m0)κ0+m0 , or, 2 (κ0 +m0)2−κ0 > v0. Note 2

(κ0 +mSI|SS

)2−κ0 = 8 − κ0. Now note 8 − κ0 > 8(2−κ0)8−3κ0 ≡v, since the LHS is at least 6 and the RHS is at most 2.Thus, dV0

dm0< 0 and utility would be maximized at mSI|SS in this segment.

13

Segment [mSS|PS,∞] Expected utility is given by case (ii) in Lemma 1 and by the expectation oflanding in PS, implying, V0 =

√κ0 (v0 −m0) v0 (κ0+m0)(1+ρ)κ0+m0+ρ and

dV0dm0

=(κ0 (v0 −m0) v0 (κ0+m0)(1+ρ)κ0+m0+ρ

)− 12

−κ0v0 (κ0+m0)(1+ρ)κ0+m0+ρ+κ0 (v0 −m0) v0 (1+ρ)(κ0+m0+ρ)−(κ0+m0)(1+ρ)(κ0+m0+ρ)2

.Note dV0

dm0< 0 whenever v0 <

(κ0+m0+ρ)(κ0+m0)ρ

+ m0. The right hand side of this expression is

increasing in m0, so the minimum is attained at m0 = mSS|PS and it equalsρ

(ρ−1)2 + 2ρρ−1 − κ0. The

highest possible value of v0, v =8(2−κ0)8−3κ0 is smaller than 2 which, in turn, is always smaller than

ρ(ρ−1)2 + 2

ρρ−1 − κ0 given that ρ < 2 (its minimum value is 4, at (κ0, ρ) = (2, 2)). Therefore the

maximum of V0 in this segment is attained at m0 = mSS|PS.

Considering all the segments together, we now show that the global maximum in this case is

m∗0 = 0. This follows from the just demonstrated fact that the maximum within each segment of the

support is at the minimum value, and from the fact that V0 is continuous. This in turn follows from

the fact that S(.) is continuous for all m0 and V0 in period t = 0 is also continuous at m̄: it is easy to

show that in t = 0, V0 in segment [0, m̄] evaluated at m̄ is 2(v0 − m̄), which is equal to V0 in segment[m̄,mSI|SS] evaluated at m̄. Thus, if v0 4κ0

( ρρ−1−κ0)4κ0− 1+ρ

ρ

, then m̄ > mSS|PS. To see this, note that from Lemma 1 and

the fact that v0S(m0)v0−m0 is increasing in m0, m̄ > mSS|PS requires

v0S(mSS|PS)

v0−mSS|PS< 4

κ0, and this follows iff

v0 >4κ0

( ρρ−1−κ0)4κ0− 1+ρ

ρ

.

Therefore everywhere in SI and SS the expected utility of the incumbent corresponds to case (i)

in Lemma 1, but in PS the expected utility switches to that corresponding to case (ii) in Lemma 1

for m0 ≥ m̄.

14

Step 3 - We now show that if v0 > v̄ ≡(

ρρ−1

)24κ0

, expected utility is increasing in m0 for as long as

m0 is in SI and SS, and that there must be an interior solution m∗0 in PS. This solution is generically

unique because the maximum stems from a concave expected utility, regardless of whether it lies in

the segment [mSS|PS, m̄] or [m̄,∞),.

Segment [0,mSI|SS] Expected utility is given by case (i) in Lemma 1 and by the expectation of

landing in SI, implying,

V0 = v0 − m0 + κ04 v0S(m0) = v0 − m0 +κ04v0(1 + κ0+m0

4

)= v0(1 +

κ04

+(κ04

)2) + m0

(κ0v016− 1).

For V0 to be increasing over [0,mSI|SS], we need v0 >16κ0

. This is guaranteed if v̄ =(

ρρ−1

)24κ0> 16

κ0⇔(

ρρ−1

)2> 4⇔ ρ

ρ−1 > 2 which holds strictly for ρ < 2 and with equality at ρ = 2.

Segment [mSI|SS,mSS|PS] Expected utility is given by case (i) in Lemma 1 and by the expectation

of landing in SS, implying, V0 = v0 −m0 + κ04 v0S(m0) = v0 −m0 +κ04v0

(2− 1

κ0+m0

).

Note this expected utility is maximized at m∗0 =√

κ0v04−κ0. But this maximum lies above mSS|PS

whenever√

κ0v04− κ0 > ρ(κ0−1)−κ01−ρ . Or, equivalently, whenever v0 >

4κ0

(ρρ−1

)2≡ v̄. Thus, the polity

must reach PS.

Segment [mSS|PS, m̄] Expected utility is given by case (i) in Lemma 1 and by the expectation of

landing in PS, implying, V0 = v0 −m0 + κ04 v0S(m0) = v0 −m0 +κ04v0

(κ0+m0)(1+ρ)κ0+m0+ρ

. This is a concave

function with a maximum at m∗0 =

√κ0v04 ρ(1 + ρ)−κ0 − ρ

, which shows that if the maximum lies in[mSS|PS, m̄], it is unique. Note if v0 >

(ρρ−1

)24κ0≡ v̄ then m∗0 > mSS|PS. To see this, rewrite the

inequality m∗0 > mSS|PS as√

κ0v04ρ(1 + ρ) − κ0 − ρ > ρρ−1 − κ0 then plug v̄ into the LHS to obtain

1 + ρ > ρ. Therefore for v0 > v̄ the polity will be in the interior of PS in period 1. Moreover, for

v0 > v̄ we have m∗0 is a strictly increasing, unbounded, function of v0.

Segment [m̄,∞) Expected utility is given by case (ii) in Lemma 1 and by the expectation of landingin PS, implying,

V0 =√κ0 (v0 −m0) v0S(m0) =

√κ0 (v0 −m0) v0 (κ0+m0)(1+ρ)κ0+m0+ρ .

The first order condition for a maximum is:

1

2

−κ0v0 (κ0+m0)(1+ρ)κ0+m0+ρ + κ0 (v0 −m0) v0(1+ρ)(κ0+m0+ρ)−(κ0+m0)(1+ρ)

(κ0+m0+ρ)2√

κ0 (v0 −m0) v0 (κ0+m0)(1+ρ)κ0+m0+ρ= 0,

15

which, after some algebra boils down the quadratic,

m20 + 2 (κ0 + ρ)m0 + κ20 + ρ (κ0 − v0) = 0,

which has only one (strictly) positive root m∗0 = − (κ0 + ρ) +√ρ (κ0 + ρ+ v0), an increasing, un-

bounded function of v0. If V0 is concave in [m̄,∞), this maximum is unique. We only need to showthat V0 is concave when evaluated at m

∗0, since V0 is continuously differentiable (see Lemma 2, point

2. in the Online Appendix C.3.2) and there is only one solution to the first order condition.

To show V0 is concave, we need to showd2V0dm20

< 0. Note that,

d2V0dm20

=1

2V 20

[(−κ0v0 dSdm0

)+ κ0v0

ddm0

((v0 −m0) dSdm0

)]V0

−(−κ0v0S (m0) + κ0 (v0 −m0) dSdm0

)dV0dm0

.The second term within curly brackets in d

2V0dm20

is zero when evaluated at m∗0, yielding,

d2V0dm20

=1

2V0

[(−κ0v0

dS

dm0

)+ κ0v0

d

dm0

((v0 −m0)

dS

dm0

)]< 0,

where the sign follows from the fact that dSdm0

= (1 + ρ) ρ > 0 and that

d

dm0

((v0 −m0)

dS

dm0

)= − dS

dm0+ (v0 −m0)

d2S

dm20< 0

because v0 −m0 ≥ 0 and d2Sdm20

= − 2ρ(1+ρ)(κ0+m0+ρ)3

< 0.

We have now proved statement 2(i) in the proposition, namely that m∗0 must lie in segments

[mSS|PS, m̄] or [m̄,∞) and in each case it is a unique solution and a positive, increasing, unboundedfunction of v0.

Proposition 2, ρ ≤ 2, part 2 (ii) We show that growth, security, and EV2 are weakly increasingin κ1, which is an increasing function of m0, and by virtue of 2(i) an increasing function of v0. For

v0 > v̄, we have EV2 =a∗1

a∗1+b∗1V2 =

a∗1√a∗1V2

V2 =√

a∗1V2V2. Since landing in PS implies a

∗1 = V2 = v1

κ1(1+ρ)κ1+ρ

from Proposition 1, and i0 = 0 from Lemma 1, we get EV2 = v0κ1(1+ρ)κ1+ρ

, which is increasing in κ1

and hence in v0. Note once in PS security is complete and invariant in v0 but prosperity increases.

The same is true when landing in PS in the case with ρ > 2 (available below in section C.3.2), while

growth is constant in v0 and security increasing when landing in PDI.

16

C.2.2 Proposition 2, ρ ≤ 2, part 3

In segment [mSS|PS, m̄] the solution is m∗0 =

√κ0v0

4ρ(1 + ρ)− κ0 − ρ, and in segment [m̄,∞) we have

the solution m∗0 = −κ0 − ρ +√ρ (κ0 + ρ+ v0). In each respective case, the v0 that is required to

produce such solution is,

v0 =4

κ0ρ(1 + ρ)(m∗0 + κ0 + ρ)

2

v0 =1

ρ(m∗0 + κ0 + ρ)

2 − (κ0 + ρ) .

Differentiating v0 with respect to ρ shows that such needed income is decreasing in ρ in each case.�

C.3 Case ρ > 2

C.3.1 Proposition 2, ρ > 2, part 1

When ρ > 2, we combine Steps 1 and 2, because the expressions for the utility function in Step 1

remain the same as in the paper (they do not change with ρ). In what follows we show that EU

is strictly decreasing in m0, so the maximum is attained at m0 = 0. Step 1 and 2 - We propose

v ≡(

16−4κ0ρ4ρ−(1+ρ)κ0

). In this case, v0 < v(κ0, ρ) is equivalent to a regime in which m̄ < mSI|PDI which

means we have to use two different EU expressions depending on whether m0 < m̄ or m0 > m̄. A

useful fact will be that v =(

16−4κ0ρ4ρ−(1+ρ)κ0

)is decreasing in both ρ and κ0. Thus, its maximum value

is at ρ = 2 and κ0 = 0. In this case, v =16−4×0×2

4×2−(1+2)×0 = 2. Let us proceed by showing that EU is

decreasing in m0 in each segment.

Segment [0, m̄] Expected utility in period 0 is

EU = v0 −m0 + κ04 v0S(m0) = v0 −m0 +κ04v0(1 + κ0+m0

4

)= v0(1 +

κ04

+(κ04

)2) +m0

(κ0v016− 1).

Since m̄ < mSI|PDI ⇐⇒ v0 < v then it follows that κ0v16 >κ0v016

. Therefore, if κ0v16

< 1, it must

follow that κ0v016− 1 < 0, implying the optimal choice is m0 = 0. To see that κ0v16 < 1, note that v is

at most 2 and κ0 < 2.

Segment [m̄,mSI|PDI] Expected utility in period 0 is

EU =√κ0v0 (v0 −m0)

(1 + κ0+m0

4

). To show this payoff is decreasing in m0 note that,

dEUdm0

=(κ0v0 (v0 −m0)

(1 + κ0+m0

4

))− 12(−κ0v0

(1 + κ0+m0

4

)+ 1

4κ0v0 (v0 −m0)

), which must be

negative because 14

(v0 −m0) < 1 + κ0+m04 ⇔ v0 < 4 + κ0 + 2m0 which must hold since v0 < v < 4. Asa result, the maximum is attained at m0 = 0 in the interval [0,mSI|PDI].

17

Segment [mSI|PDI,mPDI|PS] Expected utility in period 0 is,

EU =

√κ0v0 (v0 −m0)

(√κ0+m0

ρ(1+ρ)

2

). Marginal expected utility is,

dEUdm0

=√κ0v02

((v0 −m0)

(√κ0+m0

ρ(1+ρ)

2

))− 12

−(√

κ0+m0ρ

(1+ρ)2

)+ (v0 −m0) (1+ρ)4

(κ0+m0

ρ

)− 12 1ρ

and this is nega-tive whenever (v0 −m0) 12

(κ0+m0

ρ

)− 12 1ρ<√

κ0+m0ρ

, or, v0 < 2κ0 + 3m0. The RHS of this expression

is higher than (12/ρ)− κ0 (after re-arranging: 2κ0 + 3m0 = 3(κ0 +m0)− κ0 > 3× (4/ρ)− κ0), whichis higher than v (this last step follows from simple algebra). Since v0 < v, we conclude that

dEUdm0

< 0.

In this segment EU would be maximized at m0 = mSI|PDI.

Segment [mPDI|PS,∞] Expected utility in period 0 is,EU =

√κ0v0 (v0 −m0) (κ0+m0)(1+ρ)κ0+m0+ρ . Marginal expected utility is,

dEUdm0

=√κ0v02

((v0 −m0) (κ0+m0)(1+ρ)κ0+m0+ρ

)− 12

− (κ0+m0)(1+ρ)κ0+m0+ρ+ (v0 −m0) (1 + ρ) ρ(κ0+m0+ρ)2

. This is negativewhenever (v0 −m0) ρκ0+m0+ρ < (κ0 +m0), or whenever,

v0 <1

ρ(κ0 +m0)

2 + (κ0 +m0) +m0.

The right hand side of this expression is increasing in m0, so the minimum of this expression is attained

at m0 = mPDI|PS and it equals 3ρ− κ0. Since v0 < 2 < 3ρ− κ0 the result follows.Step 3 - The global maximum when m̄ < mSI|PDI (⇐⇒ v0 < v(κ0, ρ)) is m∗0 = 0. This follows

from the fact that the maximum within each segment of the support is at the minimum point, as we

saw above, and from EU being continuous. EU is continuous because S(.) is continuous for all m0

and hence EU in period 0 is also continuous at m̄. In period 0 in segment [0, m̄], EU evaluated at

m̄ is 2(v0 − m̄) which is equal to EU in segment [m̄,mSI|PDI] evaluated at m̄. This can be shownnoticing that κ0v0S(m̄)

4= v0− m̄, and substituting into the expression for EU in segment [0, m̄]. Thus,

the polity will stay at SI in period 1 whenever v0 < v(κ0, ρ).

C.3.2 Proof of Proposition 2, ρ > 2, part 2

Proposition 2, ρ > 2, part 2 (i): In here we show that if v0 is high enough, then the optimal choice

m∗0 is strictly positive, strictly increasing and unbounded function of v0.

Before we move on to the proof, we prove that the expected utility of the incumbent is continuous

and that its derivative is continuous in the interior of PDI ∪ PS. This result facilitates the ensuinganalysis by allowing first and second order conditions to characterize the unique global optimal m∗0

18

given (κ0, ρ, v0).

Lemma 2. For a given (κ0, ρ, v0)

1. EU is continuous in m0.

2. dEU/m0 is continuous in m0 when κ0 +m0 is in PDI ∪PS.Proof:

1. To check whether EU is continuous, we first check that S(m0) is continuous and then check

that EU = v0 − m0 + κ04 v0S(m0) equals EU =√κ0 (v0 −m0) v0S(m0) at m0 = m̄. To prove that

S(m0) is continuous, we compare its expression at both sides of the frontier of each of these regions.

At the PDI|PS, frontier, κ0 +m0 = ρ, it is direct to check that both expressions, corresponding to Sin PS and PDI equal (1 + ρ) /2. At SI|PDI, κ0 +m0 = 4/ρ, and the utilities on both sides are equalto (ρ+ 1)/ρ. It is also direct to show that S is defined for all ρ ≥ 1 as it is only undefined when ρ = 0in PDI and when ρ = −κ1 < 0 in PS. At SS|PS, κ0 +m0 = ρ/(ρ− 1), the expression in PS and theone in SS equal (ρ + 1)/ρ. At SI|SS, κ0 + m0 = 2 and the utilities on both sides of the frontier areequal to 3/2.

That both expressions of EU have the same value at m0 = m̄ follows from applying the definition

of m̄ on the EU specification:

v0 −m0 +κ04v0S(m0) =2(v0 − m̄)√

κ0 (v0 −m0) v0S(m0) =2(v0 − m̄)

2. From the expressions of EU , it is direct to see that the functions are continuously differentiable,

but proving differential continuity requires some elaboration when looking at m0 = m̄ (the point in

which EU in period 0 changes specification) and κ0 + m0 = ρ (the point at the frontier between

PDI and PS in period 1). At m0 = m̄, note that for m0 in PDI to the left of m̄, the derivative is

dEU/dm0 =v0κ0

4dSdm0− 1 and to the right of m̄

dEU

dm0=

1

2EU

[−v0κ0S(m0) + v0κ0 (v0 −m0)

dS

dm0

]where S(m0) is evaluated in PDI. At m0 = m̄, v0κ0S(m̄) = 4 (v0 − m̄) by definition, so

dEU

dm0=

1

2EU

[−4 (v0 − m̄) + v0κ0 (v0 −m0)

dS

dm0

]=

2 (v0 − m̄)EU

[−1 + v0κ0

4

dS

dm0

].

19

The term outside the square bracket is equal to 1 because EU = 2(v0 − m̄) at m̄ (see point 1. in thisLemma) and the expression in square brackets is exactly dEU/dm0 using the expression of EU to the

left of m̄, evaluated at m̄.

At the frontier PDI|PS, when κ0 + m0 = ρ, we need to keep track of whether we are using theEU to the left or right of m̄. To the left of m̄, dEU/dm0 =

v0κ04

dSdm0− 1. As a result, dEU/dm0 is

the same to the left and right of κ0 + m0 = ρ if and only ifdSdm0

evaluated in PDI is equal to dSdm0

evaluated in PS. This holds because

In PS :dS

dm0=

(1 + ρ) (κ0 +m0 + ρ)− (κ0 +m0) (1 + ρ)(κ0 +m0 + ρ)

2

=1 + ρ

4ρat κ0 +m0 = ρ

and

In PDI :dS

dm0=

(1 + ρ)

2√ρ

1

2√κ0 +m0

=1 + ρ

4ρat κ0 +m0 = ρ.

To the right of m̄, the expected utility specifications are different, but the analysis is simple because

of continuity of S and the just proven fact that dS/dm0 is continuous at κ0 +m0 = ρ. Precisely, the

derivative in region Rj, j = 1, 2 is given by

dEU

dm0=

1

2EU

[−v0κ0SRj(m0) + v0κ0 (v0 −m0)

dSRjdm0

]where SRj is S evaluated using the expression of PS or PDI. S is continuous, so SPS(m0) = SPDI(m0)

at κ0 +m0 = ρ and dSPDI/dm0 = dSPS/dm0 at κ0 +m0 = ρ (as we just showed), hence dEU/dm0 is

continuous at κ0 +m0 = ρ. �

Equipped with Lemma 2, we can now prove part 2 (i) (ρ > 2) of Proposition 2. We consider v0

high enough such that the polity invests m0 = m∗0 > 0 in defense capability in period 0. We show

below that a high enough v0 entails m̄ either in PDI or PS. For tractability, let us separate the proof

in two sub-cases depending on whether m̄ is in [mSI|PDI,mPDI|PS] (subcase A), and in [mPDI|PS,∞](subcase B).

Subcase A) m̄ in [mSI|PDI,mPDI|PS] We define v̄ =16κ0

(4+κ0)ρ(1+ρ)2

, and show that if v0 > v̄, then the

polity invests in m0 = m∗0 > 0, such that it places the polity in either PDI or PS in period 1. Similarly

20

to the case of, ρ < 2, we take 3 steps to prove this claim.

Steps 1&2 - In the frontier SI|PDI, ρ = 4/κ1 = 4/ (κ0 +m0), so mSI|PDI = 4ρ − κ0.If v0 > v̄, then m̄ > mSI|PDI. To see this, note that from Lemma 1 and the fact that

v0S(m0)v0−m0

is increasing in m0, m̄ > mSI|PDI requiresv0S(mSI|PDI)

v0−mSI|PDI< 4

κ0; this follows iff v0 >

16−4κ0ρ4ρ−(1+ρ)κ0 . This

inequality holds since 16−4κ0ρ4ρ−(1+ρ)κ0 is less than 2 (because it is decreasing in both ρ and κ0 so the maximum

is 2, see part 1); and 2 is less than v̄, because v̄ = 16κ0

(4+κ0)ρ(1+ρ)2

> 2 is equivalent to 32ρ+6ρκ0> κ0 +ρκ0ρ,

and the RHS of the latter expression is less than κ0 +4ρ because 0 ≤ ρκ0 < ρκ0(1+ρ) < 4 and κ0 < 2by assumption 1, which is in turn less than 32ρ, the first term on the LHS.

As a result, everywhere in SI, EU corresponds to case (i) in Lemma 1. In PDI or in PS, however,

the expected utility switches to that corresponding to case (ii) in Lemma 1, because m0 ≥ m̄.Step 3 - We now use the result in Lemma 2 to show that if v0 > v̄, then EU is maximal in PDI

or PS, and that there must be a unique interior solution m∗0 in PDI or PS, given (κ0, ρ, v0).

Segment [0,mSI|PDI] Expected utility in period 0 is

EU = v0 −m0 + κ04 v0S(m0) = v0 −m0 +κ04v0(1 + κ0+m0

4

)= v0(1 +

κ04

+(κ04

)2) +m0

(κ0v016− 1).

For v0 > v̄, EU can be decreasing or increasing in this interval. In particular, it is decreasing

for v0 in [min{v̄, 16/κ0}, 16/κ0] and increasing for v0 > 16/κ0. As shown in sub-case B) below, if16/κ0 < v̄, the latter condition implies that [min{v̄, 16/κ0}, 16/κ0] is empty and that both, m̄ andm∗0, lie in [mPDI|PS,∞] (because EU is not only increasing in SI, but also in PDI, and reaches itsmaximum point in PS, see subcase B) below). For now, let us focus on v0 in [v̄, 16/κ0], whenever

this set is not empty. In this case, EU is decreasing over [0,mSI|PDI], so the largest value of EU is

attained at m0 = 0.

Segment [mSI|PDI, m̄] Expected utility in period 0 is

EU = v0 −m0 + κ04 v0S(m0) = EU = v0 −m0 +κ04v0√

κ0+m0ρ

(1+ρ)2

. Differentiating and equating

to zero leads to m∗0 =(v0κ016

(1+ρ)√ρ

)2− κ0. Note that the second derivative of EU with respect to m0

is negative for all m0 in this segment, in particular at m0 = m∗0. This comes from the fact that

dEUdm0

= −1 + κ0v0(1+ρ)16ρ√

(κ0+m0)/ρis decreasing in m0. Thus, m

∗0 is strictly positive, strictly increasing and

unbounded function of v0.

Note also that since v0 > v̄, then m∗0 > mSI|PDI and EU(m

∗0) > EU(0). That m

∗0 > mSI|PDI

comes from the fact that m∗0 =(v0κ016

(1+ρ)√ρ

)2− κ0 > 4ρ − κ0 is equivalent to v0 > 32/(1 + ρ)κ0.

However, v̄ = 16κ0

(4+κ0)ρ(1+ρ)2

is greater than 32/(1 + ρ)κ0 (after simple algebra, this condition boils down

to ρ(2 + κ0) > 2, which holds for κ0 ∈ [0, 2] or ρ ∈ (2,∞)). As a result, if v0 > v̄ then the optimalpoint m∗0 lies to the right of mSI|PDI .

21

To establish that m∗0 actually characterizes the optimum investment, we must check two conditions:

1) the expected utility at m0 = m∗0 is greater than the expected utility evaluated at m0 = 0, and 2)

m∗0 lies in [mSI|PDI, m̄].

To prove that EU(m∗0) > EU(0), let us write down both of the expressions for EU :

Segment [0,mSI|PDI] : EU(0) = v0(1 +κ04

+(κ0

4

)2)

Segment [mSI|PDI, m̄] : EU(m∗0) = v0(1 + v0

(κ016

(1 + ρ)√ρ

)2) + κ0. (18)

Note that EU(m∗0) > EU(0) is implied by v0(1 + v0

(κ016

(1+ρ)√ρ

)2) > v0(1 +

κ04

+(κ04

)2), because we

exclude κ0 > 0 from EU(m∗0) in equation (18). Simplifying leads to

v0 >16

κ0

(4 + κ0)ρ

(1 + ρ)2= v̄.

which holds by definition of v̄.

We have showed that if we fix (κ0, ρ) in SI and ρ > 2 and v0 > v̄, then m0 = m∗0 lies to the right

of mSI|PDI , maximizes the expression of EU corresponding to this segment, and is associated with a

maximum value of EU that is greater than EU at m0 = 0.

The proof, however, is not complete. Nothing in the proof so far guarantees that m∗0 ∈ [mSI|PDI, m̄].It could be the case that some values of v0 > v̄ lead to m

∗0 falling to the right of [mSI|PDI, m̄]. Given

concavity of EU in [mSI|PDI, m̄], this would imply that the maximum point in this segment is m̄,

because EU is strictly increasing to the left of m∗0. But the maximum value of EU at m0 = m̄ is

lower than some values of EU to the right of m̄. This is by virtue of Lemma 2 point 2: the derivative

of EU is continuous at m̄, so EU is increasing in a neighborhood around m̄, [m̄− θ, m̄+ θ] for someθ > 0. In this case, we do not need to compare EU at m0 = m̄ with EU at m0 = 0, because there

are values of EU to the right of m̄ which are higher.

To conclude the proof in this segment, therefore, we must find the values of v0 such that m∗0 ∈

[mSI|PDI , m̄]. For m∗0 to lie in [mSI|PDI , m̄], we must define v̄

′ as the value of v0 such that m∗0 coincides

with m̄. This v̄′ would be the upper limit of the values of v0 such that the optimum lies in [mSI|PDI , m̄].

Using the equation that defines m̄, v̄′ is such that

1

4κ0v̄

′S(m∗0) = v̄′ −m∗0

22

⇐⇒ 14κ0v̄

′S

((v̄′κ016

(1 + ρ)√ρ

)2− κ0

)= v̄′ −

(v̄′κ016

(1 + ρ)√ρ

)2+ κ0.

This last equation defines a quadratic equation for v̄′. The expression looks daunting at first, but

after some algebra it becomes the much simpler quadratic equation

Av̄′2 − v̄′ − κ0 = 0,

where A = 3(κ016

(1+ρ)√ρ

)2. Solving for v̄′ and considering only the positive root:

v̄′ =1

2A

(1 +

√1 + 4Aκ0

).

We now need to make sure that v̄′ makes m∗0 such that EU at m0 = m∗0 is larger than EU at m0 = 0.

To do so, recall that EU evaluated at m̄ equals 2(v0− m̄). Plugging in the expression for m∗0 leads to

EU(m̄ = m∗0) = 2

(v̄′ − v̄′2

(κ0(1 + ρ)

16√ρ

)2+ κ0

)

If this expected utility is greater than EU at m0 = 0, we can build the set of v0, (v̄,max{v̄, v̄′}) forwhich the optimum, m∗0, lies in

[mSI|PDI , m̄

]. Comparing EU(m̄ = m∗0) and EU(m0 = 0) at (κ0, ρ, v̄

′)

leads to

EU(m̄ = m∗0) = 2

(v̄′ − v̄′2

(κ0(1 + ρ)

16√ρ

)2+ κ0

)> v̄′

(1 +

κ04

+(κ0

4

)2)= EU(m0 = 0),

which holds as long as v̄′ < (16 + 28κ0−κ20)(16ρ/ (κ20(1 + ρ)2). Straightforward algebra can show thatv̄′ = 1

2A

(1 +√

1 + 4Aκ0)< (16 + 28κ0 − κ20)(16ρ/ (κ20(1 + ρ)2) for all (κ0, ρ). To see this, we must

check that the solution to the program

maxκ0,ρ

{v̄′ − (16 + 28κ0 − κ20)(16ρ/

(κ20(1 + ρ)

2)}

subject to (κ0, ρ) ∈ PDI, is less than zero. This is the case, the maximum is negative (-10.67, andκ0 = 0 and ρ = 2), which is equivalent to v̄

′ < (16 + 28κ0− κ20)(16ρ/ (κ20(1 + ρ)2). (One direct way tocheck this is to run the following line in Wolfram Mathematica: “NMaximize[ {(16/6)* (1+(1+12*((k/16)ˆ2) * (((1+r)/(rˆ0.5))ˆ2} )* k)ˆ0.5)- 16- 28*k+k*k,k0}, {k,r} ]”).

The last consistency check is that we must show that there is a chance for v̄′ to be larger than v̄.

That is, we need to check that (16 + 28κ0 − κ20)(16ρ/ (κ20(1 + ρ)2) is greater than v̄. This is simple

23

as most of the terms simplify, which leads to the equivalent inequality 16 + 20κ0 > 3κ20, which holds

given that 0 ≤ κ0 ≤ 2, by assumption 1.We just showed that, as long as v0 is such that the local optimum, m

∗0, lies in

[mSI|PDI , m̄

],

the maximal EU in this segment is higher than EU evaluated at m0 = 0. Precisely, for v0 ∈(v̄,max{v̄, v̄′}), the polity does not choose m0 = 0. Note that we define the interval’s upper limit asmax{v̄, v̄′}. We abstain from providing a comparison between v̄ and v̄′ in this proof because we donot need to do so. In case that v̄ > v̄′, the interval is empty, but that is inconsequential. If v̄ > v̄′ the

optimum is associated with such a large v0 that m∗0 falls to the right of m̄. Such cases are analyzed

below

One final observation is in place. EU evaluated at m0 > m̄ features different specifications,

depending on whether m0 is in [m̄,mPDI|PS] or in [mPDI|PS,∞]. If that is the case, the maximal EUwhen m∗0 lies in [mSI|PDI , m̄] is increasing in v0, see equation (18). Larger values of v0 not only entail

larger values of m∗0, but also larger values of EU directly (an application of the envelope theorem).

This is important because as v0 increases the optimal value of EU , such value continues being higher

than EU at m0 = 0. Moreover, by virtue of Lemma 2 we know that EU is differentially continuous,

so if m∗0 lies past m̄, EU is increasing in [mSI|PDI , m̄] and whatever the maximum point of EU is to

the right of m̄, it must be higher than EU at m0 = 0.

Differential continuity in PDI ∪PS (Lemma 2) and the fact that EU is directly increasing in v0are key properties of EU ; they allow us to restrict attention to characterize the local optimum m0

in each segment to the right of m̄. Such a local optimal point yields an optimal value of EU that

is greater than EU(m∗0 = m̄) which in turn is greater than EU(m0 = 0). In other words, the local

optimum is global by virtue of Lemma 2 and as long as v0 > v̄.

Segment [m̄,mPDI|PS] In this segment expected utility is given by

EU =

√v0κ0 (v0 −m0)

√κ0+m0

ρ(1+ρ)

2. Differentiating with respect to m0 leads to FOC

dEUdm0

=

v0κ02EU

[−S + (v0 −m0) dSdm0

]= 0, which in turn leads to m∗0 = (v0− 2κ0)/3, which is strictly increasing,

unbounded function of v0. m∗0 is also positive because v̄ > 2κ0 (expanding v̄ generates the inequality

32ρ+ 8κ0ρ > κ20 + 2κ

20ρ+κ

20ρ

2, which holds because the LHS is higher than 64 but the RHS is smaller

than 20, as long as 0 ≤ κ0 ≤ 2, ρ > 2 and κ0ρ < 4). This critical point is a local maximum becaused2EUdm20

< 0 at m0 = m∗0. To see this note that

d2EUdm20

= v0κ02EU2

(−2 dS

dm0+ (v0 −m0) d

2Sdm20

)< 0 because

v0 −m0 ≥ 0 and S(m0) =√

κ0+m0ρ

(1+ρ)2

, so dSdm0

> 0 and d2Sdm20

< 0. If the parameters are such that m∗0

indeed lies in this segment, then by virtue of Lemma 2 and v0 > v̄, m∗0 is a global maximum point.

24

Segment [mPDI|PS,∞] By studying the EU in this segment, we reach to the same conclusionas when studying EU over [m̄,∞] in the paper. EU =

√v0κ0 (v0 −m0) (κ0+m0)(1+ρ)κ0+m0+ρ and the local

maximum would be m∗0 = −κ0 − ρ +√ρ(κ0 + ρ+ v0), which is strictly positive, strictly increasing

and unbounded in v0.

Subcase B) m̄ in [mPDI|PS,∞] Define v̄ = 16κ0 . This case is simpler because EU is increasing in m0when κ0 +m0 lies in SI, PDI and part of PS. This guarantees that the local optimum in [mPDI|PS,∞]is global. To see this, let us proceed by steps.

Steps 1&2 - In the frontier PDI|PS, ρ = κ1 = κ0 +m0, somPDI|PS satisfies mPDI|PS = ρ− κ0.It turns out that if v0 > v̄, then m̄ > mPDI|PS. To see this, note that from Lemma 1 and the

fact that v0S(m0)v0−m0 is increasing in m0, m̄ > mPDI|PS requires

v0S(mPDI|PS)

v0−mPDI|PS< 4

κ0, and this follows iff

v0 >8(ρ−κ0)

8−(1+ρ)κ0 . This inequality holds since8(ρ−κ0)

8−(1+ρ)κ0 is less than16κ0

. This latter conclusion follows

from simple algebra: 8(ρ−κ0)8−(1+ρ)κ0 <

16κ0

= v̄ is equivalent to ρκ0 + 2κ0 + 2ρκ0> 16 + κ20. The LHS of the

last expression is strictly less than 16 because 0 ≤ ρκ0 < ρκ0(1 +ρ) < 4 and κ0 < 2 under assumption1.

As a result, everywhere in SI, PDI and in part of PS expected utility of the incumbent corresponds

to case (i) in Lemma 1, but in PS the expected utility switches to that corresponding to case (ii) in

Lemma 1.

Step 3 - We now use the result in Lemma 2 to show that if v0 > v̄, EU is maximal in PS, and

that there must be a unique interior solution m∗0 in PS, given (κ0, ρ, v0).

Segment [0,mSI|PDI] Expected utility in period 0 is

EU = v0 −m0 + κ04 v0S(m0) = v0 −m0 +κ04v0(1 + κ0+m0

4

)= v0(1 +

κ04

+(κ04

)2) +m0

(κ0v016− 1).

Since v0 > v̄, it follows that EU is increasing in this interval, the optimal point is m∗0 = mSI|PDI.

Segment [mSI|PDI,mPDI|PS] The (concave) expected utility is, EU = v0 −m0 + κ0v0(1+ρ)8√

κ0+m0ρ

.

The marginal utility of m0 is: −1 + κ0v0(1+ρ)812

√ρ

κ0+m01ρ. This marginal utility may be either negative,

zero or positive in [mSI|PDI,mPDI|PS]. Given concavity, it is negative iff the value at which the

marginal utility is zero, m0 =(κ0v016

1+ρ√ρ

)2− κ0, is smaller than mSI|PDI:

m∗0 =

(κ0v016

(1 + ρ)√ρ

)2− κ0 < mSI|PDI =

4

ρ− κ0

⇐⇒ v0 <(

16κ0

)2

1+ρ. In this case, the maximum EU in this segment is at m0 = mSI|PDI.

25

The solution is interior in PDI if m∗0 is in between mSI|PDI and mPDI|PS:

4

ρ− κ0 = mSI|PDI <

(κ0v016

(1 + ρ)√ρ

)2− κ0 < mPDI|PS = ρ− κ0

⇐⇒ 16κ0

21+ρ

< v0 <16κ0

ρ1+ρ

. But since 16κ0

ρ1+ρ

< 16κ0

= v̄, then v0 >16κ0

ρ1+ρ

and the incumbent moves

to PS. The following analysis shows the incumbent moves to the interior of PS in this case.

Segment [mPDI|PS, m̄] By Lemma 2 we know the expected utility must be increasing at least at the

beginning of this interval. Expected utility is EU = v0−m0+κ04 v0S(m0) = v0−m0+κ0v0

4

((κ0+m0)(1+ρ)κ0+m0+ρ

),

and marginal utility is dEUdm0

= κ0v04

ρ(1+ρ)

(κ0+m0+ρ)2 − 1, so the unique interior optimum candidate is

m∗0 =√

κ0v04ρ(1 + ρ) − κ0 − ρ. For m∗0 to be the local, and using Lemma 2 global, optimum

it must be the case that d2EUdm20

is negative at m∗0, which follows from inspectingdEUdm0

because m0

is in the denominator. It is straightforward (using arguments analogous to the case in segment

[mSI|PDI,mPDI|PS]) to show that if16κ0

ρ(1+ρ)

< v0 then we also have m0 =√

κ0v04ρ(1 + ρ) − κ0 − ρ >

mPDI|PS. This implies that for v0 >16κ0

= v̄ the polity must move to the interior of PS in period 1.

Note that m∗0 is a strictly positive, strictly increasing and unbounded function of v0. We do not

know, however, for which parameters values, if any, it lies to the left of m̄. We are not interested

in such condition because as long as m∗0 is in PS the polity will reach civilization. That said, it is

important to check whether the optimum m∗0 is also strictly increasing and unbounded function of v0,

when v0 is such that the polity decides to invest m∗0 in [m̄,∞], which is the case we present next.

Segment [m̄,∞] This case is similar to the analysis of segment [m̄,∞] in the paper. EU =√v0κ0 (v0 −m0) (κ0+m0)(1+ρ)κ0+m0+ρ and the global maximum would be m

∗0 = −κ0 − ρ +

√ρ(κ0 + ρ+ v0),

which is strictly positive, strictly increasing and unbounded in v0.

Proposition 2, (ρ > 2), part 2 (ii): When m∗0 is in PS, the proof unfolds in the same way as in

the text of the paper by virtue of 2(i): m∗0 strictly positive, strictly increasing and unbounded in v0.

When m∗0 is in PDI, for v0 > v̄ we have that EV2 =√a∗1V2 =

v12

(1 + 1

ρ

)√ρκ1 from Proposition

1, and i0 = 0 from Lemma 1, we get EV2 =v02

(1 + 1

ρ

)√ρκ1 which is increasing in κ1 = κ0 +m

∗0 and

hence v0 by virtue of 2(i). Note once in PDI both security and prosperity increase with v0.

C.3.3 Proof of Proposition 2, (ρ > 2), part 3

When ρ > 2 and landing PS yields the same expressions as in the paper (see 2(i)).

When landing in PDI two additional expressions for m∗0 emerge. In segment [mSI|PDI, m̄] the

solution is m∗0 =(v0κ016

(1+ρ)√ρ

)2−κ0 and in segment [m̄,mPDI|PS] we have the solution m∗0 = (v0−2κ0)/3.

26

In each respective case, the v0 that is required to produce such solution is,

v0 =16√

(κ0 +m∗0) ρ

κ0(1 + ρ)

v0 = 3 (κ0 +m∗0)− κ0.

Differentiating v0 with respect to ρ shows that such needed income is decreasing in ρ in the first

expression and independent of ρ in the second expression.�

D Proof of Proposition 3

The proof of this proposition is done by generalizing our model to allow the challenger to have a

military capability κc different from 1. The next subsection lays out the formalities, and a subsequent

subsection explains the intuition, offers a graphical representation, and adds historical detail to the

collapse at the end of the Bronze Age.

D.1 A generalization of the military capability of the challenger

Given a generalized cost btκc

for the challenger the problem of the incumbent is to maximize,

L = v1 −a1κ1− i1 +

a1a1 + b1

(v1 + ρi1)

+λBC(v1 −a1κ1− i1) + λDC(κcv1 − a1 + κcρi1) + λaa1 + λii1. (19)

We characterize the solution (a1, i1, λBC , λDC , λa, λi) to this problem for each parameter combination

(ρ, κ1, κc, v1). To save on notation, let us define PS = {(κ1, κc, ρ, v1)|κ1 ≥ κc > 0, ρ > 1, v1 > 0}the parameter space we consider throughout the proof. The first order and complementary slackness

conditions that characterize the optimum are given by,

∂L∂a1

=1

2√κc

√v1 + ρi1a1

− 1κ1− λBC

κ1− λDC + λa = 0; a1 ≥ 0, λa ≥ 0, λaa1 = 0 c.s. (20)

∂L∂i1

=ρ

2√κc

√a1

v1 + ρi1− 1− λBC + λDCκcρ+ λi = 0; i1 ≥ 0, λi ≥ 0, λii1 = 0 c.s. (21)

λBC(v1 −a1κ1− i1) = 0 c.s., λDC (κcv1 − a1 + κcρi1) = 0 c.s. (22)

Since λa = 0, there are eight possible cases to be analyzed, given by whether the remaining

27

Lagrange multipliers λBC , λDC , and λi are zero or positive. We assume in each case that the conditions

defining it hold, and then determine which part if any of the parameter space supports a solution.

When the case implies conditions for the parameters that are mutually exclusive, the case is deemed

infeasible. When the case implies that the solution can be supported for knife-edge combinations of

the parameter values (i.e., combinations that, under a suitable measure, would have measure zero) we

consider the case to be non-generic and also drop it from further consideration.

1. Case λ∗BC > 0 (BC binds), λ∗DC > 0 (DC binds, consolidation), and λ

∗i = 0 (i

∗1 > 0)

Since a1 is always positive, and in this case i1 is also positive, the FOCs in (20), (21) must hold with

equality. Because this case involves binding BC and DC, constraints also hold with equality. This

means,

a1 :1

2√κc

√v1 + ρi1a1

− 1κ1− λBC

κ1− λDC = 0;

i1 :ρ

2√κc

√a1

v1 + ρi1− 1− λBC + λDCρκc = 0

DC : κcv1 − a1 + κcρi1 = 0; BC : v1 −a1κ1− i1 = 0.

BC and DC yield i∗1 = v1(κ1−κc)(κ1+κcρ)

and a∗1 = v1κ1κc(1+ρ)(κ1+κcρ)

. Then λ∗i = 0 (or i∗1 > 0) is supported by

κ1 > κc. Using a∗1 and i

∗1, the FOCs are seen to involve only λDC , λBC , and parameters κ1, κc, showing

λDC > 0 ⇔ κ1 > ρκc and λBC > 0 ⇔ ρ > κ1/(κ1 − κc). The parameter set supporting this solutionis therefore PS = {(κ1, ρ, v1) ∈ PS|ρ < κ1/κc, ρ > κ1/(κ1 − κc), κ1 > κc} and in this region there isinvestment and deterrence. The expected utility is V1 = v1

κ1(1+ρ)(κ1+κcρ)

. When κc > κ1 this case would be

infeasible, as it contradicts i1 > 0.

2. Case λ∗BC > 0 (BC binds),λ∗DC = 0 (DC does not bind, conflict), and λ

∗i = 0 (i

∗1 > 0)

The first order and complementary slackness conditions yield,

a1 :1

2√κc

√v1 + ρi1a1

− 1κ1− λBC

κ1= 0; i1 :

ρ

2√κc

√a1

v1 + ρi1− 1− λBC = 0

BC : v1 −a1κ1− i1 = 0,

Investment and army solutions are respectively i∗1 =v12

(1− 1

ρ

)and a∗1 =

κ1v12

(1 + 1

ρ

). This so-

lution is consistent with λ∗DC = 0 (DC holds with strict inequality) and λ∗i = 0 ⇔ ρ > κ1κc and ρ > 1

respectively. The solution is consistent with λ∗BC > 0⇔ ρ > 4κcκ1 (from checking λ∗BC > 0 in the FOCs).

As a result, the parameter set supporting this case is PDI = {(κ1, ρ, v1) ∈ PS|ρ > κ1/κc, ρ > 4κc/κ1 } .

28

Expected utility for the incumbent is V ∗1 =v12

(1 + 1

ρ

)√ρκ1.

3. Case λ∗BC = 0 (BC does not bind), λ∗DC = 0 (DC does not bind, conflict), and λ

∗i > 0

(i∗1 = 0) In this case a∗1 = v1κ

21/(4κc) and i

∗1 = 0. This solution is consistent with λ

∗BC = 0 and

λ∗DC = 0 ⇔ κ1 < 2κc. Also for λi > 0 we need 1 − ρκ1/(4κc) > 0 (from the FOC of i1). Thus,this holds for any triple (ρ, κ1, v1) ∈ PS such that κ1 < 2κc and ρ < 4κc/κ1. In other words,the parameter set for which this region contains the solution to the incumbent’s problem is SI =

{(κ1, ρ, v1) ∈ PS|2κc > κ1, ρ < 4κc/κ1} . Expected utility in this case is given by V ∗1 = v1(

1 + κ14κc

).

4. Case λ∗BC = 0 (BC does not bind), λ∗DC > 0 (DC binds, consolidation),and λ

∗i > 0

(i∗1 = 0) In this case the system of conditions is given by a1 :1

2κc− 1

κ1−λDC = 0; i1 : ρ2−1+λDCρκc+

λi = 0; and BC : κcv1 − a1 = 0. Since i∗1 = 0, the DC yields a∗1 = κcv1. For this to be consistentwith λ∗DC > 0, we must have from the first equation that κ1 > 2κc, and to be consistent with λ

∗i > 0

we need ρ < κ1/(κ1−κc), yielding SS = {(κ1, ρ, v1) ∈ PS|κ1 > 2κc, ρ < κ1/(κ1 − κc)} . The expectedutility is V ∗1 = v1

(2− κc

κ1

).

5. Case λ∗BC > 0 (BC binds), λ∗DC > 0 (DC binds, deterrence), and λ

∗i > 0 (i

∗1 = 0)

Because i∗1 = 0, BC binding implies that a∗1 = κ1v1, but DC binding implies that a

∗1 = κcv1, so κ1 = κc

which is non-generic.

6. Case λ∗BC > 0 (BC binds), λ∗DC = 0 (DC does not bind, conflict), and λ

∗i > 0 (i

∗1 = 0)

The BC binding and i∗1 = 0 yield a∗1 = v1κ1. The BC binds iff κ1 > 4κc. The DC not binding, however,

implies v1κc − v1κ1 > 0⇔ κc > κ1 which violates the condition for BC to bind κ1 > 4κc, making thiscase infeasible.

7. Case λ∗BC = 0 (BC does not bind), λ∗DC = 0 (DC does not bind, conflict), and λ

∗i = 0

(i∗1 > 0) Non-generic, since it is consistent for subset of the space (ρ, κ1, v1) that has (under a

suitable measure) measure zero. This follows from (20) and (21), so when λBC , λDC , λi = 0, FOCs

read a1 :1

2√κc

√v1+ρi1a1− 1

κ1= 0 and i1 :

ρ2√κc

√a1

v1+ρi1− 1 = 0. The first FOC implies

√v1+ρi1a1

=2√κc

κ1

and substituting into the second FOC, we get ρ2√κc

κ12√κc

= 1 or ρκ14κc

= 1, which implies this holds for a

non-generic parameter set.

8. Case λ∗BC = 0 (BC does not bind),λ∗DC > 0 (DC binds, consolidation), and λ

∗i = 0

(i∗1 > 0) The FOCs are,

a1 :1

2√κc

√v1 + ρi1a1

− 1κ1− λDC = 0; i1 :

ρ

2√κc

√a1

v1 + ρi1− 1 + κcρλDC = 0,

where a1 = κc (v1 + ρi1) indicating that λDC must simultaneously equal1

2κc− 1

κ1and

( ρ2−1)κcρ

, which

forces the equality ρ = κ1κc

, which is non-generic.

29

Combining these cases yields that optimal behavior by the incumbent yields a partition of the

parameter space (ρ, κ1, v1) ∈ (1,∞)× (0,∞)× R+ into four regions:Stagnant Insecurity (Region SI):{(κ1, ρ, v1)|2κc > κ1 and ρ < 4κc/κ1}In SI the solution is:

{a∗1 = v1

(κ12

)2, i∗1 = 0, V

∗1 = v1

(1 + κ1

4κc

)}Stagnant Security (Region SS):{(κ1, ρ, v1)|κ1 > 2κc, ρ < κ1/(κ1 − κc)}In SS the solution is:

{a∗1 = κcv1, i

∗1 = 0, V

∗1 = v1

(2− κc

κ1

)}.

Prosperity Despite Insecurity (Region PDI):{(κ1, ρ, v1)|ρ > κ1/κc, ρ > 4κc/κ1 and ρ > 1}In PDI the solution is:

{a∗1 =

κ1v12

(1 + 1

ρ

), i∗1 =

v12

(1− 1

ρ

), V ∗1 =

v12

(1 + 1

ρ

)√ρκ1

}Prosperous Security (Region PS): {(κ1, ρ, v1)|ρ < κ1/κc, ρ > κ1/(κ1 − κc) , κ1 > κc}In PS the solution is:

{a∗1 = v1

κ1κc(1+ρ)(κ1+κcρ)

, i∗1 = v1(κ1−κc)(κ1+κcρ)

, V ∗1 = v1κ1(1+ρ)(κ1+κcρ)

}A comparison of region by region with the case where κc = 1 yields the proposition.�

D.2 More Aggressive Challenger

The comparison of the solutions in Propositions 1 (with κc = 1) and 3 (with κc > 1) is best appreciated

in Figure D.1 where the dashed lines show the partition of the parameter space with κc = 1 and the

solid lines show the partition with κc > 1, reflecting a more aggressive challenger.

κ1

ρ

2

1

Stagnant Security (SS)

2

Prosperity Despite Insecurity (PDI)

Stagnant Insecurity (SI)

Prosperous Security (PS)

C

A

B

Figure D.1: Equilibrium partition of the parameter space with more aggressive challenger

The model helps delve deeper into the diverging fates of different regions that suffered attacks at

30

the end of the Bronze Age. The extremes of that contrast are Egypt, which managed to repel the

invasion, and cities near the Mediterranean coast of the Levant, like Ugarit, which were destroyed. In

the case of Ugarit, surviving clay tablets provide textual evidence on the threat of the Sea Peoples

and the fact that Ugarit was defenseless.5 In terms of our model, Ugarit’s vulnerability at the time of

the invasions can be interpreted as either an initial location within PS that was close to the vertex,

and thereby not too far away from SI, or within a narrow strip along the PDI/SI border on the side

of PDI. When the shocks that prompted the invasion occurred the effect was to shift the partition

boundaries landing Ugarit deep into SI. In the new location, Ugarit faced the prospects of attacks that

were too strong for the city to resist. In the model, a deep position in SI entails a lower probability

that the incumbent will prevail. In fact, for Ugarit, the attack resulted in destruction.

Egypt was also a victim of barbarian attacks but the outcome was very different. Since the end

of the Second Intermediate Period, Egypt had developed a highly professional army and a formidable

fleet. Before the shock, Egypt was likely located deep in PS so that the worst effect of the shock

could have been to relocate Egypt in a relatively safe neighborhood of PDI. Egypt became susceptible

to challenges, but it could prevail in the battlefield with high probability. Pictorial inscriptions on

the walls of the Karnak Temple attest to the threats posed by the Sea Peoples at roughly the same

time they invaded the Levant. But, in contrast to Ugarit’s tablets, the Egyptian inscriptions actually

honor king Merneptah’s success in subduing the invaders. The contrasting cases of Ugarit and Egypt

correspond respectively to the points A or C and B in Figure D.1.

Part II

Three-Period Model

We extend the model with exogenous defense capabilities from two to three periods, and show the

pattern of security-prosperity outcomes is preserved. We add a period at t = 0. In the paper we use

the expressions SI, SS, PDI, and PS to denote the regions in which the parameter space (κ1, ρ, v1)

is partitioned in equilibrium in the two-period model. We maintain the same notation to refer to

that partition in period 1, and we use the notation SI0 − PS0 to refer to the partition that obtains5In the tablets, Ammurapi, the king of Ugarit makes desperate requests to his Hittite overlord, whom he addresses

as his “father” and who was using Ugarit’s maritime fleet to defend other sections of the Hittite empire: “My father,behold, the enemy’s ships came (here); my cities(?) were burned, and they did evil things in my country. Does notmy father know that all my troops and chariots(?) are in the Land of Hatti, and all my ships are in the Land ofLukka?...Thus, the country is abandoned to itself. May my father know it: the seven ships of the enemy that camehere inflicted much damage upon us”. Letter RS 18.147 in Jean Nougaryol et al. 1968. Ugaritica V, 24: 87–90. (Note:question marks in the original.)

31

in period 0. We show that in period 0 there is a unique equilibrium that yields a partition of the

parameter space (κ0, ρ, v0) into four regions. The following proposition states this result:

Proposition D.1 There is a unique equilibrium, which yields a partition of the parameter space

(κ0, ρ, v0) ∈ (1,∞)× R+ × R+ into four regions:Stagnant Insecurity (Region SI0): SI0 = {(κ0, ρ, v0)| κ0 < 4/ρ

(1 + κ0

4

), κ0 < 2}

Secure Stagnation (Region SS0): SS0 = {(κ0, ρ, v0)| κ0 < ρ(

2− 1κ0

)/(ρ(

2− 1κ0

)− 1), κ0 > 2}.

Prosperity Despite Insecurity (Region PDI0): PDI0 = PDI∪{(κ0, ρ, v0)| κ0 > ρρ−1 , κ0 > ρ, κ0 <ρκ0(1+ρ)

κ0+ρ} ∪ {(κ0, ρ, v0)|κ0 < 4/ρ, κ0 < 2, κ0 > 4/(ρ

(1 + κ0

4

))} ∪ {(κ0, ρ, v0)|κ0 < ρ/(ρ − 1), κ0 >

2, κ0 < ρ(

2− 1κ0

)}.

Prosperous Security (Region PS0): PS0 = {(κ0, ρ, v0)| κ0 > ρ(κ0(1+ρ)κ0+ρ

), κ0 > ρ

(κ0(1+ρ)κ0+ρ

)/(ρ(κ0(1+ρ)κ0+ρ

)− 1)}.

Figure D.1: Regions in period t = 0.

Before presenting the proof, we represent graphically the regions in Proposition D.1. The solid

lines in Figure D.1 demarcate the regions in period 0 and the dashed lines the regions in period 1.

The period 0 partition is similar but not identical to the period 1 partition. PDI0 is larger than PDI,

and expands against all other period-1 regions, while PS0 occupies a portion of PS and a portion of

SS. SI and SS contract into the smaller regions SI0 and SS0, respectively. After the proof we offer

some thoughts on the intuition for the differences between the two partitions, and why they do not

alter the key empirical implications in the paper.

The proof below derives the conditions that yield the partition of the parameter space in t = 0.

Proof:

We solve the model by backward induction for a Subgame Perfect Nash Equilibrium; so the solution

is generically unique as the problem is globally concave for each (κ0, ρ, v0). Moreover, the outcomes

of period 1 in the two-period model in the paper determine the continuation value for the winner in

period 0. Note that there are four specifications for the continuation value in period 0, corresponding

32

to the outcome in each of the four regions PS, PDI, SI and SS in period 1. Precisely, in the two-

period model each point (κ0, ρ, v0) lies in only one of those regions and is associated with a unique

equilibrium. Thus, the continuation value in period 0 can be written as

V1(i0) = (v0 + ρi0)×

κ0(1+ρ)κ0+ρ

(κ0, ρ) ∈ PS√κ0ρ

(1+ρ)2

(κ0, ρ) ∈ PDI(1 + κ0

4

)(κ0, ρ) ∈ SI(

2− 1κ0

)(κ0, ρ) ∈ SS

≡ (v0 + ρi0)S.

Given the challenger’s best response, the incumbent chooses a0, i0 to maximize his expected utility,

maxa0,i0≥0

v0 −a0κ0− i0 +

a0a0 + b0(a0, i0)

V1(i0)

subject to the non-negativity constraints a0 ≥ 0, i0 ≥ 0, the budget constraint v0 − a0κ0 − i0 ≥ 0 (BC)and the deterrence constraint V1(i0)−a0 ≥ 0 (DC) for a given triple (κ0, ρ, v0). Notice the problem ofchoosing (a0, i0) is similar to the choice of (a1, i1) in the two-period model, except now the continuation

value has a different specification depending on whether (κ0, ρ) is PS, PDI, SI or SS.

The objective function is differentiable and concave in a0 and i0, the constraints are linear, so the

first order and complementary slackness conditions are necessary and sufficient for a maximum.

The problem is to maximize,

L = v0 −a0κ0− i0 +

a0a0 + b0(a0, i0)

V1(i0)

+λBC(v0 −a0κ0− i0) + λDC(V1(i0)− a0) + λaa0 + λii0.

We characterize the solution (a0, i0, λBC , λDC , λa, λi) to this problem for each parameter combination

(κ0,ρ, v0). As in the two-period model, the parameter space we consider throughout the proof is

PS = {(κ0, ρ, v0)|κ0 > 0, ρ > 1, v0 > 0}. The first order and complementary slackness conditions thatcharacterize the optimum are given by,

∂L∂a0

=1

2

√V1(i0)

a0− 1κ0− λBC

κ0− λDC + λa = 0; a0 ≥ 0, λa ≥ 0, λaa0 = 0 c.s.

∂L∂i0

=ρS

2

√a0

V1(i0)− 1− λBC + λDCρS + λi = 0; i1 ≥ 0, λi ≥ 0, λii0 = 0 c.s.

33

λBC(v0 −a0κ0− i0) = 0 c.s., λDC (V1(i0)− a0) = 0 c.s.

Note that V1 can take on four specifications, depending on whether (κ0, ρ) is in PS, PDI, SI or

SS. Moreover, since λa = 0 (the marginal benefit of army a0 at zero is infinity, it is always optimal to

have a0 > 0), there are eight possible cases to be analyzed, given by whether the remaining Lagrange

multipliers λBC , λDC , and λi are zero or positive.

In order to solve the problem, we consider (κ0, ρ) in each period 1 region separately, and find

the cases that are consistent with equilibrium under the restriction that (κ0, ρ) is in that region. To

check which cases are consistent in equilibrium, we assume in each case that the conditions defining it

hold, and then determine which subset of the period-1 region, if any, supports a solution. When the

case implies conditions for the parameters that are mutually exclusive, the case is deemed infeasible.

When the case implies that the solution can be supported for knife-edge combinations of the parameter

values (i.e., combinations that, under a suitable measure, would have measure zero) we consider the

case to be non-generic and also drop it from further consideration.

Our approach is similar to the one we use in the two-period model, with one difference. We now

have to consider 8 cases (depending on which constraints bind) for each of the four regions in period

1. We start with (κ0, ρ) in PS.

D.3 PS

We assume first that (κ0, ρ) ∈ PS in period 1. We show below that there are two cases consistent inequilibrium. One of them is when λBC > 0 (BC binds),λDC > 0 (DC binds), and λi = 0. This case

entails security and prosperity (investment), akin to PS in period 1. The other case that is consistent

in equilibrium when (κ0, ρ) ∈ PS is λBC > 0 (BC binds),λDC = 0 (DC does not bind, conflict), andλi = 0 (i0 > 0), which entails prosperity without security.

We start by studying these two cases and then we show that all the other cases are inconsistent

in equilibrium. Throughout this section we consider combinations of (κ0, ρ) such that they lie in

PS = {(κ0, ρ, v0)|κ0 > ρ, ρ > κ0/(κ0 − 1), κ0 > 1}. Note that the last condition that defines PS(κ0 > 1) is implied by the other two (this proves useful below).

1. Case λBC > 0 (BC binds),λDC > 0 (DC binds), and λi = 0

To see that this case is consistent in equilibrium, first find the condition for λDC > 0, under which

we must have,

(v0 + ρi0)S = a0,

34

and then the FOCs imply,

κ02− 1− λBC − λDCκ0 = 0

ρS

2− 1− λBC + λDCρS = 0.

Solving for λDC from the first FOC and substituting into the second we get,

1

κ0

(κ02− 1− λBC

)= λDC

ρS

2− 1− λBC +

1

κ0

(κ02− 1− λBC

)ρS = 0

λBC =

[ρS

2− 1 + 1

κ0

(κ02− 1)ρS

](1 +

ρS

κ0

)−1It follows that λDC is given by,

λDC =1

2

(κ0 − ρSκ0 + ρS

).

If λBC > 0 then,

v0 −a0κ0− i0 = 0,

and using the fact that the deterrence condition binds,

(v0 + ρi0)S = a0,

solving for a0 and i0 yields,

λDC =1

2

(κ0 − ρSκ0 + ρS

)

λBC =

[ρS

2− 1 + 1

κ0

(κ02− 1)ρS

](1 +

ρS

κ0

)−1i0 =

v0 (κ0 − S)κ0 + ρS

a0 =

(v0κ0 (1 + ρ)

κ0 + ρS

)S.

We need to establish the conditions on the parameters such that those parameter values support

35

this equilibrium. From the expression for λDC it follows directly that,

λDC =1

2

(κ0 − ρSκ0 + ρS

)> 0 ⇐⇒ κ0 > ρS.

Note that this condition is analogous to the condition for λDC > 0 in PS, namely κ0 > ρ. This pattern

remains the same throughout the proof. All conditions that involve ρ in period 1, now are written in

terms of ρS. Given that S > 1, the incumbent in period 0 faces higher returns to investment for any

fixed (κ0, ρ). This explains the shape of the partition in period 0, as we see below.

Let us check now whether the condition for λDC > 0, κ0 > ρS, is consistent with (κ0, ρ) in PS.

Replacing the expression for S in PS:

κ0 > ρκ0 (1 + ρ)

κ0 + ρ⇐⇒ κ0 > ρ2 (23)

Note that inequality (23) requires κ0 to be larger than ρ2, which is more stringent than being larger

than ρ, given ρ > 1. In words, the condition κ0 > ρ2 covers the bottom-right portion of PS.

For λBC to be strictly positive

λBC =

[ρS

2− 1 + 1

κ0

(κ02− 1)ρS

](1 +

ρS

κ0

)−1> 0

which holds generically if and only if

κ0 >ρS

ρS − 1. (24)

Note that inequality (24) is analogous to the inequality defining the border between PS and SS,

except now the condition involves ρS instead of ρ. Recall that S depends on κ0, so the condition that

λBC > 0 after plugging in the expression for S becomes

κ0 >ρκ0(1+ρ)

κ0+ρ

ρκ0(1+ρ)κ0+ρ

− 1

ρκ0 (1 + ρ)− κ0 − ρκ0 + ρ

> ρ(1 + ρ)

κ0 + ρ

distributing

ρκ0 + ρ2κ0 − κ0 − ρ > ρ+ ρ2

taking common factor κ0

κ0(ρ+ ρ2 − 1) > 2ρ+ ρ2

36

Figure D.2: Region with security and growth in t = 0 when (κ0, ρ) ∈ PS.

by adding and subtracting terms conveniently to the LHS of this expression, we reach an expression

similar to the condition for λBC > 0 in PS, κ0 > ρ/(ρ− 1),

κ0(ρ+ ρ2 − 1 + ρ− ρ+ 2− 2) > 2ρ+ ρ2

⇐⇒ κ0(2ρ+ ρ2 + 1− ρ− 2) > 2ρ+ ρ2

⇐⇒ κ0(ρ (2 + ρ)− (ρ+ 2) + 1) > 2ρ+ ρ2

⇐⇒ κ0((2 + ρ) (ρ− 1) + 1) > ρ(2 + ρ)

dividing both sides by (2 + ρ) and leaving κ0 on one side,

⇐⇒ κ0 >ρ

(ρ− 1) + (2 + ρ)−1(25)

The RHS of inequality (25) is strictly smaller than ρ/(ρ − 1) because (2 + ρ)−1 is strictly positive.Since κ0 > ρ/(ρ − 1) in PS, inequality (25) holds by transitivity. As a result, PS is fully containedby the region defined by inequality (25).

Finally, λi = 0 so

i0 =v0 (κ0 − S)κ0 + ρS

> 0 ⇐⇒ κ0 > S,

which holds in PS. In sum, this case entails,

λDC > 0 ⇐⇒ κ0 > ρS ⇐⇒ κ0 > ρ2

λBC > 0 ⇐⇒ κ0 >ρS

ρS − 1⇐⇒ κ0 >

ρ

(ρ− 1) + (2 + ρ)−1

i0 > 0 ⇐⇒ κ0 > S

a subset of PS, which we define as PS1 = {(κ0, ρ, v0)|κ0 > ρκ0(1+ρ)κ0+ρ , κ0 >ρρ−1}, and represent in

Figure D.2 (solid lines).

37

2. Case λBC > 0 (BC binds),λDC = 0 (DC does not bind, conflict), and λi = 0 (i0 > 0)

FOC

κ02

√(v0 + ρi0)S

a0− 1− λBC = 0

ρS

2

√a0

(v0 + ρi0)S− 1− λBC = 0

If λBC > 0 then

v0 −a0κ0

= i0

Equating the two FOCs after solving for λBC yieldsκ0ρS

(v0 + ρi0)S = a0, and using this expression

in the investment equation above, v0− 1ρ (v0 + ρi0) = i0, we get i0 =v0(1− 1ρ)

2which holds given ρ > 1.

The expression for i0 also yields a0 =κ0v02ρ

(1 + ρ) > 0.

The deterrence constraint not binding,λDC = 0, is equivalent to V1 > a0. Replacing terms and

simplifying yields v0 + ρv0(

1− 1ρ

)2

S > κ0v02ρ

(1 + ρ)

⇐⇒ ρS > κ0.

Note this condition is the complement to deterrence in case 1. above. As a result, this condition

includes all the (κ0, ρ) in PS that are not considered in the case 1.

The last condition to check is λBC > 0, solving from any FOC above,

λBC =κ02

√v0(

1+ρ2

)S

κ0v02ρ

(1 + ρ)− 1 = κ0

2

√ρS

κ0− 1 > 0

⇐⇒ κ0ρS > 4.

This last condition is analogous to the condition that defines the border between PDI and SI, κ0ρ > 4.

Given that S > 1, this condition moves the border to the left, so PS is fully contained in the region

defined by κ0ρS > 4. To see this, let us separate the analysis for ρ > 2 and ρ < 2, because the shape

of the boundaries differ below and above ρ = 2. When ρ > 2, we prove that κ0 > ρ implies κ0 > 4/ρS.

Note that the following sequence of inequality holds

38

Figure D.3: Region with growth but without security in t = 0 when (κ0, ρ) ∈ PS.

κ0 > ρ >4

ρ>

4

ρS,

provided that S > 1. (S > 1 follows directly from its definition: S = κ0(1+ρ)κ0+ρ

> 1 ⇐⇒ κ0+κ0ρ > κ0+ρ⇐⇒ κ0 > 1, which holds in PS). When ρ < 2, the boundary of PS is characterized by (κ0, ρ) suchthat κ0 > ρ/(ρ − 1). We must prove that ρ/(ρ − 1) is larger than 4/(ρS). Let us start by the truestatement

0 < (ρ− 2)2

⇐⇒ 0 < ρ2 − 4ρ+ 4

⇐⇒ 4ρ− 4 < ρ2

⇐�