Content uploaded by Vladimir Smirnov

Author content

All content in this area was uploaded by Vladimir Smirnov on May 21, 2018

Content may be subject to copyright.

Duplicative search∗

Alexander Matros†Vladimir Smirnov‡

June 23, 2016

Abstract

In this paper we examine the dynamic search of two rivals looking for a prize of

known value that is hidden in an unknown location, modeled as search for treasure on

an island. In every period, the players choose how much to search of the previously

unsearched portion of the island in a winner-takes-all contest. If the players cannot

coordinate so as to avoid searching the same locations, the unique equilibrium involves

complete dissipation of rents. On the other hand, if the players have some (even

limited) ability to coordinate so as to avoid duplicative search and the search area is

suﬃciently small, there is a unique equilibrium in which the full area is searched and

each player earns a positive expected return.

Keywords: R&D, search, duplication, uncertainty, coordination.

JEL classiﬁcations: D21, O32.

1 Introduction

Consider pharmaceutical ﬁrms competing in a race to discover a new drug, paparazzi

searching luxury hotels for a visiting movie star, bounty hunters pursuing a fugitive, or

researchers looking for solutions to the six Millennium Prize Problems in mathematics. All

of these scenarios have attributes of a general problem, namely, a treasure hunt for a prize

of known value in which there is uncertainty regarding the cost of search. Critically, in each

of these cases, a searching party’s incentive changes dynamically as search progresses. For

instance, a paparazzi’s incentive to search depends on the number of remaining unsearched

hotels. Similarly, pharmaceutical companies strategically adjust their eﬀort as research into

∗We would like to thank Murali Agastya, Suren Basov, Jay Pil Choi, Vincent P. Crawford, Peter Ex-

terkate, Simon Kwok, Jordi McKenzie, Mark Melatos, Jonathan Newton, Michael Paton, Suraj Prasad,

Nicolas de Roos, John Romalis, Robert Rothschild, Abhijit Sengupta, Kunal Sengupta, Rami Tabri, Rus-

sell Toth, Andrew Wait, Don Wright, Charles Zworestine, participants at the 9th Annual International

Industrial Organization Conference in Boston and, in particular, the Advisory Editor and two anonymous

referees for their helpful comments.

†Moore School of Business, University of South Carolina and Lancaster University Management School;

alexander.matros@gmail.com.

‡School of Economics, University of Sydney; vladimir.smirnov@sydney.edu.au.

1

a new medication progresses. The dynamics – and changing nature of search – is critical in

each of these problems. In this paper we analyze a dynamic non-stationary search model

for a known prize, hidden in an unknown location.

To do this, we develop a model in which two players undertake costly search for a

treasure of known value that is hidden somewhere on an island of a given area. We assume

that there is an equal probability that the treasure will be located at any given point

on the island. In each period, both players make their search decisions simultaneously.

Once the treasure is found the game ends. If the treasure is not discovered in any given

period, the game continues. In the next period, each player observes the locations that

have previously been searched. They then decide how much of the remaining area they

wish to investigate. If the treasure is simultaneously discovered – this happens if the two

players successfully search the same part of the island – both players incur their individual

costs but the treasure is destroyed. If a player discovers the treasure on their own, they

receive the full value of the treasure, while both players incur their search costs.

First, we consider the case when the duopolists are unable to coordinate to avoid search-

ing the same location. Given its dynamic nature, we analyze a game in which each state is

described by the remaining unsearched area. Following Maskin and Tirole (1988), we re-

strict our attention to symmetric Markov perfect equilibria (SMPE).1With uncoordinated

search, there is a unique equilibrium in which the potential value of the prize is completely

dissipated by excessive search. To provide some intuition for this dissipation result, con-

sider the special case when there is only one possible search period. As in the standard

price-setting Bertrand game, each player’s payoﬀ maximizing search behavior drives their

rival’s expected return to zero. The generalized game with more than one possible search

period involves the same essential tradeoﬀ, ensuring complete dissipation. This result has

far-reaching economic implications for those participating in a treasure hunt, be they phar-

maceutical ﬁrms, paparazzi, bounty hunters or pirates.

Second, if the players have some ability to coordinate search so as to avoid duplication,

the equilibrium in which all rents are dissipated no longer holds; that is, positive expected

returns are feasible even when the players can only partially coordinate their search activ-

ities. The possibility of coordination may arise between the players through observation.

For example it might be evident where miners are moving their equipment, or the direction

that a paparazzo is heading; a bounty hunter might have better contacts in one area than

another, or might only have a permit to work in a particular jurisdiction. Coordination

could also come about through some form of commitment by a player. For example, the

sort of experts a research department hires provides a (partial) commitment to the di-

rection of their search. Similarly, pharmaceutical companies might need to make public

announcements (in broad terms) about their upcoming clinical trials, and their existing

patents will highlight the most obvious place a ﬁrm will start looking for the next genera-

tion of drug. These coordination devices are, of course, only partial – these commitments

can, to a greater or lesser extent, be unwound. Importantly, our modeling suggests that

1Imposing Markov perfection not only makes our analysis simpler, while still being consistent with

rationality, but it also makes our results directly comparable to those in the previous literature. See

Maskin and Tirole (1988), Bhaskar et al. (2012) and Battaglini et al. (2014) for a general discussion of

when the use of SMPE is appropriate.

2

however imperfect, the possibility of some coordination creates an environment in which

complete dissipation of rents is not possible. This suggests that a third party (government,

industry association and so on) could improve on the decentralized market outcome, even

if its regulatory technology is quite imperfect.

Duplication with independent search also arises in the model of Fershtman and Rubin-

stein (1997). They consider a model in which two players search for a single hidden treasure

in one of a given set of labeled boxes. First, each player chooses how many costly search

units they wish to utilize. Second, each player chooses a random search strategy with which

to examine the boxes, where only one box can be examined in any given period by each

search unit. While each player knows the total number of boxes their rival can search, all

research ﬁndings are private information. Their model gives rise to some complementary

results to ours. For example, there can be search duplication in equilibrium, when players

search previously opened boxes that are empty. Almost the opposite type of coordination

failure is also possible: with an asymmetric probability that various boxes contain the

treasure, sometimes neither player searches the highest probability box ﬁrst. The ﬁrst of

these failures does not occur in our model as previous search is observable to all. The

second outcome is not possible here either due to our focus on pure strategies. Perhaps the

most important similarity is that with independent search rents are completely dissipated

in both models. The intuition is also similar: in Fershtman and Rubinstein (1997) each

player chooses the number of search units in the ﬁrst stage of the game such that the other

player has a zero expected return; in our paper each searcher chooses an area to investigate

each period that drives the other player’s return to zero.

Duplication and coordination by duopolists in a dynamic R&D search game is the

focus of Chatterjee and Evans (2004). In their model, in each period two competing

ﬁrms simultaneously choose to research one of two projects, where only one project will

eventually be successful (given it has been investigated long enough). As in our model,

previous investment is observable and it is common knowledge. They assume, however,

that ﬁrms can coordinate their search in any given period. In Chatterjee and Evans (2004),

only under certain conditions will search be eﬃcient. More typically, the outcome will be

ineﬃcient due to too much duplication (reminiscent of our dissipation result above) or too

much diversiﬁcation (in which the ﬁrms have too much incentive to investigate separate

projects). In particular, if the costs of the two projects are unequal, then there is an

equilibrium in which the high-cost project is searched too often, relative to the socially

eﬃcient choice. One important diﬀerence between their model and ours is that per-period

search costs are exogenous in their model, whereas search costs in our model endogenously

depend on the search undertaken in any given period. Critically, search endogeneity and

lack of coordination in our model leads to a complete dissipation of expected rents, which

is not observed in Chatterjee and Evans (2004).

There is a literature that models research contests as rank-order tournaments. In con-

trast to our paper, this literature analyzes the situation in which there are multiple po-

tential innovations that compete against each other. Some examples recently discussed in

the literature include: a 1992 refrigerator competition (see Taylor, 1995), an 1829 steam

locomotion tournament (see Fullerton and McAfee, 1999), a 1714 British contest for a

method of determining longitude at sea (see Che and Gale, 2003). Within this framework,

3

both Erat and Krishnan (2012) and Konrad (2014) consider the issue of duplication. In

Erat and Krishnan (2012), ﬁrms simultaneously select which area (or avenue of research)

of many clearly delineated areas to search. Each avenue has its own value. In equilibrium,

the expected return from choosing any given area is equalized, with more ﬁrms opting to

search avenues with a higher value. Coordination is facilitated in Erat and Krishnan (2012)

by the ability to commit to only search in a particular avenue. As in our model with partial

search coordination, and in a similar manner to Chatterjee and Evans (2004), while there

is duplication, not all rents are dissipated. Konrad (2014) also considers a model in which

a ﬁrm must choose a speciﬁc avenue of research, where these avenues do not overlap. Im-

portantly, if two or more ﬁrms choose the same avenue, any rents generated are dissipated,

as in Bertrand competition. Again, it is the ability of ﬁrms to commit to search only one

avenue that facilitates coordination. In equilibrium there is no duplication, as no two ﬁrms

choose the same avenue of research. In our model, a parallel result arises when duopolists

can perfectly coordinate their search patterns.

In many ways, our paper synthesizes the existing literature, from the model with in-

dependent search of Fershtman and Rubinstein (1997), to the perfect coordination case

in Konrad (2014), to the partial coordination search models in between of Chatterjee and

Evans (2004) and Erat and Krishnan (2012). One important diﬀerence between our paper

and the existing literature is that in our model ﬁrms are choosing how much to search every

period (allowing for search dynamics). On the other hand, Chatterjee and Evans (2004),

Erat and Krishnan (2012) and Konrad (2014) eﬀectively assume that the amount of search

is given exogenously, while in Fershtman and Rubinstein (1997) an equivalent choice is

made only once at the beginning at the game.

Our work extends the previous research on private-good search.2These models mostly

deal with situations that are either static or involve complete information.3One of the

few papers that considers the dynamics of investment within the private-good framework is

Reinganum (1981), who assumes that the success function is exponential and the environ-

ment is stationary.4As a result, equilibrium strategies can be represented as functions of

time only, which greatly simpliﬁes the analysis.5The memorylessness assumption, however,

is often not satisfactory; for example, when the attractive search domain, while potentially

large, is ﬁnite. As noted above, one of the contributions of this paper is ﬁnding equilibria

in a truly dynamic search model, in that players’ choices regarding eﬀort depend on the

remaining unsearched island size. Our framework allows for another important diﬀerence

with the existing literature. With a uniform distribution, investing always becomes more

attractive over time, ensuring that once ﬁrms start investing, they will continue until the

2There is also a large literature on investment dynamics in relation to public goods; for instance, Admati

and Perry (1991), Marx and Matthews (2000), Lockwood and Thomas (2002), Compte and Jehiel (2004),

Yildirim (2004, 2006), Bonatti and H¨orner (2011), Matthews (2013), Battaglini et al. (2014) and Georgiadis

(2015).

3See, for example, Loury (1979), Dasgupta and Stiglitz (1980a,b) and Lee and Wilde (1980), and

Reinganum (1989) and Long (2010) for a survey of the literature.

4In the context of our model, this would be equivalent to assuming that the present value of the treasure

is constant.

5Reinganum (1982) uses a similar framework to show how the availability of patent protection can

accelerate development of the innovation.

4

treasure is discovered (or the search area is exhausted). This is very diﬀerent to the out-

come with an exponential distribution typically used in the literature where, if there is a

positive probability that the project is unsuccessful, investing becomes less attractive over

time and it is possible that the project could be suspended.

2 The Model

In the model, two players (ﬁrms) search over an inﬁnite horizon for a treasure hidden

somewhere on an island of size x1>0. The treasure has the same value R > 0 for both

players6, and there is an equal probability that the treasure will be located at any given

point on the island.7Payoﬀs are discounted at a common factor δ.

Each period tcomprises the following sequence of events. First, both players learn the

size and location of all prior searches.8As a convention, we use the term ‘area’ to mean

the size of land searched and the term ‘region’ to mean the location of the search. Let xt

be the unsearched area of the island at the beginning of period t. Second, each player i

simultaneously plans an area Ii,t ∈[0, xt] to search at a cost of cIi,t, where c > 0.9The

regions searched by the players are partially coordinated according to a process discussed

below. Finally, search plans are implemented. If the treasure is found, the game ends. If

both players search the same region and ﬁnd the treasure simultaneously, each of them

incurs costs, but the treasure is destroyed.10

Suppose that player isearches an area Ii,t in period t, for i= 1,2. If the region of

search is chosen uniformly at random, the probability that isearches a speciﬁc point on

the island is Ii,t/xt. If both players were to choose their search region independently, then

the probability that both players search this particular point would be I1,tI2,t/x2

t, and the

area of duplicated search would be I1,tI2,t /xt.11

Under our speciﬁcation of partial coordination, the extent of duplication is given by

(1 −α)I1,tI2,t /xt, where the parameter α∈[0,1] describes the extent of coordination. If

6As players are risk neutral, in the case of random returns Rcould be replaced by E[R], requiring no

other changes to the model.

7In Appendix B we consider the case when the distribution is non-uniform.

8Using the terminology of Mailath and Samuelson (2006), we consider a stochastic game with perfect

monitoring in a dynamic setting.

9Non-linear costs are discussed in Section 4.

10This assumption is standard in the R&D literature. Intuitively, if several players discover the treasure

simultaneously, ﬁerce competition between them runs down the surplus to zero. A good example of such a

situation for just two players is Lockheed and Douglas jet development in the 1960s. For more detail, see

The Economist (1985); and Chatterjee and Evans (2004). Many examples of simultaneous discoveries in

science can be found in Merton (1973).

11 Independent search in each period of our dynamic game mimics the random search strategy of ﬁrms in

the static model of Fershtman and Rubinstein (1997). We assume that a player’s unsearched area (which

we model as the [0, xt] segment) can be partitioned into a large number of equally-sized small segments. In

this structure, from each player’s perspective the treasure is equally likely to be located in any one of the

segments. Given player ichooses Ii,t , we assume he will randomly select a portion Ii,t/xtof all unsearched

segments, where each of the unsearched segments is chosen with equal probability. For our analysis, the

area of duplicated search is deﬁned by its limit when the size of each segment goes to zero. See Appendix A

for a detailed explanation.

5

α= 1, search is perfectly coordinated and duplication is avoided; if α= 0, search is

independent; and α∈(0,1) indicates partial coordination.12

If in any period t, player 1 searches I1,t and player 2 searches I2,t, after accounting for

duplication, the players search an area of

Jt= min {xt, I1,t +I2,t −(1 −α)I1,tI2,t/xt}.(1)

Player ihas a (Jt−Ij,t)/xtprobability of ﬁnding the treasure alone and obtains the following

expected payoﬀ in period 1: Jt−Ij,t

xt

R−cIi,t.

The game ends with probability Jt/xtin period t. If the treasure is not found in period t,

which happens with probability 1 −Jt/xt, the unsearched area of the island shrinks to

xt+1 =xt−Jt,

and the game proceeds to period t+ 1.

Player i’s strategy is an inﬁnite sequence of functions specifying how much to search

each period contingent upon any possible sequence of previous searches. We assume that at

any stage, the entire history of past searches is common knowledge and can be summarized

by the ‘state’, the current unsearched area. We will consider Markov strategies in which

the past inﬂuences current play only through its eﬀect on the current unsearched area. A

pure Markov strategy for player iis a time-invariant map Ii:X→X, where X= [0, x1]

and Ii(x)∈[0, x]. Note that for ease of exposition, when convenient we skip subscript tand

indicate that Idepends on the current unsearched area x; that is, Ii,t =Ii(x). Also, when

the meaning is clear, we skip subscript iin describing the equilibrium search intensity and

the value function.

We will restrict our attention to symmetric equilibria. Therefore, the solution concept

we use is a symmetric Markov perfect equilibrium (SMPE).13 Moreover, we focus on non-

trivial SMPE; that is, equilibria in which a positive amount of search occurs somewhere

along the equilibrium path. We use the following approach to obtaining the SMPE. Player

itakes the state-contingent search plans of his rival Ij(x) as given. Given this function,

player isolves a standard optimization problem and chooses his optimal search, Ii(x).

However, given that the function Ij(x) is endogenous to the model, to obtain the symmetric

equilibrium we need to ﬁnd the function Ij(x) such that Ii(x)≡Ij(x), where Ii(x) is the

optimal search for player iwhen he takes Ijas given.14

12Diﬀerent sources of partial coordination are possible. For example, the parameter αcould describe the

degree of ﬂexibility of the players to adapt to the search plans of their rival. Alternatively, αcould represent

the proportion of information that is publicly available regarding each player’s search, whereas (1 −α) is

the proportion that cannot be credibly revealed. Additional motivation for three diﬀerent scenarios and

micro-foundations for partial coordination are provided in Section 3.

13We deﬁne equilibria in terms of the policy functions I1and I2, which describe the area but not the

region of search. An equilibrium is therefore consistent with a multiplicity of outcomes relating to the

region of search.

14We discuss both subgame perfect and asymmetric Markov perfect equilibria in Section 4.

6

Note that when multiple SMPE exist, we focus on the Superior SMPE (SSMPE); that

is, the SMPE with the highest total expected payoﬀ.15

3 Analysis of the model

In this section we consider the equilibrium outcome of the model. To do this, let us ﬁrst

establish several preliminary results for the general model that we will then use in our

subsequent analysis.

Player isolves the following Bellman equation:

V(x, c, R) = max

Ii∈[0,x]−cIi+J−Ij

xR+δ1−J

xV(x−J, c, R),(2)

where xis the part of the island which is still unsearched before the current period,

V(x, c, R) is the value function for each player (we use the symmetry assumption here),

and Jis given by equation (1). The ﬁrst term in equation (2) describes the player’s costs of

search in the current period. The second term is the player’s expected value from ﬁnding

the treasure alone in the current period. The last term is the player’s expected value from

future periods.

To simplify the analysis, we introduce the following lemma.

Lemma 1. The following identity always holds:

V(x, c, R)/R =V(cx/R, 1,1).(3)

Proof. See Appendix C.

An increase in the cost parameter cin the original problem (2) is equivalent to an

increase in the unsearched area size x. On the other hand, an increase in the treasure value

Rhas two eﬀects; it decreases the unsearched area size x, but it also directly changes the

value function. Eﬀectively, the island size is proportional to c, inversely proportional to R,

and payoﬀs are denominated in R.16

Slightly abusing notation, rename V(x) = V(x, 1,1). From Lemma 1 it follows that it

is suﬃcient to solve equation (2) when R=c= 1; that is,

V(x) = max

Ii∈[0,x]−Ii+J−Ij

x+δ1−J

xV(x−J).(4)

The following lemma allows us to distinguish between equilibria under independent and

coordinated search.

15As argued by Fudenberg and Tirole (1985), if one equilibrium Pareto dominates all others, it is the

most reasonable outcome to expect.

16For additional details and comparative statics, see our complementary paper Matros and

Smirnov (2011).

7

Lemma 2. Combined search Jdescribed by equation (1) exhibits the following properties:

i) If α= 0,I1< x, and I2< x, then J < x.

ii) If α∈(0,1], there exist I1< x and I2< x such that J=x.

Proof. See Appendix C.

Lemma 2 suggests that if α= 0, the entire island is searched in a given period only

in the case that at least one player searches the entire island. As symmetry requires each

player’s search is the same, it is not possible for the players to search the entire island

without destroying the treasure in a symmetric equilibrium. By contrast, with an element

of coordination, it is possible for the players to complete the search without full dissipation

of the treasure.

3.1 Independent search (α= 0)

Let us ﬁrst consider the extreme case when players search independently. There are many

situations when it is not possible to avoid duplication. For instance, the set of projects

may have no obvious ordering and it may be very diﬃcult to predict rival search plans;

this means that the two players might end up simultaneously searching the same area.

Duplication is more likely to arise when the players have no common background to rely

on and players cannot observe the search of their rivals. With no common expectation of

the search plans of any player, duplication could be unavoidable.17

Utilizing Lemma 2, when α= 0 equation (1) transforms to:

J=I1+I2−I1I2/x. (5)

We analyse this case in the following proposition.

Proposition 1. If α= 0, in the unique SMPE

I(x) = max{0, x −x2}and V(x)≡0.

Proof. See Appendix C.

To get some intuition, let us consider the static problem where players can only search

once and α= 0. Substituting for J, the problem in this case is

V(x) = max

Ii∈[0,x]−Ii+Ii

x1−Ij

x= max

Ii∈[0,x]Ii(x−x2−Ij)

x2.

If the search by its rival is Ij< x −x2, it is a dominant strategy for player ito choose

Ii(x) = x. If, on the other hand, search by the other player is Ij> x −x2, it is a dominant

17One example we have in mind is pharmaceutical companies trying to develop a new product, such as a

sleeping pill. Given the secrecy surrounding work in these laboratories, it is not possible to guarantee that

there is no duplication by rivals. For example, Eichler et al. (2013) discuss a documented case of duplication

in medical research. Chatterjee and Evans (2004) also provide case-study evidence of simultaneous discovery

leading to a dissipation of potential proﬁts.

8

strategy for player ito opt Ii(x) = 0. There is a unique symmetric equilibrium in which

both players search I(x) = max{0, x −x2}and each receives a value of zero. This result

mimics the outcome in the standard Bertrand price-setting game in which the best response

of each player ensures that all of the potential rents are dissipated in equilibrium.

In a similar manner, in the equilibrium of the dynamic game the island has no value

to either player for any x. To see why this is an equilibrium, note that if V(x) = 0 for

all x, then (4) reduces to the static problem above, and the unique solution has a value of

zero for both players. Because of the contraction property of the dynamic problem, this

equilibrium is unique. The economic intuition for this equivalence between the static and

dynamic problems is that there is eﬀectively no connection between the periods; there is

no beneﬁt for a player to wait until next period, as in the next period players face exactly

the same problem.

Note, not only do we have duplication in this extreme case with α= 0, the duplication

here is so egregious that the whole value of the treasure is dissipated in equilibrium. Thus,

independent search gives rise to an extreme example of the tragedy of the commons. A fea-

ture of equilibrium is that the island is never searched completely and search may continue

for arbitrarily many periods.

3.2 Coordination α∈(0,1]

We next consider the case when the parties have some ability to coordinate their search

for the treasure. As mentioned above, there are times when parties can commit to a search

pattern or to focus (or not) on a particular location. For example, pharmaceutical compa-

nies might have to lodge broad plans with the regulator regarding their upcoming clinical

trials; these plans provide some (potentially imperfect) commitment to a given search strat-

egy. Similarly, the hiring of experts in a given area of expertise, or the purchase of speciﬁc

research hardware can also be observable commitment. Another possibility is that at the

start of each period each player decides on the size of their research budget and speciﬁc

search locations, but only the choice of research budget is observable to general public

(possibly through required public announcements for a publicly listed ﬁrm). Given these

budgetary commitments (which determine the area each player searches in that period),

it is in the interests of the two rivals to coordinate in an attempt to avoid duplication.

However, if it is too costly to adjust all search locations, only partial coordination is pos-

sible.18 In the model, αindicates the proportion of a player’s search that can be adjusted,

whereas (1 −α) is the proportion of search that is suﬃciently inﬂexible so that it cannot be

adjusted.19 The generalization presented in this section allows us to capture both partial

coordination of search (0 < α < 1) and perfect coordination (α= 1).

To simplify the exposition of the model, it is convenient to introduce the following

18For example, we could assume that both parties reveal all locations of their search but only one party

can reallocate some of the locations they initially planned to search. Note that the same micro-foundations

as in the case of independent search apply here, where the area of duplication is scaled by (1 −α).

19Chatterjee and Evans (2004) and Erat and Krishnan (2012) generate partial coordination by allowing

perfect coordination with respect to project selection and independent search within a project.

9

function:

Ψ(x)≡xV (x).(6)

In terms of Ψ(x), equation (4) can be rewritten as

Ψ(x) = max

Ii∈[0,x]{J−Ij−xIi+δΨ(x−J)}.(7)

Note that equation (7) resembles the problem of extraction of exhaustable resources under

common access or, equivalently, a multi-player cake-eating problem.20

Before proceeding, we introduce the following operator Bdeﬁned by

(BΨ)(x)≡max

Ii∈[0,x]{J−Ij−xIi+δΨ(x−J)}.(8)

This allows us to consider the following lemma.

Lemma 3. If 0< α ≤1, there exists ﬁnite Tsuch that

(a) any non-trivial SMPE involves Jt>0for 0≤t≤T, and xT+1 = 0; and

(b) all SMPE can be obtained in Tsteps. Speciﬁcally,

Ψ0≡0,Ψs≡BΨs−1for s= 1,2,...,T, and Ψ = ΨT.

Proof. See Appendix C.

Lemma 3 suggests that the treasure is found in ﬁnite time in any non-trivial SMPE.

Discounting and coordination play important roles in this result. Discounting ensures

that the value of indeﬁnite search converges to zero. If search ever commences, it will be

completed in ﬁnite time. To see the role of coordination, recall from Lemma 2 that only if

search is coordinated is it possible to search the entire island and obtain positive value.

Furthermore, Lemma 3(b) reveals that the ﬁnite application of the operator Brepre-

sents an algorithm for solving the Bellman equation. From this, we derive the following

proposition for when search is completed in one period.

Proposition 2. If 0< α ≤1,∀x≤√α1−δ√α

1+√αin the unique SMPE

I(x) = x

1 + √αand V(x) = √α−x

1 + √α>0.

Proof. See Appendix C.

This Proposition shows that when players have some non-zero ability to coordinate,

the extreme dissipation result of Proposition 1 does not hold. Coordination allows for

the existence of equilibria with some search duplication and a value function that is non-

zero. Lemma 2 plays an integral role in this result. With coordination, it is possible for

20See for example Long (2011) for a recent survey of dynamics games in the economics of natural

resources.

10

a combined search to cover the whole island without each player perfectly duplicating the

locations searched by its rival; this ensures that for a suﬃciently small island, there is a

positive-payoﬀ strategy for both players.

This has potential policy implications. A government agency or regulator can improve

welfare by insisting on some publicly available information or announcement by innovating

ﬁrms (this could be a condition of receiving government subsidies or approval). Indeed,

the amount of information contained in these statements need not be onerous to help

the players avoid the extreme dissipation of value. This provides another justiﬁcation for

public announcements regarding intent to ﬁle patent applications and for the requirement

of pharmaceutical companies to register with the industry regulator information regarding

pending clinic trials.

3.3 Perfect coordination (α= 1)

We next consider perfectly coordinated search.21 There are many situations when it is

possible to avoid duplication. If, for example, players can uniquely order all potential

projects in a line, the two players have a choice to start searching from the left or the

right, avoiding duplication. (This game eﬀectively has two asymmetric equilibria, with

each player commencing their search from diﬀerent ends.) Another example is when the

treasure is located on one of two separate islands; again, the players can avoid duplication

by searching the island not occupied by their rival. Similar intuition applies when search

requires some machinery (digging equipment, for instance). Not only does this provide

observability as to the location of a rival’s search, it might also make duplication physically

impossible.22 Perfect coordination is also consistent with one of the ﬁrms being able to

adjust their search plans in the second stage of each period, as discussed at the beginning

of Section 3.2.

When α= 1 equation (1) transforms to

J= min {x, I1+I2}.(9)

Based on Lemma 3, deﬁne dto be the maximum number of search periods when α= 1.

One then can construct the sequence {Ψs}d

s=1 and ﬁnd all SMPE. This is called the value-

iteration procedure, which is equivalent to using backward induction. In general, there will

be multiple SMPE. As discussed earlier, in the case of multiplicity we focus on the SSMPE.

21An equivalent perfect coordination assumption is made in the previous R&D literature; see for example,

Cardon and Sasaki (1998) and Konrad (2014).

22Moreover, a third party like a government can often help ﬁrms avoid duplication. For example, patents

prevent rivals encroaching on a ﬁrm’s established area of expertise. Similarly, governments issue exclusive

exploration licences to mining companies.

11

We ﬁrst deﬁne the polynomials Ps(x) and critical island sizes χs, for s= 1,...,d,23

Ps(x) = −1

2(1 −x)2+1

2s(1 −x) + (4δ)s−4δ

2(4δ−1)4sfor s= 1,...,d; (10)

χs= 1 −3δ+ (4δ)s(δ−1)

2s(4δ−1) for s= 1,...,d−1 and (11)

χd= 1 + 1

2dr1−(4δ)d

1−4δ−1where d=&ln ((δ2+ 2δ)/(1 −δ)2)

ln (4δ)'.

Next we give a complete characterization of the SSMPE when α= 1 in the following

proposition.

Proposition 3. If α= 1, the non-trivial SSMPE exists if and only if x≤χdand is

described as follows. The search intensity of each ﬁrm is

I(x) =

x

2,if x≤χ1,

(1−δ)(1−x)+δ/2

2δ,if χ1< x ≤χ2,

.

.

.

(1−δ)(1−x)+δ/2s−1

2δ,if χs−1< x ≤χs,

.

.

.

(1−δ)(1−x)+δ/2d−1

2δ,if χd−1< x ≤χd;

and the value function for each ﬁrm is

V(x) = Ψ(x)/x =

P1(x)/x, if x≤χ1,

.

.

.

Pd(x)/x, if χd−1< x ≤χd.

Proof. See Appendix C.

Proposition 3 describes a ﬁnite horizon search for suﬃciently small islands. In the non-

trivial SSMPE duopolists plan to search the island for at most dperiods when x≤χd.

There are dpossibilities, depending on the island size. If x≤χ1, the duopolists search the

island in just one period. If x∈(χ1, χ2], the ﬁrms plan to search the island for at most

two periods. In general, the duopolists plan to search the island for at most speriods for

values of x∈(χs−1, χs], where s= 2,...,d.

As one can see from Proposition 3, the equilibrium search intensity I(x) is a spline of

degree one on the interval [0, χd] with knots χ1, ..., χd.24 Let us characterize the equilibrium

23Some of the expressions below are presented as ratios for expositional purposes. In particular, note that

both numerators and denominators in Ps,χs,χdand dare equal to zero when δ= 1/4. These expressions

are then deﬁned by their limits as δ→1/4.

24A spline is a special function deﬁned piecewise by polynomials, see for example Ahlberg, Nielson, and

Walsh (1967).

12

1.0

I(x)

0

0.3

x

a)

1.0

V(x)

0

0.5

b) x

Figure 1: The equilibrium search intensity and value for duopolists when δ= 0.75.

search intensity in the following corollary.

Corollary 1. If α= 1 and x≤χd, the equilibrium search intensity I(x)is a piece-wise

linear, discontinuous and quasiconcave function.

Proof. See Appendix C.

Figure 1a illustrates the equilibrium search intensity for diﬀerent island sizes in the

SSMPE for each duopolist when δ= 0.75. In this case d= 4. The equilibrium search

intensity is a discontinuous function. At each knot of the spline χs, the equilibrium strategy

of a ﬁrm switches from searching for speriods to searching for s+ 1 periods. The fact

that each ﬁrm maximizes its own value, not total surplus, leads to this discontinuity. The

discontinuities allow one to discern the intertemporal path of search by reading from right to

left. Observe that, conditional on the search being conducted in 4 periods, the equilibrium

search intensity is increasing in the ﬁrst 3 periods. In the last period the equilibrium search

intensity decreases. This is due to the fact that, while in the last period the duopolists

search all of the remaining land, this search area in the last period can in fact be smaller

than the area searched in the second-to-last period.

The following corollary characterizes the value function V(x), as derived in Proposi-

tion 3.

Corollary 2. If α= 1 and x≤χd, the value function V(x)is a continuous and monoton-

ically decreasing function.

Proof. See Appendix C.

Figure 1b illustrates the value function when δ= 0.75. Observe that the value function

is monotonically decreasing with the size of the unsearched area because it is harder to ﬁnd

the treasure on a larger island. However, the value function is not smooth; the slope of the

value function changes at knots χs, coinciding with the discontinuities in the equilibrium

search intensity.

13

3.4 The possibility of futile search when α= 1

So far we have assumed that the treasure is located on the island with certainty. To add

some extra realism, here we analyze the possibility of futile search. Speciﬁcally, we consider

that the probability that the treasure exists on an island of size x′be p∈(0,1] for the case

when α= 1.

One possible interpretation of futile search is that the treasure is located on the island

for sure but players may search only some part of the island. Extend the island to a size

x=x′/p with the convention that the treasure belongs to the island of size xwith certainty,

but only the portion x′can be searched, while the remainder is not accessible. That is, we

assume the treasure is located with uniform probability on the area of size x. Further, we

denote the portion of the island inaccessible by the players as γ=x−x′.

Using this new structure, equation (4) transforms to

V(x) = max

Ii∈[0,x−γ]−Ii+J−Ij

x+δ1−J

xV(x−J),(12)

where

J= min {x−γ, I1+I2}.(13)

The only diﬀerence from the main model is that an area γ≥0 cannot be searched. As

before, it is convenient to introduce the function Ψ(x)≡xV (x). Equation (12) transforms

to

Ψ(x) = max

Ii∈[0,x−γ]{J−Ij−xIi+δΨ(x−J)}.(14)

Next we show that the problem with γ > 0 can be transformed to the original problem

with γ= 0.

Proposition 4. Substitutions x′=x−γ

1−γand Ψ′=Ψ

(1−γ)2transform problem (14) into

problem (7).

Proof. See Appendix C.

Proposition 4 implies that the problem (14) can be derived from (7) by a linear trans-

formation. Intuitively, both (7) and (14) represent a multi-player cake eating problem. In

this game, each player’s strategy depends only on the remaining cake size (be it xor x−γ),

making both problems isomorphic. Therefore the perturbed model retains the properties

of the original. In particular, searching becomes more attractive over time and, as a result,

in any non-trivial equilibrium, players will continue to search until they either ﬁnd the

treasure or have searched the entire accessible region.

Figure 2 illustrates the equilibrium search intensity I(x) (panel a)) and the value func-

tion V(x) (panel b)) when δ= 0.75 and γ= 0.5. The substitution x′= 2x−1 transforms

Figure 2a to Figure 1a. Note that for x≤γ, search is not possible. When x > γ, because

of the transformation, search intensity appears more sensitive to island size.

On the other hand, in Figure 2b the value function is very diﬀerent from the one

represented in Figure 1b. The transformation from Ψ back to the original value function

14

1.0

I(x)

0

0.15

x

a)

x1.0

V(x)

0

0.025

b)

Figure 2: I(x), V(x) when δ= 0.75, γ= 0.5.

creates an additional eﬀect. When x≈γ, an increase in xleads to a signiﬁcant increase in

the probability that the treasure is located on the part of the island that can be searched.

This leads to a sharp increase in value, as shown in Figure 2b. However, once the island

is signiﬁcantly larger than γ, further increases in xhave a relatively small impact on the

probability that the treasure belongs to the searchable part of the island. In this case, the

increase in the cost of searching the entire island has a larger eﬀect, and value declines in

x. Consequently, there is a non-monotonic relationship between the value and island size.

As in the main model, and for the same reasons, the value function is not smooth.

3.5 Coordination and dynamics

Let us now compare investment dynamics with independent and perfectly coordinated

search. Figure 3 illustrates the equilibrium search intensity of each player, I(x) (panel a)),

and the combined search, J(x) (panel b)), when δ= 0.75, both when α= 0 and α= 1.

For small island sizes, the players search the entire island in a single period under perfect

coordination. This leads to a linear relationship between xand both I(x) and J(x) for small

x. Now consider uncoordinated search when xis small. Because of duplication, each ﬁrm

searches more than they would with coordination. However, jointly the area searched by the

two rivals is actually less with independent search, as their combined search does not cover

the whole island (Lemma 2). With larger island sizes, search intensity in any one period is

complicated by the sequencing of search over time, and it is possible for both individual and

combined search to proceed more quickly without coordination.25 Furthermore, as noted

above, the value from search with duplication is always zero, whereas with coordination

there is a positive value of search for each player provided x < χd.

4 Discussion

This paper considers duplication of search in a model with two players seeking to discover

a prize of known value that is hidden in an unknown location. We model search as a

25Speciﬁcally, when δ→1, χ2→3/4. In the neighborhood to the right of χ2when δ≈1, both individual

and combined search intensities are higher with independent search.

15

x

1.0

I(x)

0

0.3

α=0

α=1

a)

x1.0

J(x)

0

0.5

α=1

α=0

b)

Figure 3: I(x), J(x): α= 0 is concave, α= 1 is piecewise linear; δ= 0.75.

dynamic R&D race in a non-stationary environment and a ﬁnite search space. We show

that duplication can be extreme – when the players are unable to coordinate their search at

all, the unique equilibrium involves complete rent dissipation. Coordination, be it partial

or perfect, can help alleviate some of these duplication issues, allowing players to earn

positive expected returns. Consequently, this model has a clear implication for the role of

a third party (a government, industry group or regulator) in facilitating coordination in

research environments.

Several caveats relating to the model are worth commenting on. First, we use the SMPE

as our equilibrium solution concept. There also exist asymmetric Markov perfect equilibria,

where in every period each player searches a diﬀerent amount. Using a similar approach, we

consider an example of an asymmetric Markov perfect equilibrium in Appendix C. Second,

there exist other non-Markovian symmetric equilibria that rely on trigger strategies similar

to those in Compte and Jehiel (2004). In the case of small islands, the only symmetric

equilibrium that exists is the SMPE. Players can always ﬁnish the search in one period, so

trigger strategies are not useful in supporting cooperation. In contrast, in the case of large

islands with a large number of potential search periods, trigger strategies may improve

payoﬀs and additional equilibria can exist. Third, in the model we assume linear search

costs. In a complementary paper, Matros and Smirnov (2011) focus on the comparative

statics of monopoly versus duopoly and multi-player games allowing for quadratic costs.

Finally, in this paper we derive the search intensity and value generated for the players

when there is no coordination (α= 0) and when there is perfect coordination of search

(α= 1). We also derive the search intensity and value generated when there is partial

coordination (α∈(0,1]) and search is completed in one period. We have not solved the

remaining case in which there is partial coordination when it is optimal to complete search

of the island in more than one period. Deriving a complete analytical solution for the

partial coordination case is beyond the scope of this project and we leave it for future

research.

16

Appendix A

For notational simplicity, we omit time subindexes. Here we provide micro-foundations as to why

the area of duplication with independent search is I1I2

x(see footnote 11).

Without loss of generality, let us model the unsearched area at a given point in time by con-

sidering a segment [0, x]. We assume that [0, x] can be partitioned into Nequally-sized segments,

where Nis large. There are N I1

xand NI2

xsegments that are searched by players 1 and 2 re-

spectively. We consider the limit when Ngoes to inﬁnity, and therefore we can ignore issues of

discreteness.

We use the search of player 1, without loss of generality, as a reference by reordering all the

segments so that the segments searched by player 1 are closest to the left corner of the original

[0, x]. As a result, [0, x] is divided into two parts: [0, I1] where player 1 searches and [I1, x] where

player 1 does not search. Given search is independent, N I2

xsegments are randomly selected among

all Nsegments. As a consequence, search duplication is given by the amount of search the second

player undertakes in the [0, I1] segment.

Let us deﬁne kas the number of small segments that land in the [0, I1] segment, describing the

resulting number of successes. There are n=N I2

xdraws, without replacement, from a population

of size Nthat contain K=NI1

xsuccesses, wherein each draw is either a success or a failure. k

is a random variable with a hypergeometric distribution. As the size of each small segment is x

N,

the area of duplicated search is equal to A=kx

N. Consequently, search duplication is distributed

with the mean of E[A] = I1I2

xand the variance of V ar[A] = I1I2(1−I1/x)(1−I2/x)

N−1. As N→ ∞, the

distribution of Abecomes degenerate with the probability mass function equal to 1 at I1I2

x.

Appendix B

In this Appendix we modify the basic model to allow a non-uniform distribution for the location

of the treasure. We investigate the cases of independent search (α= 0) and perfect coordination

(α= 1) separately.

Consider n≥2 islands, indexed by k= 1,...,n. Assume that within each island there is an

equal probability that the treasure will be located at any given point. Each island kis described by

its size x(k)and the probability p(k)that the treasure is located on the island. The attractiveness

of island kfor search is determined by the probability of ﬁnding the treasure per unit area of the

island. Without loss of generality, we order islands according to this measure:

p(1)

x(1) >p(2)

x(2) >···>p(n)

x(n)>0 with p=

n

X

k=1

p(k)≤1,(15)

where pis the probability the treasure is located on one of the islands. By adjusting the vector

of island sizes and probabilities, we can approximate arbitrary distributions over the location of

the treasure.26

Each player inow has to choose how to structure his search across the nislands. We denote

the current state by the vectors X= (x(1),...,x(n)) and p= (p(1) ,...,p(n)) and the current value

to player ias V(X,p). Player ichooses the policy functions I(k)

i(X,p), for k= 1,...,n, which

26We follow recent papers of Erat and Krishnan (2012) and Konrad (2014) in modelling non-uniform

distributions as we partition the landscape into sites (islands) described by diﬀerent probabilities.

17

describe the extent of simultaneous search he undertakes on each of the nislands in the current

period. As in the main model, if the treasure is not found in the current period, the unsearched

area of each island kshrinks to

x(k)

t+1 =x(k)

t−J(k)

t,(16)

where J(k)

t= min nx(k)

t, I(k)

1,t +I(k)

2,t −(1 −α)I(k)

1,t I(k)

2,t /x(k)

toand J= (J(1),...,J(n)). Following the

current period search, the probability of ﬁnding the treasure in each island kupdates according

to Bayes’ rule as follows:

p(k)

t+1 =p(k)

t(1 −J(k)

t/x(k)

t)

1−Pn

m=1 p(m)

tJ(m)

t/x(m)

t

.(17)

The Bellman equation faced by each player iis then

V(X,p) = max

nI(k)

i∈[0,x(k)]on

k=1

n

X

k=1

−I(k)

i+

p(k)J(k)−I(k)

j

x(k)

(18)

+δ 1−

n

X

k=1

p(k)J(k)

x(k)!V(X′,p′)),

where transitions X→X′and p→p′are speciﬁed by relationships (16) and (17). The ﬁrst

term describes the current period costs and expected beneﬁt of search across all nislands. The

second term represents the continuation value of the remaining vector of unsearched islands. In

a similar manner as to how we deﬁned the operator Bin equation (8) from the Bellman equation

(7), here we introduce operator ¯

Bbased on equation (18).

Independent search (α= 0)

We ﬁrst reexamine the case of independent search.

Proposition 5. If α= 0, in the unique SMPE, I(k)(X,p) = max{0, x(k)−(x(k))2/p(k)}where

k∈ {1, n}and V(X,p)≡0.

Proof. See Appendix C.

Proposition 5 suggests that the extreme dissipation result of the basic model extends to

heterogeneous distributions over the location of the treasure. That is, for any distribution, when

α= 0 duplication is so egregious that the entire value of the treasure is squandered in equilibrium.

The pattern of search has the following characteristics. First, in each period only islands that have

p(k)

t> x(k)

twill be searched in equilibrium, and all of these islands will be searched simultaneously

by both players, with the intensity of search in island kincreasing in the attractiveness of search,

given by equation (15). Second, after search is completed in any period, p(k)

tupdates to p(k)

t+1

according to Bayes’ rule (17). As a result, search could expand to include new islands. This

occurs for islands kwith p(k)

t≤x(k)

tand p(k)

t+1 > x(k)

t+1. Finally, note that in a similar manner to

the uniform distribution case, in equilibrium none of the islands will ever be completely searched.

Perfect coordination (α= 1)

Now consider perfect coordination.

18

Proposition 6. If α= 1,

(1) in any SMPE, island kis searched in period tonly if islands 1,...,k −1are exhausted in

period τ, where τ≤t;

(2) there exists ﬁnite Tsuch that

(a) any non-trivial SMPE involves J(k)

t>0for 0≤t≤Tfor at least one k, and x(k)

T+1 = 0 for

all k; and

(b) all SMPE can be obtained in Tsteps. Speciﬁcally,

V0≡0, Vs≡¯

BVs−1for s= 1,2,...,T, and V=VT.

Proof. See Appendix C.

This proposition outlines the outcome when there are multiple islands that diﬀer. In this case,

part (1) states that the players will search the islands in order of their attractiveness, undertaking

an exhaustive search before moving onto the next most attractive island. This is a similar result

to Choi and Gerlach (2014) in which researchers move from the most attractive option down. This

is a signiﬁcantly diﬀerent search pattern from one we observed in the independent search case

studied earlier in this section. Part (2) extends the result in Lemma 3 with partial coordination,

showing that search will be successful in ﬁnite time when there is perfect coordination and multiple

islands. This ensures a positive expected return provided the project is close to completion. Taken

together, Propositions 5 and 6 demonstrate the key role that coordination has in facilitating that

all rents are not completely dissipated in the case of a non-uniform distribution regarding the

location of the treasure.

Appendix C

For notational simplicity, when convenient we omit ﬁrm and time subindexes.

Proof of Lemma 1

We introduce the new variables ˜x=cx/R,˜

Ii=cIi/R,˜

Ij=cIj/R and ˜

J=cJ/R and substitute

them into problem (2). This substitution results in

¯

V(˜x, 1, R) = max

˜

Ii∈[0,˜x](−˜

IiR+˜

J−˜

Ij

˜xR+δ 1−˜

J

˜x!¯

V(˜x−˜

J , 1, R)),(19)

where ¯

V(˜x, 1, R) = V(x, c, R). Next, substitute ¯

¯

V(˜x, 1,1) = ¯

V(˜x, 1, R)/R into equation (19) to

derive

¯

¯

V(˜x, 1,1) = max

˜

Ii∈[0,˜x](−˜

Ii+˜

J−˜

Ij

˜x+δ 1−˜

J

˜x!¯

¯

V(˜x−˜

J , 1,1)),

which coincides with problem (2) when R=c= 1. Consequently, V(x, c, R)/R =V(cx/R, 1,1)

proving the Lemma.

19

Proof of Lemma 2

Substitute I1=x−ε1,I2=x−ε2and α= 0 into (1) to derive

J=x−ε1ε2/x.

Part i) follows directly.

To prove part ii) assume that I1=I2. Condition J≥xtransforms to Ii≥x

1+√α. This

concludes the proof.

Proof of Proposition 1

In the proof below we follow the argument presented in the proof of Proposition 1 in Ericson and

Pakes (1995).27 Introduce the following operator ˜

Bdeﬁned by

(˜

BV )(x)≡max

I∈[0,x]−I+J−Ij

x+δ1−J

xV(x−J),(20)

where the solution has to satisfy I=Ijand J=I+Ij−I Ij/x. By the contraction property of

the operator deﬁned in equation (20), V(x) = lims→∞ Vs(x), where Vs(x)≡˜

BVs−1(x)≡˜

BsV0

and V0≡0. The remainder of the proof follows by induction. V0≡0, so an induction argument

for s= 0 is satisﬁed. Assume that Vs≡0, and let us show that Vs+1 ≡0. Substituting Vs≡0

into the right hand side of equation (20), cancels out the continuation term on the right, yielding

Vs+1(x) = ˜

BVs(x)≡max

I∈[0,x]−I+I

x1−Ij

x.(21)

First, consider when x≤1. If the search by the other player Ij< x−x2, it is a dominant strategy

for player ito choose I(x) = x. If, on the other hand, search by the other player Ij> x −x2,

it is a dominant strategy for player ito opt for I(x) = 0. Moreover, if search by the other

player Ij=x−x2, then player iis indiﬀerent among all I∈[0, x]. There is a unique symmetric

equilibrium where both players search I(x) = x−x2and the value each receives is zero; that is,

Vs+1 ≡0. Taking the limit gives V(x) = lims→∞ Vs(x)≡0.

Second, consider when x > 1. Observe that the payoﬀ in (21) is negative unless I(x) = 0.

This concludes the proof.

Proof of Lemma 3

Suppose (a) is not satisﬁed. First, if It= 0 for some t=ˆ

t, the Markovian assumption implies

search remains zero for ∀t > ˆ

t. Hence, there are ﬁnitely many search periods. Assuming that

the game lasts an inﬁnite number of periods, this means that search in some periods will take

arbitrarily small amounts, i.e. for ∀ε0>0∃ε∈(0, ε0) and ∃t∈Nsuch that It=ε≪x.

It also means that there will be an inﬁnite number of such periods. From (7) it follows Ψ(x) =

(1 −x)ε+δΨ(x−J) + o(ε), which can be rewritten as

Ψ(x−J) = Ψ(x)−(1 −x)ε+o(ε)

δ,(22)

27Also see Stokey, Lucas and Prescott (1989).

20

where J= 2ε+o(ε).

Let us prove that in this case Ψ(x) must be arbitrarily close to zero. Assume on the contrary

that Ψ(x) is equal to some positive value that is not inﬁnitely small. By iterating equation (22) a

large number of times it is possible to show that ∃ˆx∈(0, x) such that Ψ(ˆx) can be larger than any

positive value. However, by construction Ψ(ˆx) = ˆxV (ˆx)≤ˆx. We have reached a contradiction,

proving that Ψ(x) must be arbitrarily close to zero.

For values of x < 1, Ψ(x) can not be arbitrary close to zero, because each player will have

incentive to deviate by choosing I=x−ε+O(ε2) and generating Ψ(x) = (1 −x)x+O(ε)>0.

Note that this argument works even when xis very small, because a given x > 0 can not be

inﬁnitely small.

When x= 1 a similar argument applies; in the ﬁrst period a player has an incentive to deviate

by making a positive contribution, for example I= 0.5. In this case, Ψ(1) = δΨ(0.5−ε) + O(ε).

Given that Ψ(x) is not arbitrarily close to zero when x < 1, Ψ(1) is also not arbitrarily close to

zero.

When x > 1 a slightly diﬀerent proof is used. For a given size of island x1>1, each player

makes a positive search I1>0 in the ﬁrst period. Assuming that the game lasts an inﬁnite

number of periods, search has to take arbitrarily small values, i.e. for ∀ε0>0∃ε∈(0, ε0) and

∃t∈Nsuch that It=ε, where x > 1 and t > 1. It also means that there will be an inﬁnite

number of such periods. As shown above, in this case Ψ(x) is arbitrarily small. However, in the

ﬁrst period players make a positive search I1>0, which means they incur an immediate negative

payoﬀ of at least (1 −x1)I1. Subsequently, they make additional searches which add additional

negative payoﬀs. Overall, this argument implies that Ψ(x1) is negative, which is a contradiction.

Thus, Itcannot be arbitrarily close to zero, which means there exists ζ > 0 such that

It> ζ ∀t. Consequently in any non-trivial SMPE the project has to be ﬁnished in a ﬁnite

number of periods. This establishes part (a) of the Lemma.

Given that the project has to be ﬁnished in a ﬁnite number of periods, backward induction

can be applied. Namely, we assume that Ψ0≡0 and derive Ψ1≡BΨ0. This allows us to ﬁnd

all potential SMPE of the game if players could search for at most one period. Then we derive

Ψ2≡BΨ1, which allows us to ﬁnd all potential SMPE of the game if players could search for at

most two periods. We continue this process until Tis reached and ﬁnd all potential SMPE of the

game if players could search for at most Tperiods. This establishes part (b) of the Lemma and

concludes the proof.

Proof of Proposition 2

As proved in Lemma 3, the ﬁnite application of operator Brepresents the algorithm for solving

the Bellman equation. Let us start from the end period of search process. What will be the value

if players could only search for at most one period? Equation (7) transforms into

Ψ1(x) = max

I∈[0,x]{J−Ij−xI}.(23)

When xis small the beneﬁt of search is higher than the cost. In the unique SMPE both players

cumulatively search the whole island; that is, J=I+Ij−(1−α)IIj/x =x. Consequently, provided

x≤√α, each player searches

I=x

1 + √α.(24)

21

Substituting (24) into (4) and (23) results in

Ψ1(x) = xV1(x) = (√α−x)x

1 + √α.

Now we analyze the situation when players plan to search for at most two periods. Equation

(7) in this case transforms into

Ψ2(x) = max

I∈[0,x]{J−Ij−xI +δΨ1(x−J)}.(25)

The necessary condition for Ito be optimal in the interior of [0, x], provided that search lasts for

two periods, is

∂J

∂I 1−δ√α−2(x−J)

1 + √α−x= 0.(26)

Note that ∂J

∂I >0 for J≤x. This means when xis small the left hand side is positive and the

optimal solution is not interior. Consequently, in a two-period search path players search the

whole island in one period. To ﬁnd the critical value χ1for which searching the whole island in

one period stops being a dominant strategy, we use the condition that the corner solution x=J

coincides with the internal solution (26). From this we derive a unique solution

χ1=√α1−δ√α

1 + √α.

This concludes the proof.

Proof of Proposition 3

Construction of Ψ1and V1

Let us start from the end of the search process. What will be the value if players could only

search for at most one period? Equation (7) transforms into

Ψ1(x) = max

I∈[0,x]{min{x, I +Ij} − Ij−xI }.(27)

It is evident that in the unique non-trivial SMPE the equilibrium Ican be described in the

following way28 :

I=x/2,if x≤1.

If x≤1, then in the SMPE players search the whole island, 2I=x. Consequently, the

solution of (27) is

Ψ1(x) = P1(x),if x≤u1= 1; (28)

where u1= 1 is the largest positive root of polynomial P1(x) = x(1 −x)/2. For future reference

note that

P1(x) = a1

2(1 −x)2+b1

2(1 −x) + c1

2,(29)

28Note that if x= 1, then any I∈[0, x/2] is optimal. We assume that players choose I=x/2 in this

case.

22

where

a1=−1, b1= 1, c1= 0.(30)

Deﬁne the value for each player (if the players can search the island for at most kperiods) as

Vk(x)≡Ψk(x)/x, for any x≥0. From the above deﬁnition, it follows that

V1(x) = (1 −x)/2,if x≤u1= 1.

Construction of Ψ2and V2

What will be the value if players can search the island for at most two periods? In general there

could be three possibilities, depending on the island size. The ﬁrst possibility is that the players

search the whole island in just one period. Intuitively this happens for small values of xbecause

it is too costly to wait for another period when the island is very small. The second possibility

is that the players search the island for at most two periods. This happens for middle values of

x. Finally, players can ﬁnd search to be too costly. When the initial island is large (costs are

very high), there is no non-trivial SMPE (mixed, asymmetric or trivial equilibria are the only

possibilities).

We have already considered the ﬁrst possibility in the previous section. Now we analyze the

situation when players plan to search for at most two periods. The ﬁrst step is to construct Ψ2(x).

Equation (7) in this case transforms into

Ψ2(x) = max

I∈[0,x]{min{x, I +Ij} − Ij−xI +δΨ1(max{0, x −I−Ij})}.(31)

The necessary condition for Ito be optimal in the interior of [0, x], provided that search lasts for

two periods, is

(1 −x)−δΨ′

1(x−2I) = 0.(32)

In order to continue the search for the second period, the remaining island size has to satisfy

0≤x−2I≤u1.(33)

The suﬃcient condition for Ito be optimal in the interior of [0, x] is satisﬁed because

Ψ′′

1(x−2I) = a1<0.

The way to proceed is to construct the equilibrium with the help of condition (32), and then show

that the derived equilibrium satisﬁes condition (33).

From expressions (32) and (28), it follows that

(1 −x)−δ1−2(x−2I)

2= 0.

Consequently,

I=2(δ−1)x+ 2 −δ

4δ.(34)

Substituting (34) into equation (31), we obtain a spline of degree two on the interval [0, u2]:

Ψ2(x) = Ψ1(x),if 0 ≤x≤χ1,

P2(x),if χ1< x ≤u2;(35)

23

where u2>0 is the largest positive root of polynomial

P2(x) = a2

2(1 −x)2+b2

2(1 −x) + c2

2,(36)

with

a2=−1, b2=1

2,and c2=δ

4.(37)

In order to ﬁnd u2, we need to solve the quadratic equation P2(u2) = 0. It is easy to check

that

u2= 1 + √4δ+ 1 −1

4.(38)

The point x=χ1is the ﬁrst knot of the spline. When x=χ1players are indiﬀerent between

searching the island for two periods or for just one period:

Ψ1(χ1) = Ψ2(χ1).(39)

From (29) and (36), we derive a unique solution

χ1= 1 −δ

2.(40)

Note that condition (33) is satisﬁed for any x∈[χ1, u2]. Therefore, if the players can search

the island for at most two periods, the SSMPE is a spline of degree one on the interval [0, u2]:

I(x) = x

2,if x≤χ1,

2(δ−1)x+2−δ

4δ,if χ1< x ≤u2;

and the value function is

V2(x) = V1(x),if x≤χ1,

P2(x)/x, if χ1< x ≤u2.

We can describe the construction of Ψkand Vknow.

Construction of Ψkand Vk

What will be the value if players can search the whole island for at most k≥3 periods? In general

there could be k+ 1 possibilities, depending on the island size x1. The players can plan to search

the island for at most 1,2, ..., k periods in a non-trivial SMPE. It is also possible that there is no

non-trivial SMPE.

Let us construct Ψk(x). Equation (7) in this case transforms into

Ψk(x) = max

I∈[0,x]{min{x, I +Ij} − Ij−xI +δΨk−1(max{0, x −I−Ij})}.(41)

The necessary condition for Ito be the optimal value in the interior of [0, x], provided that search

lasts for kperiods, is

(1 −x) = δΨ′

k−1(x−2I).(42)

24

In order to continue search for the next period, the new value of xhas to satisfy

χk−2≤x−2I≤uk−1,(43)

where χ0= 0. The suﬃcient condition for Ito be the optimal value in the interior of [0, x] is

satisﬁed if

Ψ′′

k−1(x−2I)<0.(44)

We will use condition (42) to ﬁnd I, and then show that it satisﬁes conditions (43) and (44).

Note that if function Ψk−1(x) in (41) is a quadratic polynomial, then Ψk(x) = BΨk−1(x) has to

be a quadratic polynomial as well. Since P1(x) and P2(x) are quadratic polynomials by (29) and

(36), any Pk(x) can be represented in the following form:

Pk(x) = ak

2(1 −x)2+bk

2(1 −x) + ck

2, k ≥1.(45)

From condition (42) and expression (45), it follows that

I=−(1 −x)(1 + δak−1)

2δak−1−bk−1

4ak−1

.(46)

Deﬁne the largest root of polynomial Pk(x) as uk, and that value of xsuch that players switch

from planning to search the area for k−1 periods to planning to search the area for kperiods as

knot χk−1:

Ψk−1(χk−1) = Ψk(χk−1).(47)

For the moment, let us assume that equation (47) has a unique solution. The uniqueness of the

solution will be proved below. Substituting (46) into equation (41), we obtain a spline of degree

two on the interval [0, uk]:

Ψk(x) = Ψk−1(x),if 0 ≤x≤χk−1,

Pk(x),if χk−1< x ≤uk;(48)

where Pk(x) is deﬁned in (45). Therefore, if players plan to search the island for at most kperiods,

then I(x) is a spline of degree one on the interval [0, uk]:

I(x) =

x

2,if x≤χ1,

−(1−x)(1+δa1)

2δa1−b1

4a1,if χ1< x ≤χ2,

.

.

.

−(1−x)(1+δak−1)

2δak−1−bk−1

4ak−1,if χk−1< x ≤uk;

(49)

and the value function is

Vk(x) = Vk−1(x),if 0 ≤x≤χk−1,

Pk(x)/x, if χk−1< x ≤uk.(50)

Let us now ﬁnd ak,bk, and ckfor any k≥2. Using (41), (45) and (46), we get the following

system of diﬀerence equations:

ak=−1, bk=−bk−1

2ak−1

, ck=δ ck−1−b2

k−1

4ak−1!, k ≥2,(51)

25

where

a1=−1, b1= 1, c1= 0.(52)

Derivation of ak,bkand ck

Let us now solve the system of diﬀerence equations (51). It is straightforward to derive

ak=−1 and bk=1

2k−1.(53)

Note that ak6= 0 and the system (51) is well deﬁned.

The expression for ckin (51) can be simpliﬁed to

ck=δ(ck−1+ 1/4k−1).(54)

Introduce a new variable ek=ck4k. Equation (54) transforms to

ek= 4δ(ek−1+ 1),

where e1= 0. The solution to this linear diﬀerence equation is ek=4δ−(4δ)k

1−4δ. Substitute ck=

ek/4kto derive

ck=(4δ)k−4δ

(4δ−1)4k.(55)

Derivation of χk

To ﬁnd χk, one needs to solve the quadratic equation Pk(χk) = Pk+1(χk), namely

ak(1 −χk)2+bk(1 −χk) + ck=ak+1(1 −χk)2+bk+1(1 −χk) + ck+1, k ≥1.

From equation (53), ak=ak+1 =−1; consequently,

χk= 1 + ck+1 −ck

bk+1 −bk

.

Substitute bkand ckfrom equations (53) and (55) to derive the following indiﬀerence points

χk= 1 −3δ+ (4δ)k(δ−1)

2k(4δ−1) .(56)

Note that the solution to the quadratic equation Pk(χk) = Pk+1(χk) is always unique.

Derivation of uk

To ﬁnd uk, one needs to solve the quadratic equation Pk(uk) = 0, namely

ak(1 −uk)2+bk(1 −uk) + ck= 0, k ≥1.

Substituting ak=−1 from equation (53) and solving the above quadratic equation gives

uk= 1 + qb2

k+ 4ck−bk

2.(57)

26

Note that with the help of (53) and (55), one can simplify:

b2

k+ 4ck=(4δ)k−1

4k−1(4δ−1).(58)

Substitute equation (58) into equation (57) to get

uk= 1 + q1−(4δ)k

1−4δ−1

2k.(59)

Note that for the ease of exposition we present χd=udin the statement of the Proposition.

Derivation of d

One way to ﬁnd the maximum number of search periods is to write the condition that the largest

positive root of the quadratic polynomial Pk(x) coincides with the largest positive root of the

quadratic polynomial Pk+1(x). This condition gives a critical value of δ; for slightly larger values

of δ, there is an additional search period.

This means that kis the smallest integer such that Ψk(x)≡Ψk+1(x), or in other words

V(x)≡Vk(x). Speciﬁcally, when δis at a critical value, khas to be an integer and is the solution

to the following equation:

uk(δ, n) = tk(δ, n).(60)

Substitute tdand udfrom equations (56) and (59) into equation (60) to get

1 + q1−(4δ)d

1−4δ−1

2d= 1 −3δ+ (4δ)d(δ−1)

2d(4δ−1) .

Simplify the above expression to

s1−(4δ)d

1−4δ−1 = −3δ+ (4δ)d(δ−1)

(4δ−1) .

Further simpliﬁcations give

s(4δ)d−1

4δ−1=((4δ)d−1)(1 −δ)

(4δ−1) ,

and s(4δ)d−1

4δ−1(1 −δ) = 1.(61)

Square both sides of equation (61) to derive the maximum number of search periods:29

d=&ln ((δ2+ 2δ)/(1 −δ)2)

ln (4δ)'.(62)

29Note that when δ= 1/4 both numerator and denominator in dcontain ln (4δ) term, which means dis

not well deﬁned. The interpretation of this expression for dwhen δ= 1/4 is its limit as δ→1/4.

27

Conditions (43) and (44) are satisﬁed

Deﬁne that part of xwhich player idoes not search in the current period by y=x−Iand the

part of xno player searches in the current period by z=x−(I+Ij) = y−Ij.

Next, show that condition (43) is satisﬁed; that is, tk−2≤z≤uk−1. Let us prove the ﬁrst

part z≤uk−1by contradiction, assuming that z > uk−1. Refer to equation (41), which is written

below:

Ψk(x) = max

I∈[0,x]{min{x, I +Ij} − Ij−xI +δΨk−1(max{0, x −I−Ij})}.

Given x≥z > uk−1≥ ··· ≥ u2≥u1= 1 and I+Ij≤x, it follows that the ﬁrst term

min{x, I +Ij} − Ij−xI ≤(1 −x)Ihas to be non-positive. If z > uk−1, then the second term

δΨk−1(z) is negative. That means the whole expression on the right of equation (41) has to be

negative. Obviously that could not be an optimal choice for a player because by choosing y=x,

that is, by not searching, a player can get the value of zero. Consequently, there is a contradiction,

and condition z≤uk−1is proved.

Now let us show that χk−2≤z. From equations (46), (53) and (55), it follows that z(x) =

1−21−k−1−x

δ. It is easy to see that z(x) is a monotonically increasing function in x. Consequently,

it is suﬃcient to prove the above condition for x=χk−1. Substituting values from (56) and

simplifying gives

3δ+ (4δ)k−1(δ−1)

δ(4δ−1) ≤1 + 2δ+ 2(4δ)k−2(δ−1)

4δ−1.

Further simpliﬁcations result in (4δ)k−1≥4δ, which is satisﬁed for any k > 2.

Finally, let us show that condition (44) is satisﬁed; that is, Ψ′′

k−1(z)<0. From equations (45),

(48) and (53), it follows that Ψ′′

k−1(z) = ak−1=−1. This concludes the proof.

Proof of Corollary 1

First, I(x) presented in Proposition 3 is a piece-wise linear function. Second, let us prove that

I(x) is not a continuous function. Given that I(x) is piece-wise linear, discontinuities are only

possible in the knots of the spline. Substitute x=χsinto (49), when the project is planned to

be ﬁnished in s≥2 periods, to derive

I(χs) = 2−2δ+δ/2s−2−2(1 −δ)χs

4δ.(63)

Next, substitute x=χsinto (49), when the project is planned to be ﬁnished in s+ 1 periods, to

derive

I(χs) = 2−2δ+δ/2s−1−2(1 −δ)χs

4δ.(64)

The diﬀerence between the expressions in (63) and (64) is ∆I(χs) = 1/2s+1 >0.

Third, let us ﬁnd the slope of I(x). If the project is ﬁnished in one period then the slope

is positive, I′(x) = 1

2, otherwise it is negative, I′(x) = δ−1

2δ. This proves quasiconcavity and

concludes the proof.

28

Proof of Corollary 2

First, V(x) presented in Proposition 3 has to be continuous. Remember that knot χsis deﬁned

as an area size for which the duopolists are indiﬀerent between planning to search the area for s

periods or for s+ 1 periods.

Second, to show that the value function is monotonically decreasing we need to show that

∀s≥2 the following inequality holds:

V′

s(x) = Ps(x)

x′=as−as+bs+cs

x2<0.(65)

Substitution of as=−1 transforms (65) to

bs+cs+x2>1.

For a given s, the left hand side is increasing in x, which means it is suﬃcient to prove the

inequality for x=χs−1. Substitute bs, csand χs−1from (53),(54) and (56) to derive

1

2s−1+(4δ)s−1−1

4s−1(4δ−1)δ+1−3δ+ (4δ)s−1(δ−1)

2s−1(4δ−1) 2

>1.(66)

Let us prove inequality (66) separately for s∈[2,3] and for s∈[4, d]. When s= 2, the

inequality transforms to 2 −3δ+δ2>0, which is always satisﬁed for δ < 1. Similarly when s= 3,

the inequality transforms to 4 + 25δ−28δ2+ (3δ−4δ2)2>0, which is also always satisﬁed for

δ < 1.

From (61) when d≥4, it is the case that δ > 1/2. Simplify inequality (66) to

(2s−1−1) 2(4δ)s−1−1

(4δ−1) −1(1−δ)+