Competitive prizes: when less scrutiny induces more effort

Journal of Mathematical Economics 36 (2001) 311–336

Competitive prizes: when less scrutinyinduces more effort

Pradeep Dubeya,∗, Chien-wei Wub,1a Center for Game Theory, Department of Economics, State University of New York at Stony Brook,

Stony Brook, NY 11794-4384, USAb Department of Economics, National Chi Nan University, Puli, Nanton, Taiwan

Received 5 June 2000; received in revised form 21 May 2001; accepted 17 September 2001

Abstract

We consider a principal who is keen to induce his agents to work at their maximal effort levels.To this end, he samplesn days at random out of theT days on which they work, and awards a prizeof B dollars to the most productive agent. The principal’s policy (B, n) induces a strategic gameΓ (B, n) between the agents. We show that to implement maximal effort levels weakly (or, strongly)as a strategic equilibrium (or, as dominant strategies) inΓ (B, n), at the least costB to himself, theprincipal must choose asmallsample sizen. Thus less scrutiny by the principal induces more effortfrom the agents.

The need for reduced scrutiny becomes more pronounced when agents have information of thehistory of past plays in the game. There is an inverse relation between information and optimalsample size. As agents acquire more information (about each other), the principal, so to speak,must “undo” this by reducing his information (about them) and choosing the sample sizen evensmaller. © 2001 Elsevier Science B.V. All rights reserved.

Keywords:Competitive prizes; Principal’s policy; Strategic game

1. Introduction

The intuition behind this paper is simple. Suppose a principal writes a competitive contractbetween several agents, basing his reward on their relative performance. In many instances,he would do well to ensure that theleast skilledagent has a sufficiently good chance ofobtaining the reward, so as to be induced to work at full capacity. For then this agent generatesthe competition which spurs the next-best agent to also work at full capacity, who in turn

∗ Corresponding author. Tel.:+1-631-6327549; fax:+1-631-6327227.E-mail addresses:[email protected] (P. Dubey), [email protected] (C.-w. Wu).

1 Tel.: +886-49-2910960x4919; fax:+886-49-2914435.

0304-4068/01/$ – see front matter © 2001 Elsevier Science B.V. All rights reserved.PII: S0304-4068(01)00079-9

312 P. Dubey, C.-w. Wu / Journal of Mathematical Economics 36 (2001) 311–336

spurs the agent just above him and so on. By designing a contract which is favorable to theweakest agent, the principal can trigger competition which induces all agents to work. Webelieve that this basic principle is of quite wide application, even though for concretenesswe build here a stylized model based on the principal-agent literature. In our model, it turnsout that to favor the weak is tantamount to observing agents’ outputs on asmall sample,since all large samples would almost surely distinguish the weak from the strong and destroyincentives for the weak. In short, less scrutiny by the principal induces more effort from theagents.

The principal-agent literature is prolific and we shall make no attempt to summarizeit here (see, for example, Green and Stokey, 1983; Grossman and Hart, 1983; Lazear andRosen, 1981; Mookherjee, 1984; Nalebuff and Stiglitz, 1983; Rosen, 1986 and the referencestherein). We focus on a special scenario. There is a risk-neutral principal who values hisagents’ outputs so highly (e.g. because they fetch a very remunerative market price) thateven after compensating them for any additional effort, he makes a profit on the margin. Heis therefore keen to write a contract which will elicitmaximal effortfrom them.

Agents, on the other hand, have to be induced to work by offers of suitably largeperformance-related rewards of money, since they have a natural disutility for work. Theiroutputs are random but positively correlated with their effort levels. If an agent producessmall output, it could be because he did not work much or because he had bad luck despitehard work. Of course, the probability of small output is reduced if his effort level goes up.

We suppose that the principal is constrained to writenon-discriminatorycontracts whichare based onoutput alone. This could be because he cannot observe the inputs of ef-fort made by the agents, nor can he tell them apart in terms of their private character-istics, such as their skill (probability of being productive) or their utility for money andleisure. Alternatively, even if the principal were fully cognizant of these, it could be thatthe law requires rewards to be commensurate with performance, and to be based solelyon the outputs agents produce, with no other form of discrimination permitted. In anycase, we confine attention to contracts that are defined anonymously on the outputs of theagents.

Contracts may beindependentin that the reward to any agent depends just on his ownoutput, i.e. it is invariant of others’ outputs; or they may becompetitiverewarding agentsaccording to the rank-order of their outputs.

The focus of this paper is on non-discriminatory competitive contracts, which take theform of a prize to the most productive agent. Such contracts can often be more favorable to theprincipal than independent contracts, e.g. if agents’ productivities are sufficiently correlated(see Green and Stokey, 1983), or if they feel sufficient envy (see Dubey and Haimanko,2001). But, apart from their theoretical raison d’être, competitive prizes are simply a fact ofeveryday life. Think, for example, of a bonus to the best salesman of the year; or a promotionawarded to the branch manager who produced the highest profits, etc. The question arises:how is best performance to be measured? We point out that, even if the entire stream ofagents’ outputs is susceptible of costless observation, there are many situations in whichthe principal would be wise to deliberately create uncertainty by observing outputs only onrandom samples of small size.

Let us consider a concrete instance of our model. Suppose a government (the principal)has commissioned two manufacturers (the agents) to produceT units each of some vital

P. Dubey, C.-w. Wu / Journal of Mathematical Economics 36 (2001) 311–336 313

defence equipment,2 e.g. timed fuses for bombs. Each manufacturerα ∈ 1,2must choosean effort leveleα fromEα ≡ 1, . . . , e∗α which in turn determines the probabilitypeαα thatany one of his fuses will be of high quality. The disutility incurred byα on account of efforteα isdα(eα). We may think of the effort as being used to upgrade the production technology.It is then natural to assumepeαα > p

eαα anddα(eα) > dα(eα) whenevereα > eα.

The government buys theT fuses from 1, 2 for some previously contracted amounts.But it is evidently crucial to the government that both manufacturers commit themselvesto maximal effort. To motivate them, it announces an award to be bestowed as follows. Itwill samplen fuses at random at both production sites, and give a bonus ofB dollars to themanufacturer with more high quality pieces in the sample. In the event of a tie, it will awardB to each with probability 1/2.

A policy (B, n) induces a strategic gameΓ (B, n) between the two manufacturers. Theproblem facing the government is how to choose (B, n) so as to minimizeB while, at thesame time, ensuring that (e∗1, e

∗2) is a strategic equilibrium (SE) ofΓ (B, n). For any fixed

sample sizen it is clear that, ifB is large enough, then (e∗1, e∗2) is a strategic equilibrium

of Γ (B, n). We prove somewhat more: there exists a threshold bonusB(n) such that, ifB > B(n), then (e∗1, e

∗2) is theuniqueSE ofΓ (B, n); and ifB < B(n), then (e∗1, e

∗2) is not

an SE ofΓ (B, n).Thus the competitive contract (B, n) implements (e∗1, e

∗2) in a strong sense: not merely

as an SE ofΓ (B, n) as in much of the literature (see, for example, Mookherjee (1984) fora synopsis), but as its unique SE. Uniqueness overcomes the vexing uncertainty of howagents will coordinate upon the “chosen” candidate SE when there are many of them. Theonly uncertainty that remains is whether the agents will somehow settle down to an SE ofthe game at all. There is, in this context, the much stronger notion of implementation of(e∗1, e

∗2) as a dominant strategy equilibrium (DSE), whereby this uncertainty is considerably

reduced. But the reduction comes with a price. The thresholdB(n), needed to ensure that(e∗1, e

∗2) is a DSE ofΓ (B, n), is generally much higher thanB(n).

An optimal competitive prizeis obtained by choosing a sample size for which the thresholdbonus is minimized, i.e. (B(k), k) is optimal ifB(k) = minB(n) : 1 ≤ n ≤ T . It is easy to

see thatkmust be small. Let manufacturer 1 be more skilled than 2, i.e.pe∗11 > p

e∗22 . If kwere

big then, by the law of large numbers, 1 would almost surely producepe∗11 k high quality fuses

in any sample at (e∗1, e∗2), while his rival would almost surely produce onlyp

e∗22 k < p

e∗11 k

such fuses. To induce 2 to choosee∗2, B must be huge, compensating 2 for his extremely lowprobability of winningB. It follows that an optimal sample sizek must be small, justifyingthe title of our paper (this argument incidentally also shows that an optimal sample sizek

for implementing (e∗1, e∗2) as a DSE, i.e.B(k) = minB(n) : 1 ≤ n ≤ T is also small).

The need for reduced scrutiny comes even more to the fore when agents have informationof the history of past plays in the game. It is useful here to distinguish two scenarios. Inthe first, agents are ignorant of their rivals’ outputs,3 but may have memory of their own

2 More generally, the government could commissionT1, T2 units and samplen1, n2 units and award the bonus tothe manufacturer with the higher percentage of high quality pieces. Our main point remains intact. At least one ofn1, n2 must be small to avoid the need for huge bonuses. We takeT1 = T2 = T andn1 = n2 = n for simplicity.

3 We assume throughout that no agent can observe the inputs of effort made by his rival.


outputs. In the second, they can observe rivals’ outputs as well. Letσ ∗ denote the strategieswherein all agents exert maximal effort in all circumstances of the game. We show thatwhile all sample sizes are feasible in the first scenario (in that they implementσ ∗ for suitablylarge bonuses),optimalsample sizes are small and tend to fall as memory is refined. In thesecond scenario, evenfeasiblesample sizes must be small. More precisely, there is an upperbound on feasible sample sizes, which depends on the information pattern in the game. Asinformation is refined, this bound falls sharply. (We note, in passing, that the automaticprivilege enjoyed byσ ∗ of being the unique SE is lost in the second scenario. The bonusneeds to be raised further to bring back uniqueness.)

We know of two authors (Cowen and Glazer, 1996) who have commented on a similarphenomenon. Their analysis has been in the context of a contract between a principal anda singleagent. But to the best of our knowledge no one has considered the impact of theprincipal’s observation on agents’ behavior in acompetitiveframework.

The scope of our analysis has a natural limitation. Wefix the behaviorof agents atσ ∗(i.e. maximal effort) and examinevariable prizeswhich implementσ ∗. This is consistentwith profit-maximization for the principal if he values agents, outputs sufficientlyand ifagents’ information in the game is not too fine. Our analysis reveals that, with very fineinformation, it is not possible to implementσ ∗ without handing out huge bonuses. Then itbecomes more natural to take a complementary viewpoint: tofix the prizeand examine thevariable behaviorof agents that is induced by the prize. This is done, for the case of perfectinformation, in a companion paper (Dubey and Haimanko, 2000). An optimal policy nolonger induces maximal effort in all circumstances of the game, but it still requires samplesize to be small, and our main point remains intact.

For ease of exposition, we carry out our analysis for the case of two agents. But all ourresults, except for Theorem 3 (on uniqueness of SE) hold for any number of agents withobvious modifications in the definitions and the proofs.

2. The extensive form game (Ψ, B, n)

We present an extensive form game between the agents which involvesT time periods. Tokeep notation simple, we assume that all the data of the game, except for agents’ information,is stationary.

Each agentα ∈ 1,2 has a finite setEα of effort levels available every period, with adistinguished elemente∗α which represents his maximum effort level. Anye ∈ Eα inducesa probability distributionpeα with full support on a finite set of non-negative outputsQα ≡(q1α, . . . , q

m(α)α ) producible byα, where we have arrangedq1

α < · · · < qm(α)α . Denote the

rival of α by β, and denoteqα ≡ qm(α)α , q

α≡ q1

α. We assume, for any agentα, that:qα > q

β.

The game is played iteratively as follows. In each period both agents make a simultaneouschoice of effort levels which leads to random outputs, in accordance with the specifiedprobabilities, and brings the game into periodt + 1. Agents then choose effort levels again.

For the formal description, letΩ(t) denote the set of agents’ nodes in periodt ∈ T ≡1, . . . , T . The setΩ(1) ≡ ω∗ is a singleton, whereω∗ signifies the start of the game.


Fig. 1. The extensive game.

At anyω ∈ Ω(t), both agents 1, 2 simultaneously choose effort levels fromE1, E2. After(e1, e2) ∈ E1 × E2, there is a move of chance. Chance selects (q1, q2) ∈ Q1 ×Q2 withpositive probability4 pe11 (q1)p

e21 (q2), leading to an agents’ nodeω ≡ (ω, e1, e2, q1, q2) in

Ω(t + 1) (see Fig. 1).Thus, fort > 1,Ω(t) is isomorphic to(E1 × E2 ×Q1 ×Q2)

t−1; and anyω ∈ Ω(t) isspecified by its history(e1(τ ), e2(τ ), q1(τ ), q2(τ ))

t−1τ=1.

LetΩ ≡ t∈T Ω(t) be the set of all agents’ nodes in the game tree (where denotesdisjoint union). Each agentα has aninformation partition Iα of Ω which reflects what heknows of his own and his rival’s past history in the game. We assume throughout thatthatno agent can observe his rival’s inputs of effort. His information of the rival pertains onlyto theoutputsproduced. Thus, for any pair of nodesω, ω in Ω(t) ⊂ Ω and any agentα:

ω ≡ (eα(τ ), eβ(τ ), qα(τ ), qβ(τ ))

t−1τ=1

ω ≡ eα(τ ), eβ(τ ), qα(τ ), qβ(τ ))t−1τ=1

i.e. ω andω have the same history except perhaps for the effort ofβ

⇒ ω andω are in the same information set ofIα (1)

There is a positive correlation between effort and productivity. Our key assumption onpeαα states that when an agent exerts maximal effort, his outputs go up, in the sense of

first-order stochastic dominance:For anyα ∈ 1,2 ande ∈ Eα\e∗α∑q≥x

pe∗αα (q) >

∑q≥x

peα(q), for all x ∈ Qα\qα (2)

This completes the description of the extensive form of the game, but it still remains tospecify agents’ payoffs at each terminal nodeω ∈ Ω(T + 1).

Each agentα has a continuous, strictly monotonic utility for money given byuα : R+ →R, whereuα(B) ≡ the utility to α of receivingB dollars by way of reward at the end oftheT days. We need to assume that agents value money sufficiently. Precisely (and, for thesake of simplicity, stating it in a stronger form than necessary) we postulate:

uα(B) → ∞ asB → ∞ (3)

4 That is, agents’ probabilities are independent (for correlated probabilities, see Section 10).


The disutility for effort is given by a functiondα : (Eα)T → R+, for α ∈ 1,2. Weassume that maximal effort incurs the most disutility, i.e.

dα(e∗α, . . . , e∗α) > dα(e) (4)

for all e ∈ (Eα)T \(e∗α, . . . , e∗α).Rewards at the terminal nodes are determined by theprincipal’s policy (B, n) which

operates as follows:n days are chosen at random5 and the bonus ofB dollars is awardedto the agent with the higher total output across the days in the sample; in case of a tie, it isgiven to each agent with probability half.

For 1≤ n ≤ T , letCn ≡ I ⊂ T : |I| = n andp(n) ≡ 1/C(T , n), whereC(T , n) =T !/(T − n)!n!. Considerω ≡ (e1(τ ), e2(τ ), q1(τ ), q2(τ ))

Tτ=1 ∈ Ω(T + 1). Givenω, the

total produced byα onI ∈ Cn, isZα(ω, I) ≡ ∑τ∈Iqα(τ ). Recall thatβ denotes the rival

of α ∈ 1,2, and define6

uα(ω, I) ≡

uα(B) if Zα(ω, I) > Zβ(ω, I)

uα(0) = 0 if Zα(ω, I) < Zβ(ω, I)

12uα(B) if Zα(ω, I) = Zβ(ω, I)

Then, abusing the notationuα yet again, the expected utility from money atω isuα(ω) ≡p(n)

∑I∈Cnuα(ω, I). On the other hand, the disutility of effort atω is dα(ω) ≡

dα((eα(τ ))Tτ=1).

We assume separability, and write the payoff to agentα atω as

πα(ω) ≡ uα(ω)− dα(ω)

This completes the description of the extensive form game between the agents. Since wehold uα, dα fixed throughout, we will denote the game (Ψ , B, n) whereΨ is the extensiveform of the game, i.e. the game tree with the payoffs missing at the terminal nodes.

Remark 1. The example of “fuses”, given in Section 1, can be seen to be a special caseof our model by settingI1 = I2 = Ω (there is no loss of generality in ignoring the fixedcontracted payment to the manufacturers, since affine transformations of their utilities leavesonly the bonus term).

WhenI1 = I2 = Ω, we say that there iszero-informationin the game. In this case,our theorems hold without the full-support assumption:peα(q) > 0 for all q ∈ Qα and alle ∈ Eα.

Remark 2. WhenI1 = I2 = Ω(t) : 1 ≤ t ≤ T , we say that there islow-informationin the game. Agents now know only the time period they are in. Low-information may beinterpreted in terms of alocations model(see Dubey and Wu, 2000, for details).

5 We will assume that thesameset ofn days is sampled for each agent. But, given our stationary set-up, not muchchanges if these sets are different for the two agents.

6 W.l.o.g. (using affine transformations for each agent’s utility), we takeuα(0) = 0, to avoid writing(1/2)[uα(0)+uα(B)] in place of (1/2)uα(B).


3. The normal form gameΓ (Ψ, B, n)

We now define the game in normal (or strategic) form that arises from the extensive formgame.

A strategyσα of agentα specifies an effort levelσα(ω) in Eα for eachω ∈ Ω, and ismeasurable with respect toα’s information partitionIα.

We will define an equivalence relation on the set of all strategies of agentα, so that anytwo strategies which have the same “reduced form” get identified. To this end, we firstdefine the set of irrelevant nodes inΩ(t) for strategyσα, inductively ont. At t = 1, the setof irrelevant nodes is the empty set. Letω = (ω, e1, e2, q1, q2) be a node inΩ(t + 1) thatfollows fromω in#(t). Then defineω to be irrelevant forσα if eitherω ∈ Ω(t) is irrelevantfor σα; or if σα(ω) = eα. Denote the set of all irrelevant nodes ofσα byΩσα ⊂ Ω. We saythatσα ≈ σα if Ωσα = Ωσα andσα(ω) = σα(ω) for all ω ∈ Ω\Ωσα . It is clear that (≈) isan equivalence relation.

The set of all (reduced as above) strategies of agentα is denotedΣα. They suffice for theanalysis of the game. To see this, consider any pair of unreduced strategiesσ ≡ (σ1, σ2).Induces a probability distribution on the set of terminal nodes in the obvious manner. Indeed,given ω ≡ (ω, e1, e2, q1, q2), define the probabilitypσ (ω, ω) of reachingω from ω underσ to bepe11 (q1)p

e22 (q2) if σα(ω) = eα for bothα = 1 and 2, and to be 0 otherwise. Then the

probability7 pσ (ω) of reaching the nodeω = (e1(τ ), e2(τ ), q1(τ ), q2(τ ))tτ=1 is pσ (ω) =∏t

t=1pσ (ωt , ωt+1), whereωt+1 ≡ (ωt , e1(t), e2(t), q1(t), q2(t)) for t = 1, . . . , t (we de-

finepσ (ω∗) = 1, whereω∗ is the start of the game, i.e. whereΩ(1) = ω∗). The payoffto agentα from σ is then

∑ω∈Ω(T+1)p

σ (ω)πα(ω) ≡ ∏α(σ ).

If (σ 1,σ 2) and (σ1, σ2), are two equivalent pairs (i.e.σα ≈ σα for α ∈ 1,2), it is easyto check that they induce the same probability distribution onΩ(T + 1). Thus equivalentpairs give rise to the same expected payoffs and the normal form game, with strategy setsΣ1 andΣ2, is well-defined. We denote it byΓ (Ψ , B, n).

Let σα ∈ Σα andσβ ∈ Σβ . Thenσα is called abest replyto σβ if∏α(σα, σβ) ≥ ∏

α(σα, σβ), for all σα ∈ ΣαThe pair (σ 1, σ 2) is called a strategic equilibrium (SE) ofΓ (Ψ , B, n) if σ 1 is a best replyto σ 2, andσ 2 is a best reply toσ 1. If, moreover, both strategies are unique best replies thenthe SE is calledstrict.

Finally, we defineσα ∈ Σα to be adominant(strictly dominant) strategyof α in Γ (Ψ ,B, n) if σα is a best reply (unique best reply) toeveryσβ ∈ Σβ . If σ 1 andσ 2 are bothdominant strategies inΓ (Ψ , B, n), we say that (σ 1, σ 2) is a DSE ofΓ (Ψ , B, n).

4. Thresholds for the bonus

The strategy (modulo the equivalence relation) whereinα puts in maximal effortis denotedσ ∗α , i.e.σ ∗

α (ω) = e∗α for all ω ∈ Ω. We denote(σ ∗1 , σ

∗2 ) ≡ σ ∗.

7 It is noticed thatp(σ1,σ2)(ω) is in fact defined for arbitrary mapsσα : Ω → Eα , not necessarily just strategiesσα ∈ Σα .


First let us fix the extensive formΨ and the sample sizen, and ask how large the bonusneeds to be to implementσ ∗ either as an SE or as a DSE.

Define

B(Ψ, n) ≡ inf B : σ ∗ is an SE ofΓ (Ψ,B, n),B(Ψ, n) ≡ inf B : σ ∗ is a DSE ofΓ (Ψ,B, n)

The infimum (inf) on an empty domain is taken to be infinity. It is easy to see that if thedomain is non-empty, then the infimum is in fact achieved.

The following result is straightforward.

Theorem 1. Assume(1)–(4):

1. B > B(Ψ, n) ⇔ (σ ∗1 , σ

∗2 ) isa strict SE ofΓ (Ψ,B, n)

2. B > B(Ψ, n) ⇔ σ ∗α isa strictly dominant strategy ofΓ (Ψ,B, n) forα ∈ 1,2

This theorem justifies the use of the word “threshold” forB(Ψ , n) andB(Ψ, n).

5. Feasible and optimal sample sizes

Since the principal is interested in implementingσ ∗ with the smallest possible bonus, itis useful to make some definitions.

A sample size is called (i)SE-feasiblefor Ψ if B(Ψ, n) < ∞; (ii) DSE-feasiblefor Ψ ifB(Ψ, n) < ∞; (iii) SE-optimalfor Ψ if B(Ψ, n) = minB(Ψ, k) : 1 ≤ k ≤ T < ∞; (iv)DSE-optimalfor Ψ if B(Ψ, n) = minB(Ψ, k) : 1 ≤ k ≤ T < ∞.

6. Ignorance of the rival: memory and sample size

We shall say that agentα is ignorant of his rivalβ in Ψ if, for any pair of nodesω, ω inΩ(t) ⊂ Ω

ω ≡ (eα(τ ), eβ(τ ), qα(τ ), qβ(τ ))

t−1τ=1

ω ≡ (eα(τ ), eβ(τ ), qα(τ ), qβ(τ ))t−1τ=1

i.e. ω andω have the same history of the efforts and outputs of agentα

⇒ ω andω are in the same information set ofIαWhen each agent is ignorant of his rival,σ ∗ can be sustained as an SE for many sample

sizes with finite (albeit large) bonus. Indeed define∆ ≡ max|q1− q

2|, |q1 − q2|,∆ ≡

minq1 − q2, q2 − q

1. Then we have the following theorem for the existence of SE.

Theorem 2. Assume(1)–(4):further suppose each agent is ignorant of his rival inΨ . Then,if n ≤ (∆/∆)+ 1, n is SE- and DSE-feasible forΨ .


Corollary. Supposeq1 = q2 andq1= q

2. Then all sample sizes are SE- and DSE-feasible

for Ψ (this follows since∆ = 0).

Whenever the bonus is higher than the threshold,σ ∗ is theuniqueSE.

Theorem 3. Assume(1)–(4): further suppose each agent is ignorant of his rival inΨ . Letn be any feasible sample size forΨ . Then

B > B(Ψ, n) ⇒ σ ∗ is the unique SE ofΓ (Ψ,B, n)

In fact, σ ∗ is the unique SE in the mixed extension ofΓ (Ψ , B, n).

According to Theorem 3, whenever the principal can implementσ ∗ as an SE at all, heneed not further worry about the vexing issue of multiple equilibria: adding a penny to thethreshold will makeσ ∗ the unique SE (it is only at the thresholdB(Ψ , n) thatσ ∗ may haveto coexist with other SE of the game).

Suppose memory is refined in the game, i.e.Iα is refined toIα, still leavingα ignorantof his rival. This will tend to reduce optimal sample sizes. The intuition is clear. Supposean agent has succeeded on most days over a long past and knows so on account of refinedmemory. If the sample size is large, he will tend to shirk on the remaining few days, becausethis hardly affects his probability of winning the bonus. To get him to work there,B wouldhave to be huge. But if the sample size is small, he gives up a good chance of winningB by shirking. Thus a smallerB will sustain full effort once the sample size is lowered.Indeed, in the extreme case when the sample size is 1, the increase in the probability ofwinning through an extra day’s work remains invariant of memory. (For explicit examples,see Dubey and Wu, 2000.)

7. Information of the rival and sample size

When agents have information regarding each other’s outputs inΨ , a dramatic changeoccurs. Feasible sample sizes have an upper bound (i.e. for sample sizes that exceed thebound, no amount of bonus can sustainσ ∗ as an SE). There is an intimate inverse relationshipbetween the information pattern in the game tree and this bound. As information is refined,the bound falls. To put it another way: in order to restoreσ ∗ as an SE, the principal must“undo” the gain in agents’ information (about each other) by reducing his own information(about them).

We shall assume throughout this section that an agent has perfect recall.Furthermore, his information of past outputs (achieved by him and his rival) isinvariant of his memory of his own past efforts. To make this precise, let us defineω ∼(Iα)ω if ω andω are in the same information set inIα. Also forω ≡ (eα(τ ), eβ(τ ), qα(τ ),

qβ(τ ))t−1τ=1 and 1≤ t ′ ≤ t − 1, denote(ω|t ′) ≡ (eα(τ ), eβ(τ ), qα(τ ), qβ(τ ))

t ′τ=1. Keeping

in mind that (see (1)) thatα does not observe his rival’s efforts, we postulate thefollowing.


Perfect recallω ∼ (Iα)ω



⇒t = t1eα(τ ) = eα(τ ), for 1 ≤ τ ≤ t − 1(ω|t ′) ∼ (Iα)(ω|t ′), for 1 ≤ t ′ < t − 1

(5.1)

Invariance of output information on memory of effortThere exist partitionsJα(t) of Qt−1

α ×Qt−1β for 1< t ≤ T which characterizeIα in the

following sense:S ∈ Iα implies

∃t ∈ T , (eα(τ ))t−1τ=1 ∈ Et−1

α , K ∈ Jα(t)such thatS = (eα(τ ), eβ(τ ), qα(τ ), qβ(τ ))t−1

τ=1 : eβ(τ )

∈ Eβ, (qα(τ ), qβ(τ ))t−1τ=1 ∈ K (5.2)

We shall need one more assumption for this section.For any

α ∈ 1,2, ql+1β − qlβ < qα − q

α, for 1 ≤ l < m(β) (6)

This is not too restrictive, e.g. it automatically holds if the setsQα qα

≡ q1α, . . . , q

m(α)α ≡

qα consists of consecutive integers, forα ∈ 1,2. Its purpose is to ensure that, once anagent has averted a good or a bad scenario he will never find himself in a situation wherehis own output has no effect on his probability of winning.

We now define a positive integerfα(ω) at each nodeω ∈ Ω and for each agentα. Thefunctionsfα do not depend on the information pattern or the probabilities of production inΨ . Intuitively, fα(ω) is the largest sample size which still leaves some incentive for agentα

to put in maximal efforte∗α at the nodeω, when he knowsω perfectly.Then define, for the extensive formΨ

g(Ψ ) = minα∈1,2

minS∈Iα

maxfα(ω) : ω ∈ S

Thusg assigns a positive integer to every extensive formΨ . In contrast to the functionsfα, the functiong is highly sensitive to the information inΨ .

Pending the precise definition offα to the proof of Theorem 4 in Appendix A, we areready to state the following theorem.

Theorem 4. Assume(1)–(6),then the sample size n is SE-feasible forΨ ⇔ n ≤ g(Ψ ).

Theorem 4 makes precise our intuition that, in order to sustainσ ∗ as an SE, the principalmust lower his sample size when agents’ information is increased. Let us say thatΨ isan information-refinement ofΨ (and writeΨ Ψ ) if, for each agentα, his information


partition Iα in Ψ is a refinement of his information partitionIα in Ψ ; and other than that,the extensive formsΨ andΨ are identical. It is evident from our formula forg that

Ψ Ψ ⇒ g(Ψ ) ≤ g(Ψ )

In general,g(Ψ ) falls sharply belowg(Ψ ) when the refinement is significant. This saysthat if agents have more information about each other, then the principal must reduce hissample size in order to elicit maximal effort from them at an SE. Indeed, it is easy to checkthat if the game has three or more periods and if information is at its finest inΨ (i.e. agentsobserve everything except each other’s efforts), theng(Ψ ) = 2. The need for reducedscrutiny is then indeed obvious.

Remark 3. When agents’ information of each others’ outputs is not null, Theorem 3 isno longer true. We nevertheless can sustainσ ∗ as a unique SE for sufficiently large bonus.Indeed, for any sample sizen ≤ g(Ψ ), define

B∗(Ψ, n) ≡ inf B ∈ R+ : σ ∗ is the unique SE ofΓ (Ψ, B, n) for all B ≥ BThenB∗(Ψ, n) < ∞ as we prove in Appendix A (of course, typicallyB∗(Ψ, n) B(Ψ, n)).

8. The principal’s optimal policy

Define, forα ∈ 1,2Bα(Ψ, n) = minB : σ ∗

α is a best reply toσ ∗β in the gameΓ (Ψ,B, n)

Then

B(Ψ, n) = maxB1(Ψ, n), B2(Ψ, n)An optimal SE-policy (B(Ψ , k), k) in Ψ must satisfy

B(Ψ, k) = minB(Ψ, n) : 1 ≤ n ≤ T In the same vein, an optimal DSE-policy (B(Ψ, k), k) in Ψ must satisfyB(Ψ, k) = minB(Ψ, n) : 1 ≤ n ≤ T , whereB(Ψ, n) = maxB1(Ψ, n), B2(Ψ, n) and Bα(Ψ, n) =minB : σ ∗

α is a dominant strategy inΓ (Ψ , B, n).Consider an example withEα = Qα = 0,1 forα ∈ 1,2. Letp1

1(1) = 0.7, p01(1) = ε

andp12(1) = 0.6, p0

2(1) = ε. Suppose there are 180 periods, i.e.T = 1, . . . ,180 andlow-information, i.e.I1 = I2 = Ω(t) : 1 ≤ t ≤ 180. For any sequence of effortse ∈ 0,1180, let |e| ≡ number of 1’s in the sequence, i.e. the number of days of maximaleffort. Putd1(e) = 2|e|, d2(e) = |e|. Finally, letu1(x) = u2(x) = x. For smallε, we obtainthe computer-aided graph of Fig. 2 (the calculations were done forε = 0.001).

Notice thatB1(n) is downward-sloping, whileB2(n) has a U-shape. As was said before,raisingn increases the probability of sampling any particular day, and therefore tends tolower the bonus at which agents will work. But asn increases, the probability that the weakagent has been strongly overtaken by his rival (on the othern − 1 days of the sample) is


Fig. 2. The optimal policy.

high. This tends to raise the required bonus for the weak agent and to lower it for the strongagent. The two effects on the bonus go in the same direction for the strong agent, henceB1(n) is downward-sloping. However, they work in opposite directions for the weak agentand soB2(n) turns and starts rising, when the second effect dominates the first. Indeed theturn comes rather quickly atn = 3. (Asp2 rises, keepingp1 fixed, the turn occurs at largersample sizes). Theoptimalsample size, the intersection ofB1 andB2, is n∗ = 26.

9. Optimal sample size for long horizons

Let Ψ (1), Ψ (2), . . . be a sequence of extensive forms of length 1, 2,. . . with fixedunderlyingEα, Qα, pα for α ∈ 1,2. Let the information partitionsI1(T), I2(T) in Ψ (T)be arbitrary forT = 1,2, . . . . Denote the utility for money and the disutility for work inΨ (T) by uTα , d

Tα . We assume that the disutility of working all the time increases linearly

with T and is distributed in a non-skewed manner between the periods. Precisely, we havethe following condition.

There existsγ > 0 such that, forα ∈ 1,2 and anyT we have

dTα ((e∗α(τ ))

Tτ=1) >

T

γ(7a)

Consider(e∗α(τ ))Tτ=1 and(eα(τ ))Tτ=1. Suppose, for somet ∈ T eα(t) = e∗α = eα(t), andeα(τ ) = eα(τ ) for all τ ∈ T \t.

Then

dTα ((eα(τ ))Tτ=1)− dTα ((eα(τ ))

Tτ=1) < γ (7b)

Finally, assume that agent 1 is more skilled than agent 2∑q∈Q1

pe∗11 (q)q >

∑q∈Q2

pe∗22 (q)q (8)


that is, if both agents put in maximal effort, agent 1 produces a strictly higher averagethan 2. Define8 B(n, T ) ≡ minB:σ ∗ is an SE of the gameΓ (Ψ (T), B, n) and supposen(T ) ∈ Argmin1≤n≤T B(n, T ).

Then we have, justifying the title of this paper,

Theorem 5. Assume(7a), (7b)and(8), and that each T-period game satisfies assumptions(1)–(4).Then

n(T )

T→ 0 as T → ∞

10. Correlated probabilities

We can permit correlation in agents’ productivity without affecting many of our results.Indeed, suppose that when agents choose(e1, e2) ∈ E1 × E2 at ω ∈ Ω(t), this givesrise to a probability distributionpe1,e2 with full support onQ1 × Q2. (Notice that whilewe have dropped independence of probabilities across agents at any given time, we stillmaintain independence across time.) The assumption that more effort boosts productivitynow takes the following form. For anyα ∈ 1,2, e ∈ Eα\e∗α, q ∈ Qα\qα andqβ ∈ Qβ :∑

q∈Qα,q≥qpe∗α,e∗β (q, qβ) >

∑q∈Qα,q≥q

peα,e

∗β (q, qβ)

This says, in effect, that when agentα shifts to his maximal effort he boosts his ownproductivity, given any fixed quantity of his rival.

With this hypothesis, Theorems 1, 2 and 4 still hold. Theorem 5 holds if we assume thatagentα is more skilled thanβ in the sense∑

(x,y)∈Qα×QβPe∗α,e∗β (x, y)(x − y) > 0

The proofs require only obvious changes in notation. But Theorem 3 breaks down, exceptfor T = 1. Notice that the basic compulsions for reducing scrutiny remain intact (byTheorems 4 and 5).

The following may be worth noting. Start with a one-period model of correlatedproductivities, in which the competitive contract outperforms individual contracts (see,for example, Green and Stokey, 1983; Dubey and Wu, 2000).

Consider aT-fold repetition with low-information (see Remark 2). Then it triviallyfollows from the one-shot analysis that a competitive contract, with sample size 1, outper-forms individual contracts in theT-period game (and Theorem 5, as was said, holds in thissetting).

8 The strategyσ ∗(T), of always putting in maximal effort inΨ (T), is denotedσ ∗ without confusion.


11. Many agents

Much of our analysis clearly continues to hold when there are more than two agents. Inparticular, for any fixed sample sizen, there exist as before, finite thresholds

B(n) ≡ minB ∈ R+ : σ ∗ is a DSE ofΓ (B, n)B∗(n) ≡ inf B ∈ R+ : σ ∗ is the unique SE ofΓ (B, n) for all B > BB(n) ≡ minB ∈ R+ : σ ∗ is an SE ofΓ (B, n)

and the corresponding optimal sample sizes will again typically be much smaller thanT.We must haveB(n) ≤ B∗(n) ≤ B(n). Theorem 3 showed that, with two agents who areignorant of each other’s outputs,B∗(n) = B(n) always (though, typically,B(n) < B(n)).We have not yet fully explored the relation between these three thresholds for multipleagents. But it is curious that, ifT = 1 and if all agents can produce just two output levelsand their probabilities of being productive are independent, thenB(n) = B∗(n) = B(n) (infact, we get then a game of strategic substitutes as in Bulow et al. (1985)).

Acknowledgements

We are grateful to the Cowles Foundation for Research in Economics, Yale University,and to the Center for Economic Studies and Planning, Jawaharlal Nehru University, wherepart of this work was carried out. In particular, we wish to thank Sugato Dasgupta, OriHaimanko and Kunal Sengupta for thoughtful comments.

Appendix A. Preliminaries

If ω ≡ (ω, e1, e2, q1, q2) ∈ Ω(t+1) forω ∈ Ω(t)we say thatω is an immediate followerofω. More generally,ω is a follower ofω if there exists a sequenceω1, ω2, . . . , ωk of nodesin Ω ≡ T+1

t=1 Ω(t) = Ω Ω(T + 1) such thatω = ω1, ω = ωk and eachωl+1 is animmediate follower ofωl for 1 ≤ l ≤ k−1. DenoteF(ω) = ω ∈ Ω : ω is a follower ofω.

Next let ω ≡ (eα(τ ), eβ(τ ), qα(τ ), qβ(τ ))t−1τ=1 ∈ Ω(t) ⊂ Ω. DefineQ(α,ω) ≡

((xα(τ ))τ∈T \t, (yβ(τ ))τ∈T ) ∈ QT \tα ×QT

β : (xa(τ ), yβ(τ ))t−1τ=1 = (qα(τ ), qβ(τ ))

t−1τ=1,

i.e.Q(α, ω) lists all possible outputs of both agents, across time, that are consistent withω

and that leaveα’s output atω unspecified. For any (x, y) ∈ Q(α,ω) andI ⊂ T , with t ∈ I,put δ(x, y, I) ≡ ∑

τ∈Iy(τ)−∑τ∈I\tx(τ).

It is clear thatα wins the bonus atω, givenI and (x, y), if he produces more thanδ(x, y,I) atω; and ties for the bonus if he produces exactlyδ(x, y, I). Let

Wα = q ∈ Qα : q > δ(x, y, I),Wα = q ∈ Qα : q ≥ δ(x, y, I),Wα = q ∈ Qα : q = δ(x, y, I)


Then, conditional on the realization of (x, y) ∈ Q(α,ω) and on the days inI being sampled,the efforte ∈ Eα atω induces the probabilitypα(x, y, e,ω,I) forα to win the bonus, where9

pα(x, y, e, ω, I) = peα(Wα)+ 12(p

eα(Wα)) = 1

2(peα(Wα)+ peα(Wα))

The second equality comes from:Wα = Wα\Wα andWα ⊂ Wα. By assumption (2),

pe∗αα (V ) ≥ peα(V ) for all e ∈ Eα and setsV that are of the typeq ∈ Qα : q ≥ x orq ∈ Qα : q > x, x arbitrary; hence we immediately have

pα(x, y, e∗α, ω, I) ≥ pα(x, y, e, ω, I) (A.1)

for all e ∈ Eα. Moreover, ifqα

≤ δ(x, y, I) ≤ qα we have (again, by assumption (2) and

the fact thatpe∗αα has full support)

pα(x, y, e∗α, ω, I) > pα(x, y, e, ω, I) (A.2)

for all e ∈ Eα\e∗α.Next, supposeω ≡ (e1(τ ), e2(τ ), q1(τ ), q2(τ ))

Tτ=1 ∈ Ω(T + 1) is a terminal node and

I ⊂ T . Define

pα(ω, I) =

1 if∑τ∈Iqα(τ ) >

∑τ∈Iqβ(τ )

12 if

∑τ∈Iqα(τ ) = ∑

τ∈Iqβ(τ )0 if

∑τ∈Iqα(τ ) <

∑τ∈Iqβ(τ )

Consider two arbitrary maps10 σα, σβ fromΩ to Eα, Eβ , respectively. The probability,induced by (σα, σβ ) that the play of the game goes throughω andα wins the bonus undersampleI is given by

pα((σα, σβ)|I, ω) =∑

ω∈F(ω)∩Ω(T+1)

p(σα,σβ)(ω)pα(ω, I)

WhenI ranges over all ofCn and all samples inCn are picked with uniform probability,this event has probability

pα((σα, σβ)|n, ω) =∑I∈Cn

p(n)pα((σα, σβ)|I, ω) (A.3)

Finally, the overall probability thatα wins the bonus in the game, under (σα, σβ ) andsample sizen, ispnα(σα, σβ) = pα((σα, σβ)|n, ω∗).

It is obvious that

pnα(σα, σβ) =∑

ω∈Ω(t)pα((σα, σβ)|n, ω) (A.4)

for every 1≤ t ≤ T + 1.We now state two useful lemmas.

9 For anyS ⊂ Qα, peα(S) ≡ ∑

q∈Speα(q).10 Not necessarily strategies inΣα,Σβ .


Definition. Fix ω ∈ Ω Supposeσα and σα are two maps fromΩ to Eα which satisfy:(i) σα(ω) = σα(ω) for ω ∈ Ω\ω; (ii ) σα(ω) = σα(ω) = e∗α for ω ∈ F(ω); (iii ) σα(ω) =e∗α = σα(ω). Then we write:σα ω σα.

Lemma 1. Fix ω ∈ Ω(t) andI ⊂ T . Supposeσα ω σα. Then

1. pα((σα, σ ∗β )|I, ω) ≥ pα((σα, σ

∗β )|I, ω)

2. pα((σα, σ ∗β )|n, ω) ≥ pα((σα, σ

∗β )|n, ω)

3. pnα(σα, σ∗β ) ≥ pnα(σα, σ

∗β )

Proof. It is evident that, ift ∈ I

pα((σα, σ∗β )|I, ω) = p

(σα,σ∗β )(ω)

× ∑(x,y)∈Q(α,ω)

(T∏τ=tpe∗ββ (y(τ ))

) T∏τ=t+1

pe∗αα (x(τ ))

pα(x, y, σα(ω), ω, I)

(A.5)

for all σα, in particularσα = σα or σα. By Eqs. (A.1) and (A.5), and the fact thatp(σα,σ

∗β )(ω) = p

(σα,σ∗β )(ω)on account of (i) in the definition ofω, we getpα((σα, σ ∗

β )|I, ω)≥ pα((σα, σ

∗β )|I, ω). If t is not inI, it is obvious thatpα((σα, σ ∗

β )|I, ω) = pα((σα, σ∗β )|I, ω).

This proves (1) of Lemma 1. Now (2) is obvious from (1) and Eq. (A.3). Finally (3) is obviousfrom Eq. (A.4) and (2), and the obvious fact thatpα((σα, σ

∗β )|n, ω′) = pα((σα, σ

∗β )|n, ω′)

for all ω′ ∈ Ω(t)\ω.

Lemma 2. For any mapσα : Ω → Eα, and1 ≤ n ≤ T :

pnα(σ∗α , σ

∗β ) ≥ pnα(σα, σ

∗β )

Proof. Introduce a total order on all the nodes inΩ such that, ifω ∈ Ω(t) andω ∈ Ω(t)andt > t , thenω ω. WriteΩ ≡ ω1, ω2, . . . , ωL with ω1 ω2 · · · ωL. Go fromσα to σ ∗

α through a sequence ofL transitions:

σα ≡ σ 1α → σ 2

α → · · · → σL+1α ≡ σ ∗

α

In the lth transition, changeσ l−1α (ωl) toe∗α to and leave the rest of the choices according

to σ l−1α ; denote the resulting map byσ lα. Clearly,σ l+1

α ωl σlα.

By (3) of Lemma 1,pnα(σl+1α , σ ∗

β ) ≥ pnα(σlα, σ

∗β ) for 1 ≤ l ≤ L. Therefore,pnα(σ

∗α , σ

∗β ) ≥

pnα(σα, σ∗β ) proving Lemma 2.

A.1. Proof of fourth theorem

A.1.1. Motivation and definition of f1, f2As we said,fα(ω) is the largest sample size which still leaves some incentive for agentα

to put in maximal efforte∗α at the nodeω, when he knowsω ∈ Ω perfectly.


Let us consider what could destroy this incentive. Supposeα knows atω that his rival hasalready produced much more than him in the days sampled in the past. . . in fact so muchmore, that even if he produces his maximumq

αand his rivalβ produces his minimumqβ

on all future days remaining in the sample, he still cannot equal or beatβ. In this scenario,it would make no sense forα to work atω. We call thisthe bad scenario forα at ω, givensample size n. The good scenario is when his rival is in a bad scenario, i.e.α is so far aheadβ of that he will win without working even ifβ turns lucky and producesqβ on all futuredays while he himself produces onlyq

α. In the good scenario, too, there is no incentive for

α to work.Notice that these scenarios can only arise if two things happen at once:α knows a lot

of the past history (i.e. has refined information)andsufficiently many days of the past aresampled (i.e. the sample sizen is not too small).

With this motivation in mind, let us now do a precise calculation. Suppose the history ofoutputs that leads toω ∈ Ω(t) is (qα(t), qβ(t))

t−1τ=1 and suppose that the sample size isn.

When will a bad scenario be inescapable forα atω? If n ≤ T − t + 1 there is apositiveprobability that all then days in the sample will consist ofω andn − 1 days afterω, andthatα’s rival β will produce exactlyq

βon thesen days. By putting in efforte∗α atω, α can

boost11 the probability of producingqα atω, without lowering his chances of winning thebonus in any other event. So the bad scenario is avoided. Now supposen > T − t+1. Let usconsider an optimal way of avoiding the bad scenario forα. To this end, first letω and all theT −t days of the future be sampled. This still leavesr ≡ n−(T −t+1) = n−T +t−1 moredays to be sampled from the past ofω. Denote7(τ) ≡ qα(τ )− qβ(τ ) for 1 ≤ τ < t − 1.For any integer−∞ < j ≤ t − 1, define

h(j, ω) =

max∑τ∈I7(τ) : I ⊂ 1, . . . , t − 1, |I| = j if j > 0

0 if j ≤ 0

The subsetI of days that could be sampled from the past ofω, to best helpα avoid thebad scenario, would be aI that achievesh(r,w) and maximizesα’s lead overβ. Denoteintegers byZ and define

fα(ω) = maxk ∈ Z : 1 ≤ k ≤ T , (T − t + 1)(qa−qβ)+h(k − T+t − 1, ω) ≥ 0

It is easy to check that a bad scenario can be avoided if, and only if,n ≤ fα(ω). Arguing

in the same fashion, a good scenario forα can be avoided if, and only if,n ≤ fα(ω), where

fα(ω) = maxk ∈ Z : 1 ≤ k ≤ T , (T − t + 1)(qα

− qβ)+h(k − T + t−1, ω) ≤ 0

andh(j, ω) is defined exactly ash(j, ω) replacing “max” by “min”. Since we must avoidboth these scenarios, we put

fα(ω) = minfα(ω), fα(ω)

11 By assumption (1).


This completes the definition of the functions12 fα for α ∈ 1,2, and thereby ofg.Recall

g(Ψ ) = minα∈1,2

minS∈Iα

maxfα(ω) : ω ∈ S

A.1.2. Lemmas

Lemma 3. Leta1, . . . , aK, a, a be real numbers such thata1 ≤ a,≥ ak and|al+1 − al | ≤a − a for 1 ≤ l ≤ K − 1. Thena ≤ aj ≤ a for some1 ≤ j ≤ K.

Proof. Obvious.

Lemma 4. Assumen ≤ fα(ω) for ω ∈ Ω(t) andα ∈ 1,2. Then there existsI ∈ Cn and(x, y) ∈ Q(α,ω) such that: t ∈ I andq

α≤ δ(x, y, I) ≤ qα.

Proof. Let ω ≡ (eα(τ ), eβ(τ ), qα(τ ), qβ(τ ))t−1τ=1 ∈ Ω(t). DenoteT ∗ ≡ T \t. We shall

say that(xα(τ ), yβ(τ ))τ∈T ∗ is consistent withω if xα(τ ) = qα(τ ) andyβ(τ ) = qβ(τ ) for1 ≤ τ ≤ t − 1; andxα(τ ) ∈ Qα andyβ(τ ) ∈ Qβ for τ ∈ T ∗.

Sincefα(ω) = minfα(ω), f α(ω), n ≤ fα(ω) andn ≤ fα(ω).

Now, n ≤ fα(ω) implies13

∃I ∈ Cn and(xα(τ ), yβ(τ ))τ∈T ∗ consistent withω such that

t ∈ Iand∑

τ∈I\t(xα(τ )− yβ(τ )) ≤ qβ − q

α(A.6)

that is,ω is not a good scenario forα (he cannot clearly win if he producesqα

and his rivalproducesqβ atω); similarly, n ≤ f

α(ω) implies

∃I ∈ Cn and(xα(τ ), yβ(τ ))τ∈T ∗ consistent withω such that

t ∈ Iand∑

τ∈I\t(xα(τ )− y

β(τ )) ≥ q

β− qα (A.7)

that is,ω is not a bad scenario forα (he cannot clearly lose if he producesqα and his rivalproduces atq

βatω). Consider any sequence of samplesI ≡ I1 → I2 → · · · → Ik ≡ I

such that, in each transitionIl → Il+1, one dayτl ∈ I is removed fromIl and is replacedby a dayτ l ∈ I (which is not necessarily distinct fromτl), making sure thatIl+1 ∈ Cn.Notice t ∈ Il for 1 ≤ l ≤ k. For each suchl, define(xlα(τ ), y

lβ(τ ))τ∈T ∗ , consistent

with ω, by

(xlα(τ ), ylβ(τ )) =

(xα(τ ), yβ(τ ))

(xα(τ ), yβ(τ ))

if τ ∈ T ∗\τ1, . . . , τl−1otherwise

12 With two agentsf1(ω) = f2(ω). But the functionsfα(ω), fα(ω) can be defined, with obvious modifications,

for the case ofN agents. Then, if we defineg as above, with “minα∈1,2” replaced by “minα∈1,...,N”, Theorem 4remains intact (with the same proof).13 For τ ∈ I, τ > t we may in particular takexα(τ ) = q

α, yβ (τ ) = qβ ; and forτ ∈ I, τ > t takexα(τ ) =

qα, yβ(τ ) = q

β.


Consider7(l) ≡ ∑τ∈Il\tx

lα(τ ) − ylβ(τ ). Notice7(1) = ∑

τ∈I\txα(τ ) − yβ(τ ) ≤qβ − q

αand7(k) = ∑

τ∈I\txα(τ )− yβ(τ ) ≥ q

β− qα. Moreover,|7(l + 1)−7(l)| =

|(xα(τ l) − yβ(τ l)) − (xα(τl) − yβ(τl))| ≤ (qα − q

β) − (q

α− qβ) = (qβ − q

α) −

(qβ

− qα). Apply Lemma 3 (witha1, . . . , aK ≡ 7(1), . . . ,7(k), a ≡ qβ − qα, a ≡

qβ

− qα. We conclude that there existsI ∈ Cn and(xα(τ ), yβ(τ ))τ∈T ∗ , consistent with

ω, such thatt ∈ I andqβ

− qα ≤ ∑τ∈I\t(xα(τ ) − yβ(τ )) ≤ qβ − q

α. Rewrite this

inequality as

qβ ≥ qα

+∑

τ∈I\t(xα(τ )− yβ(τ )) ≡ A

qβ

≤ qα +∑

τ∈I\t(xα(τ )− yβ(τ )) ≡ A

Again apply Lemma 3 (witha ≡ A, a ≡ A, a1, . . . , aK ≡ qβ

≡ q1β, q

2β, . . . , q

m(β)β ≡

qβ) to conclude that there existsqlβ ∈ Qβ such thatA ≤ qlβ ≤ A. Putyβ(t) = qlβ . This

definesy ∈ QTβ (since only thetth component ofywas unspecified) and clearly (x,y) ∈ Q(α,

ω). Thenx, y, I accomplish the requirement of Lemma 4.

A.1.3. Completion of the proof of fourth theoremWe claim that

B(Ψ, n) < ∞ ⇔ pnα(σ∗α , σ

∗β ) > pnα(σα, σ

∗β ) (A.8)

for α ∈ 1,2 andσα ∈ Σα\σ ∗α . First we show (⇒). Supposepnα(σα, σ

∗β ) ≥ pnα(σ

∗α , σ

∗β )

for someσα = σ ∗α . For anyω ∈ Ω(T + 1), denoteω ≡ (eωα (τ ), e

ωβ (τ ), q

ωα (τ ), q

ωβ (τ ))

Tτ=1.

Now p(σ ∗α ,σ

∗β )(ω) > 0 for ω ∈ Ω(T + 1) implies eωα (τ ) = e∗α for τ = 1, . . . , T . Hence∑

ω∈Ω(T+1)p(σ ∗α ,σ

∗β )(ω)dα((e

ωα (τ ))

Tτ=1) = dα(e

∗α, . . . , e

∗α). On the other hand, by assump-

tion (1) and by the fact that thepeα andpeβ have full support, there existsω ∈ Ω(T + 1)

such thatp(σα,σ∗β )(ω) > 0 andeωα (τ ) = e∗α for someτ . By assumption (4), we deduce∑

ω∈Ω(T+1)p(σ ∗α ,σ

∗β )(ω)dα((e

ωα (τ ))

Tτ=1) >

∑ω∈Ω(T+1)p

(σα,σ∗β )(ω)dα((e

ωα (τ ))

Tτ=1). Then

α could switch fromσ ∗σ to σα saving disutility and still not losing on the reward, contra-

dicting that (σ ∗α , σ

∗β ) is an SE ofΓ (Ψ , B, n), for any arbitraryB. But thenB(Ψ, n) = ∞, a

contradiction, proving (⇒). Next we show (⇐). Supposepnα(σ∗α , σ


∗β ) for all

σα ∈ Σα\σ ∗α . Let

δ = minα∈1,2

minσα∈Σα\σ ∗

α (pnα(σ

∗α , σ

∗β )− pnα(σα, σ

∗β ))

SinceΣα is finite,δ > 0. By assumption (3), there exists a finiteB such thatδuα(B) ≥dα((e

∗α, . . . , e

∗α))−dα(e) for α ∈ 1,2 ande ∈ ETα \(e∗α, . . . , e∗α). But then(σ ∗

α , σ∗β )must

be an SE ofΓ (Ψ , B, n), showing thatB(Ψ, n) ≤ B < ∞. This proves (⇐) and thereby(A.8).


Thus, to prove Theorem 4, it suffices to show

n ≤ g(Ψ ) ⇔ pnα(σ∗α , σ


∗β ) (A.9)

for α ∈ 1,2 andσα ∈ Σα\σ ∗α .

Supposen > g(Ψ ). Then there exists an agentα and an information setS ∈ Iα, wheren > f (ω) for all ω ∈ S. By the definition off, any node inS is either a good scenario forα(i.e.n > fα(ω)) or a bad scenario forα (i.e.n > f

α(ω). Let S correspond to(e∗α(τ ))

t−1τ=1 and

K ∈ Jα(t) in accordance with assumption 5.2. ConsiderS that corresponds to(e∗α(τ ))t−1τ=1

andK. Then (using assumption (1)) some node inS is reached with positive probabilityunder(σ ∗

α , σ∗β ). Since the definition offα(ω) depends only on the history ofoutputsthat

lead toω, each node inS is also either a good scenario or a bad scenario. (Indeed outputhistories overS and S both yieldK). Therefore atall terminal nodes that follow from anarbitraryω ∈ S,αwins (loses) the bonus ifω is a good (bad) scenario, no matter what movesthe agents make afterω. Letα change fromσ ∗

α to σα, whereσα(ω) = e∗α for all ω ∈ S, andσα is the same asσ ∗

α at all other information sets. Thenpnα(σa, σ∗β ) = pnα(σ

∗α , σ

∗β ). This

proves (⇐) of (A.9).Now supposen ≤ g(Ψ ). Consider anyσα ∈ Σα\σ ∗

α . Let

t = maxτ : 1 ≤ τ ≤ T , σα(ω) = e∗α for someω ∈ Ω(τ)Then there existsS ∈ Iα such thatφ = S ⊂ Ω(τ), σα(ω) = e∗α for all ω ∈ S and

σα(ω) = e∗α for all ω ∈ ∪Tτ=t+1Ω(τ). First notice that there is some nodeω ∈ S which isrelevantσα for (otherwiseσα is equivalent toσ ∗

α ). By perfect recall (see assumption 5.1) allnodes inSare relevant forσα. By the definition ofg, there is a nodeω ∈ S with n ≤ fα(ω).By assumption (1) and that fact thatpα andpβ have full support, there is a nodeω′ ∈ S

which is reached with positive probability and which has the same history of outputs asω.Hencen ≤ fα(ω

′) also, sincefα depends only on the history of outputs.By Lemma 4, there exist(x′, y′) ∈ Q(α,ω′) andI′ ∈ Cn such that:t ′ ∈ I′ andq

α≤

δ(x′, y′, I′) ≤ qα. Consider any two mapsσα and σα from Ω to Eα (not necessarilystrategies inΣα) such thatσα ω′ σα. By (A.2), (A.3) and (A.5) we have

pα((σα, σ∗β )|I′, ω′) > pα((σα, σ

∗β )|I′, ω′) (A.10)

Then, by (1) of Lemma 1 and (A.10), and summing overI ∈ Cn as in (A.3), we obtain

pα((σα, σ∗β )|n, ω′) > pα((σα, σ

∗β )|n, ω′) (A.11)

Go fromσα to σ ∗α through the sequence of transitionsσα ≡ σ 1

α → · · · → σL+1α ≡ σ ∗

α

as in the proof of Lemma 2. By (3) of Lemma 1,pnα(σl+1α , σ ∗

β ) ≥ pnα(σ1α , σ

∗β ) for all l. But,

by (A.4) and (A.11), we have strict inequality at least once (at the transition involvingω′),provingpnα(σ

∗α , σ


∗β ).

Sinceσα was arbitrary element ofΣα\σ ∗α , this completes the proof.

A.2. Proof of first theorem

Consider anyω ∈ Ω(T + 1). Recallω ≡ (eωα (τ ), eωβ (τ ), q

ωα (τ ), q

ωβ (τ ))

Tτ=1. Suppose

(σ ∗α , σ

∗β ) is an SE ofΓ (Ψ , B(Ψ , n), n). Then we have


pnα(σ∗α , σ

∗β )uα(B(Ψ, n))

−∑

ω∈Ω(T+1)

p(σ ∗α ,σ

∗β )(ω)dα((e

ωα (τ ))

Tτ=1) ≥ pnα(σα, σ

∗β )uα(B(Ψ, n))

−∑

ω∈Ω(T+1)

p(σa,σ

∗β )(ω)dα((e

ωα (τ ))

Tτ=1) (A.12)

for anyσα ∈ Σ\σ ∗α .

Now, as argued in the first paragraph of Appendix A.1.3 (using only assumptions (1)–(4))∑ω∈Ω(T+1)

p(σ ∗α ,σ

∗β )(ω)dα((e

ωα (τ ))

Tτ=1) >

∑ω∈Ω(T+1)

p(σα,σ

∗β )(ω)dα((e

ωα (τ ))

Tτ=1)

Then by (A.12),pnα(σ∗α , σ


∗β ). By assumption (2),B > B(Ψ, n) implies

uα(B) > uα(B(Ψ, n)) and so all the inequalities in (A.12) become strict when we replaceB(Ψ , n) by B. This shows that(σ ∗

α , σ∗β ) is a strict SE ofΓ (Ψ,B, n), proving the “if” part

of (1).Next suppose that(σ ∗

α , σ∗β ) is a strict SE ofΓ (Ψ,B, n). Then all the inequalities of (A.12)

must be strict. By assumption (2), they will remain strict if we reduceB slightly toB− ∈,showing that(σ ∗

α , σ∗β ) is an SE ofΓ (Ψ,B− ∈, n). We conclude thatB > B(Ψ, n).

The proof of (2) is identical to the proof of (1), replacingσ ∗β by an arbitraryαβ every-

where.

A.3. Proof of second theorem

First we prove two lemmas.

Lemma 5. Letn ≤ (∆/∆)+ 1 and(x(1), . . . , x(n− 1)) ∈ Qn−1α . There exists(y(1), . . . ,

y(n)) ∈ Qnβ such that

∑n−1τ=1x(τ)+ q

α≤ ∑n

τ=1y(τ) ≤ ∑n−1τ=1x(τ)+ qα.

Proof. Consider the setY ≡ ∑nτ=1y(τ) : y(τ) ∈ Qβ, and writeY = z1, . . . , znm(β)

with z1 < · · · < znm(β). Notez1 = nqβ

andznm(β) = nqβ . Now, fromn ≤ (∆/∆) + 1,

we have∑n−1τ=1x(τ)+qα ≤ (n−1)qα+q

α≤ nqβ and

∑n−1τ=1x(τ)+ qα ≥ (n−1)q

α+ qα ≥

nqβ.

Then, by Lemma 3 (witha = ∑n−1τ=1x(τ)+ qα, a = ∑n−1

τ=1x(τ)+qα anda1, . . . , aK =z1, . . . znm(β)), there exists(y(1), . . . , y(n)) ∈ Qn

β such that∑n−1τ=1x(τ) + q

α

≤ ∑nτ=1y(τ) ≤ ∑n−1

τ=1x(τ)+ qα.

Lemma 6. Suppose neither agent can observe his rival inΨ and n ≤ (∆/∆) + 1. Fixω ∈ Ω(t), α ∈ 1,2 andσβ ∈ Σβ . Let σα and σα be two maps fromΩ to Eα such thatσα ω σα. Also assume thatp(σα,σβ)(ω) > 0. Thenpnα(σα, σβ) > pnα(σα, σβ).

Proof. Let ω be any node inΩ(t). Sinceβ cannot observeα, the probability distribution onoutput vectors thatβ can produce afterω depends only onσβ andω. Denote it byPrσβ,ω


and note that its support is(Qβ)T−t+1. Therefore, ift ∈ I ∈ Cn andσα = σα or σα, we

have

pα((σα, σβ)|I, ω) = p(σα,σβ)(ω)

× ∑(x,y)∈Q(αω)

(Prσβ,ω((y(τ ))Tτ=t ))

T∏τ=t+1

pe∗αα (x(τ ))

pα(x, y, σα(ω), ω, I)

(A.13)

Nowpnα(σα, σβ) = ∑ω∈Ω(t)

∑I∈Cnp(n)pα((σα, σβ)|I, ω). By Lemma 5 and Eq. (A.3),

pα(x, y, σα(ω), ω, I) > pα(x, y, σα(ω), ω, I); and by (A.2) we have (≥) for ω ∈ Ω(t)\ω.Moreover, it is clear thatp(σα,σβ)(ω) is independent ofσα = σα or σα for ω ∈ Ω(t) giventhe condition (i) in the definition ofω. Thus, using (A.4) and (A.13),pnα(σα, σβ) >pnα(σα, σβ).

A.3.1. Completion of the proof of the second theoremConsider anyσα ∈ Σα\σ ∗

α . Let

t = maxτ : 1 ≤ τ ≤ T , σα(ω) = e∗α for someω ∈ Ω(τ)Then there existsS ∈ Iα such thatS ∩Ω(t) = ∅, σα(ω) = e∗α for all ω ∈ S. There is somenodeω ∈ S which is relevant forσα (otherwiseσα is equivalent toσ ∗

α ). Sinceα cannotobserve his rival’s effort or output, and the probabilities of production have full supportfor every effort level, there existsω ∈ S which is reached with positive probability underσα, σ

∗β .

Go fromσα to σ ∗α through a sequence of transitionsσα ≡ σ 1

α → · · · → σL+1α ≡ σ ∗

α

as in Appendix A.1.3. By Lemma 6 we havepnα(σl+1α , σβ) > pnα(σ

lα, σβ) in the transition

at ω; and by (A.2) and (A.13), we have (≥) everywhere else. This provespnα(σ∗α , σβ) >

pnα(σα, σβ).Then the Theorem 2 follows from (A.8).

A.4. Proof of third theorem

We shall prove Theorem 3 in the context of a wider class of games thanΓ (Ψ,B, n)whichmay be of some interest in their own right.

Consider a gameΓ with players 1 and 2, who have finite pure-strategy sets denoted,w.l.o.g., byS1 ≡ 1,2, . . . ,M andS2 ≡ 1,2, . . . , N; and payoffshα : S1 × S2 → R

for α = 1,2.Payoff hypothesis: there exists a constantK and (forα = 1,2) functionsg+

α : S1×S2 → R

andg+α : Sα → R such that, for all(m, n) ∈ S1 × S2,

1. h1(m, n) = g+1 (m, n)− g−

1 (m)

2. h2(m, n) = g+2 (m, n)− g−

2 (n)

3. g+1 (m, n)+ g+

2 (m, n) = K


Let∆1 and∆2 denote the mixed-strategy sets of 1 and 2, i.e.

∆1 =(η1, . . . , ηM) : 0 ≤ ηi ≤ 1 and

M∑i=1

ηi = 1

∆2 =(ξ1, . . . , ξN) : 0 ≤ ξi ≤ 1 and

N∑i=1

ξi = 1

Then the mixed extensionΓ of the pure strategy gameΓ is defined via payoffshα(η, ξ) =∑Mi=1∑Nj=1ηiξjhα(i, j), for all (η, ξ) ∈ ∆1 ×∆2.

Lemma 7. SupposeΓ satisfies the payoff hypothesis. Then

(m, n) isa strict SE ofΓ ⇒ (m, n) is the unique SE ofΓ

Proof. Suppose, to the contrary, that(η, ξ) ∈ ∆1 × ∆2 is an SE ofΓ different from (m,n). Since (m, n) is a strict SE ofΓ , it follows thath1(m, n) > h1(η, n), i.e.

g+1 (m, n)− g−

1 (m) >

M∑i=1

ηig+1 (i, n)−

M∑i=1

ηig−1 (i) (A.14)

Similarly fromh2(m, n) > h2(m, ξ) we get

g+2 (m, n)− g−

2 (n) >

N∑j=1

ξjg+2 (m, j)−

N∑j=1

ξjg−2 (j)

which (using payoff hypothesis (3), and the fact that∑Nj=1ξj = 1) may be rewritten

K − g+1 (m, n)− g−

2 (n) > K −N∑j=1

ξig+1 (m, j)−

N∑j=1

ξjg+1 (m, j) (A.15)

Now, because(η, ξ) is an SE ofΓ we haveh1(η, ξ) > h1(m, ξ), or

M∑i=1

N∑j=1

ηiξj g+1 (i, j)−

M∑i=1

N∑j=1

ηiξj g−1 (i) ≥

N∑j=1

ξjg+1 (m, j)−

N∑j=1

ξjg−1 (m)

that isM∑i=1

N∑j=1


M∑i=1

ηig−1 (i) ≥

N∑j=1

ξjg+1 (m, j)− g−

1 (m) (A.16)

Similarly fromh2(η, ξ) ≥ h2(η, n) we get

M∑i=1

N∑j=1


N∑j=1

ξjg−2 (j) ≥

M∑i=1

ηig+2 (i, n)− g−

2 (n)


which (again using payoff hypothesis (3)) may be rewritten

K −M∑i=1

N∑j=1


N∑j=1

ξjg−2 (j) ≥ K −

M∑i=1

ηig+1 (i, n)− g−

2 (n) (A.17)

Rearranging (A.14) and (A.16), we have the following two inequalities

g+1 (m, n)−

M∑i=1

ηig+1 (i, n) > g−

1 (m)−M∑i=1

ηig−1 (i)

g−1 (m)−

M∑i=1

ηig−1 (i) ≥

N∑j=1

ξjg+1 (m, j)−

M∑i=1

N∑j=1

ηiξj g+1 (i, j)

Using the above two inequalities, we have

g+1 (m, n)−

M∑j=1

ηig+1 (i, n) >

M∑i=1

ξjg+1 (m, j)−

M∑i=1

N∑j=1

ηiξj g+1 (i, j) (A.18)

Rearranging (A.15) and (A.17), we have again the following two inequalities

N∑j=1

ξjg+1 (m, j)− g+

1 (m, n) > g−2 (n)−

N∑j=1

ξjg−2 (j)

g−2 (n)−

N∑j=1

ξjg−2 (j) ≥

M∑i=1

N∑j=1


M∑i=1

ηig+1 (i, n)

and these two inequalities imply

N∑j=1

ξjg+1 (m, j)− g+

1 (m, n) >

M∑i=1

N∑j=1


M∑i=1

ηig+1 (i, n) (A.19)

Rearranging (A.19), we have

N∑j=1

ξjg+1 (m, j)−

M∑i=1

N∑j=1

ηiξj g+1 (i, j) > g+

1 (m, n)−M∑i=1

ηig+1 (i, n)

which contradicts (A.18), proving Lemma 5.

Remark 4. The proof of Lemma 2 also shows that, if the payoff hypothesis holds, SE havethe “interchange property”: when (σ 1,σ 2) and (σ1, σ2) are SE, so are (σ1, σ2) and (σ1, σ2).However, these SE are not necessarily payoff-equivalent.


A.4.1. Completion of the proof of third theoremWhenα is ignorant of his rivalβ in Ψ , it is easy to check thatα’s expected disutility∑ω∈Ω(T+1)p

(σα,σβ)(ω)dα((eωα (τ ))

Tτ=1) is independent ofσβ . (Hereω ≡ (eωα (τ ), e

ωβ (τ ),

qωβ (τ ))Tτ=1).

Also notice that any affine transformation of von-Neumann–Morgenstern utilities leavesinvariant the set of mixed-strategy SE ofΓ . W.l.o.g., we can therefore assume thatu1(B) =u2(B) = K, a constant. Then the gameΓ (Ψ , B, n) satisfies the payoff hypothesis (sincewhenever 1 gets the bonusB with probabilityp, 2 gets it with probability 1− p). Theorem3 now follows from Lemma 7.

A.5. Proof of fifth theorem

Denoteµα ≡ ∑q∈Qαp

e∗αα (q)q andνα ≡ ∑

q∈Qαpe∗αα (q)[q − µα]2. Suppose that there

is a sequence, indexed byk, such thatn(Tk) > cTk for somec > 0. By Chebyshev’sinequality, it follows that agent 2 wins the bonus with probability at most (sinceµ1 > µ2by assumption 8):

p

∣∣∣∣∣∣n(Tk)∑τ=1

(q1(τ )− q2(τ )− n(Tk)(µ1 − µ2))

∣∣∣∣∣∣ ≥ n(Tk)(µ1 − µ2)

≤ n(Tk)(ν1 + ν2)

n(Tk)2(µ1 − µ2)2= M

n(Tk)

whereM = (ν1 + ν2)/(µ1 − µ2)2.

For σ ∗2 to be the best reply toσ ∗

1 , B must satisfy:uTk2 (B)p2(σ∗1 , σ

∗2 ) > d

Tk2 (e

∗2(Tk)) >

1/γ Tk, or uTk2 (B) > 1/γ Tk(p2(σ∗1 , σ

∗2 ))

−1 > (1/γ )Tk(n(Tk)/M), i.e.

uTk2 (B) >

c

MγT 2k (A.20)

Let pα(eα, eβ ) denote the probability thatα wins the bonus if he puts in efforteα andβputs ineβ (when both work for 1 day). Denote

δ ≡ minα∈1,2,eα∈Σα\e∗α

pα(e∗α, e

∗β)− pα(eα, e

∗β)

By assumption (2),δ > 0. We shall show that if

uTα (B) >γ

δT (A.21)

thenσ ∗α is the best reply toσ ∗

β for α ∈ 1,2, if the sample size is 1. Consider the sequence

σα ≡ σ 1α → σ 2

α → · · · → σ l+1α ≡ σ ∗

α as in the proof of Lemma 2, withσ l+1α ωl

σ lα. Let p(σα, σβ , ω) be the probability that nodeω is reached under (σα, σβ ) and notep(σ lα, σ

∗β , ωl) = p(σ l+1

α , σ ∗β , ωl). Whenα switches fromσ lα to σ l+1

α his disutility goes

up by at mostp(σ lα, σ∗β , ωl)γ by assumption (7). On the other hand,ωl is sampled with

probability p(σ lα, σ∗β , ωl)/T (since each day is sampled with probability 1/T when the


sample size is 1). Thus the gain in utility toα, when he switches fromσ lα to σ l+1α is at least

(p(σ lα, σ∗β , ωl)/T )δu

Tα (B). Hence (A.21) implies thatα profits by the switch whenever

p(σ lα, σ∗β , ωl) > 0 (and is unaffected ifp(σ lα, σ

∗β , ωl) = 0. Thenσ ∗

α is the unique best replyto σ ∗

β (since someωl are reached with positive probability). For large enoughT, there existB for which (A.21) holds while (A.20) is violated. This proves that sample size 1 is betterfor the principal than sample sizen(T) for all large enoughT, contradicting thatn(T) is anoptimal size.

A.6. Proof of third remark

By (A.9), n ≤ g(Ψ ) impliespnα(σ∗α , σ


∗β ) for σα ∈ Σα\σ ∗

α . We claimthat there is no(σ1, σ2) = (σ ∗

1 , σ∗2 ), such that (a)pn1(σ1, σ2) ≥ pn1(σ

∗1 , σ2) and (b)

pn2(σ1, σ2) ≥ pn2(σ1, σ∗2 ). Otherwise, since(σ1, σ2) = (σ ∗

1 , σ∗2 ), assume w.l.o.g.σ1 =

σ ∗1 . We havepn1(σ

∗1 , σ

∗2 ) > pn1(σ1, σ

∗2 ) ≥ pn1(σ1, σ2) ≥ pn1(σ

∗1 , σ2) ≥ pn1(σ

∗1 , σ

∗2 ) (the

first strict inequality is from (A.9) andσ1 = σ ∗1 the second inequality is from (b), the

third inequality is from (a) and the last inequality is from Lemma 2). This is a contradic-tion. Thus for every pair(σ1, σ2) = (σ ∗

1 , σ∗2 ), there existsB(σ 1, σ 2) such that, ifB >

B(σ1, σ2), at least one agentα would like to unilaterally deviate fromσα to σ ∗α . Let B =

max(σ1,σ2)∈Σ1×Σ2\(σ ∗1 ,σ

∗2 )B(σ1, σ2). Then for anyB > maxB, B(Ψ, n), (σ ∗

1 , σ∗2 ) is

the unique SE.

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Competitive prizes: when less scrutiny induces more effort