Optimal Contest Design under Reverse$Lottery

Optimal Contest Design under Reverse-Lottery Technology Jingfeng Luy

Bo Shenz

Zhewei Wangx

February 2017

Abstract In this paper, we study e¤ort-maximizing contest design under the “reverse”nested lottery contest model of Fu, Lu and Wang (2014) –which is the “mirror image”of the conventional nested lottery contest of Clark and Riis (1996). We show that under the reverse-lottery technology, a single-stage winner-take-all grand contest dominates all other feasible designs when the contest is su¢ ciently noisy. This result is in dramatic contrast to the conventional wisdom on the optimality of multistage elimination contests that is grounded under the conventional nested lottery contest technology in the literature. In the framework of a noisy-performance ranking model, the conventional and reverse models di¤er only in the noise on players’performance. Our study therefore reveals the important role that the noise term plays in modeling imperfectly discriminatory contests. JEL Nos: C72; D72; D74. Keywords: Reverse Nested Lottery Contest; Multistage Contest; E¤ort Maximization; Optimal Contest Design.

We thank Atsu Amegashie, Subhasish Chowdhury, Qiang Fu, Sang-Hyun Kim, Luisa Herbst, Dan Kovenock, Vai-Lam Mui and Aner Sela for insightful comments and suggestions. The usual disclaimer applies. y Jingfeng Lu: Department of Economics, National University of Singapore, 10 Kent Ridge Crescent, SINGAPORE, 119260; Tel: (65)65166026, Fax: (65)67752646, Email: [email protected]. z Bo Shen: Economics and Management School, Wuhan University, Luojia Hill, Wuhan, CHINA, 430072; Tel: (86)2768753199, Email: [email protected]. x Zhewei Wang: School of Economics, Shandong University, 27 Shanda Nanlu, Jinan, CHINA, 250100; Tel: (86)53188369992, Fax: (86)53188571371, Email: [email protected].

1

Introduction

Contests have been widely employed in practice to elicit productive e¤orts from agents. As pointed out by Gradstein and Konrad (1999), e¤ort maximization has long been adopted in the contest literature as one of the primary goals for optimal design. Multistage eliminations have been well studied as an e¤ective way of inducing more e¤ort from contestants.1 A well-received notion is that when the contest is su¢ ciently noisy –or, in other words, the contest is not overly discriminatory –multistage eliminations would elicit more e¤ort from contestants than single-stage competition. While adopting the multi-prize nested lottery contest technology of Clark and Riis (1996),2 Gradstein and Konrad (1999) established the optimality of a winner-take-all (i.e., only the winner receives a single prize and others get nothing) multistage elimination contest with a tree structure: In each stage, all remaining players are divided into groups of two contestants who …ght for one entry ticket to the next round of competition.3 Under the same contest technology, Fu and Lu (2012a) further showed that a winner-take-all multistage elimination contest with a pyramid structure elicits the highest total e¤ort from players.4 In this paper, we show that these wellestablished insights may vary substantially when the prevailing contest technology instead takes the form of the “reverse”nested lottery contests proposed by Fu, Lu and Wang (2014). The “conventional”model of Clark and Riis (1996) and the “reverse”model of Fu et al. (2014) can be integrated into a uni…ed framework from a perspective of noisy performance ranking. On one hand, Fu and Lu (2012b) show that the conventional model of Clark and Riis (1996) is equivalent to a noisy-performance ranking model in which players with higher perceivable outputs win higher prizes. In the equivalent noisy-performance model, an arbitrary player i’s perceivable output yi is modeled as a product of a deterministic component f (ei ) that only depends on his e¤ort level ei and an exogenous noise component "i that follows a Weibull (maximum) distribution, i.e., yi = f (ei ) "i . On the other hand, Fu et al. (2014) show that the reverse model is equivalent to the same noisy-performance ranking model, with the only di¤erence being that the exogenous noise "i instead follows a Weibull (minimum) distribution.5 1

Sisak (2009) provides a comprehensive survey on multi-prize contests. The nested lottery contest model is one of the most prominent frameworks to model multiprize contests, which has been applied by Clark and Riis (1996, 1998), Amegashie (1999, 2000), Yates and Heckelman (2001), Szymanski and Valletti (2005), Fu and Lu (2009, 2012a), Schweinzer and Segev (2012), etc. 3 Adopting a similar framework, Rosen (1986) shows that to maintain an unchanged individual e¤ort level across stages, the …nal-stage prize needs to be disproportionately large compared to early-stage prizes. 4 A pyramid structure means that in each stage all remaining players …ght together in a grand contest, and only one player is eliminated. 5 This noisy-performance model with Weibull (minimum) distribution for the shocks is …rst proposed by Hirshleifer and Riley (1992) in a two-player environment. 2

2

As illustrated by Fu and Lu (2012b) and Fu et al. (2014), the conventional (resp. reverse) model can be interpreted as a best-shot (resp. worst-shot) contest, in which a contestant’s performance is determined by the most (resp. least) favorable performance from multiple independent experiments, and one can increase (resp. decrease) the number of experiments by exerting more e¤ort. Therefore, the reverse model depicts a variety of competitive activities that are di¤erent from those captured by the conventional model. In particular, the reverse model is more suitable for depicting competitive activities in which players’ performance is judged by how closely they meet prede…ned standards or how closely they follow codes of practice. Real-world examples include acrobatics, diving, and many skilled competitions across di¤erent professions. Nevertheless, Lu and Wang (2015, 2016) show that both the conventional and reverse nested lottery contest models can be axiomatized based on two parallel (and equally elegant) sets of axioms.6 Therefore, the perspective from both noisy-performance ranking and axiomatization shows that the reverse model is as reasonable and justi…able as the conventional model in modeling multi-prize imperfectly discriminatory contests. The conventional model of Clark and Riis (1996) can be described based on the following ratio-form Contest Success Function (CSF): With I contestants, the likelihood that contestant i is selected as the winner is given by f (ei ) ; pi (e1 ; e2 ; :::; eI ) = PI f (e ) j j=1

(1)

where ej denotes contestant j’s e¤ort and the impact function f ( ) is increasing in its argument.7 Based on this CSF, the conventional model is implemented through the following (hypothetical) sequential lottery process: Players are selected one by one (from the highest ranking to the lowest ranking) in a sequence of random draws by applying CSF (1) successively among the remaining contestants and using contestants’one-shot e¤ort entries.8 In a parallel fashion, the reverse nested lottery contest model can be described based on the notion of the following ratio-form Contest Elimination Function (CEF), which gives the probability of contestant i being selected as the loser among I contestants: [f (ei )] 1 qi (e1 ; e2 ; :::; eI ) = PI j=1 [f (ej )]

1

;

(2)

where ej and f ( ) are as previously de…ned. As a result, one reduces his losing likelihood by stepping up his e¤ort. Based on this CEF, the reverse model is implemented literally 6

These axioms are adapted from those of Skaperdas (1996). The ratio-form CSF is …rst proposed by Tullock (1980) in his seminal work. 8 Note that the seemingly sequential lottery procedure is hypothetical, as each player makes a one-shot e¤ort in the entire selection process; in other words, e¤ort is not made sequentially. 7

3

through the following (hypothetical) sequential lottery process: Players are selected one by one (from the lowest ranking to the highest ranking) in a sequence of random draws by applying the CEF (2) successively among the remaining contestants and using contestants’ one-shot e¤ort entries. Chowdhury and Kim (2014) also present a variant of the Clark and Riis (1996) model by considering a multi-winner nested elimination contest in which losers are sequentially eliminated to obtain the set of winners. They show that their model is equivalent to the model of Berry (1993) with symmetric players. In this paper, we study the e¤ort-maximizing contest design under the reverse technology, while accommodating multistage eliminations, multigroup competitions within each stage, and intermediate prizes in each stage. The scope of eligible structures accommodates those of Gradstein and Konrad (1999) and Fu and Lu (2012a), as we allow subgroups of heterogeneous sizes in each stage, full ‡exibility in the number of subgroups in each stage and the numbers of those being shortlisted (who advance to the next stage) from each subgroup. Quite surprisingly, our …nding is contrary to the conventional wisdom on the optimality of multistage elimination contests, which has been established under the conventional nested lottery contest technology. We …nd instead that with the reverse technology, a single-stage winner-take-all grand contest dominates all other eligible designs, provided that the contest is su¢ ciently noisy. Our results therefore show that, in a variety of competitive activities that are better depicted by reverse technology, a single-stage winner-take-all grand contest dominates any multistage elimination contest. On the other hand, for a set of exogenously given prizes that are awarded simultaneously in a single stage, we …nd that a grand contest dominates any divided contest of a set of subcontests. This result suggests that the optimality of a grand contest among all eligible single-stage contests, as established by Fu and Lu (2009) under the conventional nested lottery technology, remains valid in the reverse contest environment.9 We …rst illustrate the intuitions behind the result of dominance of a grand contest over a divided contest consisting of a number of subcontests in a single-stage competition with a set of exogenously given prizes. Since the equilibrium e¤ort is linear in every prize, there is no loss to focus on the case in which all prizes are homogeneous.10 Suppose there are k (< I) homogeneous (strictly) positive prizes to be awarded to I players in a single-stage contest such that there are I k zero prizes in the grand contest, which includes all players and prizes. Players and prizes can be arbitrarily divided into a number of subcontests. Note 9

Chowdhury and Kim (2017) show that the result can be reversed if the contest technology of Berry (1993) or, equivalently, that of Chowdhury and Kim (2014), is adopted. 10 Note that a sequence of k (< I) heterogeneous prizes can be decomposed into k sequences of homegeneous prizes with di¤erent lengths.

4

that no matter how the subcontests are formed, in each there are at most I k zero prizes. Recall the hypothetical sequential (loser) elimination interpretation of a single-stage reverse nested lottery contest. In a grand contest, each player’s e¤ort outlay remains to be e¤ective to help the player survive the …rst I k draws for zero prizes, while in each subcontest, each player’s e¤ort outlay can remain e¤ective for at most I k draws for zero prizes. From this perspective, players’e¤ort delivers a higher return in the grand contest, which in turn leads to a higher overall e¤ort. We now reconcile the divergence of optimality of multistage elimination contests between the conventional and reverse nested lottery contest technologies. For ease of discussion, let us …x winner-take-all for prize allocation in all feasible contest structures, and compare a singlestage grand contest and a two-stage elimination contest. With conventional technology, Gradstein and Konrad (1999) and Fu and Lu (2012a) show that when the contest is not overly discriminatory, adding an additional stage of elimination would de…nitely enhance total e¤ort supply from the players. The reason is that the additional e¤ort induced from the extra stage dominates the decrease in e¤ort supply from earlier stages. What happens, then, when reverse technology is used instead? Fu et al. (2014) have shown that, due to the hypothetical loser-elimination nature of the reverse contest (compared to the hypothetical winner-selection nature of the conventional contest), it is much more e¢ cient for eliciting e¤ort than the conventional contest when the number of prizes (k) is small or the number of players (I) is large.11 This means that compared to the conventional contest, adding an extra stage would add relatively less signi…cant amount of e¤ort, but a more signi…cant drop in e¤ort supply from the …rst stage would result because of the decreased prize purse and increased number of prizes (i.e., the entry rights to the second stage). It is thus feasible –and indeed is the case – that the added e¤ort supply from a second stage would be dominated by the decrease in e¤ort supply from the …rst stage. As previously mentioned, from the noisy-performance-ranking perspective, the conventional and reverse models di¤er only in the noise on players’performance. The contrasting divergence in the optimal contest structures across the two models suggests that the noise term plays a crucial role in determining the optimal contest. Our study therefore also reveals the important role that the noise term plays in imperfectly discriminatory contests, which has so far been largely overlooked in the literature. 11

In a conventional contest with k homogeneous prizes and I contestants, one strives to improve the likelihood of being picked within the …rst k lotteries to win a prize. Only the outcomes of the …rst k lotteries a¤ect his payo¤s. By contrast, a contestant in a reverse contest strives to survive I k rounds of elimination to win a prize. His payo¤ is a¤ected by the outcomes of the I k draws. When k gets smaller and/or I gets larger, a contestant expects a higher marginal bene…t from his e¤ort in a reverse contest than his counterpart in a conventional one: One’s e¤ort in the former could remain “e¤ective”for more rounds than in the latter. Higher marginal bene…ts thus compel contestants to contribute more e¤ort.

5

2

Contest Design under Reverse Lottery Technology

In this section, we investigate the optimal contest structure in a general setup under reverse nested lottery technology. In Section 2.1, we formally propose a general framework of contest organizational structure.12 We further discuss conditions under which a pure-strategy subgame perfect equilibrium exists when the reverse model is in place in each battle throughout the entire contest. In Section 2.2, we show that with single-stage competition, a winnertake-all grand contest in which all players compete together in one contest for a single prize dominates any divided contest composed of a set of subcontests. In Section 2.3, we extend our analysis to allow full generality in our framework with multistage eliminations. We …rst show that any pyramid contest of Fu and Lu (2012a) is dominated by a single-stage winnertake-all grand contest. Finally, we establish the optimality of the single-stage winner-take-all grand contest by showing that any contest of a general structure is dominated by the pyramid contest of Fu and Lu (2012a).

2.1

Framework of Contest Structure

We now set up our general framework of contest structure with the reverse nested lottery technology in each battle. There are N ( 3) symmetric risk-neutral contestants. In this paper we focus on the case with symmetric contestants, which is analogous to Gradstein and Konrad (1999) and Fu and Lu (2012a). The organizer has a total budget of 0 for prize allocation. Let L denote the number of stages in the contest, and Nl denote the number of contestants in a stage l 2 f1; 2; :::; Lg, with N1 N . In each stage l 2 f1; 2; :::; L 1g, Nl contestants participate, and Nl+1 of them survive and proceed to the next stage. In stage l, there are sl ( 1) subcontests denoted by set sl = f1; 2; :::; sl g. Let Nlg be the number of contestants involved in a subcontest g 2 sl , and Klg ( 1) be the number of contestants in P subcontest g who will proceed to stage (l +1), where Klg Nlg . We thus have g2sl Nlg = Nl P and g2sl Klg = Nl+1 . We assume that all contestants who are shortlisted for a stage l are randomly allocated to the stage subcontests. In each stage l 2 f1; 2; :::; L 1g, in addition to the “tickets”to the next round, in each subcontest g 2 sl ; Nlg intermediate prizes Wlg;m , m 2 f1; 2; :::; Nlg g are awarded, which are nonnegative and decreasing.13 In the …nal stage L, again NL surviving 12

The proposed framework covers, as special cases, the organization structures adopted in Gradstein and Konrad (1999), Amegashi (1999), and Fu and Lu (2009, 2012a). g;N g

Intermediate prizes Wlg;m being decreasing and nonnegative implies Wlg;1 Wlg;2 Wl l 0. It is straightforward to show that for maximizing expected total e¤ort, it is never optimal to set intermediate prizes which are not decreasing. Note that with intermediate prizes being (entirely or partially) increasing, a player may receive a larger prize when he is ranked lower. 13

6

contestants are allocated into sL subcontests; and in subcontest g 2 sL , NLg contestants compete for NLg nonnegative …nal prizes WLg;m , m 2 f1; 2; :::; NLg g. In each subcontest g of each stage l, each contestant is eligible for at most one prize Wlg;m . Note that we allow some of the intermediate prizes and …nal prizes to be zero. Denote the total prizes allocated in PNlg stage l and subcontest g 2 sl by Wlg = m=1 Wlg;m , and the total prizes allocated in stage P l by Wl = g2sl Wlg . The elimination sequence of a contest is therefore represented by an L-term nonincreasing sequence fN1 ; N2 ; :::; NL g with N1 = N N2 ::: NL 1, which indicates the number of stages and the number of competing contestants in each stage. And in each stage l, the structure of subcontests is presented by sets fNlg gg2sl and fKlg gg2sl , which indicate the numbers of competing contestants and survivors who proceed to the next round in each subcontest. Therefore, the contest organizer can take control of the following instruments: number of stages (L); number of contestants in each stage (Nl ); and within each stage l, the number of subcontests (sl ), number of contestants in each subcontest (Nlg ), number of contestants who proceed to the next stage (Klg ) and intermediate prize sequence {Wlg;m } in each subcontest. The following budget constraint on prize structure is to be observed: X

l2f1;2;:::;Lg

Wl =

X

X

l2f1;2;:::;Lg

g2sl

XNlg

m=1

Wlg;m

0:

In subcontest g 2 sl of stage l, the Klg entry rights and Nlg intermediate prizes are awarded according to a reverse nested lottery technology. The reverse nested lottery contest literally employs a (hypothetical) sequence of independent lotteries to award the prizes. Contestants g in subcontest g 2 sl simultaneously exert their one-shot e¤orts eg;i l , i = 1; 2; :::; Nl . Each lottery picks one loser. This player is eliminated from the pool of candidates who are eligible for subsequent draws. The next loser is picked from the remaining pool. The procedure repeats for Nlg 1 rounds to determine the ranks of all Nlg contestants, and the players being eliminated earlier are ranked lower. Contestants are awarded the intermediate prizes and the entry rights by their ranks, which are determined by the outcomes of sequential draws. De…ne g;m to be the set of remaining contestants up for the mth draw in subcontest l g;Nlg g in stage l, with m 2 f1; 2; :::; Nlg g. Denote contestants’e¤ort vector (eg;1 ) by l ; :::; :::; el g;m g;m g el . Let #(jjej = 0, j 2 l ) be the count of players with zero e¤ort in l . And de…ne 1 f eg;j as in…nity for any j 2 f1; 2; :::; Nlg g. The conditional probability that a l = 0

7

contestant i 2

g;m l

is selected in the mth draw is given by the following CEF: 8 i 1 h g;i r > > e > l > g;i > 0; h r eg;j )= qi (egl j g;m l l j2 g;m > l > > 1 > > if eg;i : g;m l = 0; #(jjej = 0; j 2 l )

(3)

where r > 0.14 Notice that in CEF (3), the impact function f (e) = er increases with one’s own e¤ort. As a result, ceteris paribus one reduces his probability of losing by increasing his e¤ort. Analogous to that in the Tullock CSF, the magnitude of r in (3) measures the noisiness of the selection process of the contest (a higher r means that the contest is less noisy). We assume that all contestants have the same linear cost function c(ei ) = ei .15 De…ne r

1+

NP1 j=0

1 N

j

!

1

.

(4)

By (4) it can be shown that r 2 as r = 2 when N = 2 and r decreases when N increases. The conventional wisdom holds that a more discriminatory contest, i.e., a larger r, leads to a higher rate of rent dissipation. Hence, a moderate r is required to rein in contestants’ e¤ort incentives, thereby maintaining the participation constraints in the game. The logic applies in both conventional and reverse lottery contests. In a reverse contest, a pure-strategy equilibrium exists when r 2 [0; r] and the rent dissipation rate is increasing from 0 to 1 when r increases from zero to r. Notice that the rent is fully dissipated when r = r. Intuitively, a su¢ ciently large discrimination power r implies a su¢ ciently …erce competition, which leads to full rent dissipation, i.e., each player gets zero payo¤ in equilibrium. A parallel result holds in a conventional lottery contest setting: a pure-strategy equilibrium exists when r is smaller than a threshold and full rent dissipation is realized when r reaches the threshold. However, when r gets larger than the threshold, a symmetric pure-strategy equilibrium does not exist. One can reasonably guess that when r > r, mixed-strategy equilibria may exist in which the rent is also fully dissipated, as intuitively, a larger r implies …ercer competition which Note that qi (egl j g;m ) = 1= [#(jjej = 0; j 2 g;m )] if eg;i = 0 implies that if there are n ( 1) players l l l exerting zero e¤ort (with other players’e¤ort being strictly positive), then each of them has a probability of 1=n of being eliminated. 15 We can allow for a more general cost function c(e), and this setting is strategically equivalent to an alternative contest in which all contestants have the same linear cost and the impact function is h( ) = g(c 1 ( )). Following previous studies (Gradstein and Konrad, 1999; Fu and Lu, 2009, 2012a), we assume that players are symmetric. 14

8

should lower each player’s equilibrium payo¤. The following proposition partially con…rms this conjecture when r 2.16 Proposition 1 When r 2, there is an asymmetric mixed strategy equilibrium for a singlestage winner-take-all grand contest with reverse lottery technology, in which only two players are active and the rent is fully dissipated. Proof. See Appendix. When r 2 (r; 2), one can expect that a similar result should hold, though there is no existing work con…rming this due to the di¢ culties in characterizing the equilibria. In this study, we thus focus on the cases with r < r in the subsequent analysis such that a purestrategy subgame perfect equilibrium exists.17 Notice that when r < r, the rent is not fully dissipated in a single round grand contest such that the design issue is meaningful.

2.2

Optimal Design with Single-Stage Competition

In this subsection, we consider a special case of the general model: a single-stage contest, i.e., L = 1. This setup is similar to the model in Fu and Lu (2009). However, while in Fu and Lu (2009) the set of prizes is exogenously chosen, in our setting the contest organizer can endogenously choose the prize structure. Without loss of generality, we assume that N players compete for N prizes denoted by W = (W1 ; W2 ; :::; WN ); with W1 W2 ::: WN 0. Recall that Wk is the prize awarded to the player being ranked the kth highest by their noisy performance –or, in other words, the (N k + 1)th player being eliminated by the hypothetical sequential draws. We can allow all the players to compete together in a grand contest, or they can split into M 2 PM subcontests, in which Nm players compete in subcontest m, where N = m=1 Nm , and Nm 1, 8m 2 f1; :::; M g. In subcontest m, a contestant competes against Nm 1 opponents for prizes Wm , where Wm , (W1m ; W2m ; :::; WNmm ) with W1m W2m ::: WNmm 0. Every element of Wm must be in W, and every element of W must go to one and only one in Wm . From Fu et al. (2014), when N players compete in a grand contest, we can derive that, 16

In an all-pay auction setting (equivalent to a conventional Tullock contest with r ! +1), Baye, Kovenock and de Vries (1994)’s results show that with symmetric players there exists a mixed-strategy equilibrium where rent is fully dissipated. By Alcalde and Dahm (2010), it can be shown that in a conventional Tullock contest with symmetric players and r 2, an “all-pay auction equilibrium” exists where rent is also fully dissipated. Proposition 1 establishes a parallel result in our setting. 17 The conditions under which a pure-strategy subgame perfect equilibrium exists have been discussed in detail by Fu et al. (2014).

9

in the symmetric pure-strategy equilibrium, each contestant makes an e¤ort e=

r PN WN +1 N k=1

1+

k

Pk

1 j=0

1 N

(5)

:

j

It can be shown that with decreasing prizes, e is strictly positive as long as W1 > WN . The total e¤ort the N contestants make is then given by EG = r

PN

WN +1

k=1

1+

k

Pk

1 j=0

1 N

(6)

:

j

When the grand contest is split into M 2 subcontests, we denote by Em the total equilibrium e¤ort that the Nm contestants exert in subcontest m, with m 2 f1; :::; M g. Following the same procedure as above illustrated, we obtain Em = r

PNm

k=1

WNmm +1

1+

k

Pk

Nm

m

M, M

1 j=0

1 j

.

(7)

P We want to compare the total e¤ort induced from all subcontests ( M m=1 Em ) with the total e¤ort from the grand contest (EG ). The following is a key result, which facilitates this comparison. Lemma 1 Suppose N We have PM

m=1

PNm

k=1

2 and Nm

WNmm +1

k

Pk

1 j=0

1, where 1 1 Nm

j

PN

k=1

WN +1

2 and N =

k

where the strict inequality holds if and only if W1 > WN .

Pk

1 j=0

1 N

j

PM

m=1

;

Nm .

(8)

Proof. See Appendix. Following Lemma 1, we can easily obtain the following result.18 Proposition 2 The grand contest induces strictly more total e¤ort than any set of subconP m tests, i.e., EG > M 2. m=1 Em , 8fNm g; W; fW g; 8M

The above result shows that the dominance of the grand contest when a …xed set of prizes is to be awarded – which is …rst established by Fu and Lu (2009) for conventional nested lottery contests –continues to hold in reverse nested lottery contests. The contest organizer is able to further choose the prize allocation fWk g subject to P budget constraint N 0 . From (6), we know that the coe¢ cient of each prize Wk , k=1 Wk 18

The proof is omitted to save space. It is available from the authors upon request.

10

P i.e., 1 + kj=01 N1 j , is strictly increasing in k. Thus, it is optimal to allocate the entire prize purse to the …rst prize, i.e., W1 = 0 , and Wk = 0, for 8k 2 f2; :::; N g. Therefore, the single-stage winner-take-all contest dominates all other single-stage contests. We formally present this result in the following Proposition. Proposition 3 The single-stage winner-take-all grand contest dominates any divided singlestage contest that consists of a number of subcontests.19

2.3

Optimal Design Allowing Multistage Eliminations

In this subsection, we consider the general contest structure in which the contest organizer is able to fully accommodate multistage eliminations, multiple subcontests in each stage, and multiple intermediate prizes in each stage. We seek to establish that a single-stage winner-take-all grand contest remains to dominate any other contest structure. For this purpose, we …rst extend the analysis of Section 2.2. Suppose now that N players compete for two sets of prizes W and w, which are awarded independently, where w = (w1 ; w2 ; :::; wN ), with w1 w2 ::: wN 0, and W is de…ned as in Section 2.2. Thus, the total prize for the kth best performer is Wk + wk . Introducing prizes w is for the purpose of accommodating the entry rights that must be involved in multistage eliminations. Similarly, the grand contest can be split into M 2 subcontests, and in each subcontest m m = 1; :::; M , two sets of prizes Wm and wm are awarded, where wm , (w1m ; w2m ; :::; wN ) m m m m m with w1 w2 ::: wNm 0 and W is de…ned as in Section 2.2. Every element of m w must be in w, and every element of w must go to one and only one in wm . When w1 = w2 = :::wn = 0, we go back to the same setup as in Section 2.2. Similarly, we de…ne the total e¤ort the N contestants make in the grand contest by bG = r PN (WN +1 E k=1

+ wN +1 k )

k

1+

Pk

1 j=0

1 N

(9)

;

j

bm the total e¤ort the Nm contestants exert in each subcontest m, with and denote by E m 2 f1; :::; M g, where bm = r PNm E k=1

WNmm +1

k

m + wN m +1

k

1+

Pk

Following Proposition 2, we have the following corollary. bG > PM E b Corollary 1 E m=1 m , 8M 19

2.

The intuitions behind this result are provided in both Section 1.

11

1 j=0

1 Nm

j

.

(10)

To establish the optimal contest structure in a general setup, we rely on an important result: Any multistage contest is dominated by a pyramid contest in which each stage of competition is organized as a grand battle within a single group, as de…ned by Fu and Lu (2012a). This result is to be established in the following Proposition 4, using the result obtained in Corollary 1. Proposition 4 Any general contest with multiple subcontests in some stage(s) is dominated by a pyramid contest in which in each stage, all shortlisted contestants compete together in a grand contest. Proof. See Appendix. The basic idea to prove Proposition 4 is as follows: First, suppose we combine all subcontests into a grand contest in stage 1 while keeping everything else unchanged. Following Corollary 1, this would induce a higher level of total e¤ort in stage 1 without a¤ecting the total e¤ort induced in other stages. Therefore, the total e¤ort induced in the whole contest is higher. Second, …xing a grand contest in stage 1, if we combine all subcontests into a grand contest in stage 2 and keep everything else unchanged, we show that the total e¤ort induced in the …rst two stages increases, which would lead to a higher level of total e¤ort in the whole contest. Iterating this procedure until the last stage, we show that any general multistage contest with multiple subcontests in some stage(s) is dominated by a pyramid contest (in which in each stage, all remaining players compete in a grand contest). Clearly, the pyramid contest is a special case of the general structure with sl = 1 for all l 2 f1; 2; :::; Lg. For sl = 1, we can drop the superscript g from all of the notations used in the general setup for simplicity. For a pyramid contest, in each stage l 2 f1; 2; :::; L 1g, Nl contestants participate in a grand contest, and Nl+1 of them survive and proceed to the next stage. The Nl nonnegative intermediate prizes awarded in stage l are denoted by Wlm , m 2 f1; 2; :::; Nl g. We now are ready to pin down the optimal contest structure by restricting our search to all pyramid contests. Note that in Section 2.1, we discussed the existence of subgame perfect equilibrium in pure strategy. We now solve this equilibrium through backward induction. Let Vl be the conditional equilibrium expected payo¤ of a representative contestant at stage l. De…ne VL+1 0; NL+1 1. At stage l 2 f1; 2; :::; Lg, a representative contestant i chooses his e¤ort eil to maximize his expected payo¤ Nl+1 i

Vl =

X

Pm (eil ; el i )(Vl+1

+

Wlm )

m=1

+

Nl X

m=Nl+1 +1

12

Pm (eil ; el i )Wlm

eil ;

(11)

where Pm (eil ; el i ) is the probability that contestant i is selected in the (N + 1 m)th draw and thus wins the mth prize. For ease of notation, we de…ne ! ! Nl+1 NX NX Nl l m l m X X 1 1 1+ 1+ (Vl+1 + Wlm ) + Wlm : (12) l N g N g l l g=0 g=0 m=1 m=N +1 l+1

PNl m We also de…ne l m=1 Wl to be the sum of prizes awarded in stage l, and to be the entire prize purse for the contest.

PL

l=1

l

Lemma 2 In the symmetric subgame perfect Nash equilibrium of the pyramid contest, every contestant in stage l 2 f1; 2; :::; Lg exerts an e¤ort el = r Nll , where l is given by (12),20 and the payo¤ of a representative contestant at stage l is Vl =

Nl+1 Vl+1 + Nl

l

(13)

el :

Proof. See Appendix. Assume that the contest organizer chooses the optimal sequence fNl gLl=1 and prize alloP cation fWlm j m = 1; :::; Nl ; l = 1; :::; Lg to maximize the total e¤ort E = Ll=1 Nl el , subject to 0 . From (13) in Lemma 2, we obtain Nl el = Nl+1 Vl+1 +

l

(14)

Nl Vl :

Summing up (14) over the L stages gives E

L X l=1

Nl el =

L X

(Nl+1 Vl+1

Nl Vl ) +

L X

l

=

N V1 .

l=1

l=1

Thus, we obtain E=

N V1 :

(15)

Note that V1 represents the payo¤ each contestant expects in the very beginning from the entire series of competitions. Thus, N V1 represents the total surplus of all N contestants, which is equivalent to the di¤erence between the prize purse and the total e¤ort E. Therefore, from the contest organizer’s perspective, given and N , maximizing total e¤ort E is equivalent to minimizing V1 . 20

It can be shown that l 0 in the general model where the intermediate prizes Wlm are assumed to be nonnegative and decreasing, 8l 2 f1; 2; :::; Lg.

13

Lemma 3 The expression of V1 is given by 82 0 13 " Nj Nj+1 1 Nl L < Y l 1 X X 1 A5 X 1 4 @ 1+r V1 = 1 r : N Nj g g=0 m=1 l=1

r

NX l m g=0

j=1

1 Nl

g

!

Wlm

#9 = ;

:

(16)

Proof. See Appendix. Denote the weight on Wlm in (16) by Dlm , where 2 0 13 Nj Nj+1 1 l 1 Y X 1 4 @ 1 A5 Dlm 1 r 1+r N j=1 Nj g g=0

r

NX l m g=0

1 Nl

g

!

.

PNj Nj+1 1 1 Q Within stage l, coe¢ cients Dlm contain the common component N1 lj=11 1 r g=0 Nj PNl m 1 and di¤er in the other parts 1 + r r g=0 Nl g , which obviously increase in m. We thus have Dlm increases in m. Therefore, to minimize V1 , all the prize money available in stage l should be allocated to the …rst prize Wl1 at stage l, and Wlm = 0, 8m > 1. Therefore, (16) can be rewritten as 82 0 13 Nj Nj+1 1 L < Y l 1 X X 1 1 A5 4 @1 r V1 = 1+r N l=1 : j=1 N g j g=0

r

N l 1 X g=0

1 Nl

g

!

Wl1

9 = ;

:

(17)

To determine how to allocate the prize purse 0 across stages, we …rst de…ne the weight on Wl1 by l where 2 0 13 ! Nj Nj+1 1 N l 1 l 1 Y X X 1 A5 1 1 4 @ 1 r 1+r r . l N j=1 Nj g Nl g g=0 g=0

Then V1 can be rewritten as

V1 =

L X

1 l Wl ,

(18)

l=1

P where = l for all l 2 f1; 2; :::; Lg; thus, Ll=1 Wl1 = . In the Appendix, by comparing l across stages, the following result can be obtained. Wl1

Proposition 5 Within the framework of pyramid contests, to maximize the total e¤ort, the optimal prize allocation should be W11 = 0 – that is, the single-stage winner-take-all grand contest dominates all multistage pyramid contests. Proof. See Appendix. Combining Propositions 4 and 5, we therefore establish the main result of this paper: Theorem 1 A single-stage winner-take-all grand contest maximizes the total e¤ort. 14

g

,

3

Concluding Remarks

In this paper, we study the optimal contest design of multistage eliminations under the technology of reverse nested lottery contests. Contrary to previous …ndings, which are based on the conventional nested lottery contest technology, we demonstrate that a single-stage (winner-take-all) grand contest always dominates any multistage contest when a reverse nested lottery contest with su¢ cient noise is in place in every competition. Conventional and reverse nested lottery contest models depict di¤erent competitive activities, while the perspective from both noisy-performance ranking and axiomatization reveals that both are equally reasonable models for imperfectly discriminatory contests. Our study suggests that predictions from the two models can diverge dramatically. Therefore, many issues concerning contest design under conventional nested lottery contests can be re-examined in settings with reverse nested lottery contests. Our result also illustrates the importance of empirical studies on identifying prevailing contest technologies, by alerting the designer to carefully examine the contest technology in place when choosing the optimal contest structure. From this perspective, our study reemphasizes the need to conduct empirical work on identifying the contest technologies that apply in di¤erent environments.

4 4.1

Appendix Proof of Proposition 1

When there are two players, i.e. N = 2, a single-stage winner-take-all grand contest under the reverse-lottery technology exactly coincides with the conventional Tullock contest. Following Alcalde and Dahm (2010), when r 2, there exists a symmetric mixed-strategy equilibrium ( 1 ; 2 ), in which player i adopts the mixed strategy i , where i = for i = 1; 2. In this equilibrium, an arbitrary player i’s expected e¤ort is E( i ) = V =2, and the rent is fully dissipated. We seek to show that in the reverse Tullock contest setting, when N > 2, there also exists an all-pay auction equilibrium in which two players (say players 1 and 2) adopt the mixed strategy 1 and 2 , and other players exert zero e¤ort. To show the above strategies constitute an equilibrium, we only need to check that any player j > 2 can not do better than just exerting zero e¤ort and obtaining zero expected payo¤. Without loss of generality, we focus on j = 3 by symmetry. Let pi (e1 ; e2 ; : : : ; eN ) be the probability of player i winning the grand prize V .

15

We …rst show the following property (P1): for any e1 ; e2 ; e3 p2 (e1 ; e2 ; e3 ; 0; : : : ; 0) | {z }

0 and ej = 0 for j > 3,

p2 (e1 ; e2 ; 0; : : : ; 0): | {z }

N 3

N 2

Clearly, the above inequality holds when e3 = 0, or e1 e2 = 0. Next, we consider the case e1 ; e2 ; e3 > 0. Note that we have e r p2 (e1 ; e2 ; 0; : : : ; 0) = r 1 r ; | {z } e1 + e2 N 2

and

e1 r e1 r + e2 r + e3 r

p2 (e1 ; e2 ; e3 ; 0; : : : ; 0) = | {z } N 3

+

e3 r e2 r + e3 r

e3 r e1 r + e2 r + e3 r

e1 r e1 r + e2 r

:

Then we have p2 (e1 ; e2 ; 0; : : : ; 0) | {z }

p2 (e1 ; e2 ; e3 ; 0; : : : ; 0) | {z }

N 2

=

N 3

e1 r e2 r > 0: (e1 r + e2 r + e3 r )(e2 r + e3 r )

To establish the all pay auction equilibrium for r 2 in our setting, consider any pure strategy e0 chosen by player 3. In the symmetric mixed strategy equilibrium when N = 2, 0, i.e. EU2 ( 1 ; e0 ) 0. given 1 , player 2 obtains non-positive expected payo¤ for any e0 For N > 2, we must have EU2 ( 1 ; e0 ; 0; : : : ; 0) = EU2 ( 1 ; e0 ): | {z } N 2

Moreover, by symmetry,

EU3 ( 1 ; 2 ; e0 ; 0; : : : ; 0) = EU2 ( 1 ; e02 ; 2 ; 0; : : : ; 0): | {z } | {z } N 3

N 3

Further, following (P1) we have

EU2 ( 1 ; e0 ; 2 ; 0; : : : ; 0) | {z } N 3

which implies EU3 ( 1 ; 2 ; e0 ; 0; : : : ; 0) | {z }

EU2 ( 1 ; e0 ; 0; : : : ; 0) | {z }

0;

N 2

0, i.e., player 3 has no incentive to deviate.

N 3

16

4.2

Proof of Lemma 1

To prove Lemma 1, it is su¢ cient to prove inequality (8) for the case of M = 2, i.e., PN1

WN1 1 +1

k=1

PN1 +N2

k

Pk

WN1 +N2 +1

k=1

1

1 j=0

PN2

WN2 2 +1

k=1

N1

j

Pk

1 N1 + N2

1 j=0

k

+

k

Pk

1 j=0

1 N2

j

.

j

(19)

Note that (19) is equivalent to PN1

1

1

t=0

N1

PN1 +N2 t=0

1

t

PN1

k=t+1

1 N1 + N2

WN1 1 +1

t

+

k

PN1 +N2 k=t+1

PN2

1

1

t=0

N2

WN1 +N2 +1

k

t

PN2

k=t+1

WN2 2 +1

k

(20)

.

2 Consider the following three increasing sequences of nonnegative prizes: fWN2 2 +1 k gN k=1 , N1 +N2 1 fWN1 1 +1 k gN . We observe the following properties. k=1 , and fWN1 +N2 +1 k gk=1

P(i) minf P(ii) minf

1 PN2 1 PN1 1 W2 g k=1 WN1 +1 k ; N1 N2 k=1 N2 +1 k

PN 1 WN1 +N2 +1 k ; N1 + N2 k=1

PN 1 PN1 1 PN2 1 1 2 WN1 +N2 +1 k=1 WN1 +1 k ; k=1 WN2 +1 k g = N1 N2 N1 + N2 k=1 i¤

1 PN1 W1 N1 k=1 N1 +1

k

=

k

1 PN2 W2 : N2 k=1 N2 +1 k

We attempt to establish a one-to-one matching between each term on the LHS of (20) to a term on its RHS such that each component of the RHS is greater than its counterpart on the LHS. The proof is as follows. P 1 PN2 1 1 2 Without loss of generality, we assume that N11 N k=1 WN1 +1 k k=1 WN2 +1 k . By N2 the above assumption and P(i), we must have 1 PN2 W2 N2 k=1 N2 +1

PN1 +N2 1 WN1 +N2 +1 k ; N1 + N2 k=1 P 2 N2 1 2 that is, the …rst term of sequence f N21 t N is smaller than the …rst term k=t+1 WN2 +1 k gt=0 P N1 +N2 N1 +N2 1 1 in the sequence f N1 +N2 t k=t+1 WN1 +N2 +1 k gt=0 . P P 2 N N2 1 2 2 WN2 2 +1 k (second term in f N21 t N Next, we compare either N21 1 k=2 k=t+1 WN2 +1 k gt=0 ) P P P N1 N1 +N2 N1 1 1 1 1 1 1 or N11 N WN1 +N2 +1 k=1 WN1 +1 k (…rst term in f N1 t k=t+1 WN1 +1 k gt=0 ) to N1 +N2 1 k=2 P N1 +N2 N1 +N2 1 1 (second term in f N1 +N2 t k=t+1 WN1 +N2 +1 k gt=0 ). k

17

k

Note that because WN2 2 PN1 +N2 k=2

which gives 1 N1 + N2

1

PN1 +N2 k=2

WN1 +N2 , we must have

WN1 +N2 +1

WN1 +N2 +1

k

k

PN1

k=1

WN1 1 +1

1 N1 + N2

k

+

PN2

k=2

PN1

k=1

1

WN2 2 +1 k ;

WN1 1 +1

k

+

PN2

k=2

WN2 2 +1

k

: (21)

By P(i), we have minf

1 PN1 1 PN2 1 W2 g k=1 WN1 +1 k ; N1 N2 1 k=2 N2 +1 k

1 N1 + N2

1

1 N1 + N2

1

PN1

k=1

WN1 1 +1

PN1 +N2 k=2

k

+

PN2

k=2

WN2 2 +1

k

WN1 +N2 +1 k .

(22)

P 2 PN1 1 2 1 Therefore, we can match the smaller one between N21 1 N k=2 WN2 +1 k and N1 k=1 WN1 +1 k PN1 +N2 1 on the LHS of (20) to N1 +N WN1 +N2 +1 k on its RHS such that the former is no k=2 2 1 bigger than the latter. Suppose we have continued this practice for a total of n1 + n2 rounds. P m Nm 1 m Thus, each of the nm largest terms in sequence f Nm1 t N k=t+1 WNm +1 k gt=0 , m = 1, 2, has PN1 +N2 N1 +N2 1 1 . been matched to one of the n1 +n2 largest terms in sequence f N1 +N k=t+1 WN1 +N2 +1 k gt=0 2 t

We …rst consider the case nm < Nm , m = 1, 2. In this case, neither of the two sequences P m Nm 1 m f Nm t N k=t+1 WNm +1 k gt=0 , m = 1, 2, has been exhausted. Each leaves a subsequence P Nm 1 m m written as f Nm1 t N k=t+1 WNm +1 k gt=nm . The sequence on the RHS of (20) thus leaves PN1 +N2 N1 +N2 1 1 a subsequence f N1 +N k=t+1 WN1 +N2 +1 k gt=n1 +n2 . We match the smaller one between 2 t P P PN1 +N2 N1 N2 1 1 1 1 2 k=t+1 WN1 +1 k and N2 n2 k=t+1 WN2 +1 k to N1 +N2 n1 +n2 k=n1 +n2 +1 WN1 +N2 +1 k . N1 n1 We can derive 1

PN1 +N2

k=n1 +n2 +1

WN1 +N2 +1

k

which implies

1 N1 + N2

n1

n2

1 N1 + N2

n1

n2

PN1

k=n1 +1

PN1 +N2

k=n1 +n2 +1

PN1

k=n1 +1

18

WN1 1 +1

k

+

WN1 +N2 +1

WN1 1 +1

k

+

PN2

k=n2 +1

WN2 2 +1 k ,

k

PN2

k=n2 +1

WN2 2 +1

k

.

Thus, we conclude that minf

1 N1

n1 1

N1 + N 2

PN1

k=n1 +1

n1

N1 + N 2

n1

n2

N2

PN1

1 k=n1 +1 WN1 +1

n2

1

1

WN1 1 +1 k ;

PN1 +N2

k=n1 +n2 +1

n2 k

+

PN2

k=n2 +1

PN2

k=n2 +1

WN2 2 +1 k g

WN2 2 +1

k

WN1 +N2 +1 k ,

(23)

P 1 PN2 1 2 2 and we match the smaller one between N1 1 n1 N k=n1 +1 WN1 +1 k and N2 n2 k=n2 +1 WN2 +1 k P N +N 1 2 WN1 +N2 +1 k , which continues to preserve the domito the term N1 +N21 n1 n2 k=n 1 +n2 +1 nance. We iterate this procedure until one of the sequence is exhausted. Without loss of genP 1 N1 1 1 erality, we assume that after n e1 + N2 rounds, the sequence f N11 t N is k=t+1 WN1 +1 k gt=0 P N2 N2 1 1 2 not exhausted, while all the N2 terms of sequence f N2 t k=t+1 WN2 +1 k gt=0 have been P 1 N1 1 1 used. Thus, we need to match each term of the subsequence f N11 t N k=t+1 WN1 +1 k gt=e n1 PN1 +N2 N1 +N2 1 1 to a term of subsequence f N1 +N2 t k=t+1 WN1 +N2 +1 k gt=en1 +N2 . Note that the above two subsequence both have (N1 n e1 ) terms. P 1 N1 1 1 The h-th term in the sequence f N11 t N k=t+1 WN1 +1 k gt=e n1 is 1

N1

n e1

h+1

PN1

k=e n1 +h

WN1 1 +1

k

where 1

1 while the corresponding h-th term in the sequence f N1 +N 2

N1 + N2 =

1 (e n 1 + N2 + h

1 N1

n e1

h+1

1)

PN1 +N2

k=e n1 +N2 +h

PN1 +N2

k=e n1 +N2 +h

WN1 +N2 +1

k

N1

PN1 +N2 k=t+1

k=e n1 +N2 +h

n e1 ;

(24)

N1 +N2 1 WN1 +N2 +1 k gt=e n1 +N2 is

WN1 +N2 +1

k

(25)

WN1 +N2 +1 k :

P 1 1 1 1 1 Since N k=e n1 +h WN1 +1 k = W1 + W2 + ::: + WN1 +1 W1 + W2 + ::: + WN1 +1 (en1 +h) , we can see that PN1 +N2

t

h

(e n1 +h)

PN1

k=e n1 +h

and

PN1 +N2

k=e n1 +N2 +h

WN1 1 +1 k ,

WN1 +N2 +1

k

=

(26)

P 1 N1 1 1 which implies that the h-th term in the sequence f N11 t N k=t+1 WN1 +1 k gt=e n1 is smaller P N1 +N2 N1 +N2 1 1 than the corresponding h-th term in the sequence f N1 +N2 t k=t+1 WN1 +N2 +1 k gt=e n1 +N2 . We therefore prove that (20) weakly holds, i.e., the LHS is weakly smaller than the RHS. To show that (20) strictly holds when W1 > WN , it is su¢ cient to show that there exists at least one correspondence such that the term on the LHS of (20) is strictly smaller than

19

its counterpart on the RHS. Assume that the above is not true, i.e., every term on the LHS of (20) is equal to its counterpart on the RHS; then (23) and (26), respectively, imply minf =

1 N1

n1 1

N1 + N2

n1

PN1

k=n1 +1

n2

WN1 1 +1 k ;

PN1 +N2

k=n1 +n2 +1

where n1 2 f1; :::; n e1 g and n2 2 f1; :::; N2 g; PN1 +N2

k=e n1 +N2 +h

WN1 +N2 +1

k

=

1 N2

n2

PN2

k=n2 +1

WN2 2 +1 k g

WN1 +N2 +1 k ,

PN1

k=e n1 +h

(27)

WN1 1 +1 k ,

(28)

where h 2 f1; 2; :::; N1 n e1 g. We can see that when (27) and (28) hold for all n1 , n2 , and h, it must be the case that W1 = W2 = ::: = WN , which violates our initial assumption that W1 > WN , i.e., we have a contradiction. Therefore, we can safely conclude that provided W1 > WN , there must exist at least one correspondence such that the term on the LHS of (20) is strictly smaller than its counterpart on the RHS.

4.3

Proof of Proposition 4

Consider any general contest structure. Denote the total e¤ort induced in stage l by El and the expected value to each contestant by Vl . Then we have Nl Vl = Nl+1 Vl+1 +

Nl X

Wlm

El .

(29)

m=1

Given that each contestant who wins the ticket to the next stage is assumed to be randomly assigned to any subgroup, the expected value for proceeding to the next stage (Vl+1 ) is the same for all contestants. First, we want to show that if we combine all s1 subcontests in stage 1 into a grand contest while keeping everything else unchanged, the total e¤ort induced in the whole contest will be higher. Suppose that s1 2. The total e¤ort induced in subcontest g = 1; 2; :::; s1 can be written as g

E1g

=r

N1 X k=1

g;N g +1 k

"

g;N g +1 k W1 1

+

g;N g +1 k w1 1

1+

k 1 X j=0

1 N1g

g;N g +1 k

j

!#

,

(30)

where w1 1 = V2 0 for N1g +1 K1g k N1g , and w1 1 = 0 for 1 k N1g K1g . Ps1 Thus, the total e¤ort induced in stage 1 is g=1 E1g . If we combine all subcontests into a grand contest, the total e¤ort induced will be 20

E1 = r

N1 X k=1

w1N1 +1 k

"

W1N1 +1

k

+ w1N1 +1

k

1+

k 1 X

1 N1

j=0

j

!#

,

(31)

w1N1 +1 k

= 0 for 1 k N1 N2 . = V2 for N1 + 1 N2 k N2 , and where Ps1 g Following Corollary 1, it is straightforward to derive that E1 > g=1 E1 when s1 2, which implies that the total e¤ort induced in stage 1 is higher after such combination. Since the total e¤ort induced in other stages is una¤ected, the total e¤ort from the whole contest increases. We then proceed to stage 2. We have now …xed the contest structure in stage 1 as the grand contest. Since the total e¤ort induced in stage l is El = Nl+1 Vl+1 +

Nl X

Wlm

(32)

Nl Vl ;

m=1

the total e¤ort induced in the …rst two stages is 2 X

El =

l=1

2 X

Nl+1 Vl+1 +

Nl X

Wlm

Nl Vl

m=1

l=1

!

= N3 V3 +

Nl 2 X X

Wlm

N1 V1 :

(33)

l=1 m=1

Suppose we combine all s2 subcontests in stage 2 into a grand contest, again following Corollary 1, the total e¤ort induced in stage 2 will be higher (say, from E2 to E20 ). From (32), this lowers V2 . To show that it increases the total e¤ort induced in the …rst two stages, we only need to show that V1 is lower after the combination from (33). Note that E1 = r

"

N2 X

1+

m=1

NX 1 m g=0

1 N1

g

!

V2 +

N1 X

1+

m=1

NX 1 m g=0

1 N1

g

!

#

W1m :

(34)

Therefore we have

N1 V1 = N2 V2 + = N2 V2

PN1

1

m=1

r

W1m

N1 X N2 1 g=0

Since r

r

1+

PN

(35)

E1 1 N1

g

!

+

N1 X

1+r

m=1

g=0

1

1 1 g=0 N g

, it implies that

1

r

N1 X N2 1 g=0

1 N1

21

g

r

NX 1 m

> 0:

1 N1

g

!

W1m :

Therefore V1 is increasing in V2 . Combining subcontests in stage 2 into a grand contest induces a higher e¤ort in stage 2 and lowers V2 and V1 . Therefore, the total e¤ort induced in the …rst two stages is higher. Given that the total e¤ort in other stages is una¤ected, the total e¤ort induced in the whole contest is higher. Following the same idea, suppose we have combined all subcontests from stage 1 to t 1 into a grand contest, and we consider stage t. The total e¤ort induced from stage 1 to t is ! Nl t t t Nl X X X X P Wlm Nl Vl = Nt+1 Vt+1 + Wlm N1 V1 : (36) El = Nl+1 Vl+1 + l=1

l=1 m=1

m=1

l=1

Suppose we combine all st subcontests in stage t into a grand contest; the total e¤ort induced in stage t will be higher (say from Et to Et0 ). This would lower Vt . We only need to show that it also induces a lower V1 . Note that N1 V1 = Nt Vt

t 1 Y j=1

+

Since

@1

Nj Nj+1 1

r

82 0 l 1 Y 4 @1 :

t < X l=1

0

X g=0

1 Nj

Nj Nj+1 1

r

X g=0

j=1

0

@1

g

1 A 1

Nj

g

13 " A5

m=1

1

1

Nj Nj+1 1

r

X

Nj

g=0

Nl X

g

1+r

r

NX l m g=0

1 Nl

g

!

Wlm

#9 = ;

:

A>0

for 1 j L, V1 is increasing in Vt . Combining subcontests in stage t lowers Vt and V1 , and thus induces a higher total e¤ort from stage 1 to t. Therefore, the optimal structure is obtained once we have performed the same procedure until the last stage L, and the optimal structure ends up with the pyramid contest.

4.4

Proof of Lemma 2

Assuming that all other players except player i make the same e¤ort level e0l , we can derive that "N m # l 0 r Y (N 1)! (e ) (eil ) r l l Pm (eil ; el i ) = : (m 1)! k=1 (eil ) r + (Nl k)(e0l ) r (eil ) r + (m 1)(e0l ) r

22

It can be further derived that "N m l @Pm (eil ; el i ) (Nl 1)! Y = @eil (m 1)! k=1 (eil ) (Nl + (m

1)! 1)!

"N m l Y k=1

(eil ) (eil ) r + (m

r

(e0l ) r + (N l

k)(e0l )

(e0l ) (eil ) r + (Nl

r

#

r

k)(e0l )

r

( r)(eil ) [(eil )

r

# "N m l X k=1

r 1

1)(e0l ) r ]

+ (m

r

1)(e0l )

(m

r(eil ) (eil ) r + (Nl

2

r 1

k)(e0l )

r

#

r

1)(e0l )

r

:

In a symmetric equilibrium, each player’s e¤ort level is the same, eil = e0l = el , we thus obtain PNl m 1 1 r g=0 Nl g @Pm (el ; : : : ; el ) = : (37) @eil Nl el From (11), the …rst-order condition for an interior equilibrium is Nl+1

Nl X @Pm (el ; : : : ; el ) X @Pm (el ; : : : ; el ) m m (Vl+1 + Wl ) + Wl i @el @eil m=1 m=N +1

1 = 0:

(38)

l+1

Equations (37) and (38) give that if eil > 0, then it must satisfy r l =(Nl el ) Nl+1 l

X

m=1

1

NX l m g=0

1 Nl

g

!

(Vl+1 + Wlm )

Nl X

m=Nl+1 +1

1

NX l m g=0

1 = 0, where 1

Nl

g

!

Wlm :

Following this result, we have that el = r l =Nl . Each symmetric contestant has the same chance of winning each component of the total stage-award Nl+1 Vl+1 + l (including Nl intermediate prizes and Nl+1 tickets to the next stage) in a symmetric equilibrium. Thus, the equilibrium expected payo¤ of a representative contestant at stage l is given by Vl = (Nl+1 Vl+1 + l )=Nl el .

23

4.5

Proof of Lemma 3

By (13) and el = r l =Nl (in Lemma 2), we derive Nl Vl = Nl+1 Vl+1 +

l

Nl el

= Nl+1 Vl+1 +

l

r

= Nl+1 Vl+1 +

l

Nl+1

+r

X

NX l m

1

g=0

m=1

l

1 Nl

g

Nl+1

= Nl+1 Vl+1 + r

X

!

(Vl+1 + Wlm ) + r

NX l m

1

m=1

g=0

1 Nl

g

X

Nl+1 Vl+1 r

Nl

Nl Nl+1 1

= Nl+1 Vl+1

1

r

g=0

!

Vl+1 +

1

g=0

1 Nl

g

1

!

g

!

+

l

+r

Nl X

+

l

+r

Nl X

1

g=0

NX l m

1

m=1

Nl X

1+r

g=0

r

m=1

NX l m g=0

1 Nl

NX l m

m=1

1

g=0

X

NX l m

m=Nl+1 +1

Nl Nl+1 1

= Nl+1 Vl+1

Nl X

g 1

Nl 1

Nl

1 Nl

g

!

g !

g !

Wlm

!

Wlm

Wlm

Wlm

using the fact that Nl+1

X

NX l m

1

m=1

g=0

Nl

g

Nl Nl+1 1

=

Nl+1

X g=0

1 Nl

g

.

Thus, we have derived that Nl Nl+1 1

Nl Vl = Nl+1 Vl+1

1

X

r

g=0

1 Nl

g

!

+

Nl X

1+r

m=1

From the above equation, we can further derive:21 82 0 13 " Nj Nj+1 1 Nl L l 1 < X 1 X 4Y @ 1 A5 X V1 = 1 r 1+r : N N g j m=1 g=0 l=1

21

j=1

Note that we de…ne

Q0

j=1

1

r

PNj

g=0

Nj+1 1

1 Nj g

24

r

= 1.

NX l m g=0

r

NX l m g=0

g

!

g

!

1 Nl

1 Nl

Wlm :

Wlm

#9 = ;

:

4.6

Proof of Proposition 5

Now we compare

l+1

l

l

with

l+1 .

0 l Y 1 @1 = N j=1

0 l 1 Y 1 @1 N j=1 Ignoring the common part

Nj Nj+1 1

X

r

Nj

g=0

Nj Nj+1 1

X

r l 1 Y j=1

0

g

1 Nj

g=0

1 N

1

g

@1

X

r

1

r

X

1 Nl

g=0

g

!

1+r

Nl

g=0

N l 1 X g=0

Thus,

l

Nl

/

g

=

1 Nl+1

1+

g

g

X g=0

X

!

Nl

g

!

g

!

:

Nl+1

g

+

X g=0

r

N l 1 X g=0

g

Nl+1 1

1 Nl

1

1+r 1

g=0

Nl Nl+1 1

1

g=0

Nl+1 1

1

Nl+1

g=0

N l 1 X

1

1 A , Nj g

l

g=0

X

due to the fact that

r

X

r

r

1

Nl+1 1

Nl Nl+1 1

= r2

A 1+r

g=0

X

Nl+1 1

A 1+r

Nj Nj+1 1

l+1 Nl Nl+1 1

1

1

!

Nl

g

!

0

1 Nl+1

1

g

.

is increasing in l. From (15) and (18), we obtain E=

N V1 =

N

L X

1 l Wl ,

(39)

l=1

PL 1 where for all l 2 f1; 2; :::; Lg. As we have shown l is increasing in l, to l=1 Wl = P maximize E in (39), given , V1 = Ll=1 l Wl1 should be minimized when we set W11 = . Given 0 , it is obvious that to maximize ! N X1 1 E= N 1 =r 1+ , N g g=0 it should be that set W11 = 0 .

=

0.

Therefore, we …nally obtain that to maximize E, it is optimal to

25

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