Selection Efficiency in Multiple-Prize Tournaments with Sabotage

Baoting Huang; Shao-Chieh Hsueh; Min Qiang Zhao

doi:10.1515/bejeap-2024-0172

Article

Selection Efficiency in Multiple-Prize Tournaments with Sabotage

Baoting Huang , Shao-Chieh Hsueh and Min Qiang Zhao

Published/Copyright: December 16, 2024

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From the journal The B.E. Journal of Economic Analysis & Policy Volume 25 Issue 1

Abstract

Sabotage in competitive environments, like sales contests and sports, can hinder the effectiveness of rank-order tournaments in selecting the most capable individuals. Traditional winner-take-all tournaments may unintentionally level the playing field, making it difficult to distinguish the best. Despite extensive research on tournament design, the impact of sabotage on selection efficiency remains underexplored. This paper addresses this gap by investigating how the introduction of multiple-prize structures in rank-order tournaments affects selection efficiency in the presence of sabotage. Our analysis reveals that multiple prizes can improve the selection of high-performing contestants by redirecting sabotage toward weaker opponents, resulting in corner equilibria. In contrast, winner-take-all structures often result in interior equilibria, where promotion chances are equalized. By outlining the conditions under which these equilibria arise, we demonstrate that strategic prize design can enhance performance incentives, mitigate the negative impact of sabotage, and ultimately improve the selection efficiency of rank-order tournaments.

Keywords: prize design; tournament; sabotage; selection efficiency; corner equilibria

JEL Classification: C72; D23; D40

Corresponding author: Shao-Chieh Hsueh, Department of International Business, School of Economics and Management, Xiamen University of Technology, Xiamen, China, E-mail: schsuechieh@msn.com

Funding source: Natural Science Foundation of Xiamen, China

Award Identifier / Grant number: 3502Z202373068

Acknowledgments

We thank the editor and two anonymous reviewers for their helpful comments and suggestions. We are also grateful to the participants of the 10th China Meeting on Game Theory and Its Applications (July 2023) for their insightful feedback.

Research funding: Shao-Chieh Hsueh acknowledges financial support from the Natural Science Foundation of Xiamen, China (3502Z202373068), and the Social Science Foundation of Fujian, China (FJ2024B034). Baoting Huang acknowledges financial support from the Doctoral Research Start-up Foundation of Liming Vocational University (LRB202408).

Appendix A

A.1 Derivation of Probability and Expected Benefit Functions

The probability that contestant i achieves the highest rank (r _i = 1) can be computed as follows:

(A.1) P r i = 1 = Prob q i ≥ q j & q i ≥ q k = Prob f u i , θ j i + θ k i − f u j , θ i j + θ k j ≥ ε j − ε i & f u i , θ j i + θ k i − f u k , θ i k + θ j k ≥ ε k − ε i = ∫ − ∞ + ∞ ∫ − ∞ ε i + f u i , θ j i + θ k i − f u j , θ i j + θ k j g ε j d ε j × ∫ − ∞ ε i + f u i , θ j i + θ k i − f u k , θ i k + θ j k g ε k d ε k g ε i d ε i .

From equation (A.1), we can readily derive the following relationship:^[8]

(A.2) ∂ P r i = 1 ∂ u i = ∂ P r i = 1 ∂ θ i j + ∂ P r i = 1 ∂ θ i k .

Equation (A.2) demonstrates that the marginal increase in the probability of winning associated with productive effort is equivalent to the combined marginal increases in the probability of winning from sabotaging each of the other two contestants. Similarly, the probability that contestant i is ranked last (r _i = 3) can be expressed as:

(A.3) P r i = 3 = Prob q i ≤ q j & q i ≤ q k = Prob f u i , θ j i + θ k i − f u j , θ i j + θ k j ≤ ε j − ε i & f u i , θ j i + θ k i − f u k , θ i k + θ j k ≤ ε k − ε i = ∫ − ∞ + ∞ 1 − ∫ − ∞ ε i + f u i , θ j i + θ k i − f u j , θ i j + θ k j g ε j d ε j × 1 − ∫ − ∞ ε i + f u i , θ j i + θ k i − f u k , θ i k + θ j k g ε k d ε k d ε i g ε i d ε i .

The marginal winning probabilities of engaging in productive and sabotage activities can be connected in a similar way to equation (A.2):

(A.4) ∂ P r i = 3 ∂ u i = ∂ P r i = 3 ∂ θ i j + ∂ P r i = 3 ∂ θ i k .

Let p i u i , θ i j , θ i k denote the expected benefit of exerting productive and sabotage efforts in the payoff function (2):

(A.5) p i u i , θ i j , θ i k = W 1 P r i = 1 + W 2 P r i = 2 + W 3 P r i = 3 = W 2 + Δ W 1 ∫ − ∞ + ∞ G ε i − f j i G ε i − f k i g ε i d ε i − Δ W 2 ∫ − ∞ + ∞ 1 − G ε i − f j i 1 − G ε i − f k i g ε i d ε i .

A.2 Proof of Equation (A.2)

Given f _ji, as defined in Section 3.1.1, we can obtain f k i = f u k , θ i k + θ j k − f u i , θ j i + θ k i . Using the definition of P r i = 1 from equation (A.1), we can define the following equations:

(A.6) ∂ P r i = 1 ∂ u i = d f d u i ∫ − ∞ + ∞ g ε i g ε i − f j i ∫ − ∞ ε i − f k i g ε k d ε k d ε i + ∫ − ∞ + ∞ g ε i g ε i − f k i ∫ − ∞ ε i − f j i g ε j d ε j d ε i ,

(A.7) ∂ P r i = 1 ∂ θ i j = − d f d θ i j ∫ − ∞ + ∞ g ε i g ε i − f j i ∫ − ∞ ε i − f k i g ε k d ε k d ε i ,

(A.8) ∂ P r i = 1 ∂ θ i k = − d f d θ i k ∫ − ∞ + ∞ g ε i g ε i − f k i ∫ − ∞ ε i − f j i g ε j d ε j d ε i .

Since d f d u i = 1 , d f d θ i j = − 1 and d f d θ i k = − 1 , it is easy to show that

(A.9) ∂ P r i = 1 ∂ u i = ∂ P r i = 1 ∂ θ i j + ∂ P r i = 1 ∂ θ i k .

A.3 Proof of Proposition 1

The proof assumes that either ΔW ₁ or ΔW ₂ is greater than zero. According to the Kuhn–Tucker conditions, we can obtain

(A.10) Δ W 1 ∂ P r i = 1 ∂ θ i j − Δ W 2 ∂ P r i = 3 ∂ θ i j − ∂ C i ∂ θ i j ≤ 0 ,

(A.11) θ i j ≥ 0 , θ i j ∂ U i ∂ θ i j = 0 ,

where U i = W 1 P r i = 1 + W 2 P r i = 2 + W 3 P r i = 3 − C i u i , θ i j + θ i k . Since ∂ C i u i , θ i j + θ i k ∂ θ i j θ i j + θ i k = 0 = 0 , each contestant will sabotage at least one of the opponents.^[9]

Given W ₁ ≠ W ₂ or W ₂ ≠ W ₃, we can show that under each of the three conditions listed in Proposition 1, if f _j ≤ f _k, then contestant i will not sabotage opponent k. In contrast, suppose that contestant i sabotages opponent k θ i k > 0 .

Let us consider an alternative arrangement such that contestant i decreases θ _ik to 0 while increasing θ _ij by the same amount, which will not change the total cost. The corresponding benefit of such an arrangement is described by p i u i , θ i j + θ i k , 0 below:

(A.12) p i u i , θ i j + θ i k , 0 = W 2 + Δ W 1 ∫ − ∞ + ∞ G ε i − f j i + θ i k G ε i − f k i − θ i k g ε i d ε i − Δ W 2 ∫ − ∞ + ∞ 1 − G ε i − f j i + θ i k 1 − G ε i − f k i − θ i k g ε i d ε i .

Given that G ⋅ is strictly log-concave, g x G x should strictly decrease with x (An 1998). Hence, h ′ θ > 0 . Since h 0 = 0 , h θ should be positive for all θ > 0. Given equations (A.5) and (A.12), we can obtain

(A.13) p i u i , θ i j , θ i k − p i u i , θ i j + θ i k , 0 = Δ W 1 − Δ W 2 ∫ h θ i k g ε i d ε i − Δ W 2 ∫ F θ i k g ε i d ε i .

If p i u i , θ i j , θ i k ≤ p i u i , θ i j + θ i k , 0 , contestant i will not sabotage opponent k.

Condition (a). G ⋅ is convex, and ΔW ₁ − ΔW ₂ ≤ 0.

It is easy to verify condition (a) because G ⋅ is a convex function that ensures that F θ ≥ 0 .

Condition (b). G ⋅ is convex, ΔW ₁ − ΔW ₂ ≥ 0, and Δ W 1 − Δ W 2 ∫ h θ i k g ε i d ε i ≤ Δ W 2 ∫ F θ i k g ε i d ε i .

Given that h θ ≥ 0 , G ε i − f j i G ε i − f k i ≤ 1 , and ∫ g ε i d ε i = 1 , we have 0 ≤ ∫ h θ i k g ε i d ε i ≤ 1 . Next, we rearrange ∫ F θ i k g ε i d ε i as follows:

(A.14) ∫ F θ i k g ε i d ε i = ∫ ∫ ε i − f j i ε i − f j i + θ i k g ε i d ε i − ∫ ε i − f k i − θ i k ε i − f k i g ε i d ε i g ε i d ε i .

Given that F θ ≥ 0 , ∫ ε i − f j i ε i − f j i + θ i k g ε i d ε i ≤ 1 , and ∫ g ε i d ε i = 1 , we can obtain 0 ≤ ∫ F θ i k g ε i d ε i ≤ 1 . If ΔW ₂ is sufficiently large, it is easy to identify a set of multiple-prize allocations that can ensure p i u i , θ i j , θ i k ≤ p i u i , θ i j + θ i k , 0 .

Condition (c). G ⋅ is the increasing failure rate distribution. Let Y θ i k = h θ i k + F θ i k . Then, p i u i , θ i j , θ i k ≤ p i u i , θ i j + θ i k , 0 requires that Δ W 1 ∫ h θ i k g ε i d ε i ≤ Δ W 2 ∫ Y θ i k g ε i d ε i . Given that f _j ≤ f _k, G ⋅ is an increasing failure rate distribution ( g θ 1 − G θ increases with θ), which implies that:

(A.15) ∂ Y θ i k ∂ θ i k = g ε i − f j i + θ i k 1 − G ε i − f k i − θ i k − g ε i − f k i − θ i k 1 − G ε i − f j i + θ i k > 0 .

Since Y 0 = 0 , Y θ i k should be positive. If ΔW ₂ is sufficiently large and ΔW ₁ is sufficiently small, it is easy to identify a set of multiple-prize allocations that can ensure p i u i , θ i j , θ i k ≤ p i u i , θ i j + θ i k , 0 .

A.4 Proof of Lemma 1

Without loss of generality, for contestant i, we assume k is the underdog opponent such that f _ji > f _ki. Let Δ P 1 = ∂ P r i = 1 ∂ θ i j − ∂ P r i = 1 ∂ θ i k and Δ P 3 = ∂ P r i = 3 ∂ θ i j − ∂ P r i = 3 ∂ θ i k . From equation (2), we can obtain:

(A.16) ∂ U ∂ θ i j − ∂ U ∂ θ i k = Δ W 1 Δ P 1 − Δ W 2 Δ P 3 .

To prove Lemma 1, we need Lemmas 5 and 6.

Lemma 5.

If G ⋅ is strictly log-concave, then ΔP ₁ > 0.

Proof.

Let us expand ΔP ₁ as follows:

(A.17) Δ P 1 = ∂ P r i = 1 ∂ θ i j − ∂ P r i = 1 ∂ θ i k = ∫ − ∞ + ∞ g ε i g ε i − f j i ∫ − ∞ ε i − f k i g ε k d ε k d ε i − ∫ − ∞ + ∞ g ε i g ε i − f k i ∫ − ∞ ε i − f j i g ε j d ε j d ε i = ∫ − ∞ + ∞ g ε i − f j i G ε i − f k i − g ε i − f k i G ε i − f j i g ε i d ε i .

Since G ⋅ is strictly log-concave, g ( θ ) G ( θ ) should strictly decrease with θ. Given that f _ji > f _ki, it is easy to show that ∂ P r i = 1 ∂ θ i j > ∂ P r i = 1 ∂ θ i k . Q.E.D.

Lemma 6.

If G ⋅ is an increasing failure rate distribution, then ΔP ₃ > 0.

Proof.

Let us expand ΔP ₃ as follows:

(A.18) Δ P 3 = ∂ P r i = 3 ∂ θ i j − ∂ P r i = 3 ∂ θ i k = − ∫ − ∞ + ∞ g ε i g ε i − f j i 1 − ∫ − ∞ ε i − f k i g ε k d ε k d ε i + ∫ − ∞ + ∞ g ε i g ε i − f k i 1 − ∫ − ∞ ε i − f j i g ε j d ε j d ε i = ∫ − ∞ + ∞ g ε i − f k i 1 − G ε i − f j i − g ε i − f j i 1 − G ε i − f k i g ε i d ε i ,

Given that G ⋅ has an increasing failure rate , g ( θ ) 1 − G ( θ ) should increase with θ. Given that f _ji > f _ki, it is easy to show that ∂ P r i = 3 ∂ θ i j > ∂ P r i = 3 ∂ θ i k . Q.E.D.

G ⋅ is assumed to be strictly log-concave. Given that f _j > f _k, Lemma 5 implies that ΔP ₁ > 0. Let Δ = Δ P 3 − Δ P 1 = ∫ − ∞ + ∞ g ε i − f k i − g ε i − f j i g ε i d ε i . Then, ΔP ₃ = ΔP ₁ + Δ. Equation (A.16) can be rearranged as follows:

(A.19) ∂ U ∂ θ i j − ∂ U ∂ θ i k = Δ W 1 − Δ W 2 Δ P 1 − Δ W 2 Δ .

Case (1): If G ⋅ is convex ( g ′ ⋅ > 0 ), f _ji > f _ki implies that g ε i − f k i − g ε i − f j i > 0 . In other words, Δ = Δ P 3 − Δ P 1 = ∫ − ∞ + ∞ g ε i − f k i − g ε i − f j i g ε i d ε i > 0 . Given that ΔW ₁ − ΔW ₂ ≤ 0 and ΔP ₁ > 0 (from Lemma 5), it is easy to show that equation (A.19) should be negative. That is, ∂ U ∂ θ i j < ∂ U ∂ θ i k . It will be optimal to sabotage the underdog opponent.

Case (2): If G ⋅ is an increasing failure rate distribution, Lemma 6 implies that ΔP ₃ > 0. Then, if W ₁ = W ₂ and W ₂ > W ₃, equation (A.19) can be simplified as ∂ U ∂ θ i j − ∂ U ∂ θ i k = − Δ W 2 Δ P 3 < 0 . It will be optimal to sabotage the underdog opponent k.

Case (3): If W ₂ = W ₃ and W ₁ > W ₂, equation (A.19) can be simplified as ∂ U ∂ θ i j − ∂ U ∂ θ i k = Δ W 1 Δ P 1 . Since ΔW ₁ > 0, ∂ U ∂ θ i j − ∂ U ∂ θ i k > 0 . It will be optimal for contestant i to sabotage the favorite opponent j.

Case (4): If G ⋅ is concave ( g ′ ⋅ < 0 ), f _ji > f _ki implies that g ε i − f k i − g ε i − f j i < 0 . Then, Δ < 0. Given that ΔW ₁ − ΔW ₂ ≥ 0 and ΔP ₁ > 0, it is easy to show that equation (A.19) should be positive. That is ∂ U ∂ θ i j > ∂ U ∂ θ i k . It will be optimal to sabotage the favorite opponent.

A.5 Proof of Lemma 2

Suppose to the contrary that f _j ≤ f _k. If one of the conditions in Proposition 1 holds and f _j ≤ f _k, Appendix A.3 shows that p i u i , θ i j , θ i k ≤ p i u i , θ i j + θ i k , 0 . In other words, θ _ik should be zero, which contradicts with θ _ik > 0.

Furthermore, if one of the conditions in Proposition 1 holds and f _j > f _k, it is easy to show that θ _ij = 0 based on Appendix A.3.

A.6 Proof of Lemma 3

By replacing C i ⋅ with γ i C ⋅ , the objective function defined in (2) can be modified as follows:

(A.20) U i = W 2 + Δ W 1 P r i = 1 − Δ W 2 P r i = 3 − γ i C u i , θ i j + θ i k .

To evaluate the Hessian matrix of the objective function, let us first focus on its second derivatives, illustrated below:

(A.21) U u i u i = A − γ i ∂ 2 C ∂ u i 2 ,

(A.22) U u i θ i j = H − γ i ∂ 2 C ∂ u i ∂ θ i j + θ i k ,

(A.23) U u i θ i k = D − γ i ∂ 2 C ∂ u i ∂ θ i j + θ i k ,

(A.24) U θ i j θ i j = Q − γ i ∂ 2 C ∂ θ i j + θ i k 2 ,

(A.25) U θ i j θ i k = E − γ i ∂ 2 C ∂ θ i j + θ i k 2 ,

(A.26) U θ i k θ i k = F − γ i ∂ 2 C ∂ θ i j + θ i k 2 ,

where

A = ∫ g ′ ε i − f j i G ε i − f k i g ε i d ε i + 2 ∫ g ε i − f j i g ε i − f k i g ε i d ε i + ∫ g ′ ε i − f k i G ε i − f j i g ε i d ε i Δ W 1 − Δ W 2 + ∫ g ′ ε i − f j i g ε i d ε i + ∫ g ′ ε i − f k i g ε i d ε i Δ W 2 ,

H = ∫ g ε i − f j i g ε i − f k i g ε i d ε i + ∫ g ′ ε i − f j i G ε i − f k i g ε i d ε i Δ W 1 − Δ W 2 + ∫ − ∞ + ∞ g ′ ε i − f j i g ε i d ε i Δ W 2 ,

D = ∫ g ε i − f j i g ε i − f k i g ε i d ε i + ∫ g ′ ε i − f k i G ε i − f j i g ε i d ε i Δ W 1 − Δ W 2 + ∫ g ′ ε i − f k i g ε i d ε i Δ W 2 ,

Q = ∫ g ′ ε i − f j i G ε i − f k i g ε i d ε i Δ W 1 − Δ W 2 + ∫ g ′ ε i − f j i g ε i d ε i Δ W 2 ,

E = ∫ g ε i − f j i g ε i − f k i g ε i d ε i Δ W 1 − Δ W 2 ,

F = ∫ g ′ ε i − f k i G ε i − f j i g ε i d ε i Δ W 1 − Δ W 2 + ∫ g ′ ε i − f k i g ε i d ε i Δ W 2 .

Given that C ⋅ is strictly convex, we can show that ∂ 2 C ∂ u i 2 > 0 , ∂ 2 C ∂ u i 2 ∂ 2 C ∂ θ i j + θ i k 2 − ∂ 2 C ∂ u i ∂ θ i j + θ i k 2 > 0 .

To identify the conditions that ensure that the Hessian matrix of the objective function is negative semidefinite, we examine the determinants of all upper-left Hessian matrices of the objective function. First, given that C ⋅ is strictly convex, if γ _i is sufficiently large, it is easy to show that U u i u i = A − γ i ∂ 2 C ∂ u i 2 is negative and U u i u i U u i θ i j U u i θ i j U θ i j θ i j = ∂ 2 C ∂ u i 2 ∂ 2 C ∂ θ i j + θ i k 2 − ∂ 2 C ∂ u i ∂ θ i j + θ i k 2 γ i 2 − A ∂ 2 C ∂ θ i j + θ i k 2 + Q ∂ 2 C ∂ u i 2 − 2 H ∂ 2 C ∂ u i ∂ θ i j + θ i k + A B − H 2 is positive.^[10]

Next, the determinant of the whole Hessian matrix can be derived as follows:

(A.27) U u i u i U u i θ i j U u i θ i k U u i θ i j U θ i j θ i j U θ i j θ i k U u i θ i k U θ i j θ i k U θ i k θ i k = 0 · γ i 3 + I + L Δ W 1 − Δ W 2 + J + K Δ W 2 × ∂ 2 C ∂ u i 2 ∂ 2 C ∂ θ i j + θ i k 2 − ∂ 2 C ∂ u i ∂ θ i j + θ i k ∂ 2 C ∂ u i ∂ θ i j + θ i k γ i 2 + H 2 + D 2 + 2 E A − 2 H D − F A − A Q ∂ 2 C ∂ θ i j + θ i k 2 + 2 F H + 2 Q D − 2 D E − 2 H E ∂ 2 C ∂ u i ∂ θ i j + θ i k + E 2 − F Q ∂ 2 C ∂ u i 2 γ i + 2 D H E + F A Q − D 2 Q − E 2 A − H 2 F ,

where

I = ∫ − ∞ + ∞ g ′ ε i − f j i G ε i − f k i g ε i d ε i − ∫ − ∞ + ∞ g ε i − f k i g ε i − f j i g ε i d ε i

L = ∫ − ∞ + ∞ g ′ ε i − f k i G ε i − f j i g ε i d ε i − ∫ − ∞ + ∞ g ε i − f k i g ε i − f j i g ε i d ε i

J = ∫ g ′ ε i − f k i g ε i d ε i

K = ∫ g ′ ε i − f j i g ε i d ε i

If γ _i is sufficiently large, a negative sign of I + L Δ W 1 − Δ W 2 + J + K Δ W 2 can ensure that the determinant of the whole Hessian matrix is negative.

Condition (a) – G ⋅ is convex, ΔW ₁ − ΔW ₂ ≥ 0.

Given that G ⋅ is convex and ΔW ₁ − ΔW ₂ ≥ 0, it is easy to show that 0 ≤ N < J < M, 0 ≤ N < K < M, g + ∞ = B , and g − ∞ = S .

Using integration by parts, we can modify the expression of I as follows:

(A.28) I = ∫ − ∞ + ∞ G 2 ε i − f k i g ε i d g ε i − f j i G ε i − f k i = g 2 + ∞ − 2 ∫ − ∞ + ∞ g ε i − f k i g ε i − f j i g ε i d ε i − ∫ − ∞ + ∞ G ε i − f k i g ε i − f j i g ′ ε i d ε i .

Based on the upper and lower bounds of g ⋅ , it is easy to show that the second component of (A.28) is bounded as follows:

(A.29) − 2 B 2 < − 2 ∫ − ∞ + ∞ g ε i − f k i g ε i − f j i g ε i d ε i < − 2 S 2 .

The third component of (A.28) can be rearranged as follows:

(A.30) 0 < ∫ − ∞ + ∞ G ε i − f k i g ε i − f j i g ′ ε i d ε i < B ∫ − ∞ + ∞ g ′ ε i d ε i = B B − S .

Given (A.28), (A.29) and (A.30), we can obtain that

(A.31) − B 2 − B B − S < I < B 2 − 2 S 2 .

Similarly, we can also show that L is bounded between − B 2 − B B − S and B ² − 2S ². The following condition can ensure that the sign of I + L Δ W 1 − Δ W 2 + J + K Δ W 2 is negative:

(A.32) Δ W 2 Δ W 1 − Δ W 2 < 2 S 2 − B 2 M .

In other words, given that G ⋅ is convex, ΔW ₁ − ΔW ₂ ≥ 0, and Δ W 2 Δ W 1 − Δ W 2 < 2 S 2 − B 2 M , when γ _i is sufficiently large, the Hessian matrix of the objective function is negative semidefinite.

Condition (b) – G ⋅ is concave,ΔW ₁ − ΔW ₂ ≤ 0.

Given that G is concave and ΔW ₁ − ΔW ₂ ≤ 0, it is easy to show that N < J < M ≤ 0, N < K < M ≤ 0, g + ∞ = S , and g − ∞ = B .

The boundary condition is still applied to the second component of (A.28), but the third component is modified as follows:

(A.33) B S − B < ∫ − ∞ + ∞ G ε i − f k i g ε i − f j i g ′ ε i d ε i < 0 .

Given (A.28), (A.29) and (A.33), we can obtain that

(A.34) S 2 − 2 B 2 < I < − S 2 + B B − S .

Similarly, we can also show that L is bounded between S ² − 2B ² and − S 2 + B B − S . The following condition can ensure that the sign of I + L Δ W 1 − Δ W 2 + J + K Δ W 2 is negative:

(A.35) Δ W 2 Δ W 2 − Δ W 1 > B 2 − S 2 − B S M .

In other words, given that G ⋅ is concave,ΔW ₁ − ΔW ₂ ≤ 0, and Δ W 2 Δ W 2 − Δ W 1 > B 2 − S 2 − B S M , when γ _i is sufficiently large, the Hessian matrix of the objective function is negative semidefinite.

A.7 Proof of Proposition 3

Proposition 3 implies that each contestant sabotages only one opponent with a relatively lower ability level if one of the conditions in Proposition 2 is satisfied. We show that no contestant will deviate from these equilibria. Let us use contestant i as an example. Suppose to the contrary that θ _ik > 0. Then, according to Lemma 2, we have θ _ij = 0 and f _j > f _k. Let us consider an alternative arrangement such that contestant i decreases θ _ik to 0 while increasing θ _ij by the same amount, which will not change the total cost. The difference between the benefit associated with θ _ik > 0 and θ _ij = 0 (denoted by p i u i , θ i j , θ i k , see equation (3)) and the benefit under the alternative arrangement (denoted by p i u i , θ i j + θ i k , 0 , see equation (A.12)) is provided in equation (A.13).

Before we examine the difference between p i u i , θ i j , θ i k and p i u i , θ i j + θ i k , 0 , we need to show how productive and sabotage efforts change with the ability parameter γ, as summarized in Lemma 7 below.

Lemma 7.

If one of the conditions in Proposition 2 holds, both productive and sabotage efforts decrease with the ability parameter γ.

Proof.

Without loss of generality, let us focus on contestant i. We first show that ∂ 2 U i ∂ u i ∂ θ i j ≥ 0 holds if one of the conditions in Proposition 2 is satisfied.

Given (A.20), ∂ 2 U i ∂ u i ∂ θ i j can be expressed as follows:

(A.36) ∂ 2 U i ∂ u i ∂ θ i j = Δ W 1 − Δ W 2 R + Δ W 2 T − γ i ∂ 2 C ∂ u i ∂ θ i j + θ i k ,

where

(A.37) R = ∫ − ∞ + ∞ g ε i − f k i g ε i − f j i g ε i d ε i + ∫ − ∞ + ∞ g ′ ε i − f j i G ε i − f k i g ε i d ε i ,

(A.38) T = ∫ − ∞ + ∞ g ′ ε i − f j i g ε i d ε i .

According to integration by parts,

(A.39) R = ∫ − ∞ + ∞ g ε i − f k i g ε i − f j i g ε i d ε i + ∫ − ∞ + ∞ g ′ ε i − f j i G ε i − f k i g ε i d ε i = ∫ − ∞ + ∞ g ε i d G ε i − f k i g ε i − f j i .

Under Condition (a) of Proposition 2, we can obtain that g′ ≥ 0, g − ∞ = S , and g + ∞ = B . Then, R can be simplified as follows:

(A.40) R = B 2 − ∫ − ∞ + ∞ g ε i − f j i G ε i − f k i g ′ ε i d ε i .

Since S ≤ g ≤ B, 0 ≤ G ≤ 1 and g′ ≥ 0, the second component of (A.40) is bounded as follows:

(A.41) − B B − S = B − ∫ − ∞ + ∞ g ′ ε i d ε i < − ∫ − ∞ + ∞ g ε i − f j i G ε i − f k i g ′ ε i d ε i < 0 ,

Given (A.40) and (A.41), we can obtain that 0 ≤ BS < R < B ². For T, since g′ ≥ N ≥ 0, T = ∫ − ∞ + ∞ g ′ ε i − f j i g ε i d ε i > 0 . Given that ΔW ₁ − ΔW ₂ ≥ 0, R > 0, T > 0, and ∂ 2 C ∂ u i ∂ θ i j + θ i k ≤ 0 , it is easy to show that ∂ 2 U i ∂ u i ∂ θ i j ≥ 0 under Condition (a) of Proposition 2.

Under Condition (b) of Proposition 2, g′ ≤ 0, g − ∞ = B , g + ∞ = S , and equation (A.40) can be modified as follows:

(A.42) R = S 2 − ∫ − ∞ + ∞ g ε i − f j i G ε i − f k i g ′ ε i d ε i .

Similarly, we can show that 0 ≤ S ² < R < S ² + B(B − S). For T, since N ≤ g′ ≤ 0, we obtain that N = N ∫ − ∞ + ∞ g ε i d ε i < T < 0 . Given γ _i is sufficiently large (i.e., γ i ≥ Δ W 2 − Δ W 1 S 2 + B B − S + Δ W 2 − N − ∂ 2 C / ∂ u i ∂ θ i j + θ i k > 0 ) and the boundary conditions for R and T it is easy to show that ∂ 2 U i ∂ u i ∂ θ i j ≥ 0 under Condition (b) of Proposition 2.

Under Proposition 2, contestant i sabotages only one opponent. Without loss of generality, assume contestant j is sabotaged by contestant i. The resulting first-order conditions are presented below:

(A.43) ∂ U i ∂ u i = Δ W 1 ∂ P r i = 1 ∂ u i − Δ W 2 ∂ P r i = 3 ∂ u i − γ i ∂ C ∂ u i = 0 ;

(A.44) ∂ U i ∂ θ i j = Δ W 1 ∂ P r i = 1 ∂ θ i j − Δ W 2 ∂ P r i = 3 ∂ θ i j − γ i ∂ C ∂ θ i j = 0 .

To examine how γ _i affects u _i and θ _ij, we conduct a counterfactual analysis. Given a corer equilibrium, if γ _i increases slightly and all the previously determined optimal conditions are not allowed to change, then both ∂ U i ∂ u i and ∂ U i ∂ θ i j should drop below zero because ∂ C ∂ u i > 0 and ∂ C ∂ θ i j > 0 . Given that U _i is a concave function, we can obtain that ∂ 2 U i ∂ u i 2 ≤ 0 , ∂ 2 U i ∂ θ i j 2 ≤ 0 , and ∂ 2 U i ∂ u i ∂ θ i j 2 ≤ ∂ 2 U i ∂ u i 2 ⋅ ∂ 2 U i ∂ θ i j 2 . It is clear that the new equilibria cannot be obtained if u _i and θ _ij change in opposite directions because ∂ 2 U i ∂ u i 2 ≤ 0 , ∂ 2 U i ∂ θ i j 2 ≤ 0 , and ∂ 2 U i ∂ u i ∂ θ i j ≥ 0 . If both u _i and θ _ij increase, ∂ 2 U i ∂ u i ∂ θ i j 2 ≤ ∂ 2 U i ∂ u i 2 ⋅ ∂ 2 U i ∂ θ i j 2 cannot be satisfied. Therefore, both productive and sabotage efforts decrease with the ability parameter γ. Q.E.D.

Since each contestant differs only by his ability parameter γ, Lemma 7 implies that the contestant with a higher ability level should expend more productive effort u. In other words, γ _i > γ _j > γ _k implies that u _k > u _j > u _i.

Given that f _j > f _k, u _k > u _j > u _i, f _i = u _i − θ _ki − θ _ji, f _j = u _j and f _k = u _k − θ _ik, it is easy to show that ɛ _i − f _ji > ɛ _i − f _ki − θ _ik. Then, G ⋅ , which is strictly log-concave, can ensure that h θ i k > 0 for any positive value of θ _ik. According to Appendix A.3, p i u i , θ i j , θ i k − p i u i , θ i j + θ i k , 0 < 0 if one of the conditions listed in Proposition 2 is satisfied. This contradicts θ _ik > 0. Therefore, contestant i has no incentive to deviate from these equilibria, and f _j < f _k according to Lemma 2.

By the same token, neither contestant j nor contestant k has any incentive to deviate from these equilibria.

A.8 Proof of Corollary 1

If one of the conditions in Proposition 2 holds, contestant i will sabotage only one opponent. Contestant i is indifferent to choosing between j and k because γ _j = γ _k, which gives rise to two cases under Corollary 1.

Next, we show that neither contestant j nor contestant k has any incentive to deviate from those equilibria. Let us first focus on contestant j and Case (1). Suppose to the contrary that θ _jk > 0.

Lemma 2 implies that θ _ji = 0 and f _i > f _k. Next, let us consider an alternative arrangement such that contestant j decreases θ _jk to 0 while increasing θ _ji by the same amount, which will not change the total cost. The difference between the benefit associated with θ _jk > 0 and θ _ji = 0 (denoted by p j u j , θ j i , θ j k ) and the benefit under the alternative arrangement (denoted by p j u j , θ j i + θ j k , 0 ) is provided below:

(A.45) p j u j , θ j i , θ j k − p j u j , θ j i + θ j k , 0 = Δ W 1 − Δ W 2 ∫ h ̃ θ j k g ε j d ε j − Δ W 2 ∫ F ̃ θ j k g ε j d ε j ,

where h ̃ θ j k = G ε j − f i j G ε j − f k j − G ε j − f i j + θ j k G ε j − f k j − θ j k , and F ̃ θ j k = G ε j − f i j + θ j k + G ε j − f k j − θ j k − G ε j − f i j − G ε j − f k j .

Since j and k are identical contestants and act simultaneously, they must expend the same amount of productive effort, which should be greater than that of contestant i, who has lower ability according to Lemma 7. That is, u _k = u _j > u _i. Similarly, contestants j and k should expend more sabotage effort than contestant i. That is θ _ki = θ _jk > θ _ij.

Given u _k = u _j > u _i, f _i = u _i − θ _ki, f _j = u _j − θ _ij and f _k = u _k − θ _jk, we can easily show that ɛ _j − f _ij > ɛ _j − f _kj − θ _jk. Then, G ⋅ , which is strictly log-concave, can ensure that h ̃ θ j k > 0 for any positive value of θ _jk. According to Appendix A.3, under the conditions in Proposition 2, p j u j , θ j i , θ j k − p j u j , θ j i + θ j k , 0 < 0 . This contradicts θ _jk > 0. Therefore, contestant j has no incentive to deviate from this equilibrium, and f _i < f _k according to Lemma 2.

Next, let us focus on contestant j and Case (2). Again, suppose to the contrary that θ _jk > 0. Similarly, equation (A.45) is still applied. Given θ _ki = θ _jk > θ _ik, u _k = u _j > u _i, f _i = u _i − θ _ki, f _j = u _j and f _k = u _k − θ _ik − θ _jk, we can easily show that ɛ _j − f _ij > ɛ _j − f _kj − θ _jk. Then, based on the property of G ⋅ and Appendix A.3, it is easy to show that p j u j , θ j i , θ j k − p j u j , θ j i + θ j k , 0 < 0 . This again contradicts θ _jk > 0. Therefore, contestant j has no incentive to deviate from this equilibrium under case (2).

Since both j and k are identical contestants, neither contestant k will have any incentives to deviate from these equilibria under both cases, and f _i < f _j. In addition, if contestant i sabotages j, it is easy to show that f _j < f _k, and vice versa.

A.9 Proof of Corollary 2

If one of the conditions in Proposition 2 holds, contestant k will sabotage only one opponent. Contestant k is indifferent to choosing between i and j because γ _i = γ _j, which gives rise to two cases under Corollary 2.

Next, we show that neither contestant i nor contestant j has any incentive to deviate from those equilibria. Let us first focus on contestant i and Case (1). Suppose to the contrary that θ _ik > 0.

Lemma 2 implies that θ _ij = 0 and f _j > f _k. Next, let us consider an alternative arrangement such that contestant i decreases θ _ik to 0 while increasing θ _ij by the same amount, which will not change the total cost. The difference between the benefit associated with θ _ik > 0 and θ _ij = 0 (denoted by p i u i , θ i j , θ i k , see equation (3)) and the benefit under the alternative arrangement (denoted by p i u i , θ i j + θ i k , 0 , see equation (A.12)) is provided in equation (A.13).

Since i and j are identical contestants and act simultaneously, they must expend the same amount of productive effort, which should be lower than that of contestant k, who has greater ability according to Lemma 7. That is, u _k > u _j = u _i.

Given u _k > u _j = u _i, f _i = u _i − θ _ki − θ _ji, f _j = u _j and f _k = u _k − θ _ik, it is easy to show that ɛ _i − f _ji > ɛ _i − f _ki − θ _ik. Then, G ⋅ , which is strictly log-concave, can ensure that h θ i k > 0 for any positive value of θ _ik. According to Appendix A.3, p i u i , θ i j , θ i k − p i u i , θ i j + θ i k , 0 < 0 if one of the conditions listed in Proposition 2 is satisfied. This contradicts θ _ik > 0. Therefore, contestant i has no incentive to deviate from the equilibria in Case (1), and f _k > f _j according to Lemma 2.

Next, let us focus on contestant i and Case (2). Again, suppose to the contrary that θ _ik > 0. Similarly, equation (A.13) is still applied. Given u _k > u _j = u _i, f _i = u _i − θ _ji, f _j = u _j − θ _kj and f _k = u _k − θ _ik, it is easy to show that ɛ _i − f _ji > ɛ _i − f _ki − θ _ik. Then, based on the property of G ⋅ and Appendix A.3, it is easy to show that p i u i , θ i j , θ i k − p i u i , θ i j + θ i k , 0 < 0 . This again contradicts θ _ik > 0. Therefore, contestant i has no incentive to deviate from this equilibrium under case (2).

Since both i and j are identical contestants, neither contestant j will have any incentives to deviate from these equilibria under both cases, and f _k > f _i. In addition, if contestant k sabotages i, it is easy to show that f _j > f _i, and vice versa.

A.10 Proof of Lemma 4

To prove the equivalence between the two inequalities, let us first assume that u j − θ i j + θ k j > u k − θ i k + θ j k . That is, f _j > f _k. To simplify our notation, let Z = Δ W 1 ∂ P r i = 1 ∂ θ i j − Δ W 2 ∂ P r i = 3 ∂ θ i j − Δ W 1 ∂ P r i = 1 ∂ θ i k − Δ W 2 ∂ P r i = 3 ∂ θ i k . We can further expand Z as follows:

(A.46) Z = Δ W 1 − Δ W 2 ∫ − ∞ + ∞ g ε i − f j i G ε i − f k i − g ε i − f k i G ε i − f j i g ε i d ε i − Δ W 2 ∫ − ∞ + ∞ g ε i − f k i − g ε i − f j i g ε i d ε i .

Given that G ⋅ is strictly log-concave, g x G x strictly decreases with x. Therefore, f _j > f _k implies that g ε i − f j i G ε i − f k i − g ε i − f k i G ε i − f j i > 0 .

For Case (a), when ΔW ₁ > ΔW ₂ = 0, it is easy to show that Z > 0. For Case (b), given that G x is a concave function, g ′ x < 0 . In other words, g ε i − f k i − g ε i − f j i < 0 . Then, ΔW ₁ − ΔW ₂ ≥ 0 can ensure that Z > 0.

Next, let us assume that Z > 0. For Case (a), when ΔW ₁ > ΔW ₂ = 0, Z > 0 implies that ∫ − ∞ + ∞ g ε i − f j i G ε i − f k i − g ε i − f k i G ε i − f j i g ε i d ε i > 0 . Given that G ⋅ is strictly log-concave, only f _j > f _k can satisfy Z > 0, which means that u j − θ i j + θ k j > u k − θ i k + θ j k .

For Case (b), s is the opposite of f _j ≤ f _k. G x being a concave function implies that g ε i − f k i − g ( ε i − f j i ) ≥ 0 . Given that G ⋅ is strictly log-concave and f _j ≤ f _k, g ( ε i − f j i ) G ε i − f k i − g ε i − f k i G ( ε i − f j i ) ≤ 0 . Then, from (A.46), ΔW ₁ − ΔW ₂ ≥ 0 implies that Z ≤ 0, which contradicts Z > 0. Therefore, f _j > f _k. That is, u j − θ i j + θ k j > u k − θ i k + θ j k .

A.11 Proof of Proposition 4

Under interior equilibria, each contestant will sabotage all rivals. If u j − θ i j + θ k j > u k − θ i k + θ j k , Lemma 4 implies that Δ W 1 ∂ P r i = 1 ∂ θ i j − Δ W 2 ∂ P r i = 3 ∂ θ i j > Δ W 1 ∂ P r i = 1 ∂ θ i k − Δ W 2 ∂ P r i = 3 ∂ θ i k . Then, contestant i can decrease θ _ik and simultaneously increase θ _ij by the same amount, which will not change the total cost but can help increase the expected payoff. Therefore, u j − θ i j + θ k j = u k − θ i k + θ j k is required to reach interior equilibrium. The probability of winning across contestants will be equalized.

A.12 Necessary Conditions for the Existence of Interior Equilibria

To simplify the notation, let us define:

(A.47) W ̃ = Δ W 1 p 1 − Δ W 2 p 3 ,

where p 1 = ∂ P r i = 1 ∂ u i f j i = f k i = 0 = 2 ∫ − ∞ + ∞ g 2 ε G ε d ε ,

p 3 = Δ W 2 ∂ P r i = 3 ∂ u i f j i = f k i = 0 = − 2 ∫ − ∞ + ∞ g 2 ε 1 − G ε d ε .

Under the interior equilibria in which f _i = f _j = f _k, equations (4)–(6) can be modified as follows:

(A.48) u i = W ̃ γ i , u j = W ̃ γ j , u k = W ̃ γ k ,

(A.49) θ i j + θ i k = W ̃ 2 γ i , θ j i + θ j k = W ̃ 2 γ j , θ k i + θ k j = W ̃ 2 γ k

(A.50) u i − θ j i + θ k i = u j − θ i j + θ k j = u k − θ i k + θ j k

Let Θ_i denote the total sabotage effort aimed at contestant i. That is Θ_i = θ _ji + θ _ki. Then, the total amount of sabotage effort can be expressed as

(A.51) Θ i + Θ j + Θ k = W ̃ 2 r i + W ̃ 2 r j + W ̃ 2 r k .

Next, given (A.48) and (A.50), we can obtain

(A.52) W ̃ γ i − Θ i = W ̃ γ j − Θ j = W ̃ γ k − Θ k .

By combining (A.51) and (A.52), we can solve for Θ:

(A.53) Θ i = W ̃ 6 − γ i γ k + 5 γ j γ k − γ i γ j γ i γ j γ k ,

(A.54) Θ j = W ̃ 6 − γ j γ k + 5 γ i γ k − γ j γ i γ i γ j γ k ,

(A.55) Θ k = W ̃ 6 − γ k γ j + 5 γ i γ j − γ k γ i γ i γ j γ k .

Therefore, Θ_i > 0 requires that γ j + γ k < 5 γ j γ k γ i ; Θ_j > 0 requires that γ i + γ k < 5 γ i γ k γ j ; and Θ_k > 0 requires that γ i + γ j < 5 γ i γ j γ k .

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Received: 2024-05-19

Accepted: 2024-11-13

Published Online: 2024-12-16

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Articles in the same Issue

https://doi.org/10.1515/bejeap-2024-0172

Keywords for this article

prize design; tournament; sabotage; selection efficiency; corner equilibria