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Costly Rewards and Punishments

  • Frances Z. Xu Lee EMAIL logo
Published/Copyright: September 17, 2019

Abstract

To punish an agent, the principal often incurs costs. I study a principal’s least costly reward and punishment scheme for an agent whose effort the principal cannot observe. I find the principal’s cost is sometimes minimized by using both costly rewards and costly punishments because (1) the agent has an outside option, or (2) a principal without commitment ability repeatedly interacts with the agent. I also find that when an agent’s effort is better at increasing the probability of a good outcome for the principal, the agent’s payoff may decrease, because the principal replaces rewards with punishments.

JEL Classification: D23; L20; D86

Acknowledgements

The author is grateful for comments from two anonymous referees, Dan Bernhardt, Jacques Cremer, Jin Li, Albert Ma, Wing Suen and the audience at International Industrial Organization Conference

Appendix

A Proof of Lemma 1

Proof.

Suppose the optimal scheme has x1 < 0 and x2 < 0. Consider another scheme with x1=[(1pθ)x1+(p+θ)x2]/(1pθ)+ϵ and x'2 = 0. Since IR is satisfied with x1,x2 , IR must also be satisfied with x'1 and x'2 by construction for any ε > 0. Since IC is satisfied with x1,x2 , x2x1>cθ . I have x2x1=x2x1+11pθ(x2)ϵ>cθ+11pθ(x2)ϵ . Therefore, for any ϵ(0,11pθ(x2)) , the IC holds for x1,x2 . Because the new scheme x1,x2 involves lower punishment for both outcomes than scheme x1,x2 , the principal’s cost is lower under x1,x2 .

Suppose the optimal scheme has x1 > 0 and x2 > 0. Suppose IR is not binding. Consider another scheme with x'1 = 0 and x2=[(1pθ)x1+(p+θ)x2]/(p+θ)ϵ>0 . By construction, it satisfies IR for ε small enough. Since the original scheme satisfies IC, x2>x1+cθ . This implies x2x1>x1p+θ+cθϵ>cθ if ε is small enough, which satisfies IC. Then this new scheme x1,x2 reduces the cost to implement the effort, which is a contradiction. Therefore, IR has to binding.

Consider another scheme with x'1 = 0 and x2=[(1pθ)x1+(p+θ)x2]/(p+θ)>0 . By construction, it also satisfies IR. Since the original scheme satisfies IC, x2>x1+cθ . This implies x2x1>x1p+θ+cθ>cθ , so the new scheme strictly relaxes IC. Therefore, x1,x2 implement the effort and results in the same cost to the principal.

Proof of Corollary 2, It follows from rewriting Proposition 1 in the following way:

Lemma 4.

Let uA denote the agent’s expected payoff under the optimal scheme.

  1. Under low outside option t<(1p)c/λ1+λp0 ,

    1. if θ<λ1+λp , the scheme is pure reward and uA=pθc>t ;

    2. if θ>λ1+λp , the scheme is pure punishment and uA=1pθc>t .

  2. Under intermediate outside option (1p)c/λ1+λp<t<0 ,

    1. if θ<λ1+λp , the scheme is pure reward and uA=pθc>t ;

    2. if λ1+λp<θ<1ptc , the scheme uses both reward and punishment and uA = t;

    3. if θ>1ptc , the scheme is pure punishment and uA=1pθc>t .

  3. Under intermediate outside option 0<t<pc/λ1+λp ,

    1. if θ<λ1+λp , the scheme is pure reward and uA=pθc>t ;

    2. if λ1+λp<θ<ptc , the scheme uses both reward and punishment and uA = t;

    3. if θ>ptc , the scheme is pure reward and uA = t.

  4. Under high outside option t>pc/λ1+λp ,

    1. if θ<ptc , the scheme is pure reward and uA=pθc>t ;

    2. if θ>ptc , the scheme pure reward and uA = t.

Proof of Proposition 2.

Proof.

As a preparation, we solve for the kink point of CC in the ( – x1)–x2 space:

λ(1+α)(x1)+βx2=γλα(x1)+(1+β)x2=γ

This gives

x1=1λγ1+α+β=δθvλ,x2=γ1+α+β=δθv.

We consider two cases in turn.

Case 1. θ<λ1+λp . This case corresponds to Figure 1. The principal prefers rewards to punishments.

When cθγ1+β=δθv1δ(1pθ) , the left most point on IC (hitting the vertical axis) satisfies CC. Therefore, the optimal scheme is pure reward: x1 = 0 and x2=cθ .

When cθ>γ1+β=δθv1δ(1pθ) , IC intersects the upper part of CC as long as the kink point of CC is above or on IC. That is,

δθvcθ1λδθvcθ1+λλγ1+α+β=1+λλδθv.

Otherwise, the solution does not exist and the effort cannot be motivated.

Solving for the intersection of IC and the upper part of CC:

(x1)+x2=cθλα(x1)+(1+β)x2=γ

gives solution: x1=(1+β)cθγ1+βλα=(1δ(1pθ))cθδθv1(1+λ)δ(1pθ) and x2=γλαcθ1+βλα=δθvλδ(1pθ)cθ1(1+λ)δ(1pθ) .

Case 2. θ>λ1+λp . The principal prefers punishments to rewards.

When cθγλ(1+α)=δθvλ(1δ(p+θ)) , the right most point of IC (hitting the horizontal axis) satisfies CC. Therefore, the optimal scheme is pure punishment: x1=cθ and x2 = 0.

When cθ>γλ(1+α)=δθvλ(1δ(p+θ)) , IC intersects the lower part of CC as long as the kink point of CC is above or on IC. As in Case 1, this condition is:

cθ1+λλδθv.

Otherwise, the solution does not exist and the effort cannot be motivated.

Solving for the intersection of IC and the lower part of CC:

(x1)+x2=cθλ(1+α)(x1)+βx2=γ

gives solution: x1=γβcθλ(1+α)β=δθvδ(p+θ)cθλδ(1+λ)(p+θ) and x2=λ(1+α)cθγλ(1+α)β=λ(1δ)(p+θ)cθδθvλδ(1+λ)(p+θ) .

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Published Online: 2019-09-17

© 2019 Walter de Gruyter GmbH, Berlin/Boston

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