Team production with inequity-averse agents

Abstract

This paper analyzes the efficiency of team production when risk-neutral agents exhibit other-regarding preferences. It is shown that full efficiency can be sustained as an equilibrium of a budget-balancing mechanism that punishes some randomly chosen agents if output falls short of the efficient level but distributes output equally otherwise. The result depends on agents being sufficiently inequity-averse.

Appendix

Proof of Proposition 1

Recall that the efficient action e ^* is determined by:

$$\frac{\partial y(e^*)}{\partial e_i}=C_i'(e_i^*).$$

For self-interested agents, utility functions are given by (4). Given an arbitrary sharing rule (s ₁, ⋯ , s _n ) such that $\sum_i s_i=y(e^*)$, implementing e ^* means that ∀ i, $e_i^*$ is a solution to

$$\frac{\partial s_i(e_i, e^*_{-i})}{\partial e_i}=C_i'(e_i).$$

This requires $s_i(e_i, e^*_{-i})=y(e^*)$ up to a constant. This cannot be done for every agent in the team since the budget constraint $\sum_i s_i=y(e^*)$ has to be satisfied. Hence, no budget-balancing mechanism exists which implements efficiency if α _i  = β _i  = 0, ∀ i. □

Proof of Lemma 1

The optimal deviating effort $\hat e_i$ is determined by solving the maximization problem of Eq. 6. The associated Kuhn-Tucker conditions are:

$$\begin{array}{rll} \frac{\partial }{\partial e_i}u_i( \hat e_i, e^*_{-i}) =\left( \frac{1}{n}-\frac{l(\alpha_i+\beta_i)}{n(n-1)} \right) \frac{\partial y}{\partial e_i} ( \hat e_i,e_{-i}^*)-C_i'(\hat e_i)& \le & 0,\\ \hat e_i \cdot \frac{\partial }{\partial e_i}u_i(\hat e_i, e^*_{-i}) & = & 0. \end{array}$$

If α _i  + β _i  ≤ (n − 1)/l, the maximization problem has a unique interior solution due to the concavity of y and convexity of C _i , and the unique $\hat e_i$ is determined by:

$$\left( \frac{1}{n}-\frac{l(\alpha_i+ \beta_i)}{n(n-1)} \right)\frac{\partial y}{\partial e_i} (\hat e_i,e_{-i}^*) -C_i'(\hat e_i)=0.$$

(13)

The associated second order condition holds as well.

If α _i  + β _i  > (n − 1)/l, $\partial u_i/\partial e_i< 0$ for any positive e _i , the maximization problem has a corner solution, $\hat e_i=0$, and $d \hat e_i/ d \delta_i=0$. This, together with Eq. 13, implies (7). □

Proof of Proposition 2

For given ω, the derivative of Δ _i with respect to δ _i delivers:

$$\begin{array}{lll}\frac{\partial \Delta_i}{\partial \delta_i} &=&\left(-\frac{1}{n}y' +\frac{l\delta_i}{n(n-1)} y'+C_i' \right)\frac{d \hat e_i}{d \delta_i}+\frac{l}{n(n-1)} (y(\hat e_i,e^*_{-i})+n\omega)\\ &=& \frac{l}{n(n-1)} (y(\hat e_i,e^*_{-i})+n\omega)>0. \end{array}$$

The second line obtains because of (7). Δ _i is monotonically increasing in δ _i . Therefore, given ω, for each agent i, there exists a threshold $\tilde \delta_i$ such that if $\delta _i \ge \tilde \delta_i$, Δ _i  ≥ 0 and agent i has no incentive to unilaterally deviate. Efficient actions are sustained in equilibrium if this holds for every agent.

A similar argument applies to ω as

$$\frac{\partial \Delta_i}{\partial \omega} =\frac{l\delta_i}{n-1}>0.$$

□

Proof of Proposition 3

When an agent unilaterally deviates to an effort level e _i given $e^*_{-i}$, his expected utility (11) simplifies to:

$$\begin{array}{lll}u_i^h(e_i, e^*_{-i})&=& \frac{1}{n}\left(-\omega_i-\alpha_i\left( \frac{y( e_i,e^*_{-i}) +\omega_i}{n-1}+\omega_i\right)\right)+\frac{1}{n}\cdot \frac{n-1-\beta_i}{n-1}\cdot y(e_i,e^*_{-i})\\ && +\frac{1}{n}\cdot \frac{n-1-\beta_i n}{n-1}\cdot \frac{1}{n-1}\underset{j\neq i}\sum \omega_j-C_i( e_i). \end{array}$$

The optimal deviating effort $\hat e^h_i$ is determined by the associated Kuhn-Tucker conditions:

$$\frac{\partial}{\partial e_i}u\big(\hat e^h_i, e^*_{-i}\big)=\frac{n-1-\alpha_i-\beta_i}{n(n-1)}\cdot\frac{\partial y\big(\hat e^h_i,e^*_{-i}\big) }{\partial e_i}-\frac{\partial C_i\big(\hat e^h_i\big)}{\partial e_i}\le 0,$$

(14)

$$\hat e^h_i \frac{\partial}{\partial e_i}u\big(\hat e^h_i, e^*_{-i}\big)=0.$$

(15)

If α _i  + β _i  ≤ n − 1, then $\hat e^h_i$ is interior and is a decreasing function in α _i and β _i . If α _i  + β _i  > n − 1, then $\hat e^h_i=0$. Agent i will not deviate if the following is true:

$$\Delta^h_i=u_i\big(e^*_i,e^*_{-i}\big)-u^h_i\big(\hat e^h_i,e^*_{-i}\big)\ge 0,$$

where

$$u_i(e^*_i,e^*_{-i})=\frac{y\big(e^*_i,e^*_{-i}\big)}{n}-C_i(e^*_i).$$

Making use of (14) and (15), one gets:

$$\begin{array}{lll}\frac{\partial \Delta_i}{\partial \alpha_i}&=&\frac{y(\hat e^h_i,e^*_{-i})+n \omega_i}{n(n-1)}>0,\\ \frac{\partial \Delta_i}{\partial \beta_i}&=&\frac{y(\hat e^h_i,e^*_{-i})}{n(n-1)}+\frac{1}{(n-1)^2}\underset{j\neq i}{\sum}\omega_j>0. \end{array}$$

For a given vector of liability capacities (ω ₁, ⋯ , ω _n ), there exists threshold values $\tilde \alpha_i^h$ and $\tilde \beta_i^h$ such that, if $\alpha_i\ge \tilde \alpha_i^h$ and $\beta_i\ge \tilde \beta_i^h$, then $\Delta^h_i\ge 0, \ \forall i$ and efficiency is sustained in equilibrium. □

Proof of Proposition 4

Recall $\Delta^h_i$ from the proof of Proposition 3. The derivative of $\Delta^h_i$ with respect to ω _i gives:

$$\frac{\partial \Delta^h_i}{\partial \omega_i}=\frac{n-1+\alpha_i n }{n(n-1)}>0.$$

From the viewpoint of agent j, j ≠ i, we have:

$$\frac{\partial \Delta^h_j}{\partial \omega_i}=-\frac{n-1-\beta_j n}{n(n-1)^2}.$$

(16)

Note that (16) is positive if and only if $\beta_j\ge \frac{n-1}{n}$. An increase in ω _i decreases agent i’s incentive to shirk, and decreases agent j’s shirking incentive if and only if $\beta_j\ge \frac{n-1}{n}$. Therefore, if $\beta_i\ge \frac{n-1}{n}, \ \forall i$ holds, there exists a $\tilde \omega^h$ such that efficiency is obtained if $\omega_i\ge \tilde \omega^h$. □