Inequity-averse agents’ deserved concerns under the linear contract: a social network setting

Abstract

We model how inequity-averse agents’ deserved concerns generated by a social network setting impact on their behaviors and interactions under linear contract. Based on self-perception in the network, the agent will have deserved perception on the two essential elements of linear contract, i.e., deserved fixed wage and deserved output sharing, and thus generate deserved pay gap. By incorporating deserved pay gap into pay comparisons, we obtain three main findings: (1) perceived relative incentive fairness or inequity over the entire networks decides agents’ effort competition. When agents perceive over-incentivized over the networks, the heterogeneity in the network will stimulate agents to compete. While in an overall under-incentivized network, the agent tends to compete with homogenous ones, and will instead reduce effort with increasing heterogeneity in the network; (2) the normal conclusion that inequity aversion can enhance agent effort will be reversed when the agent perceives under-incentivized; and (3) wage compression remains valid, yet we provide a definitive range for the optimal incentive level.

Appendix

Proof of Proposition 1

According to Eq. 7, we have:

$$\begin{aligned} \frac{\partial ^{2}u_i }{\partial e_i \partial e_j }=p_i \left( {p_i +q_i } \right) \left( {\lambda _i -1} \right) \varphi \left( {T_i } \right) . \end{aligned}$$

(9)

The slope of agent i’s reaction function is given by:

$$\begin{aligned} \frac{de_i^*}{de_j }=-\frac{\frac{\partial ^{2}u_i }{\partial e_i \partial e_j }}{\frac{\partial ^{2}u_i }{\partial e_i^2 }}=\frac{p_i \left( {p_i +q_i } \right) \left( {\lambda _i -1} \right) \varphi \left( {T_i } \right) }{b+p_i^2 \left( {\lambda _i -1} \right) \varphi \left( {T_i } \right) }, \end{aligned}$$

(10)

where $\frac{\partial ^{2}u_i }{\partial e_i^2 }<0$ (see Eq. 8 in the text).

Because $0<\mathop \int \nolimits _{-\infty }^{T_i } \varphi \left( \varepsilon \right) d\varepsilon <1$, we incorporate it into Eq. 7 and obtain $\beta +p_i<be_i <\beta +\lambda _i p_i$ when $p_i >0$, and $\beta +\lambda _i p_i<be_i <\beta +p_i$ when $p_i <0$. In addition, $\beta +\lambda _i p_i <0$ when $p_i <0$ if $\lambda _i$ is sufficiently large. Thus, we can conclude that $e_i^*$ falls between $\frac{\beta +p_i }{b}$ and $max\left( {0,\frac{\beta +\lambda _i p_i }{b}} \right)$.

Proof of Proposition 2

Because $\frac{\partial ^{2}u_i }{\partial e_i \partial \lambda _i }=p_i \mathop \int \nolimits _{-\infty }^{T_i } \varphi \left( \varepsilon \right) d\varepsilon$, we get:

$$\begin{aligned} \frac{de_i^*}{d\lambda _i }=-\frac{\frac{\partial ^{2}u_i }{\partial e_i \partial \lambda _i }}{\frac{\partial ^{2}u_i }{\partial e_i^2 }}=\frac{p_i \mathop \int \nolimits _{-\infty }^{T_i } \varphi \left( \varepsilon \right) d\varepsilon }{b+p_i^2 \left( {\lambda _i -1} \right) \varphi \left( {T_i } \right) }. \end{aligned}$$

(11)

We can find that the sign of $\frac{de_i^*}{d\lambda _i }$ depends on $p_i$, and therefore Proposition 2 holds.

Proof of Proposition 3

Because $\frac{\partial ^{2}u_i }{\partial e_i \partial t_i }=p_i \left( {\lambda _i -1} \right) \varphi \left( {T_i } \right)$, we have:

$$\begin{aligned} \frac{de_i^*}{dt_i }=-\frac{\frac{\partial ^{2}u_i }{\partial e_i \partial t_i }}{\frac{\partial ^{2}u_i }{\partial e_i^2 }}=\frac{p_i \left( {\lambda _i -1} \right) \varphi \left( {T_i } \right) }{b+p_i^2 \left( {\lambda _i -1} \right) \varphi \left( {T_i } \right) }. \end{aligned}$$

(12)

This case also depends on the sign of $p_i$.

Proof of Proposition 4

$$\begin{aligned} \frac{de_i^*}{dq_i }=-\frac{\frac{\partial ^{2}u_i }{\partial e_i \partial q_i }}{\frac{\partial ^{2}u_i }{\partial e_i^2 }}=\frac{\frac{p_i^2 \left( {\lambda _i -1} \right) \varphi \left( {T_i } \right) }{p_i^2 +\left( {p_i +q_i } \right) ^{2}}}{b+p_i^2 \left( {\lambda _i -1} \right) \varphi \left( {T_i } \right) }\left[ {\left( {p_i +q_i } \right) e_i^*+p_i e_j -\frac{p_i +q_i }{p_i }t_i } \right] . \end{aligned}$$

(13)

First, if $p_i <0$ and $p_i +q_i \le 0$, the section in the bracket is non-positive, and thus $\frac{de_i^*}{dq_i }\le 0$ (if and only if $e_j =0$, $\frac{de_i^*}{dq_i }=0)$. Second, if $p_i >0$ and $t_i =0$, the section in the bracket is doubtlessly positive, and hence $\frac{de_i^*}{dq_i }>0$. Thus, by continuity, there must exist $\gamma >0$ to make $\frac{de_i^*}{dq_i }>0$ when $t_i \in \left[ {0,\gamma } \right]$. Third, if $p_i <0$ and $p_i +q_i >0$, the sign of the section in the bracket will mainly depend on the values of $e_i^*$ and $e_j$.

Proof of Proposition 5

The optimal output sharing is decided by $\frac{\partial {{u}_p}}{\partial {\upbeta }}=0$, where

$$\begin{aligned} \frac{\partial u_p }{\partial \upbeta }=\sum \nolimits _i^2 \left\{ {\left( {p-be_i^*} \right) \frac{de_i^*}{d\beta }+\left( {e_i^*-e_j^*} \right) \left[ {1+\left( {\lambda _i -1} \right) \int \nolimits _{-\infty }^{T_i } \varphi \left( \varepsilon \right) d\varepsilon } \right] } \right\} . \end{aligned}$$

(14)

If $\lambda _i =\lambda _j =1$, the Eq. 7 in the text is simplified as $\beta -be_i +p_i =0$, and thus the optimal effort of agent i is given by $e_i^*=\frac{\beta +p_i }{b}=\frac{2\beta -\beta _{ii} }{b}$. Similarly, the optimal effort of agent j is $e_j^*=\frac{\beta +p_j }{b}=\frac{2\beta -\beta _{jj} }{b}$. Accordingly, we get that $\frac{de_i^*}{d\beta }=\frac{de_i^*}{d\beta }=\frac{2}{b}$. Then, $\frac{\partial u_p }{\partial \upbeta }=\frac{2}{b}\left[ {2p-\left( {4\beta -\beta _{ii} -\beta _{jj} } \right) } \right] =0$. Finally, we have $\beta ^{*}=\frac{p}{2}+\frac{\beta _{ii} +\beta _{jj} }{4}$.