Other-regarding preferences in organizational hierarchies

Abstract

In this paper, we provide new theoretical insights about the role of collusion in organizational hierarchies by combining the standard principal–supervisor–agent framework with a theory of social preferences. Extending Tirole’s (J Law Econ Organ 2(2):181–214, 1986) model of hierarchy with the inclusion of Fehr and Schmidt (Q J Econ 114(3):817–868, 1999) type of other-regarding preferences, the links between inequity aversion, collusive behavior and changes in optimal contracts are studied. It turns out that other-regarding preferences do change the collusive behavior among parties depending on the nature of both agent’s and supervisor’s other-regarding preferences. The most prominent impact is on the optimal effort levels. When the agent is inequity averse, the principal can exploit this fact to make agent exert higher effort level than she would otherwise. In order to satisfy the participation constraint of the supervisor, the effort level induced for the agent becomes lower when the supervisor is status seeker, and it is higher when the supervisor is inequity averse.

Appendix

Lagrangian for the solution of the principal’s problem with other-regarding parties is:

$$\begin{aligned} L= & {} \sum _{i} p_{i} ( \theta _{i} + e_{i} - W_{i} - S_{i} ) + \nu \left( \sum _{i} p_{i} V(S_{i}-\lambda _{S}(S_{i} - W_{i})) - {\overline{V}}\right) \\&+\, \mu \left( \sum _{i} p_{i} U(W_{i} - g(e_i)-\lambda _{A}(S_{i} - W_{i})) - {\overline{U}}\right) \\&+\, \gamma ( W_3 - g(e_3) - \lambda _{A}(S_{3} - W_{3}) - W_2 + g(e_2 - \Delta \theta ) + \lambda _{A}(S_{2} - W_{2})) \\&+ \,\psi ((S_{4}-\lambda _{S}(S_{4}-W_{4})(1+2\lambda _{A})+(W_{4}-g(e_{4})-\lambda _{A}(S_{4}-W_{4}))(1-2\lambda _{S})\\&-\,(S_{3}-\lambda _{S}(S_{3}-W_{3}))(1+2\lambda _{A})-(W_{3}-g(e_{3})-\lambda _{A}(S_{3}-W_{3}))(1-2\lambda _{S}))\\&+ \,\pi ((S_{3}-\lambda _{S}(S_{3}-W_{3})(1+2\lambda _{A})+(W_{3}-g(e_{3})-\lambda _{A}(S_{3}-W_{3}))(1-2\lambda _{S})\\&-\,(S_{2}-\lambda _{S}(S_{2}-W_{2}))(1+2\lambda _{A})-(W_{2}-g(e_{2}-\Delta \theta )\\&-\,\lambda _{A}(S_{2}-W_{2}))(1-2\lambda _{S})) \end{aligned}$$

Note that we first ignore (CIC1) ; we are going to show that the solution satisfies (CIC1) .

Taking the derivatives of the Lagrangian above with respect to $S_{i},W_{i},e_{i}$ results in following FOCs:

$$\begin{aligned}&\nu V' ( S_1 - \lambda _{S}(S_{1} - W_{1}) ) = \frac{1}{1-\lambda _{S}} + \mu \frac{\lambda _{A}}{1-\lambda _{S}} U' ( W_1 - g(e_1) - \lambda _{A}(S_{1} - W_{1})) \nonumber \\\end{aligned}$$

(1)

$$\begin{aligned}&\nu V' ( S_2 - \lambda _{S}(S_{2} - W_{2}) ) = \frac{1}{1-\lambda _{S}} + \mu \frac{\lambda _{A}}{1-\lambda _{S}} U' ( W_2 - g(e_2) - \lambda _{A}(S_{2} - W_{2})) \nonumber \\&\qquad - \,\frac{\gamma \lambda _{A}}{p_2(1-\lambda _{S})} + \frac{\pi (1+\lambda _{A}-\lambda _{S})}{p_2(1-\lambda _{S})} \end{aligned}$$

(2)

$$\begin{aligned}&\nu V' ( S_3 - \lambda _{S}(S_{3} - W_{3}) ) = \frac{1}{1-\lambda _{S}} + \mu \frac{\lambda _{A}}{1-\lambda _{S}} U' ( W_3 - g(e_3) - \lambda _{A}(S_{3} - W_{3})) \nonumber \\&\qquad +\, \frac{\gamma \lambda _{A}}{p_3(1-\lambda _{S})} + \frac{(\psi - \pi ) (1+\lambda _{A}-\lambda _{S})}{p_3(1-\lambda _{S})} \end{aligned}$$

(3)

$$\begin{aligned}&\nu V' ( S_4 - \lambda _{S}(S_{4} - W_{4}) ) = \frac{1}{1-\lambda _{S}} + \mu \frac{\lambda _{A}}{1-\lambda _{S}} U' ( W_4 - g(e_4) - \lambda _{A}(S_{4} - W_{4})) \nonumber \\&\qquad -\, \frac{\psi (1+\lambda _{A}-\lambda _{S})}{p_4(1-\lambda _{S})} \end{aligned}$$

(4)

$$\begin{aligned}&\mu U' ( W_1 - g(e_1) - \lambda _{A}(S_{1} - W_{1})) = \frac{1}{1+\lambda _{A}} - \nu \frac{\lambda _S}{1+\lambda _A} V' ( S_1 - \lambda _{S}(S_{1} - W_{1}) ) \nonumber \\\end{aligned}$$

(5)

$$\begin{aligned}&\mu U' ( W_2 - g(e_2) - \lambda _{A}(S_{2} - W_{2})) = \frac{1}{1+\lambda _{A}} - \nu \frac{\lambda _S}{1+\lambda _A} V' ( S_2 - \lambda _{S}(S_{2} - W_{2}) ) \nonumber \\&\qquad +\, \frac{\gamma }{p_2}+ \frac{\pi (1+\lambda _{A}-\lambda _{S})}{p_2(1+\lambda _A)} \end{aligned}$$

(6)

$$\begin{aligned}&\mu U' ( W_3 - g(e_3) - \lambda _{A}(S_{3} - W_{3})) = \frac{1}{1+\lambda _{A}} - \nu \frac{\lambda _S}{1+\lambda _A} V' ( S_3 - \lambda _{S}(S_{3} - W_{3}) ) \nonumber \\&\qquad - \,\frac{\gamma }{p_3} + \frac{(\psi - \pi )(1+\lambda _{A}-\lambda _{S})}{p_3(1+\lambda _A)} \end{aligned}$$

(7)

$$\begin{aligned}&\mu U' ( W_4 - g(e_4) - \lambda _{A}(S_{4} - W_{4})) = \frac{1}{1+\lambda _{A}} - \nu \frac{\lambda _S}{1+\lambda _A} V' ( S_4 - \lambda _{S}(S_{4} - W_{4}) ) \nonumber \\&\qquad -\, \frac{\psi (1+\lambda _{A}-\lambda _{S})}{p_4(1+\lambda _A)} \end{aligned}$$

(8)

$$\begin{aligned}&\mu U' ( W_1 - g(e_1) - \lambda _{A}(S_{1} - W_{1})) g' (e_1) = 1\end{aligned}$$

(9)

$$\begin{aligned}&\mu U' ( W_2 - g(e_2) - \lambda _{A}(S_{2} - W_{2})) g' (e_2) -\frac{(\gamma + \pi (1-2\lambda _S))}{p_2} g'(e_2 - \Delta \theta ) =1 \nonumber \\\end{aligned}$$

(10)

$$\begin{aligned}&\mu U' ( W_3 - g(e_3) - \lambda _{A}(S_{3} - W_{3})) g' (e_3) +\frac{\gamma + (\pi - \psi )(1-2\lambda _S)}{p_3} g'(e_3) =1 \nonumber \\\end{aligned}$$

(11)

$$\begin{aligned}&\mu U' ( W_4 - g(e_4) - \lambda _{A}(S_{4} - W_{4})) g' (e_4) + \frac{\psi (1-2\lambda _S)}{p_4} g' (e_4) = 1. \end{aligned}$$

(12)

Proof of Proposition 1

Substituting (5), (6), (7), (8) into (9), (10), (11), (12) gives that $g' (e_1) = g' (e_3) = g' (e_4) = \frac{1 + \lambda _A - \lambda _S}{1-2\lambda _S}$ and $g' (e_2) < \frac{1 + \lambda _A - \lambda _S}{1-2\lambda _S}$. Since $g''(e_i) > 0$, the rank of effort levels is $e_1 = e_3 = e_4 > e_2$. Suppose $\lambda _A = -\lambda _S$. Then $g' (e_1) = g' (e_3) = g' (e_4) = 1$ and $g' (e_2) < 1$. This implies that $e_1 = e_3 = e_4 = e^{*} > e_2$.

Suppose $\lambda _A < -\lambda _S$. Then $g'(e_2)< g' (e_1) = g' (e_3) = g' (e_4) < 1$. This means that $e^{*}> e_1 = e_3 = e_4 > e_2$. Upper boundary of $g' (e_2)$ goes to $\frac{1 + \lambda _A - \lambda _S}{1-2\lambda _S}$ in our case. Thus, the principal sets $g' (e_2) = \frac{1 + \lambda _A - \lambda _S}{1-2\lambda _S} - \varepsilon$ where $\varepsilon > 0$, in order to guarantee the maximum output level. Since $\frac{1 + \lambda _A - \lambda _S}{1-2\lambda _S}- \varepsilon = g' (e_2) < g' (e_2^{B}) = 1 - \varepsilon$, the ranking of effort levels is $e_2 < e_2^{B}$.

Now suppose $\lambda _A > -\lambda _S$. Then $g' (e_1) = g' (e_3) = g' (e_4) > 1$. This implies that $e_1 = e_3 = e_4 = e^{*}$. Upper boundary of $g' (e_2)$ also increases to $\frac{1 + \lambda _A - \lambda _S}{1-2\lambda _S}$ in this case. Thus, the principal sets $g' (e_2) = \frac{1 + \lambda _A - \lambda _S}{1-2\lambda _S} - \varepsilon$ where $\varepsilon > 0$, in order to get the maximum profit. Since $g' (e_2) > g' (e_2^{B}) = 1 - \varepsilon$, we have $e_2 > e_2^{B}$. For a given $\varepsilon$ (where $\frac{\partial \varepsilon }{\partial \Delta \theta } > 0$), when we have $\frac{\lambda _A+\lambda _S}{1-2\lambda _S} = \varepsilon$ we get $g' (e_2) = g' (e^{*}) = 1$. Therefore, there exist some values for $\lambda _A$ and $\lambda _S$ where $e_2 = e^{*}$. Finally, it is easy to see that $e_2 > e^{*}$ when $\frac{\lambda _A+\lambda _S}{1-2\lambda _S} > \varepsilon$, and $e_2 < e^{*}$ when $\frac{\lambda _A+\lambda _S}{1-2\lambda _S} < \varepsilon$. $\square$

Proof of Proposition 2

First substitute (5), (6), (7) and (8) in (1), (2), (3) and (4) to get the following equations:

$$\begin{aligned} \nu V' ( S_1 - \lambda _{S}(S_{1} - W_{1}) )= & {} \frac{1+2 \lambda _{A}}{1+ \lambda _{A}-\lambda _S} \end{aligned}$$

(13)

$$\begin{aligned} \nu V' ( S_2 - \lambda _{S}(S_{2} - W_{2}) )= & {} \frac{1+2 \lambda _{A}}{1+ \lambda _{A}-\lambda _S} + \frac{\pi (1+2\lambda _{A})}{p_2} \end{aligned}$$

(14)

$$\begin{aligned} \nu V' ( S_3 - \lambda _{S}(S_{3} - W_{3}))= & {} \frac{1+2 \lambda _{A}}{1+ \lambda _{A}-\lambda _S} + \frac{(\psi - \pi ) (1+2\lambda _{A})}{p_3} \end{aligned}$$

(15)

$$\begin{aligned} \nu V' ( S_4 - \lambda _{S}(S_{4} - W_{4}))= & {} \frac{1+2 \lambda _{A}}{1+ \lambda _{A}-\lambda _S} - \frac{\psi (1+2\lambda _{A})}{p_4}. \end{aligned}$$

(16)

Now, we use (1), (2), (3) and (4) in (5), (6), (7) and (8) to get the following equations:

$$\begin{aligned} \mu U' ( W_1 - g(e_1) - \lambda _{A}(S_{1} - W_{1}))= & {} \frac{1-2 \lambda _{S}}{1+ \lambda _{A}-\lambda _S} \end{aligned}$$

(17)

$$\begin{aligned} \mu U' ( W_2 - g(e_2) - \lambda _{A}(S_{2} - W_{2}))= & {} \frac{1-2 \lambda _{S}}{1+ \lambda _{A}-\lambda _S} + \frac{\gamma }{p_2}+\frac{\pi (1-2 \lambda _{S})}{p_2} \end{aligned}$$

(18)

$$\begin{aligned} \mu U' ( W_3 - g(e_3) - \lambda _{A}(S_{3} - W_{3}))= & {} \frac{1-2 \lambda _{S}}{1+ \lambda _{A}-\lambda _S} - \frac{\gamma }{p_3} + \frac{(\psi - \pi ) (1-2 \lambda _{S})}{p_3} \nonumber \\\end{aligned}$$

(19)

$$\begin{aligned} \mu U' ( W_4 - g(e_4) - \lambda _{A}(S_{4} - W_{4}))= & {} \frac{1-2 \lambda _{S}}{1+ \lambda _{A}-\lambda _S} - \frac{\psi (1-2 \lambda _{S})}{p_4}. \end{aligned}$$

(20)

To show that (AIC) is binding, suppose $\gamma = 0$. Then, using the conditions (14), (15) and (18), (19), we get the following equality

$$\begin{aligned} \frac{V' ( S_2 - \lambda _{S}(S_{2} - W_{2}) )}{V' ( S_3 - \lambda _{S}(S_{3} - W_{3}) )} = \frac{U' ( W_2 - g(e_2) - \lambda _{A}(S_{2} - W_{2}))}{U' ( W_3 - g(e_3) - \lambda _{A}(S_{3} - W_{3}))}. \end{aligned}$$

(21)

On the other hand, (AIC) implies that

$$\begin{aligned}&W_{3}-g(e_{3})-\lambda _{A}(S_{3}-W_{3}) \ge W_{2}-g(e_{2}-\Delta \theta )-\lambda _{A}(S_{2}-W_{2}) \nonumber \\&\quad > W_{2}-g(e_{2})-\lambda _{A}(S_{2}-W_{2}). \end{aligned}$$

(22)

From (21) and (22), we get the following inequality

$$\begin{aligned} S_3 - \lambda _{S}(S_{3} - W_{3}) > S_2 - \lambda _{S}(S_{2} - W_{2}). \end{aligned}$$

(23)

Equations (22) and (23) mean that $S_3 - \lambda _{S}(S_{3} - W_{3})+W_{3}-g(e_{3})-\lambda _{A}(S_{3}-W_{3}) > S_2 - \lambda _{S}(S_{2} - W_{2}) + W_{2}-g(e_{2}-\Delta \theta )-\lambda _{A}(S_{2}-W_{2})$, i.e. (CIC3) does not bind. This implies that $\pi = 0$. Then, Eqs. (18) and (19) imply that:

$$\begin{aligned} W_{2}-g(e_{2})-\lambda _{A}(S_{2}-W_{2}) \ge W_{3}-g(e_{3})-\lambda _{A}(S_{3}-W_{3}). \end{aligned}$$

(24)

Note that (22) and (24) cannot hold at the same time. This implies that there is a contradiction which completes this part of our proof and shows that $\gamma > 0$, i.e. (AIC) is binding.

Next, suppose that the second collusion constraint is not binding, i.e. $\psi = 0$. Equations (19) and (20) imply that $W_3 - g(e_3) - \lambda _{A}(S_{3} - W_{3}) > W_4 - g(e_4) - \lambda _{A}(S_{4} - W_{4})$. Using $\lambda _{A} > 0$ we get

$$\begin{aligned} \frac{W_3 - g(e_3) - \lambda _{A}(S_{3} - W_{3})}{1+2\lambda _{A}} > \frac{W_4 - g(e_4) - \lambda _{A}(S_{4} - W_{4})}{1+2\lambda _{A}}. \end{aligned}$$

(25)

From (15) and (16), we also have $S_3 - \lambda _{S}(S_{3} - W_{3}) \ge S_4 - \lambda _{S}(S_{4} - W_{4})$. Using the assumption that $\lambda _S < 0.5$ we can write

$$\begin{aligned} \frac{S_3 - \lambda _{S}(S_{3} - W_{3})}{1-2\lambda _S} \ge \frac{S_4 - \lambda _{S}(S_{4} - W_{4})}{1-2\lambda _S}. \end{aligned}$$

(26)

Equations (25) and (26) imply that

$$\begin{aligned}&\frac{S_3 - \lambda _{S}(S_{3} - W_{3})}{1-2\lambda _S} + \frac{W_3 - g(e_3) - \lambda _{A}(S_{3} - W_{3})}{1+2\lambda _{A}} \nonumber \\&\quad > \frac{S_4 - \lambda _{S}(S_{4} - W_{4})}{1-2\lambda _S} + \frac{W_4 - g(e_4) - \lambda _{A}(S_{4} - W_{4})}{1+2\lambda _{A}}, \end{aligned}$$

(27)

which violates (CIC2) . Thus, $\psi > 0$ and (CIC2) is binding.

We now show that (CIC3) is binding. Assume to the contrary that (CIC3) is not binding, i.e. $\pi = 0$. Then, Eqs. (14) and (15) imply that $S_2 - \lambda _{S}(S_{2} - W_{2}) > S_3 - \lambda _{S}(S_{3} - W_{3})$.

We know that (AIC) is binding, so (CIC3) can now be stated as $(CIC3^{'}): S_3 - \lambda _{S}(S_{3} - W_{3}) \ge S_2 - \lambda _{S}(S_{2} - W_{2})$. However, this is a contradiction to the implication of Eqs. (14) and (15). Therefore, $\pi > 0$ and (CIC3) is binding.

With the following proof of Proposition 3, we show that (CIC1) is already satisfied with the current solution and hence (CIC1) is not binding. $\square$

Proof of Proposition 3

We know that both (AIC) and (CIC3) are binding. This fact implies that $S_2 - \lambda _{S}(S_{2} - W_{2}) = S_3 - \lambda _{S}(S_{3} - W_{3})$. Moreover, (13), (14) and (16) imply that $S_4 - \lambda _{S}(S_{4} - W_{4})> S_1 - \lambda _{S}(S_{1} - W_{1}) > S_2 - \lambda _{S}(S_{2} - W_{2})$. Therefore, the ranking of the supervisor’s utilities at different states is $S_4 - \lambda _{S}(S_{4} - W_{4})> S_1 - \lambda _{S}(S_{1} - W_{1}) > S_2 - \lambda _{S}(S_{2} - W_{2}) = S_3 - \lambda _{S}(S_{3} - W_{3})$. From Eqs. (17), (18) and (20), we have

$$\begin{aligned}&W_4 - g(e_4) - \lambda _{A}(S_{4} - W_{4})> W_1 - g(e_1) - \lambda _{A}(S_{1} - W_{1}) \nonumber \\&\quad > W_2 - g(e_2) - \lambda _{A}(S_{2} - W_{2}). \end{aligned}$$

(28)

Combining the fact that (CIC2) is binding and $g(e_3) = g(e_4)$ yields that

$$\begin{aligned}&\frac{S_4 - \lambda _{S}(S_{4} - W_{4})}{1-2\lambda _S}+\frac{W_{4}-\lambda _{A}(S_{4}-W_{4})}{1+2\lambda _{A}}\nonumber \\&\quad = \frac{S_3 - \lambda _{S}(S_{3} - W_{3})}{1-2\lambda _S}+\frac{W_{3}-\lambda _{A}(S_{3}-W_{3})}{1+2\lambda _{A}}. \end{aligned}$$

Since $S_4 - \lambda _{S}(S_{4} - W_{4}) > S_3 - \lambda _{S}(S_{3} - W_{3})$, we have $W_{3}-g(e_3)-\lambda _{A}(S_{3}-W_{3}) > W_{4}-g(e_4)-\lambda _{A}(S_{4}-W_{4})$. This completes the proof of the following ranking $W_{3}-g(e_3)-\lambda _{A}(S_{3}-W_{3})> W_4 - g(e_4) - \lambda _{A}(S_{4} - W_{4})> W_1 - g(e_1) - \lambda _{A}(S_{1} - W_{1}) > W_2 - g(e_2) - \lambda _{A}(S_{2} - W_{2})$. Now, it can be easily verified that (CIC1) is already satisfied and not binding, since $S_1 - \lambda _{S}(S_{1} - W_{1}) > S_2 - \lambda _{S}(S_{2} - W_{2})$ and $W_1 - g(e_1) - \lambda _{A}(S_{1} - W_{1}) > W_2 - g(e_2) - \lambda _{A}(S_{2} - W_{2})$. $\square$