Contract competition between hierarchies, managerial compensation and imperfectly correlated shocks

Abstract

We analyze competition through incentive contracts for managers in duopoly. Privately informed managers exert surplus enhancing effort that generates an externality on the rival. Asymmetric information on imperfectly correlated shocks creates a two-way distortion of efforts under strategic substitutability in effort and a double downward distortion under strategic complementarity in effort. In the first case, as with contracts for R&D activity or small contractual spillovers for quantity and price competition, increasing the correlation of types reduces the polarization of contracts and the differentials in managerial compensations between efficient and inefficient managers. In the second case, as with large contractual spillovers, the opposite occurs.

Appendix

Proof of Proposition 1

To prove $e_{1}^{*}\ge e_{2}^{*}$ assume, by contradiction, that the opposite holds, that is $e_{1}^{*}<e_{2}^{*}$.¹⁸ Then, since $g_{e\theta }>0$ and $g_{ee\theta }\ge 0$ by A.1., it must be that:

$$\begin{aligned} g_{e}\left( \theta _{1},e_{1}^{*}\right) <g_{e}\left( \theta _{2},e_{1}^{*}\right) \le g_{e}\left( \theta _{2},e_{2}^{*}\right) \end{aligned}$$

Moreover, since $\Pi _{11}\le 0$, it must be that:

$$\begin{aligned} g_{e}\left( \theta _{1},e_{1}^{*}\right)= & {} \lambda \Pi _{1}\left( e_{1}^{*},e_{1}^{*}\right) +\left( 1-\lambda \right) \Pi _{1}\left( e_{1}^{*},e_{2}^{*}\right) \\\ge & {} \lambda \Pi _{1}\left( e_{2}^{*},e_{1}^{*}\right) +\left( 1-\lambda \right) \Pi _{1}\left( e_{2}^{*},e_{2}^{*}\right) =g_{e}\left( \theta _{2},e_{2}^{*}\right) \end{aligned}$$

which contradicts the previous inequality.

Proof of Proposition 2

To prove $e_{1}\ge e_{2}$ assume, by contradiction, that the opposite holds, that is $e_{1}<e_{2}$. Then, since $g_{e\theta }>0$ and $g_{ee\theta }\ge 0$ by A.1, it must be that:

$$\begin{aligned} g_{e}\left( \theta _{1},e_{1}\right) <g_{e}\left( \theta _{2},e_{1}\right) \le g_{e}\left( \theta _{2},e_{2}\right) \end{aligned}$$

which implies:

$$\begin{aligned} g_{e}\left( \theta _{1},e_{1}\right) <g_{e}\left( \theta _{2},e_{2}\right) + \frac{\lambda }{1-\lambda }\Phi _{e}\left( e_{2},\theta _{1},\theta _{2}\right) \end{aligned}$$

Moreover, since $\Pi _{11}\le 0$, it must be that:

$$\begin{aligned} g_{e}\left( \theta _{1},e_{1}\right)= & {} \lambda \Pi _{1}\left( e_{1},e_{1}\right) +\left( 1-\lambda \right) \Pi _{1}\left( e_{1},e_{2}\right) \\&\quad \ge&\lambda \Pi _{1}\left( e_{2},e_{1}\right) +\left( 1-\lambda \right) \Pi _{1}\left( e_{2},e_{2}\right) =g_{e}\left( \theta _{2},e_{2}\right) + \frac{\lambda }{1-\lambda }\Phi _{e}\left( e_{2},\theta _{1},\theta _{2}\right) \end{aligned}$$

which contradicts the previous inequality.

Proof of Proposition 3

Let us consider the following system:

$$\begin{aligned} F(e_{1},e_{2})\equiv & {} \lambda \Pi _{1}\left( e_{1},e_{1}\right) +\left( 1-\lambda \right) \Pi _{1}\left( e_{1},e_{2}\right) -g_{e}\left( \theta _{1},e_{1}\right) =0, \\ H(e_{1},e_{2})\equiv & {} \lambda \Pi _{1}\left( e_{2},e_{1}\right) +\left( 1-\lambda \right) \Pi _{1}\left( e_{2},e_{2}\right) -g_{e}\left( \theta _{2},e_{2}\right) -\frac{\lambda \mu \Phi _{e}\left( e_{2},\theta _{1},\theta _{2}\right) }{1-\lambda }=0 \end{aligned}$$

which defines $\left( e_{1}^{*},e_{2}^{*}\right)$ if $\mu =0$ and $\left( e_{1},e_{2}\right)$ if $\mu =1$. Applying the Cramer rule we have:

$$\begin{aligned} \frac{de_{1}}{d\mu }=\frac{\frac{\partial F}{\partial e_{2}}\frac{\partial H }{\partial \mu }-\frac{\partial F}{\partial \mu }\frac{\partial H}{\partial e_{2}}}{\Delta }=\frac{-\lambda \Phi _{e}\left( e_{2},\theta _{1},\theta _{2}\right) \Pi _{12}\left( e_{1},e_{2}\right) }{\Delta } \end{aligned}$$

where

$$\begin{aligned} \Delta\equiv & {} \frac{\partial F}{\partial e_{1}}\frac{\partial H}{\partial e_{2}}-\frac{\partial F}{\partial e_{2}}\frac{\partial H}{\partial e_{1}}= \frac{\partial F}{\partial e_{1}}\frac{\partial H}{\partial e_{1}}\left[ f_{e}\left( e,\theta \right) -h_{e}\left( e_{2},\theta _{2},\theta _{1}\right) \right] \\ \frac{\partial F}{\partial e_{1}}= & {} \lambda \Pi _{11}\left( e_{1},e_{1}\right) +\lambda \Pi _{12}\left( e_{1},e_{1}\right) +\left( 1-\lambda \right) \Pi _{11}\left( e_{1},e_{2}\right) -g_{ee}\left( \theta _{1},e_{1}\right)<0\\ \frac{\partial H}{\partial e_{2}}= & {} \lambda \Pi _{11}\left( e_{2},e_{1}\right) +\left( 1-\lambda \right) \Pi _{11}\left( e_{2},e_{2}\right) +\left( 1-\lambda \right) \Pi _{12}\left( e_{2},e_{2}\right) \\&\quad -g_{ee}\left( \theta _{2},e_{2}\right) -\frac{\lambda \Phi _{ee}\left( e_{2},\theta _{1},\theta _{2}\right) }{1-\lambda }<0\\ \frac{\partial F}{\partial e_{2}}= & {} \left( 1-\lambda \right) \Pi _{12}\left( e_{1},e_{2}\right) \lessgtr 0\text { if }\Pi _{12}\lessgtr 0\\ \frac{\partial H}{\partial e_{1}}= & {} \lambda \Pi _{12}\left( e_{2},e_{1}\right) \lessgtr 0\text { if }\Pi _{12}\lessgtr 0 \end{aligned}$$

Notice that for any $\theta _{2}>\theta _{1}$ we have $\Phi _{e}\left( e_{2},\theta _{1},\theta _{2}\right) >0$ by A.1 and $\Delta >0$ by A.2, therefore:

$$\begin{aligned} \frac{de_{1}}{d\mu }\gtrless 0\quad \text { if }\quad \Pi _{12}\lessgtr 0. \end{aligned}$$

Analogously, we have:

$$\begin{aligned} \frac{de_{2}}{d\mu }=\frac{\frac{\partial H}{\partial e_{1}}\frac{\partial F }{\partial \mu }-\frac{\partial H}{\partial \mu }\frac{\partial F}{\partial e_{1}}}{\Delta }=\frac{\lambda \Phi _{e}\left( e_{2},\theta _{1},\theta _{2}\right) \frac{\partial F}{\partial e_{1}}}{\left( 1-\lambda \right) \Delta }<0 \end{aligned}$$

Then, the introduction of asymmetric information with $\Phi _{e}>0$ implies $e_{1}>e_{1}^{*}$ and $e_{2}<e_{2}^{*}$ (the two-way distortion) under strategic substitutability, and $\left( e_{1},e_{2}\right) <\left( e_{1},e_{2}^{*}\right)$ (double downward distortion) under strategic complementarity.

Proof of Proposition 4

To characterize the comparative statics of the equilibrium effort levels $\left( e_{1},e_{2}\right)$ with respect to $\alpha$ when $\Pi _{12}\left( e_{1},e_{2}\right) <0$ we totally differentiate the equilibrium system:

$$\begin{aligned} F(e_{1},e_{2},\alpha )\equiv & {} \left( \lambda +\frac{\alpha }{\lambda }\right) \Pi _{1}\left( e_{1},e_{1}\right) +\left( 1-\lambda -\frac{\alpha }{\lambda } \right) \Pi _{1}\left( e_{1},e_{2}\right) -g_{e}\left( \theta _{1},e_{1}\right) =0,\\ H(e_{1},e_{2},\alpha )\equiv & {} \left( \lambda -\frac{\alpha }{1-\lambda } \right) \Pi _{1}\left( e_{2},e_{1}\right) +\left( 1-\lambda +\frac{\alpha }{ 1-\lambda }\right) \Pi _{1}\left( e_{2},e_{2}\right) \\&-g_{e}\left( \theta _{2},e_{2}\right) -\frac{\lambda \Phi _{e}\left( e_{2},\theta _{1},\theta _{2}\right) }{1-\lambda }=0 \end{aligned}$$

and apply the Cramer rule as before to obtain:

$$\begin{aligned} \frac{de_{1}}{d\alpha }=\frac{\frac{\partial F}{\partial e_{2}}\frac{ \partial H}{\partial \alpha }-\frac{\partial F}{\partial \alpha }\frac{ \partial H}{\partial e_{2}}}{\Delta }<0 \end{aligned}$$

where again $\Delta >0$ by A.2. Under strategic substitutability we have:

$$\begin{aligned} \frac{\partial F}{\partial e_{2}}= & {} \left( 1-\lambda -\frac{\alpha }{\lambda } \right) \Pi _{12}\left( e_{1},e_{2}\right)<0,\\ \frac{\partial H}{\partial \alpha }= & {} \frac{\Pi _{1}\left( e_{2},e_{2}\right) -\Pi _{1}\left( e_{2},e_{1}\right) }{1-\lambda }>0,\\ \frac{\partial F}{\partial \alpha }= & {} \frac{\Pi _{1}\left( e_{1},e_{1}\right) -\Pi _{1}\left( e_{1},e_{2}\right) }{\lambda }<0,\\ \frac{\partial H}{\partial e_{2}}= & {} \left( \lambda -\frac{\alpha }{1-\lambda } \right) \Pi _{11}\left( e_{2},e_{1}\right) + \end{aligned}$$

$$\begin{aligned} \left( 1-\lambda +\frac{\alpha }{1-\lambda }\right) \left[ \Pi _{11}\left( e_{2},e_{2}\right) +\Pi _{12}\left( e_{2},e_{2}\right) \right] -g_{ee}\left( \theta _{2},e_{2}\right) -\frac{\lambda \Phi _{ee}\left( e_{2},\theta _{1},\theta _{2}\right) }{1-\lambda }<0 \end{aligned}$$

because $\Pi _{1}\left( e_{1},e_{2}\right) >\Pi _{1}\left( e_{1},e_{1}\right)$ and $\Pi _{1}\left( e_{2},e_{2}\right) >\Pi _{1}\left( e_{2},e_{1}\right)$. Analogously we have $de_{2}/d\alpha >0$.

Proof of Proposition 5

To compare extreme effort levels under the assumption $\Pi _{12}\left( e_{1},e_{2}\right) >0$, let us consider the effort of the efficient manager first. Notice that $e_{1}(0)=e_{1}$ as defined in (24):

$$\begin{aligned} \lambda \Pi _{1}\left( e_{1},e_{1}\right) +\left( 1-\lambda \right) \Pi _{1}\left( e_{1},e_{2}\right) =g_{e}\left( \theta _{1},e_{1}\right) , \end{aligned}$$

and $e_{1}(\overline{\alpha })=e_{11}$ as defined in (13 ):

$$\begin{aligned} \Pi _{1}\left( e_{11},e_{11}\right) =g_{e}\left( \theta _{1},e_{11}\right) . \end{aligned}$$

Strategic complementarity and $e_{1}>e_{2}$ imply $\Pi _{1}\left( e_{1},e_{1}\right) >\Pi _{1}\left( e_{1},e_{2}\right)$ and therefore $e_{1}( \overline{\alpha })=e_{11}>e_{1}(0)=e_{1}$.

Consider the effort of the inefficient manager now. Notice that $e_{2}(0)=e_{2}$ as defined in (25):

$$\begin{aligned} \lambda \Pi _{1}\left( e_{2},e_{1}\right) +\left( 1-\lambda \right) \Pi _{1}\left( e_{2},e_{2}\right) =g_{e}\left( \theta _{2},e_{2}\right) +\frac{ \lambda }{1-\lambda }\Phi _{e}\left( e_{2},\theta _{1},\theta _{2}\right) , \end{aligned}$$

while $e_{2}(\overline{\alpha })$ must satisfy:

$$\begin{aligned} \Pi _{1}\left( e_{2}(\overline{\alpha }),e_{2}(\overline{\alpha })\right) =g_{e}\left( \theta _{2},e_{2}(\overline{\alpha })\right) +\frac{\lambda }{ 1-\lambda }\Phi _{e}\left( e_{2}(\overline{\alpha }),\theta _{1},\theta _{2}\right) . \end{aligned}$$

For any $e_{1}>e_{2}$ strategic complementarity implies $\Pi _{1}\left( e_{2},e_{1}\right) >\Pi _{1}\left( e_{2},e_{2}\right)$, therefore $e_{2}=e_{2}(0)>e_{2}(\overline{\alpha })$.