Colluding with a conscience

Abstract

Other-regarding preferences have been documented in many strategic settings. We provide a model in which the managers of firms in an oligopoly have preferences for both consumer welfare and own income. We find that profit sharing can function as a facilitating practice. Managers must receive a sufficiently large share of profits for collusion to be sustained, and the optimal collusive price increases with the degree of profit sharing. Thus, restrictions on performance-based compensation may be consistent with the objectives of antitrust policy. We also find that an increase in industry concentration can harm consumers even if the firms were already successfully colluding.

Appendix: Proofs of the Results

Proof of Proposition 1: First observe that ${\Pi {'}}( P)=( {P-k})D {'}( P)+D(P)$ and $CS {'}( P)=-D( P)$. It follows that $\left. {\frac{du_i ( P)}{dP}} \right| _{P=k} =\frac{B{\Pi } {'}(k)}{N}+\lambda CS'( k)=\frac{B}{N}D(k)-\lambda D(k)$. Given the quasiconcavity of $u_i ( P)$, it follows that the optimal collusive price $P^C$ is greater than $k$ if and only if $\left. {\frac{du_i ( P)}{dP}} \right| _{P=k} =\left[ {\frac{B}{N}-\lambda } \right] D( k)>0$ which is true if and only if $\frac{B}{N}>\lambda$.$\square$

Proof of Lemma 1: Since $CS'( P)<0$, (4) implies ${\Pi '}( {P^U})>0$. It then follows from (2) and the quasiconcavity of ${\Pi }( P)$ that $P^{U}<P^{\uppi }$. Since $N >$ 1 and ${\Pi '}( {P^U})>0$, (4) and (3) imply $\left( {\frac{B}{N}}\right) {\Pi {'}}( {P^U})+\lambda CS'( {P^U})<0$. Thus, (3) and the quasiconcavity of $u_i ( P)$ implies $P^U>P^C$.$\square$

Proof of Lemma 2: First, it is clearly never optimal for a defecting manager to charge a price greater than the collusive price $P^*$. For collusion at $P^*>P^U$, the best defection price is $P^U$ since, by definition, $P^U$ maximizes $\left[ {B{\Pi }( P)+\lambda CS( P)} \right]$, which is the manager’s utility when all other firms charge higher prices. For any price $P<P^U$ we know that $\left[ {B{\Pi }( P)+\lambda CS( P)} \right]$ is increasing in $P$. Therefore, if the collusive price is $P^*\le P^U$, a manager that defects should charge a price just under $P^*$.

Now suppose that collusion at $P^C$ is not sustainable. We begin by showing that collusion at $P^*\in (P^C,P^U]$ cannot be sustained. Thus, for $P^*\in (P^C,P^U]$ we need to show

$$\begin{aligned} B{\Pi }( {P^*})+\lambda CS( {P^*})+\frac{\delta \lambda CS( k)}{1-\delta }>\frac{B{\Pi }(P^*)/N+\lambda CS(P^*)}{1-\delta }. \end{aligned}$$

(6)

By the definition of $P^C$, we have

$$\begin{aligned} \frac{B{\Pi }(P^C)/N+\lambda CS(P^C)}{1-\delta }>\frac{B{\Pi }(P^*)/N+\lambda CS(P^*)}{1-\delta } \end{aligned}$$

(7)

for any $P^*\ne P^C$. By (5), if $P^C$ is not sustainable, then

$$\begin{aligned} B{\Pi }(P^C)+\lambda CS(P^C)+\frac{\delta \lambda CS(k)}{1-\delta }>\frac{B{\Pi }(P^C)/N+\lambda CS(P^C)}{1-\delta }. \end{aligned}$$

(8)

The quasiconcavity of $B{\Pi }( P)+\lambda CS( P)$ implies that it is increasing in $P$ over the range $(P^C,P^U]$. It follows that

$$\begin{aligned} B{\Pi }( {P^*})+\lambda CS( {P^*})+\frac{\delta \lambda CS( k)}{1-\delta }>B{\Pi }(P^C)+\lambda CS(P^C)+\frac{\delta \lambda CS(k)}{1-\delta }, \end{aligned}$$

(9)

for $P^*\in (P^C,P^U]$. Clearly, (7), (8), (9) imply (6) which is what we needed to show. Thus, collusion cannot be sustained at $P^*\in (P^C,P^U]$.

We now show that collusion cannot be sustained at any $P^*>P^U$. For $P^*>P^U$ a defecting manager will choose $P^U$, so we need to show

$$\begin{aligned} B{\Pi }( {P^U})+\lambda CS( {P^U})+\frac{\delta \lambda CS( k)}{1-\delta }>\frac{B{\Pi }(P^*)/N+\lambda CS(P^*)}{1-\delta }. \end{aligned}$$

(10)

By the definition of $P^U$,

$$\begin{aligned} B{\Pi }( {P^U})+\lambda CS( {P^U})+\frac{\delta \lambda CS( k)}{1-\delta }>B{\Pi }(P^C)+\lambda CS(P^C)+\frac{\delta \lambda CS(k)}{1-\delta }.\quad \end{aligned}$$

(11)

since $P^C\ne P^U$, by Lemma 1. Recall that by (5), if $P^C$ is not sustainable, then

$$\begin{aligned} B{\Pi }(P^C)+\lambda CS(P^C)+\frac{\delta \lambda CS(k)}{1-\delta }>\frac{B{\Pi }(P^C)/N+\lambda CS(P^C)}{1-\delta }. \end{aligned}$$

(12)

And, by the definition of $P^C$, we know

$$\begin{aligned} \frac{B{\Pi }(P^C)/N+\lambda CS(P^C)}{1-\delta }>\frac{B{\Pi }(P^*)/N+\lambda CS(P^*)}{1-\delta } \end{aligned}$$

(13)

since $P^*\ne P^C$. Clearly, (11), (12), and (13) imply (10). Thus, collusion cannot be sustained at $P^*>P^U$.$\square$

For the remaining proofs it will be convenient to define the following slack function:

$$\begin{aligned} \sigma ( {P,\lambda ,B,N})\equiv \frac{B}{N}\left( {\frac{1-N( {1-\delta })}{\delta }}\right) {\Pi }(P)+\lambda ( {CS( P)-CS( k)}). \end{aligned}$$

By construction, $\sigma ( {P,\lambda ,B,N})$ is positive if and only if (5) holds strictly. Thus, in order for collusion to be sustained at some $P^*\le P^U$ we must have $\sigma ( {P^*,\lambda ,B,N})\ge 0$. Observe also that in order to have $\sigma ( {P,\lambda ,B,N})>0$ for $P>k$ and $\lambda \ge 0$ we must have $N( {1-\delta })<1$, and this assumption is maintained in all that follows to allow easier exposition of the results.

As discussed in the text, let $\bar{P}\equiv \hbox {Max}\left\{ {P: \sigma (P,\lambda ,B,N)\ge 0} \right\}$ denote the highest price that causes (5) to hold with equality, so that

$$\begin{aligned} \frac{B}{N}\left( {\frac{1-N( {1-\delta })}{\delta }}\right) \,{\Pi }\,\left( \bar{P})+\lambda ( {CS( {\bar{P}})-CS( k)}\right) =0. \end{aligned}$$

(14)

As discussed in footnote 11, we assume that $\bar{P}$ is finite, which be true under very mild conditions. Finally, note that we must have

$$\begin{aligned} \left. {\frac{\partial \sigma (P,\lambda ,B,N)}{\partial P}} \right| _{P={\bar{P}}} <0, \end{aligned}$$

otherwise there would exist a $P'>\bar{P}$ such that $\sigma ( {P{'},\lambda ,B,N})>0$ contradicting the definition of $\bar{P}$.

Proof of Lemma 3: Given that $CS( P)<CS( k)$ for $P>k$, it follows that $\sigma ( {P,\lambda ,B,N})<\sigma ( {P,0,B,N})$ for all $\lambda >0$ and $P>k$.$\square$

Proof of Proposition 2: We first show the “if” part. By definition of the slack function, $\sigma ( {P^*,\lambda ,B,N})\ge 0$ is sufficient for collusion to be sustainable at some $P^*\in (k,P^C]$. Since $\sigma ( {k,\lambda ,B,N})=0$, it is sufficient to show $\left. {\frac{\partial \sigma (P,\lambda ,B,N)}{\partial P}} \right| _{P=k} >0$. Using the fact that ${\Pi '}( P)=( {P-k})D{'}( P)+D(P)$ and $CS{'}( P)=-D( P)$ we have

$$\begin{aligned} \left. {\frac{\partial \sigma (P,\lambda ,B,N)}{\partial P}} \right| _{P=k} =\left[ {\frac{B}{N}\left( {\frac{1-N(1-\delta )}{\delta }}\right) -\lambda } \right] D( k). \end{aligned}$$

Therefore, $\lambda <\frac{B}{N}\left( {\frac{1-N(1-\delta )}{\delta }}\right)$ is sufficient for collusion to be sustainable at some $P^*\in (k,P^C]$.

Next we show the “only if” part. Suppose that $\lambda \ge \frac{B}{N}\left( {\frac{1-N(1-\delta )}{\delta }}\right)$. From the preceding paragraph we have

$$\begin{aligned} \left. {\frac{\partial \sigma (P,\lambda ,B,N)}{\partial P}} \right| _{P=k} =\left[ {\frac{B}{N}\left( {\frac{1-N(1-\delta )}{\delta }}\right) -\lambda } \right] D( k)<0. \end{aligned}$$

Moreover, for any $P>k$ we have

$$\begin{aligned} \frac{\partial \sigma (P,\lambda ,B,N)}{\partial P}&= \frac{B}{N}\left( {\frac{1-N(1-\delta )}{\delta }}\right) ( {P-k})D{'}( P)\nonumber \\&+\,\,\left[ {\frac{B}{N}\left( {\frac{1-N(1-\delta )}{\delta }}\right) -\lambda } \right] D(P)<0, \end{aligned}$$

since $D{'}( P)<0$. Together with $\sigma ( {k,\lambda ,B,N})=0$, it follows that $\sigma ( {P^*,\lambda ,B,N})<0$ for any $P^*>k$, which means collusion is not sustainable for any $P^*\in (k,P^C]$. Therefore, by Lemma 2, since collusion at $P^C$ cannot be sustained, collusion at $P^*>P^C$ cannot be sustained. $\square$

Proof of Proposition 4:

Part (i). We show that both $P^C$ and $\bar{P}$ are decreasing in $\lambda$. First, implicit differentiation of (3) yields

$$\begin{aligned} \frac{\partial P^C}{\partial \lambda }=\frac{-CS{'}( {P^C})}{\left( {\frac{B}{N}}\right) {\Pi }{''}( {P^C})+\lambda CS{''}( {P^C})}<0, \end{aligned}$$

where the sufficient second-order condition implies that the denominator is negative. Second, using arguments similar to the ones used in the proof of Proposition 2, we can show that $\lambda <\frac{B}{N}\left( {\frac{1-N(1-\delta )}{\delta }}\right)$ implies $\bar{P}>k$. Implicit differentiation of (14) yields

$$\begin{aligned} \frac{\partial \bar{P} }{\partial \lambda }=\frac{( {CS( k)-CS( { \bar{P} })})}{\frac{B}{N}\left( {\frac{1-N(1-\delta )}{\delta }}\right) {\Pi {'}}( {\bar{P} })+\lambda CS{'}( {\bar{P} }).}<0, \end{aligned}$$

where the sign of the denominator follows from the fact that $\sigma ({P,\lambda ,B,N})$ must be decreasing at $\bar{P}$.

Part (ii). We show that both $P^C$ and $\bar{P}$ are increasing in the bonus. Implicit differentiation of (3) yields

$$\begin{aligned} \frac{\partial P^C}{\partial B}=\frac{-\left( {\frac{{\Pi }{'}( {P^C})}{N}}\right) }{\left( {\frac{B}{N}}\right) {\Pi }{''}( {P^C})+\lambda CS{''}( {P^C})}>0, \end{aligned}$$

where the numerator is negative since $P^C<P^{\uppi }$, by Lemma 1, and the sufficient second-order condition implies that the denominator is negative.

Now let $\bar{P}( B)$ be the $\bar{P}$ corresponding to $B$, and let $B{'}>B$. We know $\sigma (P,\lambda ,B,N)$ increases in $B$ when $P^*>k$,

$$\begin{aligned} \sigma ( {\bar{P} (B),\lambda ,B',N})>\sigma ( {\bar{P} (B),\lambda ,B,N})=0. \end{aligned}$$

The fact that $\sigma ( {\bar{P} (B),\lambda ,B',N})>0$ implies $\bar{P}( {B'})>\bar{P} (B)$.

Part (iii). We show that both $P^C$ and $\bar{P}$ are decreasing in $N$. First, implicit differentiation of (3) yields

$$\begin{aligned} \frac{\partial P^C}{\partial N}=\frac{\left( {\frac{B}{N^2}}\right) {\Pi '}( {P^C})}{\left( {\frac{B}{N}}\right) {\Pi }{''}( {P^C})+\lambda CS''(P^C)}<0 \end{aligned}$$

where ${\Pi {'}}( {P^C})>0$ since $P^C<P^{\uppi }$, by Lemma 1, and the sufficient second-order condition implies that the denominator is negative. Second, the condition $\lambda <\frac{B}{N}\left( {\frac{1-N(1-\delta )}{\delta }}\right)$ implies we must have $\bar{P}>k$. Implicit differentiation of (14) yields

$$\begin{aligned} \frac{\partial \bar{P} }{\partial N}=\frac{\frac{B}{N^2\delta }{\Pi }(\bar{P} )}{\frac{B}{N}\left( {\frac{1-N(1-\delta )}{\delta }}\right) {\Pi {'}}( {\bar{P} })+\lambda CS{'}( {\bar{P} })}<0 \end{aligned}$$

where the sign of the denominator follows from the fact that $\sigma ( {P,\lambda ,B,N})$ must be decreasing at $\bar{P}$. $\square$