Optimal antitrust enforcement: information cost and deterrent effect

Abstract

This study analyzes the optimal antitrust enforcement rule and, in doing so, presents a model that illuminates two important issues. First, it compares the per se legal and illegal judicial standards to the rule of reason judicial standard in terms of information costs and general social welfare. Second, it seeks to derive the optimal judicial standard that minimizes the problems of under- and over-deterrence. These two issues are closely related because the benefit of additional information can only be measured by its deterrent effects. In this respect, this work synthesizes economic models from decision theory and the public enforcement of law. Lastly, in addition to discussing the optimal information level, we derive the optimal permissiveness of the judicial standard, the optimal burden of proof, and the optimal punishment level. We also analyze how these policy variables are interrelated .

Appendix: Proof of Lemma 1

We derive the solution by breaking down the range of $L+\sigma$.

• 1. $L+\sigma >1+ \sqrt{\frac{4\sigma }{k}}$ If $L+\sigma >1+\sqrt{\frac{4\sigma }{k} }$, then $\alpha ^{B} =1$ and

  $$\begin{aligned} W_{R} =\frac{1-\rho }{m} \int _{-m}^{-k\frac{L+\sigma }{\sigma } } L+\sigma +\frac{2\alpha }{k} \sigma \mathrm{d}\alpha -\frac{A}{\sigma } . \end{aligned}$$

  Note that $W_{R} \le 0$ and the equality hold if $L\rightarrow \infty$ and $\sigma \rightarrow \infty$. Therefore, the optimal judicial standard includes $L^{*} \rightarrow \infty$ and $\sigma ^{*} \rightarrow \infty$.

• 2. $L+\sigma \le 1+\sqrt{\frac{4\sigma }{k}}$

  First, we show that the optimal judicial standard is in the range $1\le L+\sigma \le \frac{m}{k} \sigma$. Suppose a judicial standard that is so restrictive that it will eliminate the under-deterrence problem: no $H$-type firm will engage in any transaction, $\alpha ^{H} =-m$, and $L+\sigma =\frac{m}{k} \sigma$. Once the judicial standard reaches this level of restriction, there is no marginal benefit to making the judicial standard more restrictive, because it will only aggravate the over-deterrence problem without reducing the under-deterrence problem. Therefore, the optimal judicial standard should satisfy $L+\sigma \le \frac{m}{k} \sigma$. However, there can be a judicial standard so permissive that it will allow all $B$-type firms to make their transactions, thereby eliminating the over-deterrence problem. In such a case, both $\alpha ^{B}$ and the amount of social loss caused by the over-deterrence problem will be reduced to 0. To accomplish this level of permissiveness, the judicial standard should satisfy $L+\sigma =1$. This judicial standard does not need to become any more permissive, because this would only aggravate the under-deterrence problem without reducing the over-deterrence problem. Therefore, the optimal judicial standard should satisfy $1\le L+\sigma$. Note that these results also imply that the optimal judicial standard satisfies $1\le \frac{m}{k} \sigma$.

Lemma 3

The optimal judicial standard should satisfy $1\le L+\sigma \le \frac{m}{k} \sigma$.

Since the optimal judicial standard should satisfy $1\le L+\sigma \le \frac{m}{k} \sigma$, the court now has to solve the following maximization problem:

$$\begin{aligned} \max _{\sigma ,L} \,\,W_R&= \rho \int _{\frac{k}{4\sigma }(L+\sigma -1)^2}^{1}1\,\, \mathrm {d}\alpha +\frac{1-\rho }{m}\int _{-m}^{-k\frac{L+\sigma }{\sigma }}L+\sigma +\frac{2\alpha }{k}\sigma \mathrm {d}\alpha -\frac{A}{\sigma }\\ s.t.&\quad 1\le L+\sigma \le \frac{m}{k}\sigma . \end{aligned}$$

The unconstrained solution for this maximization problem is

$$\begin{aligned} L^0+\sigma ^0 = 1+\frac{2}{k}\frac{(1-\rho )}{\rho }\sigma ^0 \\ \sigma ^0=\sqrt{\frac{Ak\rho }{(1-\rho )(\rho (1+m)-1)}}. \end{aligned}$$

It is readily verifiable that the first constraint, $L+\sigma \ge 1$, is not binding. Therefore, we only need to consider the second constraint, $L+\sigma \le \frac{m}{k} \sigma$. The second constraint is binding when $1+\frac{2m}{k} \frac{(1-\rho )}{\rho } \sigma ^{0} \ge \frac{m}{k} \sigma$, which can be reduced to

$$\begin{aligned} \rho \le \frac{2\sigma ^0}{2\sigma ^0+m\sigma ^0 -k}=\rho _0. \end{aligned}$$

Then, the optimal solution is

$$\begin{aligned} L^* +\sigma ^*=\frac{m}{k}\sigma ^*, \\ \sigma ^*=\sqrt{\frac{k(4A+k\rho )}{m^2 \rho }}. \end{aligned}$$

Lastly, we show the uniqueness of $\rho =\frac{2\sigma ^{0} }{2\sigma ^{0} +m\sigma ^{0} -k}$.

Lemma 4

There exists a unique $\rho$ that satisfies $\rho =\frac{2\sigma ^{0} }{2\sigma ^{0} +m\sigma ^{0} -k}$, where $\sigma ^{0} =\sqrt{\frac{Ak\rho }{(1-\rho )(\rho (1+m)-1)} }$.

Proof

Let $f(\rho )=\frac{2\sigma ^{0} }{2\sigma ^{0} +m\sigma ^{0} -k}$; it is easily verifiable that $f\left( \frac{1}{1+m} \right) =f(1)=2/(2+m)$ and that $f$ is a concave function ($f''<0$) that is maximized at $\rho =1/\sqrt{1+m}$. The proof is equivalent to showing that $f(\rho )-\rho =0$ at a single point of $\rho$.

Since $f''<0$, there exists a unique $\rho$ such that $f'(\rho )=1$, and at that value of $\rho$, $f(\rho )-\rho$ is maximized. Let this point of $\rho$ be $\rho ^{0}$. If $\rho ^{0} \le \frac{1}{1+m}$, $f(\rho )-\rho$ is a decreasing function in $\left( \frac{1}{1+m} ,1\right)$, $f\left( \frac{1}{1+m}\right) -\frac{1}{1+m} >0$, and $f(1)-1<0$. Therefore, there exists a single point, $\rho$, that satisfies $f(\rho )-\rho =0$. Since $f\left( \frac{1}{1+m}\right) -\frac{1}{1+m} >0$ and $f(1)-1<0$, if $\rho ^{0} >\frac{1}{1+m}$, then $\rho ^{0} <1$. Note that, for $\rho \in \left( \frac{1}{1+m} ,\rho ^{0} \right]$, $f$ is an increasing function and $f\left( \frac{1}{1+m} \right) >\frac{1}{1+m}$. Therefore, there exists no $\rho$ such that $f(\rho )-\rho =0$. Since $f(\rho )-\rho$ is a decreasing function in $(\rho ^{0} ,1)$, and $f(\rho ^{0} )-\rho ^{0} >0$ and $f(1)-1<0$, there exists a unique $\rho \in (\rho ^{0} ,1)$ that satisfies $f(\rho )=\rho$.

Lastly, because $1>f>\frac{2}{2+m}$, the unique solution $\rho _{0}$ can be found in the range $\left( \frac{2}{2+m} ,1\right)$.