Error Costs, Ratio Tests, and Patent Antitrust Law

Abstract

This paper examines the welfare tradeoff between patent and antitrust law. Since patent and antitrust law have contradictory goals, the question that naturally arises is how one should choose between the two in instances where there is a conflict. One sensible approach to choosing between two legal standards—or between proof standards with respect to evidence—is to consider the relative costs of errors. The approach in this paper is to consider the ratio of false positives to false negatives in patent antitrust. We find that the relevant error-cost ratio for patent antitrust is the proportion of the sum of the monopoly profit and the residual consumer surplus to the deadweight loss. This error-cost ratio—for a wide range of deterministic demand functions—ranges from infinity to a low of roughly three. This suggests that patent antitrust law should err on the side of protecting innovation incentives.

Appendices

Appendix 1: Model Result

1.1 The Error-Cost Ratio under Common Demand Function Assumptions

Proposition 1

The demand curve is linear$p = A {-} bq$, and the marginal cost is a constant$c$. Optimal output is$q^{{ \star }} = \frac{A - c}{2b}$and the monopoly price$p^{{ \star }} = \frac{A + c}{2}$. The competitive price is$p^{c} = c$, and the competitive output is$q^{c} = \frac{A - c}{b}$. The minimum social loss under antitrust law that deters entry is$RS + {{\uppi }} = \frac{{3\left( {A - c} \right)^{2} }}{8b}$; the maximum social loss if the firm is allowed to enter the market is$D = \frac{{\left( {A - c} \right)^{2} }}{8b}$. The error-cost ratio is therefore${{\uprho }} \equiv \frac{{RS + {{\uppi }}}}{D} = 3$.

Proposition 2

The demand curve is algebraic $p = {{\alpha }}q^{{{\upbeta }}} -$ , and the marginal cost is a constant $c,$ so the firm’s profit maximization problem is $\mathop {\hbox{max} }\limits_{q} :\left( {{{\alpha }}q^{{{\upbeta }}} - {{\upsigma }}} \right)q - cq$ . Optimal output satisfies $\alpha \left( {\beta + 1} \right)q^{\beta } = c + \sigma$ , which implies $q^{{ \star }} = \left( {\frac{{{{\upsigma }} + c}}{{{{\alpha }}\left( {{{\upbeta }} + 1} \right)}}} \right)^{{1/{{\upbeta }}}}$ and monopoly price $p^{{ \star }} = \frac{{{{\upsigma }} + c}}{{{{\upbeta }} + 1}} - {{\upsigma }}$ . The competitive output is $q^{c} = \left( {\frac{{{{\upsigma }} + c}}{{{\alpha }}}} \right)^{{1/{{\upbeta }}}}$ . The minimum social loss when antitrust law deters entry is given by

$$ \begin{aligned} RS + {{\uppi }} & = \mathop \smallint \limits_{0}^{{q^{{ \star }} }} \left[ {{{\alpha }}q^{{{\upbeta }}} - {{\upsigma }} - c} \right]dq \\ & = \frac{{{\alpha }}}{{{{\upbeta }} + 1}}q^{{{ \star }{{\upbeta }} + 1}} - \left( {{{\upsigma }} + c} \right)q^{{ \star }} \\ & = \left( {{{\upsigma }} + c} \right)\left[ {\frac{{ - {{\upbeta }}\left( {{{\upbeta }} + 2} \right)}}{{\left( {{{\upbeta }} + 1} \right)^{2} }}} \right]\left( {\frac{{{{\upsigma }} + c}}{{{{\alpha }}\left( {{{\upbeta }} + 1} \right)}}} \right)^{{1/{{\upbeta }}}} , \\ \end{aligned} $$

and the maximum social loss if the firm enters is

$$ \begin{aligned} D & = \mathop \smallint \limits_{{q^{{ \star }} }}^{{q^{c} }} \left[ {{{\alpha }}q^{{{\upbeta }}} - {{\upsigma }} - c} \right]dq \\ & = \left( {{{\upsigma }} + c} \right)\left[ {\frac{{ - {{\upbeta }}}}{{{{\upbeta }} + 1}}} \right]\left( {\frac{{{{\upsigma }} + c}}{{{\alpha }}}} \right)^{{1/{{\upbeta }}}} - \left( {{{\upsigma }} + c} \right)\left[ {\frac{{ - {{\upbeta }}\left( {{{\upbeta }} + 2} \right)}}{{\left( {{{\upbeta }} + 1} \right)^{2} }}} \right]\left( {\frac{{{{\upsigma }} + c}}{{{{\alpha }}\left( {{{\upbeta }} + 1} \right)}}} \right)^{{1/{{\upbeta }}}} . \\ \end{aligned} $$

Therefore, the error-cost ratio is

$$ \begin{aligned} \frac{{RS + {{\uppi }}}}{D} & = \frac{{\left( {{{\upsigma }} + c} \right)\left[ {\frac{{ - {{\upbeta }}\left( {{{\upbeta }} + 2} \right)}}{{\left( {{{\upbeta }} + 1} \right)^{2} }}} \right]\left( {\frac{{{{\upsigma }} + c}}{{{{\alpha }}\left( {{{\upbeta }} + 1} \right)}}} \right)^{{1/{{\upbeta }}}} }}{{\left( {{{\upsigma }} + c} \right)\left[ {\frac{{ - {{\upbeta }}}}{{{{\upbeta }} + 1}}} \right]\left( {\frac{{{{\upsigma }} + c}}{{{\alpha }}}} \right)^{{1/{{\upbeta }}}} - \left( {{{\upsigma }} + c} \right)\left[ {\frac{{ - {{\upbeta }}\left( {{{\upbeta }} + 2} \right)}}{{\left( {{{\upbeta }} + 1} \right)^{2} }}} \right]\left( {\frac{{{{\upsigma }} + c}}{{{{\alpha }}\left( {{{\upbeta }} + 1} \right)}}} \right)^{{1/{{\upbeta }}}} }} \\ & = \frac{1}{{\frac{{\left( {{{\upbeta }} + 1} \right)^{{1 + 1/{{\upbeta }}}} }}{{\left( {{{\upbeta }} + 2} \right)}} - 1}}, \\ \end{aligned} $$

with${{\upbeta }} \in \left( { - 1,0} \right)$. Because$e = \mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right)^{n}$, it follows that$\mathop {\lim }\limits_{{{{\upbeta }} \to 0}} \left( {1 + {{\upbeta }}} \right)^{{1/{{\upbeta }}}} = e$.

Therefore,

$$ \mathop {\lim }\limits_{{{{\upbeta }} \to 0}} \frac{1}{{\frac{{\left( {{{\upbeta }} + 1} \right)}}{{\left( {{{\upbeta }} + 2} \right)}} \cdot \left( {1 + {{\upbeta }}} \right)^{{\frac{1}{{{\upbeta }}}}} - 1}} = \frac{1}{{\frac{1}{2} \cdot e - 1}} \approx 2.8. $$

Proposition 4

The demand curve is exponential $q = {{\upgamma }}e^{{ - {{\upbeta }}p}}$ , and the marginal cost is a constant $c$ . The inverse demand curve is $p = {{\upbeta }}^{ - 1} \ln {{\upgamma }} - {{\upbeta }}^{ - 1} \ln q$ . Optimal output satisfies β ⁻¹ ln γ − β ⁻¹ ln q − β ⁻¹  = c, so that $q^{{ \star }} = {{\upgamma }}e^{{ - \left( {{{\upbeta }}c + 1} \right)}}$ and $p^{{ \star }} = \frac{{{{\upbeta }}c + 1}}{{{\upbeta }}}$ . The competitive price is $p^{c} = c$ , and the competitive output is $q^{c} = {{\upgamma }}e^{{ - {{\upbeta }}c}}$ . The social loss under an antitrust law that deters the firm’s entry is

$$ \begin{aligned} RS + {{\uppi }} & = \mathop \smallint \limits_{0}^{{q^{{ \star }} }} \left[ { - \frac{1}{{{\upbeta }}}\ln \left( {\frac{q}{{{\upgamma }}}} \right) - c} \right]dq \\ & = - \frac{1}{{{\upbeta }}}\left[ {q^{{ \star }} \ln \left( {\frac{{q^{{ \star }} }}{{{\upgamma }}}} \right) - q^{{ \star }} } \right] - cq^{{ \star }} \\ & = {{\upgamma }}e^{{ - \left( {{{\upbeta }}c + 1} \right)}} \frac{2}{{{\upbeta }}}, \\ \end{aligned} $$

The error cost ratio is

$$ \frac{{RS + {{\uppi }}}}{D} = \frac{{{{\upgamma }}e^{{ - \left( {{{\upbeta }}c + 1} \right)}} \frac{2}{{{\upbeta }}}}}{{{{\upgamma }}e^{{ - {{\upbeta }}c}} \left[ {\frac{1}{{{\upbeta }}}\left( {1 - \frac{2}{e}} \right)} \right]}} = \frac{{\frac{2}{e}}}{{1 - \frac{2}{e}}} \approx 2.8. $$

1.2 The Effect of Elasticity on the Error-Cost Ratio under Isoelastic Demand

From Propositions 1 and 4, we have that the error-cost ratio is a constant number when the demand function is linear or exponential. However, when the demand function is algebraic $p = {{\alpha }}q^{{{\upbeta }}} - {{\upsigma }},$ or is isoelastic $p = {{\alpha }}q^{{{\upbeta }}}$, where $\beta \in \left( { - 1,0} \right)$, the error-cost ratio is a function of demand elasticity. We claim that the ratio $\rho$ decreases with regard to demand elasticity ${\epsilon} \doteq - \frac{1}{{{\upbeta }}}$.

According to Propositions 2 and 3, the error-cost ratio is given by

$$ {{\uprho }} = \frac{1}{{\frac{{\left( {{{\upbeta }} + 1} \right)^{{1 + 1/{{\upbeta }}}} }}{{\left( {{{\upbeta }} + 2} \right)}} - 1}} $$

Since demand elasticity is given by ${\epsilon} \doteq - \frac{1}{{{\upbeta }}}$, we have

$$ {{\uprho }} = \frac{1}{{\frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}} - 1}}. $$

To study the relationship between ${{\uprho }}$ and ${\epsilon}$, we can first analyze the relationship between $\frac{1}{{{\uprho }}}$ and ${\epsilon}$ by taking derivative of $\frac{1}{{{\uprho }}}$ to ${\epsilon}$.

From $\frac{1}{{{\uprho }}} = \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}} - 1$, we have

$$ \begin{aligned} \frac{{d\left( {\frac{1}{\rho }} \right)}}{d{\epsilon}} & = \frac{{d\left( {\frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}} - 1} \right)}}{d{\epsilon}} \\ & = \frac{{d\left( {\frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}}} \right)}}{d{\epsilon}} \\ \begin{array}{l} { = \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}}\frac{{d\left( {\ln \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}}} \right)}}{d{\epsilon}}.} \\ \end{array} \\ \end{aligned} $$

(7)

To derive $\frac{{d\left( {\ln \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}}} \right)}}{d{\epsilon}}$, we first compute $\ln \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}}$ and have

$$ \ln \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}} = \left( {1 - {\epsilon}} \right)\ln \left( {1 - \frac{1}{{\epsilon}}} \right) - \ln \left( {2 - \frac{1}{{\epsilon}}} \right). $$

Then we have the derivative $\frac{{d\left( {\ln \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}}} \right)}}{d{\epsilon}}$ derived as follows

$$ \begin{array}{*{20}c} {\frac{{d\left( {\ln \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}}} \right)}}{d{\epsilon}} = \ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}.} \\ \end{array} $$

(8)

Then we have

Claim 1

$$ \forall {\epsilon} \in \left( {1, + \infty } \right), $$

$$ \ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}} > 0. $$

Proof

Since ${{\upbeta }} \in \left( { - 1,0} \right)$, we have ${\epsilon} \in \left( {1, + \infty } \right)$. Then we show that $\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}$ is monotonically decreasing with ${\epsilon}$ in $\left( {1, + \infty } \right)$.

We first derive the derivative of $\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}$ with regard to ${\epsilon}$, as follows

$$ \frac{{d\left( {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right)}}{d{\epsilon}} = \frac{{ - \frac{1}{4}}}{{{\epsilon}\left( {{\epsilon} - 1} \right)\left( {{\epsilon} - \frac{1}{2}} \right)^{2} }} < 0. $$

The last equality is from ${\epsilon} \in \left( {1, + \infty } \right).$ This proves that $\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}$ is monotonically decreasing with ${\epsilon}$ in $\left( {1, + \infty } \right)$, i.e.,$\forall {\epsilon} \in \left( {1, + \infty } \right)$, $\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}} > \mathop {\lim }\limits_{{\epsilon} \to + \infty } \left( {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right)$.

Then we have

$$ \mathop {\lim }\limits_{{\epsilon} \to + \infty } \left( {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right) = 0. $$

Thus, we have $\forall {\epsilon} \in \left( {1, + \infty } \right),\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}} > 0$.

Plugging the Eq. (8) back in Eq. (7), we have

$$ \frac{{d\left( {\frac{1}{{{\uprho }}}} \right)}}{d{\epsilon}} = \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}}\left[ {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right]. $$

From Claim 1, we have $\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}} > 0$.

Also from ${\epsilon} > 1,$ we have

$$ \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}} > 0 $$

Thus, we have

$$ \frac{{d\left( {\frac{1}{{{\uprho }}}} \right)}}{d{\epsilon}} > 0. $$

Then we have

$$ \frac{{d{{\uprho }}}}{d{\epsilon}} = - {{\uprho }}^{2} \frac{{d\left( {\frac{1}{{{\uprho }}}} \right)}}{d{\epsilon}} < 0. $$

1.3 The Negative Relationship between Elasticity and Profit

Lemma 1

(Negative relationship between elasticity and profit) The price elasticity functions for these demand representations are given by: (i) The isoelastic demand function takes a constant elasticity:

$$ {\epsilon}_{iso} = - \frac{1}{{{\upbeta }}}, $$

(ii)The price elasticity of algebraic demand function is given by

$$ {\epsilon}_{alg} \left( p \right) = - \frac{p}{{\beta \left( {p + \sigma } \right)}}, $$

(iii) The price elasticity of exponential demand function is given by

$$ {\epsilon}_{exp} \left( p \right) = \beta p. $$

Then the profit is a decreasing function in price elasticity:

$$ \frac{{\partial \pi^{*} }}{\partial {\epsilon}} < 0 $$

for the above three cases.

Proof:

1.4 Case I: Isolastic Demand Function

From the proceeding result in the Proposition 3, we have that, under isoelastic demand, the monopoly price and quantity are given by

$$ \begin{aligned} p^{*} = \frac{c}{{1 + {{\upbeta }}}}, \hfill \\ q^{{ \star }} = \left( {\frac{c}{{{{\alpha }}\left( {1 + {{\upbeta }}} \right)}}} \right)^{{1/{{\upbeta }}}} . \hfill \\ \end{aligned} $$

Then we have the monopoly profit is given by

$$ \begin{array}{*{20}c} {{{\uppi }}^{*} \doteq \left( {p^{*} - c} \right)q^{{ \star }} = \left( {\frac{c}{{1 + {{\upbeta }}}}} \right)^{{1 + \frac{1}{{{\upbeta }}}}} \left( {\frac{1}{{{\alpha }}}} \right)^{{\frac{1}{{{\upbeta }}}}} \left( { - {{\upbeta }}} \right).} \\ \end{array} $$

(9)

Plugging ${\epsilon} \doteq - \frac{1}{{{\upbeta }}}$ into Eq. (9), we have

$$ \begin{aligned} {{\uppi }}^{*} & = \left( {\frac{c}{{\epsilon} - 1}} \right)^{1 - {\epsilon}} \left( {\frac{{\epsilon}}{{{\alpha }}}} \right)^{ - {\epsilon}} \\ & = c^{1 - {\epsilon}} \left( {{\epsilon} - 1} \right)^{{\epsilon} - 1} {\epsilon}^{ - {\epsilon}} {{\alpha }}^{{\epsilon}} . \\ \end{aligned} $$

Then we have

$$ \begin{array}{*{20}c} {\ln \pi^{*} = \left( {1 - {\epsilon}} \right)\ln c + \left( {{\epsilon} - 1} \right)\ln \left( {{\epsilon} - 1} \right) - {\epsilon}\left( {\ln {\epsilon} - \ln \alpha } \right)} \\ \end{array} $$

(10)

We can back out the sign of derivative $\frac{{\partial {{\uppi }}^{*} }}{\partial {\epsilon}}$ from the sign of $\frac{{\partial \left( {{{\ln\uppi }}^{*} } \right)}}{\partial {\epsilon}}$. Formally, from $\frac{{\partial \left( {{{\ln\uppi }}^{*} } \right)}}{\partial {\epsilon}} = \frac{1}{{{{\uppi }}^{*} }}\frac{{\partial {{\uppi }}^{*} }}{\partial {\epsilon}}$ and $\pi^{*} > 0,$ we have that the two derivatives $\frac{{\partial {{\uppi }}^{*} }}{\partial {\epsilon}}$ and $\frac{{\partial \left( {{{\ln\uppi }}^{*} } \right)}}{\partial {\epsilon}}$ have the same sign.

From Eq. (10),

$$ \begin{aligned} \frac{{\partial \left( {{{\ln\uppi }}^{*} } \right)}}{\partial {\epsilon}} & = - { \ln }c + { \ln }\left( {{\epsilon} - 1} \right) - \left( {{ \ln }{\epsilon} - {{\ln\upalpha }}} \right) \\ & = { \ln }\frac{{{\alpha }}}{c} + { \ln }\left( {\frac{{\epsilon} - 1}{{\epsilon}}} \right) \\ \end{aligned} $$

Then from the model primitive ${{\upbeta }} \in \left( {0,1} \right)$, we have

$$ \begin{aligned} {{\upbeta }} & \in \left( { - 1,0} \right) \Rightarrow {\epsilon} > 1 \\ & \Rightarrow 0 < \frac{{\epsilon} - 1}{{\epsilon}} < 1 \\ & \Rightarrow { \ln }\left( {\frac{{\epsilon} - 1}{{\epsilon}}} \right) < 0. \\ \end{aligned} $$

Assume that $0 < {{\alpha }} < c$, then $\ln \frac{\alpha }{c} < 0$ then we have, $\forall {\epsilon}$

$$ \frac{{\partial \left( {{{\ln\uppi }}^{*} } \right)}}{\partial {\epsilon}} < 0 \Rightarrow \frac{{\partial {{\uppi }}^{*} }}{\partial {\epsilon}} < 0. $$

1.5 Case II: Algebraic Demand Function

From the proceeding result in the Proposition 2, we have that, $q^{{ \star }} = \left( {\frac{{{{\upsigma }} + c}}{{{{\alpha }}\left( {{{\upbeta }} + 1} \right)}}} \right)^{{1/{{\upbeta }}}}$, and monopolistic price is $p^{*} = \frac{\sigma + c}{\beta + 1} - \sigma$.

Then we have that the monopoly profit is given by

$$ \begin{aligned} {{\uppi }}^{*} & \doteq \left( {p^{*} - c} \right)q^{{ \star }} \\ & = \left( {\frac{{ - {{\upbeta \upalpha }}}}{{1 + {{\upbeta }}}}} \right)\left( {\frac{{{{\upsigma }} + c}}{{{{\alpha }}\left( {1 + {{\upbeta }}} \right)}}} \right)^{{1 + \frac{1}{{{\upbeta }}}}} \\ & = \frac{{ - {{\upbeta \upalpha }}^{{ - \frac{1}{{{\upbeta }}}}} \left( {{{\upsigma }} + c} \right)^{{1 + \frac{1}{{{\upbeta }}}}} }}{{\left( {1 + {{\upbeta }}} \right)^{{1 + \frac{1}{{{\upbeta }}}}} }}, \\ \end{aligned} $$

Plugging ${\epsilon} \doteq - \frac{1}{{{\upbeta }}}$ into the equation, we have

$$ \begin{aligned} \pi^{*} & = \frac{{\alpha^{{\epsilon}} }}{{\epsilon}}\frac{{\left( {\sigma + c} \right)^{1 - {\epsilon}} }}{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }} \\ & = \alpha^{{\epsilon}} {\epsilon}^{ - {\epsilon}} \left( {c + \sigma } \right)^{1 - {\epsilon}} \left( {{\epsilon} - 1} \right)^{{\epsilon} - 1} . \\ \end{aligned} $$

Then we have

$$ \begin{array}{*{20}c} {\ln \pi^{*} = \left( {{\epsilon} - 1} \right)\ln \left( {{\epsilon} - 1} \right) + \left( {1 - {\epsilon}} \right)\ln \left( {\sigma + c} \right) + {\epsilon}\ln \alpha - {\epsilon}\ln {\epsilon}.} \\ \end{array} $$

(11)

We can back out the sign of derivative $\frac{{\partial \pi^{*} }}{\partial {\epsilon}}$ from the sign of $\frac{{\partial \left( {{ \ln }\pi^{*} } \right)}}{\partial {\epsilon}}$. Formally, from $\frac{{\partial \left( {{ \ln }\pi^{*} } \right)}}{\partial {\epsilon}} = \frac{1}{{\pi^{*} }}\frac{{\partial \pi^{*} }}{\partial {\epsilon}}$, and $\pi^{*} > 0,$ we have that the two derivatives $\frac{{\partial \pi^{*} }}{\partial {\epsilon}}$ and $\frac{{\partial \left( {{ \ln }\pi^{*} } \right)}}{\partial {\epsilon}}$ have the same sign.

From Eq. (11)

$$ \frac{{\partial \left( {{ \ln }\pi^{*} } \right)}}{\partial {\epsilon}} = \ln \left( {{\epsilon} - 1} \right) - \ln {\epsilon} - { \ln }\left( {\sigma + c} \right) + { \ln }\alpha , $$

Then from the model primitive $\beta \in \left( { - 1,0} \right)$, we have

$$ \begin{aligned} \beta \in \left( { - 1,0} \right) \Rightarrow {\epsilon} > 1 \hfill \\ \Rightarrow { \ln }\left( {{\epsilon} - 1} \right) - { \ln }{\epsilon} < 0. \hfill \\ \end{aligned} $$

Similar to the analysis of the isoelastic demand function, if $c + \sigma > \alpha$,we have that

$$ \frac{{\partial \left( {{{\ln\uppi }}^{*} } \right)}}{\partial {\epsilon}} < 0 \Rightarrow \frac{{\partial {{\uppi }}^{*} }}{\partial {\epsilon}} < 0. $$

1.6 Case III: The Exponential Demand Function

From the previous result in the Proposition 4, the monopoly quantity, under the exponential demand function, is $q^{{ \star }} = {{\upgamma }}e^{{ - \left( {{{\upbeta }}c + 1} \right)}}$ and the monopolistic price is $p^{{ \star }} = \frac{{{{\upbeta }}c + 1}}{{{\upbeta }}}$. The monopoly profit is given by

$$ {{\uppi }}^{*} = \left( {p^{*} - c} \right)q^{{ \star }} = \frac{{{\upgamma }}}{{{\upbeta }}}e^{{\left( { - 1 - {{\upbeta }}c} \right)}} . $$

From the proceeding result, we have that the elasticity parameter ${\epsilon}_{exp}^{A} = \beta$. Thus, to analyze the effect of the market elasticity parameter on the monopoly profit, we take the partial derivative of profit with regard to ${{\upbeta }}$, i.e.,

$$ \frac{{\partial {{\uppi }}^{*} }}{{\partial {{\upbeta }}}} = - {{\upgamma }}e^{{\left( { - 1 - {{\upbeta }}c} \right)}} - \frac{{{{\upgamma }}e^{{\left( { - 1 - {{\upbeta }}c} \right)}} }}{{{{\upbeta }}^{2} }} = {{\upgamma }}e^{{\left( { - 1 - {{\upbeta }}c} \right)}} \left[ { - 1 - \frac{1}{{{{\upbeta }}^{2} }}} \right]. $$

From the demand function’s primitive setup, we have that ${{\upgamma }} > 0$, and $\beta > 0$. Thus,

$$ \frac{{\partial {{\uppi }}^{*} }}{{\partial {{\upbeta }}}} < 0. $$

1.7 Isoelastic Demand and the Uniform Distribution of Investment Cost

Assume that the R&D investment cost follows a uniform distribution on $\left( {0,\bar{K}} \right]$. To compute the adjusted error cost ratio, we first compute the probability of a firm with the firm’s entry decision in period 1. Then we have the entry probability—denoted as ${{\uptheta }}$—is given by

$$ \begin{array}{*{20}c} {\theta = \frac{{\left( {1 - v} \right){{\uppi }}^{*} }}{{\bar{K}}}.} \\ \end{array} $$

(12)

Then we have that the adjusted ratio is given by

$$ \begin{aligned} {\tilde{{\uprho }}} & = \frac{{1 - {{\uptheta }}}}{{{\uptheta }}}{{\uprho }} \\ & = \left( {\frac{{\bar{K}}}{{\left( {1 - v} \right){{\uppi }}^{ *} }} - 1} \right), \\ \end{aligned} $$

where the firm enters regardless of enforcement, if $\left( {1 - v} \right){{\uppi }}^{*} > \bar{K}$, thus,

$$ {\tilde{{\uprho }}} = \left\{ {\begin{array}{*{20}l} {\left( {\frac{{\bar{K}}}{{\left( {1 - v} \right)\pi^{*} }} - 1} \right)\rho ,} \hfill & {{\text{if}}\; \left( {1 - v} \right)\pi^{*} \le \bar{K}} \hfill \\ {0,} \hfill & {{\text{if}}\; \left( {1 - v} \right)\pi^{*} > \bar{K}} \hfill \\ \end{array} } \right. $$

When the elasticity is not small, such that $\left( {1 - v} \right){{\uppi }}^{*} \le \bar{K}$, the adjusted error cost ratio that takes into account the dynamic effect of enforcement can now be expressed as a function of the demand elasticity

$$ \begin{aligned} {\tilde{{\uprho }}} & = \frac{{1 - {{\uptheta }}}}{{{\uptheta }}}{{\uprho }} \\ & = \left( {\frac{{\bar{K}}}{{\left( {1 - v} \right){{\uppi }}^{*} }} - 1} \right)\left[ {\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} \left( {2 - \frac{1}{{\epsilon}}} \right)^{ - 1} - 1} \right]^{ - 1} . \\ \end{aligned} $$

1.8 Static and Dynamic Effects under Isoelastic Demand and the Uniform Distribution of Investment Cost

The static effect is defined by $\frac{{1 - {{\uptheta }}}}{{{\uptheta }}}\frac{{\partial {{\uprho }}}}{\partial {\epsilon}}$, and hence it is given by

$$ \frac{{1 - {{\uptheta }}}}{{{\uptheta }}}\frac{{\partial {{\uprho }}}}{\partial {\epsilon}} = \left( {\frac{{\bar{K}}}{{\left( {1 - v} \right){{\uppi }}^{*} }} - 1} \right)\frac{{\partial {{\uprho }}}}{\partial {\epsilon}} $$

From Eqs. (10) and (11), we have

$$ \begin{aligned} \frac{{d{{\uprho }}}}{d{\epsilon}} & = - {{\uprho }}^{2} \frac{{d\left( {\frac{1}{{{\uprho }}}} \right)}}{d{\epsilon}} \\ & = - {{\uprho }}^{2} \frac{{\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} }}{{2 - \frac{1}{{\epsilon}}}}\left[ {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right] \\ & = - {{\uprho }}^{2} \left( {\frac{1}{{{\uprho }}} + 1} \right)\left[ {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right] \\ & = - {{\uprho }}\left( {1 + {{\uprho }}} \right)\left[ {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right]. \\ \end{aligned} $$

where ${{\uprho }} = \left[ {\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} \left( {2 - \frac{1}{{\epsilon}}} \right)^{ - 1} - 1} \right]^{ - 1}$.

Thus, the static effect is given by

$$ \begin{array}{*{20}c} {\frac{{1 - {{\uptheta }}}}{{{\uptheta }}}\frac{{\partial {{\uprho }}}}{\partial {\epsilon}} = \left( {\frac{{\bar{K}}}{{\left( {1 - v} \right){{\uppi }}^{*} }} - 1} \right)\left[ { - {{\uprho }}\left( {1 + {{\uprho }}} \right)} \right]\left[ {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right]} \\ \end{array} $$

(13)

The dynamic effect is defined by $- \frac{{{\uprho }}}{{{{\uptheta }}^{2} }}\frac{{\partial {{\uptheta }}}}{\partial {\epsilon}}$. Plugging Eq. (12) in the expression of dynamic effect, we have

$$ \begin{aligned} - \frac{{{\uprho }}}{{{{\uptheta }}^{2} }}\frac{{\partial {{\uptheta }}}}{\partial {\epsilon}} & = - {{\uprho }}\left( {\frac{{\left( {1 - v} \right){{\uppi }}^{*} }}{{\bar{K}}}} \right)^{ - 2} \frac{{\partial {{\uptheta }}}}{{\partial {{\uppi }}^{*} }}\frac{{\partial {{\uppi }}^{*} }}{\partial {\epsilon}} \\ & = - {{\uprho }}\frac{{\frac{{\left( {1 - v} \right)}}{{\bar{K}}}}}{{\left( {\frac{{\left( {1 - v} \right){{\uppi }}^{*} }}{{\bar{K}}}} \right)^{2} }}\frac{{\partial {{\uppi }}^{*} }}{\partial {\epsilon}} \\ \begin{array}{l} { = \frac{{ - {{\uprho }}}}{{\left( {{{\uppi }}^{*} } \right)^{2} }}\frac{{\bar{K}}}{{\left( {1 - v} \right)}}\frac{{\partial {{\uppi }}^{*} }}{\partial {\epsilon}}.} \\ \end{array} \\ \end{aligned} $$

(14)

From the previous result, we have

$$ \begin{array}{*{20}c} {\frac{{\partial {{\uppi }}^{*} }}{\partial {\epsilon}} = {{\uppi }}^{*} \frac{{\partial \ln {{\uppi }}^{*} }}{\partial {\epsilon}}} \\ \end{array} $$

(15)

Combining Eqs. (15), (10), and the foregoing equation, we have

$$ \begin{array}{*{20}c} {\frac{{\partial {{\uppi }}^{*} }}{\partial {\epsilon}} = {{\uppi }}^{*} \left[ {{ \ln }\frac{{{\alpha }}}{c} + { \ln }\left( {\frac{{\epsilon} - 1}{{\epsilon}}} \right)} \right].} \\ \end{array} $$

(16)

Combining Eqs. (16) and (14), we have the dynamic effect is given by

$$ \begin{aligned} - \frac{{{\uprho }}}{{{{\uptheta }}^{2} }}\frac{{\partial {{\uptheta }}}}{\partial {\epsilon}} & = \frac{{ - {{\uprho }}}}{{\left( {{{\uppi }}^{*} } \right)^{2} }}\frac{{\bar{K}}}{{\left( {1 - v} \right)}}{{\uppi }}^{*} \left[ {{ \ln }\frac{{{\alpha }}}{c} + { \ln }\left( {\frac{{\epsilon} - 1}{{\epsilon}}} \right)} \right] \\ & \begin{array}{*{20}c} { = \frac{{ - {{\uprho }}}}{{{{\uppi }}^{*} }}\frac{{\bar{K}}}{{\left( {1 - v} \right)}}\left[ {{ \ln }\frac{{{\alpha }}}{c} + { \ln }\left( {\frac{{\epsilon} - 1}{{\epsilon}}} \right)} \right].} \\ \end{array} \\ \end{aligned} $$

(17)

We can compare the magnitude of the static and dynamic effects by the absolute value of Eqs. (13) and (17). We have that the static effect dominates dynamic effect if and only if

$$ \begin{array}{*{20}c} {\left( {\frac{{\bar{K}}}{{\left( {1 - v} \right){{\uppi }}^{*} }} - 1} \right)\left[ {{{\uprho }}\left( {1 + {{\uprho }}} \right)} \right]\left[ {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right] > \frac{{ - {{\uprho }}}}{{{{\uppi }}^{*} }}\frac{{\bar{K}}}{{\left( {1 - v} \right)}}\left[ {{ \ln }\frac{{{\alpha }}}{c} + { \ln }\left( {\frac{{\epsilon} - 1}{{\epsilon}}} \right)} \right]} \\ \end{array} $$

(18)

$$ \Leftrightarrow \begin{array}{*{20}c} {{\tilde{{\uprho }}}\left( {1 + {{\uprho }}} \right)\left[ {{ \ln }\left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right] > \frac{{\bar{K}}}{{\left( {1 - v} \right){{\uppi }}^{*} }}\rho \left[ {{ \ln }\left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - { \ln }\frac{{{\alpha }}}{c}} \right] } \\ \end{array} $$

(19)

$$ \begin{array}{*{20}c} { \Leftrightarrow \left( {\bar{K} - \left( {1 - v} \right){{\uppi }}^{*} } \right)\left( {1 + {{\uprho }}} \right)\left[ {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right] > \bar{K}\left[ {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - { \ln }\frac{{{\alpha }}}{c}} \right].} \\ \end{array} $$

(20)

From equation ${{\uprho }} = \left[ {\left( {1 - \frac{1}{{\epsilon}}} \right)^{1 - {\epsilon}} \left( {2 - \frac{1}{{\epsilon}}} \right)^{ - 1} - 1} \right]^{ - 1}$, we have $1 + {{\uprho }} = \frac{{{\epsilon}^{{\epsilon}} }}{{{\epsilon}^{{\epsilon}} - \left( {2{\epsilon} - 1} \right)\left( {{\epsilon} - 1} \right)^{{\epsilon} - 1} }}.$

Plugging this back into the comparison inequality (18), we have that the static effect dominates, if

$$ \frac{{\left( {\bar{K} - \left( {1 - v} \right){{\uppi }}^{*} } \right){\epsilon}^{{\epsilon}} }}{{{\epsilon}^{{\epsilon}} - \left( {2{\epsilon} - 1} \right)\left( {{\epsilon} - 1} \right)^{{\epsilon} - 1} }}\left[ {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - \frac{1}{{{\epsilon} - \frac{1}{2}}}} \right] > \bar{K}\left[ {\ln \left( {\frac{{\epsilon}}{{\epsilon} - 1}} \right) - { \ln }\frac{{{\alpha }}}{c}} \right]. $$

Appendix B. Simulations with Varying Enforcement Probability

In this appendix, we show how the adjusted error-cost ratio changes according to the antitrust enforcement probability. In the main text, we show the adjusted error-cost ratio and the corresponding static and dynamic effects under a modest enforcement probability: $v = 0.2.$ If the enforcement probability increases, the adjusted error-cost ratio curve shifts upward. The dynamic effect curves shift upward as well, while the change in the static effect is smaller than the change in the dynamic effect.

2.1 Medium Enforcement Probability

In Fig. 4, we assume that the enforcement probability is at the intermediate level: $v = 0.5$. The adjusted error-cost ratio is higher than that under $v = 0.2$ (see Fig. 3 in main text). The lower bound of the adjusted error-cost ratio now is about $13$, which occurs when the demand elasticity is about $1.09$

Fig. 4

Medium Enforcement Probability v = 0.5

2.2 High Enforcement Probability

In Fig. 5, we assume that the enforcement probability is high, $v = 0.8$. The adjusted error-cost ratio curve shifts up further in Fig. 5 than in Fig. 4. The lower bound of the adjusted error-cost ratio now is about $42$, which occurs when the demand elasticity is about $1.11$.

Fig. 5

High Enforcement Probability v = 0.8