Would you trust lobbies?

Abstract

We consider the regulatory problem to approve or to ban a new product/technology in the context of scientific controversy about its detrimental environmental and/or health effects. We formalize the regulator’s decision-making process as a Tullock contest (Towards a Theory of the Rent-Seeking Society, Texas A&M University Press, Texas, 1980), the contestants being an industrial lobby, representing the economic agents who have developed the new product/technology, and an environmental lobby, representing the economic agents who will be harmed by the product or technology. Assuming that the industrial lobby has private information about the unfavorable environmental and/or health effects, but can be held liable for damage ex post, we derive the equilibrium properties of the contest. In particular, we derive conditions under which it is socially preferable for the regulator to decide according to the contest’s outcome, rather than according to an ex ante cost-benefit analysis, using his prior beliefs. We find that the contest outperforms the ex ante cost-benefit analysis only if the risk of judgment-proofness is not too high.

Appendix

1.1 Existence of Nash equilibrium with full compensation of damage

We show that if $\pi \left( 0,0\right) =\alpha <1$, then no equilibrium exists when lobby E is fully compensated for damage. Indeed, if fully compensated for damage, $y=0$ is a dominant strategy for lobby E. As a result, lobby I chooses x to maximize $\pi \left( x,0\right) \left( b-\min \left( \delta ,k\right) \right) -x$. If $\alpha <1$, no equilibrium strategy exists for lobby I. Indeed, the strategy $x=0$ is dominated by any strategy x such that $0<x<\left( 1-\alpha \right) \left( b-\min \left( \delta ,k\right) \right)$; and for all $x>0$, the expected utility is strictly decreasing in x.

Proof of Property 1

For all b and k, let $\phi \left( b,k\right)$ be defined by $\phi \left( b,k\right) \equiv 8\left( b-k\right) ^{2}+\left( 1-k\right) \left( 1-2b+k^{2}\right)$.

We distinguish two cases:

• If $\delta \le k$, we show that

$${{\textstyle \frac{\partial }{\partial k}}\left( x^{*}\left( \delta \right) +y^{*}\left( \delta \right) \right) {\textstyle =-\frac{1-k}{\left( 2\left( b-k\right) +\left( 1-k\right) ^{2}\right) ^{2}}\sqrt{\frac{b-\delta }{b-k}}\frac{\phi \left( b,k\right) }{2}},}$$

$${\textstyle \frac{\partial }{\partial k}\pi \left( x^{*}\left( \delta \right) ,y^{*}\left( \delta \right) \right) }={\textstyle \frac{1-k}{\left( 2\left( b-k\right) +\left( 1-k\right) ^{2}\right) ^{2}}\sqrt{\frac{b-\delta }{b-k}}\frac{\phi \left( b,k\right) }{2\left( b-k\right) }},$$

$${\textstyle \frac{\partial }{\partial k}\left( \begin{array}{c} \pi \left( x^{*}\left( \delta \right) ,y^{*}\left( \delta \right) \right) \left( b-\delta \right) \\ -x^{*}\left( \delta \right) -y^{*}\left( \delta \right) \end{array}\right) }{\textstyle =\frac{1-k}{\left( 2\left( b-k\right) +\left( 1-k\right) ^{2}\right) ^{2}}\sqrt{\frac{b-\delta }{b-k}}\phi \left( b,k\right) .}$$

• If $\delta >k$, we show that

$${\textstyle \frac{\partial }{\partial k}\left( x^{*}\left( \delta \right) +y^{*}\left( \delta \right) \right) {\textstyle =-\frac{1-k}{\left( 2\left( b-k\right) +\left( 1-k\right) ^{2}\right) ^{2}}\left( 4\left( b-k\right) ^{2}+\left( 1-k\right) ^{3}\right) },}$$

$${\textstyle \frac{\partial }{\partial k}\pi \left( x^{*}\left( \delta \right) ,y^{*}\left( \delta \right) \right) {\textstyle =-2\frac{1-k}{\left( 2\left( b-k\right) +\left( 1-k\right) ^{2}\right) ^{2}}\left( 1-2b+k\right) },}$$

$${\textstyle \frac{\partial }{\partial k}\left( \begin{array}{c} \pi \left( x^{*}\left( \delta \right) ,y^{*}\left( \delta \right) \right) \left( b-\delta \right) \\ -x^{*}\left( \delta \right) -y^{*}\left( \delta \right) \end{array}\right) {\textstyle =\frac{1-k}{\left( 2\left( b-k\right) +\left( 1-k\right) ^{2}\right) ^{2}}\left( \begin{array}{c} \phi \left( b,k\right) \\ +2\left( 1-2b+k\right) \left( \delta -k\right) \end{array}\right) }.}$$

Property 1 follows directly, assuming that b and k are such that $\phi \left( b,k\right) >0$. To obtain the last sign, denote $\varphi \left( b,k,\delta \right) \equiv \phi \left( b,k\right) +2\left( 1-2b+k\right) \left( \delta -k\right)$. Then:

• If $1\,-\,2b\,+\,k\ge 0$, then $\varphi \left( b,k,\delta \right) >0$ directly follows from $\delta >k$ and $\phi \left( b,k\right) >0$;

• If $1-2b+k<0$, then $\varphi \left( b,k,\delta \right)$ is decreasing in $\delta$ and is bounded below by $\varphi \left( b,k,1\right)$.

Then remark that $\varphi \left( b,k,1\right) =8b^{2}-2\left( 3+5k\right) b+\left( 3-k+7k^{2}-k^{3}\right)$ and that this quadratic polynomial in b is strictly positive.¹⁹

Finally, we can check that the condition that $\phi \left( b,k\right) >0$ is not very restrictive. Rearrange $\phi \left( b,k\right)$ to write it as $\phi \left( b,k\right) =8b^{2}-2\left( 1+7k\right) b+\left( 1-k+9k^{2}-k^{3}\right)$. We have:

• If $k<7/8$, this quadratic polynomial in b is strictly positive;²⁰

• If $k\ge 7/8$, it is positive if $b\notin \left[ k+\left( 1-k\right) \left( 1-\sqrt{8k-7}\right) /8,k+\left( 1-k\right) \left( 1+\sqrt{8k-7}\right) /8\right]$.

$\square$

1.2 Calculus of the ex ante expected social surplus (private information)

The ex ante expected social surplus is

$$s^{*}\left( b,k\right) =\int \nolimits _{0}^{1}\left[ \pi \left( x^{*}\left( \delta \right) ,y^{*}\right) \left( b-\delta \right) -x^{*}\left( \delta \right) -y^{*}\right] d\delta .$$

Using (7) and (8) and integrating, we get

$$\begin{aligned}s^{*}\left( b,k\right)&=\int \nolimits _{0}^{1}\left( b-\delta \right) d\delta -\frac{\left( 1-k\right) ^{2}}{2\left( b-k\right) +\left( 1-k\right) ^{2}}\left( \begin{array}{c} 2\sqrt{b-k}\int \nolimits _{0}^{k}\sqrt{b-\delta }d\delta \\ +\int _{k}^{1}\left( 2b-\delta -k\right) d\delta \end{array}\right) \\&=\left( b-\frac{1}{2}\right) -\frac{\left( 1-k\right) ^{2}}{2\left( b-k\right) +\left( 1-k\right) ^{2}}\left( \begin{array}{c} \frac{4}{3}\sqrt{b-k}\left( b^{\frac{3}{2}}-\left( b-k\right) ^{\frac{3}{2}}\right) \\ -\frac{1}{2}\left( 1-k\right) \left( 1-4b+3k\right) \end{array}\right) .\end{aligned}$$

Rearranging, this yields

$$s^{*}\left( b,k\right) =\left( b-\frac{1}{2}\right) +\frac{\left( 1-k\right) ^{2}}{2\left( b-k\right) +\left( 1-k\right) ^{2}}\left( \frac{1}{2}-2b+k+\frac{1}{3}b^{2}f\left( \frac{k}{b}\right) \right) ,$$

where we define

$$f\left( \frac{k}{b}\right) =4\left( 1-\sqrt{1-\frac{k}{b}}\right) -2\frac{k}{b}-\frac{1}{2}\left( \frac{k}{b}\right) ^{2} {\text{for}\,\text{all}}\,0\le \frac{k}{b}\le 1.$$

Proof of Proposition 1

We first show the following lemmas.

Lemma 1

$f\left( \frac{k}{b}\right) <\frac{3}{2}\left( \frac{k}{b}\right) ^{2}$, for all $0<\frac{k}{b}<1$.

Proof

Simply observe that we can write

$$f\left( \frac{k}{b}\right) -\frac{3}{2}\left( \frac{k}{b}\right) ^{2}=2\left( \left( 1-\sqrt{1-\frac{k}{b}}\right) ^{2}-\left( \frac{k}{b}\right) ^{2}\right) .$$

This expression is negative since $1-\sqrt{1-\frac{k}{b}}<\frac{k}{b}$ for all $0<\frac{k}{b}<1$. $\square$

Lemma 2

If $\phi \left( b,k\right) >0$, the expected social surplus $s^{*}\left( b,k\right)$ is increasing in k. If $\phi \left( b,k\right) \le 0$, $s^{*}\left( b,k\right) <b-1/2\le {\mathrm{max}}\left( 0,b-1/2\right)$.

Proof

The first part follows directly from Property 1. To show the second part, first note that $\phi \left( b,k\right) \le 0$ implies $1-2b+k^{2}<0$. Then use Lemma 1 to show that

$$s^{*}\left( b,k\right) <\left( b-\frac{1}{2}\right) +\frac{\left( 1-k\right) ^{2}}{2\left( b-k\right) +\left( 1-k\right) ^{2}}\left( \frac{1}{2}-2b+k+\frac{1}{2}k^{2}\right) .$$

Rearrange the term under brackets to write

$$s^{*}\left( b,k\right) <\left( b-\frac{1}{2}\right) +\frac{\left( 1-k\right) ^{2}}{2\left( b-k\right) +\left( 1-k\right) ^{2}}\left( \frac{1}{2}\left( 1-2b+k^{2}\right) -\left( b-k\right) \right) ,$$

which proves the second part, given that $k<b$. $\square$

The rest of the proof distinguishes two cases:

• For all $b\le 1/2$, letting $\tau \left( b\right) \equiv 2\sqrt{1-b}\left( 1-\sqrt{1-b}\right) /b\in \left( 0,1\right)$, we show that there exists $a\left( b\right) \in \left( \tau \left( b\right) ,1\right)$ such that $s^{*}\left( b,a\left( b\right) b\right) =0$.

We first show that $s^{*}\left( b,\tau \left( b\right) b\right) <0$. Indeed, Lemma 1 implies that

$$s^{*}\left( b,k\right) <\left( b-\frac{1}{2}\right) +\frac{\left( 1-k\right) ^{2}}{2\left( b-k\right) +\left( 1-k\right) ^{2}}\left( \frac{1}{2}-2b+k+\frac{1}{2}k^{2}\right) .$$

For all k, remark that²¹

$$\frac{\left( 1-k\right) ^{2}}{2\left( b-k\right) +\left( 1-k\right) ^{2}}<\frac{1}{2\left( b-k\right) +1}.$$

In particular, if $k=\tau \left( b\right) b$, we obtain (after rearrangement)

$$\frac{\left( 1-k\right) ^{2}}{2\left( b-k\right) +\left( 1-k\right) ^{2}}<\frac{1}{1-2b+4\left( 1-\sqrt{1-b}\right) },$$

and

$$\frac{1}{2}-2b+k+\frac{1}{2}k^{2}=\left( \frac{1}{2}-b\right) \left( 1-2b+4\left( 1-\sqrt{1-b}\right) \right) \ge 0,$$

implying that

$$s^{*}\left( b,\tau \left( b\right) b\right) <\left( b-\frac{1}{2}\right) +\frac{1-2b+4\left( 1-\sqrt{1-b}\right) }{1-2b+4\left( 1-\sqrt{1-b}\right) }\left( \frac{1}{2}-b\right) =0.$$

We then show that $s^{*}\left( b,b\right) =b^{2}/2>0$.

From the mean-value theorem, there exists $a\left( b\right)$ such that $\tau \left( b\right)<a\left( b\right) <1$ and $s^{*}\left( b,a\left( b\right) b\right) =0$. From Lemma 2, $a\left( b\right)$ is unique and $s^{*}\left( b,k\right) >0$ for all $k/b>a\left( b\right)$.

• For all $b>1/2$, letting $\tau \left( b\right) =\left( 2\sqrt{b}-1\right) /b\in \left( 0,1\right)$, we show that there exists $a\left( b\right) \in \left( \tau \left( b\right) ,1\right)$ such that $s^{*}\left( b,a\left( b\right) b\right) =b-1/2$.

We first show that $s^{*}\left( b,\tau \left( b\right) b\right) <b-1/2$. Indeed, Lemma 1 implies that

$$s^{*}\left( b,k\right) -\left( b-\frac{1}{2}\right) <\frac{\left( 1-k\right) ^{2}}{2\left( b-k\right) +\left( 1-k\right) ^{2}}\left( \frac{1}{2}-2b+k+\frac{1}{2}k^{2}\right) .$$

In particular, if $k=\tau \left( b\right) b$, we get

$$s^{*}\left( b,\tau \left( b\right) b\right) -\left( b-\frac{1}{2}\right) <0.$$

We then show that $s^{*}\left( b,b\right) -\left( b-\frac{1}{2}\right) =\left( 1-b\right) ^{2}/2>0$.

From the mean-value theorem, there exists $a\left( b\right)$ such that $\tau \left( b\right)<a\left( b\right) <1$ and $s^{*}\left( b,a\left( b\right) b\right) =b-1/2$. From Lemma 2, $a\left( b\right)$ is unique and $s^{*}\left( b,k\right) >b-1/2$, for all $k/b>a\left( b\right)$.

1.3 Calculus of the ex ante expected social surplus (perfect information)

The ex ante expected social surplus is

$$S^{*}\left( b,k\right) =\int \nolimits _{0}^{1}\left[ \pi \left( X^{*}\left( \delta \right) ,Y^{*}\left( \delta \right) \right) \left( b-\delta \right) -X^{*}\left( \delta \right) -Y^{*}\left( \delta \right) \right] d\delta .$$

Using (14) and (15) and integrating, we get

$$\begin{aligned}S^{*}\left( b,k\right)&=\int \nolimits _{0}^{k}\left( b-\delta \right) d\delta -\left( b-k\right) \int _{k}^{1}\left( \frac{b-2\delta +k}{b+\delta -2k}\right) d\delta \\&=-2b+2k+3bk-\frac{5}{2}k^{2}+3\left( b-k\right) ^{2}\ln \left( 1+\frac{1-k}{b-k}\right) . \end{aligned}$$

observing that $\left( b-2\delta +k\right) /\left( b+\delta -2k\right)$ admits $3\left( b-k\right) \ln \left( b+\delta -2k\right) -2\delta$ as an antiderivative.

Proof of Proposition 3

We show below that $S^{*}\left( b,k\right) >\max \left( 0,b-1/2\right)$ for all $k/b\ge \tau \left( b\right)$. We need the following lemmas.

Lemma 3

$S^{*}\left( b,k\right)$ is increasing in k.

Proof

First use (14) and (15) to calculate the ex post expected social surplus:

$$\left( \begin{array}{l} \pi \left( X^{*}\left( \delta \right) ,Y^{*}\left( \delta \right) \right) \left( b-\delta \right) \\ -X^{*}\left( \delta \right) -Y^{*}\left( \delta \right) \end{array}\right) ={\left\{ \begin{array}{ll} b-\delta , & {\text{if }} \delta \le k,\\ \frac{\left( b-k\right) \left( b-2\delta +k\right) }{b+\delta -2k}, & {\text{otherwise}}. \end{array}\right. }$$

Then show by differentiation that it is increasing in k:

$$\frac{\partial }{\partial k}\left( \begin{array}{l} \pi \left( X^{*}\left( \delta \right) ,Y^{*}\left( \delta \right) \right) \left( b-\delta \right) \\ -X^{*}\left( \delta \right) -Y^{*}\left( \delta \right) \end{array}\right) ={\left\{ \begin{array}{ll} 0, & {\text{if } }\delta \le k,\\ \frac{\left( b-\delta \right) ^{2}+\left( b-k\right) ^{2}+\left( \delta -k\right) ^{2}}{\left( b+\delta -2k\right) ^{2}}, & {\text{otherwise}}. \end{array}\right. }$$

This clearly implies that the ex ante expected social surplus $S^{*}\left( b,k\right)$ is also increasing in k. $\square$

Lemma 4

$S^{*}\left( b,k\right) >-2b+2k-bk+2b^{2}-\frac{1}{2}k^{2}.$

Proof

By Assumption 1, we can show that

$$\ln \left( 1+\frac{1-k}{b-k}\right)>\ln \left( 2\right) >2/3.$$

After substitution, this directly implies that

$$S^{*}\left( b,k\right) >-2b+2k+3bk-\frac{5}{2}k^{2}+2\left( b-k\right) ^{2},$$

which gives the lemma after simplification. $\square$

The rest of the proof distinguishes two cases:

• For all $b\le 1/2$, using Lemma 4, we can write (after rearrangement)

  $$S^{*}\left( b,\tau \left( b\right) b\right) >2\sqrt{1-b}\left( 1-\sqrt{1-b}\right) ^{3},$$

  which is clearly positive.

• For all $b>1/2$, using Lemma 4, we can write (after rearrangement)

  $$S^{*}\left( b,\tau \left( b\right) b\right) -\left( b-\frac{1}{2}\right) >2\left( b+\sqrt{b}-1\right) \left( 1-\sqrt{b}\right) ^{3},$$

  which is clearly positive.