Bureaucratic Shirking, Corruption, and Firms’ Environmental Investment and Abatement

Abstract

Bureaucratic shirking and corruption are prevalent in developing countries. This paper presents a delegation model where a government authorizes an inspector to monitor a polluting firm. The inspector may shirk in monitoring and may accept bribery when discovering noncompliance. We distinguish between two types of environmentally friendly actions, emission abatement and investment to enhance abatement technology, and investigate how bureaucratic shirking and corruption affect firms’ incentives of taking these actions. A corruptible inspector exerts more effort in monitoring the firm (an effort-inducing effect) but fails to enforce environmental regulations when discovering noncompliance (a nonenforcement effect), compared to an (incorruptible) bureaucratic inspector. Moreover, the firm strategically makes more investment to reduce the corruptible inspector’s monitoring effort (a strategic effect on monitoring). We find that investment and abatement decrease when corruption becomes more widespread, only if the corruptible inspector has sufficiently small bargaining power. Moreover, the corruptible inspector’s higher bargaining power leads to higher investment and abatement.

Appendices

Appendix 1: Proofs

Proof

(Lemma 1) Differentiating the system of equations, (3) and (4), with respect to $\pi$, gives

$$\begin{aligned} c_{11}\frac{\partial a^{*}}{\partial \pi }-t\pi k\frac{\partial p^{*} }{\partial \pi }&=t\left( p^{*}k-p_{0}\right) ;\\ kt\frac{\partial a^{*}}{\partial \pi }+e^{\prime \prime }\frac{\partial p^{*}}{\partial \pi }&=0. \end{aligned}$$

Note that these are done for any given investment level I, so the derivatives are in the partial notion. Ditto for the following derivatives. In matrix notation, we have

$$\begin{aligned} \left[ \begin{array} [c]{cc} c_{11} &{} -t\pi k\\ kt &{} e^{\prime \prime } \end{array} \right] \left[ \begin{array} [c]{c} \frac{\partial a^{*}}{\partial \pi }\\ \frac{\partial p^{*}}{\partial \pi } \end{array} \right] =\left[ \begin{array} [c]{c} t\left( p^{*}k-p_{0}\right) \\ 0 \end{array} \right] . \end{aligned}$$

By the Cramer’s rule,

$$\begin{aligned} \frac{\partial a^{*}}{\partial \pi }= & {} \det \left[ \begin{array} [c]{cc} t\left( p^{*}k-p_{0}\right) &{} -t\pi k\\ 0 &{} e^{\prime \prime } \end{array} \right] /\det \left[ \begin{array} [c]{cc} c_{11} &{} -t\pi k\\ kt &{} e^{\prime \prime } \end{array} \right] ; \\ \frac{\partial p^{*}}{\partial \pi }= & {} \det \left[ \begin{array} [c]{cc} c_{11} &{} t\left( p^{*}k-p_{0}\right) \\ kt &{} 0 \end{array} \right] /\det \left[ \begin{array} [c]{cc} c_{11} &{} -t\pi k\\ kt &{} e^{\prime \prime } \end{array} \right] . \end{aligned}$$

Simplifying them gives

$$\begin{aligned} \frac{\partial a^{*}}{\partial \pi }=\frac{\left( p^{*}k-p_{0}\right) te^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}, \end{aligned}$$

(15)

and

$$\begin{aligned} \frac{\partial p^{*}}{\partial \pi }=\frac{-\left( p^{*}k-p_{0}\right) kt^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}. \end{aligned}$$

Given the curvature assumptions, $c_{11}e^{\prime \prime }+\pi t^{2}k^{2}>0$. Therefore, $sign\left( \frac{\partial a^{*}}{\partial \pi }\right) =-sign\left( \frac{\partial p^{*}}{\partial \pi }\right) =sign\left( p^{*}k-p_{0}\right)$, which proves Part 1 of the lemma.

To show Part 2, steps similar to the above establish

$$\begin{aligned} \frac{\partial p^{*}}{\partial k}=\frac{\partial p^{*}}{\partial t}=\frac{c_{11}t\left( 1-a^{*}\right) -t^{2}k\pi p^{*}}{c_{11} e^{\prime \prime }+\pi t^{2}k^{2}}. \end{aligned}$$

Then $\frac{\partial }{\partial k}\left( p^{*}k\right) =\frac{\partial p^{*}}{\partial k}k+p^{*}=\frac{kc_{11}t\left( 1-a^{*}\right) +p^{*}c_{11}e^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\ge 0$ given our curvature assumptions. Obviously $\lim _{k\rightarrow 0}p^{*}k<p_{0}$. Meanwhile from (4) we have $\lim _{k\rightarrow 1}p^{*}>p_{0}$ and so $\lim _{k\rightarrow 1}p^{*}k>p_{0}$. By these facts and the continuity of $p^{*}k$, the threshold ${{\bar{k}}}$, which depends on I, must exist.

Steps similar to the above give us

$$\begin{aligned} \frac{da^{*}}{dI}=\frac{-c_{12}e^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\ge 0, \end{aligned}$$

(16)

$$\begin{aligned} \frac{dp^{*}}{dI}=\frac{ktc_{12}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2} }\le 0, \end{aligned}$$

(17)

$$\begin{aligned} \frac{\partial a^{*}}{\partial k}=\frac{e^{\prime \prime }t\pi p^{*} +t^{2}k\pi \left( 1-a^{*}\right) }{c_{11}e^{\prime \prime }+\pi t^{2}k^{2} }\ge 0, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial a^{*}}{\partial t}=\frac{e^{\prime \prime }\left( \pi kp^{*}+\left( 1-\pi \right) p_{0}\right) +t^{2}k\pi \left( 1-a^{*}\right) }{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\ge 0, \end{aligned}$$

which prove Parts 3–4. $\square$

Proof

(Proposition 2) Define $H\left( I\right) \equiv$$\frac{dL\left( I\right) }{dI}$. With the first order condition $H\left( I\right) |_{I=I^{*}}=0$, using the implicit function theorem we have $\frac{dI^{*}}{d\pi }=-\frac{\frac{\partial H\left( I^{*}\right) }{\partial \pi }}{H^{\prime }\left( I^{*}\right) }$. Given convexity of $L\left( I\right)$, $H^{\prime }\left( I^{*}\right) \ge 0$, so $\frac{dI^{*}}{d\pi }\ge 0$ if and only if $\frac{\partial H\left( I^{*}\right) }{\partial \pi }\le 0$. From the last line of (22) in “Appendix 2”, we have

$$\begin{aligned} \frac{\partial H\left( I\right) }{\partial \pi }&=-\frac{\partial a^{*}}{\partial \pi }\left( \frac{c_{12}\pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\right) +c_{12}\frac{\partial a^{*}}{\partial \pi }+\left( 1-a^{*}\right) \frac{\partial }{\partial \pi }\left( \frac{c_{12}\pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\right) \nonumber \\&=\frac{\partial a^{*}}{\partial \pi }\underset{\le 0}{\underbrace{\left( \frac{c_{12}c_{11}e^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2} }\right) }}+\left( 1-a^{*}\right) \underset{\equiv X}{\underbrace{\frac{\partial }{\partial \pi }\left( \frac{c_{12}\pi t^{2}k^{2}}{c_{11} e^{\prime \prime }+\pi t^{2}k^{2}}\right) }} \end{aligned}$$

(18)

where the first term on the second line has sign $-sign\left( \frac{\partial a^{*}}{\partial \pi }\right)$ given the curvature assumptions; in the second term,

$$\begin{aligned} X&=\frac{\left( c_{11}e^{\prime \prime }+\pi t^{2}k^{2}\right) \left( c_{12}t^{2}k^{2}+c_{121}\pi t^{2}k^{2}\frac{\partial a^{*}}{\partial \pi }\right) -c_{12}\pi t^{2}k^{2}\left( c_{111}e^{\prime \prime }\frac{\partial a^{*}}{\partial \pi }+e^{\prime \prime \prime }c_{11}\frac{\partial p^{*} }{\partial \pi }+t^{2}k^{2}\right) }{\left( c_{11}e^{\prime \prime }+\pi t^{2}k^{2}\right) ^{2}}\\&=\frac{c_{11}e^{\prime \prime }c_{12}t^{2}k^{2}+\left( c_{11}e^{\prime \prime }+\pi t^{2}k^{2}\right) c_{121}\pi t^{2}k^{2}\frac{\partial a^{*} }{\partial \pi }}{\left( c_{11}e^{\prime \prime }+\pi t^{2}k^{2}\right) ^{2}}, \end{aligned}$$

given $c_{111}=e^{\prime \prime \prime }=0$. Equation (18) becomes

$$\begin{aligned} \frac{\partial H\left( I\right) }{\partial \pi }=\frac{\partial a^{*} }{\partial \pi }\underset{\le 0}{\underbrace{\left[ \frac{c_{12}c_{11} e^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}+\frac{\left( 1-a^{*}\right) c_{121}\pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\right] }+}\underset{\le 0}{\underbrace{\frac{\left( 1-a^{*}\right) c_{11}e^{\prime \prime }c_{12}t^{2}k^{2}}{\left( c_{11} e^{\prime \prime }+\pi t^{2}k^{2}\right) ^{2}}}}\text {,} \end{aligned}$$

(19)

where the terms in the square brackets and the second term are nonpositive due to our assumptions on c and e.

Therefore, if $k\ge {{\bar{k}}}\left( I^{*}\right)$, then $\frac{\partial a^{*}}{\partial \pi }\ge 0$ by Lemma 1, and thus $\frac{\partial H\left( I\right) }{\partial \pi }\le 0$ so $\frac{dI^{*} }{d\pi }\ge 0$; moreover, $\frac{da^{*}}{d\pi }=\frac{da^{*}}{dI} \frac{dI^{*}}{d\pi }+\frac{\partial a^{*}}{\partial \pi }\ge 0$ given $\frac{da^{*}}{dI}\ge 0$ and $\frac{\partial a^{*}}{\partial \pi }\ge 0$ because of $k\ge {{\bar{k}}}\left( I^{*}\right)$ by Lemma 1. Part 1 is thus proved.

To show Part 2, notice that when $k\rightarrow 0$, Eq. (19) becomes

$$\begin{aligned} \lim _{k\rightarrow 0}\frac{\partial H\left( I\right) }{\partial \pi } =\frac{\partial a^{*}}{\partial \pi }c_{12}\ge 0, \end{aligned}$$

where the inequality is derived from $c_{12}<0$ and $\lim _{k\rightarrow 0} \frac{\partial a^{*}}{\partial \pi }=\frac{-p_{0}t}{c_{11}}\le 0$ using Eq. (15). Consequently $\frac{dI^{*}}{d\pi }\le 0$ and thus $\frac{da^{*}}{d\pi }=\frac{da^{*}}{dI}\frac{dI^{*}}{d\pi } +\frac{\partial a^{*}}{\partial \pi }\le 0$ given $\frac{da^{*}}{dI}\ge 0$ and $\frac{\partial a^{*}}{\partial \pi }\le 0$ (for small enough k) by Lemma 1, which proves Part 2. $\square$

Proof

(Remark 1) Define

$$\begin{aligned} \phi _{FB}\left( I\right) \equiv c_{2}\left( 1,I\right) +r \end{aligned}$$

$$\begin{aligned} \phi \left( I\right)&\equiv c_{2}\left( a^{*},I\right) +r+\pi \frac{dp^{*}}{dI}b^{*}\\&=c_{2}\left( a^{*},I\right) +r+\frac{\pi k^{2}t^{2}\left( 1-a^{*}\right) c_{12}}{c_{11}e^{\prime \prime }+\pi k^{2}t^{2}}, \end{aligned}$$

where the second equality uses (17) and $b=kt\left( 1-a\right)$. We have $\frac{d\phi _{FB}}{dI}=c_{22}\ge 0$. Note that $\phi _{FB}\left( I^{FB}\right) =0$ and $\phi \left( I^{*}\right) =0$ pin down the first-best investment and the equilibrium investment respectively. Given our assumptions $c_{12}<0$ and $c_{112}\le 0$ (that is, $c_{2}\left( a,I\right)$ is a concave and decreasing function in a), we have

$$\begin{aligned} c_{2}\left( a^{*},I\right) -c_{2}\left( 1,I\right) \ge -\left( 1-a^{*}\right) c_{12}\left( a^{*},I\right) . \end{aligned}$$

(20)

For any I, we have

$$\begin{aligned}&\phi _{FB}\left( I\right) -\phi \left( I\right) \\&=c_{2}\left( 1,I\right) -c_{2}\left( a^{*},I\right) -\frac{\pi k^{2}t^{2}}{c_{11}e^{\prime \prime }+\pi k^{2}t^{2}}\left( 1-a^{*}\right) c_{12}\left( a^{*},I\right) \\&\le \left[ c_{2}\left( 1,I\right) -c_{2}\left( a^{*},I\right) \right] \left[ 1-\frac{\pi k^{2}t^{2}}{c_{11}e^{\prime \prime }+\pi k^{2} t^{2}}\right] \\&\le 0 \end{aligned}$$

where the first inequality uses (20) and the second inequality is due to $\frac{\pi k^{2}t^{2}}{c_{11}e^{\prime \prime }+\pi k^{2}t^{2}} <1$ (since $c_{11}>0$ and $e^{\prime \prime }>0$) and $c_{2}\left( 1,I\right) -c_{2}\left( a^{*},I\right) \le 0$ (since $c_{2}$ is decreasing in a by $c_{21}<0$). Since $\phi _{FB}\left( I\right) \le \phi \left( I\right)$ for all I, $\phi \left( I^{FB}\right) \ge \phi _{FB}\left( I^{FB}\right) =0$; moreover, $\phi \left( I^{*}\right) =0$. Therefore, we have $I^{*}\le I^{FB}$ given that $\phi \left( I\right)$ is an increasing function. $\square$

Proof

(Proposition 3) The implicit function theorem gives $\frac{dI^{*}}{dk}=\left. -\frac{\frac{\partial H\left( I\right) }{\partial k}}{H^{\prime }\left( I\right) }\right| _{I=I^{*}}$. Again by the convexity of L, $sign\left( \frac{dI^{*}}{dk}\right) =$$-sign\left( \frac{\partial H\left( I^{*}\right) }{\partial k}\right)$. By the last line of (22) in “Appendix 2”,

$$\begin{aligned} \frac{\partial H\left( I\right) }{\partial k}&=-\frac{\partial a^{*} }{\partial k}\left( \frac{c_{12}\pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\right) +c_{12}\frac{\partial a^{*}}{\partial k}+\left( 1-a^{*}\right) \frac{d}{dk}\left( \frac{c_{12}\pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\right) \\&=\frac{\partial a^{*}}{\partial k}\underset{\le 0}{\underbrace{\left( \frac{c_{12}c_{11}e^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2} }\right) }}+\left( 1-a^{*}\right) \underset{\equiv Y}{\underbrace{\frac{d}{dk}\left( \frac{c_{12}\pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\right) }}, \end{aligned}$$

where $\frac{\partial a^{*}}{\partial k}\ge 0$ by Lemma 1 and

$$\begin{aligned} Y&=\frac{\left( c_{11}e^{\prime \prime }+\pi t^{2}k^{2}\right) \left( 2c_{12}\pi t^{2}k+c_{121}\pi t^{2}k^{2}\frac{\partial a^{*}}{\partial k}\right) -c_{12}\pi t^{2}k^{2}\left( c_{111}e^{\prime \prime }\frac{\partial a^{*}}{\partial k}+e^{\prime \prime \prime }c_{11}\frac{\partial p^{*} }{\partial k}+2\pi t^{2}k\right) }{\left( c_{11}e^{\prime \prime }+\pi t^{2}k^{2}\right) ^{2}}\\&=\frac{2c_{11}e^{\prime \prime }c_{12}\pi t^{2}k+\left( c_{11} e^{\prime \prime }+\pi t^{2}k^{2}\right) c_{121}\pi t^{2}k^{2}\frac{\partial a^{*}}{\partial k}}{\left( c_{11}e^{\prime \prime }+\pi t^{2}k^{2}\right) ^{2}}\\&<0, \end{aligned}$$

where the second equality makes use of $c_{111}=e^{\prime \prime \prime }=0$, and the inequality makes use of $c_{12}<0$, $c_{121}\le 0$ and $\frac{\partial a^{*}}{\partial k}\ge 0$ by Lemma 1. So $\frac{\partial H\left( I\right) }{\partial k}<0$ and consequently we have $\frac{dI^{*} }{dk}>0$ and $\frac{da^{*}}{dk}=\frac{\partial a^{*}}{\partial k} +\frac{da^{*}}{dI}\frac{dI^{*}}{dk}\ge 0$ given $\frac{da^{*}}{dI}\ge 0$ and $\frac{\partial a^{*}}{\partial k}\ge 0$ by Lemma 1. This proves the first part of the statement. Exactly the same steps apply for the comparative statics with respect to t. $\square$

Proof

(Remark 2) First, Eq. (6) and the convexity of $L\left( I\right)$ imply that $\frac{dI^{*}}{dr}\le 0$. Recall from Lemma 1 that $\frac{da^{*}}{dI}\ge 0$ and $\frac{dp^{*}}{dI}\le 0$. Because $\frac{dI^{*}}{dr}\le 0$, by the chain rule we get $\frac{da^{*}}{dr}=\frac{da^{*}}{dI}\frac{dI^{*}}{dr} \le 0$ and $\frac{dp^{*}}{dr}=\frac{dp^{*}}{dI}\frac{dI^{*}}{dr} \ge 0$. $\square$

Proof

(Remark 3) Using (21) in “Appendix 2”, under the optimal $I^{*}$, we have $\frac{dL\left( I^{*}\right) }{d\pi }=\left( 1-a^{*}\right) c_{11}\frac{\partial a^{*}}{\partial \pi }$ by the envelope theorem, where $\frac{\partial a^{*}}{\partial \pi }\ge 0$ if and only if $k\ge {{\bar{k}}}\left( I^{*}\right)$ by Lemma 1. $\square$

Proof

(Proposition 4) Proposition 6 in “Appendix 4” shows $\frac{dI^{*}}{dk_{F}}\ge 0$ and $\frac{dI^{*}}{dk_{I}}\ge 0$ for all $k_{F}$ and $k_{I}$. We also have $\frac{da^{*}}{dk_{F}}=\frac{\partial a^{*}}{\partial k_{F}} +\frac{da^{*}}{dI}\frac{dI^{*}}{dk_{F}}>0$ and $\frac{da^{*}}{dk_{I} }=\frac{\partial a^{*}}{\partial k_{I}}+\frac{da^{*}}{dI}\frac{dI^{*}}{dk_{I}}>0$, using (27), (28), and (29) in the proof of Lemma 2 in “Appendix 4”. Note that $k_{I}\equiv \frac{1-\lambda \gamma _{I}}{1+\lambda \gamma _{F}}k$ is decreasing in $\lambda$, $\gamma _{I}$, and $\gamma _{F}$, while $k_{F}=k~$ is independent of $\lambda$, $\gamma _{I}$, and $\gamma _{F}$. Therefore by the chain rule, $I^{*}$ and $a^{*}$ are decreasing in $\lambda$, $\gamma _{I}$, and $\gamma _{F}$. $\square$

Proof

(Remark 4) First, derivation analogous to the proof of Proposition 3 shows $\frac{dI^{*}}{dk_{F}}>0$ and $\frac{dI^{*}}{dk_{I}}>0$ for all $k_{F}$ and $k_{I}$. We also have $\frac{da^{*}}{dk_{F}}=\frac{\partial a^{*}}{\partial k_{F}} +\frac{da^{*}}{dI}\frac{dI^{*}}{dk_{F}}>0$ and $\frac{da^{*}}{dk_{I} }=\frac{\partial a^{*}}{\partial k_{I}}+\frac{da^{*}}{dI}\frac{dI^{*}}{dk_{I}}>0$ by (27), (28), and (29) in the proof of Lemma 2 in “Appendix 4”. It remains to show how the punishment parameters $\lambda$, $\gamma _{I}$, and $\gamma _{F}$ affect $k_{I}$ and $k_{F}$ under each of the specifications.

Part 1. By the chain rule,

$$\begin{aligned} \frac{da^{*}}{d\lambda }&=\frac{da^{*}}{dk_{I}}\frac{dk_{I}}{d\lambda }+\frac{da^{*}}{dk_{F}}\frac{dk_{F}}{d\lambda },\\ \frac{dI^{*}}{d\lambda }&=\frac{dI^{*}}{dk_{I}}\frac{dk_{I}}{d\lambda }+\frac{dI^{*}}{dk_{F}}\frac{dk_{F}}{d\lambda }. \end{aligned}$$

When $k\rightarrow 1$, $k_{F}\rightarrow 1$, a constant, independent of $\lambda$, and therefore $\frac{dk_{F}}{d\lambda }\rightarrow 0$. Since $\frac{da^{*}}{dk_{F}}\frac{dk_{F}}{d\lambda }\rightarrow 0$ and $\frac{da^{*}}{dk_{I}}\frac{dk_{I}}{d\lambda }<0$ (by $\frac{da^{*} }{dk_{I}}>0$ and $\frac{dk_{I}}{d\lambda }<0$), we have $\frac{da^{*} }{d\lambda }<0$. Similarly, we can prove $\frac{dI^{*}}{d\lambda }<0$. The same argument applies for the comparative statics for $\gamma _{I}$, and $\gamma _{F}$.

When $k\rightarrow 0$, we have $k_{I}\rightarrow 0$, a constant, while $k_{F}\rightarrow$$\lambda \left( \gamma _{I}+\gamma _{F}\right)$, which is increasing in $\lambda$, $\gamma _{F}$ and $\gamma _{I}$. Following an argument similar to the above, by the chain rule, $I^{*}$ and $a^{*}$ are increasing in $\lambda$, $\gamma _{I}$, and $\gamma _{F}$.

Part 2. Given that $k_{I}\equiv k\left( 1-\lambda \gamma _{I}\right) \left( 1-\lambda \gamma _{F}\right)$ is decreasing in $\gamma _{I}$ while $k_{F}\equiv k+\left( 1-k\right) \lambda \gamma _{F}$ is independent of $\gamma _{I}$, by the chain rule, $I^{*}$ and $a^{*}$ are decreasing in $\gamma _{I}$. When $k\rightarrow 1$, we have $k_{I}\rightarrow \left( 1-\lambda \gamma _{I}\right) \left( 1-\lambda \gamma _{F}\right)$, which is decreasing in $\lambda$ and $\gamma _{F}$, while $k_{F}\rightarrow 1$, a constant. Following an argument similar to the proof of Part 1, by the chain rule, $I^{*}$ and $a^{*}$ are decreasing in $\lambda$ and $\gamma _{F}$. When $k\rightarrow 0$, we have $k_{I}\rightarrow 0$, a constant, while $k_{F}\rightarrow$$\lambda \gamma _{F}$, which is increasing in $\lambda$ and $\gamma _{F}$. Therefore by the chain rule, $I^{*}$ and $a^{*}$ are increasing in $\lambda$ and $\gamma _{F}$. $\square$

Appendix 2: Derivation of $\frac{dL\left( I\right) }{dI}$ and $\frac{d^{2}L\left( I\right) }{dI^{2}}$ in Section 2.3

Using the firm’s Stage 2 first order condition (3), (5) can be written as

$$\begin{aligned} L\left( I\right) =\left( 1-a^{*}\left( I\right) \right) c_{1}\left( a^{*}\left( I\right) ,I\right) +c\left( a^{*}\left( I\right) ,I\right) +rI. \end{aligned}$$

(21)

We have

$$\begin{aligned} \frac{dL\left( I\right) }{dI}&=\left( 1-a^{*}\right) c_{12}+\left( 1-a^{*}\right) c_{11}\frac{da^{*}}{dI}-c_{1}\frac{da^{*}}{dI} +c_{1}\frac{da^{*}}{dI}+c_{2}+r\\&=\left( 1-a^{*}\right) \left[ c_{12}+c_{11}\frac{da^{*}}{dI}\right] +c_{2}+r\nonumber \\&=\left( 1-a^{*}\right) \left( \frac{c_{12}\pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\right) +c_{2}+r,\nonumber \end{aligned}$$

(22)

where the third equality makes use of (16). By (17) in the proof of Lemma 1, (22) can be written as (6).

The second order derivative is

$$\begin{aligned} \frac{d^{2}L\left( I\right) }{dI^{2}}=-\frac{da^{*}}{dI}\left( \frac{c_{12}\pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\right) +c_{22}+\left( 1-a^{*}\right) \frac{d}{dI}\left( \frac{c_{12}\pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}\right) , \end{aligned}$$

(23)

where the last term equals

$$\begin{aligned}&\frac{\left( 1-a^{*}\right) \pi t^{2}k^{2}}{\left( c_{11} e^{\prime \prime }+\pi t^{2}k^{2}\right) ^{2}}\left[ \left( c_{11} e^{\prime \prime }+\pi t^{2}k^{2}\right) \left( c_{121}\frac{da^{*}}{dI}+c_{122}\right) -c_{12}\left( \left( c_{111}\frac{da^{*}}{dI}+c_{112}\right) e^{\prime \prime }+c_{11}e^{\prime \prime \prime } \frac{dp^{*}}{dI}\right) \right] \\&\quad =\frac{\left( 1-a^{*}\right) \pi t^{2}k^{2}}{\left( c_{11} e^{\prime \prime }+\pi t^{2}k^{2}\right) ^{2}}\left[ \left( c_{11} e^{\prime \prime }+\pi t^{2}k^{2}\right) \left( c_{121}\frac{da^{*}}{dI}+c_{122}\right) -c_{12}c_{112}e^{\prime \prime }\right] , \end{aligned}$$

by the assumption of $c_{111}=e^{\prime \prime \prime }=0$.

Substituting this back to (23) yields

$$\begin{aligned}&\frac{d^{2}L\left( I\right) }{dI^{2}}\\&\quad =-\frac{da^{*}}{dI}\left( \frac{c_{12}\pi t^{2}k^{2}}{c_{11} e^{\prime \prime }+\pi t^{2}k^{2}}\right) +c_{22}+\frac{\left( 1-a^{*}\right) \pi t^{2}k^{2}}{\left( c_{11}e^{\prime \prime }+\pi t^{2} k^{2}\right) ^{2}}\left[ \left( c_{11}e^{\prime \prime }+\pi t^{2} k^{2}\right) \left( c_{121}\frac{da^{*}}{dI}+c_{122}\right) -c_{12}c_{112}e^{\prime \prime }\right] \\&\quad =\frac{da^{*}}{dI}\left( -\frac{c_{12}\pi t^{2}k^{2}}{c_{11} e^{\prime \prime }+\pi t^{2}k^{2}}+\frac{\left( 1-a^{*}\right) \pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}c_{121}\right) +c_{22}+\frac{\left( 1-a^{*}\right) \pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2}}c_{122}-\frac{\left( 1-a^{*}\right) \pi t^{2}k^{2}}{\left( c_{11} e^{\prime \prime }+\pi t^{2}k^{2}\right) ^{2}}c_{12}c_{112}e^{\prime \prime }\\&\quad =\frac{-c_{12}e^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2} }\left( -\frac{c_{12}\pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2} }+\frac{\left( 1-a^{*}\right) \pi t^{2}k^{2}}{\left( c_{11} e^{\prime \prime }+\pi t^{2}k^{2}\right) }c_{121}\right) +c_{22}+\frac{\left( 1-a^{*}\right) \pi t^{2}k^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k^{2} }c_{122}-\frac{\left( 1-a^{*}\right) \pi t^{2}k^{2}}{\left( c_{11} e^{\prime \prime }+\pi t^{2}k^{2}\right) ^{2}}c_{12}c_{112}e^{\prime \prime }\\&\quad =\frac{\pi t^{2}k^{2}}{\left( c_{11}e^{\prime \prime }+\pi t^{2} k^{2}\right) ^{2}}\left[ e^{\prime \prime }\left( c_{12}^{2}+\left( 1-a^{*}\right) \left( c_{11}c_{122}-2c_{12}c_{121}\right) \right) +\left( 1-a^{*}\right) \pi t^{2}k^{2}c_{122}\right] +c_{22}\\&\quad \ge 0, \end{aligned}$$

where the third equality comes from (16) and the inequality comes from the assumptions $e^{\prime \prime }\ge 0$, $c_{22}\ge 0$, $c_{122}\ge 0$, and $c_{11}c_{122}\ge 2c_{12}c_{121}$.

Appendix 3: Sign Ambiguity of $\frac{da^{*}}{d\pi }$: Numerical Examples

The following numerical examples illustrate the ambiguity of the sign of $\frac{da^{*}}{d\pi }$ in Case 2 of Table 1 (where $\frac{da^{*}}{dI} \ge 0$, $\frac{dI^{*}}{d\pi }\ge 0$ and $\frac{\partial a^{*}}{\partial \pi }\le 0$), assuming $e=\frac{e_{0}}{2}\left( p-p_{0}\right) ^{2}$ and $c=\frac{c_{0}}{2I}a^{2}$. Part 3 of Lemma 1 shows that $\frac{da^{*}}{dI}\ge 0$ is always satisfied. In Fig. 2, we let $\pi =0.1$, $t=0.2$, $p_{0}=0.05$, $e_{0}=0.3$, $c_{0}=0.05$, and $r=0.001$. When $k\in \left( 0.32,0.36\right)$, we have $\frac{dI^{*} }{d\pi }\ge 0$ and $\frac{\partial a^{*}}{\partial \pi }\le 0$, while $\frac{da^{*}}{d\pi }\ge 0$. In Fig. 3, we let $\pi =0.7$, $t=0.05$, $p_{0}=0.15$, $e_{0}=0.05$, $c_{0}=0.05$, and $r=0.001$. When $k>0.22$, we have $\frac{dI^{*}}{d\pi }\ge 0$ and $\frac{\partial a^{*} }{\partial \pi }\le 0$, while $\frac{da^{*}}{d\pi }<0$.

Fig. 2

$e=\frac{e_{0}}{2}\left( p-p_{0}\right) ^{2}$, $c=\frac{c_{0}}{2I}a^{2}$, $\pi =0.1$, $t=0.2$, $p_{0}=0.05$, $e_{0}=0.3$, $c_{0}=0.05$, and $r=0.001$. The two top graphs show $\frac{\partial a}{\partial \pi }$ and $\frac{dI}{d\pi }$ respectively, while the two bottom graphs show $\frac{\partial a}{\partial I}$ and $\frac{da}{d\pi }=\frac{da}{dI}\frac{dI}{d\pi }+\frac{\partial a}{\partial \pi }$ respectively

Fig. 3

$e=\frac{e_{0}}{2}\left( p-p_{0}\right) ^{2}$, $c=\frac{c_{0}}{2I}a^{2}$, $\pi =0.7$, $t=0.05$, $p_{0}=0.15$, $e_{0}=0.05$, $c_{0}=0.05$, and $r=0.001$. The two top graphs show $\frac{\partial a}{\partial \pi }$ and $\frac{dI}{d\pi }$ respectively, while the two bottom graphs show $\frac{\partial a}{\partial I}$ and $\frac{da}{d\pi }=\frac{da}{dI}\frac{dI}{d\pi }+\frac{\partial a}{\partial \pi }$ respectively

Appendix 4: Derivation of the Model in Section 3

Applying the Cramer Rule to (11) and (12) we obtain the following lemma which is analogous to Lemma 1 in Sect. 2.

Lemma 2

For any given I , in Stage 2, we have the following.

• 1. More corruption (higher $\pi$) increases abatement,$a^{*}$, and decreases the corruptible inspector’s monitoring intensity $p^{*}$, if and only if $p^{*}k_{F}>p_{0}$.

• 2. There exists threshold ${{\bar{k}}}_{F}\left( I\right) \in \left( 0,1\right)$ such that $p^{*}k_{F}>p_{0}$ if and only if $k_{F}>{{\bar{k}}}_{F}\left( I\right)$.

• 3. Other things being equal, a higher $k_{F}$, $k_{I}$, ortinduces more abatement.

• 4. Higher investment (I)increases abatement and decreases monitoring.

Proof

(Lemma 2) Using the Cramer Rule, we obtain

$$\begin{aligned} \frac{\partial a^{*}}{\partial \pi }=\frac{\left( p^{*}k_{F} -p_{0}\right) te^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}, \end{aligned}$$

(24)

and

$$\begin{aligned} \frac{\partial p^{*}}{\partial \pi }=\frac{-\left( p^{*}k_{F} -p_{0}\right) k_{I}t^{2}}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}, \end{aligned}$$

where $sign\left( \frac{\partial a^{*}}{\partial \pi }\right) =-sign\left( \frac{\partial p^{*}}{\partial \pi }\right) =sign\left( p^{*}k_{F} -p_{0}\right)$ which proves Part 1.

To show Part 2, we first establish that

$$\begin{aligned} \frac{\partial p^{*}}{\partial k_{F}}=\frac{-t^{2}\pi k_{I}p^{*}}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\le 0, \\ \frac{\partial p^{*}}{\partial k_{I}}=\frac{c_{11}t\left( 1-a\right) }{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\ge 0, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial \left( p^{*}k_{F}\right) }{\partial k_{F}}=\frac{\partial p^{*}}{\partial k_{F}}k_{F}+p^{*}=\frac{c_{11}e^{\prime \prime }p^{*} }{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\ge 0, \end{aligned}$$

(25)

$$\begin{aligned} \frac{\partial \left( p^{*}k_{F}\right) }{\partial k_{I}}=\frac{\partial p^{*}}{\partial k_{I}}k_{F}+p^{*}\frac{\partial k_{F}}{\partial k_{I} }\ge 0, \end{aligned}$$

(26)

given $\frac{\partial p^{*}}{\partial k_{I}}\ge 0$ and $\frac{\partial k_{F}}{\partial k_{I}}=\frac{1+\gamma _{F}}{1-\gamma _{I}}>0$. Obviously $\lim _{k_{F}\rightarrow 0}p^{*}k_{F}<p_{0}$. Meanwhile from the inspector’s first-order condition (12) we get $p^{*}>p_{0}$ given $k_{I}>0$, so $\lim _{k_{F}\rightarrow 1}p^{*}k_{F}>p_{0}$. Combining these facts and $\frac{\partial \left( p^{*}k_{F}\right) }{\partial k_{F}}\ge 0$ as well as the continuity of $p^{*}k_{F}$, we conclude that there exists a threshold ${{\bar{k}}}_{F}\left( I\right) \in \left( 0,1\right)$ such that $p^{*}k_{F}>p_{0}$ if and only if $k_{F}>{{\bar{k}}}_{F}\left( I\right)$.

To show Part 3, using the Cramer Rule we have

$$\begin{aligned} \frac{\partial a^{*}}{\partial k_{F}}=\frac{e^{\prime \prime }t\pi p^{*} }{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}>0, \end{aligned}$$

(27)

$$\begin{aligned} \frac{\partial a^{*}}{\partial k_{I}}=\frac{t^{2}\pi k_{F}\left( 1-a^{*}\right) }{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}>0, \end{aligned}$$

(28)

and

$$\begin{aligned} \frac{\partial a^{*}}{\partial t}=\frac{e^{\prime \prime }\left( \pi k_{F}p^{*}+\left( 1-\pi \right) p_{0}\right) +t^{2}k_{F}k_{I}\pi \left( 1-a^{*}\right) }{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}>0. \end{aligned}$$

Finally

$$\begin{aligned} \frac{da^{*}}{dI}=\frac{-c_{12}e^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\ge 0, \end{aligned}$$

(29)

and

$$\begin{aligned} \frac{dp^{*}}{dI}=\frac{k_{I}tc_{12}}{c_{11}e^{\prime \prime }+\pi t^{2} k_{F}k_{I}}\le 0, \end{aligned}$$

which proves Part 4. $\square$

Given $k_{I}=\frac{1-\lambda \gamma _{I}}{1+\lambda \gamma _{F}}k$ and $k_{F}=k$, Parts 2 and 3 in Lemma 2 lead to the following.

Lemma 3

There exists a threshold ${{\widetilde{k}}}\left( I\right) \in \left( 0,1\right)$such that$p^{*}k_{F}>p_{0}$if and only if$k>{{\widetilde{k}}}\left( I\right)$. A higher bargaining power (k) induces more abatement.

Proof

(Lemma 3) To show the first statement, by the chain rule,

$$\begin{aligned} \frac{\partial \left( p^{*}k_{F}\right) }{\partial k}=\frac{\partial \left( p^{*}k_{F}\right) }{\partial k_{I}}\left( \frac{1-\lambda \gamma _{I}}{1+\lambda \gamma _{F}}\right) +\frac{\partial \left( p^{*} k_{F}\right) }{\partial k_{F}}\ge 0, \end{aligned}$$

by (25) and (26). Clearly $\lim _{k\rightarrow 1}k_{I} =\frac{1-\lambda \gamma _{I}}{1+\lambda \gamma _{F}}>0$, which implies $\lim _{k\rightarrow 1}p^{*}>p_{0}$ by (12). Meanwhile $\lim _{k\rightarrow 1}k_{F}=1$, which then implies $\lim _{k\rightarrow 1}p^{*}k_{F}>p_{0}$. Similarly $\lim _{k\rightarrow 0}k_{I}=\lim _{k\rightarrow 0} k_{F}=0$. From (12), this means $\lim _{k\rightarrow 0}p^{*}=p_{0}$ and so $\lim _{k\rightarrow 0}p^{*}k_{F}=p_{0}k_{F}<p_{0}$. Combining these and using $\frac{\partial \left( p^{*}k_{F}\right) }{\partial k}\ge 0$ as well as the continuity of $p^{*}k_{F}$ establish the existence of the said threshold. For the second statement of the lemma, by the chain rule $\frac{\partial a^{*}}{\partial k}=\frac{\partial a^{*}}{\partial k_{I} }\left( \frac{1-\lambda \gamma _{I}}{1+\lambda \gamma _{F}}\right) +\frac{\partial a^{*}}{\partial k_{F}}>0$ using (27) and (28). $\square$

After characterizing the subgame equilibrium in Stage 2, at Stage 1 the firm’s loss function is

$$\begin{aligned} L\left( I\right) =t\left( 1-a^{*}\right) \left( \pi p^{*} k_{F}+\left( 1-\pi \right) p_{0}\right) +c\left( a^{*},I\right) +rI. \end{aligned}$$

We have the following equation, which is the same as (22) except that $k^{2}$ is replaced by $k_{F}k_{I}$:

$$\begin{aligned} \frac{dL\left( I\right) }{dI}=\left( 1-a^{*}\right) \left( \frac{c_{12}\pi t^{2}k_{F}k_{I}}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I} }\right) +c_{2}+r. \end{aligned}$$

(30)

Steps similar to those in Sect. 2 establishes the existence of equilibrium $I^{*}$. The following result is analogous to Proposition 2 in Sect. 2, given ${{\bar{k}}}_{F}\left( I^{*}\right)$ as defined in Lemma 2.

Proposition 5

At the equilibrium investment $I^{*}$ :

• 1. If $k_{F}>{{\bar{k}}}_{F}\left( I^{*}\right)$ , a higher proportion of corruptible inspectors $(\pi )$  increases investment and abatement.

• 2. If $k_{F}\rightarrow 0$ or $k_{I}\rightarrow 0$ , a higher proportion of corruptible inspectors $(\pi )$  decreases investment and abatement.

Proof

(Proposition 5) We apply the implicit function theorem to the first-order condition $\frac{dL\left( I^{*}\right) }{dI}=0$ where $\frac{dL\left( I\right) }{dI}$ is given by Eq. (30). This yields $\frac{dI^{*}}{d\pi }\ge 0$ if and only if $\frac{\partial }{\partial \pi }\left( \frac{dL\left( I\right) }{dI}\right) \le 0$ given convexity of $L\left( I\right)$. We have

$$\begin{aligned} \frac{\partial \left( \frac{dL\left( I\right) }{dI}\right) }{\partial \pi }&=-\frac{\partial a^{*}}{\partial \pi }\left( \frac{c_{12}\pi t^{2} k_{F}k_{I}}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\right) +c_{12} \frac{\partial a^{*}}{\partial \pi }+\left( 1-a^{*}\right) \frac{d}{d\pi }\left( \frac{c_{12}\pi t^{2}k_{F}k_{I}}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\right) \\&=\frac{\partial a^{*}}{\partial \pi }\underset{\le 0}{\underbrace{\left( \frac{c_{12}c_{11}e^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F} k_{I}}\right) }}+\left( 1-a^{*}\right) \underset{\equiv Z}{\underbrace{\frac{d}{d\pi }\left( \frac{c_{12}\pi t^{2}k_{F}k_{I}}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\right) }} \end{aligned}$$

With $c_{111}=e^{\prime \prime \prime }=0$, we have

$$\begin{aligned} Z&=\frac{\left( c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}\right) \left( c_{12}t^{2}k_{F}k_{I}+c_{121}\pi t^{2}k_{F}k_{I}\frac{\partial a^{*}}{\partial \pi }\right) -\left( c_{12}\pi t^{2}k_{F}k_{I}\right) \left( t^{2}k_{F}k_{I}\right) }{\left( c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}\right) ^{2}}\\&=\frac{\left( c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}\right) c_{121}\pi t^{2}k_{F}k_{I}\frac{\partial a^{*}}{\partial \pi }+c_{11} e^{\prime \prime }c_{12}t^{2}k_{F}k_{I}}{\left( c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}\right) ^{2}}. \end{aligned}$$

Thus $\frac{\partial }{\partial \pi }\left( \frac{dL\left( I\right) }{dI}\right)$ becomes

$$\begin{aligned} \frac{\partial }{\partial \pi }\left( \frac{dL\left( I\right) }{dI}\right) =\frac{\partial a^{*}}{\partial \pi }\underset{\le 0}{\underbrace{\left[ \frac{c_{12}c_{11}e^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F} k_{I}}+\frac{\left( 1-a^{*}\right) c_{121}\pi t^{2}k_{F}k_{I}}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\right] }+}\underset{\le 0}{\underbrace{\frac{\left( 1-a^{*}\right) c_{11}e^{\prime \prime } c_{12}t^{2}k_{F}k_{I}}{\left( c_{11}e^{\prime \prime }+\pi t^{2}k_{F} k_{I}\right) ^{2}}}}. \end{aligned}$$

This equation is similar to (19) in the proof of Proposition 2. If $k_{F}>{{\bar{k}}}_{F}\left( I^{*}\right)$, by Lemma 2 we have $\frac{\partial a^{*}}{\partial \pi }\ge 0$, implying $\frac{\partial }{\partial \pi }\left( \frac{dL\left( I\right) }{dI}\right) \le 0$, and therefore $\frac{dI^{*}}{d\pi }\ge 0$. Using exactly the same argument in the proof of Proposition 2, we also have $\frac{da^{*}}{d\pi }\ge 0$.

To show Part 2, note that $\lim _{k_{F}\rightarrow 0}\frac{\partial }{\partial \pi }\left( \frac{dL\left( I\right) }{dI}\right) =c_{12}\lim _{k_{F} \rightarrow 0}\frac{\partial a^{*}}{\partial \pi }$, where $\frac{\partial a^{*}}{\partial \pi }$ is given by (24) in the proof of Lemma 2. When $k_{F}\rightarrow 0$, then clearly $\frac{\partial a^{*}}{\partial \pi }\le 0$ by Lemma 2. When $k_{I}\rightarrow 0$, then $p^{*}\rightarrow p_{0}$ by (12) and $k_{F}\rightarrow \lambda \left( \gamma _{F}+\gamma _{I}\right) <1$; therefore, $p^{*}k_{F}<p_{0}$ and thus $\frac{\partial a^{*}}{\partial \pi }\le 0$ by Lemma 2. Since $c_{12}<0$, when either $k_{F}\rightarrow 0$ or $k_{I}\rightarrow 0$, we have $\frac{\partial }{\partial \pi }\left( \frac{dL\left( I\right) }{dI}\right) \ge 0$ and thus $\frac{dI^{*}}{d\pi }\le 0$ and $\frac{da^{*}}{d\pi }\le 0$. $\square$

When $k\rightarrow 0$, then $k_{I}\rightarrow 0$. By Lemmas 2 and 3and Proposition 5, the following corollary is straightforward:

Corollary 1

At the equilibrium investment $I^{*}$:

• If $k>{{\widetilde{k}}}\left( I^{*}\right)$ , a higher proportion of corruptible inspectors $(\pi )$  increases investment and abatement.

• If $k\rightarrow 0$ , a higher proportion of corruptible inspectors $(\pi )$  decreases investment and abatement.

The following proposition corresponds to Proposition 3 in Sect. 2.

Proposition 6

With endogenous investment, a higher $k_{I}, k_{F}$ , or t increases investment and abatement.

Proof

(Proposition 6) We apply the implicit function theorem to the first-order condition $\frac{dL\left( I^{*}\right) }{dI}=0$ where $\frac{dL\left( I\right) }{dI}$ is given by Eq. (30). This yields $\frac{dI^{*}}{dk_{I}}\ge 0$ if and only if $\frac{\partial }{\partial k_{I}}\left( \frac{dL\left( I^{*}\right) }{dI}\right) \le 0$. We have

$$\begin{aligned} \frac{\partial \left( \frac{dL\left( I\right) }{dI}\right) }{\partial k_{I}}&=-\frac{\partial a^{*}}{\partial k_{I}}\left( \frac{c_{12}\pi t^{2}k_{F}k_{I}}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\right) +\left( 1-a^{*}\right) \frac{d}{dk_{I}}\left( \frac{c_{12}\pi t^{2}k_{F}k_{I} }{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\right) +c_{12}\frac{\partial a^{*}}{\partial k_{I}}\\&=\frac{\partial a^{*}}{\partial k_{I}}\left( \frac{c_{12} c_{11}e^{\prime \prime }}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\right) +\left( 1-a^{*}\right) \underset{\equiv \Psi }{\underbrace{\frac{d}{dk_{I} }\left( \frac{c_{12}\pi t^{2}k_{F}k_{I}}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}\right) }}. \end{aligned}$$

With $c_{111}=e^{\prime \prime \prime }=0$, we have

$$\begin{aligned} \Psi&=\frac{\left( c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}\right) \left( c_{12}\pi t^{2}k_{F}+c_{121}\pi t^{2}k_{F}k_{I}\frac{\partial a^{*}}{\partial k_{I}}\right) +\left( c_{12}\pi t^{2}k_{F}k_{I}\right) \left( \pi t^{2}k_{F}\right) }{\left( c_{11}e^{\prime \prime }+\pi t^{2}k_{F} k_{I}\right) ^{2}}\\&=\frac{\left( c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}\right) \left( c_{121}\pi t^{2}k_{F}k_{I}\frac{\partial a^{*}}{\partial k_{I}}\right) +c_{11}e^{\prime \prime }c_{12}\pi t^{2}k_{F}}{\left( c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}\right) ^{2}}. \end{aligned}$$

Thus $\frac{\partial \left( \frac{dL\left( I\right) }{dI}\right) }{\partial k_{I}}$ becomes

$$\begin{aligned} \frac{\partial \left( \frac{dL\left( I\right) }{dI}\right) }{\partial k_{I}}=\frac{\partial a^{*}}{\partial k_{I}}\underset{\le 0}{\underbrace{\left[ \frac{c_{12}c_{11}e^{\prime \prime }}{c_{11} e^{\prime \prime }+\pi t^{2}k_{F}k_{I}}+\frac{\left( 1-a^{*}\right) c_{121}\pi t^{2}k_{F}k_{I}}{c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I} }\right] }+}\underset{\le 0}{\underbrace{\frac{\left( 1-a^{*}\right) c_{11}e^{\prime \prime }c_{12}\pi t^{2}k_{F}}{\left( c_{11}e^{\prime \prime }+\pi t^{2}k_{F}k_{I}\right) ^{2}}}}, \end{aligned}$$

where the inequalities makes use of $c_{12}<0$ and $c_{121}\le 0$. Given $\frac{\partial a^{*}}{\partial k_{I}}\ge 0$ by (28), we conclude that $\frac{\partial \left( \frac{dL\left( I\right) }{dI}\right) }{\partial k_{I}}\le 0$ and consequently we have $\frac{dI^{*}}{dk_{I}}\ge 0$ and $\frac{da^{*}}{dk_{I}}=\frac{\partial a^{*}}{\partial k_{I}} +\frac{da^{*}}{dI}\frac{dI^{*}}{dk_{I}}\ge 0$ given $\frac{da^{*} }{dI}\ge 0$ by (29) and $\frac{\partial a^{*}}{\partial k_{I}} \ge 0$ by (28). Similar steps apply for the comparative statics with respect to $k_{F}$ and t. $\square$

Since $k_{F}$ and $k_{I}$ increase in k, Proposition 6 implies that an increase in k increases investment and abatement. Finally, it is easy to confirm that all the other comparative statics results in Section 2 hold.

4.1 Appendix 4.1: Derivation of the Model in Section 3.1

Under the two alternative specifications, the proofs of Lemma 2, Propositions 5 and 6 do not change as they only involve $k_{F}$, $k_{I}$ and t, but not k, $\gamma _{I}$ or $\gamma _{F}$.

When the punishment is proportional to the fine evaded for both the inspector and the firm, Lemma 3 still holds. The proof is as follows.

Proof

(Lemma 3) To show the first statement, by the chain rule,

$$\begin{aligned} \frac{\partial \left( p^{*}k_{F}\right) }{\partial k}=\left( \frac{\partial \left( p^{*}k_{F}\right) }{\partial k_{I}}+\frac{\partial \left( p^{*}k_{F}\right) }{\partial k_{F}}\right) \left( 1-\lambda \left( \gamma _{F}+\gamma _{I}\right) \right) \ge 0, \end{aligned}$$

by (25) and (26). Clearly $\lim _{k\rightarrow 1}k_{I} =1-\lambda \left( \gamma _{F}+\gamma _{I}\right) >0$, which implies $\lim _{k\rightarrow 1}p^{*}>p_{0}$ by (12). Meanwhile $\lim _{k\rightarrow 1}k_{F}=1$, which then implies $\lim _{k\rightarrow 1}p^{*}k_{F}>p_{0}$. Similarly $\lim _{k\rightarrow 0}k_{I}=0$ while $\lim _{k\rightarrow 0}k_{F}\rightarrow$$\lambda \left( \gamma _{F}+\gamma _{I}\right) <1$. From (12), this means $\lim _{k\rightarrow 0}p^{*}=p_{0}$ and so $\lim _{k\rightarrow 0}p^{*}k_{F}=p_{0}k_{F}<p_{0}$. Combining these and the continuity of $p^{*}k_{F}$ establishes the existence of the said threshold. For the second statement of the lemma, by the chain rule $\frac{\partial a^{*}}{\partial k}=\left( \frac{\partial a^{*}}{\partial k_{I}}+\frac{\partial a^{*}}{\partial k_{F}}\right) \left( 1-\lambda \left( \gamma _{F}+\gamma _{I}\right) \right) >0$ by (27) and (28). $\square$

When $k\rightarrow 0$, then $k_{I}\rightarrow 0$, and therefore, Corollary 1 also holds.

When the punishment of the inspector is proportional to the bribe while the punishment of the firm is proportional to the evaded fine, the proof of Lemma 3 is as follows.

Proof

(Lemma 3) To show the first statement, by the chain rule,

$$\begin{aligned} \frac{\partial \left( p^{*}k_{F}\right) }{\partial k}=\frac{\partial \left( p^{*}k_{F}\right) }{\partial k_{I}}\left( 1-\lambda \gamma _{I}\right) \left( 1-\lambda \gamma _{F}\right) +\frac{\partial \left( p^{*}k_{F}\right) }{\partial k_{F}}\left( 1-\lambda \gamma _{F}\right) \ge 0, \end{aligned}$$

by (25) and (26). Clearly $\lim _{k\rightarrow 1}k_{I}=\left( 1-\lambda \gamma _{I}\right) \left( 1-\lambda \gamma _{F}\right) >0$, which implies $\lim _{k\rightarrow 1}p^{*}>p_{0}$ by (12). Meanwhile $\lim _{k\rightarrow 1}k_{F}=1$, which then implies $\lim _{k\rightarrow 1} p^{*}k_{F}>p_{0}$. Similarly $\lim _{k\rightarrow 0}k_{I}=0$ while $\lim _{k\rightarrow 0}k_{F}=$$\lambda \gamma _{F}<1$. From (12), this means $\lim _{k\rightarrow 0}p^{*}=p_{0}$ so $\lim _{k\rightarrow 0}p^{*}k_{F}=p_{0}k_{F}<p_{0}$. Combining these and the continuity of $p^{*} k_{F}$ establishes the existence of the said threshold. For the second statement, by the chain rule $\frac{\partial a^{*}}{\partial k} =\frac{\partial a^{*}}{\partial k_{I}}\left( 1-\lambda \gamma _{I}\right) \left( 1-\lambda \gamma _{F}\right) +\frac{\partial a^{*}}{\partial k_{F} }\left( 1-\lambda \gamma _{F}\right) >0$ by (27) and (28). $\square$

When $k\rightarrow 0$, then $k_{I}\rightarrow 0$, and therefore, Corollary 1 also holds.