The strategic interaction between a company and the government surrounding disasters

Abstract

We analyze the tradeoff between safety and production. The government chooses safety effort and tax rate in the first stage, and then the company strikes a balance between safety effort and production in the second stage. The government, representing the general public, earns taxes on production. Both players’ safety efforts mitigate the negative impact of a disaster. The disaster probability is modeled as a contest between the disaster magnitude and the two players’ safety efforts. Seven propositions are developed. First, as the safety effort of one player approaches infinity, the marginal change in the other player’s safety effort, with respect to the first player’s safety effort, approaches zero. Second, an infinitely large safety effort by any player causes the disaster probability and the negative impact of the disaster to decrease toward a constant. Third, as one player’s safety effort approaches infinity, the other player’s safety effort approaches zero. Fourth, the two players’ safety efforts are strategic substitutes so that an increase in one player’s safety effort decreases the other player’s safety effort. This enables the players to free ride on each other’s safety efforts. Fifth, higher tax rate causes the company to exert higher safety effort. Sixth, with maximum tax rate the company focuses exclusively on safety effort and generates no profit. Seventh, the company’s safety effort is inverse $U$ shaped in the disaster magnitude.

Appendix: Proof of Propositions 1–7

Proof of Proposition 1

When $S = 0$, the limit $\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{dS(s)}{ds}$ is always zero so we only consider $S >0$. Differentiating the first equation in (5) gives

$$\begin{aligned}&\frac{d}{ds}\left( {F(D,S,s)\frac{\partial p(D,S,s)}{\partial S}} \right) =\frac{d}{ds}\left( {(1-\tau )\frac{\partial H(S)}{\partial S}-\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) \nonumber \\&\quad =-\frac{d}{ds}\left( {\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) \nonumber \\&\quad \Rightarrow \frac{d}{ds}\left( {F(D,S,s)\frac{\partial p(D,S,s)}{\partial S}} \right) +\frac{d}{ds}\left( {\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) =0\nonumber \\&\quad \Rightarrow \frac{d}{ds}({F(D,S,s)})\frac{\partial p(D,S,s)}{\partial S}+\frac{d}{ds}\left( {\frac{\partial p(D,S,s)}{\partial S}} \right) F(D,S,s)\nonumber \\&\qquad +\frac{d}{ds}\left( {\frac{\partial F(D,S,s)}{\partial S}} \right) p(D,S,s)+\frac{d}{ds}({p(D,S,s)})\frac{\partial F(D,S,s)}{\partial S}=0\nonumber \\&\quad \Rightarrow \left[ {\frac{\partial F(D,S,s)}{\partial s}+\frac{\partial F(D,S,s)}{\partial S}\frac{dS}{ds}} \right] \frac{\partial p(D,S,s)}{\partial S}\nonumber \\&\qquad +\left[ {\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}+\frac{\partial ^{2}p(D,S,s)}{\partial ^{2}S}\frac{dS}{ds}} \right] F(D,S,s)\nonumber \\&\qquad +\left[ {\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}+\frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}\frac{dS}{ds}} \right] p(D,S,s)\nonumber \\&\qquad +\left[ {\frac{\partial p(D,S,s)}{\partial s}+\frac{\partial p(D,S,s)}{\partial S}\frac{dS}{ds}} \right] \frac{\partial F(D,S,s)}{\partial S}=0\\&\quad \Rightarrow \frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial s}+\frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial S}\frac{dS}{ds}\nonumber \\&\qquad +F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}\nonumber \\&\qquad +F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}\frac{dS}{ds}+p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}\nonumber \\&\qquad +p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}\frac{dS}{ds}\nonumber \\&\qquad +\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial s}+\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial S}\frac{dS}{ds}=0\nonumber \\&\quad \Rightarrow \frac{dS}{ds}= \nonumber \\&-\frac{\frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial s}+F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}+p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}+\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial s}}{\frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial S}+F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}+p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}+\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial S}}\nonumber \end{aligned}$$

(11)

Since $p$ and $F$ are bounded and monotonic in s it follows that $\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial p(D,S,s)}{\partial s}=0$ and $\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial F(D,S,s)}{\partial s}=0$, and thus $\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}=0$ and $\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}=0$. Hence (11) implies $\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{dS}{ds}=0$. Although $s$ is determined before $S$, in an overall sense the players are interested in $\mathop {\lim }\nolimits _{S\rightarrow \infty } \frac{ds}{dS}=0$. In contrast to (9) we thus consider the government’s first order condition

$$\begin{aligned} \left\{ {{\begin{array}{l} {s>0\Leftrightarrow \frac{\partial u(S,s,\tau )}{\partial s}=-\frac{\partial f(D,S,s)}{\partial s}p(D,S,s)-f(D,S,s)\frac{\partial p(D,S,s)}{\partial s}-b=0}\\ {s=0\Leftrightarrow \frac{\partial u(S,s,\tau )}{\partial s}=-\frac{\partial f(D,S,s)}{\partial s}p(D,S,s)-f(D,S,s)\frac{\partial p(D,S,s)}{\partial s}-b<0}\\ \end{array} }} \right. \end{aligned}$$

(12)

which is analogous to (5) for the company. Hence analogously, $\mathop {\lim }\nolimits _{S\rightarrow \infty } \frac{ds}{dS}=0$. $\square$

Proof of Proposition 2

Follows since $\frac{\partial p}{\partial s}\le 0$ ,$\frac{\partial p}{\partial S}\le 0$ , $\frac{\partial f}{\partial s}\le$0, $\frac{\partial f}{\partial S}\le$0, $\frac{\partial F}{\partial s}\le$0, $\frac{\partial F}{\partial S}\le$0 as stated in Assumption 1, and since $p, f, F$ have lower bounds 0. $\square$

Proof of Proposition 3

Inserting $\mathop {\lim }\nolimits _{S\rightarrow \infty } \frac{\partial p(D,S,s)}{\partial s}=0$ and $\mathop {\lim }\nolimits _{S\rightarrow \infty } \frac{\partial f(D,S,s)}{\partial s}=0$ into (12) implies $s=0\Leftrightarrow 0>-b$. Inserting $\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial p(D,S,s)}{\partial S}=0$ and $\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial F(D,S,s)}{\partial S}=0$ into the second equation in (5) implies $S=0\Leftrightarrow 0>(1-\tau )\frac{\partial H(S)}{\partial S}$. $\square$

Proof of Proposition 4

When $S = 0$, $\frac{dS(s)}{ds}$ is always zero so we only consider $S >0$. Similar to the proof of Proposition 1, we have

$$\begin{aligned}&\frac{dS}{ds}=\\&-\frac{\frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial s}+F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}+p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}+\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial s}}{\frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial S}+F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}+p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}+\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial S}}\nonumber \end{aligned}$$

(13)

Since we assume $\frac{\partial p}{\partial S}\le 0$,$\frac{\partial p}{\partial s}\le 0$, $\frac{\partial F}{\partial S}\le 0$, $\frac{\partial F}{\partial s}\le 0$, $\frac{\partial ^{2}p}{\partial S^{2}}\ge 0$, and $\frac{\partial ^{2}F}{\partial S^{2}}\ge 0$, the denominator of (13) is always positive. So we have $\frac{dS}{ds}\le 0$ if and only if the numerator of (13) is positive. One sufficient condition to ensure this is: $\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}\ge 0$ and $\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}\ge 0$. In summary we have shown $\frac{dS}{ds}\le 0$ if $\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}\ge 0$ and $\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}\ge 0$. Analogously, similar to the proof of Proposition 1 and using the first order condition (11), we can show $\frac{ds}{dS}\le 0$ if $\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}\ge 0$ and $\frac{\partial ^{2}f(D,S,s)}{\partial S\partial s}\ge 0$.

Proof of Proposition 5

When $S = 0$, $\frac{dS(\tau )}{d\tau }=0$ so we only consider $S > 0$. Differentiating the first equation in (5) gives

$$\begin{aligned}&\frac{d}{d\tau }\left( {F(D,S,s)\frac{\partial p(D,S,s)}{\partial S}} \right) =\frac{d}{d\tau }\left( {(1-\tau )\frac{\partial H(S)}{\partial S}-\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) \nonumber \\&\Rightarrow \frac{\partial }{\partial S}\left( {F(D,S,s)\frac{\partial p(D,S,s)}{\partial S}} \right) \frac{dS}{d\tau }=-\frac{\partial H(S)}{\partial S}+(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}\frac{dS}{d\tau }\\&-\frac{\partial }{\partial S}\left( {\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) \frac{dS}{d\tau }\nonumber \\&\Rightarrow \frac{dS}{d\tau }=\frac{\frac{\partial H(S)}{\partial S}}{(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-\frac{\partial }{\partial S}\left( {\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) -\frac{\partial }{\partial S}\left( {F(D,S,s)\frac{\partial p(D,S,s)}{\partial S}} \right) }\nonumber \\&=\frac{\frac{\partial H(S)}{\partial S}}{(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-\frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}p(D,S,s)-2\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial S}-F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}}\nonumber \end{aligned}$$

(14)

Because of $\frac{\partial H}{\partial S}<0, \frac{\partial F}{\partial S}\le 0,\frac{\partial ^{2}p(D,S,s)}{\partial ^{2}S}\ge 0, \frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}\ge 0$, and $\frac{\partial ^{2}H(S)}{\partial S^{2}}\le 0$ as stated in Assumptions 1 and 2, we have $\frac{dS(\tau )}{d\tau }\ge 0$. $\square$

Proof of Proposition 6

Inserting tax rate $\tau =1$ into (2) gives $U(S,s,\tau =1)=-F(D,S,s)p(D,S,s)$. Since we assume $\frac{\partial p}{\partial s}\le 0$ and $\frac{\partial F}{\partial S}\le$0 in Assumption 1, S needs to be maximum to maximize $U$. Hence from (1) we have $S(\tau =1,s)=\frac{R}{B}$. $\square$

Proof of Proposition 7

When $S = 0$, $\frac{dS}{dD}=\frac{d^{2}S}{dD^{2}}=0$ is always zero so we only consider $S > 0$. From the first equation in (5) we have

$$\begin{aligned} F\frac{\partial p(D,S,s)}{\partial S}&= (1-\tau )\frac{\partial H(S)}{\partial S}\Rightarrow \nonumber \\ \frac{d}{dD}\left( {F\frac{\partial p(D,S,s)}{\partial S}} \right)&= F\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}+F\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}\frac{dS}{dD}=\frac{d}{dD}\left( {(1-\tau )\frac{\partial H(S)}{\partial S}} \right) \nonumber \\&= (1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}\frac{dS}{dD}\nonumber \\ \Rightarrow \frac{dS}{dD}&= \frac{F\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}}{(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-F\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \end{aligned}$$

(15)

where $H(S)$ does not depend on $D$ according to (1). Since we assume $\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}\ge 0$ and $\frac{\partial ^{2}H(S)}{\partial S^{2}}\le 0$ in Assumption 2 and $F\ge 0$ in (2), we have $\frac{dS}{dD}\ge 0$ if and only if $\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}\le 0$.

From (15) we have

$$\begin{aligned} \begin{array}{l} \frac{d^{2}S}{dD^{2}}=\frac{d}{dD}\left( {\frac{F\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}}{(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-F\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}}} \right) \\ =\frac{\left( {F\frac{\partial ^{3}p(D,S,s)}{\partial S\partial D^{2}}+F\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}\frac{dS}{dD}} \right) \left( {(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-F\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right) +\left( {F\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}} \right) \left( {(1-\tau )\frac{\partial ^{3}H(S)}{\partial S^{3}}\frac{dS}{dD}-F\frac{\partial ^{3}p(D,S,s)}{\partial S^{3}}\frac{dS}{dD}-F\frac{\partial ^{3}p(D,S,s)}{\partial S^{2}\partial D}} \right) }{\left( {(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-F\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right) ^{2}}\\ \end{array}\quad \end{aligned}$$

(16)

For the special case $\frac{\partial ^{2}H(S)}{\partial S^{2}}=0$ (linear production), (15) and (16) become

$$\begin{aligned} \frac{dS}{dD}&= -\frac{\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}}{\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}}\end{aligned}$$

(17)

$$\begin{aligned} \frac{d^{2}S}{dD^{2}}&= -\frac{\left( {\frac{\partial ^{3}p(D,S,s)}{\partial S\partial D^{2}}+\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}\frac{dS}{dD}} \right) \left( {\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right) +\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}} \right) \left( {\frac{\partial ^{3}p(D,S,s)}{\partial S^{3}}\frac{dS}{dD}+\frac{\partial ^{3}p(D,S,s)}{\partial S^{2}\partial D}} \right) }{\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right) ^{2}} \end{aligned}$$

(18)

Substituting (17) into (18), we have

$$\begin{aligned} \frac{d^{2}S}{dD^{2}}=-\frac{\frac{\partial ^{3}p(D,S,s)}{\partial S\partial D^{2}}\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}-\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}} \right) ^{2}-\frac{\partial ^{3}p(D,S,s)}{\partial S^{3}}\frac{\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}} \right) ^{2}}{\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}}+\frac{\partial ^{3}p(D,S,s)}{\partial S^{2}\partial D}\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}} \right) }{\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right) ^{2}} \end{aligned}$$

(19)

So we have $\frac{d^{2}S}{dD^{2}}\le 0$ if and only if (10) is satisfied. For the second part, when $D$ goes to infinity, we must have $\frac{dS}{dD}\le 0$. Since $S$ must be non-negative, the limit exists. $\square$