Strict liability, scarce generic input and duopoly competition

Abstract

Strict liability imposed on firms involved in an imperfect competition game (here duopoly) is not an obstacle to achieving a socially optimal care level of care. However, when one input becomes scarce (here a limited generic asset), the precedent result is compromised. Duopolistic competition leads to supply more goods at a lower price at the equilibrium, but also a lower care level compared to a monopoly placed in similar conditions. When damage are too high, capped strict liability coupled with an ex-ante regulation leads the firms to increase their care level. The weakest one may disappear letting the other one in monopoly situation.

Appendices

Appendix 1

1.1 1.1 Strict liability equilibrium for firms without restriction on inputs

In this economy, both firms use the generic asset without restriction. This assumption means that the firms disconnect their market activities (production and selling) from accident prevention. Under duopolistic competition this leads to the following program:

$$\underset{{q}_{1}{,x}_{1}}{\mathrm{max}}{B}_{1}\left(\left({q}_{1}{,x}_{1}\right);{q}_{2}\right)={q}_{1}\left(K- {q}_{1}-{q}_{2}\right)-{q}_{1} {c}_{1}-{q}_{1}{ x}_{1}-{q}_{1}p\left({ x}_{1}\right)D$$

(1a)

$$\underset{{q}_{2}{,x}_{2}}{\mathrm{max}}{B}_{2}\left(\left({q}_{2}{,x}_{2}\right);{q}_{1}\right)={q}_{2}\left(K- {q}_{1}-{q}_{2}\right)-{q}_{2} {c}_{2}-{q}_{2}{ x}_{2}-{q}_{2}p\left({ x}_{2}\right)D$$

(1'a)

From the first-order conditions, we get:

$$\frac{{\partial B}_{1}\left(-\right)}{\partial {q}_{1}}=0\Rightarrow\frac{{\partial B}_{1}\left(-\right)}{\partial {q}_{1}}=K-2{q}_{1}-{q}_{2}-{c}_{1}-{ x}_{1}-p\left({ x}_{1}\right)D=0$$

(2a)

$$\frac{{\partial B}_{2}\left(-\right)}{\partial {q}_{2}}=0\Rightarrow\frac{{\partial B}_{2}\left(-\right)}{\partial {q}_{2}}=K-2{q}_{2}-{q}_{1}-{c}_{2}-{ x}_{2}-p\left({ x}_{2}\right)D=0$$

(2'a)

Calculating the best response functions requires determining the care level because otherwise, they would interfere with the expression of quantity ${q}_{i}$ in relation to quantity ${q}_{j}, i\ne j, i,j=\mathrm{1,2}$. Similar to free competition, this determination is independent of the knowledge of ${q}_{i}$ for firm $i$. Then it is preferable to fix the firms’ optimum care level (here ${x}_{i}^{0}>0$), such that²⁰:

$$p{^{\prime}}{(x}_{i}^{0})=-\frac{1}{D}$$

Solving the above equations for ${q}_{1}$ and ${q}_{2}$ we can define the game Nash equilibrium $\left({q}_{1}^{*},{q}_{2}^{*}\right)$:

$${q}_{1}^{*}=\frac{1}{3}\left(K-2{c}_{1}+{c}_{2}-2{(x}_{1}^{0}+Dp\left({x}_{1}^{0}\right))+{x}_{2}^{0}+Dp\left({x}_{2}^{0}\right)\right)$$

(3a)

And,

$${q}_{2}^{*}=\frac{1}{3}\left(K-2{c}_{2}+{c}_{1}-2{({x}_{2}^{0}+D p\left({x}_{2}^{0}\right))+x}_{1}^{0}+D p\left({x}_{1}^{0}\right)\right)$$

(3'a)

1.2 1.2 The constrained monopoly choice

Let us assume that the monopoly is constrained in such a way that the program is the following

$$\underset{{q}_{m},{x}_{m}}{\mathrm{max}}{B}_{m}\left({q}_{m},{ x}_{m} \right)={q}_{m}\left(K-{q}_{m}\right)-{q}_{m}c-{q}_{m}{ x}_{m}-{q}_{m}p({ x}_{m})D$$

(7)

Under the constraints

$${q}_{m}\left(c+{x}_{m}\right)\le { Y}_{m},$$

(7a)

$$0\le p\left({ x}_{m}\right)\le 1$$

(7b)

• 1) We define the condition for which the manager may decide to comply or not with the first-best care level. Understanding this condition will involve deducing the condition for not complying to it. Using (7) and the results in 3.1, then, without constraint, the firm determines the first-best care level.

  $$\frac{{\delta B}_{m}\left({q}_{m},{ x}_{m} \right)}{{\delta x}_{m}}=-1-p\mathrm{^{\prime}}\left({ x}_{m}\right)D=0$$

  Let ${x}_{m}^{0}$ be this level that meet this condition, and the probability of an accident is known ${p(x}_{m}^{0})$. It follows that, once the optimum care level determined, the monopoly’s program becomes:

  $\underset{{q}_{m},{x}_{m}}{\mathrm{max}}{B}_{m}\left({q}_{m},{ x}_{m} \right)={q}_{m}\left(K-{q}_{m}\right)-{q}_{m}(\tilde{c }-{p(x}_{m}^{0})D)$ under the constraint: ${q}_{m}\left(\tilde{c }\right)\le { Y}_{m}$ (where $\tilde{c }=c+{x}_{m}^{0}$).

  As $\left\{\tilde{c }, {p(x}_{m}^{0}),D\right\}$ are given, the program is a classical monopoly program that expresses as:

  $$\underset{{q}_{m}\ge 0,{x}_{m}\ge 0}{\mathrm{max}}{B}_{m}\left({q}_{m} \right)\mathrm{under the constraint }{q}_{m}\left(\tilde{c }\right)\le { Y}_{m}$$

  (7')

  We determine the equilibrium quantity ${q}_{m}^{0}$ and, consequently, let ${B}_{m}\left({q}_{m}^{0}\right)$ be the equilibrium payoff of the monopoly.

• (2) If the monopolist chooses to lessen the equilibrium care level, the program is (7), (7a), (7b) and the solution is${B}_{m}\left({q}_{m}^{1} ,{x}_{m}^{1}\right)$, where $\left({q}_{m}^{1},{x}_{m}^{1}\right)$ is the equilibrium solution with${x}_{m}^{1}<{x}_{m}^{0}$.

Proposition A

Considering 1) and 2) above, the program (7), (7a, b) and the program (7’), then:

• i) If ${B}_{m}\left({q}_{m}^{0} \right)>{B}_{m}\left({q}_{m}^{1},{x}_{m}^{1}\right)$ , then, the monopolist keeps the first-best care level corresponding to the unrestricted program.

• ii) If ${B}_{m}\left({q}_{m}^{0} \right)<{B}_{m}\left({q}_{m}^{1},{x}_{m}^{1}\right)$ , then, the monopolist chooses to lessen the care level.

Proof

The proof results from the comparison of 1) and 2).

Appendix 2

2.1 2.1 Strict liability equilibrium for firms with restricted inputs

Each firm disposes of a limited but generic asset (a “generic input” or a “generic asset”) $({Y}_{i}\ne {Y}_{j}, i,j=\mathrm{1,2}, i\ne j)$ and makes a strategic tradeoff between production and care according to the other firms’ choices. The program corresponds to the Eqs. (6) to (6c) in the main-text argument. Solving this program proves the following proposition:

Proposition 1

Under the assumption that the coefficients $K$,$D$, ${c}_{i}$ combine to make the payoff function quasi-concave and since it is twice derivable, then the program (6) to (6’c) admits at least one Nash equilibrium $\left({q}_{\mathrm{i}}^{0},{x}_{\mathrm{i}}^{0}\right)$ which maximizes the firm’'s (respectively firm j’s) payoff-functions${B}_{i}\left({{ q}_{i,}x}_{i} \right), i=\mathrm{1,2}$.

Proof of proposition 1

To simplify we solve the program for firm 1; the proof is the same for firm 2. The object is to set up the best-response functions, and then to determine the game’s Nash equilibrium. The results are derived for firm 2 given the supply i.e., ${\overline{q}}_{2}$. We consider that the coefficients $D$,$K,$ ${c}_{1}\le 1$ are such that the function ${B}_{1}\left({q}_{1},{ x}_{1}\right)$ is quasi-concave. This is analyzed in part a) of the proof. Part b) defines the best response functions of each firm while part c) describes the Nash equilibrium of the game. However, first we must write the Karush–Kuhn–Tucker conditions which are:

$$\mathcal{L}\left({x}_{1}, {q}_{1},{\lambda }_{1},{\lambda }_{2}\right)={q}_{1}\left(K-{q}_{1}-{\overline{q}}_{2}\right)-{q}_{1}{c}_{1}-{q}_{1}{ x}_{1}-{q}_{1}p\left({ x}_{1}\right)D+{\lambda }_{1}\left(-{q}_{1}\left({c}_{1}+{x}_{1}\right)+{ Y}_{1}\right)+{\lambda }_{2}\left(-p\left({x}_{1}\right)+1\right){+\lambda }_{3}p\left({x}_{1}\right).$$

(6d)

$$\frac{\partial \mathcal{L}}{\partial {x}_{1}}\le 0\Rightarrow -{q}_{1}-{q}_{1}{p\left({ x}_{1}\right)}^{\mathrm{^{\prime}}}D-{\lambda }_{1}{q}_{1}-{\lambda }_{2}p\mathrm{^{\prime}}\left({x}_{1}\right)\le 0$$

(6e)

$${x}_{1}\ge 0$$

(6f)

$${x}_{1}\frac{\partial \mathcal{L}}{\partial {x}_{1}}=0\Rightarrow {x}_{1}\left(-{q}_{1}-{q}_{1}{p}^{\mathrm{^{\prime}} }\left({ x}_{1}\right)D-{\lambda }_{1}{q}_{1}-{\lambda }_{2}p\mathrm{^{\prime}}\left({x}_{1}\right)\right)=0$$

(6g)

$$\frac{\partial \mathcal{L}}{\partial {q}_{1}}\le 0\Rightarrow K-2 {q}_{1}-{\overline{q}}_{2}-{ x}_{1}-{c}_{1}-p\left({ x}_{1}\right)D-{\lambda }_{1}\left({c}_{1}+{x}_{1}\right)\le 0$$

(6h)

$${q}_{1}\ge 0$$

$${q}_{1}\frac{\partial \mathcal{L}}{\partial {q}_{1}}=0\Rightarrow {q}_{1}\left(K-2 {q}_{1}-{\overline{q}}_{2}-{ x}_{1}-{c}_{1}-p\left({ x}_{1}\right)D-{\lambda }_{1}\left({c}_{1}+{x}_{1}\right)\right)=0$$

(6i)

$$\frac{\partial \mathcal{L}}{\partial {\lambda }_{1}}\ge 0\Rightarrow {-q}_{1}\left({c}_{1}+{x}_{1}\right)+{ Y}_{1}\ge 0$$

(6j)

$${\lambda }_{1} \frac{\partial \mathcal{L}}{\partial {\lambda }_{1}}=0\Rightarrow {\lambda }_{1}\left(-{q}_{1}\left({c}_{1}+{x}_{1}\right)+{ Y}_{1}\right)=0$$

(6k)

$${\lambda }_{1}\ge 0$$

$$\frac{\partial \mathcal{L}}{\partial {\lambda }_{2}}\ge 0\Rightarrow -p\left({ x}_{1}\right)+1\ge 0$$

(6l)

$${\lambda }_{2}\ge 0.$$

(6m)

$${\lambda }_{2}\frac{\partial \mathcal{L}}{\partial {\lambda }_{2}}=0\Rightarrow {\lambda }_{2}\left(-p\left({ x}_{1}\right)+1\right)=0$$

(6n)

$$\frac{\partial \mathcal{L}}{\partial {\lambda }_{3}}\ge 0\Rightarrow p\left({ x}_{1}\right)\ge 0$$

(6o)

$${\lambda }_{3}\ge 0$$

(6p)

$${\lambda }_{3}\frac{\partial \mathcal{L}}{\partial {\lambda }_{2}}=0\Rightarrow {\lambda }_{3}\left(p\left({ x}_{1}\right)\right)=0$$

(6q)

• (a) Quasi-concavity of the program

  To simplify the process of obtaining a solution we consider strictly positive ${x}_{1}$ and ${q}_{1}$, discarding ${x}_{1}=0$ (no prevention) and ${q}_{1}=0$ (no production), and negative values (i.e., we study the “normal” state of affairs). We also consider the particular case where the constraint is binding (${\mathrm{i}.\mathrm{e}. \lambda }_{1}>0$). Thus, the firm is limited by its asset. We reject ${\lambda }_{2}>0\mathrm{ and }{\lambda }_{3}>0$ because if they are not equal to zero this means that ${x}_{1}=0$ and/or $p\left({ x}_{1}\right)=0$ which would violate our assumptions that ${x}_{1}$ is strictly positive. It is easy to verify this when ${\lambda }_{1}={\lambda }_{2}={\lambda }_{3}=0$, then, the system becomes similar to the situation without rationed asset.

  We can study the concavity or the quasi-concavity of ${B}_{1}\left({q}_{1},{ x}_{1}\right)$.

  Since the constraint is binding, we replace ${x}_{1}$ by ${x}_{1}=$ $\frac{{Y}_{1}}{{q}_{1}}-{c}_{1}$ in the payoff function:

  $${B}_{1}\left({q}_{1}\right){=q}_{1}\left(K-{q}_{1}-{\overline{q}}_{2}\right)-{ Y}_{1}-{q}_{1}p\left(\frac{{ Y}_{1}}{{q}_{1}}-{c}_{1}\right)D$$

  It is easy to see that ${q}_{1}p\left(\frac{{ Y}_{1}}{{q}_{1}}-{c}_{1}\right)D$ increases and ${q}_{1}\left(K-{q}_{1}-{\overline{q}}_{2}\right)$ decreases as ${q}_{1}$ increases in the interval $[0,{Y}_{1}]$. Then on this interval, both functions ${q}_{1}p\left(\frac{{ Y}_{1}}{{q}_{1}}-{c}_{1}\right)D$ and ${q}_{1}\left(K-{q}_{1}-{\overline{q}}_{2}\right)$ cross once. Thus, there exists ${\tilde{q }}_{1}\in [0,{Y}_{1}]$ such that:

  $${\tilde{q }}_{1}\left(K-{\tilde{q }}_{1}-{\overline{q}}_{2}\right)-{\tilde{q }}_{1}p\left(\frac{{ Y}_{1}}{{\tilde{q }}_{1}}-{c}_{1}\right)D=0$$

  It follows that: ${B}_{1}\left({\tilde{q }}_{1}\right)={B}_{1}\left(0\right)={-Y}_{1}$ and for ${\tilde{q }}_{1}$ and 0, ${B}_{1}(-)$ is negative. To be sustainable, there must be an interval $[{q}_{a},{q}_{b}]\subset [0,{Y}_{1}]$ such that for all ${q}_{1}^{a}\in [{q}_{a},{q}_{b}]$, ${B}_{1}\left({q}_{1}^{a}\right)>0$, with ${B}_{1}\left({q}_{a}\right)={B}_{1}\left({q}_{b}\right)=0$. If ${q}_{1}^{a}$ does not exist that fulfills this condition, and the program is not viable because, ${B}_{1}\left({q}_{a}\right)<0$.

Then the condition for having a viable program (i.e., a positive payoff) is that:

Condition A:

$K>\frac{{ Y}_{1}}{{q}_{1}^{a}}+{q}_{1}^{a}+{\overline{q}}_{2}+p\left({q}_{1}^{a}\right)D, {q}_{1}^{a}\in [{q}_{a},{q}_{b}]$

With this assumption, from ${B}_{1}\left({\tilde{q }}_{1}\right)={B}_{1}\left(0\right)={-Y}_{1}$, and with ${q}_{1}^{a}\in [{q}_{a},{q}_{b}]\subset [0,{Y}_{1}]$, there exists ${q}_{1}^{a}$, such that ${B}_{1}\left({q}_{1}^{a}\right)>0$, and from Rolle's lemma there exists at last one ${q}_{1}^{*}$, ${q}_{1}^{*}\in [{q}_{a},{q}_{b}]$ such that ${B}_{1}\mathrm{^{\prime}}\left({q}_{1}^{*}\right)=0$. It follows that for $[0,{Y}_{1}]$, the program admits an interval $[{q}_{a},{q}_{b}]\subset [0,{Y}_{1}]$, where for all ${q}_{1}^{a}\in [{q}_{a},{q}_{b}]$, ${B}_{1}\left({q}_{1}^{a}\right)>0$ and ${B}_{1}\left(q\right)\le 0,$ for all $q\notin [{q}_{a},{q}_{b}]$. Then the program is quasi-concave on $[{q}_{a},{q}_{b}]$.

• (b) Formation of the best-response function${x}_{1}>0$ in (6.g) implies:$-{q}_{1}-{q}_{1}{p}^{\mathrm{^{\prime}}}\left({ x}_{1}\right)D+{\lambda }_{1}{q}_{1}=0$ and consequently, ${x}_{1}^{0}>0$ such that:

  $${p}^{\mathrm{^{\prime}}}\left({x}_{1}^{0}\right)=-\left(\frac{1+{\lambda }_{1}}{D}\right)$$

  Concerning ${q}_{1}>0$, by (6.i):

  $$K-2 {q}_{1}-{\overline{q}}_{2}-{ x}_{1}-{c}_{1}-p\left({ x}_{1}\right)D-{\lambda }_{1}\left({c}_{1}+{x}_{1}\right)=0$$

  But ${\lambda }_{1}>0$, implies ${q}_{1}\left({c}_{1}+{x}_{1}\right)-{ Y}_{1}=0$.It follows that we determine a system with three unknowns ${q}_{1},{x}_{1},$ ${\lambda }_{1}$ and, three equations:

  $${q}_{1}\left({c}_{1}+{x}_{1}\right)-{ Y}_{1}=0$$

  (a1)
  $${p}^{\mathrm{^{\prime}}}\left({x}_{1}\right)=-\left(\frac{1+{\lambda }_{1}}{D}\right)$$

  (a2)
  $$K-2 {q}_{1}-{\overline{q}}_{2}-{ x}_{1}-{c}_{1}-p\left({ x}_{1}\right)D-{\lambda }_{1}\left({c}_{1}+{x}_{1}\right)=0$$

  (a3)

  To define the best response function for firm 1, it is sufficient to extract ${\lambda }_{1}$ from (a2), ${\lambda }_{1}=-1-D{p}^{\mathrm{^{\prime}}}\left({x}_{1}\right),$ and report this value in (a3) where, previously, ${q}_{1}$, from (a1) has been expressed in terms of ${x}_{1}$. Then considering these changes, from (a3) we get an equation in terms only of ${x}_{1}$:

  $$K-D p\left({x}_{1}\right)-\frac{2 {Y}_{1}}{{c}_{1}+{x}_{1}}-{\overline{q}}_{2}+D\left({c}_{1}+{x}_{1}\right){p}^{\mathrm{^{\prime}}}\left({x}_{1}\right)=0$$

  Obviously, solving this equation depends on the probability density $p\left({x}_{1}\right)$. The solution obtained depends on ${\overline{q}}_{2}$, then ${x}_{1}^{0}={x}_{1}\left( {\overline{q}}_{2}\right), {x}_{2}^{0}={x}_{2}( {\overline{q}}_{1})$. We replace all expressions in ${x}_{1}$ in (a3) to determine the best response functions or reaction functions. Then by the implicit function theorem the first order condition for firm 1 alone defines what is its best strategy given ${\overline{q}}_{2}$. A similar argument holds for firm 2:

  $${q}_{1}\left( {\overline{q}}_{2}\right)=\frac{1}{3}\left[K-2Dp\left( {x}_{1}^{0}\right)+p\left({x}_{2}^{0}\right)+2{c}_{1}D{p}^{\mathrm{^{\prime}}}{(x}_{1}^{0})+2D{x}_{1}^{0}{p}^{\mathrm{^{\prime}}}({x}_{1}^{0})-{c}_{2}Dp{^{\prime}}\left({x}_{2}^{0}\right)-D{x}_{2}^{0}{p}^{\mathrm{^{\prime}}}({x}_{2}^{0})\right]$$
  $${q}_{2}\left( {\overline{q}}_{1}\right)=\frac{1}{3}\left[K-2Dp\left({x}_{2}^{0}\right)+p\left({x}_{1}^{0}\right)+2{c}_{2}D{p}^{\mathrm{^{\prime}}}{(x}_{2}^{0})+2D{x}_{2}^{0}{p}^{\mathrm{^{\prime}}}({x}_{2}^{0})-{c}_{1}Dp{^{\prime}}\left({x}_{1}^{0}\right)-D{x}_{1}^{0}{p}^{\mathrm{^{\prime}}}({x}_{1}^{0})\right]$$

• (c) Nash equilibrium determination

  From the reaction functions we can define the Nash equilibrium of the game (see definition 1 above) which consists of determining the quantities $\left({q}_{1}^{0},{q}_{2}^{0}\right)$ which the firms will supply at equilibrium. The Nash equilibrium can be seen as occurring when each firm 1 strategy is the best response to the other firm’s strategical choice (similarly for firm 2). Substituting firm 2’s best response function into firm 1 gives:

  $${q}_{1}^{0}{=q}_{1}\left({q}_{2}\left( {\overline{q}}_{1}\right)\right)$$

  And for firm 2.

  $${q}_{2}^{0}{=q}_{2}\left({q}_{1}\left( {\overline{q}}_{2}\right)\right)$$

  Having determined ${q}_{1}^{0}$ and ${q}_{2}^{0}$, we deduce the game’s equilibrium level of care for both firms (${x}_{1}^{0}$ and ${x}_{2}^{0}$)and then ${B}_{1}\left(\left({x}_{1}^{0},{q}_{1}^{0}\right),\left({x}_{2}^{0},{q}_{2}^{0}\right)\right),{B}_{2}\left(\left({x}_{2}^{0},{q}_{2}^{0}\right),\left({x}_{1}^{0},{q}_{1}^{0}\right)\right)$ is the Nash game equilibrium.

Appendix 3

Proof of proposition 2

To demonstrate a) it is sufficient to show that ${q}_{m}^{0}$ must be such that ${{q}_{m}^{0}<q}_{1}^{0}+{q}_{2}^{0}$.

Then assume that ${q}_{m}^{0}\ge {q}_{1}^{0}+{q}_{2}^{0}$. As ${q}_{1}^{0}+{q}_{2}^{0}=\frac{2}{3}\left(K-D\left(1-c\right)\right)$ and ${q}_{m}^{0}=\frac{1}{2}\left(K-D\left(1+c\right)\right)$ (by assumption ${c}_{1}={c}_{2}=c$), with respectively $K>D\left(1-c\right)$ and $K>D(1+c)$ to guarantee positive quantities, and develop:

$$\frac{1}{2}\left(K-D\left(1+c\right)\right)>\frac{2}{3}\left(K-D\left(1-c\right)\right)$$

This inequality is true if $D\left(1-7c\right)>K$ but this contradicts the condition that

$$K>D\left(1+c\right)\mathrm{and} K>D\left(1-c\right)$$

Then it follows that:

$${{q}_{m}^{0}<q}_{1}^{0}+{q}_{2}^{0}$$

From this result, b) it follows that the duopoly equilibrium price ${\pi }_{D}^{0}$ is lower than the monopoly price ${\pi }_{m}^{0}$ then:

$${\pi }_{m}^{0}>{\pi }_{D}^{0}$$

It remains to prove c). For this, as previously, we compare the respective care levels namely: ${x}_{i}^{0}$ et ${x}_{m}^{0}$. Recall that these values are proportion that corresponds to “intensity” they are comprised between 0 and 1. We analyze successively ${x}_{i}^{0}$ and ${x}_{j}^{0}, i$ $i\ne j.$

Let us assume that ${x}_{m}^{0}>{x}_{i}^{0}$, and by taking their respective value:

$$\frac{{Y}_{m}}{\frac{1}{2}(K-D(1+c)) }-c>\frac{3{Y}_{\mathrm{i}}}{K-D(1-\mathrm{c})}-c$$

After our simplifications and as $a{Y}_{m}={Y}_{i}$

$$\frac{1}{\frac{1}{2}(K-D(1+c)) }>\frac{3\mathrm{ a}}{K-D(1-\mathrm{c})}$$

Then:

$K>D\left(1-c\left(\frac{2+3a}{2-3a}\right)\right)$

For $a\ne \frac{2}{3}$, this expression is true because by hypothesis $K>D(1+c)$ and,$D\left( {1 + c} \right) > D\left( {1 - c\left( {\frac{{2 + 3a}}{{2 - 3a}}} \right)} \right) \Rightarrow 1 > - \left( {\frac{{2 + 3a}}{{2 - 3a}}} \right)i.e.2 > - 2$

Consequently, ${x}_{m}^{0}>{x}_{i}^{0}$ i.e., the care level of the monopoly is higher than the level chosen by the duopoly firms.

Then a similar analysis is made for ${x}_{j}^{0}, i\ne j$. As by assumption ${Y}_{1}+{Y}_{2}={Y}_{m}$ with $0<a,b<1:$ $a{Y}_{m}={Y}_{i}$ et $b {Y}_{m}={Y}_{j}$ (then obviously $b {Y}_{m}=(1-a){{Y}_{m}=Y}_{j}$. By the same argument than previously:

Let us assume that ${x}_{m}^{0}>{x}_{j}^{0}$, this involves that:

$D\left( {1 + c} \right) > D\left( {1 - c\left( {\frac{{2 + 3b}}{{2 - 3b}}} \right)} \right) \Rightarrow 1 > - \left( {\frac{{2 + 3b}}{{2 - 3b}}} \right)$

Or, by replacing $b {Y}_{m}=(1-a){{Y}_{m}=Y}_{j}$

$D\left(1+c\right)>D\left(1-c\left(\frac{5- 3a}{3\mathrm{a}-1}\right)\right) \Rightarrow 1-1>-5$

for $a\ne \frac{1}{3}$

Then, ${x}_{m}^{0}>{x}_{j}^{0}$.

Consequently,

We have simultaneously, ${{q}_{m}^{0}<q}_{1}^{0}+{q}_{2}^{0}$ and ${x}_{m}^{0}>{x}_{j}^{0}$ and ${x}_{m}^{0}>{x}_{i}^{0}$

Appendix 4

Proof of proposition 3

General feature and proof of i)

The regulator introduces a standard corresponding to a minimum probability threshold $\overline{p}$ which needs ${\overline{x}}_{i}, i=\mathrm{1,2}$ as the care effort needed to achieve it. It follows that the firm uses ${q}_{i}{\overline{x}}_{i}$ as the input for prevention, then ${\overline{x}}_{i}=\overline{x} , i=\mathrm{1,2}$ (by assumption the firms use the same technology). Firm $i{^{\prime}}$ s program depends on${q}_{i}$, given${\overline{q}}_{j}$, we look for an interior solution${q}_{i}>0$:

$$\underset{{q}_{i}}{\mathrm{max}}{B}_{i}\left({q}_{i},{\overline{q}}_{j} \right)={q}_{i}\left(K-{q}_{i}-{\overline{q}}_{j}\right)-{q}_{i}\left({c}_{i}+\overline{x}\right)-{q}_{i}\overline{p}D.$$

With ${q}_{i}\in \left[0,\frac{{ Y}_{i}}{\left({c}_{i}+\overline{x}\right)}\right]$. given the asset constraint ${q}_{i}\left({c}_{i}+\overline{x}\right)\le { Y}_{1}$. Then, ${q}_{i}^{0}\left({\overline{q}}_{j}\right)$ the interior solution is such that:

$$\frac{{\partial B}_{i}\left({q}_{i},{\overline{q}}_{j} \right)}{\partial {q}_{i}}=K-\overline{p}D-\overline{x}-{c}_{i}-2{q}_{i}^{0}\left({\overline{q}}_{j}\right) -{\overline{q}}_{j}=0$$

With:

$$\frac{{\partial B^2}_{i}\left({q}_{i},{\overline{q}}_{j} \right)}{\partial^2 {q}_{i}}=-2<0$$

Then the above program is concave and admits a solution. Does this solution lie within $\left[0,\frac{{ Y}_{1}}{\left({c}_{i}+\overline{x}\right)}\right]$? To address this, we build the best response functions and determine the Nash equilibrium of the system:

$${q}_{i}^{0}=\frac{1}{3}(K-\overline{p}D-\overline{x}-2{c}_{i}+{c}_{j})$$

And for:

$${q}_{j}^{0}=\frac{1}{3}(K-\overline{p}D-\overline{x}-2{c}_{j}+{c}_{i})$$

Function ${B}_{i}\left({q}_{i}^{0},{q}_{j}^{0}\right)$ reaches a maximum at ${q}_{i}^{0}\le \frac{{ Y}_{i}}{\left({c}_{i}+\overline{x}\right)}$. As ${q}_{i}^{0}$ is determined independently of $\frac{{Y}_{i}}{\left({c}_{i}+\overline{x}\right)}$, which results in ${q}_{i}^{0}$ being higher or lower than this value. If it is lower (unsaturated constraint) the firm $i$ (respectively $j$) maximizes its profit for this value. Then, it will not be able to produce beyond $\frac{{Y}_{i}}{\left({c}_{i}+\overline{x}\right)}$ (respectively $\frac{{Y}_{j}}{{c}_{j}+\overline{x}}$). Thus, the production level is reduced.

a) Proof of ii) The equilibrium price is higher due to a lower supply,

To simplify we assume that (${c}_{i}={c}_{j}\equiv c)\mathrm{ and that}{ Y}_{j}<{ Y}_{i}.$ The following relationships result from these assumptions:

${q}_{i}^{0}={q}_{j}^{0}=\frac{1}{3}(K-\overline{p}D-\overline{x}-c)$ or also:

$${q}_{i}^{0}=\frac{1}{3}(K-\overline{p}D-\overline{x}-c)= \frac{{ Y}_{i}}{\left(c+\overline{x}\right)}$$

$${q}_{j}^{0}=\frac{1}{3}\left(K-\overline{p}D-\overline{x}-c\right)>\frac{{ Y}_{j}}{\left(c+\overline{x}\right)}$$

Hence, firm $j$ does not own enough resource to provide ${q}_{j}^{0}$ when firm $i$ supplies ${q}_{i}^{0}>\frac{{ Y}_{i}}{\left(c+\overline{x}\right)}$. The firm must then forgo this value. The question then is whether ${q}_{j}^{1}=\frac{{ Y}_{j}}{\left(c+\overline{x}\right)}$, a Nash-equilibrium may exist. As the constraint is binding for firm j, then:

$$\frac{{\partial B}_{i}\left({q}_{i},{q}_{j}^{1} \right)}{\partial {q}_{i}}=K-\overline{p}D-\overline{x}-c-2{q}_{i}-\frac{{ Y}_{j}}{\left(c+\overline{x}\right)}=0$$

We deduce:

$${q}_{i}^{1}=\frac{1}{2}\left(K-\overline{p}D-\overline{x}-c-\frac{{ Y}_{j}}{\left(c+\overline{x}\right)}\right)$$

Then (${q}_{i}^{1}, {q}_{j}^{1}$) is a Nash equilibrium for a higher price compared to if firms were not facing the limited asset. This corresponding price is ${\uppi }^{1}\left({q}_{i}^{1}, {q}_{j}^{1}\right)=K-{q}_{i}^{1}-{q}_{j}^{1}=\frac{1}{2}\left(K-\overline{p}D-\overline{x}-c\right)$ and may be compared to the price that would result from a firm j without restricted assets, i.e.

$${\pi }^{0}\left({q}_{i}^{0}, {q}_{j}^{0}\right)=K-\frac{2}{3}\left( K-\overline{p} D-\overline{x}-c\right)$$

It turns out that ${\pi }^{1}\left({q}_{i}^{1}, {q}_{j}^{1}\right)>{\pi }^{0}\left({q}_{i}^{0}, {q}_{j}^{0}\right)$ as firm $j$ restrict its offer to a maximum of $\frac{{Y}_{j}}{\left(c+\overline{x}\right)}$. Consequently, as expected, the equilibrium price is higher.

Proof of iii):The potential exit of the weakest firm

How does firm $j$ achieve a positive profit? This involves knowing the conditions for which ${B}_{j}^{1}\left({q}_{i}^{1}, {q}_{j}^{1}\right)>0$. This is true for $\frac{K}{3}> \left( \overline{p}D+\overline{x}+c\right)$.²¹ It should be emphasized that this condition is general and applies to both firms because of their symmetric costs. If, instead, we consider differentiated costs, (including unit accident costs) then the following relationship makes sense:

$${B}_{j}^{1}\left({q}_{i}^{1}, {q}_{j}^{1}\right)>0 \Rightarrow \frac{1}{2}\left(K-\overline{p}D-\overline{x}-c\right)>\left( \overline{p}{D}_{j}+{\overline{x}}_{j}+{c}_{j}\right)>0$$

Then, the equilibrium price per unit must be twice the value of the total marginal costs for the profit of firm j to be positive. For firm $i$, if the costs are lower than those of $j$, the price level at which it will make positive profits will be lower. For example, if $\left(\overline{p}{D}_{j}+{\overline{x}}_{j}+{c}_{j}\right)=2\left(\overline{p}D-\overline{x}-c\right)$(where the second member of equality designates the costs of $i$), then:

$$\frac{1}{2}\left(K-\overline{p}D-\overline{x}-c\right)>\left( \overline{p}{D}_{j}+{\overline{x}}_{j}+{c}_{j}\right)=2\left(\overline{p}D+\overline{x}+c\right)$$

Firm i, for the same price level, makes a profit if:

$\frac{1}{4}\left(K-\overline{p}D-\overline{x}-c\right)>\left(\overline{p}D+\overline{x}+c\right)$, i.e. for a lower price.

Here, if ${\uppi }^{1}\left({q}_{i}^{1}, {q}_{j}^{1}\right)\in \left] \frac{1}{4}\left(K-\overline{p}D-\overline{x}-c\right),\frac{1}{2}\left(K-\overline{p}D-\overline{x}-c\right)\right[$ then, firm i achieves a positive profit while firm j makes a loss and firm $i$ remains alone as a monopoly.