Optimal taxation in a common resource oligopoly game

Abstract

In this paper, we study the optimal tax policy in a differential oligopoly game where the competing firms share the access to a productive renewable resource. We show that, in a Feedback Nash Equilibrium of the game, a linear Markov tax, imposed on the output, and specified as an affine function of the available resource stock, leads the competing firms to produce the socially optimal quantities over time, thus overcoming the dynamic interplay between the tragedy of the commons and the firms’ market power. The optimal tax turns out to be independent from the resource stock in a monopoly, and it cannot be defined in a duopoly.

Appendices

Proof of Proposition 1

In order to get the solution of the optimal control problem in feedback form, we adopt the dynamic programming approach. The Hamilton-Jacobi-Bellman (henceforth, HJB) equation is given by:

$$\begin{aligned} rV(x)=\max _{Q\ge 0}\displaystyle \left\{ (A-Q)Q+\frac{Q^2}{2}+\frac{\partial V(x)}{\partial x}(F(x)-Q)\right\} \end{aligned}$$

where $V(x)\in C^1$ is the unknown (current) value function. The transversality condition is: $\displaystyle \lim \nolimits _{t\rightarrow \infty } e^{-rt}V(x)=0$, hence it is automatically satisfied by any bounded value function.

In what follows, we show that the value function for the considered problem is given by:

$$\begin{aligned} V(x)={\left\{ \begin{array}{ll} \displaystyle \left[ \frac{x}{{\bar{x}}_1}\right] ^{\frac{r}{\delta }}W({\bar{x}}_1) &{}\text{ if } 0\le x< {\bar{x}}_1\\ W(x) &{}\text{ if } {\bar{x}}_1\le x<{\bar{x}}_2\\ \displaystyle \frac{A^2}{2r} &{}\text{ if } x\ge {\bar{x}}_2\end{array}\right. } \end{aligned}$$

(8)

where the values for ${\bar{x}}_1$ and ${\bar{x}}_2$ are given in Proposition 1 and

$$\begin{aligned} W(x)=k_0+k_1 x+k_2 x^2 \end{aligned}$$

(9)

with:

$$\begin{aligned} k_0=\frac{A^2(r-\delta )^2}{2r\delta ^2},\quad k_1=\frac{A(2\delta -r)}{\delta }, \quad k_2=-\frac{2\delta -r}{2} \end{aligned}$$

The FOC for a maximum point of the right hand side of the HJB equation is:²⁹

$$\begin{aligned} Q=\displaystyle \max \left\{ A-\frac{\partial V}{\partial x},0\right\} \end{aligned}$$

In words, the socially optimal production is the perfect competition quantity minus $\frac{\partial V}{\partial x}$, which represents the rent associated with an additional unit of resource stock.

For our guess-function, it results:

$$\begin{aligned} \frac{\partial V}{\partial x}={\left\{ \begin{array}{ll} \displaystyle \frac{r}{\delta }\displaystyle \left[ \frac{x}{{\bar{x}}_1}\right] ^{\frac{r}{\delta }-1}W({\bar{x}}_1) &{}\text{ if } 0\le x< {\bar{x}}_1\\ \displaystyle \frac{\partial W(x)}{\partial x}=k_1+2k_2 x &{}\text{ if } {\bar{x}}_1\le x<{\bar{x}}_2\\ 0 &{}\text{ if } x\ge {\bar{x}}_2\end{array}\right. } \end{aligned}$$

(10)

Therefore, this rent is decreasing w.r.t. x, it tends to infinity when the stock approaches zero and it is zero for large values of the resource stock.

First of all, we must prove that the function V(x) is continuously differentiable. Clearly this holds true for every $x\in [0,{\bar{x}}_1)\cup ({\bar{x}}_1,{\bar{x}}_2)\cup ({\bar{x}}_2,+\infty )$. Here we prove that V(x) is also continuously differentiable at $x={\bar{x}}_1$ and $x={\bar{x}}_2$.

First we check that V(x) is continuous at $x={\bar{x}}_1$ and $x={\bar{x}}_2$:

$$\begin{aligned} \lim _{x\rightarrow {\bar{x}}_1, x<{\bar{x}}_1} V(x)= & {} W({\bar{x}}_1)=\lim _{x\rightarrow {\bar{x}}_1, x>{\bar{x}}_1} V(x)=\frac{A^2(\delta -r)}{r(2\delta -r)}\\ \lim _{x\rightarrow {\bar{x}}_2, x<{\bar{x}}_2} V(x)= & {} W({\bar{x}}_2)=\lim _{x\rightarrow {\bar{x}}_2, x>{\bar{x}}_2} V(x)=\frac{A^2}{2r} \end{aligned}$$

As far as differentiability is concerned, it amounts to check the continuity of the function $\displaystyle \frac{\partial V}{\partial x}$ at $x={\bar{x}}_1$ and $x={\bar{x}}_2$. It is easy to find:

$$\begin{aligned} \lim _{x\rightarrow {\bar{x}}_1, x<{\bar{x}}_1} \frac{\partial V}{\partial x}({\bar{x}}_1)=\frac{\partial W}{\partial x}({\bar{x}}_1)=\lim _{x\rightarrow {\bar{x}}_1, x>{\bar{x}}_1} \frac{\partial V}{\partial x}({\bar{x}}_1)=A\\ \lim _{x\rightarrow {\bar{x}}_2, x<{\bar{x}}_2} \frac{\partial V}{\partial x}({\bar{x}}_2)=\frac{\partial W}{\partial x}({\bar{x}}_2)=\lim _{x\rightarrow {\bar{x}}_2, x>{\bar{x}}_2} \frac{\partial V}{\partial x}({\bar{x}}_2)=0 \end{aligned}$$

Lastly, it can be easily proved that the considered function V(x) actually solves our optimal control problem. Indeed, for $x>{\bar{x}}_1$, the HJB equation admits the interior solution $Q^*(x)=A-\frac{\partial V}{\partial x}$, which, by considering Eq. (10), yields the linear feedback solution (for $x>{\bar{x}}_1$) given in Eq. (4). For $x\le {\bar{x}}_1$, the HJB admits the corner solution $Q^*(x)=0$. It can be easily checked that the function V(x) specified in (8) satisfies the differential equation obtained after substituting $Q^*(x)$, as specified in (4), into the HJB equation.

Proof of Corollary 1

The steady-state values of the resource stock are the solutions of the (algebraic) equation: $F(x)-Q^*(x)=0$. We must consider three cases:³⁰

• (i) For ${\bar{x}}_1\le x<\min \displaystyle \left( \frac{1}{2},{\bar{x}}_2\right) ={\bar{x}}_2$: $F(x)=\delta x, Q^*(x)=(2\delta -r)x-\frac{A(\delta -r)}{\delta }$ and $x^*_1=\frac{A}{\delta }$. As far as the stability of the equilibrium point is concerned, we have: $F(x)-Q^*(x)>0$, hence the resource stock equilibrium trajectory (monotonically) converges to $x^*_1$ for every initial value $x_0<{\bar{x}}_2=x^*_1$.

• (ii) For ${\bar{x}}_2<x<\frac{1}{2}$: $F(x)=\delta x, Q^*(x)=A$ and, again, we get $x^*_1=\frac{A}{\delta }$. Again it holds $F(x)-Q^*(x)>0$, but here this means that the basin of attraction of $x^*_1$ does not contain any initial value ${\bar{x}}_2<x_0<\frac{1}{2}$.

• (iii) For $x>\frac{1}{2}$: $F(x)=\delta (1-x), Q^*(x)=A$ and $x^*_2=1-\frac{A}{\delta }>x^*_1$. In this case we have: $F(x)-Q^*(x)>(<)0\iff x_0<(>)x^*_2$, hence, by taking into account also the results shown in the previous point, it follows that $x^*_2$ is asymptotically stable with basin of attraction $(1-\frac{A}{\delta },+\infty )$.

Proof of Proposition 3

For $\alpha =0$, the solution of the monopolist’s optimal control problem in feedback form is given by:

$$\begin{aligned} q_M(x)={\left\{ \begin{array}{ll} 0 &{}\text{ if } 0\le x< {\bar{x}}_{1,M}\\ (2\delta -r)x - \displaystyle \frac{(A-\beta )(\delta -r)}{2\delta } &{}\text{ if } {\bar{x}}_{1,M}\le x<{\bar{x}}_{2,M}\\ \displaystyle \frac{A-\beta }{2} &{}\text{ if } x\ge {\bar{x}}_{2,M}\end{array}\right. } \end{aligned}$$

where:

$$\begin{aligned} {\bar{x}}_{1,M}=\frac{(A-\beta )(\delta - r)}{2\delta (2\delta - r)},\quad {\bar{x}}_{2,M}=\frac{(A-\beta )}{2\delta } \end{aligned}$$

This result can be obtained by applying to the considered problem the same technique shown in “Appendix A”, or more easily, by considering the results shown in Proposition 2 with $N=1$ and $A-\beta$ replacing A, since the considered taxation rule plays the role of a constant marginal cost in the firm’s profit function and the FNE of the differential game, taking into account $\beta$, for $N=1$ provides the solution of the single-agent optimal control problem.

By comparing this solution with the welfare-maximizing extraction plan given in Proposition 1, it is trivial to find: $\beta ^*=-A$.

Proof of Proposition 5

Let us consider a simplified version of the regulated oligopoly game in which, as in Lambertini and Mantovani (2014) $F(x)=\delta x$ and the non-negativity constraint on the feasible output set are not taken into account.³¹ In the remainder, this game will be denoted with ${\tilde{\varGamma }}_{\alpha ,\beta }(t_0,x_0)$. In this more tractable, fully linear-quadratic framework, we are able to determine the linear FNE of the game, which will turn out to be very helpful in the computation of the first-best tax rule for the subgames $\varGamma _{\alpha ,\beta }(t_0,x_0<\frac{A}{\delta })$.

In order for the (non-degenerate) linear FNE strategies of these subgames to lead to a positive resource stock in steady-state $x^\infty _{\alpha ,\beta }$, which also constitutes an asymptotically stable equilibrium point, we must consider tax policies such that $(\alpha ,\beta )\in {\mathscr {P}}_1\cup {\mathscr {P}}_2$, where the (disjoint) sets ${\mathscr {P}}_1$ and ${\mathscr {P}}_2$ are defined as follows:

$$\begin{aligned} {\mathscr {P}}_1\equiv & {} \left\{ (\alpha ,\beta )\in {\mathbb {R}}^2\bigg | \alpha<-\frac{(N+1)^2(2\delta -r)}{2(N-1)^2}(<0) \text{ and } \beta >A\right\} \end{aligned}$$

(11)

$$\begin{aligned} {\mathscr {P}}_2\equiv & {} \left\{ (\alpha ,\beta )\in {\mathbb {R}}^2\bigg |-\frac{2\delta -r}{2}<\alpha<\frac{2\delta ^2-\delta r(N^2+3)+r^2(N^2+1)}{r(N-1)^2}(>0) \text{ and } \beta <A\right\} \nonumber \\ \end{aligned}$$

(12)

Roughly speaking if $(\alpha ,\beta )\in {\mathscr {P}}_1$, the public authority imposes a unitary tax which is decreasing with respect to the resource stock, such that the firms are discouraged to choose high production levels when the resource stock is scarce, while production is subsidized when the resource preservation is not a concern. Nevertheless, taking into account only the positivity and the stability of the state variable in steady-state, it is also possible to implement a different kind of tax policy, as far as $(\alpha ,\beta )\in {\mathscr {P}}_2$.

Lemma 1

A symmetric linear FNE of ${\tilde{\varGamma }}_{\alpha ,\beta }(t_0,x_0)$ is given by the N-tuple $({\tilde{\phi }}_{\alpha ,\beta }(x),\ldots ,{\tilde{\phi }}_{\alpha ,\beta }(x))$, where:

$$\begin{aligned} {\tilde{\phi }}_{\alpha ,\beta }(x)=\frac{1}{N+1}[H_{\alpha ,\beta }+K_{\alpha }x] \end{aligned}$$

(13)

where:

$$\begin{aligned} H_{\alpha ,\beta }\equiv & {} \frac{(A - \beta )(N + 1)[(N - 1)\sqrt{\varDelta }- 2\alpha (N-1)^2 - 2\delta (N^2 + 1) + r(3N^2 + 1)]}{2N^2[\sqrt{\varDelta } + r(N + 1)]}<0 \\ K_{\alpha }\equiv & {} \frac{(N + 1)\sqrt{\varDelta } - 2\alpha (N^2-1) + (N+1)^2(2\delta - r)}{4N^2}>0 \end{aligned}$$

with:

$$\begin{aligned} \varDelta _{\alpha }\equiv 4\alpha [(2\delta -r)(N^2+1)+\alpha (N-1)^2]+(2\delta -r)^2(N+1)^2>0 \end{aligned}$$

(14)

The resource stock in steady-state is given by:

$$\begin{aligned} x^{\infty }_{\alpha ,\beta }=\frac{(A {-} \beta )[(N {-} 1)\sqrt{\varDelta } {-} 2\alpha (N-1)^2{-} 2\delta (N^2 {+} 1) {+} r(3N^2 {+} 1)]}{(\alpha {+} \delta )(N {-} 1)\sqrt{\varDelta } {-} 2\alpha ^2(N-1)^2{+} \alpha [r(3N^2 {+} 1) {-} 4\delta (N^2 {+} 1)] {-} \delta [2\delta (N{+}1)^2{-}r(3N^2 {+} 4N {+} 1)]}>0 \nonumber \\ \end{aligned}$$

(15)

and it constitutes an asymptotically stable equilibrium point.

Proof

In order for ${\tilde{\phi }}_{i,(\alpha ,\beta )}(x)$ to be a stationary equilibrium strategy for player i, $i\in \{1,\ldots ,N\}$, it is necessary that there exists a (current) value function ${\tilde{V}}_i(x)$, continuously differentiable, such that the following HJB equation for player i is satisfied [see Dockner et al. (2000)]:

$$\begin{aligned} r{\tilde{V}}_i=\max _{{\tilde{q}}_i}\displaystyle \left\{ [A-(\alpha x+\beta )-{\tilde{q}}_i-{\tilde{\phi }}_{-i}]{\tilde{q}}_i+\frac{\partial {\tilde{V}}_i}{\partial x}(\delta x-{\tilde{q}}_i-{\tilde{\phi }}_{-i})\right\} \end{aligned}$$

(16)

where ${\tilde{\phi }}_{-i}\equiv \sum _{j\ne i}{\tilde{\phi }}_j$. The transversality condition is: $\displaystyle \lim \nolimits _{t\rightarrow \infty } e^{-rt}{\tilde{V}}_i(x)=0$. The first order conditions, which are also sufficient, for the maximum point of the right hand side of the HJB equation are:³²

$$\begin{aligned} A-(\alpha x+\beta )-2{\tilde{q}}_i-{\tilde{\phi }}_{-i}-\frac{\partial {\tilde{V}}_i}{\partial x}=0 \end{aligned}$$

Given the symmetrical structure of this linear-quadratic differential game, we look for a symmetric linear FNE, that is we impose: ${\tilde{V}}_1(x)=\ldots ={\tilde{V}}_N(x)\equiv {\tilde{V}}(x)$, ${\tilde{\phi }}_{1,(\alpha ,\beta )}(x)=\ldots {\tilde{\phi }}_{N,(\alpha ,\beta )}(x)\equiv {\tilde{\phi }}_{\alpha ,\beta }(x)$, and we guess a value function of the form:

$$\begin{aligned} {\tilde{V}}(x)=k_0+k_1 x+k_2 x^2 \end{aligned}$$

(17)

where $k_0,k_1,k_2\in {\mathbb {R}}$ have to be determined by using the undetermined coefficient technique.

As a result, the first order condition can be rewritten as follows:

$$\begin{aligned} {\tilde{\phi }}_{\alpha ,\beta }=\frac{1}{N+1}[A-k_1-\beta -(2k_2+\alpha )x] \end{aligned}$$

(18)

By substituting this expression into the HJB and by collecting the terms containing $x^2$, it follows that $k_2$ must solve the following (algebraic) equation:

$$\begin{aligned} rk_2=2\frac{1}{N+1}(-2k_2-\alpha )\displaystyle \left[ -\alpha -\frac{N(-2k_2-\alpha )}{N+1}\right] +2k_2\left[ \delta -\frac{N(-2k_2-\alpha )}{N+1}\right] \end{aligned}$$

from which we obtain two solutions:

$$\begin{aligned} k_2^{a,b}=\frac{-[2\alpha (N^2+1)+(2\delta -r)(N+1)^2]\pm (N+1)\sqrt{\varDelta }}{8N^2} \end{aligned}$$

with $\varDelta$ given by Eq. (14). By imposing $\varDelta >0$ we get:

$$\begin{aligned} \alpha <-\frac{(N+1)^2(2\delta -r)}{2(N-1)^2}\, \text{ or } \, \alpha >-\frac{2\delta -r}{2} \end{aligned}$$

(19)

which results in the first inequality which defines ${\mathscr {P}}_1$ or ${\mathscr {P}}_2$, respectively. By collecting the terms of the HJB [(after having substituted the FOC (18)] containing x, we get that $k_1$ is the solution of:

$$\begin{aligned} rk_1= & {} \frac{1}{N+1}\displaystyle \left\{ \left[ A-\beta -\frac{N(A-k_1-\beta )}{N+1}\right] (-2k_2-\alpha )-\left[ \alpha +\frac{N(-2k_2-\alpha )}{N+1}\right] (A-k_1-\beta )\right\} \\&+k_1\displaystyle \left[ \delta -\frac{N(-2k_2-\alpha )}{N+1}\right] -2k_2\left[ \frac{N(A-k_1-\beta )}{N+1}\right] \end{aligned}$$

By substituting the two obtained $k_2$ values we get:

$$\begin{aligned} k_1^{a,b}=\frac{(A-\beta )\{(N^2+1)\sqrt{\varDelta }\mp (N+1)[2\alpha (N-1)^2+(2\delta -r)(N^2+1)]\}}{2N^2[\sqrt{\varDelta }\mp r(N+1)]} \end{aligned}$$

By substituting these values into the FOC (18), it is easy to verify that, for $\alpha =\beta =0$, the solutions $k_2^{a},k_1^a$ lead to the Cournot static solution, which we disregard since it does not constitute a FNE of the general unregulated oligopoly game for the most interesting cases in which the initial resource stock endowment is not very high. We have thus considered the other pair of solutions (i.e. $k_2^{b},k_1^b$) which, for $\alpha =\beta =0$, coincides with the non-degenerate FNE strategy found in Lambertini and Mantovani (2014).

Given that we are considering a simple linear resource stock dynamic, it follows that the steady-state resource stock solves: $\delta x=N{\tilde{\phi }}_{\alpha ,\beta }(x)$, which leads to Eq. (15). This value constitutes an asymptotically stable equilibrium if and only if:

$$\begin{aligned} \delta -\frac{N}{N+1}K_{\alpha }<0 \end{aligned}$$

(20)

By imposing this condition, it turns out that this is verified for $\alpha <-\frac{2\delta (n-1)+r(n+1)}{2(n-1)}$ or $\alpha >-\frac{\delta [2\delta -r(N+1)]}{2\delta -rN}$: the latter is implied by the first inequality which defines the set ${\mathscr {P}}_2$, the former is implied by the first inequality which defines the set ${\mathscr {P}}_1$. Therefore equation (19) provides also a sufficient condition for the asymptotically stability of the equilibrium.

Finally, we notice that the steady-state resource stock can be written as:

$$\begin{aligned} x^{\infty }_{\alpha ,\beta }=\frac{NH_{\alpha ,\beta }}{(N+1)\delta -NK_{\alpha }} \end{aligned}$$

whose denominator is negative under the stability condition (20). Therefore $x^\infty _{\alpha ,\beta }>0\iff H_{\alpha ,\beta }<0$, which is verified for $\beta >A$ and $\alpha <\frac{r(3N^2+1)-2\delta (N^2+1)}{2(N-1)^2}$, i.e. for every $(\alpha ,\beta )\in {\mathscr {P}}_1$, or for every $(\alpha ,\beta )\in {\mathscr {P}}_2$. $\square$

The values for $\alpha ^*$ and $\beta ^*$ shown in Proposition 5 are computed by equating the non-degenerate linear strategy of the benevolent planner, given by Proposition 1, with the aggregate output resulting from the linear FNE of the simplified oligopoly game${\tilde{\varGamma }}_{\alpha ,\beta }(t_0,x_0)$, given in the Lemma. Notice that $(\alpha ^*,\beta ^*)\in {\mathscr {P}}_1$. In what follows, we prove that the affine linear Markov tax with coefficients $\alpha ^*$ and $\beta ^*$ leads the considered FNE of the general regulated oligopoly game to coincide with the socially optimal solution for every $x\in [0,\frac{A}{\delta }]$.

The HJB equation for player $i,i\in \{1,\ldots ,N\}$, is Eq. (16), with the considered $\alpha ^*$ and $\beta ^*$ values, which must hold with the very same transversality condition.³³ Given that we are now considering the non-negativity constraint on the output, we guess the following value function:

$$\begin{aligned} V_1(x)=\ldots = V_N(x)\equiv V(x)={\left\{ \begin{array}{ll} \displaystyle \left[ \frac{x}{{\bar{x}}_1}\right] ^{\frac{r}{\delta }}{\tilde{V}}({\bar{x}}_1) &{}\text{ if } 0\le x< {\bar{x}}_1\\ {\tilde{V}}(x) &{}\text{ if } {\bar{x}}_1\le x\le \frac{A}{\delta }\end{array}\right. } \end{aligned}$$

where the function ${\tilde{V}}(x)$ is the value function we computed for the simplified oligopoly game. By proceeding as in “Appendix A”, it can be proved that the function V(x) is continuously differentiable. Furthermore, for $x>{\bar{x}}_1$, the HJB Eq. (16) admits the interior solution given by equation (18) for the considered $\alpha ^*$ and $\beta ^*$ values, which yields the non-degenerate linear strategy of the benevolent planner. For $x<{\bar{x}}_1$, the non-negativity constraint on the production is binding, hence the HJB Eq. (16) admits the corner solution $\phi _{\alpha ^*,\beta ^*}(x)=0$. Summarizing, we have:

$$\begin{aligned} \phi _{\alpha ^*,\beta ^*}(x)={\left\{ \begin{array}{ll} 0 &{}\text{ if } 0\le x< {\bar{x}}_1\\ \displaystyle \frac{1}{N}\displaystyle \left[ (2\delta - r)x - \frac{A(\delta -r)}{\delta }\right] &{}\text{ if } {\bar{x}}_1\le x<{\bar{x}}_2=\frac{A}{\delta } \end{array}\right. } \end{aligned}$$

It can be easily checked that the function V(x) satisfies the differential equation obtained after substituting $\phi _{\alpha ^*,\beta ^*}(x)$, as specified above, into the HJB Eq. (16). In other words, we proved that, for every $x\in [0,\frac{A}{\delta }]$: $N\phi _{\alpha ^*,\beta ^*}(x)=Q^*(x)$.

Proof of Corollary 4

(a) It is trivial to find: $\tau ^*(x)=\alpha ^*x+\beta ^*>0\iff x<{\tilde{x}}$, where ${\tilde{x}}$ is given by (7). In steady state: $\tau ^*(x^*_1)=-\displaystyle \frac{A(\delta N - r)}{(N-2)[\delta (N - 2) + r]}<0$. Furthermore it holds:

$$\begin{aligned} \frac{\partial \tau ^*(x^*_1)}{\partial N}=-\frac{A(\delta N - r)}{(N-2)[\delta (N - 2) + r]}<0,\qquad \frac{\partial \tau ^*(x^*_1)}{\partial \delta }=-\frac{2Ar(N-1)}{(N - 2)[\delta (N - 2) + r]^2}<0 \end{aligned}$$

We can compute:

$$\begin{aligned} \tau ^*({\hat{x}})=-\frac{2A[\delta (2N - 1) - Nr]}{(N + 1)(N-2)[\delta (N - 2) + r]}<0 \end{aligned}$$

implying ${\tilde{x}}<{\hat{x}}$, since $\tau$ is decreasing w.r.t. x.

(b) Comparative statics provides:

$$\begin{aligned} \frac{\partial \tau ^*(x)}{\partial N}=\frac{\partial \alpha ^*}{\partial N}x+\frac{\partial \beta ^*}{\partial N} \end{aligned}$$

where:

$$\begin{aligned} \frac{\partial \alpha ^*}{\partial N}= & {} \frac{2\delta - r}{(N - 2)^2}>0 \\ \frac{\partial \beta ^*}{\partial N}= & {} \frac{-A(\delta -r)[\delta ^2(N^2 - 8N + 12) + 2\delta r(N - 3) + r^2]}{\delta (N - 2)^2[\delta (N - 2) + r]^2}>0\iff N\le 5 \end{aligned}$$

We have:

$$\begin{aligned} \frac{\partial \tau ^*(x)}{\partial N}<0\iff x<\frac{A}{\delta }\frac{\delta ^3(N^2 - 8N + 12) - \delta ^2r(N^2 - 10N + 18) + \delta r^2(7 - 2N) - r^3}{(2\delta - r)[\delta (N - 2) + r]^2}<{\bar{x}}_1 \end{aligned}$$

The last inequality follows since the difference between this threshold and ${\bar{x}}_1$ is given by:

$$\begin{aligned} \frac{2A[2\delta ^2(N - 2) + \delta r(5 - 2N) - r^2]}{(r - 2\delta )[\delta (N - 2) + r]^2} \end{aligned}$$

this value being negative, since the denominator is negative and the numerator is positive (because it is increasing in N and positive for $N=3$).

With respect to $\delta$ it results:

$$\begin{aligned} \frac{\partial \tau ^*(x)}{\partial \delta }=\frac{\partial \alpha ^*}{\partial \delta }x+\frac{\partial \beta ^*}{\partial \delta } \end{aligned}$$

where:

$$\begin{aligned} \frac{\partial \alpha ^*}{\partial \delta }= & {} -2\frac{N-1}{N-2}<0 \\ \frac{\partial \beta ^*}{\partial \delta }= & {} \frac{Ar(N - 1)[\delta ^2(N^2 - 4N + 2) + 2\delta r(N - 2) + r^2]}{\delta ^2(N - 2)[\delta (N - 2) + r)^2]}>0\iff N\ge 4 \end{aligned}$$

We have:

$$\begin{aligned} \frac{\partial \tau ^*(x)}{\partial \delta }>0\iff x< \frac{Ar[\delta ^2 (N^2 - 4 N + 2) + 2 \delta r (N - 2) + r^2]}{2 \delta ^2 [\delta (N - 2) + r]^2}<{\bar{x}}_2 \end{aligned}$$

The last inequality follows since the difference between this threshold and ${\bar{x}}_2$ is given by:

$$\begin{aligned} \frac{A[2\delta ^3(N - 2)^2 - \delta ^2 r(N^2 - 8N + 10) + 2\delta r^2(3 - N) - r^3]}{-2\delta ^2[\delta (N - 2) + r]^2} \end{aligned}$$

this value being negative, since the denominator is negative and the numerator is positive (because it is increasing in N and positive for $N=3$).

The duopoly case

Corollary 5

For any value of $\alpha$ such that $(\alpha ,\beta )\in {\mathscr {P}}_1$ (resp., $(\alpha ,\beta )\in {\mathscr {P}}_2$) we have:

$$\begin{aligned} \frac{2}{3}K_{\alpha }|_{N=2}>(<)2\delta -r \end{aligned}$$

where the LHS is a decreasing (increasing) function of $\alpha$. Therefore:

$$\begin{aligned} 2\displaystyle \frac{\partial {\tilde{\phi }}_{\alpha ,\beta }(x)|_{N=2}}{\partial x}>(<)\frac{\partial Q^*(x)}{\partial x} \end{aligned}$$

Proof

In a duopoly it holds:

$$\begin{aligned} \frac{\partial K_{\alpha }|_{N=2}}{\partial \alpha }=\frac{3}{8}\displaystyle \left[ \frac{2\alpha +5(2\delta -r)}{\sqrt{4\alpha ^2 + 20\alpha (2\delta - r) + 9(2\delta - r)^2}} -1\right] \end{aligned}$$

For any $\alpha$ satisfying the first inequality which defines the set ${\mathscr {P}}_1$, this quantity is (well defined and) negative. It follows that the infimum over $\alpha \in (-\infty ,-\frac{9(2\delta -r)}{2})$ of the values that $K_{\alpha }|_{N=2}$ can take is given by:

$$\begin{aligned} \inf _{\alpha \in \left( -\infty ,-\frac{9(2\delta -r)}{2}\right) } K_{\alpha }|_{N=2}=\lim _{\alpha \rightarrow -\frac{9(2\delta -r)}{2}}K_{\alpha }|_{N=2}=\frac{9}{4}(2\delta -r) \end{aligned}$$

therefore, in this case: $\frac{2}{3}K_{\alpha }|_{N=2}>2\delta -r$.

Analogously, for every $\alpha$ value satisfying the first inequality which defines the set ${\mathscr {P}}_2$, it holds: $\frac{\partial K_\alpha |_{N=2}}{\partial \alpha }>0$, hence:

$$\begin{aligned} \sup _{\alpha \in \left( -\frac{2\delta -r}{2},\frac{2\delta ^2-7\delta r+5r^2}{r}\right) } K_{\alpha }|_{N=2}=\lim _{\alpha \rightarrow -\frac{2\delta -r}{2}}K_{\alpha }|_{N=2}=3(\delta -r) \end{aligned}$$

therefore, in this case: $\frac{2}{3}K_{\alpha }|_{N=2}<2\delta -r$. $\square$

In a duopoly, when the subsidy per unit of resource decreases (i.e. $\alpha <0$ increases, still satisfying the first inequality which defines the set ${\mathscr {P}}_1$), firms have a less aggressive behavior, in the sense that the slope of their non-degenerate linear feedback strategy decreases, nevertheless, in order to achieve the socially-optimal extraction plan, this slope should be lower. If we consider the range of $\alpha$ values such that $(\alpha ,\beta )\in {\mathscr {P}}_2$, an increase in $\alpha$ has exactly the opposite effect on the firms’ behavior, nevertheless, given that, as far as only the state-dependent component $\alpha$ of the tax policy is concerned, production is poorly subsidized or taxed, the marginal effect on the firms’ aggregate production of an increase in the resource stock availability is lower than it would be socially optimal.³⁴