Pareto-Improving Supply Subsidy in a Simple General Oligopoly Equilibrium Model with Pollution Permits

Abstract

We introduce a competitive pollution permit market in a two-sector oligopoly equilibrium model. In this model, one commodity is inelastically supplied by one competitive trader and another one is produced by a finite set of oligopolists, using the first commodity as an input. The production of the second commodity is a polluting activity. We study both the competitive and oligopoly equilibria. We provide some conditions under which a supply subsidy given to the oligopolists that is financed by a tax on the competitive agent is Pareto-improving.

Appendices

Appendix

Competitive Equilibrium

Taking the price vector $\mathbf {p}=\left( p,1\right)$ as given, the competitive agent chooses his demands for both goods 1 and 2 by solving the problem:

$$\begin{aligned} \underset{\left( x_{1},x_{2}\right) }{\max }\alpha \ln x_{1}+(1-\alpha )\ln x_{2}\text { s.t.\ }px_{1}+x_{2}\leqslant 1. \end{aligned}$$

(52)

It is straightforward to show that his demand functions are given by $\left( x_{1},x_{2}\right) =\left( \frac{\alpha }{p},1-\alpha \right)$.

The income of trader i is equal to his profit $\Pi ^{i}\left( s^{i},e^{i}\right)$, where

$$\begin{aligned} \Pi ^{i}\left( s^{i},e^{i}\right) =p\frac{\gamma }{\beta }e^{i}-\gamma e^{i}+q\left( \lambda ^{i}E-e^{i}\right) \text {, }i\in \mathbf {\mathcal {O}}\text {.} \end{aligned}$$

(53)

Trader i takes the price vector $\left( \mathbf {p},q\right)$ as given and determines his demands for both goods 1 and 2 by solving the problem:

$$\begin{aligned} \max _{(x_{1}^{i},\,x_{2}^{i})}\text { }\alpha \ln x_{1}^{i}+(1-\alpha )\ln x_{2}^{i}\text { s.t. }px_{1}^{i}+x_{2}^{i}\leqslant \Pi ^{i}\left( s^{i},e^{i}\right) \text {, with}\,x_{1}^{i}\geqslant 0\text {, }x_{2}^{i}\geqslant 0\text {.} \end{aligned}$$

(54)

Assuming $\Pi ^{i} > 0,$ the demand functions of agent $i\in \mathbf {\mathcal {O}}$ are then given by $\left( x_{1}^{i},x_{2}^{i}\right) =\left( \frac{\alpha \Pi ^{i}}{p},(1-\alpha )\Pi ^{i}\right)$. In addition, as a producer, agent i’s demand for the input satisfies $z^{i}=\gamma e^{i}$.

The determination of the competitive equilibrium, i.e., a price vector $(\mathbf {p}^{*},q^{*})$, and an allocation $\mathcal {A}^{*}=\{\left( x_{1}^{*},x_{2}^{*}\right) ;\left( (x_{1}^{i})^{*},(x_{2}^{i})^{*}\right) _{i\in \mathbf {\mathcal {O}}}\}$ uses the market-clearing conditions. These conditions may be written as

$$\begin{aligned} \sum _{i\in \mathbf {\mathcal {O}}}\frac{\alpha \Pi ^{i}}{p}+\frac{\alpha }{p}&=\sum _{i\in \mathbf {\mathcal {O}}}\frac{\gamma }{\beta }e^{i}, \end{aligned}$$

(55)

$$\begin{aligned} \sum _{i\in \mathbf {\mathcal {O}}}(1-\alpha )\Pi ^{i}+\sum _{i\in \mathbf {\mathcal {O}}}\gamma e^{i}+1-\alpha =1, \end{aligned}$$

(56)

$$\begin{aligned} \sum _{i\in \mathbf {\mathcal {O}}}e^{i} =E\text {.} \end{aligned}$$

(57)

But as agent i’s profit ($i\in \mathbf {\mathcal {O}}$) is linear in $e^{i}$. Thus, there exists an equilibrium only if the following condition holds

$$\begin{aligned} p=\beta \left( 1+\frac{q}{\gamma }\right) \text {.} \end{aligned}$$

(58)

With such a price, the demands of agent i may be written $\left( x_{1}^{i},x_{2}^{i}\right) =\left( \alpha \frac{q}{p}\lambda ^{i}E,(1-\alpha )q\lambda ^{i}E\right)$, $i\in \mathbf {\mathcal {O}}$. Using the equilibrium condition on the permit market, the market-clearing condition for good 2 may be obtained as

$$\begin{aligned} \ \sum _{i\in \mathbf {\mathcal {O}}}(1-\alpha )q\lambda ^{i}E+\sum _{i\in \mathbf {\mathcal {O}}}\gamma E+1-\alpha =1, \end{aligned}$$

(59)

or, since $\sum _{i\in \mathbf {\mathcal {O}}}\lambda ^{i}=1$, as

$$\begin{aligned} \ (1-\alpha )qE+\gamma E+1-\alpha =1. \end{aligned}$$

(60)

The term on the left-hand side of this equation represents the aggregate demand for commodity 2, where $(1-\alpha )qE$ and $\gamma E$ are, respectively, the demands for consumption and for the input by the n agents who belong to the set $\mathbf {\mathcal {O}}$, and $1-\alpha$ is the demand by the representative consumer. The term on the right-hand side is the total supply of commodity 2. By solving (60), we obtain (11), which by using $p=\beta (1+\frac{q}{\gamma })$ yields (12). Then, equations (13)-(16) follow.

Effects on an Increase in the Emission Ceiling E

The first part of the proposition stems immediately from the expressions of (23) and (24). In addition, from (22), one easily sees that $\frac{\partial \tilde{s}^{i}}{\partial E}>0$ since $\tilde{s}^{i}=\frac{\alpha }{n}\frac{1}{\tilde{p}}$. By using the expression of (25), one can write $\tilde{e}^{i}$ as

$$\begin{aligned} \tilde{e}^{i}&= f(E) \equiv \frac{(a - b E)E}{c -d E} \end{aligned}$$

(61)

where

$$\begin{aligned} a&= \left\{ (1-\alpha ) \dfrac{n-(1-\alpha )}{n^{2}}+\lambda ^{i} \left[ \alpha \dfrac{n-(1-\alpha )}{n}\right] \right\} \\ b&= \lambda ^{i} \gamma \\ c&= \dfrac{n-(1-\alpha )}{n} \\ d&= \gamma . \end{aligned}$$

We have

$$\begin{aligned} f^{\prime }(E)&= \frac{(c-dE) (a - 2b E)+d(a-bE)E}{(c-dE)^2} \end{aligned}$$

(62)

The sign of $f^{\prime }(E)$ is the same as that of $(c-dE) (a - 2b E)+d(a-bE)E$. The latter expression can be rewritten as

$$\begin{aligned} P(E)&= db E^2 - 2 bc E + ac. \end{aligned}$$

(63)

Clearly, this polynomial takes strictly negative values only if it has 2 positive real roots (because $P(0)=ac$, $P^{\prime }(E) = 2dbE - 2 bc$, and $P^{\prime }(0) = - 2 bc < 0$). This is the case whenever the discriminant $\triangle$ is strictly positive:

$$\begin{aligned} \triangle&= 4 (bc)^2 - 4 abcd = 4bc (bc - ad)> 0. \end{aligned}$$

(64)

Now, we have

$$\begin{aligned} bc&> ad \end{aligned}$$

(65)

that is if

$$\begin{aligned} \lambda _{i} \gamma \dfrac{n-(1-\alpha )}{n} >&\gamma \left\{ (1-\alpha ) \dfrac{n-(1-\alpha )}{n^{2}}+\lambda ^{i} \left[ \alpha \dfrac{n-(1-\alpha )}{n}\right] \right\} , \end{aligned}$$

or

$$\begin{aligned} \lambda _{i} >&\frac{1}{n}. \end{aligned}$$

(66)

This condition cannot be simultaneously satisfied for all the traders. Moreover, E must be located between the two roots, which is possible, but not always possible.

Inefficiency of the OE with Pollution Permits

Let $\theta ^{i}$, where $\theta ^{i}>0$, for $i=1,\dots ,n$, with $\sum _{i}\theta ^{i}=1$, be the weight of oligopolist i in total welfare. Observe that all solutions to the problem.

$$\begin{aligned} \underset{(x_{1}^{i},x_{2}^{i},x_{1},x_{2};z^{i})}{\max } \nonumber \\ \sum _{i=1}^{n}\theta ^{i}\left( \alpha \ln x_{1}^{i}+(1-\alpha )\ln x_{2}^{i}\right) +\alpha \ln x_{1}+(1-\alpha )\ln x_{2} \nonumber \\ \text {s.t.} \sum \nolimits _{i=1}^{n}x_{1}^{i}+x_{1}=Y \nonumber \\ \sum \nolimits _{i=1}^{n}x_{2}^{i}+x_{2}+\sum \nolimits _{i=1}^{n}z_{1}^{i}=1 \end{aligned}$$

(67)

with $\frac{\sum _{i=1}^{n}z^{i}}{\beta }=Y$ and $\frac{\sum _{i=1}^{n}z^{i}}{\gamma }=E$ are Pareto-optimal.

The first-order optimality conditions for this problem are

$$\begin{aligned} \frac{\theta ^{i}}{x_{1}^{i}}=\frac{1}{\frac{\gamma }{\beta }E-\sum _{j=1}^{n}x_{1}^{j}}, \end{aligned}$$

(68)

$$\begin{aligned} \frac{\theta ^{i}}{x_{2}^{i}}=\frac{1}{1-\gamma E-\sum _{j=1}^{n}x_{2}^{j}}. \end{aligned}$$

(69)

Summing the first condition across j yields after a little algebra

$$\begin{aligned} x_{1}^{i}=\frac{\theta _{i}\frac{\gamma E}{\beta }}{1+\sum _{j=1}^{n}\theta ^{j}}. \end{aligned}$$

(70)

By the same reasoning, we also obtain

$$\begin{aligned} x_{2}^{i}=\frac{\theta ^{i}(1-\gamma E)}{1+\sum _{j=1}^{n}\theta ^{j}}. \end{aligned}$$

(71)

Using the two equations above, one gets

$$\begin{aligned} \frac{x_{2}^{i}}{x_{1}^{i}}=\frac{1-\gamma E}{\frac{\gamma }{\beta }E}. \end{aligned}$$

(72)

Using (26) to compute the ratio of the consumptions of an oligopolist in an oligopoly equilibrium, we obtain after some algebra

$$\begin{aligned} \frac{\tilde{x}_{2}^{i}}{\tilde{x}_{1}^{i}}=\frac{\frac{n-(1-\alpha )}{n}-\gamma E}{\frac{\gamma }{\beta }E}. \end{aligned}$$

(73)

Comparing (C6) with (C7), we see that

$$\begin{aligned} \frac{\tilde{x}_{2}^{i}}{\tilde{x}_{1}^{i}}<\frac{x_{2}^{i}}{x_{1}^{i}}\text {.} \end{aligned}$$

(74)

Welfare Comparison

We now compare $\tilde{U}^{i}$ with $(U^{i})^{*}$. From (15) and (28), one has

$$\begin{aligned} \tilde{U}^{i}-(U^{i})^{*}= & {} \alpha \ln (1-\gamma E)+\ln \left( \frac{\alpha (1-\alpha )}{n^{2}}+\lambda ^{i}\left[ \alpha \frac{n-(1-\alpha )}{n}-\gamma E\right] \right) \nonumber \\&-\alpha \ln \left( \frac{n-(1-\alpha )}{n}-\gamma E\right) -\ln \lambda ^{i}(\alpha -\gamma E)\text {.} \end{aligned}$$

(75)

This leads to

$$\begin{aligned} \tilde{U}^{i}-(U^{i})^{*}=\alpha \ln \left( \frac{1-\gamma E}{\frac{n-(1-\alpha )}{n}-\gamma E}\right) +\ln \left( \frac{\frac{\alpha (1-\alpha )}{n^{2}}+\lambda ^{i}\left[ \alpha \frac{n-(1-\alpha )}{n}-\gamma E\right] }{\lambda ^{i}(\alpha -\gamma E)}\right) . \end{aligned}$$

(76)

The first term in brackets is positive. The second term is positive if $\lambda ^{i}\leqslant \frac{1}{n}$. Then, for all $i\in {\mathcal {O}}$ for which $\lambda ^{i}\leqslant \frac{1}{n}$, it is certainly true that $\tilde{U}^{i}-(U^{i})^{*}>0$. Finally, one can have $\tilde{U}^{i}<(U^{i})^{*}$ for the oligopolists for which $\lambda ^{i}>\frac{1}{n}$.

We now compare $\tilde{U}$ with $U^{*}$. From (16) and (), one gets

$$\begin{aligned} \tilde{U}-U^{*}=\alpha \ln \left( \left( \frac{n-1}{n}\right) \frac{1-\gamma E}{\frac{n-(1-\alpha )}{n}-\gamma E}\right) . \end{aligned}$$

(77)

We see that this expression is negative since all the terms in the logarithm are lower than 1.

OE with a Supply Subsidy

Using equation (39), we obtain after some tedious algebra a polynomial of degree two in the strategy s

$$\begin{aligned} P(s)=as^{2}+bs-c=0\text {,} \end{aligned}$$

(78)

where $a\equiv \tau (1-\alpha )(\alpha +n)$, $b\equiv \alpha (\frac{n-(1-\alpha )}{n}-\gamma E)-\tau \frac{\gamma }{\beta }\frac{E}{n}(1-\alpha )(\left( 1-\alpha \right) n+\alpha )$, and $c\equiv \alpha (1-\alpha )\frac{\gamma }{\beta }\frac{E}{n}(1-\frac{1}{n})$.

To prove the existence of an OE with a subsidy, we must show that there is a positive solution to the above equation which satisfies: $s<\min \{\frac{\gamma }{\beta n}E,\frac{1}{\tau n}\}$. This last condition ensures that the supply of the oligopolists is feasible and that the after-tax income of the competitive agent is positive.

We first prove that there is a positive solution lower than $\gamma E/\beta n$. After some computations, we find that

$$\begin{aligned} P(0)=-\alpha (1-\alpha )\frac{\gamma }{\beta }\frac{E}{n}(1-\frac{1}{n})<0, \end{aligned}$$

(79)

$$\begin{aligned} P\left( \frac{\gamma E}{\beta n}\right) =\alpha \frac{\gamma E}{\beta n}(\alpha -\gamma E)>0. \end{aligned}$$

(80)

It follows that there is a unique positive root of the polynomial function located in $]0,\frac{\gamma E}{\beta n}[$. The value of this root is

$$\begin{aligned} s(\tau )=\frac{-\alpha (\frac{n-(1-\alpha )}{n}-\gamma E)+\tau \frac{\gamma E}{\beta n}(1-\alpha )(\left( 1-\alpha \right) n+\alpha )+\sqrt{\Omega (\tau )}}{2\tau (1-\alpha )(\alpha +n)}\text {,} \end{aligned}$$

(81)

where

$$\begin{aligned} \Omega (\tau )= & {} \left( \alpha (\frac{n-(1-\alpha )}{n}-\gamma E)-\tau \frac{\gamma }{\beta }\frac{E}{n}(1-\alpha )(\left( 1-\alpha \right) +\alpha )\right) ^{2} \nonumber \\&+4\tau \alpha (1-\alpha )^{2}(\alpha +n)\frac{\gamma E}{\beta n}(1-\frac{1}{n}). \end{aligned}$$

(82)

It remains to check that the positive root of the polynomial is lower than $1/\tau n$. We have

$$\begin{aligned} P(\frac{1}{\tau n})=\frac{1}{\tau n}\left( \frac{(1-\alpha )(1+n)}{n}+\frac{1}{\tau }(\alpha \frac{n-(1-\alpha )}{n}-\gamma E)\right) -(1-\alpha )\frac{\gamma E}{\beta n}. \end{aligned}$$

(83)

Since we will only study the case where $\tau$ is close to 0, it follows that $P(\frac{1}{\tau n})$ will always be positive.

Supply Subsidy and Supply of Good 1

Applying the implicit function theorem to equation (39), we obtain after some tedious algebra

$$\begin{aligned} \frac{\partial s}{\partial \tau }= \frac{ \frac{\gamma E}{\beta n}(\alpha +n(1-\alpha )) - s (\alpha +n)}{\tau (\alpha +n) + \frac{\gamma E}{\beta n s^2}(n-1)} \text {.} \end{aligned}$$

(84)

Since $s(\tau )$ is continuous at 0, we may define $\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}}=\lim _{\tau \rightarrow 0+}\frac{\partial s}{\partial \tau }$, where

$$\begin{aligned} \lim _{\tau \rightarrow 0+}\frac{\partial s}{\partial \tau }=\frac{\frac{\gamma E}{\beta n}(\alpha +n(1-\alpha ))-s(0)(\alpha +n)}{\tau (\alpha +n)+\frac{\gamma E}{\beta ns^{2}(0)}(n-1)}\text {.} \end{aligned}$$

(85)

Using the expression of s(0) in the above definition, we get the result.

Notice that

$$\begin{aligned} \frac{1}{s^{2}(0)}\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}}=\frac{1}{n-1}\left( n(1-\alpha )+\alpha -\frac{(\alpha +n)(1-\alpha )(1-\frac{1}{n})}{\frac{n-(1-\alpha )}{n}-\gamma E}\right) . \end{aligned}$$

(86)

Proposition 6 follows.

Effect of a Supply Subsidy on Oligopolist i’s Welfare

Differentiating $V^{i}(\tau )$ with respect to $\tau$ yields

$$\begin{aligned} \frac{\partial V^{i}(\tau )}{\partial \tau }=\frac{-\alpha }{\frac{\gamma }{\beta }\frac{E}{n}-s}\frac{\partial s}{\partial \tau }+(1-\alpha )^{2}\frac{s+\tau \frac{\partial s}{\partial \tau }}{\frac{\alpha -\gamma E}{n}+\tau s(1-\alpha )}\text {;} \end{aligned}$$

(87)

and

$$\begin{aligned} \lim _{\tau \rightarrow 0+}\frac{\partial V^{i}(\tau )}{\partial \tau }=\frac{-\alpha }{\frac{\gamma }{\beta }\frac{E}{n}-s}\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}}+(1-\alpha )^{2}\frac{s}{\frac{\alpha -\gamma E}{n}}. \end{aligned}$$

(88)

From the first-order conditions for oligopolists i’s problem, we know that

$$\begin{aligned} \frac{\alpha }{\frac{\gamma }{\beta }e^{i}-s^{i}}=\frac{(1-\alpha )\left( p+\tau -s^{i}\frac{\alpha (1-T)}{\left( \sum _{k=1}^{n}s^{k}\right) ^{2}} \right) }{(p+\tau )s^{i}-(\gamma +q)e^{i}+q\frac{E}{n}}. \end{aligned}$$

(89)

In equilibrium, we get

$$\begin{aligned} \frac{\alpha }{\frac{\gamma }{\beta }\frac{E}{n}-s}=\frac{(1-\alpha )\frac{\alpha }{ns}(1-\frac{1}{n})}{\frac{\alpha -\gamma E}{n}}. \end{aligned}$$

(90)

Using this last equality, and (G2) when $\tau$ goes to zero, we obtain

$$\begin{aligned} \lim _{\tau \rightarrow 0+}\frac{\partial V^{i}(\tau )}{\partial \tau }=-\frac{(1-\alpha )\frac{\alpha }{ns}(1-\frac{1}{n})}{\frac{\alpha -\gamma E}{n}}\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}}+(1-\alpha )^{2}\frac{s}{\frac{\alpha -\gamma E}{n}}. \end{aligned}$$

(91)

Therefore, the sign of $\lim _{\tau \rightarrow 0+}\frac{\partial V^{i}(\tau )}{\partial \tau }$ is also that of

$$\begin{aligned} \Sigma =-\frac{\alpha }{ns(0)}(1-\frac{1}{n})\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}}+(1-\alpha )s(0). \end{aligned}$$

(92)

Pareto-improving supply subsidy

Observe that the bounds $\frac{1-\alpha }{\alpha }n$ and $\frac{1-\alpha }{\alpha }\frac{n^{2}}{n-1}$ in condition (51) are both decreasing in $\alpha$, and go to $+\infty$ when $\alpha$ goes to 0, and both of them vanish when $\alpha = 1$. Now, the expression $\frac{1}{s^{2}(0)}\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}}$ is an increasing function of $\alpha$ which always takes finite values. Indeed, from equation (82) we have

$$\begin{aligned} \frac{\partial }{\partial \alpha }\left( \frac{1}{s^{2}(0)}\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}}\right)&= -1 + \frac{2 \alpha + n-1}{\frac{n+\alpha -1}{n}-\gamma E} + \frac{(\alpha +n)(1-\alpha )}{n\left( \frac{n+\alpha -1}{n}-\gamma E \right) ^2}>0. \end{aligned}$$

(93)

Moreover, we have

$$\begin{aligned} \lim _{\alpha \rightarrow 0} \frac{1}{s^{2}(0)}\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}} = \frac{n}{n-1} \frac{- \gamma E}{\frac{n-1}{n}-\gamma E}<0, \end{aligned}$$

(94)

and

$$\begin{aligned} \lim _{\alpha \rightarrow 1} \frac{1}{s^{2}(0)}\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}} = 1/(n-1). \end{aligned}$$

(95)

Therefore, the graph of this function crosses that of the two bounds (viewed as functions of $\alpha$). The value of $\underline{\alpha }$ is given by the value of $\alpha$ at which the graph of $\frac{1}{s^{2}(0)}\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}}$ crosses the lower bound $\frac{1-\alpha }{\alpha }n$. Similarly, the value of $\overline{\alpha }$ is the value of $\alpha$ at which the graph $\frac{1}{s^{2}(0)}\frac{\partial s}{\partial \tau }_{\mid _{\tau =0}}$ crosses that of the upper bound $\frac{1-\alpha }{\alpha }\frac{n^{2}}{n-1}$.

We still have to check that the values of $\alpha$ in $[\underline{\alpha }, \overline{\alpha }]$ are consistent with those ensuring the existence of an oligopoly equilibrium. First, notice that by construction of the set $[\underline{\alpha }, \overline{\alpha }]$, we have $\frac{\partial s}{\partial \tau }>0$. But Proposition 6 implies that we must then have $\alpha >\frac{\gamma nE}{1+(n-1)\gamma E}$. To ensure the existence of the oligopoly equilibrium, it is sufficient to check that $\alpha \frac{n-(1-\alpha )}{n}>\gamma E$. In turn, this last condition holds whenever

$$\begin{aligned} \frac{\alpha }{n-\alpha (n-1)}&< \alpha \frac{n-(1-\alpha )}{n}. \end{aligned}$$

(96)

But the condition above is equivalent to

$$\begin{aligned} 0&< (1-\alpha ) \left( n^2 - n (2-\alpha ) - \alpha \right) . \end{aligned}$$

(97)

This condition is always satisfied for $n\ge 2$ and $\alpha >0$.