Taxes Versus Tradable Permits Considering Public Environmental Awareness

Abstract

This paper examines the relative performance of taxes and tradable permits when public environmental awareness is taken into account in policy-making. Two sovereign regions linked by a transboundary pollutant are considered. We show when public environmental awareness in one region increases, domestic government tightens its policy setting. While for foreign government, its response is different in these two polices. Under taxes, foreign government relaxes tax rate to get a free ride; under tradable permits, it may also tighten permit supply to benefit more from the international tradable permit market. But anyway, total pollution emissions are reduced. Moreover, when public environmental awareness in one region is sufficiently large, tradable permits welfare dominates taxes. However, public environmental awareness is bounded in reality. So, we further narrow its range to match reality. It is shown for the case of global externalities, tradable permits policy is superior. While for the case of reciprocal externalities, taxes policy is superior when pollution spillover is relatively low. And with the rise of public environmental awareness, the advantage of taxes is further strengthened.

Appendix A

This appendix contains the proofs of all propositions.

Proof of Proposition 1

Rewrite the first-order conditions (2) and (3) as:

$$\frac{\partial {C}_{i}({e}_{i})}{\partial {e}_{i}}+\alpha \frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}+(1-\alpha )\frac{\partial {D}_{j}\left({E}_{j},{\theta }_{j}\right)}{\partial {E}_{j}}=0,$$

(23)

$$\frac{\partial {C}_{j}({e}_{j})}{\partial {e}_{j}}+(1-\beta )\frac{\partial {D}_{i}\left({E}_{i},{\theta }_{i}\right)}{\partial {E}_{i}}+\beta \frac{\partial {D}_{j}\left({E}_{j},{\theta }_{j}\right)}{\partial {E}_{j}}=0.$$

(24)

For ease of exposition, the left sides of Eqs. (23) and (24) are denoted by ${\Phi }_{1}(\cdot )$ and ${\Phi }_{2}(\cdot )$, respectively. Differentiating Eqs. (23) and (24) with respect to ${\theta }_{i}$, and using the implicit function theorem, we obtain:

$$\frac{\partial {e}_{i}}{\partial {\theta }_{i}}=-\frac{\left|{\Lambda }_{1,{\theta }_{i}}\right|}{\left|H\right|},\frac{\partial {e}_{j}}{\partial {\theta }_{i}}=-\frac{\left|{\Lambda }_{2,{\theta }_{i}}\right|}{\left|H\right|},$$

(25)

where $H$ is the Hessian Matrix of the system, and ${\Lambda }_{n,{\theta }_{i}}$ is the matrix which is acquired through replacing the ${n}_{th}$ column of the Hessian by the first derivatives with respect to ${\theta }_{i}$.

$$\begin{array}{c}\left|H\right|=\begin{vmatrix}\frac{\partial\Phi_1}{\partial e_i}\frac{\partial\Phi_1}{\partial e_j}\\\frac{\partial\Phi_2}{\partial e_i}\frac{\partial\Phi_2}{\partial e_j}\end{vmatrix}=\frac{\partial^2C_i}{\partial e_i^2}(\frac{\partial^2C_j}{\partial e_j^2}+\left(1-\beta\right)^2\frac{\partial^2D_i}{\partial E_i^2}+\beta^2\frac{\partial^2D_j}{\partial E_j^2})+\frac{\partial^2C_j}{\partial e_j^2}(\alpha^2\frac{\partial^2D_i}{\partial E_i^2}+\left(1-\alpha\right)^2\frac{\partial^2D_j}{\partial E_j^2})\\+\left(\alpha+\beta-1\right)^2\frac{\partial^2D_i}{\partial E_i^2}\frac{\partial^2D_j}{\partial E_j^2}>0,\end{array}$$

(26)

$$\left|{\Lambda }_{1,{\theta }_{i}}\right|=\left|\begin{array}{c}\frac{\partial {\Phi }_{1}}{\partial {\theta }_{i}}\frac{\partial {\Phi }_{1}}{\partial {e}_{j}}\\ \frac{\partial {\Phi }_{2}}{\partial {\theta }_{i}}\frac{\partial {\Phi }_{2}}{\partial {e}_{j}}\end{array}\right|=\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}(\alpha \frac{{\partial }^{2}{C}_{j}}{\partial {e}_{j}^{2}}+\beta (\alpha +\beta -1)\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}})>0,$$

(27)

$$\left|{\Lambda }_{2,{\theta }_{i}}\right|=\left|\begin{array}{c}\frac{\partial {\Phi }_{1}}{\partial {e}_{i}}\frac{\partial {\Phi }_{1}}{\partial {\theta }_{i}}\\ \frac{\partial {\Phi }_{2}}{\partial {e}_{i}}\frac{\partial {\Phi }_{2}}{\partial {\theta }_{i}}\end{array}\right|=\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}\left(\left(1-\beta \right)\frac{{\partial }^{2}{C}_{i}}{\partial {e}_{i}^{2}}+\left(1-\alpha \right)\left(1-\alpha -\beta \right)\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}\right).$$

(28)

Since $1/2\le \alpha ,\beta <1$, it is obvious that $\left|H\right|>0$ and $\left|{\Lambda }_{1,{\theta }_{i}}\right|>0$, which implies $\frac{\partial {e}_{i}}{\partial {\theta }_{i}}<0$. And if $\alpha =\beta =1/2$, then $\left|{\Lambda }_{2,{\theta }_{i}}\right|>0$ implying $\frac{\partial {e}_{j}}{\partial {\theta }_{i}}<0$; if $1/2<\alpha ,\beta <1$, $\left|{\Lambda }_{2,{\theta }_{i}}\right|>0$ is established when and only when $\beta <\frac{\frac{{\partial }^{2}{C}_{i}}{\partial {e}_{i}^{2}}+{(1-\alpha )}^{2}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}}{\frac{{\partial }^{2}{C}_{i}}{\partial {e}_{i}^{2}}+(1-\alpha )\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}}$.

Proof of Proposition 2

According to Eqs. (25)-(28), we can obtain:

$$\frac{\partial E}{\partial {\theta }_{i}}=\frac{\partial {e}_{i}}{\partial {\theta }_{i}}+\frac{\partial {e}_{j}}{\partial {\theta }_{i}}=-\frac{\left(\left(1-\beta \right)\frac{{\partial }^{2}{C}_{i}}{\partial {e}_{i}^{2}}+\alpha \frac{{\partial }^{2}{C}_{j}}{\partial {e}_{j}^{2}}+{\left(\alpha +\beta -1\right)}^{2}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}\right)\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}}{\left|H\right|}<0.$$

$$\frac{\partial {E}_{i}}{\partial {\theta }_{i}}=\alpha \frac{\partial {e}_{i}}{\partial {\theta }_{i}}+(1-\beta )\frac{\partial {e}_{j}}{\partial {\theta }_{i}}=-\frac{\left({\left(1-\beta \right)}^{2}\frac{{\partial }^{2}{C}_{i}}{\partial {e}_{i}^{2}}+{\alpha }^{2}\frac{{\partial }^{2}{C}_{j}}{\partial {e}_{j}^{2}}+{\left(\alpha +\beta -1\right)}^{2}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}\right)\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}}{\left|H\right|}<0.$$

$$\frac{\partial {E}_{j}}{\partial {\theta }_{i}}=(1-\alpha )\frac{\partial {e}_{i}}{\partial {\theta }_{i}}+\beta \frac{\partial {e}_{j}}{\partial {\theta }_{i}}=-\frac{(\beta (1-\beta )\frac{{\partial }^{2}{C}_{i}}{\partial {e}_{i}^{2}}+\alpha (1-\alpha )\frac{{\partial }^{2}{C}_{j}}{\partial {e}_{j}^{2}})\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}}{\left|H\right|}<0.$$

Proof of Proposition 3

Rewrite Eqs. (12) and (13) as:

$${t}_{i}-\alpha \frac{\partial {D}_{i}\left({E}_{i}\left({t}_{i},{t}_{j}\right),{\theta }_{i}\right)}{\partial {E}_{i}}=0,$$

(29)

$${t}_{j}-\beta \frac{\partial {D}_{j}\left({E}_{j}\left({t}_{i},{t}_{j}\right),{\theta }_{j}\right)}{\partial {E}_{j}}=0.$$

(30)

Similar to the proving process of Proposition 1, differentiate Eqs. (29) and (30) with respect to $\theta_{i}$, and use the implicit function theorem:

$$\frac{\partial {t}_{i}}{\partial {\theta }_{i}}=\frac{\alpha \frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}(1+\frac{{\beta }^{2}}{{\partial }^{2}{C}_{j}/\partial {e}_{j}^{2}}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}})}{\left|{H}_{t}\right|},$$

(31)

$$\frac{\partial {t}_{j}}{\partial {\theta }_{i}}=-\frac{\frac{\alpha \beta \left(1-\alpha \right)}{{\partial }^{2}{C}_{i}/\partial {e}_{i}^{2}}\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}}{\left|{H}_{t}\right|},$$

(32)

$$\left|{H}_{t}\right|=1+\frac{{\alpha }^{2}}{{\partial }^{2}{C}_{i}/\partial {e}_{i}^{2}}\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}^{2}}+\frac{{\beta }^{2}}{{\partial }^{2}{C}_{j}/\partial {e}_{j}^{2}}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}+\frac{\alpha \beta }{{\partial }^{2}{C}_{i}/\partial {e}_{i}^{2}}\frac{\alpha +\beta -1}{{\partial }^{2}{C}_{j}/\partial {e}_{j}^{2}}\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}^{2}}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}>0.$$

Thus, we can obtain: $\frac{\partial {t}_{i}}{\partial {\theta }_{i}}>0$ and $\frac{\partial {t}_{j}}{\partial {\theta }_{i}}<0$ for all $\frac{1}{2}\le \alpha ,\beta <1$.

Proof of Proposition 4

Based on Eq. (6) and the results of Proposition 3, we have:

$$\frac{\partial {e}_{i}}{\partial {\theta }_{i}}=\frac{\partial {e}_{i}}{\partial {t}_{i}}\frac{\partial {t}_{i}}{\partial {\theta }_{i}}=-\frac{\frac{\alpha }{{\partial }^{2}{C}_{i}/\partial {e}_{i}^{2}}\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}\left(1+\frac{{\beta }^{2}}{{\partial }^{2}{C}_{j}/\partial {e}_{j}^{2}}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}\right)}{\left|{H}_{t}\right|}<0,$$

$$\frac{\partial {e}_{j}}{\partial {\theta }_{i}}=\frac{\partial {e}_{j}}{\partial {t}_{j}}\frac{\partial {t}_{j}}{\partial {\theta }_{i}}=\frac{\frac{\alpha \beta }{{\partial }^{2}{C}_{i}/\partial {e}_{i}^{2}}\frac{1-\alpha }{{\partial }^{2}{C}_{j}/\partial {e}_{j}^{2}}\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}}{\left|{H}_{t}\right|}>0,$$

$$\frac{\partial E}{\partial {\theta }_{i}}=\frac{\partial {e}_{i}}{\partial {\theta }_{i}}+\frac{\partial {e}_{j}}{\partial {\theta }_{i}}=-\frac{\frac{\alpha }{{\partial }^{2}{C}_{i}/\partial {e}_{i}^{2}}\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}\left(1+\frac{\beta \left(\alpha +\beta -1\right)}{{\partial }^{2}{C}_{j}/\partial {e}_{j}^{2}}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}\right)}{\left|{H}_{t}\right|}<0,$$

$$\frac{\partial {E}_{i}}{\partial {\theta }_{i}}=\alpha \frac{\partial {e}_{i}}{\partial {\theta }_{i}}+(1-\beta )\frac{\partial {e}_{j}}{\partial {\theta }_{i}}=-\frac{\frac{\alpha }{{\partial }^{2}{C}_{i}/\partial {e}_{i}^{2}}\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}\left(\alpha +\frac{\beta \left(\alpha +\beta -1\right)}{{\partial }^{2}{C}_{j}/\partial {e}_{j}^{2}}\frac{{\partial }^{2}{D}_{j}}{\partial {E}_{j}^{2}}\right)}{\left|{H}_{t}\right|}<0,$$

$$\frac{\partial {E}_{j}}{\partial {\theta }_{i}}=(1-\alpha )\frac{\partial {e}_{i}}{\partial {\theta }_{i}}+\beta \frac{\partial {e}_{j}}{\partial {\theta }_{i}}=-\frac{\frac{\alpha (1-\alpha )}{{\partial }^{2}{C}_{i}/\partial {e}_{i}^{2}}\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}}{\left|{H}_{t}\right|}<0.$$

Proof of Proposition 5

Similar to the proving process of Proposition 1, differentiate Eqs. (20) and (21) with respect to $\theta_{i}$, and use the implicit function theorem:

$$\frac{\partial {a}_{i}}{\partial {\theta }_{i}}=-\frac{\frac{{\partial }^{2}S{C}_{i}}{\partial {a}_{i}\partial {\theta }_{i}}\frac{{\partial }^{2}S{C}_{j}}{\partial {a}_{j}^{2}}}{\left|{H}_{a}\right|},$$

(33)

$$\frac{\partial {a}_{j}}{\partial {\theta }_{i}}=\frac{\frac{{\partial }^{2}S{C}_{i}}{\partial {a}_{i}\partial {\theta }_{i}}\frac{{\partial }^{2}S{C}_{j}}{\partial {a}_{j}\partial {a}_{i}}}{\left|{H}_{a}\right|},$$

(34)

$$\frac{\partial S{C}_{i}}{\partial {a}_{i}\partial {\theta }_{i}}=\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}(\alpha \frac{\partial {e}_{i}}{\partial p}+(1-\beta )\frac{\partial {e}_{j}}{\partial p})\frac{\partial p}{\partial a}>0,$$

$$\begin{array}{c}\frac{\partial^2SC_j}{\partial a_j^2}=\frac{\partial^2p}{\partial a^2}(e_j-a_j)+\frac{\partial e_j}{\partial p}{(\frac{\partial p}{\partial a})}^2-2\frac{\partial p}{\partial a}+\frac{\partial^2D_j}{\partial E_j^2}{{((1-\alpha)}\frac{\partial e_i}{\partial p}+\beta\frac{\partial e_j}{\partial p})}^2{(\frac{\partial p}{\partial a})}^2\\+\frac{\partial D_j}{\partial E_j}((1-\alpha)\frac{\partial^2e_i}{\partial p^2}+\beta\frac{\partial^2e_j}{\partial p^2}){(\frac{\partial p}{\partial a})}^2+\frac{\partial D_j}{\partial E_j}((1-\alpha)\frac{\partial e_i}{\partial p}+\beta\frac{\partial e_j}{\partial p})\frac{\partial^2p}{\partial a^2}>0\end{array},$$

$$\frac{{\partial }^{2}S{C}_{j}}{\partial {a}_{j}\partial {a}_{i}}=\frac{{\partial }^{2}S{C}_{j}}{\partial {a}_{j}^{2}}+\frac{\partial p}{\partial a}=\frac{{\partial }^{2}S{C}_{j}}{\partial {a}_{j}^{2}}-\frac{\frac{{\partial }^{2}{C}_{i}}{\partial {e}_{i}^{2}}\frac{{\partial }^{2}{C}_{j}}{\partial {e}_{j}^{2}}}{\frac{{\partial }^{2}{C}_{i}}{\partial {e}_{i}^{2}}+\frac{{\partial }^{2}{C}_{j}}{\partial {e}_{j}^{2}}},$$

$$\left|{H}_{a}\right|=\frac{{\partial }^{2}S{C}_{i}}{\partial {a}_{i}^{2}}\frac{{\partial }^{2}S{C}_{j}}{\partial {a}_{j}^{2}}-\frac{{\partial }^{2}S{C}_{i}}{\partial {a}_{i}\partial {a}_{j}}\frac{{\partial }^{2}S{C}_{j}}{\partial {a}_{j}\partial {a}_{i}}>0.$$

Thus for all $\frac{1}{2}\le \alpha ,\beta <1$, we can obtain $\frac{\partial {a}_{i}}{\partial {\theta }_{i}}<0$, and $\frac{\partial {a}_{j}}{\partial {\theta }_{i}}<0$ when and only when $\frac{{\partial }^{2}S{C}_{j}}{\partial {a}_{j}^{2}}-\frac{\frac{{\partial }^{2}{C}_{i}}{\partial {e}_{i}^{2}}\frac{{\partial }^{2}{C}_{j}}{\partial {e}_{j}^{2}}}{\frac{{\partial }^{2}{C}_{i}}{\partial {e}_{i}^{2}}+\frac{{\partial }^{2}{C}_{j}}{\partial {e}_{j}^{2}}}<0$.

Proof of Proposition 6

Based on the results of Proposition 5, we have:

$$\frac{\partial a}{\partial {\theta }_{i}}=\frac{\partial {a}_{i}}{\partial {\theta }_{i}}+\frac{\partial {a}_{j}}{\partial {\theta }_{i}}=\frac{\frac{{\partial }^{2}{D}_{i}}{\partial {E}_{i}\partial {\theta }_{i}}{(\frac{\partial p}{\partial a})}^{2}(\alpha \frac{\partial {e}_{i}}{\partial p}+(1-\beta )\frac{\partial {e}_{j}}{\partial p})}{\left|{H}_{a}\right|}<0,$$

$$\frac{\partial {e}_{i}}{\partial {\theta }_{i}}=\frac{\partial {e}_{i}}{\partial p}\frac{\partial p}{\partial a}\frac{\partial a}{\partial {\theta }_{i}}<0,$$

$$\frac{\partial {e}_{j}}{\partial {\theta }_{i}}=\frac{\partial {e}_{j}}{\partial p}\frac{\partial p}{\partial a}\frac{\partial a}{\partial {\theta }_{i}}<0,$$

$$\frac{\partial E}{\partial {\theta }_{i}}=\frac{\partial a}{\partial {\theta }_{i}}<0,$$

$$\frac{\partial {E}_{i}}{\partial {\theta }_{i}}=\alpha \frac{\partial {e}_{i}}{\partial {\theta }_{i}}+(1-\beta )\frac{\partial {e}_{j}}{\partial {\theta }_{i}}<0,$$

$$\frac{\partial {E}_{j}}{\partial {\theta }_{i}}=(1-\alpha )\frac{\partial {e}_{i}}{\partial {\theta }_{i}}+\beta \frac{\partial {e}_{j}}{\partial {\theta }_{i}}<0.$$

Proof of Proposition 7

Under the optimal tax regime, it is satisfied that:

$$TSC\left({\theta }_{i},{\theta }_{j}\right)={C}_{i}\left({e}_{i}\right)+{D}_{i}\left({E}_{i},{\theta }_{i}\right)+{C}_{j}\left({e}_{j}\right)+{D}_{j}\left({E}_{j},{\theta }_{j}\right),$$

$$s.t.-\frac{\partial {C}_{i}\left({e}_{i}\right)}{\partial {e}_{i}}={t}_{i}=\alpha \frac{\partial {D}_{i}\left({E}_{i},{\theta }_{i}\right)}{\partial {E}_{i}},$$

$$-\frac{\partial {C}_{j}\left({e}_{j}\right)}{\partial {e}_{j}}={t}_{j}=\beta \frac{\partial {D}_{j}\left({E}_{j},{\theta }_{j}\right)}{\partial {E}_{j}}.$$

So it is obtained that:

$$TSC(\mathrm{0,0})=0,$$

(35)

where ${e}_{i}={e}_{i}^{\mathrm{max}}$ and ${e}_{j}={e}_{j}^{\mathrm{max}}$, since $\frac{\partial {D}_{i}({E}_{i},0)}{\partial {E}_{i}}=0=\frac{\partial {D}_{j}({E}_{j},0)}{\partial {E}_{j}}$. That is, both firms emit their unregulated pollution levels when the public in the two regions has no awareness of environmental protection.

$$\underset{{\theta }_{i}\to \infty }{\mathrm{lim}}TSC\left({\theta }_{i},{\theta }_{j}\right)=\underset{{\theta }_{i}\to \infty }{\mathrm{lim}}\left[{C}_{i}\left(0\right)+{D}_{i}\left(\left(1-\beta \right){e}_{j},{\theta }_{i}\right)+{C}_{j}\left({e}_{j}\right)+{D}_{j}\left(\beta {e}_{j},{\theta }_{j}\right)\right]=\infty ,$$

(36)

where ${e}_{i}=0$, since $\underset{{\theta }_{i}\to \infty }{\mathrm{lim}}\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}=\infty$ for ${E}_{i}>0$. That is, firm $i$ wouldn’t emit any pollution when public environmental awareness in the region is extremely high.

$$\begin{array}{lcl}\frac{\partial TSC({\theta }_{i},{\theta }_{j})}{\partial {\theta }_{i}}&=&\frac{\partial {C}_{i}}{\partial {e}_{i}}\frac{\partial {e}_{i}}{\partial {\theta }_{i}}+\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}\frac{\partial {E}_{i}}{\partial {\theta }_{i}}+\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {\theta }_{i}}+\frac{\partial {C}_{j}}{\partial {e}_{j}}\frac{\partial {e}_{j}}{\partial {\theta }_{i}}+\frac{\partial {D}_{j}({E}_{j},{\theta }_{j})}{\partial {E}_{j}}\frac{\partial {E}_{j}}{\partial {\theta }_{i}}\\ &=&\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}(\frac{\partial {E}_{i}}{\partial {\theta }_{i}}-\alpha \frac{\partial {e}_{i}}{\partial {\theta }_{i}})+\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {\theta }_{i}}+\frac{\partial {D}_{j}({E}_{j},{\theta }_{j})}{\partial {E}_{j}}(\frac{\partial {E}_{j}}{\partial {\theta }_{i}}-\beta \frac{\partial {e}_{j}}{\partial {\theta }_{i}})\\ &=&\underset{>0}{\underbrace{\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}(1-\alpha )\frac{\partial {e}_{j}}{\partial {\theta }_{i}}}}+\underset{>0}{\underbrace{\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {\theta }_{i}}}}+\underset{<0}{\underbrace{\frac{\partial {D}_{j}({E}_{j},{\theta }_{j})}{\partial {E}_{j}}(1-\beta )\frac{\partial {e}_{i}}{\partial {\theta }_{i}}}}\end{array},$$

(37)

In particular, $\frac{\partial TSC({\theta }_{i},0)}{\partial {\theta }_{i}}=\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {\theta }_{i}}>0$.

Under the optimal tradable permit regime, it is satisfied that:

$$\begin{aligned}TSC({\theta }_{i},{\theta }_{j})&={C}_{i}({e}_{i})+p({e}_{i}-{a}_{i})+{D}_{i}({E}_{i},{\theta }_{i})+{C}_{j}({e}_{j})+p({e}_{j}-{a}_{j})+{D}_{j}({E}_{j},{\theta }_{j})\\& ={C}_{i}({e}_{i})+{D}_{i}({E}_{i},{\theta }_{i})+{C}_{j}({e}_{j})+{D}_{j}({E}_{j},{\theta }_{j})\end{aligned},$$

$$\begin{array}{c}s.t.-\frac{\partial {C}_{i}({e}_{i})}{\partial {e}_{i}}=-\frac{\partial {C}_{j}({e}_{j})}{\partial {e}_{j}}=p=\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}\frac{(1-\beta )\frac{{\partial }^{2}{C}_{i}({e}_{i})}{\partial {e}_{i}^{2}}+\alpha \frac{{\partial }^{2}{C}_{j}({e}_{j})}{\partial {e}_{j}^{2}}}{2[\frac{{\partial }^{2}{C}_{i}({e}_{i})}{\partial {e}_{i}^{2}}+\frac{{\partial }^{2}{C}_{j}({e}_{j})}{\partial {e}_{j}^{2}}]}\\ +\frac{\partial {D}_{j}({E}_{j},{\theta }_{j})}{\partial {E}_{j}}\frac{\beta \frac{{\partial }^{2}{C}_{i}({e}_{i})}{\partial {e}_{i}^{2}}+(1-\alpha )\frac{{\partial }^{2}{C}_{j}({e}_{j})}{\partial {e}_{j}^{2}}}{2[\frac{{\partial }^{2}{C}_{i}({e}_{i})}{\partial {e}_{i}^{2}}+\frac{{\partial }^{2}{C}_{j}({e}_{j})}{\partial {e}_{j}^{2}}]}\end{array}.$$

So it is obtained that:

$$TSC(\mathrm{0,0})=0,$$

(38)

where ${e}_{i}={e}_{i}^{\mathrm{max}}$ and ${e}_{j}={e}_{j}^{\mathrm{max}}$, since $\frac{\partial {D}_{i}({E}_{i},0)}{\partial {E}_{i}}=0=\frac{\partial {D}_{j}({E}_{j},0)}{\partial {E}_{j}}$.

$$\underset{{\theta }_{i}\to \infty }{\mathrm{lim}}TSC\left({\theta }_{i},{\theta }_{j}\right)={C}_{i}\left(0\right)+{C}_{j}\left(0\right)<\infty ,$$

(39)

where ${e}_{i}={e}_{j}=0$. A firm’s abatement cost couldn’t be infinitely large, or the firm would choose to stop any production activities.

$$\begin{array}{lcl}\frac{\partial TSC({\theta }_{i},{\theta }_{j})}{\partial {\theta }_{i}}&=&\frac{\partial {C}_{i}}{\partial {e}_{i}}\frac{\partial {e}_{i}}{\partial {\theta }_{i}}+\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}\frac{\partial {E}_{i}}{\partial {\theta }_{i}}+\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {\theta }_{i}}+\frac{\partial {C}_{j}}{\partial {e}_{j}}\frac{\partial {e}_{j}}{\partial {\theta }_{i}}+\frac{\partial {D}_{j}({E}_{j},{\theta }_{j})}{\partial {E}_{j}}\frac{\partial {E}_{j}}{\partial {\theta }_{i}}\\ &=&\frac{\partial {C}_{i}}{\partial {e}_{i}}(\frac{\partial {e}_{i}}{\partial {\theta }_{i}}+\frac{\partial {e}_{j}}{\partial {\theta }_{i}})+\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}\frac{\partial {E}_{i}}{\partial {\theta }_{i}}+\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {\theta }_{i}}+\frac{\partial {D}_{j}({E}_{j},{\theta }_{j})}{\partial {E}_{j}}\frac{\partial {E}_{j}}{\partial {\theta }_{i}}\\ &=&-[\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}\frac{\alpha \frac{\partial {e}_{i}}{\partial a}+(1-\beta )\frac{\partial {e}_{j}}{\partial a}}{2}+\frac{\partial {D}_{j}({E}_{j},{\theta }_{j})}{\partial {E}_{j}}\frac{(1-\alpha )\frac{\partial {e}_{i}}{\partial a}+\beta \frac{\partial {e}_{j}}{\partial a}}{2}]\frac{\partial a}{\partial {\theta }_{i}}\\ &\quad+&\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}\frac{\partial {E}_{i}}{\partial {\theta }_{i}}+\frac{\partial {D}_{j}({E}_{j},{\theta }_{j})}{\partial {E}_{j}}\frac{\partial {E}_{j}}{\partial {\theta }_{i}}+\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {\theta }_{i}}\\ &=&\underset{<0}{\underbrace{\frac{1}{2}\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}\frac{\partial {E}_{i}}{\partial {\theta }_{i}}}}+\underset{<0}{\underbrace{\frac{1}{2}\frac{\partial {D}_{j}({E}_{j},{\theta }_{j})}{\partial {E}_{j}}\frac{\partial {E}_{j}}{\partial {\theta }_{i}}}}+\underset{>0}{\underbrace{\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {\theta }_{i}}}}\end{array},$$

(40)

In particular, $\frac{\partial TSC({\theta }_{i},0)}{\partial {\theta }_{i}}=\underset{<0}{\underbrace{\frac{1}{2}\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {E}_{i}}\frac{\partial {E}_{i}}{\partial {\theta }_{i}}}}+\underset{>0}{\underbrace{\frac{\partial {D}_{i}({E}_{i},{\theta }_{i})}{\partial {\theta }_{i}}}}$.

For clarity of expression, the variables under taxes and tradable permits are distinct by superscripts $t$ and $a$ below.

• (i) According to (36) and (39), it is obtained that $\underset{{\theta }_{i}\to \infty }{\mathrm{lim}}[TS{C}^{t}({\theta }_{i},{\theta }_{j})-TS{C}^{a}({\theta }_{i},{\theta }_{j})]>0$. Thus for ${\theta }_{i}$ sufficiently large, $TS{C}^{t}({\theta }_{i},{\theta }_{j})>TS{C}^{a}({\theta }_{i},{\theta }_{j})$.

• (ii) For the case of ${\theta }_{j}=0$, we have:

$$\frac{\partial (TS{C}^{t}({\theta }_{i},0)-TS{C}^{a}({\theta }_{i},0))}{\partial {\theta }_{i}}=\underset{?}{\underbrace{\frac{\partial {D}_{i}({E}_{i}^{t},{\theta }_{i})}{\partial {\theta }_{i}}-\frac{\partial {D}_{i}({E}_{i}^{a},{\theta }_{i})}{\partial {\theta }_{i}}}}\underset{>0}{\underbrace{-\frac{1}{2}\frac{\partial {D}_{i}({E}_{i}^{a},{\theta }_{i})}{\partial {E}_{i}}\frac{\partial {E}_{i}^{a}}{\partial {\theta }_{i}}}}.$$

(41)

Thus if $\frac{\partial {D}_{i}({E}_{i}^{t},{\theta }_{i})}{\partial {\theta }_{i}}-\frac{\partial {D}_{i}({E}_{i}^{a},{\theta }_{i})}{\partial {\theta }_{i}}\ge 0\iff {E}_{i}^{t}=\alpha {e}_{i}^{t}+(1-\beta ){e}_{j}^{\mathrm{max}}\ge {E}_{i}^{a}=\alpha {e}_{i}^{a}+(1-\beta ){e}_{j}^{a}$, we can conclude that the above is greater than 0.

Firstly, we prove ${e}_{i}^{t}<{e}_{i}^{a}$ by using reduction to absurdity. Assume ${e}_{i}^{t}\ge {e}_{i}^{a}$. It is obvious that:

$$-\frac{\partial {C}_{i}\left({e}_{i}^{t}\right)}{\partial {e}_{i}}\le -\frac{\partial {C}_{i}\left({e}_{i}^{a}\right)}{\partial {e}_{i}}.$$

(42)

On the other hand, we have ${E}_{i}^{t}\ge {E}_{i}^{a}$ and $\alpha >\frac{\alpha }{2}\ge \frac{(1-\beta )\frac{{\partial }^{2}{C}_{i}({e}_{i}^{a})}{\partial {e}_{i}^{2}}+\alpha \frac{{\partial }^{2}{C}_{j}({e}_{j}^{a})}{\partial {e}_{j}^{2}}}{2[\frac{{\partial }^{2}{C}_{i}({e}_{i}^{a})}{\partial {e}_{i}^{2}}+\frac{{\partial }^{2}{C}_{j}({e}_{j}^{a})}{\partial {e}_{j}^{2}}]}$. And:

$$-\frac{\partial {C}_{i}\left({e}_{i}^{t}\right)}{\partial {e}_{i}}=\alpha \frac{\partial {D}_{i}\left({E}_{i}^{t},{\theta }_{i}\right)}{\partial {E}_{i}}>\frac{\left(1-\beta \right)\frac{{\partial }^{2}{C}_{i}\left({e}_{i}^{a}\right)}{\partial {e}_{i}^{2}}+\alpha \frac{{\partial }^{2}{C}_{j}\left({e}_{j}^{a}\right)}{\partial {e}_{j}^{2}}}{2\left[\frac{{\partial }^{2}{C}_{i}\left({e}_{i}^{a}\right)}{\partial {e}_{i}^{2}}+\frac{{\partial }^{2}{C}_{j}\left({e}_{j}^{a}\right)}{\partial {e}_{j}^{2}}\right]}\frac{\partial {D}_{i}\left({E}_{i}^{a},{\theta }_{i}\right)}{\partial {E}_{i}}=-\frac{\partial {C}_{i}\left({e}_{i}^{a}\right)}{\partial {e}_{i}}=-\frac{\partial {C}_{j}\left({e}_{j}^{a}\right)}{\partial {e}_{j}}.$$

(43)

The results (42) and (43) are contradictory. So the assumption is incorrect, and it is obtained that ${e}_{i}^{t}<{e}_{i}^{a}$.

For $\forall \alpha ,\beta \in [1/2,1)$, $\exists {\stackrel{\sim }{\theta }}_{i}>0$, when ${\theta }_{i}\ge {\stackrel{\sim }{\theta }}_{i}$, we have ${E}_{i}^{t}\ge {E}_{i}^{a}$. This is because the large value of ${\theta }_{i}$ could ensure that ${e}_{i}^{a}$ and ${e}_{j}^{a}$ are small enough that $\alpha {e}_{i}^{t}+(1-\beta ){e}_{j}^{\mathrm{max}}\ge (1-\beta ){e}_{j}^{\mathrm{max}}\ge \alpha {e}_{i}^{a}+(1-\beta ){e}_{j}^{a}$. And it needs to meet the condition that $\frac{\partial {\stackrel{\sim }{\theta }}_{i}}{\partial \alpha }>0$, which states ${e}_{i}^{a}$ needs to be smaller in response to an increase in $\alpha$. In a word, for $\forall \alpha ,\beta \in [1/2,1)$, $\exists {\stackrel{\sim }{\theta }}_{i}>0$ which satisfies $\frac{\partial {\stackrel{\sim }{\theta }}_{i}}{\partial \alpha }>0$, when ${\theta }_{i}\ge {\stackrel{\sim }{\theta }}_{i}$, it is obtained that $\frac{\partial (TS{C}^{t}({\theta }_{i},0)-TS{C}^{a}({\theta }_{i},0))}{\partial {\theta }_{i}}>0$.

In addition, according to (43), we have: $-\frac{\partial {C}_{i}({e}_{i}^{t})}{\partial {e}_{i}}\ge -2\frac{\partial {C}_{i}({e}_{i}^{a})}{\partial {e}_{i}}$. Thus, with respect to quadratic abatement cost functions, we can further obtain:

$$\begin{array}{lcl}TS{C}^{t}({\theta }_{i},0)-TS{C}^{a}({\theta }_{i},0)&=&{C}_{i}({e}_{i}^{t})+{D}_{i}({E}_{i}^{t},{\theta }_{i})-{C}_{i}({e}_{i}^{a})-{C}_{j}({e}_{j}^{a})-{D}_{i}({E}_{i}^{a},{\theta }_{i})\\ &=&{C}_{i}({e}_{i}^{t})-2{C}_{i}({e}_{i}^{a})+{D}_{i}({E}_{i}^{t},{\theta }_{i})-{D}_{i}({E}_{i}^{a},{\theta }_{i})\\ &=&\underset{>0}{\underbrace{\frac{1}{2}({e}_{i}^{\mathrm{max}}-{e}_{i}^{t})\times (-\frac{\partial {C}_{i}({e}_{i}^{t})}{\partial {e}_{i}})-\frac{1}{2}({e}_{i}^{\mathrm{max}}-{e}_{i}^{a})\times (-2\frac{\partial {C}_{i}({e}_{i}^{a})}{\partial {e}_{i}})}}\\ &\quad+&\underset{\ge 0}{\underbrace{{D}_{i}({E}_{i}^{t},{\theta }_{i})-{D}_{i}({E}_{i}^{a},{\theta }_{i})}}\\ &\quad\quad>0\end{array}$$