Discussion of the Tax Scheme for Cleaner Water Use

Abstract

Resource and effluent taxes are the typical Pigovian taxes used for water environment protection. Their original purpose is to motivate innovative or green technologies. This article develops a model to examine the influence of these two types of taxes on the firm’s technology decision. In the model, the local government imposes tax policy, and in response, a profit-maximizing monopolistic firm selects technology, water consumption, production quantity and product price. By comparing the corresponding impacts on water conservation, effluent reduction and social welfare, this analysis reveals different attributes of these two types of taxes. First, neither of these taxes necessarily motivates the firm to choose the innovative technology. For the effluent tax, an initial increase may motivate the firm to leverage new technology for profit growth, but a further increase in the tax burden may induce a reverse effect. For the resource tax, the firm’s technology choice depends on downstream ecological compensation. Second, with the same tax burden, the resource tax performs more effectively than the effluent tax on both water conservation and pollutant effluent reduction, which supports the necessity of external ecological compensation. Third, the effects on social welfare improvement present two different trends, and there exists a threshold of tax rate on the gap between them. The threshold changes according to the efficiency of new technology, which implies the appropriate type of tax and the optimal range of the tax rate.

Appendices

Appendix 1

• 1.1 Proof of the existence of Region 4 $t\in \left[{t}_{2},{t}^{\mathrm{lim}}\right)$ from Proposition 1.

According to Eqs. (13) and (15),

$$\begin{array}{c}\begin{array}{c}{t}^{\mathrm{lim}}-{t}_{2}=\xi +\frac{A-\xi +2\sqrt{\xi (1-\rho )}}{\rho }-\frac{1}{\rho +1}(A+\xi \rho +\sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta })\\ =\xi +\frac{1}{\rho (\rho +1)}[A-\xi (\rho +1)-\xi {\rho }^{2}+2(\rho +1)\sqrt{\xi (1-\rho )}-\rho \sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta })]\end{array}\\ \ge \frac{1}{\rho (\rho +1)}[A-\xi (\rho +1)-\xi {\rho }^{2}-\rho \sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta })]\end{array}$$

We need to compare $A-\xi (\rho +1)-\xi {\rho }^{2}$ and $\rho \sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }$ to determine whether ${t}^{\mathrm{lim}}-{t}_{2}\ge 0$.

$${[A-\xi (\rho +1)-\xi {\rho }^{2}]}^{2}={A}^{2}-2A\xi {\rho }^{2}-2A\xi \rho -2A\xi +{\xi }^{2}{\rho }^{4}+2{\xi }^{2}{\rho }^{3}+3{\xi }^{2}{\rho }^{2}+2{\xi }^{2}\rho +{\xi }^{2}[\rho \sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }){]}^{2}={\rho }^{2}{A}^{2}-2A\xi {\rho }^{2}+{\xi }^{2}{\rho }^{2}-4\Theta {\rho }^{2}-4{\rho }^{3}\Theta$$

$${[A-\xi (\rho +1)-\xi {\rho }^{2}]}^{2}-[\rho \sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }){]}^{2}=A[(1-{\rho }^{2})A-2\xi (\rho +1)]+{\xi }^{2}{\rho }^{4}+2{\xi }^{2}{\rho }^{3}+2{\xi }^{2}{\rho }^{2}+2{\xi }^{2}\rho +{\xi }^{2}+4\Theta {\rho }^{2}+4{\rho }^{3}\Theta$$

For the assumption of $A$ and $\rho$,$(1-{\rho }^{2})A-2\xi (\rho +1)\ge 0$, then

$${[A-\xi (\rho +1)-\xi {\rho }^{2}]}^{2}-[\rho \sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }){]}^{2}\ge 0$$

This implies that ${t}^{\mathrm{lim}}-{t}_{2}\ge 0$, and Region 4 $t\in [{t}_{2},{t}^{\mathrm{lim}})$ is reasonable and exists.

• 1.1 Proof of ${t}_{1}>\xi$

  According to Eq. (14),

  $${t}_{1}-\xi =\frac{1}{\rho +1}[A+\xi \rho -\xi (\rho +1)-\sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }]$$
  $$=\frac{1}{\rho +1}[A-\xi -\sqrt{{(A-\xi )}^{2}-4\Theta (1+\rho })]\ge 0$$

  ${t}_{1}>\xi$ is proved.

Appendix 2

• 2.1 Proof of ${t}_{4}\ge 0$

According to Eq. (22), we need to compare ${\lambda }_{1}$ and $2\rho -A-\xi \rho +4$

$${\lambda }_{1}^{2}-{(2\rho -A-\xi \rho +4)}^{2}=(\rho +1)[A(6\xi -4)-4\xi +8\Theta -4\xi \rho +3\rho {\xi }^{2}-3{\rho }^{2}]$$

For $\xi \ge 2$ and the assumption of $A$, ${t}_{4}\ge 0$.

• 2.2 Proof of ${t}_{4}\le {t}^{\mathrm{lim}}$

According to Eqs. (13) and (22),

$${t}^{\mathrm{lim}}-{t}_{4}=\xi +\frac{A-\xi +2\sqrt{\xi (1-\rho )}}{\rho }-\frac{1}{\rho +1}(2\rho -A-\xi \rho +{\lambda }_{1}+4)$$

$$=(\frac{1}{\rho }+\frac{1}{\rho +1})A+\frac{-\xi +2\sqrt{\xi (1-\rho )}}{\rho }-\frac{1}{\rho +1}(2\rho -\xi \rho +\lambda +4)+\xi$$

For the assumption of $A$, $\rho$ and $\xi$, ${t}^{\mathrm{lim}}-{t}_{4}\ge 0$, then ${t}_{4}\le {t}^{\mathrm{lim}}$.

• 2.3 Proof of ${t}_{4}\ge \xi$

According to Eq. (22),${t}_{4}-\xi =\frac{1}{\rho +1}[2\rho -A-\xi \rho +{\lambda }_{1}+4-\xi (\rho +1)]$

We need to compare the ${\lambda }_{1}$ and $2\rho -A-\xi \rho +4-\xi (\rho +1)$

$${\lambda }_{1}^{2}-{(2\rho -A-2\xi \rho -\xi +4)}^{2}=4(\rho +1)[A(\xi -1)+\xi +2\Theta -{\xi }^{2}]\ge 0$$

Then, ${t}_{4}\ge \xi$ is proved.

Appendix 3

• 3. Proof of ${t}_{2}>{t}_{4}$.

According to Eqs. (15) and (22),

$$\begin{array}{c}({t}_{4}^{2}-{t}_{2}^{2}){(\rho +1)}^{2}\\ ={(2\rho -A-\xi \rho +{\lambda }_{1}+4)}^{2}-{[A+\xi \rho +\sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }]}^{2}\\ \begin{array}{c}=(8{\lambda }_{1}+4{\lambda }_{1}\rho -2{\lambda }_{1}A)+(-20A-2A\sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }+8A\xi +8A\xi \rho )-24\xi \rho \\ -2{\lambda }_{1}\xi \rho +[(32+12\Theta )(\rho +1)-2\xi \rho \sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }]-12\xi {\rho }^{2}\\ -4{\xi }^{2}-4\xi +[4(2+{\xi }^{2}){\rho }^{2}-12A\rho ]\end{array}\end{array}$$

For the assumption of $A$,

$$\begin{array}{c}8{\lambda }_{1}+4{\lambda }_{1}\rho -2{\lambda }_{1}A=2{\lambda }_{1}(4+2\rho -A)\le 0\\ -20A-2A\sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }+8A\xi +8A\xi \rho \\ \begin{array}{c}=2A(-\sqrt{{(A-\xi )}^{2}-4\Theta -4\rho \Theta }-10+4\xi +4\xi \rho )\\ \le 0(32+12\Theta )(\rho +1)-2\xi \rho \sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }\le 0\\ 4(2+{\xi }^{2}){\rho }^{2}-12A\rho \le 0\end{array}\end{array}$$

Then, with the other negative components of the equation, we can obtain

$${(2\rho -A-\xi \rho +{\lambda }_{1}+4)}^{2}-{[A+\xi \rho +\sqrt{{A}^{2}-2A\xi +{\xi }^{2}-4\Theta -4\rho \Theta }]}^{2}<0$$

Then, $({t}_{4}^{2}-{t}_{2}^{2}){(\rho +1)}^{2}>0$.

For ${t}_{2}\ge 0,{t}_{4}\ge 0$, thus, ${t}_{2}>{t}_{4}$.

Appendix 4

• 4. Proof $\Delta S{W}_{\tau G}\ge 0$ and $\Delta {\pi }_{\tau G}\ge 0$.

$$\begin{array}{c}\begin{array}{c}\Delta S{W}_{\tau G}=\Delta S{W}_{\tau }+S\\ =\frac{1-\rho }{8}(4A+4\xi -4\tau -8\Theta -6A\xi +4\xi \rho +2\xi \tau -3{\xi }^{2}\rho +3{\xi }^{2})-\frac{1-\rho }{8}(4A-6A\xi -8\Theta )\\ \begin{array}{c}=\frac{1-\rho }{8}(4\xi -4\tau +4\xi \rho +2\xi \tau -3{\xi }^{2}\rho +3{\xi }^{2})\\ =\frac{1-\rho }{8}[3{\xi }^{2}(1-\rho )+4\xi (1+\rho )+(2\xi -4)\tau ]\\ \Delta {\pi }_{\tau G}=\frac{\rho -1}{4}(4\Theta +2A\xi -2\xi \tau +{\xi }^{2}\rho -{\xi }^{2})+\frac{\rho -1}{4}(2A-3A\xi -4\Theta )\end{array}\end{array}\\ =\frac{\rho -1}{4}[4\Theta +2A\xi -2\xi \tau +{\xi }^{2}\rho -{\xi }^{2}+2A-3A\xi -4\Theta )]\\ =\frac{\rho -1}{4}[A(2-\xi )-2\xi \tau +{\xi }^{2}(\rho -1)]\end{array}$$

when $\xi \ge 2$, $\Delta S{W}_{\tau G}\ge 0$ and $\Delta {\pi }_{\tau G}\ge 0$.

Appendix 5

• 5. Proof the existence of threshold ${t}^{*}={\tau }^{*}=r$, and solve $r$.

$\Delta S{W}_{t}$ is a general quadratic equation with an upwards parabolic curve. By solving the first-order condition $\frac{\partial \Delta S{W}_{t}}{\partial t}=0$, we can find its minimum points ($t=\frac{2\rho -\xi \rho +4-A}{1+\rho }$, $\Delta S{W}_{t}^{\mathrm{min}}$). Because $\Delta S{W}_{\tau G}$ is a linear function, let $\tau =\frac{2\rho -\xi \rho +4-A}{1+\rho }$ too, $\Delta S{W}_{\tau G}\ge \Delta S{W}_{t}^{\mathrm{min}}$, then the threshold ${t}^{*}={\tau }^{*}$ must exist.

$$\begin{array}{c}\Delta S{W}_{\tau G}(\tau =\frac{2\rho -\xi \rho +4-A}{1+\rho })\\ =\frac{1-\rho }{8\rho +8}(4A+12\xi -8\rho -2A\xi +16\xi \rho +4\xi {\rho }^{2}-2{\xi }^{2}\rho +3{\xi }^{2}-3{\xi }^{2}{\rho }^{2}-16)\\ \begin{array}{c}\Delta S{W}_{t}(t=\frac{2\rho -\xi \rho +4-A}{1+\rho })=\frac{\rho -1}{8\rho +8}(16\rho -4\xi -12A+8\Theta +6A\xi -8A\rho -16\xi \rho +8\rho \Theta -8\xi {\rho }^{2}+\\ {A}^{2}-3{\xi }^{2}+4\rho {}^{2}+4{\xi }^{2}{\rho }^{2}+8A\xi \rho +16)\Delta S{W}_{\tau G}-\Delta S{W}_{t}^{\mathrm{min}}=\frac{1-\rho }{8\rho +8}[A(A+8\xi \rho -8\rho +4\xi -8)+\\ 8\rho +8\xi +8\Theta +8\rho \Theta -4\xi {\rho }^{2}-2{\xi }^{2}\rho +4{\rho }^{2}+{\xi }^{2}{\rho }^{2})]\ge 0\end{array}\end{array}$$

The threshold ${t}^{*}={\tau }^{*}$ exists.

To solve the threshold ${t}^{*}={\tau }^{*}=r$, let $\Delta S{W}_{\tau G}(\tau =r)-\Delta S{W}_{t}(t=r)=0$

That is, $\frac{\rho -1}{8}(4A-4r-8\Theta -6A\xi +2Ar-2\xi r-4r\rho +{r}^{2}\rho +{r}^{2}+2\xi r\rho )=0$

We can obtain

$$\begin{array}{c}{r}_{1}=\frac{1}{1+\rho }(\xi -A+2\rho -\xi \rho +{\lambda }_{3}+2)\\ {r}_{2}=\frac{1}{1+\rho }(\xi -A+2\rho -\xi \rho -{\lambda }_{3}+2)\\ {\lambda }_{3}=\sqrt{4\xi -8A+8\rho +8\Theta +4A\xi -8A\rho +8\rho \Theta -4\xi {\rho }^{2}-2{\xi }^{2}\rho +{A}^{2}+{\xi }^{2}+4{\rho }^{2}+{\xi }^{2}{\rho }^{2}+8A\xi \rho +4}\end{array}$$

For ${r}_{2}<0$, we focus on ${r}_{1}$.

If ${r}_{1}\ge 0$, $\lambda \ge -\xi +A-2\rho +\xi \rho -2$

For the assumption of $\xi \ge 2$,

$${\lambda }^{2}-{(-\xi +A-2\rho +\xi \rho -2)}^{2}=2(\rho +1)(4\Theta -2A+3A\xi )\ge 0$$

${r}_{1}\ge 0$ is proven, and for simplicity, we define $r={r}_{1}$.

Appendix 6

• 6. Proof $\frac{\partial r}{\partial \rho }\le 0$.

According to Eq. (34),

$$\frac{\partial r}{\partial\rho}=\frac{\lbrack4\rho-4A+4\Theta+4A\xi-4\xi\rho+\xi^2\rho-\xi^2+4+(2-\xi)\lambda_3\rbrack(1+\rho)-\lambda_3(\xi-A+2\rho-\xi\rho+\lambda_3+2)}{\lambda_3{(1+\rho)}^2}$$

Because the denominator is positive, we can judge $\frac{\partial r}{\partial \rho }\ge 0$ or $\frac{\partial r}{\partial \rho }\le 0$ by just calculating the numerator (defined as $N$),

$$N=[4\rho -4A+4\Theta +4A\xi -4\xi \rho +{\xi }^{2}\rho -{\xi }^{2}+4+(2-\xi ){\lambda }_{3}](1+\rho )-{\lambda }_{3}(\xi -A+2\rho -\xi \rho +{\lambda }_{3}+2)=4A+4A\rho -{A}^{2}-4A\xi \rho +A{\lambda }_{3}-2\xi {\lambda }_{3}-4\xi -4\Theta -4\xi \rho -4\rho \Theta +2{\xi }^{2}\rho -2{\xi }^{2}$$

If $N\ge 0$, then ${\lambda }_{3}$ should satisfy the following condition:

$${\lambda }_{3}\ge \frac{-(4A+4A\rho -{A}^{2}-4A\xi \rho -4\xi -4\Theta -4\xi \rho -4\rho \Theta +2{\xi }^{2}\rho -2{\xi }^{2})}{A-2\xi }$$

According to Eq. (33),

$$\begin{array}{c}{{\lambda }_{3}}^{2}-{[\frac{-(4A+4A\rho -{A}^{2}-4A\xi \rho -4\xi -4\Theta -4\xi \rho -4\rho \Theta +2{\xi }^{2}\rho -2{\xi }^{2})}{A-2\xi }]}^{2}\\ =-\frac{{(1+\rho )}^{2}(4\Theta -2A+3A\xi )[A(5\xi -6)+8\xi +4\Theta -4{\xi }^{2}]}{{(A-2\xi )}^{2}}\end{array}$$

For $\xi \ge 2$, $4\Theta -2A+3A\xi \ge 0$, $A(5\xi -6)+8\xi +4\Theta -4{\xi }^{2}\ge 0$, $N\le 0$, $\frac{\partial r}{\partial \rho }\le 0$.