Green Technology Transfers and Border Tax Adjustments

Abstract

We develop a two-country general equilibrium model of foreign technology transfers tied to environmental clean-up and border tax adjustments in the presence of transboundary pollution. Pollution is generated in the aid recipient as a by-product in the production of a ‘dirty’ good, which it consumes, as well as exports to the donor country. In contrast to the literature which typically treats aid as a monetary transfer, we assume that foreign aid consists of a transfer of environmental technology that lowers the cost of public clean-up. We also consider a border tax adjustment (BTA) as a secondary policy instrument used in order to internalize the residual transboundary externality. We study the environmental and welfare outcomes of the technology transfer and the BTA as well as the interaction between the two policy instruments. We derive conditions under which the additional presence of a BTA may lead to a lower transfer and less public pollution abatement in the recipient, with counter-productive consequences for both the environment and welfare. Contrary to intuition, we find that the green technology transfer and the border tax adjustment are not always complements.

Appendices

Appendix 1

1.1 Properties of the Public Abatement Cost Function $c^{g}\Big ( w\left( p,t\right) , A\Big )$

Following Hatzipanayotou and Michael (1995), the $c^{g}\Big ( w\left( p,t\right) , A\Big )$ function is concave, and homogeneous of degree one in $w$, i.e: (i) $c_{w}^{g}w=c^{g}$ ; (ii) $c_{ww}^{g}w=0$.

1.2 Properties of the Revenue Function $R\left( p,t,g\right)$

Following Abe (1992) and Hatzipanayotou and Michael (1995), and using the properties of the $c^{g}\Big ( w\left( p,t\right) , A\Big )$ function as well as our assumption that the production of good $x$ and public abatement of pollution compete for the same factors of production, the $R\left( p,t,g\right)$ function has the following properties: (i) $R_{p}=x$ ; (ii) $R_{t}=-z$ ; (iii) $R_{g}=-c^{g}$ ; (iv) $R_{pp}>0$ ; (v) $R_{pt}<0$ ; (vi) $R_{pg}=-c_{p}^{g}< 0$; (vii) $R_{tg}=-c_{t}^{g}> 0$; (viii) $R_{gg}=0$.

Appendix 2

1.1 Derivation of Expression (12)

Totally differentiating (11), and assuming that third derivatives of the revenue and expenditure functions are zero,²² we get:

$$\begin{aligned}&c_{A}^{g}dA+\left[ \theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zp}^{*}-R_{gp}\right] dp \nonumber \\&\quad +\left[ \theta \left( R_{tg}+1\right) \left( E_{z}^{*}-\tau E_{pz}^{*}\right) c_{A}^{g}-\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}-R_{gg}\right] dg \nonumber \\&\quad +\left[ \theta \left( 1-\alpha \right) \left( R_{tg}+1\right) gc_{A}^{g}E_{zp}^{*}+\alpha R_{gp}-\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{pz}^{*}\right] d\tau \nonumber \\&\quad +\,\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}dz+\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}du^{*}=0 \end{aligned}$$

(18)

Substituting (2), (9), and (11) into (18), we get the following expression:

$$\begin{aligned}&\left[ \begin{array}{c} c_{A}^{g}-\frac{\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}}{\left( 1-\tau E_{pu^{*}}^{*}\right) } \\ -\left[ \begin{array}{c} \theta \left( R_{tg}+1\right) \left( E_{z}^{*}-\tau E_{pz}^{*}\right) c_{A}^{g}-\theta ^{2}\left( R_{tg}+1\right) ^{2}gc_{A}^{g}E_{zz}^{*} \\ -R_{gg}+\frac{\theta ^{2}\left( R_{tg}+1\right) ^{2}gc_{A}^{g}E_{zu^{*}}^{*}\left( E_{z}^{*}-\tau E_{pz}^{*}\right) }{\left( 1-\tau E_{pu^{*}}^{*}\right) } \end{array} \right] \frac{gc_{A}^{g}}{c^{g}+tR_{tg}} \end{array} \right] dA \nonumber \\&\quad +\left[ \begin{array}{c} \theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zp}^{*}-R_{gp}-\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}R_{tp} \\ -\frac{\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}\left[ E_{p}^{*}-R_{p}^{*}-\tau \left( E_{pp}^{*}-R_{pp}^{*}\right) +\theta \left( \tau E_{pz}^{*}-E_{z}^{*}\right) R_{tp}\right] }{ \left( 1-\tau E_{pu^{*}}^{*}\right) } \\ -\left[ \begin{array}{c} \theta \left( R_{tg}+1\right) \left( E_{z}^{*}-\tau E_{pz}^{*}\right) c_{A}^{g}-\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}-R_{gg} \\ -\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}R_{tg} \\ +\frac{\theta ^{2}\left( R_{tg}+1\right) ^{2}gc_{A}^{g}E_{zu^{*}}^{*}\left( E_{z}^{*}-\tau E_{pz}^{*}\right) }{\left( 1-\tau E_{pu^{*}}^{*}\right) } \end{array} \right] \frac{tR_{tp}+gc_{p}^{g}}{c^{g}+tR_{tg}} \end{array} \right] dp \nonumber \\&\quad +\left[ \begin{array}{c} R_{tg}-R_{gt} \\ -\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}R_{tt} \\ +\frac{\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}\left( E_{z}^{*}-\tau E_{pz}^{*}\right) R_{tt}}{\left( 1-\tau E_{pu^{*}}^{*}\right) } \\ -\left[ \begin{array}{c} \theta \left( R_{tg}+1\right) \left( E_{z}^{*}-\tau E_{pz}^{*}\right) c_{A}^{g}-\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}-R_{gg} \\ -\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}R_{tg} \\ +\frac{\theta ^{2}\left( R_{tg}+1\right) ^{2}gc_{A}^{g}E_{zu^{*}}^{*}\left( E_{z}^{*}-\tau E_{pz}^{*}\right) }{\left( 1-\tau E_{pu^{*}}^{*}\right) } \end{array} \right] \frac{R_{t}+tR_{tt}+gc_{t}^{g}}{c^{g}+tR_{tg}} \end{array} \!\!\right] dt \nonumber \\&\quad +\left[ \begin{array}{c} \theta \left( 1-\alpha \right) \left( R_{tg}+1\right) gc_{A}^{g}E_{zp}^{*}+\alpha R_{gp}-\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{pz}^{*} \\ +\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}\alpha R_{tp} \\ +\frac{\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}\left[ \alpha \left( E_{p}^{*}-R_{p}^{*}\right) +\left( 1-\alpha \right) \tau \left( E_{pp}^{*}-R_{pp}^{*}\right) +\theta \left( \tau E_{pz}^{*}-E_{z}^{*}\right) \alpha R_{tp}\right] }{\left( 1-\tau E_{pu^{*}}^{*}\right) } \\ +\left[ \begin{array}{c} \theta \left( R_{tg}+1\right) \left( E_{z}^{*}-\tau E_{pz}^{*}\right) c_{A}^{g}-\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*} \\ -R_{gg}-\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}R_{tg} \\ +\frac{\theta ^{2}\left( R_{tg}+1\right) ^{2}gc_{A}^{g}E_{zu^{*}}^{*}\left( E_{z}^{*}-\tau E_{pz}^{*}\right) }{\left( 1-\tau E_{pu^{*}}^{*}\right) } \end{array} \right] \frac{\alpha \left( tR_{tp}+gc_{p}^{g}\right) }{c^{g}+tR_{tg}} \end{array} \right] d\tau =0\nonumber \\ \end{aligned}$$

(19)

Given that $R_{t}+tR_{tt}+gc_{t}^{g}=0$ and $tR_{tp}+gc_{p}^{g}=0$ from (1), and after some simplifications, (20) is also equivalent to

$$\begin{aligned}&\left[ \begin{array}{c} c_{A}^{g}-\frac{\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}}{\left( 1-\tau E_{pu^{*}}^{*}\right) } \nonumber \\ -\left[ \begin{array}{c} \theta \left( R_{tg}+1\right) \left( E_{z}^{*}-\tau E_{pz}^{*}\right) c_{A}^{g}-\theta ^{2}\left( R_{tg}+1\right) ^{2}gc_{A}^{g}E_{zz}^{*} \\ -R_{gg}+\frac{\theta ^{2}\left( R_{tg}+1\right) ^{2}gc_{A}^{g}E_{zu^{*}}^{*}\left( E_{z}^{*}-\tau E_{pz}^{*}\right) }{\left( 1-\tau E_{pu^{*}}^{*}\right) } \end{array} \right] \frac{gc_{A}^{g}}{c^{g}+tR_{tg}} \end{array} \right] dA \nonumber \\&\quad +\left[ \begin{array}{c} \theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zp}^{*}-R_{gp}-\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}R_{tp} \\ -\frac{\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}\left[ E_{p}^{*}-R_{p}^{*}-\tau \left( E_{pp}^{*}-R_{pp}^{*}\right) +\theta \left( \tau E_{pz}^{*}-E_{z}^{*}\right) R_{tp}\right] }{ \left( 1-\tau E_{pu^{*}}^{*}\right) } \end{array} \right] dp \nonumber \\&\quad +\left[ \begin{array}{c} \theta \left( 1-\alpha \right) \left( R_{tg}+1\right) gc_{A}^{g}E_{zp}^{*}+\alpha R_{gp}-\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{pz}^{*} \\ +\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}\alpha R_{tp} \\ +\frac{\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}\left[ \alpha \left( E_{p}^{*}-R_{p}^{*}\right) +\left( 1-\alpha \right) \tau \left( E_{pp}^{*}-R_{pp}^{*}\right) +\theta \left( \tau E_{pz}^{*}-E_{z}^{*}\right) \alpha R_{tp}\right] }{\left( 1-\tau E_{pu^{*}}^{*}\right) } \end{array} \right] d\tau \nonumber \\&\quad +\left[ \begin{array}{c} R_{tg}-R_{gt} \\ -\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}R_{tt} \\ +\frac{\theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}\left( E_{z}^{*}-\tau E_{pz}^{*}\right) R_{tt}}{\left( 1-\tau E_{pu^{*}}^{*}\right) } \end{array} \right] dt=0 \end{aligned}$$

(20)

From (20) and given that $t$ and $p$ are independent on $\tau$, we can derive the following expression of the effect of a change in $\tau$ on $A$:

$$\begin{aligned} \frac{dA}{d\tau }&= \frac{N^{A}}{D^{A}}\nonumber \\&= \frac{\left[ \begin{array}{c} -\alpha \theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zp}^{*}+\alpha R_{gp} \\ +\frac{\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}\left[ \alpha \left( E_{p}^{*}-R_{p}^{*}\right) +\left( 1-\alpha \right) \tau \left( E_{pp}^{*}-R_{pp}^{*}\right) +\theta \left( \tau E_{pz}^{*}-E_{z}^{*}\right) \alpha R_{tp}\right] }{\left( 1-\tau E_{pu^{*}}^{*}\right) } \\ +\alpha \theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}R_{tp} \end{array} \right] }{\left[ \begin{array}{c} -c_{A}^{g}+\frac{\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}}{\left( 1-\tau E_{pu^{*}}^{*}\right) } \\ +\left[ \begin{array}{c} \theta \left( R_{tg}+1\right) \left( E_{z}^{*}-\tau E_{pz}^{*}\right) c_{A}^{g}-\theta ^{2}\left( R_{tg}+1\right) ^{2}gc_{A}^{g}E_{zz}^{*} \\ -R_{gg}+\frac{\theta ^{2}\left( R_{tg}+1\right) ^{2}gc_{A}^{g}E_{zu^{*}}^{*}\left( E_{z}^{*}-\tau E_{pz}^{*}\right) }{\left( 1-\tau E_{pu^{*}}^{*}\right) } \end{array} \right] \frac{gc_{A}^{g}}{c^{g}+tR_{tg}} \end{array} \right] }\qquad \quad \end{aligned}$$

(21)

Substituting (11) in the numerator ($N^{A}$) and denominator ($D^{A}$) of the above expression, and given that $c^{g}+tR_{tg}-gR_{gg}=0$ from (1), we get expression (12) after factoring $N^{A}$ for $\alpha$.

Appendix 3

Proof of Proposition 1

To appraise the effect of the BTA on $A$, it is necessary to determine the signs of the numerator and the denominator, i.e. the signs of $N^A$ and $D^A$. From (13), we can show that $N^A=$ $\alpha -\overline{\alpha ^A }$, with

$$\begin{aligned} \overline{\alpha ^A }=\frac{\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}\tau \left( E_{pp}^{*}-R_{pp}^{*}\right) }{\left( \tau E_{pu^{*}}^{*}-1\right) \left[ \begin{array}{c} \theta ^{2}\left( R_{tg}+1\right) gc_{A}^{g}E_{zz}^{*}R_{tp}-\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zp}^{*} +tR_{tp}\left( \frac{1}{g}+\frac{\theta E_{zu^{*}}^{*}R_{tg}}{1-\tau E_{pu^{*}}^{*}}\right) \\ +\frac{\theta \left( R_{tg}+1\right) gc_{A}^{g}E_{zu^{*}}^{*}}{ \left( 1-\tau E_{pu^{*}}^{*}\right) }\left[ \left( E_{p}^{*}-R_{p}^{*}\right) -\tau \left( E_{pp}^{*}-R_{pp}^{*}\right) + \frac{R_{tp}c^{g}}{gc_{A}^{g}\left( R_{tg}+1\right) }\right] \end{array} \right] }. \end{aligned}$$

From our assumptions, we know that $R_{tg}>0,\, c_{A}<0$, $E_{zu^{*}}>0,\, R_{tp}<0,\, E_{p}^{*}-R_{p}^{*}>0$, $E_{pp}^{*}<0$, $R_{pp}^{*}>0$, $E_{zz}^{*}>0$, and $E_{zp}^{*}\geqslant 0$. When $\tau E_{pu^{*}}^{*}>1$ and $\theta >\overline{\theta }$ with $\overline{\theta }=\frac{\tau E_{pu^{*}}^{*}-1}{gE_{zu^{*}}^{*}R_{tg}}$, it directly follows that $\overline{\alpha ^A }>0$. Therefore, it is straightforward to show that $N^A\gtrless 0$ when $\alpha \gtrless \overline{\alpha ^A }$. It also follows from our assumptions that $D^A\gtrless 0$ if $\left| c_{A}^{g}\right| \lessgtr \overline{c}$, with $\overline{c}=\frac{ 3\left( c^{g}+tR_{tg}\right) }{\theta ^{2}g^{2}\left( R_{tg}+1\right) ^{2}E_{zz}^{*}}$.

$\square$

Proof of Lemma 1

From (15), we can show that $N^{e}=$ $\alpha -\overline{\alpha ^{e}}$, with

$$\begin{aligned} \overline{\alpha ^{e}}\!=\!\frac{\theta \left( R_{tg}+1\right) ^{2}g^{2}c_{A}^{g}E_{zu^{*}}^{*}\tau \left( E_{pp}^{*}-R_{pp}^{*}\right) }{\left( \tau E_{pu^{*}}^{*}\!-\!1\right) \left[ \! \begin{array}{c} -3R_{tp}\left( c^{g}+tR_{tg}\right) -\theta \left( R_{tg}+1\right) ^{2}g^{2}c_{A}^{g}E_{zp}^{*} +tR_{tp}\left( R_{tg}+1\right) \left( 1+\frac{\theta E_{zu^{*}}^{*}R_{tg}g}{1-\tau E_{pu^{*}}^{*}}\right) \\ +\frac{\theta \left( R_{tg}+1\right) ^{2}g^{2}c_{A}^{g}E_{zu^{*}}^{*} }{\left( 1-\tau E_{pu^{*}}^{*}\right) }\left[ \left( E_{p}^{*}-R_{p}^{*}\right) -\tau \left( E_{pp}^{*}-R_{pp}^{*}\right) + \frac{R_{tp}c^{g}}{gc_{A}^{g}\left( R_{tg}+1\right) }\right] \end{array} \!\right] } \end{aligned}$$

Under the assumptions highlighted in Lemma 1, it straightforward to see that $\overline{\alpha ^{e}}>0$ when $\tau E_{pu^{*}}^{*}>1$ and $\theta >\overline{\theta }$, with $\overline{\theta }=\frac{\tau E_{pu^{*}}^{*}-1}{gE_{zu^{*}}^{*}R_{tg}}$. Therefore, we have $N^{e}\gtrless 0$ when $\alpha \gtrless \overline{\alpha ^{e}}$. Under the same assumptions as in the proof of Proposition 1, it also directly follows that $D^{e}\gtrless 0$ if $\left| c_{A}^{g}\right| \gtrless \overline{c}$ with $\overline{c}= \frac{ 3\left( c^{g}+tR_{tg}\right) }{\theta ^{2}g^{2}\left( R_{tg}+1\right) ^{2}E_{zz}^{*}}$.$\square$

Proof of Proposition 2

From (14), it is straightforward to see that we always have $\frac{de}{d\tau }<0$ if $\frac{dA}{d\tau }>0$. The table below combines the conditions that determine the signs of $\frac{de}{d\tau }$, $\frac{dA}{d\tau }$ and $\frac{dg}{d\tau }$ when $\frac{dA}{d\tau }<0$ and $\frac{dg}{d\tau }<0$, as summarized in Proposition 1 and Lemma 1 respectively.

$\tau E_{pu^{*}}^{*}>1;\, \theta >\overline{\theta };\, \alpha >\overline{\alpha }$; and $\left| c_{A}^{g}\right| >\overline{c}$	$\tau E_{pu^{*}}^{*}>1;\, \theta >\overline{\theta };\, \alpha <\overline{\alpha }$; and $\left| c_{A}^{g}\right| <\overline{c}$
$\frac{dA}{d\tau }<0$ and $\frac{dg}{d\tau }<0$	$\frac{dA}{d\tau }<0$ and $\frac{dg}{d\tau }<0$
$\overline{\alpha ^{e}}>\overline{\alpha }$	$\overline{\alpha ^{e}}< \overline{\alpha }$
$\overline{\alpha ^{e}}>\alpha >\overline{\alpha }$ or $\alpha >\overline{ \alpha ^{e}}>\overline{\alpha }$	$\alpha <\overline{\alpha ^{e}}<\overline{ \alpha }$ or $\overline{\alpha ^{e}}<\alpha <\overline{\alpha }$
$\overline{\alpha ^{e}}>\alpha$ or $\alpha >\overline{\alpha ^{e}}$	$\alpha <\overline{\alpha ^{e}}$ or $\overline{\alpha ^{e}}<\alpha$
$\frac{de}{d\tau }<0$ or $\frac{de}{d\tau }>0$	$\frac{de}{d\tau }>0$ or $\frac{de}{d\tau }<0$

From this table, we can directly show that $\frac{de}{d\tau }> 0$ if $\alpha >\overline{\alpha ^{e}}>\overline{\alpha }$ or $\alpha <\overline{\alpha ^{e}}<\overline{\alpha }$, and that $\frac{de}{d\tau }< 0$ if $\overline{\alpha ^{e}}>\alpha >\overline{\alpha }$ or $\overline{\alpha ^{e}}<\alpha <\overline{\alpha }$. $\square$