International environmental policy and poverty alleviation

Abstract

We investigate how alternative national and international policies and circumstances impact the Clean Development Mechanism’s (CDM) contribution to sustainable development and the pursuit of poverty eradication goals in developing countries. In particular, we focus on the importance of technology-specific versus technology-neutral environmental regulations in the project host regions. We also consider alternative CDM benefit-sharing arrangements between the host and client regions. An analytical impure-public-good model is developed which considers CDM projects as a conditional transfer exerting price and income effects. These, in turn, induce changes in the use of environmental technologies, and with it global and local environmental protection levels. Aided by model simulations using empirical data for China and the European Union, we seek to assess conditions in which CDM transfers are more favourable towards improved environmental protection and welfare in developing regions.

Appendix

1.1 Supplementary information: derivations

1.1.1 Case 1: Fixed z

From the first-order conditions, we know that:

$$ p_{d}^{a} = {\frac{{\left( {1 - s} \right)p_{d}^{a} }}{{\bar{\alpha }_{d} }}} $$

(3)

$$ {\frac{{dq_{d} }}{{da_{d} }}} = - {\frac{1}{{\bar{\alpha }_{d} }}}. $$

(36)

1.1 Show that dq _d /ds > 0.

Rearranging Eq. (3) and obtaining the total derivative:

$$ {\frac{{dp_{d}^{q} }}{ds}} = {\frac{{p_{d}^{a} \cdot \bar{\alpha }_{d} }}{{\left( {1 - s} \right)^{2} }}} + {\frac{{\bar{\alpha }_{d} }}{{\left( {1 - s} \right)}}}\,{\frac{{dp_{d}^{a} }}{{da_{d} }}}\,{\frac{{da_{d} }}{{dq_{d} }}}\,{\frac{{dq_{d} }}{ds}}. $$

(37)

Substituting from Eq. (36) yields:

$$ {\frac{{dp_{d}^{q} }}{ds}} = {\frac{{p_{d}^{a} \cdot \bar{\alpha }_{d} }}{{\left( {1 - s} \right)^{2} }}} - {\frac{{\bar{\alpha }_{d}^{2} }}{{\left( {1 - s} \right)}}}\,{\frac{{dp_{d}^{a} }}{{da_{d} }}}\,{\frac{{dq_{d} }}{ds}}. $$

(38)

By using the chain rule we get:

$$ {\frac{{dp_{d}^{a} }}{ds}} = {\frac{{dp_{d}^{q} }}{{dq_{d} }}}\,{\frac{{dq_{d} }}{ds}}. $$

(39)

Substituting Eq. (38) into (39) yields:

$$ {\frac{{dq_{d} }}{ds}} = {\frac{{p_{d}^{a} }\cdot \bar{\alpha }_{d}}{{\left( {1 - s} \right)^{2} }}}\left[ {{\frac{{dp_{d}^{q} }}{{dq_{d} }}} + {\frac{{\bar{\alpha }_{d}^{2} }}{{\left( {1 - s} \right)}}}\,{\frac{{dp_{d}^{a} }}{{da_{d} }}}} \right]^{ - 1}. $$

(8)

Given that q and a are normal goods, dq _d /ds > 0 for all s.

1.2 Show that da _d /ds < 0.

Obtaining the total derivative of (3), and substituting in from Eq. (36):

$$ {\frac{{dp_{d}^{a} }}{ds}} = - {\frac{{\left( {1 - s} \right)}}{{\bar{\alpha }_{d}^{2} }}}\,{\frac{{dp_{d}^{q} }}{{dq_{d} }}}\,{\frac{{da_{d} }}{ds}} - {\frac{{p_{d}^{q} }}{{\bar{\alpha }_{d} }}}. $$

(40)

By applying the chain rule we obtain:

$$ {\frac{{dp_{d}^{a} }}{ds}} = {\frac{{dp_{d}^{a} }}{{da_{d} }}}\,{\frac{{da_{d} }}{ds}}. $$

(41)

Substituting Eq. (40) into (41) yields:

$$ {\frac{{da_{d} }}{ds}} = {\frac{{ - p_{d}^{a} }}{{\bar{\alpha }_{d} }}}\left[ {{\frac{{dp_{d}^{a} }}{{da_{d} }}} + {\frac{{\left( {1 - s} \right)}}{{\bar{\alpha }_{d}^{2} }}}\,{\frac{{dp_{d}^{q} }}{{dq_{d} }}}} \right]^{ - 1}. $$

(9)

Given that q and a are normal goods, we then see that da _d /ds < 0 for s < 1.

1.3 Show that dy _d /ds > 0.

According to Eq. (4) in the main text it holds that:

$$ K_{d} = \int {p_{d}^{a} \cdot da_{d} } + \left( {1 - s} \right)\int {p_{d}^{q} \cdot dq_{d} }. $$

(42)

Taking the total derivative yields:

$$ {\frac{{dK_{d} }}{ds}} = {\frac{{da_{d} }}{ds}}p_{d}^{a} + \left( {1 - s} \right)\,{\frac{{dq_{d} }}{ds}}p_{d}^{q} - \int {p_{d}^{q} \cdot dq_{d} }. $$

(43)

Application of the chain rule and substituting from Eqs. (3) and (36) yields:

$$ {\frac{{dK_{d} }}{ds}} = - \bar{\alpha }_{d} \cdot p_{d}^{a} {\frac{{dq_{d} }}{ds}} + \bar{\alpha }_{d} \cdot p_{d}^{a} {\frac{{dq_{d} }}{ds}} - \int {p_{d}^{q} \cdot dq_{d} }. $$

(44)

Hence, we get:

$$ {\frac{{dK_{d} }}{ds}} = - \int {p_{d}^{q} \cdot dq_{d} }. $$

(10)

Thus dK _d /ds < 0 for all s. Given the budget constraint, we know that dy _d /ds > 0 for all s.

1.1.2 Case 2: Fixed a

Starting with Eq. (17) of the main text and rearranging yields:

$$ p_{d}^{q} = {\frac{{\bar{\alpha }_{d} \cdot MRS_{z,y} }}{{\left( {1 - s} \right)}}}. $$

(45)

Taking the total derivative yields:

$$ {\frac{{dp_{d}^{a} }}{ds}} = {\frac{{\bar{\alpha }_{d} \cdot MRS_{z,y} }}{{\left( {1 - s} \right)^{2} }}} + {\frac{{\bar{\alpha }_{d} }}{{\left( {1 - s} \right)}}}\,{\frac{{dMRS_{z,y} }}{{dq_{d} }}}\,{\frac{{dq_{d} }}{ds}}. $$

(46)

By substituting from Eq. (39) we get:

$$ {\frac{{dq_{d} }}{ds}} = {\frac{{\bar{\alpha }_{d} \cdot MRS_{z,y} }}{{\left( {1 - s} \right)^{2} }}}\left[ {{\frac{{dp_{d}^{q} }}{{dq_{d} }}} - {\frac{{\bar{\alpha }_{d} }}{{\left( {1 - s} \right)}}}\,{\frac{{dMRS_{z,y} }}{{dq_{d} }}}} \right]^{ - 1} $$

(18)

As such, we see that dq _d /ds > 0 if dMRS _z,y /dq _d  < 0 which is the case under our assumptions for the utility function with fixed a _d.

1.1.3 Case 3: Standard optimization

1.1 Show that dq/ds > 0.

Here we can adopt Eq. (18) and its derivation from the previous case, and adjusting it for variable a and q in the utility function yields:

$$ {\frac{{dq_{d} }}{ds}} = {\frac{{\bar{\alpha }_{d} \cdot MRS_{z,y} }}{{\left( {1 - s} \right)^{2} }}}\left[ {{\frac{{dp_{d}^{q} }}{{dq_{d} }}} - {\frac{{\bar{\alpha }_{d} }}{{\left( {1 - s} \right)}}}\,{\frac{{dMRS_{z,y} }}{{dz_{d} }}}\,{\frac{{dz_{d} }}{{dq_{d} }}}} \right]^{ - 1}. $$

(27)

dq _d /ds > 0 if dMRS _z,y /dz < 0, which is the case under our assumptions for the utility function.

1.2 Derivation of da/ds.

By taking the total derivative of (3) we obtain:

$$ {\frac{{dp_{d}^{a} }}{ds}} = {\frac{1}{{\bar{\alpha }_{d} }}}\left[ {\left( {1 - s} \right){\frac{{dp_{d}^{q} }}{{dq_{d} }}}\,{\frac{{dq_{d} }}{ds}} - p_{d}^{q} } \right]. $$

(47)

Substitution in Eq. (41) and rearranging yields:

$$ {\frac{{da_{d} }}{ds}} = {\frac{{\left( {1 - s} \right){\frac{{dp_{d}^{q} }}{{dq_{d} }}}\,{\frac{{dq_{d} }}{ds}} - p_{d}^{q} }}{{\bar{\alpha }_{d} {\frac{{dp_{d}^{a} }}{da}}}}}. $$

(29)

Thus, da _d /ds < 0, if pq _d  > (1 − s)·dpq _d /dq _d ·dq _d /ds.