Environmental Liability Law and Induced Technical Change – The Role of Discounting

Abstract

We analyse the incentives of environmental liability law for inducing progress to emission abatement technology. We consider three liability rules: strict liability, a negligence rule with an emission norm as the due care standard, and a double negligence rule which combines the emission standard with an abatement technology norm. In the case of distortive discounting, i.e. where the private discount rate deviates from the social one, we show, how the level of distortion influences the ranking of liability rules, according to the criterion of generated social cost.

Appendices

Appendix A1 – Derivation of dX ₁/dI ₀, dX ₁/dr and dI ₀/dr

Total derivation of (2.2) with respect to I ₀ yields:

$$\frac{dX_1}{dI_0}=-\frac{\partial^2C_1 (X_1, I_0)/\partial X_1 \partial I_0}{\partial ^2C_1 (X_1 ,I_0)/\partial X_1^2 +D^{\prime\prime}(X_1)}\,>\, 0.$$

Total differentiation of (2.2) and (2.3) with respect to r, applying Cramer’s rule and using the first-order conditions, yields:

$$\frac{dX_1}{dr}=\frac{\partial ^2C_1 (X_1 ,I_0)/\partial X_1 \partial I_0}{\left({\partial ^2C_1 (X_1, I_0)/\partial I_0^2 \left({\partial^2C_1(X_1, I_0)/\partial X_1^2 +D^{\prime\prime}(X_1)} \right)-\left({\partial ^2C_1 (X_1,I_0)/\partial X_1 \partial I_0}\right)^2}\right)}\,<\, 0$$

Total differentiation of (2.3) with respect to r (using the first-order conditions) yields:

$$\frac{dI_0}{dr}=-\frac{\partial^2C_1 (X_1, I_0)/\partial X_1 \partial I_0 ({dX_1}/{dr})+1}{\partial^2C_1 (X_1, I_0)/\partial I_0^2}\,<\, 0$$

For a given $X_1 (=\overline{X}_1$ in case of the negligence rule) dX ₁/dr = 0 holds and hence

$$\frac{dI_0}{dr}=-\frac{1}{\partial ^2C_1 (X_1 ,I_0)/\partial I_0^2 }\,<\, 0.$$

That is, the investment level of period 0 decreases under both liability rules, when the discount rate increases.

Appendix A2 – Simple Negligence

1. $\frac{dI_0^\ast}{d\overline{X}_1}$ (if the firm complies with the standard)

Total differentiation of (7) yields $\frac{dI_0^\ast}{d\overline{X}_1}=-{\frac{\partial ^2C_1 (\overline{X}_1,I_0^\ast)}{\partial X_1\partial I_0}} \mathord{\left/ {\vphantom {{\frac{\partial ^2C_1 (\overline{X}_1, I_0^\ast)}{\partial X_1 \partial I_0}} {\frac{\partial ^2C_1 (\overline{X}_1,I_0^\ast)}{\partial I_0^2}}}} \right. \kern-\nulldelimiterspace} {\frac{\partial ^2C_1 (\overline{X}_1,I_0^\ast)}{\partial I_0^2}}\,>\, 0$

2. $I_{_0}^\ast (\overline{X}_1)+\frac{C_1 (\overline{X}_{_1},I_0^\ast (\overline{X}_1))}{1+r^\ast}\,<\, I_{_0}^{\ast {\rm SL}} +\frac{C_1 (X_{_1}^{\ast {\rm SL}}, I_0^{\ast {\rm SL}})+D(X_1^{\ast {\rm SL}})}{1+r^\ast}$ holds for $\overline{X}_1 =X_1^{\ast {\rm SL}} +\varepsilon$ with ɛ → 0

The statement follows from $\lim\limits_{\varepsilon\to 0} I_0^\ast (X_1^{\ast {\rm SL}} +\varepsilon)=I_0^{\ast {\rm SL}}$ and hence

$$\lim_{\varepsilon\to 0} \left({I_{0}^\ast (\overline{X}_1)+\frac{C_1 (\overline{X}_{1},I_0^\ast (\overline{X}_1))}{1+r^\ast}} \right)=I_{0}^{\ast {\rm SL}} +\frac{C_1 (X_{1}^{\ast {\rm SL}}, I_0^{\ast {\rm SL}})}{1+r^\ast}< I_{_0}^{\ast {\rm SL}} +\frac{C_1 (X_{1 }^{\ast {\rm SL}}, I_0^{\ast {\rm SL}})+D(X_1^{\ast {\rm SL}})}{1+r^\ast}.$$

3. $\hbox{SC}^{\rm N}(\overline{X}_1)> \hbox{SC}^{\rm N}(X_1^{\ast {\rm SL}})=\hbox{SC}^{\rm SL}(X_1^{\ast {\rm SL}})\ \forall\ \overline{X}_1< X_1^{\ast {\rm SL}}$

$$\eqalign{ \hbox{SC}^{\rm N}(X_1^{\ast {\rm SL}}, I_0^{\ast {\rm SL}})=\hbox{SC}^{\rm SL}(X_1^{\ast {\rm SL}}, I_0^{\ast {\rm SL}}) =\frac{C_1(X_{_1}^{\ast {\rm SL}}, I_0^{\ast {\rm SL}})}{(1+r^{\ast\ast})} +I_0^{\ast {\rm SL}} +\frac{D(X_1^{\ast {\rm SL}})}{(1+r^{\ast\ast})}\cr =I_0^{\ast {\rm SL}} +\frac{C_1 (X_{_1}^{\ast {\rm SL}}, I_0^{\ast {\rm SL}})}{(1+r^\ast)}+\frac{D(X_1^{\ast {\rm SL}})}{(1+r^\ast)}\cr \quad+\left( {\frac{1}{1+r^{\ast\ast }}-\frac{1}{1+r^\ast}} \right)\left( {C_1 (X_{_1}^{\ast {\rm SL}} ,I_0^{\ast {\rm SL}} )+D(X_1^{\ast {\rm SL}})} \right)\cr \quad<\left( {I_0^\ast (\overline{X}_1)+\frac{C_1 (\overline{X}_{_1},I_0^\ast (\overline{X}_1))}{(1+r^\ast)}+\frac{D(\overline{X}_1)}{(1+r^\ast)}} \right)\cr \quad+\left( {\frac{1}{1+r^{\ast\ast}}-\frac{1}{1+r^\ast}} \right)\left( {C_1 (X_{_1 }^{\ast {\rm SL}} ,I_0^{\ast {\rm SL}})+D(X_1^{\ast {\rm SL}})} \right) }$$

Hence we still have to show $\left( {C_1 (X_{_1}^{\ast {\rm SL}}, I_0^{\ast {\rm SL}} )+D(X_1^{\ast {\rm SL}})} \right)\le \left( {C_1 (\overline{X}_{_1},I_0^\ast (\overline{X}_1))+D(\overline{X}_1)} \right)$

$$\Leftrightarrow I_0^{\ast {\rm SL}} +\frac{C_1 (X_{_1}^{\ast {\rm SL}}, I_0^{\ast {\rm SL}} )+D(X_1^{\ast {\rm SL}})}{1+r^\ast}\le I_0^{\ast {\rm SL}} +\frac{C_1 (\overline{X}_{_1},I_0^\ast (\overline{X}_1))+D(\overline{X}_1)}{1+r^\ast}$$

which follows from

$$\eqalign{ I_0^{\ast {\rm SL}} +\frac{C_1 (X_{_1}^{\ast {\rm SL}} ,I_0^{\ast {\rm SL}})+D(X_1^{\ast {\rm SL}} )}{1+r^\ast}=\min_{(X_1,I_0)} \left( {I_0+\frac{C_1 (X_{1} ,I_0)+D(X_1)}{1+r^\ast}} \right)\cr \quad< I_0^{\ast {\rm SL}} +\frac{C_1 (\overline{X}_{_1},I_0^{\ast {\rm SL}})+D(\overline{X}_1)} {1+r^\ast} \,<\, I_0^{\ast {\rm SL}} +\frac{C_1 (\overline{X}_{1},I_0^\ast (\overline{X}_1))+D(\overline{X}_1)}{1+r^\ast} }$$

4. For $\overline{X}_1 =X_1^{\ast {\rm SL}} +\varepsilon$ with ɛ sufficiently small, $\frac{d{\rm SC}^{\rm N}}{d\overline{X}_1 }\,<\, 0$ holds

Because of the continuity of $\frac{d{\rm SC}^{\rm N}}{d\overline{X}_1}$, it suffices to show that $\frac{d{\rm SC}^{\rm N}}{d\overline{X}_1}\,<\, 0$ holds for $\overline{X}_1 =X_1^{\ast {\rm SL}}$.

$$\frac{d\hbox{SC}^{\rm N}}{d\overline{X}_1} =\frac{dI_0^\ast (\overline{X}_1)}{d\overline{X}_1}+\frac{1}{1+r^{\ast\ast }}\left( {\frac{\partial C_1 (\overline{X}_{_1},I_0^\ast (\overline{X}_1 ))}{\partial X_1}-(1+r^\ast)\frac{dI_0^\ast (\overline{X}_1)}{d\overline{X}_1 }+D^{\prime}(\overline{X}_1)} \right)$$

Hence, for $\overline{X}_1 =X_1^{\ast {\rm SL}}$ we have

$$\frac{d\hbox{SC}^{\rm N}}{d\overline{X}_1}(X_1^{\ast {\rm SL}})=\underbrace {\left( {1-\frac{1+r^\ast}{1+r^{\ast\ast}}} \right)}_{< 0}\underbrace {\frac{dI_0^\ast (X_1^{\ast {\rm SL}})}{d\overline{X}_1}}_{> 0}+\frac{1}{1+r^{\ast \ast}}\underbrace {\left( {\frac{\partial C_1 (X_{_1}^{\ast {\rm SL}} ,I_0^{\ast {\rm SL}})}{\partial X_1}+D^{\prime}(X_1^{\ast {\rm SL}})} \right)}_{=0}\,<\, 0.$$

5. $\overline{X}_1 =X_1^{\ast\ast} \mathop{>}\limits_{(=)} X_1^{\ast {\rm SL}} \Rightarrow I_0^{\ast {\rm SL}} \mathop{<}\limits_{(=)} I_0^\ast (\overline{X}_1)\mathop{<}\limits_{(=)} I_0^{\ast\ast}$ if $r^\ast \mathop{>}\limits_{(=)} r^{\ast\ast}$

$$I_0^\ast \mathop {<}\limits_{(=)} I_0^{\ast\ast} \hbox{ follows from } \frac{\partial C_1 (X_1^{\ast\ast} ,I_0^\ast)}{\partial I_0 }=-(1+r^\ast)\mathop{<}\limits_{(=)} -(1+r^{\ast\ast})=\frac{\partial C_1 (X_1^{\ast\ast} ,I_0^{\ast \ast})}{\partial I_0}.$$

$$I_0^{\ast {\rm SL}} \mathop{<}\limits_{(=)} I_0^\ast\hbox{ follows from } \frac{\partial C_1 (X_1^{\ast\ast}, I_0^\ast)}{\partial I_0}=-(1+r^\ast)=\frac{\partial C_1 (X_1^{\ast {\rm SL}} ,I_0^{\ast {\rm SL}})}{\partial I_0}\mathop{>}\limits_{(=)} \frac{\partial C_1 (X_1^{\ast\ast}, I_0^{\ast {\rm SL}})}{\partial I_0}.$$

6. Comparison of X ^{* N}₁ and X ^**₁

First note that if (8) is violated for $\overline{X}_1 =X_1^{\ast\ast}$ the optimal standard is smaller than X ^**₁ . Thus, we have $X_1^{\ast {\rm N}} < X_1^{\ast\ast}$.

In the following we assume that (8) holds for $\overline{X}_1 =X_1^{\ast\ast}$ and consider $\frac{d{\rm SC}^{\rm N}}{d\overline{X}_1}(X^{\ast \ast})$.

$$\eqalign{ \hbox{SC}(\overline{X}_1)=I_0^\ast (\overline{X}_1)+\frac{C_1 (\overline{X}_1 ,I_0^\ast (\overline{X}_1))+D_1 (\overline{X}_1)}{1+r^{\ast\ast}}\cr \frac{d{\rm SC}}{d\overline{X}_1}(X_1^{\ast\ast})=\underbrace {\frac{dI_0^\ast }{d\overline{X}_1}(X_1^{\ast\ast})}_{> 0}\underbrace {\left( {1+\frac{\frac{\partial C_1}{\partial I_0}(X_1^{\ast\ast} ,I_0^\ast (X_1^{\ast\ast}))}{1+r^{\ast\ast}}} \right)}_{< 0\;(\hbox{see}\;\hbox{below})}\cr \qquad\qquad\quad+\frac{1}{1+r^{\ast\ast }}\underbrace {\left( {\frac{\partial C_1}{\partial X_1}(X_1^{\ast \ast} ,I_0^\ast (X_1^{\ast\ast}))+D^{\prime}(X_1^{\ast\ast})} \right)}_{> 0\;(\hbox{see}\;\hbox{below})} }$$

We have

$$1+\frac{\frac{\partial C_1}{\partial I_0}(X_1^{\ast\ast} ,I_0^\ast (X_1^{\ast\ast})}{1+r^{\ast\ast}})> 1+\frac{\frac{\partial C_1}{\partial I_0}(X_1^{\ast\ast} ,I_0^\ast (X_1^{\ast \ast}))}{1+r^\ast }=0$$

and

$$\frac{\partial C_1}{\partial X_1}(X_1^{\ast\ast}, I_0^\ast (X_1^{\ast\ast}))+D^{\prime}(X_1^{\ast\ast})> \frac{\partial C_1}{\partial X_1 }(X_1^{\ast\ast}, I_0^{\ast \ast})+D^{\prime}(X_1^{\ast\ast})=0.$$

Thus, the first product term is negative and the second positive. Hence, the sign of $\frac{d{\rm SC}}{d\overline{X}_1}(X_1^{\ast\ast})$ cannot be determined on this general level.

Evaluation of typical functional forms reveals that either $\frac{d{\rm SC}}{d\overline{X}_1}(X_1^{\ast\ast})=0$ or $\frac{d{\rm SC}}{d\overline{X}_1}(X_1^{\ast\ast} )> 0$ , and hence either $X_1^{\ast {\rm N}} =X_1^{\ast\ast}$ or $X_1^{\ast {\rm N}} < X_1^{\ast\ast}$ , holds for the tested functions.

E.g., $\frac{d{\rm SC}}{d\overline{X}_1}(X_1^{\ast\ast})=0$ holds for $(C_1 =\frac{1}{2}\frac{\alpha X_1^2}{e^{\beta I_0}},D=\frac{\gamma}{X_1^\delta })$, $(C_1 =\frac{1}{2}\frac{\alpha X_1^3}{e^{\beta I_0}},D=\frac{\gamma }{X_1^\delta})$ or $(C_1 =e^{\alpha X_1 -\beta I_0},D=\frac{\gamma }{X_1^\delta})$ and $\frac{d{\rm SC}}{d\overline{X}_1}(X_1^{\ast\ast})> 0$ holds for $(C_1 =\frac{e^{\alpha X_1}}{(I_0 +1)^\beta},D=\frac{\gamma}{X_1})$.

Appendix A3 – Double Negligence

1. Proposition A3.1 Existence and Uniqueness of $\hat{r}^\ast$

(a) Let $\hat{r}^\ast$ be the smallest discount rate, which satisfies equation (10). Then the firm complies with both standards for all $r^{\ast\ast}< r^\ast<\hat{r}^\ast$.

(b) For all $r^\ast >\hat {r}^\ast$ the private cost of the firm is lower if it ignores the standards than if it respects them.

(c) The critical discount rate $\hat{r}^\ast$ is finite.

Proof: The difference between the firms’ costs if it ignores the standards (PC^SL), and the firms costs if it respects the socially optimal standards ( $\hbox{PC}^{\rm NN}(X_1^{\ast\ast}, I_0^{\ast\ast}))$, is given by

$$\Delta (r^\ast)=\frac{C_1 (X_1^{\ast {\rm SL}} (r^\ast),I_0^{\ast {\rm SL}} (r^\ast ))}{(1+r^\ast)}+I_0^{\ast {\rm SL}} (r^\ast)+\frac{D(X_1^{\ast {\rm SL}} (r^\ast ))}{(1+r^\ast)}-\left[ {\frac{C_1 (X_1^{\ast\ast} ,I_0^{\ast\ast})}{(1+r^\ast)}+I_0^{\ast\ast}} \right].$$

(a) Follows from Δ (r ^**) > 0 and the continuity of Δ.

(b) Total differentiation of Δ with respect to r ^* (taking (2.2) and (4) into account) yields:

$$\eqalign{\frac{d\Delta}{dr^\ast}= -\frac{C_1 (X_1^{\ast {\rm SL}}, I_0^{\ast {\rm SL}} )}{(1+r^\ast)^2}+\frac{\partial C_1 (\cdot)/\partial X_1\frac{dX_1^{\ast {\rm SL}}}{dr^\ast}+\partial C_1 (\cdot)/\partial I_0\frac{dI_0^{\ast {\rm SL}} }{dr^\ast}}{(1+r^\ast)}-\frac{D(X_1^{\ast {\rm SL}})}{(1+r^\ast )^2}\cr +\frac{D^{\prime}(X_1^{\ast {\rm SL}})\frac{dX_1^{\ast {\rm SL}}}{dr^\ast}}{(1+r^\ast )}+\frac{dI_0^{\ast {\rm SL}}}{dr^\ast}+\frac{C_1 (X_1^{\ast\ast} ,I_0^{\ast\ast})}{(1+r^\ast)^2} \cr =\frac{1}{(1+r^\ast)^2}\left( {-C_1 (X_1^{\ast {\rm SL}}, I_0^{\ast {\rm SL}} )-D(X_1^{\ast {\rm SL}})+C_1 (X_1^{\ast\ast}, I_0^{\ast\ast})} \right).}$$

Hence, we have $\frac{d\Delta}{dr^\ast}< 0\Leftrightarrow A< 0$ with $A:=C_1 (X_1^{\ast\ast}, I_0^{\ast\ast})-\left( {C_1 (X_1^{\ast {\rm SL}}, I_0^{\ast {\rm SL}})+D(X_1^{\ast {\rm SL}})} \right)$.

A <  0 means that the costs of the firm in period 1 are lower if it respects the standards than if it ignores them. Since investment costs in period 0 are higher in the case of compliance with the standard, this implies that A < 0 is a necessary condition for compliance with the standards. For $r^\ast =\hat {r}^\ast$ from (9) it follows that Δ = 0 and A < 0 holds. Hence, a marginal increase of the private discount rate ( $r^\ast =\hat {r}^\ast +\varepsilon)$ implies a breach of the standard. A further increase of the discount rate might increase Δ. However, Δ can never become positive, since the necessary condition for the acceptance of the standard (A < 0) is violated in the increasing range of Δ.

(c) Considering the limit $\lim\limits_{r^\ast \to \infty} \Delta =-I_0^{\ast\ast}$ shows that, for a sufficiently large discount rate, the firm is always better off if it violates the two standards. (q.e.d.)

2. Proposition A3.2 Optimal standards in case of $r^\ast>\hat{r}^\ast$

If$r^\ast >\hat {r}^\ast$, for the second best norms$(\overline{X}_1^{\ast\ast}, \overline{I}_0^{\ast\ast})$the following relationships hold:$\overline{X}_1^{\ast\ast}< X_1^{\ast\ast}, \overline{I}_0^{\ast\ast} < I_0^{\ast\ast}$.

Proof: (1) If $r^\ast >\hat {r}^\ast$ the regulator has to choose second best norms $(\overline{X}_{1}^{\ast\ast}, \overline{I}_0^{\ast\ast})$ which minimise social cost under the constraint

(*) $I_{_0}^\ast (\overline{X}_1^{\ast\ast})+\frac{C_1 (\overline{X}_{_1}^{\ast\ast} ,\overline{I}_0^{\ast\ast})}{1+r^\ast}=I_{_0}^{\ast {\rm SL}} +\frac{C_1 (X_{_1}^{\ast {\rm SL}} ,I_0^{\ast {\rm SL}})+D(X_1^{\ast SL})}{1+r^\ast}$.

(2) A necessary condition for (*) is given by $\left( {\overline{X}_1^{\ast\ast } < X_1^{\ast\ast} \;\hbox{or}\;\;\overline{I}_0^{\ast \ast} < I_0^{\ast\ast} } \right)$

(3) First, we show that for the equilibrium pair of standards $\overline{X}_1^{\ast\ast} < X_1^{\ast\ast}\hbox{and }\overline{I}_0^{\ast\ast} \geq I_0^{\ast\ast}$ cannot hold, since (given $\overline{X}_1^{\ast \ast})$ lowering $\overline{I}_0$ reduces private and social cost:

$$\eqalign{ \hbox{SC}^{\rm NN}(\overline{I}_0)=\overline{I}_0+\frac{C_1 (\overline{X}_{_1}^{\ast\ast}, \overline{I}_0)+D(\overline{X}_1^{\ast\ast})}{1+r^{\ast\ast}},\hbox{PC}^{\rm NN}(\overline{I}_0)=\overline{I}_0+\frac{C_1 (\overline{X}_{_1}^{\ast\ast},\overline{I}_0)}{1+r^\ast},\cr \frac{d\hbox{SC}^{\rm NN}(\overline{I}_0^{\ast\ast})}{d\overline{I}_0} =1+\frac{\frac{\partial C_1}{\partial I_0}(\overline{X}_{_1}^{\ast\ast}, \overline{I}_0^{\ast\ast})}{1+r^{\ast\ast}}> 1+\frac{\frac{\partial C_1 }{\partial I_0}(X_{_1}^{\ast\ast}, \overline{I}_0^{\ast\ast})}{1+r^{\ast\ast}}> 1+\frac{\frac{\partial C_1}{\partial I_0}(X_{_1}^{\ast\ast}, I_0^{\ast \ast})}{1+r^{\ast\ast}}=0,\cr \frac{d\hbox{PC}^{\rm NN}(\overline{I}_0^{\ast \ast})}{d\overline{I}_0 }=1+\frac{\frac{\partial C_1}{\partial I_0}(\overline{X}_{_1}^{\ast\ast} ,\overline{I}_0^{\ast \ast})}{1+r^\ast}> 1+\frac{\frac{\partial C_1}{\partial I_0}(X_{_1}^{\ast\ast} ,I_0^{\ast\ast})}{1+r^\ast }> 1+\frac{\frac{\partial C_1}{\partial I_0}(X_{_1}^{\ast\ast}, I_0^{\ast\ast})}{1+r^{\ast\ast}}=0, }$$

(4)

  (4) From (2) and (3) follows $\overline{I}_0^{\ast\ast} < I_0^{\ast\ast}$.

(5) In the following we proof $\overline{X}_1^{\ast\ast} < X_1^{\ast\ast}$ by contradiction, assuming $\overline{X}_1^{\ast\ast} \geq X_1^{\ast\ast }$.

$$(\overline{X}_1^{\ast\ast}, \overline{I}_0^{\ast\ast})=\hbox{argmin}\;\left( {I_0 +\frac{C_1 (X_1 ,I_0)+D(X_1)}{1+r^{\ast\ast}}} \right)\;\hbox{w.r.t.}\;\left( {I_0 +\frac{C_1 (X_1 ,I_0)}{1+r^\ast}=\hbox{PC}^{\rm SL}} \right).$$

Using the Lagrange-approach we get the following first order conditions:

$$\eqalign{1+\frac{\frac{\partial C_1}{\partial I_0}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast})}{1+r^{\ast\ast}}-\lambda \left( {1+\frac{\frac{\partial C_1}{\partial I_0}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast})}{1+r^\ast}} \right)=0 \hbox{ and}\cr \frac{\frac{\partial C_1}{\partial X_1}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast})+D^{\prime}(\overline{X}_1^{\ast\ast})}{1+r^{\ast\ast }}-\lambda \frac{\frac{\partial C_1}{\partial X_1}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast})}{1+r^\ast}=0 }$$

Hence, it follows that:

$$\eqalign{ \frac{1+r^{\ast\ast}+\frac{\partial C_1}{\partial I_0}(\overline{X}_1^{\ast \ast} ,\overline{I}_0^{\ast\ast})}{1+r^\ast +\frac{\partial C_1}{\partial I_0}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast} )}=\frac{\frac{\partial C_1}{\partial X_1}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast})+D^{\prime}(\overline{X}_1^{\ast\ast})}{\frac{\partial C_1 }{\partial X_1}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast})}\cr \Leftrightarrow \frac{r^\ast -r^{\ast\ast}}{1+r^\ast +\frac{\partial C_1 }{\partial I_0}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast} )}=\frac{-D^{\prime}(\overline{X}_1^\ast)}{\frac{\partial C_1}{\partial X_1}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast})} }$$

Left term:

$$\eqalign{\frac{r^\ast -r^{\ast\ast}}{1+r^\ast +\frac{\partial C_1}{\partial I_0 }(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast})}>\frac{r^\ast -r^{\ast \ast}}{1+r^\ast +\frac{\partial C_1}{\partial I_0}(\overline{X}_1^{\ast\ast } ,I_0^{\ast\ast})}>\frac{r^\ast -r^{\ast\ast}}{1+r^\ast +\frac{\partial C_1}{\partial I_0}(X_1^{\ast\ast} ,I_0^{\ast\ast} )}\cr \quad=\frac{r^\ast -r^{\ast\ast}}{1+r^\ast -(1+r^{\ast\ast})}=1}$$

Right term:

$$\frac{-D^{\prime}(\overline{X}_1^\ast)}{\frac{\partial C_1}{\partial X_1}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast\ast})}\le \frac{-D^{\prime}(X_1^{\ast\ast})}{\frac{\partial C_1}{\partial X_1}(\overline{X}_1^{\ast\ast} ,\overline{I}_0^{\ast \ast})}<\frac{-D^{\prime}(X_1^{\ast\ast} )}{\frac{\partial C_1}{\partial X_1}(X_1^{\ast\ast} ,\overline{I}_0^{\ast \ast})}<\frac{-D^{\prime}(X_1^{\ast\ast})}{\frac{\partial C_1}{\partial X_1 }(X_1^{\ast\ast} ,I_0^{\ast\ast})}=1$$

Thus, the assumption $\overline{X}_1^{\ast\ast} \geq X_1^{\ast\ast}$ leads to a contradiction and $\overline{X}_1^{\ast\ast} < X_1^{\ast \ast}$ follows. (q.e.d.)

$3.\ \hbox{SC}^{\rm NN} \,<\, \hbox{SC}^{\rm N}$

Case i: $r^\ast\le \hat{r}^\ast: \hbox{SC}^{\rm NN} =\hbox{SC}^{\ast\ast}\,<\, \hbox{SC}^{\rm N}$

Case ii: $r^\ast>\hat{r}^\ast$: Since (X ^{* NN}₁ , I ^{* N}₀ N) is the unique minimiser of SC(X ₁, I ₀) under the side constraint $A:=\frac{C_1 (X_1, I_0)}{(1+r^*)}+I_0 -\hbox{PC}^{\rm SL}\le 0$ we have to show that $(X_1^{\ast {\rm N}}, I_0^{\ast {\rm N}})\ne (X_1^{\ast {\rm NN}}, I_0^{\ast {\rm NN}})$ with $I_0^{\ast {\rm N}} =I_0^\ast (X_1^{\ast {\rm N}})$ holds.

Case ii.a: $A(X_1^{\ast {\rm N}} ,I_0^{\ast {\rm N}})<0$: Due to the strict convexity of the social cost function, the regulator may decrease SC by choosing a double norm which comes closer to the socially optimal solution than $(X_1^{\ast {\rm N}} ,I_0^{\ast {\rm N}})$. That is, $\exists\varepsilon > 0: \hbox{SC}(\tilde {X}_1 ,\tilde{I}_0)< \hbox{SC}(X_1^{\ast {\rm N}}, I_0^{\ast {\rm N}})\quad \wedge A(\tilde{X}_1, \tilde{I}_0)\le 0$ with $\tilde {X}_1 =(1-\varepsilon)X_1^{\ast {\rm N}} +\varepsilon X_1^{\ast\ast}$ and $\tilde {I}_0 =(1-\varepsilon)I_0^{\ast {\rm N}} +\varepsilon I_0^{\ast\ast}$.

Case ii.b: $A(X_1^{\ast {\rm N}} ,I_0^{\ast {\rm N}})=0$: Consider the function $X_1^A (I_0)$ implicitly defined by $A(X_1^A ,I_0)=\frac{C_1 (X_1^A ,I_0 )}{(1+r^\ast)}+I_0 -\hbox{PC}^{\rm SL}=0$. Total differentiation of A yields:

$$\frac{dX_1^A}{dI_0}=-{\left( {\frac{\partial C_1}{\partial I_0}+1+r^\ast } \right)} \mathord{\left/ {\vphantom {{\left( {\frac{\partial C_1 }{\partial I_0}+1+r^\ast} \right)} {\frac{\partial C_1}{\partial X_1}}}} \right. \kern-\nulldelimiterspace} {\frac{\partial C_1}{\partial X_1}}.$$

Hence

$$\frac{dX_1^A}{dI_0}\left| {_{I_0^{\ast {\rm N}} =I_0^\ast (X_1^{\ast {\rm N}} )}} \right.=0.$$

Further, we have

$$\frac{d\hbox{SC}(X_1^A (I_0),I_0)}{dI_0}=\left( {\frac{1}{1+r^{\ast\ast }}\frac{\partial C_1}{\partial I_0}+1} \right)+\frac{1}{1+r^{\ast \ast }}\left( {\frac{\partial C_1}{\partial X_1}+D^{\prime}} \right)\frac{dX_1^A }{dI_0}.$$

Hence

$$\frac{d\hbox{SC}(X_1^A (I_0^{\ast {\rm N}}),I_0^{\ast {\rm N}})}{dI_0}=\frac{1}{1+r^{\ast\ast}}\frac{\partial C_1}{\partial I_0}+1=-\frac{1+r^\ast}{1+r^{\ast\ast}}+1< 0. \quad \hfill\hbox{ (q.e.d.)}$$

Appendix A4 – Example

We define $C_0 (X_0)=\frac{1}{2}aX_0^2, C_1 (X_1 ,I_0)=\frac{1}{2}aX_1^2 e^{-I_0}$ and $D(X_i)=\frac{c}{X_i}$, and assume $r^{\ast\ast}\le r^\ast<<$> <$>\frac{1}{2}a^{\frac{1}{3}}c^{\frac{2}{3}}-1$ to assure inner solutions.

Evaluating the social optimum reveals $X_0^{\ast\ast} =\left({\frac{c}{a}} \right)^{\frac{1}{3}}$ (which is chosen in all scenarios), $X_1^{\ast\ast} =\frac{c}{2(1+r^{\ast\ast})}$ and $I_0^{\ast\ast} =\hbox{ln}( {\frac{ac^2}{8(1+r^{\ast \ast})^3}} )$.

The equilibrium under the strict liability rule is given by $X_1^{\ast {\rm SL}} =\frac{c}{2(1+r^\ast)}, I_0^{\ast {\rm SL}} =\ln( {\frac{ac^2}{8(1+r^\ast)^3}} )$.

For the simple negligence rule the three step procedure is illustrated through the following:

1. Let $\overline{X}_1$ be the emission norm chosen by the social planner.

Then, from $-(1+r^\ast)=\frac{\partial C_1 (\overline{X}_1, I_0)}{\partial I_0}=-\frac{a}{2}\overline{X}_1^2 e^{-I_0}$ it follows that $I_0^\ast (\overline{X}_1)=\ln({\frac{a}{2}\frac{\overline{X}_1^2}{(1+r^\ast)}} )$.

2. The representative polluter complies with the norm if the following difference is non-negative:

$$\eqalign{ \left({I_{_0}^{\ast {\rm SL}} +\frac{C_1 (X_{_1}^{\ast {\rm SL}} ,I_0^{\ast {\rm SL}} )+D(X_1^{\ast {\rm SL}})}{1+r^\ast}} \right)-\left( {I_{_0}^\ast (\overline{X}_1) +\frac{C_1 (\overline{X}_{_1},I_0^\ast (\overline{X}_1))}{1+r^\ast}} \right) \cr \quad=\ln\left({\frac{c^2}{4(1+r^\ast)^2\overline{X}_1^2}} \right)+2.}$$

Hence, the polluter complies with the norm if $\overline{X}_1\le \overline{X}_1^{\max} =\frac{ce}{2(1+r^\ast)}$.

3. For $\overline{X}_1\le \overline{X}_1^{\max} =\frac{ce}{2(1+r^\ast)}$ the social cost is

$$\hbox{SC}^{\rm N} =\ln\left({\frac{a}{2}\frac{\overline{X}_1^2}{(1+r^\ast)}} \right)+\frac{1}{1+r^{\ast\ast}}\left( {1+r^\ast +\frac{c}{\overline{X}_1}} \right)+K \hbox{ with } K=C_0 (X_0^{\ast \ast})+D(X_0^{\ast\ast})$$

$$\frac{d\hbox{SC}^{\rm N}}{d\overline{X}_1}< 0 \Leftrightarrow \overline{X}_1<\frac{1}{1+r^{\ast\ast}}\frac{c}{2}$$

Hence, the optimal standard is

$$\overline{X}_1^{\ast\ast} =\min\left\{\frac{ce}{2(1+r^\ast )},\frac{c}{2(1+r^{\ast\ast})}\right\}=\left\{\begin{array}{l} \frac{c}{2(1+r^{\ast\ast})} \hbox{if }r^\ast \le e(1+r^{\ast\ast})-1 \cr \frac{ce}{2(1+r^\ast)}\hbox{if } r^\ast > e(1+r^{\ast\ast})-1\cr \end{array}\right.$$

The corresponding investment level is

$$I_0^\ast (\overline{X}_1^{\ast\ast}) =\ln\left( {\frac{a}{2}\frac{\overline{X}_1^2}{(1+r^\ast)}} \right)=\left\{\begin{array}{l} \ln\left(\frac{ac^2}{8(1+r^\ast)(1+r^{\ast\ast})^2}\right) \hbox{if }r^\ast \le e(1+r^{\ast\ast})-1\cr \ln \left(\frac{ac^{2}e^{2}}{8(1+r^\ast)^3}\right)\hbox{if }r^\ast > e(1+r^{\ast\ast})-1\cr \end{array}\right.$$

For the double negligence rule the three step procedure is similar to the one for the simple negligence rule.

1. Let $(\overline{X}_1, \overline{I}_0)$ be the double norm chosen by the social planner.

The polluter complies with the norm if the following difference is non-negative:

$$\eqalign{ A(\overline{X}_1, \overline{I}_0)=\left(I_{0}^{\ast {\rm SL}} +\frac{C_1 (X_{1}^{\ast {\rm SL}}, I_0^{\ast {\rm SL}})+D(X_1^{\ast {\rm SL}})}{1+r^\ast} \right)-\left(\overline{I}_{0}+\frac{C_1(\overline{X}_{1},\overline{I}_0)}{1+r^\ast}\right)\cr =\left({\ln}\left(\frac{ac^2}{8(1+r^{\ast})^3}\right)+3\right)- \left(\overline{I}_{0}+\frac{a\overline{X}_1^2}{2(1+r^{\ast} )e^{\overline{I}_0}}\right). }$$

2. The critical discount rate $\hat{r}^\ast$ is implicitly defined by $\frac{C_1(X_1^{\ast\ast}, I_0^{\ast\ast})}{(1+\hat {r}^\ast)}+I_0^{\ast\ast}=\frac{C_1 (X_1^\ast, I_0^\ast)}{(1+\hat{r}^\ast )}+I_0^\ast +\frac{D(X_1^\ast)}{(1+\hat{r}^\ast)}$, which may be simplified into $A:=3+3\ln(\frac{1+r^{\ast \ast}}{1+\hat{r}^\ast})-\frac{1+r^{\ast\ast}}{1+\hat {r}^\ast}=0$. Since $B(q):=3+3\ln(q)-q$ is strictly increasing in the relevant range $0< q=\frac{1+r^{\ast\ast}}{1+r^\ast}< 1$ and $\lim\limits_{q\to 0} B(q)=-\infty, \lim\limits_{q\to 1} B(q)=2$, there exists a unique zero point of B(q) and hence a unique critical discount rate $\hat{r}^\ast$. E.g., $\hat{r}^\ast= 1.5963$ if r ^** = 0.1.

Since social cost is minimised at $(X_1^{\ast\ast}, I_0^{\ast \ast})$, and $A(X_1^{\ast\ast}, I_0^{\ast \ast})\geq0\Leftrightarrow r^\ast \le \hat {r}^\ast$ holds, the social planner will choose $(\overline{X}_1^{\ast\ast}, \overline{I}_0^{\ast\ast})=(X_1^{\ast\ast}, I_0^{\ast\ast})$ if $r^\ast\le \hat{r}^\ast.$

If $r^\ast>\hat{r}^\ast$, the social planner will choose a pair of standards $(\overline{X}_1^{\ast\ast},\overline{I}_0^{\ast\ast})$ for which $A(\overline{X}_1,\overline{I}_0)=0$ holds. That is, $(*) \overline{X}_1 =\sqrt {\frac{1}{a}\left(\ln\left(\frac{ac^2}{8(1+r^\ast)^3} \right)+3-\overline{I}_{0}\right)2(1+r^\ast)e^{\overline{I}_0}}$.

Thus, the pair of second best standards $(\overline{X}_1^{\ast\ast}, \overline{I}_0^{\ast\ast})$ is implicitly defined by (*) and

$$(**) \frac{d\hbox{SC}(\overline{X}_1 (\overline{I}_0),\overline{I}_0)}{d\overline{I}_0}=0,$$

and may be evaluated for given levels of r ^*.

Appendix A5 – Case r ^* < r ^**

For$\overline{X}_1 =X_1^{\ast {\rm SL}} -\varepsilon$with ɛ sufficiently small$\frac{d{\rm SC}^{\rm N}}{d\overline{X}_1}> 0$holds.

Because of the continuity of $\frac{d{\rm SC}^{\rm N}}{d\overline{X}_1}$ it suffices to show that $\frac{d{\rm SC}^{\rm N}}{d\overline{X}_1}> 0$ holds for $\overline{X}_1 =X_1^{\ast {\rm SL}}$.

$$\frac{d\hbox{SC}^{\rm N}}{d\overline{X}_1}(X_1^{\ast {\rm SL}})=\underbrace {\left( {1-\frac{1+r^\ast}{1+r^{\ast\ast}}} \right)}_{> 0}\underbrace {\frac{dI_0^\ast (X_1^{\ast {\rm SL}})}{d\overline{X}_1}}_{> 0}+\frac{1}{1+r^{\ast\ast}}\underbrace {\left({\frac{\partial C_1 (X_{_1}^{\ast {\rm SL}} ,I_0^{\ast {\rm SL}})}{\partial X_1}+D^{\prime}(X_1^{\ast {\rm SL}})} \right)}_{=0}> 0.$$