Windfall Profits Under Pollution Permits and Output-Based Allocation

Abstract

This paper analyzes the implementation of pollution permits. It focuses on the distributional impacts linked with the stringency of output-based allocation when two sectors are covered by the market for permits and the total cap is held constant. A new type of profit increase in sectors that are not exposed to international competition, when energy-intensive trade-exposed (EITE) sectors are granted output-based allocations, is demonstrated theoretically. The paper also illustrates a profit increase in the electricity sector in a possible fourth phase of the European Union Emission Trading Scheme, in which output-based allocation will be granted to EITE sectors, compared with the case in which all permits are auctioned.

Appendices

Appendix 1: Table of Quantities, Emissions and Prices in Each Sector

In Tables 8 and 9, we present the individual and total quantities, the emissions rates and the market prices for each sector. Sector A is a non-exposed sector, and calculations in this sector are used in the illustration to reflect the electricity sector. Sector B is an exposed sector, and calculations in this sector are used in the illustration to reflect the steel and cement sectors.

Table 9 Quantities, emissions rate and prices in sector B

Appendix 2: Condition to Ensure the Existence of Bilateral Trade

Now, we present the condition that ensures the existence of bilateral trade in sector B. We need to ensure that $q_i^{BF}(\sigma =0)>0$. Moreover, from Eq. (11), we know the following:

$$\begin{aligned} q_i^{BF}(\sigma =0)=\alpha ^B\Big (\frac{1}{\beta ^B}\Big )^{\beta ^B}\frac{((n^{BF}+n^{BH})\beta ^B-1)^{\beta ^B} \left( \beta ^B n^{BH} c^{BH}+(1-n^{BH} \beta ^B)(c^{BF}+\tau )\right) }{(n^{BH} c^{BH}+n^{BF} (c^{BF}+\tau ))^{(\beta ^B+1)}}. \end{aligned}$$

(31)

Thus, the condition that ensures the existence of bilateral trade in sector B is given as follows:

$$\begin{aligned} \alpha ^B\Big (\frac{1}{\beta ^B}\Big )^{\beta ^B}\frac{((n^{BF}+n^{BH})\beta ^B-1)^{\beta ^B} \left( \beta ^B n^{BH} c^{BH}+(1-n^{BH} \beta ^B)(c^{BF}+\tau )\right) }{(n^{BH} c^{BH}+n^{BF} (c^{BF}+\tau ))^{(\beta ^B+1)}}>0, \end{aligned}$$

(32)

From Assumption 1, $(n^{BF}+n^{BH})\beta ^B-1>0$. We thus deduce that the condition is equivalent to

$$\begin{aligned} \beta ^B n^{BH} c^{BH}+(1-n^{BH} \beta ^B)(c^{BF}+\tau )>0, \end{aligned}$$

(33)

The condition may be rewritten as follows:

$$\begin{aligned} \tau <\left( \frac{\beta ^Bn^{BH}c^{BH}}{n^{BH}\beta ^B-1} -c^{BF}\right) . \end{aligned}$$

(34)

Appendix 3: Condition to Ensure that the Optimum is a Maximum

To ensure profit maximization in sector j, we analyze the conditions in which the Hessian matrix of the maximization problem is negative.

Let us define the Hessian as follows:

$$\begin{aligned} \frac{\partial ^2 \pi ^j}{\partial q_i^{j2} } \frac{\partial ^2 \pi ^j}{\partial \mu _i^{j2}} - \left( \frac{\partial ^2 \pi ^j}{\partial q_i^j \partial \mu _i^j} \right) ^2 <0, \end{aligned}$$

(35)

Therefore, the requirement on the cost function to satisfy the second-order conditions is as follows:

$$\begin{aligned} (P''^j q_i^j+2P'^j)\gamma q_i^j +(-\sigma +\gamma ^j(\mu _0^j-\mu _i^j))^2>0. \end{aligned}$$

(36)

Appendix 4: Impact of Output-Based Allocation on Profits in the Non-exposed Sector

The profit of firm i in sector A is given as follows:

$$\begin{aligned} \pi _i^A(q_i^A,\mu _i^A)&= P^A(Q^A)q_i^A - (c_i^A + \mu _i^A\sigma )q_i^A - \gamma ^A(\mu _0^A-\mu _i^A)^2 \frac{q_i^A}{2}. \end{aligned}$$

(37)

The first-order conditions satisfy the following:

$$\begin{aligned}&\frac{\partial \pi _i^A}{\partial q_i^A}= \frac{\partial P^A}{\partial Q^A}q_i^A + P^A(Q^A) - c_i^A-\mu _i^A\sigma - \frac{\gamma ^A}{2}(\mu _0^A-\mu _i^A)^2=0. \end{aligned}$$

(38)

$$\begin{aligned}&\frac{\partial \pi _i^{A}}{\partial \mu _i^{A}} = -\sigma q_i^{A} + \gamma ^A q_i^{A} (\mu _0 ^{A} -\mu _i^{A})= 0. \end{aligned}$$

(39)

From Eq. (39), we obtain

$$\begin{aligned} \mu _i^A = \mu _0^A - \frac{\sigma }{\gamma ^A}. \end{aligned}$$

(40)

Taking the derivative of Eq. (38) with respect to $\sigma$, we obtain

$$\begin{aligned} \frac{\partial ^2 P^A}{\partial {Q^A}^2}\frac{\partial Q^A}{\partial \sigma }q_i^A+\frac{\partial P^A}{\partial q_i^A}\frac{\partial q_i^A}{\partial \sigma }+\frac{\partial P^A}{\partial Q^A} \frac{\partial Q^A}{\partial \sigma }-\mu _i^A=&0, \end{aligned}$$

(41)

$$\begin{aligned} \frac{\partial P^A}{\partial Q^A} \frac{\partial Q^A}{\partial \sigma }\left( 1+\frac{{\partial ^2 P^A}/{\partial {Q^A}^2}q_i^A}{{\partial P^A}/{\partial Q^A}}\right) +\frac{\partial P^A}{\partial q_i^A}\frac{\partial q_i^A}{\partial \sigma }-\mu _i^A=&0. \end{aligned}$$

(42)

We focus on the term $\left( 1+\frac{({\partial ^2 P^A}/{\partial {Q^A}^2})q_i^A}{{\partial P^A}/{\partial Q^A}}\right)$.

Because $P^A(Q^A)={\alpha ^A}^{\frac{1}{\beta ^A}}{Q^A}^{-\frac{1}{\beta ^A}}$, we calculate the following expression:

$$\begin{aligned}&1+\frac{({\partial ^2 P^A}/{\partial {Q^A}^2})q_i^A}{{\partial P^A}/{\partial Q^A}}=1+\frac{{\alpha ^A}^{\frac{1}{\beta ^A}} (-\frac{1}{\beta ^A})(\frac{-\beta ^A-1}{\beta ^A}) {Q^A}^{-\frac{1}{\beta ^A}-2}q_i^A}{{\alpha ^A}^{\frac{1}{\beta ^A}} (-\frac{1}{\beta ^A}) {Q^A}^{-\frac{1}{\beta ^A}-1}}, \end{aligned}$$

(43)

$$\begin{aligned}&1+\frac{({\partial ^2 P^A}/{\partial {Q^A}^2})q_i^A}{{\partial P^A}/{\partial Q^A}}=\frac{n^A\beta ^A+\beta ^A+1}{n^A\beta ^A}. \end{aligned}$$

(44)

From Eq. (42), we obtain

$$\begin{aligned} \frac{\partial P^A}{\partial q_i^A}\frac{\partial q_i^A}{\partial \sigma }=&\mu _i^A-\left( \frac{n^A\beta ^A-\beta ^A-1}{n^A\beta ^A}\right) \frac{\partial P^A}{\partial Q^A} \frac{\partial Q^A}{\partial \sigma }. \end{aligned}$$

(45)

Now, we sum the $n^A$ first-order conditions in sector A and obtain the following:

$$\begin{aligned} n^AP^A +\frac{\partial P^A}{\partial Q^A}Q^A-n^A\left( c_i^A+\mu _i^A\sigma + \frac{\gamma ^A}{2}(\mu _0^A-\mu _i^A)^2\right) =&0, \end{aligned}$$

(46)

Taking the derivative of Eq. (46) with respect to $\sigma$, we obtain the following:

$$\begin{aligned} n^A\frac{\partial P^A}{\partial Q^A}\frac{\partial Q^A}{\partial \sigma }+ \frac{\partial ^2 P^A}{\partial {Q^A}^2} \frac{\partial Q^A}{\partial \sigma }Q^A+\frac{\partial P^A}{\partial Q^A} \frac{\partial Q^A}{\partial \sigma }-n^A\mu _i^A=&0, \end{aligned}$$

(47)

$$\begin{aligned} \left( n^A+1+\frac{({\partial ^2 P^A}/{\partial {Q^A}^2}) Q^A}{{\partial P^A}/{\partial Q^A}}\right) \frac{\partial P^A}{\partial Q^A}\frac{\partial Q^A}{\partial \sigma }-n^A\mu _i^A=&0. \end{aligned}$$

(48)

Because $P^A(Q^A)={\alpha ^A}^{\frac{1}{\beta ^A}} {Q^A}^{-\frac{1}{\beta ^A}}$, we calculate the following expression:

$$\begin{aligned}&n^A+1+\frac{({\partial ^2 P^A}/{\partial {Q^A}^2}) Q^A}{{\partial P^A}/{\partial Q^A}}=n^A+1+\frac{{\alpha ^A}^{\frac{1}{\beta ^A}} (-\frac{1}{\beta ^A})(\frac{-\beta ^A-1}{\beta ^A}) {Q^A}^{-\frac{1}{\beta ^A}-2}Q^A}{{\alpha ^A}^{\frac{1}{\beta ^A}} (-\frac{1}{\beta ^A}) {Q^A}^{-\frac{1}{\beta ^A}-1}}, \end{aligned}$$

(49)

$$\begin{aligned}&n^A+1+\frac{({\partial ^2 P^A}/{\partial {Q^A}^2}) Q^A}{{\partial P^A}/{\partial Q^A}}=n^A-\frac{1}{\beta ^A}. \end{aligned}$$

(50)

The Eq. (48) may then be rewritten as follows:

$$\begin{aligned} \frac{\partial P^A}{\partial Q^A} \frac{\partial Q^A}{\partial \sigma } =\frac{n^A\mu _i^A}{(n^A-\frac{1}{\beta ^A})}. \end{aligned}$$

(51)

Because we analyze the effect of the permit price on firm profits, we study the derivative of the function $\pi _{i}^{A}$ with respect to $\sigma$. From Eq. (37), we obtain

$$\begin{aligned} \frac{\partial \pi _{i}^A}{\partial \sigma } ={q_i}^A\left( \frac{\partial P^A}{\partial Q^A} \frac{\partial Q^A}{\partial \sigma }- \frac{\partial P^A}{\partial q_i^A}\frac{\partial q_i^A}{\partial \sigma }\right) - \mu _i^A q_i^A. \end{aligned}$$

(52)

By replacing (45) and (51) in (52), we obtain

$$\begin{aligned} \frac{\partial \pi _{i}^A}{\partial \sigma } =q_i^A \mu _i^A\left( \frac{1-\beta ^A}{n^A\beta ^A-1} \right) . \end{aligned}$$

(53)

Appendix 5: Proof of Lemma 3

Let $CS^A$ be the consumer surplus in sector A:

$$\begin{aligned} CS^A&=\int _0^{Q^A}{P(y)dy}-P^A Q^A,\end{aligned}$$

(54)

$$\begin{aligned}&=\int _0^{Q^A}{{\alpha ^A}^{\frac{1}{\beta ^A}} y^{-\frac{1}{\beta ^A}} dy}-P^A Q^A,\end{aligned}$$

(55)

$$\begin{aligned}&=\left[ \left( \frac{{\alpha ^A}^{\frac{1}{\beta ^A}} \beta ^A}{\beta ^A-1}\right) {y}^{\frac{\beta ^A-1}{\beta ^A}}\right] _0^{Q^A}- {\alpha ^A}^{\frac{1}{\beta ^A}} {Q^A}^\frac{\beta ^A-1}{\beta ^A}. \end{aligned}$$

(56)

When $\beta ^A>1$, the consumer surplus is given as follows:

$$\begin{aligned} CS^A = \left( \frac{{\alpha ^A}^{\frac{1}{\beta ^A}}}{\beta ^A-1}\right) {Q^A}^{\frac{\beta ^A-1}{\beta ^A}}. \end{aligned}$$

(57)

However, when $\beta ^A<1$, the consumer surplus is given as follows:

$$\begin{aligned} CS^A =\left( \frac{{\alpha ^A}^{\frac{1}{\beta ^A}} \beta ^A}{\beta ^A-1}\right) \lim _{Q^A \rightarrow 0}\frac{1}{{Q^A}^\frac{1-\beta ^A}{\beta ^A}}+ \left( \frac{{\alpha ^A}^{\frac{1}{\beta ^A}}}{\beta ^A-1}\right) {Q^A}^{\frac{\beta ^A-1}{\beta ^A}}. \end{aligned}$$

(58)

Recall that $Q^A$ do not depend on $\omega$. In the two cases, we obtain the following:

$$\begin{aligned} \frac{\partial CS^A}{\partial \omega }&=\underbrace{\left( \frac{{\alpha ^A}^{\frac{1}{\beta ^A}}}{\beta ^A}\right) {Q^A}^{-\frac{1}{\beta ^A}}}_{(+)}\underbrace{\left[ \frac{\partial Q^A}{\partial \sigma }\frac{\partial \sigma }{\partial \omega }\right] }_{(?)}. \end{aligned}$$

(59)

We know from Lemma 2 that $\frac{\partial \sigma }{\partial \omega }>0$. Thus, the sign of $\left[ \frac{\partial Q^A}{\partial \sigma }\frac{\partial \sigma }{\partial \omega }\right]$ is given by the sign of $\frac{\partial Q^A}{\partial \sigma }$. We now calculate $\frac{\partial Q^A}{\partial \sigma }$ and obtain the following:

$$\begin{aligned} \frac{\partial Q^A}{\partial \sigma }&=-\beta ^A\left[ \frac{[{\alpha ^A}^{\frac{1}{\beta ^A}}(1-\frac{1}{n^A\beta ^A})]^{\beta ^A}}{[c_i^A+\mu _0^A\sigma -\frac{\sigma ^2}{2\gamma ^A}]^{(\beta ^A+1)}}\right] (\mu _0^A -\frac{\sigma }{\gamma ^A}). \end{aligned}$$

(60)

Knowing that $Q^A(\sigma )>0$ we can deduce that

$$\begin{aligned} \left[ \frac{[{\alpha ^A}^{\frac{1}{\beta ^A}}(1-\frac{1}{n^A\beta ^A})]^{\beta ^A}}{[c_i^A+\mu _0^A\sigma -\frac{\sigma ^2}{2\gamma ^A}]^{(\beta ^A+1)}}\right] >0. \end{aligned}$$

(61)

Moreover, we know that $\mu _i^A>0$. From Eq. (14), we obtain

$$\begin{aligned} \mu _0^A-\frac{\sigma }{\gamma ^A}>0. \end{aligned}$$

(62)

From Eqs. (60), (61) and (62), we can easily ascertain that $\frac{\partial Q^A}{\partial \sigma }<0$; finally, from Eq. (59), we find that $\frac{\partial CS^A}{\partial \omega }<0$.

Appendix 6: Proof of Lemma 4

We define the consumer surplus in sector B as follows:

$$\begin{aligned} CS^B&=\int _0^{Q^B}{P(y)dy}-P^B Q^B, \end{aligned}$$

(63)

$$\begin{aligned}&=\int _0^{Q^B}{{\alpha ^B}^{\frac{1}{\beta ^B}} y^{-\frac{1}{\beta ^B}} dy}-P^B Q^B, \end{aligned}$$

(64)

$$\begin{aligned}&=\left[ \left( \frac{{\alpha ^B}^{\frac{1}{\beta ^B}} \beta ^B}{\beta ^B-1}\right) y^{\frac{\beta ^B-1}{\beta ^B}}\right] ^{Q^B}_0-{\alpha ^B}^{\frac{1}{\beta ^B}}{Q^B}^\frac{\beta -1}{\beta }. \end{aligned}$$

(65)

When $\beta ^B>1$,

$$\begin{aligned} CS^B =\left( \frac{{\alpha ^B}^{\frac{1}{\beta ^B}}}{\beta ^B-1}\right) {Q^B}^{\frac{\beta ^B-1}{\beta ^B}}. \end{aligned}$$

(66)

However, when $\beta ^B<1$,

$$\begin{aligned} CS^B =\left( \frac{{\alpha ^B}^{\frac{1}{\beta ^B}} \beta ^B}{\beta ^B-1}\right) \lim _{Q^B \rightarrow 0}\frac{1}{{Q^B}^\frac{1-\beta ^B}{\beta ^B}}+\left( \frac{{\alpha ^B}^{\frac{1}{\beta ^B}}}{\beta ^B-1}\right) {Q^B}^{\frac{\beta ^B-1}{\beta ^B}}. \end{aligned}$$

(67)

In the two cases, we obtain

$$\begin{aligned} \frac{\partial CS^B}{\partial \omega }=\underbrace{\left( \frac{{\alpha ^B}^{\frac{1}{\beta ^B}}}{\beta ^B}\right) {Q^B}^{\frac{-1}{\beta ^B}}}_{(+)}\underbrace{\left[ \frac{\partial Q^B}{\partial \omega }\right] }_{(?)}. \end{aligned}$$

(68)

Now, we focus on and calculate the term $\frac{\partial Q^B}{\partial \omega }$.

$$\begin{aligned} \frac{\partial Q^B}{\partial \omega }= & {} -\beta ^B\frac{\left( \frac{{\alpha ^B}^{\frac{1}{\beta ^B}}}{\beta ^B}\right) ^{\beta ^B}[(n^{BF}+n^{BH})\beta ^B-1]^{\beta ^B}}{[n^{BH}(c^{BH}+(\mu _i^{BH}-\omega \mu _0^{BH})\sigma )+n^{BF}(c^{BF}+\tau )]^{\beta ^B+1}}\nonumber \\&\times \left[ \frac{\partial [(\mu _i^{BH}-\omega \mu _0^{BH})\sigma ]}{\partial \omega }\right] . \end{aligned}$$

(69)

Knowing that $Q^B(\sigma )>0$, we can deduce the following:

$$\begin{aligned} \left( \frac{\left( \frac{{\alpha ^B}^{\frac{1}{\beta ^B}}}{\beta ^B}\right) ^{\beta ^B}[(n^{BF}+n^{BH})\beta ^B-1]^{\beta ^B}}{[n^{BH}(c^{BH}+(\mu _i^{BH}-\omega \mu _0^{BH})\sigma )+n^{BF}(c^{BF}+\tau )]^{\beta ^B+1}}\right) >0. \end{aligned}$$

(70)

Now, we focus on and calculate the term $\frac{\partial [(\mu _i^{BH}-\omega \mu _0^{BH})\sigma ]}{\partial \omega }$.

$$\begin{aligned} \frac{\partial [(\mu _i^{BH}-\omega \mu _0^{BH})\sigma ]}{\partial \omega }=-\mu _0^{BH}\sigma +(1-\omega )\mu _0^{BH}\frac{\partial \sigma }{\partial \omega } -\frac{2\sigma }{\gamma ^B} \frac{\partial \sigma }{\partial \omega }. \end{aligned}$$

(71)

Assuming $\sigma >\frac{\partial \sigma }{\partial \omega }$, we obtain $\mu _0^{BH}\sigma >\mu _0^{BH}\frac{\partial \sigma }{\partial \omega }$, which implies $\mu _0^{BH}\sigma >(1-\omega )\mu _0^{BH}\frac{\partial \sigma }{\partial \omega }$. Moreover, we know that $\frac{\partial \sigma }{\partial \omega }>0$. We deduce the following:

$$\begin{aligned} \frac{\partial [(\mu _i^{BH}-\omega \mu _0^{BH})\sigma ]}{\partial \omega }&=-\mu _0^{BH}\sigma +(1-\omega )\mu _0^{BH}\frac{\partial \sigma }{\partial \omega } -\frac{2\sigma }{\gamma ^B} \frac{\partial \sigma }{\partial \omega }<0. \end{aligned}$$

(72)

From Eqs. (69), (70) and (72), we can easily ascertain that $\frac{\partial Q^B}{\partial \omega }>0$; finally, from Eq. (68), we find that $\frac{\partial CS^B }{\partial \omega }>0$.

Appendix 7: Proof of Lemma 5

We show that

$$\begin{aligned} \frac{\partial \pi _i^{BH}}{\partial \omega }>0. \end{aligned}$$

(73)

The profit for a domestic firm i in sector B is given as follows:

$$\begin{aligned} \pi _i^{BH}=\left[ P^B-c_i^{BH}-(\mu _i^{BH}-\omega _0\mu ^{BH})\sigma -\frac{\gamma ^B}{2}(\mu _0^{BH}-\mu _i^{BH})^2\right] q^{BH}_i. \end{aligned}$$

(74)

Taking the derivative with respect to $\omega$,

$$\begin{aligned} \frac{\partial \pi _i^{BH}}{\partial \omega }= & {} \left[ \frac{\partial P^B}{\partial \omega }-\frac{\partial ((\mu _i^{BH}-\omega _0\mu ^{BH})\sigma )}{\partial \omega }-\frac{\sigma }{\gamma ^B}\right] q_i^{BH} \nonumber \\&+\left[ P^B-c_i^{BH}-(\mu _i^{BH}-\omega _0\mu ^{BH})\sigma -\frac{\gamma ^B}{2}(\mu _0^{BH}-\mu _i^{BH})^2\right] \frac{\partial q_i^{BH} }{\partial \omega }. \end{aligned}$$

(75)

The mark-up is positive, and we can deduce that

$$\begin{aligned} \left[ P^B-c_i^{BH}-(\mu _i^{BH}-\omega _0\mu ^{BH})\sigma -\frac{\gamma ^B}{2}(\mu _0^{BH}-\mu _i^{BH})^2\right] >0. \end{aligned}$$

(76)

Now, we focus on the term $\frac{\partial q_i^{BH} }{\partial \omega }$.

$$\begin{aligned} \frac{\partial q_i^{BH} }{\partial \omega }=\frac{(\frac{{\alpha ^B}^{\frac{1}{\beta ^B}}}{\beta ^B})^{\beta ^B}((n^{BF}+n^{BH})\beta ^B-1)^{\beta ^B} f(\sigma )}{(n^{BH} (c^{BH}+ (\mu _i^{BH}-\omega \mu _0^{BH})\sigma )+n^{BF} (c^{BF}+\tau ))^{(\beta ^B+2)}}\frac{\partial [(\mu _i^{BH}-\omega _0\mu ^{BH})\sigma ]}{\partial \omega }, \end{aligned}$$

(77)

where $f(\sigma )$ is given by

$$\begin{aligned} f(\sigma )= & {} (c^{BH}+(\mu _i^{BH}-\omega _0\mu ^{BH})\sigma )(n^{BH}\beta ^B(n^{BF}\beta -1))+n^{BF}(c^{BF}+\tau )(1\nonumber \\&-(n^{BH}+n^{BF})\beta ^B-n^{BH}{\beta ^B}^2). \end{aligned}$$

(78)

Knowing that $q^{BH}(\sigma )>0$, we can deduce the following:

$$\begin{aligned} \frac{(\frac{{\alpha ^B}^{\frac{1}{\beta ^B}}}{\beta ^B})^{\beta ^B}((n^{BF}+n^{BH})\beta ^B-1)^{\beta ^B} f(\sigma )}{(n^{BH} (c^{BH}+ (\mu _i^{BH}-\omega \mu _0^{BH})\sigma )+n^{BF} (c^{BF}+\tau ))^{(\beta ^B+2)}}>0. \end{aligned}$$

(79)

As shown previously, we focus on and calculate the term $\frac{\partial [(\mu _i^{BH}-\omega \mu _0^{BH})\sigma ]}{\partial \omega }$.

$$\begin{aligned} \frac{\partial [(\mu _i^{BH}-\omega \mu _0^{BH})\sigma ]}{\partial \omega }&=-\mu _0^{BH}\sigma +(1-\omega )\mu _0^{BH}\frac{\partial \sigma }{\partial \omega } -\frac{2\sigma }{\gamma ^B} \frac{\partial \sigma }{\partial \omega }. \end{aligned}$$

(80)

Assuming $\sigma >\frac{\partial \sigma }{\partial \omega }$, we obtain $\mu _0^{BH}\sigma >\mu _0^{BH}\frac{\partial \sigma }{\partial \omega }$, which implies $\mu _0^{BH}\sigma >(1-\omega )\mu _0^{BH}\frac{\partial \sigma }{\partial \omega }$. Moreover, we know that $\frac{\partial \sigma }{\partial \omega }>0$. We finally deduce the following:

$$\begin{aligned} \frac{\partial [(\mu _i^{BH}-\omega \mu _0^{BH})\sigma ]}{\partial \omega }&=-\mu _0^{BH}\sigma +(1-\omega )\mu _0^{BH}\frac{\partial \sigma }{\partial \omega } -\frac{2\sigma }{\gamma ^B} \frac{\partial \sigma }{\partial \omega }<0. \end{aligned}$$

(81)

From Eqs. (77), (79) and (81), we obtain

$$\begin{aligned} \frac{\partial q^{BH}}{\partial \omega }>0. \end{aligned}$$

(82)

Now, we analyze $\frac{\partial P^B}{\partial \omega }-\frac{\partial ((\mu _i^{BH}-\omega _0\mu ^{BH})\sigma )}{\partial \omega }-\frac{\sigma }{\gamma ^B}$, which we denote with $\phi (\sigma )$.

From $P^B= \beta ^B \frac{(n^{BH}(c^{BH}+(\mu _i^{BH}-\omega \mu _0^{BH})\sigma )+ n^{BF}(c^{BF}+\tau ))}{((n^{BF}+n^{BH})\beta ^B -1)}$, we calculate

$$\begin{aligned} \frac{\partial P^B}{\partial \omega }=\frac{n^{BH}\beta ^B}{(n^{BH}+n^{BF})\beta ^B -1}\frac{\partial ((\mu _i^{BH}-\omega _0\mu ^{BH})\sigma )}{\partial \omega }. \end{aligned}$$

(83)

We can rewrite $\phi (\sigma )$ as follows:

$$\begin{aligned} \phi (\sigma )=&\left( \frac{n^{BH}\beta ^B}{(n^{BH}+n^{BF})\beta ^B -1} -1\right) \frac{\partial ((\mu _i^{BH}-\omega _0\mu ^{BH})\sigma )}{\partial \omega } -\frac{\sigma }{\gamma ^B}, \end{aligned}$$

(84)

$$\begin{aligned} \phi (\sigma )=&-\left( \frac{n^{BF}\beta ^B+1}{(n^{BH}+n^{BF})\beta ^B -1}\right) \frac{\partial ((\mu _i^{BH}-\omega _0\mu ^{BH})\sigma )}{\partial \omega } -\frac{\sigma }{\gamma ^B}. \end{aligned}$$

(85)

By replacing Eq. (80) in (85), we obtain

$$\begin{aligned} \phi (\sigma )= \lambda \left[ -\mu _0^{BH}\sigma +(1-\omega )\mu _0^{BH}\frac{\partial \sigma }{\partial \omega }-\frac{\sigma }{\gamma ^B}\frac{\partial \sigma }{\partial \omega }\right] -\lambda \frac{\sigma }{\gamma ^B}\left( \frac{\partial \sigma }{\partial \omega }+1\right) , \end{aligned}$$

(86)

where $\lambda =-\frac{n^{BF}\beta ^B+1}{(n^{BH}+n^{BF})\beta ^B -1} <0$.

From previous calculations, we know that $-\mu _0^{BH}\sigma +(1-\omega )\mu _0^{BH}\frac{\partial \sigma }{\partial \omega }-\frac{\sigma }{\gamma ^B}\frac{\partial \sigma }{\partial \omega }<0$. From $\frac{\partial \sigma }{\partial \omega }>0$, we can deduce that $\frac{\sigma }{\gamma ^B}\left( \frac{\partial \sigma }{\partial \omega }+1\right) >0$. We thus deduce the following:

$$\begin{aligned} \phi (\sigma )= \lambda \left[ -\mu _0^{BH}\sigma +(1-\omega )\mu _0^{BH}\frac{\partial \sigma }{\partial \omega }-\frac{\sigma }{\gamma ^B}\frac{\partial \sigma }{\partial \omega }\right] -\lambda \frac{\sigma }{\gamma ^B} \left( \frac{\partial \sigma }{\partial \omega }+1\right) >0. \end{aligned}$$

(87)

From Eqs. (77), (76), (82) and (87), we conclude that $\frac{\partial \pi ^{BH}}{\partial \omega }>0$.

Appendix 8: Other Scenarios and Some Variations of the Illustration

The present section analyzes the sensitivity of our illustration according to the percentage of emissions reduction and the market size in the electricity sector.

Various percentages of emissions reductions In Fig. 3, we analyze the cases in which total domestic emissions are reduced by 10 and 20%. We then show that the higher the emissions reduction is, the higher the profit increase is.

Fig. 3

Evolution of profits (M€) in electricity sector for various output-based rates and different levels of emissions reduction ($z=0.95$, $z=0.9$ and $z=0.8$)

Various market sizes in the electricity sector. We analyze the cases in which the market size in the electricity sector is equal to 1000 TWh in Table 10 and 200 TWh in Table 11. Profits in the electricity sector increase by 1.40 and 2.75% (relative to the case under which allowances are auctioned) when the market size is equal to 1000 and 200 TWh, respectively.

Table 10 Illustration of the evolution of permit prices for various output-based rates and under a 1000-TWh market size in the electricity sector

Table 11 Illustration of the evolution of permit prices for various output-based rates and under a 200-TWh market size in the electricity sector