On Emissions Trading and Market Structure: Cap-and-Trade versus Intensity Standards

Abstract

This paper examines the interdependence between imperfect competition and emissions trading. We particularly analyze the long run equilibrium in a two-sector (‘clean’ and ‘dirty’) model with Cournot competition among firms who face a fixed cost of production. The clean sector is defined as the sector with the highest long run cost margin on emissions. We compare the welfare implications of a cap-and-trade scheme with an emissions trading scheme based on relative intensity standards. It is shown that a firm’s long run equilibrium output in the clean or dirty sector does not depend on the emissions trading format, but only depends on the fixed cost of producing in the respective sector. Intensity standards can result in clean firms selling allowances to dirty firms, or dirty firms selling to clean firms. The former outcome yields higher welfare. It is demonstrated that cap-and-trade outperforms the intensity-based trading scheme in terms of long run welfare with free entry and exit. With intensity standards the size of the clean sector is too large.

Appendix A

1.1 Proofs

1.1.1 Proposition 2

Proposition 2.1 There are two solutions to Eq. (29) for each sector $i=c,d,$ (13) and (30) which we shall denote by $r$ and $\rho$. Solution $r$ is:

$$\begin{aligned} h_{c}^{r}&= \frac{-\epsilon _{d}\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) \!-\!2\epsilon _{c}(\bar{E}-L)\!+\!\epsilon _{d}\sqrt{ \left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+4L\left( \bar{E}-L\right) }}{2\left( \epsilon _{c}^{2}+\epsilon _{d}^{2}\right) }\quad \end{aligned}$$

(32a)

$$\begin{aligned} h_{d}^{r}&= \frac{\epsilon _{c}\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) -2\epsilon _{d}(\bar{E}-L)-\epsilon _{c}\sqrt{ \left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+4L\left( \bar{E}-L\right) }}{2\left( \epsilon _{c}^{2}+\epsilon _{d}^{2}\right) } \end{aligned}$$

(32b)

$$\begin{aligned} n_{c}^{r}&= \frac{2L\epsilon _{c}+\epsilon _{d}\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}+\sqrt{\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+4L\left( \bar{E}-L\right) }\right) }{ 2f_{c}\left( \epsilon _{c}^{2}+\epsilon _{d}^{2}\right) } \end{aligned}$$

(32c)

$$\begin{aligned} n_{d}^{r}&= \frac{2L\epsilon _{d}+\epsilon _{c}\left( \gamma _{d}\epsilon _{c}-\gamma _{c}\epsilon _{d}-\sqrt{\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+4L\left( \bar{E}-L\right) }\right) }{ 2f_{d}\left( \epsilon _{c}^{2}+\epsilon _{d}^{2}\right) } \end{aligned}$$

(32d)

with $\bar{E}>L$ being the unconstrained emissions given by (11) and $\gamma _{i}$ by (9). We see that $h_{d}^{r}$ in (32b) is negative, i.e., dirty firms are buying allowances so that $n_{d}^{r}<\bar{n}_{d}$ by (10) and (29). Market clearing with $n_{c}^{r},n_{d}^{r}>0$ then requires by (30) that $h_{c}^{r}>0$: clean firms are selling allowances, and $n_{c}^{r}>\bar{n}_{c}$ by (10) and (29).

Solution $\rho$ is:

$$\begin{aligned} h_{c}^{\rho }&= \frac{-\epsilon _{d}\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) -2\epsilon _{c}\left( \bar{E}-L\right) \!-\!\epsilon _{d}\sqrt{\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}\!+\!4L(\bar{E}-L)}}{2\left( \epsilon _{c}^{2}+\epsilon _{d}^{2}\right) }\quad \end{aligned}$$

(33a)

$$\begin{aligned} h_{d}^{\rho }&= \frac{\epsilon _{c}\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) -2\epsilon _{d}\left( \bar{E}-L\right) +\epsilon _{c}\sqrt{\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+4L(\bar{E}-L)}}{2\left( \epsilon _{c}^{2}+\epsilon _{d}^{2}\right) } \end{aligned}$$

(33b)

$$\begin{aligned} n_{c}^{\rho }&= \frac{2L\epsilon _{c}+\epsilon _{d}\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}-\sqrt{\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+4L\left( \bar{E}-L\right) }\right) }{2f_{c}\left( \epsilon _{c}^{2}+\epsilon _{d}^{2}\right) } \end{aligned}$$

(33c)

$$\begin{aligned} n_{d}^{\rho }&= \frac{2L\epsilon _{d}+\epsilon _{c}\left( \gamma _{d}\epsilon _{c}-\gamma _{c}\epsilon _{d}+\sqrt{\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+4L\left( \bar{E}-L\right) }\right) }{2f_{d}\left( \epsilon _{c}^{2}+\epsilon _{d}^{2}\right) }. \end{aligned}$$

(33d)

We see that $h_{c}^{\rho }$ in (33a) is negative, i.e., clean firms are buying allowances so that $n_{c}^{\rho }<\bar{n}_{c}$ by (10) and (29). Market clearing with $n_{c},n_{d}>0$ then requires by (30) that $h_{d}^{\rho }>0$: dirty firms are selling allowances so that $n_{d}^{\rho }> \bar{n}_{d}$ by (10) and (29).$\square$

Proposition 2.2 Substituting (29) and (30) into (31) gives welfare $W^{R}$ under intensity standards:

$$\begin{aligned} W^{R}=\frac{\sum _{i}\left( n_{i}f_{i}\right) ^{2}}{2}. \end{aligned}$$

(34)

Substituting (32c) and (32d) into (34) yields:

$$\begin{aligned} W^{r}=\frac{\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+2L\left( 2L-\bar{E}\right) +\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) \sqrt{\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+4L\left( \bar{E}-L\right) }}{2\left( \epsilon _{c}^{2}+\epsilon _{d}^{2}\right) }. \end{aligned}$$

Substituting (33c) and (33d) into (34) yields:

$$\begin{aligned} W^{\rho }=\frac{\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+2L\left( 2L-\bar{E}\right) -\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) \sqrt{\left( \gamma _{c}\epsilon _{d}-\gamma _{d}\epsilon _{c}\right) ^{2}+4L\left( \bar{E}-L\right) }}{ 2\left( \epsilon _{c}^{2}+\epsilon _{d}^{2}\right) }. \end{aligned}$$

By definition (12), $W^{r}>W^{\rho }$. $\square$

Proposition 2.3 Since $n_{d}^{r}< \bar{n}_{d}$ and $n_{c}^{r}>\bar{n}_{c},$ we have to ensure that $n_{d}^{r}>0$ and $P_{c}(Q_{c}^{r})>0$. From (32d), $n_{d}^{r}>0$ if and only if (14) holds. Maximizing $n_{c}^{r}$ in (32c) with respect to $L$ yields:

$$\begin{aligned} L=\frac{\bar{E}+\epsilon _{c}\sqrt{\gamma _{c}^{2}+\gamma _{d}^{2}}}{2}. \end{aligned}$$

(35)

Substituting (35) into (32c) yields, using Proposition 1:

$$\begin{aligned} Q_{c}^{\max }=n_{c}^{\max }f_{c}=\frac{\gamma _{c}+\sqrt{\gamma _{c}^{2}+\gamma _{d}^{2}}}{2}. \end{aligned}$$

Then $P_{c}(Q_{c}^{\max })>0$ if and only if (15) holds. $\square$

1.1.2 Proposition 3

Maximizing welfare (31) with respect to $n_{i}\ (i=c,d)$ yields:

$$\begin{aligned} (\alpha _{i}-c_{i}-f_{i})f_{i}-n_{i}f_{i}^{2}-\lambda \epsilon _{i}f_{i}=0. \end{aligned}$$

(36)

This is the same condition as the first-order condition under the cap-and-trade regime, substituting (7) and (16) into (4). The shadow price $\lambda$ of emissions in (36) therefore equals the allowance price $v$ in (17), and $n_{i}$ in (36) equals $n_{i}^{A}$ in (18). This means that a cap-and-trade scheme implements the welfare optimum for a given level of total emissions with $q_{i}=f_{i}$. Combining (19) and Proposition 2.1, we find $n_{c}^{A}< \bar{n}_{c}<n_{c}^{r}=n_{c}^{R}$. Combining $n_{c}^{A}<n_{c}^{R}$ with (13) yields $n_{d}^{A}>n_{d}^{R}$. $\square$

1.2 Normalization of the Slope of a Demand Function

The slope of the demand function for a good, when using conventional units for measuring the good as well as for money, is usually different from $-1$. In this appendix we show how to normalize the slopes of the demand functions for two goods (gasoline and coal) to $-1$ by changing the unit of measurement of the respective goods. We leave the money measurement intact, so that consumer surplus from the two goods can still be added together after normalization.

Suppose the inverse demand function for gasoline $(gas)$ is:

$$\begin{aligned} P_{gas}=A-BY_{gas} \end{aligned}$$

with the quantity of gasoline $Y_{gas}$ measured in gallons and its price $P_{gas}$ in dollars per gallon. Thus the units on $A$ and $B$ are “$/gallon” and “$/(gallon)$^{2}$”, respectively. Total revenue is $P_{gas}Y_{gas}$ and units for total revenue are dollars. To normalize the demand function, we first divide both sides by $b\equiv \sqrt{B}$ “$1/b$-gallons” per gallon (or equivalently, we multiply by $1/b$ gallons per “$1/b$ -gallon”). The demand function then becomes:

$$\begin{aligned} p_{gas}=\alpha _{gas}-bY_{gas} \end{aligned}$$

with $p_{gas}\equiv P_{gas}/b$ and $\alpha _{gas}\equiv A/b$. Now $p_{gas}$ and $\alpha _{gas}$ are measured in “$ per $1/b$ -gallon” and $b=B/b$ in “$\$/$(gallon $\times 1/b$-gallon)”. Finally, we introduce the quantity measure $Q_{gas}$ which is expressed in “$1/b$ -gallon” so that $Q_{gas}=bY_{gas}.\$This turns the demand function into:

$$\begin{aligned} p_{gas}=\alpha _{gas}-\beta _{gas}Q_{gas}, \end{aligned}$$

with $\beta _{gas}=\$1/(1/b$-gallon)$^{2}$. The slope of the demand function is now $-1$.

As a specific example, let us set $A=\$500/$gallon and $B=\$100/$(gallon)$^{2}$ in $P_{gas}=A-BY_{gas}$. This means the vertical intercept is $500 per gallon and the horizontal intercept is 5 gallons. Writing the unit of measurement below each parameter and variable, we have:

$$\begin{aligned} \begin{array}{ccccccc} P_{gas} &{} = &{} 500 &{} - &{} (100 &{} \times &{} Y_{gas}). \\ \frac{\$}{gallon} &{} &{} \frac{\$}{gallon} &{} &{} \frac{\$}{\left( gallon\right) ^{2}} &{} &{} gallon \end{array} \end{aligned}$$

Multiplying the left-hand side and the right-hand side by 0.1 gallon/decigallon yields:

$$\begin{aligned} \begin{array}{ccccccccccccc} (P_{gas} &{} \times &{} 0.1) &{} = &{} (500 &{} \times &{} 0.1) &{} - &{} (100 &{} \times &{} 0.1 &{} \times &{} Y_{gas}). \\ \frac{\$}{gallon} &{} &{} \frac{gallon}{decigallon} &{} &{} \frac{\$}{ gallon} &{} &{} \frac{gallon}{decigallon} &{} &{} \frac{\$}{\left( gallon\right) ^{2}} &{} &{} \frac{gallon}{decigallon} &{} &{} gallon \end{array} \end{aligned}$$

Simplifying and noting that $Y_{gas}=0.1Q_{gas}$ yields:

$$\begin{aligned} \begin{array}{ccccccccccc} p_{gas} &{} = &{} 50 &{} - &{} (100 &{} \times &{} 0.1 &{} \times &{} 0.1 &{} \times &{} Q_{gas}). \\ \frac{\$}{decigallon} &{} &{} \frac{\$}{decigallon} &{} &{} \frac{\$}{\left( gallon\right) ^{2}} &{} &{} \frac{gallon}{decigallon} &{} &{} \frac{gallon}{decigallon} &{} &{} decigallon \end{array} \end{aligned}$$

Simplifying this gives:

$$\begin{aligned} \begin{array}{ccccccc} p_{gas} &{} = &{} 50 &{} - &{} (1 &{} \times &{} Q_{gas}). \\ \frac{\$}{decigallon} &{} &{} \frac{\$}{decigallon} &{} &{} \frac{\$}{\left( decigallon\right) ^{2}} &{} &{} decigallon \end{array} \end{aligned}$$

After normalization, the vertical intercept is $50 per decigallon and the horizontal intercept is 50 decigallons.

In the same way, let the inverse demand function for coal be $P_{coal}=C-DY_{coal}$ with the quantity of coal measured in tons and its price in dollars per ton. We normalize this demand function to $p_{coal}=\alpha _{coal}-Q_{coal}$ with $p_{coal}\equiv P_{coal}/d,\ \alpha _{coal}\equiv C/d$ and$\ Q_{coal}\equiv dY_{k}$ where $d\equiv \sqrt{D}$. Now the quantity of coal is measured in “$1/d$ -tons” and its price in dollars per $1/d$-ton. As a specific example, let us set $C=\$100$ per ton and $D=\$4/$(ton)$^{2}$. This means the vertical intercept is $100 per ton and the horizontal intercept is 25 tons. We normalize this demand function by expressing the quantity of coal $Q_{coal}$ in “half tons,” with its price $p_{coal}$ expressed in dollar per half ton. After normalization the demand curve is $p_{coal}=50-Q_{coal}$. The vertical intercept is $50 per half ton and the horizontal intercept is 50 half tons.

1.3 Derivation of $p_{c}^{r}(Q_{d})$ and $p_{c}^{\rho }(Q_{d})$ Curves

Solving (29) and (30) for $Q_{c}$ yields two solutions:

$$\begin{aligned} Q_{c}&= Q_{c}^{+}(Q_{d})\equiv \frac{\gamma _{c}+\sqrt{\gamma _{c}^{2}+4Q_{d}(\gamma _{d}-Q_{d})}}{2}, \end{aligned}$$

(37)

$$\begin{aligned} Q_{c}&= Q_{c}^{-}(Q_{d})\equiv \frac{\gamma _{c}-\sqrt{\gamma _{c}^{2}+4Q_{d}(\gamma _{d}-Q_{d})}}{2}. \end{aligned}$$

(38)

The highest possible value of $Q_{d}$ in (37) and (38) is where the term under the square root is zero:¹³

$$\begin{aligned} Q_{d}^{\max }=\frac{\gamma _{d}+\sqrt{\gamma _{c}^{2}+\gamma _{d}^{2}}}{2} >\gamma _{d}. \end{aligned}$$

(39)

The $Q_{c}^{+}$ solution (37) includes the unconstrained benchmark, since $Q_{c}^{+}(\gamma _{d})=\gamma _{c}$. Substituting (37) into (13), we find that total emissions are:

$$\begin{aligned} L^{+}(Q_{d})=\epsilon _{d}+\epsilon _{c}\left( \frac{\gamma _{c}}{2}+\sqrt{ \gamma _{c}^{2}+4Q_{d}(\gamma _{d}-Q_{d})}\right) . \end{aligned}$$

(40)

The first and second derivatives are:

$$\begin{aligned} L^{+\prime }(Q_{d})=\epsilon _{d}-\frac{\epsilon _{c}\left( 2Q_{d}-\gamma _{d}\right) }{\sqrt{\gamma _{c}^{2}+4Q_{d}(\gamma _{d}-Q_{d})}},\quad L^{+\prime \prime }(Q_{d})=\frac{-2\epsilon _{c}\left( \gamma _{c}^{2}+\gamma _{d}^{2}\right) }{\left( \gamma _{c}^{2}+4Q_{d}(\gamma _{d}-Q_{d})\right) ^{\frac{3}{2}}}<0.\nonumber \\ \end{aligned}$$

(41)

From (39) and (40) we find:

$$\begin{aligned} L^{+\prime }(\gamma _{d})=\frac{\epsilon _{d}\gamma _{c}-\epsilon _{c}\gamma _{d}}{\gamma _{c}}>0,\quad \quad \underset{Q_{d}\rightarrow Q_{d}^{\max }}{\lim }L^{+\prime }(Q_{d})=-\infty . \end{aligned}$$

(42)

The inequality follows from (12). Equation (42) together with $L^{+\prime \prime }(Q_{d})<0$ from (41) implies that $L^{+}(Q_{d})$ has a unique stationary point, which is a maximum, between $\gamma _{d}$ and $Q_{d}^{\max }$. Thus $L^{+\prime }(Q_{d})>0$ for $Q_{d}\in [0,\gamma _{d}]$. From (37) it follows that $Q_{c}^{+}(Q_{d})\ge \gamma _{c}=\bar{n}_{c}f_{c}$ for $Q_{d}<\gamma _{d}= \bar{n}_{d}f_{d}$. Thus, $Q_{c}^{+}(Q_{d})$ implements solution $r$ for $Q_{d}<\gamma _{d}$. Substituting (37) into (2), the expression for $p_{c}^{r}(Q_{d})$ is then:

$$\begin{aligned} p_{c}^{r}(Q_{d})=\alpha _{c}-\frac{\gamma _{c}+\sqrt{\gamma _{c}^{2}+4Q_{d}(\gamma _{d}-Q_{d})}}{2}\ \text {for }Q_{d}\in [0,\gamma _{d}]. \end{aligned}$$

In our numerical example (21), this becomes:

$$\begin{aligned} p_{c}^{r}(Q_{d})=200-\sqrt{10000+Q_{d}(200-Q_{d})}\text { for }Q_{d}\in [0,200]. \end{aligned}$$

By (42), $L^{+}(Q_{d})>L^{+}(\gamma _{d})=\bar{E}$ for $Q_{d}$ just above $\gamma _{d}$. The other solution to $L^{+}(Q_{d})=\bar{E}$ is:

$$\begin{aligned} Q_{d}=\tilde{Q}_{d}\equiv \frac{\epsilon _{d}\left( \epsilon _{d}\gamma _{d}+\epsilon _{c}\gamma _{c}\right) }{\epsilon _{c}^{2}+\epsilon _{d}^{2}}. \end{aligned}$$

(43)

Since $L^{+}(Q_{d})$ has a unique stationary point, which is a maximum, between $\gamma _{d}$ and $Q_{d}^{\max }, L^{+\prime }(Q_{d})<0$ for $Q_{d}\in [\tilde{Q}_{d},Q_{d}^{\max }]$ and $L^{+}(Q_{d})<\bar{E}$ for $Q_{d}\in (\tilde{Q}_{d},Q_{d}^{\max }]$. With $Q_{d}\in (\tilde{Q}_{d},Q_{d}^{\max }], Q_{d}$ exceeds $\gamma _{d}$ and $L^{+}(Q_{d})<\bar{E} ,$ so that $Q_{c}$ must be below $\gamma _{c},$ which means this is part of solution $\rho$.

The other part of solution $\rho$ is found on $Q_{c}^{-}(Q_{d})$ in (38) with $Q_{c}^{-}(Q_{d}^{\max })=\gamma _{c}/2$ by (39)$,$ and $Q_{d}=0$ and $Q_{d}=\gamma _{d}$ the only solutions to $Q_{c}^{-}(Q_{d})=0$. Thus $Q_{d}\in [\gamma _{d},Q_{d}^{\max }]$. Substituting (38) into (13), total emissions are:

$$\begin{aligned} L^{-}(Q_{d})=\epsilon _{d}+\epsilon _{c}\left( \frac{\gamma _{c}}{2}-\sqrt{ \gamma _{c}^{2}+4Q_{d}(\gamma _{d}-Q_{d})}\right) \end{aligned}$$

with

$$\begin{aligned} L^{-\prime }(Q_{d})=\frac{2\epsilon _{c}(2Q_{d}-\gamma _{d})+\epsilon _{d} \sqrt{\gamma _{c}^{2}-4Q_{d}^{2}+4\gamma _{d}Q_{d}}}{\sqrt{\gamma _{c}^{2}-4Q_{d}^{2}+4\gamma _{d}Q_{d}}}>0. \end{aligned}$$

The inequality follows from $Q_{d}\ge \gamma _{d}$. The correspondence $p_{c}^{\rho }(Q_{d})$ is therefore given by:

$$\begin{aligned} p_{c}^{\rho }(Q_{d})=\left\{ \begin{array}{ccc} \alpha _{c}-\frac{\gamma _{c}+\sqrt{\gamma _{c}^{2}+4Q_{d}(\gamma _{d}-Q_{d}) }}{2} &{} &{} \text {for}Q_{d}\in [\tilde{Q}_{d},Q_{d}^{\max }] \\ &{} &{} \\ \alpha _{c}-\frac{\gamma _{c}-\sqrt{\gamma _{c}^{2}+4Q_{d}(\gamma _{d}-Q_{d}) }}{2} &{} &{} \text {for}Q_{d}\in [\frac{\gamma _{c}}{2},Q_{d}^{\max }] \end{array} \right. \end{aligned}$$

In our numerical example (21), this becomes:

$$\begin{aligned} p_{c}^{\rho }(Q_{d})=\left\{ \begin{array}{ccc} 200-\sqrt{10000+200Q_{d}-Q_{d}^{2}} &{} &{} \text {for }Q_{d}\in [240;241.42] \\ &{} &{} \\ 200+\sqrt{10000+200Q_{d}-Q_{d}^{2}} &{} &{} \text {for }Q_{d}\in [200;241.42] \end{array} \right. \end{aligned}$$