Unilateral Climate Policy with Production-Based and Consumption-Based Carbon Emission Taxes

Abstract

This paper characterizes a sub-global climate coalition’s unilateral policy of reaching a given climate damage reduction goal at minimum costs. Following Eichner and Pethig (J Environ Econ Manag, 2013) we set up a two-country two-period model in which one of the countries represents a climate coalition that implements a binding ceiling on the world’s first-period emissions. The other country is the rest of the world and refrains from taking action. The coalition can make use of production-based carbon emission taxes in both periods, as in Eichner and Pethig (J Environ Econ Manag, 2013), but here we consider consumption-based carbon emission taxes as an additional instrument. The central question is whether and how the coalition employs the consumption-based taxes along with the production-based taxes in its unilateral cost-effective ceiling policy. All cost-effective policies identified analytically and numerically consist of a mix of both types of taxes implying that there is a tax mix which is less expensive for the coalition than stand-alone consumption-based or stand-alone production-based taxes. With full cooperation both taxes are perfect substitutes (in our model), but in case of unilateral action they are imperfect substitutes, because coalition’s total welfare loss from two different but moderate distortions is smaller than that from a single but severe distortion.

Appendix

1.1 Proof of Proposition 2

Observe that (4), (5), (13) and the first-order conditions

$$\begin{aligned} X^A_{e_{A1}} (e_{A1}) = X^B_{e_{B1}} (\bar{e}_1 - e_{A1}) = p_x X^A_{e_{A2}} (e_{A2}) = p_x X^B_{e_{B2}} (\bar{e} - \bar{e}_1 - e_{A2}) \end{aligned}$$

(18)

determine $e_{A1}, e_{A2}, p_e$ and $p_x$ for given $\bar{e}_1$. Hence country $B$’s income

$$\begin{aligned} y_B := X^B(\bar{e}_1 - e_{A1}) + p_x X^B(\bar{e} - \bar{e}_1 - e_{A2}) + p_e [\alpha _B \bar{e} - (\bar{e} - e_{A1} - e_{A2})] \end{aligned}$$

is fixed and so are the levels of consumption, $x_{B1}$ and $x_{B2}$. From the market clearing condition $x_{A1}^s + x_{B1}^s = x_{A1} + x_{B1}$ therefore follows that the equilibrium value

$$\begin{aligned} x_{A1} := x_{A1}^s + x_B^s - x_{B1}, \end{aligned}$$

(19)

is uniquely determined by $\bar{e}_1$. Consider next country $A$’s demand functions for the consumption good derived from CES utility (14)

$$\begin{aligned} x_{A1}&= \frac{\bar{\gamma }_A (p_x + \tau _2 q_2)^{\sigma _A} y_A}{\bar{\gamma }_A (1+ \tau _1 q_1)(p_x +\tau _2 q_2)^{\sigma _A}+ (p_x+\tau _2 q_2)(1+\tau _1 q_1)^{\sigma _A}}, \end{aligned}$$

(20)

$$\begin{aligned} x_{A2}&= \frac{y_A}{p_x} - \frac{x_{A1}}{p_x}, \end{aligned}$$

(21)

where

$$\begin{aligned} y_A&:= \underbrace{X^A (e_{A1}) + p_x X^A (e_{A2}) + p_e (\alpha _A \bar{e} - e_{A1} - e_{A2})}_{= y_{Ao}} + \tau _1 q_1 x_{A1} + \tau _2 q_2 x_{A2},\\&= y_{Ao} + \tau _1 q_1 x_{A1} + \tau _2 q_2 x_{A2}. \end{aligned}$$

Note that the above definition of $y_A$ holds irrespective of the sign and size of $\pi _1$ and $\pi _2$. In (18) and (19) we have already accounted for all conditions of market clearing other than that of the second-period market for the consumption good. The latter is taken care of by Walras Law which is why we disregard Eq. (21). Closer inspection of (20) reveals that keeping the equilibrium values of $\pi _1, \pi _2$ and $p_x$ constant the equilibrium value $x_{A1}$ can be attained through various combinations of $\tau _1$ and $\tau _2$. Formally speaking, there is a function $F: \, \, ]-p_x, \infty [\, \, \longrightarrow \, \,] -1, \infty [$ such that the right-hand side of $\frac{U^A_{x_{A2}}}{U^A_{x_{A1}}} = \frac{p_x + \tau _2}{1 + \tau _1}$ is unchanged for all $(\tau _1, \tau _2)$ satisfying $\tau _1 = F(\tau _2)$. In order to specify the properties of that function $F$ rearrange terms in (20) to obtain

$$\begin{aligned} x_{A1} \left[1 + (p_x + \tau _2 q_2)^{1-\sigma _A} (1+\tau _1 q_1)^{\sigma _A} \bar{\gamma }_A^{-1} \right]= y_{Ao} + \tau _2 q_2 x_{A2}. \end{aligned}$$

(22)

Differentiate (22) and account for ${\mathrm {d}}x_{A1} = {\mathrm {d}}x_{A2} = {\mathrm {d}}y_{Ao}$ and ${\mathrm {d}}q_1 = {\mathrm {d}}q_2 ={\mathrm {d}}p_x = 0$ to get

$$\begin{aligned}&x_{A1} \left[(1-\sigma _A)(p_x + \tau _2 q_2)^{- \sigma _A} (1+\tau _1 q_1)^{\sigma _A} \bar{\gamma }_A^{-1} q_2 {\mathrm {d}}\tau _2 \right]\nonumber \\&\quad +\, \sigma _A (p_x + \tau _2 q_2)^{1-\sigma _A} (1+\tau _1 q_1)^{\sigma _A - 1} \bar{\gamma }_A^{-1} q_1 {\mathrm {d}}\tau _1 = x_{A2} q_2 {\mathrm {d}}\tau _2. \end{aligned}$$

(23)

Next turn (23) into

$$\begin{aligned} x_{A1} \left[ (1- \sigma _A) \left( \frac{p_x + \tau _2 q_2}{1 + \tau _1 q_1} \right) ^{-\sigma _A} q_2 {\mathrm {d}}\tau _2 + \sigma _A \left( \frac{p_x + \tau _2 q_2}{1 + \tau _1 q_1} \right) ^{1-\sigma _A} q_1 {\mathrm {d}}\tau _1 \right] = \bar{\gamma }_A x_{A2} q_2 {\mathrm {d}}\tau _2. \nonumber \\ \end{aligned}$$

(24)

Account for $\frac{x_{A1}}{x_{A2}} = \bar{\gamma }_A \left( \frac{p_x + \tau _2 q_2}{1 + \tau _1 q_1} \right) ^{\sigma _A}$ or $\left( \frac{p_x + \tau _2 q_2}{1 + \tau _1 q_1} \right) ^{-\sigma _A} = \bar{\gamma }_A \frac{x_{A2}}{x_{A1}}$ in (24) to reach

$$\begin{aligned}&x_{A1} \left[ (1-\sigma _A) \bar{\gamma }_A \frac{x_{A2}}{x_{A1}} {\mathrm {d}}\tau _2 + \sigma _A \frac{p_x+ \tau _2 q_2}{1 + \tau _1 q_1} \bar{\gamma }_A \frac{x_{A2}}{x_{A1}} \frac{q_1}{q_2} {\mathrm {d}}\tau _1 \right] = \bar{\gamma }_A x_{A2} {\mathrm {d}}\tau _2,\\&\iff (1-\sigma _A) \bar{\gamma }_A x_{A2} {\mathrm {d}}\tau _2 + \sigma _A \frac{p_x + \tau _2 q_2}{1 + \tau _1 q_1} \bar{\gamma }_A x_{A2} \frac{q_1}{q_2} {\mathrm {d}}\tau _1 = \bar{\gamma }_A x_{A2} {\mathrm {d}}\tau _2, \\&\iff - \sigma _A {\mathrm {d}}\tau _2 + \sigma _A \frac{p_x + \tau _2 q_2}{1 + \tau _1 q_1} \cdot \frac{q_1}{q_2} {\mathrm {d}}\tau _1 = 0 \quad \iff \quad \frac{{\mathrm {d}}\tau _1}{{\mathrm {d}}\tau _2} \!=\! F_{\tau _2} \!=\! \frac{1 + \tau _1 q_1}{p_x + \tau _2 q_2} \cdot \frac{q_2}{q_1} > 0. \end{aligned}$$

We conclude that $F_{\tau _2} > 0$ because $q_1>0$, $q_2>0$, $1+ \tau _1q_1 > 0$ and $p_x + \tau _2 q_2> 0$ and that

$$\begin{aligned} F_{\tau _2\tau _2} = \frac{(p_x + \tau _2 q_2) q_1 F_{\tau _2} - (1+ \tau _1 q_1) q_2}{(p_x + \tau _2 q_2)^2} \cdot \frac{q_2}{q_1} \gtreqless 0 \quad \Longleftrightarrow \quad F_{\tau _2} \gtreqless \frac{1+ \tau _1 q_1}{p_x + \tau _2 q_2} \cdot \frac{q_2}{q_1}. \end{aligned}$$

Hence $F_{\tau _2 \tau _2} = 0$.

1.2 Proof of Proposition 3

Consider first the supply-side partial equilibrium established by the equations

$$\begin{aligned} e_{A1} + e_{B1}&= \bar{e}_1, \quad \quad e_{A2} + e_{B2} = \bar{e}_2:= \bar{e} - \bar{e}_1, \end{aligned}$$

(25)

$$\begin{aligned} X_{e_{A1}}^A (e_{A1})&= p_e + \pi _{1}, \end{aligned}$$

(26)

$$\begin{aligned} X_{e_{B1}}^B (\bar{e}_1 - e_{A1})&= p_e, \end{aligned}$$

(27)

$$\begin{aligned} p_x X_{e_{A2}}^A (e_{A2})&= p_e + \pi _{2}, \end{aligned}$$

(28)

$$\begin{aligned} p_x X_{e_{B2}}^B (\bar{e}_2 - e_{A2})&= p_e. \end{aligned}$$

(29)

These equations determine $p_e$, $p_x$, $\pi _{1}$, $\pi _{2}$ and $x_{it}^s$ for $i=A,B$ and $t=1,2$ for every $(e_{A1}, e_{A2}) \in [0, \bar{e}_1] \times [0, \bar{e}_2]$. That equilibrium is partial, because the demand for good $X$ still needs to be specified in order to establish equilibrium on the commodity markets,²³

$$\begin{aligned} x_{A1} + x_{B1} = x^s_{A1} + x^s_{B1}. \end{aligned}$$

(30)

The CES utility functions yield the demands

$$\begin{aligned} x_{A1}&= \frac{\left( \frac{\gamma _1 p_x}{(1+\tau _1 q_1) \gamma _2} \right) ^\sigma (y_{Ao} + \tau _1 q_1 x_{A1})}{(1+\tau _1 q_1) \left( \frac{\gamma _1 p_x}{(1+\tau _1 q_1) \gamma _2} \right) ^\sigma +p_x} = \frac{\left( \gamma _1 p_x \right) ^\sigma y_{Ao} }{\left( \gamma _1 p_x \right) ^\sigma + (1+\tau _1 q_1)^\sigma \gamma _2^\sigma p_x}, \end{aligned}$$

(31)

$$\begin{aligned} x_{B1}&= \frac{\left( \frac{\gamma _1 p_x}{\gamma _2} \right) ^\sigma y_{B}}{ p_x + \left( \frac{\gamma _1 p_x}{\gamma _2} \right) ^\sigma }, \end{aligned}$$

(32)

where $y_{Ao}$ and $y_B$ are incomes (see (11)) with $y_{Ao}$ representing income before the tax revenue $\tau _1 x_{A1}$ is recycled to the consumer. Note that if $\sigma >0$ and $\tau _1 \in \mathbb {R}$, $x_{A1}$ in (31) and $x_{B1}$ in (32) are fully determined for every $(e_{A1}, e_{A2}) \in [0, \bar{e}_1] \times [0, \bar{e}_2]$.

We first investigate how $x_{B1}$ varies with $\sigma$. Since utility maximization implies

$$\begin{aligned} \frac{x_{B1}}{x_{B2}} = \left( \frac{\gamma _1 p_x}{\gamma _2} \right) ^\sigma , \end{aligned}$$

(33)

the straightforward conclusion from (33) and constant $y_B$ is

$$\begin{aligned} \frac{{\mathrm {d}}x_{B1}}{{\mathrm {d}}\sigma } \, \left\{ \begin{array}{l@{\quad }l} >0, \, \text{ if } \, (\gamma _1 p_x/ \gamma _2)>1 &{}(\text{ and } \, x_{B1} \rightarrow y_B,\, \text{ if } \, \sigma \rightarrow \infty ), \\ =0, \, \text{ if } \, (\gamma _1 p_x/ \gamma _2) =1, &{}\\ <0, \, \text{ if } \, (\gamma _1 p_x/ \gamma _2)<1 &{} (\text{ and } \, x_{B1} \rightarrow 0,\, \text{ if } \, \sigma \rightarrow \infty ). \end{array} \right. \end{aligned}$$

(34)

Consequently, disregarding the knife-edge case $(\gamma _1 p_x/\gamma _2)=1$ we find that for every $(e_{A1}, e_{A2})$ there is some set $S \subset \mathbb {R}$ with non-empty interior such that

$$\begin{aligned} x_{B1} \le x^s_{A1} + x^s_{B1} \quad \forall \, \sigma \in S. \end{aligned}$$

(35)

Since $x_{A1} \ge 0$, (35) is clearly a necessary condition for (30). Given (35), sufficient for (30) is $x_{A1} = x^s_{A1} + x^s_{B1} - x_{B1}$. For $\tau _1=0$ this equality will not hold, in general. But $x_{A1}$ varies with $\tau _1$. Differentiation of (31) with respect to $\tau _1$ yields

$$\begin{aligned} \frac{{\mathrm {d}}x_{A1}}{{\mathrm {d}}\tau _1} = - \frac{(\gamma _1 p_x)^\sigma y_{Ao} \sigma q_1 (1+\tau _1q_1)^{\sigma -1}}{\left[ (\gamma _1 p_x)^\sigma + (1+\tau _1 q_1)^\sigma \gamma _2^\sigma p_x \right] ^2} \quad \text{ and } \quad \lim _{\sigma \rightarrow 0} \frac{{\mathrm {d}}x_{A1}}{{\mathrm {d}}\tau _1} =0. \end{aligned}$$

(36)

It follows that if $\sigma \in S$ and $x_{A1} \ne x^s_{A1} + x^s_{B1} - x_{B1}$ for $\tau _1=0$, one can change the magnitude of $x_{A1}$ by an appropriate choice of $\tau _1$ such that (30) is satisfied. This completes the proof of Proposition 3.

1.3 Proof of Proposition 4

Based on our results in Proposition 3 we set $\tau _2 = 0$ and treat the quantities $e_{A1}$ and $e_{A2}$ as the regulator’s decision variables for analytical convenience while the tax rates $\pi _1$ and $\pi _2$ adjust endogenously in order to sustain $e_{A1}$ and $e_{A2}$. The consumption tax rate $\tau _1$ is implied by the solution of the regulator’s optimization problem. In her optimization procedure she takes into account

• the equilibrium conditions (3) for the commodity markets,

• the fuel/emission constraints²⁴

  $$\begin{aligned} e_{A1} + e_{B1} = \bar{e}_1 \quad \text{ and } \quad e_{A2} + e_{B2} = \bar{e}_2 := \bar{e} - \bar{e}_1, \end{aligned}$$

  (37)

• the input demand and output supply functions of all producers,

• and the consumption demand functions of the consumer in country $B$.

Ad (i): For the quadratic production functions (15) the supply-side partial equilibrium (25) – (29) turns into

$$\begin{aligned} p_e&= a-b(\bar{e}_1-e_{A1}), \end{aligned}$$

(38)

$$\begin{aligned} p_x&= \frac{a-b(\bar{e}_1-e_{A1})}{a-b(\bar{e}_2-e_{A2})}, \end{aligned}$$

(39)

$$\begin{aligned} \pi _{1}&= b(\bar{e}_1-2e_{A1}), \end{aligned}$$

(40)

$$\begin{aligned} \pi _{2}&= \frac{[a-b(\bar{e}_1-e_{A1})]b(\bar{e}_2-2b e_{A2})}{a-b(\bar{e}_2-e_{A2})} \end{aligned}$$

(41)

and

$$\begin{aligned} \frac{{\mathrm {d}}p_e}{{\mathrm {d}}e_{A1}} =b,&\quad \frac{{\mathrm {d}}p_e}{{\mathrm {d}}e_{A2}} =0, \end{aligned}$$

(42)

$$\begin{aligned} \frac{{\mathrm {d}}p_x}{{\mathrm {d}}e_{A1}} = \frac{b}{a-b(\bar{e}_2-e_{A2})} = \frac{b p_x}{p_e},&\quad \frac{{\mathrm {d}}p_x}{{\mathrm {d}}e_{A2}}= - \frac{p_e b}{[a-b \bar{e}_2 - e_{A2}]^2} = - \frac{b p_x^2}{p_e}. \end{aligned}$$

(43)

Next, we determine $x_{A1}$ and $x_{A2}$ with the help of country $B$’s demand

$$\begin{aligned} x_{B1} = \gamma y_B,&\quad x_{B2} = \frac{(1-\gamma )y_B}{p_x}. \end{aligned}$$

(44)

and with

$$\begin{aligned} X^A(e_{At})+X^B(e_{Bt})=x_{At}+x_{Bt} t=1,2. \end{aligned}$$

(45)

Observe that

$$\begin{aligned} x_t^s := X^A(e_{At})+X^B(e_{Bt})=a(e_{At}+e_{Bt})-\frac{b}{2} \left( e_{At}^2+e_{Bt}^2 \right) . \end{aligned}$$

(46)

From $e_{At}^2 + e_{Bt}^2=\left( e_{At}+e_{Bt} \right) ^2 - 2 e_{At}e_{Bt}$ follows $x_t^s=x_t^{s*}- b \left( \frac{\bar{e}_t^2}{4}-e_{At}e_{Bt} \right)$ with $x_t^{s*}:=a\bar{e}_t-\frac{b \bar{e}_t^2}{4}$. Combined with (45) we get

$$\begin{aligned} x_{At}=x_t^s-x_{Bt}=x_t^{s*}- \frac{b \bar{e}_t^2}{4}+ b e_{At}e_{Bt}-x_{Bt}. \end{aligned}$$

(47)

Making use of (44) and inserting (47) into the Cobb–Douglas utility function yields

$$\begin{aligned} u_A = \left[ \underbrace{\left( x_1^{s*}- \frac{b \bar{e}_1^2}{4} \right) + b e_{A1} (\bar{e}_1 - e_{A1}) -\gamma y_B}_{x_{A1}} \right] ^{\gamma } \left[ \underbrace{\left( x_2^{s*}-\frac{b \bar{e}_2^2}{4} \right) + b e_{A2} (\bar{e}_2 - e_{A2}) -\frac{(1-\gamma ) y_B}{p_x}}_{x_{A2}} \right] ^{1-\gamma }. \end{aligned}$$

(48)

We differentiate $u_A$ in (48) with respect to $e_{A1}$ and $e_{A2}$ and account for (42), (43) and

$$\begin{aligned} \frac{{\mathrm {d}}y_B}{{\mathrm {d}}e_{A1}}&= \frac{b p_x x^s_{B2}}{p_e}-b \Delta e_A, \end{aligned}$$

(49)

$$\begin{aligned} \frac{{\mathrm {d}}y_B}{{\mathrm {d}}e_{A2}}&= -\frac{b}{p_e} p_x^2 x_{B2}^s, \end{aligned}$$

(50)

to obtain, after rearrangement of terms,

$$\begin{aligned} \frac{{\mathrm {d}}u_A}{{\mathrm {d}}e_{A1}}&= \frac{b u_A}{p_e} \left[\frac{\gamma }{x_{A1}} [p_e (e_{B1}-e_{A1})- \gamma (y_B-x_{B1}^s) ]+ \frac{(1-\gamma )^2x_{B1}^s}{p_x x_{A2}} \right], \end{aligned}$$

(51)

$$\begin{aligned} \frac{{\mathrm {d}}u_A}{{\mathrm {d}}e_{A2}}&= \frac{b u_A}{p_e}\left[\frac{ \gamma ^2 p_x^2 x_{B2}^s}{x_{A1}}+ \frac{1-\gamma }{x_{A2}} [p_e(e_{B2}-e_{A2})- (1-\gamma ) (y_B-p_x x_{B2}^s) ]\right]. \end{aligned}$$

(52)

Next, we use (51), (52) and $\frac{x_{A1} (1-\gamma )}{x_{A2} \gamma }=\frac{p_x}{1+\tau _1 q_1}$ in $\frac{{\mathrm {d}}u_A}{{\mathrm {d}}e_{A1}} = 0$ and $\frac{{\mathrm {d}}u_A}{{\mathrm {d}}e_{A2}} = 0$, respectively, to get

$$\begin{aligned} \frac{{\mathrm {d}}u_A}{{\mathrm {d}}e_{A1}} = 0&\Longleftrightarrow&(1 + \tau _1 q_1) \left[ p_e (e_{B1} - e_{A1}) - \gamma (y_B - x_{B1}^s) \right] + (1 - \gamma ) x_{B1}^s = 0,\quad \end{aligned}$$

(53)

$$\begin{aligned} \frac{{\mathrm {d}}u_A}{{\mathrm {d}}e_{A2}} = 0&\Longleftrightarrow&p_e (e_{B2} - e_{A2}) \!+\! (1 \!+\! \tau _1 q_1) \gamma p_x x_{B2}^s - (1 \!-\! \gamma ) (y_B \!-\! p_x x_{B2}^s) = 0. \end{aligned}$$

(54)

From summation of (53) and (54), we obtain

$$\begin{aligned} (1+\tau _1 q_1)(e_{B1}-e_{A1})+(e_{B2}-e_{A2})=-(1+\gamma \tau _1 q_1) \Delta e_A. \end{aligned}$$

(55)

Then we make use of (38)–(41) and get

$$\begin{aligned} (1+\tau _1 q_1) \pi _{1} + \frac{\pi _{2}}{p_x} = - (1+\gamma \tau _1 q_1) b \Delta e_A. \end{aligned}$$

(56)

Observe that (45) for $t=1$, (51) and (56) determine the cost-effective policy $(e_{A1},e_{A2},\tau _1)$. Total differentiation of (56) yields

$$\begin{aligned}&- \left( 3+2 \tau _1 q_1 + \tau _1 \gamma q_1 -\frac{\tau _1 (\pi _1 +\gamma )}{2 \left( 1 -\frac{b}{2} e_{A1}\right) ^2} \right) {\mathrm {d}}e_{A1} - (3+ \tau _1 q_1 \gamma ) {\mathrm {d}}e_{A2} \nonumber \\&\quad + (\bar{e}_1-2e_{A1}+ \gamma \Delta e_A) q_1 {\mathrm {d}}\tau _1 + \tau _1 q_1 {\mathrm {d}}\bar{e}_1 = 0. \end{aligned}$$

(57)

For the laissez faire values $\tau _1=0$, $e_{A1}^o=e_{B1}^o=\bar{e}_1^o$, $\Delta e_A^o=0,$ (57) simplifies to

$$\begin{aligned} 3 {\mathrm {d}}e_{A1} + 3 {\mathrm {d}}e_{A2} = 0. \end{aligned}$$

(58)

Next we account for $x_{B1} = \gamma y_B$ and $x_{A1} = \underbrace{\frac{1}{1 + \tau _1 q_1 (1 - \gamma )}}_{=:\check{\tau }_1}\gamma \underbrace{[X(e_{A1}) + p_x X(e_{A2}) + p_e \Delta e_A ]}_{=: y_{Ao}}$ and obtain after total differentiation of the equation

$$\begin{aligned} x_{A1} + x_{B1} - X(e_{A1}) - X(\bar{e}_1 - e_{A1}) = 0 \end{aligned}$$

(59)

$$\begin{aligned}&\left[\check{\tau }_1 \gamma (X_{e_{A1}} - p_e) + \frac{b p_x x_{A2}^s}{p_e} + b \Delta e_A + \gamma \left( \frac{b p_x x_{B2}^s}{p_e} - b \Delta e_A \right) - X_{e_{A1}} + X_{e_{A1}} \right]{\mathrm {d}}e_{A1} \nonumber \\&\quad + \left[\check{\tau }_1 \gamma \left( p_x X_{e_{A2}} - p_e - \frac{b p_x^2 x_{A2}^s}{p_e}\right) - \left( \frac{\gamma b p_x^2 x_{B2}^s}{p_e} \right) \right]{\mathrm {d}}e_{A2} - \frac{(1-\gamma ) q_1}{[1 + \tau _1 q_1 (1-\gamma ) ]^2} \gamma y_{Ao} {\mathrm {d}}\tau _1 \nonumber \\&\quad + \left[ \check{\tau }_1 \gamma \left( - \frac{2 b p_x x_{A2}^s}{p_e} - \frac{b p_x^2 x_{A2}^s}{p_e} - b \Delta e_A \right) \right. \nonumber \\&\quad \left. + \gamma \left( -\frac{b p_x x_{B2}^s}{p_e}-\frac{b p_x^2 x_{B2}^s}{p_e}-b\Delta e_B\right) - X_{e_{B1}} \right] {\mathrm {d}}\bar{e}_1 = 0. \end{aligned}$$

(60)

Assessing (60) at laissez faire, which is characterized by $\tau _1=0$, $X_{e_{A1}} = p_e$, $p_x X_{e_{A2}} = p_e$, $X_{e_{A1}}= X_{e_{B1}}$, $\Delta e_A = \Delta e_B=0$, $\gamma y_A = X(e_{A1})= x_{A1}^s = x_{B1}^s$, $x_{B2}^s = x_{A2}^s$, we get

$$\begin{aligned}&\frac{2 \gamma b p_x x_{A2}^s}{p_e} {\mathrm {d}}e_{A1} - \frac{2 \gamma b p_x^2 x_{A2}^s}{p_e} {\mathrm {d}}e_{A2} - (1 - \gamma ) q_1x_{B1}^s {\mathrm {d}}\tau _1\nonumber \\&\quad = \left[ 2 \gamma \left( \frac{b p_x x_{A2}^s}{p_e} + \frac{b p_x^2 x_{A2}^s}{p_e} \right) + p_e \right] {\mathrm {d}}\bar{e}_1. \end{aligned}$$

(61)

Finally, we differentiate (53) to get

$$\begin{aligned}&\left[(\bar{e}_1 - 2 e_{A1}) \frac{{\mathrm {d}}p_e}{{\mathrm {d}}e_{A1}} - 2 p_e - \gamma \left( \frac{{\mathrm {d}}y_B}{{\mathrm {d}}e_{A1}} + X_{e_{B1}} \right) - \frac{(1-\gamma )}{1+\tau _1 q_1} X_{e_{B1}} \right]{\mathrm {d}}e_{A1} \nonumber \\&\quad + \left[(\bar{e}_1-2 e_{A1}) \frac{{\mathrm {d}}p_e}{{\mathrm {d}}e_{A2}} - \gamma \frac{{\mathrm {d}}y_B}{{\mathrm {d}}e_{A2}} \right]{\mathrm {d}}e_{A2} - \frac{(1-\gamma ) q_1 x_{B1}^s}{(1+\tau _1 q_1)^2} {\mathrm {d}}\tau _1 \nonumber \\&\quad + \left[(\bar{e}_1-2 e_{A1}) \frac{{\mathrm {d}}p_e}{{\mathrm {d}}\bar{e}_1} - \gamma \left( \frac{{\mathrm {d}}y_B}{{\mathrm {d}}\bar{e}_1} - X_{e_{B1}} \right) + \frac{(1-\gamma )}{1+\tau _1 q_1} X_{e_{B1}} \right]{\mathrm {d}}\bar{e}_1 = 0. \end{aligned}$$

(62)

Assessing (62) at laissez-faire, (62) turns into

$$\begin{aligned} \left[- 3 p_e - \frac{\gamma b p_x x_{B2}^s}{p_e} \right]{\mathrm {d}}e_{A1} + \frac{\gamma b p_x^2 x_{B2}^s}{p_e} {\mathrm {d}}e_{A2} - (1 - \gamma ) q_1 x_{B1}^s {\mathrm {d}}\tau _1 \nonumber \\ = - \left[\gamma \left( \frac{b p_x x_{B2}^s}{p_e} + \frac{p_x^2 x_{B2}^s}{p_e} \right) + p_e \right]{\mathrm {d}}\bar{e}_1. \end{aligned}$$

(63)

(58), (61) and (63) jointly determine ${\mathrm {d}}e_{A1}$, ${\mathrm {d}}e_{A2}$ and ${\mathrm {d}}\tau _1$. In matrix notation, these equations read

$$\begin{aligned} \begin{bmatrix} 3&3&0 \\ 2a&-2ap_x&-(1-\gamma ) q_1 x_{B1}^s \\ -3p_e-a&ap_x&-(1-\gamma ) q_1 x_{B1}^s \end{bmatrix} \begin{bmatrix} {\mathrm {d}}e_{A1} \\ {\mathrm {d}}e_{A2} \\ {\mathrm {d}}\tau _1 \end{bmatrix} = \begin{bmatrix} 0 \\ \left[ 2(a+ap_x) + p_e \right] {\mathrm {d}}\bar{e}_1 \\ \left( -a- a p_x-p_e \right) {\mathrm {d}}\bar{e}_1 \end{bmatrix}, \end{aligned}$$

(64)

where $a:=\frac{\gamma b p_x x_{B2}^s}{p_e}$. Solving the equation system (64) by means of Cramer’s rule yields

$$\begin{aligned} \frac{{\mathrm {d}}e_{A1}}{{\mathrm {d}}\bar{e}_1}&= \frac{3(a+ap_x)+2p_e}{3(a+ap_x p_e)}>0, \end{aligned}$$

(65)

$$\begin{aligned} \frac{{\mathrm {d}}e_{A2}}{{\mathrm {d}}\bar{e}_1}&= - \frac{3(a+ap_x)+2p_e}{3(a+ap_x p_e)}<0, \end{aligned}$$

(66)

$$\begin{aligned} \frac{{\mathrm {d}}\tau _1}{{\mathrm {d}}\bar{e}_1}&= - \frac{5p_e(a+ap_x)+3p_e}{3x_{B1}^s(1-\gamma ) q_1 (a+ap_x p_e)}<0. \end{aligned}$$

(67)

The associated changes of the prices $\pi _1$, $\pi _2$, $p_e$ and $p_x$ follow from making use of (65) and (66) in (40)–(43).

Ad (ii): In case of $\tau _1 ={\mathrm {d}}\tau _1=0$, (61) and (63) jointly determine ${\mathrm {d}}e_{A1}$ and ${\mathrm {d}}e_{A2}$. In matrix notation, these equations read

$$\begin{aligned} \begin{bmatrix} 2a&-2a p_x \\ -3p_e - a&a p_x \end{bmatrix} \begin{bmatrix} {\mathrm {d}}e_{A1} \\ {\mathrm {d}}e_{A2} \end{bmatrix} = \begin{bmatrix} \left[ 2(a+ap_x) + p_e \right] {\mathrm {d}}\bar{e}_1 \\ \left( -a- a p_x-p_e \right) {\mathrm {d}}\bar{e}_1 \end{bmatrix}. \end{aligned}$$

(68)

Solving the equation system (68), we obtain

$$\begin{aligned} \frac{{\mathrm {d}}e_{A1}}{{\mathrm {d}}\bar{e}_1}&= \frac{a p_x p_e}{6 a p_x p_e} =\frac{1}{6}>0, \end{aligned}$$

(69)

$$\begin{aligned} \frac{{\mathrm {d}}e_{A2}}{{\mathrm {d}}\bar{e}_1}&= - \frac{5a+ 6 a p_x +3 p_e}{6 a p_x}<0. \end{aligned}$$

(70)