Dynamic Climate Policy Under Firm Relocation: The Implications of Phasing Out Free Allowances

Abstract

The allocation of free allowances for firms within the European Union Emissions Trading Scheme was found to lead to substantial overcompensation, which is why some stakeholders recently called for phasing out of free allowances in the near term. This paper analyzes the consequences of phasing out free allowances in a two-period model when one country unilaterally implements climate policies. A carbon price induces firms to invest in abatement capital, but may lead to the relocation of some firms. The regulator addresses the relocation problem by offering firms transfers, i.e. free allowances, conditional on maintaining the production in the regulating country. If transfers are unrestricted in both periods, then the regulator implements the first best by equalizing the carbon price with the marginal environmental damage and using transfers to prevent any relocation. However, if transfers in the future period are restricted, the planner optimally implements a declining carbon price path with the first period price exceeding the marginal environmental damage. A high carbon price triggers investments in abatement capital and thus creates a lock-in effect. With a larger abatement capital stock, firms are less affected by carbon prices in the future and therefore less prone to relocate in the second period.

Appendix

1.1 Equivalence Between an Emissions Trading System with Free Allowances and a Price and Transfer System

Let n and N be the number of emission allowances a firm receives free of charge, then the transfers in each period are given by $g = np$ and $G = NP$. Rearranging yields

$$\begin{aligned} n&= h(p,g) \equiv {\left\{ \begin{array}{ll} p/g &{}\quad {\text {if}} \quad g > 0 \\ 0 &{}\quad {\text {if}} \quad g = 0 \end{array}\right. } \end{aligned}$$

(A.1)

$$\begin{aligned} N&= H(P,G) \equiv {\left\{ \begin{array}{ll} P/G &{}\quad {\text {if}} \quad G > 0 \\ 0 &{}\quad {\text {if}} \quad G = 0 \end{array}\right. } \end{aligned}$$

(A.2)

For the demand of certificates, let us start with the scenario with AA and AB-firms. The number of actual emissions in the first and second period reads

$$\begin{aligned} d(p,g,P,G)&= ({\bar{\theta }} - \theta _{AA}^{AB}) \cdot ({\bar{\epsilon }} - k_{AA}^*(\cdot ) - q_{AA}^*(\cdot )) + (\theta _{AA}^{AB} - {\underline{\theta }}) \cdot ({\bar{\epsilon }} - k_{AB}^*(\cdot ) - q_{AB}^*(\cdot )) \end{aligned}$$

(A.3)

$$\begin{aligned} D(p,g,P,G)&= ({\bar{\theta }} - \theta _{AA}^{AB}) \cdot ({\bar{\epsilon }} - k_{AA}^*(\cdot ) - Q_{AA}^*(\cdot )). \end{aligned}$$

(A.4)

Let ${\bar{e}}$ and ${\bar{E}}$ be the total caps of emissions in the first and second period, then the market clearing condition of the allowance market requires that

$$\begin{aligned} {\bar{e}}&= d(p,g,P,G) \quad {\text {and}} \end{aligned}$$

(A.5)

$$\begin{aligned} {\bar{E}}&= D(p,g,P,G). \end{aligned}$$

(A.6)

Define $m(p,g,P,G) = \big (h(p,g), d(p,g,P,G), H(P,G), D(p,g,P,G)\big )$, where m(p, g, P, G) is a continuously differentiable function because each of its elements is continuously differentiable. According to the implicit function theorem, m(p, g, P, G) has always an inverse function if its Jacobian matrix is invertible. It can be shown that the determinant of the Jacobian matrix is always positive, which is a sufficient condition for this matrix to be invertible. Hence, there always exists a function $f(n, {\bar{e}}, N,{\bar{E}}) = m^{-1}(n, {\bar{e}}, N,{\bar{E}})$ and we can conclude that each vector (p, g, P, G) can be uniquely determined by choosing the vector $(n, {\bar{e}}, N,{\bar{E}})$ accordingly. The same result also holds true in the case of AA and BB-firms.

1.2 Proof of Propositions

1.2.1 Proof of Proposition 2

The Kuhn–Tucker conditions for the Lagrangian from Eq. (22) read

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial p}&= \big ( W_{AB} - W_{AA} \big ) \frac{\partial \theta _{AA}^{AB}}{\partial p} + \big (\theta _{AA}^{AB} - {\underline{\theta }} \big ) \frac{\partial W_{AB}}{\partial p} + \big ( {\overline{\theta }}-\theta _{AA}^{AB} \big ) \frac{\partial W_{AA}}{\ }{\partial p} \nonumber \\&\quad +\,\lambda \frac{\partial \pi _{AB}}{\partial p} - \mu \bigg (\frac{\partial \pi _{AA}}{\partial p} - \frac{\partial \pi _{AB}}{\partial p} \bigg ) \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.7)

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial P}&= \big ( W_{AB} - W_{AA} \big ) \frac{\partial \theta _{AA}^{AB}}{\partial P} + \big ( {\overline{\theta }}-\theta _{AA}^{AB} \big ) \frac{\partial W_{AA}}{\partial P} - \mu \frac{\partial \pi _{AA}}{\partial P} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.8)

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial g}&= \lambda \frac{\partial \pi _{AB}}{\partial g} - \mu \underbrace{\bigg (\frac{\partial \pi _{AA}}{\partial g} - \frac{\partial \pi _{AB}}{\partial g} \bigg )}_{=0} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.9)

$$\begin{aligned}&\lambda (\pi _{BB} - \pi _{AB} ) + \mu (\pi _{AA} - \pi _{AB}) = 0 \end{aligned}$$

(A.10)

$$\begin{aligned}&\lambda , \mu \ge 0 \end{aligned}$$

(A.11)

Since $\frac{\partial \pi _{AB}}{\partial g} = 1 >0$, FOC (A.9) can only be satisfied for $\lambda = 0$. Using $\frac{\partial \theta _{AA}^{AB}}{\partial p} = \frac{\partial \pi _{AB}}{\partial p} - \frac{\partial \pi _{AA}}{\partial p}$ and $\frac{\partial \theta _{AA}^{AB}}{\partial P} = - \frac{\partial \pi _{AA}}{\partial P}$ immediately leads to Eqs. (23) and (24) which are the starting points for the proof of Proposition 2.

First, note that at $p=P=\psi$, we have $\frac{\partial W_{AB}}{\partial p}\big |_{p=P=\psi }=\frac{\partial W_{AA}}{\partial p}\big |_{p=P=\psi }=\frac{\partial W_{AA}}{\partial P}\big |_{p=P=\psi } =0$. Since $\frac{\partial \theta _{AA}^{AB} }{\partial P} > 0$, $\frac{\partial \theta _{AA}^{AB} }{\partial p} < 0$ and $(W_{AB}- W_{AA} - \mu )<0$, it follows that $\frac{\partial {\mathcal {L}} }{\partial p}\big |_{p=P=\psi } = \big ( W_{AB} - W_{AA} - \mu \big ) \frac{\partial \theta _{AA}^{AB}}{\partial p} > 0$ and $\frac{\partial {\mathcal {L}} }{\partial P}\big |_{p=P=\psi } = \big ( W_{AB} - W_{AA} + \mu \big ) \frac{\partial \theta _{AA}^{AB}}{\partial P} < 0$. Hence, a marginal increase (decrease) of p (P) raises the welfare at $p=P=\psi$.

Second, given that $\frac{\partial \theta _{AA}^{AB}}{\partial p}<0$ and assuming for a moment that $\big ( W_{AB} - W_{AA} + \mu \big )<0$, we must have $\frac{\partial W_{AA}}{\partial P} = Q_{AA}^*{'} (\psi - P ) + k_{AA}^*{'} (2 \psi - p - P ) > 0$ to satisfy Eq. (24). For $p \ge \psi$, this requires P to be smaller than $\psi$. As $\big ( W_{AB} - W_{AA} + \mu \big ) \frac{\partial \theta _{AA}^{AB} }{\partial p} > 0$, we must have $p > \psi$ for $P \le \psi$ to satisfy FOC (23). This leads to $p> \psi > P$. Moreover, we can exclude the case $P> \psi > p$ because $P > \psi$ requires $2\psi - p - P >0$ to satisfy Eq. (24), whereas $p < \psi$ requires $2\psi - p - P <0$ to satisfy FOC (23), leading to a contradiction. Thus, we must have $p> \psi > P$ to satisfy both FOCs.

Third, to show that $\big ( W_{AB} - W_{AA} + \mu \big ) < 0$, note that for $\mu >0$ we must have $\pi _{AA}(p,g,P,{\bar{G}}) =\pi _{AB}(p,g,{\underline{\theta }})$. If $\big ( W_{AB} - W_{AA} + \mu \big ) > 0$, then we would have $p< \psi < P$ for the same reasons as above. But then $\pi _{AA}(p< \psi ,g,P>\psi ,{\bar{G}}) =\pi _{AB}(p<\psi ,g,{\underline{\theta }}$) implies that $\pi _{AA}(\psi ,g,\psi ,{\bar{G}}) > \pi _{AB}(\psi ,g,{\underline{\theta }})$, meaning that the first best was feasible. Hence, we must have $\big ( W_{AB} - W_{AA} + \mu \big ) < 0$.

The Lagrangian for the second maximization problem of (21) is given by

$$\begin{aligned} {\mathcal {L}} = W_{AA}^{BB} - \lambda \big (\pi _{AB} - \pi _{BB} \big ) - \mu \big (\pi _{AA} - \pi _{BB} \big ) \end{aligned}$$

(A.12)

and the Kuhn-Tucker conditions read

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial p}&= \big ( W_{BB} - W_{AA} \big ) \frac{\partial \theta _{AA}^{BB}}{\partial p} + \big ( {\overline{\theta }}-\theta _{AA}^{BB} \big ) \frac{\partial W_{AA}}{\partial p} - \lambda \frac{\partial \pi _{AB}}{\partial p} - \mu \frac{\partial \pi _{AA}}{\partial p} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.13)

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial P}&= \big ( W_{BB} - W_{AA} \big ) \frac{\partial \theta _{AA}^{BB}}{\partial P} + \big ( {\overline{\theta }}-\theta _{AA}^{BB} \big ) \frac{\partial W_{AA}}{\partial P} - \mu \frac{\partial \pi _{AA}}{\partial P} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.14)

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial g}&= \big ( W_{BB} - W_{AA} \big ) \frac{\partial \theta _{AA}^{BB}}{\partial g} - \lambda \frac{\partial \pi _{AB}}{\partial g} - \mu \frac{\partial \pi _{AA}}{\partial g} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.15)

$$\begin{aligned}&\lambda (\pi _{AB} - \pi _{BB}) + \mu (\pi _{AA} - \pi _{BB}) \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.16)

$$\begin{aligned}&\lambda , \mu \ge 0 \end{aligned}$$

(A.17)

Since $\frac{\partial \pi _{AA}}{\partial g}=\frac{\partial \pi _{AB}}{\partial g} = 1$ and $\frac{\partial \theta _{AA}^{BB}}{\partial g} = - \frac{\partial \pi _{AA}}{\partial g} = -1$, it follows from Eq. (A.15) that

$$\begin{aligned} \lambda = W_{AA} - W_{BB} - \mu > 0. \end{aligned}$$

(A.18)

In order to satisfy Eq. (A.16), we must have $\pi _{AB}=\pi _{BB}$, meaning that the regulator chooses g such that firms are indifferent between relocating later or immediately. Note that $\pi _{AB}=\pi _{BB}$ implies $\theta _{AA}^{BB} = \theta _{AA}^{AB}$. Plugging in Eq. (A.18) into Eq. (A.13) and using the facts that $\frac{\partial \theta _{AA}^{BB}}{\partial p} = - \frac{\partial \pi _{AA}}{\partial p}$ and $\frac{\partial \theta _{AA}^{AB}}{\partial p} = \frac{\partial \pi _{AB}}{\partial p}- \frac{\partial \pi _{AA}}{\partial p}$ immediately leads to

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial p}&= \big ( W_{BB} - W_{AA} + \mu \big ) \frac{\partial \theta _{AA}^{AB}}{\partial p} + \big ( {\overline{\theta }}-\theta _{AA}^{AB} \big ) \frac{\partial W_{AA}}{\partial p} \,{\mathop =\limits ^{!}}\, 0. \end{aligned}$$

(A.19)

Using $\frac{\partial \theta _{AA}^{BB}}{\partial P} = -\frac{\partial \pi _{AA}}{\partial P}=\frac{\partial \theta _{AA}^{AB}}{\partial P}$ for Eq. (A.14) yields

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial P}&= \big ( W_{BB} - W_{AA} + \mu \big ) \frac{\partial \theta _{AA}^{AB}}{\partial P} + \big ( {\overline{\theta }}-\theta _{AA}^{AB} \big ) \frac{\partial W_{AA}}{\partial P} \,{\mathop =\limits ^{!}}\, 0. \end{aligned}$$

(A.20)

Equations (A.19) and (A.20) are almost equivalent to the FOCs (23) and (24). Hence, for the same reasons as above, we must have $p>\psi >P$.

1.2.2 Proof of Proposition 3

Since there are no AB-firms, the maximization problem reduces to

$$\begin{aligned} \max \left\{ \begin{array}{l l l} \max \limits _{p,P,G} W_{AA}^{AA}(p,g=0,P,G) \quad {\text {s.t.}} \quad &{}\pi _{AA}(p,g=0,P,G) &{}\ge \pi _{BB}({\underline{\theta }}) \\ &{} G\cdot ({\bar{\theta }} - {\underline{\theta }}) &{}\le T_{AA}(p,P)\cdot ({\bar{\theta }} - {\underline{\theta }}) \\ \max \limits _{p,P,G} W_{AA}^{BB}(p,g=0,P,G) \quad {\text {s.t.}} \quad &{} \pi _{AA}(p,g=0,P,G) &{}\le \pi _{BB}({\underline{\theta }}) \\ &{} G\cdot ({\bar{\theta }} - \theta _{AA}^{BB}) &{} \le T_{AA}(p,P)\cdot ({\bar{\theta }} - \theta _{AA}^{BB}) \\ \end{array} \right\} \end{aligned}$$

(A.21)

where the second line of each maximization problem represents the budget constraint with $G\cdot ({\bar{\theta }} - {\underline{\theta }})$ and $G\cdot ({\bar{\theta }} - \theta _{AA}^{AB})$ being the total transfer expenditures of the government.

The Lagrangian of the second maximization problem of (A.21) is given by

$$\begin{aligned} {\mathcal {L}} = W_{AA}^{BB} - \mu (\pi _{AA} - \pi _{BB}) - \nu (G - T_{AA}) \end{aligned}$$

(A.22)

where the term $( {\overline{\theta }}-\theta _{AA}^{BB} )$ in the budget constraint has canceled out. The Kuhn-Tucker conditions read

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial p}&= ( W_{BB} - W_{AA} ) \frac{\partial \theta _{AA}^{BB}}{\partial p} + ( {\overline{\theta }}-\theta _{AA}^{BB} ) \frac{\partial W_{AA}}{\partial p} - \mu \frac{\partial \pi _{AA}}{\partial p} + \nu \frac{\partial T_{AA}}{\partial p} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.23)

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial P}&= ( W_{BB} - W_{AA} ) \frac{\partial \theta _{AA}^{BB}}{\partial P} + ( {\overline{\theta }}-\theta _{AA}^{BB} ) \frac{\partial W_{AA}}{\partial P} - \mu \frac{\partial \pi _{AA}}{\partial P} + \nu \frac{\partial T_{AA}}{\partial P} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.24)

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial G}&= ( W_{BB} - W_{AA} ) \frac{\partial \theta _{AA}^{BB}}{\partial G} - \mu \frac{\partial \pi _{AA}}{\partial G} - \nu \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.25)

$$\begin{aligned}&\mu (\pi _{AA} - \pi _{BB}) + \nu (G - T_{AA}) \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.26)

$$\begin{aligned}&\mu , \nu \ge 0 \end{aligned}$$

(A.27)

Taking into account that $\frac{\partial \theta _{AA}^{BB}}{\partial G} = -1$ and $\frac{\partial \pi _{AA}}{\partial G} = 1$, it follows from Eq. (A.25) that $\nu = W_{AA} - W_{BB} -\mu$. Substituting $\nu$ in Eqs. (A.23) as well as (A.24) and bearing in mind that $\frac{\partial \theta _{AA}^{BB}}{\partial i}= - \frac{\partial \pi _{AA}}{\partial i}$ for $i=p$, P, G leads to

$$\begin{aligned} \frac{\partial {\mathcal {L}} }{\partial p}&= \underbrace{\big ( W_{BB} - W_{AA} + \mu \big )\cdot \bigg (\frac{\partial \theta _{AA}^{BB}}{\partial p} - \frac{\partial T_{AA}}{\partial p}\bigg )}_{<0} +\underbrace{\big ( {\overline{\theta }}-\theta _{AA}^{BB} \big )}_{>0} \frac{\partial W_{AA}}{\partial p} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.28)

$$\begin{aligned} \frac{\partial {\mathcal {L}} }{\partial P}&= \underbrace{\big ( W_{BB} - W_{AA} + \mu \big ) \cdot \bigg (\frac{\partial \theta _{AA}^{BB}}{\partial P} - \frac{\partial T_{AA}}{\partial P}\bigg )}_{<0} +\underbrace{\big ( {\overline{\theta }}-\theta _{AA}^{BB} \big )}_{>0} \frac{\partial W_{AA}}{\partial P} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.29)

where $\mu$ is the Lagrangian multiplier for the constraint $\pi _{AA}(\cdot ) \le \pi _{BB}({\underline{\theta }})$. Moreover, note that

$$\begin{aligned} \frac{\partial \theta _{AA}^{BB}}{\partial p} - \frac{\partial T_{AA}}{\partial p} = p (q_{AA}^{*'} + k_{AA}^{*'}) + P (Q_{AA}^{*'} + k_{AA}^{*'})>0. \end{aligned}$$

(A.30)

For the first part of Proposition 3, note that if the regulator sets the highest possible transfer $G=T_{AA}(p,P)$, then the profit of an AA firm reads $\pi _{AA}^*(p,P,g=0, G=T_{AA}(p,P)) = 2 - \kappa (k_{AA}^*(p + P)) - \gamma (q_{AA}^*(p)) - \gamma (Q_{AA}^*(P))$. Thus, there is no relocation for first-best prices as long as ${\underline{\theta }} \ge \kappa (k_{AA}^*(2 \psi )) + \gamma (q_{AA}^*(\psi )) + \gamma (Q_{AA}^*(\psi ))$ and the regulator can implement the first-best.

For the second part, if ${\underline{\theta }} < \kappa (k_{AA}^*(2 \psi )) + \gamma (q_{AA}^*(\psi )) + \gamma (Q_{AA}^*(\psi ))$, the regulator optimally reduces the carbon prices. At $p=P=\psi$, we have $\frac{\partial W_{AA}}{\partial p}\big |_{p=P=\psi }=\frac{\partial W_{AA}}{\partial P}\big |_{p=P=\psi }=0$, meaning that $\frac{\partial {\mathcal {L}}}{\partial p}\big |_{p=P=\psi } = \frac{\partial {\mathcal {L}}}{\partial P}\big |_{p=P=\psi } = \big ( W_{BB} - W_{AA} + \mu \big ) \frac{\partial \theta _{AA}^{BB}}{\partial i} < 0$ for $i = p$, P. Since $\frac{\partial \theta _{AA}^{BB}}{\partial p} >0$ as well as $\frac{\partial \theta _{AA}^{BB}}{\partial P} >0$ and $W_{BB} - W_{AA} + \mu < 0$, we must have $\frac{\partial W_{AA}}{\partial p}>0$ and $\frac{\partial W_{AA}}{\partial P}>0$ to satisfy the FOCs (A.28) and (A.29). This requires $p+P<2 \psi$. Since the FOCs (A.28) and (A.29) are symmetric, there is a unique welfare maximum with $p=P<\psi$.

1.2.3 Proof of Proposition 4

The Lagrangian for the first optimization problem in (27) reads

$$\begin{aligned} {\mathcal {L}} = W_{AA}^{AB} - \lambda (\pi _{BB}-\pi _{AB}) - \mu (\pi _{AA} - \pi _{AB}) - \nu (g ({\bar{\theta }} - {\underline{\theta }}) - T) \end{aligned}$$

(A.31)

and the Kuhn-Tucker conditions are given by

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial p}&= ( W_{AB} - W_{AA} ) \frac{\partial \theta _{AA}^{AB}}{\partial p} + (\theta _{AA}^{AB} - {\underline{\theta }}) \frac{\partial W_{AB}}{\partial p} + ( {\overline{\theta }}-\theta _{AA}^{AB} ) \frac{\partial W_{AA}}{\partial p} \nonumber \\&\quad +\,\lambda \frac{\partial \pi _{AB}}{\partial p} - \mu \bigg (\frac{\partial \pi _{AA}}{\partial p}- \frac{\partial \pi _{AB}}{\partial p} \bigg ) + \nu \frac{\partial T}{\partial p} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.32)

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial P}&= ( W_{AB} - W_{AA} ) \frac{\partial \theta _{AA}^{AB}}{\partial P} + ( {\overline{\theta }}-\theta _{AA}^{AB} ) \frac{\partial W_{AA}}{\partial P} - \mu \frac{\partial \pi _{AA}}{\partial P} + \nu \frac{\partial T}{\partial P} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.33)

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial g}&= \lambda \frac{\partial \pi _{AB}}{\partial g} - \mu \bigg (\frac{\partial \pi _{AA}}{\partial g}- \frac{\partial \pi _{AB}}{\partial g} \bigg ) - \nu ({\bar{\theta }} - {\underline{\theta }}) \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.34)

$$\begin{aligned}&\lambda (\pi _{BB} - \pi _{AB}) + \mu (\pi _{AA} - \pi _{AB})+\nu (g ({\bar{\theta }} - {\underline{\theta }}) - T) \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.35)

$$\begin{aligned}&\lambda , \mu , \nu \ge 0 \end{aligned}$$

(A.36)

Taking into account that $\frac{\partial \pi _{AB}}{\partial g}=1$ and that $\frac{\partial \pi _{AA}}{\partial g}- \frac{\partial \pi _{AB}}{\partial g} =0$, Eq. (A.34) can be reduced to $\lambda = \nu ({\bar{\theta }} - {\underline{\theta }})$. Plugging this in into Eq. (A.32) and performing the same transformations as in the proof of Proposition 2 leads to Eq. (28) from the text. For the first line in Table 2 from Proposition 4 note that the sign of the last term of Eq. (28) is indeterminate. Hence, the third best p can be either above or below $\psi$. For the last entry in the first line, the last term of Eq. (29) is certainly negative, implying the term $\frac{\partial W_{AA}}{\partial P}$ to be positive which holds only true for $P < \psi$ for the same reasons as pointed out in the proof of Proposition 2. However, if $\frac{\partial T(p,P)}{\partial P} > 0$, then P can be below or above $\psi$.

The Lagrangian for the second optimization problem in (27) reads

$$\begin{aligned} {\mathcal {L}} = W_{AA}^{BB} - \lambda (\pi _{AB}-\pi _{BB}) - \mu (\pi _{AA} - \pi _{BB}) - \nu (g - T_{AA}) \end{aligned}$$

(A.37)

and the Kuhn-Tucker conditions are given by

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial p}&= ( W_{BB} - W_{AA} ) \frac{\partial \theta _{AA}^{BB}}{\partial p} + ( {\overline{\theta }}-\theta _{AA}^{BB} ) \frac{\partial W_{AA}}{\partial p} - \lambda \frac{\partial \pi _{AB}}{\partial p} - \mu \frac{\partial \pi _{AA}}{\partial p} + \nu \frac{\partial T_{AA}}{\partial p} \,\,{\mathop =\limits ^{!}}\,\, 0 \end{aligned}$$

(A.38)

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial P}&= ( W_{BB} - W_{AA} ) \frac{\partial \theta _{AA}^{BB}}{\partial P} + ( {\overline{\theta }}-\theta _{AA}^{BB} ) \frac{\partial W_{AA}}{\partial P} - \mu \frac{\partial \pi _{AA}}{\partial P} + \nu \frac{\partial T_{AA}}{\partial P} \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.39)

$$\begin{aligned} \frac{\partial {\mathcal {L}}}{\partial g}&= ( W_{BB} - W_{AA} ) \frac{\partial \theta _{AA}^{BB}}{\partial g} - \lambda \frac{\partial \pi _{AB}}{\partial g} - \mu \frac{\partial \pi _{AB}}{\partial g} + \nu \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.40)

$$\begin{aligned}&\lambda (\pi _{BB} - \pi _{AB}) + \mu (\pi _{AA} - \pi _{BB}) + \nu (g - T_{AA}) \,{\mathop =\limits ^{!}}\, 0 \end{aligned}$$

(A.41)

$$\begin{aligned}&\lambda , \mu , \nu \ge 0 \end{aligned}$$

(A.42)

Taking into account that $\frac{\partial \theta _{AA}^{BB}}{\partial g} = -1$ and $\frac{\partial \pi _{AB}}{\partial g} =1$, Eq. (A.40) reduces to $\nu = W_{AA} - W_{BB} - \lambda - \mu$. Plugging this into Eqs. (A.38) and (A.39) leads to Eqs. (31) and (32) from the text. If $\lambda >0$, then p can be above or below $\psi$ because the sign of the last term in Eq. (31) is indeterminate which proofs the first entry in the second line of Table 2. If $\frac{\partial T_{AA}}{\partial P} > 0$, then the last term of Eq. (32) is negative and P must be below $\psi$. If the opposite holds true, then P can be above or below $\psi$ which proofs the other entries of the second line. For the last line, if $\lambda =0$, then the FOCs (31) and (32) are equivalent to (A.28) and (A.29) and we have $p=P< \psi$ for the same reasons as outlined in the proof of Proposition 3.