Auctioning Versus Grandfathering in Cap-and-Trade Systems with Market Power and Incomplete Information

Abstract

We compare auctioning and grandfathering as allocation mechanisms of emission permits when there is a secondary market with market power and firms have private information on their own abatement technologies. Based on real-life cases such as the EU ETS, we consider a multi-unit, multi-bid uniform auction. At the auction, each firm anticipates its role in the secondary market, either as a leader or a follower. This role affects each firms’ valuation of the permits (which are not common across firms) as well as their bidding strategies and it precludes the auction from generating a cost-effective allocation of permits, as it occurs in simpler auction models. Auctioning tends to be more cost-effective than grandfathering when the firms’ abatement cost functions are sufficiently different from one another, especially if the follower has lower abatement costs than the leader and the dispersion of the marginal costs is large enough.

Appendices

Appendix 1: Notations

Fundamentals
$\bar{Q}$	Total amount of permits to be shared, exogenous
$i,j\in \left\{ F,L\right\}$	Firms: $F$ and $L$ denote follower and leader, respectively
$MC _{i}\left( e\right) =\alpha _{i}-\beta e$	Marginal abatement cost of firm $i$, where $e$ is the emission level
$\alpha _{i}$, $\varvec{\alpha }=\left( \alpha _{F},\alpha _{L}\right)$	Firms’ types, random. The support of $\varvec{\alpha }$ belongs to $\Omega := \left[ \theta ,\theta +\sigma \right] ^{2}$
$D$	Diagonal of $\Omega$, i.e., $D:=\left\{ \left( \alpha , \alpha \right) \in \Omega \right\}$
$\kappa :=\frac{\beta \bar{Q}}{\sigma }$	Ratio market size to across-firm heterogeneity
$E\{\alpha _{i}\mid \alpha _{j}\}=\mu _{j}+\lambda _{j}\alpha _{j}$	Expectation of $\alpha _{i}$ conditional on $\alpha _{j}$, assumed linear with coeffs. $\mu _{j},\lambda _{j}$
$e_{i}^{ BAU }$, $e_{i}^{CE}$	Business as usual and cost-effective emissions of $i$, respectively
$\xi ,\Theta ,\Theta _{L},\Theta _{F}$	Auxiliary coefficients
Secondary market
$q_{i1},$ $p_{1}$	Post-market holdings of permits for firm $i$ and market price, respectively
$\Pi _{i}\left( q_{i1},p_{1},\varvec{\alpha }\right)$	Realized profit of firm $i$
$\pi _{i}\left( q_{i0},\varvec{\alpha }\right) :=\Pi _{i}\left( q_{i1}^{*},p_{1}^{*},\varvec{\alpha }\right)$	Realized profit of firm $i$ (under eq. in the secondary market) (the asterisk denotes equilibrium value)
$v_{i}\left( q_{i0},\alpha _{i}\right)$	Value function for firm $i$ (valuation of initially getting $q_{i0}$ permits)
Primary market (auctioning or grandfathering)
$q_{i0}$	Initial allocation of permits for firm $i$ (under grand. or auctioning)
$p_{0}$	Price of permits at the auction
$\delta :=\frac{q_{F0}}{\bar{Q}}$	$F^{\prime }$s share of permits ($\delta ^{A}$, $\delta ^{G}$ under auctioning or grandf., respectively)
$h\left( \delta ,\varvec{\alpha }\right)$	Monotone transformation of total abatement cost (incorporates secondary-market behavior)
$q_{i}^{b}\left( \alpha _{i},p_{0}\right)$	Firm $i$’s bid (mapping from support of $\alpha _{i}$ to set of demand funct.)
$\Omega _{l},$ $l=1,2,3$	Set of types’ values such that type-$l$ equilibrium occurs
$\left( p_{L},p_{Fu},p_{Fd}\right)$	Auxiliary variables to define equilibrium strategies at the auction

Appendix 2: Proofs

1.1 Proof of Proposition 1

Let us initially assume that the solution is interior (which we check below). The optimal behavior of the follower is derived by maximizing (9) for $i=F$ while taking $p_{1}$ as given. From the first-order condition (FOC), we get the follower’s net demand for permits:

$$\begin{aligned} q_{F1}^{d}\left( p_{1}\right) =\frac{1}{\beta }\left( \alpha _{F}-p_{1}\right) , \end{aligned}$$

(23)

where $d$ stands for “demand”. The leader’s problem is to maximize (9) for $i=L$ including the market clearance condition, $q_{L1}+q_{F1}=\bar{Q}$, and the follower’s demand, (23), as constraints. The (interior) solution to this problem is given by (10). Using the market clearing condition and (23), we get the equilibrium price in the secondary market,

$$\begin{aligned} p_{1}^{*}=\frac{\alpha _{L}+2\alpha _{F}-\beta \left( 2\bar{Q} -q_{L0}\right) }{3}, \end{aligned}$$

(24)

and using (24) in (23) we obtain the equilibrium number of permits for the follower, given by (11). Using (10), (11) and (24) in the expressions for $F$’s and $L$’s profits and rearranging, we get (12) and (13), where $\Theta _{L}$ and $\Theta _{F}$ include the terms that do not depend on the initial allocation.

To show that the solution satisfies $q_{i1}\in [0,\min \{\bar{Q} ,e_{i}^{ BAU }\}]$ w.p.1, note first that, by construction, $0\le q_{F1}\le \bar{Q}$ ensures $0\le q_{L1}\le \bar{Q}$ and using (23) we have that $0\le q_{F1}\le \bar{Q}\iff \alpha _{F}\ge p_{1}\ge \alpha _{F}-\beta \bar{Q}$, which, using (24), can be written as $\frac{\alpha _{F}-\alpha _{L}}{\beta } +2\bar{Q}\ge q_{L0}\ge \frac{(\alpha _{F}-\alpha _{L})}{\beta }-\bar{Q}$. As $q_{L0}\in [0,\bar{Q}]$, the previous inequalities are ensured for any initial allocation iff $\frac{\alpha _{F}-\alpha _{L}}{\beta }+2\bar{Q} \ge \bar{Q}$ and$\ 0\ge \frac{(\alpha _{F}-\alpha _{L})}{\beta }-\bar{Q}$ hold at the same time. The intersection of these inequalities is: $\bar{Q} \ge \frac{1}{\beta }|\alpha _{L}-\alpha _{F}|$, which is guaranteed if

$$\begin{aligned} \beta \bar{Q}\ge \sigma \end{aligned}$$

(25)

as the maximum value of $|\alpha _{L}-\alpha _{F}|$ is $\sigma$.

Firm $i$ makes a strictly positive abatement effort iff $q_{i1}\le e_{i}^{ BAU }=\frac{\alpha _{i}}{\beta }$. If we particularize this inequality for $i=F$, using (23), we have $p_{1}\ge 0$ , which, using (24), can be written as $\alpha _{L}+2\alpha _{F}\ge \beta (2\bar{Q}-q_{L0})$. As $q_{L0}\in [0,\bar{Q}]$ and the minimum value for each type is $\theta$, the latter equality is guaranteed under (4). For $i=L$, using (10), $q_{L1}\le \frac{\alpha _{L}}{\beta }$ can be written as

$$\begin{aligned} 2\alpha _{L}+\alpha _{F}\ge \beta (\bar{Q}+q_{L0}). \end{aligned}$$

(26)

The most adverse values for the fulfilment of (26) are $\alpha _{L}=\alpha _{F}=\theta$, $q_{L0}= \bar{Q}$. To ensure that (26) holds w.p.1, we plug these values to conclude $\theta \ge \frac{2}{3}\beta \bar{Q}$. Combining this inequality with (25), we get (4), which holds by assumption.

1.2 Proof of Corollary 1

Total cost can be computed by inserting (10) and (11) in (1).²⁶ Substituting $q_{L0}$ by $\bar{Q} -q_{F0}$ and rearranging, we get (14), where $\Theta$ includes the terms that do not depend on the initial allocation.

1.3 Proof of Proposition 2

Using (13), $v_{F}\left( q_{F0},\alpha _{F}\right)$ can be written as

$$\begin{aligned} v_{F}\left( q_{F0},\alpha _{F}\right) =E\left\{ \Theta _{F}\mid \alpha _{F}\right\} +\frac{1}{9}\left( 2E\{\alpha _{L}\mid \alpha _{F}\}+7\alpha _{F}-2\beta \bar{Q}\right) q_{F0}-\frac{5}{18}\beta q_{F0}^{2}. \end{aligned}$$

As $v_{F}$ is strictly concave in $q_{F0}$ for every $\alpha _{F}$, an interior solution to $F$’s problem (if it exists) follows easily from the first-order condition, which can be solved to get

$$\begin{aligned} q_{F}^{b}=\tau _{F}\left( q_{F0},\alpha _{F}\right) :=\frac{2E\{\alpha _{L}\mid \alpha _{F}\}+7\alpha _{F}-2\beta \bar{Q}-9p_{0}}{5\beta }, \end{aligned}$$

(27)

where, for simplicity of exposition, we have denoted as $\tau _{F}\left( q_{F0},\alpha _{F}\right)$, the right-hand side of the previous equality. Since the submitted demand must lie in $[0,\bar{Q}]$, and given the concavity of $v_{F}$ on $q_{F0}$, we conclude that $q_{F}^{b}=0$ and $q_{F}^{b}=\bar{Q}$ whenever $\tau _{F}\left( p_{0},\alpha _{F}\right) \le 0$ and $\tau _{F}\left( p_{0},\alpha _{F}\right) \ge \bar{Q}$, respectively. Note also that $\tau _{F}\left( p_{0},\alpha _{F}\right)$ is strictly decreasing in $p_{0}$ for every $\alpha _{F}$, hence $p_{Fu}$ and $p_{Fd}$ are defined by $\tau _{F}(p_{Fu},\alpha _{F})=0$ and $\tau _{F}(p_{Fd},\alpha _{F})=\bar{Q}$, respectively. Thus, we have

$$\begin{aligned} p_{Fu}=\frac{2E\{\alpha _{L}\mid \alpha _{F}\}+7\alpha _{F}-2\beta \bar{Q}}{9 }\ ,\quad p_{Fd}=\frac{2E\{\alpha _{L}\mid \alpha _{F}\}+7\alpha _{F}-7\beta \bar{Q}}{9}. \end{aligned}$$

(28)

Let us now consider $L$’s problem, which is $max_{q_{L0}}\{v_{L}\left( q_{L0},\alpha _{L}\right) -p_{0}q_{L0}\}$ where

$$\begin{aligned} v_{L}\left( q_{L0},\alpha _{L}\right)&= E\{\pi _{L}\left( q_{L0},\varvec{ \varvec{\alpha }}\right) \mid \alpha _{L},q_{L0}\}=E\left( \Theta _{L}\mid \alpha _{L}\right) \\&+\,\frac{\alpha _{L}+2E\{\alpha _{F}\mid \alpha _{L}\}-2\beta \bar{Q}}{3}q_{L0}\,+\,\frac{\beta }{6}q_{L0}^{2}. \end{aligned}$$

Clearly $v_{L}\left( q_{L0},\alpha _{L}\right)$ is strictly convex in $q_{L0}$ for every $\alpha _{L}$, and therefore the leader’s problem only has corner solutions, i.e., $q_{L}^{b}\in \{0,\bar{Q}\}$ with $q_{L}^{b}=\bar{Q} \iff v_{L}\left( \bar{Q},\alpha _{L}\right) \ge v_{L}\left( 0,\alpha _{L}\right)$. Given that $v_{L}\left( 0,\alpha _{L}\right) =E\left\{ \Theta _{L}\mid \alpha _{L}\right\}$, $L$ chooses $q_{L0}=\bar{Q}$ whenever $v_{L}\left( \bar{Q},\alpha _{L}\right) \ge E\left\{ \Theta _{L}\mid \alpha _{L}\right\} +p_{0}\bar{Q}$ holds. Let us denote as $p_{L}$ the maximum value of $p_{0}$ such that this inequality holds, i.e., $p_{L}=\frac{ v_{L}\left( \bar{Q},\alpha _{L}\right) -E\left\{ \Theta _{L}\mid \alpha _{L}\right\} }{\bar{Q}}$. Using the expression for $v_{L}$ we have that

$$\begin{aligned} p_{L}:=\frac{\alpha _{L}+2E\{\alpha _{F}\mid \alpha _{L}\}}{3}\,-\,\frac{1}{2} \beta \bar{Q}. \end{aligned}$$

(29)

1.4 Proof of Proposition 3

Given that the support of $\alpha _{L}$ and $\alpha _{F}$ is $\left[ \theta ,\theta +\sigma \right]$, from (29) we know $p_{L}>\theta -\frac{1}{2 }\beta \bar{Q}>0$, where the latter inequality follows from (4). Thus, there exists at least one positive price ($p_{L}$) at which the demanded quantity is not smaller than supply, which guarantees the existence of an equilibrium in the auction. Uniqueness follows from the fact that each firm’s demand is non-increasing in $p_{0}$ for every possible pair $\left( \alpha _{F},\alpha _{L}\right)$. Seeing as $p_{Fu}>p_{Fd}$ and $p_{L}>0$ hold w.p.1, there are just three possible cases.

• First case: $p_{L}<p_{Fd}$. In this case, the stop-out price cannot be $p_{0}>p_{Fd}$ because there would be excess supply (i.e., some units would not be awarded). As all the permits would be awarded for any $p_{0}\le p_{Fd}$, the opt-out price must be $p_{Fd}$. At this price, $F$ bids $\bar{Q}$, $L$ bids zero and hence $F$ gets all the permits and $L$ gets nothing. This corresponds to type-1 equilibrium.

• Second case: $p_{Fu}<p_{L}$. In this case, for any $p_{0}>p_{L}$ there would be excess supply and all the permits would be awarded for any $p_{0}\le p_{L}$. Therefore, the opt-out price must be $p_{L}$. At this price, $L$ bids $\bar{Q}$ and $F$ bids zero. Thus, $L$ gets all the permits, which corresponds to type-2 equilibrium.

• The third case is $p_{Fd}\le p_{L}\le p_{Fu}$, which corresponds to type-3 equilibrium, in which permits are shared between $F$ and $L$. The equilibrium price is $p_{L}$, at which $F$ and $L$ demand $\tau _{F}(p_{L},\alpha _{F})\in \left[ 0,\bar{Q}\right]$ and $\bar{Q}$, respectively, and rationing is required. Using (28) and (29), straightforward algebra leads to $p_{L}<p_{Fd}\iff \xi <-1$, $p_{Fu}<p_{L}\iff \xi >1$ and $p_{Fd}\le p_{L}\le p_{Fu}\iff -1\le \xi \le 1$.

1.5 Proof of Proposition 4

We split the proof into the following parts:

Part i)

Under independent types, using (5), (6), (21) and the definition of $\kappa$, the conditions given in Proposition 3 for type-1 and type-2 equilibria collapse to

$$\begin{aligned} \text {type 1:}&{\qquad }\xi <-1\Leftrightarrow 3\alpha _{L}-7\alpha _{F}+4\theta +2\sigma +\frac{5\sigma \kappa }{2}<0, \end{aligned}$$

(30)

$$\begin{aligned} \text {type 2:}&{\qquad }\xi >1\Leftrightarrow 7\alpha _{F}-3\alpha _{L}-4\theta -2\sigma +\frac{5\sigma \kappa }{2}<0. \end{aligned}$$

(31)

For any pair in $D$, (30) and (31) become, respectively

$$\begin{aligned} 5\kappa <8a-4, \quad 5\kappa <4-8a, \end{aligned}$$

none of which, under (4), holds for any admissible values of $\kappa$ and $a$. Thus, we only have type-3 equilibria in $D$, which proves the first statement in the proposition.

Parts ii) and iii)

The results in these parts straightforwardly follow from the fact that conditions (30) and (31) define two convex sets, that $\left( \alpha _{F},\alpha _{L}\right) =\left( \theta +\sigma ,\theta \right)$, is the most favorable pair to satisfy (30) and that $\left( \alpha _{F},\alpha _{L}\right) =\left( \theta ,\theta +\sigma \right)$, is the most favorable pair to satisfy (31).

Part iv)

To compute the probabilities of type-1 and type-2 equilibria, notice first that the pair $\left( \alpha _{F},\alpha _{L}\right) =\left( \theta +\sigma ,\theta \right)$, which is the most favorable pair to satisfy (30), satisfies it if and only if $\kappa <2$ holds. Similarly, the pair $\varvec{\alpha }=\left( \theta ,\theta +\sigma \right)$ , which is the most favorable pair to satisfy (31), satisfies it if and only if $\kappa <2$ holds. Contrarily, if $\kappa \ge 2$ holds, (30) and (31) never hold and hence $\Pr (\Omega _{1})=\Pr (\Omega _{2})=0$.

Let us now consider the $\kappa <2$ case. Under the uniform distribution assumption, the probability of type-$i$ equilibria is proportional to the area of $\Omega _{i}$ (for $i=1,2$).

Regarding $\Omega _{1}$, as shown in Fig. 2, this area is a triangle in the bottom right corner of $\Omega$. To compute the size of this triangle, we first compute the infimum value for $\alpha _{F}$ in $\Omega _{1}$, which can be obtained by taking $\alpha _{L}=\theta$ in (30) and solving for $\alpha _{F}$ to obtain

$$\begin{aligned} \alpha _{F}^{Inf}:=\theta +n_{F}\sigma , \quad \text { where }\quad n_{F}:= \frac{4+5\kappa }{14}. \end{aligned}$$

Similarly, the supremum value for $\alpha _{L}$ in $\Omega _{1}$ is obtained by taking $\alpha _{F}=\theta +\sigma$ in (30) and solving to obtain

$$\begin{aligned} \alpha _{L}^{Sup}:=\theta +\frac{5\left( 2-\kappa \right) }{6}\sigma . \end{aligned}$$

Using these values, we can obtain the area of $\Omega _{1}$:

$$\begin{aligned} A(\Omega _{1}):=\frac{1}{2}\left( \theta +\sigma -\alpha _{F}^{Inf}\right) \left( \alpha _{L}^{Sup}-\theta \right) =\frac{25\sigma ^{2}}{42}\left( 1-\frac{\kappa }{2} \right) ^{2} \end{aligned}$$

and bearing in mind that the size of the square $\Omega$ is equal to $\sigma ^{2}$, the probability of type-1 equilibrium is $\Pr (\Omega _{1})= \frac{1}{\sigma ^{2}}A(\Omega _{1})$.

Regarding $\Omega _{2}$, as shown in Fig. 2, this region is a triangle in the top left corner of $\Omega$. Proceeding as above, from (30) we can compute the infimum value of $\alpha _{L}$ and the supremum value of $\alpha _{F}$ within $\Omega _{2}$:

$$\begin{aligned} \alpha _{L}^{Inf}=\theta +\frac{5\kappa -4}{6}\sigma \text {, }\qquad \alpha _{F}^{Sup}=\theta +\frac{5\left( \kappa -2\right) }{14}\sigma . \end{aligned}$$

Using the relevant expressions, it is straightforward to check that $\theta +\sigma -\alpha _{F}^{Inf}=\alpha _{F}^{Sup}-\theta$ and $\alpha _{L}^{Sup}-\theta =\theta +\sigma -\alpha _{L}^{Inf}$, i.e., $\Omega _{2}$ is simply a $180 {{}^o}$-rotation of $\Omega _{1}$ and, therefore, both areas are identical, which, jointly with the uniform distribution assumption, implies $\Pr (\Omega _{1})=\Pr (\Omega _{2})$.

Part v)

Notice that, if both types are independent, under grandfathering we always have

$$\begin{aligned} \delta ^{G}=\frac{q_{F0}^{G}}{\bar{Q}}=\frac{E\{e_{F}^{*}\}}{ E\{e_{F}^{*}\}+E\{e_{L}^{*}\}}=\frac{1}{2}. \end{aligned}$$

Let us first consider type-1 equilibria. In this case we know that the follower gets all the permits and, therefore, $\delta ^{A}=1$. Using $\kappa \sigma =\beta \bar{Q}$ and the values of $\delta ^{G}$ and $\delta ^{A}$ in (15), we have that

$$\begin{aligned}&E\{h(\delta ^{G},{\varvec{\alpha }}) \mid \Omega _{1}\}=E\{h(1/2,{\varvec{\alpha }})\mid \Omega _{1}\}=\frac{1}{2} E\{\alpha _{L}-\alpha _{F}\mid \Omega _{1}\}-\frac{\kappa \sigma }{4},\\&E\{h(\delta ^{A},{\varvec{\alpha }}) \mid \Omega _{1}\}=E\{h(1, {\varvec{\alpha }})\mid \Omega _{1}\}=E\{\alpha _{L}-\alpha _{F}\mid \Omega _{1}\}, \end{aligned}$$

and direct comparison reveals that

$$\begin{aligned} E\{h(\delta ^{A},{\varvec{\alpha }})\mid \Omega _{1}\}<E\{h(\delta ^{G},{\varvec{\alpha }})\mid \Omega _{1}\}\iff E\{\alpha _{L}-\alpha _{F}\mid \Omega _{1}\}<-\frac{\kappa \sigma }{2}. \end{aligned}$$

(32)

Therefore, to determine the relative expected costs we need to compute $E\left\{ \alpha _{L}-\alpha _{F}\mid \Omega _{1}\right\}$. The first step is to obtain the conditional distributions of $\alpha _{F}\mid \Omega _{1}$ and $\alpha _{L}\mid \Omega _{1}$. As regards $\alpha _{F}\mid \Omega _{1}$, note that the relevant support is the interval $[\alpha _{F}^{Inf},\theta +\sigma ]$. For any arbitrary value of $\alpha _{F}$ in that interval, say $\alpha$, according to (30) we know that $\alpha _{L}$ ranges (within $\Omega _{1}$) from $\theta$ to $\hat{\alpha }_{L}(\alpha ):= \frac{7\alpha -4\theta }{3}-\frac{\left( 4+5\kappa \right) \sigma }{6}$. Thus, the joint probability of the events “$\alpha _{F}=\alpha$” and “type-1 equilibrium occurring” is²⁷

$$\begin{aligned} \Pr (\alpha _{F}=\alpha \cap \Omega _{1})=\frac{1}{\sigma ^{2}}\int _{\theta }^{\hat{\alpha }_{L}(\alpha )}dz=\frac{1}{\sigma ^{2}}\left( \hat{\alpha } _{L}(\alpha )-\theta \right) . \end{aligned}$$

We now apply Bayes’ rule to obtain the conditional density

$$\begin{aligned} \Pr (\alpha _{F}=\alpha \mid \Omega _{1})=\frac{\Pr (\alpha _{F}=\alpha \cap \Omega _{1})}{\Pr (\Omega _{1})}=7k_{0}^{F}\left( \alpha -\theta -n_{F}\sigma \right) \end{aligned}$$

where $k_{0}^{F}=\left( 3A(\Omega _{1})\right) ^{-1}$. The corresponding conditional expectation is

$$\begin{aligned} E\{\alpha _{F}\mid \Omega _{1}\}=\int _{\alpha _{F}^{Inf}}^{\theta +\sigma }\alpha \Pr (\alpha _{F}=\alpha \mid \Omega _{1})d\alpha =7k_{0}^{F}\int _{\alpha _{F}^{Inf}}^{\theta +\sigma }\left\{ \alpha ^{2}- \left[ \theta +n_{F}\sigma \right] \alpha \right\} d\alpha \end{aligned}$$

which, after some straight-forward algebra, results in

$$\begin{aligned} E\{\alpha _{F}\mid \Omega _{1}\}=\theta +\left( \frac{5\kappa +32}{42} \right) \sigma . \end{aligned}$$

(33)

We proceed analogously with $\alpha _{L}\mid \Omega _{1}$. Its support is $[\theta ,\alpha _{L}^{Sup}]$. For any arbitrary $\alpha$ in that interval, from (30) we conclude that $\alpha _{F}$ ranges (within $\Omega _{1}$) from $\hat{\alpha }_{F}(\alpha ):=\frac{3}{7}\alpha + \frac{4}{7}\theta +\frac{1}{7}\left( 2+\frac{5}{2}\kappa \right) \sigma$ to $\theta +\sigma$.

The joint probability of the events “$\alpha _{L}=\alpha$” and “type-1 equilibrium occurring” is

$$\begin{aligned} \Pr (\alpha _{L}=\alpha \cap \Omega _{1})=\frac{1}{\sigma ^{2}}\int _{\hat{ \alpha }_{F}(\alpha )}^{\theta +\sigma }dz=\frac{1}{\sigma ^{2}}\left( \theta +\sigma -\hat{\alpha }_{F}(\alpha )\right) \end{aligned}$$

and using Bayes’ rule,

$$\begin{aligned} \Pr (\alpha _{L}=\alpha \mid \Omega _{1})=k_{0}^{L}\left( \theta -\alpha + \frac{5}{3}\left( 1-\frac{1}{2}\kappa \right) \sigma \right) \end{aligned}$$

where $k_{0}^{L}=3\left( 7\sigma ^{2}\Pr (\Omega _{1})\right) ^{-1}$. The corresponding conditional expectation is

$$\begin{aligned} E\{\alpha _{L}\mid \Omega _{1}\}=\int _{\theta }^{\alpha _{L}^{Sup}}\alpha \Pr (\alpha _{L}=\alpha \mid \Omega _{1})d\alpha =\theta +\frac{5\left( 2-\kappa \right) }{18}\sigma . \end{aligned}$$

(34)

Using (33) and (34), we have that

$$\begin{aligned} E\{\alpha _{L}-\alpha _{F}\mid \Omega _{1}\}=-\frac{13+25\kappa }{63}\sigma \end{aligned}$$

(35)

and using (35) in (32) we conclude

$$\begin{aligned} E\{h(\delta ^{A},\varvec{{\alpha }})\mid \Omega _{1}\}\le E\{h(\delta ^{G},\varvec{{\alpha }})\mid \Omega _{1}\}\iff \kappa \le 2 \end{aligned}$$

where the latter inequality is trivially true if $\Omega _{1}$ is nonempty (which requires $\kappa <2$).

Under type-2 equilibrium, we know that the leader obtains all the permits and, therefore, $\delta ^{A}=0$. Using this value and $\delta ^{G}=\frac{1}{2 }$ in (15), we have that

$$\begin{aligned}&E\{h(\delta ^{A},{\varvec{\alpha }}) \mid \Omega _{2}\}=E\{h(0,{\varvec{\alpha }})\mid \Omega _{2}\}=0, \\&E\{h(\delta ^{G},{\varvec{\alpha }}) \mid \Omega _{2}\}=E\{h(1/2,{\varvec{\alpha }})\mid \Omega _{2}\}=\frac{1}{2} E\{\alpha _{L}-\alpha _{F}\mid \Omega _{2}\}-\frac{1}{4}\kappa \sigma \end{aligned}$$

and comparing both expressions,

$$\begin{aligned} E\{h(\delta ^{A},{\varvec{\alpha }})\mid \Omega _{2}\}<E\{h(\delta ^{G},{\varvec{\alpha }})\mid \Omega _{2}\}\Leftrightarrow E\{\alpha _{L}-\alpha _{F}\mid \Omega _{2}\}>\frac{ \kappa \sigma }{2}. \end{aligned}$$

(36)

To compute $E\{\alpha _{L}-\alpha _{F}\mid \Omega _{2}\}$, we can proceed as we did for equilibrium one (obtaining the conditional distributions and then computing the conditional expectations of $\alpha _{L}$ and $\alpha _{F}$ ). However, it is possible to exploit the rotational symmetry property between areas $\Omega _{1}$ and $\Omega _{2}$ (see part iii) above) to show ²⁸

$$\begin{aligned} E\{\alpha _{L}-\alpha _{F}\mid \Omega _{2}\}=-E\{\alpha _{L}-\alpha _{F}\mid \Omega _{1}\}=\frac{7+31\kappa }{63}\sigma \end{aligned}$$

and using this value in (36), we conclude that

$$\begin{aligned} E\{h(\delta ^{A},{\varvec{\alpha }})\mid \Omega _{2}\}<E\{h(\delta ^{G},{\varvec{\alpha }})\mid \Omega _{2}\}\Leftrightarrow E\{\alpha _{L}-\alpha _{F}\mid \Omega _{2}\}>\frac{ \kappa \sigma }{2}\Leftrightarrow \kappa >-14 \end{aligned}$$

and the last inequality is trivially true because $\kappa$ is positive by construction.

1.6 Proof of Corollary 2

Our strategy to prove the corollary is to show that, provided that $\kappa \ge 2$ holds, for any arbitrary realization $\varvec{\alpha }:=(\alpha _{F},\alpha _{L})$, we have $\delta ^{A}<\delta ^{CE}=\frac{\alpha _{F}\,-\,\alpha _{L}\,+\,\beta \bar{Q}}{2\beta \bar{Q}}$, where the last equality comes directly from (3).

Under type-3 equilibrium, the auction allocation is given by $\delta ^{A}:= \frac{\tau _{F}\left( q_{F0},\alpha _{F}\right) }{\tau _{F}\left( q_{F0},\alpha _{F}\right) \,+\,\overline{Q}}$, where $\tau _{F}\left( q_{F0},\alpha _{F}\right)$ is defined in (27). Using this expression for $\delta ^{A}$, we have that

$$\begin{aligned} \delta ^{A}<\delta ^{CE}\iff \beta \tau _{F}\left( q_{F0},\alpha _{F}\right) \le \frac{\delta ^{CE}}{1-\delta ^{CE}}\beta \bar{Q}. \end{aligned}$$

(37)

We now evaluate $\tau _{F}\left( q_{F0},\alpha _{F}\right)$ in the equilibrium price $p_{0}^{*}$, which, under type-3 equilibrium is equal to $p_{L}$, where $p_{L}$ is given by (29). Using the expression for $\delta ^{CE}$ together with (5) and, using (6), (37) can be written as

$$\begin{aligned} \delta ^{A}<\delta ^{CE}\iff \frac{7\alpha _{F}-3\alpha _{L}-4\theta -2\sigma }{5}<\frac{3(\alpha _{F}-\alpha _{L})+\beta \overline{Q}}{2(\beta \overline{Q}+\alpha _{L}-\alpha _{F})}\beta \overline{Q}. \end{aligned}$$

Notice that we can write $\alpha _{i}=\theta +r_{i}\sigma$ for $i\in \{F,L\}$, where $r_{i}$ is uniformly distributed in $[0,1]$, and there is a one-to-one mapping between $\alpha _{i}$ and $r_{i}$. Additionally, we use the definition of $\kappa$ to write $\beta \overline{Q}=\kappa \sigma$. Making these substitutions in the latter inequality, we get rid of $\theta$ and $\sigma$ to obtain

$$\begin{aligned} \delta ^{A}<\delta ^{CE}\iff 7r_{F}-3r_{L}-2<\frac{3(r_{F}-r_{L})+\kappa }{ \kappa -(r_{F}-r_{L})}\cdot \frac{5}{2}\kappa . \end{aligned}$$

(38)

To check that this inequality holds for any pair $(\alpha _{F},\alpha _{L})$ or, equivalently, for any pair $(r_{F},r_{L})$, we distinguish between two cases. Let us first consider first $r_{F}\ge r_{L}$. Notice that $7r_{F}-3r_{L}-2\le 3(r_{F}-r_{L})+2$ and that

$$\begin{aligned} \frac{3(r_{F}-r_{L})+\kappa }{\kappa -(r_{F}-r_{L})}\ge \frac{ 3(r_{F}-r_{L})+2}{\kappa -(r_{F}-r_{L})} \end{aligned}$$

where we have used $\kappa \ge 2$. Thus, a sufficient condition for (38) to hold is

$$\begin{aligned} 3(r_{F}-r_{L})+2<\frac{3(r_{F}-r_{L})+2}{\kappa -(r_{F}-r_{L})}\frac{5}{2} \kappa . \end{aligned}$$

(39)

If $r_{F}\ge r_{L}$, the left-hand side of (39) is positive and we can divide both sides by $3(r_{F}-r_{L})+2$ and rearrange to get $\frac{3}{2}\kappa >r_{L}-r_{F}$, which trivially holds when $\kappa \ge 2$ and $r_{F}\ge r_{L}$.

Let us now consider $r_{F}<r_{L}$. Notice that the right-hand side in (38) increases with $\kappa$, so it suffices to prove the inequality for $\kappa =2$. Introducing this value of $\kappa$ and rearranging, we have that

$$\begin{aligned} -7r_{F}^{2}-3r_{L}^{2}+r_{F}+7r_{L}+10r_{F}r_{L}-14<0. \end{aligned}$$

(40)

A sufficient condition for (40) to hold is that both of the following inequalities hold:

$$\begin{aligned} r_{F}+7r_{L}-8&\le 0 \end{aligned}$$

(41)

$$\begin{aligned} -7r_{F}^{2}-3r_{L}^{2}+10r_{F}r_{L}-6&< 0 \end{aligned}$$

(42)

given that, if both (41) and (42) hold, (40) follows from the sum of them both. Clearly, (41) holds as $r_{F}$, $r_{L}\in [0,1]$. To prove ( 42), we define $s:=\frac{r_{F}}{r_{L}}$, where $s\in [0,1]$ and rewrite (42) as $(-7s^{2}-3+10s)r_{L}^{2}-6<0$. For any $r_{L}$, the expression inside the parenthesis attains its maximum at $s=5/7$. For this value of $s$, the latter inequality becomes $\frac{4}{7}r_{L}^{2}-6<0$, which clearly holds true for any $r_{L}\in [0,1]$.

1.7 Proof of Proposition 5

As shown in Proposition 3, the condition for type-1 and type-2 equilibria to arise are, respectively, $\xi <-1$ and $\xi >1$. Using (21), (5), (7) and the definition of $\xi$ we obtain, respectively:

$$\begin{aligned} \text {type 1:}&{\qquad }2\alpha _{L}-\frac{8}{3}\alpha _{F}+\frac{2}{3} \theta +\sigma +\frac{5}{6}\beta \bar{Q}<0, \end{aligned}$$

(43)

$$\begin{aligned} \text {type 2:}&{\qquad }-2\alpha _{L}+\frac{8}{3}\alpha _{F}-\frac{2}{3 }\theta -\sigma +\frac{5}{6}\beta \bar{Q}<0. \end{aligned}$$

(44)

As for type-1 equilibria, particularizing (43) in the diagonal, we have $5\kappa \le 4a-6$, which clearly does not hold for any admissible value of $a$. In addition, the most favorable pair of realizations for type-1 equilibrium is $\alpha _{F}=\theta +\sigma$ and $\alpha _{L}=\theta$, under which the condition $\xi <-1$ becomes $\kappa <2$. Thus, there is a positive probability of obtaining type-1 equilibrium if and only if this latter inequality holds and, if it holds, the pair $\left( \alpha _{F}=\theta +\sigma \text {, }\alpha _{L}=\theta \right)$ leads to type-1 equilibrium. This proves part (i), whereas part (ii) follows from the convexity of the half-space defined by (43) whenever it is nonempty.

Notice that any pair $\varvec{\alpha }$ can be written as $\left( \theta +r_{F}\sigma ,\theta +r_{L}\sigma \right)$, with $r_{F}$, $r_{L}\in \left[ 0,1\right]$. Moreover, as $\alpha _{L}\le \alpha _{F}$, we must have $r_{L}\le r_{F}$. Using this notation, (44) becomes

$$\begin{aligned} \left( \frac{8}{3}r_{F}-2r_{L}-1\right) \sigma +\frac{5}{6}\beta \bar{Q}\le 0. \end{aligned}$$

(45)

To identify the feasible value of $\varvec{\alpha }$ that is more favorable for type-2 equilibrium, we minimize the left-hand side of (45) subject to $0\le r_{L}\le r_{F}\le 1$ ²⁹ and obtain $(r_{F},r_{L})=(0,0)$, or, equivalently, $\varvec{\alpha }=(\theta ,\theta )$. Under this value, (45) becomes $\kappa \le \frac{6}{5}$. This proves (iii), whereas part (iv) follows from the convexity of the half-space defined by (44) whenever it is nonempty. To prove part (v) it suffices to note that, under (4), at $\varvec{\alpha }=(\theta +\sigma ,\theta +\sigma )$ we have $\xi =\frac{2\sigma }{5\beta \bar{Q}}$, which belongs to $\left( -1,1\right)$ under (4) and, by continuity, this is true for some non-empty interval around $\left( \theta +\sigma ,\theta +\sigma \right)$.

1.8 Proof of Proposition 6

In case 3, using (5) and (8) we conclude that the condition for type-1 and type-2 equilibria to occur are given by

$$\begin{aligned} \text {type 1:}&{\qquad }2\alpha _{L}-\frac{8}{3}\alpha _{F}+\frac{2}{3} \theta -\frac{1}{3}\sigma +\frac{5}{6}\beta \bar{Q}<0, \end{aligned}$$

(46)

$$\begin{aligned} \text {type 2:}&{\qquad }-2\alpha _{L}+\frac{8}{3}\alpha _{F}-\frac{2}{3 }\theta +\frac{1}{3}\sigma +\frac{5}{6}\beta \bar{Q}<0. \end{aligned}$$

(47)

Consider (46). We use the same notation as in the proof of proposition 5 to write any pair $\varvec{\alpha }$ as $(\theta +r_{L}\sigma ,\theta +r_{F}\sigma )$ with $0\le r_{F}\le r_{L}\le 1$. Using this notation, (46) becomes

$$\begin{aligned} \left( 2r_{L}-\frac{8}{3}r_{F}-\frac{1}{3}\right) \sigma +\frac{5}{6}\beta \bar{Q}<\sigma . \end{aligned}$$

(48)

For (46), the most favorable feasible value is $(r_{F},r_{L})=(1,1)$,³⁰ or, equivalently, $\varvec{\alpha }=(\theta +\sigma ,\theta +\sigma )$. Under this value, (48) becomes $\ \kappa \le \frac{6}{ 5}$. This proves (i), whereas part (ii) follows from the convexity of the half-space defined by (46) whenever it is nonempty. Now consider (47). Using the previous notation, it can be written as

$$\begin{aligned} \left( -2r_{L}+\frac{8}{3}r_{F}+\frac{1}{3}\right) \sigma +\frac{5}{6}\beta \bar{Q}<\sigma . \end{aligned}$$

(49)

Notice that any $\varvec{\alpha }$ in $D$ is characterized by $r_{F}=r_{L}=r$ with $r\in [0,1]$, and (49) fails to hold for any of these values. In addition, the most favorable value of the pair $(r_{F},r_{L})$ for (49) within $0\le r_{F}\le r_{L}\le 1$ is $(r_{F},r_{L})=(0,1)$ or, equivalently, $\varvec{\alpha } =(\theta ,\theta +\sigma )$, under which (49) becomes $\kappa <2$. This proves (iii), whereas part (iv) follows from the convexity of the half-space defined by (47) whenever it is nonempty. To prove part (v) it suffices to note that, at $\varvec{\alpha }=(\theta ,\theta )$, we have $\xi =\frac{2\sigma }{5\beta \bar{Q}}$, which belongs to $\left( -1,1\right)$ under (4) and, by continuity, this is true for some non-empty interval around $\left( \theta ,\theta \right)$.