Assigning pollution permits: are uniform auctions efficient?

Abstract

We study the efficiency of the uniform auction as an allocation mechanism for emission permits among polluting firms. In our model, firms have private information about their abatement costs, which differ across firms and across units, and bidders’ demands are linear. We show that there is a continuum of interior Bayesian Nash equilibria, and only one is efficient, minimizing abatement costs. We find that the existence of many bidders is not a sufficient condition to guarantee an efficient equilibrium in the uniform auction. Additionally, bidders’ types have to be uncorrelated.

Appendix: Proofs

The following notation is convenient for several of the proofs, while it is omitted from the main text to ease the overall exposition.

$$\begin{aligned} \mu :=(1-\lambda )E \lbrace \tilde{\alpha }_i \rbrace \end{aligned}$$

(17)

And (8) can be rewritten as

$$\begin{aligned} E \lbrace \tilde{\alpha }_j \mid \alpha _i \rbrace = \mu + \lambda \alpha _i \end{aligned}$$

Proof of Proposition 1

Assume that all bidders but i follow some—and the same—linear strategy, as in (3). The residual supply for bidder i at price p is

$$\begin{aligned} S_{-i}(p) = Q- \sum _{j \ne i} \tau ( \alpha _j) + (I-1)\delta p \end{aligned}$$

(18)

where $\sum _{j \ne i} \tau ( \alpha _j)$ depends on the types of all bidders but the i. For an arbitrary stop-out price p, bidder i’s realized cost is $C(S_{-i}(p),\alpha _i )$. After some algebra from (1), (2) and (18), we have

$$\begin{aligned} E \lbrace C(S_{-i}(p),\alpha _i ) \mid \ \varvec{\gamma }_{-i} \rbrace&= \left( -\alpha _i (I-1)\delta + (Q-\hat{\rho }(\alpha _i))(1+\beta (I-1)\delta ) \right) p\nonumber \\&\quad + \left( \frac{1}{2}\beta (I-1)\delta + 1 \right) (I-1)\delta p^2 + \theta \end{aligned}$$

(19)

where $\theta$ contains terms that do not depend on p and $\hat{\rho }(\alpha _i):=E \lbrace \sum _{j \ne i} \tau ( \alpha _j) \mid \alpha _i \rbrace$. It is

$$\begin{aligned} \theta = \alpha _i+\left( Q-\hat{\rho }(\alpha _i)\right) -\dfrac{\beta }{2}\left( e^*+\left( Q-\hat{\rho }(\alpha _i)\right) ^2\right) \end{aligned}$$

(20)

Under the uniform format, bidder i’s cost depends only on the stop-out price and on the quantity demanded at that price and, given a residual supply, there is a one-to-one mapping between that price and that quantity. Thus, given a residual supply and $\alpha _i$, choosing the stop-out price is equivalent to choosing the quantity demanded at that price. Given $\alpha _i$ and the rivals’ strategy, the stop-out price that minimizes bidder i’s expected cost is

$$\begin{aligned} \min _{p} E \lbrace C(S_{-i}(p),\alpha _i) \mid \varvec{\gamma }_{-i} \rbrace \end{aligned}$$

Denote by $p^{*}(\alpha _i,\varvec{\gamma }_{-i})$ the solution to that problem. From (19), first-order conditions of bidder’s i minimization problem imply that

$$\begin{aligned} p^{*}(\alpha _i,\varvec{\gamma }_{-i}) = \frac{\alpha _i(I-1)\delta -(Q-\hat{\rho }(\alpha _i))(1+\beta (I-1)\delta )}{(2+\beta (I-1)\delta )(I-1)\delta } \end{aligned}$$

(21)

Assuming that all bidders (including i) play $\gamma$ defined in (3), bidder i has no incentives to deviate if and only if, for each $\alpha _i \in \Omega$, the expected stop-out price when all bidders are playing $\gamma$, conditional on $\alpha _i$, is precisely $p^{*}(\alpha _i,\varvec{\gamma }_{-i})$.

On the other hand, if all bidders follow $\gamma$ defined in (3), the stop-out price can be characterized as the solution in p to

$$\begin{aligned} S_{-i}(p) = \tau (\alpha _i)-\delta p \end{aligned}$$

Solving this latter equation in p, denoting the solution by $p_o$, and taking expectations conditional on $\alpha _i$, we get

$$\begin{aligned} E \lbrace p_o \mid \alpha _i, \gamma \rbrace = \frac{1}{\delta I}\left( \hat{\rho }(\alpha _i) + \tau (\alpha _i)-Q\right) \end{aligned}$$

(22)

Thus, $\gamma$ is an equilibrium if and only if $(\tau ,\delta )$ satisfy

$$\begin{aligned} p^{*}(\alpha _i,\varvec{\gamma }_{-i}) = E \lbrace p_o \mid \alpha _i, \gamma \rbrace \qquad \forall \alpha _i \in \Omega \end{aligned}$$

(23)

Using (21) and (22), we can rewrite (23) as

$$\begin{aligned} \alpha _i = \frac{\xi }{I} \times \frac{\tau (\alpha _i)}{\delta } + \left( 1- \frac{\xi }{I} \right) \times \frac{1}{I-1}\frac{1}{\delta } \left( (\hat{\rho }(\alpha _i) -Q\right) \end{aligned}$$

(24)

where $\xi := 2 + \beta (I-1)\delta$. Consider the equation $S_{-i}(p)=0$, where $S_{-i}(p)$ is given by (18), and denote its solution by $p_{-i}$. It is

$$\begin{aligned} E \lbrace p_{-i} \mid \gamma , \; \alpha _i \rbrace = \frac{1}{I-1}\frac{1}{\delta } ((\hat{\rho }(\alpha _i) -Q) \end{aligned}$$

which substituted in (24) leads to

$$\begin{aligned} \alpha _i = \frac{\xi }{I} \times \frac{\tau (\alpha _i)}{\delta } + \left( 1- \frac{\xi }{I} \right) \times E \lbrace p_{-i} \mid \gamma , \; \alpha _i \rbrace \end{aligned}$$

This latter equality can be easily rewritten as in the statement in the proposition. $\square$

Proof of Lemma 1

Assuming that all players demand some positive quantity at the stop-out price, the market clearing condition is

$$\begin{aligned} \sum _i \gamma (\alpha _i,p) = Q \iff p = \frac{1}{\delta } \left( \kappa _0 + \kappa _1 \frac{1}{I}\sum _i \alpha _i - \frac{Q}{I} \right) \end{aligned}$$

where the last equality characterizes the auction’s stop-out price. Recall that the support of $\tilde{\alpha }_i$ is $[\underline{\alpha },\overline{\alpha }]$. The auction’s stop-out price is positive with probability 1 if and only if¹⁷

$$\begin{aligned} \kappa _0 + \kappa _1 \underline{\alpha } - \frac{Q}{I} > 0 \end{aligned}$$

(25)

The quantity demanded by bidder i at the stop-out price is

$$\begin{aligned} \begin{aligned} \gamma _i(\alpha _i,p)&= \kappa _0 + \kappa _1 \alpha _i - \left( \kappa _0 + \kappa _1 \frac{1}{I}\sum _i \alpha _i - \frac{Q}{I} \right) \\&= \kappa _1 \left( \alpha _i - \frac{1}{I}\sum _i \alpha _i \right) + \frac{Q}{I}\\&= \frac{I-1}{I} \kappa _1 \left( \alpha _i - \frac{1}{I-1}\sum _{j \ne i} \alpha _j \right) + \frac{Q}{I} \end{aligned} \end{aligned}$$

Thus

$$\begin{aligned} \gamma _i> 0 \iff (I-1)\kappa _1 \left( \alpha _i - \frac{1}{I-1}\sum _{j \ne i} \alpha _j \right) + Q > 0 \end{aligned}$$

The latter equality holds with probability 1 iff

$$\begin{aligned} (I-1)\kappa _1 (\underline{\alpha }-\overline{\alpha }) + Q > 0 \end{aligned}$$

(26)

Combining (25) and (26) we obtain (6).

Next, we give conditions for a nonnegative marginal abatement cost. It is

$$\begin{aligned} \phi (e;\alpha _i) \ge 0 \iff \alpha _i - \beta (\kappa _0+\kappa _1 \alpha _i - \delta p) \ge 0 \end{aligned}$$

Substituting p from the market clearing condition for interior solution and collecting terms in $\alpha$’s, the latter inequality becomes

$$\begin{aligned} \left( 1 - \beta \kappa _1 \left( 1 - \frac{1}{I} \right) \right) \alpha _i + \beta \kappa _1 \frac{1}{I} \sum _{j \ne i} \alpha _j \ge \frac{\beta Q}{I} \end{aligned}$$

The most adverse case on the second term on the left is $\alpha _j=\underline{\alpha }$ for all $j \ne i$, so that the inequality becomes

$$\begin{aligned} \left( 1 - \beta \kappa _1 \left( 1 - \frac{1}{I} \right) \right) \alpha _i + \beta \kappa _1 \frac{I-1}{I}\underline{\alpha } \ge \frac{\beta Q}{I} \end{aligned}$$

(27)

The coefficient of $\alpha _i$ in (27) is nonnegative if and only if

$$\begin{aligned} \beta \kappa _1 \le \frac{I}{I-1} \end{aligned}$$

(28)

Thus, if (28) holds, the most adverse case of $\alpha _i$ for inequality (27) is $\alpha _i = \underline{\alpha }$. Substituting this value of $\alpha _i$, the inequality (27) becomes

$$\begin{aligned} \underline{\alpha } \ge \frac{\beta Q}{I} \end{aligned}$$

On the other hand, if the inequality in (28) does not hold, the most adverse case of $\alpha _i$ for inequality (27) is $\alpha _i = \overline{\alpha }$. Substituting this value of $\alpha _i$, the inequality (27) becomes

$$\begin{aligned} \overline{\alpha }-\beta \kappa _1 \frac{I-1}{I}\left( \overline{\alpha }-\underline{\alpha }\right) \ge \frac{\beta Q}{I} \end{aligned}$$

Inequality (7) summarizes these latter two inequalities. $\square$

Proof of Proposition 2

Assume that $\tau$ is linear, i.e., we restrict to equilibria in which $\gamma$ is linear in both arguments, $\alpha _i$ and p:

$$\begin{aligned} \tau (\alpha ) = \kappa _0 + \kappa _1 \alpha \end{aligned}$$

(29)

Using (29), (8) and (17), we can write

$$\begin{aligned} \hat{\rho }(\alpha _i) = (I-1)(\kappa _0+\kappa _1\mu ) + (I-1)\kappa _1 \lambda \alpha _i \end{aligned}$$

In turn, with the above expression for $\hat{\rho }(\alpha _i)$, both sides of (23) are linear on $\alpha _i$. Specifically, substituting in (21) we have

$$\begin{aligned} \begin{aligned} p^{*}(\alpha _i,\varvec{\gamma }_{-i})&= \frac{1}{2+\beta (I-1)\delta } \left( 1 + (1+\beta (I-1)\delta )\frac{\lambda \kappa _1}{\delta } \right) \alpha _i\\&\quad + \frac{1+\beta (I-1)\delta }{(2+\beta (I-1)\delta )(I-1)\delta } \left( (I-1)(\kappa _0+\kappa _1\mu )- Q\right) \end{aligned} \end{aligned}$$

and from (22)

$$\begin{aligned} E \lbrace p_o \mid \alpha _i, \gamma \rbrace = \frac{1}{I\delta } (\lambda (I-1)+1)\kappa _1\alpha _i + \frac{1}{I\delta }((I-1)(\kappa _0+\kappa _1\mu )+\kappa _0 - Q) \end{aligned}$$

Using these expressions, from (23), the coefficients of $\alpha _i$ and the intercept on both expressions have to be equal. Equalizing the coefficients of $\alpha _i$ we obtain the following equation

$$\begin{aligned} \frac{1}{\beta (I-1)} - \left( \frac{2}{\beta (I-1)} + \lambda \kappa _1 \right) \frac{1}{\xi } = \frac{1}{I}(1-\lambda )\kappa _1 \end{aligned}$$

(30)

where $\xi := 2 + \beta (I-1) \delta$, as in the proof of Proposition 1. Notice that there is a one-to-one mapping between $\xi$ (or $\xi ^{-1}$) and $\delta$. Equalizing the intercepts (the terms that do not depend on $\alpha _i$), we obtain the following equation

$$\begin{aligned} \left( \frac{1}{I-1}Q -(\kappa _0+\kappa _1 \mu ) \right) \frac{1}{\xi } + \frac{1}{I}\kappa _1 \mu = \frac{1}{I(I-1)}Q \end{aligned}$$

(31)

Equations (30) and (31) characterize the equilibrium parameters for any linear strategy $\gamma$. The unknowns are $\kappa _0$, $\kappa _1$ and $\xi ^{-1}$. From (30), we have that

$$\begin{aligned} \frac{1}{\xi } = m(\kappa _1) \end{aligned}$$

(32)

where we have denoted

$$\begin{aligned} m(\kappa _1) := \frac{I-\beta (I-1)(1-\lambda )\kappa _1}{2I+\beta (I-1)I\lambda \kappa _1} \end{aligned}$$

(33)

It follows that

$$\begin{aligned} \delta =g_{\delta }\left( \kappa _1 \right) = \left( \frac{1}{m(\kappa _1)}-2 \right) \frac{1}{\beta (I-1)} \end{aligned}$$

(34)

Clearly, m is differentiable. Straightforward computations show that $m'(\kappa _1) <0$ if $\lambda \ge 0$ and $m(0)=\frac{1}{2}$. The properties of $g_{\delta }$ follow from the properties of m.

Substituting (32) into (31) and solving for $\kappa _0$, we get

$$\begin{aligned} \kappa _0 = g_{0}\left( \kappa _1 \right) = \left( 1 - \frac{1}{Im(\kappa _1)}\right) \left( \frac{1}{I-1}Q-\kappa _1\mu \right) \end{aligned}$$

(35)

The right-hand side of Eq. (35) defines $g_{0}$. $\square$

Proof of Proposition 3

Consider the first inequality in (6), substitute the expression for $\kappa _0$ given in (35) and reorder terms to obtain

$$\begin{aligned} \left( 1-\frac{1}{m(\kappa _1)}\right) \frac{1}{I-1}Q + \left( \underline{\alpha } + \mu \left( \frac{1}{Im(\kappa _1)}-1\right) \right) I\kappa _1 \ge 0 \end{aligned}$$

(36)

Step 1. We show that (36) neither holds for $\kappa _1=0$ nor for $\kappa _1 \rightarrow \infty$. If $\kappa _1=0$, using that $m(0)=1/2$, (36) collapses to

$$\begin{aligned} -\frac{1}{I-1} Q \ge 0 \end{aligned}$$

which does not hold. Furthermore, as $\kappa _1 \rightarrow \infty$, we have $m(\kappa _1) \rightarrow -\frac{1-\lambda }{I\lambda }$. For $\lambda \ne 1$, the sign of the left-hand side in (36) as $\kappa _1 \rightarrow \infty$ is given by the coefficient of $\kappa _1$ with $m(\kappa _1)$ at its limiting value:

$$\begin{aligned} \underline{\alpha } - \mu \frac{1}{1-\lambda } = \underline{\alpha } - E \lbrace \tilde{\alpha } \rbrace \end{aligned}$$

(37)

where we have used (17). The expression on the right-hand side of (37) is clearly negative.

Step 2. Consider $\kappa _1^p$, defined by $\beta \kappa _1(I-1)=I$. It is

$$\begin{aligned} m \left( \kappa _1^p \right) = \frac{\lambda }{2 + \lambda I} \end{aligned}$$

For $\kappa _1=\kappa _1^p$, taking limits in (36) as $\lambda \rightarrow 0$ we have

$$\begin{aligned} -\frac{1}{I-1}Q + \mu \kappa _1^p \ge 0 \end{aligned}$$

where $\mu \rightarrow E \lbrace \tilde{\alpha }_i \rbrace$ as $\lambda \rightarrow 0$. Using the expression for $\kappa _1^p$, the latter inequality becomes

$$\begin{aligned} \frac{\beta Q}{I} \le E \lbrace \tilde{\alpha }_i \rbrace \end{aligned}$$

(38)

Now consider the second inequality in (6) for $\kappa _1=\kappa _1^p$. It is

$$\begin{aligned} (I-1)\kappa _1^p (\overline{\alpha }-\underline{\alpha }) \le Q \iff \overline{\alpha }-\underline{\alpha } \le \frac{\beta Q}{I} \end{aligned}$$

(39)

Finally, condition (7) for $\kappa _1=\kappa _1^p$ becomes

$$\begin{aligned} \frac{\beta Q }{I} \le \underline{\alpha } \end{aligned}$$

(40)

which is more restrictive than (38). Combining (39) and (40) we obtain (9). Using a continuity argument, it follows the existence of an interval containing $\kappa _1^p$ and some non-empty interval in $[0,\lambda ^u)$. $\square$

Proof of Lemma 2

• 1. We solve the problem that characterizes the efficient allocation for a given vector of type realizations, say $\varvec{\alpha }=\left( \alpha _1,\dots ,\alpha _I \right)$. We define the Lagrangian

  $$\begin{aligned} \mathcal {L} = \sum _{i=1}^{I} \int _{q_i}^{e^{*}_i(\alpha _i)} \phi (e;\alpha _i)\mathrm{d}e + \theta \left( \sum _{i=1}^{I} q_i- Q \right) \end{aligned}$$

  where nonnegativity constraints are omitted for sake of simplicity and $\theta$ denotes the multiplier. The Kuhn–Tucker conditions are

  $$\begin{aligned} \phi (q_i;\alpha _i)= & {} \theta \qquad i \in \lbrace 1,\dots , I \rbrace \nonumber \\&\qquad \theta \left( \sum _{i=1}^{I} q_i- Q \right) = 0 \qquad \theta \ge 0 \end{aligned}$$

  (41)

  An interior allocation occurs when $\phi (q_i;\alpha _i) >0$ holds, which implies $\theta >0$ and thus

  $$\begin{aligned} \sum _{i=1}^{I} q_i- Q = 0 \end{aligned}$$

  (42)

  Next, we solve in q’s the set of equations given by (41) and (42). These equations conform a system of linear equations which can be solved using standard linear algebra. Alternatively, consider any $i \ne 1$ and use (41) to write

  $$\begin{aligned} q_1 - q_i = \frac{1}{\beta } (\alpha _1-\alpha _i) \end{aligned}$$

  (43)

  Substituting in (42) for any $i \ne 1$ and then solving (42) for $q_1$ we have

  $$\begin{aligned} q_1 = \frac{Q}{I} - \frac{1}{\beta I} \left( \sum _{i \ne 1}\alpha _i - (I-1)\alpha _1 \right) \end{aligned}$$

  Substituting back in (43) and solving for $q_i$ we obtain

  $$\begin{aligned} q_i = \frac{Q}{I} - \frac{1}{\beta I} \left( \sum _{j \ne i}\alpha _j - (I-1)\alpha _i \right) \end{aligned}$$

  (44)

  for any $i \in \lbrace 1,\dots ,I \rbrace$. Using (44), the nonnegativity requirement can be written

  $$\begin{aligned} q_i \ge 0 \iff Q \ge \frac{1}{\beta } \left( \sum _{j \ne i} \alpha _j - (I-1)\alpha _i\right) \end{aligned}$$

  Considering the most adverse realizations, the latter inequality becomes

  $$\begin{aligned} Q \ge \frac{1}{\beta }(I-1) \left( \overline{\alpha } - \underline{\alpha } \right) \end{aligned}$$

  (45)

  Analogously, using (44), the nonnegativity of the marginal abatement cost (equivalently, the condition for $\theta >0$) can be written

  $$\begin{aligned} \phi (q_i;\alpha _i) \ge 0 \iff Q \le \frac{1}{\beta } \sum _i \alpha _i \end{aligned}$$

  Considering the most adverse realizations, the latter inequality becomes

  $$\begin{aligned} Q \le \frac{1}{\beta } I \underline{\alpha } \end{aligned}$$

  (46)

  This part of the Lemma follows trivially from the combination of (45) and (46).

• 2. If the efficient allocation is interior, from (41), for any $i,j \in \lbrace 1,\dots , I \rbrace$, it is

  $$\begin{aligned} \alpha _i - \alpha _j = \beta (q_i-q_j) \end{aligned}$$

  Now assume that q’s in this latter equality come from an equilibrium in the auction in which all bidders play a strategy as in (3) and that leads to an interior allocation, that is, for any $h \in \lbrace 1,\dots , I \rbrace$ it is

  $$\begin{aligned} q_h = \tau (\alpha _h) - \delta p^{*} \end{aligned}$$

  where $p^{*}$ is the stop-out price in the auction. Combining the previous two equalities, we have

  $$\begin{aligned} \frac{1}{\beta } = \frac{\tau (\alpha _i)-\tau (\alpha _j)}{\alpha _i-\alpha _j} \end{aligned}$$

  This latter equality holds for any arbitrary pair of realizations $\alpha _i$ and $\alpha _j$ in $[\underline{\alpha },\overline{\alpha }]$ if and only if it is $\tau (\alpha )= \kappa _0 + \beta ^{-1}\alpha$, where $\kappa _0$ is an arbitrary parameter. $\square$

Proof of Proposition 4

Write the inequalities that characterize interior equilibria for $\lambda = 0$. First, take $\lambda = 0$ in (33) and substitute into (36) to obtain

$$\begin{aligned} \left( 1 - \frac{2}{1-x} \right) \frac{\beta Q}{I} + \left( I \underline{\alpha } + \mu \left( \frac{2}{1-x} - I \right) \right) x \ge 0 \end{aligned}$$

(47)

where we have denoted $x := \beta \frac{I-1}{I}\kappa _1$. From Proposition 2, interior equilibria are indexed by $\kappa _1$, so they are equivalently indexed by x. Recall that this latter inequality is the first inequality in (6). Using the definition of x, the second inequality in (6) becomes:

$$\begin{aligned} \frac{\beta Q}{I} \ge x (\overline{\alpha }-\underline{\alpha }) \end{aligned}$$

(48)

Finally, the condition (7) can be written

$$\begin{aligned} \frac{\beta Q }{I } \le {\left\{ \begin{array}{ll} \underline{\alpha } &{} \text { if }x \le 1 \\ \overline{\alpha }- x\left( \overline{\alpha } - \underline{\alpha } \right) &{} \text { otherwise} \\ \end{array}\right. } \end{aligned}$$

(49)

Thus, for $\lambda =0$, x defines an interior equilibrium if and only if it satisfies (47) to (49).

Rewrite (47) as

$$\begin{aligned} \frac{\beta Q}{I} + I (\underline{\alpha }-\mu )x + \left( \mu x - \frac{\beta Q}{I} \right) \frac{2}{1-x} \ge 0 \end{aligned}$$

(50)

where we must recall that $x := \beta \frac{I-1}{I}\kappa _1$. Notice that for $x = 1$ this latter inequality collapses to

$$\begin{aligned} \mu \ge \frac{\beta Q}{I} \end{aligned}$$

which is implied by (9). In addition, for $x=1$ (48) and (49) collapse to (9). Thus, under (9) there is an interior equilibrium for $x=1$ or, equivalently

$$\begin{aligned} \kappa _1 = \frac{1}{\beta } \frac{I}{I-1} \end{aligned}$$

Now consider $1-x >0$. Write (50) as

$$\begin{aligned} I (\mu - \underline{\alpha })x^2 + \left( 2 \mu - I(\mu - \underline{\alpha }) - \frac{\beta Q}{I} \right) x - \frac{\beta Q}{I} \ge 0 \end{aligned}$$

(51)

Let us denote by h a function of x such that (51) is $h(x) \ge 0$. Clearly, h is quadratic, convex since $\mu - \underline{\alpha } >0$ holds and it satisfies $h(0)<0$ and $h(1)=2\left( \mu - \frac{\beta Q}{I} \right) >0$. Thus, there is a unique value of x in (0, 1), say $x^{*}$, such that (51) holds if and only if $x \in [x^{*},1]$. More concretely, $h(x)=0$ has necessarily two real roots and $x^{*}$ is the largest (and the only positive) root. Notice also that (48) and (49) are implied by (9) for any $x<1$.

If $1-x <0$, then from (50) we obtain $h(x) \le 0$, where h is still the left-hand side of (51). Thus, this latter inequality cannot hold for $x>1$. Therefore, the set of interior equilibria are characterized by $x \in [x^{*},1]$. Denoting $\kappa _1^{*}$ such that $x^{*} = \beta \frac{I-1}{I}\kappa _1^{*}$, the interior equilibria are characterized equivalently by $\kappa _1 \in \left( \kappa _1^{*}, \frac{1}{\beta }\frac{I}{I-1}\right)$.

The sensitivity analysis of $\kappa _1^{*}$ is equivalent to $x^{*}$. Consider first a variation in $\mu$ keeping all other parameters constant. Taking total differential in the latter equality, we have

$$\begin{aligned} \left( 2 + I(x^{*}-1) \right) x^{*} d \mu + \left( 2 \mu + I(\mu - \underline{\alpha })(2x^{*}-1) - \frac{\beta Q}{I} \right) dx^{*} = 0 \end{aligned}$$

(52)

Since $x^{*} <1$ holds, the coefficient of $d\mu$ is negative if I is large enough. To analyze the sign of the coefficient of dx we must notice first that (9) implies $2\mu - \frac{\beta Q}{I} >0$, so a sufficient condition for that coefficient to be positive is $2x^{*}-1 >0$ or, equivalently, $x^{*} > \frac{1}{2}$. It is

$$\begin{aligned} h \left( \frac{1}{2} \right) = \mu - \frac{1}{4} I (\mu - \underline{\alpha }) - \frac{3}{2} \frac{\beta Q }{I} \end{aligned}$$

Thus, $h \left( \frac{1}{2} \right) <0$ holds if I is large enough, which in turn implies that $x^{*} > \frac{1}{2}$. Therefore, for I large enough, the coefficient of $d \mu$ is negative whereas the coefficient of dx is positive, which implies $d \mu$ and $dx^{*}$ must have the same sign.

Let $s:= \mu - \underline{\alpha }$. Consider a variation in s keeping all other parameters (in particular $\mu$) constant. Taking total differential in $h(x^{*})=0$ we have

$$\begin{aligned} Ix^{*}(x^{*}-1)ds + \left( 2 \mu + Is(2x^{*}-1)-\frac{\beta Q}{I} \right) dx^{*} = 0 \end{aligned}$$

The coefficient of ds in the previous expression is negative since $x^{*}<1$, whereas, following an argument as above, the coefficient of $dx^{*}$ is positive if I is large enough. Thus, for I large enough ds and $dx^{*}$ must have the same sign.

Consider a variation in Q keeping all other parameters constant. Taking total differential in $h(x^{*})=0$ we have

$$\begin{aligned} - \frac{1}{I}(x^{*}+1) \beta dQ + \left( 2 \mu - \frac{\beta Q}{I} + I (\mu - \underline{\alpha })(2x^{*}-1) \right) dx^{*} = 0 \end{aligned}$$

Using an analogous reasoning, when I is large enough we obtain that dQ and $dx^{*}$ must have the same sign.

It rests to prove that the efficient equilibrium, $\beta \kappa _1 = 1$, belongs to the set of interior equilibrium for I large enough. Notice that $\beta \kappa _1 = 1$ implies $x = \frac{I-1}{I}$. It conforms an interior equilibrium if and only if it satisfies (47) to (49). Substituting in (47), we have

$$\begin{aligned} \frac{I-1}{2I-1} (\underline{\alpha } + \mu ) \ge \frac{\beta Q}{I} \end{aligned}$$

(53)

whereas substituting in (48) and (49) we have

$$\begin{aligned} \underline{\alpha } \ge \frac{\beta Q}{I} \ge \frac{I-1}{I} (\overline{\alpha }-\underline{\alpha }) \end{aligned}$$

(54)

For any I finite, (54) is implied by (9). In addition, (53) is also implied by (9) if

$$\begin{aligned} \frac{I-1}{2I-1} (\underline{\alpha } + \mu ) \ge \underline{\alpha } \end{aligned}$$

The coefficient of $\underline{\alpha } + \mu$ in this latter inequality is strictly increasing and continuous in I. It converges to $\frac{1}{2}$ as $I \rightarrow \infty$. The inequality clearly holds at the limiting value of that coefficient; thus, it must hold for any I larger than some finite threshold. $\square$

Proof of Proposition 5

• 1. The necessary and sufficient conditions for $\kappa _1$ to constitute an interior equilibrium are (47) to (49), in Proposition 4, where $x=\beta \frac{I-1}{I}\kappa _1$. Taking $\beta \kappa _1=1$, (47) becomes

  $$\begin{aligned} \underline{\alpha } + \mu \ge \left( 1 + \frac{I}{I-1} \right) \frac{\beta Q}{I} \end{aligned}$$

  (55)

  Also (48) and (49) become

  $$\begin{aligned} \overline{\alpha } - \underline{\alpha } \le \frac{\beta Q}{I-1} \le \frac{I}{I-1} \underline{\alpha } \end{aligned}$$

  (56)

  The inequalities (55) and (56) are implied by (10) just noting that $\lambda =0$ implies $\mu = E \lbrace \tilde{\alpha } \rbrace > \underline{\alpha }$. The efficient equilibrium strategy follows from taking $\kappa _1 = \beta ^{-1}$ and $\lambda = 0$ in (32) to (35), in the proof of Proposition 2. To show that the efficient strategy satisfies (11) is left to the part 4 of this proof.

• 2. Take again the characterization of equilibrium as (47) to (49). We write $I \rightarrow \infty$ to represent: $I \rightarrow \infty$ and $Q \rightarrow \infty$ while $\frac{Q}{I}$ remains constant. If $I \rightarrow \infty$ for $\beta \kappa _1 \ne 1$, then (47) converges to

  $$\begin{aligned} \left( \underline{\alpha } - \mu \right) x \ge 0 \end{aligned}$$

  which cannot hold: $\underline{\alpha } - \mu <0$ holds for any non-degenerated distribution of types, while $x>0 \iff \kappa _1>0$ and, from Proposition 2, this latter equality must hold at any equilibrium under which firms submit downward sloping demand functions. In addition, for $\beta \kappa _1 = 1$, the conditions for an interior equilibrium are (55) and (56), which hold in the limiting case $I \rightarrow \infty$ under (10).

• 3. Take $I=2$ and let $\beta \kappa _1$ be arbitrary. Using (32) to (35), in the proof of Proposition 2, we have

  $$\begin{aligned} m(\kappa _1) = \frac{1}{4} (2-\beta \kappa _1); \; \kappa _0 = \frac{\beta \kappa _1}{2 - \beta \kappa _1} \left( \kappa _1 \mu - Q \right) ; \; \delta = \frac{2 \kappa _1}{2 - \beta \kappa _1} \end{aligned}$$

  (57)

  From the latter equality, we have a downward sloping demand if and only if $\beta \kappa _1 \in (0,2)$. The conditions for an interior equilibrium in Lemma 1, for $I=2$, can be written as follows. (6) is

  $$\begin{aligned} \beta \kappa _0+ \beta \kappa _1 \underline{\alpha }> \frac{\beta Q}{2} > \frac{1}{2}\beta \kappa _1 \left( \overline{\alpha }-\underline{\alpha } \right) \end{aligned}$$

  (58)

  In addition, since $\beta \kappa _1 \le 2$, the condition (7) is

  $$\begin{aligned} \frac{\beta Q}{2} \le \underline{\alpha } \end{aligned}$$

  (59)

  The second equality in (58) and (59) is implied by (10). Use the expression for $\kappa _0$ above to write the first inequality in (58) as

  $$\begin{aligned} \beta \kappa _1 \mu + (2-\beta \kappa _1) \underline{\alpha } \ge \left( \frac{2}{\beta \kappa _1} + 1 \right) \frac{\beta Q}{2} \end{aligned}$$

  The left-hand side of the previous inequality is continuous and strictly increasing in $\beta \kappa _1$ as $\mu > \underline{\alpha }$ holds. The right-hand side is continuous and strictly decreasing. Clearly, the inequality fails to hold as $\beta \kappa _1 \rightarrow 0$, whereas it is implied by (10) at $\beta \kappa _1 = 1$. Thus, under (10), there must exist a unique $b^{0} \in (0,1)$ such that for any $\beta \kappa _1 \in (0,2)$ the inequality holds iff $\beta \kappa _1 \in (b^{0},2)$.

• 4. We prove (11) for $I=2$ and $\beta \kappa _1 \in (0,2)$. For any linear strategy, the stop-out price in the auction is defined by the market clearing condition

  $$\begin{aligned} 2 \kappa _0 + \kappa _1 \sum _i \alpha _i -2 \delta p^{*} = Q \end{aligned}$$

  Using the expressions for $\kappa _0$ and $\delta$ in (57), we have

  $$\begin{aligned} p^{*}(\hat{\alpha }) = \frac{1}{\beta \delta } \left( \beta \kappa _0 - \frac{1}{2} \beta Q + \beta \kappa _1 \hat{\alpha } \right) \end{aligned}$$

  (60)

  where $\hat{\alpha }$ denotes the sample mean of types and the notation emphasizes that the stop-out price depends on it. Denoting by $q^{*}_i$ the auction assignment for bidder i, his marginal saving on abatement cost is

  $$\begin{aligned} \phi _i(q^{*}_i, \alpha _i) = \alpha _i - \beta q^{*}_i = \alpha _i - \beta \left( \kappa _0 + \kappa _1 \alpha _i - \delta p^{*}(\hat{\alpha }) \right) \end{aligned}$$

  (61)

  Taking expectations in (61), substituting in (11) and re-arranging terms, (11) is equivalent to

  $$\begin{aligned} (1-\beta \delta ) E \lbrace p^{*}(\hat{\alpha }) \mid \alpha _i \rbrace \le (1 - \beta \kappa _1) \alpha _i - \beta \kappa _0 \end{aligned}$$

  (62)

  Taking expectations in (60) and substituting into (62), we can rewrite (62) as

  $$\begin{aligned} \left( \frac{1}{\beta \delta }-1 \right) \left( \beta \kappa _0 - \frac{1}{2}\beta Q + \frac{1}{2}\beta \kappa _1 \left( \alpha _i + E \lbrace \tilde{\alpha }_i \rbrace \right) \right) \le (1 - \beta \kappa _1) \alpha _i - \beta \kappa _0 \end{aligned}$$

  where we have used that, since bidder i only observes his own type and types are independent, it is

  $$\begin{aligned} E \lbrace \hat{\alpha } \mid \alpha _i \rbrace = \frac{1}{2} \left( \alpha _i + E \lbrace \tilde{\alpha }_i \rbrace \right) \end{aligned}$$

  The latter inequality is equivalent to

  $$\begin{aligned}&\frac{1}{\beta \delta } \beta \kappa _0 -\frac{1}{2} \left( \frac{1}{\beta \delta } -1 \right) \beta Q + \left( \left( \frac{1}{\beta \delta } -1 \right) \frac{1}{2} \beta \kappa _1 - 1 + \beta \kappa _1 \right) \alpha _i \\&\qquad + \frac{1}{2}\left( \frac{1}{\beta \delta } -1 \right) \beta \kappa _1 E \lbrace \tilde{\alpha }_i \rbrace \le 0 \end{aligned}$$

  Using the expression for $\delta$ in this latter inequality yields

  $$\begin{aligned} \frac{2-\beta \kappa _1}{2 \beta \kappa _1} \beta \kappa _0 - \frac{2-3 \beta \kappa _1}{4 \beta \kappa _1} \beta Q - \frac{1}{4}(2-\beta \kappa _1)\alpha _i + \frac{1}{4} \left( 2-3 \beta \kappa _1 \right) E \lbrace \tilde{\alpha }_i \rbrace \le 0 \end{aligned}$$

  Finally, using the expression for $\kappa _0$ in this latter inequality yields

  $$\begin{aligned} - \frac{2 - \beta \kappa _1}{4} \left( \frac{\beta Q}{\beta \kappa _1} + \alpha _i - E \lbrace \tilde{\alpha }_i \rbrace \right) \le 0 \end{aligned}$$

  since $\beta \kappa _1 \in (0,2)$, the latter inequality is equivalent to

  $$\begin{aligned} \beta Q \ge \beta \kappa _1 \left( E \lbrace \tilde{\alpha }_i \rbrace - \alpha _i \right) \end{aligned}$$

  (63)

  But (63) is implied by the first inequality in (10) since $\beta \kappa _1 \le 2$ and $\overline{\alpha } > E \lbrace \tilde{\alpha }_i \rbrace$. Finally, we prove (11) for $I > 2$ and $\beta \kappa _1 = 1$. Notice that in this case (11) can still be written as (62). The efficient strategy satisfies $\beta \kappa _1 =1$. In addition, using the value for $\kappa _0$ and $\delta$ for the efficient strategy, in the part 1 of the proposition, we can rewrite (62) as

  $$\begin{aligned} E \lbrace p^{*} (\hat{\alpha }) \mid \alpha _i \rbrace \ge E \lbrace \tilde{\alpha }_i \rbrace - \frac{I}{I-1} \frac{\beta Q}{I} \end{aligned}$$

  (64)

  To obtain the stop-out price, we consider the market clearing condition

  $$\begin{aligned} \sum _i \gamma (\alpha _i,p) = Q \end{aligned}$$

  Use the expression for $\gamma$ in the part 1 of the proposition, solve for p, to obtain

  $$\begin{aligned} p^{*}(\hat{\alpha }) = \frac{1}{2} \left( E \lbrace \tilde{\alpha }_i \rbrace + \hat{\alpha } \right) - \frac{1}{2} \left( \frac{I}{I-1}+1 \right) \frac{\beta Q}{I} \end{aligned}$$

  where, as before, $\hat{\alpha }$ denotes the sample mean of types. Analogously to the two-bidder case, notice that

  $$\begin{aligned} E \lbrace \hat{\alpha } \mid \alpha _i \rbrace = \frac{I-1}{I} E \lbrace \tilde{\alpha }_i \rbrace + \frac{1}{I} \alpha _i \end{aligned}$$

  Taking conditional expectations on the previous expression for $p^{*}(\hat{\alpha })$, substituting in (64) and re-arranging terms, (64) is equivalent to

  $$\begin{aligned} \frac{\beta Q}{I-1} > E \lbrace \tilde{\alpha }_i \rbrace - \alpha _i \end{aligned}$$

  This latter inequality is clearly implied by the first inequality in (10) just noting that it is

  $$\begin{aligned} E \lbrace \tilde{\alpha }_i \rbrace - \alpha _i \le \overline{\alpha } - \underline{\alpha } \end{aligned}$$

$\square$

Proof of Proposition 6

• 1. We write the inequalities that characterize interior equilibria for $\lambda = 1/2$. Note that $\lambda =1/2$ implies $\mu = \frac{E\lbrace \tilde{\alpha _i}\rbrace }{2}$. First, (33) is

  $$\begin{aligned} m(\kappa _1) = \frac{1-x}{2+Ix} \end{aligned}$$

  where $x := \beta \frac{I-1}{2I}\kappa _1$. From Proposition 2, interior equilibria are indexed by $\kappa _1$, so they are equivalently indexed by x. Substitute the expressions for $m(\kappa _1)$ and $\mu$ into (36) to obtain

  $$\begin{aligned} \left( 1-\frac{2+Ix}{1-x}\right) \frac{\beta Q}{2I} + \left( I\underline{\alpha } + \frac{E \lbrace \tilde{\alpha _i} \rbrace }{2} \left( \frac{ 2+Ix}{1-x}-I\right) \right) x \ge 0 \end{aligned}$$

  (65)

  Recall that this latter inequality is the first inequality in (6). Using the definition of x, the second inequality in (6) becomes:

  $$\begin{aligned} \frac{\beta Q}{I} \ge 2 x (\overline{\alpha }-\underline{\alpha }) \end{aligned}$$

  (66)

  Finally, condition (7) can be written

  $$\begin{aligned} \frac{\beta Q }{I } \le {\left\{ \begin{array}{ll} \underline{\alpha } &{} \text { if }x \le 1/2 \\ \overline{\alpha }- 2x\left( \overline{\alpha } - \underline{\alpha } \right) &{} \text { otherwise} \\ \end{array}\right. } \end{aligned}$$

  (67)

  Thus, for $\lambda =1/2$, x defines an interior equilibrium if and only if it satisfies (65) to (67). Take $\beta \kappa _1 = 1$. Then

  $$\begin{aligned} \frac{2+Ix}{1-x} = \frac{I(I+3)}{I+1} \end{aligned}$$

  Thus, (65) can be written

  $$\begin{aligned} \underline{\alpha } + \frac{1}{2} E \lbrace \tilde{\alpha }_i \rbrace \left( \frac{I+3}{I+1}-1 \right) \ge \left( \frac{I+3}{I+1}- \frac{1}{I} \right) \frac{\beta Q}{I-1} \end{aligned}$$

  or, equivalently,

  $$\begin{aligned} \underline{\alpha } + \frac{1}{I+1} E \lbrace \tilde{\alpha }_i \rbrace \ge \frac{\beta Q}{I} + \frac{2}{I+1} \frac{\beta Q}{I-1} \end{aligned}$$

  Using (10), it suffices to add the first inequality in (14) for this latter inequality to hold. In addition, (66) and (67), for $\beta \kappa _1 = 1$, collapse to

  $$\begin{aligned} \frac{I-1}{I} \left( \overline{\alpha } - \underline{\alpha } \right) \le \frac{\beta Q}{I} \le \underline{\alpha } \end{aligned}$$

  These latter two inequalities are implied by (10). The efficient equilibrium strategy follows from taking $\kappa _1 = \beta ^{-1}$ and $\lambda = \frac{1}{2}$ in (32) to (35), in the proof of Proposition 2. As in the proof of Proposition 5, we can write (11) as (62). For the case $\lambda = \frac{1}{2}$, the efficient equilibrium strategy is presented in the part 1 of this proposition, in particular

  $$\begin{aligned} \kappa _0 = \frac{1}{\beta }\frac{1}{I+1} \left( E \lbrace \tilde{\alpha }_i \rbrace - 2 \frac{I}{I-1} \frac{\beta Q}{I} \right) ; \qquad \kappa _1 = \frac{1}{\beta }: \qquad \delta = \frac{1}{\beta } \frac{I+2}{I+1} \end{aligned}$$

  Substituting in (62) and re-arranging terms, it is

  $$\begin{aligned} E \lbrace p^{*} \left( \hat{\alpha }\right) \mid \alpha _i \rbrace \ge E \lbrace \tilde{\alpha }_i \rbrace -2 \frac{I}{I-1} \frac{\beta Q}{I} \end{aligned}$$

  (68)

  In the other hand, the stop-out price follows from the usual market clearing condition, as in the proof of Proposition 5, leading to

  $$\begin{aligned} p^{*} (\hat{\alpha }) = \frac{1}{\beta \delta } \left( \beta \kappa _0 + \beta \kappa _1 \hat{\alpha } - \frac{\beta Q}{I} \right) \end{aligned}$$

  (69)

  where $\hat{\alpha }$ denotes the sample mean of the types. Under $\lambda = \frac{1}{2}$, it is

  $$\begin{aligned} E \lbrace \hat{\alpha } \mid \alpha _i \rbrace = \frac{I-1}{I} E \lbrace \tilde{\alpha }_j \mid \alpha _i \rbrace + \frac{1}{I} \alpha _i = \frac{I-1}{2I} E \lbrace \tilde{\alpha }_i \rbrace + \frac{I+1}{2I} \alpha _i \end{aligned}$$

  (70)

  where $j \ne i$. Taking the conditional expectation on the stop-out price, using the parameter values of the equilibrium strategy and substituting into (68), we can rewrite it as

  $$\begin{aligned}&\left( \frac{2I}{I-1}{-} \frac{I+1}{I+2} \left( \frac{1}{I+1} \frac{2I}{I-1} + 1 \right) \right) \frac{\beta Q}{I} \ge \left( 1{-} \frac{I+1}{I+2} \left( \frac{1}{I+1}{+}\frac{I-1}{2I} \right) \right) \\&\qquad E \lbrace \tilde{\alpha }_i \rbrace - \frac{I+1}{I+2}\frac{I+1}{2I}\alpha _i \end{aligned}$$

  or, equivalently

  $$\begin{aligned} \frac{\beta Q}{I-1} \ge \frac{1}{2} \left( E \lbrace \tilde{\alpha }_i \rbrace - \alpha _i \right) \end{aligned}$$

  which is implied for all $\alpha _i \in \Omega$ by the first inequality in (10).

• 2. In the remainder of the proof, we write $I \rightarrow \infty$ to represent: $I \rightarrow \infty$ and $Q \rightarrow \infty$ while $\frac{\beta Q}{I}$ remains constant. Taking $I \rightarrow \infty$, $x \rightarrow \frac{\beta \kappa _1}{2}$. Additionally, considering $x\ne 0$, equation (65) converges to

  $$\begin{aligned} \underline{\alpha } + \frac{1}{2}v(x) \ge 0 \end{aligned}$$

  (71)

  where we have denoted

  $$\begin{aligned} v(x) := \frac{1}{1-x} \left( E\lbrace \tilde{\alpha _i}\rbrace \left( 2x-1 \right) - \frac{\beta Q}{I}\right) \end{aligned}$$

  Note that v is unbounded at $x=1$. Under (9), it is

  $$\begin{aligned} E\lbrace \tilde{\alpha _i}\rbrace - \frac{\beta Q}{I}> \underline{\alpha } - \frac{\beta Q}{I} > 0 \end{aligned}$$

  Thus, as $x \rightarrow 1$ from the left and from the right, it is $\lim _{x \rightarrow 1^{-}}v(x)= \infty$ and $\lim _{x \rightarrow 1^{+}}v(x)= -\infty$, respectively. Furthermore

  $$\begin{aligned} v'(x) = \frac{1}{(1-x)^2} \left( E\lbrace \tilde{\alpha _i}\rbrace - \frac{\beta Q}{I}\right) \end{aligned}$$

  which is positive from (9). In turn, (9) is implied by (10). In fact, in the remainder of this part of the proof it will suffice to use (9) instead of (10). Note that $\lim _{x \rightarrow \infty } v(x) = -2 E \lbrace \tilde{\alpha }_i \rbrace$. Using this latter limit in (71) and taking into account that v is strictly increasing, (71) cannot hold for any $x > 1$. Equivalently, there cannot be interior equilibrium for any $\kappa _1$ such that $\beta \kappa _1 >2$ holds as $I \rightarrow \infty$. Note that $v(1/2)= -2 \frac{\beta Q}{I}$. Thus, (71) holds with strict inequality under (9) for $x=1/2$. Taking into account that v is strictly increasing, (71) holds for any $x \in [1/2,1]$. Notice that (66) and (67) are unaffected by taking $I \rightarrow \infty$. For $x=1/2$, (66) and (67) are implied by (9), so that $x=1/2$, or equivalently $\beta \kappa _1=1$, that is, the efficient allocation, constitutes an interior equilibrium. Within $x \in (1/2,1]$, the most restrictive case for (66) and (67) to hold is at $x=1$. For that value of x, (66) and (67) are equivalent to the second and third inequalities in (14). Now consider $x \in \left( 0, \frac{1}{2}\right)$. (66) and (67) are implied by (9). Under (9), (71) holds with strict inequality for $x=1/2$. Since v is continuous and increasing at any $x \le 1/2$, (71) must also hold if and only if $x \in (x^{*},1/2)$, for some $x^{*} \in (0,1/2)$. The relation between $x^{*}$ and $b^1$ follows from $x \rightarrow \frac{\beta \kappa _1}{2}$ as $I \rightarrow \infty$. As in the proof of Proposition 5, we can write (11) as (62). Combining with the expression of the stop-out price, in (69), we can write (11) as

  $$\begin{aligned} \frac{1}{\beta \delta } \beta \kappa _0 + \left( \frac{1}{\beta \delta } -1 \right) \left( \beta \kappa _1 E \lbrace \hat{\alpha } \mid \alpha _i \rbrace - \frac{\beta Q}{I} \right) \le (1-\beta \kappa _1) \alpha _i \end{aligned}$$

  Taking limits as $I \rightarrow \infty$ in (70) and substituting in the later inequality, we can rewrite it as

  $$\begin{aligned} \frac{1}{\beta \delta } \beta \kappa _0 + \left( \frac{1}{\beta \delta } -1 \right) \left( \frac{1}{2} \beta \kappa _1 E \lbrace \tilde{\alpha }_i \rbrace - \frac{\beta Q}{I} \right) \le \left( 1-\beta \kappa _1 - \left( \frac{1}{\beta \delta } -1 \right) \frac{1}{2} \beta \kappa _1 \right) \alpha _i \end{aligned}$$

  (72)

  Next, we use the expressions in (32) to (35), in the proof of Proposition 2. Taking $\lambda = \frac{1}{2}$ and $I \rightarrow \infty$, it is

  $$\begin{aligned} \lim _{I \rightarrow \infty } \frac{I}{\xi } = \lim _{I \rightarrow \infty } I m(\kappa _1) \iff \frac{1}{\beta \delta } = \frac{2-\beta \kappa _1}{\beta \kappa _1}; \end{aligned}$$

  and

  $$\begin{aligned} \lim _{I \rightarrow \infty } \beta \kappa _0 = \left( 1- \frac{\beta \kappa _1}{2-\beta \kappa _1} \right) \left( \frac{\beta Q}{I} - \frac{1}{2} E \lbrace \tilde{\alpha }_i \rbrace \beta \kappa _1 \right) \end{aligned}$$

  Using these limit values into (72), we have $0 \le 0$ for any $\beta \kappa _1 \ne 0$.

$\square$

Proof of Corollary 1

With $\beta \kappa _1 = \frac{3}{2}$, x, as defined in the proof of Proposition (6) is $x = \frac{3(I-1)}{4I}$, so that $m(\kappa _1) = \frac{I+3}{I(3I+5)}$. From the definition of $\xi$ in the proof of Proposition 2, substituting in Eq. (32)

$$\begin{aligned} \delta = \frac{I(3I+5)-2(I+3)}{\beta (I+3)(I-1)} \end{aligned}$$

(73)

that tends to $\frac{3}{\beta }$ as $I \rightarrow \infty$. From (35)

$$\begin{aligned} \kappa _0 = \left( 1-\frac{3I+5}{I+3} \right) \left( \frac{1}{I-1}Q-\frac{3}{2\beta }\frac{E \lbrace \tilde{\alpha }_i \rbrace }{2} \right) \end{aligned}$$

(74)

that tends to $\frac{3}{2\beta }E \lbrace \tilde{\alpha }_i \rbrace$ as $I \rightarrow \infty$. $\square$

Proof of Corollary 2

It follows from the definition of $\xi$ in the proof of the Proposition 2, substitution of the corresponding value of $\lambda$ and $\kappa _1=1/\beta$ in Eqs. (32) to (35), and then taking limits as $I \rightarrow \infty$, $Q \rightarrow \infty$ and $\frac{Q}{I}$ stays constant. $\square$