Moral Hazard and Tradeable Pollution Emission Permits

Proofs

Proof of Proposition 1. (i). We use a second-order Taylor’s expansion for the exponential function: expδx≃1+δx+12δ2x2. The objective function of the problem that characterizes the efficient allocation of the economy is

Using eq. [1], the i.i.d. assumption of the shocks and zi’s, we can write

In addition, it is

Take mi=1−γ(zi), differentiate J with respect to zi and set the derivative equal to zero to obtain

where

The expression in part (i) of the proposition is straightforward from the previous expression. To prove part (ii), note that the right-hand side of eq. [4] is constant across i’s whereas its left-hand side increases monotonically with zio and decreases monotonically with λi. ∎

Proof of Proposition 2. (i). Using zi as the unique decision variable and taking the second-order Taylor expansion for the exponential function, the first-order condition of firm i’s problem is

where hr=Emax0,1−γzi+εi−λizi+qi∗r. It is

Using Leibniz’s rule, we have

and

Substituting in the previous first-order condition and re-arranging terms, we obtain the desired expression. (ii). (ii.1) Differentiating the first-order condition obtained in Proposition 2 with respect to zi and θi, and collecting terms we obtain

Since the terms multiplying dzi and dθi are both positive (provided f′′<0, γ′′≥0, γ′>0, and f′>0), then we must have dzidθi>0, and, therefore, we can conclude that the abatement effort of firm i increases with θi (since both g and γ are increasing functions). Differentiating the first-order condition in Proposition 2 with respect to zi and qi∗ and collecting terms, we obtain

Since the terms multiplying dzi and dqi∗ have different signs, then dzidqi∗<0. (ii.2) Comparing Proposition 1 to part (i), an efficiency-inducing policy (qi∗,θi) must satisfy

Take Pr(ti<εi)=0, i.e. Pr(εi<ti)=1. In this case, the right-hand side of eq. [16] becomes zero, while its left-hand side is strictly positive, so we must have that Pr(ti<εi)>0, even when each firm exerts the Parero Optimal equilibrium effort. (ii.3) Any efficiency-inducing environmental policy (qi∗,θi) satisfies eq. [16] with ti=λizio+qi∗−1−γzio. Then, given qi∗ there exists a unique value of θi,Υ(qi∗) that satisfies the first equality. Differentiability (and thus continuity) of Υ follows trivially. Differentiating eq. [16] with respect to qi∗ and θi and collecting terms leads to

Since the terms multiplying dqi∗ and dθi in the above expression have different signs, then Υ is increasing. ∎

Proof of Proposition 3. (i). The first-order condition with respect to zi mimics the corresponding one in Proposition 2 just replacing qi∗ by qi and ti by ui. Similarly, the first-order condition with respect to qi can be written as

where we have denoted hˆr=Emax0,1−γzi+εi−λizi+qir. It is

Using Leibniz’s rule, we have that

and

Substituting in the first-order condition and re-arranging terms, we obtain the expression in the proposition. (ii) Necessity (only if part) follows trivially from the comparison of eq. [4] to the first equality in eq. [5] and the fact that the left side of eq. [4] is monotone in zio. To prove sufficiency (if part), consider an allocation z=z1,…,zI and a market price p such that eqs [5] and [6] hold. Then z satisfies eq. [4]. ∎

Proof of Proposition 4. (i) Differentiating the first equality in eq. [5] with respect to zi and p and collecting terms we obtain

Use eq. [5] to observe that

Then, we have that

The expressions multiplying dzi and dp are both positive provided f′′<0 and γ′′≥0. Therefore dzidp>0 holds. (ii) Consider both equalities in eq. [5]. When p increases, zi also increases, and, hence, ui also increases. Moreover, as ui increases, the integrals in eq. [5] decrease. In summary, if qi remains constant as p increases, the term on the right in eq. [5] decreases. On the other hand, this term decreases with qi. Therefore, p and qi must move in the opposite direction to satisfy eq. [5]. (iii) It follows trivially from the fact that the term on the left in eq. [5] decreases monotonically with λi whereas it increases monotonically with zi. (iv) From (iii), we have that, given p, an increase of λi conveys an increase in zi as well, and thus an increase in ui in the rightest term of eq. [5]. Consequently, for the second equality of eq. [5] to hold, we must decrease qi. (v) It follows trivially from (iv). ∎

Proof of Proposition 5. (i) Condition [7] is equivalent to ti<εinf. Thus, under condition [7] the integrals in the left-hand side of Proposition 2 become: ∫ti∞ϕεdε=1 and ∫ti∞εϕεdε=μ. Thus, the optimal decision of the firm is given by

where ti=λizi+qi∗−1−γzi. Part (i) follows trivially from the direct comparison of eqs [4]–[17] and monotonicity in zio of the left-hand side of eq. [4]. (ii) For any arbitrary z=z1,…,zI, it is Eeˉ|z>Eei|zi=mi+μ−λizi>mi+μ−λizi−qi∗=μ−ti, where we have used eq. [2] for the first inequality. Therefore, for eq. [8] to hold, it must be the case that δ<ρθi holds. (iii) The larger λi, the larger the distance between Eeˉ|z and μ−ti, and therefore, the larger the value of θi. ∎

Proof of Proposition 6. (i) Condition [9] implies ui<εinf, where ui is defined in Proposition 3. Thus, under eq. [9] the integrals in the rightest term of eq. [5] are ∫ui∞ϕεdε=1 and ∫ui∞εϕεdε=μ. Consequently, under eqs [9] and [10], eq. [5] becomes

The latter two equalities together with the definition of ui constitute a system of three equations on three unknowns: ui, qi and zi. The unique solution is

(ii) Since ui<εinf holds under eq. [9], the latter solution in qi constitutes the individual demand function of permits. The equilibrium price equals aggregate demand to total supply. Then, the equilibrium price is a solution in p to

Multiplying by p and collecting terms, we obtain

The left-hand side of this equation is a quadratic concave function which is strictly positive at p=0. Thus, there exists a unique positive value of p which solves it. (iii) po is characterized in the part (ii) of Proposition 3. Using eq. [10] and the solution to zi given in part (i), after some algebra, the equation in part (ii) of Proposition 3 becomes

where M=Iμ−1−2∑i∈χλi. The left-hand side of the previous equality is a quadratic convex function strictly negative at p=0. This implies that there exists a unique positive value of p which solves it. (iv) By direct comparison of eqs [18] and [19], we obtain trivially that eq. [11] is a sufficient condition. (v) From the second equality in eq. [11], it follows trivially that it must be δ2ρ2θi2<1 for all i∈χ, or, equivalently, the statement in the proposition. (vi) Both eqs [18] and [19] have the same positive solution in p if and only if

where k0=δ1ρ∑i∈χ1θi−Q∗ and k1=δ2ρ2∑i∈χ1(θi)2. Eq. [20] allows for k0+δMk1>1+δM together with 1k1<1. Assume first that M>0 holds. Then the previous inequalities together imply that k0>k1>1 must hold, or, equivalently, ∑i∈χδρθi−δQ∗>∑i∈χδρθi2>1. This latter chain of inequalities can hold if δρθi>1 holds for all i∈χ∖j, being j arbitrary in χ, which is part (a) of (vi). If M<0 holds, we can still have k0>k1, and the subsequent argument follows. To prove part (b) of (vi) it suffices to let r=θi for all i∈χ and to show that eq. [20] has a positive solution in r. The left-hand side of eq. [20] is strictly positive (regardless the sign of M). The right-hand side of eq. [20] is a continuous function of r that tends to zero as r tends to ∞ whereas it tends to ∞ as r tends to zero. ∎

Proof of Proposition 7. (i–iii) It is convenient to introduce the following notation: mˉ=∑i∈χ1−γzio and sˉ=∑i∈χλizio. Thus Eeˉ|zo=mˉ+Iμ−sˉ, and eq. [4] becomes

for all i∈0,1, where we must recall that the subscript i refers to type and not to a specific firm. Using eq. [10] and the fact that the left-hand side of eq. [21] must be equal across types, we have

Note that eq. [22] together with λ0<λ1 implies part (iii) of the proposition. Using eq. [22] we can write

and

After some algebra, these two previous expressions lead to

Using this expression, eq. [21] for type 1 firms can be written

The left-hand side of eq. [23] increases with z1 whereas the right-hand side decreases with z1. Thus, if there exists a solution in z1 to eq. [23], it is unique. In addition, such solution exists and lies in the interval (0,1) if and only if: (a) the left-hand side of eq. [23] is smaller than the right-hand side at z1=0, and thus the solution is larger than zero, and (b) the left-hand side of eq. [23] is larger than the right-hand side at z1=1, and thus the solution is smaller than 1. Condition (b) is

or, equivalently

Notice that, since z0<z1 holds, eq. [24] ensures both z1<1 and z0<1. Condition (b) ensures z1>0 but not z0>0. Thus, we need to replace (a) with a stronger condition that ensures z0>0. Using eq. [22], we can write z0>0 as z1>zˆ, where zˆ=21−1+λ01+λ1. Since we have that zˆ>0, z1>zˆ is stronger than condition (a). Therefore, replacing 0 with zˆ in condition (a) leads to the following condition: (a′) the left-hand side of eq. [23] is smaller than the right-hand side at zˆ, and thus the solution is larger than zˆ. Condition (a′) ensures that z0>0 and consequently z1>0 holds. We can write (a′) as

Using (2−zˆ)(1+λ1)=2(1+λ0) within the left-hand side of the latter inequality and using (1+λ1)zˆ=2(λ1−λ0) within the right-hand side, we can write it as

Conditions [24] and [25] characterize the subset of parameter values for which the efficient interior solutions exist and are unique. Let us denote that subset by Ω. Next, we analyze the non-emptiness of Ω. First, since the interior efficient solution z1, defined by the solution to eq. [23] in z1, must belong to the interval (zˆ,1), it must be the case that zˆ<1. The latter inequality is equivalent to

This condition is thus necessary for Ω to be non-empty. We show that it is also sufficient. The previous assumptions imply that the right-hand side of eq. [24] is strictly positive. Thus, there exists a unique value of α, say α∗, such that eq. [24] holds with equality. Replacing α by α∗, we write eq. [25] as

where we have denoted T1=2I0I(λ1−λ0)+μ−λ1 and T2=1−2I1I(λ1−λ0)+μ. Condition [26] implies that 121+λ11+λ0<1. Therefore, for the previous inequality to hold, it suffices that δ(1+δIT1)<δ(1+δIT2) or, equivalently, T1<T2, but this latter inequality is equivalent to eq. [26]. The non-emptiness of Ω follows now trivially using a continuity argument on α. Furthermore, if, for every α satisfying both eqs [24] and [25], we use a continuity argument on the rest of the parameters, we obtain that Ω has a non-zero measure along any dimension in α,I0,I1,λ0,λ1,δ,μ, i.e. Ω has a non-empty interior. (iv and v) Write eq. [23] as

where

Thus, defining z˜=(1+λ1)(2−z1o) and taking into account the definition of k, eq. [23] can be written as δ2Iz˜2+kz˜−α=0, which is a quadratic equation in z˜. More specifically, the left-hand side of this equation is convex and strictly negative at z˜=0, thus it has exactly one positive root. Once z˜ is obtained, we have that z1o=2−11+λ1z˜ and z0o=2−11+λ0z˜, where we have used eq. [22] for the latter equality. The signs of the comparative statics analysis are straightforward: when a parameter changes, first look at the changes in z˜ and then at the changes in zio. ∎

Proof of Proposition 8. (i) Consider the expressions in Proposition 2 taking zi=zio. It is ti=qi∗+(1+λi)zio−1=μ, where we have used eq. [13]. Using in addition that zio=2−11+λiz˜ holds for every i∈0,1, which is shown in the proof of Proposition 7, the expression in Proposition 2 becomes

Clearly, the solution in θi to this latter equation is type-independent. (ii) It follows trivially if we note that Prε−μ>0=∫μ∞ϕ(ε)dε and that both z0o and z1o decrease with z˜. (iii) Notice that σ>0 is necessary and sufficient for the integrals on the right-hand side of the above equation to be strictly positive, which in turn is a necessary and sufficient condition for the above equation to have a (positive) solution in θ. Using the uniform distribution, we have ∫μ∞ϕ(ε)dε=12 and ∫μ∞(ε−μ)ϕ(ε)dε=σ8, which substituted in the above equation leads to the desired expression. ∎

Proof of Proposition 9. (i) Using eq. [10], eq. [5] becomes

where κ(ui)=ρθ∫ui∞ϕεdε+ρθ∫ui∞ε−uiϕεdε and ui is defined in Proposition 3. Under eq. [10], we have ui=qi−1+(1+λi)zi. It is κ(ui)≥0 for all ui, and the strict inequality holds iff ui<εsup. Thus, for any p>0, from the second equality in eq. [28], ui<εsup must hold. Also, κ′(ui)=−ρθϕui+ρθ∫ui∞ϕεdε, thus it is κ′(ui)≤0 for all ui, and the strict inequality holds iff ui<εsup. Thus, under the latter inequality, the inverse of κ, denoted κ−1, is well defined, and the second equality in eq. [28] is ui=κ−1(p). Using this latter equality together with the first equality in eq. [28] and the definition of ui, we have

The latter expression is the individual demand of permits for any p>0, and it is continuous and strictly decreasing in p. Notice that ui<εsup is equivalent to Prei>qi|zi>0. (ii) Consider the first equality in eq. [28] with zi=zio and use the definition of z˜ given in Proposition 8. The unique price that restores efficiency is p=α/z˜. Of course, for it to be an equilibrium price, the market must clear at that price. The aggregate demand at that price is

whereas the aggregate supply (at any price) is given by eq. [15]. Note that

where, for the latter equality, we have used the definition of z˜. Using the latter expression into eq. [15] to obtain the aggregate supply, the market clearing condition, after canceling out common terms on both sides, becomes κ−1(α/z˜)=μ or, equivalently

but the latter equation is exactly eq. [27].