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Private-Sector Solutions to Negative Externalities. 5 . 2. The Solution. Coase Theorem (Part I) When there are well-defined property rights and costless bargaining, then negotiations between the party creating the externality and the party affected

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slide1
Private-Sector Solutions to Negative Externalities

5 . 2

The Solution

Coase Theorem (Part I) When there are well-defined property rights and costless bargaining, then negotiations between the party creating the externality and the party affected

by the externality can bring about the socially optimal market quantity.

Coase Theorem (Part II) The efficient solution to an externality does not depend on which party is assigned the property rights, as long as someone is assigned those rights.

slide2

5 . 2

Example I

Net Benefit to the factory associated with marginal production = $1.0

Net Cost to the Laundromat associated with the firm’s marginal production = $1.20

*Efficient outcome?

Case (i): Factory has the property right.

Case (ii): Laundromat has the property right

slide3

5 . 2

Example II

Net Benefit to the factory associated with marginal production = $1.20

Net Cost to the Laundromat associated with the firm’s marginal production = $1.0

*Efficient outcome?

Case (i): Factory has the property right

Case (ii): Laundromat has the property right

slide4

5 . 2

The problem of the Common

Example: 1000 identical persons who can do nothing but fish. Each can catch 4 fish on shore.

*

* *

slide5
Distinctions Between Price and Quantity Approaches to Addressing Externalities

5 . 4

Basic Model

slide6
Abatement: Algebraic Illustration

Ē = firm’s pollution without abatement

X = abatement

E = Ē-X = pollution

C(X) = abatement cost

D(E) = D(Ē–X) = pollution damage

C’(X) = marginal abatement cost

D’(E) = marginal damage of pollution

slide7
1. Optimal abatement: Choose X to

Minimize C(X) + D(E) = C(X) + D(Ē-X)

=> C’(X) - D’(Ē-X)=0.

Or, C’(X) = D’(E).

slide8
2. Optimal solution for a firm in the presence of a tax:

Minimize C(X) + t E = C(X) + t Ē – t X

(x)

=> t= C’(x)

To attain social optimum then, set t= D’(E).

slide9
Distinctions Between Price and Quantity Approaches to Addressing Externalities

5 . 4

Multiple Plants with Different Reduction Costs

slide10
Example with Multiple Firms

Ē1, Ē2;

X1, X2;

E1 = Ē1 - X1;

E2 = Ē2 - X2

Pollution damage = D(E1+E2) =D(Ē1 + Ē2 - X1 - X2)

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* Optimal abatement:

Minimize C1(X1) + C2(X2) + D(Ē1 + Ē2 - X1 - X2)

C1’ (X1) = C2’(X2) = D’(E).

* Firm’s solution: Minimizes Ci(Xi) + t (Ēi - Xi)

=> Ci’(Xi) = t.

=> Set: t = D’(E)

slide12
Example

Assume:

D(E) =10 E => D’(E) =10

C1(X1)=F + 1/10 (X1)2 => C1’(X1) =1/5 (X1)

C2(X2)=F + 1/30 (X2)2 => C2’(X2) =1/15 (X2)

Setting C1’(X1) = C2’(X2) = D’(E)

=> X1=50; X2=150

slide13
Equal pollution Reduction: Ask each firm to reduce pollution by 100.

Same benefit of damage reduction as with the Pigouvian solution.

Costs:

C1 = F + 1/10 (100)2

C2 = F + 1/30 (100)2

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Total cost of abatement=

C1 + C2 = 2F + (100)2 [1/10 + 1/30] = 2F + 4000/3

Versus the total cost for the Pigouvian solution:

C1 = F + 1/10 (50)2

C2 = F + 1/30 (150)2

=> C1 +C2 = 2F + 1000.

slide15
Market for Permits

Suppose Ē1 + Ē2 = 500.

Want 200 reduction

Issue 300 permits (150 each)

Firm i’s pollution level is

Ei = Ēi - Xi = 150 + ni

ni denotes the number of extra permits purchased.

If ni is negative, it will be the number of permits sold.

slide16
Price of a permit= p

Cost of polluting Ei = Ci (Xi) + ni p

Or Ci (Xi) + (Ē - Xi – 150) p

Minimizing costs yields

Ci’(Xi)=p.

C1’(X1)= C2’(X2)

If p=t, we will have the Pigouvian solution.