1. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 1 of 33 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.

2. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 2 of 33 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

3. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 3 of 33 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

4. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 4 of 33 5. 2 The problem of the Common Example: 1000 identical persons who can do nothing but fish. Each can catch 4 fish on shore. No of Men Total Catch on Board MP (on board) AP (on board) Net Social MP (on board) Social Total 000004000+0=4000 16+6623396+6=4002 216+10863392+16=4008 324+8844012 430+67.524014 534+46.804014 636+26-24012 73605.14-44008 832-44-84000 927-53-93991 1021-621-103981 * *

5. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 5 of 33 Distinctions Between Price and Quantity Approaches to Addressing Externalities 5. 4 Basic Model

6. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 6 of 33 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

7. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 7 of 33 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).

8. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 8 of 33 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).

9. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 9 of 33 Distinctions Between Price and Quantity Approaches to Addressing Externalities 5. 4 Multiple Plants with Different Reduction Costs

10. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 10 of 33 Example with Multiple Firms Ē1, Ē2; X1, X2; E1 = Ē1 - X1; E2 = Ē2 - X2 Pollution damage = D(E1+E2) =D(Ē1 + Ē2 - X1 - X2)

11. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 11 of 33 * 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)

12. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 12 of 33 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

13. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 13 of 33 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

14. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 14 of 33 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.

15. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 15 of 33 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.

16. Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 16 of 33 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.