Notes on Chapter 5

7/28/2019 Notes on Chapter 5

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© 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 doesnot depend on which party is assigned

the property rights, as long as someone

is assigned those rights.



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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



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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



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5 . 2

The problem of the Common

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

No ofMen

Total Catchon Board

MP

(on board)

AP

(on board)

Net Social

MP (on board)

Social Total

0 0 0 0 0 4000+0=4000

1 6 +6 6 2 3396+6=4002

2 16 +10 8 6 3392+16=4008

3 24 +8 8 4 4012

4 30 +6 7.5 2 4014

5 34 +4 6.8 0 4014

6 36 +2 6 -2 4012

7 36 0 5.14 -4 4008

8 32 -4 4 -8 4000

9 27 -5 3 -9 3991

10 21 -6 21 -10 3981

*

* *



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Distinctions Between Price and QuantityApproaches to Addressing Externalities

5 . 4

Basic Model



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Abatement: Algebraic Illustration

Ē = firm’s pollution without abatement

X = abatementE = 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



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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).



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

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

• => t= C’(x)

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



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Distinctions Between Price and QuantityApproaches to Addressing Externalities

5 . 4

Multiple Plants with Different Reduction Costs



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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)



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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



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Equal pollution Reduction: Ask each firm to reducepollution by 100.

• Same benefit of damage reduction as with thePigouvian 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.



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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 extrapermits purchased.

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



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© 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.

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Notes on Chapter 5