Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public...
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Transcript of Chapter 5 Externalities: Problems and Solutions © 2007 Worth Publishers Public Finance and Public...
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© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
Private-Sector Solutions to Negative Externalities5 . 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 affectedby 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.
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© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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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© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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
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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© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
Distinctions Between Price and Quantity Approaches to Addressing Externalities
5 . 4
Basic Model
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© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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
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© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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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© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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).
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© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
Distinctions Between Price and Quantity Approaches 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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ExampleAssume:
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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© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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.
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© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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.
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• 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.