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7/28/2019 Notes on Chapter 5
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 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.
7/28/2019 Notes on Chapter 5
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 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
7/28/2019 Notes on Chapter 5
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 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
7/28/2019 Notes on Chapter 5
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 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 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
*
* *
7/28/2019 Notes on Chapter 5
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 5 of 33
Distinctions Between Price and QuantityApproaches to Addressing Externalities
5 . 4
Basic Model
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 6 of 33
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
7/28/2019 Notes on Chapter 5
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 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).
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 8 of 33
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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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 9 of 33
Distinctions Between Price and QuantityApproaches to Addressing Externalities
5 . 4
Multiple Plants with Different Reduction Costs
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 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)
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 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)
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C h a p t e r 5
E x t e r n a l i t i e s : P r
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t i o n s
© 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
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C h a p t e r 5
E x t e r n a l i t i e s : P r
o b l e m s a n d S o l u
t i o n s
© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e 13 of 33
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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C h a p t e r 5
E x t e r n a l i t i e s : P r o b l e m s a n d S o l u
t i o n s
© 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.
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C h a p t e r 5
E x t e r n a l i t i e s : P r o b l e m s a n d S o l u
t i o n s
© 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 extrapermits purchased.
• If ni is negative, it will be the number of permits sold.
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E x t e r n a l i t i e s : P r o b l e m s a n d S o l u
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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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