Microeconomics 2: Problem Set 1

Problem Set 4: Externalities
Matthew Robson
University of York
Microeconomics 2
1
Question 1
Consider a two-commodity (π‘₯ and 𝑦), two-consumer (𝐴 and 𝐡)
pure-exchange economy.
Suppose consumer 𝐴’s utility function is π‘ˆπ΄ π‘₯𝐴 , 𝑦𝐴 , π‘₯𝐡 =
π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅βˆ’1 and she is endowed with 100 units of π‘₯ and zero units
of 𝑦.
Suppose consumer B’s utility function is π‘ˆπ΅ π‘₯𝐡 , 𝑦𝐡 = π‘₯𝐡0.5 𝑦𝐡0.5 and
she is endowed with zero units of x and 100 units of y.
Derive the conditions for Pareto optimality in the above
economy(notice that an externality is present).
2
Question 1
Choose an allocation π‘₯𝐴 , 𝑦𝐴 , π‘₯𝐡 , 𝑦𝐡 to maximise:
π‘₯𝐡0.5 𝑦𝐡0.5
subject to:
π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅βˆ’1 = π‘ˆπ΄
π‘₯𝐴 + π‘₯𝐡 = 100
𝑦𝐴 + 𝑦𝐡 = 100
Feasible Allocations
3
Question 1
Form the Lagrangian:
β„’ = π‘₯𝐡0.5 𝑦𝐡0.5 + πœ† π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅βˆ’1 βˆ’ π‘ˆπ΄ + 𝛼 100 βˆ’ π‘₯𝐴 βˆ’ π‘₯𝐡 + 𝛾 100 βˆ’ 𝑦𝐴 βˆ’ 𝑦𝐡
Where πœ†, 𝛼 and 𝛾 are Lagrange multipliers.
Derive the FOC’s:
π‘₯𝐴 : πœ†0.5π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅βˆ’1 βˆ’ 𝛼 = 0,
(1)
𝑦𝐴 : πœ†0.5π‘₯𝐴0.5 π‘¦π΄βˆ’0.5 π‘₯π΅βˆ’1 βˆ’ 𝛾 = 0
π‘₯𝐡 : 0.5π‘₯π΅βˆ’0.5 𝑦𝐡0.5 βˆ’ πœ†π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅βˆ’2 βˆ’ 𝛼 = 0, (3) 𝑦𝐡 : 0.5π‘₯𝐡0.5 π‘¦π΅βˆ’0.5 βˆ’ 𝛾 = 0
πœ†: π‘₯𝐴0.5 𝑦𝐴0.5 βˆ’ π‘ˆπ΄ = 0,
(5) 𝛼: 100 βˆ’ π‘₯𝐴 βˆ’ π‘₯𝐡 = 0
𝛾: 100 βˆ’ 𝑦𝐴 βˆ’ 𝑦𝐡 = 0
(2)
(4)
(6)
(7)
4
Question 1
(1)=(3)
πœ†0.5π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅βˆ’1 = 𝛼 = 0.5π‘₯π΅βˆ’0.5 𝑦𝐡0.5 βˆ’ πœ†π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅βˆ’2
0.5π‘₯π΅βˆ’0.5 𝑦𝐡0.5
πœ†=
0.5π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅βˆ’1 + π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅βˆ’2
(2)=(4)
πœ†0.5π‘₯𝐴0.5 π‘¦π΄βˆ’0.5 π‘₯π΅βˆ’1 = 𝛾 = 0.5π‘₯𝐡0.5 π‘¦π΅βˆ’0.5
π‘₯𝐡0.5 π‘¦π΅βˆ’0.5
πœ† = 0.5 βˆ’0.5 βˆ’1
π‘₯𝐴 𝑦𝐴 π‘₯𝐡
0.5π‘₯π΅βˆ’0.5 𝑦𝐡0.5
π‘₯𝐡0.5 π‘¦π΅βˆ’0.5
βˆ’0.5 0.5 βˆ’1
0.5 0.5 βˆ’2 = πœ† = 0.5 βˆ’0.5 βˆ’1
0.5π‘₯𝐴 𝑦𝐴 π‘₯𝐡 + π‘₯𝐴 𝑦𝐴 π‘₯𝐡
π‘₯𝐴 𝑦𝐴 π‘₯𝐡
0.5π‘₯𝐴0.5 π‘¦π΄βˆ’0.5
π‘₯𝐡1.5 π‘¦π΅βˆ’0.5
βˆ’0.5 0.5 βˆ’1
0.5 0.5 βˆ’2 = βˆ’0.5 0.5
0.5π‘₯𝐴 𝑦𝐴 π‘₯𝐡 + π‘₯𝐴 𝑦𝐴 π‘₯𝐡
π‘₯𝐡 𝑦𝐡
5
Question 1
0.5π‘₯𝐴0.5 π‘¦π΄βˆ’0.5
π‘₯𝐡2
=
0.5π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅βˆ’1 + π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅βˆ’2 𝑦𝐡
π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅βˆ’1 π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅βˆ’2 𝑦𝐡
0.5 βˆ’0.5 +
0.5 βˆ’0.5 = π‘₯ 2
π‘₯𝐴 𝑦𝐴
0.5π‘₯𝐴 𝑦𝐴
𝐡
𝑦𝐴
𝑦𝐴
𝑦𝐡
+
2 = 2
0.5π‘₯
π‘₯𝐡
π‘₯𝐴 π‘₯𝐡
𝐡
𝑦𝐴
π‘₯𝐴
+
2𝑦𝐴
π‘₯𝐡
=
𝑦𝐡
π‘₯𝐡
Condition required for Pareto Optimality
6
Question 1 (Extension)
Generalise the Utility Function
π‘ˆπ΄ π‘₯𝐴 , 𝑦𝐴 , π‘₯𝐡 = π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿ
π‘ˆπ΅ π‘₯𝐡 , 𝑦𝐡 = π‘₯𝐡0.5 𝑦𝐡0.5
β€’ Use π‘Ÿ as a parameter to enable differences in the preference for π‘₯𝐡 by 𝐴.
β€’ We can then see no externalities, positive externalities and negative externalities.
7
Question 1
Form the Lagrangian:
β„’ = π‘₯𝐡0.5 𝑦𝐡0.5 + πœ† π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿ βˆ’ π‘ˆπ΄ + 𝛼 100 βˆ’ π‘₯𝐴 βˆ’ π‘₯𝐡 + 𝛾 100 βˆ’ 𝑦𝐴 βˆ’ 𝑦𝐡
Where πœ†, 𝛼 and 𝛾 are Lagrange multipliers.
Derive the FOC’s:
π‘₯𝐴 : πœ†0.5π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿ βˆ’ 𝛼 = 0,
(1)
π‘₯𝐡 : 0.5π‘₯π΅βˆ’0.5 𝑦𝐡0.5 + πœ†π‘Ÿπ‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿβˆ’1 βˆ’ 𝛼 = 0, (3)
πœ†: π‘₯𝐴0.5 𝑦𝐴0.5 βˆ’ π‘ˆπ΄ = 0,
𝑦𝐴 : πœ†0.5π‘₯𝐴0.5 π‘¦π΄βˆ’0.5 π‘₯π΅π‘Ÿ βˆ’ 𝛾 = 0 (2)
𝑦𝐡 : 0.5π‘₯𝐡0.5 π‘¦π΅βˆ’0.5 βˆ’ 𝛾 = 0
(5) 𝛼: 100 βˆ’ π‘₯𝐴 βˆ’ π‘₯𝐡 = 0
𝛾: 100 βˆ’ 𝑦𝐴 βˆ’ 𝑦𝐡 = 0 (7)
(4)
(6)
8
Question 1
(1)=(3)
πœ†0.5π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿ = 𝛼 = 0.5π‘₯π΅βˆ’0.5 𝑦𝐡0.5 + πœ†π‘Ÿπ‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿβˆ’1
0.5π‘₯π΅βˆ’0.5 𝑦𝐡0.5
πœ†=
0.5π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿ βˆ’ π‘Ÿπ‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿβˆ’1
(2)=(4)
πœ†0.5π‘₯𝐴0.5 π‘¦π΄βˆ’0.5 π‘₯π΅π‘Ÿ = 𝛾 = 0.5π‘₯𝐡0.5 π‘¦π΅βˆ’0.5
π‘₯𝐡0.5 π‘¦π΅βˆ’0.5
πœ† = 0.5 βˆ’0.5 π‘Ÿ
π‘₯𝐴 𝑦𝐴 π‘₯𝐡
0.5π‘₯π΅βˆ’0.5 𝑦𝐡0.5
π‘₯𝐡0.5 π‘¦π΅βˆ’0.5
βˆ’0.5 0.5 π‘Ÿ
0.5 0.5 π‘Ÿβˆ’1 = πœ† = 0.5 βˆ’0.5 π‘Ÿ
0.5π‘₯𝐴 𝑦𝐴 π‘₯𝐡 βˆ’ π‘Ÿπ‘₯𝐴 𝑦𝐴 π‘₯𝐡
π‘₯𝐴 𝑦𝐴 π‘₯𝐡
0.5π‘₯𝐴0.5 π‘¦π΄βˆ’0.5
π‘₯𝐡0.5βˆ’π‘Ÿ π‘¦π΅βˆ’0.5
βˆ’0.5 0.5 π‘Ÿ
0.5 0.5 π‘Ÿβˆ’1 =
0.5π‘₯𝐴 𝑦𝐴 π‘₯𝐡 βˆ’ π‘Ÿπ‘₯𝐴 𝑦𝐴 π‘₯𝐡
π‘₯π΅βˆ’0.5 𝑦𝐡0.5
9
Question 1
0.5π‘₯𝐴0.5 π‘¦π΄βˆ’0.5
π‘₯𝐡1βˆ’π‘Ÿ
βˆ’0.5 0.5 π‘Ÿ
0.5 0.5 π‘Ÿβˆ’1 =
0.5π‘₯𝐴 𝑦𝐴 π‘₯𝐡 βˆ’ π‘Ÿπ‘₯𝐴 𝑦𝐴 π‘₯𝐡
𝑦𝐡
π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿ π‘Ÿπ‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿβˆ’1
𝑦𝐡
βˆ’
=
π‘₯𝐡1βˆ’π‘Ÿ
π‘₯𝐴0.5 π‘¦π΄βˆ’0.5
0.5π‘₯𝐴0.5 π‘¦π΄βˆ’0.5
𝑦𝐴 π‘₯π΅π‘Ÿ
π‘₯𝐴
π‘Ÿπ‘¦π΄ π‘Ÿβˆ’1
𝑦𝐡
βˆ’
π‘₯𝐡 = 1βˆ’π‘Ÿ
0.5
π‘₯𝐡
𝑦𝐴
π‘Ÿπ‘¦π΄ βˆ’1
𝑦𝐡
βˆ’
π‘₯𝐡 = 1βˆ’π‘Ÿ π‘Ÿ
0.5
π‘₯𝐡 π‘₯𝐡
π‘₯𝐴
𝑦𝐡
=
𝑦𝐴
βˆ’
2π‘Ÿπ‘¦π΄
π‘₯𝐡
π‘₯𝐴
π‘₯𝐡
Condition required for Pareto Optimality (take this further to get 𝑦𝐴 = β‹―)
10
Question 1
From (6):
100 βˆ’ π‘₯𝐴 βˆ’ π‘₯𝐡 = 0
π‘₯𝐡 = 100 βˆ’ xA
From (7):
100 βˆ’ 𝑦𝐴 βˆ’ 𝑦𝐡 = 0
𝑦𝐡 = 100 βˆ’ yA
Plug into:
𝑦𝐡
π‘₯𝐡
=
𝑦𝐴
π‘₯𝐴
100 βˆ’ yA 𝑦𝐴
2π‘Ÿπ‘¦π΄
=
βˆ’
100 βˆ’ xA π‘₯𝐴
100 βˆ’ xA
100
𝑦𝐴
βˆ’1=
100
π‘₯𝐴
βˆ’
2π‘Ÿπ‘¦π΄
π‘₯𝐡
100 βˆ’ yA
∴
βˆ’ 1 βˆ’ 2π‘Ÿ
1
𝑦𝐴
∴
2π‘Ÿ
𝑦𝐴 =
βˆ’
100
π‘₯𝐴
1
𝑦𝐴
=
100 βˆ’ xA
π‘₯𝐴
βˆ’ 2π‘Ÿ
1
2π‘Ÿ
=
βˆ’
100
π‘₯𝐴
βˆ’1
11
Question 2
For the economy in Question 1, derive the conditions that prevail in
a general competitive equilibrium, and thus demonstrate that the
general competitive equilibrium allocation is not Pareto optimal.
12
Question 2
General Competitive Equilibrium
β€’ Both consumers are maximising utility subject to their budget
constraints
β€’ max π‘ˆπ΄ (π‘₯𝐴 , 𝑦𝐴 ) subject to 𝑝π‘₯ π‘₯𝐴 + 𝑝𝑦 𝑦𝐴 = π‘šπ΄ = 𝑝π‘₯ πœ”π‘₯𝐴 + 𝑝𝑦 πœ”π‘¦π΄
β€’ max π‘ˆπ΅ (π‘₯𝐡 , 𝑦𝐡 ) subject to 𝑝π‘₯ π‘₯𝐡 + 𝑝𝑦 𝑦𝐡 = π‘šπ΅ = 𝑝π‘₯ πœ”π‘₯𝐡 + 𝑝𝑦 πœ”π‘¦π΅
β€’ Market Clear (Demand = Supply)
β€’ π‘₯𝐴 + π‘₯𝐡 = πœ”π‘₯𝐴 + πœ”π‘₯𝐡
β€’ 𝑦𝐴 + 𝑦𝐡 = πœ”π‘¦π΄ + πœ”π‘¦π΅
13
Question 2
For Consumer A:
β„’ = π‘₯𝐴0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿ + πœ† 100𝑝π‘₯ βˆ’ 𝑝π‘₯ π‘₯𝐴 βˆ’ 𝑝𝑦 𝑦𝐴
FOC:
(1)=(2)
π‘₯𝐴 : 0.5π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿ βˆ’ 𝑝π‘₯ Ξ» = 0
(1)
𝑦𝐴 : 0.5π‘¦π΄βˆ’0.5 π‘₯𝐴0.5 π‘₯π΅π‘Ÿ βˆ’ 𝑝𝑦 Ξ» = 0
(2)
πœ†: 100𝑝π‘₯ βˆ’ 𝑝π‘₯ π‘₯𝐴 βˆ’ 𝑝𝑦 𝑦𝐴 = 0
(3)
0.5π‘₯π΄βˆ’0.5 𝑦𝐴0.5 π‘₯π΅π‘Ÿ
0.5π‘¦π΄βˆ’0.5 π‘₯𝐴0.5 π‘₯π΅π‘Ÿ
=πœ†=
𝑝π‘₯
𝑝𝑦
𝑝π‘₯
𝑦𝐴 = π‘₯𝐴 ,
𝑝𝑦
𝑝𝑦
π‘₯𝐴 = 𝑦𝐴 ,
𝑝π‘₯
∴
𝑦𝐴 𝑝π‘₯
=
π‘₯𝐴 𝑝𝑦
14
Question 2
For Consumer B:
β„’ = π‘₯𝐡0.5 𝑦𝐡0.5 + πœ† 100𝑝𝑦 βˆ’ 𝑝π‘₯ π‘₯𝐡 βˆ’ 𝑝𝑦 𝑦𝐡
FOC:
(1)=(2)
π‘₯𝐡 : 0.5π‘₯π΅βˆ’0.5 𝑦𝐡0.5 βˆ’ 𝑝π‘₯ Ξ» = 0
(1)
𝑦𝐡 : 0.5π‘¦π΅βˆ’0.5 π‘₯𝐡0.5 βˆ’ 𝑝𝑦 Ξ» = 0
(2)
πœ†: 100𝑝𝑦 βˆ’ 𝑝π‘₯ π‘₯𝐡 βˆ’ 𝑝𝑦 𝑦𝐡 = 0
(3)
0.5π‘₯π΅βˆ’0.5 𝑦𝐡0.5
0.5π‘¦π΅βˆ’0.5 π‘₯𝐡0.5
=πœ†=
𝑝π‘₯
𝑝𝑦
𝑝π‘₯
𝑦𝐡 = π‘₯𝐡 ,
𝑝𝑦
𝑝𝑦
π‘₯𝐡 = 𝑦𝐡 ,
𝑝π‘₯
∴
𝑦𝐡 𝑝π‘₯
=
π‘₯𝐡 𝑝𝑦
15
Question 2
β€’ Therefore, in General Competitive Equilibrium, we have:
𝑦𝐴 𝑝π‘₯ 𝑦𝐡
=
=
π‘₯𝐴 𝑝𝑦 π‘₯𝐡
β€’ However, for Pareto Optimality our (generalised) condition is:
𝑦𝐡
π‘₯𝐡
=
𝑦𝐴
π‘₯𝐴
βˆ’
2π‘Ÿπ‘¦π΄
π‘₯𝐡
β€’ Then, GE = PO when π‘Ÿ = 0 (i.e. no externalities)
16