8. Example: Consider again the Edgeworth box economy in Question 7. Can the allocation (x",2) = ((1,1), (3,3) be supported as a price equilibrium with transfers? If yes, find the supporting prices and transfers. If no, explain. Solution: Yes. Allocation (r',x²) is feasible and Parento Optimal. MRS' x =1 MRS? - MRS' MRS So, by the Second Welfare Theorem, we can support this allocation as an equi- librium with transfers. The supporting prices are given by the MRS of the consumers at this allocation. That is, 2 = 1. With our normalization, we get P1 =1, P2 = 1. So, for transfers, T1 = (p1, p2)-(z)- (p1,p2)-(e,e) = (1,1)-(1,1)-(1,1)-(1,3) = -2 T2 = -T1=2

ENGR.ECONOMIC ANALYSIS
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Author:NEWNAN
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Chapter1: Making Economics Decisions
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8. Example: Consider again the Edgeworth box economy in Question 7. Can
the allocation (x',²) = ((1,1), (3,3)) be supported as a price equilibrium with
transfers? If yes, find the supporting prices and transfers. If no, explain.
Solution:
Yes. Allocation (x', x²) is feasible and Parento Optimal.
MRS'
3.
MRS =
3.
- MRS' = MRS?
So, by the Second Welfare Theorem, we can support this allocation as an equi-
librium with transfers. The supporting prices are given by the MRS of the
consumers at this allocation. That is, 2 = 1. With our normalization, we get
P1 = 1, P2 = 1. So, for transfers,
Pa
T: = (p1, p2) (x}, x) – (p1,p2)·(e},e5)
= (1,1) (1,1)- (1,1) (1,3) = -2
T2 = -T1 =2
Transcribed Image Text:8. Example: Consider again the Edgeworth box economy in Question 7. Can the allocation (x',²) = ((1,1), (3,3)) be supported as a price equilibrium with transfers? If yes, find the supporting prices and transfers. If no, explain. Solution: Yes. Allocation (x', x²) is feasible and Parento Optimal. MRS' 3. MRS = 3. - MRS' = MRS? So, by the Second Welfare Theorem, we can support this allocation as an equi- librium with transfers. The supporting prices are given by the MRS of the consumers at this allocation. That is, 2 = 1. With our normalization, we get P1 = 1, P2 = 1. So, for transfers, Pa T: = (p1, p2) (x}, x) – (p1,p2)·(e},e5) = (1,1) (1,1)- (1,1) (1,3) = -2 T2 = -T1 =2
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