hw6_soln

hw6_soln - Professor Faingold Solution prepared by Anne...

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Professor Faingold Solution prepared by Anne Guan Economics 121b: Intermediate Microeconomics Solution to Problem Set 6 1. Let’s transform these utility functions using logs to make the algebra easier: u A x A 1 ,x A 2 = log x A 1 +(1 ) log x A 2 u B x B 1 ,x B 2 = 1 2 log x B 1 + 1 2 log x B 2 a. A feasible allocation is a consumption vector (consumption of each good by each consumer) that satis±es the resource constraints (total consumption of good 1 equals total endowment of good 1, and the same for good 2). The set of feasible allocations is:  x A 1 ,x A 2 ,x B 1 ,x B 2 : x A 1 + x B 1 = A 1 + B 1 AND x A 2 + x B 2 = A 2 + B 2 Using the simplifying notation, this becomes:  x A 1 ,x A 2 ,x B 1 ,x B 2 : x A 1 + x B 1 = 1 AND x A 2 + x B 2 = 2 b. 1
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c. A Pareto e cient allocation is one where no consumer can be made better o without making another worse o . d. You can either do this formally or intuitively (only one is needed for full credit, but you should understand both). Answer 1 is the formal proof using Lagrangians and answer 2 is the intuitive answer. You can think of "in general" as "if preferences are monotonic, convex and smooth". Answer 1 (Formal, also given in lecture 12) Pareto e ciency is equivalent to maximizing A’s utility subject to B getting a certain utility level and subject to the feasibility constraints. Note that we are choosing both consumers’ consumption, this is not the same as consumer A’s utility maximization problem. The problem can be written as follows [ ( x 1 ,x 2 ) represent consumer A’s consumption, ( x 1 ,x 2 ) represent consumer B’s consump- tion; we do this to distinguish free variables from the solution]: max ( x 1 ,x 2 ,x 1 ,x 2 ) u A ( x 1 ,x 2 ) s.t. u B ( x 1 ,x 2 )= u B (Lagrange mult. μ ) x 1 + x 1 = 1 (Lagrange mult. 1 ) x 2 + x 2 = 2 (Lagrange mult. 2 ) The Lagrangian is: L ( x 1 ,x 2 ,x 1 ,x 2 ,μ, 1 , 2 )= u A ( x 1 ,x 2 )+ μ u B ( x 1 ,x 2 ) u B + 1 [ 1 x 1 x 1 ]+ 2 [ 2 x 2 x 2 ] The FOCs are: L x 1 =0 MU A 1 1 =0 MU A 1 = 1 L x 2 =0 MU A 2 2 =0 MU A 2 = 2 L x 1 =0 MU B 1 1 =0 MU B 1 = 1 L x 2 =0 MU B 2 2 =0 MU B 2 = 2
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hw6_soln - Professor Faingold Solution prepared by Anne...

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