Question

2) Consider an Exchange economy composed of two individuals A and B and two goodsx1 and x2. Individual A has an endowment of W(3,5) and individual B has an endowment of Wa^- (3,3). A's utility function is given byUA Xx2. Suppose that B is neutral about x1 (neither increasing nor decreasing the amount of x1 affects her utility) and she prefers more of x2 to less. Specifv a utility function for B. Eind the equilibrium price and allocations. 3) Consider an Exchange economy composed of five individuals and two goods x1 and x2 Each individual has the utility function of xix2 and an endowment of (2,4). Find the equilibriumecise and allocations. We have onlỵ done Exchange with two consumers but the idea easily extends to any number of individuals. Hence. if you understand how to find an eauilibrium with 2 individuals, you should be able to solve this problem) 4) (10 points) Consider an exchange economy,composed of two individuals A and B and two goods xi and x2. 's utility function is given by UA-2xi +X2. Individual B's utility function is given by usxix2. In the economy the total endowment of xı is 2 and the total endowment of xz is 1. Normalize p2 to 1. We know that, in this economy equilibrium price is givenbypi-1. a. (6 points) Find the equilibrium allocation. b. (4 points) Write a possible endowment forA and B. That is, with the endowment you have written the equilibrium price should be 1. (Note that there might be multiple values for endowments that would lead to an eauilibrium price of 1. You are.onlỵ required tRindone of these)

Answer 1

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W_A = (3, 5) w_B = (3, 3) v_A = X_1, X_2 Now utility funcn for B, u_B (X_1, X_2) = X_2 Now demand funcn of X_1 = X_1^A = M_A/P_1, Let P_2 = 1, P_1 = P so M_A = 3P_1 + 5P_2 = 3P + 5 X^A_2 = M_A/2P_2 M_B = 3P_1 + 3P_2 = 3P + 3, thus x^A_1 = 3P + 5/2P X^A_2 = 3P + 5/2, X^B_2 = M_B/P_2 = 3P + 3/1 thus at eqn X^A_2 + X^B_2 = 8 3P + 5/2 + 3P + 3 = 8 implies 3P + 5 + 6P + 5 = 16 P*_1 = 5/9 P_2 = 1 thus X^A_1 = 3(5/9) + 5/2P* X^A_1 = 5/3 + 5/2 times 5/9 = 5 + 15/3[10] times 9 = 60/10 = 6, X^B_2 = 3 times 5/9 + 3 = 14/3 X^A_2 = 8 - 14/3 = 10/3, X^B_1 = 6 - 6 = 0 Thus (X_1, X_2)^A = (6, 10/3) (X_1, X_2)^B