Micro II Problem Set 1 Solutions PDF

Micro II Solutions to exercises Problem Set 1: Solutions Chapter 1: Preliminaries Exercise 1.1: Direct Proof Theorem 1: Direct Proof Let A be x2 − 8x + 15 = 0 and B =“x is either 3 or 5.” Using a direct proof,...

Micro II Solutions to exercises Problem Set 1: Solutions Chapter 1: Preliminaries Exercise 1.1: Direct Proof Theorem 1: Direct Proof Let A be x2 − 8x + 15 = 0 and B =“x is either 3 or 5.” Using a direct proof, show that A ⇒ B. We build a chain of implications from the statement A to the statement B. A: x2 − 8x + 15 = 0 ⇓ S1 : x2 − 3x − 5x + 15 = 0 ⇓ S2 : x(x − 3) − 5(x − 3) = 0 ⇓ S3 : (x − 5)(x − 3) = 0 ⇓ B: x is either 3 or 5 Exercise 1.2: Proof by Contraposition: Infintiy of prime numbers Theorem 2: Proof by contraposition Prove that there exists an infinity of prime numbers. Let us translate the text of the exercise into our language about statements. Let P be the set of all prime numbers. Then, we can define the statements A and B as A: P is the set of all prime numbers B: P does not have a maximum We need to prove A ⇒ B. Instead, as we know from class, we can equivalently prove ¬B ⇒ ¬A. The statement ¬B is “P does have a maximum.” Call that maximum x. Furthermore, define the number y as the product of all numbers in the set P plus 1. That is: y = (2 · 3 · 5 ·... · x) + 1 1 Micro II Solutions to exercises Observe that y is not divisible by any number in the set P (it leaves the remainder of 1). Furthermore, as any non-prime number can be written as a product of primes, then the number y is also not divisible by any non-prime number. But then, y is a prime number, and since y > x, it holds y ∈ P. Thus, P is not the set of all prime numbers, or ¬A. Exercise 1.3: Proof by Induction Theorem 3: Proof by Induction Prove that the sum of first n odd numbers equals n2. We can prove this statement directly for any fixed k odd numbers. For example, if k = 4, the proof would be as follows: S(2) : 1 + 3 = 22 S(3) : 1 + 3 + 5 = 32 S(4) : 1 + 3 + 5 + 7 = 16 = 42. Having directly verified that the sums of first 2, 3 and 4 odd numbers equals 22 , 32 and 42 , we have proven the statement for k = 4, but we don’t know anything about k = 5, 6, · · ·. It is clear that the pattern supports the statement, but we need to prove that S(n) is true for any n ∈ N. This can be done with a proof by induction. A proof by induction consists of a basis step, where we directly show that the statement holds for n0 (the smallest n for which the statement needs be true), and of an inductive step, where we show that for any k ∈ N, k ≥ n0 : S(k) ⇒ S(k + 1). In words, we need to show that whenever S(k) is true, then S(k + 1) also must be true. Next, we show that the basis step and the inductive step hold. Basis step: S(2) is true since 1 + 3 = 22. Inductive step: Suppose S(k) is true (inductive hypothesis). That is, the sum of the first k odd numbers is k 2. We need to calculate the sum of the first k + 1 odd numbers. For this we need to figure out what is the (k + 1)-th odd number. Since each odd number is by 2 larger than the previous one, we can write the k-th odd number as 2k − 1. 2k − 1 : 2·1−1=1 2·2−1=3 2·3−1=5 2·4−1=7 So that the (k + 1)-th odd number is odd numbers is 2(k + 1) − 1. Then the sum of the first k + 1 odd numbers is: k2  + 2(k + 1) − 1    sum of first k numbers (k+1)−th number 2 Micro II Solutions to exercises where the k 2 comes from the inductive hypothesis. After some simple algebra, we can demonstrate that the inductive step is fulfilled: k 2 + 2(k + 1) − 1 = k 2 + 2k + 2 − 1 = k 2 + 2k + 1 = (k + 1)2. 3 Micro II Solutions to exercises Chapter 2: Pure Exchange Economy Exercise 2.1: Edgeworth Box Theorem 4: Edgeworth Box Consider the following exchange economy. The agent 1 is endowed with ω11 = 3 of good 1 and ω12 = 1 of good 2. The agent 2 is endowed with ω21 = 1 of good 1 and ω22 = 2 of good  2. The utility function of both agents is the same and is given by ui (x1i , x2i ) = x1i x2i. Represent this economy in an Edgeworth box. Mark the endowment point. What is the set of feasible allocations? Sketch the indifference curves passing through the endowment point. Mark the set of allocations that both agents would prefer to their initial endowments. The economy is characterized by the initial endowments and the utility function: w11 = 3, w12 = 1 w21 = 1, w22 = 2  ui (x1i , x2i ) = x1i · x2i 4 Micro II Solutions to exercises Exercise 2.2: Walrasian Equilibrium Theorem 5: Walrasian Equilibrium Consider the economy as in Exercise 2.1. The agent 1 is endowed with ω11 = 3 of good 1 and ω12 = 1 of good 2. The agent 2 is endowed with ω21 = 1 of good 1 and ω22 = 2 of good  2. The utility function of both agents is the same and is given by ui (x1i , x2i ) = x1i x2i. 1. Show that p1 = p2 = 1 does not constitute an equilibrium. 2. Find the equilibrium price and allocation. 3. Depict the equilibrium allocation, the initial endowment and the budget sets in the Edgeworth box. The economy is characterized by the initial endowments and the utility function: ω11 = 3, ω12 = 1 ω21 = 1, ω22 = 2  ui (x1i , x2i ) = x1i · x2i This gives us the following optimization problem subject to the budget constraint:  1 2 max ui (xi , xi ) = x1i · x2i x11 ,x21 s.t. p1 x1i + p2 x2i = mi A solution to the optimization can be found via use of Lagrangian,  Li = x1i x2i + λ(p1 x1i + p2 x2i − mi ) Observe that the problem is identical for both agents. Thus, ignore the subscript i for the moment and solve it generally. The first oder conditions are given by  1 1 x2 ! FOC x : + λp1 = 0 (1) 2  x1 1 x1 ! x2 : + λp2 = 0 (2) 2 x2 ! λ: p1 x 1 + p 2 x 2 − m = 0 (3) 5 Micro II Solutions to exercises Solve (1) and (2) for λ and combine them   1 x2 1 1 x1 1 ⇐⇒ · = · 2 x 1 p1 2 x 2 p2 p2 x1 = p1 x2 p2 x 2 ⇐⇒ x1 = 1 p We can substitute x1 into (3), our initial budget constraint, and solve for x2 p2 x 2 p1 + p2 x 2 = m p1 m x2 = 2 2p from above, we get p2 x 2 p2 m m x1 = 1 = 1 2 = 1 p 2p p 2p We have now the equilibrium demand for both goods in terms of the budget. However, we want it in terms of prices. For that, analyze the rest of the problem for each agent, i.e. put the subscripts back since their endowments differ. Because we have a pure exchange economy, the budget m is mi = p1 ωi1 + p2 ωi2 m1 = p1 ω11 + p2 ω12 = 3p1 + 1p2 m2 = p1 ω21 + p2 ω22 = 1p1 + 2p2 Now we can derive the demands in terms of prices m1 3p1 + p2 x11 = = 2p1 2p1 m2 p1 + 2p2 x12 = 1 = 2p 2p1 m1 3p1 + p2 x21 = 2 = 2p 2p2 m2 p1 + 2p2 x22 = 2 = 2p 2p2 Now we did all the necessary preparation to answer the exercise questions. 6 Micro II Solutions to exercises Part 1: By contradiction. Suppose that p1 = p2 = 1. Then demand for good 2 would be: 3·1+1 4 x21 = = 2·1 2 2 1+2·1 3 x2 = = 2 2 7 x21 + x22 = 2 But supply of good 2 is given by 6 ω12 + ω22 = 1 + 2 = , 2 which is strictly lower than 72 , the demand of good 2, i.e. x2 > ω 2 which is not feasible. Hence, p1 = p2 = 1 does not constitute an equilibrium. Part 2: Let us find prices that would clear market 2, i.e. x2 = ω 2 : 3p1 + p2 p1 + 2p2 + =3 2p2 2p2 3p1 + p2 + p1 + 2p2 = 6p2 4p1 + 3p2 = 6p2 4p1 = 3p2 p1 3 2 = p 4 Thus, we can only determine relative prices. Setting p1 = 3, p2 = 4, 3p1 + p2 9+4 13 x21 = = = 2p2 8 8 1 p + 2p 2 3+8 11 x22 = 2 = = 2p 8 8 2 2 2 2 x 1 + x 2 = 3 = ω 1 + ω2 such that the market for good 2 clears and for good 1, 3p1 + p2 9+4 13 x11 = 1 = = 2p 6 6 p 1 + 2p2 3 + 8 11 x12 = = = 2p1 6 6 1 1 1 1 x 1 + x 2 = 4 = ω 1 + ω2 the market also clears. 7 Micro II Solutions to exercises Part 3: Note that the ratio of prices determines the slope of the black price line. Exercise 2.3: A Cobb-Douglas Economy Theorem 6: A Cobb-Douglas Economy Let agent 1 have utility function u1 (x11 , x21 ) = (x11 )a (x21 )1−a and endowment ω1 = (1, 0). Let agent 2 have utility function u2 (x12 , x22 ) = (x12 )b (x22 )1−b and endowment ω2 = (0, 1). Find the equilibrium allocation and prices. Denote the budget of agent i with mi. Then m1 = p1 · 1 + p2 · 0 m2 = p1 · 0 + p2 · 1 The problem of agent 1 is max(x11 )a (x21 )1−a x11 ,x21 s.t. p1 x11 + p2 x21 = m1 8 Micro II Solutions to exercises and we can set up the Lagrangian: L= (x11 )a (x21 )1−a + λ(p1 x11 + p2 x21 − m1 ) ! FOC x11 : a(x11 )a−1 (x21 )1−a + λp1 = 0 ! x21 : (1 − a)(x11 )a (x21 )−a + λp2 = 0  a−1 x11 a x21 = −λ p1  a x11 (1 − a) x21 = −λ p2 Equating the two expression for −λ and multiplying with p1 p2 , a−1 a x11 x11 a p2 = (1 − a) p1 x21 x21 x11 ap2 = (1 − a) p1 x21 ap2 x21 = (1 − a)p1 x11 (∗) Now substitute p2 x21 from the budget constraint, a(m1 − p1 x11 ) = (1 − a)p1 x11 am1 − ap1 x11 = x11 p1 − ap1 x11 am1 x11 =. p1 Substitute m1 , and we get ap1 x11 (p1 , p2 ) = =a p1 Using the same approach for agent 2 we would obtain bm2 x12 =. p1 Substituting m2 = p2 , we get bp2. x12 = p1 In equilibrium, total demand has to equal total supply for all goods (since goods are desirable!). But, by Walras’ law, we only need to make sure that k − 1 markets clear. In this exercise, that means we only need to make sure one market clears. Thus, x11 + x12 = ω11 + ω21 bp2 a+ =1+0 p1 bp2 =1−a p1 p2 1−a = p1 b 9 Micro II Solutions to exercises Like in exercise 2.2, we have found the relative prices. Thus, we can set p1 = b and p2 = 1 − a and calculate demand: 1−a x11 = a x12 = b =1−a b The same can be done for good 2, agent 1, 1 x11 p1 + x21 p2 = m1 = p1 · 1 p1 p2 x11 + x21 =1 p1 1−a a + x21 =1 b x21 = b and agent 2, 1 x12 p1 + x22 p2 = ω2 = p2 · 1 p2 p1 x12 + x22 = 1 p2 b (1 − a) + x22 = 1 1−a x22 = 1 − b 10