ANSWER KEY University of California, Davis Date: August 20, 2015 Department of Economics Time: 5 hours Microeconomic Theory Reading Time: 20 minutes PRELIMINARY EXAMINATION FOR THE Ph.D. DEGREE Please answer four questions (out of five) Question 1. Throughout this question we assume that all the optimization problems considered have unique solutions which are interior and characterized by the first-order equalities. We also assume that all decision makers take all prices as given, and that the Lagrange multipliers are nonzero. 1.1. This part covers standard price-taking production theory, and we minimally adapt the notation used in class and in Chapter 5 of Mas-Colell et al. We consider a price-taking firm that produces one output by using N inputs, according to a direct production function f : R N + R :(z 1,...,z N )! f (z 1,...,z N ), where (z 1,...,z N ) z denotes the input vector. We denote by q the amount of output, and in order to avoid notational confusion with Part 1.2 below we use capital letters for prices, i. e., P R ++ denotes the price of the output, and W (W 1,...,W N ) R N ++ the vector of input prices. We consider the following optimization problems. Problem INPUTPROFITMAX[W, P]. Given (W, P), choose (z 1,...,z N ) in order to maximize Pf (z) W i z. Denote by!z(w,p) and by Π(W,P) the solution and the value, respectively, of this problem. Problem COSTMIN[W, q]. Given (W, q), choose (z 1,...,z N ) in order to minimize W i z subject to f (z) q. Denote by ẑ(w,q) and by c(w,q) the solution and the value, respectively, of this problem, and by β(w,q)its Lagrange multiplier.
2 Problem OUTPUTPROFITMAX[W, P]. Given (W, P), choose q order to maximize Pq c(w,q). 1.1(a). Prove that z* solves INPUTPROFITMAX[W, P] if and only if, defining q* = f(z*): (i) z* solves COSTMIN[W, q*], and (ii) q* solves OUTPUTPROFITMAX[W, P]. Remark. It is here preferable that you write direct proofs by contradiction rather that manipulating first order conditions. ANSWER. Proof. The statement is intuitive, and each of its parts can easily be proven by contradiction. (1) We first show that if z* solves INPUTPROFITMAX[W, P], then z* solves COSTMIN[W, q*] (for q* = f(z*)), i. e., the input vector that maximizes profits also minimizes the cost of producing the profit-maximizing output. Suppose not, i. e., assume that there is a cheaper way, say z to produce. Then Pq W i z > Pq W i!z(w,p), contradicting the assumption that!z(w,p) solves OUTPUTPROFITMAX[W, P]. (2) Now we show that if z* solves INPUTPROFITMAX[W, P], then q* f(z*) solves OUTPUTPROFITMAX[W, P]. Suppose not. Let q satisfy: Pq c(w,q) > Pq * c(w,q*). (A.1) Let z solve COSTMIN[W, q ], i. e., c(w,q) = W i z, and q = f (z ). Then the LHS of (A.1) can be written Pf (z ) W i z. On the other hand, because f(z*) = q*, it must be the case that c(w,q*) W i z *. Hence, Pq * c(w,q*) Pf (z)* W i z *. Introducing these weak inequalities in (A.1), we obtain Pf (z ) W i z = Pq c(w,q) > Pq * c(w,q*) Pf (z*) W i s*, contradicting the fact that z* solves INPUTPROFITMAX[W, P]. (3) Now we prove the other direction, i. e., if z* solves COSTMIN[W, q*] and q* solves OUTPUTPROFITMAX[W, P], then z* solves INPUTPROFITMAX[W, P]. Suppose not: let z satisfy: Pf (z ) W i z > Pf (z*) W i z *. (A.2)
3 Write q f (z ). By the definition of the cost function, c(w,q) W i z, and hence Pq c(w,q) Pf (z ) W i z. Because z* solves COSTMIN[W, q*], it must be the case that q* = f(z*) and that c(w, q*) = W i z *. Using (A.2), we have that: Pq c(w,q) Pf (z ) W i z > Pf (z*) W i z* = Pq * c(w,q*), contradicting the assumption that q* solves OUTPUTPROFITMAX[W, P]. 1.1(b). Interpret β(w,q). ANSWER. By the envelope theorem, the Lagrange multiplier is the marginal value of relaxing the constraint, i. e., β(w,q) = c (W,q). q (A.3) In words, β(w,q)is the rate at which cost increases as output increases, i.e., the marginal cost of producing output. 1.1(c). Argue that, if z* solves INPUTPROFITMAX[W, P], and q* = f(z*), then P =β(w,q*). ANSWER. The first order equality of problem COSTMIN[W, q ] is P = c (W,q), q which together with (A.3) yields the desired result. 1.2. We now consider a consumer with utility function!u :R + N R :(x 1,..., x N ) "!u(x 1,..., x N ), which is in particular assumed to be locally nonsatiated, where (x 1,...,x N ) x denotes her consumption vector. We denote by p (p 1,..., p N ) R N ++ the price vector for the consumption goods, and by w > 0 the wealth of the consumer. We consider the following optimization problems. Problem UMAX[p, w]. Given (p, w), choose x (x 1,...,x N ) in order to maximize!u(x)subject to p i x w. Denote by!x(p,w) and by v(p,w) the solution and the value, respectively, of this problem, and by λ(p,w)its Lagrange multiplier. Problem EMIN[p, u]. Given (p, u), choose x (x 1,...,x N ) in order to minimize p i x subject to!u(x) u.
4 Denote by h(p,u) and by e(p,u) the solution and the value, respectively, of this problem, and by µ(p,u)its Lagrange multiplier. Now a new one: Problem FRISCH[p, r]. Given p R N ++ and r R ++, choose x (x 1,...,x N ) in order to maximize r!u(x) p i x. Denote by ϕ(p,r) and by Ω(p,r) the solution and the value, respectively, of this problem. 1.2(a). Interpret Problem FRISCH[p, r] and the parameter r. ANSWER. Problem FRISCH[p, r] is a reinterpretation of Problem OUTPUTPROFITMAX[W, P] in terms of the decision of a consumer, who faces a price r for utility and the usual prices p for the consumer goods that yield utility. In problem FRISCH[p, r] the consumer, instead of maximizing utility under constraints, unconstrainedly maximizes the utility revenue minus the utility cost. To repeat, parameter r is then visualized as the price of utility. 1.2(b). What can you say about Ω p j (p,r)? Justify your answer. ANSWER. In a manner parallel to Hotelling s and Shephard s lemmas, Ω p j (p,r) = ϕ j (p,r), i. e., the partial derivative of the value function with respect to the price p j is the Frischian demand for good j with a minus sign (note that Ω p j (p,r) < 0 ) This can be argued most simply from the envelope theorem, since the (unconstrained) N objective function is r!u(x 1,..., x N ) p k x k, and its derivative with respect to p j is just - x j. k=1 1.2(c). Some of the optimization problems in this part (Part 1.2) are formally (mathematically) identical to some of the ones in Part 1.1 above. Which ones are those? Explain. ANSWER. Problem INPUTPROFITMAX[W, P] is formally identical to Problem FRISCH[p, r]. Problem COSTMIN[W, q ] is formally identical to Problem EMIN[p, u]. Of course, the interpretations vary: in 1.1, we have N-dimensional vectors z of production inputs, with associated price vector W, while in 1.2 we have N-dimensional vectors of consumption goods, with associated price vector p. The notational correspondence is as follows.
5 Part 1.1. Above Here (Part 1.2) z x W p P r q u f!u c e Π Ω Problem UMAX[p, w] of Part 1.2 does not have a formal parallel in Part 1.1, but it is of course related, by duality, to problem EMIN[p, u], which does have a counterpart in Part 1.1. On the other hand, Problem OUTPUTPROFITMAX[W, P] of Part 1.1 does not have a formal parallel in Part 1.2, but as seen in 1.1(a), it is intimately related to problems INPUTPROFITMAX[W, P] and COSTMIN[W, q], which do have formal parallels in Part 1.2. 1.2(d). Interpret the Lagrange multipliers λ(p,w) and µ(p,u). ANSWER. By the envelope theorem, λ(p,w) = v w (p,w), i. e., λ(p,w) is the marginal utility of wealth. (A.4) Similarly, and in a manner parallel to 1.1(b) above, µ(p,u) = e u (p,u), (A.5) i. e., µ(p,u)is the marginal cost of utility, the rate at which the required expenditure increases as the required level of utility increases. 1.2(e). Show that if u* = v(p,w*), then λ( p,w*) = Interpret. (Hint. Use duality.) 1 µ(p,u*). ANSWER. By duality e(p, u) = v v(p, w) = u, i. e., e(p, v(p, w)) = w.
6 Differentiating both sides with respect to w yields, by the chain rule, e v (p,v(p,w)) u w (p,w) = 1, which for w = w* and u* = v(p, w*), can be written, using (A.5) and (A.4), µ(p,u*)λ(p,w*) = 1. The assumption of nonzero multipliers yields the desired result. In words, the marginal utility of wealth is the reciprocal of the marginal cost of utility. 1.2(f). Let x* = ϕ(p,r) and u* =!u(x*). Show that r = µ(p,u*) and interpret. ANSWER. We saw in 1.1(b) that β(w,q) = c (W,q), i. e., the Lagrange multiplier of the q COSTMIN[W, q] problem is the marginal cost. On the other hand, from 1.1(c) the price equals the marginal cost, yielding P = β(w,q). The translation of this result to the consumer interpretation is r = µ(p,u) = e u (p,u), i. e., when the consumer solves the FRISCH[p, r] problem, the price of utility equals the marginal cost of utility. interpret. 1.2(g). Let x* = ϕ(p,r), u* =!u(x*)and w* = e(p,u*). Show that r = ANSWER. Because of 1.2(d)-(f) we have that r = 1 λ(p,w*) = 1, v w (p,u*) i. e., the price of utility equals the reciprocal of the marginal utility of wealth. 1 λ(p,w*) and 1.3. Comment on the results of parts 1.1-1.2 above. ANSWER. The results in parts 1.1 and 1.2 above have provided interesting interpretations of the Lagrange multipliers associated with the various optimization problems. First, by using the duality between problems UMAX[p, w] and EMIN[p, u], we have established that the marginal utility of wealth is the reciprocal of the marginal cost of utility. Second, by appealing the to the formal identity between the INPUTPROFITMAX[W, P] problem (production theory) and the FRISCH[p, r] Problem (interpreted as a consumer
7 maximization problem), as well as to the formal identity between the COSTMIN[W, q] problem (production theory) and the EMIN[p, u] problem (c0nsumer theory), we have introduced the idea of price of utility (r), which in parallelism with competitive production theory equals the marginal cost of utility. Putting together these two notions, we also find that the price of utility equals the reciprocal of the marginal utility of wealth.
Microeconomics Prelim August 2015 - Answer Keys for Questions 4 and 5 ANSWER KEYS FOR QUESTION 4 (a) If the firm offers w < S 0 its expected profit is zero, because nobody will apply. If the firm offers w > S 0 then everybody will apply and the firm s monthly expected profit per worker is q R + (1 q ) R w. H H H L (b) qh RH + (1 qh ) RL < S0 (in order to attract any workers at all, I-Start would have to offer w S 0 but then its expected profit would be qh RH + (1 qh ) RL w < 0 ). (c) Given the values of the parameters we have that qh RH + (1 qh ) RL = 4,500 < S0 = 4,800. Thus if the firm offers a monthly wage that attracts any workers (that is, a wage w 4,800) then the firm makes negative profits. Hence the best option is to get zero profits by not hiring anybody (equivalently, by offering a wage less than 4,800 with the consequence that nobody will apply). Note that the value of n is irrelevant. (d) Suppose first that the values of the parameters are such that every type of worker applies. Then during the probationary period the firm gets an expected monthly revenue of q R + (1 q ) R and pays a monthly salary of d, so that the total expected profit for the H H H L first m months is 50 m[ q R (1 q ) R d] 50(1 p) + 50 p qh audited workers not audited audited and retained +. The expected number of retained workers is H H H L. Retained workers are paid $c. The expected revenue from the first group of retained workers [whose number is 50(1 p )] is q R + (1 q ) R, while the Page 1 of 5 H H H L expected revenue from the second group [whose number is 50 pq ] is expected profits are 50 f ( RH, RL, qh, c, d, m, n, p, k ) where f ( R, R, q, c, d, m, n, p, k) H L H [ ] ( ) H R. Thus total m qhrh + (1 qh ) RL d + ( n m) (1 p) qhrh + (1 qh ) RL c + pqh ( RH c) kp Suppose now that the values of the parameters are such that only type H workers apply. Then total expected profits are 50 g( R, c, d, m, n, k ) where H ( ) ( ) g( R, c, d, m, n, k) = m R d + ( n m) R c = nr md ( n m) c kp. H H H H (e) Replacing the given values in the function f given in part (d), we get f ( m, p ) = 2,500m + 40,500 p 1,125mp 36, 000. The given inequality corresponds to f ( m, p ) < 0 : if the firm were to choose values of m and p that satisfy this inequality and if for such values of m and p all workers applied, then the firm s profits would be negative. On the other hand, replacing the given values in the function g given in part (d), we get g( m) = 18,000 + 2,500m, which is always positive; thus if the firm chooses values of m and p such that only the H types apply then the firm will make positive profits. (f) We only need to see under what conditions type H workers apply and type L workers do not. A worker who stays at her current job gets a total remuneration of = H
Microeconomics Prelim August 2015 - Answer Keys for Questions 4 and 5 ns 0 = 36(4,800) = 172,800 (recall that there is no discounting). A worker of type L who applies gets a total expected remuneration of (recall that if the worker is audited and laid off then she can go back to her earlier job that pays S 0 ) L( m, p) = md + ( n m) [ ps0 + (1 p) c] = 198, 000 + 700mp 25, 200 p 2,500m. Thus a type L worker will not apply if 198, 000 + 700mp 25, 200 p 2, 500m 172,800. This is the first inequality. A worker of type H who applies gets a total remuneration of H ( m) = md + ( n m) c = 198, 000 2,500m. Thus a type H worker will apply if 198, 000 2,500m 172,800. This is the second inequality. The third inequality, for positive profits, is (using the function g of part (e)) 18, 000 + 2,500m pk > 0. Max π (g) ( ) m {0,1,...,36} p [0,1] ( m, p; k) = 50 18,000 + 2,500m kp50 subject to (1) 198, 000 2,500m 172,800 (this constraint says that the H people are going to apply), and (2) 198, 000 + 700mp 25, 200 p 2, 500m 172,800 (this constraint says that the L people are not going to apply). (h) The first constraint is equivalent to m 10 (given the integer constraint). Since the maximand is decreasing in p, the second constraint must be satisfied as an equality and thus can be 252 25m written as p =. Replacing this expression in the maximand we get 252 7m 252 25m 252 25m Π( m; k) = π ( m, 252 7m ; k ) = 50 18,000 + 2,500m k. Thus the maximization 252 7m problem reduces to 252 25m Max Π( m; k) = 50 18, 000 + 2,500m k m {0,1,...,10} 252 7m. 252 25m The maximand is increasing in m (because is decreasing in m: its derivative is 252 7m 648 ); thus the maximum is achieved at m = 10. Hence the solution to the firm s 2 7(36 m) 252 25(10) 1 maximization problem is m = 10 and p = = = 0.01099, provided that 252 7(10) 91 Π (10, k) > 0, that is, that 489,125 k < 142 = 3, 444.542. (i) If the firm is not committed, then it will be tempted to not audit the workers and retain them all, because it figures out that they must all be H workers. However, if this change of policy is anticipated by the workers, then also the L workers will apply ( L (10, 0) = 173, 000 > 172,800 ) and the actual profits of the firm would be negative, because inequality (*) holds: f (10, 0) = 11, 000. Page 2 of 5
Microeconomics Prelim August 2015 - Answer Keys for Questions 4 and 5 ANSWER KEYS FOR QUESTION 5 1 (a) For the US customs: U ( I ) = 1, U ( I ) =, U ( P ) = U ( P ) = 0. L S 2 L S 1 For the drug kingpins: U ( PL ) = 1, U ( PS ) = 2, U ( I L) = U ( IS ) = 0. From now on we will write USC instead of US customs and DK instead of drug kingpins (b) If the drugs are on the Cartagena flight then, since there are 100 passengers and the 7 officers can inspect a total of 70 passengers, the probability of finding the drugs is 70 = 7. If the drugs 100 10 are on the Bogota flight then, since there are 200 passengers and the 8 officers can inspect a total of 80 passengers, the probability of finding the drugs is 80 = 2. 200 5 (c) Assigning 10 officers to the Cartagena flight guarantees that, if the drug shipment is there, it will be intercepted, because all the passengers will be inspected. Assigning more than 10 officers to the Cartagena flight does not increase the probability of intercepting the drug shipment if it is on the Cartagena flight (that probability is already 1 with 10 officers) but decreases the probability of intercepting the drug shipment if it is on the Bogota flight. Thus, letting n be the number of officers assigned to the Cartagena flight, every n such that 10 < n 15 is weakly dominated by n = 10. (d) Let n be the number of officers assigned to the Cartagena flight, B be DK s pure strategy of sending the shipment on the Bogota flight and C be DK s pure strategy of sending the shipment on the Cartagena flight. The unique best response to B is n = 0, but the unique best response to n = 0 is C. Similarly, a best response to C is any n 10 (and no other) but the unique best response to any n 10 is B. Thus there is no pure-strategy Nash equilibrium. (e) Fix an arbitrary n with 0 n 10 (we might as well ignore the dominated strategies). Then the expected payoffs are as follows: that is, US Drug B kingpins 10(15 n) 10(15 n) 10n 1 10n 1 customs n 1, 1 1, 1 200 200 100 2 100 2 US Drug B kingpins 75 5n 25 + 5n 5n 50 5n customs n,, 100 100 100 100 B C In order for the strategy profile n, p 1 p with 0 < p < 1 to be a Nash equilibrium it is necessary that DK be indifferent between playing B for sure and playing C for sure, that is, that C C Page 3 of 5
Microeconomics Prelim August 2015 - Answer Keys for Questions 4 and 5 25 + 5n 50 5n = which gives n = 2.5, not an integer (cutting a customs officer into two equal 100 100 parts was declared unconstitutional by the Supreme Court several years ago in a split 5-4 vote). (f) While n = 2.5 is not a possible choice, it is possible to obtain it as an average, that is, the USC 2 3 could play the following mixed strategy: 1 1. Can this be part of a Nash equilibrium? 2 2 2 3 1 35 1 40 75 Against 1 1 B yields an expected payoff of + =, while C yields an 2 2 2 100 2 100 200 1 40 1 35 75 expected payoff of + =. Thus any mixture of B and C is a best reply to 2 100 2 100 200 2 3 B C 1 1. Now let s see what value of p in the mixed strategy makes the USC 2 2 p 1 p indifferent between playing n = 2 and playing n = 3. The expected payoff from playing n = 2 is 65 10 10 + 55 p p + (1 p) =, while the expected payoff from playing n = 3 is 100 100 100 60 15 15 + 45 p 1 p + (1 p) =. The two are equal if and only if p =, in which case USC s 100 100 100 2 expected payoff is 75. Thus we have found a candidate for a Nash equilibrium, namely 200 2 3 B C 1 1, 1 1. To complete the proof that it is indeed a Nash equilibrium we need to show 2 2 2 2 that the USC cannot get a payoff higher than 75 200 against B C 1 1 with a value of n different 2 2 B C 1 75 5n 1 5n 75 from 2 and 3. Now, playing n against 1 1 yields + = ; thus the USC 2 2 2 100 2 100 200 2 3 is indifferent among all its undominated strategies and therefore 1 1 is indeed a best reply to 2 2 B C 1 1 (as would be any other mixed strategy). 2 2 Page 4 of 5
Microeconomics Prelim August 2015 - Answer Keys for Questions 4 and 5 (g) (g.1) The extensive game is as follows: Nature B DK large C p 1 p small B DK C US customs n n n n 75 5n 100 25 + 5n 100 10n 100 100 10n 100 75 5n 200 25 + 5n 200 10n 200 100 10n 200 2 3 (g.2) Assume that the USC uses the mixed strategy 1 1. Then the probability that a shipment 2 2 1 130 1 120 125 on the Bogota flight is intercepted is + = (yielding DK an expected payoff 2 200 2 200 200 125 75 75 1 75 of 1 = with a large shipment and = with a small shipment), while the 200 200 200 2 400 1 20 1 30 50 probability that a shipment on the Cartagena flight is intercepted is + = 2 100 2 100 200 50 150 150 1 150 (yielding DK an expected payoff of 1 = with a large shipment and = 200 200 200 2 400 with a small shipment); thus, no matter what the size of the shipment is, it is best for DK to 2 3 choose C that is, the best reply to 1 1 is (C,C). 2 2 2 3 (g.3) No: 1 1 is not a best reply (C,C): a best reply is to choose n 10 (since the drug 2 2 shipment is for sure on the Cartagena flight). Page 5 of 5