University of California, Davis Date: June 24, PRELIMINARY EXAMINATION FOR THE Ph.D. DEGREE. Answer four questions (out of five)

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1 University of California, Davis Date: June 4, 03 Department of Economics Time: 5 hours Microeconomics Reading Time: 0 minutes ANSWER KEY PREIMINARY EXAMINATION FOR TE Ph.D. DEGREE Answer four questions (out of five) Question. Complements and substitutes. A consumer consumes consumption goods, with quantities denoted (x,, x ), and her preferences are represented by a twice differentiable, strictly quasiconcave and locally nonsatiated utility function u : R+ R: x ( x,..., x) ux ( ). ut these consumption goods are not sold in the market. Instead, consumption good ( =,, ) is home-produced by using N[] marketed goods, with quantities denoted (z,, z N[], ), according to the eontief technology x z min z, z,..., N[ ], = a a an[ ], where all denominators are positive. (The first subscript indicates the marketed good, and the second one the consumption good.) A double index (k, ) labels the kth marketed good in the list of marketed goods used in the production of consumption good, =,,, k =,, N[]. ence, there are altogether N =, N[ ] marketed goods, N[] of which are used in the home production of consumption good. Denote by π k the price of the (k, ) marketed good, =,,, k =,, N[]. (a). A vector π ( π,..., πn[],, π,..., πn[],,..., π,..., πn[ ], ) of prices of marketed goods induces, via the home production technology, a vector p (p,, p ) of (implicit) prices for the consumption goods. For =,,, define the function p that expresses the price of consumption good in terms of the prices of the marketed goods. ANSWER. For =,,, in order to home-produce one unit of consumption good, the consumer needs to buy a k units of marketed good (k, ), =,,, k =,, N[]. We accordingly define:

2 p : R R: p ( π ) = a π + a π a π. (.) N ++ N[ ], N[ ], (b). The EMIN[p, u] Problem is defined by min x px subect to ux ( ) u, with solution function the icksian demand function h for consumer goods. Show that S(p, u)p = 0, where S(p, u) is the Slutsky matrix. ANSWER. The solution to the EMIN[p, u] problem is the same as that of the EMIN[tp, u] problem, for any t > 0. ence, icskian demand is homogeneous of degree zero in prices. p h h Euler s theorem then yields... 0 p p =, =,,. p (c). Define the icksian demand for marketed good (k, ) by N ξ : R U R: ξ ( π, u) = ah( p ( π), u), k ++ k k where U is the relevant domain of utility levels. at (π, u)? (π, u)? at (π, u) if When can we say that marketed good (m, i) is a (net) complement of marketed good (k, ) When can we say that marketed good (m, i) is a (net) substitute of marketed good (k, ) at ANSWER. Adapting the usual definitions, we say that (m, i) is a (net) complement of (k, ) ξk ( π, u) 0, and that it is a (net) substitute of (k, ) at (π, u) if π mi ξk ( π, u) 0. π (d). Coffee and cream are popular textbook examples of complements. ut Paul Samuelson suggested that, when a consumer uses cream both in her coffee and in her tea, cream and coffee may actually behave as substitutes. In order to analyze this somewhat paradoxical result, we specialize the previous model to the case of two home-produced consumer goods: coffee (good ) and tea (good ). Coffee requires coffee beans (marketed good (, )) and cream for coffee (marketed good (, )), whereas tea requires tea leaves (marketed good (, )) and cream for tea (marketed good (, )). (The prices of marketed goods (, ) and (, ) may well be the same, but this does not play any role here.) (d)(i). Is marketed good (, ) (coffee beans) a complement or a substitute for marketed good (, ) (cream for coffee)? Argue your answer. mi

3 3 ANSWER. In this simplified model, the price functions (.) for the consumer goods are as follows. Good (coffee): p ( π ) = aπ + aπ, (.) Good (tea): p ( π ) = aπ + aπ, (.3) and the icksian demand for marketed good (k, ) can be written ξ ( π, u) = ah( p ( π), p ( π), u). k k Applying the chain rule, we compute p From (.), π = a ξ ( π, u) = a h p h + p p, and from (.3) π π p π p π = 0.Writing s i. (.4) hi ( pu, ) for the (i, ) entry of the p Slutsky matrix of (b) above, (.4) becomes ξ( π, u) = a s a π, (.5) which must be less than or equal to zero by the negative semidefiniteness of the Slutsky matrix. ence, as is intuitively plausible, coffee beans are a complement of cream for coffee. (d)(ii). Is marketed good (, ) a complement or a substitute for marketed good (, ) (cream for tea)? Argue your answer. ANSWER. Now we compute y (b) above, ξ ( π, u) = a h p h + p = a s a. π p π p π s s p 0 s s = p 0. (.6) ence, s > 0, and therefore ξ( π, u) = asa 0. π (.7)

4 4 It follows that coffee beans are a substitute of cream for tea. (d)(iii). Define the demand for cream as the sum of the demand for marketed goods (, ) and (, ). What can you say about the complementarity or substitutability of cream and marketed good (, )? Argue and discuss your answer. ANSWER. The derivative of the icksian demand for cream with respect to the price of coffee beans is given by the sum of the nonpositive term (.5) and the nonnegative term (.7). Its sign will then depend on the relative strength of these two terms. We can compute the sum as follows. ξ( π, u) ξ( π, u) + = a s a + a s a π π. (.8) y (.6), s p + s p = 0, which given the symmetry of the Slutsky matrix implies that s p + s p = 0. ence (.8) can be written ξ ( π, u) ξ ( π, u) p a p + = a s a a s a = a a s π π p a p. (.9) oth definitions in (c) are satisfied if s = 0. So let s < 0. The paradoxical result of coffee beans being a substitute for cream appears a p a when (.9) is positive, i. e., when < 0. This requires that a p a p consumer uses a lot more cream in her tea than in her coffee) in relation to p be relatively small (the, the relative price of coffee. An increase in the price of coffee beans then induces a large substitution of tea for coffee, which may result in an overall increase in the demand for cream. Question. Capping emissions. A firm that behaves in a perfectly competitive manner in all markets produces one output by using - inputs according to a direct production function f R R z z z f z. : + : (,..., ) ( ) assumed to be differentiable with a strictly positive gradient on R ++ and concave.

5 5 Denote by p > 0 the price of the output, and by w (w,, w - ) input prices. Assume that the cost function R ++ the vector of : ++ + : (,..., ; ) (,..., ; ) c R R R w w q c w w q is differentiable and convex in q, and that at the profit maximizing solution the quantities of the inputs and of output are positive. (a). What is the relation between the output price and the marginal cost at a profitmaximizing solution? Prove your answer. ANSWER. Write the profit maximization problem as: Given w, choose q > 0 in order to maximize p q c(w, q). The first-order condition is: cwq (, ) p, with equality if q > 0, as assumed, i. e., the q output price equals the marginal cost. (b). ow does an increase in an input price affect the marginal cost at a profitmaximizing solution? Prove your answer. ANSWER. The PRICE = MARGINA COST equality is maintained before and after an increase in the price of any input. This necessitates an adustment in the amount of output, because an increase in an input price increases the marginal cost at any given level of output, as can be seen as follows. The cost-optimization problem can equivalently be written in the minimization or maximization form. Its maximization form is as follows. Given (w, q) choose (z,, z - ) in order to maximize w z subect to q < f(z). The agrangian of the problem is w z - λ[ q - f(z)]. The first-order conditions are:. w +λ 0, w +λ z = 0, =,, -, q < f(z), [q - f(z)] λ = 0. The value function of this problem is c(w, q), and by the envelope theorem, ( cwq (, )) = λ 0. ence, q cwq (, ) =λ 0, i. e., the marginal cost equals λ. q Implicitly differentiating the FO equality w +λ = 0, we obtain dλ dw = > 0. ence, as long as the solution is interior the marginal cost is increasing in every input price.

6 6 (c). In order to limit greenhouse gas emissions, the public authority imposes a fixed cap or quota k on the CO emissions of the firm. All inputs may contribute to emissions: more precisely, the firm s emissions are a convex, differentiable function with nonnegative partial derivatives. η R R z z η z z : + : (,..., ) (,..., ) (c)(i). Is it necessarily true that the production of a larger amount of output requires emissions to increase? Discuss it in the simpler two-dimensional case. ANSWER. The answer is in general NO. Assume that, at some point, the slope of the isoquant, f η η is different from the slope of the iso-emissions curve, say >. η η η ε et (ε, ε ) be small and satisfy > > ε η and consider increasing z by ε while decreasing z by ε. The FO approximation to the change in output is ε ε > 0, whereas η η that of emissions is ε ε < 0, i. e., output increases, while emissions decrease. See Figure. η If, on the other hand <, then decrease z by ε while increasing z by ε, for η η ε < < ε η. In that case the FO approximation to the change in output is η η ε+ ε > 0, whereas that of emissions is ε+ ε < 0. See Figure.. ence, it is often technologically possible to increase output while decreasing emissions.

7 7 z Figure. Isoquant (+ ε, - ε ) Iso-emissions Z z Figure. (- ε, + ε ) Isoquant Iso-emissions z

8 8 (c)(ii). Write the Kuhn-Tucker conditions of the profit-maximizing problem of the firm that faces an emission cap, and discuss the implications of the size of the emission cap on the agrange multiplier ANSWER. The problem of the firm when facing the emissions constraint is: Given p, w and k, choose z in order to maximize pf(z) w.z subect to η(z) < k, with agrangian pf(z) w.z - µ[η(z) k]. Its KT conditions are: η η p w µ 0, p w z = 0, =,...,, η( z) k 0, µ [ η( z) k] = 0. The multiplier µ is nonnegative. If k is large enough, the firm ust chooses the same input vector as when its emissions are unconstrained and emits an amount less than k. The last KT condition then implies than µ is zero. If, on the other hand, the emissions constraint is binding, then µ is typically positive. (c)(iii). Compare the Kuhn-Tucker conditions of the profit-maximizing problem of the firm under the emissions constraint with those of the standard profit-maximizing problem (without the emissions constraint). ANSWER. Under the assumption that the solution is interior, the first KT conditions can be written: f No emissions constraint: p = w, =,...,, η Emissions constraint: p = w +µ, =,...,. (.) (.) We observe that the RS of (.) is never lower than that of (.), and is higher than η (.) unless µ or is zero. (c)(iv). Argue that the profit-maximizing output of the firm cannot be higher under the emissions constraint than in the absence of such a constraint.

9 9 ANSWER. We see in (c)(iii) that the emissions constraint has an effect on the profit maximizing solution of the firm formally comparable to a (weak) increase in input prices. From (b) we know that an increase in input prices will induce an increase in the marginal cost, and because marginal costs are nondecreasing in q (by the assumed convexity of the cost function with respect to q), the PRICE = MARGINA COST equality of (a) above cannot be satisfied at a higher level of output. Therefore the profit-maximizing output of the firm cannot be higher under the emissions constraint than in the absence of such a constraint. If the emissions constraint is binding, then the firm will typically choose to produce a smaller amount of output.

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13 C RC r (d) gives h = βγ + i. γ r +γ i C 4 If i N, then h i β[ ω i + rhi] βωi = =, since h i = 0. r r Supply = Demand gives γ C RC + r β β i, i C i r + C r ω= γ +γ i N β R C β [ +γc β] α+γc + RN = = =. r +γ C +γ C +γ C +γc Thus β [ RC + ( +γc) RN] r =. [ α+γ ] (f) Substituting r in the expression found for y gives C γc [ RC ( C) RN] C [ C] RC ( C) RN] y β + +γ R γ α+β+γ +β +γ = C + =, +γc α+γ C +γc α+γc γc i.e., y = [ RC +βrn]. α+γ C

14 Micro Prelim June 4, 03 Answer Keys for Question 4 4(a) The game is as follows: pa pa pb pa pb pb pa pa pb pa pb pb pa pa pb pa pb pb pa pa pb pb pa pb 4(b) There are four subgames in the second stage. In the subgame where they both choose, by ertrand s theorem there is only one Nash equilibrium given by p = p = where both firms make zero profits. In the subgame where they both choose, by ertrand s theorem there is only one Nash equilibrium given by p = p = 0 where both firms make zero profits. Now consider the situation where one firm has chosen and the other. Then the profit π π functions are π = ( p ) D and π = pd. Solving = 0 and = 0 we get p p p = 8, p = 4 with corresponding profits of π = 69 and π = 49. Thus the game can be reduced to: ence there is a unique subgame-perfect equilibrium given by (, p =, p = 8, p = 4, p = 0 ),(,, p =, p = 4, p = 8, p = 0). The subgame- strategy of Firm strategy of Firm perfect equilibrium play is: Firm chooses, Firm follows with and then Firm chooses p = 8 and Firm chooses p = 4 with corresponding profits of π = 69 and π = 49.

15 4(c) y Part (b) the problem reduces to finding the Nash equilibria of a simultaneous auction where the winner gets $(69 her bid) at the loser gets $49. There is only one Nash equilibrium of this game where both players bid $0. et b A be the bid of Player A and b the bid of Player. Proof: first of all, there cannot be a Nash equilibrium where ba > b because the winner (in this case, Player A) can increase her payoff by reducing her bid slightly. Similarly, there cannot be a Nash equilibrium where b > ba. Thus the only candidates for Nash equilibrium are pairs ( ba, b ) where ba = b. Call this common bid b. If b > 0 then the winner (Player A) gets a payoff of 69 b < 8 and she can increase her payoff to 49 by switching to ba < b. It cannot be that b < 0, because Player s payoff is 49 and he can increase it to (69 b ε ) by switching to b = b + ε with a sufficiently small ε > 0. Finally we show that (0,0) is indeed a Nash equilibrium. The payoff of each player is 49. Player A is the winner; if she increases her bid to any b A > 0, then her payoff becomes 69 b A < 49 and if she switches to any b A < 0, then her payoff remains 49. If Player increases his bid to any b > 0, then his payoff becomes 69 b < 49 and if he switches to any b < 0, then his payoff remains 49. Thus there is only one subgame-perfect equilibrium of the entire game given by the bids of 0 together with the strategies determined in Part (b). 4(d) The game can be reduced as in Part (c). In this case there are many Nash equilibria. First note that ba = b = b requires b = 0 (if b > 0, the winner Player A prefers to become the loser and if b < 0 then Player prefers to become the winner). Secondly, ba b requires ba 0 (otherwise Player wants to become the winner) and b 0 (otherwise Player A wants to become the loser). Finally, b > ba requires b 0 (otherwise Player A wants to become the winner) and also ba 0 (otherwise Player wants to become the loser). The Nash equilibria are: () every pair ( ba, b ) with ba b and ba 0 and b 0 (Player s payoff is 49 and he cannot increase it by changing his bid, while Player A s payoff is at least 49 and she cannot increase it by changing her bid), () every pair ( ba, b ) with b > ba and b 0 and ba 0. 4(e) In this case there are no Nash equilibria. Nash equilibrium requires the loser s bid to be zero and the winner s bid to be as low as possible; thus the only candidate would be ba = b = 0, but this is not a NE because Player can increase his payoff by slightly increasing his bid. ere is a more detailed argument: () ba b > 0 is not a NE because Player can increase his payoff by reducing his bid, () b ba > 0 is not a NE because Player A can increase her payoff by reducing her bid, (3) b > b = 0 is not a NE because Player A can increase her payoff by reducing her bid, A (4) b > b = 0 is not a NE because Player can increase his payoff by reducing her bid. A

16 Answer Keys for Question 5 5(a) The game is as follows (where A = { a, b}, c = { c}, A = { a, c}, b = { b} ) 5(b.) The normal form is as follows (where the strategy ( x, y ) for Player means x if { a, b } and y if {c} and the strategy ( z, w ) for Player means z if { a, c } and w if {b}). Inside each cell the corresponding sets of outcomes are given: Player ({ a, c},{ a, c} ) ({ a, c},{ b} ) ({ b},{ a, c} ) ({ b},{ b} ) ({ a, b},{ a, b} ) {(5,5),(5,0),(0,5)} {(5,5),(4,),(0,5) } {(4,6),(5,0),(,4) } {(4,6),(4,),(, 4) } ( a b c ) { } { } { } { )} ( c a b ) { } { } { } { } ({ c},{ c} ) {(6,4),(4,),(, 4) } {(6,4),(4,),(,4) } {(5,5),(4,),(,4) } {(5,5),(4,),(, 4) } Pl {, },{ } (5,5),(5,0),(,4) (5,5),(4,),(,4) (4,6),(5,0),(,4) (4,6),(4,),(, 4 { },{, } (6,4),(4,),(0,5) (6,4),(4,),(0,5) (5,5),(4,),(,4) (5,5),(4,),(, 4) Taking as payoffs the smallest sum of money in each cell (for the corresponding player) the game can be written as follows: ( a b a b ) ( a b c ) ( c a b ) ({ c c ) Player ({ a, c},{ a, c} ) ({ a, c},{ b} ) ({ b},{ a, c} ) ({ b},{ b} ) {, },{, } 0, 0 0,, 0 Pl {, },{ }, 0,, 0, { },{, } 0, 0,,,, },{ },,,, 5(b.) There are 9 Nash equilibria which are highlighted in red. ( ) 5(b.3) Truth telling is represented by the strategy profile ({, },{ }),({, },{ }) of the Nash equilibria. a b c a c b and it is one

17 5(c.) No. If the state is b then it is a good idea for Player to report truthfully because { a, c } yields her 0 while {b} yields her. ut if the state is either a or c then, by ayes rule, Player must assign probability to the left-most node and probability to the right-most node of her information set; thus her expected payoff from reporting { a, c } is = 4.5 while the expected payoff from reporting {b} is = 5. ( ) 5(c.) Always lie corresponds to the strategy profile ({ },{, }),({ },{, }) the corresponding beliefs must be: for Player (, ) and for Player ( ) 3 3 c a b b a c. y ayes rule 0,,,0 at the top information set and (0,) at the bottom information set. Sequential rationality is then satisfied at every information set: for Player at the information set on the left {c} gives an expected payoff of = 3 while {a,b} gives = 3 and at the node on the right {a,b} gives and so does {c}; for Player at the top information set {b} gives an expected payoff of = and {a,c} gives = 4.5 and at the bottom information set both {a,c} and {b} give.