Final Exam - Math Camp August 27, 2014

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1 Final Exam - Math Camp August 27, 2014 You will have three hours to complete this exam. Please write your solution to question one in blue book 1 and your solutions to the subsequent questions in blue book 2. Write your name on each blue book. Read each question carefully before answering. Good luck! 1. Consider the problem of maximizing utility over two goods, x 1 and x 2, where utility is given by the function U(x 1, x 2 ) kx α 1 x β 2, where α, β (0, 1). (a) Under what conditions is U( ) homogenous of degree one in (x 1, x 2 )? U( ) is homogeneous of degree one if U(cx 1, cx 2 ) cu(x 1, x 2 ) for c > 0. Therefore, we need or α + β 1. k (cx 1 ) α (cx 2 ) β kc α+β x 1 x 2 kcx 1 x 2, The consumer has exogenous income given by y > 0, and prices for x 1 and x 2 are p 1 > 0 and p 2 > 0 respectively. There are no other goods in this economy, nor are there opportunities for saving. Recall that any increasing transformation of the objective function will yield the same optimal choices of x 1 and x 2. Take the increasing transformation ln U(x 1, x 2 ) ln k before attempting parts (b)-(f). (b) Write down the constrained optimization problem of maximizing the consumer s utility subject to a budget constraint as carefully as possible. The problem is to choose x 1 and x 2 to solve maximize x 1,x 2 α ln x 1 + β ln x 2 subject to p 1 x 1 + p 2 x 2 y, x 1 0, x 2 0. (c) Prove that there exists a unique solution to the problem that you wrote down in part (b) without first solving the problem. Proof. We know that if the objective function is strictly quasiconcave and continuous, the constraint function is concave, and the feasible set if convex and nonempty then there exists a unique solution to the KKT problem given in part (b). To show that the objective function is strictly quasiconcave, it is sufficient to show that it is strictly concave. 1

2 Let λ (0, 1) and let (ˆx 1, ˆx 2 ) and ( x 1, x 2 ) be two points in R 2. Since the natural log is a strictly concave function, it is true that ln[λˆx 1 + (1 λ) x 1 ] > λ ln ˆx 1 + (1 λ) ln x 1 and ln[λˆx 2 + (1 λ) x 2 ] > λ ln ˆx 2 + (1 λ) ln x 2. Thus, α ln[λˆx 1 + (1 λ) x 1 ]+β ln[λˆx 2 + (1 λ) x 2 ] > λ [α ln ˆx 1 + β ln ˆx 2 ] + (1 λ) [α ln x 1 + β ln x 2 ], and the objective is strictly concave. Furthermore, the natural log is a continuous function and the linear combination of continuous functions is continuous. The constraint function is linear and therefore both convex and concave. Since y > 0, the constraint set is trivially nonempty, since it contains the origin. The constraint set is a subset of R 2, so it is sufficient to show that it is closed and bounded to show that it is compact by the Heine-Borel Theorem. Let ɛ 2y/ p p 2 2. Then every point in the feasible set lies within the ball B ɛ (0). To prove this, show that the solution to the problem of maximizing the Euclidean norm of any point in the constraint set is x 1 p 1 p p 2 2 y, x 2 p 1 p p 2 2 y, and thus the maximum value of the Euclidean norm of any feasible point is y/ p p 2 2. To show that the constraint set is closed, it is straightforward to show that it s complement is open. Its complement is the set {x R 2 x 1 < 0} {x R 2 x 2 < 0} {x R 2 p 1 x 1 + p 2 x 2 > y}. It is straightforward to show by contradiction that for each point in the three sets above, there exists an ɛ > 0 such that the ɛ-ball around the point is fully contained in the set. Furthermore, the union of open sets is open. Thus, a unique solution to the KKT problem must exist. (d) Write down the first-order necessary conditions and complementary slackness conditions that result from this consumer s problem. Provide an argument for which conditions will bind and which will be slack. The Lagrangean is L(x 1, x 2, λ) α ln x 1 + β ln x 2 + λ(y p 1 x 1 p 2 x 2 ). 2

3 The first order necessary conditions and corresponding complementary slackness conditions are ( ) α α x 1 p 1 λ 0, x 1 x 1 p 1 λ 0, ( ) β β x 2 p 2 λ 0, x 2 x 2 p 2 λ 0, y p 1 x 1 p 2 x 2 0, λ (y p 1 x 1 p 2 x 2 0). If x 1 0, then the FOC with respect to x 1 yields λ, which is impossible. The same is true if x 2 0. If λ 0, then α, β 0, which contradicts the assumptions of the model. Therefore, the budget constraint must bind and the non-negativity constraints are slack. (e) Write the demand functions x 1 (p, y) and x 2 (p, y) that solve the constrained utility maximization problem and prove that the demand functions are homogeneous of degree zero. Solving the first-order conditions yields x 1 (p, y) α y, α + β p 1 We see that for c > 0, x 1 (cp, cy) x 2 (cp, cy) x 2 (p, y) Thus, x 1 (p, y) and x 2 (p, y) are h.o.d. 0. β α + β y p 2. α cy α y x 1 (p, y), α + β cp 1 α + β p 1 β cy β y x 2 (p, y). α + β cp 2 α + β p 2 (f) Write the optimal (transformed) utility of the consumer as a function of prices and income using the transformed utility function and your solution to part (e). Take the third- and fourth-order leading principal minors of the bordered Hessian of the transformed utility function. What can you say about the quasiconcavity/quasiconvexity of this function? (In other words, is the function quasiconcave, quasiconvex, possibly quasiconcave, possibly quasiconvex, none of the above, some combination of the above?) Evaluating the transformed utility at x 1 (p, y) and x 2 (p, y) yields the indirect utility function v(p, y) (α + β) ln y α ln x 1 β ln x 2 + α ln[α/(α + β)] + β ln[β/(α + β)] This yields the bordered Hessian H 0 α/p 1 β/p 2 (α + β)/y α/p 1 α/p β/p 2 0 β/p (α + β)/y 0 0 (α + β)/y 2 3.

4 The third-order leading principal minor is the determinant of the matrix comprised of the first 3 rows and columns of H, 0 α/p 1 β/p 2 α/p 1 α/p αβ(α + β) < 0. β/p 2 0 β/p 2 p 1 p 2 2 The fourth-order leading principal minor is the determinant of H. One way to compute that is H α p 1 M 21 + α p 2 1 M 22, where and M 22 M 21 α/p 1 β/p 2 (α + β)/y 0 β/p (α + β)/y 2 0 β/p 2 (α + β)/y β/p 2 β/p (α + β)/y 0 (α + β)/y 2 αβ(α + β) p 1 p 2 y 2, β(α + β)2 p 2 2y 2 + β2 (α + β) p 2 2y 2. Therefore, H 0. Since the third- and fourth-order leading principal minors are both nonpositive, it is possible that the function is quasiconvex. However, the weak inequality is a necessary condition, not a sufficient condition. To be certain that the function is quasiconvex, we would need H < 0. Since the third-order leading principal minor is strictly negative, this function cannot be quasiconcave as it violates the necessary condition. In ECON602, you will learn that under general conditions, the indirect utility function is indeed quasiconvex. 2. Prove the following proposition using the envelope theorem: Proposition 1 (Roy s Identity). Suppose that u : R n R is continuous and strictly quasiconcave. Let v(p, y) satisfy: v(p, y) maximize x R n u(x) subject to p x y x i 0, i 1, 2,..., n, where p R n is a vector of non-negative prices. Furthermore assume that for some point ( p, ȳ) R n+1 where p i > 0 i 1, 2,..., n, ȳ > 0, v( p, ȳ) is continuously differentiable. Then for all i 1, 2,..., n, x i ( p, ȳ) v( p, ȳ)/ p i v( p, ȳ)/ y. 4

5 Proof. v(p, y) is the indirect utility function, which is the maximand of the constrained optimization problem given above. The Lagrangian is given by By the envelope theorem, L(x, λ; p, y) u(x) + λ(y p x) v(p, y) p i for all i 1, 2,..., n. Similarly, we have v( p, ȳ) y Combining both equation yields L(x, λ; p, y) p i λx i xx( p,ȳ); λλ( p,ȳ) λ( p, ȳ)x i ( p, ȳ) L(x, λ; p, ȳ) y λ( p, ȳ) Rearranging yields our desired result xx(p,ȳ); λλ( p,ȳ) xx( p,ȳ); λλ( p,ȳ) v( p, ȳ) v( p, ȳ) x i ( p, ȳ) p i y v( p,ȳ) p x i ( p, ȳ) i v( p,ȳ) y 3. Parts (a)-(d) are unrelated (a) Prove that every finite set is compact. (Note: This set is not necessarily a subset of R n.) Proof. We denote the finite set as S, and without loss of generality, we can label the finite elements of the initial set as s 1, s 2,..., s n. Let i S i be an open cover of the finite set S. Since the set S is a subset of the open cover, we can find a set S i1 such that s 1 S i1. We repeat this process for s 2, s 3,..., s n. By this methodology we find that we can write S n j1s j. Therefore, any open cover of S emits a finite subcover. The following proof assumes that the set is a subset of R n and therefore the Heine-Borel Theorem is applicable. This proof is worth 3 points. 5

6 Proof. Let S be a finite set with a finite number of elements s 1, s 2,..., s n where n <. To show that S is compact, we need show that it is bounded and closed. Define S max{ s 1, s 2,..., s n }, where x is any arbitrary norm. Thus, S is bounded since s i S i. To prove that S is closed, notice that we can rewrite S as the union of n singleton sets each consisting of distinct elements in S. Let S i {s i } for all i 1, 2,... n and for all s i S. Hence we can write S as S n S i. From Thm in Simon and Blume, a finite union of closed sets is closed. It is easy to see that a singleton set is closed, hence by Thm 12.10, S is closed. (b) An point x is said an accumulation point of the set S if every open ball around x contains elements of S other than x. Note that x does not need to be an element of S. Prove that if S is closed, any accumulation point of S lies within S. Proof. We prove by contradiction. Suppose that if S is closed, x an accumulation point of S that does not lie within S, i.e. x S c where S c denotes the complement of S. Since S is closed, S c is open. Hence ɛ > 0 such that the open ball with radius ɛ centered around x lies entirely within S c, i.e. B ɛ (x) S c. This means that S B ɛ (x), contradicting the definition of an accumulation point. (c) Is the function 2 ln x + 3 ln y a homothetic function? Why or why not? To show that it is a homothetic function, we need to verify that the given function is a strictly increasing transformation of a homogenous function. Define h(x, y) x 2 y 3 and v(x) ln(x). Notice that h(x, y) is HOD 5 since for any c > 0, h(cx, cy) (cx) 2 (cy) 3 c 5 x 2 y 3 c 5 h(x, y). The natural log function is strictly increasing so v(x) is strictly increasing. Finally, we can write f(x, y) 2 ln x + 3 ln y as the composition of v and h, i.e. f(x, y) v(h(x, y)), so f(x, y) is a homothetic function. (d) Prove that any intersection of convex sets is convex. Proof. Let s and s be elements of i S i where each S i is a convex set, then it must also be the case that s, s S i for all i. Since each S i is a convex set, for any α [0, 1], αs + (α 1)s S i for all i. It follows that αs + (α 1)s i S i so i S i is a convex set. 4. Consider the problem where consumers live for T periods and maximize their utility as a function of consumption c t and healthcare consumption h t. Consumers receive an initial endowment of wealth a 1 in period t 1, and can invest in an one-period i1 6

7 asset a t in each subsequent period that yields return (1 + r). The price of consumption is set equal to one, while healthcare consumption costs p t per unit in each period. This problem can be written as the following dynamic optimization problem: maximize {c t,h t} T t0 T t0 1 α βt (h α t + c α t ) subject to a t (1 + r)a t 1 c t p t h t, a T 0, a 1 given, c t 0, h t 0, where α (, 1) and p t is assumed to be positive. Furthermore, assume that r > 0 and (1 + r)β 1. (a) Specify state and control variables and write down the Bellman equation. State: a t 1 Control: c t, h t, a t (choose any two) The Bellman equation can be written as: for all t 0, 1,..., T. 1 V t (a t 1 ) max h t,c t α (hα t + c α t ) + βv t+1 [(1 + r)a t 1 c t p t h t ] (b) Write down the first-order conditions and envelope conditions for the Bellman equation that you wrote down in part (a). First order conditions: h t : h t βp t V t+1(a t ) (1) Envelope condition: c t : c t βv t+1(a t ) (2) V t (a t 1 ) V t+1(a t ) (3) (c) Derive the Euler equation relating c t to c t+1, showing that c t is some constant c over time. Additionally, derive the Euler equation relating h t to c t. Using (2) and (3), bring the equation one period forward and we obtain: c t+1 βv t+1(a t ) Plug this back into (2) to get the Euler equation for consumption: c t c t+1 7

8 or c t c t+1 t (4) Hence consumption is constant over time at some value c. To derive the intratemporal Euler equation, combine (1) and (2): h t p t c t (5) Using (4) and rearraging terms, we can write h t in terms of the constant consumption amount c and the p t as follows: c h t p 1 t. (6) (d) Use your solution to rewrite the household s budget constraint as a first-order non-autonomous difference equation in a. Use the boundary conditions to solve for c as a function of model parameters and the sequence of prices. (Note: there are other ways to solve for c as a function of model parameters and prices. You must solve the difference equation to receive full credit.) Using the results in (c), the asset accumulation equation becomes a first-order non-autonomous difference equation a t (1 + r)a t 1 c(1 + p α t ). (7) To find the complementary solution to a t, iterate backwards on the homogenous difference equation a C t (1 + r)a C t 1 (1 + r) 2 a C t 2... (1 + r) t+1 k where k is some constant. For the particular solution of a t, iterate (7) backwards: a P t (1 + r)a P t 1 c(1 + p α t ) (1 + r) 2 a P t 2 c(1 + r)(1 + p α t 1 ) c(1 + p α. s 1 (1 + r) s a t s c t j ). t ) Since a 1 is the logical starting point for the sequence, let s t + 1. This yields a P t (1 + r) t+1 a 1 c t t j ). 8

9 The general solution to a t is then a t a C t + a P t (1 + r) t+1 (k + a 1 ) c t t j ). To find k and c, we use both boundary conditions. We obtain k 0 by evaluating the general solution at time t 1. Finally, we can solve for c by evaluating the general solution at time t T : 0 (1 + r) T +1 a 1 c T T j ) Rearranging, we get c (1 + r) T +1 a 1. (8) T T j ) (e) How would a change in price in period t impact the consumption decision c in period zero? How does this relationship depend on α? We take the derivative of (??) with respect to p t to obtain c α α 1 (1 + r)t t p (2)/() (1 + r) T +1 a 1 t [ T ] 2 T j ) Since r > 0, a 1 and p t > 0 for all t, it is clear that (1 + r) T t p 1/() t (1 + r) T +1 a 1 [ T ] 2 > 0. T j ) Thus, if α (0, 1), α/(α 1) > 0 and thus an increase in p t causes c to rise. If α < 0, then α/(α 1) < 0 and thus an increase in p t causes c to fall. If α 0, the utility function is undefined. 9