LONDON SCHOOL OF ECONOMICS Professor Leonardo Felli Department of Economics S.478; x7525 EC400 2010/11 Math for Microeconomics September Course, Part II Problem Set 1 with Solutions 1. Show that the general quadratic form of a 11 x 2 1 + a 12 x 1 x 2 + a 22 x 2 2 ( ) ( ) a11 a 12 x1 can be written as ( x 1 x 2 ). 0 a 22 Solution: multiply the vector and the matrices and show that it yields the quadratic form. x 2 2. List all the principal minors of a general (3 3) matrix and denote which are the three leading principal submatrices. Solution: for a general 3 3 matrix, A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 there is one 3rd order principal minor, det(a). There are three 2nd order principal minors: ( ) a11 a 12 (1) det formed by deleting column 3 and row 3 from A, a 21 a 22
( ) a11 a 13 (2) det formed by deleting column 2 and row 2 from A, a 31 a 33 ( ) a22 a 23 (3) det deleting column 1 and row 1 from A. a 21 a 33 Then, there are three 1st order principal minors: det(a 11 ) formed by deleting columns and rows 2,3, det(a 22 ) formed by deleting columns and rows 1,3, and det(a 33 ) formed by deleting columns and rows 1,2. Finally, the leading principal minors are det(a 11 ), det ( a11 a 12 a 21 a 22 ) and det(a). 3. Let C = ( 0 0 0 c ), and determine the definiteness of C. Solution: Note that det(c 1 )=0 and that det(c 2 ) = 0. The definiteness of C depends completely on c; C is positive semidefinite if c 0 and negative semidefinite if c 0. This is especially obvious when we look at the corresponding quadratic form cx 2 2. 4. Determine the definiteness of the following symmetric matrices: ( ) ( ) 1 2 0 2 1 3 4 a) b) c) 1 1 4 6 2 4 5 0 5 6 Solution: a) positive definite; b) negative definite; c) indefinite; 5. Approximate e x at x = 0 with a Taylor polynomial of order three and four. Then compute the values of these approximation at h =.2 and at h = 1 and compare with the actual values. 2
Solution: e h 1 + h + 1 2 h2 + 1 6 h3 ; e h 1 + h + 1 2 h2 + 1 6 h3 + 1 24 h4. e.2 1.22133...; e.2 1.2214...; e.2 1.22140... e 1 2.666..; e 1 2.70833...; e 1 2.71828... 3
LONDON SCHOOL OF ECONOMICS Professor Leonardo Felli Department of Economics S.478; x7525 EC400 2009/10 Math for Microeconomics September Course, Part II Problem Set 2 with Solutions 1. For each of the following functions, find the critical points and classify these as local max, local min, or can t tell : a) x 4 + x 2 6xy + 3y 2, b) x 2 6xy + 2y 2 + 10x + 2y 5 c) xy 2 + x 3 y xy Solution: a) (0,0) is neither a local max nor a local min; (1,1) is a local min (-1,-1) is a local min. b) (13/7,16/7) is neither a local max nor a local min. c) (0,0); (0,1); (1,0) and (-1,0) are neither local max nor local min (this is called a saddle point). (1/ 5, 2/5) is a local min; (-1 5/, 2/5) is a local max. 2. Let S R n be an open set and f : S R be a twice differentiable function. Suppose that Df(x ) = 0. State the weakest sufficient conditions the relevant points, corresponding to the Hessian of f must, satisfy for: (i) x to be a local max. Solution: negative definite at x. (ii) x to be a strict local min. Solution: positive definite at x. 4
3. Which of the critical points found in Problem 1 are also global maxima or global minima? Solution: a) global min at (1,1) and (-1,-1). To see this, we use a two-stage process. First, we use the Hessian to determine that the function is convex for the open sets U + = {(x, y) x > 1/ 3} and U = {(x, y) x < 1/ 3}. Second, for the remaining region, we either draw or compute the values of the function to see that x 4 + x 2 6xy + 3y 2 1. Solution: b) no local max or min. c) (1/ 5, 2/5) is a local min, not a global min; (-1 5/, 2/5) is a local max, not a global max. 4. Check whether f(x, y) = x 4 + x 2 y 2 + y 4 3x 8y is concave or convex using its Hessian. Solution: The Hessian is ( 12x 2 + 2y 2 4xy 4xy 2x 2 + 12y 2 ) For (x, y) (0, 0), the two leading principal minors, 12x 2 +2y 2, and 24x 4 +132x 2 y 2 + 24y 4 are both positive, so f is a convex function on all R n. 5
LONDON SCHOOL OF ECONOMICS Professor Leonardo Felli Department of Economics S.478; x7525 EC400 2009/10 Math for Microeconomics September Course, Part II Problem Set 3 with Solutions 1. A commonly used production or utility function is f(x, y) = xy. Check whether it is concave or convex using its Hessian. Solution: the Hessian is simply ( 0 1 D 2 F (x, y) = 1 0 ) and the second order principal minor is 1. Since this is negative, then D 2 F is indefinite, and F is neither concave nor convex. 2. Prove that the sum of two concave functions is a concave function as well. Solution: trivial. 3. Let f be a function defined on a convex set U in R n. Prove that the following statements are equivalent: (i) f is a quasiconcave function on U (ii) For all x, y U and t [0, 1], f(x) f(y) f(tx + (1 t)y) f(y) 6
(iii) For all x, y U and t [0, 1], f(tx + (1 t)y) min{f(x), f(y)} Solution: If f is quasiconcave, and f(x) f(y), since the set {z f(z) f(y)} is convex, and y and x belong to this set, then for any t, f(tx + (1 t)y) f(y). Similarly, if (ii) holds, then trivially (iii) holds. Suppose that (iii) holds. Assume without loss of generality that f(x) f(y). Then (iii) implies that for all t, the set {z f(z) f(y)} is convex, and hence f is quasiconcave. 4. State the corresponding theorem for quasiconvex functions. Solution: Let f be a function defined on a convex set U in R n. Then, the following statements are equivalent: (i) f is a quasiconvex function on U (ii) For all x, y U and t [0, 1], f(x) f(y) f(tx + (1 t)y) f(x) (iii) For all x, y U and t [0, 1], f(tx + (1 t)y) max{f(x), f(y)} 5. For each of the following functions on R 1, determine whether they are quasiconcave, quasiconvex, both, or neither: 7
a) e x ; b) ln x; c) x 3 x. Solution: a) both (this is an increasing function in R. Any increasing function will satisfy both quasiconcavity and quasiconvexity). b) both; c) neither (graph the function: it has a strict interior local min and a strict interior local max). 8
LONDON SCHOOL OF ECONOMICS Professor Leonardo Felli Department of Economics S.478; x7525 EC400 2009/10 Math for Microeconomics September Course, Part II Problem Set 4 with Solutions 1. For the following program subject to min f(x) = x x (x 2 ) 0, find the optimal solution. Solution: this problem is designed to illustrate when the constraint qualification can fail. In this problem the constraint qualification requirement does not hold, i.e., we cannot find x > 0 which satisfies the constraint since for all x > 0, x 2 < 0. Alternatively, one can see that the optimal x (which is 0) is actually a critical point of the constraint function. This means that we cannot use Lagrange to characterize all solutions. However, it is trivial in this problem that x = 0 is the solution (and the only value of x which satisfies the constraint). 2. Solve the following problem: max x 1,x 2 f(x 1, x 2 ) = x 2 1x 2 subject to 2x 2 1 + x 2 2 = 3. Solution: see the Simon and Blume book, p.418. 9
3. Solve the following problem: subject to max x,y x2 + y 2 ax + y = 1 when a [ 1 2, 3 2 ]. Solution: here the objective function is convex and so will be the Lagrangian function. This means that we cannot use first order conditions - these will give us minimum and not maximum of the objective function (see also lecture 6). Hence, graphically, one can see that this problem results in a corner solution in which we either choose x = 1 a or we choose y = 1. When a = 1 2, then x2 = 4 > y 2 = 1. On the other hand, when a = 3, x < 1 and hence we prefer choosing y = 1. Thus, for all a < 1 2 the solution is x = 1, y = 0, and for all a 1 the solution is x = 0, y = 1. a Alternatively, one can use Lagrange and find the set of critical points which satisfy the relevant conditions (complementary slackness, etc.). One can then compute which one maximizes the Lagrangian. This one will also solve the constrained optimization problem. 4. Consider the following problem: max f(x) x subject to g(x) a x X Let X be a convex subset of R n, f : X R a concave function, g : X R m a convex function, a is a vector in R m. What is the Largrangian for this problem? prove it is a concave function of the choice variable x on X. 10
Solution: the Lagrangian is L(x, q, a) = f(x) + q(a g(x)) We will show that it is concave in x. Take x and x. Then: tf(x) + (1 t)f(x ) f(tx + (1 t)x ) tg(x) + (1 t)g(x ) g(tx + (1 t)x ) and thus with positive q, we have: f(tx + (1 t)x ) + qa qg(tx + (1 t)x ) tf(x) + (1 t)f(x ) + qa q(tg(x) + (1 t)g(x )) which proves that the Lagrangian is concave. 11
LONDON SCHOOL OF ECONOMICS Professor Leonardo Felli Department of Economics S.478; x7525 EC400 2009/10 Math for Microeconomics September Course, Part II Problem Set 5 with Solutions 1. Assume that the utility function of the consumer is u(x, y) = x + y The consumer has a positive income I > 0 and faces positive prices p x > 0, p y > 0. The consumer cannot buy negative amounts of any of the goods. a) Use Kuhn-Tucker to solve the consumer s problem. b) Show how the optimal value of u,depends on I. Solution: a) this is a quasi linear utility function which is heavily used in economic theory. The Lagrangian is: L(x, y, q, I) = x + y + q 0 (I p x x p y y) + q 1 x + q 2 y The FOCs are 1 q 0 p x + q 1 = 0 1 2 y q 0p y + q 2 = 0 12
Obviously the budget constraint will bind. Hence, q 0 > 0. Then assume that q 1 = q 2 = 0. We then find the interior solution so that 2 y = p x y = p2 x, x = I p x ; q p y 4p 2 0 = 1 y p x 4p y p x This solution exists when 4p y I p 2 x > 0. Now consider the solution where q 1 > 0 and q 2 = 0. Hence, x = 0 and then y = I p y. From the FOCs, this implies that q 0 = 1 and therefore q 1 = 1 p x 1 > 0 when 2 Ip y 2 Ip y p x > 2 Ip y or when p 2 x > 4p y I. Thus, these two solutions exhaust all possibilities (can you explain why there is no solution such that q 2 > 0 and q 1 = 0?). Solution: b) Let us take first the solution when q 1 = q 2 = 0, i.e., both goods are bought. Then:y = p2 x, x = I 4p 2 y p x px 4p y. Hence: u(x, y ) = I p x p x 4p y + p x 2p y Hence, taking a derivative w.r.t. I, we get: u(x, y ) I = 1 p x = q 0. Now consider a corner solution, for example when q 1 > 0. Then y = I p y u(x, y I ) = p y and hence and then: u(x, y ) I = 1 2p y I p y = q 0. 13
2. Solve the following problem: max(min{x, y} x 2 y 2 ) Solution: Here differentiability fails and we cannot use first order conditions. One has to set x = y at the optimal solution which gives x = y = 1 4. 14
LONDON SCHOOL OF ECONOMICS Professor Leonardo Felli Department of Economics S.478; x7525 EC400 2009/10 Math for Microeconomics September Course, Part II Problem Set 6 with Solutions 1. Consider the problem of maximizing xyz subject to x + y + z 1, x 0, y 0 and z 0. Obviously, the three latter constraints do not bind, and we can then concentrate only on the first constraint (x + y + z 1). Find the solution and the Lagrange multiplier, and show how the optimal value would change if instead the constraint is x + y + z.9. Solution: it is easy to see that the solution is x = y = z = 1 and the Lagrange 3 multiplier is 1 1. The optimal value is. Given the new constraint, the optimal value 9 27 is now: 1 27 + 1 9 ( 1 10 ).0259 2. Consider the problem of maximizing xy subject to x 2 + ay 2 1. What happens to the optimal value when we change a = 1 to a = 1.1? Solution: the Lagrangian for the problem is L(x, y, λ; a) = xy + λ(1 x 2 ay 2 ) + λ 1 x+λ 2 y. Note however that it cannot be the case that the maximizer will have either x = 0 or y = 0. Thus, the solution is interior when a = 1, and the solution is x = y = 1/ 2, λ = 1/2. The Envelope Theorem tells us that as a changes from 1 to 1.1,the optimal value of f changes by approximately L( 1 1 a 2, 2, 1 ; 1) (0.1). Then: 2 L a = λy2 = 1 2 ( 1 2 ) 2 = 1 4 15
Hence the optimal value will decrease from.5 to.475. 3. Consider Problem 1 in Problem set 5. Set the first order conditions, and for the case of an interior solution use comparative statics to find changes in the endogenous variables when I and p x change (one at a time), i.e., find (i) (ii) x I, x, p x y I, q 0 I ; y q 0,. p x p x Solution: The first conditions of this problem are: I p x x p y y = 0 1 q 0 p x + q 1 = 0 1 2 y q 0p y + q 2 = 0 Consider J : det 0 p x p y p x 0 0 p y 0 1 4y 3 2 = p2 x 1 4y 3 2 > 0. This is not surprising, since the original Lagrangian is concave; at the unique interior optimum (y > 0), the bordered Hessian, H, is negative definite (satisfying sufficient second order conditions) and hence H 2 = det H > 0. Using total differentiation, and assuming an interior solution (i.e., q 1 = q 2 = 0), we have: di p x dx p y dy = 0 p x dq 0 = 0 16
1 dy p 4y 3 y dq 0 = 0 2 It is obvious that dq 0 = 0 and hence we also have dy = 0. Thus, x I = 1 p x, y I = q 0 I = 0. Now consider a change in p x : or: p x dx xdp x p y dy = 0 p x dq 0 q 0 dp x = 0 1 dy p 4y 3 y dq 0 2 = 0 and in matrix notation: det system. 0 p x p y p x 0 0 p y 0 1 4y 3 2 p x dx p y dy = xdp x p x dq 0 = q 0 dp x 1 dy p 4y 3 y dq 0 2 = 0 0 p x p y p x 0 0 p y 0 1 4y 3 2 q 0 p x x p x y p x = x q 0 0 > 0 and hence one can use Cramer s rule to solve this 17