Microeconomics I. September, c Leopold Sögner

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1 Microeconomics I c Leopold Sögner Department of Economics and Finance Institute for Advanced Studies Stumpergasse Wien Tel: soegner@ihs.ac.at soegner September, 2012

2 Motivation (1) Motivation: Sufficient conditions for maximum for Kuhn Tucker problem: Suppose that there are no nonlinear equality constraints and each inequality constraint is given by a quasiconvex function. Suppose that the objective function satisfies f(x)(x x) > 0 for any x and x with f(x ) > f(x). If x satisfied the Kuhn-Tucker conditions, then x is a global maximizer. (see Mas-Colell, Theorem [M.K.3]). Jehle, Reny: Chapter A 1.4, 1.4. Mas-Colell, Chapter M.C 1

3 Concave Functions (1) Consider a convex subset A of R n. Definition - Concave Function: A function f : A R is concave if f(νx + (1 ν)x) νf(x ) + (1 ν)f(x), ν [0, 1]. If strict > holds then f is strictly concave; ν (0, 1) and x x. This last equation can be rewritten with z = x x and α = ν: f(x + αz) αf(x ) + (1 α)f(x). If f is (strictly) concave then f is (strictly) convex. 2

4 Concave Functions (2) Theorem - Tangents and Concave Functions: If f is continuously differentiable and concave, then f(x ) f(x) + f(x)) (x x) (and vice versa). < holds if f is strict concave for all x x. [Theorem M.C.1] For the univariate case this implies that the tangent line is above the function graph of f(x); strictly for x x with strict concave functions. 3

5 Concave Functions (3) : For α (0, 1] the definition of a concave function implies: f(x ) = f(x + z) f(x) + f(x + αz) f(x) α If f is differentiable the limit of the last term exists such that f(x + z) f(x) + f(x) z 4

6 Concave Functions (4) : Suppose that f(x + z) f(x) f(x) z for any non-concave function. Since f(.) is not concave f(x + z) f(x) > f(x + αz) f(x) α for some x, z and α (0, 1]. Taking the limit results in f(x + z) f(x) > f(x) z, i.e. we arrive at a contradiction. 5

7 Concave Functions (5) Theorem - Hessian and Concave Functions: If f is twice continuously differentiable and concave, then the Hessian matrix D 2 f(x) is negative semidefinite; negative definite for strict concave functions (and vice versa). [Theorem M.C.2] 6

8 Concave Functions (6) : A Taylor expansion of f(x ) around the point α = 0 results in f(x + αz) = f(x) + f(x) (αz) + α2 2 (z D 2 (f(x + β(α)z))z) By the former theorem we know that f(x + αz) f(x) f(x) (αz) 0 for concave functions z D 2 (f(x + β(α)z))z 0. For arbitrary small α we get z D 2 (f(x))z 0. 7

9 Concave Functions (7) : If the right hand side of f(x + αz) f(x) f(x) (αz) = 0.5α 2 (z D 2 (f(x + β(α)z))z) is 0 then the left hand side. By the former theorem f is concave. 8

10 Quasiconcave Functions (1) Definition - Quasiconcave Function: A function f : A R is quasiconcave if f(νx + (1 ν)x) min{f(x ), f(x)}, ν [0, 1]. If > holds it is said to be strict quasiconcave; ν (0, 1) and x x. Quasiconvex is defined by f(νx + (1 ν)x) max{f(x ), f(x)}. If f is quasiconcave than f is quasiconvex. If f is concave then f is quasiconcave but not vice versa. E.g. f(x) = x for x > 0 is concave and also quasiconcave. x 3 is quasiconcave but not concave. 9

11 Quasiconcave Functions (2) Transformation property: Positive monotone transformations of quasiconcave functions result in a quasiconcave function. Definition - Superior Set: S(x) := {x A f(x ) f(x)} is called superior set of x (upper contour set of x). Note that if f(x ν ) min{f(x ), f(x )}, then if f(x ) t and f(x ) t this implies that f(x ν ) t; where t = f(x). 10

12 Quasiconcave Functions (3) Theorem - Quasiconcave Function and Convex Sets: The function f is quasiconcave if and only if S(x) is convex for all x A. 11

13 Quasiconcave Functions (4) Sufficient condition : If f is quasiconcave then S(x) is convex. Consider x 1 and x 2 in S(x). We need to show that f(x ν ) in S(x); f(x) = t. Since f(x 1 ) t and f(x 2 ) t, the quasiconcave f implies f(x ν ) min{f(x 1 ), f(x 2 )} t. Therefore f(x ν ) S(x); i.e. the set S(x) is convex. 12

14 Quasiconcave Functions (5) Necessary condition : If S(x) is convex then f(x) has to be quasiconcave. W.l.g. assume that f(x 1 ) f(x 2 ), x 1 and x 2 in A. By assumption S(x) is convex, such that S(x 2 ) is convex. Since f(x 1 ) f(x 2 ), we get x 1 S(x 2 ) and x ν S(x 2 ). From the definition of S(x 2 ) we conclude that f(x ν ) f(x 2 ) = min{f(x 1 ), f(x 2 )}. Therefore f(x) has to be quasiconcave. 13

15 Quasiconcave Functions (6) Theorem - Gradients and Quasiconcave Functions: If f is continuously differentiable and quasiconcave, then f(x) (x x) 0 whenever f(x ) f(x) (and vice versa). [Theorem M.C.3] If f(x) (x x) > 0 whenever f(x ) f(x) and x x then f(x) is strictly quasiconcave. If f(x) is strictly quasiconcave and if f(x) 0 for all x A, then f(x) (x x) > 0 whenever f(x ) f(x) and x x. 14

16 Quasiconcave Functions (7) : For f(x ) f(x) and α (0, 1] the definition of a quasiconcave function implies: f(x + α(x x)) f(x) α 0 If f is differentiable, then the limit exists such that f(x) z 0 15

17 Quasiconcave Functions (8) : Suppose that f(x) z 0 holds but f is not quasiconcave. Then f(x + αz) f(x) < 0 for some x, z and α (0, 1]. Such that (f(x + αz) f(x))/α < 0. Taking the limit results in a contradiction. 16

18 Quasiconcave Functions (9) Theorem - Hessian Matrix and Quasiconcave Functions: Suppose f is twice continuously differentiable. f(x) is quasiconcave if and only if D 2 (f(x)) is negative semidefinite in the subspace {z f(x) z = 0}. I.e. z D 2 (f(x))z 0 whenever f(x) z = 0. [Theorem M.C.4] If the Hessian D 2 (f(x)) is negative definite in the subspace {z f(x) z = 0} for every x A then f(x) is strictly quasiconcave. 17

19 Quasiconcave Functions (10) : If f is quasiconcave then whenever f(x ν ) f(x), so f(x) (αz) 0 has to hold. Thus f(x 1 ) f(x) 0 and the above theorem imply: f(x) (z) 0, where z = x 1 x. A first order Taylor series expansion of f in α (at α = 0) results in f(x + αz) = f(x) + f(x)αz + α2 2 (z D 2 f(x + β(α)z)z ). 18

20 Quasiconcave Functions (11) Apply this to x 1, x with f(x 1 ) f(x): f(x + αz) f(x) f(x)αz = α2 2 z D 2 f(x + β(α)z)z. If z = x 1 x fulfills f(x)(x 1 x) = 0 the above inequality still has to hold. This implies α 2 /2z D 2 f(x + β(α)z)z 0. 19

21 Quasiconcave Functions (12) To fulfill this requirement on the subspace {z f(x) z = 0}, where f(x)αz = 0, this requires a negative definite Hessian of f(x). : In the above equation a negative semidefinite Hessian implies that

22 Envelope Theorem (1) Consider f(x; q), x are variables in R N and q are parameters in R S. We look at the constrained maximization problem max x f(x; q) s.t.g m (x; q) b m m = 1,..., M. Assume that the solution of this optimization problem x = x(q) is at least locally differentiable function (in a neighborhood of a q considered). v(q) = f(x(q); q) is the maximum value function associated with this problem. 21

23 Envelope Theorem (2) With no constraints (M = 0) and S, N = 1 the chain rule yields: d f(x( q); q) x( q) v( q) = dq x q + f(x( q); q). q With an unconstrained maximization problem the first order = 0 results in condition f(x( q); q) x d f(x( q); q) v( q) =. dq q 22

24 Envelope Theorem (3) [T. M.L.1] Consider the value function v(q) for the above constrained maximization problem. Assume that v(q) is differentiable at q R S and (λ 1,..., λ M ) are the Lagrange multipliers associated with the maximizer solution x(q) at q. In addition the inequality constraints are remain unaltered in a neighborhood of q. Then v( q) q s = f(x( q); q) q s M m=1 λ m g m (x( q); q) q s. For s = 1,..., S. 23

25 Envelope Theorem (4) With no constraints (M = 0) and S, N = 1 the chain rule yields: v( q) dq s = N n=1 f(x( q); q) x n ( q) x n q s + f(x( q); q) q s. The first order conditions tell us f(x( q); q) x n = M m=1 λ m g m (x( q); q) x n. 24

26 Envelope Theorem (5) In addition we observe N n=1 g m (x( q); q) x n ( q) x n q s + g m( q) q s = 0. if a constraint is binding; if not the multiplier λ m is zero. 25

27 Envelope Theorem (6) Plugging in and changing the order of summation results in : v( q) dq s = M m=1 λ m N n=1 g m (x( q); q) x n ( q) x n q s + f(x( q); q) q s. and v( q) dq s = M m=1 λ m g m (x( q); q) q s + f(x( q); q) q s. Remark: remember that the Lagrangian of the problem is L(x, λ; q) = f(x; q) m λ mg m (x; q). Hence we get v( q) dq s by means of the partial derivative of the Lagrangian with respect to q l, evaluated at q. 26