PS_8_Solution_Key

PS_8_Solution_Key - TA Yankun Wang Economics 617 Problem...

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TA: Yankun Wang Economics 617 Problem Set 8 Solution Key 1. [Unconstrained Optimization: Sufficient Conditions for a Global Maximum] Suppose A is an open convex set in R n , and f : A R is twice continuously differentiable and quasi-concave on A . Suppose there is x ̄ in A satisfying: (i) f x ̄ 0 , and (ii) H f x ̄ is negative definite. Show that x ̄ is the unique point of global maximum of f on A . Solution: We will prove this result via two steps: first we prove that x ̄ is a strict local maximum; second we prove that this strict local maximum has to be the unique global maximum given f is quasi-concave. Claim1: x ̄ is a strict local maximum of f on A . Proof: In the Lecture Notes, we are given that if f : A R is twice continuously differentiable and x ̄ in A satisfies: (i) f x ̄ 0 , and (ii) H f x ̄ is negative definite, then x ̄ is a local maximum of f on A . However, this result is not strong enough for our purpose. We actually need to prove that x ̄ is a strict local maximum of f on A .The main idea is to use Taylor’s Theorem. However, before using it, we need to prove one thing: if H f x ̄ is negative definite and f is twice continuously differentiable, then H f x is negative definite for all the x in a small neighborhood around x ̄ . We are given that f is twice continuously differentiable, which means D ij f : A R is a continuous function for i , j 1,. ., n . We will also utilize the fact that the summation and multiplication of continuous functions defined on A is continuous. Let F r : A R , r 1,. ., n denote the corresponding leading principal minors (lpm) of H f . Note here that each F r x is a continuous function from A to R .S ince H f x ̄ is negative definite, lpm of degree 1; F 1 x ̄ D 11 f x ̄ 0 For  F 1 x ̄ 0, 1 0: whenever d x , x ̄ 1 , we have | F 1 x ̄ F 1 x |  , which means D 11 f x 0. Since A is an open set, 1 0: B x ̄ , 1 A . Set 1 min 1 , 1 , so that we have B x ̄ , 1 A , with D 11 f x 0 for all x B x ̄ , 1 . lpm of degree 2; F 2 x ̄ D 11 f x ̄ D 22 f x ̄ D 12 f x ̄ D 21 f x ̄ 0 For  F 2 x ̄ 0, 2 0: whenever d x , x ̄ 2 , we have | F 2 x ̄ F 2 x |  , which means D 11 f x D 22 f x D 12 f x D 21 f x 0. Since A is an open set, 2 0: B x ̄ , 2 A . Set 2 min 2 , 2 , so that we have B x ̄ , 2 A , with D 11 f x D 22 f x D 12 f x D 21 f x 0 for all x B x ̄ , 1 . ...

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This note was uploaded on 08/29/2009 for the course ECON 617 taught by Professor Staff during the Fall '08 term at Cornell.

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PS_8_Solution_Key - TA Yankun Wang Economics 617 Problem...

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