{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PS9_Solution_Key

# PS9_Solution_Key - TA Yankun Wang Economics 617 Problem Set...

This preview shows pages 1–3. Sign up to view the full content.

TA: Yankun Wang Economics 617 Problem Set 9 Solution Key 1. [On Slater’s Condition] Let X = R 2 + , and f, g 1 and g 2 be functions from X to R , de fi ned by: f ( x 1 , x 2 ) = 2 x 1 + x 2 g 1 ( x 1 , x 2 ) = 1 ( x 1 + x 2 ) g 2 ( x 1 , x 2 ) = ( x 1 + x 2 ) 1 Consider the following optimization problem: Max f ( x 1 , x 2 ) subject to g j ( x 1 , x 2 ) 0 for j { 1 , 2 } and x X ( P ) (a) Show that ˆ x = (1 , 0) solves problem ( P ) . Solution: In order to have g j ( x 1 , x 2 ) 0 for j = 1 , 2 at the same time, we must have ( x 1 + x 2 ) 1 = 0 . Thus x 1 = 1 x 2 ; f ( x 1 , x 2 ) = 2 x 1 + x 2 = 2 x 2 . To maximize f , we want x 2 to be as small as possible, and the minimum of x 2 is 0 . When x 2 = 0 , x 1 = 1 x 2 = 1 . Remark: The problem can be view at a utility maximization problem with linear utility function. Think about the economic explanation as well as the graphical illustration for this problem. Just notice that neither the graph nor the intuition serves as the formal proof. (b) Note that X is a convex set in R 2 , and that f, g 1 and g 2 are concave functions from X to R . Show that Slater’s Condition is not satis fi ed. Solution: If we know what the slater’s condition is, this is kind of obvious: we can never have a point x X such that 1 ( x 1 + x 2 ) > 0 as well as ( x 1 + x 2 ) 1 > 0 , i.e., x 1 + x 2 < 1 and x 1 + x 2 > 1 . Remark: Make sure you know how to use the de fi nition to prove that linear functions are both concave and convex. (c) Does there exist ˆ λ R 2 + such that x, ˆ λ ) is a saddle point? Explain. Solution: We want to know if we can fi nd ˆ λ R 2 + such that: 2 x 1 + x 2 + ( ˆ λ 1 ˆ λ 2 )(1 ( x 1 + x 2 )) 2 2 , for all x X, X = R 2 + . where the de fi nition of saddle point has been applied. If we let ˆ λ 1 ˆ λ 2 = 2 , the above inequality becomes: 2 x 2 2 , and this is always true for x 2 0 . Thus by de fi nition we have fi nd that x, ˆ λ ) is a saddle point as long as: ˆ x = (1 , 0) , ˆ λ 1 ˆ λ 2 = 2 , and ˆ λ R 2 + . Remark: The Lecture Notes has an example (right after Theorem 45 ) in 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
which Slater’s Condition is violated, and we can’t fi nd a saddle point. Try to understand the di ff erence between that example and this problem. 2. [Application of the theorem: “GM SP”] Let a (0 , 1) and b (0 , 1) be given parameters. Let A be the set de fi ned by: A = { ( x, y, z ) R 3 + : y x, z bx and [ y + a ( z bx )] 1 } Consider the following constrained optimization problem: Maximize y subject to ( x, y, z ) A and z x ( B ) Problem ( B ) has a solution (you do not have to show this); denote an arbitrary solution of problem ( B ) by x, ¯ y, ¯ z ) . Show that there is ¯ p > 0 such that: y + ¯ pz ¯ px ¯ y for all ( x, y, z ) A Proof: The title of the problem serves as a hint: use Theorem 45 in the Lecture Notes. It is given that x, ¯ y, ¯ z ) is a constrained global maximum. Let g ( x, y, z ) = z x, f ( x, y, z ) = y, X = A, we need to check fi rst that A is convex, f, g are concave. All of these can be done by checking the de fi nitions of convex sets and concave functions. Is the Slater’s Condition satis fi ed? Yes, since (0 , 0 , 1) in an element in A, and g (0 , 0 , 1) = 1 > 0 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

PS9_Solution_Key - TA Yankun Wang Economics 617 Problem Set...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online