Solution Jehle Reny

Only available on StudyMode
  • Topic: Convex function, Hicksian demand function, Consumer theory
  • Pages : 41 (10800 words )
  • Download(s) : 837
  • Published : June 22, 2012
Open Document
Text Preview
Solutions to selected exercises from Jehle and Reny (2001): Advanced Microeconomic Theory Thomas Herzfeld September 2010 Contents
1 Mathematical Appendix 1.1 Chapter A1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Chapter A2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Consumer Theory 2.1 Preferences and Utility . . . . . . 2.2 The Consumer’s Problem . . . . . 2.3 Indirect Utility and Expenditure . 2.4 Properties of Consumer Demand 2.5 Equilibrium and Welfare . . . . . 3 Producer Theory 3.1 Production . . . . . 3.2 Cost . . . . . . . . . 3.3 Duality in production 3.4 The competitive firm 2 2 6 12 12 14 16 18 20 23 23 26 28 30

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

1

1 Mathematical Appendix

1 Mathematical Appendix
1.1 Chapter A1
A1.7 Graph each of the following sets. If the set is convex, give a proof. If it is not convex, give a counterexample. Answer (a) (x, y)|y = ex This set is not convex. Any combination of points would be outside the set. For example, (0, 1) and 1 / (1, e) ∈ (x, y)|y = ex , but combination of the two vectors with t = 2 not: ( 1 , e+1 ) ∈ 2 2 x (x, y)|y = e . (b) (x, y)|y ≥ ex This set is convex. Proof: Let (x1 , y1 ), (x2 , y2 ) ∈ S = (x, y)|y ≥ ex . Since y = ex is a continuous function, it is sufficient to show that (tx1 + (1 − t)x2 , ty1 + (1 − t)y2 ) ∈ S for any particular t ∈ (0, 1). Set t = 1 . Our task is to show that 1 (x1 + x2 ), 1 (y1 + y2 ) ∈ 2 2 2 1 S. 2 (y1 + y2 ) ≥ 1 (ex1 + ex2 ), since yi ≥ ex1 for i = 1, 2. Also, 2 x2 x1 1 1 x1 (e + ex2 ) ≥ e 2 (x1 +x2 = e 2 · e 2 2 x2 x1 ⇔ ex1 + ex2 ≥ 2e 2 · e 2

⇔ ex1 − 2e 2 · e 2 + ex2 ≥ 0 ⇔ (ex1 − ex2 )2 ≥ 0. (c) (x, y)|y ≥ 2x − x2 ; x > 0, y > 0 This set is not convex. 9 1 For example, 10 , 1 , 1 10 , 1 ∈ S = (x, y)|y ≥ 2x − x2 ; x > 0, y > 0. However, 2 2 1 1 1 9 1, 2 = 2 10 , 1 + 1 1 10 , 1 ∈ S / 2 2 2 (d) (x, y)|xy ≥ 1; x > 0, y > 0 This set is convex. Proof: Consider any (x1 , y1 ), (x2 , y2 ) ∈ S = (x, y)|xy ≥ 1; x > 0, y > 0. For any t ∈ [0, 1], (tx1 + (1 − t)x2 )(ty1 + (1 − t)y2 ) = t2 x1 y1 + t(1 − t)(x1 y2 + x2 y1 ) + (1 − t)2 x2 y2 > t2 + (1 − t)2 + t(1 − t)(x1 y2 + x2 y1 ), since xi yi > 1. = 1 + 2t2 − 2t + t(1 − t)(x1 y2 + x2 y1 ) = 1 + 2t(t − 1) + t(1 − t)(x1 y2 + x2 y1 ) = 1 + t(1 − t)(x1 y2 + x2 y1 − 2) ≥ 1 iff x1 y2 + x2 y1 ≥ 0. y1 y2 y1 y2 + −2≥0 x1 y2 + x2 y1 = x1 y1 + x2 y2 − 2 ≥ y1 y2 y1 y2 y − 1 − 2y1 y2 + y2 ≥ 0 (y1 − y2 )2 ≥ 0,

x1

x2

2

1 Mathematical Appendix which is always true and therefore, (tx1 + (1 − t)x2 , ty1 + (1 − t)y2 ) ∈ S which is convex. (e) (x, y)|y ≤ ln(x) This set is convex. 1 Proof. Let (x1 , y1 ) + (x2 , y2 ) ∈ S. Then 2 (y1 + y2 ) ≤ (ln(x1 ) + ln(x2 )). S is convex ⇒ if

1 2

1 1 ln(x1 ) + ln(x2 ) ≤ ln( x1 + x2 ) 2 2 1 1 1 ⇔ ln(x1 x2 ) ≤ ln( x1 + x2 ) 2 2 2 1 1 1/2 ⇔ (x1 x2 ) ≤ ( x1 + x2 ) 2 2 ⇔ x1 − 2(x1 x2 )1/2 + x2 ≥ 0 ⇔ x1 + x2 1/2 1/2 2

≥0

which is always true. A1.40 Sketch a few level sets for the following functions: y = x1 x2 , y = x1 + x2 and y = min[x1 , x2 ]. Answer x2 6

x2
6 @ @@ @ @@@ @ @ @

x2
6

(a) y = x1 x2

x1

x1

-

x1

(b) y = x1 + x2

(c) y = min(x1 , x2 )

Figure 1: Sets to Exercise A1.40 A1.42 Let D = [−2, 2] and f : D → R be y = 4 − x2 . Carefully sketch this function. Using the definition of a concave function, prove that f is concave. Demonstrate that the set A is a convex set. Answer Proof of concavity: Derive the...
tracking img