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Solution Jehle Reny

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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

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The second order partial derivative is always less than zero. Therefore, the function is concave. Proof of convexity: The area below a concave function forms a convex set (Theorem

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1 Mathematical Appendix A1.13). Alternatively, from the definition of convexity the following inequality should hold 4 − (tx1 + (1 − t)x2 )2 ≥ t(4 − (x1 )2 ) + (1 − t)(4 − (x2 )2 ). Multiply out to get 4 − (tx1 + x2 − tx2 )2 ≥ 4 − x2 + t[(x1 )2 − (x2 )2 ]. Again, the area below the function forms 2 a convex set. y
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Figure 2: Graph to Exercise A1.42 A1.46 Consider any linear function f (x) = a · x + b for a ∈ Rn and b ∈ R. (a) Show that every linear function is both concave and convex, though neither is strictly concave nor strictly convex. Answer The statement is true iff, for any x1 , x2 ∈ Rn , t ∈ [0, 1], it is true that f (tx1 + (1 − t)x2 ) = tf (x1 ) + (1 − t)f (x2 ). Substituting any linear equation in this statement gives f (tx1 +(1−t)x2 ) = a[tx1 +(1−t)x2 ]+b = tax1 +(1−t)ax2 +tb+(1−t)b = tf (x1 )+(1−t)f (x2 ) for all x1 , x2 ∈ Rn , t ∈ [0, 1]. (b) Show that every linear function is both quasiconcave and quasiconvex and, for n > 1, neither strictly so. (There is a slight inaccuracy in the book.) Answer As it is shown in (a) that a linear function is concave and convex, it must also be quasiconcave and quasiconvex (Theorem A1.19). More formally, the statement is true iff, for any x1 , x2 ∈ Rn (x1 = x2 ) and t ∈ [0,

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