JEHLE RENY Solutions to Selected Exercises

August 20, 2017 | Author: Syed Yusuf Saadat | Category: Utility, Demand, Derivative, Economic Theories, Functions And Mappings
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Solutions to selected exercises from Jehle and Reny (2001): Advanced Microeconomic Theory Thomas Herzfeld Autumn 2011 Remark: Some answers might not be the most elegant ones from a mathematical perspective. Any comment and suggestion, also in case of obscurities, are highly welcome. The graphs are not always perfect.

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

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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, e) ∈ (x, y)|y = ex , but combination of the two vectors with t = 12 not: 1 1 1 e+1 (0, 1) + (1, e) = ( , )∈ / (x, y)|y = ex . 2 2 2 2 (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 = 21 . Our task is to show that 12 (x1 + x2 ), 12 (y1 + y2 ) ∈ S. 21 (y1 + y2 ) ≥ 12 (ex1 + ex2 ), since yi ≥ ex1 for i = 1, 2. Also, x2 x1 1 1 x1 (e + ex2 ) ≥ e 2 (x1 +x2 = e 2 · e 2 2 x2 x1 ⇔ ex1 + ex2 ≥ 2e 2 · e 2 x1

x2

⇔ 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. The graph shows the at (0,0). To provide a formal  that  function has a maximum  9 1 1 1 2 , , 1 , ∈ S = (x, y)|y ≥ 2x − x ; x > 0, y > 0. However, 1, 21 = example, 10 2 10 2   1 1 1 9 1 , + 12 1 10 ,2 ∈ /S 2 10 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.

2

1 Mathematical Appendix x1 y 2 + x2 y 1 = x1 y 1

y1 y2 y1 y2 + x2 y2 − 2 ≥ + −2≥0 y1 y2 y1 y2 y − 1 − 2y1 y2 + y2 ≥ 0 (y1 − y2 )2 ≥ 0,

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. Proof. Let (x1 , y1 ) + (x2 , y2 ) ∈ S. Then 21 (y1 + y2 ) ≤ (ln(x1 ) + ln(x2 )). S is convex   1 1 if 1 ⇒ ln(x1 ) + ln(x2 ) ≤ ln( x1 + x2 ) 2 2 2 1 1 1 ⇔ ln(x1 x2 ) ≤ ln( x1 + x2 ) 2 2 2 1 1 ⇔ (x1 x2 )1/2 ≤ ( x1 + x2 ) 2 2 1/2 ⇔ x1 − 2(x1 x2 ) + x2 ≥ 0  2 1/2 1/2 ⇔ x1 + x2 ≥0 which is always true.

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−30 −6

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(a) (x, y)|y = ex

−4

−2

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(b) (x, y)|y ≥ ex 4

−4

1 y

y

3 0.5 0 −0.5 1

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(d) (x, y)|xy ≥ 1; x > 0, y > 0

(e) (x, y)|y ≤ ln(x)

Figure 1: Sets to Exercise A1.7

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0 x

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4

6

(c) (x, y)|y ≥ 2x − x2 ; x > 0, y > 0

1.5

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1 Mathematical Appendix A1.40 Sketch a few level sets for the following functions: y = x1 x2 , y = x1 + x2 and y = min[x1 , x2 ]. Answer x2

x2 6

(a) y = x1 x2

x1

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

6

x1

-

(b) y = x1 + x2

x1

(c) y = min(x1 , x2 )

Figure 2: 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 first and second order partial derivative: ∂ 2y = −2 ∂x2

∂y = −2x ∂x

The first derivative is strictly positive for values x < 0 and negative for values x > 0. 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 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 − x22 + t[(x1 )2 − (x2 )2 ]. Again, the area below the function forms a convex set. y 6

-x

Figure 3: 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 ).

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1 Mathematical Appendix 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 6= x2 ) and t ∈ [0, 1], we have f (tx1 + (1 − t)x2 ) ≥ min[f (x1 ), f (x2 )](quasiconcavity) f (tx1 + (1 − t)x2 ) ≤ max[f (x1 ), f (x2 )](quasiconvexity) Again by substituting the equation into the definition, we get tf (x1 ) + (1 − t)f (x2 ) ≥ min[f (x1 ), f (x2 )] tf (x1 ) + (1 − t)f (x2 ) ≤ max[f (x1 ), f (x2 )]

∀t ∈ [0, 1]

A1.47 Let f (x) be a concave (convex) real-valued function. Let g(t) be an increasing concave (convex) function of a single variable. Show that the composite function, h(x) = g(f (x)) is a concave (convex) function. Answer The composition with an affine function preserves concavity (convexity). Assume that both functions are twice differentiable. Then the second order partial derivative of the composite function, applying chain rule and product rule, is defined as h00 (x) = g 00 (f (x)) f 0 (x)2 + g 0 (f (x)) f 00 (x)2 For any concave function, ∇2 f (x) ≤ 0, ∇2 g(x) ≤ 0, it should hold ∇2 h(x) ≤ 0. In the case the two functions are convex: ∇2 f (x) ≥ 0 and ∇2 g(x) ≥ 0, it should hold ∇2 h(x) ≥ 0. A1.48 Let f (x1 , x2 ) = −(x1 − 5)2 − (x2 − 5)2 . Prove that f is quasiconcave. Answer Proof: f is concave iff H(x) is negative semidefinite and it is strictly concave if the Hessian is negative definite.   −2 0 H= 0 −2 zT H(x)z = −2z12 − 2z22 < 0, for z = (z1 , z2 ) 6= 0 Alternatively, we can check the leading principal minors of H: H1 (x) = −2 < 0 and H2 (x) = 4 > 0. The determinants of the Hessian alternate in sign beginning with a negative value. Therefore, the function is even strictly concave. Since f is concave, it is also quasiconcave.

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1 Mathematical Appendix A1.49 Answer each of the following questions “yes” or ”no“, and justify your answer. (a) Suppose f (x) is an increasing function of one variable. Is f (x) quasiconcave? Answer Yes, an increasing function of one variable is quasiconcave. Any convex combination of two points on this function will be at least as large as the smallest of the two points. Using the differential-based approach, f is quasiconcave, if for any x0 and x1 , f (x1 ) ≥ f (x0 ) ⇒ ∂f (x0 )/∂x(x1 − x0 ) ≥ 0. This must be true for any increasing function. (b) Suppose f (x) is a decreasing function of one variable. Is f (x) quasiconcave? Answer Yes, a decreasing function of one variable is quasiconcave. Similarly to (a), f is quasiconcave if for any x0 , x1 and t ∈ [0, 1], it is true that f (tx0 + (1 − t)x1 ) ≥ min[f (x0 ), f (x1 )]. (c) Suppose f (x) is a function of one variable and there is a real number b such that f (x) is decreasing on the interval (− inf, b] and increasing on [b, + inf). Is f (x) quasiconcave? Answer No, if f is decreasing on (− inf, b] and increasing on [b, + inf) then f (x) is not quasiconcave. c−b ∈ [0, 1], tb a + (1 − tb )c = b. Given the nature Proof: Let a < b < c, and let tb = c−a of f , f (b) < min[f (a), f (c)]. Then f (tb a + (1 − tb )c) < min[f (a), f (c)], so f is not quasiconcave. (d) Suppose f (x) is a function of one variable and there is a real number b such that f (x) is increasing on the interval (− inf, b] and decreasing on [b, + inf). Is f (x) quasiconcave? Answer Yes. Proof: Let a < b < c, for x ∈ [a, b], f (x) ≥ f (a) and for x ∈ [b, c], f (x) ≥ f (c). Hence, for any x ∈ [a, c], f (x) ≥ min[f (a), f (c)]. (e) You should now be able to come up with a characterization of quasiconcave functions of one variable involving the words “increasing” and “decreasing”. Answer Any function of one variable f (x) is quasiconcave if and only if is either continuously increasing, continuously decreasing or first increasing and later decreasing.

1.2 Chapter A2 A2.1 Differentiate the following functions. State whether the function is increasing, decreasing, or constant at the point x = 2. Classify each as locally concave, convex, or linear at the point x = 2. (a) f (x) = 11x3 − 6x + 8 f1 = 33x2 − 6 increasing locally convex (b) f (x) = (3x2 − x)(6x + 1) f1 = 54x2 − 6x − 1 increasing locally convex

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1 Mathematical Appendix (c) f (x) = x2 −

1 x3

3 x4 increasing locally concave f1 = 2x +

(d) f (x) = (x2 + 2x)3

f1 = (6x + 6)(x2 + 2x)2 increasing locally convex

x3 − 3x2 + 1 (x3 + 1)3 increasing locally concave   3 4 1 8 1 2 4 (f) f (x) = [(1/x + 2) − (1/x − 2)] f1 = − − +4 x2 x3 x2 x increasing locally convex Z 1 2 2 et dt f1 = −ex (g) f (x) = (e) f (x) = [3x/(x3 + 1)]2

f1 = 18x

x

decreasing locally convex A2.2 Find all first-order partial derivatives. (a) f (x1 , x2 ) = 2x1 − x21 − x22 f1 = 2 − 2x1 = 2(1 − x1 )

f2 = −2x2

(b) f (x1 , x2 ) = x21 + 2x22 − 4x2 f1 = 2x1 f2 = 4x2 − 4 (c) f (x1 , x2 ) = x31 − x22 − 2x2 f1 = 3x1 f2 = −2(x2 + 1) (d) f (x1 , x2 ) = 4x1 + 2x2 − x21 + x1 x2 − x22 f1 = 4 − 2x1 + x2 f2 = 2 − 2x2 + x1 (e) f (x1 , x2 ) = x31 − 6x1 x2 + x32 f1 = 3x21 − 6x2 f2 = 3x22 − 6x1 (f) f (x1 , x2 ) = 3x21 − x1 x2 + x2 f1 = 6x1 − x2 f 2 = 1 − x1

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1 Mathematical Appendix (g) g(x1 , x2 , x3 ) = ln x21 − x2 x3 − x23 2x1 g1 = 2 x1 − x2 x3 − x23 −x2 − 2x3 g3 = 2 x1 − x2 x3 − x23

 g2 =

x21

−x3 − x2 x3 − x23

A2.4 Show that y = x21 x2 + x22 x3 + x23 x1 satisfies the equation ∂y ∂y ∂y + + = (x1 + x2 + x3 )2 . ∂x1 ∂x2 ∂x3 The first-order partial derivatives are: ∂y/∂x1 = 2x1 x2 + x23 , ∂y/∂x2 = x21 + 2x2 x3 , and ∂y/∂x3 = x22 + 2x3 x1 . Summing them up gives ∂y ∂y ∂y + + = x21 + x22 + x23 + 2x1 x2 + 2x1 x3 + 2x2 x3 = (x1 + x2 + x3 )2 . ∂x1 ∂x2 ∂x3 A2.5 Find the Hessian matrix and construct the quadratic form, zT H(x)z, when (a) y = 2x1 − x21 − x22 

 −2 0 H= 0 −2 zT H(x)z = −2z12 + 2 · 0z1 z2 − 2z22 (b) y = x21 + 2x22 − 4x2 

2 0 H= 0 4



zT H(x)z = 2z12 + 2 · 0z1 z2 + 4z22 (c) y = x31 − x22 + 2x2 

 6x1 0 H= 0 −2 zT H(x)z = 6x1 z12 − 2z22

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1 Mathematical Appendix (d) y = 4x1 + 2x2 − x21 + x1 x2 − x22 

 −2 1 H= 1 −2 zT H(x)z = −2z12 + 2z1 z2 − 2z22 (e) y = x31 − 6x1 x2 − x32 

6x1 −6 H= −6 6x2



zT H(x)z = 6x1 z12 − 12z1 z2 + 6x2 z22 A2.6 Prove that the second–order own partial derivatives of a convex function must always be nonnegative. Answer A convex function must be increasing over their whole domain, i.e. might not have a maximum. Therefore, the second–order own partial derivative can not change from nonnegative to negative. The second-oorder own partial derivatives represent the diagonal elements of the Hessian. Convexity of the function requires the Hessian to be positive semidefinite. A2.7 Complete Example A2.4 for the partial with respect to x2 . Answer The partial with respect to x2 is ∂f (x1 , x2 ) x1 α = βAxα1 xβ−1 = (1 − α)A 2 ∂x2 x2 After multiplication by the factor t, we obtain: x1 α ∂f (x1 , x2 ) tx1 α ∂f (tx1 , tx2 ) = tα /tα (1 − α)A = (1 − α)A = ∂x2 tx2 x2 ∂x2 p A2.8 Suppose f (x1 , x2 ) = x21 + x22 . (a) Show that f (x1 , x2 ) is homogeneous of degree 1. q q p 2 2 2 2 2 f (tx1 , tx2 ) = (tx1 ) + (tx2 ) = t (x1 + x2 ) = t x21 + x22 (b) According to Euler’s theorem, we should have f (x1 , x2 ) = (∂f /∂x1 ) x1 +(∂f /∂x2 ) x2 . Verify this. x21 + x22 1 · f (x1 , x2 ) = p 2 x x = 1+ p 2 = p x1 + x22 x21 + x22 x21 + x22 x1

x2

9

q x21 + x22

1 Mathematical Appendix A2.9 Suppose f (x1 , x2 ) = (x1 x2 )2 and g(x1 , x2 ) = (x21 x2 )3 . (a) f (x1 , x2 ) is homogeneous. What is its degree? f (tx1 , tx2 ) = t4 (x1 x2 )2 k = 4 (b) g(x1 , x2 ) is homogeneous. What is its degree? g(tx1 , tx2 ) = t9 (x21 x2 )3 k = 9 (c) h(x1 , x2 ) = f (x1 , x2 )g(x1 , x2 ) is homogeneous. What is its degree? h(x1 , x2 ) = x81 x52 h(tx1 , tx2 ) = t13 (x81 x52 k = 13 Obviously, whenever two functions are homogeneous of degree m and n, their product must be homogeneous of degree m + n. (d) k(x1 , x2 ) = g (f (x1 , x2 ), f (x1 , x2 )) is homogeneous. What is its degree? k(tx1 , tx2 ) = t36 (x1 x2 )18 k = 36 (e) Prove that whenever f (x1 , x2 ) is homogeneous of degree m and g(x1 , x2 ) is homogeneous of degree n, then k(x1 , x2 ) = g (f (x1 , x2 ), f (x1 , x2 )) is homogeneous of degree mn. k(tx1 , tx2 ) = [tm (f (x1 , x2 ), f (x1 , x2 ))]n k = mn

A2.15 Check the calculations in Example A2.6 by using the substitution method to solve the system of first–order partials. Then evaluate the function at x∗1 = 3/7 and x∗2 = 8/7 and find y ∗ . Verify what we found in Example A2.7 by evaluating the function at any other point and comparing to y ∗ . Answer The system of first–order partials is ∂f (x1 , x2 ) = −8x1 + 3x2 ∂x1 ∂f (x1 , x2 ) = 1 + 3x1 − 2x2 ∂x2 Re–arrange the first partial gives: x1 = 3/8x2 . Substitute into the second and solve for x2 gives: x∗2 = 8/7 and x∗1 = 3/7. This is the same result as found by using the matrix method. The value of the function at the critical point is y ∗ (3/7, 8/7) = 4/7. A2.16 Find the critical points when (a) f (x1 , x2 ) = 2x1 − x21 − x22 . x1 = 1 and x2 = 0 (b) f (x1 , x2 ) = x21 + 2x21 − 4x2 . x1 = 0 and x2 = 1

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1 Mathematical Appendix (c) f (x1 , x2 ) = x31 − x22 + 2x2 . x1 = 0 and x2 = 1 (d) f (x1 , x2 ) = 4x1 + 2x2 − x21 + x1 x2 − x22 . x1 = 10/3 and x2 = 8/3 (e) f (x1 , x2 ) = x31 − 6x1 x2 + x32 . x1 = 0 and x2 = 0 A2.18 Let f (x) be a real-valued function defined  0 f1 · · ·  f1 f11 · · ·  H∗ =  .. .. . . . . . fn fn1 · · ·

on Rn+ , and consider the matrix  fn f1n   ..  . .  fnn

This is a different sort of bordered Hessian than we considered in the text. Here, the matrix of second-order partials is bordered by the first–order partials and a zero to complete the square matrix. The principal minors of this matrix are the determinants 0 f1 f2 0 f1 , D3 = f1 f11 f12 , . . . , Dn = |H∗ |. D2 = f1 f11 f2 f21 f22 Arrow & Enthoven (1961) use the sign pattern of these principal minors to establish the following useful results: (i) If f (x) is quasiconcave, these principal minors alternate in sign as follows: D2 ≤ 0, D3 ≥ 0, . . . . (ii) If for all x ≥ 0, these principal minors (which depend on x) alternate in sign beginning with strictly negative: D2 < 0, D3 > 0, . . . , then f (x) is quasiconcave on the nonnegative orthant. Further, it can be shown that if, for all x  0, we have this same alternating sign pattern on those principal minors, then f (x) is strictly quasiconcave on the (strictly) positive orthant. (a) The function f (x1 , x2 ) = x1 x2 + x1 is quasiconcave on R2+ . Verify that its principal minors alternate in sign as in (ii). Answer The bordered Hessian is   0 x2 + 1 x1 0 1 . H ∗ =  x2 + 1 x1 1 0

The two principal minors are D2 = −(x2 +1)2 < 0 and D3 = 2x1 x2 +2x1 ≥ 0. Which shows that the function will be quasiconcave and will be strictly quasiconcave for all x1 , x2 > 0.

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1 Mathematical Appendix (b) Let f (x1 , x2 ) = a ln(x1 + x2 ) + b, where a > 0. Is this function strictly quasiconcave for x  0? It is quasiconcave? How about for x ≥ 0? Justify. Answer The bordered Hessian is   a a 0 x1 +x2 x1 +x2   a −a −a H∗ =  x1 +x2 (x1 +x2 )2 (x1 +x2 )2  . a x1 +x2

−a (x1 +x2 )2

−a (x1 +x2 )2

a The two principal minors are D2 = −( x1 +x )2 < 0 for x1 , x2 > 0 and D3 = 0. 2 Which shows that the function can not be strictly quasiconcave. However, it can be quasiconcave following (i). For x1 = x2 = 0 the function is not defined. Therefore, curvature can not be checked in this point.

A2.19 Let f (x1 , x2 ) = (x1 x2 )2 . Is f (x) concave on R2+ ? Is it quasiconcave on R2+ ? Answer In order to prove concavity we can check whether the Hesssian is negative semidefinite for all x. The Hessian is   2x22 4x1 x2 H= . 4x1 x2 2x21 The principal minors of the Hessian are not consistent with a concave curvature: D1 = 2x22 > 0 and D2 = −12x21 x22 < 0. The bordered Hessian is   0 2x1 x22 2x21 x2 H∗ = 2x1 x22 2x22 4x1 x2  . 2x21 x2 4x1 x2 2x21 The two principal minors of the bordered Hessian are D2 = −(2x1 x2 )2 < 0 and D3 = 16x41 x42 > 0. Which shows that the function will be strictly quasiconcave. Note that this function is not homogeneous of degree one. A2.25 Solve the following problems. State the optimised value of the function at the solution. (a) minx1 ,x2 = x21 + x22 s.t. x1 x2 = 1 x1 = 1 and x2 = 1 or x1 = −1 and x2 = −1, optimised value= 2 2 2 (b) minx1 ,x p2 = x1 x2 s.t. x1 + px2 = 1 p p x1 = 1/2 and x2 = − 1/2 or x1 = − 1/2 and x2 = 1/2, optimised value= −1/2

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1 Mathematical Appendix 2 2 2 2 2 (c) maxx1p ,x2 = x1 x2 s.t. x 1 /a + x2 /b = 1 p p x1 = a2 /3 and x2 = 2b2 /3 or x2 = − 2b2 /3, optimised value=

2ab2 33 /2

(d) maxx1p x41 + x42 = 1 ,x2 = x1 + x2 s.t. √ p 4 4 4 x1 = 1/2 and x2 = 1/2, optimised value= 23 = 23/4 (e) maxx1 ,x2 ,x3 = x1 x22 x33 s.t. x1 + x2 + x3 = 1 x1 = 1/6 and x2 = 1/3 = 2/6 and x3 = 1/2 = 3/6, optimised value= 1/432 = 108/66 A2.26 Graph f (x) = 6 − x2 − 4x. Find the point where the function achieves its unconstrained (global) maximum and calculate the value of the function at that point. Compare this to the value it achieves when maximized subject to the nonnegativity constraint x ≥ 0. Answer This function has a global optimum at x = −2. It is a maximum as the secondorder partial derivative is less than zero. The value of the function at that point is f (−2) = 10. Obviously, the global maximum is not a solution in the presence of a nonnegativity constraint. The constrained maximization problem is L(x, z, λ) = 6 − x2 − 4x + λ(x − z) The first–order conditions are: ∂L = −2x − 4 + λ = 0 ∂x ∂L = −λ ≤ 0 ∂z ∂L =x−z =0 ∂λ

(i) (ii) (iii)

Since we do not know a priori whether or not the constraint will be binding, the partial derivative of L with respect to the slack variable is not equal to zero. Either the constraint is binding, i.e. the slack variable redundant z = 0, or the constraint is not binding, i.e. the shadow value of the constraint λ = 0. Such a condition can be summarised in stating that the product of the two variables must be zero: zLz = z(−λ) = 0 If λ = 0, then x = −2 would solve the problem. However, it does not satisfy the nonnegativity constraint. If λ 6= 0, then z = 0 and due to the equality of z and x derived from (iii) x = 0. As the function is continuously decreasing for all values x ≥ 0, it is the only maximizer in this range.

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1 Mathematical Appendix y 6

x

-

Figure 4: Graph to Exercise A2.26

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2 Consumer Theory

2 Consumer Theory 2.1 Preferences and Utility 1.6 Cite a credible example were the preferences of an ‘ordinary consumer’ would be unlikely to satisfy the axiom of convexity. Answer : Indifference curves representing satiated preferences don’t satisfy the axiom of convexity. That is, reducing consumption would result in a higher utility level. Negative utility from consumption of ‘bads’ (too much alcohol, drugs, unhealthy food etc.) would rather result in concave preferences. 1.8 Sketch a map of indifference sets that are parallel, negatively sloped straight lines, with preference increasing northeasterly. We know that preferences such as these satisfy Axioms 1, 2, 3, and 4. Prove the they also satisfy Axiom 50 . Prove that they do not satisfy Axiom 5. Answer : Definition of convexity (Axiom 50 ): If x1 % x0 , then tx1 + (1 − t)x0 % x0 for all t ∈ [0, 1]. Strict convexity (Axiom 5) requires that, if x1 6= x0 and x1 % x0 , then tx1 + (1 − t)x0  x0 for all t ∈ [0, 1]. The map of indifference sets in Figure 5 represents perfect substitues. We know that those preferences are convex but not stricly convex. Intuitively, all combinations of two randomly chosen bundles from one indifference curve will necessarily lie on the same indifference curve. Additionally, the marginal rate of substitution does not change by moving from x0 to x1 . To prove the statement more formally, define xt as convex combination of bundles x0 to x1 : xt = tx0 + (1 − t)x1 . Re-writing in terms of single commodities gives us:  xt = (tx01 , tx02 ) + (1 − t)x11 , (1 − t)x12 . A little rearrangement and equalising the two definitions results in the equality tx0 + (1 − t)x1 = (tx01 + (1 − t)x11 ), (tx02 + (1 − t)x12 ). That is, the consumer is indifferent with respect to the convex combination and the original bundles, a clear violation of strict convexity.

15

2 Consumer Theory x2 6

4

2 x1 1

2

3

4

5

6

7

Figure 5: Indifference sets to Exercise 1.8 1.9 Sketch a map of indifference sets that are parallel right angles that “kink” on the line x1 = x2 . If preference increases northeasterly, these preferences will satisfy Axioms 1, 2, 3, and 4’. Prove that they also satisfy Axiom 5’. Do they also satisfy Axiom 4? Do they satisfy Axiom 5? Answer : Convexity (Axiom 50 ) requires that, if x1 % x0 , then tx1 + (1 − t)x0 % x0 for all t ∈ [0, 1]. Take any two vectors x0 , x1 such that x0 ∼ x1 . Given the nature of these preferences, it must be true that min[x01 , x02 ] = min[x11 , x12 ]. For any t ∈ [0, 1] consider the point tx1 + (1 − t)x2 . If we can show that min[tx01 + (1 − t)x02 , tx11 + (1 − t)x12 ] ≥ min[x01 , x02 ] = min[x11 , x12 ], then we have shown that these preferences are convex. min[tx01 + (1 − t)x02 , tx11 + (1 − t)x12 ] ≥ min[tx01 , tx11 ] + min[(1 − t)x02 , +(1 − t)x12 ] = min[x02 , x12 ] + t[min(x01 , x11 ) − min(x02 , x12 )] = min[x02 , x12 ] Definition of strict monotonicity (Axiom 4): For all x0 , x1 ∈ Rn+ , if x0 ≥ x1 , then x0 % x1 , while if x0  x1 , then x0  x1 . The map of indifference sets in Figure 6 represents perfect complements. Take two points x0 , x1 along one indifference curve. If x0  x1 , “preferences increase northeasterly”, then x0  x1 . For any two vectors on the same indifference curve, that is x0 ≥ x1 , it follows x0 % x1 . Therefore, the definition of strict monotonicity is satisfied for these indifference sets. Strict convexity (Axiom 5) requires that, if x1 6= x0 and x1 % x0 , then tx1 +(1−t)x0  x0 for all t ∈ [0, 1]. Take any two points along the horizontal or vertical part of an indifference curve such as (x01 , x02 ) and (x01 , x12 ), where x02 > x12 . Any convex combination xt = x01 , tx02 + (1 − t)x12 lies on the same indifference curve as x1 and x0 . Therefore, it is not possible that

16

2 Consumer Theory xt  tx0 + (1 − t)x1 . That is, the consumer is indifferent with respect to the convex combination and the original bundles, a clear violation of strict convexity. x2 6 0

px1 px -

x1

Figure 6: Indifference sets to Exercise 1.9 1.12 Suppose u(x1 , x2 ) and v(x1 , x2 ) are utility functions. (a) Prove that if u(x1 , x2 ) and v(x1 , x2 ) are both homogeneous of degree r, then s(x1 , x2 ) ≡ u(x1 , x2 ) + v(x1 , x2 ) is homogeneous of degree r. Answer : Whenever it holds that tr u(x1 , x2 ) = u(tx1 , tx2 ) and tr v(x1 , x2 ) = v(tx1 , tx2 ) for all r > 0, it must also hold that tr s(x1 , x2 ) ≡ u(tx1 , tx2 ) + v(tx1 , tx2 ) = tr u(x1 , x2 ) + tr v(x1 , x2 ). (b) Prove that if u(x1 , x2 ) and v(x1 , x2 ) are quasiconcave, then m(x1 , x2 ) ≡ min{u(x1 , x2 ), v(x1 , x2 )} is also quasiconcave. Answer : Forming a convex combination of the two functions u and v and comparing with m(xt ) satisfies the definition of quasiconcavity:  When u(xt ) ≥ min tu(x1 ) + (1 − t)u(x2 ) and  v(xt ) ≥ min tv(x1 ) + (1 − t)v(x2 ) so  m(xt ) ≥ min u(xt ), v(xt ) .

2.2 The Consumer’s Problem 1.20 Suppose preferences are represented by the Cobb-Douglas utility function, u(x1 , x2 ) = Axα1 x1−α , 0 < α < 1, and A > 0. Assuming an interior solution, solve for the Marshal2 lian demand functions. Answer : Use either the Lagrangian or the equality of Marginal Rate of Substitution and price ratio. The Lagrangian is L = Axα1 x1−α + λ(y − p1 x1 − p2 x2 ). The first-order conditions (FOC) are 2 ∂L − λp1 = 0 = αAxα−1 x1−α 1 2 ∂x1 ∂L = (1 − α)Axα1 x−α 2 − λp2 = 0 ∂x2 ∂L = y − p 1 x1 + p 2 x2 = 0 ∂λ

17

2 Consumer Theory By dividing first and second FOC and some rearrangement, we get either x1 = x2 =

(1−α)p1 x1 . αp2

αx2 p2 (1−α)p1

or

Substituting one of these expressions into the budget constraint, results

in the Marshallian demand functions: x1 =

αy p1

and x2 =

(1−α)y . p2

1.21 We’ve noted that u(x) is invariant to positive monotonic transforms. One common transformation is the logarithmic transform, ln(u(x)). Take the logarithmic transform of the utility function in 1.20; then, using that as the utility function, derive the Marshallian demand functions and verify that they are identical to those derived in the preceding exercise (1.20). Answer : Either the Lagrangian is used or the equality of Marginal Rate of Substitution with the price ratio. The Lagrangian is L = ln(A) + α ln(x1 ) + (1 − α) ln(x2 ) + λ(y − p1 x1 − p2 x2 ). The FOC are α ∂L = − λp1 = 0 ∂x1 x1 ∂L (1 − α) = − λp2 = 0 ∂x2 x2 ∂L = y − p 1 x1 + p 2 x2 = 0 ∂λ The Marshallian demand functions are: x1 = αy and x2 = (1−α)y . They are exactly p1 p2 identical to the demand functions derived in the preceding exercise. 1.22 We can generalise further the result of the preceding exercise. Suppose that preferences are represented by the utility function u(x). Assuming an interior solution, the consumer’s demand functions, x(p, y), are determined implicitly by the conditions in (1.10). Now consider the utility function f (u(x)), where f 0 > 0, and show that the first–order conditions characterising the solution to the consumer’s problem in both cases can be reduced to the same set of equations. Conclude from this that the consumer’s demand behavior is invariant to positive monotonic transforms of the utility function. Answer : The set of first–order conditions for a utility function and its positive monotone transform, assuming a linear budget constraint, are as follows: ∂u ∂L ∂f (u) ∂u(x) ∂L = − λpi = − λpi ∂xi ∂xi ∂xi ∂u ∂xi ∂L ∂u ∂L ∂f (u) ∂u(x) = − λpj = − λpj ∂xj ∂xj ∂xj ∂u ∂xj ∂u/∂xi pi ∂f (u)/∂u ∂u/∂xi pi = = ∂u/∂xj pj ∂f (u)/∂u ∂u/∂xj pj After forming the Marginal Rate of Substitution, the outer derivative cancels out. Therefore, the demand function should be unaffected by the positive monotonic transformation of the utility function.

18

2 Consumer Theory 1.24 Let u(x) represent some consumer’s monotonic preferences over x ∈ Rn+ . For each of the functions F (x) that follow, state whether or not f also represents the preferences of this consumer. In each case, be sure to justify your answer with either an argument or a counterexample. Answer : (a) f (x) = u(x) + (u(x))3 Yes, all arguments of the function u are transformed equally by the third power. Checking the first- and second-order partial derivatives reveals that, although the second-order partial is not zero, the sign of the derivatives is always invariant and positive. 2 ∂ 2u ∂u ∂ 2f 2∂ u = + 6(u(x)) + 3(u(x) ∂x2i ∂x2i ∂xi ∂x2i

Thus, f represents a monotonic transformation of u. (b) f (x) = u(x) − (u(x))2 No, function f is decreasing with increasing consumption for any u(x) < (u(x))2 . Therefore, it can not represent the preferences of the consumer. It could do so if the minus sign is replaced by a plus sign. P (c) f (x) = u(x) + ni=1 xi Yes, the transformation is a linear one, as the first partial is a positive constant, here one, and the second partial of the transforming function ∂f ∂u is zero. Checking the partial derivatives proves this statement: ∂x = ∂x + 1 and i i ∂2f ∂x2i

=

∂2u . ∂x2i

1.28 An infinitely lived agent owns 1 unit of a commodity that she consumes over her lifetime. The commodity is perfect storable and she will receive no more than she has now. Consumption of the commodity in period t is denoted xt , and her lifetime utility function is given by ∞ X β t ln(xt ), where 0 < β < 1. u(x0 , x1 , x2 , . . .) = t=0

Calculate her optimal level of consumption in each period. Answer : Establish a geometric series to calculate her lifetime utility: u = β 0 ln(x0 ) + β ln(x1 ) + β 2 ln(x2 ) + . . . + β t ln(xt ) As β is less than one, this series approaches a finite value. To find the solution, multiply the expression by β and subtract from the original equation [(1)-(2)]. βu = β 1 ln(x0 ) + β 2 ln(x1 ) + β 3 ln(x2 ) + . . . + β t+1 ln(xt ) u − βu = (1 − β)u = ln(x0 ) − β t+1 ln(xt ) ln(x0 ) − β t+1 ln(xt ) u= = ln(x0 ) 1−β Thus, the consumer’s utility maximising consumption will be constant in every period.

19

2 Consumer Theory

2.3 Indirect Utility and Expenditure 1.29 In the two-ggod case, the level sets of the indirect utility function in price space are sets of the form {(p1 , p2 )|v(p1 , p2 , y) = v 0 } for v 0 ∈ R. These are sometimes called price–indifference curves. Sketch a possible map of price–indifference curves. Give separate arguments to support your claims as to their slope, curvature, and the direction of increasing utility. Answer : Figure 7 presents a map of price–indifference curves. Two budget constraints represent different price vectors holding income and utility constant. The combination of prices can be recovered by moving along the price–indifference curve. Because v is decreasing in prices, utility increases by moving further towards the origin. The price– indifference curve must be stricly convex towards the origin. The price–indifference curves must be negatively sloped, i.e. one of the prices must decrease whenever the other price increases holding utility constant. In order to illustrate this fact, re-arrange a homothetic indirect utility function in a two–commodity world v(p, y) = φ(y)v(f (p1 , p2 , 1) in terms of p1 and differentiate with respect to p2 . p1 = vφ(y)−1 f (p1 )−1 ∂f (p1 ) ∂p1 = −vφ(y)−1 f (p1 )−2 ∂p2 ∂p1

Figure 7: Map of price–indifference curves 1.30 Show that the indirect utility function in Example 1.2 is a quasi-convex function of prices and income.

20

2 Consumer Theory Answer : The indirect utility function corresponding to CES preferences is: v(p, y) = y (pr1 + pr2 )−1/r , where r ≡ ρ/(ρ − 1). There are several ways. First, using the inequality relationship, let pt = tp0 + (1 − t)p1 and y t = ty 0 + (1 − t)y 1 . We need to show that the indirect utility function fulfills the inequality    tr −1/r 0r 0r −1/r 1r 1r −1/r y ptr + p ≤ max[y p + p , y p + p ] 1 2 1 2 1 2 which gives:    0r r 1r 1r −1/r 0r −1/r 1r −1/r y tr (p0r ≤ max[y p0r , y p1r ]. 1 + p2 ) + (1 − t) (p1 + p2 ) 1 + p2 1 + p2 Remember that the inverse of concave function is convex. Therefore, it is sufficient to prove that, for any v ∈ R and y > 0, the set {p ∈ R2 : v(p, y) ≤ v} is convex. Define f (p) = (pr1 + pr2 )1/r . If r ∈ (0, 1), then f (p)r = pr1 + pr2 is a concave function. Hence f (p) = (f (p)1/r ) is convex. Since v(p, y) = y/f (p), this implies that the indirect utility funtion is convex for every y and p. 1.37 Verify that the expenditure function obtained from the CES direct utility function in Example 1.3 satisfies all the properties given in Theorem 1.7. Answer : The expenditure function in a two–commodity world is e(p, u) = u (pr1 + pr2 )1/r where r ≡ ρ/(ρ − 1). 1. Zero when u takes on the lowest level of utility in U . The lowest value in U is u((0)) because the utility function is strictly increasing. Consequently, 0(pr1 + pr2 )1/r = 0. 2. Continuous on its domain Rn++ × U . This property follows from the Theorem of Maximum. As the CES direct utility function satisfies the axiom of continuity, the derived expenditure function will be continuous too. 3. For all p >> 0, strictly increasing and unbounded above in u. Take the first partial derivative of the expenditure function with respect to utility: ∂e/∂u = (pr1 +pr2 )1/r . For all strictly positive prices, this expression will be positive. Alternatively, by the Envelope theorem it is shown that the partial derivative of the minimum-value function e with respect to u is equal to the partial derivative of the Lagrangian with respect to u, evaluated at (x∗ , λ∗ ), what equals λ. Unboundness above follows from the functional form of u. 4. Increasing in p. 1 Again, take all first partial derivatives with respect to prices: ∂e/∂pi = upr−1 (pr1 + pr2 ) r −1 , i what is, obviously, positive. 5. Homogeneous of degree 1 in p. e(tp, u) = u ((tp1 )r + (tp2 )r )1/r = t1 u (pr1 + pr2 )1/r

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2 Consumer Theory 6. Concave in p. The definition of concavity in prices requires h i h i r r 1/r r r 1/r t u p 0 1 + p0 2 + (1 − t) u p1 1 + p1 2 ≤ e(pt , u) for pt = tp0 + (1 − t)p1 . Plugging in the definition of the price vector into e(pt , u) yields the relationship h i h i r r 1/r r r 1/r t u p0 1 + p0 2 + (1 − t) u p1 1 + p1 2 ≤ r r r r 1/r u t(p0 1 + p0 2 ) + (1 − t)(p1 1 + p1 2 ) . Alternatively, we can check the negative semidefiniteness of the associated Hessian matrix of all second-order partial derivatives of the expenditure function. A third possibility is to check (product rule!)  ∂ 2e r−2 r r−1 r r 1/r−1 r 1/r−2 < 0 by r < 0. = u (r − 1)p (p + p ) − rp (p + p ) 1 2 1 2 i i ∂p2i 7. Shephard’s lemma r 1/r−1 r what is exactly the definition of a CES-type Hicksian ∂e/∂p1 = upr−1 1 (p1 + p2 ) demand function. 1.38 Complete the proof of Theorem 1.9 by showing that xh (p, u) = x (p, e(p, u)). Answer : We know that at the solution of the utility maximisation or expenditure minimisation problem e(p, u) = y and u = v(p, y). Substitute the indirect utility function v into the Hicksian demand function gives xh (p, v(p, y)). As the new function is a function of prices and income only, it is identical to the Marshallian demand function. Furthermore, by replacing income by the expenditure function we get the expression x (p, e(p, u)).

2.4 Properties of Consumer Demand 1.40 Prove that Hicksian demands are homogeneous of degree zero in prices. Answer : We know that the expenditure function must be homogeneous of degree one in prices. Because any Hicksian demand function equals, due to Shephard’s lemma, the first partial derivative of the expenditure function and, additionally, we know that the derivative’s degree of homogeneity is k − 1. The Hicksian demand functions must be homogeneous of degree 1 − 1 = 0 in prices. 1.42 For expositional purposes, we derived Theorems 1.14 and 1.15 separately, but really the second one implies the first. Show that when the substitution matrix σ(p, u) is negative semidefinite, all own–substitution terms will be nonpositive.

22

2 Consumer Theory Answer : Theorem 1.15 implies Theorem 1.14 because all elements of the substitution matrix represent second–order partial derivatives of the expenditure function. Therefore, the second–order cross-partial derivatives must be equal due to Young’s theorem which results in a symmetric matrix of second–order partial derivatives. The diagonal elements of this matrix must be negative because the expenditure function is required to be concave in prices. Subsequently, any compensated demand function is required to be non–increasing in its own price. Thus, the leading principal minors will be alternating in sign, starting with negative: ∂xh1 (p, u) ∂ 2 e(p, u) = ≤0 ∂p21 ∂p1  2 2 ∂ 2 e(p, u) ∂ 2 e(p, u) ∂ e(p, u) 2 (−1) D2 = − ≥0 ∂p21 ∂p22 ∂p1 ∂p2 (−1)3 D3 ≤ 0 (−1)1 D1 =

1.43 In a two-good case, show that if one good is inferior, the other good must be normal. Answer : The Engel-aggregation in a two-good case is the product of the income elasticity and the repsective expenditure share s1 η1 + s2 η2 = 1. An inferior good is characterised by a negative income elasticity, thus, one of the two summands will be less than zero. Therefore, to secure this aggregation, the other summand must be positive (even larger one) and the other commodity must be a normal good (even a luxury item). 1.55 What restrictions must the αi , f (y), w(p1 , p2 ), and z(p1 , p2 ) satisfy if each of the following is to be a legitimate indirect utility function? Answer : (a) v(p1 , p2 , p3 , y) = f (y)pα1 1 pα2 2 pα3 3 The function P f (y) must be continuous, strictly increasing and homogeneous of degree 0 − αi . Each of the exponents αi has to be less than zero to satisfy v decreasing in prices. Furthermore, negative partial derivatives of v with respect to each price are required to get positive Marshallian demand functions by using Roy’s identity. (b) v(p1 , p2 , y) = w(p1 , p2 )+z(p1 , p2 )/y Both functions, w and z, must be continuous and decreasing in prices. Function z has to be homogeneous of degree one and function w has to be homogeneous of degree zero: v(tp1 , tp2 , ty) = t0 w(p1 , p2 ) + (t1 z(p1 , p2 ))/(ty) = t0 (w(p1 , p2 ) + z(p1 , p2 )/y) . To satisfy v increasing in income, function z must be < 0. 1.60 Show that the Slutsky relation can be expressed in elasticity form as ij = hij − sj ηi , where hij is the elasticity of the Hicksian demand for xi with respect to price pj ,

23

2 Consumer Theory and all other terms are as defined in Definition 1.6. Answer : The Slutsky relation is given by ∂xhi ∂xi ∂xi = − xj . ∂pj ∂pj ∂y Multiplying the total expression with y/y and pj gives ∂xhi pj xj ∂xi ∂xi pj = pj − y. ∂pj ∂pj y ∂y By assuming that xhi = xi before the price change occurs, we can divide all three terms by xi . The result of this operation is ∂xi pj ∂xhi pj ∂xi y = − sj = ij = hij − sj ηi ∂pj xi ∂pj xi ∂y xi 1.62 The substitution  matrix of a utility–maximising consumer’s demand system at a b prices (8, p) is 2 −1/2 . Find a, b, and p. Answer : Whenever demand is generated from utility maximisation, the substitution matrix must possess the followign three properties: negative semidefiniteness, symmetry, and satisfy σ(p, u)p = 0 (known as Hick’s Third Law). Putting things together, due to symmetry b = 2. From the second row we get p2 = 32 because 2p1 − 1/2p2 = 0. Subsequently, a = −8. Checking the first and second leading principal minor to see whether the matrix is negative semidefinite, gives D1 = −8 < 0 and D2 = 4 − 4 = 0. Thus, the matrix is negative semidefinite. Additional exercise Relationship between utility maximisation and expenditure minimisation Let’s explore the relationship with an example of a concrete utility function. A con1/2 1/2 sumer’s utility function is u = x1 x2 . For the derived functions see 1

2.5 Equilibrium and Welfare 4.19 A consumer has preferences over the single good x and all other goods m represented by the utility function, u(x, m) = ln(x) + m. Let the price of x be p, the price of m be unity, and let income be y. (a) Derive the Marshallian demands for x and m. Answer The equality of marginal rate of substitution and price ratio gives 1/x = p. Thus, the Marshallian demand for x is x = 1/p. The uncompensated demand for m separates into two cases depending on the amount of income available: ( 0 when y ≤ 1 m= y − 1 when y > 1.

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2 Consumer Theory Start from the utility function Minimise expenditures s.t. u to find the Hicksian demand function and derive the Marshallian demand for x1 x1 = y/2p1 xh1 = u (p2 /p1 )1/2 Plug in the respective demand functions to get the indirect utility function expenditure function e = u(4p1 p2 )1/2 v = y/(4p1 p2 )1/2 Substitute the expenditure function Substitute the indirect utility function into the Hicksian demand function into the Marshallian demand function to derive the Hicksian demand function to derive the Marshallian demand function x1 = (u(4p1 p2 )1/2 )/2p1 = u(p2 /p1 )1/2 xh1 = (p2 /p1 )1/2 y/(4p1 p2 )1/2 = y/2p1 Invert v, re–arrange and replace v by u Invert e, re–arrange and replace e by y to get the expenditure function to get the indirect utility function 1/2 v = u = y(4p1 p2 )−1/2 e = y = u(4p1 p2 ) Check Roy’s identity Check Shephard’s lemma ∂v/∂p1 2y(p1 p2 )1/2 ∂e 1/2 2 − ∂v/∂y = 4(p3 p2 )1/2 = y/2p1 = 2(4pu4p 1/2 = u(p2 /p1 ) ∂p1 1 p2 ) 1 Establish the Slutsky equation ∂x1 = 2(p1 pu2 )1/2 − 2py2 · 2p11 ∂p2 substitute u = v(p, y) into the substitution effect ∂x1 = 4py1 p2 − 4py1 p2 = 0 ∂p2 Table 1: Relationship between UMP and EMP (b) Derive the indirect utility function, v(p, y). Answer Again, depending on the amount of income available there will be two indirect utility functions: (   when m ≤ 1 ln p1 v(p, y) = y − 1 − ln p when m > 1. (c) Use the Slutsky equation to decompose the effect of an own-price change on the demand for x into an income and substitution effect. Interpret your result briefly. Answer A well-known property of any demand function derived from a quasi-linear utility function is the absence of the income effect. Which can be easily seen in the application of the Slutsky equation: ∂x ∂xh ∂x ∂x = +x ∂p ∂p ∂y ∂p

=−

1 1 ∂xh + 0 · = p2 p ∂p

Therefore, the effect of an own-price change on the demand for x consists of the substitution effect only, the partial derivative of the compensated demand function with respect to price. (d) Suppose that the price of x rises from p0 to p1 > p0 . Show that the consumer surplus area between p0 and p1 gives an exact measure of the effect of the price change on

25

2 Consumer Theory

Figure 8: Graph to 4.19 consumer welfare. Answer The consumer surplus area can be calculated by integrating over the inverse uncompensated demand function of x: Z

p0

CS = p1

1 dx = ln p0 − ln p1 . x

Calculating the change in utility induced by a price change gives: ∆v = v 1 (p1 , y 0 ) − v 0 (p0 , y 0 ) = y − 1 − ln p1 − (y − 1 − ln p0 ) = − ln p1 + ln p0 . As the two expressions are equal, the consumer surplus area gives an exact measure of the effect of the price change on consumer welfare in the case of quasi-linear preferences. (e) Carefully illustrate your findings with a set of two diagrams: one giving the indifference curves and budget constraints on top, and the other giving the Marshallian and Hicksian demands below. Be certain that your diagrams reflect all qualitative information on preferences and demands that you’ve uncovered. Be sure to consider the two prices p0 and p1 , and identify the Hicksian and Marshallian demands. Answer See Figure 8. Please note, that Hicksian and Marshallian demands are identical here.

26

3 Producer Theory

3 Producer Theory 3.1 Production i (x) · APxii(x) . Show that this is 3.1 The elasticity of average product is defined as ∂AP ∂xi equal to µi (x) − 1. Show that average product is increasing, constant, or decreasing as marginal product exceeds, is equal to, or less than average product. Answer : In order to know the first part of the elasticity, which is at the same time the slope of the average product, we need to take the first partial derivative of AP = f (x)/y. Applying quotient rule to get the first partial derivative of the average product gives:

∂APi (x) xi ∂f (x)/∂xi − f (x) MP AP M P − AP = = − = 2 ∂xi xi xi xi xi Multiply this term with the right part of the definition (xi /AP ) gives M P/AP − 1 what xi = µi (x) − 1. is exactly ∂f∂x(x) y i The first part of the above definition equals the slope of the average product: (M P − AP )/xi . It is straightforward to show that whenever marginal product exceeds the average product the slope has to be positive. The average product reaches a maximum when the marginal product equals average product. Finally, whenever M P < AP average product is sloping downwards. That shows, the marginal product function intersects the average product function always in the maximum. 3.3 Prove that when the production function is homogeneous of degree one, it may be P written as the sum f (x) = M Pi (x)xi , where M Pi (x) is the marginal product of input i. Answer : The answer to this exercise gives a nice application of Euler’s Theorem. The sum of the partial differentials of a function multiplied with the level of the respective inputs is equal to the function times the degree of homogeneity k. The sum of all marginal products multiplied with input levels gives the production function times k = 1. 3.7 Goldman & Uzawa (1964) have shown that the production function is weakly separable with respect to the partition {N1 , . . . , NS } if and only if it can be written in the form  f (x) = g f 1 (x(1) ), . . . , f S (x(S) ) , where g is some function of S variables, and, for each i, f i (x(i) ) is a function of the subvector x(i) of inputs from group i alone. They have also shown that the production function will be strongly separable if and only if it is of the form  f (x) = G f 1 (x(1) ) + · · · + f S (x(S) ) , where G is a strictly increasing function of one variable, and the same conditions on the subfunctions and subvectors apply. Verify their results by showing that each is separable as they claim.

27

3 Producer Theory Answer To show that the first equation is weakly separable with respect to the partitions, ∂[f (x)/f (x)] = 0 ∀i, j ∈ NS and k ∈ / NS . Calculate the marginal we need to show that i ∂xk j products of the first equation for two arbitrary inputs i and j: ∂g ∂f S fi (x) = ∂f S ∂xi

∂g ∂f S fj (x) = . ∂f S ∂xj

The marginal rate of technical substitution between these two inputs is fi (x) = fj (x)

∂f S ∂xi ∂f S ∂xj

This expression is independent of any other input which is not in the same partition N S and, therefore, the production function is weakly separable. ∂(fi /fj ) = 0 for k ∈ / NS ∂xk To show that the second equation is strongly separable we have to perform the same exercise, however, assuming that the three inputs are elements of three different partitions i ∈ NS , j ∈ NT and k ∈ / NS ∪ NT . The marginal products of the two inputs i and j are:   S T x(S) x(T ) 0 ∂f 0 ∂f fi (x) = G fj (x) = G . ∂xi ∂xj The MRTS is:

∂f S /∂xi fi (x) = . fj (x) ∂f T /∂xj

It follows for k ∈ / NS ∪ NT ∂(fi /fj ) = 0. ∂xk 3.8 A Leontief production function has the form y = min {αx1 , βx2 } for α > 0 and β > 0. Carefully sketch the isoquant map for this technology and verify that the elasticity of substitution σ = 0, where defined. Answer : Taking the total differential of the log of the factor ratio gives d ln (x2 /x1 ) = 1/x2 dx2 − 1/x1 dx1 . However, the MRTS is not defined in the kinks as the function is discontinuous. Along all other segments of the isoquants the MRTS is zero. Therefore, the elasticity of substitution is only defined when the input ratio remains constant. In this case, σ = 0. 3.9 Calculate σ for the Cobb-Douglas production function y = Axα1 xβ2 , where A > 0, α > 0 and β > 0. Answer : The total differential of the log of the factor ratio gives

28

3 Producer Theory

x2 6

-

x1

Figure 9: Isoquant map of Leontief technology d ln(x2 /x1 ) = 1/x2 dx2 −1/x1 dx1 . The marginal rate of technical substitution is M RT S = αx2 . Write in logs and take the total differential results in βx1 ln M RT S = ln α + ln x2 − ln β + ln x1 d ln M RT S = (1/x1 dx1 − 1/x2 dx2 ) Putting both parts together results in σ=

1/x2 dx2 − 1/x1 dx1 =1 1/x1 dx1 − 1/x2 dx2

P P 1/ρ 3.14 Let y = ( ni=1 αi xρi ) , where i αi = 1 and 0 6= ρ < 1. Verify that σij = 1/(1 − ρ) for all i 6= j. Answer Apply the definition of the elasticity of substitution. d (ln(xj ) − ln(xi )) d ln (fi (x)/fj (x)) 1 dxj − x1i dxi xj   = P αi xiρ−1 ( i αi xρi )1/ρ−1 d ln α xρ−1 (P α xρ )1/ρ−1 j j i i i   1 1 − xi dxi − xj dxj   = 1 1 ρ − 1 xi dxi − xj dxj

σij =

=

−1 1 = ρ−1 1−ρ

29

3 Producer Theory Another way to calculate the elasticity of subsitution starts with the definition of the MRTS:  ρ−1 x1 M RT S = x2 x2 = M RT S 1/(1−ρ) x1 x2 1 ln = ln M RT S x1 1−ρ 1 ln(x2 /x1 ) = σ= ln M RT S 1−ρ Note that assuming perfect competition, the elasticity of substitution can be expressed in terms of the input price ratio in place of the M RT S: σ=

d ln(xj /xi ) wi /wj d(xj /xi ) = . d ln(wi /wj ) xj /xi d(wi /wj )

Alternatively, the elasticity of substitution can also be expressed in terms of the elasticity of input demand and the inverse of the cost share (assuming perfect competition): σ=

∂xi wj wj xj . ∂wj xi c

3.15 For the generalised CES production function, prove the following claims made in the text. y=

n X

!1/ρ αi xρi

, where

i=1

n X

αi = 1 and 0 6= ρ < 1

i=1

1. lim y =

ρ→0

n Y

xαi i

i=1

P Answer : Write the log of the CES production function ln y = 1/ρ ln αi xρi . At ρ = 0, the value of the function is indeterminate. However, using L’H`opital’s rule we can write P αi xρi ln xi P lim ln y = . ρ→0 αi xρi P P At ρ = 0 this expression turns into ln y = αi ln xi / αi . Because the denominator isQdefined to be one, we can write the CES production function at this point as y = xαi i , what is exactly the generalised Cobb-Douglas form. 2. lim y = min {x1 , . . . , xn }

ρ→−∞

30

3 Producer Theory Answer : Let us assume that αi = αj . Then the CES production function has the P P form y = ( i xρi )1/ρ . Let us suppose that x1 = min( xi ) and ρ < 0. We want P to show that x1 = limρ→−∞ ( xρi )1/ρ all commodities xi are required be P. Since P to ρ ρ ρ 1/ρ nonnegative, we canP establish x1 ≤ xi . Thus, P it must hold that x1 ≥ ( xi ) . On the other hand, P xρi ≤ n · xρ1 . Hence ( xρi )1/ρ ≥ n1/ρ · x1 . Letting ρ → −∞, we obtain limρ→−∞ ( xρi )1/ρ = x1 , because limρ→−∞ n1/ρ · x1 = x1 .

3.2 Cost 3.19 What restrictions must there be on the parameters of the Cobb-Douglas form in Example 3.4 in order that it be a legitimate cost function: c(w, y) = Aw1α w2β y? Answer : The parameters A, w1 , w2 and y are required to be larger than zero. A cost function is required to be increasing in input prices. Therefore, the exponents α and β must be larger zero. To fulfill the property of homogeneity of degree one in input prices, the exponents have to add up to one: α + β = 1. To secure concavity in input prices, the Hesse matrix of all second–order partial derivatives must be negative semidefinite. As a necessary condition, the second–order own–partial derivatives cannot be positive. Thus, each of the exponents can not be larger than one (α ≤ 1, β ≤ 1). 3.24 Calculate the cost function and conditional input demands for the Leontief production function in Exercise 3.8. Answer This problem is identical to the expenditure function and compensated demand functions in the case of perfect complements in consumer theory. Because the production is a min-function, set the inside terms equal to find the optimal relationship between x1 and x2 . In other words, αx1 = βx2 . For a given level of output y, we must have y = αx1 = βx2 . Rearrange this expression to derive the conditional input demands: y y x2 (w, y) = . x1 (w, y) = α β The cost function is obtained by substituting the two conditional demands into the definition of cost: c(w, y) = w1 x1 (w, y) + w2 x2 (w, y) =

w1 y w2 y + . α β

3.27 In Fig. 3.85, the cost functions of firms A and B are graphed against the input price w1 for fixed values of w2 and y. (a) At wage rate w10 , which firm uses more of input 1? At w10 ? Explain? Answer : Input demand can be obtained by using Shephard’s lemma, represented by the slope of the cost function. Therefore, at w10 firm B demands more of factor 1 and at wage rate w10 firm A has a higher demand of that input. (b) Which firm’s production function has the higher elasticity of substitution? Explain. Answer : The first-order conditions for cost minimisation imply that the marginal rate of technical substitution between input i and j equals the ratio of factor prices wi /wj . In

31

3 Producer Theory the two input case, we can re-write the original definition of the elasticity of substitution as d ln(x2 /x1 ) xˆ2 − xˆ1 d ln(x2 /x1 ) = = , σ= d ln(f1 /f2 ) d ln(w1 /w2 ) wˆ1 − wˆ2 where the circumflex denotes percentage change in input levels and input prices, respectively. Because w ˆ2 = 0, the denominator reduces to wˆ1 , which is assumed to be the same for both firms. In (a) we established that input demand at w10 is larger for firm B compared to firm A. It follows that the numerator will be larger for B and, subsequently, firm A’s production function shows the higher elasticity of substitution at w10 . 3.29 The output elasticity of demand for input xi is defined as iy (w, y) ≡

y ∂xi (w, y) . ∂y xi (w, y)

(a) Show that iy (w, y) = φ(y)iy (w, 1) when the production function is homothetic. Given a homothetic production function, the cost function can be written as c(w, y) = φ(y)c(w, 1). Shephard’s lemma states that the first order partial derivative with respect to the price of input i gives demand of xi and to obtain the elasticity we need to take take the second-order cross-partial derivative of the cost function with respect to output. However, by Young’s theorem it is known that the order of differentiation does not matter. Therefore, the following partial derivatives should be equal: ∂xi ∂mc ∂ 2 c(w, y) = = . ∂wi ∂y ∂wi ∂y Putting everything together gives: ∂φ(y) y 1 ∂ 2c y ∂wi = xi (w, 1) = 0 iy (w, 1). iy (w, y) = ∂y∂wi ∂c ∂y φ(y)xi (w, 1) φ (y) Unfortunately, this is not the result we should get. (b) Show that iy = 1, for i = 1, . . . , n, when the production function has constant returns to scale. Answer For any production function with constant returns to scale, the conditional input demand xi is linear in output level y (see Theorem 3.4). More formally, the conditional input demand of a production function homogeneous of degree α > 0 can be written as xi (w, y) = y 1/α xi (w, 1). By definition, a constant returns to scale technology requires a production function homogeneous of degree 1. Therefore, the conditional input demand reduces to xi (w, y) = yxi (w, 1). Calculating the output elasticity of demand for input xi results in: iy (w, y) ≡

∂xi (w, y) y y = xi (w, 1) = 1. ∂y xi (w, y) yxi (w, 1)

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3 Producer Theory 3.33 Calculate the costPfunction and the conditional input demands for the linear production function y = ni=1 αi xi . Answer Because the production function is linear, the inputs can be substituted for another. The most efficient input (i.e. input with the greatest marginal product/ price) will be used and the other inputs will not be used. (y α if wαii > wjj ∀j 6= i, j ∈ {1, . . . , n} αi xi (w, y) = α 0 if wαii < wjj for at least one j 6= i, j ∈ {1, . . . , n}. The cost function is then c(w, y) =

wi y , αi

where i is the input where

αi αj > ∀j 6= i, j ∈ 1, . . . , n. wi wj

3.3 Duality in production Additional exercise (Varian (1992) 1.6) For the following “cost functions” indicate which if any of properties of the cost function fails; e.g. homogeneity, concavity, monotonicity, or continuity. Where possible derive a production function. (a) c(w, y) = y 1/2 (w1 w2 )3/4  Homogeneity: c(tw, y) = y 1/2 (tw1 tw2 )3/4 = t3/2 y 1/2 (w1 w2 )3/4 The function is not homogeneous of degree one. Monotonicity: ∂c(w, y) ∂c(w, y) −1/4 3/4 3/4 −1/4 >0 = 3/4y 1/2 w1 w2 > 0 = 3/4y 1/2 w1 w2 ∂w1 ∂w2 The function is monotonically increasing in input prices. Concavity: " # 3 1/2 −5/4 3/4 9 1/2 −1/4 −1/4 − y w w2 y w1 w2 16 H = 9 161/2 − 1 −1/4 3 1/2 3/4 −5/4 y w 1/41 w2 − 16 y w1 w2 16 |H1 | < 0 |H2 | = −

72 y ∂w1 2 w1 2 w1  r  r r ∂c 1 w1 1 w1 w2 =y 1− or 2 > positive for 1 > ∂w2 2 w2 2 w2 w1

The function is not concave as the first partial derivatives with respect to both input prices are negative and the second-order partial derivatives are positive. The determinants of the Hessian matrix are |H1 | > 0 and |H2 | = 0. Thus, the function is convex. √ (f) c(w, y) = (y + 1/y) w1 w2 The function satisfies all properties, except continuity in y = 0. 3.40 We have seen that every Cobb-Douglas production function, y = Axα1 x21−α , gives rise to a Cobb-Douglas cost function, c(w, y) = yAw1α w21−α , and every CES production function, y = A (xρ1 + xρ2 )1/ρ , gives rise to a CES cost function, c(w, y) = yA (xr1 + xr2 )1/r . For each pair of functions, show that the converse is also true. That is, starting with the respective cost functions, “work backward” to the underlying production function and show that it is of the indicated form. Justify your approach. Answer : Using Shephard’s lemma we can derive the conditional input demand functions. The first step to solve this exercise for a Cobb-Douglas cost function is to derive Shephard’s lemma and to rearrange all input demands in such a way to isolate the ratio of input prices on one side, i.e. left-hand side of the expression. On the right-hand side we have the quantity of input(s) and output. Second, equalise the two expressions and

34

3 Producer Theory solve for y. The result will be the corresponding production function. ∂c(w, y) = αyA x1 = ∂w1



w2 w1

1−α

1/(1−α) x1 Aαy  −1/α w2 x2 = w1 A(1 − α)y

w2 = w1

 −α ∂c(w, y) w2 x2 = = (1 − α)yA ∂w2 w1 1−α α (Aαy) x2 = α x1 (A(1 − α)y)1−α



xα1 x1−α 2 y= α = axα1 x1−α where a = (Aαα (1 − α)1−α )−1 2 1−α α (1 − α) For the CES cost function a short-cut is used: Derive the conditional input demand functions and substitute them into the production function. 1 ∂c(w, y) = yAw1r−1 (w1r + w2r ) r −1 ∂w1 1 ∂c(w, y) = yAw2r−1 (w1r + w2r ) r −1 x2 = ∂w2  −r −r 1/ρ ρ w1 + w2 y = (Ay) −r = Ay w1 + w2−r

x1 =

3.4 The competitive firm Additional exercise (Varian (1992) 1.21) Given the following production function 1/2 1/4 y = 100x1 x2 . (a) Find c(w1 , w2 , y). Answer : Starting from the equality of MRTS and ratio of factor prices, we get w1 /w2 = 2x2 /x1 . Solving for one of the inputs, substituting back in the production function and rearranging, we derive the conditional input demand functions:  y 4/3  2w 1/3 2 and x1 = 100 w1  y 4/3  2w −2/3 2 x2 = . 100 w1 Substituting the two functions in the definition of costs, the resulting cost function is:  y 4/3  y 4/3 2/3 1/3 2/3 1/3 c(w1 , w2 , y) = w1 (2w2 ) + w2 w1 2−2/3 100 100  y 4/3  2/3 1/3 = 21/3 + 2−2/3 w1 w2 . 100

35

3 Producer Theory (b) Find the effect of an increase in output on marginal cost, and verify that λ = marginal cost. Answer : Marginal costs are  2/3 1/3 1 (y/100)1/3 w1 w2 21/3 + 2−2/3 . Marginal costs are increasing M C = ∂c/∂y = 75 with output which is shown by  2/3 1/3 ∂M C ∂2c 1 2 = ∂y (y/100)−2/3 w1 w2 21/3 + 2−2/3 . From the FOC of the La2 = ∂y 150 grangian we can derive that λ∗ =

1/2

w1 x1

1/4 50x2

3/4

=

w2 x2

1/2 25x1

. Substituting the conditional input

demand functions into one of those expressions gives 1/2 −1/4 w1 (y/100)4/3 (2w2 /w1 )1/3 (y/100)4/3 (2w2 /w1 )−2/3 λ∗ = 50 21/3 y 1/3 2/3 1/3 = ( ) w1 w2 50 100 When you solve the ratios, this expression will be equal to the marginal cost function. (c) Given p = price of output, find x1 (w, p), x2 (w, p) and π(w, p). Use Hotelling’s lemma to derive the supply function y(w, p). Answer : By maximising π = py − c(w, y) the first-order condition is ∂π 1  y 1/3 2/3 1/3 1/3 =p − w1 w2 (2 + 2−2/3 ) = 0 ∂y 75 100 !3  3 75 p y =100 1/3 2/3 1/3 2 + 2−2/3 w1 w2 The first expression affirms the equality of price and marginal cost as the profit maximum for any competitive firm. The last expression gives already the profit maximising supply function. Furthermore, the two following unconditional demand functions emerge as solution of this optimisation problem: !4/3   3 1/3  4 1/3 4 75p 75 2w2 2 p −2 −1 x1 = w1 w2 = 1/3 −2/3 1/3 −2/3 2 +2 w1 2 +2 w13 w2 !4/3   3 2/3 75p w1 75 p4 −2 −1 w w = . x2 = 1 2 21/3 + 2−2/3 2w2 21/3 + 2−2/3 22/3 w12 w22 The profit function is    3 4  1/3 75 100p4 75p 2 w1 w2 − + 2/3 2 2 π = 1/3 2 + 2−2/3 w12 w2 21/3 + 2−2/3 w13 w2 2 w1 w2  3 4 75 p = (100 − 75) 1/3 −2/3 2 +2 w12 w2  3 4 75 p = 25 1/3 . 2 + 2−2/3 w12 w2

36

3 Producer Theory Hotelling’s lemma confirms the output supply function shown above. (d) Derive the unconditional input demand functions from the conditional input demands. Answer One, among several, way is to substitute the conditional input demands into the definition of cost to obtain the cost function. Calculating marginal cost and equalising with output price, gives, after re-arrangement, the output supply function. Substitution of the output supply function into the conditional input demands results in the unconditional input demand functions. Using the example at hand, and starting from the equality ∂c/∂y = p gives:   y 1/3 2/3 1/3 1 1/3 −2/3 2 +2 w1 w2 p= 75 100  −3 3 −2 −1 y = 753 21/3 + 2−2/3 p w1 w2 · 100 4  75 p4 w1−3 w2−1 x1 (w, p) = 1/3 −2/3 2 +2 (e) Verify that the production function is homothetic. Answer : The cost function is a factor of a function of output and input prices. Similarly, the conditional input demand functions are products of a function of y and input prices. Therefore, the possibility to separate the two functions multiplicatively and following Theorem 3.4 shows that the production function has to be a homothetic function. (f) Show that the profit function is convex. Answer : Inorder to simply this step, I write the constant part of the profit function  3

75 Calculating the second-order partial derivatives of the profit as K = 25 21/3 +2 −2/3 function with respect to all prices gives the following Hessian matrix. Be aware of the doubble sign change after each derivative with respect to the input price (check Theorem 3.8).  

H=

12Kp2 2 1 w2  w−8Kp − 3 3  w1 w23 − −4Kp w12 w22

−8Kp3 w13 w2 4 − −6Kp w14 w2 4 − −2Kp w13 w22

−4Kp3 w12 w22 4  − −2Kp w13 w22  4 − −2Kp w12 w23

The own supply effect is positive, the own demand effects are negative and all crossprice effects are symmmetric. Checking the determinants becomes quite tedious. Intuituively, it should become clear that they all have to be positive. (g) Assume x2 as a fixed factor in the short run and calculate short-run total cost, short-run marginal cost, short-run average cost and short-run profit function. Short-run total cost are obtained by re-arranging the production function to get x1 on the left-hand side and plugging in into the definition of cost c(w, y) = 1/2 (y/100)2 w1 /x2 + w2 x2 . The first-partial derivative gives the short-run marginal

37

References cost function smc = 1/2 y w /x2 + w2yx2 . 1002 1

1/2 1 y w /x2 . 50 100 1

The short-run average costs are equal to sac =

References Arrow, K. J. & Enthoven, A. C. (1961), ‘Quasi-concave programming’, Econometrica 29(4), 779–800. Goldman, S. M. & Uzawa, H. (1964), ‘A note on separability in demand analysis’, Econometrica 32(3), 387–398. Varian, H. R. (1992), Microeconomic Analysis, 3rd edn, W. W. Norton and Company, New York.

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