1 Mathematical Appendix
1 Mathematical Appendix 1.1 Chapter A1
Solutions to selected exercises from Jehle and Reny (2001): Advanced Answer (a) (x, y)|y = e Microeconomic Theory This set is not convex.
A1.7 Graph each of the following sets. If the set is convex, give a pro of. If it is not convex, give a counterexample. counterexample.
x
Any combination of points would be outside the set. For example, (0, 1) and (1, e) (x, y)|y = e (x, y)|y = e x . (b) (x, y)|y = e
x
, but combThomas ination of theHerzfeld two vectors with t =
2
, y2 ) S = (x, y)|y = e
x
. Since y = e
function, it is su cient to show that (tx + (1 - t)x 1 1 . Our task is to show particular t (0, 1). Set t = show that Contents 1 2
1 2
, e+1 ) / 2
September 2010
x
This set is convex. Proof: Let (x , y1 ), (x 1
S.
not: (
1 2
(y 1 + y 2 ) =
1 2
(e x 1 + e
2
x2
), since y
i
=e
x1
2
, ty
1
x1
- 2e
x1 2
·e
x2 2
is a continuous
+ (1 - t)y 2 ) S for any 1 (x + x 2 ), 1 (y 1 + y 2 ) 1 2
2
for i = 1, 2. Also,
1 x 1 + e x 2 ) = e 1 (x 1 +x 2 = e 1 Mathematical Appendix 2 2 (e 1.1 Chapter A1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 x x 1.2 Chapter A2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .e. . . . 1. 6+ e 2 = 2e e
x
+ e x2 = 0 (e
-e
x1
x1 2
x1
x2
2
·e
x2
·e
x2
2
2
2
) 2 = 0.
2 Consumer Theory 2; x > 0 . ,. y . .>. .0. . . . . . . . . . . . . . . . . . . . . . 12 (c) (2.1 x, yPreferences )|y = 2x - x and Utility 2.2 The Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Problem This setConsumer’s is not convex.
12
1 , Expenditure 1 2.3 and . . (.x., .y.)|.y. = . .2.x. .- .x. . . . . 16 ForIndirect exampleUtility , , 1 9 , 1 . . .S. = 10 2 10 2 1 1 Consumer 9 , 1nd S 2.41,Properties Dema Demand ./ . . . . . . . . . . . . . . . . . . . . . 18 = 1 of ,1 + 1 1 2
2
10
2
2
10
2;
x > 0, y > 0. However,
2
2.5 Equilibrium and Welfare . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 (d) (x, y)|xy = 1; x > 0, y > 0 3 ProThis duceset r Thiseoconvex. ry ProPro of: C onsider. a. n. y. .(.x. . . . . .1 ., .y.1 .)., .(x. . 2. ,. y |xy = 1; x > 0, y > 0. For any 3.1 duction . 2. .) . .S. =. .(x . ,. y . .)23 t [0, 1],. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.2 Cost
23
3.3 Duality in production . . . . . . . . . . . . . . . .2. . . . . . . . . . . . . 28 (tx 1 + (1 - t)x 2 )(ty 1 + (1 - t)y 2 ) = t x1 y1 + t(1 - t)(x y + x 2 y1 ) + (1 - t) 2 x 2 y2 3.4 The competitive firm firm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 1 2 > t 2 + (1 - t) 2 + t(1 - t)(x y + x 2 y1 ), since x i yi > 1. 1 2 = 1 + 2t
2
- 2t + t(1 - t)(x
1
= 1 + 2t(t - 1) + t(1 - t)(x = 1 + t(1 - t)(x x 1 y2 + x
y =x 2 1
y 1 1
y2 y1
+ x 2 y2
y1 y2
1
1
y2 + x 2 y1 )
y2 + x 2 y1 - 2) = 1 i x
-2=y
2
+y 1 -2=0 y1 y2
y - 1 - 2y
1
y2 + y
(y 1 - y
1 2
y2 + x 2 y1 )
2
2
=0
)2 = 0,
1
y2 + x 2 y1 = 0.
1 Mathematical Appendix which is always true and therefore, (tx
+ (1 - t)x
1
2
, ty 1 + (1 - t)y
2
) S which is
convex. (e) (x, y)|y = ln(x) This set is convex. Proof. Let (x S is convex
1
, y1 ) + (x
1if
2
, y2 ) S. Then
2 ln(x
1
) + ln(x
1
(y 1 + y 2 ) = (ln(x
1 2
2
(x
1
x
1
2 x1
) + ln(x
2 x1
+1 2) 2x
x 2 )1/2 = (1
2 x1
+1 2) 2x
- 2(x
1
x2 )1/2 + x
1/2 1
x
+x
2
)).
+1 2) 2x
x 2 ) = ln(1
1
2 ln(x
) = ln(1
1
1/2 2
2 2
=0 =0
which is always true. A1.40 Sketch a few level sets for the following functions: y = x y = min[x Answer
1
1
x2 , y = x
1
+x
, x 2 ]. x2
x2 6
x2 6
6
@ @ (a) y = x
1
@
@ @@
x1
x2
@ @ @@ -
(b) y = x
1
-
x1
+x
(c) y = min(x
2
x1 1
, x2 )
Figure 1: Sets to Exercise A1.40
2 . Carefully sketch this function. A1.42 Let D = [-2, 2] and f : D R be y = 4 - x 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: y
2y
x = -2x
x
2
= -2
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
3
2
and
1 Mathematical Appendix A1.13). Alternatively, from the definition of convexity the following inequality should 1 + (1 - t)x 2 ) 2 = t(4 - (x 1 )2 ) + (1 - t)(4 - (x 2 ) 2 ). Multiply out to get hold 4 - (tx 4 - (tx 1 + x 2 - tx 2 )2 = 4 - x 2 + t[(x 1 ) 2 - (x 2 )2 ]. Again, the area below the function forms 2 a convex set. y
6
x
-
Figure 2: Graph to Exercise A1.42
A1.46 Consider any linear function f (x) = a · x + b for a R
and b R.
n
(a) Show that every linear function is both concave and convex, though neither is strictly concave nor strictly convex. Answer The statement is true i , for any x f (tx
1
+ (1 - t)x
1
2
, x2 R
) = tf (x
1
n
, t [0, 1], it is true that
) + (1 - t)f (x
2
).
Substituting any linear equation in this statement gives f (tx
1
+(1-t)x
for all x
1
2)
, x2 R
= a[tx n
1
+(1-t)x
2 ]+b
= tax
1
+(1-t)ax
2 +tb+(1-t)b
= tf (x
, t [0, 1].
(b) Show that every linear function is both quasiconcave 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 quasiconcave and quasiconvex quasiconvex (Theorem (Theorem A1.19). More formally, the statement statement 1 , x2 R n is true i , for any x (x 1 = x 2 ) and t [0, 1], we have f (tx f (tx
1 1
+ (1 - t)x + (1 - t)x
2) 2
= min[f (x
1
), f (x 2 )](quasiconcavity)
) = max[f (x
1
), f (x 2 )](quasiconvexity)
Again by substituting the equation into the definition, we get tf (x
1
) + (1 - t)f (x
2
) = min[f (x
1 ),
tf (x
1
) + (1 - t)f (x
2
) = max[f (x
1
f (x
2
)]
), f (x 2 )] 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 a ne function preserves concavity (convexity). (convexity). Assume that both functions are twice di erentiable. Then the second order partial derivative of the composite function, applying chain rule and product rule, is defined as h (x) = g (f (x)) f (x)
2
4
+ g (f (x)) f (x)
2
1
)+(1-t)f (x
2
)
1 Mathematical Appendix For any concave function,
2f
(x) = 0,
g(x) = 0, it should hold
2
the case the two functions are convex: 2 h(x) = 0.
2
f (x) = 0 and
2
2
h(x) = 0. In
g(x) = 0, it should hold
A1.48 Let f (x , x 2 ) = -(x 1 - 5) 2 - (x 2 - 5) 2 . Prove that f is quasiconcave. 1 Answer Proof: f is concave concave i H(x) is negative semidefinite semidefinite and it is strictly concave concave if the Hessian is negative definite. H = -2 0
0 -2
z T H(x)z = -2z
- 2z
2 1
2 2
< 0, for z = (z
1
, z2 ) = 0
Alternatively, we can check the leading principal minors of H: H (x) = -2 < 0 and 1 H 2 (x) = 4 > 0. The determina determinants nts of the Hessian Hessian alter alternate nate in sign sign beginning beginning with a negative value. Therefore, Therefore, the function is even strictly concave. Since f is concave, it is also quasiconcave. quasiconcave. 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 di erential-based approach, approach, f is quasiconcave, if for any x 0 and x 1 , f (x 1 ) = f (x increasing function.
0
)
f (x
0
)/ x(x
1
-x
0)
= 0. This must be true for any
(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. quasiconcave. Similarly to (a), f is quasiconcave if for any x min[f (x 0 ), f (x 1 )].
0
,x
1
and t [0, 1], it is true that f (tx
0
+ (1 - t)x
(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. quasiconcave. Proof: Let a < b < c, and let t = b of f , f (b) < min[f (a), f (c)]. Then f (t
c-b c-a
[0, 1], t b a + (1 - t b )c = b. Given the nature a + (1 - t b )c) < min[f (a), f (c)], so f is not b
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)].
5
1
)=
1 Mathematical Appendix (e) You should now be able to come up with a characterization of quasiconcave funcfunctions 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 Di erentiate 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) = 11x
- 6x + 8 f
3
1
= 33x
2
-6
increasing locally convex (b) f (x) = (3x
2
- x)(6x + 1) f
1
= 54x
2
- 6x - 1
increasing locally convex (c) f (x) = x
-1
2
(d) f (x) = (x
2
f 1 = 2x + 3
x3
x4 increasing locally concave
+ 2x)
f 1 = (6x + 6)(x
3
2
+ 2x)
2
increasing lo cally convex - 3x
3
(e) f (x) = [3x/(x
3
+ 1)]
f 1 = 18x x
2
2
(x 3 + 1)
+1 3
increasing locally concave (f ) f (x) = [(1/x
2
+ 2) - (1/x - 2)]
f1 = 4
4
x2
-8
1 x3
x2
increasing lo cally convex 1
e t2 dt f
(g) f (x) =
1
x2
= -e
x
decreasing locally convex A2.2 Find all first-order partial derivatives. derivatives.
(a) f (x
1
, x 2 ) = 2x
1
-x
2 1
f 1 = 2 - 2x (b) f (x
1
, x2 ) = x
2 1
+ 2x
f 1 = 2x
1
-x 1
2 2
2 2
= 2(1 - x
- 4x
1
)f
2
= -2x
2
f 2 = 4x
2
-4
6
2
3
-1
x+4
1 Mathematical Appendix (c) f (x
1
, x2 ) = x
3 1
-x
f 1 = 3x (d) f (x
1
, x 2 ) = 4x
1
- 2x
2 2
1
, x2 ) = x
3 1
+ 2x
- 6x
f 1 = 3x (f ) f (x
1
, x 2 ) = 3x
2 1
(g) g(x
1
-x
2
1
1
2
, x 2 , x 3 ) = ln x
+ x 1 x2 - x
2 2
f 2 = 2 - 2x
f 2 = 3x
2
g1 = 2x x2 - x 1
x3 - x 1
x3 - x - 2x 3
2
x
1
x
1
2
=x
2 1
2 3
+ y x
+ y x
+x
2 2
3
x3 - x
2
2 3
x1 satisfies the equation
2
2
1
+ y x
1
2 3
The first-order partial derivatives are y / x y/ x = x 2 + 2x 2 x 3 , and y/ x =x 2 3 + y x
1
1
x2 + x 2 x3 + x y
y
- 6x
g2 = -x x2 - x
2 3
1
2 1
1
2 3
2 g3 = -x 2 x - x 2 x3 - x
A2.4 Show that y = x
2 2
f2 = 1 - x 2
+x
2
2
-x
2
3 2
x2 + x
2 1
+ 1)
2
x2 + x
-x
1
2 1
+x
1
- 6x
2 1
-x
f 1 = 6x
f 2 = -2(x
1
f 1 = 4 - 2x (e) f (x
2
+x
2 3
= (x
1
+x
+ x 3 )2 .
2
3
= 2x 1 x2 + x 2 , 3 + 2x 3 x 1 . Summing them up gives 1
2 2
+ 2x
1
x2 + 2x
x3 + 2x
1
2
x3 = (x
1
-x
2 1
-x
0 -2
z T H(x)z = -2z 2 1
+ 2x
2 2
- 4x
T
2 2
H = -2 0
(b) y = x
+x
2
+ x 3 )2 .
3
A2.5 Find the Hessian matrix and construct the quadratic form, z
(a) y = 2x
1
+ 2 * 0z
2 1
1
z2 - 2z
2 2
2
H=20 zT H(x)z = 2z
04 2 1
7
+ 2 * 0z
1
z2 + 4z
2 2
H(x)z, when
1 Mathematical Appendix (c) y = x
3 1
-x
+ 2x
2 2
2
0 -2
z T H(x)z = 6x (d) y = 4x
1
+ 2x
-x
2
2 1
+ x 1 x2 - x
- 6x
1
x2 - x
1
2 2
+ 2z 1 z2 - 2z
2 1
2 2
3 2
H = 6x
-6
1
-6 6x
zT H(x)z = 6x A2.8 Suppose f (x (a) Show that f (x
z 2 - 2z
1 -2
z T H(x)z = -2z 3 1
1
2 2
H = -2 1
(e) y = x
0
1
H = 6x
1
2
z 2 - 12z 1
1
z2 + 6x
2
z2
2
2 + x 2 . , x2 ) = x 1 2 , x ) is homogeneous of degree 1. 1 2 1
2 (x 2 + x 2 ) = t x f (tx 1 , tx 2 ) = (tx )2 + (tx 2 )2 = t 1 1 2 (b) According to Euler’s theorem, we should have f (x
+x2. 2 , x ) = (f/x 1 2 2 1
1
) x 1 +( f / x
Verify this. 1 · f (x
, x2 ) = x 1
1
x2 + x 1
A2.9 Suppose f (x
(a) (a) f (x
1
1
, x2 ) = (x
1
2 2
x1 + x
2
x2 + x
x 2 )2 and g(x
1
1
2 2
x2 = x
, x 2 ) = (x
2 1
+x
2 2
x2 + x 1
2 1
x2 ) 3 .
, x 2 ) is is hom homog ogen eneo eous us.. Wha Whatt is is its its degr degree ee??
f (tx 1 , tx 2 ) = t
4
(x 1 x 2 )2
k=4
(b) (b) g(x g(x 1 , x2 ) is is hom homog ogen eneo eous us.. Wha Whatt is is its its degr degree ee?? g(tx 1 , tx 2 ) = t 9 (x 2 x 2 )3 k = 9 1
(c) h(x 1 , x 2 ) = f (x 1 , x 2 )g(x 1 , x 2 ) is homogeneous. What is its degree? h(x 1 , x 2 ) = (x 3 x2 ) 5 h(tx 1 , tx 2 ) = t 25 (x 3 x 2 )5 k = 25 1
2
1
8
2
=x 2 2
2 1
+x
2 2
2
) x2 .
1 Mathematical Appendix (d) k(x k(tx
1
, x 2 ) = g (f (x 1
, tx 2 ) = t
36
1
, x 2 ), f (x 1 , x2 )) is homogeneous. What is its degree?
(x 1 x 2 )18
(e) Prove that whenever f (x neous of degree n, then k(x mn. k(tx
1
, tx 2 ) = [t
m
k = 36
1
, x 2 ) is homogeneous of degree m and g(x , x 2 ) is homoge1 , x 2 ) = g (f (x 1 , x 2 ), f (x 1 , x 2 )) is homogeneous of degree 1
(f (x 1 , x 2 ), f (x 1 , x 2 ))] n
k = mn
A2.18 Let f (x) be a real-valued fu function defined on R
n +
, and co consider the matrix
0f 1 ···f n f 1 f 11 · · · f 1n . .. . .. . .. .. . . f n f n1 · · · f nn
H* =
This is a di erent 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 0f 1
D2 = 0 f f1
f 11
, D3
=
f1 f2
1
f 11 f 21
f 2 f 12 , . . . , D n = |H f 22
*
|.
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: D D 3 = 0, . . . .
2
(ii) If for all x = 0, these principal minors (which (which depend on x) alternate in sign beginning with strictly strictly negative: D < 0, D 3 > 0, . . . , then then f (x) is quasiconcave quasiconcave 2 on the nonnegative orthant. Further, 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 (x
1
, x2 ) = x
1
x2 + x
1
is quasiconcave on R
2 +
minors alternate in sign as in (ii). Answer The bordered Hessian is
H
*
=
0x x2 + 1 0 1 x1
2
+1x 10
9
1
.
. Verify that its principal
= 0,
1 Mathematical Appendix The two principal minors are D
= -(x
2
2
+1)
< 0 and D
2
3
= 2x
1
x2 +2x
1
= 0. Which
shows that the function will be quasiconcave and will be strictly quasiconcave for all x 1 , x 2 > 0. (b) Le Let f (x
1
, x2 ) = a ln ln(x
+ x 2 ) + b, wh where a > 0. 0. Is Is th this fu function st strictly qu quasiconcave
1
for x 0? It is quasiconcave? How about for x = 0? Justify. Answer The bordered Hessian is 0 H* =
a +x +x a +x +x
x1 x1
a x 1 +x -a (x 1 +x +x -a (x 1 +x +x
2
2
a x 1 +x 2 -a (x 1 +x 2 )2 -a (x 1 +x 2 )2
2
2
)2
2
)2
.
a The two principal minors are D = -( ) 2 < 0 for x , x2 > 0 and D 2 1 x 1 +x 2 Which shows that the function can not be strictly quasiconcave. However, However, it can be
quasiconcave fo following (i (i). Fo For x =x 1 curvature can not be checked in this point. A2.19 Let f (x
1
, x 2 ) = (x
1
2
3
= 0.
= 0 th the fu function is is no not de defined. Th Therefore,
x2 ) 2 . Is f (x) concave on R
2 +
? Is it quasiconcave on R
2 +
Answer The bordered Hessian is
H* =
0 2x 2x 1 x 2
2x 2
2x 2 x 2
4x 1 x2
2
1
1
x2
2x 2 x 2 1 4x 1 x 2 .
2
2
2x 2 1
The two principal minors are D = -(2x 1 x 2 )2 < 0 and D = 16x 2 3 shows that the function will be strictly quasiconcave. Strict quasiconcavity quasiconcavity implies
4 1
x4 = 0. Which 2
quasioncavity. A2.25 Solve the following problems. State the optimised value of the function at the solution. (a) min x ,x = x 2 + x 2 s.t. x 1 x2 = 1 1 2 1 2 x 1 = 1 and x 2 = 1 or x 1 = -1 and x (b) min
=x
x 1 ,x 2
1
x 2 s.t. x
x 1 = 1/2 and x (c) max x1 = a
2 /3
(d) max x1 =
x 1 ,x 2
4
x1 ,x 2
1
x 22 s.t. x
and x =x
1/2 and x
2 1
= 2b
+x 2
+x
2 2
=
2 4
2 1
/a
= - 1/2 and x
2
+x
2 /3
or x
s.t. x
= -1, optimised value= 2
=1
= - 1/2 or x
2
=x
2 1
2
4 1
+x
1 2 /b 2 2 2 4 2
= - 2b
2 /3,
=1
2
= 1/2, optimised value= -1/2
=1 optimised value=
2ab 2 3 3 /2
v2
1/2, optimised value=
(e) max x ,x ,x = x 1 x 2 x 3 s.t. x 1 + x 2 3 1 2 3 x 1 = 1/6 and x 2 = 1/3 = 2/6 and x
2
4
3
= 2 3/4
+x3 =1 = 1/2 = 3/6, optimised value= 1/432 = 108/6 3
10
6
?
1 Mathematical Appendix A2.26 Graph f (x) = 6 - x
2
- 4x. Find the point where the function ac 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. Obviously, the global maximum is not a solution in the presence of a nonnegativity constraint. The constrained maximization problem is L(x, z, ) = 6 - x
2
- 4x + (x ( x - z)
The first order conditions and derived equations are: L x = -2x - 4 + = 0 L
z=- =0 L
=x-z=0
= x - zz = x x = 0 If = 0, then x = -2 would solve the problem. problem. However, However, it does not satisfy satisfy the nonnegativity constraint. constraint. If = 0, then x = 0. As the function is continuously decreasing decreasing for all values x = 0, it is the only maximizer in this range. y 6
-
Figure 3: Graph to Exercise A2.26
11
x
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 : Indi erence curves representing 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 etc.) would rather result in concave preferences. 1.8 Sketch a map of indi erence sets that are parallel, negatively sloped straight lines, with preference increasing northeasterly. northeasterly. We know that preferences such as these satisfy Axioms 1, 2, 3, and 4. Prove the they also satisfy Axiom 5 . Prove that they do not satisfy Axiom 5. Answer : Definition of convexity (Axiom 5 ): If x
1
x 0 , then tx
all t [0, 1]. Strict convexity (Axiom 5) requires that, if x tx 1 + (1 - t)x 0 x 0 for all t [0, 1].
1
1
=x
+ (1 - t)x 0
and x
x 0 for
0
x 0 , then
1
The map of indi erence sets in the figure below represent perfect substitues. We know that those preferences are convex but not stricly convex. Intuitively, all combinations of two randomly chosen bundles from one indi erence curve will necessarily necessarily lie on the same indi erence curve. Additionally, the marginal rate of substitution does not change 0 to x 1 . To prove the statement more formally, define by moving from x define x combination of bundles x commodities gives us:
0
to x
1
:x
1 , (1 - t)x x t = (tx 0 , tx 0 ) + ((1 - t)x 1 2 1 definitions results in the equality
t
= tx
1 ). 2
0
+ (1 - t)x
1
t
as convex
. Re-writing in terms of single
A little rearrangement and equalising the two
tx 0 + (1 - t)x 1 = (tx 0 + (1 - t)x 1 ), tx 0 + (1 - t)x 1 ). That is, the consumer is indi erent 1 1 2 2 with respect to the convex combination and the original bundles, a clear violation of strict convexity.
12
2 Consumer Theory
x2
6 HHHHHHHHHHHHHHHHHHHHHHH HHHHHHHHHHHHHHHHH
x0
HHHHHHHHHHHH HHr HHHHHHHHHH
xt
r HHHHHH
x1
r
-
x1 Figure 4: Indi erence sets to Exercise 1.8
1.9 Sketch a map of indi erence sets that are parallel right angles that “kink” on the line line x 1 = x 2 . If pre prefer ferenc encee incre increase asess northe northeast asterl erly, y, these these prefer preferenc ences es will will sati satisfy sfy Axio Axioms ms 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 5 ) requires that, if x all t [0, 1]. Take any two vectors x must be true that min[x
x 0 , then tx
1
1
+ (1 - t)x
x 0 for
0
, x 1 such that x 0 ~ x 1 . Given the nature of these preferences, it 0 , x 0 ] = min[x 1 , x 1 ]. For any t [0, 1] consider the point tx +(1 1
0
1
2
1
2
0 + (1 - t)x 0 , tx 1 + (1 - t)x 1 ] = min[x 0 , x 0 = min[x t)x 2 . If we can show that min[tx 1 2 1 2 1 2 0 + (1 - t)x 0 , tx 1 + (1 - t)x then we shown that these preferences are convex. min[tx 1
0 , +(1 - t)x 1 ] = min[x min[tx 0 , tx 1 ] + min[(1 - t)x 1 1 2 2 min[x 0 , x1 ] 2 2 Definition of strict monotonicity (Axiom 4): For all x
0 2
2
, x 1 ] + t[min(x 2
0
,x
1
R
0 1
1
, x 1 ) - min(x 1
n +
1 1
, if x
0
=x
0, 2
, x 1 ], 2 1] = 2
x 1 )] = 2
1,
then
x0 x 1 , while if x 0 x 1 , then x 0 x1 . The map of indi erence sets in the figure below represents perfect complements. Take 0 two points x 0 , x1 along one indi erence curve. If x x 1 , “preferences increase north0 easterly”, th then x x 1 . Fo For an any tw two ve vectors on on th the sa same indi er erence cu curve, th that is is
x 0 = x 1 , it it fo follows x 0 for these indi erence sets.
x 1 . Th Therefore, th the de definition of of st strict mo monotonicity is is sa satisfied
1 = x 0 and x 1 Strict convexity (Axiom 5) requires that, if x x 0 , then tx for all t [0, 1]. Take any two points along the horizontal or vertical part of an indi erence curve such
as (x 0 , x 0 ) and (x 0 , x1 ), where x 0 > x 1 2 1 2 2 lies on the same indi erence curve as x
1 2
1
+(1-t)x
0
t = x 0 , tx 0 + (1 - t)x . Any convex combination x 1 2 1 and x 0 . Therefore, it is not possible that
1 . That is xt tx 0 + (1 - t) t)x is, the consumer is is indi er erent wi with respect to th the convex combination and the original bundles, a clear violation of strict convexity.
13
x0
1 2
2 Consumer Theory
x2
6
x0
r
x r
1
-
x1 Figure 5: Indi erence sets to Exercise 1.9
1.12 Suppose u(x
1
, x 2 ) and v(x
1
, x 2 ) are utility functions.
(a) Prove that if u(x , x 2 ) and v(x 1 , x 2 ) are both homogeneous of degree r, then s(x , x2 ) = 1 1 u(x 1 , x 2 ) + v (x 1 , x 2 ) is homogen geneous of degre gree r. r u(x , x ) = u(tx Answer : Whenever it holds that t , tx 2 ) and t r v(x 1 , x 2 ) = v(tx 1 , tx 2 ) 1 2 1 for all r > 0, it must also hold that t tr u(x 1 , x2 ) + t r v(x 1 , x 2 ).
r
(b) Prove that if u(x , x 2 ) and v(x 1 v(x v(x 1 , x2 ) is is als also o qua quasi sico conc ncav ave. e.
1
s(x 1 , x2 ) = u(tx
1
, tx 2 ) + v(tx
, x 2 ) are quasiconcave, then m(x
1
1
, x 2 ) = u(x
, tx 2 ) =
1
, x 2) +
Answer : Forming a convex combination of the two functions u and v and comparing with with m(x t ) satis satisfie fiess the defini definitio tion n of qua quasic siconc oncavi avity: ty: When u(x
t
) = min tu(x
1
v(x t ) = min tv(x
) + (1 - t)u(x 1
) + (1 - t)v(x
m(x t ) = min u(x = t(u(x
2
t 1
) + v(x
1 ))
+ (1 - t)(u(x
2)
) and 2)
so
) + v(x
+ v(x
2
t
)
)) .
2.2 The Consumer’s Problem 1.20 Suppose preferences are represented by the Cobb-Douglas utility function, u(x Ax a x1-a , 0 < a < 1, 1, and and A > 0. Ass Assum umin ing g an an inte interi rior or solu soluti tion on,, sol solve ve for for the the Mars Marsha hall1
2
lian demand functions. Answer : Use either the Lagrangian or the equality of Marginal Rate of Substitution and price ratio. The The Lagrangian Lagrangian is
14
1
, x2 ) =
2 Consumer Theory L = Ax
a 1
x 1-a 2
+ (y - p
1
x1 - p
2
L x
= aAx
x 1-a
a-1 1
- p
2
1
=0
1
L x L
x2 ). The first-order conditions (FOC) are
= (1 - a)Ax
a 1
x-a
- p
2
2
=0
2
1
=y-p
x1 + p 2 x 2 = 0
By dividing first and second FOC and some rearrangement, we get either x x2 =
(1-a)p ap
1
x1
1
=
ax
2
p2
(1-a)p
or 1
. Substituting one of these expressions into the budget constraint, results
2
in the Marshallian demand functions: x
1
=
and x
ay p1
2
=
(1-a)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 exercise (1.20). (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) + a ln(x ) + (1 - a) ln(x 1
2
L x L
) + (y (y - p
=a 1
x 2 L
x1
1
- p
1
= (1 - a) x2
=y-p
x1 - p
x 2 ). The FOC are
=0
- p
1
2
=0
2
x1 + p 2 x 2 = 0
The Marshallian demand functions are: x = ay and x 1 p1 identical to the demand functions derived in the preceding exercise.
2
=
(1-a)y p2
. They are exactly
1.24 Let u(x) represent some consumer’s monotonic preferences over x R n . For For each each of the functions functions F (x) (x) that follow, follow, state state wheth whether er or not f also also represen represents ts + the preferences preferences of this consumer. In each case, be sure to justify your answer with either an argument or a counterexample. counterexample. Answer : (a) f (x) = u(x) + (u(x))
3
Yes, all arguments of the function u are transformed equally
by the third power. power. Checking the first- and second-order second-order partial derivatives derivatives reveals 2 2 u f u that, although the second-order partial = +6(u(x))( ) 2 is not zero, the sign x
2 i
x
2 i
x
i
of the derivatives is always invariant and positive. Thus, f represents a monotonic transformation transformation of u.
15
2 Consumer Theory 2
(b) f (x) = u(x) - (u(x))
No, function f is decreasing with increasing consumption for
2
any u(x) < (u(x)) . Therefore, it can not represent the preferences of the consumer. It could do so if the minus sign is replaced by a plus sign.
n (c) f (x) = u(x) + x i Yes, the the transf transforma ormation tion is a linear linear one, one, as the first partial partial i=1 is a positive constant, here one, and the second partial of the transforming function f
is zero. Checking the partial derivatives proves this statement: 2
x
f 2 i
=
2
x
u 2 i
x
= i
u x
+ 1 and i
.
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 x function is given by
, and her lifetime utility
t
8
u(x 0 , x 1 , x 2 , . . .) =
ß t ln(x t ), where 0 < ß < 1. t=0
Calculate her optimal level of consumption in each perio d. Answer : Establish a geometric series to calculate her lifetime utility: u=ß
0
ln(x 0 ) + ß ln(x
1
)+ß
2
ln(x 2 ) + . . . + ß
t
ln(x t )
As ß is less than one, this series approaches a finite value. To find the solution, multiply the expression by ß and subtract from the original equation [(1)-(2)]. ßu = ß
1
ln(x 0 ) + ß
ln(x 1 ) + ß
2
u - ßu = (1 - ß)u = ln(x u = ln(x
0
)-ß
0 t+1
)-ß
t+1
ln(x 2 ) + . . . + ß
3
t+1
ln(x t )
ln(x t )
ln(x t ) 0
1 - ß = ln(x
)
Thus, the consumer’s utility maximising consumption will be constant in every perio d.
2.3 Indirect Utility and Expenditure 1.30 Show that the indirect utility function in Example 1.2 is a quasi-convex function of prices and income. Answer : The indirect utility function corresponding to CES preferences is: v(p, y) = y (p r + p r )-1/r 1
2
, where r = /( - 1).
There are several ways. First, using the inequality relationship, let p We need to show that the indirect utility function fulfills the inequality yp
tr 1
+p
tr 2
-1/r
= max[y p
0r 1
+p
16
0r 2
-1/r
,yp
t
1r 1
+p
1r 2
= tp
-1/r
]
0
+ (1 - t)p
1
.
2 Consumer Theory which gives: yt
r
(p 0r + p 1
0r 2
) + (1 - t)
r
(p1r + p 1r ) 1
-1/r
= max[y p
2
+ p 0r
0r 1
-1/r
2
,yp
+p
1r 1
-1/r
1r 2
].
Second, the bordered Hessian can be derived and their determinants checked. The determinants will be all negative. 0p H=
p -1/r -p -p
r-1 1 r-1 2
, where p = (p
0p
p -1/r-1
yp
-1/r
p -1/r-1
yp
-1/r
r 1
-p
-1/r
p -1/r-1
r-1 1
y -p
r-1 2
-1/r
yp -1/r -1
pr-2 1
(1 + r)p
((1 - r) - rp
r 1
pr2-1 p -1/r-2
r-1 1
p -1 ) (1 + r)p
y yp
p -1/r-1 p -1/r
r-1 1
pr2-2
-1/r-1
y
pr-1 p -1/r-2
y
2
((1 - r) - rp
r 2
-1 ) p-1
+ p r ). 2
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 for two commodities is e(p, u) = u (p
r 1
+ p r ) 1/r where 2
r = /( - 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(p
r 1
+ p r )1/1/r = 0. 2
n 2. Continuous on its domain R ×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: r +p r ) 1/r . For all stric e/ u = (p trictl tly y posi positi tiv ve pric pricees, thi this expre pressio ssion n will will be posit ositiv ive. e. 1
2
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 boundness above follows from the functional form form of u.
*,
*
), what equals . Un-
4. Increasing in p. Again, take all first partial derivatives with respect to prices: e/ p what is, obviously, positive.
5. Homogeneous of degree 1 in p. e(tp, u) = u ((tp
1
) r + (tp
2
) r ) 1/r = t 1 u (p r + p r ) 1/r 1
17
2
i
= up
r-1 i
(p r + p r )(1/r)-1 1
2
,
2 Consumer Theory 6. Concave in p. The definition of concavity in prices requires tup for p
t
= tp
0
0r 1
+p
+ (1 - t)p
1
1/ r
0r 2
+ (1 - t) u p
1r 1
+ p 1r
1/r
= e (p
2
, u)
t
. Plugging in the definition of the price vector into e(p
t
, u)
yields the relationship tup
0r 1
+ p 0r
1/r
+ (1 - t) u p
2
u t(p
0r 1
+p
0r 2
1r 1
) + (1 - t)(p
+p
1r 1
1/r
1r 2
+p
1/r
)
1r 2
=
.
Alternatively, we can check the negative semidefiniteness semidefiniteness of the associated Hessian matrix of all second-order second-order partial derivatives of the expenditure function. A third possibility is to check check (product rule!) rule!) 2
p
e 2 i
= u (r - 1)p
r i
(p r + p r )1/r 1
2
p2 (p r + p r ) - r p 1
1
2
2r i
(p r + p r )1/r
1 p2 (p r i 1
+
2 pr 2
)2
< 0 by r < 0.
Homogeneity of degree one, together with Euler’s theorem, implies that
2
0. Hence the diagonal elements of the Hessian matrix must be zero and the matrix will be negative semidefinite.
7. Shephard’s lemma e/ u = (p (p function.
r 1
+p
r )1/ 1/r 2
what
is is ex exactly th the de definition of of a CES-type Hi Hicksian de demand
1.38 Complete the pro of of Theorem 1.9 by showing that x h (p, (p, u) u) = x (p, (p, e(p e(p,, u)) 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 h (p, v(p, y)). As the new function is a v into the Hicksian demand function gives x function of prices and income only, it is identical to the Marshallian demand function.
Furthermore, 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.
18
e/ p
2p i i
=
2 Consumer Theory 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 s + s 2 2 = 1. An inferior go od is characterised 1 1 by a negative income elasticity, elasticity, thus, one of the two summands will will be less than zero. zero. Therefore, Therefore, to secure this aggregation, the other summand must be positive (even larger one) and the other commo dity must be a normal good (even a luxury item). 1.55 What restrictions must the a , f (y), w(p i following is to be a legitimate indirect utility function? Answer : (a) v(p 1 , p 2 , p3 , y ) = f (y)p
a1 1
1
, p 2 ), and z(p
1
, p2 ) satisfy if each of the
pa 2 pa 3 The function f (y) must be continuous, strictly in2
3
creasing and homogeneous of degree 0 - a . Each of the exponents a i than zero to satisfy v decreasing in prices. Furthermore, negative partial derivatives derivatives of
i
has to be less
v with respect to each price are required to get positive Marshallian demand functions by using Roy’s identity. identity. (b) v( v(p 1 , p2 , y) y) = w(p 1 , p2 )+z(p 1 , p2 )/y Th The fu functions w and z must be be continuous an and de decreasing in prices. Function z has to be homogeneous of degree one and w homogeneous 0 (w(p , p ) + z(p of degree zero: v(tp 1 , tp 2 , ty) = t 0 w(p 1 , p 2 )+(t 1 z(p 1 , p2 ))/(ty) = t 1 2
1
, p2 )/y).
To satisfy v increasing in income, z must be < 0. 1.60 Show that the Slutsky relation can be expressed in elasticity form as h s j i , where is the elasticity of the Hicksian demand for x ij
= with respect to price p ij
i
and all other terms are as defined in Definition 1.6. Answer : The Slutsky relation is given by h x i x i = x i - x j p j p j y. Multiplying the total expression with y/y and p x
i
p = x p j j p By assu assumi ming ng that that x
h i
=x
gives
j h i
p j - p
j
j
xj x y
i
y y.
befo before re the the pric pricee chan change ge occu occurs rs,, we we can can div divid idee all thre threee term termss
i
by x i . The result result of this operation operation is x p
i j
pj
= x xi p
h i
p j
j
xi
-s
x j
y
i
y xi
=
ij
=
h ij
-s
j
i
Additional exercise Relationship between utility maximisation and expenditure minimisation Let’s explore the relationship with an example of a concrete utility function. A consumer’s utility function is u = x
1/2 1
x1/2 . For the derived functions see 1 2
19
h ij
j
,
2 Consumer Theory Start from the utility function Minimise expenditures s.t. u and derive the Marshallian demand for x to find the Hicksian demand function 1 x 1 = y/2p
x h = u (p
1
1
2
/p 1 )1/2
1
p2 )1/2
Plug in the respective demand functions to get the indirect utility function expenditure function v = y/(4p 1 p2 )1/2
e = u(4p
Substitute the expenditure function Substituting the indirect utility function into the Marshallian demand function into the Hicksian demand function to derive the Hicksian demand function to derive the Marshallian demand function x1 = (u(4p 1 p2 )1/2 )/2p 1 = u(p 2 /p 1 ) 1/2 xh = (p 1 Invert v and replace y by u Invert e and replace u by v
2
/p 1 ) 1/2 y/(4p
1
p2 )1/2 = y/2p
to get the expenditure function to get the indirect utility function v-1 = u(4p 1 p2 )1/2 Check Roy’s identity Check Shephard’s lemma -
v/ p v/ y
1
=
2y(p
1
p2 )1 / 2
3
4(p p 2 )1 /2 1
= y/2p
e-1
1 e
p
= 1
= y(4p u4p
2(4p
1
2
p2 )1 /2
1
p2 )-1/2
= u(p
Establish the Slutsky equation x p
=
1
u 2(p
2
1
p2 )1/ 2
-
y 2p 2
·
1 2p 1
substitute u = v(p, y ) into the substitution e ect y y x 1 = =0 p
2
4p 1 p2
4p 1 p2
Table 1: Relationship between UMP and EMP
2.5 Equilibrium and Welfare 4.19 A consumer has preferences over the single good x and all other goo ds 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:
m = 0 when y = 1 y - 1 when y > 1. (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:
v (p, y) = ln
when m = 1
1 p
y - 1 - ln p when m > 1.
20
2
/p 1 )1/2
1
2 Consumer Theory (c) Use the Slutsky equation to decompose the e ect of an own-price change on the demand for x into an income and substitution e ect. Interpret your result brie y. Answer A well-known property of any demand function derived from a quasi-linear utility function is the absence of the income e ect. Which can be easily seen in the application of the Slutsky equation: x
h
p = x
p+x x y
x
h
p = - 1
p2
+0·1
p= xp.
Therefore, the e ect of an own-price change on the demand for x equals the substitution e ect. 0 to p 1 > p 0 . Show that the consumer surplus (d) Suppose that the price of x rises from p area area betw betwee een n p 0 and and p 1 give givess an an exa exact ct meas measur uree of of the the e ect ect of the the pric pricee cha chang ngee on
consumer welfare. Answer The consumer surplus area can be calculated by integrating over the inverse demand function of x: p1
CS = p0
1
1
x dx = ln(p
-p
0
).
Calculating the change in utility induced by a price change gives: v=v
1
(p1 , y1 ) - v
0
(p 1 , y0 ) = y - 1 - ln p
1
- (y - 1 - ln p
0
) = ln(p
As the two expressions are equal, the consumer surplus area gives an exact measure of the e ect of the price change on consumer welfare in the case of quasi-linear preferences. 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 re ect all qualitative information on preferences and demands that you’ve uncovered. Be sure to consider the the two two pric prices es p 0 and p 1 , and and iden identi tify fy the the Hic Hicks ksia ian n and and Mars Marsha hall llia ian n dem deman ands ds.. Answer See Figure 6. Please note, that Hicksian and Marshallian demands are identical here.
21
1
-p
0
).
2 Consumer Theory
Figure 6: Graph to 4.19
22
3 Pro ducer Theory
3 Producer Theory 3.1 Production AP
3.1 The elasticity of average product is defined as equal to µ
i
(x)
i
x
·
i
xi AP
i
(x)
. Show that this is
(x) - 1. Show that average product is increasing, constant, or decreasing decreasing as
marginal product exceeds, is equal to, or less than average product. Answer : Applying quotient rule to get the first partial derivative of the average product gives: AP i (x) f (x)/ x - f (x) i =x i = MP - AP = MP - AP x i x2 xi xi xi i
Multiply this term with the right part of the definition (x /AP ) gives M P/AP - 1 what i is exactly µ i (x) - 1. The first part of the above definition equals the slope of the average product: (MP AP )/x i . It is straightf straightforwa orward rd to show show that whene whenever ver margina marginall pro duct duct exceeds exceeds the average pro duct the slope has to be positive. The average product reaches a maximum when the marginal product equals average product. Finally, whenever MP < AP average pro duct is sloping downwards. 3.3 Prove that when the production function is homogeneous of degree one, it may be written as the sum f (x) = MP i.
i
(x)x i , where MP
i
(x) is the marginal product of input
Answer : The answer to this exercise gives a nice application of Euler’s Theorem. The sum of the partial di erentials 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 pro duction function times k = 1. 3.7 Goldman & Uzawa (1964) have shown that the pro duction function is weakly separable with respect to th the partition {N the form
1
f (x) = g f
1
, . . . , N S } if and on only if it ca can be written in
(x (1) ), . . . , f
S
(x (S) ) ,
i (x (i) ) is a function of the where g is some function of S variables, and, for each i, f subvec subvector tor x (i) of inputs inputs from from grou group p i alone. alone. They They have have also also show shown n that that the pro pro duct duction ion 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. Answer To show that the first equation is weakly separable with respect to the partitions, [f i (x)/f j (x)] we need to show that = 0 i, i, j N and k / N S . Calculate the marginal S x
k
23
3 Pro ducer Theory pro ducts of the first equation for for two arbitrary arbitrary inputs i and j: f i (x) = g
f f
f j (x) = g
x
S
f
S
i
f
S
S
.
x
j
The marginal rate of technical substitution between these two inputs is f i (x)
f x
S
f j (x) =
f
S
x
j
i
This expression is independent of any other input which is not in the same partition N and, therefore, the production function is weakly separable. (f
i
x
/f j )
= 0 for k / N
S
S
k
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 di erent partitions i N
S
,j N
T
and k / N
S
N
T
. The marginal products of the two inputs i and j are: x (S)
S
f i (x) = G f
x
x (T )
T
f j (x) = G f
x
i
.
j
The MRTS is:
It follows for k /
N
S
N
f i (x)
S
/x
f j (x) = f f
T
/x
T
(f
i
x
/f j )
i
.
j
= 0.
k
3.8 A Leontief pro duction function has the form y = min {ax
1
, ßx 2 } for a > 0 and
ß > 0. Carefully sketch the isoquant map for this technology and verify that the elasticity of substitution s = 0, where defined. Answer : Taking the total di erential of the log of the factor ratio gives d ln (ß x
2
/ax
1
ß /x /x 2 dx 2 - a/ a/x 1 dx 1 . How Howeever, ver, the the MRT MRTS S is is not not def define ined in in the the kinks inks as the the func unction tion 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, s = 0. 3.9 Calculate s for the Cobb-Douglas production function y = Ax 0, a > 0 and ß > 0.
a 1
x ß , where A > 2
Answer : The total di erential of the log of the factor ratio gives d ln(x 2 /x 1 ) = ß/x 2 dx 2 -a/x 1 dx 1 . Th The total di di erential of the marginal ra rate of technical substitution gives a-1 x ß 1 2 d ln Aax Aß xa x ß-1 1
= a/ß (dx
2
24
1
/x
1
- dx
2
/x 2 )
)=
3 Pro ducer Theory
x2 6
-
x1 Figure 7: Isoquant map of Leontief technology
Putting both parts together results in 2 s = ß /x a/ß (dx
3.14 Let y = (
n i=1
a i x )1/
dx 2 - a/x /x
1
, where
i
1
dx 1
1
- dx
2
/x 2 ) = 1
a i = 1 and 0 = < 1. Verify that s
i
1/(1 - ) for all all i = j. Answer Apply the definition of the elasticity of substitution. ) - ln(x
j
s ij = (ln(x ln (f
i
(x)/f
j
x
-
1 x j
=
j
-1
a i xi
ln
=
1 xi
-1
x
i
x
1 xi
= -1 -1=1
1 xi i
(sum
j
-
(x))
(
a j x- 1
i
))
i
1 xj
-
x
i
a i x i ) 1/ i
-1
a i x )1 / - 1 i
x 1 xj
j
x
j
1-
3.15 For the generalised CES production function, prove the following claims made in the text. 1/
n
y=
aix
i
n
, where
a i = 1 and 0 = < 1
i=1
i=1
25
ij
=
3 Pro ducer Theory (a) n
xa i
lim y =
i
0 i=1
Answer : Write the log of the CES production function ln y = 1/ ln a the value of the function is indeterminate. However, However, using L’H` x ln x i
i
lim ln y = a
.
i
aix
0
x . At = 0, i opital’s rule we can write i
i
At = 0 this expression turns into ln y = a ln x i / a . Because the denominator is i i a i , what is defined to be one, we can write the CES production at this point as y = x i
exactly the generalised Cobb-Douglas form. (b) lim y = min {x -8
Answer : Let us us as assume th that a y = (x that x
1 1
i
=a
j
1
2
(x
-8
i
we can establish x Hence ( x ) 1/
=x = n 1/
because lim
n 1/
i
-8
1
, . . . , xn }
. The Then n th the CE CES pr production fu function ha has th the fo form
+ x )1/ . Let us suppose that x = lim
1
= min( x
)1/ . Sin Since ce all comm ommodit oditie iess x . Thus, x * x 1 . Letting i
1
i i
) and < 0. We want to show are re requir quireed to to be be non nonn negati gativ ve,
=(x )1/ . On the other hand, x i -8, we obtain lim -8
(x
i i
=n*x =x
) 1/
* x1 = x 1.
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? Answ Answeer : The The pa param rameter ters A, A, w , w2 and y ar are re requir uired to to be be la larger ger th than ze zero. ro. A cost ost 1 function is required to be increasing in input prices. Therefore, Therefore, the exponents a and ß must be larger zero. To fulfill the property of homogeneity of degree one in input prices, the exponents have to add up to one. To secure concavity in input prices, the second order partials should be less than zero. Thus, each of the exponents can not be larger one. 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 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 x and x 2 . In other words, ax = ßx 2 . For a given level of output 1 1 y , we must ha have y = ax ax = ßx ßx 2 . Re Rearrange th this ex expre pression to to de derive the co conditional 1 input demands: x 1 (w, y) = y
2
ax
26
(w, y) = y
ß.
. , 1 1
3 Pro ducer Theory The cost function is obtained by substituting the two conditional demands into the definition of cost: c(w, y ) = w
x (w, y) + w 1 1
x (w, y) = w 2 2
1
y
a+w
2
y
ß.
3.27 In Fig. 3.85, the cost functions of firms A and B are graphed against the input price w 1 for fixed values values of w 2 and y. 0 , which firm uses more of input 1? At w (a) At wage rate w ? Explain? 1 1 Answer : Input demand can be obtained by using Shephard’s lemma, represented represented by the 0 firm B demands more of factor 1 and at slope of the cost function. Therefore, at w 1 wage rate w firm A has a higher demand of that input. 1
(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 w the two input case, we can re-write the original definition of the elasticity of substitution as /x ) /x 1 ) x2 - ˆ x1 2 s = d ln(x 2 1 , d ln(f 1 /f 2 ) = d ldn(ln x(w 1 /w 2 ) = ˆ w ˆ 1 - ˆ w2
i
/w j . In
where the circum ex denotes percentage change in input levels and input prices, respectively. Be Because ˆ w 2 = 0, the denominator reduces to to ˆ w1 , wh which is as assumed to be be the 0 is larger for firm B same for both firms. In (a) we established that input demand at w 1
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 w 3.29 The output elasticity of demand for input x
iy
(a) Show that
iy
(w, y) = f(y)
(w, y) = x
iy
i
i
(w, y )
0 1
.
is defined as y
y
x i (w, y) .
(w, 1) when the production function is homothetic.
Given a homothetic production function, the cost function can be written as c(w, y) = f(y)c(w, 1). Shephard’s lemma states that the first order partial derivative with respec spectt to to the the pric pricee of of inpu inputt i give givess dem deman and d of of x and and to to obt obtai ain n the the elas elasti tici city ty we need need to i take take the second-order 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 di erentiation does not matter. Therefore, Therefore, the following partial derivatives should be equal: 2
c(w, y)
w
i
y = mc w
i
= x i y.
Putting everything together gives: 2 iy
(w, y ) =
yw
c
y i
cw
i
= f(y) yx
i
Unfortunately, this is not the result we should get.
27
(w, 1) y f(y)x
(w, 1) = 1 f (y) i
iy
(w, 1).
3 Pro ducer Theory (b) Show how tha thatt
= 1, 1, for for i = 1, . . . , n, wh when the the produ roducction ion fun funcction tion has con const staant
iy
returns to scale. Answer For any production function with constant returns to scale, the conditional inpu inputt dema demand nd x is lin linea earr in out outpu putt leve levell y (see (see The Theor orem em 3.4 3.4). ). Mor Moree form formal ally ly,, the the i conditional input demand of a production function homogeneous of degree a > 0 can be be written as x (w, y) = y 1/a x i (w, 1) 1). By de definition, a constant re return urns to to scale i technology requires a pro duction function homogeneous of degree 1. Therefore, Therefore, the conditional in input de demand re reduces to to x (w, y) y) = yx i (w, 1) 1). Ca Calculating th the output i elasticity of demand fo for input x
iy
results in:
i
i
(w, y) = x
(w, y)
y
y
xi (w ( w, y) = x
i
(w, 1) y y xi (w, (w, 1) 1) = 1.
3.33 Calculate the cost function and the conditional input demands for the linear n pro duction function function y = a i xi . i=1 Answer Because the production function is linear, the inputs can be substituted for
another. The most e cient input (i.e. input with the greatest marginal product/ price) will be used and the other inputs will not be used. y
x i (w, y) y) =
ai
if
0 if
>
ai wi ai wi
<
a j wj a j
j = i, j {1, . . . , n} for at least one j = i, j {1, . . . , n}.
wj wi y
The cost function is then c(w, y) =
ai
ai wi
>a
j
wj
, where i is the input where
j = i, j 1, . . . , n.
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 pro duction function. 1/2 (w w ) 3/4 (a) c(w, y) = y 1 2 Homogeneity: c(tw, y) = y
1/2
(tw 1 tw 2 ) 3/4 = t
3/2
y1/2 (w 1 w 2 ) 3/4
The function is not
homogeneous of degree one. Monotonicity: c(w, y) w
1
= 3/4y
1/2
w-1/4 1
w3/4 > 0 c(w, y) 2 w 2
The function is monotonically increasing in input prices.
28
= 3/4y
1/2
w 3/4 w -1/4 1
2
>0
3 Pro ducer Theory Concavity: w -5/4
3 y1/2 16 9 y1/2 w 16
H=-
1
1/4
w3/4
2 -1/4 w 2 1
9 16
-
y1/2 w-1/4
w -1/4
1
3 16
y1/2
w 3/4 1
2 w-5/4 2
|H 1 | < 0 |H 2 | = - 72
vw
256
y 1
w2
<0
The function is not concave in input prices. Continuity: Yes 1/2 (b) c(w, y) = vy(2w w21/2 ) 1 The function satisfies all properties. The underlying technology is represented by y = x 1 x2 .
(c) c(w, y) = y (w + vw 1 w2 + w 2 ) 1 The function satisfies all properties. The underlying technology is represented by y = 2/3 (x
1
(d) c(w, y) = y (w
+x
1
2
e -w
)+x 1
2 1
-x
1
x2 + x
2 2
.
+ w 2)
The function is not homogeneous of degree one. Using Euler’s Theorem we get c the result wi = y(w 1 e-w 1 + w 2 e -w 1 + w 2 ) what is clearly not equal to the w
1
i
original cost function. Alternatively, it becomes clear from the expression c(tw, y) = ty (w 1 e-tw 1 + w 2 ). The The func unction tion is not mon monoton otonic icaally lly inc incrreasing sing in inpu inputt pr price ices as as the the fir firsst pa partia tial de deriva ivative tive with with respe spect to to w is only only posi posittive ive for for pric pricees le less tha than one one:: 1 c/ w 1 = ye -w 1 (1 -w 1 ). Furthermore the function is only concave for prices w what can be seen from the first determinant of the Hessian matrix: |H 2)e -w
1
< 2, | = y(w 1 1 1
.
(e) c(w, y) = y (w
- vw
1
1
w2 + w 2 )
Monotonicity of the cost function holds only for a narrow set of input prices with the characteristics 1/4w < w 1 < 4 4 . The conclusion can be derived from the first 2 partial derivatives derivatives and a combination combination of the two inequalities. c w
=y1-1 1
c w
=y1-1 2
w2 2
w1 w1
2
w2
positive for 1 > 1 positive for 1 > 1
w2 2
w1 w1
2
w2
or 2 > w
2
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 |H is convex.
1
29
| > 0 and |H
2
| = 0. Thus, the function
3 Pro ducer Theory (f ) c(w, y) = (y + 1/y)vw
1
w2
The function satisfies all properties, except continuity in y = 0. 3.40 We have seen that every Cobb-Douglas pro duction function, y = Ax a x1-a , give givess rise ise to a Cob Cobb-Do b-Doug ugla lass cost func unction tion,, 1
2
c(w, y) = yAw
a 1
w 1-a , and every CES production function, y = A (x 2
1
+ x ) 1/ , gives rise 2
r + x r ) 1/r . For each pair of functions, show that to a CES cost function, c(w, y) = yA (x 1 2 the converse is also true. That is, starting with the respective cost functions, “work
backward” to the underlying underlying production function function and show that that it is of the indicated 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 solve for y. The result will be the corresponding production function. 1-a
x 1 = c(w, y) w 1
= ayA w
2
w2
w1
w1
1/(1-a)
=x
-a
x 2 = c( c(w, y) w 2 (Aay) xa
1
y=x
= (1 - a)yA w
a
2
w2
w1
w1
1
Aay -1/a 2
=x
A(1 - a)y
1 -a 2
=x (A(1 - a)y) a 1
x 1-a 2
a a (1 - a)
1 -a
= ax 1-a
a 1
x 1-a 2
a
where a = (Aa
(1 - a)
1-a
For the CES cost function a short-cut is used: Derive the conditional input demand functions and substitute them into the production function. 1
x 1 = c(w, y) w 1
= yAw
r-1 1
(w r + w r ) r -1
x 2 = c(w, y) w 2
= yAw
r-1 2
(w r + w r ) r -1
y = (Ay)
w-r 1 w-r 1
1
1
1
+ w -r
1/
+w
2 -r 2
2
2
= Ay
3.4 The competitive firm Additional exercise (Varian (1992) 1.21) Given the following production function y = 100x
1/2 1
x1/4 . 2
(a) Find c(w 1 , w2 , y). Answer : Starting from the equality of M RTS and ratio of factor prices, we get
30
)-1
3 Pro ducer Theory w1 /w
2
= 2x 2x
2
/x 1 . Sol Solvi ving ng for for one one of the the inp input uts, s, subs substi titu tuti ting ng back back in the the pro pro duct ductio ion n
function and rearranging, we derive the conditional input demand functions:
100
and
w1 -2/3
2w2
4/3
x2 = y
1/3
2w2
4/3
x1 = y
100
.
w1
Substituting the two functions in the definition of costs, the resulting cost function is: c(w 1 , w 2 , y) = y =2
4/3
4/3
w 2/3 (2w 2 )1/3 + y
100
100
1
+2
1/3
y
-2/3
4/3
w 1/3 w2/3 2-2/3 2
1
w2/3 w1/3 . 1
100
2
(b) Find the e ect of an increase increase in output output on marginal cost, cost, and verify that that = marginal cost. Answer : Marginal costs are 1 (y/100) M C = c/ y =
w2/3 w1/3
1/3
1
75
21/3 + 2
2
. Marginal costs are increasing
-2/3
with output which is shown by MC y
=
2
y
c 2
=
1 150
2
(y/100)
w2/3 w1/3
-2/3
1
grangian we can derive that
*
1/2
w1 x 1
=
50x
21/3 + 2
2
1/4
3/ 4
w2 x 2
=
25x
2
demand functions into one of those expressions gives *
=w 1 50 (y/100)
. Substituting the conditional input
1 /2 1
(2w 2 /w 1 ) 1/3 1/2
4/3
. From the FOC of the La-
-2/3
(y/100)
4/3
(2w 2 /w 1 )-2/3 -1/4
1/3
=2
1/3
50 ( y100 )
w2/3 w1/3 1
2
When you solve the ratios, this expression will be equal to the marginal cost function.
(c) Given p = price of output, find x (w, p), x 1 lemma to derive the supply function y(w, p).
2
(w, p) and p(w, p). Use Hotelling’s
Answer : By maximising p = py - c(w, y) the first-order condition condition is p
y
y =p - 1
75
1/3
100
w2/3 w 1/3 (2 1/3 + 2 -2/3 1
2
3
y =100 75
21/3 + 2 -2/3
3
p 2/3
w
1
)=0
1/3
w
2
The first expression a rms the equality of price and marginal cost as the profit maximum for any competitive firm. The last expression gives already the profit
31
3 Pro ducer Theory maximising supply function. Furthermore, the two following unconditional demand functions emerge as solution of this optimisation problem: 4/3
3
x1 = 75p
21/3 + 2
w-2 1
-2/3
4/3
21/3 + 2
1
-2/3
21/3 p4 w 3 w2
-2/3
= 75 21/3 + 2 -2/3
2w2
2
21/3 + 2
1
2/3
w1
w-2 w-1
4
= 75
w1
3
x2 = 75p
1/3
2w2
w-1 2
p4 22/3 w2 w2 1
The profit function is 3
p = 75
21/3 + 2 -2/3
= 75
21/3 + 2
4
100p 4 w2 w 2 1 3
p4 w2 w2
-2/3
21/3 + 2
w3 w2 1
2 +w 22/3 w2 w 2 1
(100 - 75)
1
3
= 25 75
21/3 w1
- 75p 21/3 + 2 -2/3
p4
.
w2 w2
-2/3
1
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, 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 from the equality c/ y = p gives: p = 1
1/3
75 2
y = 75
3
+2
y
-2/3
1/3
1
100
21/3 + 2 -2/3 -3
w2/3 w1/3 2
p3 w -2 w -1 · 100 1
2
4
x1 (w, p) = 75
21/3
+2
-2/3
p4 w -3 w -1 1
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 : In order to simply this step, I write the constant part of the profit function
32
2
2
.
References
as K = 25
2 1/ 3
75 +2 - 2 /3
3
Calculating the second-order partial derivatives of the profit
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). 12Kp
2
-8Kp
w12
H=
3
3
-4Kp
w31
w 12
w22
-
w2 -8Kp 3 w 3 w2
-
w2 -6Kp 4 w 4 w2
-
-2Kp 4 w3 w 2
-
-4Kp 3 w2 w 2
-
-2Kp 4 w 3 w2
-
-2Kp 4 w2 w 3
1
1
1
2
1
2
1
2
1
2
The own supply e ect is positive, the own demand e ects are negative and all cross price e ects ects are symmmetric. symmmetric. Checking the determinants becomes becomes quite tedious. tedious. Intuituively, it should become clear that they all have to be positive. (g) (g) Assu Assume me x
2
as a fixe fixed d fact factor or in in the the shor shortt run run and and calc calcul ulat atee shor shortt-ru run n tota totall cost 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 x 1 on the left-han left-hand d side and pluggi plugging ng in into the the definiti definition on of cost c(w, y) = (y/100)
2
w1 /x 1/2 + w 2 x 2 . The firstfirst-par partia tiall deriva derivativ tivee gives gives the the shor short-r t-run un margi marginal nal 2
cost function smc = y 100 2
w 1 /x
1/2 2
+
w2 x 2 y
1 y 50 100
w1 /x 1/2 . The short-run average costs are equal to sac = 2
.
Final 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.
References Arrow, K. J. & Enthoven, A. C. (1961), ‘Quasi-concave programming’, Econometrica 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 & Company, New York.
33
