Econ 211 Spring 2016 Final Exam
Instructions: NO electronic aids of any type are permitted. Answer eight (8)
of the following ten (10) questions. All questions are of equal weight. Indicate
clearly on the first page of your examination which questions you want marked.
1. Let, X, the set for the following questions, be <n (note: n may exceed 2). Using the
mathematical notation developed in class, carefully define:
(a) the Euclidean distance function, d (a, b) , a ∈ X, b ∈ X;
(b) an epsilon neighbourhood of a ∈ X, N (a);
(c) let {ai }∞
i=1 ⊂ X; ‘a is the limit of ai as i goes to infinity’;
(d) the convex combination of a ∈ X and b ∈ X;
(e) the interior of any set A ⊂ X.
ANSWER (a) Given that X = <n , let a = (a1 , . . . , an ) and b = (b1 , . . . , bn ). Then
1/2
d (a, b) = (a1 − b1 )2 + · · · + (an − bn )2
,
where we take the positive root.
(b)
N (a) = {x ∈ X : d (a, x) < , > 0}
(c) It means that for any > 0, there is a natural number K which depends on such that
{an }∞
n=K ⊂ N (a) .
(d) The convex combination of a ∈ X and b ∈ X is {c ∈ X : c = ta + (1 − t)b, 0 ≤ t ≤ 1}.
(e) The interior of set A is {a ∈ A : for some > 0, N (a) ⊂ A}.
2. Answer both parts of this question.
(a) Let u : <2+ → < be the utility function for some consumer who always prefers more
to less of either good. Define what it means to say that u is “strictly concave”. Let x0 ∈ <2+ .
Use this utility function to define the upper contour set (or better set) of x0 . Prove that if u
1
is strictly concave the upper contour set of x0 must be strictly convex, that is, u is strictly
quasi-concave.
(b) Let
2 2
if x1 ≥ 0 and 0 ≤ x2 ≤ 2
x1 x2
3
u (x1 , x2 ) =
2 x2 +2
x1 3
if x1 ≥ 0 and x2 > 2
For each of the following, calculate the marginal utility of good 2, that is, u2 (x1 , x2 ), if it exists, or, if it doesn’t exist, explain why this partial derivative doesn’t exist.
(i) (x1 , x2 ) = (1, 1); (ii) (x1 , x2 ) = (2, 2); (iii) (x1 , x2 ) = (3, 3).
ANSWER
(a) Let a, b ∈ <2+ . u is strictly concave if for t ∈ (0, 1)
u (ta + (1 − t)b) > tu(a) + (1 − t)u(b).
A ⊂ <2+ is strictly convex if for all a, b ∈ A and all 0 < t < 1, ta + (1 − t)b is in the interior
of A. The upper contour set (or better set) of x0 , call this B (x0 ), for this function is
B x0 = {x ∈ <2+ : u(x) ≥ u x0 }
Let a and b be any two elements of B (x0 ). Given that u is strictly concave we know that
for t ∈ (0, 1)
u (ta + (1 − t)b) > tu(a) + (1 − t)u(b).
But a, b ∈ B (x0 ) so
tu(a) + (1 − t)u(b) ≥ tu(x0 ) + (1 − t)u(x0 ) = u(x0 ),
and therefore
u (ta + (1 − t)b) > tu(a)+(1−t)u(b) ≥ tu(x0 )+(1−t)u(x0 ) = u(x0 ) or u (ta + (1 − t)b) > u(x0 ).
The boundary of B (x0 ) is given by {x ∈ <2+ : u(x) = u (x0 )} so ta + (1 − t)b must lie in the
interior of the better set. Thus the better set must be strictly convex.
(b) (i) At (x1 , x2 ) = (1, 1), u2 (x1 , x2 ) = 2x2 x21 so u2 (1, 1) = 2.
(ii) u (2, 2) = 22 22 = 16. The limit of u (2, x2 ) as x2 approaches 2 from the right side is
22 = 4 times the limit of (x32 + 2)/3 as x2 approaches 2. This equals (4)(10/3) = 40/3, which
is different from 16. Thus the function is not continuous in x2 at x2 = 2 and thus u2 (2, 2)
does not exist.
(iii) At (x1 , x2 ) = (3, 3), u2 (x1 , x2 ) = x21 x22 so u2 (3, 3) = 81.
3. (a) Derive a formula for the sum of the first n terms of the following geometric series.
2
−
6000
500
1
300
4000
1
+
.
−
+
2
6
2
8
(1 + i)
(1 + i)
(1 + i) i
(1 + i) i
(c) Let 1/s = i/n or cross-multiplying n = is. Then
limn→∞
i
1+
n
n
is
1
= lims→∞ 1 +
s
s i
1
= lims→∞ 1 +
s
i
= e,
where
s
1
e ≡ lims→∞ 1 +
.
s
4. (a) Derive the ordinary demand functions (x1 (p1 , p2 , w) , x2 (p1 , p2 , w)) and the value
function V (p1 , p2 , w) for a price-taking consumer who maximizes the utility function
u (x1 , x2 ) = x1 x22 ,
x1 ≥ 0, x2 ≥ 0
subject to the budget constraint
p1 x1 + p2 x2 = w.
In your derivation use the method of Lagrange multipliers.
(b) Verify that the demand functions are homogeneous of degree zero in prices and wealth.
(c) Is
x1 (p1 , p2 , w) = −
∂V (p1 , p2 , w) /∂p1
?
∂V (p1 , p2 , w) /∂w
ANSWER
Form the Lagrangian
L (x1 , x2 , λ, p1 , p2 , w) = x1 x22 + λ (w − p1 x1 − p2 x2 ) .
The first-order conditions for an interior solution are
∂L
= x22 − λp1 = 0
∂x1
∂L
= 2x1 x2 − λp2 = 0
∂x2
∂L
= w − p1 x1 − p2 x2 = 0.
∂λ
4
These three equations can be solved for (x1 , x2 , λ) as functions of (p1 , p2 , w). In the first two
equations take the λ terms to the right-hand side and divide the first equation equation by
the second.
λp1
p1
x22
=
=
or
2x1 x2
λp2
p2
x2
p1
=
or
2x1
p2
p2 x2 = 2p1 x1 .
Substituting this into the third equation
p1 x1 + 2p1 x1 = w
3p1 x1 = w
w
x1 =
and using the budget constraint
3p1
2w
.
x2 =
3p2
Then the value function for this problem, which is usually called the indirect utility function,
is
V (p1 , p2 , w) ≡ u (x1 (p1 , p2 , w) , x2 (p1 , p2 , w))
2
2w
w
=
3p1
3p2
4 w3
=
.
27 p1 p22
(b) Verification that the demand functions are homogeneous of degree zero in prices and
wealth: for any α > 0
w
αw
=
= x1 (p1 , p2 , w)
3αp1
3p1
α2w
2w
x2 (αp1 , αp2 , αw) =
=
= x2 (p1 , p2 , w)
3αp2
3p2
x1 (αp1 , αp2 , αw) =
(c) By the envelope theorem this must be true. Checking
5
∂V (p1 , p2 , w)
4 w3
= −
∂p1
27 p21 p22
∂V (p1 , p2 , w)
4 3w2
=
∂w
27 p1 p22
∂V (p1 , p2 , w) /∂p1
4 w3 27 p1 p22
−
=
∂V (p1 , p2 , w) /∂w
27 p21 p22 4 3w2
w
= x1 (p1 , p2 , w) .
=
3p1
5. (a) Derive Hicksian demand functions (h1 (p1 , p2 , u0 ) , h2 (p1 , p2 , u0 )) and the expenditure (or cost) function e (p1 , p2 , u0 ) for a price-taking consumer whose utility function is
u (h1 , h2 ) = h1 h22 ,
h1 ≥ 0, h2 ≥ 0.
In your derivation use the method of Lagrange multipliers.
(b) Verify that the Hicksian demand functions are homogeneous of degree zero in prices.
(c) Is
h1 (p1 , p2 , u0 ) =
∂e (p1 , p2 , u0 )
?
∂p1
ANSWER
Form the Lagrangian
L (h1 , h2 , µ, p1 , p2 , u0 ) = p1 h1 + p2 h2 + µ u0 − h1 h22 .
The first-order conditions for an interior solution are
∂L
= p1 − µh22 = 0
∂h1
∂L
= p2 − 2µh1 h2 = 0
∂h2
∂L
= u0 − h1 h22 = 0.
∂λ
These three equations can be solved for (h1 , h2 , µ) as functions of (p1 , p2 , u0 ). In the first two
equations take the µ terms to the right-hand side and divide the first equation equation by
the second.
h2
p1
=
or
p2
2h1
2p1 h1
h2 =
.
p2
6
Substituting this into the third equation and simplifying
h1
2
2p1 h1
= u0
p2
4p2
h31 21 = u0
p2
−2/3 2/3 1/3
−2/3 2/3 1/3
p 2 u0
h1 = 4−1/3 p1 p2 u0 = 2−2/3 p1
2p1 h1
1/3 −1/3 1/3
= 21/3 p1 p2 u0
h2 =
p2
and then
Then the value function for this problem, which is the expenditure or cost function, is
e (p1 , p2 , w) ≡ p1 h1 + p2 h2
1/3 2/3 1/3
1/3 2/3 1/3
= 2−2/3 p1 p2 u0 + 21/3 p1 p2 u0
1/3 2/3 1/3
= 2−2/3 p1 p2 u0 (1 + 2)
1/3
27 2
=
p1 p2 u 0
4
(b) Check that the demand functions are homogeneous of degree zero in prices. For any
α>0
2/3
2/3
p2
αp2
1/3
1/3
−2/3
u0 = 2
u0 = h1 (p1 , p2 , u0 )
h1 (αp1 , αp2 , u0 ) = 2
αp1
p1
1/3
1/3
αp1
p1
1/3
1/3
h2 (αp1 , αp2 , u0 ) = 21/3
u0 = 21/3
u0 = h2 (p1 , p2 , u0 )
αp2
p2
−2/3
(c) This must be true by the envelope theorem. Checking
1/3 1/3
27 2
27 2
1/3
e (p1 , p2 , w) =
p1 p2 u0
=
p2 u0
p1
4
4
1/3
∂e (p1 , p2 , u0 )
1 27 2
−2/3
=
p u0
p1
= h1 (αp1 , αp2 , u0 ) .
∂p1
3 4 2
6. Consider a price-taking, cost-minimizing firm in a one-output, two-input world. Write
the production function as q = f (z1 , z2 ) and denote the prices of inputs by w1 , w2 . Assume
the production function is strictly quasi-concave and differentiable, and assume an interior
solution for optimal input use.
(a) Describe how one would obtain input demand functions zic (w1 , w2 , q) , i = 1, 2, and
show these functions must be homogeneous of degree zero in (w1 , w2 ).
7
(b) Using the cost function c (w1 , w2 , q) and the envelope theorem sketch a proof that
∂zic (w1 , w2 , q)
≤ 0, for i = 1, 2
∂wi
ANSWER
(a) In this context the cost-minimizing firm will minimize total cost, which equals w1 z1 +
w2 z2 , subject to producing q = f (z1 , z2 ) units of output. Assuming the production function
is strictly quasi-concave and differentiable, and assuming an interior solution for optimal
input use, optimal input demands may be obtained by setting the derivatives of the following
Lagrangian expression with respect to z1 , z2 , λ equal to zero and solving for these variables.
w1 z1 + w2 z2 + λ (q − f (z1 , z2 ))
We obtain
∂f (z1 , z2 )
∂f (z1 , z2 )
= 0 or w1 = λ
∂z1
∂z1
∂f (z1 , z2 )
∂f (z1 , z2 )
= 0 or w2 = λ
w2 − λ
∂z2
∂z2
q − f (z1 , z2 ) = 0.
w1 − λ
The ratio of the first two equations together with the third give us the following two
equations to solve for zic (w1 , w2 , q) , i = 1, 2.
w1
∂f (z1 , z2 )
=
w2
∂z1
q = f (z1 , z2 ) .
∂f (z1 , z2 )
∂z2
Note that the zi s that solve these equations would be the same if (w1 , w2 ) were scaled by some
positive number α. So zic (w1 , w2 , q) , i = 1, 2 are homogeneous of degree zero in (w1 , w2 ).
(b) The definition of the cost function is
C (w1 , w2 , q) =
Opt
w z + w2 z2 + λ (q − f (z1 , z2 )) .
z1 , z2 , λ 1 1
Since this function must be concave in (w1 , w2 ) the Hessian of the cost function must be negative semi-definite, which, among other things, means its main diagonal must be nonpositive.
Applying the envelope theorem to the definition of C (w1 , w2 , q) we obtain
∂ 2 C (w1 , w2 , q)
∂ ∂C (w1 , w2 , q)
∂zic (w1 , w2 , q)
≡
=
≤ 0 for i = 1, 2.
∂wi2
∂wi
∂wi
∂wi
8
7.


−1 2 1
A =  1 −1 2 
1
0 2
 
2
B= 3 
5


2 2 1
C =  3 −1 2 
5 0 2


−1 1 1
D= 2 3 5 
1 2 2
Calculate: (i) A−1 , (ii) A−1 A, (iii) A−1 B, (iv) det (C) / det (A), and (v) det (D) / det (A).
ANSWER
For any n by n matrix Z, if det(Z) 6= 0, the inverse of Z is the transpose of the matrix
of cofactors of Z divided by det(Z). For matrix A,


−2 0
1
(cij ) =  −4 −3 2  ,
5
3 −1
and expanding along the third row of A, det(A) = (1)(5) + (2)(−1) = 3. So


−2 −4 5
1
A−1 =  0 −3 3  .
3
1
2 −1
Checking




 

−2 −4 5
−1 2 1
3 0 0
1 0 0
1
1
A−1 A =  0 −3 3   1 −1 2  =  0 3 0  =  0 1 0  .
3
3
1
2 −1
1
0 2
0 0 3
0 0 1
For (iii)

   
 
−2 −4 5
2
9
3
1
1   
−1



0 −3 3
3
6
2
A B=
=
=
3
3
1
2 −1
5
3
1
(iv) Note that C is A with the first column replaced by B. Therefore Cramer’s Rule tells
us det (C) / det (A) must the first element of the column vector answer for (iii), that is, 3.
9
(v) Note D> is A with the second column replaced by B. Using the fact that the determinant of a matrix equals the determinant of its transpose and Cramer’s Rule det (D) /det (A)
must the second element of the column vector answer for (iii), that is, 2.
8. (a) State whether the following matrices are positive definite, positive semi-definite,
negative definite, negative semi-definite, or indefinite, and justify your answers carefully.






1 3 0
1 0 1
1 0 1
(i)  1 4 2  (ii)  0 1 1  (iii)  0 1 1 
0 2 3
1 1 2
1 0 2
(b) Consider the following utility function:
u (x1 , x2 ) = xa1 xb2 , a > 0, b > 0, x1 > 0, x2 > 0.
For what values of a and b is u (x1 , x2 ) strictly concave? Justify your answer carefully.
ANSWER
(a) (i) The first step is to make the matrix symmetric about the main diagonal by
replacing the off-diagonal elements by (aij + aji )/2. This yields


1 2 0
 2 4 2 
0 2 3
With a positive main diagonal this matrix is a candidate for positive definite or positive
semi-definite or indefinite. But, using the first column, the determinant of the whole matrix
is (1)(8) + 2(−6) = −4. This rules out positive definite or positive semi-definite and so the
matrix is indefinite.
(ii) Expanding down the first column the determinant of this three by three matrix is
1(1) + 1(−1) = 0. So the matrix cannot be positive definite but with positive elements on
its main diagonal it could be positive semi-definite. We must check aii ajj − a2ij for all i not
equal to j. Checking we see that
a11 a22 − a212 = 1
a11 a33 − a213 = 1
a22 a33 − a223 = 1,
so yes it is positive semi-definite.
(iii) This matrix is not symmetric but we know its “definiteness” is the same as that of
the symmetric matrix B formed by averaging A’s off-diagonal elements. Thus


1 0
1
B =  0 1 1/2 
1 1/2 2
10
Expanding down the first column the determinant of B is (1)(7/4) + (1)(−1) = 3/4 > 0.
Since the other two leading principal minors are positive (both equal 1), the matrix is positive
definite.
(b) The matrix of second-order partial derivatives of this function, which is called the
Hessian of u (x1 , x2 ), can be written as
u11 (x1 , x2 ) u12 (x1 , x2 )
H=
.
u21 (x1 , x2 ) u22 (x1 , x2 )
Calculating the first and second-order partial derivatives yields
au
x1
bu
=
x2
b
u1 = axa−1
1 x2 =
u2 = bxa1 xb−1
2
b
u11 = a(a − 1)xa−2
1 x2 =
a(a − 1)u
x21
abu
= u21
x1 x2
b(b − 1)u
= b(b − 1)xa1 xb−2
=
2
x22
b−1
u12 = abxa−1
=
1 x2
u22
H is negative definite if and only if u11 < 0 and u11 u22 − u212 > 0. The first condition requires
a − 1 < 0. The second condition requires
u11 u22 −
u212
a(a − 1)u b(b − 1)u
−
=
x21
x22
abu
x1 x2
2
=
abu2
((a − 1)(b − 1) − ab) > 0.
x21 x22
This is equivalent to a + b < 1 which, since a and b are positive numbers, implies the first
condition that a < 1. This argument shows that H is negative definite (or u is strictly
concave) for x1 , x2 > 0 if and only if a + b < 1.
9. Let


a11 a12 a13
A =  a21 a22 a23 
a31 a32 a33
(a) Define the determinant of A, det (A), by expanding down the first column of A. Now
let B = A except that in B the second and third columns of A are interchanged. Write out
det (B) by expanding down the first column of B. What is the relation between det (A) and
det (B)?
(b) Define the quadratic form of A. What does it mean to say that A is negative definite?
Under what conditions will A be negative definite?
11
ANSWER
(a) Let Z = [zij ] be any n by n matrix. Let [cij ] be the matrix of cofactors of Z where
cij is (−1)i+j times the determinant of the matrix obtained by deleting the ith row and j
column of Z. The determinant of Z may be obtained by expanding along any row or any
column of Z, that is,
det(Z) =
j=n
X
zij cij for i = 1, . . . , n =
j=1
i=n
X
zij cij for j = 1, . . . , n.
i=1
Setting Z = A and j = 3 we have
det(A) = a11 (a22 a33 − a23 a32 ) + a21 (−a12 a33 + a13 a32 ) + a31 (a12 a23 − a13 a22 ) .
Now

a11 a13 a12
B =  a21 a23 a22 
a31 a33 a32

Setting Z = B and j = 3 we have
det(B) = a11 (a23 a32 − a22 a33 ) + a21 (−a13 a32 + a12 a33 ) + a31 (a13 a22 − a12 a23 ) ,
from which it is clear that det(B) = − det(A).
(b) The quadratic form of matrix A which is 3 by 3 is
QF ≡ x> Ax,
where x is any 3 by 1 column vector of real numbers. A is negative definite if QF < 0 for all
x where at least one element of x is nonzero. In case A is not symmetric use B = (A + A> )/2
which must be symmetric. Necessary and sufficient conditions for A to be negative definite
are that the leading principal minors of B alternate in sign with the first negative, that is,
b11 < 0, b11 b22 − b212 > 0, det(B) < 0.
10. Consider a profit-maximizing monopolist who faces the following inverse demand
curve
p = a − bY, p ≥ 0, Y ≥ 0, a > 0, b > 0,
and whose total cost function is
C (Y ) = cY, a > c > 0.
Assume this monopolist sells every unit of output at the same price so that its price is equal
to average revenue.
12
(a) Prove marginal revenue equals a − 2bY , that is, marginal revenue has the same priceaxis intercept as average revenue, and the slope of marginal revenue is twice the slope of
average revenue.
(b) Derive functions for profit-maximizing output, price and profits as functions of the
parameters a, b, c; that is, Y (a, b, c), p(a, b, c), π(a, b, c).
(c) Why is the following relationship true?
∂π (a, b, c)
= −Y (a, b, c)
∂c
ANSWER
(a) Observe that
p = average revenue = a − bY
total revenue = pY = aY − bY 2
d(total revenue)
= a − 2bY
marginal revenue ≡
dY
(b)
Profits ≡
=
=
=
dProfits
=
dY
d2 Profits
=
dY 2
Revenue - Cost
pY − C (Y )
(a − bY ) Y − cY
(a − c) Y − bY 2
a − c − 2bY
−2b < 0.
Since the second-order conditions for a maximum are met for any Y > 0 set the first-order
condition to zero to get optimal output.
dProfits
= a − c − 2bY = 0
dY
a−c
Y (a, b, c) =
> 0 because we assumed a > c
2b
a−c
a+c
p(a, b, c) = a − bY ∗ = a − b
=
2b
2
(a − c)2
π(a, b, c) = (a − c) Y ∗ − b (Y ∗ )2 =
4b
13
(c) By definition
π(a, b, c) =
Max
Max
Profits =
(a − c) Y − bY 2
Y
Y
Applying the envelope theorem
∂π (a, b, c)
= −Y (a, b, c) .
∂c
14