General Equilibrium Models1
Matthew Hoelle
www.matthew-hoelle.com
Spring 2014
1 This
manuscript is dedicated to Dave Cass, both for his immeasurable contributions to general
equilibrium theory and for his inspirational work in graduate education and advising.
ii
Contents
1 Mathematical Prerequisites
1
1.1 Farkas Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1.1
Hyperplane Theorems . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1.2
Farkas Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.1.3
Application: Duality . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2 Kuhn-Tucker Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.2.1
De…nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.2.2
Constraint Quali…cation . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.2.3
Kuhn-Tucker Theorem . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.3 Correspondences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
1.3.1
De…nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
1.3.2
Berge’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
1.3.3
Kakutani’s Fixed Point Theorem . . . . . . . . . . . . . . . . . . . .
14
1.3.4
Application: Dynamic Programming . . . . . . . . . . . . . . . . . .
14
1.4 Di¤erential Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
1.4.1
Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
1.4.2
Properness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
1.4.3
Regular vs. Critical values . . . . . . . . . . . . . . . . . . . . . . . .
18
1.4.4
Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
1.4.5
Finite Local Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . .
20
1.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
2 Arrow-Debreu Model
25
2.1 The Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
2.2 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
2.3 First Basic Welfare Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
iii
iv
CONTENTS
2.4 Second Basic Welfare Theorem . . . . . . . . . . . . . . . . . . . . . . . . .
31
2.5 Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
2.6 Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
2.6.1
Proof of Theorem 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
2.6.2
Proof of Theorem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
2.6.3
Proof of Theorem 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
2.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
3 General Financial Model
47
3.1 The Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
3.1.1
Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
3.1.2
Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
3.1.3
No Arbitrage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
3.1.4
No Redundancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58
3.2 Complete Markets
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
3.3 Incomplete Markets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
3.3.1
Pareto suboptimal . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
3.3.2
Constrained Pareto suboptimal . . . . . . . . . . . . . . . . . . . . .
64
3.4 Real Assets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
3.4.1
Nonexistence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
3.4.2
Generic existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
70
3.5 Proof of Theorem 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71
h T
6= 0 8h 2 H . . . . . . . . . . . . . . . . . . . . . . .
74
3.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
79
3.5.1
3.5.2
Case I:
x
Case II:
x
h T
= 0 for some h 2 H . . . . . . . . . . . . . . . . . .
4 Incomplete Markets and Money
4.1 The Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1
Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Complete Markets
78
85
86
88
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
4.3 Incomplete Markets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
A Solutions to the Exercises
A.1 Mathematical Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
97
CONTENTS
v
A.2 Arrow-Debreu Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
A.3 General Financial Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
A.4 Incomplete Markets and Money . . . . . . . . . . . . . . . . . . . . . . . . . 122
vi
CONTENTS
Preface
These notes have been prepared to serve as the theoretical foundation for graduate courses
in theoretical economics. These notes have previously been used as the preliminary material
for a 2nd year Ph.D. course on monetary theory (Purdue University, Spring 2012). They
are perhaps best suited for a reference for any theorist wanting a quick refresher in general
equilibrium theory.
The models of general equilibrium have been adopted by macroeconomics (Real Business
Cycle school of macroeconomics), …nance, and other …elds. Given the importance of these
models, it is unfortunate that a concise (and precise) source does not exist to summarize
the predictions of properties of general equilibrium theory. That is, such a source does not
exist until now. The present manuscript introduces the two fundamental general equilibrium
models (the static Arrow-Debreu Model and the dynamic General Financial Model).
The material is a compilation of (i) the course material from the general equilibrium
PhD course taught by Dave Cass, (ii) the major results in the …eld of general equilibrium in
the past 30 years (yes, we’ve advanced beyond an Arrow-Debreu or Walrasian equilibrium
setting), and (iii) the insights from the author about the use of these models in applications.
This manuscript is organized as follows. Chapter 1 contains the mathematical prerequisites for the course. Students are presumed to have a su¢ cient grasp of proof-based
mathematics (the logic of a mathematical proof) and a working knowledge of real analysis
and matrix algebra. To check if you are equipped to handle the material in this manuscript,
verify that you can complete each of the following tasks:
De…ne what it means for a set to be convex.
De…ne what it means for a set to be open (and closed).
De…ne what it means for a set to be bounded.
vii
viii
CONTENTS
In Euclidean space, does the de…nition of compactness imply closed and bounded?
Vice versa?
De…ne what it means for a function to be concave (and strictly concave).
De…ne what it means for a function to be continuous.
State and prove the Extreme Value Theorem.
De…ne what it means for a function to be di¤erentiable.
State and prove the Mean Value Theorem.
For a refresher on any of these mathematical tasks, I encourage you to review the notes
on real analysis on my teaching website: http://www.matthew-hoelle.com/teaching.html.
Chapter 2 introduces the Arrow-Debreu model, the static general equilibrium model
of pure exchange where the markets are anonymous and perfectly competitive. This is
the canonical model of an economic market. I demonstrate that an equilibrium of the
model satis…es three properties: existence, optimality, and regularity. Chapter 3 introduces
the general …nancial model, an extension of the Arrow-Debreu model to a dynamic and
stochastic framework. Facing uncertainty in the future periods, households are permitted to
trade …nancial assets. I analyze how the predictions of the model compare under the settings
of complete …nancial markets and incomplete …nancial markets.
Chapter 4 adds money to the general …nancial model in Chapter 3, by specifying that the
assets pay out in the unit of account (e.g., the currency of a country). The exogenous money
supply pins down the price level, using the Quantity Theory of Money. I show that with
complete …nancial markets, monetary policy is neutral, whereas with incomplete …nancial
markets, monetary policy may induce a change in the real allocation.
Each chapter contains its own reference list if readers wish to review a particular topic in
greater depth by reading the primary sources. Any proofs that are too long or complicated
to justify being placed in the heart of a chapter are placed in the penultimate section of a
chapter. The longer proofs are included to provide a complete account of the theory, but (as
they are quite technical in nature) can typically be skipped by all but the most courageous
readers.
In the ultimate section of each chapter, I have included a handful of exercises. These
exercises not only review the main theoretical contributions, but also o¤er important appli-
CONTENTS
ix
cations of that theory. Solutions to the exercises can be found in the Appendix. Use this
resource as a complement to learning, instead of a replacement.
Notation The following is a list of mathematical terminology and notation utilized throughout this manuscript:
8 is the universal quanti…er; it is the symbol "for all"
9 is the existential quanti…er; it is the symbol "there exists"
9! is the unique existence quanti…er; it is the symbol "there exists a unique"
’i¤’refers to ’if and only if’; it is used when two statements are necessary and su¢ cient
for each other
’wlog’refers to ’without loss of generality’
Unless otherwise indicated, x 2 Rn is a column vector. Its tranpose xT is a row vector.
The value kxk denotes the Euclidean norm of the vector x 2 Rn :
The space Rm;n refers to the space of m
For x; y 2 Rn ; x
y i¤ xi
For x; y 2 Rn ; x > y i¤ x
n real matrices.
yi for i = 1; :::; n:
y and x 6= y:
For x; y 2 Rn ; x >> y i¤ xi > yi for i = 1; :::; n:
The set AnB = fx 2 A : x 2
= Bg :
The set N (x) is the open neighborhood of radius
around x:
The set N (x) is the deleted open neighborhood of radius around x; that is, N (x) =
N (x) nfxg:
f is C 0 if it is continuous and is C n (for any n 2 N) if its nth derivative exists and is
continuous.
The set intA denotes the interior of A :
intA = fx 2 A : 9 > 0 such that N (x)
Ag :
x
CONTENTS
The set A0 denotes the set of limit points of A :
A0 = fx : 8 > 0; N (x) \ A 6= ;g :
The set clA denotes the closure of A :
clA = A [ A0 :
n 1
is the (n
1) dimensional simplex de…ned as
n 1
= fx 2 Rn+ :
Pn
i=1
xi = 1g:
Chapter 1
Mathematical Prerequisites
The topics to be covered in this chapter are (1) Farkas Lemma (Section 1.1) with Duality as
an important application, (2) Kuhn-Tucker theorem (Section 1.2), an important economic
result that uses Farkas Lemma in its proof, (3) correspondences (Section 1.3) as used to prove
equilibrium existence, and (4) di¤erential topology (Section 1.4) as used to prove equilibrium
regularity.
1.1
Farkas Lemma
Before stating the Farkas Lemma, I …rst state and prove the Separating Hyperplane Theorem.
As a corollary of this, I state the Supporting Hyperplane Theorem. Interestingly enough, the
Supporting Hyperplane Theorem is used in the proof of the Second Basic Welfare Theorem
(Section 2.4). Finally, I state the Farkas Lemma and then apply it to the problem of Duality.
1.1.1
Hyperplane Theorems
The following theorem, the Separating Hyperplane Theorem, is used to prove (a) Supporting
Hyperplane Theorem and (b) Farkas Lemma.
Theorem 1.1 Separating Hyperplane Theorem
Suppose Z Rn is nonempty, closed, and convex, and that y 2 Rn ; but y 2
= Z:
n
T
T
Then there exists q 2 R nf0g such that q y > q z 8z 2 Z:
The Separating Hyperplane Theorem can be depicted in Figure 1.1 in the separate ’Figures’document. In Figure 1.1, the boundary of the set is a solid line, indicating that the set
1
2
CHAPTER 1. MATHEMATICAL PREREQUISITES
Z is closed.
Proof. Pick any z 0 2 Z and de…ne Z 0 = fz 2 Z : kz yk kz 0 ykg : The set Z 0 is compact
(as it is closed and bounded). The function kz yk is a continuous function of the variable
z:
Applying the Extreme Value Theorem, 9z = arg min kz
then kz
yk > 0: Thus kz
yk : For any z 2 Z; as y 2
= Z;
z2Z 0
yk > 0:
Take any z 2 Z and de…ne the real-valued function g : [0; 1] ! R as follows:
g ( ) = ((1
)z + z
y)T ((1
)z + z
y) :
The function is minimized at 0; namely g ( )
g (0) 8 2 [0; 1] : Using the de…nition of
derivative, Dg (0) 0: We can rewrite the function
g ( ) = ( (z
z )+z
y)T ( (z
z )+z
y) :
Taking the derivative yields
Dg (0) = (z
Using algebra, (y
z )T (z
z )
0 or (y
y)T (z
z )
0:
z )T z
(y
z )T z
De…ne the separating hyperplane as q = (y
T
8z 2 Z:
z ) : Therefore, q z
q T z 8z 2 Z:
As kz
yk2 > 0; then (z
y)T (z
y) > 0 or (y z )T (y z ) > 0: This implies
q T y > q T z : From above, q T y > q T z
q T z 8z 2 Z: This …nishes the argument.
As a straightforward corollary of the Separating Hyperplane Theorem, we can now state
the Supporting Hyperplane Theorem. As previously mentioned, this theorem is used in the
proof of the Second Basic Welfare Theorem (Section 2.4).
Corollary 1.1 Supporting Hyperplane Theorem
Suppose Z
Rn is nonempty and convex, and that y 2 Z; but y 2
= intZ:
Then there exists a q 2 Rn nf0g such that q T y
q T z 8z 2 Z:
The Supporting Hyperplane Theorem can be depicted in Figure 1.2. The boundary of
the set is a dashed line, indicating that the set Z need not be closed.
1.1. FARKAS LEMMA
1.1.2
3
Farkas Lemma
We are now prepared to state the Farkas Lemma (which as a result of its importance in
economics has been promoted to a ’Theorem’).
Theorem 1.2 Farkas Lemma
Let ai 2 Rn nf0g for i = 1; :::; m; matrix A = [a1 :::am ] 2 Rn;m ; and
Z = z 2 Rn : z = A for some
2 Rm
+ :
For z 2 Rn ; exactly one of the following conditions holds:
(i) z 2 Z:
(ii) 9q 2 Rn nf0g such that q T z > 0
q T z 8z 2 Z:
The theorem is illustrated in the two-panel Figure 1.3. In the left panel labeled Farkas
(i), the element z 2 Z: In the right panel labeled Farkas (ii), z 2
= Z and 9q 2 Rn nf0g such
that q T z > 0 q T z 8z 2 Z:
1.1.3
Application: Duality
As an application of Farkas Lemma, let us state and prove the Duality Theorem. The
Duality Theorem in this manuscript will be stated in terms of a linear objective function
and linear constraints (I will refer to this as the "Linear Duality Theorem"). In general, a
Duality Theorem shows the equivalence between a maximization problem and the related
minimization problem (which is called the "dual" of the maximization problem). In order
to obtain a Duality Theorem for nonlinear objective functions and nonlinear constraint
functions, we must take derivatives of all nonlinear functions. Recall that a derivative is a
linear mapping. Thus, a Duality Theorem for an objective function and constraint functions
that are di¤erentiable and concave (not necessarily linear) can be equivalently represented
in the form of a Linear Duality Theorem (after taking derivatives).
We begin by writing down the maximization problem:
maximize
x 0
xT
subject to xT A
T
:
4
CHAPTER 1. MATHEMATICAL PREREQUISITES
The dual minimization problem is then given by:
yT
subject to Ay
minimize
y 0
:
De…nition 1.1 A linear program is feasible if there exists a vector satisfying the constraints.
De…nition 1.2 A feasible vector is an optimal vector if it maximizes or minimizes the linear
form. The value of this max or min is called the value.
Theorem 1.3 Duality
If both the maximization problem and its dual are feasible, then both have optimal vectors
and the values of the two are the same.
The following Lemma is used in the proof of Theorem 1.3.
Lemma 1.1 To prove Theorem 1.3, it su¢ ces to …nd (x; y)
three equations:
T
xT A
:
Ay
xT
(1.1)
(1.2)
:
yT
0 that satis…es the following
(1.3)
0:
T
Proof. From (1.1), xT Ay
y: From (1.2), xT Ay
xT : The previous two inequalities
T
together imply xT
y: If (1.3) is satis…ed, then xT = T y:
Suppose, for contradiction, that (x; y) are both feasible, but at least one is not optimal,
wlog x: Then there exists feasible z such that
zT
zT
> xT :
T
y:
T
Using the equality xT = T y; then z T
y = xT ; which is a contradiction.
We are now prepared to prove the Duality Theorem.
Proof. Suppose, for contradiction, there does not exist (x; y) 0 satisfying (1.1), (1.2), and
(1.3). The equations can be rewritten as:
= AT x + Iv
= Ay Iw
T
0 = Tx
y
for v
for w
u for u
0:
0:
0:
1.1. FARKAS LEMMA
5
0
B
B
B
=B
B
B
@
Thus, there does not exist
0
1
x
v
y
w
u
1
C
C
C
C
C
C
A
2
AT I
B C 6
@ A=4 0 0
T
0
0
0 such that
0
A
T
0
3B
0 B
7B
0 5B
B
B
1 @
0
I
0
x
v
y
w
u
1
C
C
C
C:
C
C
A
Let’s apply the Farkas Lemma with the following relations between the statement of the
Farkas Lemma and this proof:
Farkas Lemma
This proof
q
()
z
()
A
()
()
0
2 T
A I
6
40 0
T
0
q
1
B
@
0
0
A
C
A
T
0
B
B
B
B
B
B
@
x
v
y
w
u
1
C
C
C
C
C
C
A
0
I
0
3
0
7
05 :
1
6
CHAPTER 1. MATHEMATICAL PREREQUISITES
De…ne
Z=
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
z
2 T
A I
6
:z=4 0 0
T
0
0
A
T
0
3B
0 B
7B
0 5B
B
B
1 @
0
I
0
x
v
y
w
u
1
0
C
B
C
B
C
B
C for some B
C
B
C
B
A
@
0
B
The implications of our proof by contradiction is that @
1
0
x
v
y
w
u
1
C
C
C
C
C
C
A
9
>
>
>
>
>
>
=
0 :
>
>
>
>
>
>
;
C
= Z: This says that part (i) of
A2
0
1
q1
B
C
6 0
the Farkas Lemma does not hold, meaning that part (ii) must. Thus, there is q = @ q2 A =
q3
0 1
such that q
written as:
T
B
@
0
C
A>0
q T z 8z 2 Z: The strict inequality on the left-hand side can be
q1T + q2T > 0:
2 T
3
A I
0
0
0
6
7
As q T z
0 8z 2 Z; then q1T ; q2T ; q3 4 0 0 A
I 05
T
T
0
0
1
!
! !
1
0
determined by the unit vectors from = !
to =
):
0
1
q1T AT + q3
q1T
q2T A
T
0:
T
0 (letting z 2 Z be
(1.5)
(1.6)
0:
q3
(1.4)
0:
(1.7)
q2T
0:
(1.8)
q3
0:
(1.9)
Two steps remain. The …rst is to show that q3 > 0: If not, q3 = 0; then (1.5) and (1.1)
imply q1T
0 (recall that q1T
0). Additionally, from (1.7) and (1.2), q2T
0: The last
1.2. KUHN-TUCKER THEOREM
7
two inequalities contradict (1.4).
Now with q3 > 0; from (1.5):
q1T
q3
AT
q2
q3
T
q2
q3
(1.10)
:
From (1.7):
q2T
q3
A
q1
q3
T
q1
q3
:
(1.11)
After taking the transpose of (1.11) and adding it to (1.10), I obtain T qq23 + T qq13
0;
which is a contradiction of (1.4). This is a contradiction, allowing us to conclude that there
must exist (x; y) 0 satisfying (1.1), (1.2), and (1.3).
1.2
1.2.1
Kuhn-Tucker Theorem
De…nitions
Prior to the statement of the theorem, I introduce two de…nitions. The …rst, of a concave
function, should be familiar to the students. The second may be new. The convention
throughout the entire manuscript is that for a function f : Rn ! R; the values of the
derivative mapping Df : Rn ! R are row vectors. Thus, Df (x) is a 1 n row vector.
Extending this idea, for g : Rn ! Rm ; the values of the derivative mapping are m n
Jacobian matrices. Thus, Dg (x) is a m n matrix.
De…nition 1.3 Given X
Df (x ) (x x ) + f (x )
Rn is convex, a di¤erentiable function f : X ! R is concave i¤
f (x):
De…nition 1.4 Given X
Rn is convex, a di¤erentiable function f : X ! R is quasiconcave i¤ f (x) f (x ) 0 =) Df (x ) (x x ) 0:
A concave function is quasi-concave, but not vice-versa. Figure 1.4 provides an example
of a function that is quasi-concave, but not concave.
The following equivalent de…nitions of concave and quasi-concave may prove useful in
the sequel and highlight why concavity is a stronger assumption.
De…nition 1.5 A function f is concave i¤ 8x; y 2 X and 8 2 [0; 1] ;
f ( x + (1
) y)
f (x) + (1
) f (y) :
8
CHAPTER 1. MATHEMATICAL PREREQUISITES
De…nition 1.6 A function f is quasi-concave i¤ 8x; y 2 X and 8 2 [0; 1] ;
f ( x + (1
1.2.2
) y)
min ff (x) ; f (y)g :
Constraint Quali…cation
De…ne the programming problem (P ) as:
maximize
(P )
with multipliers
f (x)
:
subject to gj (x)
0 j = 1; :::; m
j
The Lagrange multipliers
= ( ; :::; ) are taken to be a row vector. The vector-valued
0 11 m
g1
B
C
function g is de…ned as g = @ : A : To state the Kuhn-Tucker Theorem, we must assume
gm
that the so-called "Constraint Quali…cation" holds. Examples of the Constraint Quali…cation
are:
The constraint functions gj are linear functions of x:
If x is an optimal solution to (P ) ; then rankDg (x ) = m ; where m is the number
of constraints that bind at the optimal solution (wlog, gj (x ) = 0 for j = 1; :::; m ).
Basically, the Constraint Quali…cation is any su¢ cient condition for the following statement:
x is an optimal solution to (P ) =) x is an optimal solution to (LP ) ;
where the linear programming problem (LP ) is:
(LP )
maximize
with multipliers
Df (x )x
;
subject to Dgj (x ) (x
x)
0 j = 1; :::; m
j
and as above, the …rst m constraints bind at the optimal solution x :
Proof. Assume that the …rst suggested Constraint Quali…cation holds. The following argument shows that if x is an optimal solution to (P ); then x is an optimal solution to (LP ):
Similar methods can be used to show that the second suggested Constraint Quali…cation
yields the same implication.
1.2. KUHN-TUCKER THEOREM
9
The proof only requires the additional assumption that X is a convex set.
Suppose that x is not an optimal solution to (LP ): Then 9y 2 X such that Df (x )y >
Df (x )x and Dgj (x ) (y x ) 0 for j = 1; :::; m : Since the functions gj are linear, the
dot products Dgj (x )y = gj (y) and Dgj (x )x = gj (x ) (the derivative vector Dgj (x ) is
independent of x ). This implies that gj (y) gj (x )
0 or simply that gj (y)
0 since
gj (x ) = 0 for j = 1; :::; m ; by de…nition.
De…ne z = y + (1
)x for any 2 (0; 1) : Since X is convex, z 2 X: Since gj (y)
gj (x ) = 0 for j = 1; :::; m ; and gj are linear functions, then for any 2 (0; 1) ; gj (z )
for j = 1; :::; m :
0;
0
By de…nition, gj (x ) > 0 for j = m + 1; ::::; m: There exists values of near 0 such that
gj (z ) > 0 for j = m + 1; ::::; m: All told, for values of near 0; gj (z ) 0 for j = 1; :::; m:
Observe that if Df (x ) z
derivative,
f z
lim
!0
x
> 0 for any z 2 Rn ; then by the de…nition of the
f (x )
jz
Df (x ) z
xj
x
= 0:
Since the supposition is that Df (x )y > Df (x )x ; then for any 2 (0; 1) ; Df (x ) z
x >
0: This implies that for values of near 0; f z
f (x ) > 0; or f z > f (x ) : Combined
with the facts that for values of near 0; z 2 X and gj (z ) 0 for j = 1; :::; m; the vector
x is not an optimal solution to (P ): This …nishes the proof by contraposition.
1.2.3
Kuhn-Tucker Theorem
The Kuhn-Tucker Theorem can now be stated.
Theorem 1.4 Given X Rn is convex, assume that f : X ! R is di¤erentiable and concave. Further assume that gj : X ! R are di¤erentiable and quasi-concave 8j = 1; :::; m; and
the Constraint Quali…cation is satis…ed. Then x is an optimal solution to the programming
problem (P ) i¤ x satis…es the Kuhn-Tucker conditions:
First Order Conditions
Df (x ) + Dg(x ) = 01
n:
Complimentary Slackness Conditions
g(x ) = 01 1 ; with
0 and g (x)
0:
10
CHAPTER 1. MATHEMATICAL PREREQUISITES
The proof proceeds in two steps: Su¢ ciency and Necessity. The easier of the two steps
is to show that the Kuhn-Tucker conditions are su¢ cient for an optimal solution to (P ) ; so
we begin there.
Proof. Su¢ ciency of Kuhn-Tucker conditions
We utilize a proof by contradiction. Suppose that x satis…es the Kuhn-Tucker conditions,
but is not an optimal solution to (P ) : Then 9y such that gj (y) 0 8j = 1; :::; m and f (y) >
f (x ) : Suppose that the …rst m constraints bind at x (wlog). Then gj (y) gj (x ) = 0
8j = 1; :::; m : Using the quasi-concavity of gj ; then Dgj (x ) (y x )
0 8j = 1; :::; m :
From the concavity of f; Df (x ) (y x ) > 0: Then for any values 1 ; :::; m
0; the
equation
m
X
Df (x ) (y x ) +
x ) > 0:
j Dgj (x ) (y
j=1
As the constraints gj (x ) > 0 for j > m ; the Complimentary Slackness Conditions imply
j = 0 for j > m : Thus, we have (in vector notation):
(Df (x ) + Dg(x )) (y
x ) > 0:
This contradicts the First Order Conditions: Df (x ) + Dg(x ) = 0:
To show that the Kuhn-Tucker conditions are necessary for an optimal solution to (P ) ;
I make use of the Farkas Lemma. The proof is as follows.
Proof. Necessity of Kuhn-Tucker conditions
As the Constraint Quali…cation is satis…ed by assumption, then x is an optimal solution
to (LP ) : For simplicity, I de…ne cT = Df (x ) 2 Rn ; b = Dg(x )x 2 Rm ; and A = Dg(x ) 2
Rm;n : The linear programming problem (LP ) is then equivalent to the simpler linear problem:
cT x
maximize
(LP 2)
with multipliers
:
subject to Aj x
j
bj
0 j = 1; :::; m
Suppose, wlog, that the …rst m constraints bind, Aj x
= 0 for j > m :
j
bj = 0 for j = 1; :::; m : De…ne
As x is an optimal solution to (LP 2) ; there does not exist x such that cT x cT x > 0
and Aj x bj Aj x
bj = 0 8j = 1; :::; m : Thus, there does not exist x 2 Rn such that:
(x
x )T c > 0
(x
x )T ( Aj )T 8j = 1; :::; m :
1.2. KUHN-TUCKER THEOREM
11
Let’s apply the Farkas Lemma with the following relations between the statement of the
Farkas Lemma and this proof:
Farkas Lemma
n
This proof
q 2 R nf0g
z 2 Rn
()
()
2 Rm
+
()
A 2 Rn;m
()
x ) 2 Rn nf0g
c 2 Rn
h
i
T
T
(A1 ) ::::: (Am )
(x
T
2 Rm
+
De…ning
n
h
Z = z 2 Rn : z =
(A1 )T :::::
(Am )T
i
T
T
for some
2 Rm
+
o
;
then ( Aj )T 2 Z 8j = 1; :::; m : As there does not exist (x x ) such that (ii) of Farkas
Lemma holds (given the implication (x x )T c > 0 (x x )T z 8z 2 Z), then (i) must
hold (c 2 Z). Rewriting Z as
Z=
8
>
>
<
>
>
:
z 2 Rn : z =
0
B
@
2
31T
A1
6
7C
4 : 5A for some
Am
T
2 Rm
+
2
9
>
>
=
>
>
;
;
3
A1
6
7
T
then c 2 Z implies that there exists T 2 Rm
4 : 5 : From above,
+ such that c =
Am
T
recall that we de…ned j = 0 for j > m : Thus, we have c + A = 0: These are the
Kuhn-Tucker First Order Conditions:
Df (x ) + Dg(x ) = cT + A = 0:
Additionally, we have that 8j = 1; :::; m : j 0 and j = 0 if Aj x
Complimentary Slackness Conditions. This …nishes the argument.
bj > 0: These are the
12
CHAPTER 1. MATHEMATICAL PREREQUISITES
1.3
Correspondences
A correspondence is a multi-valued mapping. If a correspondence is single-valued over its
entire domain, then it is a function. Correspondences are crucial in economics, because
demand and best responses may not always be single-valued. Speci…cally, there may exist
multiple variables that comprise the demand or the best response.
The value of a correspondence is a set. When dealing with functions, the value of the
function is also a set, albeit a singleton.
Over the entire domain, if the value of a correspondence is compact, then the correspondence is said to be compact-valued.
Over the entire domain, if the value of a correspondence is convex, then the correspondence is said to be convex-valued.
Over the entire domain, if the value of a correspondence is nonempty, then the correspondence is said to be well-de…ned (or nonempty-valued).
Some more advanced de…nitions are provided in the next subsection. With these, I can
discuss the main results related to correspondences.
1.3.1
Let U
:U
De…nitions
Rm be the domain and W
W:
Rn be the codomain of the correspondence ; denoted
De…nition 1.7 The correspondence : U
W is upper hemi-continuous (uhc) i¤ for any
u 2 U; w 2 (u ) for 2 N such that lim u = u 2 U; there is a subsequence, wlog the
!1
original sequence, such that lim w = w 2 (u):
!1
If W is a compact set (as will often be the case in economics), we know that there exists
a subsequence, wlog the original sequence, such that lim w = w: All that remains to be
!1
shown for the de…nition of uhc is that w 2 (u):
Consider any u 2 U: We can certaintly specify a sequence u 2 U such that u = u 8 2
N: By the de…nition of uhc, all sequences w 2 (u ) = (u) have a convergent subsequence
(wlog the original sequence) lim w = w 2 (u): As all sequences w 2 (u ) = (u)
!1
have a convergent subsequence, then the set (u) is compact. Thus, is a compact-valued
correspondence. So uhc implies compact-valued.1
1
In some textbooks, compact-valued is assumed (redundantly) along with uhc.
1.3. CORRESPONDENCES
13
Additionally, for a correspondence that takes all values in the codomain, i.e., (u) = W
8u 2 U; if W is compact, then is uhc.
De…nition 1.8 The correspondence : U
W is lower hemi-continuous (lhc) i¤ for any
u 2 U for 2 N such that lim u = u 2 U and w 2 (u); there is a sequence w 2 (u )
!1
for 2 N such that lim w = w:
!1
If the correspondence
is lhc.
takes all values in the codomain, i.e., (u) = W 8u 2 U; then
De…nition 1.9 The correspondence
and lower hemi-continuous.
:U
W is continuous i¤ it is upper hemi-continuous
Figure 1.5 illustrates the distinction between uhc and lhc. The left panel shows a continuous correspondence (both uhc and lhc). The middle panel shows a correspondence that
is uhc, but not lhc. The right panel shows a correspondence that is lhc, but not uhc.
1.3.2
Berge’s Theorem
There are two major results associated with correspondences: Berge’s Maximum Theorem
and Kakutani’s Fixed Point Theorem. The proof of Kakutani is not included in this manuscript.
Theorem 1.5 Berge’s Maximum Theorem
Suppose X
Rn ; A Rm ; f : X A ! R is continuous, X f : A
and continuous, and
Xo : A
X s.t. 7 ! arg maxf (x; ) :
X is well-de…ned
x2X f ( )
Then X o is uhc.
Proof. Suppose that X o is not uhc. Then for some sequence
2 A; x 2 X o ( ) for
2 N such that lim
= 2 A; it is not true that lim x = x 2 X o ( ) : As X f is uhc,
!1
!1
there is a subsequence, wlog the original sequence, such that lim x = x 2 X f ( ) : Then, it
!1
must be that x 2
= X o ( ) ; even though x 2 X o ( ) for 2 N: As X f is lhc, then for every
x0 2 X f ( ) ; in particular x0 2 X f ( ) such that f (x; ) < f (x0 ; ) ; there exists a sequence
x0 2 X f ( ) for 2 N such that lim x0 = x0 : Since f is continuous,
!1
lim f (x ;
!1
) = f (x; ) < f (x0 ; ) = lim f (x0 ;
!1
):
14
CHAPTER 1. MATHEMATICAL PREREQUISITES
As x0 2 X f (
) for
2 N; then this contradicts that x 2 X o (
Corollary 1.2 Suppose X
Rn ; A
well-de…ned and continuous, and
Rm ; f : X
V : A ! X s.t.
) for
2 N:
A ! R is continuous, X f : A
X is
7 ! max f (x; ) :
x2X f ( )
Then V is continuous.
Proof. Since X o is uhc, then for some sequence
2 A; x 2 X o ( ) for 2 N such that
lim
= 2 A; there is a subsequence, wlog the original sequence, such that lim x =
!1
!1
o
x 2 X ( ) : Since (x ; ) ! (x; ) and f is continuous, then f (x ; ) ! f (x; ) : Since
x 2 X o ( ) and x 2 X o ( ) ; then this implies that V ( ) ! V ( ) : This completes the
argument.
1.3.3
Kakutani’s Fixed Point Theorem
By the Extreme Value Theorem, X o is well-de…ned. Further, if f is quasi-concave (De…nition
1.4), then X o is convex-valued (details are left to the reader).
Theorem 1.6 Kakutani’s Fixed Point Theorem
Let : X
X be a well-de…ned, convex-valued, and uhc correspondence and X be a
compact, convex, and nonempty set. Then 9x 2 X such that x 2 (x) :
1.3.4
Application: Dynamic Programming
This application requires the use of some basic results related to metric spaces. If the only
metric space you have worked with is the Euclidean space, then this material might be out
of reach. This digression will be brief, so please bear with me as I take the steps to introduce
dynamic programming. The mathematics behind dynamic programming includes Berge’s
Maximum Theorem.
The standard dynamic optimization problem is:
Given x0 2 X; max
(xn )n2N
f (x0 ; x1 ) +
8n 2 N:
The following assumptions are made:
1
X
i=1
i
f (xi ; xi+1 ) such that xn 2
(xn 1 )
1.3. CORRESPONDENCES
15
Assumption 1: f is continuous and bounded.
Assumption 2:
is continuous (uhc and lhc).
Assumption 3:
2 (0; 1) :
Assumption 4: X is compact.
Rather than …nding an entire optimal sequence all at once, we can …nd a sequence of
optimal solutions. To do this, we de…ne the recursive programming problem as follows:
For all n 2 N; given xn
(xn 1 ) :
1
2 X; max f (xn 1 ; xn ) + V (xn ) such that xn 2
xn
We need to verify that the function V : X ! R satis…es
V (xn 1 ) = max f (xn 1 ; xn ) + V (xn )
xn
for all n 2 N such that xn 2
(xn 1 ) :
De…ne the mapping T such that
(T V ) (x) = max ff (x; y) + V (y)g for all x 2 X:
y2 (x)
Assume that V 2 CB(X); where CB(X) is the set of bounded and continuous functions
with compact domain X: This means that V is a continuous function. Using Assumptions
1 and 2, the Corollary to Berge’s Theorem dictates that T V is a continuous function. Since
X is compact, then T V is also a bounded function. Consequently, T V 2 CB(X):
So the mapping T is a self-map CB(X) ! CB(X):
The space CB(X) is a complete metric space. A complete metric space is one in which
all Cauchy sequences in the set converge in the set.
We want to claim that the mapping T is a contraction.
De…nition 1.10 A self map : K ! K is a contraction if there exists 2 (0; 1) such that
d ( (x) ; (y))
d (x; y) for all x; y 2 K; where d ( ) is the metric for the metric space K:
Rather than verify directly that T is a contraction, we will use the Blackwell su¢ cient
conditions.
16
CHAPTER 1. MATHEMATICAL PREREQUISITES
Lemma 1.2 If
: K ! K is increasing and there exists
(f + )
for all (f; ) 2 K
R+ ; then
2 (0; 1) such that
(f ) +
is a contraction.
Using these su¢ cient conditions, is the mapping T a contraction? Let’s consider each
condition in turn:
1. Increasing
If V
W; where V; W 2 CB (X) ; this implies V (x)
W (x) 8x 2 X: So, if
V (x) W (x) 8x 2 X; then T (V ) (x) T (W ) (x) 8x 2 X: This is by the de…nition
of T (V ) (x) = max ff (x; y) + V (y)g as any y = arg max ff (x; y) + W (y)g can
y2 (x)
y2 (x)
also be selected for arg max ff (x; y) + V (y)g (if not selected, then a higher max is
y2 (x)
achieved for T (V ) (x)). Thus, T is increasing.
2. 9 2 (0; 1) such that
(f + )
(f ) +
By de…nition, T (V + ) (x) = max ff (x; y) + V (y) +
y2 (x)
nition, max ff (x; y) + V (y) +
y2 (x)
g = T (V ) (x) +
: As
g for all x 2 X: By de…< 1 by Assumption 3,
then the condition is satis…ed.
So, T is a contraction. How does this help us? We can apply a di¤erent …xed point
theorem, called the Contraction Mapping Theorem.
Theorem 1.7 Contraction Mapping Theorem
If
: K ! K is a contraction and K is a complete metric space, then there exists a
unique k 2 K such that (k ) = k :
As T is a contraction and CB(X) is a complete metric space, then the Contraction
Mapping Theorem guarantees that there exists a unique …xed point V 2 CB (X) such that
T (V ) = V:
This results in the so-called Bellman equation:
V (x) = max ff (x; y) + V (y)g for all x 2 X:
y2 (x)
1.4. DIFFERENTIAL TOPOLOGY
17
Not only does a value function V : X ! R exist that is consistent with the Bellman equation,
but this value function is unique from the statement of the Contraction Mapping Theorem.
This implies that any solution to standard dynamic optimization problem, stated as
Given x0 2 X; max
(xn )n2N
f (x0 ; x1 ) +
1
X
i=1
8n 2 N;
i
f (xi ; xi+1 ) such that xn 2
(xn 1 )
is also a solution to the Bellman equation, and vice versa.
1.4
Di¤erential Topology
I …rst introduce the concept of measure and properness before moving into the basic results
for the analysis of di¤erentiable equations.
1.4.1
Measure
The set X 0 Rn has zero measure if for every > 0; X 0 can be covered by n dimensional
cubes whose total volume is less than : For any open set X Rn ; the subset X
X is a
generic subset of X i¤:
1. X is open.
2. The complement relative to X; (X )C
X = XnX has zero measure.
If a property holds over a generic subset, the property is said to hold generically.
1.4.2
Properness
For X
Rn and Y
Rm ; the function f : X ! Y is proper i¤:
1. f is continuous.
2. For any Y 0
Y compact, the set f
1
(Y 0 ) is compact.
Using the de…nition of properness, we obtain the following property (of great value in
the coming sections).
18
CHAPTER 1. MATHEMATICAL PREREQUISITES
Lemma 1.3 Properness Property
Suppose that for X Rn and Y
Rm ; the function f : X ! Y is proper. If X 0
closed (relative to X), then f (X 0 ) Y is closed (relative to Y ).
X is
Proof. Consider a sequence y 2 f (X 0 ) for 2 N such that lim y = y 0 : As y 2 f (X 0 )
!1
for 2 N; then there exists (x ) 2N such that x 2 X 0 and y = f (x ) 8 2 N: De…ne the
set Y 0 = (y ) 2N ; y 0 : The set Y 0 is compact. As f is proper, then f 1 (Y 0 ) is compact.
As x 2 f 1 (Y 0 ) 8 2 N; then there exists a subsequence, wlog the original sequence, such
that lim x = x0 2 f 1 (Y 0 ) : As X 0 is closed, then x0 2 X 0 : As f is continuous, y 0 = f (x0 ) ;
!1
so y 0 2 f (X 0 ) : This veri…es that f (X 0 ) is closed.
1.4.3
Regular vs. Critical values
The following framework is utilized when analyzing di¤erentiable equations:
RJ is the set of variables; it is open with typical element :
RK is the set of parameters; it is open with typical element :
H=
is the set of variables and parameters, with typical element
= ( ; ):
: H ! RL is the system of equations characterizing equilibria; it is C 1 :
The set of solutions are 2 H such that ( ) = 0: The results that we will obtain will
depend upon whether (i) J = L (equal number of variables and equations) or (ii) J < L
(more equations than variables). For the case of J = L (as discussed in the remainder of
the section), the results that we obtain (provided the assumptions are met) can be viewed
as ’positive.’ For the case of J < L (as considered in Exercise 3 at the end of the chapter),
the results that we obtain (provided the exact same assumptions are met) can be viewed as
’negative.’
The following are important mathematical de…nitions:
M =f 2H:
( ) = 0g ; the set of solutions.
is the projection of M onto
:
2 M is a critical point if rankD
( ) < L; a regular point if rankD
( ) = L:
1.4. DIFFERENTIAL TOPOLOGY
2 is a critical value if
set of critical values.
1
19
c
( ) contains at least one critical point. De…ne
Any 2 that is not a critical value must be a regular value. De…ne
the set of regular values.
r
=
as the
n
c
as
The terms critical/regular point and critical/regular value are with respect to the projection : Notice that if no solution exists for some parameter ; that is, 1 ( ) = ;; then
is a regular value (any value that is not a critical value is a regular value, by de…nition).
Figure 1.6 illustrates the concepts of critical point, regular point, critical value, and regular
value.
1.4.4
Results
This section contains the required mathematical results. The results are stated for the case
of J = L: The following section combines the results and provides a simple routine in order
to prove the important regularity result known as FLU (…nite and locally unique).
Theorem 1.8 Closedness Theorem
If
is proper, then
c
is closed, so that
r
is open.
1
Proof. The set f0g is closed and the function is continuous, so
(0) = M is closed
(relative to H). With J = L; we can de…ne the determinant det : M ! R on the matrices
D ( ) : The determinant is continuous and the set f0g is closed, so the set of critical points,
those 2 M such that 2 det 1 (0) ; is closed. As is proper, then by the Properness
Property, the set of critical values c is closed.
Concerning notation, the derivative matrix D ( ) considers the derivatives with respect
to the variables : As J = L (number of variables equals number of equations), then D ( )
is a square matrix. The derivative matrix D ( ) considers the derivatives with respect to
both the variables and the parameters : This matrix could be more accurately written as
follows: D ( ) = D ; ( ) : This matrix now has more columns than rows (as the number
of variables + the number of parameters is greater than the number of equations).
Theorem 1.9 Transversality Theorem
If, for every
measure.
2 M; rankD ( ) = L; then
c
has zero measure, so that
r
has full
20
CHAPTER 1. MATHEMATICAL PREREQUISITES
The proof of the theorem is too involved to include in these notes, but it is basically
an application of Sard’s Theorem. Consider what the theorem says. The set of parameters
that are both (a) critical (that is, rankD ( ) < L) and (b) satisfy the rank condition
rankD ( ) = L is a set of measure zero. This brings us to the aptly named Stack of Records
Theorem.
Theorem 1.10 Stack of Records Theorem
If is proper, 0 2 r ; and 1 ( 0 ) 6= ;; then
1
1.
( 0 ) = f 0i : i = 1; :::; I with I < 1g :
2. There is an open neighborhood 0 around 0 and open neighborhoods Hi0 around
C 1 mappings i : 0 ! Hi0 \ M 8i = 1; :::; I such that for any 2 0 and any 2
there exists a unique i such that = i ( ) :
0
i
1
and
( );
Figure 1.7 illustrates the result.
Proof. Consider any 0 2 1 ( 0 ) : Since rankD (n) = L and J = L; then we can apply
the Implicit Function Theorem. Thus, there exists a neighborhood 0 of and a neighbohood
H 0 of 0 and a C 1 mapping : 0 ! H 0 such that for any 2 0 ; 2 H 0 \ M i¤ = ( ) :
Thus, 1 ( 0 )
[ 0 H 0 ; an open covering. Since is proper and f 0 g is compact, then
1
02
0
1(
)
( ) is compact, so the cover
02
[
H 0 must have a …nite subcover
1( 0)
[ Hi0 : That is,
i=1;:::;I
( 0 ) = f 0i : i = 1; :::; I with I < 1g :
By taking the neighborhood 0 to be small enough, then Hi0 \ Hj0 = ; 8i 6= j: Thus, for
any 2 0 ; each 2 1 ( ) has a unique i :
1
1.4.5
Finite Local Uniqueness
For any system of equations
( ) = 0; we seek to verify the following two properties:
is proper.
For every
2 M; rankD ( ) = L (called the rank condition).
Then over a generic subset of parameters (the set of regular values r ), there are a …nite
1
number of solutions (that is,
( ) is a …nite set 8 2 r ) and the solutions are locally
unique. By locally unique, we mean that for any 2 r with a solution 2 1 ( ) ; there is
only one solution 0 2 1 ( 0 ) in a neighborhood around ; where 0 lies in a neighborhood
around :
1.5. EXERCISES
21
This is the result Finite Local Uniqueness (FLU). It implies that small changes in the
parameters do not change the number of solutions nor do they allow for any "jumps".
1.5
Exercises
1. (Applying Farkas Lemma)
Let R be a m
implication.
n matrix and
There does not exist
2 Rn : Use Farkas Lemma to prove the following
such that R
> 0
+
2 Rm
++ s.t.
9
T
R = 0:
(Note: The other implication is a simple one-line proof. Make sure that you are proving
the correct implication [i.e., the multi-page behemoth of a proof]).
2. (Applying Kakutani’s Fixed Point Theorem)
We will use the theory of correspondences to prove the existence of a Nash equilibrium. Consider a game with I players. Each player i has a …nite number of actions
ai1 ; ::::; aiJi : The set of strategies for each player i is then the simplex Ji 1 of dimension Ji 1: The strategies are simply the probabilities that each player assigns to
each of the …nite number of actions. For simplicity, denote the strategy set for each
player as S i with element si : These sets are nonempty, compact, and convex.
If player i0 s payo¤ value for the action pro…le a = (a1 ; :::; ai ; :::; aI ) is pi (a); then the
objective function for each player is (using the convention s = (s1 ; :::; si ; :::; sI )):
ui (s) =
X
a
pi (a)
Y
i
s i ai ;
where si (ai ) is the probability that player i selects the action ai : This objective function
is quasi-concave in s (as it’s linear).
Denote s i = (s1 ; :::; si 1 ; si+1 ; :::; sI ) as the vector of strategies (not including the
strategy for player i) and S j = S 1 ::: S i 1 S i+1 ::: S I as the Cartesian product
j6=i
22
CHAPTER 1. MATHEMATICAL PREREQUISITES
of the strategy sets (not including the strategy set for player i). De…ne F i :
Sj
Si
j6=i
as the set of "feasible" strategies. Notice that F i (s i ) = S i 8s
De…ne BRi :
Sj
i
2
Sj :
j6=i
S i as the best response correspondence for player i: Use the
j6=i
Berge’s Maximum Theorem and Kakutani’s Fixed Point Theorem to prove that the
self-map BR1 ; :::; BRI : S i
S i has a …xed point. By de…nition, a …xed point is
i
i
a Nash equilibrium.
3. (A Di¤erent Application of Di¤erential Topology)
Let
RJ be the set of variables (with typical element ),
RK be the set of
parameters (with typical element ), H =
(so = ( ; )), and : H ! RL be the
C 1 system of equations. Assume that J < L: Suppose that the projection is proper
and the rank condition holds (that is, rankD ( ) = L). Using the mathematical
results from Section 1.4, what conclusions can be drawn? The conclusions will be of
the form, "over a generic subset of parameters (the set of regular values r ), then...".
Bibliography
[1] Berge, Claude (1997, reprint): Topological Spaces (Dover Publications: Mineola, New
York).
[2] Cass, David (2005): Basic Results for the Analysis of Smooth Equations (mimeo, University of Pennsylvania).
[3] Cass, David (2005): Hemi-continuity of Correspondences and Berge’s Maximum Thereom
(mimeo, University of Pennsylvania).
[4] Milnor, John (1997, reprint): Topology from the Di¤erentiable Viewpoint (Princeton
University Press: Princeton, New Jersey).
[5] Ok, Efe A. (2007): Real Analysis with Economic Applications (Princeton University Press:
Princeton and Oxford).
23
24
BIBLIOGRAPHY
Chapter 2
Arrow-Debreu Model
This chapter will introduce the canonical general equilibrium model of pure-exchange in the
static setting. The equilibrium is commonly called a Walrasian equilibrium. However, the
existence of the equilibrium was not shown until the early 1950s in joint work by the Nobel
laureates Kenneth Arrow and Gerard Debreu. For this reason, I use the term "Arrow-Debreu
equilibrium."
After introducing the model and de…ning an equilibrium, I can prove the fundamental
properties of an Arrow-Debreu equilibrium. First, I prove existence (using the results for
correspondences from Section 1.3). Second, I prove the First Basic Welfare Theorem. Next,
I prove the Second Basic Welfare Theorem (using the Supporting Hyperplane Theorem from
Section 1.1). Finally, I will verify the regularity of the Arrow-Debreu equilibria, namely that
the equilibria are …nite and locally unique (using the FLU result from Section 1.4).
Some proofs are relegated to the end of the chapter. I prefer to spend the main sections
of the chapter motivating the results and discussing why certain assumptions are required
and what is required to carry out the proofs.
2.1
The Model
I will be fairly pedantic about the notation used in the model. The reason is that the
fundamentals of the model will be used throughout the manuscript, so proper notation can
help to minimize confusion.
The …rst component of the environment is households. Let H = f1; :::; Hg be the set
of households, where the number of households is …nite (H < 1): A typical element of the
set H will be h: Households trade and consume G goods. Let G = f1; :::; Gg be the set
25
26
CHAPTER 2. ARROW-DEBREU MODEL
of goods, with typical element g: Denote xhg as the consumption of good g by household h:
Then xh = xhg g2G is the vector of consumption by household h: Finally, x = xh h2H is
the vector of consumption by all households. This is also referred to as the allocation of the
economy.
This model is one of pure exchange. Production is not considered in this manuscript.
Each household begins with a vector of endowments. As with consumption, let ehg denote
the endowment of good g for household h; where eh = ehg g2G is the vector of endowments
of household h and e = eh h2H is the vector of all endowments.
Each household must select consumption within its consumption set. The consumption
set is denoted X h and must be a subset of the nonnegative orthant, X h RG
+ : We assume that
h
h
h
the endowments e belong to the consumption set, e 2 X : This allows for the possibility
of a no-trade equilibrium (also called autarchy).
The objective function of a household is called the utility function uh : X h ! R: These
household primitives (parameters of the model) are listed below.
h
X h RG
+ is the consumption set. We typically assume that X is closed (relative to
the set RG
+ ) and convex.
uh : X h ! R is the utility function. We typically assume that uh is continuous, locally
non-satiated, and quasi-concave.
eh 2 X h is the endowment vector. We typically assume that eh >> 0:
De…nition 2.1 The function uh : X h ! R is locally non-satiated if 8x 2 X h and 8 > 0;
there exists y 2 X h \ N (x) such that uh (y) > uh (x) :
The second component of the environment is the market institutions. In the ArrowDebreu model, markets are perfectly competitive, meaning that all households take prices
as given and do not account for how their actions a¤ect the market prices. Each good has
a market price pg : The vector of all prices is p = (pg )g2G 2 RG nf0g: The convention is that
the price vector p is a row vector. As can be shown (see Exercise 1), the assumptions on
utility guarantee that p > 0: An equilibrium, as will be evident from the de…nition, exhibits
nominal price indeterminacy. To remove this, we normalize the prices so that p 2 G 1 :
We are now prepared to de…ne an Arrow-Debreu equilibrium.
De…nition 2.2 An Arrow-Debreu equilibrium is
xh
h2H
; p such that
2.2. EXISTENCE
27
1. 8h 2 H; given p; xh is an optimal solution to the household problem (HP )
maximize uh xh
subject to xh 2 X h
p e h xh
(HP )
2. Markets clear
2.2
X
h2H
xhg =
X
h2H
ehg
:
0
8g 2 G:
Existence
Which assumptions are required to guarantee that an Arrow-Debreu equilibrium exists for
all parameters eh ; uh h2H ? These assumptions are listed below. After the statement of the
theorem, I will discuss why the third assumption is required.
Assumption E1:
X h = RG
+ 8h 2 H:
Assumption E2:
concave 8h 2 H:
uh : X h ! R is continuous, locally non-satiated, and quasi-
Assumption E3:
eh >> 0 8h 2 H:
Assumption E4:
For some household h0 ; uh is non-decreasing.
0
Theorem 2.1 Under Assumptions E1-E4, an Arrow-Debreu equilibrium
ists.
xh
h2H
; p ex-
Proof. For the complete proof, see Section 2.6.
The proof method proceeds as follows. We begin by de…ning the budget correspondence
bh : G 1
X h 8h 2 H such that bh (p) = xh 2 X h : p eh xh
0 : Using this
h
G 1
h
de…nition, the demand correspondence can be de…ned as d :
X such that dh (p) =
arg max uh xh : If the conditions of Berge’s Maximum Theorem are satis…ed, namely that
xh 2bh (p)
h
b is well-de…ned and continuous, then the correspondence dh is upper hemi-continuous.
Using Assumption E2 (continuous and quasi-concave), the correspondence dh is also wellde…ned and convex-valued. Thus, we can apply the Kakutani’s Fixed Point Theorem to
show that a …xed point exists. Further details are left to Section 2.6.
28
CHAPTER 2. ARROW-DEBREU MODEL
From the outline of the proof, the key step is to verify that the conditions of Berge’s
Maximum Theorem are satis…ed. It is straightforward to show that the budget correspondence bh is well-de…ned and upper hemi-continuous. But the conditions of Berge’s Maximum
Theorem require that the correspondence is also lower hemi-continuous. In order for this to
be achieved, we must satisfy the equilibrium condition peh > 0: Since p > 0; an assumption
of eh > 0 does not su¢ ce as leaves open the possibility that peh = 0 (namely, any good with
a strictly positive price [pg > 0] may have a zero endowment ehg = 0 ). So we require that
eh >> 0; which is Assumption E3.
What is so special about requiring peh > 0? Suppose otherwise. That is, suppose
(p1 ; p2 ) = (0; 1) : This is depicted in Figure 2.1 in the companion ’Figures’document. In both
panels, the initial endowment is labeled as the point e (we drop the household superscript for
convenience). Given the prices (p1 ; p2 ) = (0; 1) ; the choice x is budget feasible, x 2 bh (p) :
Consider any sequence p ! p; speci…cally p = 1 ; 1 1 for 2 N: Equilibrium prices
(see Exercise 1) can never contain a negative element: p > 0: The de…nition of lower hemicontinuity requires that a sequence x exists such that (i) x ! x and (ii) x 2 bh (p ) for
2 N: However, as can be seen in the right panel of Figure 2.1, no such sequence can be
found (budget feasible choices are those lying to the lower-left of the dotted budget lines).
Thus, the correspondence bh is not lower-hemicontinuous unless peh > 0:1
2.3
First Basic Welfare Theorem
The First Basic Welfare Theorem states that all Arrow-Debreu equilibria are Pareto optimal.
Let’s de…ne a Pareto optimal allocation.
De…nition 2.3 A feasible allocation xh
h2H
is such that
1. xh 2 X h 8h 2 H and
1
A weaker assumption can be made in order to guarantee peh > 0: This weaker assumption was derived
by Lionel McKenzie and termed irreducibility. Formally, his assumption contains two parts:
P
1. eh > 0 8h 2 H and h2H eh >> 0
2. 8H1 ; H2 6= ; such that H1 \ H2 = ; and H1 [ H2 = H and any allocation xh h2H ; there exP
P
~h h2H2
ists y h h2H1
0 and x
0 such that (a) h2H1 ygh = 0 whenever h2H1 ehg = 0; (b)
P
P
P
h
h
~h
+ h2H2 eh ; and (c) uh x
~h h2H2 > uh xh h2H2 :
h2H2 x
h2H1 y + e
Basically, the second condition says that each household has a strictly positive endowment of at least one
good that is "desired" by some other household (forcing the price of that good to be strictly positive).
2.3. FIRST BASIC WELFARE THEOREM
2.
X
h2H
xhg =
X
h2H
ehg
29
8g 2 G:
For simplicity, de…ne the set of feasible allocations as F A:
De…nition 2.4 A Pareto optimal allocation xh h2H is such that there does not exist a
0
0
0
0
uh xh 8h 2 H and uh y h > uh xh for
feasible allocation y h h2H where uh y h
some h0 :
To …nd the Pareto optimal allocations, we have two options. First, we can assume that
u is strictly increasing for some household h0 and 8h 6= h0 ; uh is continuous. Then xh h2H
is a Pareto optimal allocation i¤ xh h2H is an optimal solution to the following nonlinear
programming problem:
h0
0
0
(P Oh0 ) maximize uh xh
subject to uh xh
uh xh
xh h2H 2 F A
8h 6= h0 :
The proof is in Section 2.6.
The second option employs concavity. If xh h2H is a Pareto optimal allocation and
uh is concave 8h 2 H; then there exists h h2H 2 H 1 such that xh h2H is an optimal
solution to the following nonlinear programming problem:
(P O) maximize
subject to
X
xh
h
h2H
h2H
u h xh
2 FA
:
Conversely, if there exists h h2H 2 int H 1 such that xh h2H is an optimal solution
to (P O) ; then it is a Pareto optimal allocation. The proof is in Section 2.6.
The assumptions required to prove the First Basic Welfare Theorem are:
Assumption F1:
uh : X h ! R is locally non-satiated 8h 2 H:
Theorem 2.2 Under Assumption F1, all Arrow-Debreu equilibrium allocations xh
Pareto optimal.
h2H
are
Proof. See Section 2.6.
The assumption of locally non-satiated utility implies that all budget constraints hold
with equality: p eh xh = 0 8h 2 H: In this case, why do we de…ne an Arrow-Debreu
30
CHAPTER 2. ARROW-DEBREU MODEL
equilibrium with inequalities in the budget constraints? The reason is that the First Basic Welfare Theorem no longer holds when we de…ne an Arrow-Debreu equilibrium with
equalities in the budget constraint.
De…nition 2.5 An Arrow-Debreu equilibrium with equalities in the budget constraint is
xh h2H ; p such that
1. 8h 2 H; given p; xh is an optimal solution to the household problem (HP =)
maximize uh xh
:
subject to xh 2 X h
h
h
p e
x =0
(HP =)
2. Markets clear
X
h2H
xhg =
X
h2H
ehg
8g 2 G:
Why does the First Basic Welfare Theorem not hold for an Arrow-Debreu equilibrium
with equalities in the budget constraint? Consider the following counterexample. Assume
the economy has 2 households and 2 goods, so we can illustrate the economy in an Edgeworth
box. The consumption sets for both households are given by:
X1 =
x1 2 R2+ : x12
1
x11
:
X 2 = R2+ :
These consumption sets are closed and convex (satisfying Assumption S1 below, which are
typical assumptions for X h h2H ). The utility functions for both households are given by:
u 1 x1
=
x11 :
u2 x2
= x21 :
Both utility functions are locally non-satiated (satisfying Assumption F1). The endowments
are labeled e in Figure 2.2. The shaded region above the curve x12 = x11 denotes X 1 : There
1
are two price vectors that comprise Arrow Debreu equilibria with equalities in the budget
constraints: (i) p >> 0 and (ii) p = (1; 0) : Given the equilibrium price vector (i) p >> 0;
the equilibrium allocation is x near the top-left corner of the Edgeworth box. Given the
equilibrium price vector (ii) p = (1; 0) ; the equilibrium allocation lies anywhere along the
2.4. SECOND BASIC WELFARE THEOREM
31
darkened vertical line through the endowment e: In both cases, the noted equilibrium allocations are the optimal solutions to the household maximization problem (HP =) (equalities
in the budget constraints).
The First Basic Welfare Theorem requires that each of these equilibrium allocations is
Pareto optimal. Yet, the only Pareto optimal allocation is x : All of the allocations along the
darkened vertical line can be Pareto improved (both households made better o¤) by moving
to the left in the Edgeworth box.
2.4
Second Basic Welfare Theorem
The Second Basic Welfare Theorem serves as a partial converse to the First Basic Welfare
Theorem. The Second Basic Welfare Theorem states that for any Pareto optimal allocation,
equilibrium prices can be found so that that allocation is an Arrow-Debreu equilibrium
allocation.
The assumptions required to prove the Second Basic Welfare Theorem are:
Assumption S1:
Xh
RG
+ is closed and convex 8h 2 H:
Assumption S2:
uh : X h ! R is continuous and quasi-concave 8h 2 H:
Assumption S3:
For some household h0 ; wlog h0 = 1; X 1 is unbounded above
and u1 is increasing.
Theorem 2.3 Under Assumptions S1-S3, if eh
h2H
= xh
h2H
is a Pareto optimal al-
location and xh 2 intX h 8h 2 H; then there is p > 0 such that
Arrow-Debreu equilibrium.
xh
h2H
;p
is an
Proof. See Section 2.6.
If we want to consider a Pareto optimal allocation xh h2H that di¤ers from the endowment eh h2H ; then the statement of the theorem includes "an imposed tax/subsidy scheme
P
h
with h2H h = 0 such that h = p xh
p eh 8h 2 H:" If h > 0; the transfer is
h2H
a subsidy and that amount is added to the household’s income. If h < 0; the transfer is a
tax and the amount is subtracted from the household’s income.
Notice that in the statement of Theorem 2.3, we require xh 2 intX h 8h 2 H: Why
is this? Why can’t the Second Basic Welfare Theorem be valid for an allocation on the
boundary of some household’s consumption set? Consider the following counter-example.
32
CHAPTER 2. ARROW-DEBREU MODEL
The economy contains 2 households and 2 goods, so the allocations can be depicted in the
Edgeworth box. The consumption sets for both households are given by:
X1 =
x1 2 R2+ : x12
1
x11 :
X 2 = R2+ :
These consumption sets are closed and convex (satisfying Assumption S1). The utility
functions for both households are given by:
u1 (x1 ) = min fx11 ; x12 g :
u2 (x2 ) is "smooth" (C 2 ; strictly concave, strictly increasing):
Both utility functions are continuous and quasi-concave (satisfying Assumption S2). Consider Figure 2.3, where X 1 is the shaded region above the curve x12 = 1 x11 : Figure 2.3
contains three indi¤erence curves for h = 1 and two for h = 2:
The allocation E is Pareto optimal. Why? We cannot …nd a feasible allocation that
makes both households better o¤ than they currently are at E:
But does there exist a price p > 0 such that the allocation E is an Arrow-Debreu
equilibrium allocation? Well, the only possible equilibrium price is p as drawn in Figure 2.3
(tangent to the indi¤erence curve for household h = 2). But given this equilibrium price p ;
the optimal consumption choice by household h = 1 would be E 0 ; not E:
Thus, the Second Basic Welfare Theorem relies importantly on the assumption that
x 2 intX h 8h 2 H:
h
2.5
Regularity
When we state that the Arrow-Debreu equilibria are regular, what we mean is that the number of equilibria is …nite and that each is locally unique. Recall the Finite Local Uniqueness
result from Section 1.4.
The assumptions required to prove Regularity are:
Assumption R1:
X h = RG
++ 8h 2 H:
Assumption R2:
uh : X h ! R is C 2 ; di¤erentiably strictly increasing (meaning that Duh xh >> 0), di¤erentiably strictly concave (meaning that D2 uh xh
2.5. REGULARITY
33
is a negative de…nite matrix), and satis…es the boundary condition (meaning that
h
8x 2 RG
uh (x)
RG
++ ; cl y : u (y)
++ ) 8h 2 H:
Assumption R3:
eh >> 0 8h 2 H:
Theorem 2.4 Under Assumptions R1-R3, all Arrow-Debreu equilibria satisfy Finite Local
Uniqueness.
The remainder of the section walks through the proof of Theorem 2.4, with useful properties of the model sprinkled in.
Recall that the household problem (HP ) is given by
maximize uh xh
subject to xh 2 X h
p eh x h
(HP )
:
0
Under Assumption R2 (namely, the boundary condition), the condition xh 2 X h never binds.
Applying the Kuhn-Tucker Theorem, the following conditions are necessary and su¢ cient
conditions for an optimal solution to (HP ) :
Duh xh
p eh
h
xh
p = 0
= 0;
where h 2 R is the Lagrange multiplier associated with the budget constraint p eh xh
0 in the problem (HP ) : The convention is that Duh xh is a row vector (as is the case with
p). Under Assumption R2 (namely, Duh xh >> 0), the budget constraint binds and both
h
> 0 and p >> 0:
If
xh
h2H
;p
satis…es the de…nition of an Arrow-Debreu equilibrium, then so does
xh h2H ; p for any scalar > 0: This is called nominal indeterminacy, as a continuum
of prices exist that satisfy the equilibrium de…nition. In the household problem (HP ) ; only
relative prices matter, not the absolute price level. To remove this nominal indeterminacy, we
impose the price normalization pG = 1:2 Thus, the only remaining price variables are pnG =
1
(p1 ; :::; pG 1 ) 2 RG
++ : What we show with regularity is that there is no real determinacy;
namely that the Arrow-Debreu equilibria are determinate (not a continuum).
2
This is a di¤erent price normalization than previously used in this chapter.
34
CHAPTER 2. ARROW-DEBREU MODEL
Walras’Law (obtained by summing over all households’budget constraints) states:
P
h2H
p eh
xh = 0:
P
P
As p >> 0; if h2H ehg xhg = 0 8g < G; then it must also be the case that h2H ehG xhG =
P
0: Thus, the market clearing condition h2H ehG xhG = 0 is redundant given that the market clearing conditions hold 8g < G: I use the notation
for the (G
P
h2H
ehnG
xhnG =
P
h2H
:
h2H
xh ;
xh
h2H
xh1 ; :::;
1) dimensional column vector.
De…ne the system of equations
where xh ;
eh1
h
h2H
H(G+1)
h
h2H
2 R++
0
B
;p = B
@
RG+1
++
P
ehG
h2H
1
H(G+1)+G
RG
++ ! R
Duh xh
p eh
P
h2H
h
p
T
xh
ehnG
xhnG
1
and p 2 RG
++ : By de…nition,
!
T
xhG
1
h2H
1
1
as:
1
C
C;
A
xh ;
h
h2H
;p
= 0 i¤
; p is an Arrow-Debreu equilibrium.
Recalling the proof method discussed in Section 1.4.5, Finite Local Uniqueness is proven
if we can show that (i) is proper and (ii) rankD
xh ; h h2H ; p = H (G + 1) + G 1:
The …rst condition is left to Exercise 6. We will now walk through the second condition.
The derivative D
h
h
xh ;
h
; p means that we are taking derivatives with respect
h2H
h
e h2H
to x h2H ;
; p; and
(all variables and all parameters). If we show that the
h2H
rank condition holds by only taking derivatives with respect to xh h2H ; h h2H ; p; and
e1 ; then we have …nished the argument.
The derivative matrix M = D(xh ; h ) ;p;e1
xh ; h h2H ; p has [H (G + 1) + G 1]
h2H
rows and [H (G + 1) + G 1 + G] columns. The rows of M correspond to the equations in
xh ; h h2H ; p ; while the columns correspond to the variables (or parameters) that we
are taking derivatives with respect to. The derivative matrix is given by:
2.5. REGULARITY
35
2
1
2 1
1
6 D u (x )
6
6
6
p
6
6
6
0
6
6
0
M =6
6
6
6
0
6
6
6
6
6
0
4
IG 1 0
pT
0
0
0
0
0
0
0
0
0
0
0
::: :::
::: :::
0
0
0
0
IG
0
e1nG
x1nG
:
:
H
0
0
0
0
0
0
0
::: :::
D2 uH xH
IG
IG
0
pT
p
eH
nG
0
0
0
1
1
1
xH
nG
!
3
0
T
p
0
0
!
0
T
0
0
IG
1
0
7
7
7
7
7
7
7
7
7
7:
7
7
7
7
7
7
7
7
5
To show that the matrix M has full rank (there are more columns than rows, so we have to
show that the matrix M has full row rank), we set T M = 0 and must verify T = 0; where
2 RH(G+1)+G 1 corresponds to equations in :
0
x1
T
1
1
F OC1
BC1
:
F OCh
BCh
:
MC
CB
CB
CB
B
: C
CB
CB
xh C B
CB
h CB
CB
CB
: A@
p
B
B
B
B
B
B
=B
B
B
B
B
@
The equations
10
C
C
C
C
C
C
C:
C
C
C
C
A
M = 0 are given by:
T
1
( x1 ) D2 u1 (x1 )
pT
p
T
( x1 ) p T = 0
h
:
xh
T
h
D2 uh xh
xh
T
p
pT = 0
pT
IG
h
0
1
IG
1
i
0
=0
(A:1:a)
(A:1:b)
i
=0
(A:1:c)
(A:1:d)
:
P
h
h2H
1
p+
pT
h
T
xhnG
IG
P
i h2H
0 =0
+
1
h
ehnG
xhnG
T
=0
(A:1:e)
(A:1:f )
:
(2.1)
36
CHAPTER 2. ARROW-DEBREU MODEL
Let’s show that
T
= 0 in three steps:
1. (A:1:f ) can be written as
1
1
As p >> 0; then
(p1 ; :::; pG 1 ; 1) + ( p1 ; :::; pG 1 ; 0) = 0:
= 0: Consequently,
pT = 0:
2. Use (A:1:f ) to simplify (A:1:a) and then postmultiply the equation by
T
x1
D2 u1 x1
x1 :
x1 = 0:
As u1 is di¤erentiably strictly concave, the Hessian matrix D2 u1 (x1 ) is negative de…T
nite. Thus, ( x1 ) = 0:
3. For any household h > 1; postmultiply (A:1:c) by
xh
From (A:1:d); p xh =
T
D 2 u h xh
xh
T
T
pT
xh
xh :
h
p xh = 0:
= 0: As uh is di¤erentiably strictly concave,
the Hessian matrix D2 uh xh is negative de…nite. Thus,
h
= 0:
Thus,
2.6
2.6.1
T
T
= ( x1 ) ;
1
; :::;
xh
T
;
h
; :::; pT
xh
T
= 0: From (A:1:c);
= 0; …nishing the argument.
Proofs
Proof of Theorem 2.1
The existence proof is divided into 6 parts.
Part 1: Price and consumption space
The price space G 1 is compact, convex, and nonempty. De…ne the bounded consumption
P
h
h
set X h = xh 2 X h : xhg 2
h2H eg 8g 2 G : The set X is compact, convex, and nonempty. We show in Part 6 that equilibrium consumption xh is an optimal solution to (HP )
2.6. PROOFS
37
i¤ xh is an optimal solution to:
maximize uh xh
subject to xh 2 X h
p eh x h
(HP )
:
0
Part 2: Demand correspondence
De…ne the budget correspondence bh :
G 1
X h 8h 2 H such that
bh (p) = xh 2 X h : p eh
xh
0 :
The correspondence is well-de…ned (consider an element eh 2 bh (p)).
Claim 2.1 bh is upper hemi-continuous.
Proof. Consider sequences p and x such that x 2 bh (p ) 8 2 N: Let p ! p and x ! x:
To prove upper hemi-continuity, we must prove that x 2 bh (p) : Suppose otherwise, that is,
x2
= bh (p) : Then, either xg < 0 for some good g or p eh x < 0: Then, by continuity, for
some ; either xg < 0 for some good g or p eh x < 0: This contradicts that x 2 bh (p )
8 2 N:
Claim 2.2 bh is lower hemi-continuous.
Proof. Consider a sequence p such that p ! p and x 2 bh (p) : To prove lower hemicontinuity, we must …nd a sequence x such that x 2 bh (p ) for 2 N and x ! x: There
are two cases to consider.
Case I : p eh x > 0: Then 9 such that 8
8
: Then, x ! x and x 2 bh (p ) 8
:
Case II : p eh
x = 0: I will de…ne x =
=
As p ! p and p eh
pv eh
! 1: Thus, 8
pv x
pv eh
pv x
1 and p x =
(
pv eh
pv x
1
if
pv eh
pv x
; p
x for
x > 0: De…ne x = x
2 N; where
< 1 and pv x 6= 0
otherwise
eh
)
:
x = 0; then 9v such that 8
; p x > 0: As p ! p; then
;x =
x ! x: By de…nition, p x =
p x = 1 pv x pv eh if
p x=
pv eh
pv x
pv x = pv eh if
pv eh
pv x
< 1: Thus, x 2 bh (p ) 8
:
38
CHAPTER 2. ARROW-DEBREU MODEL
Thus, bh is upper hemi-continuous and lower hemi-continuous. De…ne the demand correspondence dh : G 1
X h such that
dh (p) = arg max uh xh :
xh 2bh (p)
From Berge’s Maximum Theorem, dh is upper hemi-continuous. As uh is continuous and
X h is compact, the Extreme Value Theorem implies that dh is well-de…ned. As uh is quasiconcave and X h is convex, dh is convex-valued. To see this, let x; y 2 dh (p) : Then x; y 2 X h
and both p eh x
0 and p eh y
0: Therefore, 8 2 [0; 1] ; the choice x+(1
)y 2
X h (convexity) and p eh ( x + (1
) y)
0 (budget constraints are linear). Further,
h
h
h
h
u ( x + (1
) y) min u (x) ; u (y) = u (x) = uh (y) (de…nition of quasi-concavity).
Thus, x + (1
) y 2 dh (p) :
Part 3: Price correspondence
G 1
De…ne the price correspondence :
Xh
such that z = z 1 ; :::; z H 7! arg max
h2H
p2 G 1
P
h
h
p
e : The correspondence is well-de…ned, using the Extreme Value Theorem.
h2H z
The correspondence is convex-valued as the objective function is quasi-concave (in p) and
G 1
is convex.
is upper hemi-continuous.
Claim 2.3
Proof. Consider sequences z and p such that p 2 (z ) for 2 N: Let z ! z and p ! p:
To prove upper hemi-continuity, we must show that p 2 (z) : Suppose otherwise, that is,
P
P
h
h
p2
= (z) : Then, there exists p^ 2 G 1 such that p^
eh > p
eh :
h2H z
h2H z
P
P
h
h
Then for some ; p^
eh > p
eh : This contradicts that p 2 (z )
h2H z
h2H z
for 2 N:
Part 4: Fixed point
De…ne
Xh
:
h2H
Xh
set
G 1
Xh
h2H
G 1
G 1
as the Cartesian product of dh
h2H
and : The
is compact, convex, and nonempty. As the correspondences dh (for any h)
h2H
and are upper hemi-continuous, convex-valued, and well-de…ned, then so is the Cartesian
product.
From Kakutani’s Fixed Point Theorem, there exists a …xed point
xh h2H ; p
2
xh
h2H
;p
:
2.6. PROOFS
39
Part 5: Market clearing
As uh h2H are locally non-satiated, then p xh eh = 0 8h 2 H: Walras’ Law is then
P
xh eh = 0: This together with the de…nition of the price correspondence
given by p
P h2Hh
P
implies h2H xg ehg
0 8g 2 G: Otherwise, if h2H xhg0 ehg0 > 0 for some g 0 ; then
P
h
eh > 0 by setting pg0 = 1:
the maximum of the price correspondence is p
h2H z
P
If h2H xhg ehg < 0; then pg = 0 (see Walras’Law). For all households h 6= h0 ; de…ne
P
0
0
0
h
eh : As uh is non-decreasing (Assumption
xh = xh : For h0 ; de…ne xh = xh
h2H x
4), then xh h2H are optimal solutions to the household problems (HP ) : Further, market
clearing is satis…ed as:
P
0
h
=
h2H x
P
h6=h0
h 0
xh + xh
P
h2H
xh
eh
i
=
P
h2H
eh :
Part 6: Innocuous to bound consumption
I want to show that equilibrium consumption xh is an optimal solution to (HP ) i¤ xh is an
optimal solution to:
(HP )
maximize uh xh
subject to xh 2 X h
p eh x h
:
0
To do this, I use the assumptions that uh is quasi-concave and locally non-satiated.
If xh is an optimal solution to (HP ) ; then the market clearing requirement implies that
xh is an optimal solution to (HP ):
For the other direction, suppose that xh is an equilibrium consumption choice under
(HP ); but 9y 0 2 X h nX h such that p eh y 0
0 and uh (y 0 ) > uh xh : The market
clearing requirement implies xh 2 intX h and p eh xh = 0: By continuity, 9y 2 X h nX h
such that p eh y > 0 and uh (y) > uh xh : As uh is locally non-satiated, then there
exists x 2 intX h such that uh (x ) > uh xh : As uh is quasi-concave, then 8 2 [0; 1] ;
uh ( y + (1
) x ) > uh xh : Further, provided that is small enough, y + (1
)x 2
h
h
h
X : As y satis…es p e
y > 0; then p e
( y + (1
) x ) > 0: This contradicts
that xh is an equilibrium consumption choice under (HP ); as the allocation y + (1
)x
satis…es the constraints and provides strictly higher utility.
40
2.6.2
CHAPTER 2. ARROW-DEBREU MODEL
Proof of Theorem 2.2
Before proving the First Basic Welfare Theorem, I prove the two statements about how to
…nd Pareto optimal allocations as optimal solutions to programming problems.
Optimal solution to (P Oh0 )
Suppose that xh h2H is not an optimal solution to the programming problem (P Oh0 ) : Then
0
0
0
0
9 y h h2H such that uh y h > uh xh ; uh y h
uh xh
8h 6= h0 ; and y h h2H 2 F A:
As y h h2H is feasible, then xh h2H is not Pareto optimal.
Suppose that xh h2H is not Pareto optimal. Then 9 y h h2H 2 F A such that uh y h
uh xh 8h 2 H and uk y k > uk xk for some k: There are two cases to consider: (i) k = h0
and (ii) k 6= h0 : In case (i) k = h0 ; the allocation xh h2H would not be an optimal solution
0
to (P Oh0 ) by de…nition. In case (ii) k 6= h0 ; as uh is strictly increasing and uk is continuous,
0
0
then for some > 0; de…ne a new allocation y^h = y h 8h 2
= fh0 ; kg ; y^h = y h + ( ; :::; ) ; and
0
0
0
0
0
0
y^k = y k ( ; :::; ) : Then y^h h2H is such that uh y^h > uh y h = uh xh ; uk y^k >
uk xk ; and uh y^h
uh xh
8h 2
= fh0 ; kg : Thus, xh h2H is not an optimal solution
to the programming problem (P Oh0 ) :
Optimal solution to (P O)
Suppose that xh h2H is not Pareto optimal. Then 9 y h h2H 2 F A such that uh y h
0
0
0
0
uh xh 8h 2 H and uh y h > uh xh for some k: Therefore, for any h h2H >> 0;
P
P
h
uh y h > h2H h uh xh : Thus, xh h2H is not an optimal solution to the
h2H
programming problem (P O) :
Suppose that xh h2H is Pareto optimal. De…ne the set U = fu 2 RH : uh
uh xh
8h 2 H; xh h2H 2 F Ag: The point u = u1 (x1 ) ; ::::; uH xH
2 U; but u 2
= intU: The
h
h
h
set U is nonempty. Select any points v; w 2 U; where v
u x 8h 2 H and wh uh y h
8h 2 H for xh h2H ; y h h2H 2 F A: Then as uh is concave 8h 2 H; any convex combination
v + (1
) w 2 U as xh h2H + (1
) y h h2H 2 F A and
uh
xh + (1
) yh
uh xh + (1
) uh y h
v h + (1
) wh :
Thus, the set U is convex. From the Supporting Hyperplane Theorem (Corollary 1.1. in
Section 1.1), there exists a q 2 RH nf0g such that q T u
q T u 8u 2 U: We need to show that
there exists a q > 0 such that q T u
q T u 8u 2 U: Suppose not, that is, q h < 0 for some
2.6. PROOFS
41
0
0
0
h: Then, we can specify a new vector uh = uh xh 8h0 6= h and uh < uh xh : This new
q
vector is an element of U; yet q T u < q T u: Thus, q > 0 and we de…ne h h2H = kqk
2 H 1:
So, there exists h h2H 2 H 1 such that xh h2H is an optimal solution to (P O) :
Proof of First Basic Welfare Theorem
Suppose that
xh
h2H
; p is an Arrow-Debreu equilibrium, but xh
h2H
is not a Pareto
optimal allocation. Then there exists a feasible allocation y h h2H such that uh y h
0
0
0
0
uh xh 8h 2 H; with uh y h > uh xh for some h0 :
Then p eh y h
p eh xh = 0 8h 2 H: Why? If p eh y h > p eh xh = 0;
then as uh is locally non-satiated, there exists y^h such that p eh y^h > p eh xh = 0
and uh y^h > uh xh : This contradicts that xh is an optimal solution to (HP ) :
Likewise, if uh y h > uh xh for some h; then p eh y h < p eh xh = 0: Otherwise,
p eh xh = 0 and xh is not an optimal solution to (HP ) :
p eh y h
Summing the budget constraints over all households:
p
X
h2H
eh
yh < p
X
ehg
As p > 0; this implies that
h2H
y h h2H is a feasible allocation.
2.6.3
X
h2H
eh
xh = 0:
ygh < 0 for some good g: This contradicts that
Proof of Theorem 2.3
The proof proceeds in two steps: …rst an application of the Supporting Hyperplane Theorem
(Corollary 1.1 in Section 1.1) and second an application of the Duality Theorem (Theorem
1.3 in Section 1.1).
Part 1: Supporting Hyperplane Theorem
De…ne the set
(
Z=
z 2 RG : z
P
h2H
eh
xh
where xh
h2H
is such that
(
xh 2 X h and
u h xh
u h xh
)
)
8h 2 H :
P
h
The set Z is convex. Why? Consider any a; b 2 Z: Then a
xh and uh xh
h2H e
P
h
uh xh 8h 2 H where xh 2 X h 8h 2 H; and b
y h and uh y h
h2H e
uh xh 8h 2 H where y h 2 X h 8h 2 H: As uh is quasi-concave, then 8 2 [0; 1] ;
42
CHAPTER 2. ARROW-DEBREU MODEL
)b
uh xh and a + (1
)y h
uh xh + (1
ther, xh + (1
)y h 2 X h 8h 2 H: Thus, a + (1
P
h2H
eh
xh + (1
)y h
: Fur-
) b 2 Z:
Considering the allocation xh h2H = xh h2H ; we know that 0 2 Z; but 0 2
= intZ: To
prove the latter, suppose that 0 2 intZ: Then there exists z >> 0 such that z 2 Z: The
allocation xh h2H such that x1 = x1 +z and xh = xh 8h 2 is the allocation corresponding
uh xh 8h 2 and u1 (x1 ) > u1 (x1 ) (Assumption S3). This
to z: This implies uh xh
contradicts that xh h2H is a Pareto optimal allocation.
Applying the Supporting Hyperplane Theorem, there exists p 2 RG nf0g such that
0 p z 8z 2 Z: Further, p > 0: Why? Suppose not, that is pg < 0 for some g: Consider
the element z such that zg = 1 and zg0 = 0 8g 0 6= g: Obviously, z 2 Z; yet p z > 0: This
contradiction veri…es that p > 0:
Part 2: Duality Theorem
The conclusion from Part 1 is that for any allocation such that xh 2 X h and uh xh
P
h
uh xh 8h 2 H; 0 p
eh : In particular, for any household h; we can de…ne
h2H x
an allocation y h
h2H
= xh ; xh
h6=h0
: If xh 2 X h and uh xh
u h xh
; it must be
0 p xh eh : As this holds for all households h 2 H; then xh is an optimal solution to
the following Expenditure Minimization problem 8h 2 H :
minimize p xh
subject to xh 2 X h
uh xh
u h xh
(ExpM in)
:
0
I wish to show that this implies xh is an optimal solution to the Utility Maximization
problem (HP ) 8h 2 H: Suppose not. Then for some household h; there exists y h such that
p eh y h
0; y h 2 X h ; and uh y h > uh xh : Consider the consumption xh for some
< 1: As xh 2 intX h ; then for close to 1; xh 2 X h : Further, p eh
xh > 0: Then,
de…ning the convex combination y h + (1
) x h ; p eh
y h + (1
) xh
>0
h
h
8 2 (0; 1): By de…nition, p e = p x ; implying
p
y h + (1
)
xh
< p xh :
By the continuity of uh ; for some close to 1; uh y h + (1
) xh
> uh xh : Thus,
xh is not an optimal solution to the problem (ExpM in) : This completes the contrapositive
2.7. EXERCISES
43
argument.
2.7
Exercises
1. Show that if uh is locally non-satiated 8h 2 H; then the Arrow-Debreu equilibrium
prices satisfy p > 0:
2. Assume that uh is di¤erentiable, di¤erentiably strictly increasing, and concave 8h 2
H: Using the results from the programming problems in this chapter, show that if
xh h2H ; p is an Arrow-Debreu equilibrium, then the allocation xh h2H is an optimal solution to the problem (P Oh0 ) for any h0 : From Section 2.3, the allocation xh h2H
is then Pareto optimal. As this was the …rst proof of the First Basic Welfare Theorem,
it is often called the "classical proof."
3. Consider an economy with two households and two goods. Both utility functions are
di¤erentiable, strictly increasing, and strictly concave, but there is a missing market
for the second good. That is, each household can consume no more than its initial
endowment of the second good. Using an Edgeworth box, show that the equilibrium
allocations are typically Pareto suboptimal.
4. Consider an economy with two households and two goods. The two households have
utility functions
u 1 x1
u 2 x2
1
1
log x11 + x21 + log x12
2
2
1
1
=
log x21 + log x22
2
2
=
and endowments e1 = (1; 2) and e2 = (2; 1) : Notice that household 1 cares about
household 2’s consumption of the …rst good (x21 ). Using the programming problems
from this chapter, characterize the Pareto optimal allocations and then characterize
the equilibrium allocations. From this, can you claim that the First Basic Welfare
Theorem holds? Use an Edgeworth box to illustrate your argument.
5. Consider an economy with two households and two goods. Prove that if the utility
functions are Cobb-Douglas (of the form uh xh = 1 log xh1 + 2 log xh2 ), then
there is a unique Arrow-Debreu equilibrium. Does this property hold for an economy
with more than two households and more than two goods?
44
CHAPTER 2. ARROW-DEBREU MODEL
H(G+1)+G 1
6. Show that the projection is proper. is the projection : R++
which maps xh h2H ; p; eh h2H 7! eh h2H such that
xh
HG
RHG
++ ! R++
; p; eh h2H = 0:
h2H
7. Prove that for any endowments eh h2H within an open set of allocations around
a Pareto optimal allocation, the resulting Arrow-Debreu equilibrium is unique. To
do this, you must verify that the matrix D(xh ; h ) ;p
xh ; h h2H ; p has full rank
h2H
(square matrix) given that the initial endowment is a Pareto optimal allocation eh h2H =
xh h2H : Notice that the derivative matrix only contains derivatives with respect to
variables, not the endowment e1 :
8. Using "A Di¤erent Application of Di¤erential Topology" from Exercise 3 in Chapter
1, prove that over a generic subset of endowments, if G > 1; then p1 6= p2 :
9. As a general result of the ideas in Exercise 3 above, show that over a generic subset
of endowments, if there is a missing market for one of the goods, then the allocation
xh h2H is not Pareto optimal. To attack this problem, use "A Di¤erent Application of
Di¤erential Topology" from Exercise 3 in Chapter 1 and consider that a necessary con0
0
Dg uh xh
Dg uh (xh )
dition for Pareto optimality is D uh xh = D uh0 xh0 8 (h; h0 ; g; g 0 ) 2 H H G G:
( )
( )
g0
g0
Bibliography
[1] Arrow, Kenneth and Gerard Debreu (1954): "Existence of an Equilibrium for a Competitive Economy," Econometrica 22:3, 265-290.
[2] Debreu, Gerard (1970): "Economies with a Finite Set of Equilibria," Econometrica 38:3,
387-392.
[3] Debreu, Gerard (1972): "Smooth Preferences," Econometrica 40:4, 603-615.
[4] Geanakoplos, John and Herakles Polemarchakis (1986): "Existence, regularity, and constrainted suboptimality of competitive allocations when the asset market is incomplete."
In: Uncertainty, Information, and Communication: Essays in Honor of K.J. Arrow, Vol.3,
edited by W. Heller, R. Starr, and D. Starrett (Cambridge University Press: Cambridge,
pgs. 65-95).
[5] Magill, Michael and Martine Quinzii (1996): Theory of Incomplete Markets, Vol. 1 (MIT
Press: Cambridge, MA).
[6] Mas-Colell, Andreu, Michael Whinston, and Jerry Green (1995): Microeconomic Theory
(Oxford University Press: Oxford).
[7] McKenzie, Lionel (1959): "On the Existence of General Equilibrium for a Competitive
Market," Econometrica 27, 54-71.
[8] Stokey, Nancy L. and Robert E. Lucas (1989): Recursive Methods in Economic Dynamics
(Harvard University Press: Cambridge, MA).
[9] Villanacci, Antonio, Laura Carosi, Pierluigi Benevieri, and Andrea Battinelli (2002):
Di¤erential Topology and General Equilibrium with Complete and Incomplete Markets
(Kluwer Academic Publishers: Boston).
45
46
BIBLIOGRAPHY
Chapter 3
General Financial Model
This chapter explores a pure-exchange general equilibrium model in a dynamic setting. For
simplicity, there are only two time periods, with a …nite number of states of uncertainty
in the …nal period. Models with a longer, but still …nite, time horizon can be reduced to
the two-period model. Available to the households are …nancial markets that allow for the
transfer of wealth between the states of uncertainty. This model in which households trade
not only commodities, but also …nancial assets, is labeled the general …nancial model. We
will introduce the concept of incomplete markets in this chapter. Given the importance of
this topic, theorists have taken to calling this model the GEI model, where GEI stands for
"general equilibrium with incomplete markets". The equilibrium concept, typically called a
GEI equilibrium or Radner equilibrium (the latter in recognition of the work of Roy Radner),
will be referred to as a general …nancial equilibrium in this manuscript.
In this chapter, I …rst introduce the …nancial model with numeraire assets. Then I discuss
the equilibrium properties under two conditions: complete markets and incomplete markets.
I next consider how the results di¤er when considering a more general class of assets called
real assets. One large proof is relegated to Section 3.5.
3.1
The Model
The model is dynamic with two time periods, t = 0 and t = 1: In the …nal period, one of S
possible states of uncertainty is realized, where S < 1 is …nite. Thus, the total number of
states, including the initial period, is S + 1: Denote the set of states as S = f0; 1; :::; Sg with
typical element s: The uncertainty tree is depicted in Figure 3.1.
There are a …nite number of households. Denote the set of households as H= f1; :::; Hg
47
48
CHAPTER 3. GENERAL FINANCIAL MODEL
with typical element h: In each state, households trade and consumer L commodities. De…ne
the commodity space as L = f1; :::; Lg : The number of commodities is …nite. By de…nition,
one commodity in a state is a di¤erent good than the same commodity in a di¤erent state
(an avocado may be ripe today, but spoiled tomorrow). Thus, the total number of goods
is G = L (S + 1) : The convention is that xhl (s) is the consumption by household h of good
(l; s) ; or the lth commodity in state s; xh (s) = xhl (s) l2L is the vector of consumption by
household h of all commodities in state s; and xh = xh (s) s2S is the vector of consumption
by household h of all goods.
As with the static model, the household primitives are given below.
X h = RG
+ is the consumption set.
uh : X h ! R is the utility function. We typically assume that the function is continuous, locally non-satiated, and quasi-concave.
eh 2 X h is the endowment. We typically assume that eh >> 0:
Each good (l; s) has a market price pl (s) : The vector of all prices is p = (pl (s))(l;s)2L S 2
RG nf0g: The convention is that
2 the price vector p (s)
3 is a row vector 8s 2 S: I de…ne the
p(0) 0
0
0
6
7
6 0 p(1) 0
7
0
7:
(S + 1) G price matrix P = 6
6 0
0 :::
0 7
4
5
0
0
0 p(S)
Now, we consider the …nancial elements of the model. There are J …nancial assets that
a household can choose to buy or sell in the initial period.1 The number of assets is …nite
(J < 1). Denote the set of assets as J = f1; :::; Jg ; with typical element j: The decision to
buy or sell an asset in the initial period is in‡uenced by the price of that asset. Denote the
price of an asset j in the initial period as qj : Denote all asset prices as q = (qj )j2J : These
prices are variables in the equilibrium and will be determined to satisfy market clearing
conditions.
In the state of uncertainty s in the …nal period, the asset j has a …xed payout rj (s): The
payouts of all assets are in terms of the same physical commodity, de…ned as the numeraire
commodity, in all states. Without loss of generality, we label the numeraire commodity as
0
commodity l = L: We typically assume that for some household h0 ; the utility uh is strictly
1
Both buying and selling the same asset is redundant in this model without transaction costs and without
default.
3.1. THE MODEL
49
increasing in the commodity l = L in all states s 2 S: This way, the commodity price of the
numeraire commodity is pL (s) > 0 8s 2 S: To remove the nominal indeterminacy, we make
the price normalization pL (s) = 1 8s 2 S:
Further, the assets are assumed to have nonnegative payouts rj (s)
0 8(j; s) 2
J Snf0g; where rj = (rj (s))s>0 > 0 8j 2 J : Let’s collect the payouts of all assets in
all states s > 0 into2one payout matrix
3 R: This payout matrix has S rows and J columns:
r1 (1) ::: rJ (1)
h
i
6
7
R = r1 ::: rJ = 4 :
:::
: 5:
r1 (S) ::: rJ (S)
The lone …nancial primitive is the matrix of asset payouts.
R 2 RS;J
+ is the payout matrix. We typically assume that R is nonnegative and has
full column rank (though the …nal assumption is innocuous with respect to the real
variables as we will see in Section 3.1.4).
Given R and q; households determine their positions on all assets (also called their portfolio). As with the static general equilibrium model, the trading of assets is perfectly competitive and anonymous. That is, all households can select any asset positions that they
wish to. There are no investment constraints and households do not take into consideration
market clearing conditions when making their optimal choice.
Household h chooses portfolio z h 2 RJ ; where the position for a particular asset j is
denoted zjh 2 R: If zjh < 0; then the household has sold the asset (often called short-selling).
In this case, the asset has negative payout in the states s > 0 : rj (s) zjh 0 (the household
has borrowed). If zjh > 0; then the household has purchased the asset. In this case, the
asset has positive payout in the states s > 0 : rj (s) zjh 0 (the household has saved). We
typically assume that the economy is closed, so that the assets are in zero net supply. That
simply means that the sum of all assets sold equals the sum of all assets purchased.
3.1.1
Existence
We are now prepared to de…ne a general …nancial equilibrium, which is the equilibrium
concept in this …nancial model.
De…nition 3.1 A general …nancial equilibrium is
xh ; z h
h2H
; p; q such that
50
CHAPTER 3. GENERAL FINANCIAL MODEL
1. 8h 2 H; given (p; q) ; xh ; z h is an optimal solution to the household problem (HP )
maximize uh xh
subject to xh 2 X h
z h 2 RJ
(HP )
P eh
xh +
q
R
!
:
zh
0
2. Markets clear
X
h2H
X
xhl (s) =
h2H
X
h2H
zjh = 0
ehl (s) 8 (l; s) 2 L
S:
8j 2 J:
It is important to verify, under roughly the same conditions as the static model, that a
…nancial equilibrium exists. The assumptions su¢ cient for existence are given by:
Assumption E1:
X h = RG
+ 8h 2 H:
Assumption E2:
uh : X h ! R is continuous, non-decreasing, strictly increasing in the commodity l = L in all states s 2 S; and quasi-concave 8h 2 H:
Assumption E3:
eh >> 0 8h 2 H:
Assumption E4 (No Redundancy): The payout matrix R has full column rank
(that is, the assets are linearly independent).
Notice that Assumption E4 implies that J
S:
Theorem 3.1 Under Assumptions E1-E4, a …nancial equilibrium
xh ; z h
h2H
; p; q exists.
Proof. The arguments from the static model apply across the board. All that remains is to
show that households’portfolios are bounded, so that households’budget sets are compact.
Given that consumption is bounded (using the argument in Subsection 2.6.1, Part 1), the
budget constraints imply that the portfolio payouts Rz h are bounded. Given Assumption
E4, the portfolio z h must also be bounded. We discuss the validity of Assumption E4 in
Section 3.1.4. In particular, we can show that it is innocuous to make this assumption. By
innocuous, I mean that making this assumption does not a¤ect the real equilibrium variables.
3.1. THE MODEL
3.1.2
51
Regularity
The assumptions required to prove Regularity are:
Assumption R1:
X h = RG
++ 8h 2 H:
Assumption R2:
uh : X h ! R is C 2 ; di¤erentiably strictly increasing (meaning that Duh xh >> 0), di¤erentiably strictly concave (meaning that D2 uh xh
is a negative de…nite matrix), and satis…es the boundary condition (meaning that
h
uh (x)
RG
8x 2 RG
++ ) 8h 2 H:
++ ; cl y : u (y)
Assumption R3:
eh >> 0 8h 2 H:
Assumption R4 (No Redundancy): The payout matrix R has full column rank
(that is, the assets are linearly independent).
Theorem 3.2 Under Assumptions R1-R4, over a generic subset of endowments, all …nancial equilibria satisfy Finite Local Uniqueness.
The Lagrange multipliers for the budget constraints are the elements of the (S + 1) dimensional
row vector h : Applying the Kuhn-Tucker Theorem, the following conditions are necessary
and su¢ cient conditions for an optimal solution to (HP ) :
h
Duh xh
P eh
q
R
xh +
h
P = 0;
!
z h = 0;
q
R
= 0;
!
where the third set of equations contains the …rst order conditions with respect to the
portfolio z h : From this third set (knowing that h >> 0 from the …rst order conditions with
respect to consumption and rj > 0 8j 2 J by Assumption R4), q >> 0:
52
CHAPTER 3. GENERAL FINANCIAL MODEL
De…ne the system of equations
:
h2H
RG+S+1
++
0 0
B
B
B
B
B
B
B
h
h
h
x ; ; z h2H ; p; q = B
B
B
B
B
B
@
Duh xh
B
B
B P e h xh
B
B
B
"
B
@
h
X
G (S+1)
RJ
R++
T
h
P !
q
+
zh
R
!#T
q
R
h2H
X
ehnG
h2H
zh
xhnG
RJ++ ! Rn as:
1
1
C
C
C
C
C
C
C
A
h2H
C
C
C
C
C
C
C
C;
C
C
C
C
C
A
where n = H (G + S + 1 + J) + G (S + 1) + J: Allow me to verify that the equilibrium
h
h
J
2 RS+1
8h 2 H: Further,
variables belong to open sets: xh 2 X h = RG
++ ; and z 2 R
++ ;
G (S+1)
J
p 2 R++
following the price normalization pL (s) = 1 8s 2 S and q 2 R++ : By de…nition,
h
xh ; ; z h h2H ; p; q = 0 i¤ xh ; h ; z h h2H ; p; q is a general …nancial equilibrium.
Recalling the proof method discussed in Section 1.4.5 (in Chapter 1), Finite Local Uniqueness is proven if we can show that (i) is proper and
xh ;
(ii) rankD
h
; zh
h2H
; p; q = n:
(3.1)
The …rst condition is shown in the same fashion as Exercise 6 in Chapter 2 (the details are
left to the intrepid reader). We will now walk through the second condition.
xh ;
The derivative D
h
; zh
h2H
; p; q means that we are taking derivatives with re-
h
spect to xh h2H ;
; z h h2H ; p; q; and eh h2H (both variables and parameters). If
h2H
we show that the rank condition holds by only taking derivatives with respect to xh h2H ;
h
; z h h2H ; and e1 ; then we have …nished the argument.
h2H
The derivative matrix
M = D(xh ;
h
;z h )
h2H
;e1
xh ;
h
; zh
h2H
; p; q
has H (G + S + 1 + J) + G (S + 1) + J rows and H (G + S + 1 + J) + G columns. The
rows of M will correspond to the equations in ; while the columns will correspond to
the variables (or parameters) that we are taking derivatives with respect to. For simplicity,
3.1. THE MODEL
de…ne
=
q
R
!
matrix is given by:
53
and
2
6
=6
4
2 2 1 1
D u (x )
6
6
P
6
6
0
6
6
6
6
6
6
6
M =6
6
6
6
6
6
6
6
6
6
4
0
IL
0
0
0
:::
0
0
0
0
1
PT
0
0
T
0
0
IL
0
1
7
7 2 RG
5
0
0
:: :: ::
:: :: ::
:: :: ::
0
0
D2 uH xH
P
0
0
0
0
3
0 :: :: ::
IJ :: :: ::
PT
0
0
T
0
0
IJ
0
0
0
(S+1);G
: The derivative
3
0
7
P7
7
07
7
7
07
7
07
7
7
07:
7
07
7
7
07
7
07
7
7
5
0
To show that the matrix M has full rank (there are more columns than rows, so we have to
show that the matrix M has full row rank), we set T M = 0 and must verify T = 0; where
2 Rn corresponds to equations in :
0
B
B
B
B
B
B
B
B
B
B
=B
B
B
B
B
B
B
B
B
@
x1
1
z1
:
xh
h
zh
:
p
q
10
CB
CB
CB
CB
CB
CB
CB
CB
CB
CB
CB
CB
CB
CB
CB
CB
CB
CB
CB
A@
F OCx1
BC1
F OCz1
:
F OCxh
BCh
F OCzh
:
M Cx
M Cz
1
C
C
C
C
C
C
C
C
C
C
C:
C
C
C
C
C
C
C
C
A
54
CHAPTER 3. GENERAL FINANCIAL MODEL
The equations
T
M = 0 are given by:
1 T
T
( x1 ) D2 u1 (x1 )
T
T
( x1 ) P T + ( z 1 )
1 T
T
P
=0
pT
=0
(3:2:a)
(3:2:b)
qT = 0
+
(3:2:c)
:
h T
T
xh D 2 u h xh
T
xh P T + z h
h T
T
T
pT
P
=0
=0
qT = 0
+
(3.2)
(3:2:d)
(3:2:e)
(3:2:f )
:
1
Let’s show that
T
pT
P
=0
(3:2:g)
= 0 in four steps:
1
1. From (3:2:g); by the de…nition of ;
= 0: Consequently,
pT = 0:
x1 :
2. Postmultiply (3:2:a) by
x1
T
D2 u1 x1
x1 = 0:
As u1 is di¤erentiably strictly concave, the Hessian matrix D2 u1 (x1 ) is negative de…T
nite. Thus, ( x1 ) = 0:
!
q
3. From (3:2:b); as =
and R has full column rank (Assumption R4), then T
R
T
has full row ranking leading to the implication: ( z 1 )
(3:2:c); q T = 0:
T
D 2 u h xh
From (3:2:e); after postmultiplying by
h T
T
xh
h
P xh =
T
T
= 0 =) ( z 1 ) = 0: From
xh :
4. For any household h > 1; postmultiply (3:2:d) by
xh
T
h T
P xh = 0:
and then taking the transpose:
h T
zh:
h
From (3:2:f );
= 0: Thus,
xh D2 uh xh xh = 0: As uh is di¤erentiably
T
strictly concave, the Hessian matrix D2 uh xh is negative de…nite. Thus,
xh = 0:
3.1. THE MODEL
55
h
From (3:2:d);
= 0: From (3:2:e); as R has full column rank (Assumption R4), then
T
T
T
zh
= 0 implies
z h = 0:
1 T
T
Thus, T = ( x1 ) ;
0; …nishing the argument.
3.1.3
T
; ( z1)
; :::;
xh
T
;
h T
zh
;
T
; :::; pT ; q T
No Arbitrage
!
q
Given R and q; an arbitrage opportunity exists if 9z 2 RJ such that
z > 0: This
R
is an arbitrage opportunity, because the household could hold the portfolio z as ! 1;
and provide itself with unbounded wealth in some states (without having unbounded debt
in other states).
Thus, the No Arbitrage condition is given by:
No Arbitrage:
q
R
There does not exist z 2 RJ such that
!
z > 0:
This condition is quite intuitive, but di¢ cult to work with mathematically. What we
would like to do is prove that the No Arbitrage condition is equivalent to the following
condition involving state prices ^ 2 RS+1
++ :
No Arbitrage II:
There exists ^ 2 RS+1
++ so that ^
q
R
!
= 0:
State prices are ’price-like’variables, so the convention is that they are written as row
vectors (similar to prices p and q). As mentioned above, we believe this No Arbitrage II
condition is equivalent to No Arbitrage I. Let’s verify this.
1. No Arbitrage II implies No Arbitrage I
q
R
!
Suppose that No Arbitrage I does not hold. Then 9z 2 RJ so that
z > 0:
!
q
This implies that for any ^ 2 RS+1
z > 0: This is a contradiction, as No
++ ; ^
R
!
q
Arbitrage II requires that for some ^ 2 RS+1
z = 0:
++ ; ^
R
=
56
CHAPTER 3. GENERAL FINANCIAL MODEL
2. No Arbitrage I implies No Arbitrage II
Let us recall something that we veri…ed in Chapter 1 (speci…cally, Exercise 1):
!
q
Let
be a (S + 1) J matrix and z 2 RJ : The Farkas Lemma can be used
R
to prove the following implication.
There does not exist z such that
q
R
!
z > 0
+
9^ 2 RS+1
++ s.t. ^
q
R
!
= 0:
This tells us exactly what we need to …nish the argument.
You are asked to verify in Exercise 1 that No Arbitrage is a necessary condition of
equilibrium.
Let’s use the No Arbitrage II condition to see whether asset prices can be "No Arbitrage
prices" or not. The method that I am set to introduce was developed by Thorsten Hens,
and is henceforth known as the Hens method. The method is valid for economies with either
J = 2 or J = 3 assets.
Hens method: Two assets
Consider an economy with S
R is given by:
2 states of uncertainty and J = 2 assets. The payout matrix
2
3
r1 (1) r2 (1)
7
6
R=4 :
: 5:
r1 (S) r2 (S)
S
From No Arbitrage II, the asset prices are given by q = R for
! some 2 R++ : To see this,
q
take the No Arbitrage II de…nition, which says that ^
= 0 for some ^ >> 0: Then
R
de…ning
steps.
=
^ (1)
; :::; ^^(S)
^ (0)
(0)
; we obtain
q + R = 0: The Hens method proceeds in two
1. Plot the pairs (r1 (s) ; r2 (s)) in R2+ for all s > 0: As you can see in Figure 3.2, I
3.1. THE MODEL
57
2
3
0 1
6
7
have plotted the pairs corresponding to the payout matrix R = 44 25 (the x-axis
3 3
corresponds to j = 1 and the y-axis to j = 2).
2. The convex cone of this set f(r1 (1) ; r2 (1)) ; :::; (r1 (S) ; r2 (S))g can then be found. In
Figure 3.2, the convex cone is the shaded area.
3. If the asset prices q = (q1 ; q2 ) belong to the interior of this convex cone, then they are
No Arbitrage prices. Otherwise, they are not No Arbitrage prices.
It is important to notice that Step 3 requires the asset prices to lie in the interior of the
cone.
The No Arbitrage condition is a necessary condition for a general …nancial equilibrium
(Exercise 1). Consequently, any equilibrium asset prices must be No Arbitrage prices (the
converse is not true). Exercise 2 provides practice with the Hens method with two assets.
Hens method: Three assets
Consider an economy with S 3 states of uncertainty and J = 3 assets. To apply the Hens
method, we must take an initial step involving linear operations on the payout matrix R:
1. Given the payout matrix
! R; we need to perform appropriate linear operations on the
q
column space of
so that the equivalent matrix R has one asset (wlog asset
R
j = 1) with constant payouts of 1 in every state s > 0:
The !
equivalent asset price vector and payout matrix following the linear operations are
q
; where
R
2
3
1 r2 (1) r3 (1)
6
7
R = 41
:
: 5:
1 r2 (S) r3 (S)
2
3
2 0 1
6
7
As an example, suppose that R = 43 4 25 : Then the linear operation that I will perform
4 3 3
will be C1 7! C1 C3; meaning that I replace the 1st column with the di¤erence (1st
58
CHAPTER 3. GENERAL FINANCIAL MODEL
column) - (3rd column). This results in the equivalent asset price vector and payout matrix:
q
R
!
2
6
6
=6
6
4
(q1 q3 )
1
1
1
q2
0
4
3
q3
1
2
3
3
7
7
7
7
5
From No Arbitrage, the asset prices are given by q = R for some 2 RS++ : This
P
!
means that q1 =
1 or q1 = s>0 (s) : The Hens method continues with the following
steps.
2. Plot the pairs (r2 (s) ; r3 (s)) in R2+ for all s > 0: As you can see in Figure 3.3, I have
2
3
1 0 1
6
7
plotted the pairs corresponding to the payout matrix R = 41 4 25 :
1 3 3
3. The convex hull of this set f(r2 (1) ; r3 (1)) ; :::; (r2 (S) ; r3 (S))g can then be found. In
Figure 3.3, the convex hull is the shaded area inside the triangle.
q
q
4. If the asset prices q2 ; q3 belong to the interior of this convex hull, then they are
1
1
No Arbitrage prices. Otherwise, they are not No Arbitrage prices. Continuing with
my example, as q1 = q1 q3 ; the asset price pairs that I will actually be plotting are
q2
; q3
: Obviously, any prices such that q1 q3
0 automatically fail the No
q1 q3 q1 q3
Arbitrage requirement.
Notice two elements of Step 4. First, we require that the asset prices are all divided by
q1 : Second, we again require that the asset prices lie in the interior of the convex hull.
Exercise 3 provides practice with the Hens method and three assets.
3.1.4
No Redundancy
The No Redundancy condition assumes that the payout matrix R has full column rank
(that is, the assets are linearly independent). You are asked to verify in Exercise 4 that No
Redundancy is an innocuous assumption (in terms of the real variables) to make.
3.2. COMPLETE MARKETS
3.2
59
Complete Markets
Throughout this section, we assume that the No Arbitrage condition and the No Redundancy
condition both hold. The assumption of "Complete Markets" states that J = S (equal
number of assets and states of uncertainty). The payout matrix R is now a full rank square
matrix (hence, invertible). Let’s see what this gets us.
Examples of payout matrices with complete markets are given by:
2
3
1 0 0
6
7
R = 40 1 05
0 0 1
2
3
1 2 2
6
7
R = 41 1 25
1 1 1
I de…ne the column span of any payout matrix as
hRi = y 2 RS : y = Rz for some z 2 RJ :
(3.3)
For instance, under the assumption of complete markets, hRi = RS :
Given complete markets and the equilibrium asset prices q 2 RJ++ ; there exists a unique
vector of state prices. Recall the No Arbitrage pricing conditions:
q = R;
; :::; ^^(S)
: Given that R has full rank, then the state prices are uniquely
where = ^^ (1)
(0)
(0)
1
de…ned as = qR :
Suppose that the market clearing conditions hold for all commodities:
X
h2H
eh
xh = 0:
!
X
q
Walras’Law states that P
e h xh +
z h = 0 (sum of budget conh2H
h2H
R
straints equals zero). Given commodity market clearing and full column rank R; then
X
z h = 0: Thus, the market clearing conditions for assets are guaranteed to hold when
h2H
the market clearing conditions for commodities hold.
X
The following theorem (called Arrow’s equivalency theorem) proves that a general …nancial equilibrium with complete markets is allocation-equivalent to an Arrow-Debreu equilibrium.
60
CHAPTER 3. GENERAL FINANCIAL MODEL
Theorem 3.3 Arrow’s Equivalency Theorem
Under Assumptions E1-E4, if J = S and xh ; z h
librium, then
xh
h2H
;
any full rank payout matrix R; 9 z h
xh ; z h
h2H
; p; q is a general …nancial equi= (1; qR 1 ) P:
is an Arrow-Debreu equilibrium where
Under Assumptions E1-E3, if
vector
h2H
xh
h2H
h2H
;
is an Arrow-Debreu equilibrium, then for
such that for
(s) =
L (s)
L (0)
8s > 0 and q = R; the
; p; q is a general …nancial equilibrium.
Proof. General …nancial =) Arrow-Debreu
As the markets are complete, 9! state prices = qR 1 as discussed above. Let 1 2 RL
be the L dimensional row vector with 1 for each element. This will be the state price
for s 2= 0: Then the state3prices for all s 2 S are (1; ) : Recall that the price matrix
p(0) 0
0
0
6
7
6 0 p(1) 0
7
0
7 is a (S + 1) G matrix. Premultiply P by the state prices to
P =6
6 0
7
0
:::
0
4
5
0
0
0 p(S)
obtain the Arrow-Debreu prices 2 RG
+ :
= (1; ) P:
In particular, l (0) = pl (0) 8l 2 L and l (s) = (s) pl (s) 8(l; s) 2 L Snf0g:
The proof requires two steps. First, let xh ; z h be budget feasible under the general
…nancial equilibrium budget constraints:
xh ; z h 2 B h (p; q) =
(
(x; z) : P eh
x +
q
R
!
z
)
0 :
Premultiply the budget constraints by the state prices (1; ) : Then the budget constraints
reduce to
!
q
h
(1; ) P e
x + (1; )
z 0:
R
!
q
From No Arbitrage II, (1; )
= 0: Thus, we have eh x
0: So xh 2 bh ( ) =
R
x : eh x
0 :
For the second step, suppose that y h is budget feasible under the Arrow-Debreu equilib(s)
rium budget set bh ( ) : De…ne (s) = L(0)
8s > 0 and pl (s) = l (s)
8(l; s) 2 L S: For any
(s)
L
L
3.3. INCOMPLETE MARKETS
61
full rank R; de…ne q = R: There exists
0
p(1) eh (1)
B
zh = R 1 @
:
h
p(S) e (S)
xh (1)
xh (S)
1
C
A:
By construction, xh ; z h 2 B h (p; q) for the selected full rank payout matrix R:
To conclude, we pose the question: if xh ; z h 2 arg max uh (x) ; then since xh 2 bh ( ) ;
(x;z)2B h (p;q)
is it also true that x
h
h
2 arg max u (x)? Suppose not, that is, 9y 2 bh ( ) such that
x2bh ( )
h
h
h
h
u (y) > u x : If y 2 b ( ) ; then 9^
z 2 RJ such that (y; z^) 2 B h (p; q) : As uh (y) > uh xh
and (y; z^) 2 B h (p; q) ; then this contradicts that xh ; z h 2 arg max uh (x) :
(x;z)2B h (p;q)
Arrow-Debreu =) General …nancial
Choose any full rank payout matrix R: The proof requires three steps.
If xh is budget feasible under the Arrow-Debreu equilibrium budget set bh ( ) ; then
(s)
8s > 0; (ii) pl (s) = l (s)
using arguments from above, 9z h such that for (i) (s) = L(0)
L
L (s)
h h
h
8(l; s) 2 L S; and (iii) q = R; x ; z 2 B (p; q) :
If xh 2 arg max uh (x) ; then since 9z h such that xh ; z h 2 B h (p; q) ; is it also true that
x2bh ( )
h
x ;z
h
2 arg max uh (x)? Suppose not, that is, 9 (y; z^) 2 B h (p; q) such that uh (y) >
(x;z)2B h (p;q)
h
h
u x : If (y; z^) 2 B h (p; q) ; then y 2 bh ( ) (citing arguments from above). But this
contradicts that xh 2 arg max uh (x) :
x2bh ( )
For the third step, we use the previously stated fact that the market clearing conditions
for assets are redundant given that the market clearing conditions for commodities hold.
Given Arrow’s Equivalency Theorem, when the complete markets assumption holds, the
First Basic Welfare Theorem is applicable. That is, if J = S; all general …nancial equilibrium
allocations are Pareto optimal.
3.3
Incomplete Markets
Throughout this section, we assume that the No Arbitrage condition and the No Redundancy
condition both hold. "Incomplete Markets" means that J < S (fewer assets than states of
uncertainty). The payout matrix R is no longer invertible. How does this change things?
Consider any two payout matrices R0 ; R00 with the same column span: hR0 i = hR00 i (recall
62
CHAPTER 3. GENERAL FINANCIAL MODEL
the de…nition of the span in (3.3)). These two matrices are then equivalent up to a change of
basis. For our purposes in the general …nancial model, if a household h has portfolio payouts
R0 z h ; then there exists z^h 2 RJ such that R0 z h = R00 z^h : Thus, given any payout matrix with
0
full column "
rank R
# (No Redundancy condition holds), we consider the equivalent payout
R1
matrix R =
such that R1 2 RJ;J has full rank.
R2
With incomplete markets, given the equilibrium asset prices q 2 RJ++ ; there no longer
exists a unique vector of state prices. In fact, the set of state prices
: q = R is an
(S J) dimensional space.
3.3.1
Pareto suboptimal
Consider the characterization of Pareto optimal allocations using either the problem (P Oh0 )
or the problem (P O) from Chapter 2. A necessary condition for a Pareto optimal allocation
is:
8h 2 H; 1 = h h for some scalar h > 0:
Recall that h is the vector of Lagrange multipliers corresponding to the budget constraints
for household h:
1
2
(1)
Let’s add the constraint 1 (0)
= 2 (1)
to the system of equations : Now the number of
(0)
equations is one greater than the number of unknowns. Let’s apply "A Di¤erent Application
of Di¤erential Topology" from Exercise 3 in Chapter 1 (similar to Exercises 8 and 9 in
2
1
= 2 (1)
Chapter 2) to show that over a generic subset of endowments, the equation 1 (1)
(0)
(0)
cannot hold. If the equation cannot hold, as it is a necessary condition for a Pareto optimal
allocation, then the general …nancial equilibrium allocations are Pareto suboptimal (over a
generic subset of endowments).
Theorem 3.4 Under Assumption R1-R4, if J < S; then over a generic subset of endowments, the general …nancial equilibrium allocations are Pareto suboptimal.
Proof. Exercise 5.
Let’s consider an example that illustrates why Theorem 3.4 only holds over a generic
subset of household endowments (rather than over the entire set of household endowments).
In the following example, we have incomplete markets and an equilibrium allocation that is
Pareto optimal.
3.3. INCOMPLETE MARKETS
63
Let’s introduce some machinery 2that will be useful in
3 the example. De…ne the S G price
0 p(1) 0
0
6
7
matrix (for states s > 0) as Pn0 = 40 0 :::
0 5 (this is the matrix P with the …rst
0 0
0 p(S)
an any vector of state prices satisfying q = R: The
row removed). De…ne 2
! budget
q
zh
0:
constraints in a general …nancial equilibrium are given by P eh xh +
R
This is equivalent to the following two conditions:
RS++
1. Single budget constraint
(1; ) P eh
xh
0:
2. Span condition
Pn0 xh
eh 2 hRi :
The single budget constraint is satis…ed by both Pareto optimal and Pareto suboptimal
allocations. The key condition to determine Pareto optimality is the span condition.
X
xh (s)
Example 3.1 Let L = 1: All households have identical utility uh xh =
s2S
for some > 0: All households have endowments such that eh (s) s>0 2 hRi :
Since the utility is homothetic and identical across households, then there is a unique
general …nancial equilibrium. Let’s …rst …nd the Arrow-Debreu equilibrium and then show
that this is also the general …nancial equilibrium. Proceeding in this manner, we will obtain
the Pareto optimality as a byproduct.
You are asked to verify in Exercise 6 that the Arrow-Debreu equilibrium allocation is such
that:
X
xh (s) = h
eh (s) 8h 2 H
h2H
for some household-speci…c fractions
h
:
Given that L = 1; the span condition is equivalent to xh (s) eh (s) s>0 2 hRi : Since
eh (s) s>0 2 hRi 8h 2 H and the space hRi is linear, the span condition reduces to
xh (s) s>0 2 hRi 8h 2 H:
Do the household choices (in the Arrow-Debreu equilibrium) satisfy the span condition?
P
h
Well, since eh (s) s>0 2 hRi 8h 2 H (by assumption), then
h2H e (s) s>0 2 hRi : As
X
eh (s) and the space hRi is linear, then in fact xh (s) s>0 2 hRi 8h 2 H:
xh (s) = h
h2H
Thus, the Arrow-Debreu equilibrium allocation satis…es the span condition and therefore
64
CHAPTER 3. GENERAL FINANCIAL MODEL
satis…es the general …nancial equilibrium budget constraints. Consequently, the Arrow-Debreu
equilibrium is also a general …nancial equilibrium.
So we have a setting with incomplete markets (the example is valid for any J < S) and
yet, the general …nancial equilibrium allocation is Pareto optimal. Why? The endowments
eh (s) s>0 2 hRi 8h 2 H: Whenever J < S; the space hRi is a measure zero subset of RS :
Thus, the endowments in this example do not belong to a generic subset. We can see now
why the result in Theorem 3.4 holds only over a generic subset of endowments.
3.3.2
Constrained Pareto suboptimal
With incomplete markets, it seems unfair to compare the general …nancial equilibrium allocations with Pareto optimal allocations. The set of Pareto optimal allocations implicitly allows
the planner to make transfers between all households and all states of uncertainty. This is
not possible in a general …nancial equilibrium. In a general …nancial equilibrium, the possible
allocations are restricted by the …xed asset structure (transfers can only be made according
to the payout matrix R). In this section, we introduce the concept of constrained Pareto
optimality, where constrained Pareto optimal allocations are those that respect the …xed
asset structure. We now have a fair comparison: general …nancial equilibrium allocations vs.
constrained Pareto optimal allocations.
De…nition 3.2 The allocation xh h2H is constrained Pareto suboptimal if there exists equilibrium prices p~ and a feasible allocation x~h h2H 2 F A such that 8h 2 H;
x~h 2 arg max uh (x)
x:P~n0 (xh eh )2hRi
and uh x~h
h2H
> uh xh
h2H
:2
We will see shortly that this de…nition is equivalent to a planner …xing the asset choices
of households and making a transfer to each household in the initial state. The planner is
X
h
choosing h h2H for h 2 RJ+1 such that
= 0; where h (1) is the transfer (in
h2H
units of account) in state s = 0 and h (j + 1) for j 2 J is the holding of asset j: Given these
2
Equilibrium prices p~ simply means that
X
h2H
x
~h
eh = 0; which is a requirement of the set F A:
3.3. INCOMPLETE MARKETS
65
…xed transfers by the planner, the new budget constraints for each household become:
P eh
xh +
1 0
0 R
!
h
0:
The main result that we show in this section is that the general …nancial equilibrium
allocations (under some conditions) are not constrained Pareto optimal. To motivate this,
we …rst consider two examples. The examples dictate what conditions are required to obtain
this main result.
In the …rst example, we consider any economy with only one physical commodity traded
in each state (L = 1) :
Example 3.2 Consider any economy with L = 1: Suppose that the general …nancial equilibrium allocation xh h2H is constrained Pareto suboptimal. Then there exists x~h h2H such
that:
1. x~h 2
2.
X
arg max uh (x) 8h 2 H;
x:(xh (s) eh (s))
2hRi
s>0
h2H
3. uh x~h
x~h
h2H
eh = 0; and
> uh xh
h2H
:
But then xh h2H cannot be optimal choices for all households (as x~h is chosen under
the exact same budget constraints and some households strictly prefer x~h ). Thus, xh h2H
cannot be a general …nancial equilibrium allocation. This …nishes the argument that all
general …nancial equilibrium allocations (when L = 1) are constrained Pareto optimal.
Thus, we know that the main result (general …nancial equilibrium allocations are constrained Pareto suboptimal) requires L > 1:
In the second example, we consider an economy with L > 1 and a special form of utility
(namely, identical and homothetic [speci…cally, Cobb-Douglas] utility).
X
P
h
Example 3.3 Suppose that the utility functions are uh xh =
l2L l (s) log xl (s)
s2S
8h 2 H: The Cobb-Douglas utility has some nice properties. For each household, the consumption is given by:
X
xhl (s) = h (s)
ehl (s)
h2H
66
CHAPTER 3. GENERAL FINANCIAL MODEL
where
H 1
h
(s) 2 (0; 1) is determined according to a household’s endowments and h (s) h2H 2
P
eh (s)
l (s)
8(l; s) 2 L S: You
8s 2 S: The prices are given such that pl (s) = L (s) Ph2H Lh
e
h2H l
(s)
are asked to verify these previous two facts in Exercise 7.
In each state s 2 S the household choices can be reduced to simply the fraction h (s) 2
(0; 1) for each h 2 H (rather than the commodity vector xh (s) 2 RL+ ). Suppose that the
general …nancial equilibrium allocation xh h2H is constrained Pareto suboptimal. Then there
exists x~h h2H such that:
1. x~h 2
2.
X
arg max uh (x)
8h 2 H;
h
h
~
x: Pn0 (x (s) e (s))
2hRi
s>0
h2H
3. uh x~h
x~h
h2H
Thus, 9 ~ h (s)
eh = 0; and
> uh xh
(h;s)2H S
h2H
:
such that
X
x~hl (s) = ~ h (s)
h2H
X
~ h (s) = 1 8s 2 S and
h2H
ehl (s)
(h;l;s)2H L S
are the maximizers of the household problem (HP ) and uh x~h h2H > uh xh h2H : The
commodity prices don’t change with changes in ~ h (s) (h;s)2H S ; so the choices ~ h (s) (h;s)2H
lie in the budget sets for all households. This contradicts that xh h2H is a general …nancial
equilibrium allocation as some household is not optimally solving (HP ) :
With this second example, we know that the main result (general …nancial equilibrium
allocations are constrained Pareto optimal) is not valid for all utility functions. Rather it is
only valid for a generic subset of utility functions. The set of utility functions, unlike the set of
endowments, belongs to an in…nite-dimensional space. What does it mean to consider a full
measure subset of an in…nite-dimensional space? Well, this concept is not de…ned. Therefore,
we need to rede…ne the set of utility functions as belonging to a …nite-dimensional space.
This is taken up in the proof in Section 3.5.
Theorem 3.5 Under Assumption R1-R4 (strengthening R2 such that uh is C 3 ), if J < S
and L > 1; then over a generic subset of endowments and utility functions, the general
…nancial equilibrium allocations are constrained Pareto suboptimal.
Proof. See Section 3.5.
S
3.4. REAL ASSETS
3.4
67
Real Assets
In the discussion so far in this chapter, the assets were numeraire assets, meaning that the
assets paid out in the numeraire commodity l = L: In this section, the assets will be real
assets, meaning that they will pay o¤ in the vector of all commodities. By this speci…cation,
numeraire assets are a special case of real assets. With real assets, the asset j in state s > 0
has the vector of payouts yj (s) 2 RL+ : This vector is often called the "yields." The yields
vector is a column vector.
The value of asset j payouts in state s > 0 is then given by p (s) yj (s)
payout matrix is then the S J matrix:
0: The entire
2
3
p (1) y1 (1) ::: p (1) yJ (1)
6
7
R (p) = 4
:
:
5:
p (S) y1 (S) ::: p (S) yJ (S)
With numeraire assets, we can go ahead and assume that the payout matrix has full column
rank (our No Redundancy condition). With real assets, we can no longer make that same
assumption. The rank of the payout matrix R (p) is endogenously determined as a function
of the commodity prices p:
The de…nition of a general …nancial equilibrium remains the same as in Section 3.1.1.
In the next section, I show that a …nancial equilibrium may not exist when real assets are
considered. The section that follows will show how the approach developed by Du¢ e and
Shafer (1985) is able to prove the generic existence of a general …nancial equilibrium with
real assets.
3.4.1
Nonexistence
The following example was provided by Oliver Hart (1975). I present the Hart example in
its original form. I do this not because I am too lazy to construct my own example, but
because the original Hart example is beautiful in its simplicity.
In the example, Hart does not allow for consumption in the initial period. Such a
modeling choice was fashionable at the time. This does not change our model at all, except
that the utility function is now only de…ned over consumption in the states of uncertainty
s > 0:
Example 3.4 Consider an economy with two households (H = 2), two states of uncertainty
68
CHAPTER 3. GENERAL FINANCIAL MODEL
in the …nal period (S = 2), and two physical commodities traded in each state (L = 2).
Consumption only takes place in the …nal period, so the household is only endowed with
commodities in the …nal period. The endowments are:
e1 (1) =
e2 (1) =
5
;
2
1
;
2
50
21
13
21
e1 (2) =
e2 (2) =
13
;
21
50
;
21
1
2
5
2
:
The utility functions are given by:
u
1
x
1
u 2 x2
q
q
q
q
1:5
1
1
1
= 2
x1 (1) + x2 (1) + 2
x1 (2) + x12 (2)
q
q
q
q
1:5
1:5
2
2
2
=
x1 (1) + 2
x2 (1) + x1 (2) + 2
x22 (2)
1:5
There are two real assets that are traded in the initial period and pay out in the …nal
period (J = 2). The asset yields are:
y1 (1) =
y1 (2) =
!
1
y2 (1) =
0 !
1
y2 (2) =
0
!
0
1 !
:
0
1
We will show that a general …nancial equilibrium does not exist for this economy by
considering two cases.
Case I: The payout matrix R (p) has full column rank.
With a full rank payout matrix, using Arrow’s Equivalency Theorem (Theorem 3.3), the
general …nancial equilibrium allocation and commodity prices are in fact an Arrow-Debreu
equilibrium. The Arrow-Debreu budget constraints are:
X
(l;s)2L S
l
(s) xhl (s)
X
l
(s) ehl (s)
(l;s)2L S
for both h = 1; 2: The equilibrium market clearing conditions are:
x1l (s) + x2l (s) = e1l (s) + e2l (s) = 3 8 (l; s) 2 L
S:
Uniqueness of an Arrow-Debreu equilibrium is guaranteed from the utility functions. It is
3.4. REAL ASSETS
69
easy to show (Exercise 8) that the Arrow-Debreu equilibrium is given by:
x1 (1) = 38 ; 31
x2 (1) = 13 ; 38
(1) = (1; 1)
x1 (2) = 83 ; 31
x2 (2) = 13 ; 38 :
(2) = (1; 1)
The corresponding general …nancial equilibrium allocation is the same, with the commodity
prices given by p(1) = (1; 1) and p(2) = (1; 1) : The payout matrix is given by:
2
6p (1)
6
R (p) = 6
6
4
p (2)
!
1
p (1)
0 !
1
p (2)
0
!3
0
#
7 "
1 !7
1
1
7=
:
1 1
0 7
5
1
This payout matrix does not have full column rank as required for Case I.
Case II: The payout matrix R (p) does not have full column rank.
The real assets are then redundant, so the general …nancial equilibrium is found as if
only a single asset existed. The initial period budget constraint (recall, no consumption in
this period) is given by qz h
0 for both h = 1; 2: To satisfy market clearing for the lone
h
asset, then z = 0 for both h = 1; 2: Thus, the general …nancial equilibrium does not have any
…nancial transfers, meaning that the allocation in state s = 1 is the Arrow-Debreu equilibrium
allocation of that state considered in isolation (and likewise for state s = 2). Again, we know
that there is a unique Arrow-Debreu equilibrium in both states. It is easy to show (Exercise
9) that the general …nancial equilibrium is given by:
; 31
x1 (1) = 62
21 21
1 32
x2 (1) = 21
; 21
p (1) = (2; 1)
x1 (2) =
x2 (2) =
p (2) =
32 1
;
21 21
31 62
;
21 21
1
;1
2
:
The payout matrix is given by:
2
6p (1)
6
R (p) = 6
6
4
p (2)
!
1
p (1)
0 !
1
p (2)
0
!3
0
#
7 "
1 !7
2
1
7=
:
1
1
0 7
5
2
1
70
CHAPTER 3. GENERAL FINANCIAL MODEL
This payout matrix does have full column rank, contradicting that we are in Case II.
3.4.2
Generic existence
Consider the Hart (1975) example above. Is there something special about the example?
Well, the endowments and asset yields are specially chosen to arrive at contradictions in
both cases. Do we expect similar results to hold for arbitrarily chosen endowments and asset
yields? Well no, but how do we know that by restricting our attention to a generic subset
of endowments and asset yields we can guarantee the existence of an equilibrium? We know
this because Du¢ e and Shafer (1985) came along and introduced the method required to
prove such a claim.
I state the result of Du¢ e and Shafer and then brie‡y sketch the proof method. The
complete details of the proof have no place in a text like this, but the interested reader is
directed toward the well-written primary source.
Theorem 3.6 Under Assumption R1-R3, over a generic subset of endowments and asset
yields, a general …nancial equilibrium exists.
The proof method proceeds as follows.
Step 1: Let L be a J dimensional linear subspace of RS (think of this as the column
span of a payout matrix). De…ne a pseudo-equilibrium as xh h2H ; p; L such that
1. 8h 2 H; given p; xh is an optimal solution to the household problem (HP )
maximize uh xh
:
subject to xh 2 X h
h
h
Pn0 x
e 2L
(HP )
2. Markets clear
X
h2H
xhl (s) =
X
h2H
ehl (s) 8(l; s) 2 L
S:
Show that a pseudo-equilibrium always exists.
Step 2: If
xh
h2H
; p; L
is a pseudo-equilibrium, then we can show that, over a generic
subset of endowments and asset yields, there exists z h h2H ; q such that xh ; z h h2H ; p; q
is a general …nancial equilibrium with real assets. When we refer to a generic subset of asset
yields, we are looking at a generic subset of a Grassmanian manifold. This is because the
3.5. PROOF OF THEOREM 3.5
71
space of J dimensional linear subspaces of RS (the column span of a payout matrix) has
the structure of a topological manifold called the "Grassmanian manifold."
3.5
Proof of Theorem 3.5
This section contains the rather long proof of Theorem 3.5. The original proof is due to
Geanakoplos and Polemarchakis (1986), while the general method (which can be applied to
all sorts of Pareto-improving policies) is due to Citanna, Kajii, and Villanacci (1998).
The equilibrium variables are
=
xh ;
h
; zh
; p; q and the planner transfers were
h2H
h
previously introduced as =
: The principal task will be to show that the vector of
h2H
household utility functions U ( ; ) = u1 (x1 ); ::; uH (xH ) is a submersion (i.e., its derivative
mapping is surjective).
xh ;
Take as given a general …nancial equilibrium
= eh ; uh
h2H
; the variables ^ =
x^h ; ^
h
0
for F OCx =
P
h
h2H (enL (s)
h
^ P^
Duh (^
xh )
iT
h2H
; p; q : Given parameters
; p^ and transfers
constitute a constrained
B
B
(^; ; ) = B
B
@
F OCx
BC
M Cx
MC
P^ eh
; BC =
h2H
x^hnL (s))
; and M C
s2S
(S + 1) + J + 1 equations.
; zh
h2H
feasible vector i¤ (^; ; ) = 0; where
h
h
=
P
h2H
h
:
1
C
C3
C
C
A
x^h +
1 0
0 R
!
h
!
; M Cx =
h2H
has m = H (G + S + 1) + G
Picking a vector of parameters = eh ; uh h2H such that eh h2H belongs to a generic
subset of RHG
: In
++ ; then all resulting general …nancial equilibria are regular values of
0
particular, this means that there exists an open set
around such that for any parameters
2 0 ; the resulting equilibria satisfy the rank condition of (3.1). Let xh h2H be one such
equilibrium allocation given the parameters : Then, the set of allocations xh h2H in a local
neighborhood around xh h2H such that u1 (x1 ); ::; uH (xH ) >> u1 (x1 ); ::; uH (xH ) is open.
3
See the discussion in Section 3.3.2.
72
CHAPTER 3. GENERAL FINANCIAL MODEL
If for some planner transfers
= h h2H ; the resulting constrained feasible allocation is
Pareto superior, then all constrained feasible allocations for in an open neighborhood
around
are Pareto superior as well.
!
!
= 0 : By de…nition, if (^; 0 ; ) = 0
The analysis is conducted evaluating equations at
and ( ; ) = 0; then ^ = xh ; h h2H ; p :
De…ne the (H + m)
(m0 + H (J + 1)) matrix
0
=
0
:
!
0
;
(^; ; )
Dx; ;p U (^; )
Dx; ;p (^; ; ) D
where m0 = H (G + S + 1) + G (S + 1) is the number of variables in the vector (x; ; p) =
xh ; h h2H ; p and H (J + 1) is the number of planner transfers h h2H : From Citanna,
Kajii, and Villanacci (2002), if
0
has full row rank, 9^ =
6
xh ;
h
h2H
; p s.t. ^ satis…es
(^; ; ) = 0 (for some ) and U x^h h2H > U xh h2H : For full row rank, I must
ensure that there are fewer rows than columns, so I need the following inequality to hold:
H +m
m0 + H (J + 1) :
(3.4)
(3.4) reduces to
H +J +1
(3.5)
H (J + 1) :
(3.5) is always satis…ed as H
2: The matrix 0 is square if H + J + 1 = H (J + 1) ; but
if H + J + 1 < H (J + 1) ; then there are more columns than rows and I must remove some
columns (it does not matter which) in order to obtain a square matrix : This matrix
!
does not have full rank i¤ 9 2 RH+m s.t. 0 (^; 0 ; ; ) = 0 where
0
!
(^; 0 ; ; ) =
T
T
=2
1
!
:
For simplicity, I divide the vector T into subvectors that each represent a certain equation
T
T
in : De…ne T =
u T ; xT ;
; pT ;
2 RH+m where each subvector corresponds
3.5. PROOF OF THEOREM 3.5
sensibly to an equation (row) in
73
as follows:
uT () Dx;
^; )
;p U (
xT () F OCx
T
() BC
pT () M Cx
T
A subset of the equations T
to (xh ; h )h2H in that order):
uh Duh (xh ) +
x
h T
xh
T
() M C :
= 0 are given by (corresponding to derivatives with respect
h T
D2 uh (xh )
P
pT
= 0: (3:6:a)
h2H
PT
= 0:
(3:6:b)
(3.6)
h2H
where
2
6
=6
4
IL
1
0
0
j 0
0
0
:::
0
0
IL
1
3
j 0
7
7 as in Subsection 3.1.2.
5
The proof is complete if I can show that for a generic choice of
exist ( ; ) s.t.
2
; there does not
(3.7)
( ; ) = 0
0
!
((x; ; p) ; 0 ; ; ) = 0:
Counting equations and unknowns, (3.7) has n equations in ; n variables ; H + m +
1 equations in 0 ; and only H + m variables : I must show that over a generic subset
of parameters!(exactly which generic subset will be discussed next), the derivative matrix
Dx;
;p;
0
has full row rank. I will reference the (N D) condition of Citanna, Kajii, and
!
Villanacci (1998), which is a su¢ cient condition for the full row rank of Dx;
;p;
0
:
74
CHAPTER 3. GENERAL FINANCIAL MODEL
The condition states that for
!
= 0 and ^ =
T
T
where
!
D
0
xh ;
!
h
h2H
; p ; the matrix
has full row rank,
(3.8)
are the parameters on which the genericity statement is made.
T
For simplicity, I break up the analysis into two cases: Case I:
xh 6= 0 8h 2 H and
T
Case II:
xh = 0 for some h 2 H: In Case I, I show that (3.8) holds over a generic subset
of parameters. In Case II, I show that (3.7) will generically not have any solution.
3.5.1
Case I:
xh
T
6= 0 8h 2 H
!
_h
d
A
0
Lemma 3.1 For
=
where d A_ h
!
0
corresponds to the rows for derivatives with respect to xh h2H :
!
= 0 ; then Du
has full row rank and
Proof. The set U is in…nite-dimensional and is endowed with the C 3 uniform convergence
topology on compact sets. This means that a sequence of functions fu g converges uniformly
to u i¤ fDu g; fD2 u g; and fD3 u g uniformly converge to Du; D2 u; and D3 u, respectively.
Additionally, any subspace of U is endowed with the subspace topology of the topology of
U: I will use the regularity result from Theorem 3.2 to de…ne utility functions as locally
belonging to the …nite-dimensional subset A U:
Using Theorem 3.2, pick a regular value : For that , there exist …nitely many equilibria
0h
0
h
0h
and A0h
i ; i = 1; :::; I: Further, there exist open sets
i s.t. xi 2 Ai ; the sets Ai are disjoint
0h
across i; and 8 2 0 ; 9! equilibrium allocation xhi 2 A0h
i : Choose Ai such that the closure
0h
~0h
clA0h
clA0h
A~0h
i is compact and there exist disjoint open sets Ai s.t. Ai
i
i :
For each household, de…ne a bump function h : X h ! [0; 1] with I bumps as h = 1 on
h
c
h
A0h
= 0 on (A~0h
G symmetric matrix Ah by:
i and
i ) : Now, I de…ne u in terms of a G
uh (xh ; Ah ) = uh (xh ) +
1
2
h
(xh )
X
i
(xh
xhi )T Ah (xh
xhi ) :
Thus, the space of symmetric matrices (denote this as A) is a …nite dimensional subspace
of U: Since A has the subspace topology of U; then uh ( ; A ) ! uh ( ; A) i¤ A ! A: This
can be seen by taking derivatives and noting that the function u stays …xed at the regular
value.
3.5. PROOF OF THEOREM 3.5
75
Taking derivatives with respect to xh 2 A0h
i yields:
Dx uh (xh ; Ah ) = Duh (xh ) + Ah (xh
xhi )
2
Dxx
uh (xh ; Ah ) = D2 uh (xh ) + Ah :
A is a G(G + 1)=2 dimensional space, so write Ah as the vector
(Ahi;i )i=1;::;G ; (Ahi;j )i<j;i=1;::;G
1
:
xh :
2
by
Postmultiply Dxx
2
Dxx
uh (xh ; Ah ) xh = D2 uh (xh ) xh + Ah xh :
Taking derivatives with respect to the parameter uh is equivalent to taking derivatives
with respect to Ah :
2
Du Dxx
uh (xh ; Ah ) xh = DA Ah xh
1
0
0
xh1 0
C
B
= @ 0 :::
0
(1) :::
(G 1) A 2 RG;G(G+1)=2
0
0
xhG
where the submatrix
(i) is de…ned as
0
0 2 Ri 1;G
B 0 h
B
x
:::
B B i+1
h
(i) = B
xi
0
B B
B B
B
:::
@ @ 0
0
0
Thus, since
xh 6= 0 (without loss of generality
i
1
xhG
C
0 C
C
0 C
A
h
xi
1
C
C
C
C 2 RG;G i :
C
C
A
xh1 6= 0), then
2
uh (xh ; Ah ) xh = G:
rankDAh Dxx
Out of all the rows
T
(3.9)
; the utility function uh only appears in the row for derivatives
76
CHAPTER 3. GENERAL FINANCIAL MODEL
with respect to xh : This row in
U ( ; ^)
Duh (xh )
T
for household h is given by (as in (3:6:a)):
F OCx
D2 uh (xh )
T
DAh
Thus, taking the derivative DAh 0 =
0
h
the rows for derivatives with respect to x :
DAh
Duh (xh )
= DAh
T
BC M Cx M C
T
PT
0
T
!
; the only nonzero terms correspond to
uh + D2 uh (xh ) xh
Duh (xh ; Ah )
T
:
PT
h
T
p
uh + DAh D2 uh (xh ; Ah ) xh :
From the …rst derivative, Dx uh (xh ; Ah ) = Du(xh ) + Ah (xh xhi ) = Du(xh ) evaluated at
!
= 0 (since xh = xhi ). Thus DAh (Dx uh (xh ; Ah ) uh ) = 0: Using (3.9),
d A_ h
2
3
DA1 (D2 u1 (x1 ; A1 ) x1 ) 0
0
6
7
=4
0
:::
0
5
2 H
H
H
H
0
0 DAH D u (x ; A ) x
is a full row rank matrix of size HG
T
The matrix
T
!
DA
0
!
HG(G + 1)=2:
is given below (where the rows correspond to the equi-
librium variables (xh ; h )h2H ; p ; planner transfers ( ); and vector v T in that order). Recall
from (3.8) that I aim to show that this matrix has full row rank. I use the convention
0
0 1
1
1
M1
M 0
0
B
C
B
C
c M h = @ : A ; r M h = M 1 ::: M H ; and d M h = @ 0 :::
A ; where
MH
0 0 MH
c stands for column, r stands for row, and d stands for diagonal. For simplicity, de…ne
3.5. PROOF OF THEOREM 3.5
=
!
1 0
: The matrix
0 R
0
B
B
B
B
B
B
B
@
77
!
T
T
!
0
DA
is given by:
d Duh (xh )T
d D 2 uh
d( P T )
0
d( P)
0
0
h T
)
2
r(
0
0
h
r( u )
r(
x
Zh
r
T
T
d
h T
T
c(
)
d A_ h
0
0
0
0
0
0
0
0
h T
) r(
)
p
c (IS+1 )
T
0
T
0
1
C
C
C
C
C
C
C
A
where I de…ne the following two submatrices:
h
2
Z
h
Lemma 3.2
2
6
6
6
= 6
6
6
4
2
6
6
= 6
4
h
(0) IL
0
!
1
0
0
0
:::
0
0
0
h
ehnL
uh
xhnL
(0)
h2H
(S) IL
0
1
T
(0)
7
7
7
7 2 RG;G
!7
7
5
0
0
0
:::
0
0
0
:
3
xhnL (S)
ehnL (S)
(S+1)
7
7
7 2 RS+1;G
T5
(S+1)
:
; p 6= 0:
Proof. Suppose not, that is
uh ; p = 0 for some h: From (3:6:a);
xh
Post-multiply (3.10) by
3
T
D2 uh (xh )
h T
(3.10)
P = 0:
xh and use (3:6:b) to yield:
xh
From (3.11) and Assumption R2, then
T
D2 uh (xh ) xh = 0:
(3.11)
xh = 0: But this contradicts that we are in Case I.
T
From Lemmas 3.1 and 3.2, the …rst and last row blocks of
T
!
DA
0
!
are lin-
78
CHAPTER 3. GENERAL FINANCIAL MODEL
h
2;
early independent from the others. By the de…nition of
of size [H(S + 1) + G
0
B
B
the submatrix B
B
@
P
0
0
T
( 12 )
HG is a full rank matrix.
d T
! The submatrix
!
(S + 1)]
0
0
::
0
0
P
H
::
2
also has
T
full rank (as R has full column rank). Thus,
DA
T
0
has full rank and this
concludes the proof under Case I.
3.5.2
Case II:
T
xh
= 0 for some h 2 H
I will show that over a generic subset of endowments, (3.7) has no solution. Suppose 9h0 2 H
0 T
such that
xh
= 0: From (3:6:a) and ; I obtain
0
0
T
h0
0
uh Duh (xh )
P
h0
pT
= 0
h0
P = 0
h0
Du (x )
T
h0
pT = 0 and
which together imply that
=
0
uh
h0
:
For all other h 6= h0 ; postmultiply uh Duh (xh ) by xh and use the …rst order condition
(consumption) in and (3:6:b) to get uh Duh (xh ) xh = 0: Next, postmultiply (3:6:a) by
xh and use (3:6:b) to arrive at the equation:
xh
xh
By Assumption R2,
uh h 8h 2 H:
T
T
D2 uh (xh ) xh = 0:
= 0 8h 2 H: From what remains of (3:6:a) and
h
From the column for derivatives with respect to
h T
As
=
!
1 0
; then (3.13) implies
0 R
The terms
uh ;
(3.12)
h T
;
h2H
+
h
T
T
h2H
1
h T
=
;
= 0 8h 2 H:
(0) =
;
(3.13)
(0) 8h 2 H:
are the only nonzero elements of
: Further,
1
T
C
C
C
C
A
3.6. EXERCISES
uh
h
(0) =
u1
79
1
(0) 8h 2 H: This yields:
1
h
u =
u
1
h
(0)
8h 2 H:
(0)
(3.14)
The equation from T = 0 corresponding to derivatives with respect to p (s) (for any
h
s > 0) is given as follows (after replacing
(s) with uh h (s)):
X
h2H
h
uh
(s) ehnL (s)
xhnL (s)
T
= 0:
(3.15)
For the analysis to hold at this point, I must use the assumption that L > 1 (as there
are no commodity price variables when L = 1). In (3.15), only consider the …rst physical
commodity l = 1 and use (3.14):
u1
1
(0)
X
h
h2H
h
(s) h
e1 (s)
(0)
xh1 (s)
T
= 0:
As with Theorem 3.4 and "A Di¤erent Application of Di¤erential Topology" from Exercise
3 in Chapter 1, we can show that over a generic subset of endowments, the equation
X
h2H
h
h
(s) h
e1 (s)
(0)
xh1 (s)
T
6= 0:
T
h
This implies that u1 = 0; with (3.14) implying
uh h2H = 0: Consequently,
=0
T
T
0
8h 2 H: From (3.13),
= 0: In conclusion,
= 0; which cannot satisfy
as this
system contains the equation T =2 = 1: This completes the argument that (3.7) cannot
hold (generically), meaning that Case II is not possible (generically).
3.6
Exercises
1. Show that No Arbitrage is a necessary condition of equilibrium. That is, show that if
No Arbitrage does not hold, then a general …nancial equilibrium does not exist.
80
CHAPTER 3. GENERAL FINANCIAL MODEL
2. Suppose that there are S = 4 states and J = 2 assets with payouts
2
2
6
61
R=6
60
4
3
3
3
7
17
7:
17
5
2
Which of the following asset prices satisfy No Arbitrage (hint: Hens method): (i)
q = (0; 2) ; (ii) q = (2; 1:75) ; and (iii) q = (2; 1:25)?
3. Suppose that there are S = 4 states and J = 3 assets with payouts
2
3
1 0 0
6
7
60 1 17
7
R=6
62 1 17 :
43
35
1
2
2
3
1
2
Which of the following asset prices satisfy No Arbitrage: (i) q = (1; 4; 2) ; (ii) q =
(2; 1; 1) ; and (iii) q = (3; 2; 1)?
4. Show that it is innocuous to assume No Redundancy. In other words, show that if
xh ; h ; z h h2H ; p; q is a general …nancial equilibrium in which No Redundancy does
not hold, then there exists z^h h2H ; q^ such that xh ; h ; z^h
…nancial equilibrium in which No Redundancy does hold.
h2H
; p; q^ is a general
5. Prove Theorem 3.4.
6. In Example 3.1, verify that the Arrow-Debreu equilibrium allocation is such that 8h 2
X
H : xh (s) = h
eh (s) for some h :
h2H
7. In Example 3.3, verify that the general …nancial equilibrium consumption is given by:
xhl (s) =
for some
h
h
(s)
X
h2H
ehl (s)
(s) 2 (0; 1) and the prices are given such that pl (s) =
l (s)
L (s)
8(l; s) 2 L
S:
8. In Example 3.4, verify that the Arrow-Debreu equilibrium allocation and commodity
3.6. EXERCISES
81
prices in Case I are given by:
x1 (1) = 83 ; 13
x2 (1) = 13 ; 83
(1) = (1; 1)
x1 (2) = 83 ; 31
x2 (2) = 13 ; 38 :
(2) = (1; 1)
9. In Example 3.4, verify that the general …nancial equilibrium allocation and commodity
prices in Case II are given by:
62 31
x1 (1) = 21
; 21
1 32
2
x (1) = 21 ; 21
p (1) = (2; 1)
x1 (2) =
x2 (2) =
p (2) =
32 1
;
21 21
31 62
;
21 21
1
;1
2
:
82
CHAPTER 3. GENERAL FINANCIAL MODEL
Bibliography
[1] Citanna, Alessandro, Atsushi Kajii, and Antonio Villanacci (1998): "Constrained Suboptimality in Incomplete Markets," Economic Theory 11, 495-521.
[2] Du¢ e, Darrell and Wayne Shafer (1985): "Equilibrium in Incomplete Markets I: A Basic
Model of Generic Existence," Journal of Mathematical Economics 14, 285-300.
[3] Geanakoplos, John and Herakles Polemarchakis (1986): "Existence, Regularity, and Constrainted Suboptimality of Competitive Allocations when the Asset Market is Incomplete,"
in Uncertainty, Information, and Communication: Essays in Honor of K.J. Arrow, Vol.
3, ed. by Walter P. Heller, Ross M. Starr, and David A. Starrett (Cambridge University
Press: Cambridge).
[4] Hart, Oliver D. (1975): "On the Optimality of Equilibrium when the Market Structure
is Unique," Journal of Economic Theory 11, 418-443.
[5] Magill, Michael and Martine Quinzii (1996): Theory of Incomplete Markets, Vol. 1 (MIT
Press: Cambridge, MA).
[6] Villanacci, Antonio, Laura Carosi Carosi, Pierluigi Benevieri, and Andrea Battinelli
(2002): Di¤erential Topology and General Equilibrium with Complete and Incomplete Markets (Kluwer Academic Publishers: Boston).
83
84
BIBLIOGRAPHY
Chapter 4
Incomplete Markets and Money
In the previous chapter, we made the following price normalizations in all states s 2 S :
pL (s) = 1: Does this choice matter? Would a di¤erent price normalization a¤ect the equilibrium consumption? Consider the budget constraints in the previous chapter, when we have
numeraire assets:
p (s) eh (s)
xh (s) + pL (s)
P
j2J
rj (s) zjh
0 for s > 0:
Thus, any other price normalization (pL (s) = k for some k > 0 or p (s) 2 L 1 ) does not
change the budget constraint. Price normalizations are neutral; they have no real e¤ects. The
conclusion with numeraire assets (as well as with real assets, the generalization of numeraire
assets) is that nominal indeterminacy does not imply real indeterminacy.
That conclusion does not hold when we look at a third type of assets: nominal assets.
Nominal assets are those that pay o¤ in the unit of account in all states s > 0: The unit of
account can be thought of as the currency of the economy. Assets paying o¤ in the unit of
account change the wealth of households, but not through the promise of payment in a real
commodity.
This chapter introduces a two-period general equilibrium model with uncertainty and
nominal assets. In principle, we are free to make any price normalizations that we wish.
Yet, in this context, it makes sense to include an institution that will pin down the price
normalizations. That institution is a monetary exchange that issues money. The supplies of
money will determine the nominal price levels.
In Section 4.1, I introduce the model. In Section 4.2, I consider an asset structure
with complete markets. With complete markets, nominal indeterminacy does not imply
85
86
CHAPTER 4. INCOMPLETE MARKETS AND MONEY
real indeterminacy (this is the same result obtained for the numeraire/real asset case). In
Section 4.3, I consider an asset structure with incomplete markets. With incomplete markets,
nominal indeterminacy can imply real indeterminacy (the main result is Theorem 4.5). This
is the problem (or the reality) with nominal assets.
4.1
The Model
The uncertainty of the dynamic model remains unchanged from Chapter 3. Additionally,
the real side of the model remains the same. The household primitives are:
X h = RG
+ is the consumption set.
uh : X h ! R is the utility function. We assume that the function is continuous, locally
non-satiated, and quasi-concave.
eh 2 X h is the endowment.
Each good (l; s) has a market price pl (s) : The vector of all prices is p = (pl (s))(l;s)2L S 2
R nf0g: The convention is that 2the price vector p (s) 3is a row vector 8s 2 S: I de…ne the
p(0) 0
0
0
6
7
6 0 p(1) 0
0 7
6
7 : The price normalizations will be
(S + 1) G price matrix P = 6
7
0
0
:::
0
4
5
0
0
0 p(S)
determined as a function of the money supplies. The money supplies are Ms > 0 for all
states s 2 S: The vector M = (Ms )s2S is a parameter of the model.
G
There are J nominal assets that are traded in the initial period. Denote the price of an
asset j in the initial period as qj : Denote all asset prices as q = (qj )j2J : These prices are
variables in the equilibrium and will be determined to satisfy market clearing conditions.
In state s > 0; the asset j has a …xed payout rj (s): The payouts are made in terms
of the unit of account. The payouts are assumed to be nonnegative rj (s)
0 8 (j; s) 2
J Snf0g; where rj = (rj (s))s>0 > 0 8j 2 J : Let’s collect the payouts of all assets in all
states s > 0 into one
2 payout matrix 3R: This payout matrix has S rows and J columns:
r1 (1) ::: rJ (1)
h
i
6
7
R = r1 ::: rJ = 4 :
:::
: 5:
r1 (S) ::: rJ (S)
4.1. THE MODEL
87
Household h chooses a portfolio z h 2 RJ in the initial period, where the position for a
particular asset j is denoted zjh 2 R:
The timing of the model is as follows. There are two distinct entities: households and the
Monetary Exchange. There are three substages in the initial period. In the …rst substage, all
households turn over their endowments to the Monetary Exchange. In return, the Monetary
Exchange provides each household with mh (0) = p (0) eh (0) units of account (view this as
X
currency), so that
mh (0) = M0 : In the second substage, households trade the nominal
h2H
assets. Following this trade, each household has available the following amount of the unit
of account: m
^ h (0) = mh (0) qz h : Given this amount of the unit of account, the households
then enter the third substage of the initial period, in which they purchase commodities for
consumption with the monetary resources that they have available:
p (0) xh (0)
m
^ h (0) :
After the purchase of commodities, the money in the market
to the Monetary Exchange.
X
h2H
m
^ h (0) = M0 is returned
A similar story unfolds in each state s > 0: Three substages take place. In the …rst,
households sell their endowments to the Monetary Exchange and receive mh (s) = p (s) eh (s)
X
mh (s) = Ms : In the second substage, the payouts of the
units of account, so that
h2H
nominal assets are accounted for, so each household now has available the following amount
of the unit of account: m
^ h (s) = mh (s) + r (s) z h : In the third substage, households use this
money to pay for the purchase of commodities:
p (s) xh (s)
m
^ h (s) :
The money in the market from the sale of commodities, of size
returned to the Monetary Exchange.
X
h2H
m
^ h (s) = Ms ; is
The story may seem involved, but it is a simple means to introduce a cash-in-advance
constraint (also called a Clower constraint in recognition of Robert Clower’s work on monetary models). In this setup, the velocity of money is set equal to 1 by de…nition (a single
unit of account can only be used in one purchase). The monetary elements of the model are
summarized succinctly in the de…nition of a …nancial equilibrium with money.
88
CHAPTER 4. INCOMPLETE MARKETS AND MONEY
4.1.1
Existence
De…nition 4.1 A …nancial equilibrium with money is
xh ; z h
h2H
; p; q such that
1. 8h 2 H; given (p; q) ; xh ; z h is an optimal solution to the household problem (HP )
maximize uh xh
subject to xh 2 X h
z h 2 RJ
(HP )
P eh
xh +
q
R
!
:
zh
0
2. Markets clear
X
h2H
X
xhl (s) =
h2H
X
zjh = 0
h2H
ehl (s) 8(l; s) 2 L
S:
8j 2 J :
3. Monetary supply condition
p (s)
X
h2H
xh (s) = Ms 8s 2 S:
It is important to verify that a …nancial equilibrium with money exists. The assumptions
for this are given by:
Assumption E1:
X h = RG
+ 8h 2 H:
Assumption E2:
uh : X h ! R is C 1 ; increasing, and quasi-concave 8h 2 H:
Assumption E3:
eh >> 0 8h 2 H:
Assumption E4 (No Redundancy): The payout matrix R has full column rank
(that is, the assets are linearly independent).
Assumption E5:
M >> 0:
Theorem 4.1 Under Assumptions E1-E5, a …nancial equilibrium with money
exists.
xh ; z h
h2H
; p; q
Proof. The proof is identical to the proof of Theorem 3.1, which is simply an extension of
the …rst existence proof showing the existence of an Arrow-Debreu equilibrium (see Theorem
2.1 and the proof in Section 2.6).
4.2. COMPLETE MARKETS
4.2
89
Complete Markets
Under Assumption E4 (No Redundancy), complete markets means simply that J = S: Under
this assumption, the Arrow’s Equivalency Theorem (Theorem 3.3) is still valid. Thus, all
…nancial equilibrium with money allocations are equivalent to Arrow-Debreu equilibrium
allocations. Therefore, the allocations are Pareto optimal (First Basic Welfare Theorem; see
Theorem 2.2). Additionally, all equilibria satisfy Finite Local Uniqueness (see Theorem 2.4).
This leads us to the following "immunity" result stating that nominal indeterminacy does
not imply real indeterminacy.
Theorem 4.2 Assume E1-E5 and J = S: With the money supply M = (M0 ; :::; MS ) ; the
…nancial equilibrium with money is xh ; z h h2H ; p; q : Consider a change in money supply
to M 0 = (M00 ; :::; MS0 ) : Then the new …nancial equilibrium allocation is still xh
h2H
:
This result can equivalently be called the "neutrality of monetary policy."
4.3
Incomplete Markets
This section shows that with incomplete markets, nominal indeterminacy can imply real
indeterminacy. This result that I’ll show can equivalently be called the "non-neutrality of
monetary policy." Much of the material in this section is borrowed from Chapter 7 of the
Magill and Quinzii (1996) text.
: De…ne
Let’s de…ne the purchasing power of money for all states s 2 S : s = P 1
p (s)
l2L l
2
3
0
1 0
6
7
= ( s )s>0 and the diagonal matrix [ ] = 4 0 :: 0 5 :
0 0 S
We scale the payout matrix by the purchasing power of money. The budget constraints
for states s > 0 are given by:
[p] xh eh = [ ] Rz;
2
3
0 P p(1)
0
0
p (1)
6
7
l2L l
6
7
where I de…ne [p] = 60
0
:::
0
7 2 RS;G : Recall that the span of the
4
5
0
0
0 P p(S)
l2L
S
pl (S)
payout matrix is h[ ] Ri = x 2 R : x = [ ] Rz for some z 2 RJ : The budget constraints
are equivalently expressed by the following two conditions.
90
CHAPTER 4. INCOMPLETE MARKETS AND MONEY
1. Single budget constraint
(1; ) [p] eh
for any vector of state prices
xh
0
2 RS++ :
2. Span condition
[p] xh
eh 2 h[ ] Ri :
We make the following assumptions.
Assumption M1:
2
Assumption M2:
3
1
6 7
4 : 5 2 hRi :
1
J < S:
Assumption M3:
J
H:
Theorem 4.3 Under Assumptions E1-E5 and M1-M3, there is a subspace
RS with
dim
S J such that if a change in the monetary policy from M to M 0 implies a change
from to 0 with 0
2 ; then h[ ] Ri =
6 h[ 0 ] Ri :
Before we walk through the proof of this theorem, let’s see what changes of monetary
policy do not satisfy the theorem’s conditions. For M = (M0 ; M1 ; :::; MS ) ; consider the
change to M 0 = (M00 ; M10 ; :::; MS0 ) such that Ms0 = Ms 8s > 0 and for some > 0: Then
[ 0 ] = 1 [ ] ; so the span remains unchanged: h[ ] Ri = h[ 0 ] Ri : Thus, for this proportional
monetary policy, there are no real e¤ects (the policy is neutral).
Now, let’s consider the proof of the theorem.
Proof. The span of the payout matrix h[ ] Ri has dimension equal to J (from Assumption
E4). Thus, the space that is orthogonal to h[ ] Ri ; denoted h[ ] Ri? ; has dimension equal
to S J: De…ne
= h[ ] Ri? n f0g : From the statement of the theorem, 0
2 : It
?
0
0
0
must be that 2
= h[ ] Ri (if not, then
2 h[ ] Ri and
2 h[ ] Ri ; a contradiction
0
0
0
of
6= ). Under Assumption M1,
2 h[ ] Ri (as there exists z = (1; 0; :::; 0) such that
0
0
0
0
= [ ] Rz). If 2
= h[ ] Ri and 2 h[ 0 ] Ri ; it can only be the case that h[ ] Ri =
6 h[ 0 ] Ri :
Theorem 4.3 is nice, but if we make one additional assumption, then we are able to prove
a much stronger result. Speci…cally, the subspace of possible policy changes can be shown
to have dimension S 1 (bigger than S J).
4.3. INCOMPLETE MARKETS
Assumption M4:
has rank J:
91
R is in general position, meaning that any J
J submatrix
Theorem 4.4 Under Assumptions E1-E5 and M1-M4, there is a subspace
RS with
dim = S 1 such that if a change in the monetary policy from M to M 0 implies a change
from to 0 with 0
2 ; then h[ ] Ri =
6 h[ 0 ] Ri :
The proof of Theorem 4.4 is more challenging than the proof of Theorem 4.3, but our
e¤orts are rewarded with the knowledge that the asset span will change for any policy changes
in a (S 1) dimensional subset :
Proof. Suppose that h[ ] Ri = h[ 0 ] Ri : This implies that for any j 2 J; the vector [ 0 ] rj is
a linear combination of the vectors [ ] r1 ; :::; [ ] rJ : We can express these linear combinations
using the J J matrix C such that:
[ 0 ] R = [ ] RC:
(4.1)
(4.1) written for any state s > 0 is equivalent to the following form:
C T [r(s)]T =
0
s
[r(s)]T :
(4.2)
s
0
If we recall our lessons on eigenvalues, (4.2) implies that ss is an eigenvalue, with associated
eigenvector [r(s)]T ; of the matrix C T : Notice that the same can be said for any state s > 0:
Now this is where it gets interesting. Further recalling our lessons on eigenvalues (a good
reference is Strang, 2006), if two eigenvalues are distinct, then their corresponding eigenvectors must be linearly independent. Assume that the eigenvalue for state s is di¤erent from
0
0
the eigenvalue for state : ss 6= : Then this implies that [r(s)]T is linearly independent
from [r( )]T : With J < S; all vectors [r(s)]T cannot be linearly independent from the vectors
all other states 6= s: Thus, we are able to …nd a collection of J or fewer vectors
n for o
T
[r(s)]
such that the number of states s 2 S is greater than the rank of the submatrix
s2S
[r(s)]s2S : This is a contradiction of the general position assumption (Assumption M4).1
0
0
We conclude that 8s; > 0 : ss =
=
that do not change the span are of the form
> 0: Thus, the only changes from to 0
0
s 8s > 0: These changes lie in the
s =
1
Adding a row vector r(s) to a matrix increases the rank by at most 1. Add states to the submatrix
[r(s)]s2S until there are J rows. The rank of the new submatrix is then at most J 1: General position
requires that a submatrix with any J rows of R has full rank.
92
CHAPTER 4. INCOMPLETE MARKETS AND MONEY
one-dimensional subspace de…ned by the vector
0
All other changes, those in the set
change the span: h[ ] Ri =
6 h[ 0 ] Ri :
0
:
2 h i:
2 h i? ; where h i? has dimension S
1; will
The following lemma is crucial for proving the main result of this chapter (Theorem 4.5).
You will get the opportunity to verify Lemma 4.1 through a series of exercises (Exercises
1-4) at the end of the chapter.
Lemma 4.1 Under Assumptions E1-E5 and M1-M3, over a generic subset of household
endowments and money supplies, the vectors ah h2H span h[ ] Ri ; where ah 2 RS is de…ned
by:
ah = [p] xh eh 8h 2 H:
Equivalently, this lemma states that the collection of J vectors ah h2J are linearly independent. This brings us to the main result of the chapter (which can be equivalently called
the "non-neutrality of monetary policy").
Theorem 4.5 Under Assumptions E1-E5 and M1-M4, over a generic subset of household
endowments and money supplies, there is a subspace
RS with dim = S 1 such
that all changes in the monetary policy from M to M 0 that induce a change from to 0
with 0
2 ; are non-neutral. That is, the …nancial equilibrium with money allocation
changes: x = xh h2H 6= x0 = x0h h2H :
Proof. From Theorem 4.4, h[ ] Ri =
6 h[ 0 ] Ri : As the dimensions dim h[ ] Ri = dim h[ 0 ] Ri =
J (Assumption E4), then dim (h[ ] Ri \ h[ 0 ] Ri) < J: By Lemma 4.1, the vectors ah h2H
span h[ ] Ri ; so some of the vectors cannot be elements of the set h[ ] Ri \ h[ 0 ] Ri : Since
the vectors a0h = [p0 ] x0h eh 2 h[ 0 ] Ri 8h 2 H by de…nition, then ah h2H 6= a0h h2H :
h
By the de…nition of ah = [p] xh eh and …rst order conditions Duh xh
P = 0; then
h
0h
this implies x h2H 6= x h2H ; …nishing the argument.
4.4
Exercises
Exercises 1-4 walk through all steps required to prove Lemma 4.1.
4.4. EXERCISES
93
1. Show that the vectors ah
are linearly independent.
h2J
are linearly independent i¤ the portfolio vectors z h
h2J
2. Write down the system of equilibrium equations :
! Rn for n = H (G + S + 1 + J)+
G + J so that =
xh ; h ; z h h2H ; p; q is a …nancial equilibrium with money i¤
( ) = 0: Here
=
eh
h2H
;M 2
:
3. Use the mathematical tools from Section 1.4 and the results in Theorems 2.4 and 3.2
to show that, for a generic selection of households endowments e = eh h2H and money
supplies M = (M0 ; M1 ; :::; MS ) ; the …nancial equilibria with money satisfy Finite Local
Uniqueness. This requires proving that (i) is proper and (ii) if ( ; ) = 0; then
rankD ( ; ) = n (the derivative is taken with respect to the variables and the
parameters = eh h2H ; M ). The proof of (i) is straightforward, so only prove (ii).
4. Given the outcome of Exercise 3 above, show that rankD ( ; ) = n + J + 1; where
0
1
( )
B
C
( ) = @
z 1 ; :::; z J A (again taking derivatives with respect to the variables
T
=
xh ;
h
; zh
=2
1
h2H
; p; q;
2 Rn+J and the parameters
=
eh
h2H
; M ). The
additional equations
z 1 ; :::; z J ; where 2 RJ nf0g (nonzero from T =2 = 1),
imply that the portfolios vectors z h h2J are linearly dependent. Using "A Di¤erent
Application of Di¤erential Topology" from Exercise 3 in Chapter 1, we can then conclude that for a generic selection of = eh h2H ; M ; the portfolio vectors z h h2J
are linearly independent.
94
CHAPTER 4. INCOMPLETE MARKETS AND MONEY
Bibliography
[1] Cass, David (1992): "Incomplete Financial Markets and Indeterminacy of Competitive
Equilibrium," in: Advances in Economic Theory; Sixth World Congress, Volume II, edited
by J.J. La¤ont (Cambridge: Cambridge University Press).
[2] Magill, Michael and Martine Quinzii (1996): Theory of Incomplete Markets, Vol. 1 (MIT
Press: Cambridge, MA).
[3] Strang, Gilbert (2006): Linear Algebra and its Applications, Fourth Edition (Thomson:
Belmont, CA).
[4] Villanacci, Antonio, Laura Carosi Carosi, Pierluigi Benevieri, and Andrea Battinelli
(2002): Di¤erential Topology and General Equilibrium with Complete and Incomplete Markets (Kluwer Academic Publishers: Boston).
95
96
BIBLIOGRAPHY
Appendix A
Solutions to the Exercises
A.1
Mathematical Prerequisites
1. (Applying Farkas Lemma)
Let R be a m
implication.
n matrix and
There does not exist
2 Rn : Use Farkas Lemma to prove the following
such that R
> 0
+
9
2 Rm
++ s.t.
T
R = 0:
(Note: The other implication is a simple one-line proof. Make sure that you are proving
the correct implication [i.e., the multi-page behemoth of a proof]).
Solution:
2
3
R1
6
7
De…ne R = 4 : 5 : Consider the 1st row of the matrix R: The following process
Rm
will be repeated for all m rows of R: The statement "There does not exist such that
R > 0" implies that @ such that R1 > 0 and Ri
0 8i = 2; :::; m: Thus, @ such
that R1 > 0
Ri 8i = 2; :::; m: De…ne
Z(1) = z 2 Rn : z =
T
R2T ; :::; Rm
97
(1) for (1) 2 Rm
+
1
:
98
APPENDIX A. SOLUTIONS TO THE EXERCISES
T
z 8z 2 Z(1): Use the following table
With this de…nition, @ such that T R1T > 0
to connect the statement of Farkas Lemma to the proof at hand.
Farkas Lemma
n
q 2 R nf0g
z 2 Rn
A 2 Rn;m 1
1
2 Rm
+
This proof
2 Rn nf0g
R1T 2 Rn
T
R2T ; :::; Rm
1
(1) 2 Rm
+
()
()
()
()
As condition (ii) of Farkas Lemma does not hold, then condition (i) must. This means
1
that R1T 2 Z(1); so 9 (1) 2 Rm
such that
+
R1T =
T
R2T ; :::; Rm
(1):
Rearranging and taking transposes, we have:
R1 +
m
X1
i (1)
Ri+1 = 0:
i=1
~ 1 (1) = 1 and ~ i (1) =
De…ne ~ (1) 2 Rm
+ such that
T
~ (1)
R = 0:
i 1 (1)
for i = 2; :::; m: Then
Repeating the entire process for the remaining rows i = 2; :::; m; then we obtain
T
~ (2); :::; ~ (m) 2 Rm such that ~ i (i) = 1 8i = 2; :::; m: In all cases, ~ (i)
R=0
+
8i = 2; :::; m:
De…ne
2
argument.
Rm
++
as
=
m
X
~ (i): By construction,
T
R = 0: This completes the
i=1
2. (Applying Kakutani’s Fixed Point Theorem)
We will use the theory of correspondences to prove the existence of a Nash equilibrium. Consider a game with I players. Each player i has a …nite number of actions
ai1 ; ::::; aiJi : The set of strategies for each player i is then the simplex Ji 1 of dimension Ji 1: The strategies are simply the probabilities that each player assigns to
each of the …nite number of actions. For simplicity, denote the strategy set for each
player as S i with element si : These sets are nonempty, compact, and convex.
A.1. MATHEMATICAL PREREQUISITES
99
If player i0 s payo¤ value for the action pro…le a = (a1 ; :::; ai ; :::; aI ) is pi (a); then the
objective function for each player is (using the convention s = (s1 ; :::; si ; :::; sI )):
ui (s) =
X
a
pi (a)
Y
i
s i ai ;
where si (ai ) is the probability that player i selects the action ai : This objective function
is quasi-concave in s (as it’s linear).
Denote s i = (s1 ; :::; si 1 ; si+1 ; :::; sI ) as the vector of strategies (not including the
strategy for player i) and S j = S 1 ::: S i 1 S i+1 ::: S I as the Cartesian product
j6=i
of the strategy sets (not including the strategy set for player i). De…ne F i :
Sj
Si
j6=i
as the set of "feasible" strategies. Notice that F i (s i ) = S i 8s
De…ne BRi :
Sj
i
2
Sj :
j6=i
S i as the best response correspondence for player i: Use the
j6=i
Berge’s Maximum Theorem and Kakutani’s Fixed Point Theorem to prove that the
S i has a …xed point. By de…nition, a …xed point is
self-map BR1 ; :::; BRI : S i
i
i
a Nash equilibrium.
Solution:
By de…nition, S i = Ji 1 is compact, convex, and nonempty. These properties are
maintained through the Cartesian product operator, meaning that S i is compact,
i
convex, and nonempty.
The objective function is continuous and the strategy set is compact. Using the Extreme Value Theorem, the best response correspondence BRi is well-de…ned.
The objective function is quasi-concave (as it’s linear) and the strategy set is convex.
Consider any two x; y 2 BRi (s i ) for some s i 2
S j : By convexity of S i ; then
j6=i
i
i
x + (1
)y 2 S : By quasi-concavity, u ( x + (1
)y) ui (x) = ui (y): This implies
that x + (1
)y 2 BRi (s i ) ; meaning that BRi is convex-valued.
De…ne the constraint correspondence C i :
Sj
S i : The correspondence is de…ned
j6=i
such that C i (s i ) = S i 8s
i
2
S j : Given this de…nition, the best response correj6=i
spondence is formally de…ned as:
BRi (s i ) = arg max ui (s) :
si 2C i (s
i)
100
APPENDIX A. SOLUTIONS TO THE EXERCISES
The objective function ui (s) is continuous. Given that the codomain S i is compact
and C i maps into the entire set S i ; the correspondence C i is both uhc and lhc. From
Berge’s Maximum Theorem, BRi is a uhc correspondence.
The vector of correspondences BR1 ; :::; BRI has the same properties as each of its
elements: well-de…ned, convex-valued, and uhc. Applying Kakutani’s Fixed Point TheS i guarantees the existence of a …xed
orem to the self-map BR1 ; :::; BRI : S i
i
i
point.
3. (A Di¤erent Application of Di¤erential Topology)
Let
RJ be the set of variables (with typical element ),
RK be the set of
parameters (with typical element ), H =
(so = ( ; )), and : H ! RL be the
C 1 system of equations. Assume that J < L: Suppose that the projection is proper
and the rank condition holds (that is, rankD ( ) = L). Using the mathematical
results from Section 1.4, what conclusions can be drawn? The conclusions will be of
the form, "over a generic subset of parameters (the set of regular values r ), then...".
Solution:
With J < L; then all solutions 2 M are critical points, as they must necessarily satisfy
rankD ( ) < L: Thus, (M ) are critical values. Let’s suppose that we are able to
verify the two key properties of regularity: (i) is proper and (ii) rankD ( ) = L
8 2 M: Then the Closedness Theorem and the Transversality Theorem allow us to
conclude that the set of critical values is a closed subset of measure zero (equivalently,
the set of regular values is an open subset of full measure). Thus, (M ) is a closed
subset of measure zero, where (M ) = f : 9 such that ( ; ) = 0g by de…nition.
In conclusion, over a generic subset of parameters (the set of regular values r =
n (M )), there does not exist a solution to the system of equations ( ; ) = 0:
A.2
Arrow-Debreu Model
1. Show that if uh is locally non-satiated 8h 2 H; then the Arrow-Debreu equilibrium
prices satisfy p > 0:
Solution:
To begin, I show that in any Arrow-Debreu equilibrium xh h2H ; p ; an optimal
solution to the household problem (HP ) must satisfy pxh = peh 8h 2 H: Suppose
A.2. ARROW-DEBREU MODEL
101
otherwise, that pxh < peh for some h: By local non-satiation, 9y h 2 N xh \ X h such
that uh y h > uh xh : The value for can be made small enough so that py h < peh ;
contradicting that xh is an optimal solution to the household problem (HP ):
The initial price space is p 2 RG nf0g: To prove the claim, suppose that pg < 0 for some
g 2 G: Suppose that xh is the equilibrium consumption choice for household h: From
the previous paragraph, pxh = peh : De…ne x~h such that x~hg = xhg + and x~hg0 = xhg0
8g 0 6= g: By linearity, p~
xh < peh : As x~h is not optimal, it must be that uh xh >
uh x~h : By local non-satiation, 9y h 2 N xh \ X h such that uh y h > uh xh : For
xh to be optimal, it must be that py h > peh : De…ne y~h such that y~gh = ygh + and
y~gh0 = ygh0 8g 0 6= g: There exists such that 8
; p~
y h peh : For xh to be optimal,
it must be that uh xh
uh y~h : We have uh y h > uh xh
uh y~h : Letting
( ; ) ! 0; we see that uh is strictly decreasing with xhg : This implies that the optimal
solution to (HP ) lies on the boundary: xh 2
= intX h : This violates local non-satiation
as @z h 2 N xh \ X h such that uh z h > uh xh :
2. Assume that uh is di¤erentiable, di¤erentiably strictly increasing, and concave 8h 2
H: Using the results from the programming problems in this chapter, show that if
xh h2H ; p is an Arrow-Debreu equilibrium, then the allocation xh h2H is an optimal solution to the problem (P Oh0 ) for any h0 : From Section 2.3, the allocation xh h2H
is then Pareto optimal. As this was the …rst proof of the First Basic Welfare Theorem,
it is often called the "classical proof."
Solution:
The following are necessary and su¢ cient conditions for an Arrow-Debreu equilibrium:
h
Duh (xh )
p = 0 8h 2 H (F OC)
X
ehg xhg = 0 8g 2 G
(M C) :
h2H
The following are necessary and su¢ cient conditions for an optimal solution to (P Oh0 )
h
for household h0 = 1 (without loss of generality). Let ~
be the Lagrange multipliers for the constraints uh (xh )
uh (x h ) and
h>1
g g2G
be the Lagrange multipliers
102
APPENDIX A. SOLUTIONS TO THE EXERCISES
for the constraints xh
h2H
2 F A; namely
X
h2H
ehg
xhg = 0 8g 2 G :
Du1 (x1 )
=0
(F OC1)
h
h
h
~ Du (x )
= 0 8h > 1 (F OCh) :
X
ehg xhg = 0 8g 2 G (F A)
h2H
To show that both systems of equations yield the same optimal allocation, de…ne
h
1
= 1 p >> 0 and ~ = h > 0:
3. Consider an economy with two households and two goods. Both utility functions are
di¤erentiable, strictly increasing, and strictly concave, but there is a missing market
for the second good. That is, each household can consume no more than its initial
endowment of the second good. Using an Edgeworth box, show that the equilibrium
allocations are typically Pareto suboptimal.
Solution:
The Edgeworth box in Figure A.1 shows that the equilibrium allocation occurs at the
endowment point. As the utility functions are strictly increasing, equilibrium prices
must satisfy p >> 0: For any such prices, the only allocation that lies in both the
budget set for h = 1 and the budget set for h = 2 is the endowment point e: This is
an autarchic equilibrium.
The Edgeworth box in Figure A.2 shows that the set of Pareto optimal allocations
(called the Pareto set) will typically not pass through the endowment point e: The
Pareto set is de…ned (given the assumptions on the utility functions) as points where
an indi¤erence curve for h = 1 is tangent to an indi¤erence curve for h = 2: Thus, the
equilibrium allocation will be Pareto suboptimal.
4. Consider an economy with two households and two goods. The two households have
utility functions
u 1 x1
u 2 x2
1
1
log x11 + x21 + log x12
2
2
1
1
=
log x21 + log x22
2
2
=
and endowments e1 = (1; 2) and e2 = (2; 1) : Notice that household 1 cares about
A.2. ARROW-DEBREU MODEL
103
household 2’s consumption of the …rst good (x21 ). Using the programming problems
from this chapter, characterize the Pareto optimal allocations and then characterize
the equilibrium allocations. From this, can you claim that the First Basic Welfare
Theorem holds? Use an Edgeworth box to illustrate your argument.
Solution:
The Pareto optimal allocations are x1 = (0; ) and x2 = (3; 3
This is illustrated in the Edgeworth box below.
) for any
2 [0; 3] :
To …nd the Arrow-Debreu equilibrium, normalize p2 = 1: The …rst order conditions for
household h = 2 are given by:
0:5
x21
0:5
x22
Solving for x2 as a function of
(x22 1) = 0 yields:
2
1
2
p1 = 0:
2
= 0:
and plugging into the budget constraint p1 (x21
0:5
2
Thus,
2
+
0:5
2
2) +
= 2p1 + 1:
= 2p1 + 1; so the demand functions are:
x21 (p1 ) =
p1 + 0:5
and x22 (p1 ) = p1 + 0:5:
p1
The …rst order conditions for household h = 1 are given by:
0:5
x11 + x21
0:5
x12
1
p1 = 0:
1
= 0:
This means that x12 = 0:51 and x11 + x21 = 0:5
: By market clearing, this implies that
1
p1
0:5
= 3; so 1 = 6p11 : This means that x12 = 3p1 :
1
p1
Market clearing requires that x12 +x22 = 3: From the demand functions, 3p1 +p1 +0:5 = 3;
104
APPENDIX A. SOLUTIONS TO THE EXERCISES
meaning that p1 = 58 : This implies an equilibrium allocation given by:
x1 = (1:2; 1:875) :
x2 = (1:8; 1:125) :
This allocation is plotted in the Edgeworth box in Figure A.3. The equilibrium allocation does not lie in the Pareto set, which is shown in red. This is a violation of the
First Basic Welfare Theorem.
5. Consider an economy with two households and two goods. Prove that if the utility
functions are Cobb-Douglas (of the form uh xh = 1 log xh1 + 2 log xh2 ), then
there is a unique Arrow-Debreu equilibrium. Does this property hold for an economy
with more than two households and more than two goods?
Solution:
For the two good economy, we will solve for the demand functions. Normalize the price
p2 = 1: For any household h; the …rst order conditions are given by:
1
h
xh1
2
h
x2
h
Solving for xh as a function of
xh2 eh2 = 0 yields:
xh1 (p1 ) =
1
h
=
1
(
(
1+
2)
2
h
+
and xh2 (p1 ) =
X
X
xh1 (p1 ) =
h2H
household independent, then
h2H
eh1 +
= p1 eh1 + eh2 :
p1 eh1 + eh2
p1
Market clearing requires that
X
= 0:
p1 eh1 + eh2 : The demand functions are given by:
1+ 2)
1
h
and plugging into the budget constraint p1 xh1
1
h
This means that
p1 = 0:
xh1 (p1 ) =
(
1+
2)
p1 eh1 + eh2 :
eh1 : Given that the coe¢ cients are
h2H
1
p1 (
2
1
+
2)
p1
X
h2H
eh1 +
X
h2H
!
eh2 :
A.2. ARROW-DEBREU MODEL
105
The market clearing condition
X
xh1 (p1 ) =
h2H
X
eh1 is then a linear function of p1 ;
h2H
meaning that there exists a unique price vector, and subsequently a unique equilibrium
allocation. The market clearing condition allows us to write a closed form expression
for the price p1 :
p1
1
p1 (
1+ 2)
X
eh1 +
h2H
(
p1
eh1 +
2)
X h2H
1+
p1 =
2
1
(
eh
2
1
h2H
X
eh2
h2H
X
1
X
1+ 2)
!
=
X
h2H
X
eh1
eh2
= p1
h2H
!
X
eh1
h2H
:
eh
1
h2H
If we repeat this procedure for an economy with an arbitrary number of goods and an
arbitrary number of households, the demand functions are of the form:
xhg pnG = P
g
g2G
g
peh
pg
for g < G and xhG pnG = P
G
peh :
g2G
g
Here we have normalized the price pG = 1 and only consider the prices pnG = (p1 ; :::; pG 1 ):
The market clearing conditions for the goods g < G yield equations of the following
form:
!
X
X
P g
ehg 8g < G:
eh = pg
p
g2G
g
h2H
h2H
These linear equations (G 1 in total) have a unique solution for pnG : Thus, CobbDouglas utility always leads to a unique equilibrium allocation.
H(G+1)+G 1
6. Show that the projection is proper. is the projection : R++
which maps xh ; h h2H ; p; eh h2H !
7
eh h2H such that
xh ;
0:
h
HG
RHG
++ ! R++
; p; eh h2H =
h2H
Solution:
A projection is by de…nition a continuous mapping. Consider a compact set of en1
dowments 0 : We have left to show that
( 0 ) is a compact set. As consumption
X
is nonnegative and the endowments
ehg are bounded 8g 2 G; then the equilibrium
h2H
106
APPENDIX A. SOLUTIONS TO THE EXERCISES
consumption xh h2H also belong to a compact set. From the …rst order conditions
with respect to the good G; we have:
DxG uh xh
h
= 0:
Recall that we have normalized the price pG = 1: As DxG uh xh is bounded (Duh is
continuous), then h h2H belongs to a compact set. From the …rst order conditions
with respect to all other goods g < G :
Dxg uh xh
h
pg = 0 8g < G:
As all terms are bounded, then (pg )g<G belongs to a compact set.
Thus, the set
1
( 0 ) is compact, …nishing the argument that
is proper.
7. Prove that for any endowments eh h2H within an open set of allocations around
a Pareto optimal allocation, the resulting Arrow-Debreu equilibrium is unique. To
do this, you must verify that the matrix D(xh ; h ) ;p
xh ; h h2H ; p has full rank
h2H
(square matrix) given that the initial endowment is a Pareto optimal allocation eh h2H =
xh h2H : Notice that the derivative matrix only contains derivatives with respect to
variables, not the endowment e1 :
Solution:
I will show that the matrix M = D(xh ; h ) ;p
xh ; h h2H ; p has full row rank
h2H
when eh h2H = xh h2H ; a Pareto optimal allocation. The FLU result then implies
that there exists an open neighborhood around the Pareto optimal allocation such that
for any endowment in this open neighborhood the equilibia are locally unique. This
means that the number of equilibria for any endowment within this set must be equal
to the number of equilibria at the endowment eh h2H = xh h2H (in other words, 1
equilibrium). This veri…es uniqueness.
The rows of M correspond to the equations in
xh ; h h2H ; p ; while the columns
correspond to the variables that we are taking derivatives with respect to. The deriv-
A.2. ARROW-DEBREU MODEL
107
ative matrix is given by:
2
:::
6
6:::
6
6
60
6
6
6
M =6
60
6
6
60
6
60
4
:::
:::
:::
0
0
0
D 2 u h xh
0
0
0
0
0
0
pT
0
0
0
0
0
0
0
::: :::
::: :::
:
:
0
::: :::
0
h
0
p
0
0
0
0
:::
IG
1
0
3
:
:
IG
0
1
xhnG
ehnG
7
7
!7
7
7
7
7
7
T 7:
7
7
7
7
7
7
5
To show that the matrix M has full rank, we set T M = 0 and must verify
where 2 RH(G+1)+G 1 corresponds to equations in :
0
10
:
:
h CB
x C B F OCh
CB
h CB
C B BCh
CB
: A@
:
p
MC
B
B
B
=B
B
B
@
T
The equations
T
x
= 0;
1
C
C
C
C:
C
C
A
M = 0 are given by:
:
xh
T
h
D2 uh xh
h T
p
h
pT
pT = 0
IG
0
1
i
=0
(A:1:a)
(A:1:b) :
(A.1)
:
P
h
h2H
xhnG
T
+
P
h2H
As eh h2H = xh h2H ; then the term
show that T = 0 in three steps:
(a) Postmultiply (A:1:a) by
xh
T
D 2 u h xh
h
ehnG
P
h2H
xhnG
h
ehnG
T
=0
xhnG
(A:1:c)
T
= 0 in (A:1:c): Let’s
xh :
xh
1
p xh
( p1 ; :::; pG 1 ; 0) xh = 0:
108
APPENDIX A. SOLUTIONS TO THE EXERCISES
From (A:1:b); p xh = 0: This implies
h
xh
T
D2 uh xh
xh =
h
X
pg xhg ;
(A.2)
g<G
where
h
> 0 is multiplied by both sides.
(b) From (A:1:c) :
P
h
xh1 ; :::; xhG
1
= 0:
h2H
(c) Add the equation (A:2) over all households h 2 H: Using (A:2); this implies
P
h
xh
T
D 2 u h xh
h2H
xh =
X
pg
g<G
P
h
xhg
= 0:
(A.3)
h2H
As uh is di¤erentiably strictly concave, the Hessian matrix D2 uh xh is negative
T
de…nite. This means h
xh D 2 u h xh xh
0 8h 2 H; with equality i¤
T
xh h2H = 0: From (A:1:a);
xh = 0: (A:3) implies
h
h
As p >> 0; then
Thus,
T
= :::;
xh
(p1 ; :::; pG 1 ; 1) + ( p1 ; :::; pG 1 ; 0) = 0:
h2H
T
;
h
= 0: Consequently,
; :::; pT
pT = 0:
= 0; …nishing the argument.
8. Using "A Di¤erent Application of Di¤erential Topology" from Exercise 3 in Chapter
1, prove that over a generic subset of endowments, if G > 1; then p1 6= p2 :
Solution:
!
( )
We utilize Assumptions R1-R3. Consider the system of equations ( ) =
:
p1 p2
As L = J + 1 > J; if I can show that is proper (same proof as Exercise 6) and
rankD ( ) = L 8 :
( ) = 0; then using the result from Exercise 3 in Chapter 1,
over a generic subset of endowments, there does not exist a solution to the system
of equations
( ; ) = 0: This implies that any Arrow-Debreu equilibrium (de…ned
such that ( ; ) = 0) must be such that p1 p2 6= 0:
To show that rankD ( ) = L; repeat the exact same three steps as in Section
2.5. Notice that these steps do not involve using the columns for the derivatives with
A.2. ARROW-DEBREU MODEL
109
respect to the price variables. This shows that the …rst J equations of
are linearly
independent. For the …nal equation of
; use the column for the derivatives with
respect to p1 or p2 to show that the …nal row is linearly independent from the rest.
This veri…es that the matrix D ( ) has full row rank L:
9. As a general result of the ideas in Exercise 3 above, show that over a generic subset
of endowments, if there is a missing market for one of the goods, then the allocation
xh h2H is not Pareto optimal. To attack this problem, use "A Di¤erent Application of
Di¤erential Topology" from Exercise 3 in Chapter 1 and consider that a necessary con0
0
Dg uh xh
Dg uh (xh )
dition for Pareto optimality is D uh xh = D uh0 xh0 8 (h; h0 ; g; g 0 ) 2 H H G G:
( )
( )
g0
g0
Solution:
We utilize Assumptions R1-R3. Suppose wlog that the missing market is for good
g = 1: Then in equilibrium, xh1 = eh1 8h 2 H: The equilibrium variables are =
xh ; h ; h h2H ; p and the equilibrium system of equations is
0 0
B
B
B
B
B
( )=B
B
B
B
B
@
h
h
h
h
Dx1 u x
p1 = 0
B
h
h
h
pg = 0 1<g<G
B D xg u x
B
h
B
=0
DxG uh xh
B
B
h
h
e1 x1
@
p eh xh
P
h
xhnG
h2H enG
1
C
C
C
C
C
C
A
h2H
1
C
C
C
C
C
C:
C
C
C
C
A
The number of equations in equals the number of variables in : From the …rst order
0
0
Dx1 uh xh
h
h
Dx uh (xh )
Dx uh (xh )
conditions, D 1 uh xh = +h pg p1 8g > 1: This means that D 1 uh xh = D uh0 xh0 i¤
xg
( )
xg
( )
xg
( )
h
h0
h =
h0 :
!
( )
Consider the system of equations
( ) =
: As L = J + 1 > J;
1 2
2 1
if I can show that
is proper (same proof as Exercise 6) and rankD ( ) = L
8 :
( ) = 0; then using the result from Exercise 3 in Chapter 1, over a generic
subset of endowments, there does not exist a solution to the system of equations
( ; ) = 0: This implies that any equilibrium (de…ned such that ( ; ) = 0) must
1
2
2 1
be such that 1 2
6= 0: Without 1 = 2 ; the equilibrium allocation must be
Pareto suboptimal.
110
APPENDIX A. SOLUTIONS TO THE EXERCISES
The rows of M = D ( ) correspond to the equations in
; while the columns
correspond to the variables and parameters (endowment vectors e1 and e2 ) that we are
taking derivatives with respect to. To show that the matrix M has full rank, we set
T
M = 0 and must verify T = 0; where 2 RH(G+2)+G 1 corresponds to equations in
:
0
10
1
:
:
B
CB
C
B
C
xh C B
F OCh
B
CB
C
h CB
h
h
B
C
x
e
1
1
B
CB
C
B
C
B
C
h
=B
BCh
CB
C:
B
CB
C
B : CB
C
:
B
CB
C
B
CB
C
p A@
MC
@
A
2
1
1
2
EE
Given that xh1 = eh1 ; the budget constraint is written only for the remaining G
goods.
1
From the columns corresponding to derivatives with respect to (e1 ; e2 ) ; we obtain (as
1
2
1
2
in Section 2.5)
;
=
;
= 0 and pT = 0: For all households h 2 H;
the columns corresponding to derivatives with respect to xh imply:
x
h T
2 h
D u
x
h
+
h
1
h
;
! !
0
= 0:
p
(A.4)
For the households h > 2; the columns corresponding to derivatives with respect to
h
; h imply:
! !T
1
0
T
xh
= 0:
p
T
The previous two equations imply that
xh D2 uh xh xh = 0 8h 2 H; and thus
T
xh = 0 8h 2 H using the assumption that uh is di¤erentiably strictly concave.
h
h
From (A.4),
;
= 0 8h 2 H:
For the column for the derivatives with respect to
Thus,
T
= 0; …nishing the argument.
1
; then
EE = 0 (as
2
> 0).
A.3. GENERAL FINANCIAL MODEL
A.3
111
General Financial Model
1. Show that No Arbitrage is a necessary condition of equilibrium. That is, show that if
No Arbitrage does not hold, then a general …nancial equilibrium does not exist.
Solution:
!
q
Assume that No Arbitrage does not hold. Then 9z h such that
z h > 0: Such
R
h
J
h
h
a portfolio
z for > 1 such that
! z 2 R and
! there always exists a portfolio z~ =
q
q
z~h >
z h : As the asset set RJ is unbounded, then there will never
R
R
exist a solution to the household problem (HP ):
2. Suppose that there are S = 4 states and J = 2 assets with payouts
2
2
6
61
R=6
60
4
3
3
3
7
17
7:
17
5
2
Which of the following asset prices satisfy No Arbitrage (hint: Hens method): (i)
q = (0; 2) ; (ii) q = (2; 1:75) ; and (iii) q = (2; 1:25)?
Solution:
To solve this problem, let’s use the Hens method for two assets. I plot the points
(r1 (1); r2 (1)) = (2; 3) ; (r1 (2); r2 (2)) = (1; 1) ; (r1 (3); r2 (3)) = (0; 1) ; and (r1 (4); r2 (4)) =
(3; 2) in Figure A.4. The x-axis corresponds to asset j = 1 and the y-axis to asset j = 2:
The prices q satisfy No Arbitrage i¤ they lie in the interior of the cone generated by
the 4 pairs of plotted points.
As indicated in Figure A.4, the asset prices q = (0; 2) lie on the boundary of the
cone, the asset prices q = (2; 1:75) lie in the interior of the cone, and the asset prices
q = (2; 1:25) lie outside the cone. Thus, the only prices that satisfy No Arbitrage are
q = (2; 1:75) :
112
APPENDIX A. SOLUTIONS TO THE EXERCISES
3. Suppose that there are S = 4 states and J = 3 assets with payouts
2
3
1 0 0
6
7
60 1 17
7
R=6
62 1 17 :
43
35
1
2
2
3
1
2
Which of the following asset prices satisfy No Arbitrage: (i) q = (1; 4; 2) ; (ii) q =
(2; 1; 1) ; and (iii) q = (3; 2; 1)?
Solution:
!
q
We need to …nd a linear operation for the matrix
such that the new payout
R
matrix has
! the risk-free payout of 1 in all states. To do this, simply replace Column 1
q
of
with the sum of Columns 1 and 3:
R
!
q
R
2
6
6
6
=6
6
6
4
q1
q3
3
q3
7
0 7
7
1 7
7:
1 7
3 5
q2
0
1
1
1
1
1
1
2
3
1
2
We are now ready to plot the points (r2 (1); r3 (1)) = (0; 0) ; (r2 (2); r3 (2)) = (1; 1) ;
(r2 (3); r3 (3)) = 1; 31 ; and (r2 (4); r3 (4)) = 32 ; 12 : Figure A.5 shows these plotted
points and the convex hull that they generate. The prices q satisfy No Arbitrage i¤
q2 q3
2
;q
= q1q+q
; q3
lies in the interior of the convex hull. For the three asset
q
3 q1 +q3
1
1
prices in the question, we have (i)
and (iii)
q2
; q3
q1 +q3 q1 +q3
=
1 1
;
2 4
q2
; q3
q1 +q3 q1 +q3
=
4 2
;
3 3
; (ii)
q2
; q3
q1 +q3 q1 +q3
=
1 1
;
3 3
;
:
As indicated in Figure A.5, the pair 43 ; 32 lies outside the convex hull, the pair 13 ; 13
lies on the boundary of the convex hull, and the pair 12 ; 14 lies in the interior of the
convex hull. Thus, the only prices that satisfy No Arbitrage are q = (3; 2; 1) :
4. Show that it is innocuous to assume No Redundancy. In other words, show that if
xh ; h ; z h h2H ; p; q is a general …nancial equilibrium in which No Redundancy does
not hold, then there exists
z^h
h2H
; q^ such that
xh ;
h
; z^h
h2H
; p; q^ is a general
A.3. GENERAL FINANCIAL MODEL
113
…nancial equilibrium in which No Redundancy does hold.
Solution:
Consider a payout matrix R that does not satisfy No Redundancy. For this payout
matrix, the general …nancial equilibrium is xh ; h ; z h h2H ; p; q : From the …rst order
!
q
h
= 0 8h 2 H: As R does not satisfy No
conditions with respect to z h ; then
R
Redundancy, then rankR = J < J: Without loss of generality,
h assume that
i the …rst J
columns of R are linearly independent, so we can write R = R j R
such that R
J submatrix with rankR = J : I can de…ne asset holdings z^h = z^jh j=1;:::;J
!
!
q
q
^
zh =
z^h : Each of the columns in R is a linear combination
such that
R
R
0
1
is an S
1
of the columns in R ; meaning that rj = R
2
6
where rj = 4
B
C
= @ : A 2 Rj and 8j > j ;
for some
j
3
rj (1)
7
: 5 : Thus, de…ne the new asset holdings as:
rj (S)
z^jh = zjh +
j
X
zkh :
k>j
Since
q
R
!
zh =
q^
R
!
z^h ; the real variables
The new asset prices satisfy q^j = qj 8j
j as
xh ;
h
q^
R
h
h2H
!
; p remain unchanged.
= 0 8h 2 H:
Thus, for the payout matrix R (satisfying No Redundancy as R has full column
rank), xh ; h ; z^h h2H ; p; q^ is a general …nancial equilibrium.
5. Prove Theorem 3.4.
Solution:
This proof will use "A Di¤erent Application of Di¤erential Topology" from Exercise 3
in Chapter 1. That is, we will write down a system of equations that must be satis…ed
by a general …nancial equilibrium that is also Pareto optimal. This system of equations
will satisfy J < L (more equations than variables). If we can then show that (i) is
114
APPENDIX A. SOLUTIONS TO THE EXERCISES
proper and (ii) rankD ( ) = L 8 2 M; then we can apply "A Di¤erent Application
of Di¤erential Topology" conclusion from Exercise 3 to conclude that over a generic
subset of endowments, there does not exist variables = xh ; h ; z h h2H ; p; q such
that xh ; h ; z h h2H ; p; q is a general …nancial equilibrium and xh
optimal allocation.
h2H
is a Pareto
The condition " is proper" can be veri…ed in the same fashion as Exercise 6 in Chapter
2 (the details are left to the intrepid reader).
As in Exercise 9 in Chapter 2, a necessary and su¢ cient condition for Pareto optimality
0
0
D(l;s) uh xh
D
uh (xh )
is D (l;s) uh xh = D
for any pair of households (h; h0 ) and any commodity0
h
h0
( )
(l0 ;s0 )
(l0 ;s0 ) u (x )
state pairs (l; s): Using the …rst order condition with respect to consumption, this
0
h0
(1);:::; h (S)
( h (1);:::; h (S))
=
condition is equivalent to
for any pair of households
h
h0
(0)
(0)
(h; h0 ); that is, the Lagrange multipliers are proportional.
De…ne the system of equations
:
h2H
RG+S+1
++
0 0
B
B
B
B
B
B
B
B
xh ; h ; z h h2H ; p; q = B
B
B
B
B
B
B
@
Du
h
x
B
B
B P eh xh
B
B
B
"
B
@
h
X
1
G (S+1)
RJ
h
R++
h
T
P !
q
+
zh
R
!#T
q
R
h2H
X
ehnG
h2H
(1) 2 (0)
zh
1
xhnG
RJ++ ! Rn+1 as:
1
C
C
C
C
C
C
C
A
1
h2H
(0) 2 (1)
where n = H (G + S + 1 + J) + G (S + 1) + J: The additional equation
1
(0) 2 (1) = 0 is a necessary condition for Pareto optimality.
xh ;
I will show that rankD
h
h
h
; zh
h2H
C
C
C
C
C
C
C
C
C;
C
C
C
C
C
C
A
1
(1) 2 (0)
; p; q = n+1; taking derivatives with respect
h
to the variables x ; ; z h2H and the endowment e1 : By assumption, J < S: This
means that R has full column rank J (by No Redundancy) and 9s > 0 such that the
submatrix R (ns) de…ned by removing the row for state s also has full column rank
2
3
r(2)
6
7
J: Without loss of generality, let s = 1: This means that R (n1) = 4 : 5 has full
r(S)
A.3. GENERAL FINANCIAL MODEL
115
column rank J:
T
= :::;
xh
T
h T
;
h
xh ;
Premultiply the derivative matrix D
; zh
zh
;
T
h2H
; p; q by
; :::; pT ; q T ; EE ;
1
where EE 2 R corresponds to the extra equation 1 (1) 2 (0)
(0) 2 (1) = 0: The
proof proceeds as in the proof of Theorem 3.2, with one exception. In Step 3, the
columns corresponding to the derivatives with respect to 1 are given by:
z
1 T
T
+
EE
2
(1);
2
(0); 0; :::; 0
= 0
q T r(1)T R(n1)T +
EE
2
(1);
2
(0); 0; :::; 0
= 0
z1
T
(A.5)
0
1
q
B
C
T
As = @ r(1) A and R(n1) has full column rank, then (A.5) implies that ( z 1 ) =
R(n1)
0 and EE = 0 (as both 2 (0) > 0 and 2 (1) > 0). The rest of the proof remains
unchanged, ultimately resulting in T = 0: Thus D
xh ; h ; z h h2H ; p; q has full
row rank n + 1; …nishing the argument.
6. In Example 3.1, verify that the Arrow-Debreu equilibrium allocation is such that 8h 2
X
H : xh (s) = h
eh (s) for some h :
h2H
Solution:
X
With utility function utility uh xh =
xh (s)
s2S
with respect to consumption are given by:
xh (s)
( +1)
h
; the …rst order conditions
(s) = 0 8s 2 S:
(A.6)
Recall that in an Arrow-Debreu equilibrium, there is a single budget constraint and
thus a single Lagrange multiplier h > 0: I denote the Arrow-Debreu prices as ( (s))s2S
such that (0) = 1: Equation (A.6) yields:
h
xh (s) =
1
+1
(s)
1
+1
:
(A.7)
116
APPENDIX A. SOLUTIONS TO THE EXERCISES
From the Arrow-Debreu budget constraint:
X
(s)xh (s) =
s2S
X
(s)eh (s) :
s2S
Using the expression for xh (s) from (A.7), the budget constraint becomes:
1
+1
h
X
(s)
+1
=
s2S
This implies that
(s)eh (s) :
s2S
X
1
+1
h
X
(s)eh (s)
s2S
= X
(s)
8h 2 H:
+1
(A.8)
s2S
The market clearing condition
X
X
xh (s) =
eh (s) for any state s 2 S implies
h2H
h2H
1
X
P h
(s) +1 X
X
eh (s) 8s 2 S:
e (s) =
(s)
+1
h2H
(s)
s2S
h2H
s2S
1
+1
Solving for (s)
yields
1
+1
(s)
=
X
s2S
(s)
X
+1
(s)
De…ne
h
= Xs2S
P
eh (s)
!
eh (s)
h2H
eh (s)
(s)eh (s)
(s)
s2S
P
X
8h 2 H: Notice that
h2H
P
h
h
X
h2H
This completes the argument.
(A.9)
= 1; by de…nition. Insert (A.8)
h2H
and (A.9) into the expression for xh (s) from (A.7) yields:
xh (s) =
8s 2 S:
h2H
s2S
X
!
!
eh (s) :
A.3. GENERAL FINANCIAL MODEL
117
7. In Example 3.3, verify that the general …nancial equilibrium consumption is given by:
xhl (s) =
for some
h
h
X
(s)
h2H
ehl (s)
(s) 2 (0; 1) and the prices are given such that pl (s) =
l (s)
L (s)
8(l; s) 2 L
S:
Solution:
X
P
h
With utility function utility uh xh =
l2L l (s) log xl (s)
s2S
conditions with respect to consumption are given by:
l (s)
h
xl (s)
h
(s)pl (s) = 0 8 (l; s) 2 L
; the …rst order
S:
Solving for xhl (s) yields
xhl (s) =
(s)
8 (l; s) 2 L
(s)pl (s)
l
h
(A.10)
S:
The budget constraint for any household in any state s 2 S is given by:
X
pl (s)xhl (s) =
l2L
X
pl (s)ehl (s) :
l2L
Using the expression for xhl (s) from (A.10) yields:
1 X
h
(s) l2L
l (s) =
X
pl (s)ehl (s) :
l2L
Thus, the expression for xhl (s) can be updated as:
1P
0
pl (s)ehl (s)
(s)
l
A l2L
xhl (s) = @ P
8 (l; s) 2 L
pl (s)
(s)
l
S:
(A.11)
l2L
Consider the market clearing conditions
X
h2H
0
1
xhl (s) =
X
h2H
ehl (s) for any (l; s) 2 L
X
P h
1 @ l (s) A X
P
pl (s)
el (s) =
ehl (s) :
pl (s)
l (s)
h2H
l2L
h2H
l2L
S:
118
APPENDIX A. SOLUTIONS TO THE EXERCISES
This implies that pl (s) can be expressed as:
0
pl (s) = @ P
l
l2L
1
(s) A
l (s)
X
P
pl (s)
l2L
X
ehl (s)
h2H
(A.12)
:
ehl (s)
h2H
Given the price normalization pL (s) = 1 8s 2 S; then
0
1
1
1
A
= @P
L (s)
l (s)
X
l2L
l2L
by de…nition. For any (l; s) 2 L
pl (s)
P
ehl (s)
h2H
X
;
ehL (s)
h2H
S; equation (A.12) veri…es that
P
h
(s)
h2H eL (s)
P
:
pl (s) =
h
L (s)
h2H el (s)
l
Using the expression (A.12) for pl (s); then the expression (A.11) for xhl (s) can be
updated as:
P
pl (s)ehl (s)
l2L
xhl (s) = X
pl (s)
h2H
l2L
De…ne
h
P
(s) = X
ehl
(s)
pl (s)
P
h2H
and notice that
eh
l (s)
h2H
!
ehl (s)
pl (s)eh
l (s)
l2L
l2L
P
X
X
h2H
h
8 (l; s) 2 L
S:
(s) = 1 8s 2 S; by de…nition.
This completes the second part of the claim that xhl (s) =
L S:
h
(s)
X
eh
h2H l
(s) 8 (l; s) 2
8. In Example 3.4, verify that the Arrow-Debreu equilibrium allocation and commodity
prices in Case I are given by:
x1 (1) = 38 ; 13
x2 (1) = 13 ; 83
(1) = (1; 1)
x1 (2) = 83 ; 31
x2 (2) = 13 ; 38 :
(2) = (1; 1)
A.3. GENERAL FINANCIAL MODEL
119
Solution:
To solve for the Arrow-Debreu equilibrium, I make the price normalization 2 (2) = 1:
The equilibrium variables are (x1 (1) ; x1 (2) ; x2 (1) ; x2 (2)) (8 consumption variables),
1
; 2 (2 Lagrange multipliers for the 2 budget constraints, 1 for each household),
and ( 1 (1); 2 (1); 1 (2)) (3 price variables). The equations are the …rst order conditions
with respect to consumption (8 of these), the 2 budget constraints (1) xh (1) + (2)
xh (2) = (1) eh (1) + (2) eh (2) for h = 1; 2; and the 3 market clearing conditions
x11 (1) + x21 (1) = 3; x12 (1) + x22 (1) = 3; and x11 (2) + x21 (2) = 3:
The …rst order conditions for household h = 1 are given by:
20:5 (x11 (1))
20:5 (x11 (2))
0:5
1
0:5
1
1 (1)
=0
1 (2) = 0
2
2
1
(x12 (1))
1
(x12 (2))
0:5
1
0:5
1
2 (1)
=0
=0
:
The expressions for consumption are given by:
x11 (1) =
x11 (2) =
(
1
(
1
2
2
1 (1))
2
2
1 (2))
x12 (1) =
x12 (2) =
1
4(
1
)
2 (1)
2
1
4(
(A.13)
:
1 2
)
Inserting these expressions into the budget constraint yields:
1
1
2
2
1
+
+
4 2 (1)
1 (1)
2
1
+
4
1 (2)
=
5
2
1 (1)
+
50
21
2 (1)
+
13
21
1 (2)
1
+ :
2
The expressions for consumption are then given by:
x11 (1) =
2
2
1 (1)
5
50
13
1
(1) + 21
2 (1) + 21 1 (2) + 2
2 1
2
+ 4 1(1) + 2(2) + 41
1 (1)
2
1
1
x12 (1) =
4 ( 2 (1))2
x11 (2) =
1
x12 (2) =
4
2
2
1 (2)
!
5
13
1
(1) + 50
(1) + 21
1 (2) + 2
2 1
21 2
2
+ 4 1(1) + 2(2) + 41
1 (1)
2
1
5
50
13
1
(1) + 21
2 (1) + 21 1 (2) + 2
2 1
2
+ 4 1(1) + 2(2) + 41
1 (1)
2
1
5
50
13
1
(1) + 21
2 (1) + 21 1 (2) + 2
2 1
2
+ 4 1(1) + 2(2) + 14
1 (1)
2
1
!
!
(A.14)
!
Let’s make use of the symmetry assumed in this economy. The commodity (l; s) = (1; 1)
120
APPENDIX A. SOLUTIONS TO THE EXERCISES
for household h = 1 is identical to the commodity (l; s) = (2; 2) for household h = 2:
Additionally, the commodity (l; s) = (1; 2) for household h = 1 is identical to the
commodity (l; s) = (2; 1) for household h = 2: This means that (1) = (2) = ( ; 1)
and we only need to consider the market clearing condition for the commodity (l; s) =
(1; 1):
The market clearing condition x11 (1) + x21 (1) = 3 is then given by:
5
2
2
2
+ 50
+
21
1
2
+4+
13
21
2
+
+
1
4
1
2
!
1
+ 2
2
1
2
Distributing the 2 in the …rst term and the
1
2
The value of
(A.13):
5
2
+ 50
+
21
1
1
+8+
13
21
1
+
+
1
8
1
2
!
+
1
1
2
8
3
8
3
50
21
1
4
+ 52
+2
!
= 3:
in the second term yields:
1
2
2
= 1 is a solution to (A.15). With
x11 (1) =
x11 (2) =
13
+ 21
+
1
+2+
4
+ 13
+
21
1
+4+
2
50
21
1
2
+ 52
+4
1
= 1; then
x12 (1) =
x12 (2) =
1
3
1
3
1
!
2
(A.15)
= 3:
=
6
4:5
=
4
3
and from
:
The consumptions for household h = 2 are found using the market clearing conditions.
9. In Example 3.4, verify that the general …nancial equilibrium allocation and commodity
prices in Case II are given by:
62 31
x1 (1) = 21
; 21
1 32
2
x (1) = 21 ; 21
p (1) = (2; 1)
x1 (2) =
x2 (2) =
p (2) =
32 1
;
21 21
31 62
;
21 21
1
;1
2
:
Solution:
To solve for the general …nancial equilibrium, I consider each state s = 1; 2 as its own
Arrow-Debreu economy.
In state s = 1; I make the price normalization p2 (1) = 1: The equilibrium variables
are (x1 (1) ; x2 (1)) (4 consumption variables), 1 ; 2 (2 Lagrange multipliers for the 2
budget constraints, 1 for each household), and p1 (1) (1 price variable). The equations
are the …rst order conditions with respect to consumption (4 of these), the 2 budget
A.3. GENERAL FINANCIAL MODEL
121
constraints p (1) xh (1) = p (1) eh (1) for h = 1; 2; and 1 market clearing condition
x11 (1) + x21 (1) = 3:
The …rst order conditions for household h = 1 are given by:
20:5 (x11 (1))
0:5
1
p1 (1) = 0
2
1
(x12 (1))
0:5
1
=0 :
The expressions for consumption are given by:
x11 (1) =
(
1
2
2
p1 (1))
x12 (1) =
1
4(
(A.16)
) :
1 2
Inserting these expressions into the budget constraint yields:
1
1
2
1
2
+
p1 (1) 4
50
5
= p1 (1) + :
2
21
The expressions for consumption are then given by:
50
(1) + 21
2
+ 14
p1 (1)
!
5
50
p
(1)
+
1
2
21
2
1
+
p1 (1)
4
2
x11 (1) =
p1 (1)2
x12 (1) =
1
4
5
p
2 1
!
Similarly, the expressions for consumption of household h = 2 can be found as:
x21
x12 (1) = 2
1
p
2 1
13
(1) + 21
1
+2
4p1 (1)
!
1
13
p (1) + 21
2 1
1
+2
4p1 (1)
1
(1) =
4p1 (1)2
!
The market clearing condition x11 (1) + x21 (1) = 3 is then given by:
2
p1 (1)2
5
p
2 1
(1) + 50
21
2
1
+
p1 (1)
4
!
1
+
4p1 (1)2
1
p
2 1
13
(1) + 21
1
+2
4p1 (1)
!
= 3:
The value p1 (1) = 2 is the solution to this equation. With p1 (1) = 2; then
1
1
2
=
124
21
122
APPENDIX A. SOLUTIONS TO THE EXERCISES
and from (A.16):
x11 (1) =
124 2
21 (2)2
=
62
21
x12 (1) =
124 1
21 4
=
31
21
:
The consumptions for household h = 2 are found using the market clearing conditions.
For state s = 2; the equilibrium variables are equal to those from state s = 1; with
the commodity l = 1 switched with commodity l = 2: For instance, the relative price
= pp21 (1)
= 21 :
is now pp12 (2)
(2)
(1)
A.4
Incomplete Markets and Money
Exercises 1-4 walk through all steps required to prove Lemma 4.1.
1. Show that the vectors ah
are linearly independent.
h2J
are linearly independent i¤ the portfolio vectors z h
h2J
Solution:
The vectors ah h2J are linearly independent provided that the matrix a1 ; :::; aJ has
full column rank. By de…nition, ah = [ ] Rz h 2 RS ; so a1 ; :::; aJ = [ ] R z 1 ; :::; z J :
The matrix [ ] R 2 RS has full column rank by de…nition. Thus, [ ] R z 1 ; :::; z J has
full rank i¤ z 1 ; :::; z J has full rank.
2. Write down the system of equilibrium equations :
! Rn for n = H (G + S + 1 + J)+
G + J so that =
xh ; h ; z h h2H ; p; q is a …nancial equilibrium with money i¤
( ) = 0: Here
=
eh
h2H
;M 2
:
Solution:
The price vectors are now p(s) = (p1 (s); :::; pL (s)) 2 RL++ : There are no price normalizations.
A.4. INCOMPLETE MARKETS AND MONEY
De…ne the system of equations
:
h2H
RG+S+1
++
0 0
B
B
B
B
B
B
B
B
h h
h
x ; ; z h2H ; p; q = B
B
B
B
B
B
B
B
@
123
Duh xh
B
B
B P e h xh
B
B
B
"
B
@
h
X
p (s)
X
RJ++ ! Rn as:
RG
++
RJ
T
h
P !
q
+
zh
R
!#T
q
R
h
h2H
X
h2H
e
h2H
h
x
h
1
1
C
C
C
C
C
C
C
A
h2H
zh
x (s)
Ms
s2S
C
C
C
C
C
C
C
C
C:
C
C
C
C
C
C
C
A
3. Use the mathematical tools from Section 1.4 and the results in Theorems 2.4 and 3.2
to show that, for a generic selection of households endowments e = eh h2H and money
supplies M = (M0 ; M1 ; :::; MS ) ; the …nancial equilibria with money satisfy Finite Local
Uniqueness. This requires proving that (i) is proper and (ii) if ( ; ) = 0; then
rankD ( ; ) = n (the derivative is taken with respect to the variables and the
parameters = eh h2H ; M ). The proof of (i) is straightforward, so only prove (ii).
Solution:
We are tasked with showing rankD ( ; ) = n; i.e., that D ( ; ) has full row rank.
The easiest way is to …rst consider the columns of D ( ; ) corresponding to the derivatives with respect to the money supplies M = (M0 ; M1 ; :::; MS ) : These parameters
X
only appear in one equation each, p (s)
xh (s) Ms = 0: Thus, these …nal S + 1
h2H
equations are linearly independent from all other equations.
To show that the remaining equations are linearly independent, proceed using the exact
same steps as in the proof of Theorem 3.2.
4. Given the outcome of Exercise 3 above, show that rankD ( ; ) = n + J + 1; where
0
1
( )
B
C
( ) = @
z 1 ; :::; z J A (again taking derivatives with respect to the variables
T
=
xh ;
h
; zh
=2
1
h2H
; p; q;
2 Rn+J and the parameters
=
eh
h2H
; M ). The
additional equations
z 1 ; :::; z J ; where 2 RJ nf0g (nonzero from T =2 = 1),
imply that the portfolios vectors z h h2J are linearly dependent. Using "A Di¤erent
124
APPENDIX A. SOLUTIONS TO THE EXERCISES
Application of Di¤erential Topology" from Exercise 3 in Chapter 1, we can then conclude that for a generic selection of = eh h2H ; M ; the portfolio vectors z h h2J
are linearly independent.
Solution:
We are tasked with showing rankD ( ; ) = n + J + 1; i.e., that D ( ; ) has full
row rank. The easiest way is to …rst consider the columns of D ( ; ) corresponding
to the derivatives with respect to the money supplies M = (M0 ; M1 ; :::; MS ) : These
X
parameters only appear in one equation each, p (s)
xh (s) Ms = 0: Thus, these
h2H
…nal S + 1 equations are linearly independent from all other equations.
Set
T
D
( ; ) = 0; where
0
B
B
B
B
B
B
B
B
B
B
=B
B
B
B
B
B
B
B
B
@
2 Rn+J+1 corresponds to equations in
10
:
CB
B
xh C
CB
h CB
CB
CB
h CB
z CB
B
: C
CB
CB
B
p C
CB
B
q C
CB
CB
B
M C
CB
CB
A@
EE
p (s)
X
Full row rank is veri…ed upon showing that
that M = 0:
1
:
F OCxh
BCh
F OCzh
:
M Cx
M Cz
h2H
:
xh (s)
Ms
z 1 ; :::; z J
T
=2 1
s2S
C
C
C
C
C
C
C
C
C
C
C:
C
C
C
C
C
C
C
C
A
= 0: The preceding paragraph speci…es
Recall Steps 1-3 in the proof of Theorem 3.2. These steps are used to show that
T
T
1 T
( x1 ) ;
; ( z 1 ) ; pT ; q T = 0; by using the columns of D ( ; ) corresponding to the derivatives with respect to x1 ; 1 ; z 1 ; e1 : For households h > 1; using
the columns corresponding to derivatives with respect to xh ; h ; eh ; the same Steps
xh
1-3 can be used to show that
T
;
h T
;
zh
T
= 0 8h > 1:
The vector 6= 0; so wlog suppose that 1 6= 0: Then the columns corresponding to
derivatives with respect to z11 ; :::; z1J yield:
j
1
= 0 8j = 1; :::J:
A.4. INCOMPLETE MARKETS AND MONEY
125
T
Thus,
= 0: Derivatives with respect to
EE = 0 (as 6= 0).
then imply that
All told,
( ; ) = n + J + 1:
= 0; …nishing the proof that rankD
EE
= 0; so