General Equilibrium Models1General Equilibrium Models1 Matthew Hoelle Spring 2014 1This manuscript...

General Equilibrium Models1

Matthew Hoelle

www.matthew-hoelle.com

Spring 2014

1This manuscript is dedicated to Dave Cass, both for his immeasurable contributions to generalequilibrium theory and for his inspirational work in graduate education and advising.

Contents

1 Mathematical Prerequisites 11.1 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 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.2 Constraint Qualification . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.3 Kuhn-Tucker Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Correspondences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 Differential 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 252.1 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

3.5.1 Case I:(∆xh

)T 6= 0 ∀h ∈ H . . . . . . . . . . . . . . . . . . . . . . . 74

3.5.2 Case II:(∆xh

)T= 0 for some h ∈ H . . . . . . . . . . . . . . . . . . 78

3.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4 Incomplete Markets and Money 85

4.1 The Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

4.1.1 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.2 Complete Markets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

4.3 Incomplete Markets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

A Solutions to the Exercises 97

A.1 Mathematical Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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), finance, and other fields. 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 field 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 prereq-

uisites for the course. Students are presumed to have a suffi 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:

• Define what it means for a set to be convex.

• Define what it means for a set to be open (and closed).

• Define what it means for a set to be bounded.

vii

viii CONTENTS

• In Euclidean space, does the definition of compactness imply closed and bounded?Vice versa?

• Define what it means for a function to be concave (and strictly concave).

• Define what it means for a function to be continuous.

• State and prove the Extreme Value Theorem.

• Define what it means for a function to be differentiable.

• 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 satisfies three properties: existence, optimality, and regularity. Chapter 3 introduces

the general financial 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 financial assets. I analyze how the predictions of the model compare under the settings

of complete financial markets and incomplete financial markets.

Chapter 4 adds money to the general financial 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 financial markets, monetary policy is neutral, whereas with incomplete financial

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 offer 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 through-

out this manuscript:

• ∀ is the universal quantifier; it is the symbol "for all"

• ∃ is the existential quantifier; it is the symbol "there exists"

• ∃! is the unique existence quantifier; it is the symbol "there exists a unique"

• ’iff’refers to ’if and only if’; it is used when two statements are necessary and suffi cientfor each other

• ’wlog’refers to ’without loss of generality’

• Unless otherwise indicated, x ∈ Rn is a column vector. Its tranpose xT is a row vector.

• The value ‖x‖ denotes the Euclidean norm of the vector x ∈ Rn.

• The space Rm,n refers to the space of m× n real matrices.

• For x, y ∈ Rn, x ≥ y iff xi ≥ yi for i = 1, ..., n.

• For x, y ∈ Rn, x > y iff x ≥ y and x 6= y.

• For x, y ∈ Rn, x >> y iff xi > yi for i = 1, ..., n.

• The set A\B = x ∈ A : x /∈ B .

• The set Nε (x) is the open neighborhood of radius ε around x.

• The set N∗ε (x) is the deleted open neighborhood of radius ε around x, that is, N∗ε (x) =

Nε (x) \x.

• f is C0 if it is continuous and is Cn (for any n ∈ N) if its nth derivative exists and iscontinuous.

• The set intA denotes the interior of A :

intA = x ∈ A : ∃ε > 0 such that Nε (x) ⊂ A .

x CONTENTS

• The set A′ denotes the set of limit points of A :

A′ = x : ∀ε > 0, N∗ε (x) ∩ A 6= ∅ .

• The set clA denotes the closure of A :

clA = A ∪ A′.

• ∆n−1 is the (n− 1)−dimensional simplex defined as

∆n−1 = x ∈ Rn+ :∑n

i=1 xi = 1.

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) differential topology (Section 1.4) as used to prove equilibrium

regularity.

1.1 Farkas Lemma

Before stating the Farkas Lemma, I first 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 TheoremSuppose Z ⊆ Rn is nonempty, closed, and convex, and that y ∈ Rn, but y /∈ Z.Then there exists q ∈ Rn\0 such that qTy > qT z ∀z ∈ Z.

The Separating Hyperplane Theorem can be depicted in Figure 1.1 in the separate ’Fig-

ures’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′ ∈ Z and define Z ′ = z ∈ Z : ‖z − y‖ ≤ ‖z′ − y‖ . The set Z ′ is compact(as it is closed and bounded). The function ‖z − y‖ is a continuous function of the variablez.

Applying the Extreme Value Theorem, ∃z∗ = arg minz∈Z′

‖z − y‖ . For any z ∈ Z, as y /∈ Z,

then ‖z − y‖ > 0. Thus ‖z∗ − y‖ > 0.

Take any z ∈ Z and define the real-valued function g : [0, 1]→ R as follows:

g (θ) = ((1− θ) z∗ + θz − y)T ((1− θ) z∗ + θz − y) .

The function is minimized at 0, namely g (θ) ≥ g (0) ∀θ ∈ [0, 1] . Using the definition of

derivative, Dg (0) ≥ 0. We can rewrite the function

g (θ) = (θ (z − z∗) + z∗ − y)T (θ (z − z∗) + z∗ − y) .

Taking the derivative yields

Dg (0) = (z∗ − y)T (z − z∗) ≥ 0.

Using algebra, (y − z∗)T (z − z∗) ≤ 0 or (y − z∗)T z ≤ (y − z∗)T z∗ ∀z ∈ Z.Define the separating hyperplane as q = (y − z∗) . Therefore, qT z∗ ≥ qT z ∀z ∈ Z.As ‖z∗ − y‖2 > 0, then (z∗ − y)T (z∗ − y) > 0 or (y − z∗)T (y − z∗) > 0. This implies

qTy > qT z∗. From above, qTy > qT z∗ ≥ qT z ∀z ∈ Z. This finishes 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 ∈ Z, but y /∈ intZ.Then there exists a q ∈ Rn\0 such that qTy ≥ qT z ∀z ∈ 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 3

1.1.2 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 ∈ Rn\0 for i = 1, ...,m, matrix A = [a1...am] ∈ Rn,m, and

Z =z ∈ Rn : z = Aα for some α ∈ Rm+

.

For z∗ ∈ Rn, exactly one of the following conditions holds:(i) z∗ ∈ Z.(ii) ∃q∗ ∈ Rn\0 such that q∗T z∗ > 0 ≥ q∗T z ∀z ∈ Z.

The theorem is illustrated in the two-panel Figure 1.3. In the left panel labeled Farkas

(i), the element z∗ ∈ Z. In the right panel labeled Farkas (ii), z∗ /∈ Z and ∃q∗ ∈ Rn\0 suchthat q∗T z∗ > 0 ≥ q∗T z ∀z ∈ 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 differentiable 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 xTγ

x ≥ 0 subject to xTA ≤ βT.


The dual minimization problem is then given by:

minimize yTβ

y ≥ 0 subject to Ay ≥ γ.

Definition 1.1 A linear program is feasible if there exists a vector satisfying the constraints.

Definition 1.2 A feasible vector is an optimal vector if it maximizes or minimizes the linearform. The value of this max or min is called the value.

Theorem 1.3 DualityIf 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 suffi ces to find (x, y) ≥ 0 that satisfies the following

three equations:

xTA ≤ βT . (1.1)

Ay ≥ γ. (1.2)

xTγ − yTβ ≥ 0. (1.3)

Proof. From (1.1), xTAy ≤ βTy. From (1.2), xTAy ≥ xTγ. The previous two inequalities

together imply xTγ ≤ βTy. If (1.3) is satisfied, then xTγ = βTy.

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

zTγ > xTγ.

zTγ ≤ βTy.

Using the equality xTγ = βTy, then zTγ ≤ βTy = xTγ, 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:

β = ATx+ Iv for v ≥ 0.

γ = Ay − Iw for w ≥ 0.

0 = γTx− βTy − u for u ≥ 0.

1.1. FARKAS LEMMA 5

Thus, there does not exist α =

x

v

y

w

u

≥ 0 such that

β

γ

0

=

AT I 0 0 0

0 0 A −I 0

γT 0 −βT 0 −1

x

v

y

w

u

.

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∗ ⇐⇒ q∗

z∗ ⇐⇒

β

γ

0

A ⇐⇒

AT I 0 0 0

0 0 A −I 0

γT 0 −βT 0 −1

α ⇐⇒

x

v

y

w

u

.


Define

Z =

z : z =

AT I 0 0 0

0 0 A −I 0

γT 0 −βT 0 −1

x

v

y

w

u

for some

x

v

y

w

u

≥ 0

.

The implications of our proof by contradiction is that

β

γ

0

/∈ Z. This says that part (i) of

the Farkas Lemma does not hold, meaning that part (ii) must. Thus, there is q =

q1

q2

q3

6= 0

such that q∗T

β

γ

0

> 0 ≥ q∗T z ∀z ∈ Z. The strict inequality on the left-hand side can be

written as:

qT1 β + qT2 γ > 0. (1.4)

As q∗T z ≤ 0 ∀z ∈ Z, then(qT1 , q

T2 , q3

)AT I 0 0 0

0 0 A −I 0

γT 0 −βT 0 −1

≤ 0 (letting z ∈ Z be

determined by the unit vectors from α =

(1−→0

)to α =

( −→0

1

)):

qT1 AT + q3γ

T ≤ 0. (1.5)

qT1 ≤ 0. (1.6)

qT2 A− q3βT ≤ 0. (1.7)

−qT2 ≤ 0. (1.8)

−q3 ≤ 0. (1.9)

Two steps remain. The first is to show that q3 > 0. If not, q3 = 0, then (1.5) and (1.1)

imply qT1 β ≤ 0 (recall that qT1 ≤ 0). Additionally, from (1.7) and (1.2), qT2 γ ≤ 0. The last

1.2. KUHN-TUCKER THEOREM 7

two inequalities contradict (1.4).

Now with q3 > 0, from (1.5):(−q

T1

q3

)AT(q2

q3

)≥ γT

(q2

q3

). (1.10)

From (1.7): (−q

T2

q3

)A

(−q1

q3

)≤ βT

(−q1

q3

). (1.11)

After taking the transpose of (1.11) and adding it to (1.10), I obtain γT(q2q3

)+βT

(q1q3

)≤ 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 Kuhn-Tucker Theorem

1.2.1 Definitions

Prior to the statement of the theorem, I introduce two definitions. The first, 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 thederivative 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.

Definition 1.3 Given X ⊆ Rn is convex, a differentiable function f : X → R is concave iffDf (x∗) (x− x∗) + f (x∗) ≥ f(x).

Definition 1.4 Given X ⊆ Rn is convex, a differentiable function f : X → R is quasi-

concave iff 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 definitions of concave and quasi-concave may prove useful in

the sequel and highlight why concavity is a stronger assumption.

Definition 1.5 A function f is concave iff ∀x, y ∈ X and ∀θ ∈ [0, 1] ,

f (θx+ (1− θ) y) ≥ θf (x) + (1− θ) f (y) .


Definition 1.6 A function f is quasi-concave iff ∀x, y ∈ X and ∀θ ∈ [0, 1] ,

f (θx+ (1− θ) y) ≥ min f (x) , f (y) .

1.2.2 Constraint Qualification

Define the programming problem (P ) as:

(P ) maximize f(x) with multipliers

subject to gj(x) ≥ 0 j = 1, ...,m λj.

The Lagrange multipliers λ = (λ1, ..., λm) are taken to be a row vector. The vector-valued

function g is defined as g =

g1

:

gm

. To state the Kuhn-Tucker Theorem, we must assume

that the so-called "Constraint Qualification" holds. Examples of the Constraint Qualification

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 Qualification is any suffi cient condition for the following state-

ment:

x∗ is an optimal solution to (P ) =⇒ x∗ is an optimal solution to (LP ) ,

where the linear programming problem (LP ) is:

(LP ) maximize Df(x∗)x with multipliers

subject to Dgj(x∗) (x− x∗) ≥ 0 j = 1, ...,m∗ λj

,

and as above, the first m∗ constraints bind at the optimal solution x∗.

Proof. Assume that the first suggested Constraint Qualification holds. The following argu-ment 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 Qualification

yields the same implication.


The proof only requires the additional assumption that X is a convex set.

Suppose that x∗ is not an optimal solution to (LP ). Then ∃y ∈ 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 definition.

Define zθ = θy+ (1− θ)x∗ for any θ ∈ (0, 1) . Since X is convex, zθ ∈ X. Since gj(y) ≥ 0,

gj(x∗) = 0 for j = 1, ...,m∗, and gj are linear functions, then for any θ ∈ (0, 1) , gj(z

θ) ≥ 0

for j = 1, ...,m∗.

By definition, gj(x∗) > 0 for j = m∗ + 1, ....,m. There exists values of θ near 0 such that

gj(zθ) > 0 for j = m∗ + 1, ....,m. All told, for values of θ near 0, gj(z

θ) ≥ 0 for j = 1, ...,m.

Observe that if Df(x∗)(zθ − x∗

)> 0 for any zθ ∈ Rn, then by the definition of the

derivative,

limθ→0

∣∣f (zθ)− f (x∗)−Df (x∗)(zθ − x∗

)∣∣|zθ − x∗| = 0.

Since the supposition is thatDf(x∗)y > Df(x∗)x∗, then for any θ ∈ (0, 1) , Df (x∗)(zθ − x∗

)>

0. This implies that for values of θ near 0, f(zθ)− f (x∗) > 0, or f

(zθ)> f (x∗) . Combined

with the facts that for values of θ near 0, zθ ∈ X and gj(zθ) ≥ 0 for j = 1, ...,m, the vector

x∗ is not an optimal solution to (P ). This finishes 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 differentiable and con-cave. Further assume that gj : X → R are differentiable and quasi-concave ∀j = 1, ...,m, and

the Constraint Qualification is satisfied. Then x∗ is an optimal solution to the programming

problem (P ) iff x∗ satisfies the Kuhn-Tucker conditions:

First Order Conditions

Df(x∗) + λDg(x∗) = 01×n.

Complimentary Slackness Conditions

λg(x∗) = 01×1, with λ ≥ 0 and g (x) ≥ 0.


The proof proceeds in two steps: Suffi ciency and Necessity. The easier of the two steps

is to show that the Kuhn-Tucker conditions are suffi cient for an optimal solution to (P ) , so

we begin there.

Proof. Suffi ciency of Kuhn-Tucker conditions

We utilize a proof by contradiction. Suppose that x∗ satisfies the Kuhn-Tucker conditions,

but is not an optimal solution to (P ) . Then ∃y such that gj (y) ≥ 0 ∀j = 1, ...,m and f (y) >

f (x∗) . Suppose that the first m∗ constraints bind at x∗ (wlog). Then gj (y) ≥ gj (x∗) = 0

∀j = 1, ...,m∗. Using the quasi-concavity of gj, then Dgj (x∗) (y − x∗) ≥ 0 ∀j = 1, ...,m∗.

From the concavity of f, Df (x∗) (y − x∗) > 0. Then for any values λ1, ..., λm∗ ≥ 0, the

equation

Df (x∗) (y − x∗) +m∗∑j=1

λj ·Dgj (x∗) (y − x∗) > 0.

As the constraints gj (x∗) > 0 for j > m∗, the Complimentary Slackness Conditions imply

λj = 0 for j > m∗. Thus, we have (in vector notation):

(Df(x∗) + λDg(x∗)) (y − x∗) > 0.

This contradicts the First Order Conditions: Df(x∗) + λDg(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 Qualification is satisfied by assumption, then x∗ is an optimal solution

to (LP ) . For simplicity, I define cT = Df(x∗) ∈ Rn, b = Dg(x∗)x∗ ∈ Rm, and A = Dg(x∗) ∈Rm,n. The linear programming problem (LP ) is then equivalent to the simpler linear problem:

(LP2) maximize cTx with multipliers

subject to Ajx− bj ≥ 0 j = 1, ...,m λj.

Suppose, wlog, that the first m∗ constraints bind, Ajx∗ − bj = 0 for j = 1, ...,m∗. Define

λj = 0 for j > m∗.

As x∗ is an optimal solution to (LP2) , there does not exist x such that cTx− cTx∗ > 0

and Ajx− bj ≥ Ajx∗ − bj = 0 ∀j = 1, ...,m∗. Thus, there does not exist x ∈ Rn such that:

(x− x∗)T c > 0 ≥ (x− x∗)T (−Aj)T ∀j = 1, ...,m∗.


Let’s apply the Farkas Lemma with the following relations between the statement of the

Farkas Lemma and this proof:


q∗ ∈ Rn\0 ⇐⇒ (x− x∗) ∈ Rn\0z∗ ∈ Rn ⇐⇒ c ∈ Rn

A ∈ Rn,m∗ ⇐⇒[− (A1)T .....− (Am∗)

T]

α ∈ Rm∗+ ⇐⇒ λT ∈ Rm∗+

Defining

Z =z ∈ Rn : z =

[− (A1)T .....− (Am∗)

T]· λT for some λT ∈ Rm∗+

,

then (−Aj)T ∈ Z ∀j = 1, ...,m∗. As there does not exist (x− x∗) such that (ii) of FarkasLemma holds (given the implication (x− x∗)T c > 0 ≥ (x− x∗)T z ∀z ∈ Z), then (i) musthold (c ∈ Z). Rewriting Z as

Z =

z ∈ Rn : z = −

λ · A1

:

Am∗

T

for some λT ∈ Rm∗+

,

then c ∈ Z implies that there exists λT ∈ Rm∗+ such that cT = −λ

A1

:

Am∗

. From above,

recall that we defined λj = 0 for j > m∗. Thus, we have cT + λA = 0. These are the

Kuhn-Tucker First Order Conditions:

Df(x∗) + λDg(x∗) = cT + λA = 0.

Additionally, we have that ∀j = 1, ...,m : λj ≥ 0 and λj = 0 if Ajx∗ − bj > 0. These are the

Complimentary Slackness Conditions. This finishes the argument.


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. Specifically, 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 correspon-

dence is said to be compact-valued.

Over the entire domain, if the value of a correspondence is convex, then the correspon-

dence is said to be convex-valued.

Over the entire domain, if the value of a correspondence is nonempty, then the corre-

spondence is said to be well-defined (or nonempty-valued).

Some more advanced definitions are provided in the next subsection. With these, I can

discuss the main results related to correspondences.

1.3.1 Definitions

Let U ⊆ Rm be the domain and W ⊆ Rn be the codomain of the correspondence φ, denotedφ : U ⇒ W.

Definition 1.7 The correspondence φ : U ⇒ W is upper hemi-continuous (uhc) iff for any

uν ∈ U, wν ∈ φ (uν) for ν ∈ N such that limν→∞

uν = u ∈ U, there is a subsequence, wlog theoriginal sequence, such that lim

ν→∞wν = w ∈ φ(u).

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 limν→∞

wν = w. All that remains to be

shown for the definition of uhc is that w ∈ φ(u).

Consider any u ∈ U.We can certaintly specify a sequence uν ∈ U such that uν = u ∀ν ∈N. By the definition of uhc, all sequences wν ∈ φ (uν) = φ (u) have a convergent subsequence

(wlog the original sequence) limν→∞

wν = w ∈ φ(u). As all sequences wν ∈ φ (uν) = φ (u)

have a convergent subsequence, then the set φ (u) is compact. Thus, φ is a compact-valued

correspondence. So uhc implies compact-valued.1

1In 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., φ(u) = W

∀u ∈ U, if W is compact, then φ is uhc.

Definition 1.8 The correspondence φ : U ⇒ W is lower hemi-continuous (lhc) iff for any

uν ∈ U for ν ∈ N such that limν→∞

uν = u ∈ U and w ∈ φ(u), there is a sequence wν ∈ φ (uν)

for ν ∈ N such that limν→∞

wν = w.

If the correspondence φ takes all values in the codomain, i.e., φ(u) = W ∀u ∈ U, then φis lhc.

Definition 1.9 The correspondence φ : U ⇒ W is continuous iff it is upper hemi-continuous

and lower hemi-continuous.

Figure 1.5 illustrates the distinction between uhc and lhc. The left panel shows a con-

tinuous 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 manu-

script.

Theorem 1.5 Berge’s Maximum Theorem

Suppose X ⊆ Rn, A ⊆ Rm, f : X × A → R is continuous, Xf : A ⇒ X is well-defined

and continuous, and

Xo : A⇒ X s.t. α 7−→ arg maxx∈Xf (α)

f (x, α) .

Then Xo is uhc.

Proof. Suppose that Xo is not uhc. Then for some sequence αν ∈ A, xν ∈ Xo (αν) for

ν ∈ N such that limν→∞

αν = α ∈ A, it is not true that limν→∞

xν = x ∈ Xo (α) . As Xf is uhc,

there is a subsequence, wlog the original sequence, such that limν→∞

xν = x ∈ Xf (α) . Then, it

must be that x /∈ Xo (α) , even though xν ∈ Xo (αν) for ν ∈ N. As Xf is lhc, then for every

x′ ∈ Xf (α) , in particular x′ ∈ Xf (α) such that f (x, α) < f (x′, α) , there exists a sequence

x′ν ∈ Xf (αν) for ν ∈ N such that limν→∞

x′ν = x′. Since f is continuous,

limν→∞

f (xν , αν) = f (x, α) < f (x′, α) = limν→∞

f (x′ν , αν) .


As x′ν ∈ Xf (αν) for ν ∈ N, then this contradicts that xν ∈ Xo (αν) for ν ∈ N.

Corollary 1.2 Suppose X ⊆ Rn, A ⊆ Rm, f : X × A → R is continuous, Xf : A ⇒ X is

well-defined and continuous, and

V : A→ X s.t. α 7−→ maxx∈Xf (α)

f (x, α) .

Then V is continuous.

Proof. Since Xo is uhc, then for some sequence αν ∈ A, xν ∈ Xo (αν) for ν ∈ N such thatlimν→∞

αν = α ∈ A, there is a subsequence, wlog the original sequence, such that limν→∞

xν =

x ∈ Xo (α) . Since (xν , αν) → (x, α) and f is continuous, then f (xν , αν) → f (x, α) . Since

xν ∈ Xo (αν) and x ∈ Xo (α) , then this implies that V (αν) → V (α) . This completes the

argument.

1.3.3 Kakutani’s Fixed Point Theorem

By the Extreme Value Theorem, Xo is well-defined. Further, if f is quasi-concave (Definition

1.4), then Xo is convex-valued (details are left to the reader).

Theorem 1.6 Kakutani’s Fixed Point TheoremLet Γ : X ⇒ X be a well-defined, convex-valued, and uhc correspondence and X be a

compact, convex, and nonempty set. Then ∃x ∈ X such that x ∈ Γ (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 ∈ X, max(xn)n∈N

f (x0, x1) +∞∑i=1

βif (xi, xi+1) such that xn ∈ Γ (xn−1)

∀n ∈ N.

The following assumptions are made:

1.3. CORRESPONDENCES 15

Assumption 1: f is continuous and bounded.

Assumption 2: Γ is continuous (uhc and lhc).

Assumption 3: β ∈ (0, 1) .

Assumption 4: X is compact.

Rather than finding an entire optimal sequence all at once, we can find a sequence of

optimal solutions. To do this, we define the recursive programming problem as follows:

For all n ∈ N, given xn−1 ∈ X, maxxn

f (xn−1, xn) + βV (xn) such that xn ∈Γ (xn−1) .

We need to verify that the function V : X → R satisfies

V (xn−1) = maxxn

f (xn−1, xn) + βV (xn)

for all n ∈ N such that xn ∈ Γ (xn−1) .

Define the mapping T such that

(TV ) (x) = maxy∈Γ(x)

f (x, y) + βV (y) for all x ∈ X.

Assume that V ∈ 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 TV is a continuous function. Since

X is compact, then TV is also a bounded function. Consequently, TV ∈ 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.

Definition 1.10 A self map Φ : K → K is a contraction if there exists δ ∈ (0, 1) such that

d (Φ (x) ,Φ (y)) ≤ δd (x, y) for all x, y ∈ 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 suffi cient

conditions.


Lemma 1.2 If Φ : K → K is increasing and there exists δ ∈ (0, 1) such that

Φ (f + α) ≤ Φ (f) + δα

for all (f, α) ∈ K × R+, then Φ is a contraction.

Using these suffi cient conditions, is the mapping T a contraction? Let’s consider each

condition in turn:

1. Increasing

If V ≥ W, where V,W ∈ CB (X) , this implies V (x) ≥ W (x) ∀x ∈ X. So, if

V (x) ≥ W (x) ∀x ∈ X, then T (V ) (x) ≥ T (W ) (x) ∀x ∈ X. This is by the definitionof T (V ) (x) = max

y∈Γ(x)f (x, y) + βV (y) as any y∗ = arg max

y∈Γ(x)

f (x, y) + βW (y) can

also be selected for arg maxy∈Γ(x)

f (x, y) + βV (y) (if not selected, then a higher max is

achieved for T (V ) (x)). Thus, T is increasing.

2. ∃δ ∈ (0, 1) such that Φ (f + α) ≤ Φ (f) + δα

By definition, T (V + α) (x) = maxy∈Γ(x)

f (x, y) + βV (y) + βα for all x ∈ X. By defi-

nition, maxy∈Γ(x)

f (x, y) + βV (y) + βα = T (V ) (x) + βα. As β < 1 by Assumption 3,

then the condition is satisfied.

So, T is a contraction. How does this help us? We can apply a different fixed point

theorem, called the Contraction Mapping Theorem.

Theorem 1.7 Contraction Mapping Theorem

If Φ : K → K is a contraction and K is a complete metric space, then there exists a

unique k∗ ∈ K such that Φ (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 fixed point V ∈ CB (X) such that

T (V ) = V.

This results in the so-called Bellman equation:

V (x) = maxy∈Γ(x)

f (x, y) + βV (y) for all x ∈ 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 ∈ X, max(xn)n∈N

f (x0, x1) +

∞∑i=1

βif (xi, xi+1) such that xn ∈ Γ (xn−1)

∀n ∈ N,

is also a solution to the Bellman equation, and vice versa.

1.4 Differential Topology

I first introduce the concept of measure and properness before moving into the basic results

for the analysis of differentiable equations.

1.4.1 Measure

The set X0 ⊆ Rn has zero measure if for every ε > 0, X0 can be covered by n−dimensionalcubes whose total volume is less than ε. For any open set X ⊆ Rn, the subset X∗ ⊆ X is a

generic subset of X iff:

1. X∗ is open.

2. The complement relative to X, (X∗)CX = X\X∗ 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 iff:

1. f is continuous.

2. For any Y ′ ⊆ Y compact, the set f−1 (Y ′) is compact.

Using the definition of properness, we obtain the following property (of great value in

the coming sections).


Lemma 1.3 Properness Property

Suppose that for X ⊆ Rn and Y ⊆ Rm, the function f : X → Y is proper. If X ′ ⊆ X is

closed (relative to X), then f (X ′) ⊆ Y is closed (relative to Y ).

Proof. Consider a sequence yν ∈ f (X ′) for ν ∈ N such that limν→∞

yν = y′. As yν ∈ f (X ′)

for ν ∈ N, then there exists (xν)ν∈N such that xν ∈ X ′ and yν = f (xν) ∀ν ∈ N. Define the

set Y ′ =

(yν)ν∈N , y′ . The set Y ′ is compact. As f is proper, then f−1 (Y ′) is compact.

As xν ∈ f−1 (Y ′) ∀ν ∈ N, then there exists a subsequence, wlog the original sequence, suchthat lim

ν→∞xν = x′ ∈ f−1 (Y ′) . As X ′ is closed, then x′ ∈ X ′. As f is continuous, y′ = f (x′) ,

so y′ ∈ f (X ′) . This verifies that f (X ′) is closed.

1.4.3 Regular vs. Critical values

The following framework is utilized when analyzing differentiable equations:

• Ξ ⊂ RJ is the set of variables; it is open with typical element ξ.

• Θ ⊂ RK is the set of parameters; it is open with typical element θ.

• H = Ξ×Θ is the set of variables and parameters, with typical element η = (ξ, θ) .

• Φ : H → RL is the system of equations characterizing equilibria; it is C1.

The set of solutions are η ∈ H such that Φ (η) = 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 definitions:

• M = η ∈ H : Φ (η) = 0 , the set of solutions.

• π is the projection of M onto Θ.

• η ∈M is a critical point if rankDξΦ (η) < L, a regular point if rankDξΦ (η) = L.

1.4. DIFFERENTIAL TOPOLOGY 19

• θ ∈ Θ is a critical value if π−1 (θ) contains at least one critical point. Define Θc as the

set of critical values.

• Any θ ∈ Θ that is not a critical value must be a regular value. Define Θr = Θ\Θc as

the set of regular values.

The terms critical/regular point and critical/regular value are with respect to the pro-

jection π. Notice that if no solution exists for some parameter θ∗, that is, π−1 (θ∗) = ∅, thenθ∗ is a regular value (any value that is not a critical value is a regular value, by definition).

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 (finite and locally unique).

Theorem 1.8 Closedness TheoremIf π is proper, then Θc is closed, so that Θr is open.

Proof. The set 0 is closed and the function Φ is continuous, so Φ−1 (0) = M is closed

(relative to H). With J = L, we can define the determinant det : M → R on the matricesDξΦ (η) . The determinant is continuous and the set 0 is closed, so the set of critical points,those η ∈ M such that η ∈ det−1 (0) , is closed. As π is proper, then by the Properness

Property, the set of critical values Θc is closed.

Concerning notation, the derivative matrixDξΦ (η) considers the derivatives with respect

to the variables ξ. As J = L (number of variables equals number of equations), then DξΦ (η)

is a square matrix. The derivative matrix DΦ (η) considers the derivatives with respect to

both the variables ξ and the parameters θ. This matrix could be more accurately written as

follows: DΦ (η) = Dξ,θΦ (η) . 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 TheoremIf, for every η ∈ M, rankDΦ (η) = L, then Θc has zero measure, so that Θr has full

measure.


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

θ that are both (a) critical (that is, rankDξΦ (η) < L) and (b) satisfy the rank condition

rankDΦ (η) = L is a set of measure zero. This brings us to the aptly named Stack of Records

Theorem.

Theorem 1.10 Stack of Records TheoremIf π is proper, θ′ ∈ Θr, and π−1 (θ′) 6= ∅, then1. π−1 (θ′) = η′i : i = 1, ..., I with I <∞ .2. There is an open neighborhood Θ′ around θ′ and open neighborhoods H ′i around η

′i and

C1 mappings φi : Θ′ → H ′i ∩M ∀i = 1, ..., I such that for any θ ∈ Θ′ and any η ∈ π−1 (θ) ,

there exists a unique i such that η = φi (θ) .

Figure 1.7 illustrates the result.

Proof. Consider any η′ ∈ π−1 (θ′) . Since rankDξΦ (n) = L and J = L, then we can apply

the Implicit Function Theorem. Thus, there exists a neighborhoodΘ′ of θ and a neighbohood

H ′ of η′ and a C1 mapping φ : Θ′ → H ′ such that for any θ ∈ Θ′, η ∈ H ′ ∩M iff η = φ (θ) .

Thus, π−1 (θ′) ⊂ ∪η′∈π−1(θ′)

H ′, an open covering. Since π is proper and θ′ is compact, then

π−1 (θ′) is compact, so the cover ∪η′∈π−1(θ′)

H ′ must have a finite subcover ∪i=1,...,I

H ′i. That is,

π−1 (θ′) = η′i : i = 1, ..., I with I <∞ .By taking the neighborhood Θ′ to be small enough, then H ′i ∩H ′j = ∅ ∀i 6= j. Thus, for

any θ ∈ Θ′, each η ∈ π−1 (θ) has a unique φi.

1.4.5 Finite Local Uniqueness

For any system of equations Φ (η) = 0, we seek to verify the following two properties:

• π is proper.

• For every η ∈M, rankDΦ (η) = L (called the rank condition).

Then over a generic subset of parameters (the set of regular values Θr), there are a finite

number of solutions (that is, π−1 (θ) is a finite set ∀θ ∈ Θr) and the solutions are locally

unique. By locally unique, we mean that for any θ ∈ Θr with a solution η ∈ π−1 (θ) , there is

only one solution η′ ∈ π−1 (θ′) in a neighborhood around η, where θ′ lies in a neighborhood

around θ.

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 × n matrix and θ ∈ Rn. Use Farkas Lemma to prove the followingimplication.

There does not exist θ such that Rθ > 0

⇓∃λ ∈ Rm++ s.t. λ

TR = 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 equilib-

rium. Consider a game with I players. Each player i has a finite number of actionsai1, ...., a

iJi

. The set of strategies for each player i is then the simplex ∆Ji−1 of di-

mension Ji − 1. The strategies are simply the probabilities that each player assigns to

each of the finite number of actions. For simplicity, denote the strategy set for each

player as Si with element si. These sets are nonempty, compact, and convex.

If player i′s payoff value for the action profile a = (a1, ..., ai, ..., aI) is pi(a), then the

objective function for each player is (using the convention s = (s1, ..., si, ..., sI)):

ui (s) =∑

api(a) ·

∏isi(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 ×j 6=iSj = S1×...×Si−1×Si+1×...×SI as the Cartesian product


of the strategy sets (not including the strategy set for player i). Define F i : ×j 6=iSj ⇒ Si

as the set of "feasible" strategies. Notice that F i (s−i) = Si ∀s−i ∈ ×j 6=iSj.

Define BRi : ×j 6=iSj ⇒ Si as the best response correspondence for player i. Use the

Berge’s Maximum Theorem and Kakutani’s Fixed Point Theorem to prove that the

self-map(BR1, ..., BRI

): ×iSi ⇒ ×

iSi has a fixed point. By definition, a fixed point is

a Nash equilibrium.

3. (A Different Application of Differential Topology)

Let Ξ ⊂ RJ be the set of variables (with typical element ξ), Θ ⊂ RK be the set of

parameters (with typical element θ), H = Ξ×Θ (so η = (ξ, θ)), and Φ : H → RL be theC1 system of equations. Assume that J < L. Suppose that the projection π is proper

and the rank condition holds (that is, rankDΦ (η) = 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 Θr), 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, Uni-

versity 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 Differentiable 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 defining 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 finite 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 first component of the environment is households. Let H = 1, ..., H be the setof households, where the number of households is finite (H < ∞). A typical element of the

set H will be h. Households trade and consume G goods. Let G = 1, ..., G 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)g∈G is the vector of consumption by household h. Finally, x =

(xh)h∈H 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)g∈G is the vector of endowments

of household h and e =(eh)h∈H is the vector of all endowments.

Each household must select consumption within its consumption set. The consumption

set is denotedXh and must be a subset of the nonnegative orthant,Xh ⊆ RG+.We assume thatthe endowments eh belong to the consumption set, eh ∈ Xh. 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 : Xh → R. Thesehousehold primitives (parameters of the model) are listed below.

• Xh ⊆ RG+ is the consumption set. We typically assume that Xh is closed (relative to

the set RG+) and convex.

• uh : Xh → R is the utility function. We typically assume that uh is continuous, locallynon-satiated, and quasi-concave.

• eh ∈ Xh is the endowment vector. We typically assume that eh >> 0.

Definition 2.1 The function uh : Xh → R is locally non-satiated if ∀x ∈ Xh and ∀ε > 0,

there exists y ∈ Xh ∩Nε (x) such that uh (y) > uh (x) .

The second component of the environment is the market institutions. In the Arrow-

Debreu model, markets are perfectly competitive, meaning that all households take prices

as given and do not account for how their actions affect the market prices. Each good has

a market price pg. The vector of all prices is p = (pg)g∈G ∈ RG\0. The convention is thatthe 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 definition, exhibits

nominal price indeterminacy. To remove this, we normalize the prices so that p ∈ ∆G−1.

We are now prepared to define an Arrow-Debreu equilibrium.

Definition 2.2 An Arrow-Debreu equilibrium is((xh)h∈H , p

)such that

2.2. EXISTENCE 27

1. ∀h ∈ H, given p, xh is an optimal solution to the household problem (HP )

(HP ) maximize uh(xh)

subject to xh ∈ Xh

p(eh − xh

)≥ 0

.

2. Markets clear ∑h∈H

xhg =∑

h∈Hehg ∀g ∈ G.

2.2 Existence

Which assumptions are required to guarantee that an Arrow-Debreu equilibrium exists for

all parameters(eh, uh

)h∈H? These assumptions are listed below. After the statement of the

theorem, I will discuss why the third assumption is required.

Assumption E1: Xh = RG+ ∀h ∈ H.

Assumption E2: uh : Xh → R is continuous, locally non-satiated, and quasi-concave ∀h ∈ H.

Assumption E3: eh >> 0 ∀h ∈ H.

Assumption E4: For some household h′, uh′is non-decreasing.

Theorem 2.1 Under Assumptions E1-E4, an Arrow-Debreu equilibrium((xh)h∈H , p

)ex-

ists.

Proof. For the complete proof, see Section 2.6.

The proof method proceeds as follows. We begin by defining the budget correspondence

bh : ∆G−1 ⇒ Xh ∀h ∈ H such that bh (p) =xh ∈ Xh : p

(eh − xh

)≥ 0. Using this

definition, the demand correspondence can be defined as dh : ∆G−1 ⇒ Xh such that dh (p) =

arg maxxh∈bh(p)

uh(xh). If the conditions of Berge’s Maximum Theorem are satisfied, namely that

bh is well-defined and continuous, then the correspondence dh is upper hemi-continuous.

Using Assumption E2 (continuous and quasi-concave), the correspondence dh is also well-

defined and convex-valued. Thus, we can apply the Kakutani’s Fixed Point Theorem to

show that a fixed point exists. Further details are left to Section 2.6.


From the outline of the proof, the key step is to verify that the conditions of Berge’s

Maximum Theorem are satisfied. It is straightforward to show that the budget correspon-

dence bh is well-defined 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 suffi 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 ∈ bh (p) .

Consider any sequence pν → p, specifically pν =(

1ν, 1− 1

ν

)for ν ∈ N. Equilibrium prices

(see Exercise 1) can never contain a negative element: pν > 0. The definition of lower hemi-

continuity requires that a sequence xν exists such that (i) xν → x and (ii) xν ∈ bh (pν) for

ν ∈ N. However, as can be seen in the right panel of Figure 2.1, no such sequence can befound (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 define a Pareto optimal allocation.

Definition 2.3 A feasible allocation(xh)h∈H is such that

1. xh ∈ Xh ∀h ∈ H and1A 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:

1. eh > 0 ∀h ∈ H and∑

h∈H eh >> 0

2. ∀H1,H2 6= ∅ such that H1 ∩ H2 = ∅ and H1 ∪ H2 = H and any allocation(xh)h∈H , there ex-

ists(yh)h∈H1

≥ 0 and(xh)h∈H2

≥ 0 such that (a)∑

h∈H1yhg = 0 whenever

∑h∈H1

ehg = 0, (b)∑h∈H2

xh ≤∑

h∈H1

(yh + eh

)+∑

h∈H2eh, and (c)

(uh(xh))h∈H2

>(uh(xh))h∈H2

.

Basically, the second condition says that each household has a strictly positive endowment of at least onegood that is "desired" by some other household (forcing the price of that good to be strictly positive).

2.3. FIRST BASIC WELFARE THEOREM 29

2.∑

h∈Hxhg =

∑h∈H

ehg ∀g ∈ G.

For simplicity, define the set of feasible allocations as FA.

Definition 2.4 A Pareto optimal allocation(xh)h∈H is such that there does not exist a

feasible allocation(yh)h∈H where uh

(yh)≥ uh

(xh)∀h ∈ H and uh

′ (yh′)> uh

′ (xh′)for

some h′.

To find the Pareto optimal allocations, we have two options. First, we can assume that

uh′is strictly increasing for some household h′ and ∀h 6= h′, uh is continuous. Then

(xh∗)h∈H

is a Pareto optimal allocation iff(xh∗)h∈H is an optimal solution to the following nonlinear

programming problem:

(POh′) maximize uh′ (xh′)

subject to uh(xh)≥ uh

(xh∗)∀h 6= h′(

xh)h∈H ∈ FA

.

The proof is in Section 2.6.

The second option employs concavity. If(xh∗)h∈H is a Pareto optimal allocation and

uh is concave ∀h ∈ H, then there exists(µh)h∈H ∈ ∆H−1 such that

(xh∗)h∈H is an optimal

solution to the following nonlinear programming problem:

(PO) maximize∑

h∈Hµh · uh

(xh)

subject to(xh)h∈H ∈ FA

.

Conversely, if there exists(µh)h∈H ∈ int

(∆H−1

)such that

(xh∗)h∈H is an optimal solution

to (PO) , 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 : Xh → R is locally non-satiated ∀h ∈ H.

Theorem 2.2 Under Assumption F1, all Arrow-Debreu equilibrium allocations(xh)h∈H are

Pareto optimal.

Proof. See Section 2.6.The assumption of locally non-satiated utility implies that all budget constraints hold

with equality: p(eh − xh

)= 0 ∀h ∈ H. In this case, why do we define an Arrow-Debreu


equilibrium with inequalities in the budget constraints? The reason is that the First Ba-

sic Welfare Theorem no longer holds when we define an Arrow-Debreu equilibrium with

equalities in the budget constraint.

Definition 2.5 An Arrow-Debreu equilibrium with equalities in the budget constraint is((xh)h∈H , p

)such that

1. ∀h ∈ H, given p, xh is an optimal solution to the household problem (HP =)

(HP =) maximize uh(xh)


p(eh − xh

)= 0

.


xhg =∑

h∈Hehg ∀g ∈ 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 ∈ R2

+ : x12 ≥

1

x11

.

X2 = R2+.

These consumption sets are closed and convex (satisfying Assumption S1 below, which are

typical assumptions for(Xh)h∈H). The utility functions for both households are given by:

u1(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 = 1

x11denotes X1. There

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 alloca-

tions 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 off) 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 ∀h ∈ H.

Assumption S2: uh : Xh → R is continuous and quasi-concave ∀h ∈ H.

Assumption S3: For some household h′, wlog h′ = 1, X1 is unbounded above

and u1 is increasing.

Theorem 2.3 Under Assumptions S1-S3, if(eh)h∈H =

(xh∗)h∈H is a Pareto optimal al-

location and xh∗ ∈ intXh ∀h ∈ H, then there is p∗ > 0 such that((xh∗)h∈H , p

∗)is an

Arrow-Debreu equilibrium.

Proof. See Section 2.6.

If we want to consider a Pareto optimal allocation(xh∗)h∈H that differs from the endow-

ment(eh)h∈H , then the statement of the theorem includes "an imposed tax/subsidy scheme(

τh)h∈H with

∑h∈H τ

h = 0 such that τh = p∗xh∗ − p∗eh ∀h ∈ H." If τh > 0, the transfer is

a subsidy and that amount is added to the household’s income. If τh < 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∗ ∈ intXh ∀h ∈ H. Whyis 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.


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 ∈ R2

+ : x12 ≥ 1− x1

1

.

X2 = R2+.

These consumption sets are closed and convex (satisfying Assumption S1). The utility

functions for both households are given by:

u1 (x1) = min x11, x

12 .

u2 (x2) is "smooth" (C2, strictly concave, strictly increasing).

Both utility functions are continuous and quasi-concave (satisfying Assumption S2). Con-

sider Figure 2.3, where X1 is the shaded region above the curve x12 = 1 − x1

1. Figure 2.3

contains three indifference curves for h = 1 and two for h = 2.

The allocation E is Pareto optimal. Why? We cannot find a feasible allocation that

makes both households better off 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 indifference curve for household h = 2). But given this equilibrium price p∗,

the optimal consumption choice by household h = 1 would be E ′, not E.

Thus, the Second Basic Welfare Theorem relies importantly on the assumption that

xh∗ ∈ intXh ∀h ∈ H.

2.5 Regularity

When we state that the Arrow-Debreu equilibria are regular, what we mean is that the num-

ber of equilibria is finite 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: Xh = RG++ ∀h ∈ H.

Assumption R2: uh : Xh → R is C2, differentiably strictly increasing (mean-

ing thatDuh(xh)>> 0), differentiably strictly concave (meaning thatD2uh

(xh)

2.5. REGULARITY 33

is a negative definite matrix), and satisfies the boundary condition (meaning that

∀x ∈ RG++, cly : uh(y) ≥ uh(x)

⊆ RG++) ∀h ∈ H.

Assumption R3: eh >> 0 ∀h ∈ H.

Theorem 2.4 Under Assumptions R1-R3, all Arrow-Debreu equilibria satisfy Finite LocalUniqueness.

The remainder of the section walks through the proof of Theorem 2.4, with useful prop-

erties of the model sprinkled in.

Recall that the household problem (HP ) is given by



p(eh − xh

)≥ 0

.

Under Assumption R2 (namely, the boundary condition), the condition xh ∈ Xh never binds.

Applying the Kuhn-Tucker Theorem, the following conditions are necessary and suffi cient

conditions for an optimal solution to (HP ) :

Duh(xh)− λhp = 0

p(eh − xh

)= 0,

where λh ∈ 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

λh > 0 and p >> 0.

If((xh)h∈H , p

)satisfies the definition of an Arrow-Debreu equilibrium, then so does((

xh)h∈H , κp

)for any scalar κ > 0. This is called nominal indeterminacy, as a continuum

of prices exist that satisfy the equilibrium definition. 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 p\G =

(p1, ..., pG−1) ∈ RG−1++ . What we show with regularity is that there is no real determinacy;

namely that the Arrow-Debreu equilibria are determinate (not a continuum).

2This is a different price normalization than previously used in this chapter.


Walras’Law (obtained by summing over all households’budget constraints) states:

∑h∈H p

(eh − xh

)= 0.

As p >> 0, if∑

h∈H(ehg − xhg

)= 0 ∀g < G, then it must also be the case that

∑h∈H

(ehG − xhG

)=

0. Thus, the market clearing condition∑

h∈H(ehG − xhG

)= 0 is redundant given that the mar-

ket clearing conditions hold ∀g < G. I use the notation

∑h∈H

(eh\G − xh\G

)=(∑

h∈H(eh1 − xh1

), ...,

∑h∈H

(ehG−1 − xhG−1

))Tfor the (G− 1)−dimensional column vector.

Define the system of equations Φ : ×h∈H

(RG+1

++

)× RG−1

++ → RH(G+1)+G−1 as:

Φ((xh, λh

)h∈H , p

)=

( [

Duh(xh)− λhp

]Tp(eh − xh

) )h∈H∑

h∈H

(eh\G − xh\G

) ,

where(xh, λh

)h∈H ∈ RH(G+1)

++ and p ∈ RG−1++ . By definition, Φ

((xh, λh

)h∈H , p

)= 0 iff((

xh)h∈H , 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) π is proper and (ii) rankDΦ((xh, λh

)h∈H , p

)= H (G+ 1) +G− 1.

The first condition is left to Exercise 6. We will now walk through the second condition.

The derivative DΦ((xh, λh

)h∈H , p

)means that we are taking derivatives with respect

to(xh)h∈H ,

(λh)h∈H , p, and

(eh)h∈H (all variables and all parameters). If we show that the

rank condition holds by only taking derivatives with respect to(xh)h∈H ,

(λh)h∈H , p, and

e1, then we have finished the argument.

The derivative matrix M = D(xh,λh)h∈H

,p,e1Φ((xh, λh

)h∈H , p

)has [H (G+ 1) +G− 1]

rows and [H (G+ 1) +G− 1 +G] columns. The rows of M correspond to the equations in

Φ((xh, λh

)h∈H , 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

M =

D2u1 (x1) −pT 0 0 0 0

(−λ1IG−1

0

)0

−p 0 0 0 0 0(e1\G − x1

\G

)Tp

0 0 ... ... 0 0 : 0

0 0 ... ... 0 0 : 0

0 0 0 0 D2uH(xH)−pT

(−λHIG−1

0

)0

0 0 0 0 −p 0(eH\G − xH\G

)T0(

−IG−1 0)

0 ... ...(−IG−1 0

)0 0

(IG−1 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 νTM = 0 and must verify νT = 0, where

ν ∈ RH(G+1)+G−1 corresponds to equations in Φ :

ν =

∆x1

∆λ1

:

∆xh

∆λh

:

∆p

FOC1

BC1

:

FOCh

BCh

:

MC

.

The equations νTM = 0 are given by:

(∆x1)TD2u1 (x1)−∆λ1p−∆pT

[IG−1 0

]= 0 (A.1.a)

− (∆x1)TpT = 0 (A.1.b)

:(∆xh

)TD2uh

(xh)−∆λhp−∆pT

[IG−1 0

]= 0 (A.1.c)

−(∆xh

)TpT = 0 (A.1.d)

:

−∑

h∈H λh(

∆xh\G

)T+∑

h∈H ∆λh(eh\G − xh\G

)T= 0 (A.1.e)

∆λ1p+ ∆pT[IG−1 0

]= 0 (A.1.f)

. (2.1)


Let’s show that νT = 0 in three steps:

1. (A.1.f) can be written as

∆λ1 (p1, ..., pG−1, 1) + (∆p1, ...,∆pG−1, 0) = 0.

As p >> 0, then ∆λ1 = 0. Consequently, ∆pT = 0.

2. Use (A.1.f) to simplify (A.1.a) and then postmultiply the equation by ∆x1 :

(∆x1

)TD2u1

(x1)

∆x1 = 0.

As u1 is differentiably strictly concave, the Hessian matrix D2u1 (x1) is negative defi-

nite. Thus, (∆x1)T

= 0.

3. For any household h > 1, postmultiply (A.1.c) by ∆xh :

(∆xh

)TD2uh

(xh)

∆xh −∆λhp∆xh = 0.

From (A.1.d), p∆xh =((

∆xh)TpT)T

= 0. As uh is differentiably strictly concave,

the Hessian matrix D2uh(xh)is negative definite. Thus,

(∆xh

)T= 0. From (A.1.c),

∆λh = 0.

Thus, νT =(

(∆x1)T,∆λ1, ...,

(∆xh

)T,∆λh, ...,∆pT

)= 0, finishing the argument.

2.6 Proofs

2.6.1 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. Define the bounded consumption

set Xh =xh ∈ Xh : xhg ≤ 2 ·

∑h∈H e

hg ∀g ∈ G

. The set Xh is compact, convex, and non-

empty. We show in Part 6 that equilibrium consumption xh is an optimal solution to (HP )

2.6. PROOFS 37

iff xh is an optimal solution to:



p(eh − xh

)≥ 0

.

Part 2: Demand correspondence

Define the budget correspondence bh : ∆G−1 ⇒ Xh ∀h ∈ H such that

bh (p) =xh ∈ Xh : p

(eh − xh

)≥ 0.

The correspondence is well-defined (consider an element eh ∈ bh (p)).

Claim 2.1 bh is upper hemi-continuous.

Proof. Consider sequences pν and xν such that xν ∈ bh (pν) ∀ν ∈ N. Let pν → p and xν → x.

To prove upper hemi-continuity, we must prove that x ∈ bh (p) . Suppose otherwise, that is,

x /∈ bh (p) . Then, either xg < 0 for some good g or p(eh − x

)< 0. Then, by continuity, for

some ν, either xνg < 0 for some good g or pν(eh − xν

)< 0. This contradicts that xν ∈ bh (pν)

∀ν ∈ N.

Claim 2.2 bh is lower hemi-continuous.

Proof. Consider a sequence pν such that pν → p and x ∈ bh (p) . To prove lower hemi-

continuity, we must find a sequence xν such that xν ∈ bh (pν) for ν ∈ N and xν → x. There

are two cases to consider.

Case I : p(eh − x

)> 0. Then ∃ν∗ such that ∀ν ≥ ν∗, pν

(eh − x

)> 0. Define xν = x

∀ν ≥ ν∗. Then, xν → x and xν ∈ bh (pν) ∀ν ≥ ν∗.

Case II : p(eh − x

)= 0. I will define xν = θν · x for ν ∈ N, where

θν =

pveh

pvxif p

veh

pvx< 1 and pvx 6= 0

1 otherwise

.

As pν → p and p(eh − x

)= 0, then ∃v∗ such that ∀ν ≥ ν∗, pνx > 0. As pν → p, then

pveh

pvx→ 1. Thus, ∀ν ≥ ν∗, xν = θν · x→ x. By definition, pνxν = θν · pνx = 1 · pvx ≤ pveh if

pveh

pvx≥ 1 and pνxν = θν · pνx =

(pveh

pvx

)pvx = pveh if p

veh

pvx< 1. Thus, xν ∈ bh (pν) ∀ν ≥ ν∗.


Thus, bh is upper hemi-continuous and lower hemi-continuous. Define the demand cor-

respondence dh : ∆G−1 ⇒ Xh such that

dh (p) = arg maxxh∈bh(p)

uh(xh).

From Berge’s Maximum Theorem, dh is upper hemi-continuous. As uh is continuous and

Xh is compact, the Extreme Value Theorem implies that dh is well-defined. As uh is quasi-

concave and Xh is convex, dh is convex-valued. To see this, let x, y ∈ dh (p) . Then x, y ∈ Xh

and both p(eh − x

)≥ 0 and p

(eh − y

)≥ 0. Therefore, ∀λ ∈ [0, 1] , the choice λx+(1− λ) y ∈

Xh (convexity) and p(eh − (λx+ (1− λ) y)

)≥ 0 (budget constraints are linear). Further,

uh (λx+ (1− λ) y) ≥ minuh (x) , uh (y)

= uh (x) = uh (y) (definition of quasi-concavity).

Thus, λx+ (1− λ) y ∈ dh (p) .

Part 3: Price correspondence

Define the price correspondence ρ : ×h∈H

Xh ⇒ ∆G−1 such that z =(z1, ..., zH

)7→ arg max

p∈∆G−1

p ·∑

h∈H(zh − eh

). The correspondence is well-defined, using the Extreme Value Theorem.

The correspondence is convex-valued as the objective function is quasi-concave (in p) and

∆G−1 is convex.

Claim 2.3 ρ is upper hemi-continuous.

Proof. Consider sequences zν and pν such that pν ∈ ρ (zν) for ν ∈ N. Let zν → z and pν → p.

To prove upper hemi-continuity, we must show that p ∈ ρ (z) . Suppose otherwise, that is,

p /∈ ρ (z) . Then, there exists p ∈ ∆G−1 such that p ·∑

h∈H(zh − eh

)> p ·

∑h∈H

(zh − eh

).

Then for some ν, pν ·∑

h∈H(zh − eh

)> pν ·

∑h∈H

(zh − eh

). This contradicts that pν ∈ ρ (zν)

for ν ∈ N.

Part 4: Fixed point

Define Γ : ×h∈H

Xh×∆G−1 ⇒ ×h∈H

Xh×∆G−1 as the Cartesian product of(dh)h∈H and ρ. The

set ×h∈H

Xh×∆G−1 is compact, convex, and nonempty. As the correspondences dh (for any h)

and ρ are upper hemi-continuous, convex-valued, and well-defined, then so is the Cartesian

product.

From Kakutani’s Fixed Point Theorem, there exists a fixed point((xh)h∈H , p

∗)∈

Γ((xh)h∈H , p

∗).

2.6. PROOFS 39

Part 5: Market clearing

As(uh)h∈H are locally non-satiated, then p∗

(xh − eh

)= 0 ∀h ∈ H. Walras’Law is then

given by p∗∑

h∈H(xh − eh

)= 0. This together with the definition of the price correspondence

implies∑

h∈H(xhg − ehg

)≤ 0 ∀g ∈ G. Otherwise, if

∑h∈H

(xhg′ − ehg′

)> 0 for some g′, then

the maximum of the price correspondence is p∗∑

h∈H(zh − eh

)> 0 by setting p∗g′ = 1.

If∑

h∈H(xhg − ehg

)< 0, then p∗g = 0 (see Walras’Law). For all households h 6= h′, define

xh∗ = xh. For h′, define xh′∗ = xh

′ −∑

h∈H(xh − eh

). As uh

′is non-decreasing (Assumption

4), then(xh∗)h∈H are optimal solutions to the household problems (HP ) . Further, market

clearing is satisfied as:

∑h∈H x

h′∗ =∑

h6=h′ xh +

[xh′ −∑

h∈H(xh − eh

)]=∑

h∈H eh.

Part 6: Innocuous to bound consumption

I want to show that equilibrium consumption xh is an optimal solution to (HP ) iff xh is an

optimal solution to:



p(eh − xh

)≥ 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 ∃y′ ∈ Xh\Xh such that p(eh − y′

)≥ 0 and uh (y′) > uh

(xh). The market

clearing requirement implies xh ∈ intXh and p(eh − xh

)= 0. By continuity, ∃y ∈ Xh\Xh

such that p(eh − y

)> 0 and uh (y) > uh

(xh). As uh is locally non-satiated, then there

exists x∗ ∈ intXh such that uh (x∗) > uh(xh). As uh is quasi-concave, then ∀λ ∈ [0, 1] ,

uh (λy + (1− λ)x∗) > uh(xh). Further, provided that λ is small enough, λy + (1− λ)x∗ ∈

Xh. As y satisfies p(eh − y

)> 0, then p

(eh − (λy + (1− λ)x∗)

)> 0. This contradicts

that xh is an equilibrium consumption choice under (HP ), as the allocation λy + (1− λ)x∗

satisfies the constraints and provides strictly higher utility.



Before proving the First Basic Welfare Theorem, I prove the two statements about how to

find Pareto optimal allocations as optimal solutions to programming problems.

Optimal solution to (POh′)

Suppose that(xh∗)h∈H is not an optimal solution to the programming problem (POh′) . Then

∃(yh)h∈H such that u

h′(yh′)> uh

′ (xh′∗) , uh (yh) ≥ uh

(xh∗)∀h 6= h′, and

(yh)h∈H ∈ FA.

As(yh)h∈H is feasible, then

(xh∗)h∈H is not Pareto optimal.

Suppose that(xh∗)h∈H is not Pareto optimal. Then ∃

(yh)h∈H ∈ FA such that u

h(yh)≥

uh(xh∗)∀h ∈ H and uk

(yk)> uk

(xk∗)for some k. There are two cases to consider: (i) k = h′

and (ii) k 6= h′. In case (i) k = h′, the allocation(xh∗)h∈H would not be an optimal solution

to (POh′) by definition. In case (ii) k 6= h′, as uh′is strictly increasing and uk is continuous,

then for some ε > 0, define a new allocation yh = yh ∀h /∈ h′, k , yh′ = yh′+ (ε, ..., ε) , and

yk = yk − (ε, ..., ε) . Then(yh)h∈H is such that u

h′(yh′)> uh

′ (yh′)

= uh′ (xh′∗) , uk (yk) >

uk(xk∗), and uh

(yh)≥ uh

(xh∗)∀h /∈ h′, k . Thus,

(xh∗)h∈H is not an optimal solution

to the programming problem (POh′) .

Optimal solution to (PO)

Suppose that(xh∗)h∈H is not Pareto optimal. Then ∃

(yh)h∈H ∈ FA such that uh

(yh)≥

uh(xh∗)∀h ∈ H and uh

′ (yh′)> uh

′ (xh′∗) for some k. Therefore, for any (µh)

h∈H >> 0,∑h∈H µ

h · uh(yh)>∑

h∈H µh · uh

(xh∗). Thus,

(xh∗)h∈H is not an optimal solution to the

programming problem (PO) .

Suppose that(xh∗)h∈H is Pareto optimal. Define the set U = u ∈ RH : uh ≤ uh

(xh)

∀h ∈ H,(xh)h∈H ∈ FA. The point u

∗ =(u1 (x1∗) , ...., uH

(xH∗

))∈ U, but u∗ /∈ intU. The

set U is nonempty. Select any points v, w ∈ U, where vh ≤ uh(xh)∀h ∈ H and wh ≤ uh

(yh)

∀h ∈ H for(xh)h∈H ,

(yh)h∈H ∈ FA. Then as u

h is concave ∀h ∈ H, any convex combinationθv + (1− θ)w ∈ U as θ

(xh)h∈H + (1− θ)

(yh)h∈H ∈ FA and

uh(θxh + (1− θ) yh

)≥ θuh

(xh)

+ (1− θ)uh(yh)≥ θvh + (1− θ)wh.

Thus, the set U is convex. From the Supporting Hyperplane Theorem (Corollary 1.1. in

Section 1.1), there exists a q ∈ RH\0 such that qTu∗ ≥ qTu ∀u ∈ U.We need to show thatthere exists a q > 0 such that qTu∗ ≥ qTu ∀u ∈ U. Suppose not, that is, qh < 0 for some

2.6. PROOFS 41

h. Then, we can specify a new vector uh′

= uh′ (xh′∗) ∀h′ 6= h and uh < uh

(xh∗). This new

vector is an element of U, yet qTu∗ < qTu. Thus, q > 0 and we define(µh)h∈H = q

‖q‖ ∈ ∆H−1.

So, there exists(µh)h∈H ∈ ∆H−1 such that

(xh∗)h∈H is an optimal solution to (PO) .

Proof of First Basic Welfare Theorem

Suppose that((xh)h∈H , p

)is an Arrow-Debreu equilibrium, but

(xh)h∈H is not a Pareto

optimal allocation. Then there exists a feasible allocation(yh)h∈H such that uh

(yh)≥

uh(xh)∀h ∈ H, with uh′

(yh′)> uh

′ (xh′)for some h′.

Then p(eh − yh

)≤ p

(eh − xh

)= 0 ∀h ∈ H. Why? If p

(eh − yh

)> p

(eh − xh

)= 0,

then as uh is locally non-satiated, there exists yh such that p(eh − yh

)> p

(eh − xh

)= 0

and uh(yh)> uh

(xh). This contradicts that xh is an optimal solution to (HP ) .

Likewise, if uh(yh)> uh

(xh)for some h, then p

(eh − yh

)< p

(eh − xh

)= 0. Otherwise,

p(eh − yh

)≥ p

(eh − xh

)= 0 and xh is not an optimal solution to (HP ) .

Summing the budget constraints over all households:

p∑

h∈H

(eh − yh

)< p

∑h∈H

(eh − xh

)= 0.

As p > 0, this implies that∑

h∈H

(ehg − yhg

)< 0 for some good g. This contradicts that(

yh)h∈H is a feasible allocation.


The proof proceeds in two steps: first 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

Define the set

Z =

z ∈ RG : z ≤

∑h∈H

(eh − xh

)where

(xh)h∈H is such that

xh ∈ Xh and

uh(xh)≥ uh

(xh∗) ∀h ∈ H .

The set Z is convex. Why? Consider any a, b ∈ Z. Then a ≤∑

h∈H(eh − xh

)and uh

(xh)≥

uh(xh∗)∀h ∈ H where xh ∈ Xh ∀h ∈ H, and b ≤

∑h∈H

(eh − yh

)and uh

(yh)≥

uh(xh∗)∀h ∈ H where yh ∈ Xh ∀h ∈ H. As uh is quasi-concave, then ∀θ ∈ [0, 1] ,


uh(θxh + (1− θ)yh

)≥ uh

(xh∗)and θa + (1− θ) b ≤

∑h∈H

(eh −

[θxh + (1− θ)yh

]). Fur-

ther, θxh + (1− θ)yh ∈ Xh ∀h ∈ H. Thus, θa+ (1− θ) b ∈ Z.Considering the allocation

(xh)h∈H =

(xh∗)h∈H , we know that 0 ∈ Z, but 0 /∈ intZ. To

prove the latter, suppose that 0 ∈ intZ. Then there exists z >> 0 such that z ∈ Z. The

allocation(xh)h∈H such that x

1 = x1∗+z and xh = xh∗ ∀h ≥ 2 is the allocation corresponding

to z. This implies uh(xh)≥ uh

(xh∗)∀h ≥ 2 and u1 (x1) > u1 (x1∗) (Assumption S3). This

contradicts that(xh∗)h∈H is a Pareto optimal allocation.

Applying the Supporting Hyperplane Theorem, there exists p∗ ∈ RG\0 such that0 ≥ p∗z ∀z ∈ Z. Further, p∗ > 0. Why? Suppose not, that is pg < 0 for some g. Consider

the element z such that zg = −1 and zg′ = 0 ∀g′ 6= g. Obviously, z ∈ Z, yet p∗z > 0. This

contradiction verifies that p∗ > 0.

Part 2: Duality Theorem

The conclusion from Part 1 is that for any allocation such that xh ∈ Xh and uh(xh)≥

uh(xh∗)∀h ∈ H, 0 ≥ p∗

∑h∈H

(xh − eh

). In particular, for any household h, we can define

an allocation(yh)h∈H =

(xh,(xh∗)h6=h′

). If xh ∈ X h and uh

(xh)≥ uh

(xh∗), it must be

0 ≥ p∗(xh − eh

). As this holds for all households h ∈ H, then xh∗ is an optimal solution to

the following Expenditure Minimization problem ∀h ∈ H :

(ExpMin) minimize p∗xh


uh(xh)− uh

(xh∗)≥ 0

.

I wish to show that this implies xh∗ is an optimal solution to the Utility Maximization

problem (HP ) ∀h ∈ H. Suppose not. Then for some household h, there exists yh such thatp∗(eh − yh

)≥ 0, yh ∈ Xh, and uh

(yh)> uh

(xh∗). Consider the consumption βxh∗ for some

β < 1. As xh∗ ∈ intXh, then for β close to 1, βxh∗ ∈ Xh. Further, p∗(eh − βxh∗

)> 0. Then,

defining the convex combination θyh + (1− θ)(βxh∗

), p∗

(eh −

[θyh + (1− θ)

(βxh∗

)])> 0

∀θ ∈ (0, 1). By definition, p∗eh = p∗xh∗, implying

p∗(θyh + (1− θ)

(βxh∗

))< p∗xh∗.

By the continuity of uh, for some θ close to 1, uh(θyh + (1− θ)

(βxh∗

))> uh

(xh∗). Thus,

xh∗ is not an optimal solution to the problem (ExpMin) . This completes the contrapositive

2.7. EXERCISES 43

argument.

2.7 Exercises

1. Show that if uh is locally non-satiated ∀h ∈ H, then the Arrow-Debreu equilibriumprices satisfy p > 0.

2. Assume that uh is differentiable, differentiably strictly increasing, and concave ∀h ∈H. Using the results from the programming problems in this chapter, show that if((xh)h∈H , p

)is an Arrow-Debreu equilibrium, then the allocation

(xh)h∈H is an opti-

mal solution to the problem (POh′) for any h′. From Section 2.3, the allocation(xh)h∈H

is then Pareto optimal. As this was the first 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

differentiable, 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

u1(x1)

=1

2log(x1

1 + x21

)+

1

2log(x1

2

)u2(x2)

=1

2log(x2

1

)+

1

2log(x2

2

)and endowments e1 = (1, 2) and e2 = (2, 1) . Notice that household 1 cares about

household 2’s consumption of the first 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)

= α1 log(xh1)

+ α2 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?


6. Show that the projection π is proper. π is the projection π : RH(G+1)+G−1++ ×RHG++ → RHG++

which maps((xh)h∈H , p,

(eh)h∈H

)7→(eh)h∈H such that Φ

((xh)h∈H , p;

(eh)h∈H

)= 0.

7. Prove that for any endowments(eh)h∈H 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,λh)h∈H

,pΦ((xh, λh

)h∈H , p

)has full rank

(square matrix) given that the initial endowment is a Pareto optimal allocation(eh)h∈H =(

xh∗)h∈H . Notice that the derivative matrix only contains derivatives with respect to

variables, not the endowment e1.

8. Using "A Different Application of Differential 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)h∈H is not Pareto optimal. To attack this problem, use "A Different Application of

Differential Topology" from Exercise 3 in Chapter 1 and consider that a necessary con-

dition for Pareto optimality isDguh(xh)Dg′u

h(xh)=

Dguh′(xh′)

Dg′uh′(xh′)

∀ (h, h′, g, g′) ∈ H×H×G×G.

Bibliography

[1] Arrow, Kenneth and Gerard Debreu (1954): "Existence of an Equilibrium for a Compet-

itive 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 con-

strainted 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):

Differential 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 finite number of states of uncertainty

in the final period. Models with a longer, but still finite, time horizon can be reduced to

the two-period model. Available to the households are financial markets that allow for the

transfer of wealth between the states of uncertainty. This model in which households trade

not only commodities, but also financial assets, is labeled the general financial 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 financial equilibrium in this manuscript.

In this chapter, I first introduce the financial model with numeraire assets. Then I discuss

the equilibrium properties under two conditions: complete markets and incomplete markets.

I next consider how the results differ 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 final period, one of S

possible states of uncertainty is realized, where S < ∞ is finite. Thus, the total number of

states, including the initial period, is S+ 1. Denote the set of states as S = 0, 1, ..., S withtypical element s. The uncertainty tree is depicted in Figure 3.1.

There are a finite number of households. Denote the set of households as H= 1, ..., H

47

48 CHAPTER 3. GENERAL FINANCIAL MODEL

with typical element h. In each state, households trade and consumer L commodities. Define

the commodity space as L = 1, ..., L . The number of commodities is finite. By definition,one commodity in a state is a different good than the same commodity in a different 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)

)l∈L is the vector of consumption by

household h of all commodities in state s, and xh =(xh(s)

)s∈S is the vector of consumption

by household h of all goods.

As with the static model, the household primitives are given below.

• Xh = RG+ is the consumption set.

• uh : Xh → R is the utility function. We typically assume that the function is continu-ous, locally non-satiated, and quasi-concave.

• eh ∈ Xh 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)∈L×S ∈RG\0. The convention is that the price vector p (s) is a row vector ∀s ∈ S. I define the

(S + 1)×G price matrix P =

p(0) 0 0 0

0 p(1) 0 0

0 0 ... 0

0 0 0 p(S)

.Now, we consider the financial elements of the model. There are J financial assets that

a household can choose to buy or sell in the initial period.1 The number of assets is finite

(J <∞). Denote the set of assets as J = 1, ..., J , with typical element j. The decision tobuy or sell an asset in the initial period is influenced 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)j∈J . These

prices are variables in the equilibrium and will be determined to satisfy market clearing

conditions.

In the state of uncertainty s in the final period, the asset j has a fixed payout rj(s). The

payouts of all assets are in terms of the same physical commodity, defined as the numeraire

commodity, in all states. Without loss of generality, we label the numeraire commodity as

commodity l = L. We typically assume that for some household h′, the utility uh′is strictly

1Both buying and selling the same asset is redundant in this model without transaction costs and withoutdefault.

3.1. THE MODEL 49

increasing in the commodity l = L in all states s ∈ S. This way, the commodity price of thenumeraire commodity is pL (s) > 0 ∀s ∈ S. To remove the nominal indeterminacy, we makethe price normalization pL (s) = 1 ∀s ∈ S.

Further, the assets are assumed to have nonnegative payouts rj(s) ≥ 0 ∀(j, s) ∈J× S\0, where rj = (rj(s))s>0 > 0 ∀j ∈ J . Let’s collect the payouts of all assets in

all states s > 0 into one payout matrix R. This payout matrix has S rows and J columns:

R =[r1 ... rJ

]=

r1(1) ... rJ(1)

: ... :

r1(S) ... rJ(S)

.The lone financial primitive is the matrix of asset payouts.

• R ∈ RS,J+ is the payout matrix. We typically assume that R is nonnegative and has

full column rank (though the final 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 port-

folio). As with the static general equilibrium model, the trading of assets is perfectly com-

petitive 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 zh ∈ RJ , where the position for a particular asset j isdenoted zhj ∈ R. If zhj < 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) zhj ≤ 0 (the household

has borrowed). If zhj > 0, then the household has purchased the asset. In this case, the

asset has positive payout in the states s > 0 : rj (s) zhj ≥ 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 define a general financial equilibrium, which is the equilibrium

concept in this financial model.

Definition 3.1 A general financial equilibrium is((xh, zh

)h∈H , p, q

)such that


1. ∀h ∈ H, given (p, q) ,(xh, zh

)is an optimal solution to the household problem (HP )



zh ∈ RJ

P(eh − xh

)+

(−qR

)zh ≥ 0

.


xhl (s) =∑

h∈Hehl (s) ∀ (l, s) ∈ L× S.∑

h∈Hzhj = 0 ∀j ∈ J.

It is important to verify, under roughly the same conditions as the static model, that a

financial equilibrium exists. The assumptions suffi cient for existence are given by:


Assumption E2: uh : Xh → R is continuous, non-decreasing, strictly increas-ing in the commodity l = L in all states s ∈ S, and quasi-concave ∀h ∈ 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 financial equilibrium((xh, zh

)h∈H , p, q

)exists.

Proof. The arguments from the static model apply across the board. All that remains is toshow 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 Rzh are bounded. Given Assumption

E4, the portfolio zh 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 affect the real equilibrium variables.

3.1. THE MODEL 51

3.1.2 Regularity

The assumptions required to prove Regularity are:

Assumption R1: Xh = RG++ ∀h ∈ H.

Assumption R2: uh : Xh → R is C2, differentiably strictly increasing (mean-

ing thatDuh(xh)>> 0), differentiably strictly concave (meaning thatD2uh

(xh)

is a negative definite matrix), and satisfies the boundary condition (meaning that

∀x ∈ RG++, cly : uh(y) ≥ uh(x)

⊆ RG++) ∀h ∈ H.

Assumption R3: eh >> 0 ∀h ∈ H.

Assumption R4 (No Redundancy): The payout matrix R has full column rank


Theorem 3.2 Under Assumptions R1-R4, over a generic subset of endowments, all finan-cial equilibria satisfy Finite Local Uniqueness.

The Lagrange multipliers for the budget constraints are the elements of the (S + 1)−dimensionalrow vector λh. Applying the Kuhn-Tucker Theorem, the following conditions are necessary

and suffi cient conditions for an optimal solution to (HP ) :

Duh(xh)− λhP = 0,

P(eh − xh

)+

(−qR

)zh = 0,

λh

(−qR

)= 0,

where the third set of equations contains the first order conditions with respect to the

portfolio zh. From this third set (knowing that λh >> 0 from the first order conditions with

respect to consumption and rj > 0 ∀j ∈ J by Assumption R4), q >> 0.



(RG+S+1

++ × RJ)× RG−(S+1)

++ × RJ++ → Rn as:

Φ((xh, λh, zh

)h∈H , p, q

)=

[Duh

(xh)− λhP

]TP(eh − xh

)+

(−qR

)zh[

λh

(−qR

)]T

h∈H∑

h∈H

(eh\G − xh\G

)∑h∈H

zh

,

where n = H (G+ S + 1 + J) + G − (S + 1) + J. Allow me to verify that the equilibrium

variables belong to open sets: xh ∈ Xh = RG++, λh ∈ RS+1

++ , and zh ∈ RJ ∀h ∈ H. Further,p ∈ RG−(S+1)

++ following the price normalization pL (s) = 1 ∀s ∈ S and q ∈ RJ++. By definition,

Φ((xh, λh, zh

)h∈H , p, q

)= 0 iff

((xh, λh, zh

)h∈H , p, q

)is a general financial equilibrium.

Recalling the proof method discussed in Section 1.4.5 (in Chapter 1), Finite Local Unique-

ness is proven if we can show that (i) π is proper and

(ii) rankDΦ((xh, λh, zh

)h∈H , p, q

)= n. (3.1)

The first 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.

The derivative DΦ((xh, λh, zh

)h∈H , p, q

)means that we are taking derivatives with re-

spect to(xh)h∈H ,

(λh)h∈H ,

(zh)h∈H , p, q, and

(eh)h∈H (both variables and parameters). If

we show that the rank condition holds by only taking derivatives with respect to(xh)h∈H ,(

λh)h∈H ,

(zh)h∈H , and e

1, then we have finished the argument.

The derivative matrix

M = D(xh,λh,zh)h∈H

,e1Φ((xh, λh, zh

)h∈H , 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 Φ, while the columns will correspond to

the variables (or parameters) that we are taking derivatives with respect to. For simplicity,

3.1. THE MODEL 53

define Ψ =

(−qR

)and Λ =

(IL−1 0

)0 0

0 ... 0

0 0(IL−1 0

) ∈ RG−(S+1),G. The derivative

matrix is given by:

M =

D2u1 (x1) −P T 0 0

−P 0 Ψ 0 0 P

0 ΨT 0 0

.. .. .. 0

0 .. .. .. 0 0

.. .. .. 0

D2uH(xH)−P T 0 0

0 0 −P 0 Ψ 0

0 ΨT 0 0

−Λ 0 0 .. .. .. −Λ 0 0 Λ

0 0 IJ .. .. .. 0 0 IJ 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 νTM = 0 and must verify νT = 0, where

ν ∈ Rn corresponds to equations in Φ :

ν =

∆x1

∆λ1

∆z1

:

∆xh

∆λh

∆zh

:

∆p

∆q

FOCx1

BC1

FOCz1

:

FOCxh

BCh

FOCzh

:

MCx

MCz

.



(∆x1)TD2u1 (x1)−

(∆λ1

)TP −∆pTΛ = 0 (3.2.a)

− (∆x1)TP T + (∆z1)

TΨT = 0 (3.2.b)(

∆λ1)T

Ψ + ∆qT = 0 (3.2.c)

:(∆xh

)TD2uh

(xh)−(∆λh

)TP −∆pTΛ = 0 (3.2.d)

−(∆xh

)TP T +

(∆zh

)TΨT = 0 (3.2.e)(

∆λh)T

Ψ + ∆qT = 0 (3.2.f)

:

∆λ1P −∆pTΛ = 0 (3.2.g)

(3.2)

Let’s show that νT = 0 in four steps:

1. From (3.2.g), by the definition of Λ, ∆λ1 = 0. Consequently, ∆pT = 0.

2. Postmultiply (3.2.a) by ∆x1 :

(∆x1

)TD2u1

(x1)

∆x1 = 0.

As u1 is differentiably strictly concave, the Hessian matrix D2u1 (x1) is negative defi-

nite. Thus, (∆x1)T

= 0.

3. From (3.2.b), as Ψ =

(−qR

)and R has full column rank (Assumption R4), then ΨT

has full row ranking leading to the implication: (∆z1)T

ΨT = 0 =⇒ (∆z1)T

= 0. From

(3.2.c), ∆qT = 0.

4. For any household h > 1, postmultiply (3.2.d) by ∆xh :

(∆xh

)TD2uh

(xh)

∆xh −(∆λh

)TP∆xh = 0.

From (3.2.e), after postmultiplying by ∆λh and then taking the transpose:

(∆λh

)TP∆xh =

(∆λh

)TΨ∆zh.

From (3.2.f),(∆λh

)TΨ = 0. Thus,

(∆xh

)TD2uh

(xh)

∆xh = 0. As uh is differentiably

strictly concave, the Hessian matrix D2uh(xh)is negative definite. Thus,

(∆xh

)T= 0.

3.1. THE MODEL 55

From (3.2.d), ∆λh = 0. From (3.2.e), as R has full column rank (Assumption R4), then(∆zh

)TΨT = 0 implies

(∆zh

)T= 0.

Thus, νT =((

(∆x1)T,(∆λ1

)T, (∆z1)

T), ...,

((∆xh

)T,(∆λh

)T,(∆zh

)T), ...,∆pT ,∆qT

)=

0, finishing the argument.

3.1.3 No Arbitrage

Given R and q, an arbitrage opportunity exists if ∃z ∈ RJ such that(−qR

)z > 0. This

is an arbitrage opportunity, because the household could hold the portfolio κz as κ → ∞,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: There does not exist z ∈ RJ such that(−qR

)z > 0.

This condition is quite intuitive, but diffi 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 α ∈ RS+1++ :

No Arbitrage II: There exists α ∈ RS+1++ so that α

(−qR

)= 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

Suppose that No Arbitrage I does not hold. Then ∃z ∈ RJ so that(−qR

)z > 0.

This implies that for any α ∈ RS+1++ , α

(−qR

)z > 0. This is a contradiction, as No

Arbitrage II requires that for some α ∈ RS+1++ , α

(−qR

)z = 0.


2. No Arbitrage I implies No Arbitrage II

Let us recall something that we verified in Chapter 1 (specifically, Exercise 1):

Let

(−qR

)be a (S + 1)×J matrix and z ∈ RJ . The Farkas Lemma can be used

to prove the following implication.

There does not exist z such that

(−qR

)z > 0

⇓

∃α ∈ RS+1++ s.t. α

(−qR

)= 0.

This tells us exactly what we need to finish 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 ≥ 2 states of uncertainty and J = 2 assets. The payout matrix

R is given by:

R =

r1 (1) r2 (1)

: :

r1 (S) r2 (S)

.From No Arbitrage II, the asset prices are given by q = αR for some α ∈ RS++. To see this,

take the No Arbitrage II definition, which says that α

(−qR

)= 0 for some α >> 0. Then

defining α =(α(1)α(0)

, ..., α(S)α(0)

), we obtain −q + αR = 0. The Hens method proceeds in two

steps.

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

have plotted the pairs corresponding to the payout matrix R =

0 1

4 2

3 3

(the x-axiscorresponds to j = 1 and the y-axis to j = 2).

2. The convex cone of this set (r1 (1) , r2 (1)) , ..., (r1 (S) , r2 (S)) can then be found. InFigure 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 thecone.

The No Arbitrage condition is a necessary condition for a general financial 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

column space of

(−qR

)so that the equivalent matrix R∗ has one asset (wlog asset

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∗

R∗

), where

R∗ =

1 r∗2 (1) r∗3 (1)

1 : :

1 r∗2 (S) r∗3 (S)

.

As an example, suppose that R =

2 0 1

3 4 2

4 3 3

. Then the linear operation that I will performwill be C1 7→ C1 − C3, meaning that I replace the 1st column with the difference (1st


column) - (3rd column). This results in the equivalent asset price vector and payout matrix:

(−q∗

R∗

)=

− (q1 − q3) −q2 −q3

1 0 1

1 4 2

1 3 3

From No Arbitrage, the asset prices are given by q∗ = αR∗ for some α ∈ RS++. This

means that q∗1 = α · −→1 or q∗1 =∑

s>0 α (s) . The Hens method continues with the following

steps.

2. Plot the pairs (r∗2 (s) , r∗3 (s)) in R2+ for all s > 0. As you can see in Figure 3.3, I have

plotted the pairs corresponding to the payout matrix R∗ =

1 0 1

1 4 2

1 3 3

.3. The convex hull of this set (r∗2 (1) , r∗3 (1)) , ..., (r∗2 (S) , r∗3 (S)) can then be found. InFigure 3.3, the convex hull is the shaded area inside the triangle.

4. If the asset prices(q∗2q∗1,q∗3q∗1

)belong to the interior of this convex hull, then they are

No Arbitrage prices. Otherwise, they are not No Arbitrage prices. Continuing with

my example, as q∗1 = q1 − q3, the asset price pairs that I will actually be plotting are(q2

q1−q3 ,q3

q1−q3

). Obviously, any prices such that q1 − q3 ≤ 0 automatically fail the No

Arbitrage requirement.

Notice two elements of Step 4. First, we require that the asset prices are all divided by

q∗1. 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 59

3.2 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:

R =

1 0 0

0 1 0

0 0 1

R =

1 2 2

1 1 2

1 1 1

I define the column span of any payout matrix as

〈R〉 =y ∈ RS : y = Rz for some z ∈ RJ

. (3.3)

For instance, under the assumption of complete markets, 〈R〉 = RS.Given complete markets and the equilibrium asset prices q ∈ RJ++, there exists a unique

vector of state prices. Recall the No Arbitrage pricing conditions:

q = αR,

where α =(α(1)α(0)

, ..., α(S)α(0)

). Given that R has full rank, then the state prices are uniquely

defined as α = qR−1.

Suppose that the market clearing conditions hold for all commodities:∑h∈H

(eh − xh

)= 0.

Walras’Law states that P∑

h∈H

(eh − xh

)+

(−qR

)∑h∈H

zh = 0 (sum of budget con-

straints equals zero). Given commodity market clearing and full column rank R, then∑h∈H

zh = 0. Thus, the market clearing conditions for assets are guaranteed to hold when

the market clearing conditions for commodities hold.

The following theorem (called Arrow’s equivalency theorem) proves that a general finan-

cial equilibrium with complete markets is allocation-equivalent to an Arrow-Debreu equilib-

rium.


Theorem 3.3 Arrow’s Equivalency TheoremUnder Assumptions E1-E4, if J = S and

((xh, zh

)h∈H , p, q

)is a general financial equi-

librium, then((xh)h∈H , ρ

)is an Arrow-Debreu equilibrium where ρ = (1, qR−1) · P.

Under Assumptions E1-E3, if((xh)h∈H , ρ

)is an Arrow-Debreu equilibrium, then for

any full rank payout matrix R, ∃(zh)h∈H such that for α (s) = ρL(s)

ρL(0)∀s > 0 and q = αR, the

vector((xh, zh

)h∈H , p, q


Proof. General financial =⇒ Arrow-DebreuAs the markets are complete, ∃! state prices α = qR−1 as discussed above. Let 1 ∈ RL

be the L−dimensional row vector with 1 for each element. This will be the state price

for s = 0. Then the state prices for all s ∈ S are (1, α) . Recall that the price matrix

P =

p(0) 0 0 0

0 p(1) 0 0

0 0 ... 0

0 0 0 p(S)

is a (S + 1) × G matrix. Premultiply P by the state prices to

obtain the Arrow-Debreu prices ρ ∈ RG+ :

ρ = (1, α) · P.

In particular, ρl (0) = pl(0) ∀l ∈ L and ρl (s) = α (s) · pl(s) ∀(l, s) ∈ L× S\0.The proof requires two steps. First, let

(xh, zh

)be budget feasible under the general

financial equilibrium budget constraints:

(xh, zh

)∈ Bh (p, q) =

(x, z) : P

(eh − x

)+

(−qR

)z ≥ 0

.

Premultiply the budget constraints by the state prices (1, α) . Then the budget constraints

reduce to

(1, α)P(eh − x

)+ (1, α)

(−qR

)z ≥ 0.

From No Arbitrage II, (1, α)

(−qR

)= 0. Thus, we have ρ

(eh − x

)≥ 0. So xh ∈ bh (ρ) =

x : ρ(eh − x

)≥ 0.

For the second step, suppose that yh is budget feasible under the Arrow-Debreu equilib-

rium budget set bh (ρ) . Define α (s) = ρL(s)ρL(0)

∀s > 0 and pl(s) = ρl(s)ρL(s)

∀(l, s) ∈ L× S. For any

3.3. INCOMPLETE MARKETS 61

full rank R, define q = αR. There exists

zh = R−1

p(1)(eh(1)− xh(1)

):

p(S)(eh(S)− xh(S)

) .

By construction,(xh, zh

)∈ Bh (p, q) for the selected full rank payout matrix R.

To conclude, we pose the question: if(xh, zh

)∈ arg max

(x,z)∈Bh(p,q)

uh (x) , then since xh ∈ bh (ρ) ,

is it also true that xh ∈ arg maxx∈bh(ρ)

uh (x)? Suppose not, that is, ∃y ∈ bh (ρ) such that

uh (y) > uh(xh). If y ∈ bh (ρ) , then ∃z ∈ RJ such that (y, z) ∈ Bh (p, q) . As uh (y) > uh

(xh)

and (y, z) ∈ Bh (p, q) , then this contradicts that(xh, zh

)∈ arg max

(x,z)∈Bh(p,q)

uh (x) .

Arrow-Debreu =⇒ General financial

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 (ρ) , then

using arguments from above, ∃zh such that for (i) α (s) = ρL(s)ρL(0)

∀s > 0, (ii) pl(s) = ρl(s)ρL(s)

∀(l, s) ∈ L× S, and (iii) q = αR,(xh, zh

)∈ Bh (p, q) .

If xh ∈ arg maxx∈bh(ρ)

uh (x) , then since ∃zh such that(xh, zh

)∈ Bh (p, q) , is it also true that(

xh, zh)∈ arg max

(x,z)∈Bh(p,q)

uh (x)? Suppose not, that is, ∃ (y, z) ∈ Bh (p, q) such that uh (y) >

uh(xh). If (y, z) ∈ Bh (p, q) , then y ∈ bh (ρ) (citing arguments from above). But this

contradicts that xh ∈ arg maxx∈bh(ρ)

uh (x) .

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 financial 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 R′, R′′ with the same column span: 〈R′〉 = 〈R′′〉 (recall


the definition of the span in (3.3)). These two matrices are then equivalent up to a change of

basis. For our purposes in the general financial model, if a household h has portfolio payouts

R′zh, then there exists zh ∈ RJ such that R′zh = R′′zh. Thus, given any payout matrix with

full column rank R′ (No Redundancy condition holds), we consider the equivalent payout

matrix R =

[R1

R2

]such that R1 ∈ RJ,J has full rank.

With incomplete markets, given the equilibrium asset prices q ∈ RJ++, there no longer

exists a unique vector of state prices. In fact, the set of state prices α : q = αR is an

(S − J)−dimensional space.

3.3.1 Pareto suboptimal

Consider the characterization of Pareto optimal allocations using either the problem (POh′)

or the problem (PO) from Chapter 2. A necessary condition for a Pareto optimal allocation

is:

∀h ∈ H, λ1 = κhλh for some scalar κh > 0.

Recall that λh is the vector of Lagrange multipliers corresponding to the budget constraints

for household h.

Let’s add the constraint λ1(1)

λ1(0)= λ2(1)

λ2(0)to the system of equations Φ. Now the number of

equations is one greater than the number of unknowns. Let’s apply "A Different Application

of Differential Topology" from Exercise 3 in Chapter 1 (similar to Exercises 8 and 9 in

Chapter 2) to show that over a generic subset of endowments, the equation λ1(1)

λ1(0)= λ2(1)

λ2(0)

cannot hold. If the equation cannot hold, as it is a necessary condition for a Pareto optimal

allocation, then the general financial 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 endow-

ments, the general financial 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.


Let’s introduce some machinery that will be useful in the example. Define the S×G price

matrix (for states s > 0) as P\0 =

0 p(1) 0 0

0 0 ... 0

0 0 0 p(S)

(this is the matrix P with the first

row removed). Define α ∈ RS++ an any vector of state prices satisfying q = αR. The budget

constraints in a general financial equilibrium are given by P(eh − xh

)+

(−qR

)zh ≥ 0.

This is equivalent to the following two conditions:

1. Single budget constraint

(1, α)P(eh − xh

)≥ 0.

2. Span condition

P\0(xh − eh

)∈ 〈R〉 .

The single budget constraint is satisfied by both Pareto optimal and Pareto suboptimal

allocations. The key condition to determine Pareto optimality is the span condition.

Example 3.1 Let L = 1. All households have identical utility uh(xh)

= −∑

s∈S

(xh (s)

)−γfor some γ > 0. All households have endowments such that

(eh (s)

)s>0∈ 〈R〉 .

Since the utility is homothetic and identical across households, then there is a unique

general financial equilibrium. Let’s first find the Arrow-Debreu equilibrium and then show

that this is also the general financial 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:

xh (s) = θh∑

h∈Heh (s) ∀h ∈ H

for some household-specific fractions θh.

Given that L = 1, the span condition is equivalent to(xh (s)− eh (s)

)s>0∈ 〈R〉 . Since(

eh (s))s>0∈ 〈R〉 ∀h ∈ H and the space 〈R〉 is linear, the span condition reduces to(

xh (s))s>0∈ 〈R〉 ∀h ∈ H.

Do the household choices (in the Arrow-Debreu equilibrium) satisfy the span condition?

Well, since(eh (s)

)s>0∈ 〈R〉 ∀h ∈ H (by assumption), then

(∑h∈H e

h (s))s>0∈ 〈R〉 . As

xh (s) = θh∑

h∈Heh (s) and the space 〈R〉 is linear, then in fact

(xh (s)

)s>0∈ 〈R〉 ∀h ∈ H.

Thus, the Arrow-Debreu equilibrium allocation satisfies the span condition and therefore


satisfies the general financial equilibrium budget constraints. Consequently, the Arrow-Debreu

equilibrium is also a general financial equilibrium.

So we have a setting with incomplete markets (the example is valid for any J < S) and

yet, the general financial equilibrium allocation is Pareto optimal. Why? The endowments(eh (s)

)s>0∈ 〈R〉 ∀h ∈ H. Whenever J < S, the space 〈R〉 is a measure zero subset of RS.

Thus, the endowments in this example do not belong to a generic subset. We can see nowwhy 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 financial equilibrium alloca-

tions 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 financial equilibrium. In a general financial equilibrium, the possible

allocations are restricted by the fixed 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 fixed

asset structure. We now have a fair comparison: general financial equilibrium allocations vs.

constrained Pareto optimal allocations.

Definition 3.2 The allocation(xh)h∈H is constrained Pareto suboptimal if there exists equi-

librium prices p and a feasible allocation(xh)h∈H ∈ FA such that ∀h ∈ H,

xh ∈ arg max uh (x)x:P\0(xh−eh)∈〈R〉

and(uh(xh))h∈H >

(uh(xh))h∈H .

2

We will see shortly that this definition is equivalent to a planner fixing the asset choices

of households and making a transfer to each household in the initial state. The planner is

choosing(τh)h∈H for τh ∈ RJ+1 such that

∑h∈H

τh = 0, where τh(1) is the transfer (in

units of account) in state s = 0 and τh(j + 1) for j ∈ J is the holding of asset j. Given these

2Equilibrium prices p simply means that∑

h∈H

(xh − eh

)= 0, which is a requirement of the set FA.


fixed transfers by the planner, the new budget constraints for each household become:

P(eh − xh

)+

(1 0

0 R

)τh ≥ 0.

The main result that we show in this section is that the general financial equilibrium

allocations (under some conditions) are not constrained Pareto optimal. To motivate this,

we first consider two examples. The examples dictate what conditions are required to obtain

this main result.

In the first 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 financial equilib-

rium allocation(xh)h∈H is constrained Pareto suboptimal. Then there exists

(xh)h∈H such

that:

1. xh ∈ arg max uh (x)x:(xh(s)−eh(s))

s>0∈〈R〉

∀h ∈ H,

2.∑

h∈H

(xh − eh

)= 0, and

3.(uh(xh))h∈H >

(uh(xh))h∈H .

But then(xh)h∈H cannot be optimal choices for all households (as xh is chosen under

the exact same budget constraints and some households strictly prefer xh). Thus,(xh)h∈H

cannot be a general financial equilibrium allocation. This finishes the argument that all

general financial equilibrium allocations (when L = 1) are constrained Pareto optimal.

Thus, we know that the main result (general financial equilibrium allocations are con-

strained 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 [specifically, Cobb-Douglas] utility).

Example 3.3 Suppose that the utility functions are uh(xh)

=∑

s∈S

(∑l∈L θl (s) · log

(xhl (s)

))∀h ∈ H. The Cobb-Douglas utility has some nice properties. For each household, the con-sumption is given by:

xhl (s) = αh (s) ·∑

h∈Hehl (s)


where αh (s) ∈ (0, 1) is determined according to a household’s endowments and(αh (s)

)h∈H ∈

∆H−1 ∀s ∈ S. The prices are given such that pl (s) = θl(s)θL(s)·∑

h∈H ehL(s)∑h∈H ehl (s)

∀(l, s) ∈ L× S. Youare asked to verify these previous two facts in Exercise 7.

In each state s ∈ S the household choices can be reduced to simply the fraction αh (s) ∈(0, 1) for each h ∈ H (rather than the commodity vector xh (s) ∈ RL+). Suppose that thegeneral financial equilibrium allocation

(xh)h∈H is constrained Pareto suboptimal. Then there

exists(xh)h∈H such that:

1. xh ∈ arg max uh (x)x: P\0(xh(s)−eh(s))

s>0∈〈R〉

∀h ∈ H,

2.∑

h∈H

(xh − eh

)= 0, and

3.(uh(xh))h∈H >

(uh(xh))h∈H .

Thus, ∃(αh (s)

)(h,s)∈H×S such that

∑h∈H

αh (s) = 1 ∀s ∈ S and

(xhl (s) = αh (s) ·

∑h∈H

ehl (s))

(h,l,s)∈H×L×S

are the maximizers of the household problem (HP ) and(uh(xh))h∈H >

(uh(xh))h∈H . The

commodity prices don’t change with changes in(αh (s)

)(h,s)∈H×S , so the choices

(αh (s)

)(h,s)∈H×S

lie in the budget sets for all households. This contradicts that(xh)h∈H is a general financial

equilibrium allocation as some household is not optimally solving (HP ) .

With this second example, we know that the main result (general financial 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 infinite-dimensional space. What does it mean to consider a full

measure subset of an infinite-dimensional space? Well, this concept is not defined. Therefore,

we need to redefine the set of utility functions as belonging to a finite-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 C3), if J < S

and L > 1, then over a generic subset of endowments and utility functions, the general

financial equilibrium allocations are constrained Pareto suboptimal.

Proof. See Section 3.5.

3.4. REAL ASSETS 67

3.4 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 off in the vector of all commodities. By this specification,

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) ∈ RL+. This vector is often called the "yields." The yieldsvector is a column vector.

The value of asset j payouts in state s > 0 is then given by p (s) · yj (s) ≥ 0. The entire

payout matrix is then the S × J matrix:

R (p) =

p (1) · y1 (1) ... p (1) · yJ (1)

: :

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 definition of a general financial equilibrium remains the same as in Section 3.1.1.

In the next section, I show that a financial equilibrium may not exist when real assets are

considered. The section that follows will show how the approach developed by Duffi e and

Shafer (1985) is able to prove the generic existence of a general financial 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 defined over consumption in the states of uncertainty

s > 0.

Example 3.4 Consider an economy with two households (H = 2), two states of uncertainty


in the final period (S = 2), and two physical commodities traded in each state (L = 2).

Consumption only takes place in the final period, so the household is only endowed with

commodities in the final period. The endowments are:

e1 (1) =(

52, 50

21

)e1 (2) =

(1321, 1

2

)e2 (1) =

(12, 13

21

)e2 (2) =

(5021, 5

2

) .The utility functions are given by:

u1(x1)

= 21.5√x1

1 (1) +√x1

2 (1) + 21.5√x1

1 (2) +√x1

2 (2)

u2(x2)

=√x2

1 (1) + 21.5√x2

2 (1) +√x2

1 (2) + 21.5√x2

2 (2)

There are two real assets that are traded in the initial period and pay out in the final

period (J = 2). The asset yields are:

y1 (1) =

(1

0

)y2 (1) =

(0

1

)

y1 (2) =

(1

0

)y2 (2) =

(0

1

) .

We will show that a general financial 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 financial equilibrium allocation and commodity prices are in fact an Arrow-Debreu

equilibrium. The Arrow-Debreu budget constraints are:∑(l,s)∈L×S

ρl (s)xhl (s) ≤

∑(l,s)∈L×S

ρl (s) ehl (s)

for both h = 1, 2. The equilibrium market clearing conditions are:

x1l (s) + x2

l (s) = e1l (s) + e2

l (s) = 3 ∀ (l, s) ∈ 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) =(

83, 1

3

)x1 (2) =

(83, 1

3

)x2 (1) =

(13, 8

3

)x2 (2) =

(13, 8

3

)ρ (1) = (1, 1) ρ (2) = (1, 1)

.

The corresponding general financial 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:

R (p) =

p (1)

(1

0

)p (1)

(0

1

)

p (2)

(1

0

)p (2)

(0

1

) =

[1 1

1 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 financial equilibrium is found as if

only a single asset existed. The initial period budget constraint (recall, no consumption in

this period) is given by qzh ≤ 0 for both h = 1, 2. To satisfy market clearing for the lone

asset, then zh = 0 for both h = 1, 2. Thus, the general financial equilibrium does not have any

financial 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 financial equilibrium is given by:

x1 (1) =(

6221, 31

21

)x1 (2) =

(3221, 1

21

)x2 (1) =

(121, 32

21

)x2 (2) =

(3121, 62

21

)p (1) = (2, 1) p (2) =

(12, 1) .

The payout matrix is given by:

R (p) =

p (1)

(1

0

)p (1)

(0

1

)

p (2)

(1

0

)p (2)

(0

1

) =

[2 112

1

].


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 Duffi e and Shafer (1985) came along and introduced the method required to

prove such a claim.

I state the result of Duffi e and Shafer and then briefly 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 assetyields, a general financial 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 columnspan of a payout matrix). Define a pseudo-equilibrium as

((xh)h∈H , p, L

∗)such that

1. ∀h ∈ H, given p, xh is an optimal solution to the household problem (HP )



P\0(xh − eh

)∈ L∗

.


xhl (s) =∑

h∈Hehl (s) ∀(l, s) ∈ L× S.

Show that a pseudo-equilibrium always exists.

Step 2: If((xh)h∈H , p, L

∗)is a pseudo-equilibrium, then we can show that, over a generic

subset of endowments and asset yields, there exists((zh)h∈H , q

)such that

((xh, zh

)h∈H , p, q

)is a general financial 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) hasthe 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 ξ =((xh, λh, zh

)h∈H , p, q

)and the planner transfers were

previously introduced as τ =(τh)h∈H . The principal task will be to show that the vector of

household utility functions U (ξ, τ) =(u1(x1), .., uH(xH)

)is a submersion (i.e., its derivative

mapping is surjective).

Take as given a general financial equilibrium((xh, λh, zh

)h∈H , p, q

). Given parameters

θ =(eh, uh

)h∈H , the variables ξ =

((xh, λ

h)h∈H

, p)and transfers τ constitute a constrained

feasible vector iff Γ(ξ, τ , θ) = 0, where

Γ(ξ, τ , θ) =

FOCx

BC

MCx

MCτ

3

for FOCx =[Duh(xh)− λhP

]Th∈H

, BC =

(P(eh − xh

)+

(1 0

0 R

)τh

)h∈H

, MCx =(∑h∈H(eh\L(s)− xh\L(s))

)s∈S

, and MCτ =∑

h∈H τh. Γ has m = H (G+ S + 1) + G −

(S + 1) + J + 1 equations.

Picking a vector of parameters θ =(eh, uh

)h∈H such that

(eh)h∈H belongs to a generic

subset of RHG++ , then all resulting general financial equilibria are regular values of Φ. In

particular, this means that there exists an open set Θ′ around θ such that for any parameters

θ ∈ Θ′, the resulting equilibria satisfy the rank condition of (3.1). Let(xh)h∈H be one such

equilibrium allocation given the parameters θ. Then, the set of allocations(xh)h∈H in a local

neighborhood around(xh)h∈H such that

(u1(x1), .., uH(xH)

)>>

(u1(x1), .., uH(xH)

)is open.

3See the discussion in Section 3.3.2.


If for some planner transfers τ ∗ =(τh∗)h∈H , the resulting constrained feasible allocation is

Pareto superior, then all constrained feasible allocations for τ in an open neighborhood

around τ ∗ are Pareto superior as well.

The analysis is conducted evaluating equations at τ =−→0 . By definition, if Γ(ξ,

−→0 , θ) = 0

and Φ(ξ, θ) = 0, then ξ =((xh, λh

)h∈H , p

).

Define the (H +m)× (m′ +H (J + 1)) matrix Ψ0 :

Ψ0 =

(Dx,λ,pU(ξ, τ) 0

Dx,λ,pΓ(ξ, τ , θ) DτΓ(ξ, τ , θ)

),

where m′ = H (G+ S + 1) +G− (S + 1) is the number of variables in the vector (x, λ, p) =((xh, λh

)h∈H , p

)and H (J + 1) is the number of planner transfers

(τh)h∈H . From Citanna,

Kajii, and Villanacci (2002), if Ψ0 has full row rank, ∃ξ 6=((xh, λh

)h∈H , p

)s.t. ξ satisfies

Γ(ξ, τ , θ) = 0 (for some τ) and U((xh)h∈H

)> U

((xh)h∈H

). For full row rank, I must

ensure that there are fewer rows than columns, so I need the following inequality to hold:

H +m ≤ m′ +H (J + 1) . (3.4)

(3.4) reduces to

H + J + 1 ≤ H (J + 1) . (3.5)

(3.5) is always satisfied as H ≥ 2. The matrix Ψ0 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 Ψ. This matrix Ψ

does not have full rank iff ∃ν ∈ RH+m s.t. Φ′(ξ,−→0 , ν, θ) = 0 where

Φ′(ξ,−→0 , ν, θ) =

(ΨTν

νTν/2− 1

).

For simplicity, I divide the vector νT into subvectors that each represent a certain equation

in Ψ. Define νT =(∆uT ,∆xT ,∆λT ,∆pT ,∆τT

)∈ RH+m where each subvector corresponds


sensibly to an equation (row) in Ψ as follows:

∆uT ⇐⇒ Dx,λ,pU(ξ, τ)

∆xT ⇐⇒ FOCx

∆λT ⇐⇒ BC

∆pT ⇐⇒ MCx

∆τT ⇐⇒ MCτ.

A subset of the equations νTΨ = 0 are given by (corresponding to derivatives with respect

to((xh, λh)h∈H

)in that order):(

∆uhDuh(xh) +(∆xh

)TD2uh(xh)−

(∆λh

)TP −∆pTΛ

)h∈H

= 0. (3.6.a)(−(∆xh

)TP T)h∈H

= 0. (3.6.b)(3.6)

where Λ =

(IL−1 | 0

)0 0

0 ... 0

0 0(IL−1 | 0

) as in Subsection 3.1.2.

The proof is complete if I can show that for a generic choice of θ ∈ Θ, there does not

exist (ξ, ν) s.t.

Φ(ξ, θ) = 0 (3.7)

Φ′((x, λ, p) ,−→0 , ν, θ) = 0.

Counting equations and unknowns, (3.7) has n equations in Φ, n variables ξ, H + m +

1 equations in Φ′, and only H + m variables ν. I must show that over a generic subset

of parameters (exactly which generic subset will be discussed next), the derivative matrix

Dx,λ,p,ν

(Φ

Φ′

)has full row rank. I will reference the (ND) condition of Citanna, Kajii, and

Villanacci (1998), which is a suffi cient condition for the full row rank of Dx,λ,p,ν

(Φ

Φ′

).


The condition states that for τ =−→0 and ξ =

((xh, λh

)h∈H , p

), the matrix

( (ΨT

νT

)DθΦ

′

)has full row rank, (3.8)

where θ are the parameters on which the genericity statement is made.

For simplicity, I break up the analysis into two cases: Case I:(∆xh

)T 6= 0 ∀h ∈ H and

Case II:(∆xh

)T= 0 for some h ∈ 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 ∀h ∈ H

Lemma 3.1 For τ =−→0 , then DuΦ

′ =

(d(Ah)

−→0

)where d

(Ah)has full row rank and

corresponds to the rows for derivatives with respect to(xh)h∈H .

Proof. The set U is infinite-dimensional and is endowed with the C3 uniform convergence

topology on compact sets. This means that a sequence of functions uν converges uniformlyto u iff Duν, D2uν, and D3uν uniformly converge to Du, D2u, and D3u, respectively.

Additionally, any subspace of U is endowed with the subspace topology of the topology ofU . I will use the regularity result from Theorem 3.2 to define utility functions as locally

belonging to the finite-dimensional subset A ⊆ U .Using Theorem 3.2, pick a regular value θ. For that θ, there exist finitely many equilibria

ξi, i = 1, ..., I. Further, there exist open sets Σ′ and A′hi s.t. xhi ∈ A′hi , the sets A′hi are disjoint

across i, and ∀θ ∈ Σ′, ∃! equilibrium allocation xhi ∈ A′hi . Choose A′hi such that the closureclA′hi is compact and there exist disjoint open sets A

′hi s.t. A

′hi ⊂ clA′hi ⊂ A′hi .

For each household, define a bump function δh : Xh → [0, 1] with I bumps as δh = 1 on

A′hi and δh = 0 on (A′hi )c. Now, I define uh in terms of a G×G symmetric matrix Ah by:

uh(xh;Ah) = uh(xh) +1

2δh(xh)

∑i

[(xh − xhi )TAh(xh − xhi )

].

Thus, the space of symmetric matrices (denote this as A) is a finite dimensional subspaceof U . Since A has the subspace topology of U , then uh(·;Aν) → uh(·;A) iff Aν → A. This

can be seen by taking derivatives and noting that the function u stays fixed at the regular

value.


Taking derivatives with respect to xh ∈ A′hi yields:

Dxuh(xh;Ah) = Duh(xh) + Ah(xh − xhi )

D2xxu

h(xh;Ah) = D2uh(xh) + Ah.

A is a G(G+ 1)/2 dimensional space, so write Ah as the vector

((Ahi,i)i=1,..,G, (A

hi,j)i<j,i=1,..,G−1

).

Postmultiply D2xx by ∆xh :

D2xxu

h(xh;Ah)∆xh = D2uh(xh)∆xh + Ah∆xh.

Taking derivatives with respect to the parameter uh is equivalent to taking derivatives

with respect to Ah :

Du

(D2xxu

h(xh;Ah)∆xh)

= DA

(Ah∆xh

)=

∆xh1 0 0

0 ... 0

0 0 ∆xhG

Σ(1) ... Σ(G− 1)

∈ RG,G(G+1)/2

where the submatrix Σ(i) is defined as

Σ(i) =

0 ∈ Ri−1,G−i

∆xhi+1 ... ∆xhG∆xhi 0 0

0 ... 0

0 0 ∆xhi

∈ RG,G−i.

Thus, since ∆xh 6= 0 (without loss of generality ∆xh1 6= 0), then

rankDAh(D2xxu

h(xh;Ah)∆xh)

= G. (3.9)

Out of all the rows ΨT , the utility function uh only appears in the row for derivatives


with respect to xh. This row in ΨT for household h is given by (as in (3.6.a)):

U(ξ, ξ) FOCx BC MCx MCτ(Duh(xh)

)TD2uh(xh) −P T −ΛT 0

.

Thus, taking the derivative DAhΦ′ =

(DAhΨTν

0

), the only nonzero terms correspond to

the rows for derivatives with respect to xh :

DAh

((Duh(xh)

)T∆uh +D2uh(xh)∆xh − P T∆λh − ΛT∆p

)= DAh

((Duh(xh;Ah)

)T∆uh

)+DAh

(D2uh(xh;Ah)∆xh

).

From the first derivative, Dxuh(xh;Ah) = Du(xh) + Ah(xh − xhi ) = Du(xh) evaluated at

τ =−→0 (since xh = xhi ). Thus DAh(Dxu

h(xh;Ah)∆uh) = 0. Using (3.9),

d(Ah)

=

DA1 (D2u1(x1;A1)∆x1) 0 0

0 ... 0

0 0 DAH(D2uH(xH ;AH)∆xH

)

is a full row rank matrix of size HG×HG(G+ 1)/2.

The matrix

( (ΨT

νT

)DAΦ′

)is given below (where the rows correspond to the equi-

librium variables((xh, λh)h∈H, p

), planner transfers (τ), and vector vT in that order). Recall

from (3.8) that I aim to show that this matrix has full row rank. I use the convention

c(Mh)

=

M1

:

MH

, r(Mh)

=(M1 ... MH

), and d

(Mh)

=

M1 0 0

0 ...

0 0 MH

, where

c stands for column, r stands for row, and d stands for diagonal. For simplicity, define


Ω =

(1 0

0 R

). The matrix

( (ΨT

νT

)DAΦ′

)is given by:

d(Duh(xh)T

)d(D2uh

)d(−P T ) c(−ΛT ) 0 d

(Ah)

0 d (−P ) 0 0 0 0

0 r(−(Λh

2

)T) r

((Zh)T)

0 0 0

0 0 d(ΩT)

0 c (IS+1) 0

r(∆uh) r((∆xh

)T) r(

(∆λh

)T) ∆pT ∆τT 0

where I define the following two submatrices:

Λh2 =

(λh (0) IL−1

0

)0 0

0 ... 0

0 0

(λh (S) IL−1

0

) ∈ R

G,G−(S+1).

Zh =

(eh\L (0)− xh\L (0)

)T0 0

0 ... 0

0 0(eh\L (S)− xh\L (S)

)T ∈ RS+1,G−(S+1).

Lemma 3.2((

∆uh)h∈H , p

)6= 0.

Proof. Suppose not, that is(∆uh, p

)= 0 for some h. From (3.6.a),

(∆xh

)TD2uh(xh)−

(∆λh

)TP = 0. (3.10)

Post-multiply (3.10) by ∆xh and use (3.6.b) to yield:

(∆xh

)TD2uh(xh)∆xh = 0. (3.11)

From (3.11) and Assumption R2, then ∆xh = 0. But this contradicts that we are in Case I.

From Lemmas 3.1 and 3.2, the first and last row blocks of

( (ΨT

νT

)DAΦ′

)are lin-


early independent from the others. By the definition of Λh2 , the submatrix

−P 0 0

0 .. 0

0 0 −P− (Λ1

2)T

.. −(ΛH

2

)T

of size [H(S + 1) +G− (S + 1)]×HG is a full rank matrix. The submatrix d

(ΩT)also has

full rank (as R has full column rank). Thus,

( (ΨT

νT

)DAΦ′

)has full rank and this

concludes the proof under Case I.

3.5.2 Case II:(∆xh

)T= 0 for some h ∈ H

I will show that over a generic subset of endowments, (3.7) has no solution. Suppose ∃h′ ∈ Hsuch that

(∆xh

′)T= 0. From (3.6.a) and Φ, I obtain

∆uh′Duh

′(xh

′)−

(∆λh

′)T

P −∆pTΛ = 0

Duh′(xh

′)− λh′P = 0

which together imply that ∆pT = 0 and(

∆λh′)T

= ∆uh′λh′.

For all other h 6= h′, postmultiply ∆uhDuh(xh) by ∆xh and use the first order condition

(consumption) in Φ and (3.6.b) to get ∆uhDuh(xh)∆xh = 0. Next, postmultiply (3.6.a) by

∆xh and use (3.6.b) to arrive at the equation:

(∆xh

)TD2uh(xh)∆xh = 0. (3.12)

By Assumption R2,(∆xh

)T= 0 ∀h ∈ H. From what remains of (3.6.a) and Φ,

(∆λh

)T=

∆uhλh ∀h ∈ H.

From the column for derivatives with respect to(τh)h∈H ,(

∆λh)T

Ω + ∆τT = 0 ∀h ∈ H. (3.13)

As Ω =

(1 0

0 R

), then (3.13) implies ∆λh (0) = ∆λ1 (0) ∀h ∈ H.

The terms((

∆uh,(∆λh

)T)h∈H

,∆τT)are the only nonzero elements of ν. Further,

3.6. EXERCISES 79

∆uhλh (0) = ∆u1λ1 (0) ∀h ∈ H. This yields:

∆uh = ∆u1λ1 (0)

λh (0)∀h ∈ H. (3.14)

The equation from νTΨ = 0 corresponding to derivatives with respect to p (s) (for any

s > 0) is given as follows (after replacing ∆λh(s) with ∆uhλh(s)):∑h∈H

∆uhλh(s)(eh\L(s)− xh\L(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 first physical

commodity l = 1 and use (3.14):

∆u1λ1 (0)∑

h∈H

λh (s)

λh (0)

(eh1(s)− xh1(s)

)T= 0.

As with Theorem 3.4 and "A Different Application of Differential Topology" from Exercise

3 in Chapter 1, we can show that over a generic subset of endowments, the equation

∑h∈H

λh (s)

λh (0)

(eh1(s)− xh1(s)

)T 6= 0.

This implies that ∆u1 = 0, with (3.14) implying(∆uh

)h∈H = 0. Consequently,

(∆λh

)T= 0

∀h ∈ H. From (3.13), ∆τT = 0. In conclusion, νT = 0, which cannot satisfy Φ′ as this

system contains the equation νTν/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 financial equilibrium does not exist.


2. Suppose that there are S = 4 states and J = 2 assets with payouts

R =

2 3

1 1

0 1

3 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)?


R =

1 0 0

0 1 123

1 13

12

23

12

.

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, λh, zh

)h∈H , p, q

)is a general financial equilibrium in which No Redundancy does

not hold, then there exists((zh)h∈H , q

)such that

((xh, λh, zh

)h∈H , p, q

)is a general

financial equilibrium in which No Redundancy does hold.

5. Prove Theorem 3.4.

6. In Example 3.1, verify that the Arrow-Debreu equilibrium allocation is such that ∀h ∈H : xh (s) = θh

∑h∈H

eh (s) for some θh.

7. In Example 3.3, verify that the general financial equilibrium consumption is given by:

xhl (s) = αh (s) ·∑

h∈Hehl (s)

for some αh (s) ∈ (0, 1) and the prices are given such that pl (s) = θl(s)θL(s)

∀(l, s) ∈ 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, 1

3

)x1 (2) =

(83, 1

3

)x2 (1) =

(13, 8

3

)x2 (2) =

(13, 8

3

)ρ (1) = (1, 1) ρ (2) = (1, 1)

.

9. In Example 3.4, verify that the general financial equilibrium allocation and commodity

prices in Case II are given by:

x1 (1) =(

6221, 31

21

)x1 (2) =

(3221, 1

21

)x2 (1) =

(121, 32

21

)x2 (2) =

(3121, 62

21

)p (1) = (2, 1) p (2) =

(12, 1) .

Bibliography

[1] Citanna, Alessandro, Atsushi Kajii, and Antonio Villanacci (1998): "Constrained Sub-

optimality in Incomplete Markets," Economic Theory 11, 495-521.

[2] Duffi 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 Con-

strainted 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.



[6] Villanacci, Antonio, Laura Carosi Carosi, Pierluigi Benevieri, and Andrea Battinelli

(2002): Differential Topology and General Equilibrium with Complete and Incomplete Mar-

kets (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 ∈ S :

pL (s) = 1. Does this choice matter? Would a different price normalization affect the equilib-

rium consumption? Consider the budget constraints in the previous chapter, when we have

numeraire assets:

p (s)(eh (s)− xh (s)

)+ pL (s)

∑j∈J rj (s) zhj ≥ 0 for s > 0.

Thus, any other price normalization (pL (s) = k for some k > 0 or p (s) ∈ ∆L−1) does not

change the budget constraint. Price normalizations are neutral; they have no real effects. 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 off 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 off 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:

• Xh = RG+ is the consumption set.

• uh : Xh → R is the utility function. We assume that the function is continuous, locallynon-satiated, and quasi-concave.

• eh ∈ Xh is the endowment.

Each good (l, s) has a market price pl (s) . The vector of all prices is p = (pl (s))(l,s)∈L×S ∈RG\0. The convention is that the price vector p (s) is a row vector ∀s ∈ S. I define the

(S + 1) × G price matrix P =

p(0) 0 0 0

0 p(1) 0 0

0 0 ... 0

0 0 0 p(S)

. The price normalizations will bedetermined as a function of the money supplies. The money supplies are Ms > 0 for all

states s ∈ S. The vector M = (Ms)s∈S is a parameter of the model.

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)j∈J . 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 fixed payout rj(s). The payouts are made in terms

of the unit of account. The payouts are assumed to be nonnegative rj(s) ≥ 0 ∀ (j, s) ∈J× S\0, where rj = (rj(s))s>0 > 0 ∀j ∈ J . Let’s collect the payouts of all assets in allstates s > 0 into one payout matrix R. This payout matrix has S rows and J columns:

R =[r1 ... rJ

]=

r1(1) ... rJ(1)

: ... :

r1(S) ... rJ(S)

.

4.1. THE MODEL 87

Household h chooses a portfolio zh ∈ RJ in the initial period, where the position for aparticular asset j is denoted zhj ∈ 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 first 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

currency), so that∑

h∈Hmh (0) = M0. In the second substage, households trade the nominal

assets. Following this trade, each household has available the following amount of the unit

of account: mh (0) = mh (0)− qzh. Given this amount of the unit of account, the householdsthen 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) ≤ mh (0) .

After the purchase of commodities, the money in the market∑

h∈Hmh (0) = M0 is returned

to the Monetary Exchange.

A similar story unfolds in each state s > 0. Three substages take place. In the first,

households sell their endowments to the Monetary Exchange and receivemh (s) = p (s) eh (s)

units of account, so that∑

h∈Hmh (s) = Ms. In the second substage, the payouts of the

nominal assets are accounted for, so each household now has available the following amount

of the unit of account: mh (s) = mh (s) + r (s) zh. In the third substage, households use this

money to pay for the purchase of commodities:

p (s)xh (s) ≤ mh (s) .

The money in the market from the sale of commodities, of size∑

h∈Hmh (s) = Ms, is

returned to the Monetary Exchange.

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 mon-

etary models). In this setup, the velocity of money is set equal to 1 by definition (a single

unit of account can only be used in one purchase). The monetary elements of the model are

summarized succinctly in the definition of a financial equilibrium with money.


4.1.1 Existence

Definition 4.1 A financial equilibrium with money is((xh, zh

)h∈H , p, q

)such that

1. ∀h ∈ H, given (p, q) ,(xh, zh

)is an optimal solution to the household problem (HP )



zh ∈ RJ

P(eh − xh

)+

(−qR

)zh ≥ 0

.


xhl (s) =∑

h∈Hehl (s) ∀(l, s) ∈ L× S.∑

h∈Hzhj = 0 ∀j ∈ J .

3. Monetary supply condition

p (s)∑

h∈Hxh (s) = Ms ∀s ∈ S.

It is important to verify that a financial equilibrium with money exists. The assumptions

for this are given by:


Assumption E2: uh : Xh → R is C1, increasing, and quasi-concave ∀h ∈ H.


Assumption E4 (No Redundancy): The payout matrix R has full column rank


Assumption E5: M >> 0.

Theorem 4.1 Under Assumptions E1-E5, a financial equilibrium with money((xh, zh

)h∈H , p, q

)exists.

Proof. The proof is identical to the proof of Theorem 3.1, which is simply an extension of

the first 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 89

4.2 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

financial 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

financial equilibrium with money is((xh, zh

)h∈H , p, q

). Consider a change in money supply

to M ′ = (M ′0, ...,M

′S) . Then the new financial equilibrium allocation is still

(xh)h∈H .

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.

Let’s define the purchasing power of money for all states s ∈ S : νs = 1∑l∈L pl(s)

. Define

ν = (νs)s>0 and the diagonal matrix [ν] =

ν1 0 0

0 .. 0

0 0 νS

.We scale the payout matrix by the purchasing power of money. The budget constraints

for states s > 0 are given by:

[p](xh − eh

)= [ν]Rz,

where I define [p] =

0 p(1)∑

l∈L pl(1)0 0

0 0 ... 0

0 0 0 p(S)∑l∈L pl(S)

∈ RS,G. Recall that the span of thepayout matrix is 〈[ν]R〉 =

x ∈ RS : x = [ν]Rz for some z ∈ RJ

. The budget constraints

are equivalently expressed by the following two conditions.


1. Single budget constraint

(1, α) [p](eh − xh

)≥ 0

for any vector of state prices α ∈ RS++.

2. Span condition

[p](xh − eh

)∈ 〈[ν]R〉 .

We make the following assumptions.

Assumption M1:

1

:

1

∈ 〈R〉 .Assumption M2: J < S.

Assumption M3: J ≤ H.

Theorem 4.3 Under Assumptions E1-E5 and M1-M3, there is a subspace ∆ ⊂ RS withdim ∆ ≥ S− J such that if a change in the monetary policy from M to M ′ implies a change

from ν to ν ′ with ν ′ − ν ∈ ∆, then 〈[ν]R〉 6= 〈[ν ′]R〉 .

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 ′ = (M ′0,M

′1, ...,M

′S) such that M ′

s = βMs ∀s > 0 and for some β > 0. Then

[ν ′] = 1β

[ν] , so the span remains unchanged: 〈[ν]R〉 = 〈[ν ′]R〉 . Thus, for this proportionalmonetary policy, there are no real effects (the policy is neutral).

Now, let’s consider the proof of the theorem.

Proof. The span of the payout matrix 〈[ν]R〉 has dimension equal to J (from Assumption

E4). Thus, the space that is orthogonal to 〈[ν]R〉 , denoted 〈[ν]R〉⊥ , has dimension equalto S − J. Define ∆ = 〈[ν]R〉⊥ \ 0 . From the statement of the theorem, ν ′ − ν ∈ ∆. It

must be that ν ′ /∈ 〈[ν]R〉 (if not, then ν ′− ν ∈ 〈[ν]R〉⊥ and ν ′− ν ∈ 〈[ν]R〉 , a contradictionof ν ′ 6= ν). Under Assumption M1, ν ′ ∈ 〈[ν ′]R〉 (as there exists z = (1, 0, ..., 0) such that

ν ′ = [ν ′]Rz). If ν ′ /∈ 〈[ν]R〉 and ν ′ ∈ 〈[ν ′]R〉 , it can only be the case that 〈[ν]R〉 6= 〈[ν ′]R〉 .

Theorem 4.3 is nice, but if we make one additional assumption, then we are able to prove

a much stronger result. Specifically, the subspace ∆ of possible policy changes can be shown

to have dimension S − 1 (bigger than S − J).


Assumption M4: R is in general position, meaning that any J × J submatrixhas rank J.

Theorem 4.4 Under Assumptions E1-E5 and M1-M4, there is a subspace ∆ ⊂ RS withdim ∆ = S − 1 such that if a change in the monetary policy from M to M ′ implies a change

from ν to ν ′ with ν ′ − ν ∈ ∆, then 〈[ν]R〉 6= 〈[ν ′]R〉 .

The proof of Theorem 4.4 is more challenging than the proof of Theorem 4.3, but our

efforts are rewarded with the knowledge that the asset span will change for any policy changes

in a (S − 1)−dimensional subset ∆.

Proof. Suppose that 〈[ν]R〉 = 〈[ν ′]R〉 . This implies that for any j ∈ J, the vector [ν ′] rj is

a linear combination of the vectors [ν] r1, ..., [ν] rJ .We can express these linear combinations

using the J × J matrix C such that:

[ν ′]R = [ν]RC. (4.1)

(4.1) written for any state s > 0 is equivalent to the following form:

CT [r(s)]T =ν ′sνs

[r(s)]T . (4.2)

If we recall our lessons on eigenvalues, (4.2) implies that ν′sνsis an eigenvalue, with associated

eigenvector [r(s)]T , of the matrix CT . 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 eigenvec-

tors must be linearly independent. Assume that the eigenvalue for state s is different from

the eigenvalue for state σ : ν′sνs6= ν′σ

νσ. Then this implies that [r(s)]T is linearly independent

from [r(σ)]T . With J < S, all vectors [r(s)]T cannot be linearly independent from the vec-

tors for all other states σ 6= s. Thus, we are able to find a collection of J or fewer vectors[r(s)]T

s∈S∗

such that the number of states s ∈ S∗ is greater than the rank of the submatrix[r(s)]s∈S∗ . This is a contradiction of the general position assumption (Assumption M4).

1

We conclude that ∀s, σ > 0 : ν′sνs

= ν′σνσ

= β > 0. Thus, the only changes from ν to ν ′

that do not change the span are of the form ν ′s = βνs ∀s > 0. These changes lie in the

1Adding a row vector r(s) to a matrix increases the rank by at most 1. Add states to the submatrix[r(s)]s∈S∗ until there are J rows. The rank of the new submatrix is then at most J − 1. General positionrequires that a submatrix with any J rows of R has full rank.


one-dimensional subspace defined by the vector ν :

ν ′ − ν ∈ 〈ν〉 .

All other changes, those in the set ν ′ − ν ∈ 〈ν〉⊥ , where 〈ν〉⊥ has dimension S − 1, will

change the span: 〈[ν]R〉 6= 〈[ν ′]R〉 .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 householdendowments and money supplies, the vectors

(ah)h∈H span 〈[ν]R〉 , where ah ∈ RS is defined

by:

ah = [p](xh − eh

)∀h ∈ H.

Equivalently, this lemma states that the collection of J vectors(ah)h∈J are linearly inde-

pendent. 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 householdendowments and money supplies, there is a subspace ∆ ⊂ RS with dim ∆ = S − 1 such

that all changes in the monetary policy from M to M ′ that induce a change from ν to ν ′

with ν ′ − ν ∈ ∆, are non-neutral. That is, the financial equilibrium with money allocation

changes: x =(xh)h∈H 6= x′ =

(x′h)h∈H .

Proof. From Theorem 4.4, 〈[ν]R〉 6= 〈[ν ′]R〉 . As the dimensions dim 〈[ν]R〉 = dim 〈[ν ′]R〉 =

J (Assumption E4), then dim (〈[ν]R〉 ∩ 〈[ν ′]R〉) < J. By Lemma 4.1, the vectors(ah)h∈H

span 〈[ν]R〉 , so some of the vectors cannot be elements of the set 〈[ν]R〉 ∩ 〈[ν ′]R〉 . Sincethe vectors a′h = [p′]

(x′h − eh

)∈ 〈[ν ′]R〉 ∀h ∈ H by definition, then

(ah)h∈H 6=

(a′h)h∈H .

By the definition of ah = [p](xh − eh

)and first order conditions Duh

(xh)− λhP = 0, then

this implies(xh)h∈H 6=

(x′h)h∈H , finishing 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)h∈J are linearly independent iff the portfolio vectors

(zh)h∈J

are linearly independent.

2. Write down the system of equilibrium equationsΦ : Ξ×Θ→ Rn for n = H (G+ S + 1 + J)+

G + J so that ξ =((xh, λh, zh

)h∈H , p, q

)is a financial equilibrium with money iff

Φ (ξ) = 0. Here θ =((eh)h∈H ,M

)∈ Θ.

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)h∈H and money

suppliesM = (M0,M1, ...,MS) , the financial equilibria with money satisfy Finite Local

Uniqueness. This requires proving that (i) π is proper and (ii) if Φ (ξ, θ) = 0, then

rankDΦ (ξ, θ) = n (the derivative is taken with respect to the variables ξ and the

parameters θ =((eh)h∈H ,M

)). The proof of (i) is straightforward, so only prove (ii).

4. Given the outcome of Exercise 3 above, show that rankDΦ∗ (ξ∗, θ) = n+ J + 1, where

Φ∗ (η) =

Φ (η)

µ ·(z1, ..., zJ

)µTµ/2− 1

(again taking derivatives with respect to the variables

ξ∗ =((xh, λh, zh

)h∈H , p, q, µ

)∈ Rn+J and the parameters θ =

((eh)h∈H ,M

)). The

additional equations µ ·(z1, ..., zJ

), where µ ∈ RJ\0 (nonzero from µTµ/2 = 1),

imply that the portfolios vectors(zh)h∈J are linearly dependent. Using "A Different

Application of Differential Topology" from Exercise 3 in Chapter 1, we can then con-

clude that for a generic selection of θ =((eh)h∈H ,M

), the portfolio vectors

(zh)h∈J


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Equilibrium," in: Advances in Economic Theory; Sixth World Congress, Volume II, edited

by J.J. Laffont (Cambridge: Cambridge University Press).



[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): Differential Topology and General Equilibrium with Complete and Incomplete Mar-

kets (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 × n matrix and θ ∈ Rn. Use Farkas Lemma to prove the followingimplication.

There does not exist θ such that Rθ > 0

⇓∃λ ∈ Rm++ s.t. λ

TR = 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:

Define R =

R1

:

Rm

. Consider the 1st row of the matrix R. The following process

will be repeated for all m rows of R. The statement "There does not exist θ such that

Rθ > 0" implies that @θ such that R1θ > 0 and Riθ ≥ 0 ∀i = 2, ...,m. Thus, @θ suchthat R1θ > 0 ≥ −Riθ ∀i = 2, ...,m. Define

Z(1) =z ∈ Rn : z =

[−RT

2 , ..., RTm

]· λ(1) for λ(1) ∈ Rm−1

+

.

97

98 APPENDIX A. SOLUTIONS TO THE EXERCISES

With this definition, @θ such that θTRT1 > 0 ≥ θT z ∀z ∈ Z(1). Use the following table

to connect the statement of Farkas Lemma to the proof at hand.


q∗ ∈ Rn\0 ⇐⇒ θ ∈ Rn\0z∗ ∈ Rn ⇐⇒ RT

1 ∈ Rn

A ∈ Rn,m−1 ⇐⇒[−RT

2 , ..., RTm

]α ∈ Rm−1

+ ⇐⇒ λ(1) ∈ Rm−1+

As condition (ii) of Farkas Lemma does not hold, then condition (i) must. This means

that RT1 ∈ Z(1), so ∃λ(1) ∈ Rm−1

+ such that

RT1 =

[−RT

2 , ..., RTm

]· λ(1).

Rearranging and taking transposes, we have:

R1 +m−1∑i=1

λi(1) ·Ri+1 = 0.

Define λ(1) ∈ Rm+ such that λ1(1) = 1 and λi(1) = λi−1(1) for i = 2, ...,m. Then(λ(1)

)T·R = 0.

Repeating the entire process for the remaining rows i = 2, ...,m, then we obtain

λ(2), ..., λ(m) ∈ Rm+ such that λi(i) = 1 ∀i = 2, ...,m. In all cases,(λ(i)

)T· R = 0

∀i = 2, ...,m.

Define λ ∈ Rm++ as λ =m∑i=1

λ(i). By construction, λT · R = 0. This completes the

argument.

2. (Applying Kakutani’s Fixed Point Theorem)

We will use the theory of correspondences to prove the existence of a Nash equilib-

rium. Consider a game with I players. Each player i has a finite number of actionsai1, ...., a

iJi

. The set of strategies for each player i is then the simplex ∆Ji−1 of di-

mension Ji − 1. The strategies are simply the probabilities that each player assigns to

each of the finite number of actions. For simplicity, denote the strategy set for each

player as Si with element si. These sets are nonempty, compact, and convex.

A.1. MATHEMATICAL PREREQUISITES 99

If player i′s payoff value for the action profile a = (a1, ..., ai, ..., aI) is pi(a), then the

objective function for each player is (using the convention s = (s1, ..., si, ..., sI)):

ui (s) =∑

api(a) ·

∏isi(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 ×j 6=iSj = S1×...×Si−1×Si+1×...×SI as the Cartesian product

of the strategy sets (not including the strategy set for player i). Define F i : ×j 6=iSj ⇒ Si

as the set of "feasible" strategies. Notice that F i (s−i) = Si ∀s−i ∈ ×j 6=iSj.

Define BRi : ×j 6=iSj ⇒ Si as the best response correspondence for player i. Use the

Berge’s Maximum Theorem and Kakutani’s Fixed Point Theorem to prove that the

self-map(BR1, ..., BRI

): ×iSi ⇒ ×

iSi has a fixed point. By definition, a fixed point is

a Nash equilibrium.

Solution:

By definition, Si = ∆Ji−1 is compact, convex, and nonempty. These properties are

maintained through the Cartesian product operator, meaning that ×iSi is compact,

convex, and nonempty.

The objective function is continuous and the strategy set is compact. Using the Ex-

treme Value Theorem, the best response correspondence BRi is well-defined.

The objective function is quasi-concave (as it’s linear) and the strategy set is convex.

Consider any two x, y ∈ BRi (s−i) for some s−i ∈ ×j 6=iSj. By convexity of Si, then

θx+(1−θ)y ∈ Si. By quasi-concavity, ui (θx+ (1− θ)y) ≥ ui(x) = ui(y). This implies

that θx+ (1− θ)y ∈ BRi (s−i) , meaning that BRi is convex-valued.

Define the constraint correspondence Ci : ×j 6=iSj ⇒ Si. The correspondence is defined

such that Ci (s−i) = Si ∀s−i ∈ ×j 6=iSj. Given this definition, the best response corre-

spondence is formally defined as:

BRi (s−i) = arg maxsi∈Ci(s−i)

ui (s) .


The objective function ui (s) is continuous. Given that the codomain Si is compact

and Ci maps into the entire set Si, the correspondence Ci 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-defined, convex-valued, and uhc. Applying Kakutani’s Fixed Point The-

orem to the self-map(BR1, ..., BRI

): ×iSi ⇒ ×

iSi guarantees the existence of a fixed

point.

3. (A Different Application of Differential Topology)

Let Ξ ⊂ RJ be the set of variables (with typical element ξ), Θ ⊂ RK be the set of

parameters (with typical element θ), H = Ξ×Θ (so η = (ξ, θ)), and Φ : H → RL be theC1 system of equations. Assume that J < L. Suppose that the projection π is proper

and the rank condition holds (that is, rankDΦ (η) = 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 Θr), then...".

Solution:

With J < L, then all solutions η ∈M are critical points, as they must necessarily satisfy

rankDξΦ (η) < L. Thus, π (M) are critical values. Let’s suppose that we are able to

verify the two key properties of regularity: (i) π is proper and (ii) rankDΦ (η) = L

∀η ∈ 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, π (M) is a closed

subset of measure zero, where π (M) = θ : ∃ξ such that Φ (ξ, θ) = 0 by definition.

In conclusion, over a generic subset of parameters (the set of regular values Θr =

Θ\π (M)), there does not exist a solution ξ to the system of equations Φ (ξ, θ) = 0.

A.2 Arrow-Debreu Model

1. Show that if uh is locally non-satiated ∀h ∈ H, then the Arrow-Debreu equilibriumprices satisfy p > 0.

Solution:

To begin, I show that in any Arrow-Debreu equilibrium((xh)h∈H , p

), an optimal

solution to the household problem (HP ) must satisfy pxh = peh ∀h ∈ H. Suppose

A.2. ARROW-DEBREU MODEL 101

otherwise, that pxh < peh for some h. By local non-satiation, ∃yh ∈ Nε

(xh)∩Xh such

that uh(yh)> uh

(xh). The value for ε can be made small enough so that pyh < peh,

contradicting that xh is an optimal solution to the household problem (HP ).

The initial price space is p ∈ RG\0. To prove the claim, suppose that pg < 0 for some

g ∈ G. Suppose that xh is the equilibrium consumption choice for household h. From

the previous paragraph, pxh = peh. Define xh such that xhg = xhg + δ and xhg′ = xhg′

∀g′ 6= g. By linearity, pxh < peh. As xh is not optimal, it must be that uh(xh)>

uh(xh). By local non-satiation, ∃yh ∈ Nε

(xh)∩Xh such that uh

(yh)> uh

(xh). For

xh to be optimal, it must be that pyh > peh. Define yh such that yhg = yhg + δ and

yhg′ = yhg′ ∀g′ 6= g. There exists δ such that ∀δ ≥ δ, pyh ≤ peh. For xh to be optimal,

it must be that uh(xh)≥ uh

(yh). We have uh

(yh)> uh

(xh)≥ uh

(yh). Letting

(ε, δ)→ 0, we see that uh is strictly decreasing with xhg . This implies that the optimal

solution to (HP ) lies on the boundary: xh /∈ intXh. This violates local non-satiation

as @zh ∈ Nε

(xh)∩Xh such that uh

(zh)> uh

(xh).

2. Assume that uh is differentiable, differentiably strictly increasing, and concave ∀h ∈H. Using the results from the programming problems in this chapter, show that if((xh)h∈H , p

)is an Arrow-Debreu equilibrium, then the allocation

(xh)h∈H is an opti-

mal solution to the problem (POh′) for any h′. From Section 2.3, the allocation(xh)h∈H

is then Pareto optimal. As this was the first proof of the First Basic Welfare Theorem,

it is often called the "classical proof."

Solution:

The following are necessary and suffi cient conditions for an Arrow-Debreu equilibrium:

Duh(xh)− λhp = 0 ∀h ∈ H (FOC)∑h∈H

(ehg − xhg

)= 0 ∀g ∈ G (MC) .

The following are necessary and suffi cient conditions for an optimal solution to (POh′)

for household h′ = 1 (without loss of generality). Let(λh)h>1

be the Lagrange mul-

tipliers for the constraints uh(xh) ≥ uh(x∗h) and(µg)g∈G be the Lagrange multipliers


for the constraints(xh)h∈H ∈ FA, namely

∑h∈H

(ehg − xhg

)= 0 ∀g ∈ G :

Du1(x1)− µ = 0 (FOC1)

λhDuh(xh)− µ = 0 ∀h > 1 (FOCh)∑

h∈H

(ehg − xhg

)= 0 ∀g ∈ G (FA)

.

To show that both systems of equations yield the same optimal allocation, define

µ = λ1p >> 0 and λh

= λ1

λh> 0.

3. Consider an economy with two households and two goods. Both utility functions are

differentiable, 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 defined (given the assumptions on the utility functions) as points where

an indifference curve for h = 1 is tangent to an indifference 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

u1(x1)

=1

2log(x1

1 + x21

)+

1

2log(x1

2

)u2(x2)

=1

2log(x2

1

)+

1

2log(x2

2

)and endowments e1 = (1, 2) and e2 = (2, 1) . Notice that household 1 cares about


household 2’s consumption of the first 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, θ) and x2 = (3, 3− θ) for any θ ∈ [0, 3] .

This is illustrated in the Edgeworth box below.

To find the Arrow-Debreu equilibrium, normalize p2 = 1. The first order conditions for

household h = 2 are given by:

0.5

x21

− λ2p1 = 0.

0.5

x22

− λ2 = 0.

Solving for x2 as a function of λ2 and plugging into the budget constraint p1 (x21 − 2) +

(x22 − 1) = 0 yields:

0.5

λ2 +0.5

λ2 = 2p1 + 1.

Thus, 1λ2

= 2p1 + 1, so the demand functions are:

x21 (p1) =

p1 + 0.5

p1

and x22 (p1) = p1 + 0.5.

The first order conditions for household h = 1 are given by:

0.5

x11 + x2

1

− λ1p1 = 0.

0.5

x12

− λ1 = 0.

This means that x12 = 0.5

λ1and x1

1 + x21 = 0.5

λ1p1. By market clearing, this implies that

0.5λ1p1

= 3, so λ1 = 16p1. This means that x1

2 = 3p1.

Market clearing requires that x12+x2

2 = 3. From the demand functions, 3p1+p1+0.5 = 3,


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 allo-

cation 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)

= α1 log(xh1)

+ α2 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 first order conditions are given by:

α1

xh1− λhp1 = 0.

α2

xh2− λh = 0.

Solving for xh as a function of λh and plugging into the budget constraint p1

(xh1 − eh1

)+(

xh2 − eh2)

= 0 yields:α1

λh+α2

λh= p1e

h1 + eh2 .

This means that 1λh

= 1(α1+α2)

(p1e

h1 + eh2

). The demand functions are given by:

xh1 (p1) =α1

(α1 + α2)

(p1e

h1 + eh2p1

)and xh2 (p1) =

α2

(α1 + α2)

(p1e

h1 + eh2

).

Market clearing requires that∑h∈H

xh1 (p1) =∑h∈H

eh1 . Given that the coeffi cients are

household independent, then

∑h∈H

xh1 (p1) =α1

p1 (α1 + α2)·(p1

∑h∈H

eh1 +∑h∈H

eh2

).


The market clearing condition∑h∈H

xh1 (p1) =∑h∈H

eh1 is then a linear function of p1,

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 :

α1p1(α1+α2)

·(p1

∑h∈H

eh1 +∑h∈H

eh2

)=∑h∈H

eh1(α1

(α1+α2)· p1

∑h∈H

eh1 + α1(α1+α2)

·∑h∈H

eh2

)= p1

∑h∈H

eh1

p1 =

α1·∑h∈H

eh2

α2·∑h∈H

eh1

.

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(p\G)

=αg∑g∈G αg

(peh

pg

)for g < G and xhG

(p\G)

=αG∑g∈G αg

(peh).

Here we have normalized the price pG = 1 and only consider the prices p\G = (p1, ..., pG−1).

The market clearing conditions for the goods g < G yield equations of the following

form:αg∑g∈G αg

(p∑h∈H

eh

)= pg

∑h∈H

ehg ∀g < G.

These linear equations (G − 1 in total) have a unique solution for p\G. Thus, Cobb-

Douglas utility always leads to a unique equilibrium allocation.

6. Show that the projection π is proper. π is the projection π : RH(G+1)+G−1++ ×RHG++ → RHG++

which maps((xh, λh

)h∈H , p,

(eh)h∈H

)7→(eh)h∈H such thatΦ

((xh, λh

)h∈H , p;

(eh)h∈H

)=

0.

Solution:

A projection is by definition a continuous mapping. Consider a compact set of en-

dowments Θ′. We have left to show that π−1 (Θ′) is a compact set. As consumption

is nonnegative and the endowments∑h∈H

ehg are bounded ∀g ∈ G, then the equilibrium


consumption(xh)h∈H also belong to a compact set. From the first order conditions

with respect to the good G, we have:

DxGuh(xh)− λh = 0.

Recall that we have normalized the price pG = 1. As DxGuh(xh)is bounded (Duh is

continuous), then(λh)h∈H belongs to a compact set. From the first order conditions

with respect to all other goods g < G :

Dxguh(xh)− λhpg = 0 ∀g < G.

As all terms are bounded, then (pg)g<G belongs to a compact set.

Thus, the set π−1 (Θ′) is compact, finishing the argument that π is proper.

7. Prove that for any endowments(eh)h∈H 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,λh)h∈H

,pΦ((xh, λh

)h∈H , p

)has full rank

(square matrix) given that the initial endowment is a Pareto optimal allocation(eh)h∈H =(

xh∗)h∈H . 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,λh)h∈H

,pΦ((xh, λh

)h∈H , p

)has full row rank

when(eh)h∈H =

(xh∗)h∈H , 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)h∈H =

(xh∗)h∈H (in other words, 1

equilibrium). This verifies uniqueness.

The rows of M correspond to the equations in Φ((xh, λh

)h∈H , p

), while the columns

correspond to the variables that we are taking derivatives with respect to. The deriv-


ative matrix is given by:

M =

... ... 0 0 0 0 :

... ... 0 0 0 0 :

0 0 D2uh(xh)

−pT 0 0

(−λhIG−1

0

)0 0 −p 0 0 0

(eh\G − xh\G

)T0 0 0 0 ... ... :

0 0 0 0 ... ... :

... ...(−IG−1 0

)0 ... ... 0

.

To show that the matrix M has full rank, we set νTM = 0 and must verify νT = 0,

where ν ∈ RH(G+1)+G−1 corresponds to equations in Φ :

ν =

:

∆xh

∆λh

:

∆p

:

FOCh

BCh

:

MC

.


:(∆xh

)TD2uh

(xh)−∆λhp−∆pT

[IG−1 0

]= 0 (A.1.a)

−(∆xh

)TpT = 0 (A.1.b)

:

−∑

h∈H λh(

∆xh\G

)T+∑

h∈H ∆λh(eh\G − xh\G

)T= 0 (A.1.c)

. (A.1)

As(eh)h∈H =

(xh∗)h∈H , then the term

∑h∈H ∆λh

(eh\G − xh\G

)T= 0 in (A.1.c). Let’s

show that νT = 0 in three steps:

(a) Postmultiply (A.1.a) by ∆xh :

(∆xh

)TD2uh

(xh)

∆xh −∆λ1p∆xh − (∆p1, ...,∆pG−1, 0) ∆xh = 0.


From (A.1.b), p∆xh = 0. This implies

λh ·(∆xh

)TD2uh

(xh)

∆xh = λh ·∑g<G

∆pg∆xhg , (A.2)

where λh > 0 is multiplied by both sides.

(b) From (A.1.c) : ∑h∈H

λh ·(∆xh1 , ...,∆x

hG−1

)= 0.

(c) Add the equation (A.2) over all households h ∈ H. Using (A.2), this implies

∑h∈H

λh ·(∆xh

)TD2uh

(xh)

∆xh =∑g<G

∆pg

(∑h∈H

λh ·∆xhg)

= 0. (A.3)

As uh is differentiably strictly concave, the Hessian matrix D2uh(xh)is negative

definite. This means λh ·(∆xh

)TD2uh

(xh)

∆xh ≤ 0 ∀h ∈ H, with equality iff(∆xh

)T= 0. (A.3) implies

(∆xh

)h∈H = 0. From (A.1.a),

∆λh (p1, ..., pG−1, 1) + (∆p1, ...,∆pG−1, 0) = 0.

As p >> 0, then(∆λh

)h∈H = 0. Consequently, ∆pT = 0.

Thus, νT =(...,(∆xh

)T,∆λh, ...,∆pT

)= 0, finishing the argument.

8. Using "A Different Application of Differential 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 equationsΦ∗ (η) =

(Φ (η)

p1 − p2

).

As L = J + 1 > J, if I can show that π is proper (same proof as Exercise 6) and

rankDΦ∗ (η) = L ∀η : Φ∗ (η) = 0, then using the result from Exercise 3 in Chapter 1,

over a generic subset of endowments, there does not exist a solution ξ to the system

of equations Φ∗ (ξ, θ) = 0. This implies that any Arrow-Debreu equilibrium ξ (defined

such that Φ (ξ, θ) = 0) must be such that p1 − p2 6= 0.

To show that rankDΦ∗ (η) = 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


respect to the price variables. This shows that the first J equations of Φ∗ are linearly

independent. For the final equation of Φ∗, use the column for the derivatives with

respect to p1 or p2 to show that the final row is linearly independent from the rest.

This verifies that the matrix DΦ∗ (η) 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)h∈H is not Pareto optimal. To attack this problem, use "A Different Application of

Differential Topology" from Exercise 3 in Chapter 1 and consider that a necessary con-

dition for Pareto optimality isDguh(xh)Dg′u

h(xh)=

Dguh′(xh′)

Dg′uh′(xh′)

∀ (h, h′, g, g′) ∈ H×H×G×G.

Solution:

We utilize Assumptions R1-R3. Suppose wlog that the missing market is for good

g = 1. Then in equilibrium, xh1 = eh1 ∀h ∈ H. The equilibrium variables are ξ =((xh, µh, λh

)h∈H , p

)and the equilibrium system of equations is

Φ (η) =

Dx1u

h(xh)− µh − λhp1 = 0(

Dxguh(xh)− λhpg = 0

)1<g<G

DxGuh(xh)− λh = 0

eh1 − xh1p(eh − xh

)

h∈H∑

h∈H

(eh\G − xh\G

)

.

The number of equations in Φ equals the number of variables in ξ. From the first order

conditions,Dx1u

h(xh)Dxgu

h(xh)= µh+λhp1

λhpg∀g > 1. This means that

Dx1uh(xh)

Dxguh(xh)

=Dx1u

h′(xh′)

Dxguh′(xh′)

iff

µh

λh= µh

′

λh′ .

Consider the system of equations Φ∗ (η) =

(Φ (η)

µ1λ2 − µ2λ1

). As L = J + 1 > J,

if I can show that π is proper (same proof as Exercise 6) and rankDΦ∗ (η) = L

∀η : Φ∗ (η) = 0, then using the result from Exercise 3 in Chapter 1, over a generic

subset of endowments, there does not exist a solution ξ to the system of equations

Φ∗ (ξ, θ) = 0. This implies that any equilibrium ξ (defined such that Φ (ξ, θ) = 0) must

be such that µ1λ2 − µ2λ1 6= 0. Without µ1

λ1= µ2

λ2, the equilibrium allocation must be

Pareto suboptimal.


The rows of M = DΦ∗ (η) correspond to the equations in Φ∗, 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

νTM = 0 and must verify νT = 0, where ν ∈ RH(G+2)+G−1 corresponds to equations in

Φ :

ν =

:

∆xh

∆µh

∆λh

:

∆p

∆EE

:

FOCh

eh1 − xh1BCh

:

MC

µ1λ2 − µ2λ1

.

Given that xh1 = eh1 , the budget constraint is written only for the remaining G − 1

goods.

From the columns corresponding to derivatives with respect to (e1, e2) , we obtain (as

in Section 2.5)(∆µ1,∆λ1

)=(∆µ2,∆λ2

)= 0 and ∆pT = 0. For all households h ∈ H,

the columns corresponding to derivatives with respect to xh imply:

(∆xh

)TD2uh

(xh)

+(∆µh,∆λh

)( −1−→0

−p

)= 0. (A.4)

For the households h > 2, the columns corresponding to derivatives with respect to(µh, λh

)imply: (

∆xh)T ( −1

−→0

−p

)T

= 0.

The previous two equations imply that(∆xh

)TD2uh

(xh)

∆xh = 0 ∀h ∈ H, and thus(∆xh

)T= 0 ∀h ∈ H using the assumption that uh is differentiably strictly concave.

From (A.4),(∆µh,∆λh

)= 0 ∀h ∈ H.

For the column for the derivatives with respect to µ1, then ∆EE = 0 (as λ2 > 0).

Thus, νT = 0, finishing the argument.

A.3. GENERAL FINANCIAL MODEL 111

A.3 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 financial equilibrium does not exist.

Solution:

Assume that No Arbitrage does not hold. Then ∃zh such that(−qR

)zh > 0. Such

a portfolio zh ∈ RJ and there always exists a portfolio zh = κ · zh for κ > 1 such that(−qR

)zh >

(−qR

)zh. As the asset set RJ is unbounded, then there will never

exist a solution to the household problem (HP ).


R =

2 3

1 1

0 1

3 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 iff 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) .



R =

1 0 0

0 1 123

1 13

12

23

12

.

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:

We need to find a linear operation for the matrix

(−qR

)such that the new payout

matrix has the risk-free payout of 1 in all states. To do this, simply replace Column 1

of

(−qR

)with the sum of Columns 1 and 3:

(−q∗

R∗

)=

−q1 − q3 −q2 −q3

1 0 0

1 1 1

1 1 13

1 23

12

.

We are now ready to plot the points (r∗2(1), r∗3(1)) = (0, 0) , (r∗2(2), r∗3(2)) = (1, 1) ,

(r∗2(3), r∗3(3)) =(1, 1

3

), and (r∗2(4), r∗3(4)) =

(23, 1

2

). Figure A.5 shows these plotted

points and the convex hull that they generate. The prices q satisfy No Arbitrage iff(q∗2q∗1,q∗3q∗1

)=(

q2q1+q3

, q3q1+q3

)lies in the interior of the convex hull. For the three asset

prices in the question, we have (i)(

q2q1+q3

, q3q1+q3

)=(

43, 2

3

), (ii)

(q2

q1+q3, q3q1+q3

)=(

13, 1

3

),

and (iii)(

q2q1+q3

, q3q1+q3

)=(

12, 1

4

).

As indicated in Figure A.5, the pair(

43, 2

3

)lies outside the convex hull, the pair

(13, 1

3

)lies on the boundary of the convex hull, and the pair

(12, 1

4

)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, λh, zh

)h∈H , p, q

)is a general financial equilibrium in which No Redundancy does

not hold, then there exists((zh)h∈H , q

)such that

((xh, λh, zh

)h∈H , p, q

)is a general


financial 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 financial equilibrium is((xh, λh, zh

)h∈H , p, q

). From the first order

conditions with respect to zh, then λh(−qR

)= 0 ∀h ∈ H. As R does not satisfy No

Redundancy, then rankR = J∗ < J.Without loss of generality, assume that the first J∗

columns of R are linearly independent, so we can write R =[R∗ | R∗∗

]such that R∗

is an S×J∗ submatrix with rankR∗ = J∗. I can define asset holdings zh =(zhj)j=1,...,J∗

such that

(−qR

)zh =

(−qR∗

)zh. Each of the columns in R∗∗ is a linear combination

of the columns in R∗, meaning that rj = R∗θ for some θ =

θ1

:

θj∗

∈ Rj∗ and ∀j > j∗,

where rj =

rj(1)

:

rj(S)

. Thus, define the new asset holdings as:zhj = zhj + θj

∑k>j∗

zhk .

Since

(−qR

)zh =

(−qR∗

)zh, the real variables

((xh, λh

)h∈H , p

)remain unchanged.

The new asset prices satisfy qj = qj ∀j ≤ j∗ as λh(−qR∗

)= 0 ∀h ∈ H.

Thus, for the payout matrix R∗ (satisfying No Redundancy as R∗ has full column

rank),((xh, λh, zh

)h∈H , p, q


5. Prove Theorem 3.4.

Solution:

This proof will use "A Different Application of Differential Topology" from Exercise 3

in Chapter 1. That is, we will write down a system of equations that must be satisfied

by a general financial 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) π is


proper and (ii) rankDΦ (η) = L ∀η ∈M, then we can apply "A Different Application

of Differential Topology" conclusion from Exercise 3 to conclude that over a generic

subset of endowments, there does not exist variables ξ =((xh, λh, zh

)h∈H , p, q

)such

that((xh, λh, zh

)h∈H , p, q

)is a general financial equilibrium and

(xh)h∈H is a Pareto

optimal allocation.

The condition "π is proper" can be verified 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 suffi cient condition for Pareto optimality

isD(l,s)u

h(xh)D(l′,s′)u

h(xh)=

D(l,s)uh′(xh′)

D(l′,s′)uh′(xh′)

for any pair of households (h, h′) and any commodity-

state pairs (l, s). Using the first order condition with respect to consumption, this

condition is equivalent to (λh(1),...,λh(S))λh(0)

=

(λh′(1),...,λh

′(S)

)λh′(0)

for any pair of households

(h, h′), that is, the Lagrange multipliers are proportional.


(RG+S+1

++ × RJ)× RG−(S+1)

++ × RJ++ → Rn+1 as:

Φ((xh, λh, zh

)h∈H , p, q

)=

[Duh

(xh)− λhP

]TP(eh − xh

)+

(−qR

)zh[

λh

(−qR

)]T

h∈H∑

h∈H

(eh\G − xh\G

)∑h∈H

zh

λ1(1)λ2(0)− λ1(0)λ2(1)

,

where n = H (G+ S + 1 + J) +G− (S+ 1) +J. The additional equation λ1(1)λ2(0)−λ1(0)λ2(1) = 0 is a necessary condition for Pareto optimality.

I will show that rankDΦ((xh, λh, zh

)h∈H , p, q

)= n+1, taking derivatives with respect

to the variables(xh, λh, zh

)h∈H and the endowment e

1. By assumption, J < S. This

means that R has full column rank J (by No Redundancy) and ∃s > 0 such that the

submatrix R (\s) defined by removing the row for state s also has full column rank

J. Without loss of generality, let s = 1. This means that R (\1) =

r(2)

:

r(S)

has full


column rank J.

Premultiply the derivative matrix DΦ((xh, λh, zh

)h∈H , p, q

)by

νT =(...,((

∆xh)T,(∆λh

)T,(∆zh

)T), ...,∆pT ,∆qT ,∆EE

),

where ∆EE ∈ R corresponds to the extra equation λ1(1)λ2(0)− λ1(0)λ2(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 λ1 are given by:

(∆z1

)TΨT + ∆EE

(−λ2(1), λ2(0), 0, ..., 0

)= 0 (A.5)(

∆z1)T (−qT r(1)T R(\1)T

)+ ∆EE

(−λ2(1), λ2(0), 0, ..., 0

)= 0

As Ψ =

−qr(1)

R(\1)

and R(\1) has full column rank, then (A.5) implies that (∆z1)T

=

0 and ∆EE = 0 (as both λ2(0) > 0 and λ2(1) > 0). The rest of the proof remains

unchanged, ultimately resulting in νT = 0. Thus DΦ((xh, λh, zh

)h∈H , p, q

)has full

row rank n+ 1, finishing the argument.

6. In Example 3.1, verify that the Arrow-Debreu equilibrium allocation is such that ∀h ∈H : xh (s) = θh

∑h∈H

eh (s) for some θh.

Solution:

With utility function utility uh(xh)

= −∑

s∈S

(xh (s)

)−γ, the first order conditions

with respect to consumption are given by:

γ(xh (s)

)−(γ+1) − λhρ(s) = 0 ∀s ∈ S. (A.6)

Recall that in an Arrow-Debreu equilibrium, there is a single budget constraint and

thus a single Lagrange multiplier λh > 0. I denote the Arrow-Debreu prices as (ρ(s))s∈Ssuch that ρ(0) = 1. Equation (A.6) yields:

xh (s) =

(λh

γ

)− 1γ+1

ρ(s)−1

γ+1 . (A.7)


From the Arrow-Debreu budget constraint:∑s∈S

ρ(s)xh (s) =∑s∈S

ρ(s)eh (s) .

Using the expression for xh(s) from (A.7), the budget constraint becomes:

(λh

γ

)− 1γ+1 ∑

s∈S

ρ(s)γγ+1 =

∑s∈S

ρ(s)eh (s) .

This implies that (λh

γ

)− 1γ+1

=

∑s∈S

ρ(s)eh (s)∑s∈S

ρ(s)γγ+1

∀h ∈ H. (A.8)

The market clearing condition∑h∈H

xh (s) =∑h∈H

eh (s) for any state s ∈ S implies

ρ(s)−1

γ+1∑s∈S

ρ(s)γγ+1

∑s∈S

ρ(s)∑h∈H

eh (s) =∑h∈H

eh (s) ∀s ∈ S.

Solving for ρ(s)−1

γ+1 yields

ρ(s)−1

γ+1 =

(∑s∈S

ρ(s)γγ+1

)(∑h∈H

eh (s)

)∑s∈S

ρ(s)∑h∈H

eh (s)∀s ∈ S. (A.9)

Define θh =

∑s∈S

ρ(s)eh(s)∑s∈S

ρ(s)∑h∈H

eh(s)

∀h ∈ H. Notice that∑h∈H

θh = 1, by definition. Insert (A.8)

and (A.9) into the expression for xh(s) from (A.7) yields:

xh (s) = θh

(∑h∈H

eh (s)

).

This completes the argument.


7. In Example 3.3, verify that the general financial equilibrium consumption is given by:

xhl (s) = αh (s) ·∑

h∈Hehl (s)

for some αh (s) ∈ (0, 1) and the prices are given such that pl (s) = θl(s)θL(s)

∀(l, s) ∈ L× S.

Solution:

With utility function utility uh(xh)

=∑

s∈S

(∑l∈L θl (s) · log

(xhl (s)

)), the first order

conditions with respect to consumption are given by:

θl (s)

xhl (s)− λh(s)pl(s) = 0 ∀ (l, s) ∈ L× S.

Solving for xhl (s) yields

xhl (s) =θl (s)

λh(s)pl(s)∀ (l, s) ∈ L× S. (A.10)

The budget constraint for any household in any state s ∈ S is given by:∑l∈L

pl(s)xhl (s) =

∑l∈L

pl(s)ehl (s) .

Using the expression for xhl (s) from (A.10) yields:

1

λh(s)

∑l∈L

θl (s) =∑l∈L

pl(s)ehl (s) .

Thus, the expression for xhl (s) can be updated as:

xhl (s) =

θl (s)∑l∈L

θl (s)

∑l∈L

pl(s)ehl (s)

pl(s)∀ (l, s) ∈ L× S. (A.11)

Consider the market clearing conditions∑h∈H

xhl (s) =∑h∈H

ehl (s) for any (l, s) ∈ L× S :

1

pl(s)

θl (s)∑l∈L

θl (s)

∑l∈L

pl(s)∑h∈H

ehl (s) =∑h∈H

ehl (s) .


This implies that pl(s) can be expressed as:

pl(s) =

θl (s)∑l∈L

θl (s)

∑l∈L

pl(s)∑h∈H

ehl (s)∑h∈H

ehl (s). (A.12)

Given the price normalization pL(s) = 1 ∀s ∈ S, then

1

θL (s)=

1∑l∈L

θl (s)

∑l∈L

pl(s)∑h∈H

ehl (s)∑h∈H

ehL (s),

by definition. For any (l, s) ∈ L× S, equation (A.12) verifies that

pl (s) =θl (s)

θL (s)·∑

h∈H ehL (s)∑

h∈H ehl (s)

.

Using the expression (A.12) for pl(s), then the expression (A.11) for xhl (s) can be

updated as:

xhl (s) =

∑l∈L

pl(s)ehl (s)∑

l∈L

pl(s)∑h∈H

ehl (s)

(∑h∈H

ehl (s)

)∀ (l, s) ∈ L× S.

Define αh (s) =

∑l∈L

pl(s)ehl (s)∑

l∈L

pl(s)∑h∈H

ehl (s)

and notice that∑h∈H

αh (s) = 1 ∀s ∈ S, by definition.

This completes the second part of the claim that xhl (s) = αh (s)·∑

h∈Hehl (s) ∀ (l, s) ∈

L× S.

8. In Example 3.4, verify that the Arrow-Debreu equilibrium allocation and commodity

prices in Case I are given by:

x1 (1) =(

83, 1

3

)x1 (2) =

(83, 1

3

)x2 (1) =

(13, 8

3

)x2 (2) =

(13, 8

3

)ρ (1) = (1, 1) ρ (2) = (1, 1)

.


Solution:

To solve for the Arrow-Debreu equilibrium, I make the price normalization ρ2 (2) = 1.

The equilibrium variables are (x1 (1) , x1 (2) , x2 (1) , x2 (2)) (8 consumption variables),(λ1, λ2

)(2 Lagrange multipliers for the 2 budget constraints, 1 for each household),

and (ρ1(1), ρ2(1), ρ1(2)) (3 price variables). The equations are the first order conditions

with respect to consumption (8 of these), the 2 budget constraints ρ (1) ·xh (1) +ρ (2) ·xh (2) = ρ (1) · eh (1) + ρ (2) · eh (2) for h = 1, 2, and the 3 market clearing conditions

x11 (1) + x2

1 (1) = 3, x12 (1) + x2

2 (1) = 3, and x11 (2) + x2

1 (2) = 3.


20.5 (x11 (1))

−0.5 − λ1ρ1(1) = 0 2−1 (x12 (1))

−0.5 − λ1ρ2(1) = 0

20.5 (x11 (2))

−0.5 − λ1ρ1(2) = 0 2−1 (x12 (2))

−0.5 − λ1 = 0.

The expressions for consumption are given by:

x11 (1) = 2

(λ1ρ1(1))2 x1

2 (1) = 1

4(λ1ρ2(1))2

x11 (2) = 2

(λ1ρ1(2))2 x1

2 (2) = 1

4(λ1)2

. (A.13)

Inserting these expressions into the budget constraint yields:(1

λ1

)2(2

ρ1(1)+

1

4ρ2(1)+

2

ρ1(2)+

1

4

)=

5

2ρ1(1) +

50

21ρ2(1) +

13

21ρ1(2) +

1

2.

The expressions for consumption are then given by:

x11 (1) =

2

ρ1(1)2

(52ρ1(1) + 50

21ρ2(1) + 13

21ρ1(2) + 1

22

ρ1(1)+ 1

4ρ2(1)+ 2

ρ1(2)+ 1

4

)(A.14)

x12 (1) =

1

4 (ρ2(1))2

(52ρ1(1) + 50

21ρ2(1) + 13

21ρ1(2) + 1

22

ρ1(1)+ 1

4ρ2(1)+ 2

ρ1(2)+ 1

4

)

x11 (2) =

2

ρ1(2)2

(52ρ1(1) + 50

21ρ2(1) + 13

21ρ1(2) + 1

22

ρ1(1)+ 1

4ρ2(1)+ 2

ρ1(2)+ 1

4

)

x12 (2) =

1

4

(52ρ1(1) + 50

21ρ2(1) + 13

21ρ1(2) + 1

22

ρ1(1)+ 1

4ρ2(1)+ 2

ρ1(2)+ 1

4

)

Let’s make use of the symmetry assumed in this economy. The commodity (l, s) = (1, 1)


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 ρ (1) = ρ(2) = (ρ, 1)

and we only need to consider the market clearing condition for the commodity (l, s) =

(1, 1).

The market clearing condition x11 (1) + x2

1 (1) = 3 is then given by:

2

ρ2

(52ρ+ 50

21+ 13

21ρ+ 1

22ρ

+ 14

+ 2ρ

+ 14

)+

1

2ρ2

(12ρ+ 13

21+ 50

21ρ+ 5

214ρ

+ 2 + 14ρ

+ 2

)= 3.

Distributing the 2 in the first term and the 12in the second term yields:

1

ρ2

(52ρ+ 50

21+ 13

21ρ+ 1

21ρ

+ 18

+ 1ρ

+ 18

)+

1

ρ2

(12ρ+ 13

21+ 50

21ρ+ 5

212ρ

+ 4 + 12ρ

+ 4

)= 3. (A.15)

The value of ρ = 1 is a solution to (A.15). With ρ = 1, then(

1λ1

)2= 6

4.5= 4

3and from

(A.13):x1

1 (1) = 83

x12 (1) = 1

3

x11 (2) = 8

3x1

2 (2) = 13

.

The consumptions for household h = 2 are found using the market clearing conditions.

9. In Example 3.4, verify that the general financial equilibrium allocation and commodity

prices in Case II are given by:

x1 (1) =(

6221, 31

21

)x1 (2) =

(3221, 1

21

)x2 (1) =

(121, 32

21

)x2 (2) =

(3121, 62

21

)p (1) = (2, 1) p (2) =

(12, 1) .

Solution:

To solve for the general financial 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),(λ1, λ2

)(2 Lagrange multipliers for the 2

budget constraints, 1 for each household), and p1 (1) (1 price variable). The equations

are the first order conditions with respect to consumption (4 of these), the 2 budget


constraints p (1) · xh (1) = p (1) · eh (1) for h = 1, 2, and 1 market clearing condition

x11 (1) + x2

1 (1) = 3.


20.5 (x11 (1))

−0.5 − λ1p1 (1) = 0 2−1 (x12 (1))

−0.5 − λ1 = 0 .

The expressions for consumption are given by:

x11 (1) = 2

(λ1p1(1))2 x1

2 (1) = 1

4(λ1)2 . (A.16)

Inserting these expressions into the budget constraint yields:(1

λ1

)2(2

p1 (1)+

1

4

)=

5

2p1 (1) +

50

21.

The expressions for consumption are then given by:

x11 (1) =

2

p1 (1)2

(52p1 (1) + 50

212

p1(1)+ 1

4

)

x12 (1) =

1

4

(52p1 (1) + 50

212

p1(1)+ 1

4

)

Similarly, the expressions for consumption of household h = 2 can be found as:

x21 (1) =

1

4p1 (1)2

(12p1 (1) + 13

211

4p1(1)+ 2

)

x12 (1) = 2

(12p1 (1) + 13

211

4p1(1)+ 2

)

The market clearing condition x11 (1) + x2

1 (1) = 3 is then given by:

2

p1 (1)2

(52p1 (1) + 50

212

p1(1)+ 1

4

)+

1

4p1 (1)2

(12p1 (1) + 13

211

4p1(1)+ 2

)= 3.

The value p1 (1) = 2 is the solution to this equation. With p1 (1) = 2, then(

1λ1

)2= 124

21


and from (A.16):

x11 (1) = 124

212

(2)2= 62

21x1

2 (1) = 12421

14

= 3121 .

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

is now p1(2)p2(2)

= p2(1)p1(1)

= 12.

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)h∈J are linearly independent iff the portfolio vectors

(zh)h∈J


Solution:

The vectors(ah)h∈J are linearly independent provided that the matrix

(a1, ..., aJ

)has

full column rank. By definition, ah = [ν]Rzh ∈ RS, so(a1, ..., aJ

)= [ν]R

(z1, ..., zJ

).

The matrix [ν]R ∈ RS has full column rank by definition. Thus, [ν]R(z1, ..., zJ

)has

full rank iff(z1, ..., zJ

)has full rank.

2. Write down the system of equilibrium equationsΦ : Ξ×Θ→ Rn for n = H (G+ S + 1 + J)+

G + J so that ξ =((xh, λh, zh

)h∈H , p, q

)is a financial equilibrium with money iff

Φ (ξ) = 0. Here θ =((eh)h∈H ,M

)∈ Θ.

Solution:

The price vectors are now p(s) = (p1(s), ..., pL(s)) ∈ RL++. There are no price normal-

izations.

A.4. INCOMPLETE MARKETS AND MONEY 123


(RG+S+1

++ × RJ)× RG++ × RJ++ → Rn as:

Φ((xh, λh, zh

)h∈H , p, q

)=

[Duh

(xh)− λhP

]TP(eh − xh

)+

(−qR

)zh[

λh

(−qR

)]T

h∈H∑

h∈H

(eh − xh

)∑h∈H

zh(p (s)

∑h∈H

xh (s)−Ms

)s∈S

.

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)h∈H and money

suppliesM = (M0,M1, ...,MS) , the financial equilibria with money satisfy Finite Local

Uniqueness. This requires proving that (i) π is proper and (ii) if Φ (ξ, θ) = 0, then

rankDΦ (ξ, θ) = n (the derivative is taken with respect to the variables ξ and the

parameters θ =((eh)h∈H ,M

)). The proof of (i) is straightforward, so only prove (ii).

Solution:

We are tasked with showing rankDΦ (ξ, θ) = n, i.e., that DΦ (ξ, θ) has full row rank.

The easiest way is to first consider the columns of DΦ (ξ, θ) corresponding to the deriv-

atives with respect to the money supplies M = (M0,M1, ...,MS) . These parameters

only appear in one equation each, p (s)∑

h∈Hxh (s)−Ms = 0. Thus, these final S + 1

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 rankDΦ∗ (ξ∗, θ) = n+ J + 1, where

Φ∗ (η) =

Φ (η)

µ ·(z1, ..., zJ

)µTµ/2− 1

(again taking derivatives with respect to the variables

ξ∗ =((xh, λh, zh

)h∈H , p, q, µ

)∈ Rn+J and the parameters θ =

((eh)h∈H ,M

)). The

additional equations µ ·(z1, ..., zJ

), where µ ∈ RJ\0 (nonzero from µTµ/2 = 1),

imply that the portfolios vectors(zh)h∈J are linearly dependent. Using "A Different


Application of Differential Topology" from Exercise 3 in Chapter 1, we can then con-

clude that for a generic selection of θ =((eh)h∈H ,M

), the portfolio vectors

(zh)h∈J


Solution:

We are tasked with showing rankDΦ∗ (ξ∗, θ) = n+J + 1, i.e., that DΦ∗ (ξ∗, θ) has full

row rank. The easiest way is to first consider the columns of DΦ∗ (ξ∗, θ) corresponding

to the derivatives with respect to the money supplies M = (M0,M1, ...,MS) . These

parameters only appear in one equation each, p (s)∑

h∈Hxh (s)−Ms = 0. Thus, these

final S + 1 equations are linearly independent from all other equations.

Set νTDΦ∗ (ξ∗, θ) = 0, where ν ∈ Rn+J+1 corresponds to equations in Φ∗ :

ν =

:

∆xh

∆λh

∆zh

:

∆p

∆q

∆M

∆µ

∆EE

:

FOCxh

BCh

FOCzh

:

MCx

MCz(p (s)

∑h∈H

xh (s)−Ms

)s∈S

µ ·(z1, ..., zJ

)µTµ/2− 1

.

Full row rank is verified upon showing that ν = 0. The preceding paragraph specifies

that ∆M = 0.

Recall Steps 1-3 in the proof of Theorem 3.2. These steps are used to show that((∆x1)

T,(∆λ1

)T, (∆z1)

T,∆pT ,∆qT

)= 0, by using the columns of DΦ∗ (ξ∗, θ) corre-

sponding to the derivatives with respect to(x1, λ1, z1, e1

). For households h > 1, using

the columns corresponding to derivatives with respect to(xh, λh, eh

), the same Steps

1-3 can be used to show that((

∆xh)T,(∆λh

)T,(∆zh

)T)= 0 ∀h > 1.

The vector µ 6= 0, so wlog suppose that µ1 6= 0. Then the columns corresponding to

derivatives with respect to(z1

1 , ..., zJ1

)yield:

∆µj · µ1 = 0 ∀j = 1, ...J.

A.4. INCOMPLETE MARKETS AND MONEY 125

Thus, ∆µT = 0. Derivatives with respect to µ then imply that ∆EE · µ = 0, so

∆EE = 0 (as µ 6= 0).

All told, ν = 0, finishing the proof that rankDΦ∗ (ξ∗, θ) = n+ J + 1.

General Equilibrium Models1General Equilibrium Models1 Matthew Hoelle Spring 2014 1This manuscript...

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