Business School
WORKING PAPER SERIES
Working Paper
2014-122
Arbitrage and asset market equilibrium
in finite dimensional economies with
short-selling and risk-averse expected
utilities
Thai Ha-Huy
Cuong Le Van
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Arbitrage and asset market equilibrium in finite
dimensional economies with short-selling and
risk-averse expected utilities
Thai Ha-Huy∗, Cuong Le Van†
February 21, 2014
Abstract
We consider a model with an finite number of states of nature where
short sells are allowed.
Keywords: asset market equilibrium, individually rational attainable allocations, individually rational utility set, no-arbitrage prices, no-arbitrage
condition.
JEL Classification: C62, D50, D81,D84,G1.
∗
University of Evry-Val-D’Essonne. E-mail: [email protected]
IPAG Business School, CNRS, Paris School of Economics, VCREME. E-mail: [email protected]
†
1
1
Introduction
Various conditions have been given for the existence of a general equilibrium in
an exchange economy. There are different approaches. We can classify them in
three main categories.
The first category based on conditions on net trade, for example Hart (1974),
Page (1987), Nielsen (1989, Page and Wooders (1996), Allouch (1999), Page,
Wooders and Monteiro (2000). We define Individual arbitrage opportunity as
the set of directions which agent wants to trade infinitely large quantities. Hart
(1974) is the first who considers the problem of existence of general equilibrium
of assets market with short-sales. In the case the agents disagree too much,
some agents can make an arbitrage, which is ”An opportunity with a mutually compatible set of net trades which are utility non-decreasing and, at most,
costless to make.” Taking the fact that this opportunity can be repeated indefinitely, equilibrium may do not exist. The Weak-no-market-arbitrage (WNMA)
requires that that all mutually compatible net trades which are non-decreasing
be useless. Page (1987) proposes the no-unbounded-arbitrage (NUBA), a situation in which no group of agents can make mutually compatible, unbounded
and utility increasing trades. In 2000, Page Wooders and Monteiro, use Inconsequential arbitrage condition to ensure the existence of an equilibrium.
The second category is based on conditions on prices, for example Green
(1973), Grandmont (1972, 1977), Hammond (1983) and Werner (1987) . These
authors define Non-arbitrage-price as the strictly positive dual of the set of
useful vectors. If the intersection of No-arbitrage-price sets of all the agents is
non empty (existence of No-arbitrage-price-system NAPS ), then there exists a
general equilibrium.
Other authors, like Brown and Werner (1995), Dana, Le Van, Magnien
(1999) propose, as no-arbitrage condition, the compactness of attainable utility
set in order to ensure the existence of an equilibrium.
When the utility functions are quasi-concave, Allouch, Le Van and Page
(2002) , Ha-Huy and Nguyen (2009) by different approaches, prove the equivalence between the existence of No-arbitrage-price-system (NAPS) and NUBA
or WNMA. Allouch, Le Van and Page(2002) extend the definition of no arbitrage price system (NAPS) introduced by Werner (1987) to the case where some
agent in the economy has only useless vectors. They show that an extension of
NAPS condition of Werner (1987) is actually equivalent to the weak no market
arbitrage (WNMA) condition introduced by Hart (1974). They mention that
this result implies the one given by Page and Wooders (1996) who prove that
no unbounded arbitrage (NUBA) condition, a special case of Page (1987), is
equivalent to NAPS when no agent has non-null useless vectors. The first part
2
of this paper is to show that when the statement NUBA implies NAPS (Page
and Wooders, 1996) is true then we have WNMA implies NAPS (Allouch, Le
Van and Page 2002). But it is obvious that if the second statement holds then
the first one also holds. The novelty of the result of this note is that the results
are self-contained. While Allouch, Le Van and Page (2002) prove WNMA implies NAPS by using a difficult Lemma in Rockafellar (1970) and then deduce
that NUBA implies NAPS when no agent has non-null useless vectors, we show
that these conditions are actually circular. These conditions let us think of the
circularty of Brouwer and Kakutani fixed-point theorems. Moreover, the proofs
are simple and elementary.
In 2010, Dana and Le Van , consider the relationships between the agents
beliefs and risk where there is ambiguity, and propose to use the set of derivatives of utility function as no-arbitrage-prices set. With their definition of noarbitrage prices, they give an equivalence between the non-emptiness of the intersection of the interiors of no-arbitrage-price sets and NUBA condition. More
generally, they prove an equivalence between the non-emptiness of the intersection of the relative interior of no-arbitrage-price sets and WNMA condition.
Hence, if this intersection is non empty, the existence of general equilibrium is
ensured.
We can prove that the existence of equilibrium implies the non emptiness
of the intersection of weak no-arbitrage prices sets (with the definition given by
Dana and Le Van ), but does not imply the non-emptyness of the intersection of
the interiors of no-arbitrage sets. Actually, NUBA and WNMA conditions are
quite strong. In this paper, we will give an example where NUBA and WNMA
are violated, and, however, an equilibrium exists.
We give in this paper conditions based on the closedness of the gradients set and
on the existence of maximal value on affine subspace. Under realistic conditions,
the existence of equilibrium is equivalent with the existence of a common weak
no-arbitrage price.
In this paper, we reconsider the equilibrium theory of assets with shortselling when there is risk and ambiguity. Using the notion of weak no-arbitrage
price proposed by Dana and Le Van, we prove the equivalence between existence
of general equilibrium and non-emptiness of intersection of no-arbitrage-price
cones.
Our paper is organized as follows. The Section 2 gives the circular results
concerning NUBA, WNMA and NAPS. The Section 3 presents important results in the literature when utility functions are concave. The Sections 4 and 5
treat the problem of general equilibrium in the model which has utility functions
satisfies no half-line condition, closed gradient condition or maximal condition.
In the Section 5, we consider the problem of the existence of equilibrium in a
3
model with risk and ambiguity.
2
The model
We have an economy E with m agents. Each agent is characterized by a consumption set X i ⊂ RS , an endowment ei and a utility function U i : RS → RS .
We impose those conditions on utility functions:
Definition 1 An equilibrium is a list (x∗i )i=1,...,m , p∗ ) such that x∗i ∈ X i for
every i and p∗ ∈ RS+ \ {0} and
(a) For any i, U i (x) > U i (x∗i ) ⇒ p · x > p · ei
P
Pm i
∗i
(b) m
i=1 x =
i=1 e .
Definition 2 A quasi-equilibrium is is a list (x∗i )i=1,...,m , p∗ ) such that x∗i ∈
X i for every i and p∗ ∈ RS+ \ {0} and
(a) For any i, U i (x) > U i (x∗i ) ⇒ p · x ≥ p · ei
Pm i
P
∗i
(b) m
i=1 e .
i=1 x =
Since short-sales are allowed, we have here the result of Geistdorfer-Florenzano
(1982):
Proposition 1 If the initial endownments belong to the interior of consumption, then any quasi-equilibrium is an equilibrium.
1. The individually rational attainable allocations set A is de-
Definition 3
fined by
i
S m
A = {(x ) ∈ (R ) |
m
X
i=1
i
x =
m
X
ei and U i (xi ) ≥ U i (ei ) for all i}.
i=1
2. The individually rational utility set U is defined by
U = {(v 1 , v 2 , ..., v m ) ∈ Rm | ∃ x ∈ A such that U i (ei ) ≤ v i ≤ U i (xi ) for all i}.
Denote by Ri the set of useful vectors and by Li the set of useless vectors.
By definition, the set of useless vectors of agent i is the biggest linear sub space
included in Ri :
Li = Ri ∩ (−Ri ).
We have Ri is the positive dual cone of P i , i.e. Ri = −(P i )o . Observe that
Ri has no empty interior since RS+ ⊆ Ri .
Definition 4 The vector p ∈ RS is a no-arbitrage price of agent i if for any
w ∈ Ri \ Li we have p · w > 0, and for w ∈ Li , p · w = 0.
4
Denote by S i the set of no-arbitrage prices of agent i. In [8], Dana and
Le Van propose to use the derivative of utility function like no-arbitrage price.
Their idea comes from the next Lemma, which gives us an important property
of useful vector.
Lemma 1 The vector w is useful vector for agent i if and only if for all x, we
have: U i0 (x) · w ≥ 0.
Proof : (⇒): From the concavity of U we have:
0 ≥ U i (x) − U i (x + λw) ≥ U i0 (x) · (−λw) for all λ ≥ 0.
Hence U i0 (x) · w ≥ 0.
(⇐): Observe that for all λ ≥ 0:
U i (x + λw) − U i (x) ≥ U i0 (x + λw) · (λw) ≥ 0.
Hence U i (xi + λw) ≥ U i (x) ∀ x.
From Lemma 1, we now define the set of weak no-arbitrage prices for agent
i which is the cone
P i := {p ∈ RS : ∃ x ∈ RS , λ > 0 such that p = λU i0 (x)}.
Evidently, the set of no-arbitrage prices is included in the convex hull of
the set of weak no-arbitrage prices: S i ⊂ convexP i . We have also S i =
ri(convexP i ), the relative interior of P i .
Hart [?] proposed the WNMA condition, and Page [?] proposed the NUBA
condition, which ensure the existence of equilibrium. The definitions of which
are presented in the Section 2. Observe that in the case there is no half-line in
indifferent surfaces of utility functions, NUBA and WNMA are equivalents.
The Inconsequential arbitrage is proposed by Page, Monteiro and Wooders
in [19].
Definition 5 The economy satisfies Inconsequential arbitrage ( IC) condition
P
if for all (w1 , w2 , . . . , wm ) with wi ∈ Ri and i wi = 0, sequence {xn } ⊂ A,
positif sequence λn → 0 such that limn λn xin = wi , there exists > 0 such that
for N big enough we have U i (xin − wi ) ≥ U i (xin ), ∀n ≥ N .
We present here the Theorem of Page, Wooders and Monteiro in [19]. Their
proof are applied for the case the utility functions are quasi-concaves. In this
paper, using the concavity of utility functions, we can give a more simple proof.
Theorem 1 Suppose that the economy E satisfies IC condition. The attainable
utility set U is compact.
5
Proof : Assume that (v 1 , v 2 , . . . , v m ) belongs to the boundary of U. For all
n ∈ N, define the closed set
An = {x ∈ A such that U i (xi ) ≥ v i −
1
for all i}.
n
By definition of v, the set An is no empty for all n. For each n, take xn ∈ An
which has property:
m
m
X
X
kxin k = min
kxi k.
i=1
x∈An
i=1
P
We will prove that {xn } is bounded. Indeed, suppose the contrary, limn i kxin k
−1
P
i
converges to infinity. Define λn =
. Without loss of generality,
i kxn k
i
i
i
assume that for all i, limn λn xn = w . Since U (xin ) ≥ v i − 1, by the concavity
of U i , we have wi is a useful vector of agent i.
Since the economy satisfies IC condition, we can take 0 < < 1 such that
for n big enough, we have U i (xin − wi ) ≥ U i (xin ) ≥ v i − n1 for all i. For
n big enough, we have for all s ∈ S+ (wi ), 0 < xin,s − wsi < xin,s and for
s ∈ S− (wi ), 0 > xin,s − wsi > xin,s . That implies if wi 6= 0, kxin − wi k < kxin k
P
for n sufficiently big. Since the sum i kwi k = 1, there exists wi 6= 0. Hence
P
P
i
i
i
i kxn k: a contradiction with the definition of xn .
i kxn − w k <
The set {xn } is bounded, so we can extract a sub sequence {xnk } which
P i
P i
i i
i
converges to x. Easily, we can verify that
i e and U (x ) ≥ v .
ix =
Hence v ∈ U.
Theorem 2 The compactness of U implies the existence of equilibrium.
Proof : See Dana, Le Van and Magnien in [?].
With the Theorem 1 and the Theorem 2, we have a direct proof for the
existence of equilibrium under NUBA condition or WNMA condition.
Theorem 3 Assume that the economy E satisfies NUBA condition or WNMA
condition. Then the attainable utility set U is compact and this implies the
existence of equilibrium.
Proof : We can verify easily that NUBA or WNMA imply IC condition. By the
Theorem 1, we have U is compact. Using the Theorem 2, we have the existence
of equilibrium.
In the Theorem 4, we see that the existence of equilibrium implies the
existence of common weak no-arbitrage price.
6
Theorem 4
Equilibrium exists ⇒
m
\
P i 6= ∅.
i=1
Proof : Denote by (p∗ , (x∗i )i ) the equilibrium. The equilibrium allocation (x∗i )
solves the problems:
max U i (xi )
s.t. p∗ · xi = p∗ · ei .
From Theorem V.3.1, page 91, in Arrow-Hurwicz-Uzawa (1958) [2], for any i,
there exists ζi > 0 s.t.
U i (x∗i ) − ζi p∗ · x∗i ≥ U i (xi ) − ζi p∗ · xi
for any xi ∈ RS . Hence U i0 (x∗i ) = ζi p∗ , ∀ i. Let λi =
λi U i0 (x∗i ) ∈ P i for all i. Hence ∩i P i 6= ∅.
3
1
ζi ,
we have p∗ =
The closed gradient condition and the maximal
condition
In this section, if there is not confusion, we will use U (x) instead of U i (x).
Firstly, we present some preparation results about limits of utility function
along a useful direction.
Definition 6 Given a concave function U , and w ∈ R. Define V [U ] : RS ×R →
R:
V [U ](x, w) = sup U (x + λw) = lim U (x + λw).
λ→+∞
λ∈R
We will present here some properties of the function V [U ]. From now on,
when there exists no confusion, we will write V (x, w) instead of V [U ](x, w).
Lemma 2 (a) V is concave.
(b) If there exists x such that V (x, w) = +∞ then V (x, w) = +∞, ∀x ∈ RS .
(c) If V (x, w) < +∞ ∀x ∈ RS , V (·, w) is continuous on RS .
Proof : (a) We consider t ∈ [0, 1] and a couple (x, w), (y, w0 ) in RS × R. For
all λ ≥ 0, by the concavity of U we have:
V ((1 − t)x + ty, (1 − t)w + tw0 )) =
≥
lim U ((1 − t)x + ty + λ((1 − t)w + tw0 ))
lim (1 − t)U (x + λw) + tU (y + λw0 )
λ→+∞
λ→+∞
= (1 − t)V (x, w) + tV (y, w0 )
7
because U (x + λw) and U (y + λw0 ) are increasing in λ.
(b) and (c): They are direct consequences of the concavity of V (·, w).
Lemma 3 If w ∈ intR then V (x, w) = +∞ for all x ∈ RS .
Proof : Fix x. Denote 1 = (1, 1, . . . , 1) ∈ RS . We will prove that V (x, 1) =
+∞. By the monotonicity of U , we have 1 ∈ intR. From the concavity of
V (x, ·), if V (x, 1) = +∞, then V (x, w) = +∞ ∀ w ∈ intR.
Indeed, for all y ∈ RS , there exists λ > 0 such that x + λ1 >> y. Then
U (x + λ1) > U (y). We have supy U (y) = +∞, so supλ U (x + λ1) = +∞.
Remark 1 From Lemma 3, the necessary condition to have V (x, w) < +∞ is
w ∈ ∂R.
Lemma 4 Given w ∈ ∂R, x ∈ RS . Suppose that there exists λ > 0 such that
U (x + λw) = U (x + λw) for all λ ≥ λ. Then V is differentiable at x and
V 0 (x, w) = U 0 (x + λw).
Proof :
Take p ∈ ∂V (x, w). We have p · w = 0. Indeed, observe that w is an useless
vector of V (·, w), then for all λ ∈ R, by the concavity of V we have:
0 = V (x + λw) − V (x, w) ≤ λp · w.
Hence p · w = 0.
By assumption, U (x + λw) = U (x + λw) ∀ λ ≥ λ, so V (x, w) = U (x + λw).
For all y ∈ RS we have:
U (y) − U (x + λw) ≤ V (y, w) − V (x, w) ≤ p · (y − x) = p · (y − (x + λw)).
Then p = U 0 (x + λw).
Take p = U 0 (x + λw). Observe that p · w = 0. We have:
V (y, w) − V (x, w) =
≤
=
lim [U (y + λw) − U (x + λw)]
λ→+∞
lim p · (y − x + λw − λw)
λ→+∞
lim p · (y − x) = p · (y − x).
λ→+∞
Then p ∈ ∂V (x, w).
8
Lemma 5 Given w ∈ ∂R such that V (x, w) ∈ R, a useful vector of U is also
a useful vector of V (·, w).
Proof : Indeed, take r ∈ R. For all λ, µ ≥ 0 we have:
U (x + µr + λw) ≥ U (x + λw).
This implies supλ U (x + µr + λw) ≥ supλ U (x + λw).
We say that w ∈ RS is a half-line direction if there exists x ∈ RS such that
U (x + λw) = U (x), ∀ λ ≥ 0. Let HL denote the set of half-line directions.
Given an affine space H of RS such that supx∈H U (x) = +∞, define the
0 the projection of
function UH : H → R as the restriction of U on H. Define UH
gradient of UH on H. Let σα denote an indifference surface, σα = {x | U (x) =
α} with α ∈ R. We set the following assumptions.
C (Closed gradient condition): For any α ∈ R, any H affine space such that
supH U (x) = +∞, define σαH as the indifference surface of U on H. Then the
0 (x)/kU 0 (x)k is closed.
map x ∈ σαH → UH
H
M condition: For all affine space H of RS , only one of the two following
cases holds:
(i) supH U (x) = +∞.
(ii) maxH U (x) exists.
Proposition 2 (a) HL ⊂ ∂R.
U 0 (x)
(b) Assume that for α ∈ R, the set { kU
0 (x)k }x∈σα is closed, then HL = ∂R.
Proof :
(a) Suppose that w is HL, by using Lemma 3, we have w ∈ ∂R.
(b) With α ∈ R, let σα = {x | U (x) = α}.
We will prove that if w ∈ ∂R, then there exists x
b ∈ σα such that U (b
x+λw) =
U (b
x) = α for all λ ≥ 0. If w = 0, obviously, the claim is true. We assume now
that w ∈ ∂R\{0}.
Take some x ∈ σα . If U (x + λw) = U (x) ∀ λ ≥ 0, the claim is true. Assume
that there exists µ > 0 such that U (x + µw) > U (x). If µ > 1, the concavity of
U implies that U (x + w) > U (x). If µ ≤ 1, then U (x + w) ≥ U (x + µw) > U (x).
This implies that U (x + w) > U (x). From the continuity of U , there exists
ρ > 0 such that for all y ∈ B(x + w, ρ), a ball center x + w and radius ρ, we
have U (y) > U (x). Since w ∈ ∂R, we can take a sequence wn converging to
w such that x + wn ∈ B(x + w, ρ) and wn ∈
/ R. We have U (x + wn ) > U (x).
9
Therefore, for any n, there exists λn > 1 such that U (x + λn wn ) = U (x) and
U (x + λwn ) < U (x) for all λ > λn .
Define xn := x + λn wn . We claim that λn → +∞. Indeed, in the contrary,
λn → λ ≥ 1. We have U (x + λw) ≥ U (x + w) > U (x): a contradiction because
U (x + λw) = limn U (xn ) = U (x).
Since λn → ∞, we have kxn k → ∞.
By the concavity of U we have:
0 = U (xn ) − U (x) ≥ U 0 (xn ) · (xn − x).
This implies:
xn
U 0 (xn )
x
U 0 (xn )
·
≤
· n .
0
n
n
0
n
kU (x )k kx k
kU (x )k kx k
Observe that when n → ∞, kxn k → +∞, and
of
U 0 (x)
{ kU
0 (x)k }x∈σα
xn
kxn k
→ w. From the closeness
we have:
U 0 (xn )
U 0 (x̂)
→
kU 0 (xn )k
kU 0 (x̂)k
where x
b ∈ σα . We obtain that:
U 0 (b
x)
· w ≤ 0.
0
kU (b
x)k
So U 0 (b
x) · w = 0 and that implies U (b
x + λw) = U (b
x) ∀ λ ≥ 0.
Proposition 3 Let U satisfy C condition. Let w ∈ ∂R.Then for all x, the
function λ 7→ U (x + λw) has a maximum.
Proof :
From Proposition 2, there exist x̂ such that U (x̂ + λw) = U (x̂), ∀ λ ≥ 0.
Hence V (x̂, w) < +∞. By Lemma 2, we have V (x, w) < +∞ ∀ x ∈ RS , or
limλ→+∞ U (x + λw) < +∞.
Suppose that there exists x such that U (x + λw) has no maximum.
Take any p ∈ ∂V (x, w). Take any z in interior of R. Let Π denote the plane
spanned by (w, z) and Πx = Π + {x}. We will prove that p is orthogonal to Π
and then orthogonal to z.
By the way of choosing z and Lemma 2, we have supλ U (x + λz) = +∞, so
supΠx U (x) = +∞. Let α := limλ→∞ U (x + λw) = V (x, w). Since U (x + λw) <
α and U (x + λz) → +∞ when λ → ∞, there exists y ∈ Πx which satisfies
U (y) = α. In other words, Cα := Πx ∩ σα 6= ∅.
10
By using the same argument as in the proof of Proposition 2 and the closed
gradient condition for the affine space Πx , we can prove that there exists x̂ ∈ Cα
such that
U (x̂) = U (x̂ + λw) = α, ∀ λ ≥ 0.
We have V (x, w) = V (x̂, w) = α, so
0 = V (x, w) − V (x̂, w) ≥ p · (x − x̂)
We obtained p · (x − x̂) ≤ 0.
By Lemma 4, we have V 0 (x̂, w) = U 0 (x̂), and then V 0 (x̂, w) · w = 0.
Denote q = V 0 (x̂, w). Let p1 , q1 respectively the orthogonal projections of p
and q on Π. Since p · w = q · w = 0, then p1 and q1 are orthogonal to w, and so
p1 = µq1 with µ ∈ R. Since z ∈ int(R), z is also a useful vector of V (·, w), and
V 0 (x, w) · z ≥ 0 which implies p1 · z ≥ 0. We have also U 0 (x̂) · z > 0 ⇒ q1 · z > 0.
These inequalities imply µ ≥ 0.
Observe that:
U (x̂) − U (x + λw) ≥ U 0 (x̂) · (x̂ − x) − λU 0 (x̂) · w = U 0 (x̂) · (x̂ − x) = q1 · (x̂ − x).
Let λ → +∞. We obtain q1 · (x̂ − x) ≤ 0. And then p1 · (x̂ − x) ≤ 0. Remember
that p · (x − x̂) ≤ 0 or p1 · (x − x̂) ≤ 0. Then, we have p1 · (x − x̂) = 0.
By the assumption of the non existence of a maximum of U (x + λw), x̂ does
not belong to the line {x + λw}, so w and x − x̂ are linearly independent and
orthogonal to p1 , so p is orthogonal to Π, or p · z = 0.
Since z is arbitrarily chosen, p is orthogonal to all vectors who belongs to
intR, and hence to all vectors in RS+ . This implies p = 0. Then V (x, w) ≥
V (y, w) ≥ U (y) for all y, or supy U (y) < +∞: a contradiction.
Proposition 4 Suppose w ∈ R is such that for all x, ∃ maxλ≥0 U (x + λw).
Define V (x, w) = maxλ≥0 U (x + λw). If U satisfies the C condition, then so is
V (·, w).
Proof : By Proposition 3, for all x ∈ RS , there exists λ ≥ 0 big enough such
that V 0 (x, w) = U 0 (x + λw).
Suppose that H is an affine space of RS , and supx∈H V (x, w) = +∞. We
consider two cases:
1. H is parallel to w: ∀ x ∈ H, x + w ∈ H.
2. H is not parallel to w.
Denote σαV,H := {x ∈ H | V (x, w) = α} and consider the sequence {xn }n ⊂
V 0 (x ,w)
σαw,H such that limn kVH0 (xnn ,w)k = p. For each n, there exists λn ≥ 0 such
H
that V (xn , w) = U (xn + λn w), or xn + λn w ∈ σα , and using Lemma 4,
11
U 0 (x +λ w)
limn kUH0 (xnn +λnn w)k = p.
H
For any affine subspace F , πF denotes the projection on F .
Consider the first case. In this case, xn +λn w ∈ H for all n, and supx∈H U (x) =
supx∈H V (x, w) = +∞. By applying the closed gradients condition for the
U 0 (x̂)
affine space H, there exists x̂ ∈ σαV,H such that kUH0 (x̂)k = p. Observe that
H
U 0 (xn + λn w) · w = 0 ⇒ πH [U 0 (xn + λn w)] · w = 0 ⇒ πH [U 0 (x̂)] · w = 0 ⇒
U 0 (x̂) · w = 0 and so V (b
x, w) = U (b
x) = α, or U (b
x + λw) = U (b
x) = α for all
0
0
λ ≥ 0, and V (b
x, w) = U (b
x). So we have
0 (b
VH0 (x̂, w)
UH
x)
πH (V 0 (b
x, w))
πH (U 0 (b
x))
=
=
=
= p.
0 (b
kVH0 (b
x, w)k
kπH (V 0 (b
x, w))k
kπH (U 0 (b
x)))k
kUH
x)k
Consider the second case. Let G := H + {λw}λ∈R . Then G is parallel to w.
Observe that supx∈G U (x) = supx∈H V (x, w) = +∞.
V 0 (x ,w)
Suppose that kVH0 (xnn ,w)k = p with xn ∈ H and V (xn , w) = α for all n.
H
Without loss of generality, we can suppose that
0 (x +λ w)
UG
n
n
0 (x +λnw)k
kUG
n
q ∈ RS .
→ p. Observe
that H ⊂ G, so πH (πG (q)) = πH (q) for all vector
Firstly, we prove that p is not orthogonal to H. Using the same argument
as the first case, applying closed gradient condition for affine space G, we know
that there exists x̂ ∈ G such that
UG0 (x̂)
= p and U 0 (x̂) · w = 0
kUG0 (x̂)k
or we can write
VG0 (x̂, w)
= p.
kVG0 (x̂, w)k
H is not parallel to w, so there exists λ ∈ R satisfies x̂ + λw ∈ H. Denote
x = x̂ + λw. Observe that V (x, w) = V (x̂, w) and V 0 (x, w) = V 0 (b
x, w).
0
If p is orthogonal to H then V (x, w) is orthogonal to H, then for all y ∈ H
we have:
V (y, w) − V (x, w) ≤ V 0 (x, w) · (y − x) = 0.
Thus V (x, w) = supy∈H V (y, w) = +∞: a contradiction.
Since p is not orthogonal to H, we have πH (p) 6= 0.
(p)
Secondly, we prove that kππH
= p. Indeed, for all n we have
H (p)k
0
(U (xn +λn ))
πH ( kππG
)
0
πH (πG (U 0 (xn + λn w))
G (U (xn +λn ))k
=
.
0
(U (xn +λn ))
kπH (πG (U 0 (xn + λn w))k
kπH ( kππG
0 (x +λ ))k )k
(U
n
n
G
And we let n → ∞, the left hand side tends to p, and the right hand side
πH (p)
H (p)
tends to kππH
(p)k . Hence, kπH (p)k = p.
12
Finally we have
0
G (V (x̂,w))
πH ( kππG
VH0 (x, w)
πH (V 0 (b
x, w))
πH (πG (V 0 (x̂, w)))
(V 0 (x̂,w))k )
=
=
))
=
0 x,w))
G (V (b
kVH0 (x, w)k
kπH (V 0 (x̂, w))k
kπH (πG (V 0 (b
x, w)))k
kπH ( kππG
(V 0 (b
x,w))k )k
⇒
VH0 (x, w)
πH (p)
=
= p.
0
kVH (x, w)k
kπH (p)k
V (., w) satisfies C condition.
Proposition 5 Suppose that U satisfies M condition. Given a vector w ∈ R
such that for an x, supλ≥0 U (x + λw) < +∞, then the function V (x, w) =
supλ≥0 U (x + λw) also satisfies M.
Proof :
Firstly, observe that from property of condition M, if supλ U (x+λw) < +∞
for some x, then ∃ λ such that U (x + λw) = supλ U (x + λw) = V (x, w).
Suppose that supH V (x, w) < ∞. Define the affine subspace G = H +
{λw}λ∈R . We can verify easily that supG U (x) = supH V (x, w) < +∞, so there
exists x ∈ G such that U (x) = supG U (x). Observe that V (x, w) = supλ U (x +
λw) ≥ U (x) = supG U (x) = supH V (x, w), so V (x, w) = supH V (x, w). Since
x ∈ H + {λw}λ , there exists x̂ ∈ H, λ ∈ R such that x = x̂ + λw. We have
V (x̂, w) = supH V (x, w).
Definition 7 For each vector w ∈ R, define S+ (w) = {s such that ws > 0},
S− (w) = {s such that ws < 0}.
The Proposition gives us an example of utility function which satisfies C
and M conditions.
Proposition 6 We consider the case utility function is defined as U (x) =
Pm
s=1 πs u(xs ), with u : R → R is a concave, strictly increasing, differentiable
function, π is a probability measure on {1, . . . , S}, with πs > 0, ∀ s. Define
a = u0 (+∞), b = u0 (−∞). Suppose that 0 < a ≤ b < +∞ and there exists
z > 0 such that u0 (x) = a ∀ x > z and u0 (x) = b ∀ x < −z. Then we have:
(a) U satisfies C condition.
(b) U satisfies M condition.
(c) There exists M vectors a1 , a2 , . . . , aM in RS such that the cone of weak
no-arbitrage prices P is the convex cone generated by {a1 , a2 , . . . , aM }.
13
Proof : (a) Fix a affine sub space H. Denote by σαH the set {x ∈ H such that U (x) =
U 0 (x)
α}. Suppose that p is limit point of { kU
H . For each n, define
0 (x)k }σα
An = {x ∈ σαH such that kp −
U 0 (x)
1
k ≤ }.
0
kU (x)k
n
Denote by xn the element of An which has the smallest norm: kxn k ≤ kxk for
all x ∈ An . We prove that the set {xn } is bounded.
Suppose the contrary, limn kxn k = ∞. Without loss of generality, we can
suppose that limn kxxnn k = w. Observe that w is parallel to H. Since U (xn ) = α
for all n, the vector w is a useful vector. We have also p · w = 0. Indeed, denote
by 0 the vector (0, 0, . . . , 0). We have U (xn ) − U (0) ≥ U 0 (xn ) · xn . Hence
α − U (0)
xn
≥ U 0 (xn ) ·
.
kxn k
kxn k
Let n converges to infinity, the LHS converges to 0. Observe that since 0 < a ≤
b < +∞, we have {U 0 (x)} is bounded. Hence without loss of generality, assume
q
that U 0 (xn ) converges to a vector q, with kqk
= p. The inequality above implies
that p · w ≤ 0, that implies p · w = q · w = 0.
Dana and Le Van in [8] prove that for all x we have
X
X
U 0 (x) · w ≥ a
πs ws + b
πs ws ≥ 0.
s∈S+ (w)
Since limn U 0 (xn ) · w = 0, we have a
s∈S− (w)
P
s∈S+ (w) πs ws
+b
P
s∈S− (w) πs ws
= 0.
Fix 0 < < 1. Take N big enough such that for all n ≥ N we have
xn,s − ws > z for s ∈ S+ (w) and xn,s − ws < −z for s ∈ S− (w). We will prove
that U (xn − w) = α ∀ n ≥ N . Indeed,
U (xn − w) − U (xn ) ≥ U 0 (xn ) · w
=
S
X
πs u0 (xn,s − ws )ws
s=1
X
=
s∈S+ (w)
X
= a
X
πs u0 (xn,s − ws )ws +
πs u0 (xn,s − ws )ws
s∈S− (w)
π s ws + b
s∈S+ (w)
X
πs ws
s∈S− (w)
= 0.
Hence U (xn − w) ≥ U (xn ) ≥ U (xn − w). This implies U (xn − w) = α.
Observe that for s ∈ S+ ∪ S− , we have u0 (xn,s − ws ) = u0 (xn,s ), which is
equal to a is s ∈ S+ (w), and to b if s ∈ S− , so U 0 (xn − w) = U 0 (xn ). Since
14
w is parallel to H, xn − w ∈ H. We have also kxn − wk < kxn k. This is a
contradiction for the definition of xn .
The set {xn } is bounded. Take a limit point x of this set, we have x ∈ H,
U 0 (x)
U (x) = α, and kU
0 (x)k = p.
(b) Fix affine space H such that supH U (x) = a < +∞. For each n, define
An = {x ∈ H such that U (x) ≥ a −
1
}.
n
Take xn ∈ An such that kxn k ≤ kxk for all x ∈ An . We prove that the
set {xn } is bounded. Suppose the contrary. We can suppose that limn kxxnn k =
w. Since {U (xn )} is bounded below, we have w is a useful vector. We will
P
P
prove that a s∈S+ (w) πs ws + b s∈S− (w) πs ws = 0. Indeed, suppose that
P
P
a s∈S+ (w) πs ws + b s∈S− (w) πs ws > 0. Fix x ∈ H. For λ > 0 big enough
we have xs + λws > z, ∀s ∈ S+ (w), and xs + λws < −z, ∀s ∈ S− . Hence
U (x + λw) − U (x) ≥ λU 0 (x + λw) · w


X
X
= λ a
πs ws + b
πs ws  .
s∈S+ (w)
s∈S− (w)
Let λ converges to infinity, the RHS tends to +∞. Observe that since w is parallel to H, x+λw ∈ H for all λ. This implies supH U (x) = +∞, a contradiction.
P
P
We have a s∈S+ (w) πs ws + b s∈S− (w) πs ws = 0. By using the same arguments in the part (a), we have U (xn − w) ≥ U (xn ) ≥ a − n1 , and kxn − wk <
kxn k, a contradiction. Hence the set {xn } is bounded. Since limn U (xn ) = a,
there exists maxH U (x).
(c) By induction, we prove that the weak no-arbitrage prices cone can be
generated by 2S vectors a1 , a2 , . . . a2S who satisfy ∀k, s, ak,s = πs a or ak,s = πs b.
It is easy to verify the claim with S = 1. Suppose that we have proved the
claim with S − 1.
Take any vector p ∈ P , with ps = λπs u0 (xs ). Since a ≤ u0 (xS ) ≤ b, there
exists 0 ≤ t ≤ b such that u0 (xS ) = ta + (1 − t)b. Hence we have
p = tλ(π1 u0 (x1 ), . . . , πS−1 u0 (xS−1 ), πS a)+(1−t)λ(π1 u0 (x1 ), . . . , πS−1 u0 (xS−1 ), πS b).
By the hypothesis of induction, there exists t1 , . . . , t2S−1 positives such that
P
(π1 u0 (x1 ), . . . , πS−1 u0 (xS−1 )) = k tk ak . Hence we have
p = tλ
S−1
2X
tk (ak , πS a) + (1 − t)λ
k=1
S−1
2X
k=1
15
tk (ak , πS b).
The claim has been proved.
It is easy to verify that a1 , a2 , . . . , a2S belong to P . The proof has finished.
4
Equivalence between the existence of general equilibrium and the existence of common weak noarbitrage price
Let E = {U i , ei , X i } be the initial economy, with X i = RS . Denote
1
2
m
1
2
W = {(w , w , . . . , w ) ∈ R × R × · · · × R
m
such that
m
X
wi = 0}.
i=1
4.1
Existence of equilibrium under C and M conditions
Lemma 6 Suppose that ∩i P i 6= ∅. Then U is bounded.
Proof : Take any p ∈ ∩i P i . For each i, there exists λi > 0, xi ∈ RS such that
p = λi U i0 (xi ). For (x1 , x2 , . . . , xm ) ∈ A, we have:
m
X
i
i
i
i
λi [U (x ) − U (x )] ≤
m
X
p · (xi − xi )
i=1
i=1
m
X
= p·(
xi − e)
i=1
then for all i:
m
X
X
X
λi U (e ) ≤ λi U (x ) ≤ p · (
xi − e) −
λj U j (ej ) +
λj U j (xj ).
i
i
i
i
i=1
j6=i
j6=i
Fix a vector w = (w1 , w2 , . . . wm ) ∈ W , like in Definition 6, define V i (xi , wi ) =
V [U i ](xi , wi ). We define E V the economy with m agents, each agent represented
by utility function V i , endowment ei and consumption set X i = RS . The attainable allocation set AV is defined as:
V
1
2
m
S m
A = {(x , x , . . . , x ) ∈ (R )
|
m
X
i
x =
i=1
m
X
ei and V i (xi , wi ) ≥ U i (ei )}.
i=1
The attainable utility set U V is defined as:
U V = {(v 1 , v 2 , . . . , v m ) ∈ Rm | ∃x ∈ AV : U i (ei ) ≤ v i ≤ V i (xi , wi ), ∀i}.
We have the result:
16
Proposition 7 Suppose that for all (x1 , x2 , . . . , xm ) ∈ A, there exists maxλ≥0 U i (xi +
λwi ), ∀ i. Then U V = U.
Proof : Evidently, since U i (xi ) ≤ V i (xi , wi ), U ⊂ U V .
Take (v 1 , v 2 , . . . , v m ) ∈ U V . There exists (x1 , x2 , . . . , xm ) ∈ U V such that
U i (ei ) ≤ v i ≤ V i (xi , wi ), ∀i. By assumption, there exists λ ≥ 0 big enough such
that V i (xi , wi ) = U i (xi + λwi ). Observe that (x1 + λw1 , x2 + λw2 , . . . , xm +
λwm ) ∈ A. This implies (v 1 , v 2 , . . . , v m ) ∈ U.
Theorem 5 Given economy E. Suppose that for all i, U i satisfy C condition
or No-half-line condition. Then:
m
\
P i 6= ∅ ⇔ U is compact ⇔ there exists General Equilibrium.
i=1
Proof : Suppose there exists equilibrium. By the Theorem 4, we have ∩i P i 6= ∅.
Suppose that ∩i P i 6= ∅. We will prove that U is compact, and this implies
the existence of equilibrium. Take any p ∈ ∩i P i . There exists λi > 0, xi ∈ RS
such that p ∈ λi U i0 (xi ), ∀ i.
Suppose that (w1 , w2 , . . . , wm ) ∈ W . From the Lemma 1, we know that for
all i, p · w ≥ 0, so we have:
0=p·
m
X
wi ≥ 0.
i=1
wi
This implies p ·
= 0 for all i.
Define L(W ) := W ∩ (−W ), the biggest sub linear space included in W . If
W is a linear subsspace, then the WNMA condition is satisfied. By the Theorem
3, U is compact and there general equilibrium exists.
Suppose that W 6= L(W ), or there exists w = (w1 , w2 , . . . , wm ) ∈ W such
that −w ∈
/ W . Denote U1i (xi ) = V [U i ](xi , wi ) as in Definition 6.
If U i satisfies no-half-line condition, then wi = 0. Indeed, p · wi = 0, this
implies U i0 (xi ) · wi = 0, so U i (xi + λwi ) = U i (xi ). If wi 6= 0, this direction will
be half-line of i, a contradiction. Hence if U i has no half-line, wi = 0 and this
implies U1i = U i . Evidently, U1i satisfies no half-line property.
If U i satisfies C condition, by Proposition 4, U1i also satisfies C condition.
Now we define E1 the economy who is characterized by {(U1i , ei , X i )}. Like
above, denote by W1 the cone of mutual opportunities of arbitrage.
Observe that any useful vector of U i is also a useful vector of U1i . So,
W ⊂ W1 . From the very definition of V [U i ], we have that, for all i, wi is a useless
vector of U1i . Thus, (w1 , w2 , . . . , wm ) ∈ L(W1 ). Hence we have dimL(W1 ) ≥
dimL(W ) + 1.
17
We will prove that the utility set U1 = U. If U i satisfies no half-line condition, then wi = 0, and evidently ∃ maxλ U i (xi +λwi ). If U i satisfies C condition,
by the Proposition 3, ∃ maxλ U i (xi + λwi ). By applying the Proposition 7, we
have U1 = U.
If we denote by P1i , the set of weak no-arbitrage prices of agent i, we have
p ∈ P1i for all i. Indeed, p · wi = 0, this implies U i (xi ) = V i (xi , wi ). Hence by
Lemma 4, p = λi V i0 (xi , wi ).
The economy E1 satisfies ∩i P1i 6= ∅, utility functions satisfy no half-line
condition or C condition. By induction and the same arguments, suppose that
we have constructed the economy Et , with t ≥ 1. If Wt is not a linear space,
we can construct a new economy Et+1 , with dimL(Wt+1 ) ≥ dimL(Wt ) + 1. Our
economies satisfy the utility sets are all the same Ut = U ∀ t. For all t, the
cone of mutual directions Wt ⊂ RS×m , so we have to stop at step T . The
economy ET has WT is a linear space, or we can say that the economy ET
satisfies WNMA condition. By the Theorem 3, we have UT is compact. This
implies U is compact and our initial economy has equilibrium.
Theorem 6 Suppose that for all agent i, U i is differentiable, and satisfies Nohalf-line condition or M condition. Then:
m
\
P i 6= ∅ ⇔ U is compact ⇔ ∃ general equilibrium.
i=1
Proof : Suppose that the exists equilibrium. We use the Theorem 4 to prove
that ∩i P i 6= ∅.
Suppose that the exists common weak no-arbitrage price, i.e. ∩i P i 6= ∅.
Take p ∈ ∩i P i . There exists λi > 0, xi ∈ RS such that p = λi U i0 (xi ). Suppose
that (w1 , w2 , . . . , wm ) ∈ W . Denote V i (xi , wi ) = supλ U i (xi + λwi ). We have
for all λ > 0, U i (xi + λwi ) = U i (xi ), or V i (xi , wi ) < +∞. By Lemma 2, we
have V i (xi , wi ) < +∞ for all xi ∈ RS .
If U i satisfies No-half-line condition, then wi = 0, and V i (xi , wi ) = U i (xi ).
If U i satisfies M condition, then by Lemma 5, V i is also satisfies this condition.
Observe that, by Lemma 4, we have V i (·, wi ) is differentiable, and if we denote
PVi = {p ∈ RS | ∃ λ > 0, x ∈ RS such that p = λV i0 (xi , wi )}
we have ∩i PVi 6= ∅, since p ∈ P i ∩ {wi }⊥ = PVi .
We have constructed a new economy with m agents, utility functions V i (·, wi )
have the same properties of U i , and ∩i PVi 6= ∅. By using the same argument in
the proof of Theorem 5, we have U is compact and there exists equilibrium.
18
4.2
When the cone P i is generated by a finite number of vectors
Theorem 7 Suppose that for all i, either P i satisfies the no-half-line condition,
or there exists M i vectors a1 , a2 , . . . , aM i in RS such that P i is the convex cone
generated by {a1 , a2 , . . . , aM i }. Suppose also that for all (w1 , w2 , . . . , wm ) ∈ W ,
there exists maxλ≥0 U i (xi + λwi ). Then U is compact and hence there exists
equilibrium.
Proof : If W is linear space, then the condition WNMA is satisfied. This implies
the compactness of U and the existence of equilibrium.
Suppose that W is not a linear space. Take w ∈ W which satisfies −w 6∈ W .
Define the function V i (x, wi ) = V [U ](xi , wi ) as in Definition 6. Since for any
i, there exists maxλ U i (xi +λwi ), by Lemma 4, we have V i (·, wi ) is differentiable.
Denote by PVi the cone generated by the gradients of V i (., wi ).
PVi = {p ∈ RS | ∃λ > 0, ∃x ∈ RS such that p = λV 0 (x, wi )}.
We will prove two claims.
1. For all i, either PVi satisfies no-half-line condition, or there exists a subset
D ⊂ {a1 , a2 , . . . , aM i } such that PVi = convex cone{D}.
2. Let (u1 , u2 , . . . , um ) satisfy: for any i, ui is a useful vector of V i , ∀ i, and
P i
i i
i
i
i
S
i u = 0. Then there exists maxλ≥0 V (x + λu , w ), ∀ x ∈ R , ∀ i.
We prove the first claim. If P i satisfies the no-half-line condition, then
wi = 0, and PVi = P i . Now we consider second case. Without loss of generality,
we can assume that for an 1 ≤ N i ≤ M i , a1 , a2 , . . . , aN i are orthogonal to wi ,
and for all N + 1 ≤ k ≤ M i , the scalar product ak · wi is strictly positif. We
actually have:
PVi = P i ∩ {wi }⊥ = cone{a1 , a2 , . . . , aN i }.
Indeed, observe that if p ∈ PVi , so ∃ µ > 0, x ∈ RS such that p ∈ µ∂V i (x, wi ) ⊂
µ∂U i (x + λwi ) with λ ≥ 0 big enough. From Lemma 4 we know that p · wi = 0.
So PVi ⊂ P i ∩ {wi }⊥ .
Given p ∈ P i ∩ {wi }⊥ , by the very definition of P i , p ∈ µ∂U i (x) with µ > 0,
x ∈ RS . Since p · wi = 0, we have U i (x + λwi ) = U i (x) for all λ ≥ 0. Indeed,
U i (x) − U i (x + λwi ) ≥ −λ µ1 p · wi = 0. Then U i (x) ≥ U i (x + λwi ), ∀ λ ≥ 0.
Then V i (x, wi ) = U i (x). By Lemma 4, we have p ∈ PVi . So P i ∩ {wi }⊥ ⊂ P i .
Evidently, cone{a1 , a2 , . . . , aN i } ⊂ P i ∩ {wi }⊥ . Take any p ∈ P i ∩ {wi }⊥ ,
P i
i
we can write p as p = M
i=1 λk ak . If there exists k ≥ N + 1 such that λk is
i
strictly positif, then p · w > 0: contradiction. Hence λk = 0 ∀k ≥ N i + 1, and
p belongs to the convex cone generated by {a1 , a2 , . . . , aN i }.
19
We now prove the second claim. Denote by RVi the set of useful vectors of
V i . We will prove the follow assertion:
RVi = Ri + {λwi }λ∈RS .
Since Ri ⊂ RVi and wi is a useless vector of V i , we have Ri + λwi λ∈RS ⊂ RVi .
Take any vector u ∈ RVi . Observe that for all 1 ≤ k ≤ N i , ak · u ≥ 0. Consider
the vector u + λwi , with λ ≥ 0. For 1 ≤ k ≤ N i , ak · (u + λwi ) ≥ 0. For
k ≥ N i + 1, since ak · wi > 0, we have ak · (u + λwi ) = ak · u + λak · wi > 0 with
λ big enough. With this λ, we have ak · (u + λwi ) ≥ 0 for all k, or u + λwi ∈ Ri .
We have proved that RVi ⊂ Ri + {λwi }λ∈RS .
P
Suppose that (u1 , u2 , . . . , um ) such that ui ∈ RVi for all i, and i ui = 0.
For each i, we can write ui = ri + µi wi , with ri ∈ Ri , µi ∈ R. Choose µ > 0
big enough such that for all i, λi = µ + µi > 0. We have:
m
X
i=1
(ri + λi wi ) =
m
X
(ri + µi wi ) + µ
i=1
m
X
wi = 0.
i=1
For all i, λi > 0, so ri + λi wi ∈ Ri . Observe that since wi is a useless
vector of V i , we have V i (xi + λui , wi ) = V (xi + λri , wi ). From the property
of utility function U i , we have there exists maxλ≥0 U i (x + λ(ri + λi wi )). Then
there exists λ1 ≥ 0 such that for all λ ≥ λ1 we have:
U i (x + λ(ri + λi wi )) = U i (x + λ1 (ri + λi wi )) ≤ V i (x + λ1 ri , wi ).
Fix any λ2 ≥ λ1 . For all λ > λ2 , we have:
U i (x + λ2 ri + λλi wi ) ≤ U i (x + λ(ri + λi wi )) ≤ V i (x + λ1 ri , wi ).
Let λ → +∞ the left hand side U i (x + λ2 ri + λλi wi ) tends to V i (x + λ2 ri , wi )
so
V i (x + λ2 ri , wi ) ≤ V i (x + λ1 ri , wi ).
This inequality is true for any λ2 ≥ λ1 . Since ri is also a useful vector of V i ,
there exists maxλ≥0 V i (x + λri , wi ), or equivalently there exists maxλ≥0 V i (x +
λui , wi ).
By the two claims that we have just proved, the utility functions {V i } has
the same properties of the utility functions {U i }. Define U1i (xi ) = V i (xi , wi ).
Define a new economy E1 who is characterized by {(U1i , ei , X i )} with X i =
RS .
Observe that since U i0 (xi ) = V i0 (xi , wi ) for all i, we have ∩i P1i 6= ∅.
By Lemma 5, any useful vector of U i is also a useful vector of U1i . So,
W ⊂ W1 . From the very definition of V [U i ], we have that, for all i, wi is a useless
20
vector of U1i . Thus, (w1 , w2 , . . . , wm ) ∈ L(W1 ). Hence we have dimL(W1 ) ≥
dimL(W ) + 1.
The economy E1 satisfies ∩i P1i 6= ∅, utility functions satisfy the same condition of utility functions of the economy E. By induction and the same arguments, suppose that we have the economy Et , with t ≥ 1. If Wt is not a linear
space, we can construct a new economy Et+1 , with dimL(Wt+1 ) ≥ dimL(Wt )+1.
Observe that the utility sets are all the same Ut = U ∀ t. For all t, the cone of
mutual directions Wt ⊂ RS×m , so there exists T such that WT is linear space,
or we can say that the economy ET satisfies WNMA condition. Then UT is
compact. Hence U is compact and our initial economy has equilibrium.
4.3
Existence of a General equilibrium with separable expected
utilities
We consider now the economy E with m agents i = 1, 2, . . . , m, a standard
model of Arrow-Debreu of complete contingent security markets. There are
two dates, 0 and 1. At date 0, there is uncertainty about which state s from
a state space {1, 2, . . . , S} will occur at date 1. At date 0, agents who are
uncertain about their future endowments trade contingent claims for date 1.
Agent i is characterized by an endowment ei ∈ RS of contingent claims, a
consumption set RS and a probability π i = (π1i , π2i , . . . , πSi ) which represents
P
his belief about future, πsi > 0 ∀s and s πsi = 1. For each contingent claim
(x1 , x2 , . . . , xS ) ∈ RS , the utility of agent i is:
i
U (x) =
S
X
πsi ui (xs )
s=1
with ui : R → R is a concave, strictly increasing differentiable function.
Observe that U i0 (x) = (πsi ui0 (xs ))Ss=1 . Denote by ai and bi the limits of
derivation of ui at infinity:
ai = lim ui0 (x) and bi = lim ui0 (x).
x→+∞
x→−∞
Proposition 8 If ai < ui0 (x) ∀ x, or bi > ui0 (x) ∀ x, then P i is open.
Proof : See Dana and Le Van [8].
Proposition 9 Suppose that ai < ui0 (x) ∀ x, or bi > ui0 (x) ∀ x, then we have:
m
\
P i 6= ∅ ⇔ NUBA condition ⇔ U is compact ⇔ there exists equilibrium.
i=1
21
Proof : Easy.
Now we consider the case utility function ui is linear at far, i.e. there exists
z i such that
ui0 (x) = z i for x > z i and ui0 (x) = bi for x < −z i .
We present here an example in which the weak no-arbitrage prices cones are
closed, their intersection is no empty, the model does not satisfy NUBA or
WNMA conditions, but there exists equilibrium.
Example 1 We consider an economy with two agents, the number of states
is S = 2. The belief of agent 1 is represented by probability π11 = π21 = 12 . The
one of agent 2 is π12 = 13 , π12 = 32 .


 log(x) if x ∈ [1/3, 1/2]
1
u (x) =
2x − 1 − log 2 if x ≥ 1/2


3x − 1 − log 3 if x ≤ 1/3


 log(x) if x ∈ [1/3, 4/9]
2
9
4
4
u (x) =
4 x − 1 + log( 9 ) if x ≥ 9


3x − 1 − log 3 if x ≤ 13
We have u10 (+∞) = 2, u10 (−∞) = 3 and u20 (+∞) = 94 , u20 (−∞) = 3.
Therefore, the cone of no-arbitrage prices of agent 1 is P 1 = {λ(ζ1 , ζ2 )}λ>0
with 1 ≤ ζ1 ≤ 23 , 1 ≤ ζ2 ≤ 23 . The one of agent 2 is P 2 = {λ(ζ1 , ζ2 )}λ>0 with
3
3
4 ≤ ζ1 ≤ 1, 2 ≤ ζ2 ≤ 2.
The common no-arbitrage prices is the intersection of two cones P 1 ∩ P 2 =
{λ(1, 32 )}λ>0 . Denote S 1 , S 2 the interiors of P 1 and P 2 . We have S 1 ∩S 2 = ∅. If
we consider useful vector w1 = (1, − 32 ) of agent 1, the useful vector w2 = (−1, 23 )
of agent 2, observe that w1 + w2 = 0. Whenever −w1 , −w2 are not useful
vectors of agent 1 and agent 2. Our economy does not satisfies NUBA or
WNMA conditions. In the next of this subsection, we will prove the existence
of equilibrium un such model.
Example 2 We consider an economy with two agents, the number of states
is S = 2. The belief of agent 1 is represented by probability π11 = π21 = 12 . The
one of agent 2 is π12 = 13 , π12 = 32 .



5
2x
− 1 if x ∈ [−1, 1]
1
u (x) =
2x − 2x
if x ≥ 1


1
3x + 2x if x ≤ −1

11

 4 x − 1 if x ∈ [−1, 1]
2
9
1
u (x) =
4 x − 2x if x ≥ 1


1
3x + 4x − 21 if x ≤ −1
1
22
We have u1 , u2 are continuous, differentiable, and for all x ∈ R, 2 < u10 (x) <
3, 49 < u20 (x) < 3. Therefore, the cone of no-arbitrage prices of agent 1 is
P 1 = {λ(ζ1 , ζ2 )}λ>0 with 1 < ζ1 < 32 , 1 < ζ2 < 32 . The one of agent 2 is
P 2 = {λ(ζ1 , ζ2 )}λ>0 with 43 < ζ1 < 1, 32 < ζ2 < 2. The weak no-arbitrage prices
cones are open, and coincide with the Werner’s no-arbitrage cones, S 1 and S 2 .
Since P 1 ∩ P 2 = ∅, equilibrium does not exists.
The agent 1 has w1 = (1, − 23 ) like useful vector. And the agent 2 has
w2 = (−1, 32 ) like useful vector. We can verify easily that limλ→+∞ U 1 (λw1 ) <
1
2
+∞ and limλ→+∞ U 2 (λw2 ) < +∞. The intersection of closures P ∩ P =
{λ(1, 23 )}λ>0 6= ∅. The limited arbitrage condition hence is satisfied. But the
equilibrium does not exists.
Theorem 8 In the model with separable utility, we have:
m
\
P i 6= ∅ ⇔ U is compact ⇔ there exists equilibrium.
i=1
Proof : Suppose that ∩i P i 6= ∅. We use the Proposition 6, and one of the
Theorems 5, 6, 7 to prove the compactness of utility set U and existence of
equilibrium.
For the applying of the Theorem 7, we prove now that if (w1 , . . . , wm )
P
satisfies i wi = 0 and wi ∈ Ri for all i, for all xi , there exists λ ≥ 0 such that
U i (xi + λwi ) = U i (xi + λwi ), ∀λ ≥ λ.
Take p ∈ ∩i P i , there exists λi > 0, xi such that ps = λi ui0 (xis ). Easily, we
can prove that p · wi = 0 ∀i. This implies U i0 (xi ) · wi = 0. Hence for all i,
P
P
ai S+ πsi wsi + bi S− πsi wsi = 0.
For all i, fix xi . If ui0 (z) > ai for all z or ui0 (z) < bi for all z, then P i is
open (see Dana and Le Van [8]). In this case, there is no half line in indifferent
surfaces, this implies wi = 0. And the claim is evidently true.
Suppose that there exists z i such that ui0 (z) = ai for z > z i and ui0 (z) = bi
for z < −z i . Take λ ≥ 0 big enough such that for s ∈ S+ (wi ), xis + λwsi > z i ,
and for s ∈ S− (wi ), xis + λwsi < −z i . For all λ ≥ λ we have
U i (xi + λwi ) − U i (xi + λwi ) ≤ (λ − λ)
S
X
πsi ui0 (xis + λwsi )wsi
s=1

= (λ − λ) ai

X
s∈S+ (wi )
πsi wsi + bi
X
πsi wsi 
s∈S− (wi )
= 0.
Since wi is a useful vector of i, U i (xi + λwi ) = U i (xi + λwi ), for all λ ≥ λ.
23
If there exists general equilibrium, we use the Theorem 4 to prove that
∩i P i 6= ∅.
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