Walrasian Equilibrium: Proof of Existence
Ram Singh
Lecture 6
September 27, 2017
Market Equilibrium Price I
Definition
Walrasian Equilibrium Price: A price vectorp∗is equilibrium price vector, if for allj =1, ...,M,
zj(p∗) =
N
X
i=1
xji(p∗,p∗.ei)−
N
X
i=1
eij =0, i.e., if
z(p∗) =0= (0, ...,0).
Proposition
Ifp∗is equilibrium price vector, thenp0=tp∗, t >0, is also an equilibrium price vector
Ifp∗is equilibrium price vector, thenp06=tp∗,t >0, may or may not be an equilibrium price vector
WE: General Case I
References: Jehle and Reny (2008).
Also see: Arrow and Debreu (1954), and McKenzie (2008); Arrow and Hahn (1971).
Recall, we have
For alli =1,2, ...,N, we have:xi(tp) =xi(p), for allt >0.
z(tp) =z(p), for allt>0.
Let
pˆ= (ˆp1, ...,ˆpM)>>(0, ...,0)be a price vector t =pˆ 1
1,...,ˆpM
WE: General Case II
Thenp=tpˆis such that
pj ≥>0 and
M
X
j=1
pj =1
Without any loss of generality, we can restrict attention to the following set
P=
p= (p1, ...,pM)|
M
X
j=1
pj =1 andpj ≥ 1+2M
,
where≥0.
Remark
Note thatPis non-empty, bounded, closed and convex set for all∈(0,1).
WE: General Case III
Theorem
Suppose ui(.)satisfies the above assumptions, ande>>0. Let{ps}be a sequence of price vectors inRM++, such that
{ps}converges to¯p, where
p¯∈RM+,p¯6=0, but for some j,p¯j =0.
Then, for some good k with¯pk =0, the sequence of excess demand (associated with{ps}), say{zk(ps)}, is unbounded above.
Theorem
Under the above assumptions on ui, there exists a price vectorp∗>>0, such thatz(p∗) =0.
WE: Proof I
Let,
z¯j(p) = min{zj(p),1}. (1) Note
¯zj(p) = min{zj(p),1} ≤1.Therefore, 0≤max{¯zj(p),0} ≤1.
Next, let’s define the functionf = (f1(p), ...,fM(p))such that: Forj=1, ..,M, fj(p) = +pj+ max{¯zj(p),0}
M+1+PM
j=1max{¯zj(p),0} =Nj(p) D(p),
Note thatPM
j=1fj(p) =1. Moreover, we have fj(p)≥ Nj(p)
M+1+M.1 ≥
M+1+M.1 ≥
1+2M.
WE: Proof II
Therefore,
f(p) :P7→P.
SinceD(p)≥1>0, the above function is well defined and continuous over a compact and convex domain. Therefore, there existspsuch that
f(p) =p,i.e.,
fj(p) =pj, for allj =1, ..,M.
That is, for allj=1, ..,M,
Nj(p)
D(p) = pj,i.e.,
pj[M+
M
X
j=1
max{¯zj(p),0}] = + max{¯zj(p),0}. (2)
Next, we let→0. Consider the sequence of price vectors{p}, as→0.
WE: Proof III
Sequence{p}, as→0, has a convergent subsequence, say{p0}.
Why?
Let{p0}converge top∗, as0→0.
Clearly,p∗≥0. Why?
Suppose,p∗k =0. Recall, we have
pk0
M0+
M
X
j=1
max{¯zj(p0),0}
=0+ max{¯zk(p0),0}. (3)
as0 →0:
The LHS converges to 0, sincelim0→0pk0 =0 and term [M0+PM
j=1max{¯zj(p0),0}]on LHS is bounded.
However, the RHS takes value 1 infinitely many times. Why?
WE: Proof IV
This is a contradiction, because the equality in (3) holds for all values of0. Therefore,pj∗>0 for allj =1, ..,M. That is,
p∗>>0,i.e.,
lim
→0p=p∗>>0.
In view of continuity ofz¯(p)overRM++, from (2) we get (by taking limit→0):
WE: Proof V
For allj=1, ..,M
pj∗
M
X
j=1
max{¯zj(p∗),0} = max{z¯j(p∗),0},i.e.,
zj(p∗)pj∗
M
X
j=1
max{¯zj(p∗),0}
= zj(p∗) max{¯zj(p∗),0},i.e.,
M
X
j=1
zj(p∗)pj∗
M
X
j=1
max{¯zj(p∗),0}
=
M
X
j=1
zj(p∗) max{¯zj(p∗),0},i.e.,
M
X
j=1
zj(p∗) max{¯zj(p∗),0}=0. (4)
WE: Proof VI
Recall,z¯j(p) = min{zj(p),1}.Therefore,
zj(p∗)>0 ⇒ max{¯zj(p∗),0}>0;
zj(p∗)≤0 ⇒ max{¯zj(p∗),0}=0.
Therefore,
zj(p∗)>0 ⇒ zj(p∗) max{¯zj(p∗),0}>0;
zj(p∗)≤0 ⇒ zj(p∗) max{z¯j(p∗),0}=0.
Suppose, for somej, we havezj(p∗)>0, then we will have
M
X
j=1
zj(p∗) max{¯zj(p∗),0}>0. (5)
But, this is a contradiction in view of (4).
WE: Proof VII
Therefore:
For anyj=1, ..,M, we have
zj(p∗)≤0. (6)
Suppose,zk(p∗)<0 for somek. From (4), we know
p∗1z1(p∗) +...+p∗kzk(p∗) +...+p∗MzM(p∗) =0.
Sincepj∗>0 for allj =1, ..,M.
zk(p∗)<0 impliesp∗kzk(p∗)<0
Therefore, there must existk0, such that
WE: Proof VIII
zk0(p∗)>0, (7)
which is a contradiction in view of (6). Therefore,
For all j =1, ..,M, we have:zj(p∗) = 0,i.e., z(p∗) = 0.
That is,
p∗is a WE price vector.
