General Equilibrium Analysis: Lecture 8
Ram Singh
Course 001
October 1, 2014
Questions
Is Competitive/Walrasian equilibrium unique?
Why a unique equilibrium helpful?
If WE is not unique, how many WE can be there?
What are the conditions, for a unique WE?
Do these conditions hold in the real world?
Readings: Arrow and Hahn. (1971), Jehle and Reny (2008), MWG*(1995)
Local Uniqueness of WE I
For a 2×2 economy: Let,p= (p,1), wherep>0. Note:
Lemma
Price vectorp= (p,1)is equilibrium price vector of a2×2economy iff z1(p) =0. Why?
For anyN−1×M−1 economy, WLOG we can consider vectors in the set
P={p|p∈RM++, andpM =1}
Let
pvM= (p1, ...,pM−1)and zvM = (z1(p), ...,zM−1(p))
Local Uniqueness of WE II
Proposition
A price vectorp= (p1, ...,pM−1,1)is an equilibrium price vector iff it solves the M−1×M−1systemzvM(p) =0, i.e., iff it solves the system:
z1(p) = 0 ... = ... zM−1(p) = 0.
Define theM−1×M−1 matrix of first order derivatives:
DzvM(p) =
∂z1(p)
∂p1
∂z1(p)
∂p2 · · · ∂∂zp1(p)
M−1
... ... . .. ...
∂zM−1(p)
∂p1
∂zM−1(p)
∂p2 · · · ∂z∂M−1p (p)
M−1
Local Uniqueness of WE III
Definition
A price vectorp= (p1, ...,pM−1,1)isregularif theM−1×M−1 matrix, DzvM(.), is non-singular atp= (p1, ...,pM−1,1).
Definition
An economy is regular if everyequilibriumprice vectorp= (p1, ...,pM−1,1) is is regular.
Theorem
A regular equilibrium price vectorp= (p1, ...,pM−1,1)is locally unique. That is, there exists an >0such that: for everyp0= (p10, ...,p0M−1,1),p06=p, and kp0−pk< , we have
z(p0)6=0.
Local Uniqueness of WE IV
ProofSuppose,p= (p1, ...,pM−1,1)is an equilibrium price vector, i.e., z(p) =0.
Now, consider an infinitesimal change inp, saydp6=0. Let dp= (dp1, ...,dpM−1,0), and
p0 =p+dp= (p1+dp1, ...,pM−1+dpM−1,1) Since,DzvM(p)is non-singular, we have
DzvM(p)dpvM=
∂z1(p)
∂p1
∂z1(p)
∂p2 · · · ∂p∂z1(p)
M−1
... ... . .. ...
∂zM−1(p)
∂p1
∂zM−1(p)
∂p2 · · · ∂z∂pM−1(p)
M−1
dp1
dp2
... dpM−1
(1)
DzvM(p)dpvM 6=0. Why?
Local Uniqueness of WE V
zvM(p0)≈zvM(p) +DzvM(p)dpvM 6=0.
Therefore,
zvM(p0) 6= 0,i.e., z(p0) 6= 0.
That is,p0 is not WE.
Number of a WE I
Let
E={p|p∈P, andz(p) =0}.
Remark
Note:E⊂⊂P⊆RM++. Also, if an economy is regular, the setEis discrete.
Theorem
If an economy is regular, there are only finitely many equilibrium prices.
Proof: Suppose,p= (p1, ...,pM−1,1)is an equilibrium price vector, i.e., z(p) =0. In view of the ‘boundary conditions’,pj cannot be arbitrarily close to zero.
For a two goods Economy: Boundary conditions onz(.)imply that z1(.)>0 for very smallp1
z1(.)<0 for very highp1
Number of a WE II
That is, if the ‘boundary conditions’ hold, then there existsr >0 such that:
For allj=1, ..,J
1
r <pj <r.
Therefore, the setEis bounded.
Exercise
Assuming thatzis continuous inp, show thatEis closed.
Hint: Consider a sequence of prices inEand use continuity ofz.
Next, use the following result:
Theorem
If a set is compact and discrete, then it has to be finite.
SinceEis bounded, closed and discrete, it is a finite set.
Number of a WE III
Theorem
If an economy is regular and the ‘boundary conditions’ onz(P)holds, then Either there will be a unique equilibrium
The number of equilibria will be odd.
Unique WE: Conditions? I
For an individual consumer. Let
u(.)is a continuous, strictly increasing and strictly quasi-concave utility function
u∗=v(p,I), and
x:RM+1++ 7→RM+ be the (Marshallian) demand function generated byu(.).
xH :RM+1++ 7→RM+be the associated Hicksian demand function.
∂xj(p,I)
∂pk = ∂x
H j (p,u∗)
∂pk −xk(p,I)∂xj∂I(p,I), i.e.,
∂xjH(p,u∗)
∂pk =∂x∂pj(p,I)
k +xk(p,I)∂xj∂I(p,I).
Unique WE: Conditions? II
We know that
∂xjH(p,u∗)
∂pj
=
∂xj(p,u∗)
∂pj
du=0
=∂xj(p,I)
∂pj
+xj(p,I)∂xj(p,I)
∂I <0.
Let,
D(xH) =
∂x1H(p,u)
∂p1 · · · ∂x1∂pH(p,u) .. M
. · · · ...
∂xMH(p,u)
∂p1 · · · ∂xM∂pH(p,u)
M
=
∂2E(p,u)
∂p21 · · · ∂∂p2E(p,u)
M∂p1
... · · · ...
∂2E(p,u)
∂p1 · · · ∂2E(p,u)
∂p2M
We can write
D(xH) =
∂x1(p,I)
∂p1 +x1(p,I)∂x1∂I(p,I) · · · ∂x∂p1(p,I)
M +xM(p,I)∂x1∂I(p,I)
... . .. ...
∂xM(p,I)
∂p1 +x1(p,I)∂xM∂I(p,I) · · · ∂x∂pM(p,I)
M +xM(p,I)∂xM∂I(p,I)
Unique WE: Conditions? III
We know that
MatrixD(xH)is negative semi-definite. Why?
Moreover, for every pair(p,I), there is unique ‘equilibrium’ for the consumer. Why
Theorem
Ifx:RM+1++ 7→RM+, is is a demand function generated by a continuous, strictly increasing and strictly quasi-concave utility function,then xsatisfies budget balancedness, symmetry and negative semi-definiteness.
Theorem
If a function,x, satisfies budget balancedness, symmetry and negative semi-definiteness, then it is a demand functionx:RM+1++ 7→RM+is generated by some continuous, strictly increasing and strictly quasi-concave utility function.
Unique WE: Conditions? IV
That is, if matrix
D{x}=
∂x1(p,I)
∂p1 +x1(p,I)∂x1∂I(p,I) · · · ∂x∂p1(p,I)
M +xM(p,I)∂x1∂I(p,I)
... . .. ...
∂xM(p,I)
∂p1 +x1(p,I)∂xM∂I(p,I) · · · ∂x∂pM(p,I)
M +xM(p,I)∂xM∂I(p,I)
is symmetric and negative semi-definite, then
There is a continuous, strictly increasing and strictly quasi-concave utility functionu(p)that induces demand functionx.
The ‘equilibrium of the consumer’ is unique.
Moreover, if goodj is normal, then ∂x∂jp(p)
j <0.
Unique WE: Conditions? V
Question
Does a similar result hold for the aggregate demand function? That is, does a similar result hold at the aggregate level?
Let
D{z(p)}=
∂zj(p)
∂pk
, j,k =1, ...,J−1.
At aggregate level, we have:
Theorem
WE is unique, if the matrix
D{z(.)}is negative semi-definite at allp∈E, i.e., D{−z(.)}has a positive determinant at allp∈E.
Unique WE: Conditions? VI
Note, now Slutsky equation is:
∂xji(p∗)
∂pk
!
dui=0
= ∂xji(p∗)
∂pk + xki(p∗)−eik ∂xji(p∗)
∂I
!
dp=0
We have seen that even for 2×2 economy, Remark
D{z(.)}is not necessarily negative semi-definite, even ifD(xH)is negative semi-definite.
Moreover, even if the goods are all normal, ∂z∂pj(p)
j <0 may not hold.
