Conditions for Unique Walrasian Equilibrium

Conditions for Unique Walrasian Equilibrium

Ram Singh

Lecture 8

Gross Substitutes I

Suppose,

There are two goods

Consider three price vectors:p= (p1,p2) = (2,1),p⁰ = (p⁰₁,p₂⁰) = (3,1) andp¯= (¯p1,¯p2) = (3,2).

Let,xⁱ(p), be the demand function for individuali.

Question

Suppose, the above goods are ‘gross substitutes’ for individual i.

How will x₂ⁱ(p⁰)compare with x₂ⁱ(p)?

How will x₂ⁱ(¯p)compare with x₂ⁱ(p)?

Letλ= max_j{^¯p_p^j

j},j =1,2.

Note hereλ=^p¯2

p2 =2

Gross Substitutes II

Also,λp≥¯p. Since(4,2)≥(3,2).

Question

What can we say about the individual demand for the two goods at these two price vectorsλp= (4,2)andp¯= (3,2)?

Question

What can we say about the individual demand for the two goods at the price vectorsp= (2,1)andλp= (4,2)?

Gross Substitutes III

Next, consider two price vectors

p= (p1,p2,p3) = (3,2,1)and¯p= (¯p1,p¯2,¯p3) = (5,1,4)

Question

What can we say about the excess demand at these two price vectors?

Letλ= max_j{^¯p_p^j

j},j =1, ..,3.

Note hereλ= max{⁵₃,¹₂,⁴₁}= ^¯p_p³

3 =4

Also,λp≥¯p. Since(12,8,4)≥(5,1,4).

Remark

z(λp) =z(p), i.e.,z(4p) =z(p).

Gross Substitutes IV

Consider the following price vectors

p= (p1,p2,p3) = (3,2,1),pˆ= (ˆp1,pˆ2,ˆp3) = (12,8,4)and p¯= (¯p1,p¯2,p¯3) = (5,1,4).

Question

What can we say about the excess demand for 3rd good at pricespˆ and p? That is,¯

How is z3(ˆp)expected to compare with zj(¯p)?

Note:

pˆ=λpandpˆ ≥p¯ pˆ3=λp3= ¯p3.

GS and No of WE I

Question

What are the assumptions needed for the aggregate demand function to exist?

Aggregate demand functionx(.)will exist iff if the individual demand function, i.e.,xⁱ(.), exists for alli =1, ..,N.

xⁱ(.)exists if the underlying utility function satisfies the assumption of continuity, strong monotonicity and strict quasi-concavity.

GS and No of WE II

Definition

Aggregate demand function,z(.), satisfies condition of ‘Gross Substitutes’

(GS) if for allp,ˆ p¯∈R^M++, such thatpˆ≥p¯and ˆp6= ¯p:

pˆj = ¯pj ⇒zj(ˆp)>zj(¯p).

Theorem

If Z(.)satisfies condition of GS, then there is unique WE.

WLOG, we can consider vectors in the set

P={p|p∈R^M++, andpM =1}.

Proof: Suppose, WE is not unique. If possible, supposep,p⁰∈E. Moreover,p6=p⁰.

GS and No of WE III

Let

λ = max

j

p´j

p_j

forj =1, ..,M.

= max p´1

p1

,p´2

p2

, ...,´pM

pM

Suppose, ^p_p^´k

k ≥ ^´p_p^j

j for allj =1, ..,M.That is, λ= ´p_k

pk

Clearly,λp≥p⁰, andp_kλ= ´p_k. Letp¯=λp.

This meansp¯ ≥p⁰andp¯k = ´pk. Hence

zk(¯p)>zk(p⁰).

GS and No of WE IV

Butzk(p⁰) =0. Therefore,

z_k(¯p=λp)>0, which is a contradiction. Why?

Sincep∈E, therefore

zk(p) =0.

Sincep¯=λp,

z_k(¯p) =z_k(p) =0.

WARP and no of WE I

Let,

x=x(p)denote the bundle demanded at pricep; wherex= (x1, ...xm).

x⁰ =x(p⁰)denote the bundle demanded at pricep⁰; -x⁰= (x₁⁰, ...x_m⁰).

Therefore,

p.x=p.x(p)is the expenditure incurred at pricep.

p⁰.x⁰=p⁰.x(p⁰)is the expenditure incurred at pricep⁰.

The demand satisfies Weak Axiom of Revealed Preference (WARP), if p.x⁰≤p.x⇒p⁰.x⁰ <p⁰.x.

p.x⁰≤p.ximplies that the bundlex⁰ was affordable at pricep.

p⁰.x>p⁰.x⁰implies that the bundlex=x(p)is strictly more expensive (thanx⁰=x(p⁰)) at pricep⁰.

WARP and no of WE II

Restating: The demand satisfies WARP, if

p.x⁰≤p.x ⇒ p⁰.x⁰<p⁰.x.

p(x(p⁰)−x(p))≤0 ⇒ p⁰(x(p⁰)−x(p))<0.

Theorem

If the aggregate demand function satisfies the WARP, then the WE is unique.

WARP and no of WE III

Define (aggregate) vectors:

x = (

N

X

i=1

x₁ⁱ,

N

X

i=1

x₂ⁱ, ...,

N

X

i=1

x_Mⁱ )

e = (

N

X

i=1

eⁱ₁,

N

X

i=1

eⁱ₂, ...,

N

X

i=1

e_Mⁱ )

z = x−e= (

N

X

i=1

(x₁ⁱ −e₁ⁱ), ...,

N

X

i=1

(x_Mⁱ −e_Mⁱ ))

z = x−e= (z1, ...,zM)

WARP and no of WE IV

Proof: Suppose, WE is not unique. If possible, supposep,p⁰∈E, andp6=p⁰. Note for any price vectorspandp⁰, we have:

p.z(p) = 0,i.e.,

p.(x(p)−e) = 0. (1)

Sincep⁰is an equi. price vector,z(p⁰) =x(p⁰)−e=0, i.e.,x(p⁰) =e.

Therefore, the assumptionp⁰ ∈Egives us

p.(x(p)−x(p⁰)) = 0,i.e.,

p.(x(p⁰)−x(p)) = 0. (2) From WARP, we know that

p.(x(p⁰)−x(p))≤0⇒p⁰.(x(p⁰)−x(p))<0. (3) (2) and (3) give us,

p⁰.(x(p⁰)−x(p))<0. (4)

WARP and no of WE V

Similarly, we get:

p⁰.z(p⁰) = 0,i.e., p⁰.(x(p⁰)−e) = 0

p⁰.(x(p⁰)−x(p)) = 0 (5)

which is a contradiction, since in view of (4), we have p⁰(x(p⁰)−x(p))<0.

The assumption that there are two price vectorsp,p⁰ ∈Eleads to a contradiction.

There cannot be two or more equilibrium price vectors.