General Equilibrium Analysis: Lecture 7

General Equilibrium Analysis: Lecture 7

Ram Singh

Course 001

September 29, 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)

Multiple WE: Example

Example Two consumers:

u¹(.) =x₁¹−¹₈ ¹

x₂¹ andu²(.) =−¹₈ ¹

x₁² +x₂² e¹= (2,r)ande²= (r,2);r =2⁸⁹ −2¹⁹ The equilibria are solution to

p2

p1

⁻¹₉

+2+r p2

p1

− p2

p1

⁸₉

=2+r,i.e.,

there are three equilibria:

p₂ p1

= 1

2,1, and 2.

Multiple WE: Example I

Consider

Two goods: food and cloth Let(p_f,p_c)be the price vector.

We know that for allt >0:z(tp) =z(p).

Therefore, we can work with p= (pf

pc

,1) = (p,1).

From Walras’s Law, we have

pz_f(p) +zc(p) =0.

Assumptions for existence of WE:

zi(p)is continuous for allp>>p, i.e., for allp>0.

Multiple WE: Example II

there exists smallp= >0 s.t.z_f(,1)>0, and there exists anotherp⁰> ¹ s.t.z_c(p⁰,1)>0.

Additional assumption for uniqueness of WE:

z_f⁰(p)<0 for allp>0.

Let

E={p|p∈P, andz(p) =0}.

Later on, we will show that:

Theorem

For an N×M economy, under suitable conditions, Either there will be a unique equilibrium The number of equilibria will be odd.

WARP and no of WE I

Let,

x(p)denote the demanded bundle at pricep.

Definition

The demand satisfies WARP, if

px(´p)≤px(p) ⇒ px(´´ p)<´px(p)).

p(x(´p)−x(p))≤0 ⇒ p(x(´´ p)−x(p))<0.

Theorem

If for the aggregate demand WARP holds, then there is unique WE.

WARP and no of WE II

Proof: Suppose, WE is not unique. If possible, supposep,p´∈E, andp6= ´p.

Note

p.z(p) = 0,i.e., p.(x(p)−e) = 0,i.e., p.(x(p)−x(´p)) = 0,i.e.,

p.(x(´p)−x(p)) = 0. (1) From WARP, we know that

p(x(´p)−x(p))≤0⇒p(x(´´ p)−x(p))<0. (2) (1) and (2) give us,

p.(x(´´ p)−x(p))<0. (3)

WARP and no of WE III

But

p.z(´´ p) = 0,i.e., p.(x(´´ p)−e) = 0

p.(x(´´ p)−x(p)) = 0 (4) which is a contradiction, since in view of (3), we have

p(x(´´ p)−x(p))<0.

Gross Substitutes I

Suppose,

There are two goods

Consider two price vectors:p= (p₁,p₂) = (2,1)and p¯= (¯p₁,p¯₂) = (4,1).

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

Question

If the above goods are ‘gross substitutes’ for individual i, how will x₂ⁱ(p) compare with x₂ⁱ(¯p)?

Gross Substitutes II

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=maxj{^¯p_p^j

j},j=1, ..,M,J =3.

Note hereλ=^p_p^¯3

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 III

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 z_j(¯p)?

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).

GS and No of WE I

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,p´∈E. Moreover,p6= ´p.

Let

λ=max

j {´pj

p_j}forj=1, ..,M. Suppose,

λ= ´pk

pk

GS and No of WE II

Clearly,p¯=λp≥p, and´ p¯k =pkλ= ´pk. Therefore, z_k(¯p=λp)>z_k(´p),

That is,

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

Normal Goods I

In case of two goods, we have

pz_f(p) +z_c(p) =0.

Assumptions:

z_i(p)is continuous for allp>>0, i.e., for allp>0.

there exists smallp= >0 s.t.z_f(,1)>0 and anotherp⁰ >¹ s.t.

z_c(p⁰,1)>0.

Note

zf(p) = (z_f¹)(p) + (z_f²)(p) Since endowments are fixed, we get

Normal Goods II

∂z_f(p)

∂p =

∂x_f¹(p)

∂p

du¹=0

−(x_f¹(p)−e_f¹)

∂x_f¹(p)

∂I

+

∂x_f²(p)

∂p

du²=0

−(x_f²(p)−e_f²)

∂x_f²(p)

∂I

WLOG assumex_f¹(p)−e_f¹>0. Using the fact x_f²(p)−e²_f =−[x_f¹(p)−e¹_f],

we get

∂z_f(p)

∂p =

∂x_f¹(p)

∂p

du¹=0

+

∂x_f²(p)

∂p

du¹=0

+ (x_f¹(p)−e_f¹)

∂x_f²(p)

∂I −∂x_f¹(p)

∂I

,

Normal Goods III

Now, even if goods are normal,

2 might have large income effect that can offset the negative substitution effect.

∂zf(p)

∂p <0 might not hold.