General Equilibrium Analysis: Lecture 5

General Equilibrium Analysis: Lecture 5

Ram Singh

Course 001

September 24, 2014

Individual UMP I

Take a price vectorp= (p₁, ...,p_M)∈R^M++. Assume(p₁, ...,p_M)>(0, ...,0).

The consumeri solves:

max

x∈R^J+

uⁱ(x) s.t. p.x≤p.eⁱ

Assumption

For alli∈I,uⁱ is continuous, strongly increasing, and strictly quasi-concave onR^M+

Definition

uⁱ is strongly increasing if for any two bundlesxandx⁰ x⁰ ≥x⇒uⁱ(x⁰)>uⁱ(x).

Individual UMP II

In view of monotonicity, for givenp= (p₁, ...,p_M)>>(0, ...,0), consumeri solves:

max

x∈R^M+

uⁱ(x) s.t. p.x=p.eⁱ (1)

Theorem

Under the above assumptions on uⁱ(.), for every(p₁, ...,p_M)>(0, ...,0), (1) has a unique solution, sayxⁱ(p,p.eⁱ).

Note: For eachi =1, ...,N,

xⁱ(p,p.eⁱ) :R^M++7→R^M+;

xⁱ(p,p.eⁱ) = (x₁ⁱ(p,p.eⁱ,x_Mⁱ (p,p.eⁱ).

Individual UMP III

Theorem

Under the above assumptions on uⁱ(.), for every(p1, ...,pM)>(0, ...,0), xⁱ(p,p.eⁱ)is continuous inpoverR^M+.

For all i=1,2, ...,N, we have:xⁱ(tp) =xⁱ(p), for all t>0. That is, demand of each good j by individual i satisfies the following property:

x_jⁱ(tp) =x_jⁱ(p)for all t >0.

Question

Given that uⁱ(.)is strongly increasing, isxⁱ(p)continuous overR^M+?

is the demand function x_jⁱ(p)defined at pj =0?

Excess Demand Function I

Definition

The excess demand forjth good by theith individual is give by:

zⁱ_j(p) =x_jⁱ(p,p.eⁱ)−eⁱ_j. The aggregate excess demand forjth good is give by:

z_j(p) =

N

X

i=1

x_jⁱ(p,p.eⁱ)−

N

X

i=1

e_jⁱ.

So, Aggregate Excess Demand Function is:

z(p) = (z₁(p), ...,z_M(p)),

Excess Demand Function II

Theorem

Under the above assumptions on uⁱ(.), for anyp>>0, z(.)is continuous inp

z(tp) =z(p), for all t>0 p.z(p) =0. (the Walras’ Law) For any given price vectorp, we have

p.xⁱ(p,p.eⁱ)−p.eⁱ = 0,i.e.,

M

X

j=1

pj[x_jⁱ(p,p.eⁱ)−eⁱ_j] = 0.

This gives:

Excess Demand Function III

N

X

i=1 M

X

j=1

p_j[x_jⁱ(p,p.eⁱ)−eⁱ_j] = 0,i.e.,

M

X

j=1 N

X

i=1

pj[x_jⁱ(p,p.eⁱ)−eⁱ_j] = 0,i.e.,

M

X

j=1

p_j

" _N X

i=1

x_jⁱ(p,p.eⁱ)−

N

X

i=1

e_jⁱ

#

= 0

That is,

M

X

j=1

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

Excess Demand Function IV

So,

p1z1(p) +p2z2(p) +...+pj−1zj−1(p) +pj+1zj+1(p) + +p_MzM(p) =−p_jzj(p) For a price vectorp>>0,

ifzj⁰(p) =0 for allj⁰ 6=j, thenzj(p) =0 For two goods case,

p₁z₁(p) =−p₂z₂(p).

So,

z₁(p)>0⇒z₂(p)<0; andz₁(p) =0⇒z₂(p) =0

Walrasian Equilibrium I

Definition

Walrasian Equilibrium Price: A price vectorp^∗is equilibrium price vector, if for allj =1, ...,J,

z_j(p^∗) =

N

X

i=1

x_jⁱ(p^∗,p^∗.eⁱ)−

N

X

i=1

eⁱ_j =0, i.e., if

z(p^∗) =0= (0, ...,0).

Two goods:food and cloth

Let(pf,pc)be the price vector. We can work withp= (^p_p^f

c,1) = (p,1). Since, we know that for allt>0:

z(tp) =z(p)

Walrasian Equilibrium II

Therefore, we have

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

Assumptions:

zi(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_f(p⁰,1)<<0.

WE: General Case I

References: Arrow and Debreu (1954), and McKenzie (2008); Arrow and Hahn (1971). Jehle and Reny (2008).

Recall, we have

For alli =1,2, ...,N, we have:xⁱ(tp) =xⁱ(p), for allt >0.

z(tp) =z(p), for allt>0.

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 II

Theorem

Suppose uⁱ(.)satisfies the above assumptions, ande>>0. Let{p^s}be a sequence of price vectors inR^M++, such that

{p^s}converges to¯p, where

p¯∈R^M+,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{p^s}), say{z_k(p^s)}, is unbounded above.

Theorem

Under the above assumptions on uⁱ, there exists a price vectorp^∗>>0, such thatz(p^∗) =0.

WE: Proof I

Let,

¯zj(p) =min{z_j(p),1}. (2) Note

z¯j(p) =min{z_j(p),1} ≤1.Therefore, 0≤max{¯z_j(p),0} ≤1.

Next, let’s define the functionf :P7→Psuch that: Forj =1, ..,M, f_j(p) = +pj+max{¯zj(p),0}

M+1+PM

j=1max{¯zj(p),0} =Nj(p) D(p),

Note thatPM

j=1f_j(p) =1. Moreover, we have f_j(p)≥ Nj(p)

M+1+M.1 ≥

M+1+M.1 ≥

1+2M.

WE: Proof II

Therefore,

f(p) = (f1(p), ...,fM(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., f_j(p) =p_j, for allj =1, ..,M.

That is, for allj=1, ..,M,

N_j(p)

D(p) = p_j,i.e., p_j[M+

M

X

j=1

max{z¯_j(p),0}] = +max{z¯_j(p),0}. (3)

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^∗, as→0.

Clearly,p^∗≥0. Why?

Suppose,p^∗_k =0. Recall, we have

p_k⁰



M⁰+

M

X

j=1

max{¯zj(p⁰),0}



=⁰+max{¯zk(p⁰),0}. (4)

as⁰ →0 while the LHS converges to 0, since lim⁰→0p_k⁰ =0 and term [M⁰+PM

j=1max{¯zj(p⁰),0}]on LHS is bounded.

WE: Proof IV

However, the RHS takes value 1 infinitely many times. Why? This is a contradiction, because the equality in (4) holds for all values of⁰. Therefore, p^∗_j >0 for allj=1, ..,M. That is,

p^∗>>0,i.e.,

→0limp=p^∗>>0.

In view of continuity ofz¯(p)overR^M++, from (4) we get (by taking limit→0):

WE: Proof V

For allj=1, ..,M

p_j^∗

M

X

j=1

max{¯zj(p^∗),0} = max{z¯j(p^∗),0},i.e.,

zj(p^∗)p^∗_j





M

X

j=1

max{¯zj(p^∗),0}



 = zj(p^∗)max{¯zj(p^∗),0},i.e.,

M

X

j=1

zj(p^∗)p^∗_j





M

X

j=1

max{¯zj(p^∗),0}



 =

M

X

j=1

zj(p^∗)max{¯zj(p^∗),0},i.e.,

M

X

j=1

z_j(p^∗)max{¯z_j(p^∗),0}=0. (5)

WE: Proof VI

You can verify that, given the definition ofz¯_j(p^∗):

zj(p^∗)>0 ⇒ max{¯zj(p^∗),0}>0;

z_j(p^∗)≤0 ⇒ max{¯z_j(p^∗),0}=0.

Suppose, for somej, we havezj(p^∗)>0, then we will have

M

X

j=1

zj(p^∗)max{z¯j(p^∗),0}>0. (6) But, this is a contradiction in view of (5). Therefore:

For anyj=1, ..,M, we have

zj(p^∗)≤0. (7)

WE: Proof VII

Suppose,z_k(p^∗)<0 for somek. We know

p^∗₁z1(p^∗) +...+p^∗_kzk(p^∗) +...+p^∗_MzM(p^∗) =0.

Sincep_j^∗>0 for allj =1, ..,M.

z_k(p^∗)<0 implies: There existsk⁰, such that

zk⁰(p^∗)>0, (8)

which is a contradiction in view of (7). Therefore,

For all j =1, ..,M, we have:zj(p^∗) = 0,i.e., z(p^∗) = 0.