The existence of walrasian equilibrium Assumption3: market clearing Desirability: z(p)>0 for p=0 lemma: free goods: if p"" is a walrasian equilibrium price, andz (p)<0, then p=0 lemma: Equivalent of demand and supply: all the commodities are desirable and p is a walrasian equilibrium price, then, (p)=0
The existence of Walrasian equilibrium • Assumption3: market clearing • Desirability: • lemma1: free goods: if p * is a walrasian equilibrium price, and ,then • lemma2: Equivalent of demand and supply: all the commodities are desirable and p * is a walrasian equilibrium price, then ( ) 0 for 0 i z p p = ( ) 0 j z p 0 j p = ( ) 0 j z p =
The existence of walrasian equilibrium Brouwer's fixed-point theorem: continuous function f: s">s",(s is a unit simplex) and there exist anx. that x=f(x) Scarf. 1973 when n=1; let g(x)=f(x)-x g(0)=f(0)-0≥0g(1)=f(1)-1≤0 彐x,g(x)=0=f(x)-x
The existence of Walrasian equilibrium • Brouwer’s fixed-point theorem: continuous function , ( is a unit simplex) and there exist an x, that: • Scarf,1973. • when n=1;let g(x)=f(x)-x, : n n f s s → x x = f ( ) g f g f (0) (0) 0 0; (1) (1) 1 0 = − = − = = − x g x f x x , ( ) 0 ( ) n s
The existence of walrasian equilibrium Ifzsk->k is a continuous function witch satisfied Walras'law, pz(p)=0, then there exist some p in sk-I such that =(p)<0 Construct a unit simplex: p K ∑ s={pin9:),p;=1 k-1 Defined 8.S>S g, (p) P1+max(0,2=(p) max(0,二,(p)
The existence of Walrasian equilibrium • If is a continuous function witch satisfied Walras’ law, , then there exist some in such that . – Construct a unit simplex: – Defined 1 : k k z s − → p p z( ) 0 = p k 1 s − z( ) 0 p 1 i i K j j p p p = = 1 1 { in : 1 k k k i i s p − = = = p 1 1 : k k g s s − − → 1 max(0, ( )) ( ) 1 max(0, ( ) i i i K j j p z g z = + = + p p p
The existence of walrasian equilibrium By brouwer's fixed-point theorem, there is a p, such that p=g(p) p=g(p) p1+max(0,2(p) 1+∑max(0,=,(p) So p'max(O, =(p)1=max(0,=(p) for i=1.k
The existence of Walrasian equilibrium – By brouwer’s fixed-point theorem, there is a p * , such that – So g( ) p p = 1 max(0, ( )) ( ) 1 max(0, ( ) i i i i K j j p z p g p z = + = = + p p 1 [ max(0, ( )] max(0, ( )) for 1 K i j i j p z z i k = p p = =