Economics 201B–Second HalfLecture 2 3/11/10, Revised 3/17/2010, Revisionsnoted by ** and Sticky NotesTwo Graphical “Proofs” of the Existence of WalrasianEquilibrium in the Edgeworth BoxDemand: Di(p)={x ∈ Bi(p):∀y∈Bi(p)x iy}Walrasian Equilibrium (in the Edgeworth Box) is a pair (p, x)where• x is an exact allocation• xi∈ Di(p)(i =1, 2)In the following Edgeworth Box Diagram, we give a graphical rep-resentation of Walrasian Equilibrium. In fact, there are (at least)three Walrasian Equilibria in the drawing, and there is nothingapparently pathological in the preferences of the two agents. Notethat if the demands of the two agents at a single price p are rep-resented by the same point in the Edgeworth Box, it indicatesthat the sum of the demands equals the total supply, so we haveWalrasian Equilibrium; on the other hand, if the demands of thetwo agents at a price p are represented by different points in theEdgeworth Box, the sum of the demands does not equal the totalsupply; p is not an equilibrium price.Why the quotes on “Proofs”? Why the Proofs inside the quotes?• graphical arguments prone to introduction of tacit assumptions• these arguments can be turned into proofs; our real proof laterfollows the first of the two “proofs”Price Normalization: p ∈ Δ0= {p ∈ R2++: p1+ p2=1};1Δ={p ∈ R2+: p1+ p2=1}Notation:• D(p)=D1(p)+D2(p)MarketDemand• Ei(p)=Di(p) − ωiExcess Demand of i• E(p)=E1(p)+E2(p)=D(p) − ¯ω Market Excess Demand• Offer Curve:– OCi= {x : ∃p∈Δ0x ∈ Di(p)} This is a curve in the Edge-worth Box Diagram; OC1measured from O1, OC2fromO2.– OC = {x : ∃p∈Δ0x ∈ E(p)} This is a curve in R2.– 0 ∈ OC ⇔ there is a Walrasian Equilibrium: straightfor-ward.– (OC1∩OC2)\{ω} = ∅⇒there is a Walrasian Equilibrium;we’ll see why.ItemsCommontothetwo“Proofs:”• Lemma 1 If pn∈ Δ0and pn → 0 as n →∞,then|Di(pn)|→∞.This follows from strong monotonicity, and was likely provedin 201A. We’ll prove later in a more general case.• Walras’ Law:– p · Di(p) ≤ p · ωi. Comes from definition, with no assump-tions on preferences.– By strong monotonicity, can’t have p · Di(p) <p· ωi,sop · Di(p)=p · ωi,sop · Ei(p)=0,sop · E(p)=0. In2particular, ∃p∈Δ0(Di(p) <ωi∨ Di(p) >ωi) ∃p∈Δ0(E(p) < 0 ∨ E(p) > 0)(1)• Observe that this is where we use the fact that Di(p) ≥ 0,equivalently Ei(p) ≥−ωi:(pn)2D(pn)2≤ pn· D(pn)= pn· ¯ω≤ max{¯ω1, ¯ω2}If pn1→ 0, pn2→ 1, so for n sufficienly large, D(pn)2≤2max{¯ω1, ¯ω2},soD(pn)2→∞. Therefore,pn1→ 0 ⇒ D(pn)1→∞⇒E(pn)1→∞pn2→ 0 ⇒ D(pn)2→∞⇒E(pn)2→∞(2)• Given p ∈ Δ0, D1(p),D2(p)andE(p) each consist of a singleelement. In other words, every ray through the origin withnegative slope intersects OC in exactly one point other thanzero. In the Edgeworth Box diagram, each ray through ω withnegative slope intersects OC1and OC2in exactly one point,other than ω,each. Givenapointx ∈ OC, x = 0, there is aunique p ∈ Δ0such that x ∈ E(p); p is the perpendicular tothe ray through 0 and x.Givenapointx ∈ OCi, x = ω,thereis a unique p ∈ Δ0such that x ∈ Di(p); p is the perpendicularto the ray through ω and x.“Proof 1:” (In Consumption Space, using OC)•0 ∈ OC ⇔∃p∈Δ0E(p)=0⇔ Walrasian Equilibrium exists3Hence, it suffices to show that 0 ∈ OC•E(p)=D(p) − ¯ω ≥−¯ω (3)• In the following diagram, Equations (2) and (3) tell us thatOC goes from the region (A)(whenp1is small) to the region(B)(whenp2is small).• Equation (1) tells us that OC avoids the first (northeast) andthird (southwest) quadrants, so OC must pass through zero,so Walrasian Equilibrium exists!• However, it appears that OC may go through the origin morethan once, reinforcing the earlier conclusion that WalrasianEquilibrium need not be unique.“Proof 2” (uses OC1and OC2as in diagrams in MWG, assumespreferences are smooth)• Suppose x ∈ OC1∩ OC2, x = ω.Thenxi= Di(pi)forsome pi∈ Δ0(i =1, 2), so xi≥ 0, and hence x lies in theEdgeworth Box; although each offer curve can go outside theEdgeworth Box, any intersection of the offer curves must liein the Edgeworth Box. There is a unique ray going through xand ω,andp1and p2are both perpendicular to it, so p1= p2.Since x is a point in the Edgeworth Box, x1+ x2=¯ω,sop1is a Walrasian Equilibrium Price. In other words, it suffices toshow that OC1∩ OC2contains at least one x = ω.•ω ∈ OC1∩ OC2(4)4To see this, let pibe the “support price” to iat ωi.Inotherwords,y iωi⇒ pi· y ≥ pi· ωiWe’ll explain more carefully later why the support price exists.Then ωi= Di(pi)soωi∈ OCi,soω ∈ OC1∩ OC2.• If preferences are smooth **“smooth preferences” is a termused in the literature. Smooth preferences produce C1demandfunctions**, thenpi· (Di(p) − ωi)= p · (Di(p) − ωi)+(pi− p) · (Di(p) − Di(pi))= 0 (by Walras’ Law) + O(|pi− p|2)which shows that piis **perpendicular to the** tangent toOCiat ωi.• If it turns out that p1= p2, then this common price is aWalrasian Equilibrium Price and ω is a Walrasian Equilibriumallocation. If p1= p2,then– OC1and OC2cross at ω.– By Equation (1), OC1∪ OC2cannot enter the quadrantnortheast of ω or the quadrant southwest of ω.– By Equation (2), as the price of the first good moves from0to1,OC1and OC2travel from (A)to(B). Notice thatOC1at (A) lies northeast of OC2at (B), and OC1at (B)lies northeast of OC2at (A). Thus, OC1and OC2“must”cross an even number of times, hence they cross at somex = ω, so Walrasian Equilibrium