Economics 201B–Second HalfLecture 2 3/11/10Two Graphical “Proofs” of the Existence of Walrasian Equilibrium 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 representation of Walrasian Equilibrium.In fact, there are (at least) three Walrasian Equilibria in the drawing, and there is nothing apparentlypathological in the preferences of the two agents. Note that if the demands of the two agents at a singleprice p are represented by the same point in the Edgeworth Box, it indicates that the sum of the demandsequals the total supply, so we have Walrasian Equilibrium; on the other hand, if the demands of the twoagents at a price p are represented by different points in the Edgeworth Box, the sum of the demands doesnot equal the total supply; 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 later follows the first of the two “proofs”Price Normalization: p ∈ Δ0= {p ∈ R2++: p1+ p2=1} ;Δ={p ∈ R2+: p1+ p2=1}1Notation:• 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 Edgeworth Box Diagram; OC1measuredfrom O1, OC2from O2.– OC = {x : ∃p∈Δ0x ∈ E(p)} This is a curve in R2.– 0 ∈ OC ⇔ there is a Walrasian Equilibrium: straightforward.– (OC1∩ OC2) \{ω} = ∅⇒there is a Walrasian Equilibrium; we’ll see why.Items Common to the two “Proofs:”• Lemma 1 If pn∈ Δ0and pn→ 0 as n →∞,then|Di(pn)|→∞.This follows from strong monotonicity, and was likely proved in 201A. We’ll prove later in a moregeneral case.• Walras’ Law:– p · Di(p) ≤ p · ωi. Comes from definition, with no assumptions 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. In particular, ∃p∈Δ0(Di(p) <ωi∨ Di(p) >ωi) ∃p∈Δ0(E(p) < 0 ∨ E(p) > 0)(1)2• 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 single element. In other words, every raythrough the origin with negative slope intersects OC in exactly one point other than zero. In theEdgeworth Box diagram, each ray through ω with negative slope intersects OC1and OC2in exactlyone point, other than ω,each. Givenapointx ∈ OC, x = 0, there is a unique p ∈ Δ0such thatx ∈ E(p); p is the perpendicular to the ray through 0 and x.Givenapointx ∈ OCi, x = ω,thereisa unique p ∈ Δ0such that x ∈ Di(p); p is the perpendicular to the ray through ω and x.“Proof 1:” (In Consumption Space, using OC)•0 ∈ OC ⇔∃p∈Δ0E(p)=0⇔ Walrasian Equilibrium existsHence, it suffices to show that 0 ∈ OC•E(p)=D(p) − ¯ω ≥−¯ω (3)3• In the following diagram, Equations (2) and (3) tell us that OC goes from the region (A)(whenp1is small) to the region (B)(whenp2is small).• Equation (1) tells us that OC avoids the first (northeast) and third (southwest) quadrants, so OCmust pass through zero, so Walrasian Equilibrium exists!• However, it appears that OC may go through the origin more than once, reinforcing the earlierconclusion that Walrasian Equilibrium need not be unique.“Proof 2” (uses OC1and OC2as in diagrams in MWG, assumes preferences are smooth)• Suppose x ∈ OC1∩ OC2, x = ω.Thenxi= Di(pi)forsomepi∈ Δ0(i =1, 2), so xi≥ 0, andhence x lies in the Edgeworth Box; although each offer curve can go outside the Edgeworth Box, anyintersection of the offer curves must lie in the Edgeworth Box. There is a unique ray going throughx and ω,andp1and p2are both perpendicular to it, so p1= p2.Sincex is a point in the EdgeworthBox, x1+ x2=¯ω,sop1is a Walrasian Equilibrium Price. In other words, it suffices to show thatOC1∩ OC2contains at least one x = ω.•ω ∈ OC1∩ OC2(4)To see this, let pibe the “support price” to iat ωi. In other words,y iωi⇒ pi· y ≥ pi· ωiWe’ll explain more carefully later why the support price exists. Then ωi= Di(pi)soωi∈ OCi,soω ∈ OC1∩ OC2.4• If preferences are smooth, 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 tangent to OCiat ωi.• If it turns out that p1= p2, then this common price is a Walrasian Equilibrium Price and ω is aWalrasian Equilibrium allocation. If p1= p2,then– OC1and OC2cross at ω.– By Equation (1), OC1∪OC2cannot enter the quadrant northeast of ω or the quadrant southwestof ω.– By Equation (2), as the price of the first good moves from 0 to 1, OC1and OC2travel from (A)to (B). Notice that OC1at (A) lies northeast of OC2at (B), and OC1at (B) lies northeast ofOC2at (A). Thus, OC1and OC2“must” cross an even number of times, hence they cross atsome x = ω, so Walrasian Equilibrium