Economics 201B–Second HalfLecture 6, 4/1/10Revised 4/6/10, Revisions Marked by ** and StickyNotesThe Second Welfare Theorem in the Arrow DebreuEconomyTheorem 1 (Second Welfare Theorem) (Pure ExchangeCase) If x∗is Pareto Optimal in a pure exchange economy,with strongly monotone, continuous, convex preferences, thereexists a price vector p∗and an income transfer T such that(p∗,x∗,T) is a Walrasian Equilibrium with Transfers.Outline of Proof:• LetAi= {x i− x∗i: x iix∗i}A =Ii=1Ai= {a1+ ···+ aI: ai∈ Ai}Then 0 ∈ A (if it were, we’d have a Pareto improvement).• By Minkowski’s Theorem, find p∗=0suchthatinf p∗· A ≥ 0• Show RL+\{0}⊂ Aiand hence p∗≥ 0.• Show inf p∗· Ai=0foreachi.• Define T to make x∗iaffordable at p∗:Ti= p∗· x∗i− p∗· ωiShowIi=1Ti=0andx∗i∈ Qi(p∗,T)1• Use strong monotonicity to show that p∗ 0.• Showp∗ 0 ⇒ Qi(p∗,T)=Di(p∗,T)Now, for the details:• LetAi= {x i− x∗i: x iix∗i}A =Ii=1Ai= {a1+ ···+ aI: ai∈ Ai}Claim:0 ∈ AIf 0 ∈ A, there exists ai∈ Aisuch thatIi=1ai=0Letx i= x∗i+ aiSince x i− x∗i= ai∈ Ai,wehavex iix∗iIi=1x i=Ii=1(x∗i+ ai)=Ii=1x∗i+Ii=1ai=Ii=1x∗i=¯ωTherefore, x is an exact** allocation, x Pareto improves x∗,so x∗is not Pareto Optimal, contradiction. Therefore, 0 ∈ A.2•∃p∗=0inf p∗· A ≥ 0Aiis convex, so A is convex (easy exercise). By Minkowski’sTheorem, there exists p∗=0suchthat0=p∗· 0 ≤ inf p∗· A =Ii=1inf p∗· AiThe fact that inf p∗· A =Ii=1inf p∗· Aiis an exercise; onceyou figure out what you have to prove, it is obvious.• We claim that p∗≥ 0.Suppose not, so p∗ < 0forsome ,WLOGp∗1< 0. Letx i= x∗i+⎛⎜⎜⎝−1p∗1, 0,...,0⎞⎟⎟⎠By strong monotonicity, x iix∗i,so⎛⎜⎜⎝−1p∗1, 0,...,0⎞⎟⎟⎠∈ AiSoinf p∗· Ai≤ p∗·⎛⎜⎜⎝−1p∗1, 0,...,0⎞⎟⎟⎠= −1 < 0inf p∗· A =Ii=1inf p∗· Ai≤−I< 0a contradiction that shows p∗≥ 0.• We claim that inf p∗· Ai=0foreachi:Suppose ε>0. By strong monotonicity,x∗i+(ε,...,ε) ix∗i3so(ε,...,ε) ∈ Aisoinf p∗· Ai≤ p∗· (ε,...,ε)Since ε is an arbitrary positive number, inf p∗· Aiis less thanevery positive number, soinf p∗· Ai≤ 0SinceIi=1inf p∗· Ai≥ 0,inf p∗· Ai=0(i =1,...,I)• Define T to make x∗iaffordable at p∗. We claim that T is anincome transfer andx∗i∈ Qi(p∗,T)LetTi= p∗· x∗i− p∗· ωiIi=1Ti=Ii=1(p∗· x∗i− p∗· ωi)= p∗·⎛⎜⎝Ii=1x∗i−Ii=1ωi⎞⎟⎠= p∗· (¯ω − ¯ω)=0so T is an income transfer.p∗· x∗i= p∗· (ωi+(x∗i− ωi))= p∗· ωi+ p∗· (x∗i− ωi)= p∗· ωi+ Ti4sox∗i∈ Bi(p∗,T)If x iix∗i,thenx i− x∗i∈ Ai,sop∗· x i= p∗· (x∗i+(x i− x∗i))= p∗· x∗i+ p∗· (x i− x∗i)≥ p∗· x∗i+infp∗· Ai= p∗· x∗i= p∗· ωi+ Tisox∗i∈ Qi(p∗,T) ∗∗• Use strong monotonicity to show that p∗ 0.Lemma 2 If iis continuous and complete, and x iy,then there exists ε>0 such that(B(x, ε) ∩ Xi) iyProof: **If (B(x, ε) ∩ Xi)={x} for some ε>0, i.e. x isan isolated point in Xi, then the lemma is true, since x iy.If x is not an isolated point in Xi,thenwecanfindxn→ x,xn∈ Xi, xniy; by completeness, we have y ixnfor eachn.Sinceiis continuous, y ix,sox iy, a contradictionwhich proves the lemma.Since p∗≥ 0andp∗=0,p∗> 0; since in addition ¯ω  0,p∗· ¯ω>0, sop∗· ωi+ Ti> 0forsomeiIf p∗ =0forsome (WLOG = 1), letx i= x∗i+(1, 0,...,0)5By strong monotonicity, x iix∗i.p∗· x i= p∗· x∗i= p∗· ωi+ Ti> 0Find (WLOG = 2) such thatp∗ > 0,x 2i> 0Since x iix∗i,letε>0 be chosen to satisfy the conclusionof the Lemma. If necessary, we may make ε smaller to ensurethat ε ≤ 2x 2i.Letx i= x i− (0,ε/2, 0,...,0)Since Xi= RL+, x i∈ Xi, so by the Lemma, x iix∗i.Butp∗· x i<p· x i= p∗· ωi+ Ti, which shows that x∗i∈ Qi(p∗,T),a contradiction which proves that p∗ 0.• Showp∗ 0 ⇒ Qi(p∗,T)=Di(p∗,T)– Case 1: p∗· ωi+ Ti=0. Sincep∗ 0, Bi(p∗,T)={0},soQi(p∗,T)=Di(p∗,T)={0}– Case 2: p∗· ωi+ Ti> 0Suppose x ∈ Qi(p∗,T) but x ∈ Di(p∗,T). Then thereexists z ix such that z ∈ Bi(p∗,T), hence p∗· z ≤p∗· ωi+ Ti.Sincex ∈ Qi(p∗,T), p∗· z ≥ p∗· ωi+ Ti,sop∗· z = p∗· ωi+ Ti> 0By Lemma 2, there exists ε>0 such that|z − z| <ε,z ∈ RL+⇒ z  x6Letz = z⎛⎜⎜⎝1 −ε2|z|⎞⎟⎟⎠Since z ∈ RL+, z ∈ RL+.|z − z| =εz2|z|=ε2<εso z  x.p∗· z = p∗· z⎛⎜⎜⎝1 −ε2|z|⎞⎟⎟⎠=(p∗· ωi+ Ti)⎛⎜⎜⎝1 −ε2|z|⎞⎟⎟⎠<p∗· ωi+ Tiwhich contradicts the assumption that x ∈ Qi(p∗,T). Thisshows Qi(p∗,T) ⊂ Di(p∗,T); since clearly Di(p∗,T) ⊂Qi(p∗,T), Qi(p∗,T)=Di(p∗,T).What if preferences are not convex?• Second Welfare Theorem may fail if preferences are nonconvex.• Diagram gives an economy with two goods and two agents,and a Pareto optimum x∗so that so that the utility levels ofx∗cannot be approximated by a Walrasian Equilibrium withTransfers.• If p∗is the price which locally supports x∗,andT is the in-come transfer which makes x affordable with respect to theprices p∗, there is a unique Walrasian equilibrium with trans-fers (z∗,q∗,T); z∗is much more favorable to agent I and muchless favorable to agent II than x∗is.7• This is the worst that can happen under standard assumptionson preferences. Given a Pareto optimum x∗, there is a Wal-rasian quasiequilibrium with transfers (z∗,p∗,T) such that allbut L people are indifferent between x∗and z∗.ThoseL peo-ple are treated quite harshly (they get zero consumption). Onecould be less harsh and give these L people carefully chosenconsumption bundles in the convex hull of their quasidemandsets, but one would then have to forbid them from trading,a prohibition that would in practice be difficult to