A feasible allocation \(\{x^i,y^j\}\) Pareto dominates a feasible allocation \(\{\bar x^i,\bar y^j\}\) if $$u^i(x^i)\ \ge\ u^i(\bar x^i)\ \text{ for all } i\in I,\qquad u^{i'}(x^{i'})\ >\ u^{i'}(\bar x^{i'})\ \text{ for some } i'\in I .$$ A feasible allocation \(\{\bar x^i,\bar y^j\}\) is Pareto optimal (PO) if no feasible allocation Pareto dominates it.
Three remarks that are easy to forget under exam pressure:
\(\bar x\) is weakly PO if there is no feasible \(x\) with \(u^i(x^i)>u^i(\bar x^i)\) for every \(i\). PO ⇒ weakly PO. The converse can fail when some agent is satiated. You will see in Lesson 4 that every CE is weakly PO with no assumptions at all; LNS is what upgrades this to full PO.
Use the Edgeworth box from Lesson 2: total endowment \((1,1)\), \(u^A=a\log x_1+(1-a)\log x_2\), \(u^B=b\log x_1+(1-b)\log x_2\). Pick any allocation, A's bundle \(\bar x=(\bar x_1,\bar x_2)\) and B's \(\mathbf 1-\bar x\), and draw both indifference curves through it.
"MRSs equal" characterizes PO only for interior allocations with smooth, convex preferences. The definition is "no feasible Pareto improvement". At corners, with kinks (Leontief), or with non-convexities you must go back to the definition. Exercise 3.3 is a typical exam case.
\((u^i,X^i)\) satisfies local non-satiation if for every \(x\in X^i\) and every \(\varepsilon>0\) there is an \(\hat x\in B_\varepsilon(x)\cap X^i\) with \(u^i(\hat x)>u^i(x)\), where $$B_\varepsilon(x)=\{x'\in X^i:\ \|x-x'\|\lt \varepsilon\},\qquad \|x\|^2=\sum_{k=1}^m x_k^2 .$$
In words: however close you insist on staying, there is always something slightly better nearby. Note that LNS is a joint property of \(u^i\) and \(X^i\): the better bundle must be consumable. Two pictures show how it fails:
Let \((u^i,X^i)\) satisfy LNS, let \(p\) be a price vector, and let \(\bar x\) maximize \(u^i\) on \(\{x\in X^i: p\cdot x\le M\}\). Then
Parts 2 and 3 are, word for word, equations (1) and (2) in the proof of the First Welfare Theorem (Lesson 4). Part 1 is what makes Walras' law hold (Lesson 2).
The proof only ever perturbs a bundle by a tiny amount to make a budget argument. So we need "better is available nearby", not "more is better". LNS allows goods that are bads (\(u=x_2-x_1^2\) satisfies it: just raise \(x_2\)), and it says nothing about convexity or continuity. It is about the weakest assumption that makes "money left on the table is wasted" true.
Fix the vocabulary first; the answers depend on it. For bundles in \(\mathbb{R}^m\), write \(x'\ge x\) for "every coordinate weakly larger" and \(x'\gg x\) for "every coordinate strictly larger".
| Name | Condition |
|---|---|
| increasing (weakly) | \(x'\ge x\ \Rightarrow\ u(x')\ge u(x)\) |
| monotone (MWG) | \(x'\gg x\ \Rightarrow\ u(x')>u(x)\) |
| strictly increasing | \(x'\ge x,\ x'\neq x\ \Rightarrow\ u(x')>u(x)\) |
Throughout, take \(X^i=\mathbb{R}^m_+\) (or any set such that \(x\in X^i\) implies \(x+t\,\mathbf e_k\in X^i\) for small \(t>0\)). Try the three questions as Exercise 3.1 before opening the solutions.
On exams, Pareto optimality and LNS show up in two ways: as true/false traps about exact hypotheses, and inside longer questions that ask you to characterize the PO allocations of a specific economy, usually with Pareto weights (the planner approach is developed in Lesson 21). The exercises below practise both.
For each statement, answer True or False and justify in at most four lines.
False. Interior PO allocations satisfy \(\lambda^i\pi^i(s)v^{i\prime}(c^i(s))=\gamma(s)\), so \(\frac{v^{i\prime}(c^i(s))}{v^{j\prime}(c^j(s))}=\frac{\lambda^j\pi^j(s)}{\lambda^i\pi^i(s)}\). With different beliefs the likelihood ratio \(\pi^j(s)/\pi^i(s)\) varies across states, so two states with the same aggregate endowment can give \(i\) different consumption: agents bet on their beliefs. (Exercise 3.6.)
True. The probabilities cancel from the ratio above, so \(v^{i\prime}(c^i(s))/v^{j\prime}(c^j(s))=\lambda^j/\lambda^i\) is the same in every state. Together with \(\sum_ic^i(s)=Y(s)\), this pins down each \(c^i(s)\) as an increasing function of \(Y(s)\) alone: perfect risk sharing.
False. With strictly increasing utilities, giving the whole aggregate endowment to one agent is PO: any other feasible allocation takes something from her and makes her strictly worse off (Exercise 3.4). PO is about waste, not distribution.
False without convexity. Tangency is necessary for an interior PO, but sufficient only with quasi-concave utilities. Counterexample: total endowment \((1,1)\), \(u^A=u^B=x_1^2+x_2^2\). At \(x^A=x^B=(\frac12,\frac12)\) both MRSs equal 1, yet giving A \((1,0)\) and B \((0,1)\) raises each utility from \(\frac12\) to 1.
True, and continuity is not even needed. Given \(x\) and \(\varepsilon\), the bundle \(x+\frac\varepsilon2\mathbf e_1\) is within \(\varepsilon\) of \(x\) and strictly preferred by strict monotonicity (Exercise 3.1).
False. \(u(x_1,x_2)=x_1-x_2\) treats good 2 as a bad but satisfies LNS: from any \(x\), increase \(x_1\) slightly. LNS only requires some better bundle nearby, in any direction.
True. At an interior point \(x\) of the thick set, a small enough ball around \(x\) lies inside the set, so every nearby bundle is indifferent to \(x\) and none is strictly better.
For (a), construct \(\hat x\) explicitly and check its distance from \(x\). Then ask which hypotheses your construction actually used. For (b), think about the most boring increasing function there is.
(a) Yes. Fix \(x\in X\) and \(\varepsilon>0\). Let \(\hat x = x+\tfrac{\varepsilon}{2}\mathbf e_1\). Then \(\hat x\in X\) (we may increase a coordinate), \(\|\hat x-x\|=\varepsilon/2\lt \varepsilon\), and \(\hat x\ge x\) with \(\hat x\ne x\), so strict monotonicity gives \(u(\hat x)>u(x)\). ∎
(b) Not in general. A constant function \(u\equiv 0\) is continuous and (weakly) increasing but nothing is ever strictly better. A less trivial example is the thick-band utility in the figure: continuous, increasing, and violating LNS inside the band. Caveat: if "increasing" is read as MWG's "monotone" (\(x'\gg x\Rightarrow u(x')>u(x)\)), then the answer is yes: take \(\hat x=x+\frac{\varepsilon}{2\sqrt m}\mathbf 1\), which has \(\|\hat x-x\|=\varepsilon/2\) and \(\hat x\gg x\). On an exam, state which definition you use.
(c) Yes. The proof of (a) never used continuity. Continuity is irrelevant for LNS; it matters for existence of optima and for the Second Welfare Theorem (Arrow's remark, Lesson 7). The slide pairs (a) with (c) to make exactly this point.
Use the Edgeworth box of Exercise 2.2: \(u^A=a\log x_1+(1-a)\log x_2\), \(u^B=b\log x_1+(1-b)\log x_2\), total endowment \((1,1)\). Write \(k_a=\frac{a}{1-a}\) and \(k_b=\frac{b}{1-b}\).
(a) \(MRS^A=k_a\frac{x_2}{x_1}\), \(MRS^B=k_b\frac{1-x_2}{1-x_1}\).
(b) Setting them equal: \(k_ax_2(1-x_1)=k_b(1-x_2)x_1\), so \(\phi(x_1)=\dfrac{k_bx_1}{k_a(1-x_1)+k_bx_1}\).
(c) \(k_a(1-a)=a\), so \(\phi(a)=\frac{k_ba}{a+k_ba}=\frac{k_b}{1+k_b}=b\) ✓.
(d) In a CE each consumer's interior optimum sets \(MRS^i=p_1/p_2\). Both face the same price ratio, so \(MRS^A=MRS^B\): tangency, which (with these convex preferences) is the contract curve.
(e) When \(a=b\): then \(\phi(x_1)=x_1\). Identical homothetic preferences give both agents the same expenditure shares, so efficiency requires the same ratio \(x_2/x_1\) as the aggregate endowment.
Total endowment \((1,1)\). \(u^A=x_1+x_2\) (perfect substitutes) and \(u^B=\min\{x_1,x_2\}\) (Leontief). Characterize the set of all PO allocations.
If B's bundle is off its diagonal, B has some "useless" units. What happens if you give those to A? Then write A's utility in terms of B's bundle: \(u^A=2-(x_1^B+x_2^B)\).
Claim: PO ⇔ B's bundle is on the diagonal, \(x^B=(t,t)\) for some \(t\in[0,1]\), and A gets \((1-t,1-t)\).
(⇒) If \(x^B_1\ne x^B_2\), say \(x^B_1>x^B_2\), move \(x^B_1-x^B_2\) units of good 1 from B to A. B's min is unchanged; A gains strictly. So the original allocation was not PO.
(⇐) At \(x^B=(t,t)\): \(u^B=t\), \(u^A=2-2t\). Any feasible allocation has \(u^B=\min\{x^B_1,x^B_2\}\le\frac{x^B_1+x^B_2}{2}=\frac{2-u^A}{2}\), i.e. \(u^A+2u^B\le 2\), with equality on the diagonal. Raising one agent's utility therefore strictly lowers the other's bound, so no Pareto improvement exists. Note that MRS^B is undefined at the kink, so tangency would not have helped.
Pure exchange, \(I\) agents, each \(u^i\) strictly increasing on \(\mathbb{R}^m_+\). Show that the allocation giving the entire aggregate endowment to agent 1 is PO.
Let \(\bar x^1=\sum_i e^i\) and \(\bar x^i=0\) for \(i\ne1\). Take any feasible \(x\ne\bar x\). By feasibility \(x^1=\sum_i e^i-\sum_{i\ne1}x^i\le\bar x^1\), and since \(x\ne\bar x\), some \(x^i\ne0\) for \(i\ne1\), so \(x^1\le\bar x^1\) with \(x^1\ne\bar x^1\). Strict monotonicity gives \(u^1(x^1)\lt u^1(\bar x^1)\): agent 1 is strictly worse. Hence no Pareto improvement exists. ∎
Two goods, two agents, each endowed with \((1,1)\). \(u^A=\min\{x_1+x_2,\,1\}\) and \(u^B=x_1+x_2\). Show that \(p=(1,1)\) with no trade is a CE, and that this CE is not PO. Which part of the Lemma fails for A?
At \(p=(1,1)\), A's budget is \(x_1+x_2\le 2\). Every bundle with \(x_1+x_2\ge1\) gives the maximum utility 1, so \(x^A=(1,1)\) is optimal. B maximizes \(x_1+x_2\) subject to \(x_1+x_2\le2\), so \(x^B=(1,1)\) is optimal. Markets clear: this is a CE.
Not PO: give A \((\tfrac12,\tfrac12)\) and B \((\tfrac32,\tfrac32)\). A still gets 1; B gets 3 instead of 2.
For A, \((\tfrac12,\tfrac12)\) is weakly preferred to \((1,1)\) yet strictly cheaper. That is a failure of Part 2 of the Lemma, which is equation (1) in the 1WT proof. Notice that A's chosen bundle does exhaust its budget; what fails is Part 2, not Part 1. A could have spent less and been just as happy, and that wasted value is what B could have used.
One good, \(S\) states of nature, so a bundle is \(c^i=(c^i(1),\dots,c^i(S))\). Agent \(i\) has expected utility \(u^i(c^i)=\sum_s\pi^i(s)v^i(c^i(s))\), with \(v^i\) strictly increasing, strictly concave and satisfying Inada conditions, and beliefs \(\pi^i(s)>0\). The aggregate endowment in state \(s\) is \(Y(s)>0\).
Divide the FOC by \(\pi(s)\) and define \(\tilde\gamma(s)=\gamma(s)/\pi(s)\). Then each \(c^i(s)\) is a decreasing function of \(\tilde\gamma(s)\) alone. What pins down \(\tilde\gamma(s)\)?
(a) \(\max\sum_i\lambda^i\sum_s\pi^i(s)v^i(c^i(s))\) s.t. \(\sum_ic^i(s)=Y(s)\) for each \(s\). FOC (interior by Inada): \(\lambda^i\pi^i(s)v^{i\prime}(c^i(s))=\gamma(s)\).
(b) With \(\pi^i=\pi\): \(v^{i\prime}(c^i(s))=\tilde\gamma(s)/\lambda^i\), so \(c^i(s)=h^i(\tilde\gamma(s))\) with \(h^i=(v^{i\prime})^{-1}(\cdot/\lambda^i)\) strictly decreasing. Resource balance \(\sum_ih^i(\tilde\gamma)=Y(s)\) has a left side strictly decreasing in \(\tilde\gamma\), so it defines \(\tilde\gamma=\Gamma(Y(s))\) with \(\Gamma\) decreasing. Hence \(c^i(s)=h^i(\Gamma(Y(s)))\equiv g^i(Y(s))\), increasing in \(Y\).
(c) \(c^{-\sigma}=\tilde\gamma/\lambda^i\) gives \(c^i=(\lambda^i/\tilde\gamma)^{1/\sigma}\). Summing: \(\tilde\gamma^{-1/\sigma}\sum_j(\lambda^j)^{1/\sigma}=Y\). Substituting back, \(c^i=\omega^iY\).
(d) From the FOCs of agents 1 and 2: \(\frac{v^{1\prime}(c^1(s))}{v^{2\prime}(c^2(s))}=\frac{\lambda^2}{\lambda^1}\cdot\frac{\pi^2(s)}{\pi^1(s)}\). Take states \(s,s'\) with \(Y(s)=Y(s')\) and \(\frac{\pi^1(s)}{\pi^2(s)}>\frac{\pi^1(s')}{\pi^2(s')}\). If \(c^1(s)\le c^1(s')\), then \(c^2(s)\ge c^2(s')\) by feasibility (with two agents), so the left side would be weakly larger in \(s\) than in \(s'\), while the right side is strictly smaller: contradiction. So \(c^1(s)>c^1(s')\).
(e) By the 1WT, every complete-markets CE is Pareto optimal, so it is one of the allocations in (a), with weights \(\lambda^i=1/\mu^i\) (Exercise 2.5). With common beliefs it therefore features perfect risk sharing: \(c^i(s)=g^i(Y(s))\), and individual endowment risk is fully insured.