Theory of Income · Part I: Welfare Theorems · Lesson 7 of 25

Arrow's Remark, Decentralization, and Non-Convexities

Covers Slides 1, pp. 31–35 · About 40 min · Prerequisite: Lesson 6

By the end you can:

prove Arrow's remark: expenditure minimization + "not the cheapest point" ⇒ utility maximization;
exhibit the classic counterexample where a PO allocation is a quasi-equilibrium but not a CE, and state assumptions that rule it out;
design endowments and ownership shares that decentralize a given PO allocation, and explain why they are not unique;
draw the two pictures from slide 35 where non-convexities make the 2WT fail.

Primary source Alvarez, Slides 1, pp. 31–35. Mas-Colell, Whinston & Green §16.D, which introduces price quasi-equilibrium with transfers, gives the condition under which it is a full equilibrium, and draws the "exceptional case". The exceptional case goes back to Arrow's 1951 Berkeley Symposium paper, "An Extension of the Basic Theorems of Classical Welfare Economics".

1. Where Lesson 6 left us

The 2WT delivered a price $p\neq0$ at which firms maximize profits and each household's $\bar x^i$ minimizes expenditure over $\{x\in X^i: u^i(x)\ge u^i(\bar x^i)\}$. Give household $i$ exactly enough wealth, $M^i=p\,\bar x^i$, to buy its PO bundle. Is $\bar x^i$ then its utility-maximizing choice in $\{x\in X^i: p\,x\le M^i\}$? These are dual problems, and duality can fail at exactly one kind of point.

2. Arrow's remark S1 p.31

Arrow's remark

Let $X^i$ be convex and $u^i$ continuous (the slides also assume strict quasi-concavity). Suppose $\bar x^i$ solves $\min_{x\in X^i}p\,x$ s.t. $u^i(x)\ge u^i(\bar x^i)$, and suppose $\bar x^i$ is not a cheapest point of $X^i$: there is $\tilde x^i\in X^i$ with $p\,\tilde x^i\lt p\,\bar x^i$. Then $\bar x^i$ solves $\max_{x\in X^i}u^i(x)$ s.t. $p\,x\le p\,\bar x^i$.

Consequently, if this holds for every $i$, the PO allocation and $p$ form a competitive equilibrium for any ownership structure that gives each $i$ wealth $p\,\bar x^i$.

Proof S1 pp.32–33

Suppose not. Then there is $\hat x\in X^i$ with $p\,\hat x\le p\,\bar x^i$ and $u^i(\hat x)>u^i(\bar x^i)$.
Mix toward the cheap point. For $\theta\in(0,1)$ let $x^\theta=\theta\tilde x^i+(1-\theta)\hat x$, which lies in $X^i$ by convexity.
Still good enough. As $\theta\to0$, $x^\theta\to\hat x$. Since $u^i(\hat x)>u^i(\bar x^i)$ and $u^i$ is continuous, $u^i(x^\theta)>u^i(\bar x^i)$ for all $\theta$ small enough. (Only continuity is needed here. Strict quasi-concavity is not.)
Strictly cheaper. $p\,x^\theta=\theta\,p\,\tilde x^i+(1-\theta)\,p\,\hat x\lt \theta\,p\,\bar x^i+(1-\theta)\,p\,\bar x^i=p\,\bar x^i$. The strict inequality comes from $p\,\tilde x^i\lt p\,\bar x^i$ and $\theta>0$.
Contradiction. $x^\theta$ reaches at least utility $u^i(\bar x^i)$ at strictly lower cost than $\bar x^i$. That contradicts expenditure minimization.

∎

Intuition: why we need room to economize

A better bundle $\hat x$ that costs no more than $\bar x^i$ is not itself a contradiction, because expenditure minimization only rules out strictly cheaper ways to reach $u^i(\bar x^i)$. We need to shave a little off $\hat x$'s cost without losing too much utility. Mixing in a strictly cheaper bundle does exactly that. If $\bar x^i$ is already the cheapest point in $X^i$, there is nothing to mix with, and the argument collapses.

3. The exceptional case: a PO allocation that is not a CE S1 p.34

Slide 34 asks for an example with $m=2$ where $\bar x$ minimizes expenditure but the hint warns that "the cost minimizing choice must be the cheapest point". Here is the canonical one. One household, no firms, $X=\mathbb{R}^2_+$, endowment $e=(0,1)$, and $$u(x_1,x_2)=\sqrt{x_1}+x_2 .$$ $u$ is continuous, increasing and concave (so quasi-concave). Good 2 has constant marginal utility, which is the "no Inada condition" of the hint. Good 1 has infinite marginal utility at $x_1=0$.

PO. The only feasible allocation (up to disposal) is $\bar x=e=(0,1)$, so it is trivially PO.
The only supporting price is $p=(1,0)$. The indifference curve through $(0,1)$, $x_2=1-\sqrt{x_1}$, has slope $-1/(2\sqrt{x_1})\to-\infty$ at $x_1=0$: it is vertical at $\bar x$. The only line through $\bar x$ with $A$ on one side is the vertical axis $x_1=0$, so $p\propto(1,0)$. At this $p$, $p\,\bar x=0=\min_{x\ge0}p\,x$: $\bar x$ is a cheapest point.
Quasi-equilibrium, yes. Expenditure $x_1$ is minimized at $x_1=0$, and $(0,1)$ reaches the PO utility. ✓
CE, no. At $p=(1,0)$ the budget set is $\{x_1=0,\ x_2\ \text{anything}\}$: good 2 is free, so utility is unbounded and there is no demand. At any $p$ with $p_2>0$, demand is $x_1=(p_2/2p_1)^2>0=e_1$, so good 1 is in excess demand. No CE exists, even though the 2WT's quasi-equilibrium does.

Interactive. $q=p_1/(p_1+p_2)$ is the weight on good 1. The black line is the budget through the endowment $e=(0,1)$; the blue curve is the indifference curve through $e$; the orange dot is the household's demand. For every $q<1$ the household wants positive $x_1$, and there is none to give. At $q=1$ the budget line becomes the vertical axis, tangent to the indifference curve at $e$, but then good 2 is free and demand runs off to infinity.

Ruling the exceptional case out S1 p.34

Slide 34's last question: what assumptions on $u^i$, $X^i$ guarantee that no PO bundle is a cheapest point? A clean sufficient set:

$X^i=\mathbb{R}^m_+$, so $0\in X^i$ and $p\cdot0=0$.
Preferences are monotone, so $p\ge0$ (Lesson 6, Exercise 6.5). With $p\neq0$, some $p_k>0$.
Interiority: every PO bundle is strictly positive, $\bar x^i\gg0$. This is guaranteed if bundles on the boundary of $\mathbb{R}^m_+$ are worse than every strictly positive bundle, e.g. Cobb–Douglas $\prod_kx_k^{a_k}$, or any $u$ satisfying Inada conditions in every good, as long as each agent gets positive utility at the PO allocation.

Then $p\,\bar x^i\ge p_k\bar x^i_k>0=p\cdot0$, so $\tilde x^i=0$ is strictly cheaper and Arrow's remark applies. Alternative routes: assume $p\gg0$ (e.g. strictly monotone preferences together with an interiority condition) and $\bar x^i\neq0$; or assume each household's wealth exceeds the minimum cost of $X^i$ (a "minimum wealth" or survival condition). In the counterexample, (3) fails: $\bar x_1=0$, and the household has no wealth to spare at $p=(1,0)$.

4. Decentralization: who must own what? S1 p.34

For $(p,\bar x,\bar y)$ to be a CE we need each household's wealth to be exactly the cost of its PO bundle: $$p\,e^i+\sum_j\theta^i_j\,\pi^j\ =\ p\,\bar x^i\quad\text{for all }i,\qquad \pi^j=p\,\bar y^j .$$ If wealth were higher, $i$ would buy more than $\bar x^i$. If lower, $i$ could not afford it.

A decentralizing ownership structure

Pick any shares $\theta^i_j\ge0$ with $\sum_i\theta^i_j=1$ (e.g. $\theta^i_j=1/I$) and set $$e^i\ \equiv\ \bar x^i-\sum_j\theta^i_j\,\bar y^j .$$ Then (i) $\sum_ie^i=\sum_i\bar x^i-\sum_j\bar y^j$ equals the economy's aggregate endowment, by feasibility; and (ii) $p\,e^i+\sum_j\theta^i_j\pi^j=p\,\bar x^i-\sum_j\theta^i_jp\,\bar y^j+\sum_j\theta^i_jp\,\bar y^j=p\,\bar x^i$. ✓

Is it unique? No, far from it. Only the value $p\,e^i+\sum_j\theta^i_j\pi^j$ matters. (i) Any shares $\theta$ work, with $e^i$ adjusted. (ii) The composition of $e^i$ can change in any direction $d^i$ with $p\cdot d^i=0$ and $\sum_id^i=0$. (iii) Equivalently, starting from any initial endowments, lump-sum transfers $T^i=p\,\bar x^i-p\,e^i-\sum_j\theta^i_j\pi^j$ (which sum to zero) implement the allocation. This is the precise sense in which "any efficient allocation can be reached by markets plus lump-sum transfers".

Why this is less useful in practice than it sounds

The transfers depend on $\bar x^i$ and $p$, i.e. on everyone's preferences. They must also be lump-sum: independent of anything the household chooses. Real tax systems condition on income or consumption, which distorts MRS ≠ MRT (Lesson 5, Exercise 5.1, item 9). The 2WT is a benchmark, not a policy recipe.

5. Non-convexities: the two failures on slide 35 S1 p.35

(a) A non-quasi-concave household in the Edgeworth box

Total endowment $(1,1)$. A has $u^A=x_1^2+x_2^2$: strictly increasing (so LNS holds) but convex, with indifference curves that bow outward (quarter circles). B has Leontief preferences $u^B=\min\{x^B_1,x^B_2\}$. Consider $\bar x^A=(t,t)$ and $\bar x^B=(1-t,1-t)$.

It is PO. B is at least as well off only at $x^A\in\{x_1\le t,\ x_2\le t\}$, a square whose farthest point from A's origin is $(t,t)$. So A can't gain without hurting B.
It is not a CE for any prices or transfers. On any budget line through $(t,t)$, A's convex utility is maximized at an endpoint of the line, never in the middle. So A never demands $(t,t)$.

Interactive. Move $t$ along the diagonal (every such point is PO) and tilt the price line through it. The orange L is B's indifference curve, with B's better set shaded below-left. The blue arc is A's indifference curve through $(t,t)$. However you tilt the line, A's demand (violet) jumps to an intercept, which is often outside the box, where A gets strictly higher utility. The PO point is never chosen.

(b) A fixed cost in production

One consumer, one firm, two goods: leisure $\ell$ and consumption $c$. The consumer has one unit of time and utility $\log c+b\log\ell$. The firm needs a fixed labour input $n_0$ before producing anything, and then has increasing marginal cost: $$c=\begin{cases}0 & n\le n_0\\ (n-n_0)^\gamma & n>n_0\end{cases},\qquad 0\lt \gamma<1 .$$ The production set has a flat segment from $n=0$ to $n=n_0$ followed by a concave piece, so it is not convex. The PO allocation is interior (because $\log c\to-\infty$ at $c=0$), at $\bar n=(\gamma+bn_0)/(\gamma+b)$. The only candidate price is the slope there, $w=\gamma(\bar n-n_0)^{\gamma-1}$. At that price the firm's profit from the PO plan is $$\bar\pi=(\bar n-n_0)^{\gamma}-w\bar n=(\bar n-n_0)^{\gamma-1}\big[(1-\gamma)\bar n-n_0\big],$$ which is negative iff $n_0>(1-\gamma)\bar n$, i.e. iff $n_0>\frac{1-\gamma}{1+b}$. In that case the firm prefers to shut down ($n=0$, $\pi=0$), and the PO allocation cannot be decentralized.

Interactive. Blue: the producible set (the flat segment at $c=0$ between $\ell=1-n_0$ and $\ell=1$ is the fixed cost). Orange: the PO indifference curve. Black: the supporting line at the PO point. The violet dot is the "no production" plan $(\ell,c)=(1,0)$. When it lies above the price line, that plan is worth more than the PO plan at the supporting price, so the firm shuts down. Raise $n_0$ past $(1-\gamma)/(1+b)$ and watch it happen.

6. Check your understanding

1. Where does the proof of Arrow's remark use the cheaper bundle $\tilde x^i$?

To show the better bundle is at least as costly as x̄
To make a mixture strictly cheaper than the bundle x̄
To show that the price vector p is strictly positive
To keep the mixture inside the upper contour set of x̄

$p\,x^\theta=\theta p\tilde x+(1-\theta)p\hat x\lt p\bar x$ needs $p\tilde x\lt p\bar x$. Staying above $u(\bar x)$ comes from continuity and $\theta$ small.

2. In the example $u=\sqrt{x_1}+x_2$, $e=(0,1)$, why is there no CE?

Because u is not continuous at the endowment point e
Because the allocation (0,1) is not Pareto optimal
Because u is not quasi-concave on the consumption set
Because good 1 is in excess demand unless p₂ equals 0

For $p_2>0$, demand $x_1=(p_2/2p_1)^2>0$. At $p_2=0$, good 2 is free and demand doesn't exist. $u$ is continuous and concave, and $(0,1)$ is trivially PO.

3. Which ownership structures decentralize a PO allocation at price $p$?

Any with each household's wealth equal to p times its bundle
Only the one giving every household equal shares of each firm
Only the one where endowments equal the PO consumption bundles
Any with endowments proportional to the aggregate endowment

The condition is $p\,e^i+\sum_j\theta^i_j\pi^j=p\,\bar x^i$. Many $(e,\theta)$ satisfy it, so the structure is not unique.

4. With a fixed cost, when does the PO allocation fail to be decentralizable?

Whenever the fixed labour requirement is strictly positive
Whenever the marginal cost of production is increasing
When the firm's profit at the supporting price is negative
When the household's utility function is not quasi-concave

A small fixed cost can be harmless: the firm still prefers the PO plan to shutting down. Failure happens when $\bar\pi<0$, i.e. $n_0>(1-\gamma)/(1+b)$ in the example.

7. Exercises

These follow the format of Alvarez's exams. The true/false block mixes past midterm statements with traps from student review notes. Exercises 7.2–7.6 are the questions on slides 34–35, and Exercise 7.7 is built on a 2025 midterm item.

Exercise 7.1 Past exams True or false? Justify.

One to four lines each.

"Increasing returns to scale in production imply that the First Welfare Theorem fails." student review drill bank; cf. Fall 2025 Midterm 1, I.2
Show answer
False. IRS breaks the convexity assumption of the 2WT, as in §5(b), where a PO allocation cannot be decentralized. The 1WT needs no convexity: any CE that exists is PO.
"Consider a pure endowment economy with aggregate endowment $\bar e$. The competitive equilibrium with equal endowments $e^i=\bar e/I$ is envy-free: no agent prefers another agent's bundle to its own." Fall 2025 Midterm 1, I.6
Show answer
True. Every agent faces the same budget set $\{x:p\cdot x\le p\cdot\bar e/I\}$. So $x^j$ is affordable to $i$, and $u^i(x^i)\ge u^i(x^j)$ by optimality. Exercise 7.7 builds on this.
"Envy-freeness of the equal-endowment CE requires all agents to have the same preferences." student review drill bank
Show answer
False. The argument in item 2 uses only equal budget sets and optimality. It puts no restriction on $u^i$.
"If $u^i$ is strictly increasing, Arrow's condition holds automatically." exam-inspired
Show answer
False. $u=\sqrt{x_1}+x_2$ is strictly increasing, yet $\bar x=(0,1)$ is the cheapest point at $p=(1,0)$ (§3). Monotonicity gives $p\ge0$, not $p\gg0$, and not interiority.
"A small fixed cost of production always makes the 2WT fail." exam-inspired
Show answer
False. In §5(b), decentralization survives as long as $n_0\le(1-\gamma)/(1+b)$ (Exercise 7.6). Non-convexity matters only when it binds at the supporting price, i.e. when "shut down" beats the PO plan.
"A PO allocation supported as a quasi-equilibrium with $p\gg0$, $X^i=\mathbb{R}^m_+$ and every $\bar x^i\ne0$ is a CE with transfers (given continuous $u^i$)." exam-inspired
Show answer
True. With $p\gg0$ and $\bar x^i\ge0$, $\bar x^i\ne0$, we get $p\cdot\bar x^i>0=p\cdot0$. So $\tilde x^i=0$ is strictly cheaper, and Arrow's remark turns expenditure minimization into utility maximization.
"Under the 2WT assumptions and Arrow's condition, the endowments and shares that decentralize a given PO allocation are unique." S1 p.34
Show answer
False. Only each household's wealth $p\cdot\bar x^i$ is pinned down. Any $d^i$ with $p\cdot d^i=0$ and $\sum_id^i=0$ can be added to endowments, and shares can be reshuffled with offsetting endowment changes (Exercise 7.2).
"A planner can decentralize any PO allocation equally well with a proportional tax on labour income, rebated lump-sum, instead of lump-sum transfers." exam-inspired
Show answer
False. The labour tax drives a wedge: households face $(1-t)w$ while firms face $w$, so MRS $\neq$ MRT and the resulting CE is not PO (cf. Exercise 5.1, item 9). The 2WT's transfers must be lump-sum, i.e. independent of anything the agent chooses.

Exercise 7.2 Slides p.34 Ownership structure

Let $\{\bar x^i,\bar y^j\}$ be PO and $p$ the price from the 2WT, with Arrow's condition holding for every $i$.

Describe endowments and shares $(e^i,\theta^i_j)$ for which $(p,\bar x,\bar y)$ is a CE. (10 points)
Is the structure unique? (5 points)
Starting instead from given endowments $\hat e^i$ and shares $\hat\theta^i_j$, what lump-sum transfers work? Show that they sum to zero. (10 points)

Show solution

(a) Any $\theta$ with $\sum_i\theta^i_j=1$, and $e^i=\bar x^i-\sum_j\theta^i_j\bar y^j$ (§4). Firms maximize profit at $p$ by the 2WT. Households have wealth exactly $p\cdot\bar x^i$ and maximize utility by Arrow's remark. Markets clear because the allocation is feasible.
(b) No. Any $(e,\theta)$ with the right wealth values works, e.g. perturb $e^i$ by $d^i$ with $p\cdot d^i=0$ and $\sum_i d^i=0$.
(c) $T^i=p\cdot\bar x^i-p\cdot\hat e^i-\sum_j\hat\theta^i_j\, p\cdot\bar y^j$. Then $\sum_iT^i=p\cdot\big(\sum_i\bar x^i-\sum_i\hat e^i-\sum_j\bar y^j\big)=0$ by feasibility and $\sum_i\hat\theta^i_j=1$.

Exercise 7.3 Slides p.34 The cheapest point

Construct an example with $m=2$ and quasi-concave $u$ in which $\bar x$ minimizes expenditure at given $p$ but does not maximize utility on $\{x:p\cdot x\le p\cdot\bar x\}$. Draw indifference curves and the budget line, and verify every claim.

Hint

The slide's hint: one good with zero price, $u$ increasing, no Inada condition. Make the free good one with constant marginal utility.

Show solution

$u=\sqrt{x_1}+x_2$, $p=(1,0)$, $\bar x=(0,1)$ (§3 and the figure at $q=1$). Expenditure $p\cdot x=x_1\ge0=p\cdot\bar x$, so $\bar x$ minimizes expenditure, and it is the cheapest point of $\mathbb{R}^2_+$. The budget set $\{x_1\le0\}=\{(0,x_2)\}$ contains $(0,2)$ with $u=2>1$, so $\bar x$ is not a utility maximizer. The budget line is the vertical axis, tangent to the indifference curve $x_2=1-\sqrt{x_1}$ at $\bar x$.

Exercise 7.4 Slides p.34 Proof Ruling out the cheapest point

Give assumptions on $u^i$ and $X^i$ ensuring that, at every PO allocation and its supporting $p$, no $\bar x^i$ is a cheapest point. Prove your claim, including that the relevant prices are non-negative.

Show solution

Take $X^i=\mathbb{R}^m_+$, and let $u^i$ be continuous, quasi-concave and increasing, with strictly positive bundles strictly preferred to all boundary bundles (e.g. Cobb–Douglas). Assume also that each agent gets a utility above the boundary level at the PO allocation. Then $\bar x^i\gg0$. Monotonicity gives $p\ge0$ (Exercise 6.5: otherwise $\bar x+te_k\in A$ and $p\cdot(\bar x+te_k)\to-\infty$). Since $p\neq0$, some $p_k>0$, so $p\cdot\bar x^i\ge p_k\bar x^i_k>0=p\cdot0$. Hence $\tilde x^i=0$ is strictly cheaper.

Exercise 7.5 Slides p.35 Non-quasi-concave Edgeworth box

Sketch an Edgeworth box with one agent whose utility is not quasi-concave, in which some PO allocation cannot be decentralized. Prove that it is PO and that no price supports it.

Show solution

Use §5(a): $u^A=x_1^2+x_2^2$, $u^B=\min\{x_1^B,x_2^B\}$, $\bar x^A=(t,t)$. PO: any allocation with $u^B\ge 1-t$ has $x^A\le(t,t)$ componentwise, so $u^A(x^A)\le 2t^2=u^A(\bar x^A)$. Not decentralizable: on any budget line $L$ through $(t,t)$ with $p\ge0$, $p\ne0$, $u^A$ is strictly convex along $L$. So its maximum over the segment $L\cap\mathbb{R}^2_+$ is attained only at an endpoint. Since $(t,t)$ is interior to that segment, it is not demanded. (If $p_1$ or $p_2$ is 0, demand is unbounded.)

Exercise 7.6 Slides p.35 Exam-inspired A fixed cost

The economy of §5(b): one unit of time, technology $c\le (n-n_0)^\gamma$ for $n\ge n_0$ (and $c=0$ otherwise) with $\gamma\in(0,1)$ and fixed cost $n_0\in(0,1)$, utility $\log c+b\log\ell$.

Show that the PO labour input (when production is worthwhile) is $\bar n=(\gamma+bn_0)/(\gamma+b)$. (5 points)
Let $w$ be the slope of the frontier at $\bar n$. Show that the firm's profit at $\bar n$ is negative iff $n_0>(1-\gamma)\bar n$. (10 points)
Show that the PO allocation can be decentralized iff $n_0\le n_0^*\equiv\dfrac{1-\gamma}{1+b}$. (10 points)
Sign $\partial n_0^*/\partial b$ and $\partial n_0^*/\partial\gamma$, and interpret each in one line. (10 points)
Sketch the production possibility set and explain which step of the 2WT proof fails when $n_0>n_0^*$. (5 points)

Show solution

(a) Maximize $\gamma\log(n-n_0)+b\log(1-n)$. FOC: $\gamma(1-n)=b(n-n_0)$, so $n=(\gamma+bn_0)/(\gamma+b)$.
(b) $w=\gamma(\bar n-n_0)^{\gamma-1}$, so $\pi(\bar n)=(\bar n-n_0)^\gamma-\gamma(\bar n-n_0)^{\gamma-1}\bar n=(\bar n-n_0)^{\gamma-1}\big[(\bar n-n_0)-\gamma\bar n\big]$. This is negative iff $n_0>(1-\gamma)\bar n$.
(c) On $n\ge n_0$, profit is concave, so at $w$ the best producing plan is $\bar n$. The only alternative is shutting down, with $\pi=0$. Decentralization therefore works iff $\pi(\bar n)\ge0$, i.e. iff $n_0\le(1-\gamma)\bar n$. Substitute (a): $n_0(\gamma+b)\le(1-\gamma)(\gamma+bn_0)\iff n_0\,\gamma(1+b)\le\gamma(1-\gamma)\iff n_0\le\frac{1-\gamma}{1+b}$.
(d) $\partial n_0^*/\partial b=-\frac{1-\gamma}{(1+b)^2}<0$. A household that values leisure more works less, so the fixed cost is spread over less output and losses come sooner. $\partial n_0^*/\partial\gamma=-\frac{1}{1+b}<0$. With $\gamma$ closer to 1, marginal product falls more slowly, so the wage absorbs more of the revenue and there is less inframarginal profit to cover the fixed cost.
(e) The set is flat between $\ell=1-n_0$ and $\ell=1$, then rises concavely, so it is not convex. Assumption FF fails, and Step 6 of the 2WT (firms maximize profits at $p$) breaks: the plan "no production" lies above the supporting hyperplane.

Exercise 7.7 Past exam: Fall 2025 Midterm 1, I.6 Efficiency and fairness

A pure endowment economy with $I$ agents, $m$ goods and aggregate endowment $\bar e$. An allocation is envy-free if $u^i(x^i)\ge u^i(x^j)$ for all $i,j$.

Prove that the CE allocation with $e^i=\bar e/I$ is envy-free. Your answer should take at most three lines. (10 points)
Assume LNS. Conclude that this allocation is both PO and envy-free. (5 points)
Give an example of a PO allocation that is not envy-free. (5 points)
Show by example that with unequal endowments the CE need not be envy-free. (5 points)
Now add firms and give every household equal endowments and equal shares $\theta^i_j=1/I$. Is the CE still envy-free? (5 points)
Connect to the 2WT. Under its assumptions plus Arrow's condition, which PO allocations can be reached starting from equal endowments, and with which transfers? (10 points)

Show solution

(a) All agents have wealth $p\cdot\bar e/I$, so they face the same budget set. Each $x^j$ is affordable to $i$. If $u^i(x^j)>u^i(x^i)$, then $x^i$ would not be optimal for $i$.
(b) By the 1WT, a CE allocation under LNS is PO. Together with (a), the equal-division CE is "fair": efficient and envy-free.
(c) Give everything to agent 1: $x^1=\bar e$ and $x^j=0$ for $j\neq1$. With strictly increasing utilities this is PO (anyone else's gain costs agent 1), but every $j\ne1$ envies agent 1.
(d) Two agents with identical preferences, Cobb–Douglas with equal weights, and $e^1=(1,1)$, $e^2=(0,0)$. The CE has no trade, and agent 2 envies agent 1.
(e) Yes. Wealth is $p\cdot\bar e/I+\sum_j\pi^j/I$, the same for everyone, so the argument of (a) goes through.
(f) Every PO allocation can be reached. Take $p$ from the 2WT and transfers $T^i=p\cdot\bar x^i-p\cdot\bar e/I$, which sum to zero by feasibility. From equal division, the equal-division CE needs no transfers at all. Every other PO allocation requires moving wealth away from equality, so $T^i\neq0$ for some agent.

Key takeaways

2WT ⇒ quasi-equilibrium. Arrow's remark (a strictly cheaper bundle exists, plus continuity) ⇒ full CE with transfers.
The exceptional case: a zero-wealth agent at a boundary with a vertical indifference curve ($\sqrt{x_1}+x_2$, $e=(0,1)$). There is a quasi-equilibrium but no CE.
Interiority + monotonicity + $X=\mathbb{R}^m_+$ rule it out.
Decentralizing wealth: $p\,e^i+\sum_j\theta^i_j\pi^j=p\,\bar x^i$. Infinitely many $(e,\theta)$ do it, or equivalently zero-sum lump-sum transfers.
Non-convex preferences make households go to corners; non-convex technology (IRS, fixed costs) makes firms shut down or go to corners. Either way the 2WT fails.

💬 Stuck or curious? Ask me (your Claude teacher). Try "why does a continuum of agents convexify the economy (Aumann)?" or "can a two-part tariff decentralize the fixed-cost example?"

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