Theory of Income · Part I: Welfare Theorems · Lesson 6 of 25

The Second Welfare Theorem

Covers Slides 1, pp. 24–30 · About 40 min · Prerequisites: Lesson 3, Lesson 4

By the end you can:

state Assumptions HH and FF and the 2WT exactly, including what it does not conclude (utility maximization);
reproduce the proof: the two sets $A$ and $B$, why $A\cap\operatorname{Int}(B)=\emptyset$, the separating hyperplane, and the two "construct a violating point" arguments the slides leave implicit;
say which hypothesis each step uses, and draw the picture where it fails.

Primary source Alvarez, Slides 1, pp. 24–30. Textbook: Mas-Colell, Whinston & Green §16.D (same theorem, separating the strictly preferred set instead) and Mathematical Appendix M.G (separating hyperplane theorems). Debreu, Theory of Value, ch. 6.

1. The question: can any efficient allocation be a market outcome?

The 1WT says CE ⇒ PO. The 2WT asks the converse. Pick any PO allocation $\{\bar x^i,\bar y^j\}$, however unequal or equal you like. Is there a price vector $p$ at which every firm wants to produce $\bar y^j$ and every household wants $\bar x^i$? If so, the allocation can be decentralized: a planner only needs to choose the distribution of wealth (endowments and ownership shares, i.e. lump-sum transfers), and markets do the rest. That is the theoretical basis for separating efficiency (leave it to markets) from equity (handle it with lump-sum transfers).

The answer is yes, under convexity. The tool is the separating hyperplane theorem: prices are the normal vector of a hyperplane that separates "what people would like" from "what the economy can produce".

2. The assumptions S1 p.24

Assumption HH (households)

For every $i$: $X^i$ is convex, and $u^i$ is continuous with strictly convex upper contour sets S1 p.25: for all $x,x'\in X^i$, $$u^i(x')>u^i(x)\ \Longrightarrow\ u^i\big(\theta x+(1-\theta)x'\big)>u^i(x)\quad\text{for all }\theta\in(0,1).$$

Assumption FF (firms)

The aggregate production set is convex: $$Y=\Big\{y\in L:\ y=\textstyle\sum_{j\in J}y^j,\ y^j\in Y^j\ \text{for all }j\Big\}\ \text{is convex.}$$

Reading the assumptions

What "strictly convex" means here. If you mix a bundle with a strictly better one, you are strictly better off than with the original. This rules out "dents" in indifference curves; it is a form of strict quasi-concavity. It implies the upper contour sets $\{x\in X^i: u^i(x)\ge u^i(\bar x)\}$ are convex, which is the property the proof actually uses most. (With continuity, the slide's condition implies ordinary quasi-concavity.)
FF is about the aggregate. Only $Y=\sum_jY^j$ must be convex. If every $Y^j$ is convex, so is the sum, but a sum can be convex even when individual sets aren't.
The 1WT used neither assumption. Convexity is the price of going backwards from PO to CE.

3. The theorem S1 p.26

Second Welfare Theorem (main part)

Assume HH and FF. Let $\{\bar x^i,\bar y^j\}$ be a PO allocation, and suppose some household $h$ is not satiated at its bundle: there is $\tilde x^h\in X^h$ with $u^h(\tilde x^h)>u^h(\bar x^h)$. Then there is a price vector $p\ne 0$ such that

(profit maximization) $p\,\bar y^j\ \ge\ p\,y$ for all $y\in Y^j$ and all $j$;
(expenditure minimization) for each $i$, $\bar x^i$ solves $\min_{x\in X^i} p\,x$ s.t. $u^i(x)\ge u^i(\bar x^i)$.

Read the conclusion carefully

Part 2 says households minimize expenditure to reach their PO utility. It does not say they maximize utility subject to the budget $p\,x\le p\,\bar x^i$. The pair $(p,\{\bar x^i,\bar y^j\})$ is a quasi-equilibrium (with transfers). Upgrading it to a full CE needs one more condition, which is Arrow's remark (Lesson 7). There is a genuine counterexample when that condition fails.

4. The tool: separating hyperplanes

Separating hyperplane theorem (Minkowski)

Let $A, B\subset\mathbb{R}^m$ be convex, with $\operatorname{Int}(B)\ne\emptyset$ and $A\cap\operatorname{Int}(B)=\emptyset$. Then there is $p\in\mathbb{R}^m$, $p\neq 0$, such that $$p\cdot x\ \ge\ p\cdot z\qquad\text{for all }x\in A,\ z\in B .$$ Geometrically, the hyperplane $\{v:p\cdot v=c\}$, for any $c$ between $\sup_B p\cdot z$ and $\inf_A p\cdot x$, has $A$ on one side and $B$ on the other.

Both convexity and the non-empty interior matter. Two interlocking crescents can be disjoint with no line between them, and that is exactly the failure of the 2WT under non-convexities. A set with empty interior (a single point, say) touches everything in the sense of "$A\cap\operatorname{Int}B=\emptyset$" trivially, so that hypothesis would carry no information.

5. The proof, step by step S1 pp.27–30

Proof

Define the two sets. $A$ is everything the economy would need in aggregate to make everyone at least as well off as at the PO allocation. $B$ is everything the economy can make available: $$A=\Big\{\textstyle\sum_i x^i:\ x^i\in X^i,\ u^i(x^i)\ge u^i(\bar x^i)\ \forall i\Big\},\qquad B=Y+\Big\{\textstyle\sum_i e^i\Big\}.$$
Both sets are convex. Each upper contour set $U^i=\{x\in X^i:u^i(x)\ge u^i(\bar x^i)\}$ is convex by HH. $A=U^1+\dots+U^I$ is a set-sum of convex sets, hence convex: if $a=\sum_i x^i$ and $a'=\sum_i x'^i$, then $\theta a+(1-\theta)a'=\sum_i[\theta x^i+(1-\theta)x'^i]$, and each bracket lies in $U^i$. $B$ is a translate of $Y$, which is convex by FF.
$A$ and the interior of $B$ don't meet: $A\cap\operatorname{Int}(B)=\emptyset$. Suppose $x=\sum_ix^i\in A\cap\operatorname{Int}(B)$. Take the non-satiated household $h$ and its $\tilde x^h$, and for $\theta\in(0,1)$ set $x^{h,\theta}=(1-\theta)x^h+\theta\tilde x^h\in X^h$ (by convexity of $X^h$).
- $h$ strictly gains. We know $u^h(x^h)\ge u^h(\bar x^h)$ and $u^h(\tilde x^h)>u^h(\bar x^h)$. If $u^h(\tilde x^h)>u^h(x^h)$, HH gives $u^h(x^{h,\theta})>u^h(x^h)\ge u^h(\bar x^h)$. If $u^h(x^h)>u^h(\tilde x^h)$, HH (with roles swapped) gives $u^h(x^{h,\theta})>u^h(\tilde x^h)>u^h(\bar x^h)$. If they are equal, both exceed $u^h(\bar x^h)$ strictly, and continuity gives $u^h(x^{h,\theta})>u^h(\bar x^h)$ for $\theta$ small. (The slides do this in one line; these are the cases hiding inside it.)
- It's still producible. The new aggregate is $x+\theta(\tilde x^h-x^h)$. Since $x$ is in the interior of $B$, a small enough move stays inside $B$. So there are $y^j\in Y^j$ with $x+\theta(\tilde x^h-x^h)=\sum_jy^j+\sum_ie^i$.
Give everyone $i\ne h$ the bundle $x^i$ and give $h$ the bundle $x^{h,\theta}$. This allocation is feasible, everyone is weakly better off than at $\bar x$, and $h$ is strictly better off. That contradicts Pareto optimality.
Separate. By Steps 2–3 (and $\operatorname{Int}(B)\neq\emptyset$; see the remark below), there is $p\ne0$ with $$p\,x\ \ge\ p\,z\qquad\forall\,x\in A,\ z\in B. \tag{6}$$ Why insist on $p\ne0$? $p=0$ satisfies (6) trivially and says nothing. With zero prices every bundle is free, so no budget set means anything.
The hyperplane passes through the PO aggregate. Let $\bar x\equiv\sum_i\bar x^i$ and $\bar z\equiv\sum_j\bar y^j+\sum_ie^i$. Trivially $u^i(\bar x^i)\ge u^i(\bar x^i)$, so $\bar x\in A$. And $\bar z\in B$. Feasibility says $\bar x=\bar z$, so $$p\sum_i\bar x^i = p\sum_j\bar y^j+p\sum_ie^i. \tag{7}$$ Combining with (6): $p\,\bar x=\min_{x\in A}p\,x=\max_{z\in B}p\,z$. The PO aggregate is simultaneously the cheapest point of $A$ and the most valuable point of $B$. Steps 6–7 split each of these statements into its individual pieces.
Firms maximize profits (the slides' "one can construct a $z$"). Suppose some firm $j$ has $\hat y\in Y^j$ with $p\,\hat y>p\,\bar y^j$. Replace only firm $j$'s plan: $$\hat z\ \equiv\ \hat y+\sum_{k\neq j}\bar y^k+\sum_ie^i\ \in B .$$ Then $p\,\hat z = p\,\bar z + (p\,\hat y-p\,\bar y^j) > p\,\bar z \overset{(7)}{=} p\,\bar x$. But $\bar x\in A$, so (6) requires $p\,\bar x\ge p\,\hat z$. Contradiction. Hence $p\,\bar y^j\ge p\,y$ for all $y\in Y^j$.
Households minimize expenditure (the slides' "one can construct an $x$"). Suppose some $i$ has $\hat x\in X^i$ with $u^i(\hat x)\ge u^i(\bar x^i)$ and $p\,\hat x\lt p\,\bar x^i$. Replace only household $i$'s bundle: $$\hat a\ \equiv\ \hat x+\sum_{k\neq i}\bar x^k\ \in A .$$ Then $p\,\hat a = p\,\bar x-(p\,\bar x^i-p\,\hat x)\lt p\,\bar x\overset{(7)}{=}p\,\bar z$. But $\bar z\in B$, so (6) requires $p\,\hat a\ge p\,\bar z$. Contradiction. Hence $\bar x^i$ minimizes $p\,x$ over $\{x\in X^i: u^i(x)\ge u^i(\bar x^i)\}$.

∎

A remark the slides skip: $\operatorname{Int}(B)\ne\emptyset$

In a pure exchange economy, $Y=\{0\}$ and $B=\{\sum_ie^i\}$ is a single point with empty interior. Then Step 3 is vacuous and Step 4 has nothing to separate. Two standard fixes:

(i) Assume free disposal, $-\mathbb{R}^m_+\subseteq Y$. Then $B\supseteq\sum_ie^i-\mathbb{R}^m_+$, which has an interior.
(ii) Follow MWG and separate the convex set of strictly preferred aggregates, $A^{\succ}=\sum_i\{x:u^i(x)>u^i(\bar x^i)\}$, from $B$. These are disjoint by PO, and the separating theorem for disjoint convex sets needs no interior. LNS then extends the inequality to weakly preferred bundles.

6. See it: Robinson Crusoe

One household and one firm. The goods are leisure $\ell$ and consumption $c$. The endowment is one unit of time, $e=(1,0)$. The firm turns labour $n$ into output $c\le n^\gamma$, so $B=\{(\ell,c):\ \ell\le1,\ c\le(1-\ell)^\gamma\}$ (with free disposal). Utility is $u=\log c+b\log\ell$. The PO allocation maximizes $\gamma\log n+b\log(1-n)$, so $\bar n=\gamma/(\gamma+b)$. The candidate supporting price of leisure (in units of $c$) is the slope of the frontier at $\bar n$: $w=\gamma\bar n^{\gamma-1}$.

Interactive. Blue: the producible set $B$. Orange: the set $A$ of bundles weakly preferred to the PO point. Black: the price line through the PO point with slope $-w$. With $\gamma\le1$ (convex $B$) the line separates the sets: $A$ lies above it and $B$ below. That gives a supporting price at which the firm maximizes profit. Push $\gamma$ above 1 (increasing returns): $B$ becomes non-convex, part of it pokes above the line (violet), and at the "tangency" price the firm would rather produce at a corner. The 2WT fails.

7. What each hypothesis buys

Hypothesis	Used in	If it fails…
Some $h$ not satiated ($\tilde x^h$)	Step 3	If everyone is at a bliss point with resources to spare, $A$ can sit inside $B$. Then no hyperplane separates them, and only $p=0$ "supports" the allocation (Exercise 6.3).
Convex upper contour sets (HH)	Steps 2–3	$A$ is non-convex. The PO tangency may not be a household optimum at the tangency price (Lesson 7, Edgeworth box).
$X^i$ convex, $u^i$ continuous	Step 3	The mixture $x^{h,\theta}$ may not be consumable, or may not beat $\bar x^h$.
Aggregate $Y$ convex (FF)	Steps 2, 6	Increasing returns or fixed costs: firms don't maximize profit at the supporting price (figure above; Lesson 7).
$\operatorname{Int}(B)\neq\emptyset$	Steps 3–4	The proof is vacuous. Use free disposal or MWG's strict version.

Intuition: prices as shadow values

$p$ is the vector of shadow prices of the planner's problem: the Lagrange multipliers on the resource constraints when maximizing a weighted sum of utilities. In the smooth interior case, separation boils down to "every MRS equals every MRT equals the price ratio". The hyperplane argument gets this with no derivatives, so it covers kinks and corners too. Convexity is exactly what makes these first-order "tangency" conditions global optimality conditions for each agent.

8. Check your understanding

1. In the 2WT proof, why is the set $A$ convex?

Because the aggregate production set Y is assumed convex
Because the PO allocation maximizes a weighted sum of utility
Because sums of convex upper contour sets remain convex sets
Because prices are linear, so every budget set is convex

HH makes each $U^i=\{x\in X^i:u^i(x)\ge u^i(\bar x^i)\}$ convex, and $A=\sum_iU^i$. FF is what makes $B$ convex. Prices haven't been found yet at that stage of the proof.

2. Where in the proof is the non-satiated household $h$ used?

To show that A does not intersect the interior of B
To show that firms maximize profits at the price p
To show that the separating price vector is nonzero
To show that the set B has a nonempty interior here

Without $h$, an interior point of $B$ could lie in $A$: everyone satiated, with leftover resources. $p\neq0$ comes from the separation theorem itself.

3. What does the 2WT (main part) conclude about households?

Each maximizes utility on the budget set at price p
Each receives the same value of wealth at price p
Each is at an interior bundle of its consumption set
Each minimizes the cost of reaching its PO utility

The conclusion is expenditure minimization, i.e. a quasi-equilibrium. Utility maximization needs Arrow's extra condition (Lesson 7).

4. In Step 6, firm $j$ could earn more with $\hat y$. What point violates the separation inequality?

The aggregate demand with household j's bundle replaced
The aggregate supply with only firm j's plan replaced by ŷ
The aggregate endowment with every firm's plan set to zero
The aggregate supply with every firm's plan replaced by ŷ

$\hat z=\hat y+\sum_{k\neq j}\bar y^k+\sum e^i\in B$ has $p\hat z>p\bar z=p\bar x$, contradicting $p\bar x\ge p\hat z$ with $\bar x\in A$.

9. Exercises

These follow the format of Alvarez's exams. The true/false items mirror the assumption-checking traps on past first midterms. The derivations show how the 2WT gets used on exams: through planner weights and first-order conditions, and through "which step fails".

Exercise 6.1 Past exams True or false? Justify.

One to four lines each. Name the hypothesis or the step of the proof (§5) that settles it.

"If there are increasing returns to scale, some Pareto-optimal allocations may not be supportable as competitive equilibria." the flip side of Fall 2025 Midterm 1, I.2
Show answer
True. IRS makes $Y$ non-convex, which violates Assumption FF. Then $B$ need not be convex and the separating hyperplane can fail to exist, so at the tangency price the firm prefers another plan (§6 with $\gamma>1$; Exercise 6.6). Contrast with the 1WT, which IRS does not affect.
"The Second Welfare Theorem requires convex preferences." student review drill bank
Show answer
True (in our version). HH asks for convex $X^i$ and continuous, strictly quasi-concave $u^i$, so that each upper contour set and their sum $A$ are convex. Without it, a PO point can sit where $A$ curves around the price line (Lesson 7, §5a). It also needs a convex aggregate $Y$ (FF) and one non-satiated household.
"If a production function is convex, the corresponding production set is convex." student review drill bank
Show answer
False. The language flips. A concave $f$ gives a convex set $\{(-n,c):c\le f(n)\}$. A convex $f$ (e.g. $n^\gamma$, $\gamma>1$) gives a non-convex set.
"Let $\mu_i$ be agent $i$'s marginal utility of income and $p_s$ the price of good $s$. If we set the planner's weights equal to $\mu_i$, and the multipliers on the resource constraints equal to $p_s$, the planner's allocation coincides with the CE allocation." Fall 2005 Midterm, Ex. 1 Q5
Show answer
False. The CE FOC is $u^i_s=\mu_ip_s$ and the planner FOC is $\alpha_iu^i_s=\gamma_s$. They coincide when $\gamma=p$ and $\alpha_i=1/\mu_i$: the weights are the inverse of the marginal utility of income (Negishi weights; see Lesson 23). Rich agents (low $\mu_i$) get high weight.
"Assume all utility functions are strictly concave, strictly increasing, differentiable and satisfy Inada conditions, in a pure endowment economy. Then in a CE, the price of each good is proportional to a weighted average across agents of the marginal utility of that good, with the same weights for all goods." Fall 2025 Midterm 1, I.4
Show answer
True. Inada conditions give interior demands, so $u^i_\ell=\mu_ip_\ell$ for every $i,\ell$. Hence $p_\ell=\frac1I\sum_i\frac{1}{\mu_i}u^i_\ell$. The weight $1/\mu_i$ is agent-specific but the same for every good, because each agent has one budget constraint and so one multiplier. Exercise 6.4 develops this.
"Under the 2WT assumptions, any PO allocation is a CE for the original endowments." exam-inspired
Show answer
False. It is a quasi-equilibrium with transfers: wealth must be redistributed so that each $i$ can afford exactly $p\cdot\bar x^i$ (Lesson 7). Even then, utility maximization needs Arrow's condition.
"The 2WT requires every firm's production set to be convex." exam-inspired
Show answer
False. Assumption FF is on the aggregate $Y=\sum_jY^j$. (In practice one usually assumes each $Y^j$ is convex, which is sufficient.)
"The supporting prices in the 2WT are unique." exam-inspired
Show answer
False in general. At kinks (Leontief preferences, linear technologies with corners), many hyperplanes separate $A$ and $B$. With smooth indifference curves at an interior point, relative prices are pinned down by the common MRS.
"In a pure exchange economy, the slides' argument goes through as written: $B=\{\sum_ie^i\}$ and we separate $A$ from $\operatorname{Int}B$." exam-inspired (a gap in the slides)
Show answer
False. Without production, $B$ is a single point, so $\operatorname{Int}B=\emptyset$. Separating $A$ from the empty set says nothing. Fixes (§5): assume free disposal, so $B=\sum_ie^i-\mathbb{R}^m_+$ has an interior; or separate the strictly preferred set from $B$, as in MWG.

Exercise 6.2 Proof The proof from a blank page

Reproduce the 2WT proof from a blank page, in this order: (i) define $A$ and $B$; (ii) prove convexity; (iii) prove $A\cap\operatorname{Int}B=\emptyset$; (iv) separate; (v) derive (7); (vi) profit maximization; (vii) expenditure minimization. Then check yourself against §5. For each step, name the hypothesis it uses.

Show checklist

(ii) HH gives convex $U^i$; $A$ is their set-sum; FF makes $B$ convex. (iii) Non-satiated $h$, convex $X^h$, HH or continuity, $x$ interior to $B$, and PO. (iv) Minkowski with $\operatorname{Int}B\ne\emptyset$. (v) Feasibility of the PO allocation. (vi) Build $\hat z$ by changing only firm $j$. (vii) Build $\hat a$ by changing only household $i$.

Exercise 6.3 Proof Why we need $h$

One household, two goods, no firms but free disposal: $Y=-\mathbb{R}^2_+$. The endowment is $e\gg(1,1)$ and $u(x)=-\|x-(1,1)\|^2$ (bliss at $(1,1)$). Show that $\bar x=(1,1)$, with $\bar y=(1,1)-e$ (disposing of the excess), is PO; that no household satisfies the non-satiation hypothesis; and that no $p\neq0$ satisfies the 2WT's conclusions.

Show solution

Nothing beats the bliss point, so $\bar x$ is PO and no $\tilde x$ exists. $A=\{(1,1)\}$ and $B=e-\mathbb{R}^2_+$. Since $(1,1)\in\operatorname{Int}B$, Step 3's conclusion fails. Suppose $p\neq0$ supported it. Profit maximization over $Y=-\mathbb{R}^2_+$ requires $p\ge0$ (otherwise disposing of more of a negatively priced good raises profit without bound). It also requires $p\cdot\bar y\ge p\cdot0=0$, i.e. $p\cdot(e-(1,1))\le0$. With $e-(1,1)\gg0$, these force $p=0$. Contradiction: a satiated economy has only zero shadow prices.

Exercise 6.4 Past exam: Fall 2025 Midterm 1, I.4 Problem set 2022 (PS3 P3) Planner weights and prices

A pure endowment economy with $I$ agents and $m$ goods, aggregate endowment $\bar e\gg0$. Each $u^i$ is strictly increasing, strictly concave, differentiable, and satisfies Inada conditions.

Write agent $i$'s FOC for good $\ell$ in a CE, with budget multiplier $\mu_i$. Why does it hold with equality? (5 points)
Write the planner's problem $\max\sum_i\lambda_iu^i(x^i)$ subject to $\sum_ix^i_\ell\le\bar e_\ell$ (multipliers $\gamma_\ell$) and its FOC. (5 points)
Show that the CE allocation solves the planner's problem with $\lambda_i=1/\mu_i$ and $\gamma=p$. Which welfare theorem is this a proof of, under these assumptions? (10 points)
Deduce the statement of Exercise 6.1, item 5. Which agents get a high weight? (10 points)
Conversely, take any weights $\lambda\gg0$ and the planner's solution $\bar x(\lambda)$. Show that it is a CE once endowments are redistributed, and give the lump-sum transfer $T^i$ in terms of $p$, $\bar x^i$ and $e^i$. Show that $\sum_iT^i=0$. (10 points)
Let $u^i(x)=\sum_\ell\pi_\ell\frac{x_\ell^{1-\sigma}}{1-\sigma}$ for all $i$, with $\sigma>0$. Show that $\bar x^i_\ell(\lambda)=\dfrac{\lambda_i^{1/\sigma}}{\sum_j\lambda_j^{1/\sigma}}\,\bar e_\ell$. Interpret: what do the weights control, and what don't they control? (10 points)

Hint

For (c), the planner's problem is concave, so its FOCs are sufficient. For (f), use the planner FOC to write $x^i_\ell$ in terms of $\lambda_i$ and $\gamma_\ell$, then sum over $i$ to eliminate $\gamma_\ell$.

Show solution

(a) $u^i_\ell(\bar x^i)=\mu_ip_\ell$. Inada conditions make $u^i_\ell\to\infty$ at $x_\ell=0$, so demands are interior and the non-negativity constraints are slack. LNS makes the budget bind, so $\mu_i>0$.
(b) FOC: $\lambda_iu^i_\ell(x^i)=\gamma_\ell$ for all $i,\ell$.
(c) With $\lambda_i=1/\mu_i$ and $\gamma=p$, the planner FOC becomes $u^i_\ell/\mu_i=p_\ell$, which is the CE FOC. Feasibility holds because CE allocations are feasible. The planner's objective is strictly concave and the constraints are linear, so the FOCs are sufficient and $\bar x$ is the planner's solution. Every solution of a weighted planner with positive weights is PO, so this proves the 1WT by duality (under smoothness and concavity, which the general proof does not need).
(d) From (a), $p_\ell=u^i_\ell/\mu_i$ for every $i$. Averaging, $p_\ell=\frac1I\sum_i\frac{1}{\mu_i}u^i_\ell$. The weights $1/\mu_i$ do not depend on $\ell$. Agents with a low marginal utility of income, i.e. the rich, get high weight.
(e) Set $p=\gamma$ and $\mu_i=1/\lambda_i$. Then each agent's FOC holds at $\bar x^i$. With strictly concave $u^i$ and a budget $p\cdot x\le p\cdot\bar x^i$, the FOC is sufficient, so $\bar x^i$ is demanded. This is the 2WT. The transfer is $T^i=p\cdot\bar x^i-p\cdot e^i$, and $\sum_iT^i=p\cdot(\sum_i\bar x^i-\bar e)=0$ by feasibility (binding, since $u$ is increasing).
(f) FOC: $\lambda_i\pi_\ell(x^i_\ell)^{-\sigma}=\gamma_\ell$, so $x^i_\ell=(\lambda_i\pi_\ell/\gamma_\ell)^{1/\sigma}$. Summing over $i$: $\bar e_\ell=(\pi_\ell/\gamma_\ell)^{1/\sigma}\sum_j\lambda_j^{1/\sigma}$. Dividing gives the share formula. The weights control how big a slice each agent gets: the same share of every good. They do not affect relative prices, $\gamma_\ell/\gamma_k=(\pi_\ell/\pi_k)(\bar e_k/\bar e_\ell)^{\sigma}$, which depend only on aggregates.

Exercise 6.5 Proof Prices are non-negative

Suppose $X^i=\mathbb{R}^m_+$ and some household's $u^i$ is (weakly) increasing in every good. Show that the separating $p$ from the 2WT satisfies $p\ge0$.

Hint

If $p_k<0$, what happens to $p\cdot(\bar x+t\,e_k)$ as $t\to\infty$? Is $\bar x+t\,e_k$ in $A$?

Show solution

Give household $i$ the extra $t\,e_k$, where $e_k$ is the $k$-th unit vector. By monotonicity it is weakly better off, and others are unchanged, so $\bar x+t\,e_k\in A$ for all $t\ge0$. By (6), $p\cdot\bar x+t\,p_k\ge p\cdot\bar z$ for all $t$. If $p_k<0$, the left side goes to $-\infty$. Contradiction. Hence $p_k\ge0$ for every $k$.

Exercise 6.6 Exam-inspired Robinson Crusoe: when do supporting prices exist?

The economy of §6 with general $\gamma>0$ and $b>0$: one unit of time, technology $c\le n^\gamma$, utility $\log c+b\log\ell$.

Find the PO labour input $\bar n$. Your answer should be one equation. (5 points)
Find the candidate supporting wage $w$ (price of leisure in units of $c$) as the slope of the frontier at $\bar n$. (5 points)
Show that the firm's profit at $\bar n$ is $\pi=(1-\gamma)\bar n^\gamma$. (5 points)
Let $\gamma\le1$. Show that at $w$ the firm maximizes profit at $\bar n$, and that the household with income $w+\pi$ demands exactly $(\bar\ell,\bar c)$. (15 points)
Let $\gamma>1$. Show that at $w$, $\bar n$ is a profit minimum on $[0,1]$ and the firm prefers $n=0$. Which step of the 2WT proof fails, and which assumption is violated? (10 points)

Show solution

(a) Maximize $\gamma\log n+b\log(1-n)$: $\gamma/n=b/(1-n)$, so $\bar n=\gamma/(\gamma+b)$.
(b) $w=\gamma\bar n^{\gamma-1}$.
(c) $\pi=\bar n^\gamma-w\bar n=\bar n^\gamma-\gamma\bar n^\gamma=(1-\gamma)\bar n^\gamma$.
(d) For $\gamma\le1$, $n^\gamma-wn$ is concave, so the FOC $\gamma n^{\gamma-1}=w$ identifies the maximum, at $\bar n$. The household maximizes $\log c+b\log\ell$ subject to $c+w\ell=w+\pi$, so $c=\frac{w+\pi}{1+b}$. Now $w+\pi=\bar n^{\gamma-1}\big(\gamma+(1-\gamma)\bar n\big)$ and $\gamma+(1-\gamma)\bar n=\frac{\gamma(1+b)}{\gamma+b}=(1+b)\bar n$. So $w+\pi=(1+b)\bar n^\gamma$ and $c=\bar n^\gamma=\bar c$. Then $w\ell=b\bar n^\gamma$ gives $\ell=b\bar n/\gamma=1-\bar n=\bar\ell$. ✓
(e) For $\gamma>1$, $n^\gamma-wn$ is strictly convex, so the point where the FOC holds is a minimum. Indeed $\pi(\bar n)=(1-\gamma)\bar n^\gamma<0=\pi(0)$. Step 6 (firms maximize profits at $p$) fails, because Assumption FF (convex $Y$) is violated: part of $B$ lies above the price line through the PO point.

Exercise 6.7 Exam-inspired Decentralizing an Edgeworth-box optimum

Two agents with $u^A=a\log x_1+(1-a)\log x_2$ and $u^B=b\log x_1+(1-b)\log x_2$, total endowment $(1,1)$, initial endowments $e^A$ and $e^B=(1,1)-e^A$. Write $k_A=a/(1-a)$ and $k_B=b/(1-b)$.

Derive the contract curve $x^A_2=\phi(x^A_1)$. (10 points)
For the PO allocation with $x^A_1=t\in(0,1)$, find the supporting relative price $p_1/p_2$ in terms of $t$. (5 points)
With $p_2=1$, find the lump-sum transfer $T^A$ that decentralizes it. Show that $T^B=-T^A$. (10 points)
Verify that with wealth $p\cdot e^A+T^A$, agent A demands $(t,\phi(t))$. (5 points)

Show solution

(a) Equal MRS: $k_A\,x_2/x_1=k_B\,(1-x_2)/(1-x_1)$. Solving, $\phi(x_1)=\dfrac{k_Bx_1}{k_A(1-x_1)+k_Bx_1}$.
(b) $p_1/p_2=MRS^A=k_A\,\phi(t)/t=\dfrac{k_Ak_B}{k_A(1-t)+k_Bt}$.
(c) $T^A=p\cdot\bar x^A-p\cdot e^A=p_1t+\phi(t)-p_1e^A_1-e^A_2$. Feasibility gives $\bar x^B=(1,1)-\bar x^A$ and $e^B=(1,1)-e^A$, so $T^B=p\cdot(\bar x^B-e^B)=-p\cdot(\bar x^A-e^A)=-T^A$.
(d) With wealth $M=p\cdot\bar x^A=p_1t+\phi(t)$, Cobb–Douglas demand is $x_1=aM/p_1$. Now $p_1t=k_A\phi(t)$, so $M=(1+k_A)\phi(t)=\phi(t)/(1-a)$. Then $x_1=a\phi(t)/((1-a)p_1)=k_A\phi(t)/p_1=t$ and $x_2=(1-a)M=\phi(t)$. ✓

Key takeaways

$A$ = aggregates making everyone at least as well off; $B$ = producible aggregates. PO + a non-satiated $h$ ⇒ $A\cap\operatorname{Int}B=\emptyset$.
Convexity (HH, FF) ⇒ a separating $p\neq0$ exists. The PO aggregate is the cheapest point of $A$ and the most valuable point of $B$.
Change one firm's plan to get profit maximization; change one household's bundle to get expenditure minimization.
The conclusion is a quasi-equilibrium with transfers, not yet a CE (Lesson 7).
Non-convexities (IRS, fixed costs, non-convex preferences) are where decentralization breaks.

💬 Stuck or curious? Ask me (your Claude teacher). Try "prove the separating hyperplane theorem in $\mathbb{R}^2$" or "show me how the Lagrange multipliers of the planner's problem equal $p$."

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Hypothesis	Used in	If it fails…
Some \(h\) not satiated (\(\tilde x^h\))	Step 3	If everyone is at a bliss point with resources to spare, \(A\) can sit inside \(B\). Then no hyperplane separates them, and only \(p=0\) "supports" the allocation (Exercise 6.3).
Convex upper contour sets (HH)	Steps 2–3	\(A\) is non-convex. The PO tangency may not be a household optimum at the tangency price (Lesson 7, Edgeworth box).
\(X^i\) convex, \(u^i\) continuous	Step 3	The mixture \(x^{h,\theta}\) may not be consumable, or may not beat \(\bar x^h\).
Aggregate \(Y\) convex (FF)	Steps 2, 6	Increasing returns or fixed costs: firms don't maximize profit at the supporting price (figure above; Lesson 7).
\(\operatorname{Int}(B)\neq\emptyset\)	Steps 3–4	The proof is vacuous. Use free disposal or MWG's strict version.