The two welfare theorems connect competitive equilibria with Pareto optima. This lesson adds a third object: the solution to a planner's problem that maximizes a weighted sum of utilities, $$W(\lambda):\qquad \max_{\{x^i,y^j\}}\ \sum_{i\in I}\lambda_i\,u^i(x^i)\quad\text{subject to feasibility.}$$ The main theorem says that, under convexity assumptions, the Pareto optima are exactly the solutions of \(W(\lambda)\) as \(\lambda\) ranges over non-negative weight vectors N3 s.5. That is what licenses the "representative agent" of macroeconomics: a single maximization problem whose objective is \(\sum_i\lambda_iu^i\) N3 s.1. Lesson 23 will show that the CE weights are \(\lambda_i=1/\mu_i\), the reciprocal of each household's marginal utility of income.
Pareto optimality is a statement about a set: no feasible point is better for everyone. A planner with weights \(\lambda\) instead maximizes a single number, "move as far as possible in direction \(\lambda\)". If the set of attainable utility vectors is convex, every point on its outer edge is the farthest point in some direction. So choosing \(\lambda\) is the same as choosing where on the Pareto frontier to be, i.e. choosing a distribution of welfare.
$$U=\Big\{u\in\mathbb R^I:\ u_i\le u^i(x^i)\ \text{for all } i\in I,\ \text{for some feasible allocation }\{x^i,y^j\}\Big\}.$$
Feasible means \(x^i\in X^i\), \(y^j\in Y^j\) and \(\sum_ix^i\le\sum_jy^j+\sum_ie^i\) (the slides' exchange economy, eq. (1), allows disposal: N3 s.14).
Three features of this definition do real work later.
The theorem's key step separates \(U\) with a hyperplane, which needs \(U\) to be convex. We get there in two steps: convexity of the set of feasible allocations, then convexity of \(U\).
If the aggregate production set \(Y=\sum_jY^j\) and every \(X^i\) are convex (Assumptions FF and the first half of HH from Lesson 6), then the set \(\mathcal F\) of feasible allocations is convex.
Convexity of \(\mathcal F\) is not enough for convexity of \(U\): the map from allocations to utilities can bend a convex set into a non-convex one. We need a property of the functions \(u^i\) themselves.
Every \(u^i\) is concave: \(u^i(tx+(1-t)x')\ge tu^i(x)+(1-t)u^i(x')\) for all \(x,x'\in X^i\), \(t\in[0,1]\).
An ordinal property survives every strictly increasing transformation \(f\), since \(f\circ u\) represents the same preferences and generates the same demands. Concavity does not survive.
So CC is an assumption about the utility numbers we feed the planner, not about behaviour. This matters in §7.
Step 3 is the only place CC enters. With merely quasi-concave \(u^i\), we only know \(u^i(x^i_t)\ge\min\{u^i(x^i),u^i(\hat x^i)\}\), which bounds \(x_t\)'s utility below by the worse endpoint, not by the average. That is too weak to put \(tu+(1-t)\hat u\) in \(U\).
Assume the \(u^i\) are strictly increasing, that HH, CC and FF hold, and that \(U\) has a nonempty interior.
Notice the asymmetry: strictly positive weights in part 1, merely non-negative weights in part 2. Both halves are sharp, as §6 shows.
What this used: only \(\lambda\gg0\). No convexity, no concavity, no monotonicity. If \(\lambda_k=0\), step 3 fails: the agent who gains gets no weight, so \(\hat x\) only ties with \(\bar x\).
| Assumption | Where the proof uses it |
|---|---|
| \(\lambda\gg0\) | Sufficiency, step 3. Nothing else is needed for that direction. |
| \(X^i\), \(Y\) convex (HH, FF) | Feasible allocations are convex (Step A), feeding into Step B. |
| \(u^i\) concave (CC) | \(U\) is convex (Step B, step 3), so the supporting hyperplane exists. |
| "\(\le\)" in the definition of \(U\) | Step B, step 4; and the sign of the weights (necessity, step 5). |
| Nonempty interior of \(U\) | Rules out a degenerate \(U\) that lies inside a hyperplane, where "support" would be vacuous. With \(U\) defined by "\(\le\)" it holds automatically whenever a feasible allocation exists, since \(U\supseteq\bar u-\mathbb R^I_+\). (Compare Lesson 6, where \(\operatorname{Int}B\ne\emptyset\) was a genuine restriction.) |
| \(u^i\) strictly increasing | Not used in either argument. It keeps the frontier downward-sloping and matters for the zero-weight discussion and the first-order conditions of Lessons 22–23. |
In Lesson 6 we separated the "better than" set from the production set in commodity space; the normal was a price vector. Here we support the utility possibility set in utility space; the normal is a vector of Pareto weights. In both cases convexity is what guarantees that a linear functional (prices, or weights) can "see" the optimum. In both cases the sign of the normal comes from free disposal or comprehensiveness: you can always throw goods (or utility) away.
Two households share an aggregate endowment \(\bar e=(1,1)\) of two goods and have identical preferences \(u^i(x_1,x_2)=(x_1x_2)^k\), \(k>0\). The preferences are Cobb–Douglas for every \(k\), so they are strictly quasi-concave and every Pareto optimum lies on the diagonal of the Edgeworth box: agent 1 gets \(s\bar e\) and agent 2 gets \((1-s)\bar e\) (Exercise 21.5 proves this). Utilities are then \(u_1=s^{2k}\) and \(u_2=(1-s)^{2k}\), so the Pareto frontier is $$u_1^{1/(2k)}+u_2^{1/(2k)}=1 .$$ \(u^i\) is concave exactly when \(2k\le1\), and that is also exactly when \(U\) is convex.
The theorem's two halves use different weight sets, and the slides warn that this is not cosmetic N3 s.9. Two examples make both points.
One good, aggregate endowment \(\bar e\), two households with \(u^i(x)=x\). Household 1 can consume at most \(a\lt\bar e\): \(X^1=[0,a]\), \(X^2=\mathbb R_+\). Feasibility is \(x^1+x^2\le\bar e\). All assumptions of the theorem hold (the \(X^i\) are convex intervals, the \(u^i\) linear, hence concave and strictly increasing).
A rescue. If the solution of \(W(\lambda)\), \(\lambda\ge0\), is unique, it is Pareto optimal. Proof: a Pareto-dominating \(\hat x\) would satisfy \(\sum_i\lambda_iu^i(\hat x^i)\ge\sum_i\lambda_iu^i(\bar x^i)\) (weights are non-negative), so \(\hat x\) would be another maximizer. And \(\hat x\ne\bar x\), since someone is strictly better off under \(\hat x\). That contradicts uniqueness.
One good, \(\bar e=1\), \(u^1(x)=\sqrt x\), \(u^2(x)=\sqrt x\). The frontier is the quarter circle \(u_1^2+u_2^2=1\). The allocation "household 1 gets everything", \(\bar u=(1,0)\), is Pareto optimal. But the circle is vertical there, so its only supporting normal is \(\lambda\propto(1,0)\). With \(\lambda_2>0\) the planner always gives household 2 something, because \(u^{2\prime}(0)=\infty\). So in the necessity half, \(\mathbb R^I_+\) cannot be replaced by \(\mathbb R^I_{++}\).
Every Pareto optimum is a planner solution, but some need zero weights; and once zero weights are allowed, a planner solution need not be Pareto optimal. For the smooth interior allocations of Lessons 22–25 neither problem arises, so from now on we work with \(\lambda\gg0\).
N3 s.10 asks three drawing questions. The figure in §5 lets you do all of them.
Draw a frontier that bows inward (towards the origin), like the §5 figure with \(k>\tfrac12\), and pick an interior point \(\bar u\). Any line through \(\bar u\) with non-negative normal either crosses into \(U\) on one side or is the tangent. In the tangent case, the frontier lies above the line on both sides, so \(\bar u\) is a local minimum of \(\lambda\cdot u\) along the frontier. For every \(\lambda\ge0\), some other point of \(U\), a corner here, gives a higher \(\lambda\cdot u\). So \(\bar u\) is Pareto optimal but is not the solution to any planner problem.
Yes. The §5 economy is the example: \(u^i=(x_1x_2)^k\) is strictly quasi-concave for every \(k>0\) (its upper contour sets are those of Cobb–Douglas), but for \(k>\tfrac12\) it is not concave and the frontier \(u_1^{1/(2k)}+u_2^{1/(2k)}=1\) bows inward. An even simpler example: one good, \(u^i(x)=x^2\). With one good, any strictly increasing function is strictly quasi-concave, and the frontier \(\sqrt{u_1}+\sqrt{u_2}=\bar e\) bows inward.
The deeper point. The set of Pareto-optimal allocations is ordinal: it depends only on preferences. In the §5 economy it is the diagonal of the box for every \(k\). Whether every Pareto optimum is reachable as a solution of \(W\) depends on which utility numbers we use. Replace \(u^i=(x_1x_2)^k\) by the equivalent \(v^i=(u^i)^{1/(2k)}=\sqrt{x_1x_2}\), which is concave. The new frontier is the straight line \(v_1+v_2=1\), and every Pareto optimum solves \(W\) (with equal weights; the planner is indifferent along the line). So when preferences admit a concave representation, the planner approach can trace out every Pareto optimum: choose that representation. This is why the theorem assumes CC, a cardinal assumption, even though everything it describes is ordinal.
Fix \(\lambda\gg0\) and suppose there is no production. Then \(W(\lambda)\) is a problem of splitting a given aggregate bundle \(x\) among households. Its value, $$u(x;\lambda)=\max_{\{x^i\}}\ \sum_i\lambda_iu^i(x^i)\quad\text{s.t.}\quad\sum_ix^i=x,$$ is the utility function of a single representative agent N3 s.11. By this lesson's theorem and the welfare theorems, every CE corresponds to some \(\lambda\). Lesson 22 studies \(u(\cdot;\lambda)\) and the planner's first-order conditions. Lesson 23 finds the CE's \(\lambda\). Lessons 24–25 ask when the representative agent's prices do not depend on \(\lambda\) at all (aggregation).
No past exam has asked directly about \(U\) or this theorem; the exam questions on Class Note 3 start from the planner's first-order conditions (Lessons 22–25). These exercises cover the in-slide questions and use Alvarez's formats: true/false with short justifications, and proofs broken into small graded steps. Knowing exactly which assumption does what is how Midterm 1 true/false items about the welfare theorems are won.
Answer True or False and justify in at most four lines. Unless stated otherwise, HH, FF and CC hold and the \(u^i\) are strictly increasing.
False. The theorem only delivers \(\lambda\in\mathbb R^I_+\setminus\{0\}\). Counterexample (§6b): one good, \(u^i=\sqrt x\). The corner allocation giving everything to household 1 is PO, but the frontier \(u_1^2+u_2^2=1\) is vertical there, so the only supporting weights are \(\lambda\propto(1,0)\).
True. The sufficiency proof uses only \(\lambda\gg0\): a Pareto-dominating feasible allocation would raise \(\sum_i\lambda_iu^i\) strictly, since the agent who gains has positive weight. No convexity is involved.
False. With \(\lambda_k=0\), a Pareto improvement that only helps agent \(k\) leaves \(W\) unchanged, so a non-PO allocation can tie for the maximum. Example (§6a): \(X^1=[0,a]\), \(u^i=x\), disposal allowed, \(\lambda=(1,0)\): every allocation with \(x^1=a\) maximizes \(W\), but only \(x^2=\bar e-a\) is PO. It becomes true if the maximizer is unique.
False. Two households, \(\bar e=(1,1)\), \(u^i=(x_1x_2)^k\) with \(k>\tfrac12\): strictly quasi-concave, not concave. The frontier \(u_1^{1/(2k)}+u_2^{1/(2k)}=1\) bows inward, so \(U\) is not convex and interior PO points solve \(W(\lambda)\) for no \(\lambda\): \(W\) always picks a corner.
False. \(\log x\) is concave, but the equivalent representation \(e^{2\log x}=x^2\) is convex. Concavity is cardinal; its ordinal counterpart is quasi-concavity (convex upper contour sets are preserved by increasing transformations).
False. "Some agent strictly better off and nobody worse off" only compares bundles, so it is invariant to strictly increasing transformations. What does depend on the representation is the shape of \(U\), and hence whether every PO allocation can be obtained from \(W\) (§7(iii)).
False. It lies on the boundary. If \(\bar u\) were interior, \(\bar u+\varepsilon\mathbf 1\in U\) for small \(\varepsilon>0\), so a feasible allocation would make everyone strictly better off.
False. At a kink of the frontier, many supporting lines exist. In §6(a) the point \((a,\bar e-a)\) is supported by every \(\lambda=(\lambda_1,\lambda_2)\) with \(\lambda_1\ge\lambda_2\ge0\). Weights are unique up to scale where the frontier is smooth.
True. The sum \(Y=\sum_jY^j\) of convex sets is convex. Convex combinations of feasible allocations keep each \(x^i\in X^i\) and \(y\in Y\), and the resource constraint is linear (Step A).
Try one good, two households, \(\bar e=1\), \(u^i(x)=\sqrt x\), and no disposal (\(x^1+x^2=1\)). What shape is \(U'\)?
(a) Step A in §3: averaging keeps each piece in its convex set, and adding \(t\times\) and \((1-t)\times\) the two resource constraints gives the resource constraint for the average.
(b) Cardinal. \(\log x\) is concave, \(x^2=e^{2\log x}\) represents the same preferences and is convex.
(c) Step B in §3: \(u^i(x^i_t)\ge tu^i(x^i)+(1-t)u^i(\hat x^i)\ge tu_i+(1-t)\hat u_i\), and \(x_t\) is feasible.
(d) Step 4: the averaged vector is only weakly below the utilities of \(x_t\), and "\(\le\)" is what puts it in \(U\). \(U'\) need not be convex. In the hint's example, \(U'=\{(\sqrt t,\sqrt{1-t}):t\in[0,1]\}\) is a quarter circle (a curve), and the midpoint of \((1,0)\) and \((0,1)\), namely \((\tfrac12,\tfrac12)\), is not on it.
(a) §4, sufficiency. With \(\lambda_k=0\), step 3 fails: a dominating \(\hat x\) that is strictly better only for \(k\) gives the same weighted sum.
(b) §4, necessity step 2.
(c) If \(C\subset\mathbb R^I\) is convex and \(\bar u\) is not in its interior, there is \(\lambda^*\ne0\) with \(\lambda^*\cdot\bar u\ge\lambda^*\cdot u\) for all \(u\in C\). Apply it to \(C=U\), which is convex by HH (convex \(X^i\)), FF (convex \(Y\)) and CC (concave \(u^i\)). For any feasible \(x\), \((u^i(x^i))_i\in U\), so \(\sum\lambda^*_iu^i(x^i)\le\sum\lambda^*_iu^i(\bar x^i)\).
(d) §4, necessity step 5: if \(\lambda^*_i\lt0\), then \(\bar u-\varepsilon e_i\in U\) and \(\lambda^*\cdot(\bar u-\varepsilon e_i)>\lambda^*\cdot\bar u\), a contradiction.
(e) If \(x\) is feasible, \(u(x)\in U\), so the whole orthant \(u(x)-\mathbb R^I_+\subseteq U\). It contains the open set \(\{u: u_i\lt u^i(x^i)\ \forall i\}\).
One good with aggregate endowment \(\bar e>0\), two households, \(u^i(x)=x^a\) with \(a>0\), \(x^i\ge0\), \(x^1+x^2=\bar e\).
For (e), the objective \(\lambda_1x^a+\lambda_2(\bar e-x)^a\) is strictly convex in \(x\) when \(a>1\). Where does a strictly convex function on an interval attain its maximum?
(a) One good and strictly increasing utilities: any reallocation that helps one household hurts the other, and nothing is wasted under \(x^1+x^2=\bar e\). With \(u_i=(x^i)^a\), \(x^i=u_i^{1/a}\), and the resource constraint gives the frontier.
(b) The frontier is \(u_2=(\bar e-u_1^{1/a})^a\). \(U\) is convex iff this is a concave function of \(u_1\). That holds iff \(a\le1\), which is exactly when \(x^a\) is concave (CC). For \(a>1\) it bows inward. (For \(a=1\) it is the line \(u_1+u_2=\bar e\).)
(c) FOC: \(\lambda_1a(x^1)^{a-1}=\lambda_2a(x^2)^{a-1}\), so \((x^1/x^2)^{1-a}=\lambda_1/\lambda_2\), i.e. \(x^1/x^2=(\lambda_1/\lambda_2)^{1/(1-a)}\). The objective is strictly concave, so the FOC characterizes the maximum; combine with \(x^1+x^2=\bar e\). Yes, since \(1/(1-a)>0\): more weight, more consumption.
(d) Invert: \(\lambda_1/\lambda_2=\big(s/(1-s)\big)^{1-a}\in(0,\infty)\). Every interior share is reached by some \(\lambda\gg0\). (The corners \(s\in\{0,1\}\) need \(\lambda_2=0\) or \(\lambda_1=0\), because \(u^{i\prime}(0)=\infty\).)
(e) A strictly convex function on \([0,\bar e]\) attains its maximum only at endpoints. So the planner gives everything to household 1 if \(\lambda_1\bar e^a>\lambda_2\bar e^a\), i.e. \(\lambda_1>\lambda_2\), to household 2 if \(\lambda_1\lt\lambda_2\), and either corner if they are equal. Only the two corner allocations are planner solutions, though every allocation is PO.
(f) No: with one good, every \(a>0\) represents "more is better". Preferences, and hence the PO set, are identical for all \(a\). CC is a restriction on the utility numbers, and with \(a>1\) the planner approach fails purely because of the representation. Use \(v^i=(u^i)^{1/a}=x\) instead and every PO point is a planner solution.
Two households, two goods, \(\bar e=(1,1)\), identical utilities \(u^i(x_1,x_2)=(x_1x_2)^k\), \(k>0\). No disposal.
(a) \(k\log x_1+k\log x_2\) is an increasing transformation of \(u^i\). A Cobb–Douglas function \(x_1^{b}x_2^{c}\) is concave iff \(b+c\le1\). (d) Along the frontier the weighted sum is \(\lambda_1s^{2k}+\lambda_2(1-s)^{2k}\); look at its second derivative in \(s\).
(a) \(\log u^i=k(\log x_1+\log x_2)\) is strictly concave, so its upper contour sets are strictly convex. Upper contour sets are ordinal, so \(u^i\) is strictly quasi-concave. \(x_1^kx_2^k\) is Cobb–Douglas with exponent sum \(2k\): concave iff \(2k\le1\). (Along the diagonal, \(u^i(t,t)=t^{2k}\), which is convex for \(2k>1\), so concavity fails.)
(b) At an interior PO, if MRSs differed, a small trade would make both better off. \(MRS^i=x^i_2/x^i_1\) (the \(k\) cancels). Equality: \(\frac{x^1_2}{x^1_1}=\frac{1-x^1_2}{1-x^1_1}\Rightarrow x^1_2=x^1_1\). So \(x^1=(s,s)\) and \(x^2=(1-s,1-s)\).
(c) \(u_1=s^{2k}\), \(u_2=(1-s)^{2k}\), so \(u_1^{1/(2k)}+u_2^{1/(2k)}=s+(1-s)=1\). It bows inward iff \(2k>1\).
(d) Suppose an interior PO allocation \(x^1=(s,s)\), \(s\in(0,1)\), solved \(W(\lambda)\). Then in particular it would beat every other diagonal allocation, i.e. \(s\) would maximize \(f(s)=\lambda_1s^{2k}+\lambda_2(1-s)^{2k}\) on \([0,1]\). For \(2k>1\) and \(\lambda\ge0\), \(\lambda\ne0\), \(f''(s)=2k(2k-1)[\lambda_1s^{2k-2}+\lambda_2(1-s)^{2k-2}]>0\) on \((0,1)\), so \(f\) is strictly convex and \(f(s)\lt\max\{f(0),f(1)\}\): a contradiction. Interior PO points are never chosen; only the two corners are.
(e) \(v^i=(u^i)^{1/(2k)}=\sqrt{x_1x_2}\), which is concave (exponents sum to 1). The frontier becomes \(v_1+v_2=1\), and with \(\lambda=(1,1)\) every point on it, i.e. every PO allocation, solves \(W\). The PO set never changed; only the utility numbers did.
One good, aggregate endowment \(\bar e\), \(u^1(x)=u^2(x)=x\), \(X^1=[0,a]\) with \(0\lt a\lt\bar e\), \(X^2=\mathbb R_+\), and feasibility \(x^1+x^2\le\bar e\).
(a) \(X^1,X^2\) are intervals (convex); there is no production, so the aggregate set \(-\mathbb R_+\) (disposal) is convex; linear \(u^i\) are concave and strictly increasing. \(U\cap\mathbb R^2_+=\{u_1\le a,\ u_1+u_2\le\bar e\}\) (figure in §6).
(b) Maximize \(x^1\): \(x^1=a\), and any \(x^2\in[0,\bar e-a]\). Only \(x^2=\bar e-a\) is PO.
(c) The objective is \(\lambda_1x^1+\lambda_2x^2\); at an optimum nothing is disposed (\(\lambda_2>0\)), so \(x^2=\bar e-x^1\) and the objective is \(\lambda_2\bar e+(\lambda_1-\lambda_2)x^1\). If \(\lambda_1>\lambda_2\): \(x^1=a\). If \(\lambda_1\lt\lambda_2\): \(x^1=0\). If equal: any \(x^1\in[0,a]\). All of these are PO, as the theorem's sufficiency half promises.
(d) From (c), it is a solution whenever \(\lambda_1\ge\lambda_2\ge0\) (with \(\lambda\ne0\)). Not unique: the frontier has a kink there.
(e) §6, "a rescue".
(f) Then \(x^2=\bar e-x^1\) is forced; \(\lambda=(1,0)\) has the unique solution \((a,\bar e-a)\), which is PO, consistent with (e). The non-PO maximizers in (b) exist only because the planner can waste goods it gives zero weight to.