Lesson 21 proved that, under convexity and concavity, every Pareto optimum maximizes a weighted sum of utilities \(\sum_i\lambda_iu^i\) for some \(\lambda\ge0\), \(\lambda\ne0\). This lesson turns that weighted sum into the utility function of a single fictitious consumer CN3 p.11.
For weights \(\lambda\in\mathbb R^I_+\) and an aggregate consumption vector \(x\in\mathbb R^m\), $$u(x;\lambda)\;=\;\max_{\{x^i\}}\;\sum_{i\in I}\lambda_i\,u^i(x^i)\qquad\text{s.t.}\qquad \sum_{i\in I}x^i=x,\quad x^i\in X^i\ \text{for all } i .$$ So \(u:\big(\sum_{i}X^i\big)\times\mathbb R^I_+\to\mathbb R\).
Read it as a two-stage problem. Someone hands the economy a bundle \(x\), and \(u(x;\lambda)\) is the best weighted welfare the planner can get by splitting \(x\) among the households. Four features of the definition:
Let \(Y=\sum_jY^j\) and \(\bar e=\sum_ie^i\). Then the weighted planner's problem \(W\) of Lesson 21 equals $$\max_{y\in Y}\;u(\bar e+y;\lambda).$$ The planner first picks aggregate production, treating the representative agent as the only consumer. Then it divides the resulting bundle optimally among households.
Averaging the two optimal splits is one way to share the averaged aggregate bundle. Concave households like averages, so even this naive split does at least as well as the average of the two welfare levels, and the optimal split can only do better. Concavity of \(u(\cdot;\lambda)\) is what makes the representative agent's problem well behaved: its first-order conditions are sufficient, and its demand is single-valued under strict concavity. Note what the proof needs: concave \(u^i\), which is cardinal (Lesson 21), and non-negative weights. A negative \(\lambda_i\) would flip Step 4.
Slide 13 compresses a chain of results into four bullets CN3 p.13. Here it is with every link labelled.
A representative agent always exists in this weak sense, but its utility \(u(\cdot;\lambda^*)\) depends on the weights, and the weights depend on who owns what. Change the distribution of endowments (a tax reform, a transfer, a shock that hits some agents and not others) and \(\lambda^*\) changes, so a different representative agent describes the new equilibrium. You cannot do comparative statics on the distribution with a fixed \(u(\cdot;\lambda)\). Lessons 24–25 ask when \(u(\cdot;\lambda)\) gives the same prices for every \(\lambda\). That is the aggregation result macroeconomists rely on.
The aggregate production set is the negative orthant, \(Y=-\mathbb R^m_+\). Goods can be thrown away (free disposal) but not produced. With \(y\le0\), feasibility \(\sum_ix^i=\bar e+y\) becomes $$\sum_{i\in I}x^i\;\le\;\sum_{i\in I}e^i\equiv\bar e,\qquad x^i\in X^i\ \text{for all } i, \tag{1}$$ where \(\bar e\in\mathbb R^m_+\) is the aggregate endowment.
By Section 1, nothing about the representative agent is lost by dropping firms. The rest of the note uses smooth (differentiable), concave utility functions. Then planner optima and CE can be characterized by first-order conditions, and the welfare theorems can be proved a second way, by matching FOCs (Lesson 23). Slide 15 asks you to keep track of which direction each argument uses CN3 p.15:
Take \(X^i=\mathbb R^m_+\) and differentiable \(u^i\). The planner's Lagrangian, with multiplier \(\gamma_l\) on the resource constraint (1) for good \(l\), is
$$\mathcal L=\sum_i\lambda_i\,u^i(x^i)+\sum_{l=1}^m\gamma_l\Big(\bar e_l-\sum_ix^i_l\Big).$$For every agent \(i\) and good \(l\): $$\lambda_i\frac{\partial u^i(\bar x^i)}{\partial x_l}\;\le\;\gamma_l,\qquad\text{with equality if } \bar x^i_l>0;$$ $$\gamma_l\ge0,\qquad \gamma_l\Big(\bar e_l-\sum_i\bar x^i_l\Big)=0 .$$ At an interior solution, the first line is the slides' equation (2) CN3 p.16: $$\lambda_i\,\frac{\partial u^i(\bar x^i)}{\partial x_l}=\gamma_l\qquad\text{for all } i,\ l. \tag{2}$$
| What it needs | Why | |
|---|---|---|
| Necessary | \(u^i\) differentiable at \(\bar x^i\) | All constraints (1) and \(x^i\ge0\) are linear, so a constraint qualification holds automatically, and KKT conditions hold at any local maximum. |
| Sufficient | Each \(u^i\) concave (CC) and \(\lambda\ge0\) | The objective \(\sum_i\lambda_iu^i\) is then concave and the feasible set is convex (linear constraints). For a concave program, a KKT point is a global maximum. |
| Interior form (2) | \(\bar x^i\gg0\) for all \(i\); in practice, Inada conditions | Slack non-negativity constraints have zero multipliers, so the inequality holds with equality. |
| \(\lambda\gg0\) | Implied by interiority plus strictly increasing \(u^i\) | If some agent has \(\lambda_j=0\), then (2) for \(j\) would say \(\gamma_l=0\). Another agent \(k\) with \(\lambda_k>0\) and \(u^k_l>0\) forces \(\gamma_l>0\). So zero-weight agents sit at a corner (they get nothing), and the interior case requires strictly positive weights. |
Slides 16–17 state the conditions under HH and CC "for an interior solution with strictly positive Pareto weights" CN3 p.16. The table explains each qualifier. Note also that \(\gamma_l=\lambda_iu^i_l>0\) when utility is strictly increasing, so complementary slackness forces (1) to bind: the planner throws nothing away.
Equation (2) carries two separate pieces of information. Dividing it for good \(l\) by itself for good \(k\), the \(\lambda_i\) cancels:
$$\underbrace{\frac{\partial u^i(\bar x^i)/\partial x_l}{\partial u^i(\bar x^i)/\partial x_k}}_{MRS^i_{lk}}=\frac{\gamma_l}{\gamma_k}=\underbrace{\frac{\partial u^j(\bar x^j)/\partial x_l}{\partial u^j(\bar x^j)/\partial x_k}}_{MRS^j_{lk}}\qquad\text{for all } i,j,\ l,k .$$Writing (2) for agents \(i\) and \(j\) and the same good \(l\), the \(\gamma_l\) cancels instead CN3 p.16:
$$\frac{\partial u^i(\bar x^i)/\partial x_l}{\partial u^j(\bar x^j)/\partial x_l}=\frac{\lambda_j}{\lambda_i}\qquad\text{for all } i,j,\ l .$$The condition \(MRS^i=MRS^j\) has no \(\lambda\) in it. The common value \(\gamma_l/\gamma_k\) is evaluated at the allocation the weights select, so in general it changes with \(\lambda\). The figure below shows this. Slide 17 says the same thing: \(\gamma\) is a function of \(\lambda\) and of all the utility functions CN3 p.17. The exceptions, where \(\gamma_l/\gamma_k\) does not depend on \(\lambda\), are the subject of Lessons 24–25.
Two agents, two goods, one unit of each in aggregate. A has \(u^A=a\log x_1+(1-a)\log x_2\) and B has \(u^B=b\log x_1+(1-b)\log x_2\). Using (2) and feasibility, \(\lambda_Aa/x^A_1=\gamma_1=\lambda_Bb/(1-x^A_1)\). Adding \(x^A_1=\lambda_Aa/\gamma_1\) and \(x^B_1=\lambda_Bb/\gamma_1\) and setting the sum to 1 gives a closed form:
$$\gamma_1=\lambda_Aa+\lambda_Bb,\qquad \gamma_2=\lambda_A(1-a)+\lambda_B(1-b),\qquad x^A_l=\frac{\lambda_A\beta^A_l}{\gamma_l},$$with \(\beta^A=(a,1-a)\) and \(\beta^B=(b,1-b)\). The shadow-price ratio \(\gamma_1/\gamma_2\) is a weighted average of the two agents' tastes, weighted by \(\lambda\).
Slide 17 says \(\gamma_l\) "has the interpretation of the marginal value for the planner (or rep-agent) of an extra unit of the aggregate endowment" CN3 p.17. Here is the precise statement and a proof that needs no differentiability of the optimal allocation.
Let \(X^i\) be convex, \(u^i\) concave and differentiable, and \(\lambda\ge0\). Let \(\{\bar x^i\}\) be an interior solution of the planner's problem at aggregate endowment \(\bar e\), with multipliers \(\gamma\) satisfying (2). Then $$u(x;\lambda)\;\le\;u(\bar e;\lambda)+\gamma\cdot(x-\bar e)\qquad\text{for every } x\in\textstyle\sum_iX^i,$$ so \(\gamma\) is a supergradient of the concave function \(u(\cdot;\lambda)\) at \(\bar e\). If \(u(\cdot;\lambda)\) is differentiable at \(\bar e\), then $$\frac{\partial u(\bar e;\lambda)}{\partial\bar e_l}=\gamma_l=\lambda_i\frac{\partial u^i(\bar x^i)}{\partial x_l}\qquad\text{for every agent } i .$$
You may know the textbook envelope argument: differentiate \(\mathcal L\) at the optimum with respect to \(\bar e_l\). The terms multiplying \(\partial x^i_k/\partial\bar e_l\) vanish by (2), the terms multiplying \(\partial\gamma_k/\partial\bar e_l\) vanish because the constraint binds, and what remains is \(\gamma_l\). That argument assumes the optimal allocation is differentiable in \(\bar e\). The proof above does not.
On Alvarez's exams this material shows up in two forms. The first is a long guided derivation, typically: planner FOC → solve for \(x^i_l\) → sum over agents → \(\gamma_l(\lambda)\) → compare with a representative agent. The second is true/false statements about weights and multipliers. The derivations below are the front half of that chain. Lessons 24–25 finish it.
For each statement, answer True or False and justify in at most four lines.
False. By the 1WT the CE is PO, so it solves a planner problem. With \(u^i=\sum_s\pi^i_sv^i(x_s)\), the FOC is \(\lambda_i\pi^i_sv^{i\prime}(x^i_s)=\gamma_s\). For two agents, \(\dfrac{v^{1\prime}(x^1_s)}{v^{2\prime}(\bar e_s-x^1_s)}=\dfrac{\lambda_2\pi^2_s}{\lambda_1\pi^1_s}\). So \(x^1_s\) depends on \(\bar e_s\) and on the belief ratio \(\pi^2_s/\pi^1_s\). Two states with the same \(\bar e_s\) but different belief ratios get different allocations. With common beliefs the \(\pi_s\) cancel, and \(x^i_s\) is a function of \(\bar e_s\) alone. (See Exercise 22.5.)
False. The condition holds at every PO allocation, but it is evaluated at the allocation that \(\lambda\) selects. In the two-agent Cobb–Douglas box, \(\gamma_1/\gamma_2=\frac{\lambda_Aa+\lambda_Bb}{\lambda_A(1-a)+\lambda_B(1-b)}\), which varies with \(\lambda_A/\lambda_B\) unless \(a=b\). Slide 17: \(\gamma\) is a function of \(\lambda\).
False in general. (2) gives \(u^i_l(\bar x^i)\lt u^j_l(\bar x^j)\) for every \(l\): a lower marginal utility, not necessarily more consumption. If \(u^i=a\log x_1+(1-a)\log x_2\) with \(a\) close to 0, agent \(i\) consumes very little good 1 even with a large weight. The statement is true under extra structure. One case is identical, additively separable, strictly concave utility: then \(u_l\) depends only on \(x_l\) and is decreasing, so a lower marginal utility means more of good \(l\). Another is identical homothetic preferences, where every agent consumes a multiple of the same vector.
False. The proof needs concavity, which is cardinal. Counterexample: one good, \(u^i(x)=x^2\) on \(\mathbb R_+\) (strictly increasing, so quasi-concave). Then \(u(x;\lambda)=\max_{x_1\in[0,x]}\lambda_1x_1^2+\lambda_2(x-x_1)^2=\max(\lambda_1,\lambda_2)\,x^2\), which is convex. The planner gives everything to one agent, because splitting is wasteful when utility is convex.
True. The objective \(\sum_ik\lambda_iu^i\) is a positive multiple of the original, so the argmax is unchanged. If \((\bar x,\gamma)\) satisfies \(\lambda_iu^i_l=\gamma_l\), then \((\bar x,k\gamma)\) satisfies \(k\lambda_iu^i_l=k\gamma_l\). Equivalently \(u(\bar e;k\lambda)=k\,u(\bar e;\lambda)\), so its gradient scales by \(k\). Only ratios of \(\gamma\) are meaningful.
False. Interior KKT for \(j\) would read \(0=\lambda_ju^j_l=\gamma_l\). For an agent \(k\) with \(\lambda_k>0\) we have \(\gamma_l=\lambda_ku^k_l>0\), a contradiction. Directly: anything given to \(j\) adds nothing to the objective and could raise it if given to \(k\), so \(j\) gets \(\bar x^j=0\) (a corner).
False. \(u(x;\lambda)\) only splits a given aggregate bundle among households, using \(\lambda\), \(u^i\) and \(X^i\). Technology enters afterwards, when the representative agent chooses \(y\in Y\) in \(\max_yu(\bar e+y;\lambda)\). That is why the note can study pure exchange without loss for this purpose.
There are \(m\) goods and \(I\) agents with \(u^i(c)=H(c-\theta^i)\), where \(\theta^i\in\mathbb R^m\). \(H\) is homogeneous of degree one, strictly increasing and strictly quasi-concave. Assume consumption must satisfy \(c^i\ge\theta^i\) (each agent's \(\theta^i\) is a subsistence vector) and \(\bar e\ge\bar\theta\equiv\sum_i\theta^i\). Fix weights \(\lambda\in\mathbb R^I_+\) and an aggregate endowment \(\bar e\in\mathbb R^m_+\).
Change variables to \(z^i=c^i-\theta^i\ge0\). The constraint becomes \(\sum_iz^i=\bar e-\bar\theta\). A function that is HD1 and quasi-concave, and positive on the relevant cone, is concave. Concave and HD1 together imply superadditivity: \(H(a+b)=2H\big(\tfrac{a+b}{2}\big)\ge H(a)+H(b)\).
(a) \(\displaystyle U(\bar e;\lambda)=\max_{\{c^i\}}\sum_{i=1}^I\lambda_iH(c^i-\theta^i)\quad\text{s.t.}\quad\sum_{i=1}^Ic^i=\bar e\) (good by good), with \(c^i\ge\theta^i\).
(b) With \(z^i=c^i-\theta^i\) and \(Z\equiv\bar e-\bar\theta\), the problem is \(\max\lambda\sum_iH(z^i)\) s.t. \(\sum_iz^i=Z\), \(z^i\ge0\). By superadditivity (applied repeatedly), \(\sum_iH(z^i)\le H(\sum_iz^i)=H(Z)\). Any split along the ray, \(z^i=s_iZ\) with \(s_i\ge0\) and \(\sum_is_i=1\), attains the bound, since \(\sum_iH(s_iZ)=\sum_is_iH(Z)=H(Z)\). So $$U(\bar e;\lambda)=\lambda\,H(\bar e-\bar\theta),\qquad\bar\theta=\sum_i\theta^i .$$ The "simple transformation" is the sum of the subsistence vectors. Equal weights are exactly the case that matters for the CE: with HD1 \(H\), every agent's marginal utility of income equals \(1/b(p)\), where \(b(p)\) is the cost of one unit of \(H\). So the Negishi weights \(1/\mu_i\) supporting any CE are equal across agents (Lessons 23 and 25).
(c) (i) With one agent, the constraint forces \(c^1=\bar e\), so \(U=\lambda_1H(\bar e-\theta^1)\). (ii) Scaling every weight by \(k\) scales the objective by \(k\) without changing the constraint. The candidate \(H(\bar e-\sum_i\lambda_i\theta^i)\) fails both checks. With \(I=1\) it gives \(H(\bar e-\lambda_1\theta^1)\ne\lambda_1H(\bar e-\theta^1)\) unless \(\lambda_1=1\), and scaling \(\lambda\) moves the argument of \(H\) instead of scaling the value. A commonly circulated answer to this exam question is exactly this candidate, and it is wrong. It comes from the step "\(\lambda_iH(z^i)=H(\lambda_iz^i)\), so the weights multiply the \(\theta^i\)". That step is valid, but the resource constraint restricts \(\sum_iz^i\), not \(\sum_i\lambda_iz^i\), so the weights cannot be pushed inside.
(d) With \(z^i\ge0\) and \(H\ge0\) on the cone, \(\sum_i\lambda_iH(z^i)\le(\max_i\lambda_i)\sum_iH(z^i)\le(\max_i\lambda_i)H(Z)\). Giving all of \(Z\) to an agent with the largest weight attains it. Because \(H\) is linear along rays, the planner does not care about spreading consumption, so any surplus above subsistence goes to the top-weight agent(s). This is a corner, unlike the interior allocations of Section 5. (Without the restriction \(c^i\ge\theta^i\), the problem can even be unbounded. For instance, with a Leontief \(H(z)=\min_lz_l\) and \(\lambda_1>\lambda_2\), give agent 1 \(Z+kv\) and agent 2 \(-kv\) with \(v=(1,\dots,1)\). Welfare rises like \((\lambda_1-\lambda_2)k\).)
(e) In both cases \(U=\kappa(\lambda)H(\bar e-\bar\theta)\) with \(\kappa(\lambda)>0\), so \(\nabla_{\bar e}U=\kappa(\lambda)\nabla H(\bar e-\bar\theta)\). By the envelope property, \(\gamma=\nabla_{\bar e}U\), so \(\gamma_l/\gamma_k=H_l(\bar e-\bar\theta)/H_k(\bar e-\bar\theta)\), independent of \(\lambda\). The shadow prices do not depend on the distribution: this is aggregation, which Lesson 25 develops. Unlike the Cobb–Douglas box of Section 6, here the representative agent's MRS is the same for every \(\lambda\).
Pure exchange with \(m\) goods and \(I\) consumers, with $$u^i(x^i)=-\sum_{l=1}^m\pi_l\frac{(\theta^i_l-x^i_l)^{1+\sigma}}{1+\sigma},\qquad X^i=\{x\in\mathbb R^m: x_l\le\theta^i_l\ \text{for all } l\},$$ where \(\pi_l>0\), \(\sigma>0\) and \(\theta^i_l>0\). The planner uses weights \(\lambda_i>0\) and feasibility \(\sum_ix^i_l=\bar e_l\). Let \(\bar\theta_l\equiv\sum_i\theta^i_l\), and assume \(\bar e_l\lt\bar\theta_l\) for every \(l\). On the exam each part had a line limit (1–4 lines). Keep to it.
Work with one term \(v(c)=-\pi_l(\theta-c)^{1+\sigma}/(1+\sigma)\) at a time. In part 6, raise both sides of part 5's answer to the power you need so that the sum over \(i\) factors.
1. Each term \(v(c)=-\pi_l(\theta-c)^{1+\sigma}/(1+\sigma)\) has \(v'(c)=\pi_l(\theta-c)^\sigma>0\) for \(c\lt\theta\), and \(v''(c)=-\sigma\pi_l(\theta-c)^{\sigma-1}\lt0\). The Hessian of \(u^i\) is diagonal with negative entries, hence negative definite: \(u^i\) is strictly concave and strictly increasing in each argument. (The derivative vanishes only at the boundary point \(c=\theta\).) Note: the official answer key writes \(v''=\sigma(\theta-c)^{\sigma-1}\lt0\), dropping the minus sign and \(\pi_l\). The expression above is the correct one.
2. \(\sigma=1\). 3. \(\theta^i_l\) is \(i\)'s satiation (bliss) point for good \(l\): marginal utility falls to zero there.
4. \(\lambda_i\pi_l(\theta^i_l-x^i_l)^\sigma=\gamma_l\).
5. \(x^i_l-\theta^i_l=-\big(\gamma_l/\pi_l\big)^{1/\sigma}\big(1/\lambda_i\big)^{1/\sigma}\).
6. Sum over \(i\): \(\bar e_l-\bar\theta_l=-(\gamma_l/\pi_l)^{1/\sigma}\sum_j(1/\lambda_j)^{1/\sigma}\), so $$\gamma_l=\pi_l\,\frac{(\bar\theta_l-\bar e_l)^\sigma}{\Big(\sum_j(1/\lambda_j)^{1/\sigma}\Big)^\sigma}.$$
7. \(\displaystyle x^i_l-\theta^i_l=(\bar e_l-\bar\theta_l)\,\frac{(1/\lambda_i)^{1/\sigma}}{\sum_j(1/\lambda_j)^{1/\sigma}}\).
8. \(\delta^i=\dfrac{(1/\lambda_i)^{1/\sigma}}{\sum_j(1/\lambda_j)^{1/\sigma}}\), and the shares obviously sum to one. Write \(\delta^i=a_i/(a_i+A_{-i})\) with \(a_i=(1/\lambda_i)^{1/\sigma}\). Then \(\delta^i\) increases with \(a_i\), and \(a_i\) falls as \(\lambda_i\) rises, so \(\delta^i\) falls with \(\lambda_i\). Since \(\bar e_l-\bar\theta_l\lt0\), the shortfall from satiation \(\theta^i_l-x^i_l=\delta^i(\bar\theta_l-\bar e_l)\) shrinks, so \(x^i_l\) rises with \(\lambda_i\). A higher weight moves \(i\) closer to bliss in every good. (The official key just says "\((1/\lambda_i)^{1/\sigma}\) decreases". The complete argument is about the ratio \(\delta^i\).)
9. Replacing \(\lambda_j\) by \(k\lambda_j\) multiplies \(\sum_j(1/\lambda_j)^{1/\sigma}\) by \(k^{-1/\sigma}\), and its \(\sigma\)-th power by \(k^{-1}\), so \(\gamma\to k\gamma\): only relative shadow values mean anything. \(\gamma_l\) rises with \(\bar\theta_l-\bar e_l\): the scarcer good \(l\) is relative to collective satiation, the more an extra unit is worth. (Feasibility needs \(\bar e_l\lt\bar\theta_l\). The exam uses this without stating it.) Notice that \(\lambda\) enters \(\gamma_l\) through a factor common to all goods. That is why \(\gamma_l/\gamma_k\) will turn out to be independent of \(\lambda\) in parts 9–11 of the exam, which are done in Lesson 25.
Pure exchange with \(I\) agents, \(m\) goods and \(u^i(x^i)=\sum_l\beta^i_l\log x^i_l\), where \(\beta^i_l>0\) and \(\sum_l\beta^i_l=1\). Aggregate endowment \(\bar e\gg0\), weights \(\lambda\gg0\).
In (c), write \(x^i_l=s^i_l\bar e_l\) with \(s^i_l=\lambda_i\beta^i_l/\sum_j\lambda_j\beta^j_l\). These shares do not depend on \(\bar e\).
(a) \(\lambda_i\beta^i_l/x^i_l=\gamma_l\), so \(x^i_l=\lambda_i\beta^i_l/\gamma_l\).
(b) \(\sum_ix^i_l=\bar e_l\Rightarrow\gamma_l=\sum_i\lambda_i\beta^i_l/\bar e_l\).
(c) \(x^i_l=s^i_l\bar e_l\) with the shares in the hint, so $$u(\bar e;\lambda)=\sum_i\lambda_i\sum_l\beta^i_l\big[\log s^i_l+\log\bar e_l\big]=\text{const}(\lambda)+\sum_l\Big(\sum_i\lambda_i\beta^i_l\Big)\log\bar e_l .$$ Then \(\partial u/\partial\bar e_l=\sum_i\lambda_i\beta^i_l/\bar e_l=\gamma_l\) ✓.
(d) \(\dfrac{\gamma_l}{\gamma_k}=\dfrac{\bar e_k}{\bar e_l}\,\dfrac{\sum_i\lambda_i\beta^i_l}{\sum_i\lambda_i\beta^i_k}\). It is independent of \(\lambda\) for all \(\lambda\) if and only if \(\beta^i\) is the same for all \(i\) (identical preferences). This is slide 31's example (Lesson 25), and the figure in Section 6 with \(I=m=2\).
(e) Yes: up to a constant, \(u(\bar e;\lambda)=\sum_lB_l(\lambda)\log\bar e_l\) with \(B_l(\lambda)=\sum_i\lambda_i\beta^i_l\). The representative agent's taste for good \(l\) is the \(\lambda\)-weighted average of individual tastes. This is exactly why different weights give different representative agents, and different prices.
For fixed \(x\), \(u(x;\lambda)\) is the maximum over feasible splits of a function that is linear in \(\lambda\). In (b), compare the split that is optimal at \(\lambda\) with the one that is optimal at \(\lambda'\).
(a) HD1: the objective at \(k\lambda\) is \(k\) times the objective at \(\lambda\), with the same feasible set. Convexity: for each feasible split \(\{x^i\}\), the map \(\lambda\mapsto\sum_i\lambda_iu^i(x^i)\) is linear. A pointwise maximum of linear functions is convex.
(b) Let \(v_i(\lambda)\equiv u^i(x^i(\lambda))\). For any \(\lambda'\), the split optimal at \(\lambda\) is feasible at \(\lambda'\), so \(u(x;\lambda')\ge\sum_i\lambda'_iv_i(\lambda)=u(x;\lambda)+\sum_i(\lambda'_i-\lambda_i)v_i(\lambda)\). So \(v(\lambda)\) is a subgradient of the convex function \(u(x;\cdot)\) at \(\lambda\). Where \(u(x;\cdot)\) is differentiable, the subgradient is the gradient: \(\partial u/\partial\lambda_i=v_i(\lambda)\). Monotonicity: swap the roles of \(\lambda\) and \(\lambda'\) in the same inequality and add the two. This gives \(\big(v(\lambda')-v(\lambda)\big)\cdot(\lambda'-\lambda)\ge0\). With \(\lambda'=\lambda+te_i\), \(t>0\), this reads \(t\,[v_i(\lambda')-v_i(\lambda)]\ge0\). So agent \(i\)'s utility is weakly increasing in its own weight, as the figure shows. This step needs no differentiability.
(c) Let \(\hat x\ne\bar x\) with optimal splits \(\{\hat x^i\},\{\bar x^i\}\). Since \(\sum_i\hat x^i\ne\sum_i\bar x^i\), some agent \(k\) has \(\hat x^k\ne\bar x^k\). In Step 4 of the concavity proof, \(k\)'s inequality is strict (strict concavity), and \(\lambda_k>0\) keeps it strict after weighting. Hence \(u(\rho\hat x+(1-\rho)\bar x;\lambda)>\rho u(\hat x;\lambda)+(1-\rho)u(\bar x;\lambda)\).
One good, \(S\) states, two agents. Agent \(i\) has expected utility \(\sum_s\pi^i_s\,\dfrac{(x^i_s)^{1-\rho}}{1-\rho}\) with \(\rho>0\) and subjective probabilities \(\pi^i_s>0\). The aggregate endowment in state \(s\) is \(\bar e_s>0\). Use planner weights \(\lambda_1,\lambda_2>0\).
(a) \(\lambda_i\pi^i_s(x^i_s)^{-\rho}=\gamma_s\). Dividing agent 1's condition by agent 2's: \(\big(x^1_s/x^2_s\big)^{\rho}=\dfrac{\lambda_1\pi^1_s}{\lambda_2\pi^2_s}\). Let \(k_s\equiv\big(\lambda_1\pi^1_s/(\lambda_2\pi^2_s)\big)^{1/\rho}\). With \(x^1_s+x^2_s=\bar e_s\), \(x^1_s/\bar e_s=k_s/(1+k_s)\).
(b) With \(\pi^1_s=\pi^2_s\), \(k_s=(\lambda_1/\lambda_2)^{1/\rho}\) is the same in every state, so agent 1's share is constant, and it rises with \(\lambda_1\). Consumption depends only on \(\bar e_s\) (linear sharing rules: mutual insurance of idiosyncratic risk).
(c) \(k_s\) rises with the belief ratio \(\pi^1_s/\pi^2_s\). Agent 1 gets a larger share in the states that agent 1 considers relatively likely: each agent effectively bets on its own beliefs. So \(x^1_s\) is not a function of \(\bar e_s\) alone, which is why the Fall 2005 statement is false. The CE has this form because, by the 1WT, it is PO, and by Lesson 21 every PO allocation solves \(W\) for some \(\lambda\).
(d) \(x^i_s=(\lambda_i\pi^i_s/\gamma_s)^{1/\rho}\), and summing, \(\gamma_s=\Big(\dfrac{\sum_i(\lambda_i\pi^i_s)^{1/\rho}}{\bar e_s}\Big)^{\rho}\). With common beliefs \(\pi_s\), \(\gamma_s=\pi_s\,\bar e_s^{-\rho}\Big(\sum_i\lambda_i^{1/\rho}\Big)^\rho\), so \(\gamma_s/\gamma_{s'}=\frac{\pi_s}{\pi_{s'}}\big(\bar e_s/\bar e_{s'}\big)^{-\rho}\), independent of \(\lambda\). The representative agent has the same CRRA utility and prices states by its marginal utility at the aggregate endowment. With heterogeneous beliefs this independence fails.