Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 36 (Mathematical Methods in General Equilibrium Theory)

Let $B(p,w)=\{z\in {\mathbb{R}}^L:p\cdot z\le w\}$ be the set allowed by the budget constraint. A demand correspondence is understood to choose only feasible bundles: $x(p,w)\subseteq B(p,w)$. Define the direct revealed preference relation byThe indirect revealed preference relation is its reflexive transitive closure. Thus $yRz$ precisely when there is a finite chain $y=y_0,y_1,\dots ,y_m=z$ whose successive comparisons satisfy $y_j{R}_D{y}_{j+1}$. Chains of length zero are included. This makes $R$ a preorder on all of ${\mathbb{R}}^L$, including alternatives that are never chosen. The associated strict preference isIf a convention defines indirect revelation using only positive-length chains, one must add the diagonal before the intersection assertion in part (c) can hold: every complete weak preference relation contains the diagonal.

The congruence axiom requires that an affordable alternative indirectly revealed at least as desirable as a chosen alternative is itself chosen. In the notation of part (a),Equivalently, a chosen bundle $x$ and an affordable but unchosen bundle $y$ satisfy $\neg (yRx)$. Since $x{R}_{D}y$ already holds, this means $xPy$. The axiom therefore distinguishes a genuine strict preference for a chosen bundle from indifference between two chosen bundles; it applies to the entire demand correspondence, including budgets with several chosen alternatives.

First pass from the preorder $R$ to a partial order. Define the indifference relation $y\sim z$ by $yRz$ and $zRy$. Reflexivity and transitivity make this an equivalence relation. On the quotient set $C={\mathbb{R}}^L/\sim$, putTransitivity makes this independent of the representatives, and the quotient relation is antisymmetric, hence a partial order.

We need its order-extension property, including freedom to order incomparable classes. Here is the argument. Order all partial order extensions of $⊵$ by inclusion. The union of a chain of extensions is again a partial order: any finite list of comparisons needed to check an axiom occurs together in one member of the chain. Zorn's lemma therefore supplies a maximal extension. If two classes $A,B$ remain incomparable, adjoin $A⊵B$ and take the transitive closure of a relation. The only possible new comparisons have the form $X⊵A⊵B⊵Y$. A cycle would require a previously existing path $B⊵A$, contradicting incomparability. The enlarged relation is still a partial order, contradicting maximality. Consequently the maximal extension is a total order. This also proves the Szpilrajn extension theorem used here.

Lift such a total order to the original bundles by comparing their equivalence classes. The lifted preference relation $R^{\prime }$ is complete and transitive, contains $R$, and preserves its strict preference: if $yPz$, their distinct classes satisfy $[y]⊵[z]$, and antisymmetry of the extended order prevents the reverse comparison. In particular, the class $\mathcal{P}$ is nonempty and $R\subseteq {\bigcap }_{R^{\prime }\in \mathcal{P}}{R}^{\prime }$.

For the reverse inclusion, take a pair with $\neg (yRz)$. If $zRy$, then $zPy$, so every member of $\mathcal{P}$ strictly ranks $z$ above $y$ and excludes $y{R}^{\prime }z$. If neither comparison holds, their classes are incomparable. First adjoin $[z]⊵[y]$, which creates no cycle by the preceding argument, and then extend to a total order. Its lift belongs to $\mathcal{P}$ and again excludes $y{R}^{\prime }z$. Thus every pair outside $R$ is outside at least one member of $\mathcal{P}$, provingThe intersection proof is an order-theoretic fact about any preorder. The congruence axiom additionally ensures exact rationalization of the observed demand correspondence: every chosen bundle weakly outranks every affordable bundle, while an affordable unchosen bundle is strictly worse by part (b). Hence each completion's budget-maximizing set is exactly the observed chosen set.

Use the exact-resource-balance convention for an exchange economy. If $\omega$ is the integrable initial-endowment profile, a feasible consumption allocation is a measurable integrable function $f:A\to {\mathbb{R}}_{+}^L$ withIt is Pareto efficient if there is no other feasible consumption allocation $g$ satisfying $g(a){⪰}_{a}f(a)$ for almost every consumer and $g(a){\succ }_{a}f(a)$ on a set of positive measure. A Pareto improvement makes nobody worse off, up to a null set, and makes a positive-measure group strictly better off. Positive measure is essential: reallocating to a single consumer of zero mass does not constitute an aggregate Pareto improvement.

The support statement uses a price quasi-equilibrium with transfers. A pair $(f,p)$ consists of an exactly feasible consumption allocation and a nonzero price vector $p\in {\mathbb{R}}^L$ such that, outside one null set of consumers,The allocation is on the consumer's transfer-adjusted budget boundary, at wealth $m(a)=p\cdot f(a)$. Relative to the original endowment, the lump-sum transfer is $\tau (a)=p\cdot (f(a)-\omega (a))$; exact feasibility gives ${\int }_A\tau d\mu =0$. A price quasi-equilibrium rules out cheaper strict improvements, but may allow a strictly preferred bundle of exactly equal cost. That is why it is weaker than maximizing the preference relation over the entire weak budget set.

Two conventions matter. The conclusion allows redistribution; it does not assert support at each consumer's original wealth $p\cdot \omega (a)$. Also, no monotonicity or free-disposal hypothesis is given, so the nonzero supporting vector is not asserted to have nonnegative coordinates. Those stronger economic conventions require additional hypotheses.

The measurable structure of an exchange economy includes measurability of consumers' preference relations: in particular, for each rational vector $q$, the setis measurable. Continuity of each individual's preference relation alone would not imply this across-consumer requirement. We use this usual measurability convention explicitly.

There are only countably many vectors in ${\mathbb{Q}}^L$. Remove all the null sets $E_q$ that have zero measure, obtainingLet $V=\{q\in {\mathbb{Q}}^L:\mu (E_q)>0\}$. Equivalently, $V$ is the union over $a\in D$ of the rational strict-improvement directions available to consumer $a$. This choice of $D$ ensures that a direction available to even one consumer in $D$ is available to a positive-measure group.

We claim that $0\notin \mathrm{co}V$. Otherwise, take finitely many $q_1,\dots ,q_m\in V$ and positive weights ${\lambda }_i$ with ${\sum }_i{\lambda }_i=1$ and ${\sum }_i{\lambda }_i{q}_i=0$. Choose $t>0$ so small that $2t<{\min }_i\mu (E_{q_i})$. We can choose disjoint measurable sets $F_i\subseteq E_{q_i}$ with $\mu (F_i)=t{\lambda }_i$. Indeed, after the earlier choices remove at most $t$ mass, the remaining portion of $E_{q_i}$ has mass greater than $t$. The scalar case of the Lyapunov convexity theorem, applied to its restricted atomless measure, realizes every mass between zero and its total, including $t{\lambda }_i$.

Now change $f(a)$ to $f(a)+q_i$ on $F_i$ and leave it unchanged elsewhere. Every changed bundle is a feasible consumption bundle and is strictly preferred by its recipient. This new consumption allocation $g$ remains integrable, andThus $g$ is exactly feasible and improves a group of total mass $t>0$, contradicting Pareto efficiency. Notice that no individual's consumption bundles were averaged: an atomless measure allows different small cohorts to make different improvements. Individual convexity of the preference relations is unnecessary.

If $V$ is nonempty, separate its convex hull from zero using the supporting hyperplane theorem. There exists $p\ne 0$ with $p\cdot q\ge 0$ for every $q\in V$. Only weak separation is needed, so it is harmless if zero belongs to the closure of this convex hull.

Fix any $a\in D$ and any $y{\succ }_{a}f(a)$. Completeness and continuity make the strictly preferred set relatively open in ${\mathbb{R}}_{+}^L$. Approximate $y-f(a)$ by rational vectors $q_n$ for which $f(a)+q_n$ is feasible and still strictly preferred. At a zero consumption coordinate, choose the rational approximations from the feasible side. Then $a\in E_{q_n}$ implies $q_n\in V$, and passage to the limit gives $p\cdot (y-f(a))\ge 0$. The same full-measure set $D$ works for all such bundles $y$, not merely for each bundle separately.

If $V$ is empty, the same rational approximation argument says that consumers in $D$ have no strictly preferred feasible bundles at all. In that case any nonzero $p$ satisfies the required implication. In either case,To see why transfers cannot simply be omitted, give all consumers endowment $(1,1)$ and utility function $u(x_1,x_2)=x_1+x_2$. Allocate $(3/2,3/2)$ to half of them and $(1/2,1/2)$ to the other half. This is Pareto efficient, since any Pareto improvement would increase the integral of utility while aggregate resources fix it. At these interior bundles the support inequalities force $p_1=p_2>0$: directions arbitrarily close to $(1,-1)$ force equality, and strict increases force positivity. The poorer half then receives less wealth than the original endowment supplies and has a cheaper strict improvement within its original budget. The support theorem concerns the redistributed budget, as defined in part (b).

For $v\in Z$, write $M(v)={\max }_i{v}_i$. Maximizing a linear functional on the price simplex givesThus $g(v)$ is a nonempty closed convex face of the compact price simplex. Its graph of a set-valued mapping is closed: if $v_n\to v$, $p_n\to p$ and $p_n\in g(v_n)$, then, for every fixed $q\in P$, the inequalities $p_n\cdot v_n\ge q\cdot v_n$ pass to $p\cdot v\ge q\cdot v$.

Each value $z(p)$ is closed in $Z$ by the closed-graph hypothesis, hence compact; it is nonempty and convex by assumption. Therefore $F(p,v)=g(v)\times z(p)$ has nonempty closed convex values in the nonempty compact convex set $K=P\times Z$.

Its graph of a set-valued mapping is also closed. If $(p_n,v_n)\to (p,v)$ and $(q_n,w_n)\to (q,w)$ with $q_n\in g(v_n)$ and $w_n\in z(p_n)$, the two closed-graph properties give $q\in g(v)$ and $w\in z(p)$. Finally, this closed graph implies upper hemicontinuity because the whole output space $K$ is compact. If an open set containing $F(p,v)$ failed to contain the nearby values, select outputs outside it along a convergent input sequence. A subsequence of those outputs converges in $K$; closedness puts its limit in $F(p,v)$, whereas closedness of the open set's complement puts it outside. This contradiction proves upper hemicontinuity.

Consequently all the hypotheses of the Kakutani fixed-point theorem hold: the domain is nonempty, compact and convex, and the set-valued mapping maps it into itself with nonempty closed convex values and upper hemicontinuity.

By the Kakutani fixed-point theorem, choose $(p^{\ast },v^{\ast })\in P\times Z$ with $p^{\ast }\in g(v^{\ast })$ and $v^{\ast }\in z(p^{\ast })$. The maximizing property of $g$ and the assumed Walras law giveEvery coordinate of $v^{\ast }$ is therefore nonpositive. Since $v^{\ast }\in z(p^{\ast })$, we obtainThis step does not yet imply that the entire vector is zero: a zero price can coexist with negative excess demand in its coordinate.

Use the same Kakutani fixed-point theorem pair as in part (b). If any coordinate of $p^{\ast }$ vanished, the additional boundary condition would force some coordinate of every member of $z(p^{\ast })$, including $v^{\ast }$, to be positive. This contradicts $v^{\ast }\le 0$. Hence $p_i^{\ast }>0$ for every $i$.

Now Walras law reads ${\sum }_i{p}_i^{\ast }{v}_i^{\ast }=0$, a sum of nonpositive terms with strictly positive coefficients. Every term must vanish, soThe boundary hypothesis first excludes boundary prices; strict positivity of all prices then turns nonpositive excess demand into exact market clearing.

Normalize the second price to one and write $p>0$ for the first price and $t>0$ for the varying endowment. Let $Z(p,t)$ be first-good aggregate excess demand. Under Walras law, the second excess-demand coordinate equals $-pZ(p,t)$, so one market-clearing equation suffices. A regular exchange economy at fixed $t$ has $Z_p(p,t)\ne 0$ at every equilibrium.

A sufficient joint condition is that zero be a regular value of the smooth map $Z:(0,\infty {)}^2\to \mathbb{R}$:To prove the almost-everywhere conclusion, the regular level set theorem makes $M=Z^{-1}(0)$ a smooth one-dimensional manifold. Project $M$ to the parameter axis by $\pi (p,t)=t$. Its tangent space consists of vectors $(h,k)$ with $Z_{p}h+Z_{t}k=0$. If $Z_p\ne 0$, every $k$ is possible, so the derivative of $\pi$ is surjective. If $Z_p=0$, the joint condition gives $Z_t\ne 0$, forcing $k=0$; the point is critical for $\pi$.

Thus the parameters with a singular equilibrium are exactly the critical values of this projection. Sard theorem makes that set a null set. Therefore almost every parameter gives a regular economy. No claim of existence at every parameter is needed for this sufficient regularity criterion.

Write the first endowment as $(a,t)$. The budget constraint at normalized prices is $px+y\le pa+t$. Because the quasilinear utility is increasing in $y$, the budget binds, and its maximization reduces to maximizing $\log x-px+pa+t$ over $x>0$. Its derivative is $1/x-p$ and its second derivative is $-1/x^2<0$, giving the unique demand correspondenceThe first-good demand has no wealth effect. Since the second consumer's endowment and preference relation are fixed, $Z_t=0$ everywhere. Thus the joint condition of part (a) becomes $Z_p\ne 0$ at equilibrium; it is not automatically satisfied.

Here is a complete smooth counterexample, with the second consumer's consumption set specified explicitly. The question does not impose the first consumer's consumption set on the second consumer. Take endowments $(a,t)=(1/10,t)$ and $(b,c)=(2,1)$. Give the second consumer consumption set $X_2=(-\infty ,A)\times \mathbb{R}$, where $A=43/20$, and put $B=31/20$. Its utility function isThis defines complete, transitive, continuous preference relations. It is in fact smooth on $X_2$: writing $r=A-x>0$ and $s=y-B>0$, the positive branch is $\exp (-r^2/(4s)-r/2-s/4)$; at $s=0$ all derivatives tend to zero locally uniformly where $r>0$. The endowment belongs to $X_2$. Allowing negative holdings is explicit here, just as the first consumer's second coordinate is unrestricted in the printed question.

We now derive, rather than posit, the second consumer's demand correspondence. For $v>0$, the upper contour $U_2\ge e^{-v}$ is described by $(r+s{)}^2\le 4vs$, with $r,s>0$. At a fixed $s$, its cheapest first coordinate uses $r=2\sqrt{vs}-s$. Minimizing total expenditure over $s$ therefore means minimizingThe unique minimum occurs at $\sqrt{s}=p\sqrt{v}/(1+p)$, with $r=vp(p+2)/(1+p{)}^2>0$. The expenditure necessary to reach this contour is consequentlySet $\delta =A-b=3/20$ and $\epsilon =B-c=11/20$. Wealth is $pb+c$, so the best affordable contour hasAll affordable positive-utility bundles have $v\ge v_{\ast }$, and the minimum-expenditure bundle on that contour is unique and spends precisely the budget. Zero-utility bundles cannot be optimal because this positive-utility bundle is available. Substitution yields globally smooth unique demands for every $p>0$:The aggregate excess demand is nowFor every $t>0$, the unique equilibrium price is $p=1$, and $Z_p(1,t)=Z_t(1,t)=0$. At this price the allocations are $(1,t-9/10)$ and $(11/10,19/10)$, whose sum equals the aggregate endowment $(21/10,t+1)$. Second-good excess demand is $-pZ$, so it too is smooth and clears. Hence every economy in this smooth parameterized family is singular. This demonstrates why the first consumer's missing wealth effect alone gives no transversality guarantee; the stronger conclusion in part (c) comes from the second consumer's additional demand restriction.

When both consumers have the same quasilinear utility on $(0,\infty )\times \mathbb{R}$, the same strictly concave one-variable maximization gives $x_1=x_2=1/p$, independently of their wealth. If their first-good endowments are $a,b$, thenThus the joint regular-value condition always holds. With positive total first-good endowment, the equilibrium isBoth first-good markets and second-good markets clear: second-good demands sum to $p^{\ast }(a+b)+t+c-2=t+c$. If a nonstandard endowment choice has $a+b\le 0$, there is no positive-price zero, so the regular value condition is vacuous. In the usual positive-endowment case every parameter has the displayed unique regular exchange economy.

Put $X={\mathbb{R}}_{+}^L$ and identify a continuous preference relation $R$ with its closed graph $G_R=\{(x,y)\in X\times X:xRy\}$. The closed convergence of preference relations is the topology induced by the Fell topology on these closed subsets of $X\times X$.

Concretely, a subbasis consists of the setsFinite intersections of these requirements form a basis, restricted to the specified complete, transitive, continuous preference relations. Thus a neighborhood may require some comparisons in prescribed open regions and forbid any comparison in prescribed compact regions. The use of compact sets in the avoidance condition is what makes this the Fell topology, rather than a stronger hyperspace topology on an unbounded consumption space.

There are two equivalent graph-limit requirements. Both are necessary. The upper condition is that, for every subsequence $n_j$ and convergent pairs $x_j\to x$, $y_j\to y$,The lower condition is that every limiting comparison can be approximated: for every $x⪰y$, there exist $x_n\to x$ and $y_n\to y$ withThe upper condition excludes spurious limiting comparisons; the lower condition prevents genuine limiting comparisons from disappearing. The subsequence quantifier in the upper condition is important: a violation need occur along only some of the indices. On the locally compact separable metric space $X\times X$, these two conditions are exactly closed convergence of preference relations.

No: uniform convergence of utilities on bounded sets does not imply closed convergence of preferences. It does imply the upper condition. Indeed, suppose $x_j\to x$, $y_j\to y$ and $u_{n_j}(x_j)\ge u_{n_j}(y_j)$. These bundles all lie in one bounded ball. Uniform convergence there, together with continuity of $u$, givesso $x⪰y$.

For a strict limiting comparison $u(x)>u(y)$, the fixed pair itself satisfies $u_n(x)>u_n(y)$ eventually. The obstruction is the lower condition at an indifference relation, where the strict approximating rankings need not disappear.

For a counterexample valid in every dimension $L\ge 1$, takeOn $\Vert x\Vert \le k$, $|u_n(x)-u(x)|\le k/n\to 0$, exactly as required. Every $u_n$ induces $x{⪰}_{n}y$ if and only if $x_1\ge y_1$, while $u$ makes every pair indifferent. In particular, $0⪰e_1$ holds in the limit. Any sequences $x_n\to 0$, $y_n\to e_1$ eventually have $(x_n{)}_1<(y_n{)}_1$, so they cannot satisfy $x_n{⪰}_n{y}_n$. The lower graph-limit condition fails.

Equivalently, an open neighborhood of $(0,e_1)$ sufficiently small to keep the first coordinate of its first bundle below that of its second bundle meets the limiting graph and none of the approximating graphs. The corresponding hit-open neighborhood in the Fell topology proves failure of convergence directly. HenceThe uniform utility convergence need not preserve closed preference convergence phenomenon reflects the loss of ranking information when the limiting utility has a flat indifference region.