6. Exchange Economy

Jun He June 01, 2026

中文English 交换经济Exchange Economy 瓦尔拉斯均衡Walrasian Equilibrium 超额需求Excess Demand 福利定理Welfare Theorems 核Core

6. Exchange Economy

Note

本章导读本章研究交换经济与一般均衡：先定义瓦尔拉斯（竞争）均衡 $p^\star$ 及其配置（WEA）；在 Assumption 1（连续、强单调、严格拟凹）下证明个体需求存在唯一且连续（Thm 6.1，借 Weierstrass 与最大值定理），定义超额需求 $\mathbf{z}(p)$ 并给出其连续性、零次齐次、Walras 律三性质（Thm 6.4），从而 $\mathbf{z}(p^\star)=\mathbf{0}$ 为均衡的等价定义。随后用 Brouwer 不动点定理证明均衡存在（Thm 6.5、6.7）；最后讨论 WEA 的效率：核（core）、Pareto 效率、第一福利定理（Thm 6.8/6.9）、无交易定理与第二福利定理（Thm 6.10/6.11）。

6. Exchange Economy

Note

Overview This chapter studies the exchange economy and general equilibrium: we define the Walrasian (competitive) equilibrium $p^\star$ and its allocation (WEA); under Assumption 1 (continuous, strongly increasing, strictly quasi-concave) we prove individual demand exists, is unique and continuous (Thm 6.1, via Weierstrass and the Theory of Maximum), define excess demand $\mathbf{z}(p)$ with its three properties — continuity, homogeneity of degree zero, Walras' law (Thm 6.4) — so that $\mathbf{z}(p^\star)=\mathbf{0}$ is an equivalent definition of equilibrium. We then prove existence via Brouwer's Fixed Point Theorem (Thm 6.5, 6.7); finally we discuss the efficiency of WEA: the core, Pareto efficiency, the First Welfare Theorem (Thm 6.8/6.9), the No-Trade theorem, and the Second Welfare Theorem (Thm 6.10/6.11).

6.1 Set-up

交换经济由 $I$ 个消费者组成。令 $\mathcal{I}=\{1,2,\dots,I\}$ 为消费者集合；
每个消费者 $i\in\mathcal{I}$ 有禀赋向量 $\mathbf{e}^i=(e_1^i,e_2^i,\dots,e_n^i)\in\mathbb{R}_+^n$；
每个消费者 $i\in\mathcal{I}$ 的偏好由效用函数 $u^i:\mathbb{R}_+^n\to\mathbb{R}$ 表示；
交换经济定义为 $\mathcal{E}=(u^i,\mathbf{e}^i)_{i\in\mathcal{I}}$。

6.2 Definition of Walrasian equilibrium

非平凡的问题是：是否存在一个市场出清的价格向量。

6.1 Set-up

An exchange economy consists of $I$ consumers. Let $\mathcal{I}=\{1,2,\dots,I\}$ be the set of consumers;
Each consumer $i\in\mathcal{I}$ has an endowment vector $\mathbf{e}^i=(e_1^i,e_2^i,\dots,e_n^i)\in\mathbb{R}_+^n$;
Each consumer $i\in\mathcal{I}$ has preferences represented by a utility function $u^i:\mathbb{R}_+^n\to\mathbb{R}$;
An exchange economy is defined by $\mathcal{E}=(u^i,\mathbf{e}^i)_{i\in\mathcal{I}}$.

6.2 Definition of Walrasian equilibrium

The non-trivial question is whether we have a market-clearing price vector.

Important

Definition 6.1（Walrasian equilibrium for exchange economy）称 $\mathbf{p}^\star\in\mathbb{R}_+^n$ 为 $\mathcal{E}=(u^i,\mathbf{e}^i)_{i\in\mathcal{I}}$ 的瓦尔拉斯均衡（竞争均衡），当且仅当 $\exists\hat{\mathbf{x}}^1,\hat{\mathbf{x}}^2,\dots,\hat{\mathbf{x}}^I\in\mathbb{R}_+^n$ 使：(1) 效用最大化：$\hat{\mathbf{x}}^i$ 在 $\mathbf{p}^\star\cdot\mathbf{x}^i\le\mathbf{p}^\star\cdot\mathbf{e}^i=\sum_{k=1}^n p_k^\star e_k^i$ 下最大化 $u^i(\mathbf{x}^i)$，$\forall i\in\mathcal{I}$；(2) 市场出清：$\sum_{i\in\mathcal{I}}\hat{\mathbf{x}}^i=\sum_{i\in\mathcal{I}}\mathbf{e}^i$。称 $\hat{\mathbf{x}}=\{\hat{\mathbf{x}}^i\}_{i\in\mathcal{I}}$ 为瓦尔拉斯均衡配置（WEA）。We say $\mathbf{p}^\star\in\mathbb{R}_+^n$ is a Walrasian equilibrium (competitive equilibrium) of $\mathcal{E}=(u^i,\mathbf{e}^i)_{i\in\mathcal{I}}$ if and only if $\exists\hat{\mathbf{x}}^1,\hat{\mathbf{x}}^2,\dots,\hat{\mathbf{x}}^I\in\mathbb{R}_+^n$ such that: (1) utility maximization: $\hat{\mathbf{x}}^i$ maximizes $u^i(\mathbf{x}^i)$ subject to $\mathbf{p}^\star\cdot\mathbf{x}^i\le\mathbf{p}^\star\cdot\mathbf{e}^i=\sum_{k=1}^n p_k^\star e_k^i$ for $\forall i\in\mathcal{I}$; (2) market clearing: $\sum_{i\in\mathcal{I}}\hat{\mathbf{x}}^i=\sum_{i\in\mathcal{I}}\mathbf{e}^i$. And $\hat{\mathbf{x}}=\{\hat{\mathbf{x}}^i\}_{i\in\mathcal{I}}$ is a Walrasian equilibrium allocation (WEA).

下面作一些假设以保证瓦尔拉斯均衡 $\mathbf{p}^\star$ 存在。

6.3 Useful assumptions and theorems

Now let's make some assumptions such that the Walrasian equilibrium $\mathbf{p}^\star$ exists.

6.3 Useful assumptions and theorems

Important

Assumption 1 每个 $u^i:\mathbb{R}_+^n\to\mathbb{R}$ 都是连续、强单调（strongly increasing）且严格拟凹（strictly quasi-concave）的。Each $u^i:\mathbb{R}_+^n\to\mathbb{R}$ is continuous, strongly increasing, and strictly quasi-concave.

Important

Definition 6.2（strongly increasing）& Definition 6.3（strict quasi-concavity） Def 6.2：$u^i$ 强单调，当且仅当 $\forall\mathbf{x}\ge\mathbf{y}$ 且 $\mathbf{x}\ne\mathbf{y}\Rightarrow u^i(\mathbf{x})>u^i(\mathbf{y})$（$\mathbf{x}\ge\mathbf{y}$ 为逐分量不小于，$\mathbf{x}\ne\mathbf{y}$ 指至少一个分量严格更大）。事实上强单调排除了 $p_k=0$（$\forall k$）。
Def 6.3：$u^i$ 严格拟凹，若对 $\mathbf{x}\ne\mathbf{y}$ 及 $\forall t\in(0,1)$，$u^i(t\mathbf{x}+(1-t)\mathbf{y})>\min\{u^i(\mathbf{x}),u^i(\mathbf{y})\}$。Def 6.2: $u^i$ is strongly increasing iff $\forall\mathbf{x}\ge\mathbf{y}$ and $\mathbf{x}\ne\mathbf{y}\Rightarrow u^i(\mathbf{x})>u^i(\mathbf{y})$ ($\mathbf{x}\ge\mathbf{y}$ element-wise, $\mathbf{x}\ne\mathbf{y}$ means at least one element strictly larger). In fact strong increasingness rules out $p_k=0$ for $\forall k$.
Def 6.3: $u^i$ is strictly quasi-concave if for $\mathbf{x}\ne\mathbf{y}$ and $\forall t\in(0,1)$, $u^i(t\mathbf{x}+(1-t)\mathbf{y})>\min\{u^i(\mathbf{x}),u^i(\mathbf{y})\}$.

Important

Theorem 6.1 若 $u^i$ 满足 Assumption 1，则对 $\forall\mathbf{p}\in\mathbb{R}_{++}^n$，消费者的效用最大化问题 $\max_{\mathbf{x}^i\in\mathbb{R}_+^n}u^i(\mathbf{x}^i)$ s.t. $\mathbf{p}\cdot\mathbf{x}^i\le\mathbf{p}\cdot\mathbf{e}^i$ 有唯一解 $\mathbf{x}^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)$，且 $\mathbf{x}^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)$ 在 $\mathbb{R}_{++}^n$ 上关于 $\mathbf{p}$ 连续。If $u^i$ satisfies Assumption 1, then for $\forall\mathbf{p}\in\mathbb{R}_{++}^n$ the consumer's utility maximization problem $\max_{\mathbf{x}^i\in\mathbb{R}_+^n}u^i(\mathbf{x}^i)$ s.t. $\mathbf{p}\cdot\mathbf{x}^i\le\mathbf{p}\cdot\mathbf{e}^i$ has a unique solution $\mathbf{x}^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)$. In addition, $\mathbf{x}^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)$ is continuous in $\mathbf{p}$ on $\mathbb{R}_{++}^n$.

Note

证明 / Proof (Theorem 6.1)

因 $\mathbf{p}\in\mathbb{R}_{++}^n$，预算集可定义为对应 $\Gamma(\mathbf{p})=\{\mathbf{x}^i\in\mathbb{R}_+^n:\mathbf{p}\cdot\mathbf{x}^i\le\mathbf{p}\cdot\mathbf{e}^i\}$，它有界；又因预算约束不等式为弱不等式，$\Gamma(\mathbf{p})$ 亦闭。由 $\Gamma(\mathbf{p})\subseteq\mathbb{R}_+^n$，$\Gamma(\mathbf{p})$ 紧。因 $u^i$ 连续，由 Weierstrass 定理（Thm 6.2） 保证最大值存在。由 $u^i$ 严格拟凹可证解唯一：设有两个不同解 $\tilde{\mathbf{x}}^i$ 与 $\hat{\mathbf{x}}^i$，取凸组合，由严格拟凹 $u^i(t\tilde{\mathbf{x}}^i+(1-t)\hat{\mathbf{x}}^i)>\min\{u^i(\tilde{\mathbf{x}}^i),u^i(\hat{\mathbf{x}}^i)\}$，与二者为解矛盾。连续性由最大值定理（Thm 6.3）得：$\Gamma(\mathbf{p})$ 既上半连续又下半连续，故作为对应连续。定义 $\mathbf{p}$ 处最大化效用 $h(\mathbf{p})=\max_{\mathbf{x}^i\in\Gamma(\mathbf{p})}u^i(\mathbf{x}^i)$，因 $u^i(\mathbf{x}^i)$ 关于 $\mathbf{p}$ 与 $\mathbf{x}^i$ 连续、$\Gamma(\mathbf{p})$ 紧且连续，最大值定理蕴含 $h(\mathbf{p})$ 连续。又 $h(\mathbf{p})=u^i(\mathbf{x}^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i))$，因 $u^i$ 连续单调，可定义其逆 $u^{i(-1)}$（亦连续单调），故 $\mathbf{x}^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)=u^{i(-1)}(h(\mathbf{p}))$ 连续。$\blacksquare$

Since $\mathbf{p}\in\mathbb{R}_{++}^n$, the budget set can be defined as a correspondence $\Gamma(\mathbf{p})=\{\mathbf{x}^i\in\mathbb{R}_+^n:\mathbf{p}\cdot\mathbf{x}^i\le\mathbf{p}\cdot\mathbf{e}^i\}$, which is bounded; and since the inequality in the budget constraint is weak, $\Gamma(\mathbf{p})$ is also closed. Since $\Gamma(\mathbf{p})\subseteq\mathbb{R}_+^n$, $\Gamma(\mathbf{p})$ is compact. Since $u^i$ is continuous, the Weierstrass theorem (Thm 6.2) guarantees existence of a maximum. By strict quasi-concavity of $u^i$ the solution is unique: suppose two distinct solutions $\tilde{\mathbf{x}}^i$ and $\hat{\mathbf{x}}^i$; taking a convex combination, by strict quasi-concavity $u^i(t\tilde{\mathbf{x}}^i+(1-t)\hat{\mathbf{x}}^i)>\min\{u^i(\tilde{\mathbf{x}}^i),u^i(\hat{\mathbf{x}}^i)\}$, contradicting that both are solutions. Continuity follows from the Theory of Maximum (Thm 6.3): $\Gamma(\mathbf{p})$ is both upper and lower hemi-continuous, hence continuous as a correspondence. Define the maximized utility at $\mathbf{p}$ as $h(\mathbf{p})=\max_{\mathbf{x}^i\in\Gamma(\mathbf{p})}u^i(\mathbf{x}^i)$; since $u^i(\mathbf{x}^i)$ is continuous in both $\mathbf{p}$ and $\mathbf{x}^i$, and $\Gamma(\mathbf{p})$ is compact and continuous, the Theory of Maximum implies $h(\mathbf{p})$ is continuous. Recall $h(\mathbf{p})=u^i(\mathbf{x}^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i))$; since $u^i$ is continuous and monotone, we can define the inverse $u^{i(-1)}$ (also continuous and monotone), so $\mathbf{x}^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)=u^{i(-1)}(h(\mathbf{p}))$ is continuous in $\mathbf{p}$. $\blacksquare$

Important

Theorem 6.2（Weierstrass）& Theorem 6.3（Theory of Maximum） Thm 6.2：对连续函数 $f:S\to\mathbb{R}$，若 $S$ 为 $\mathbb{R}^m$ 的紧子集，则 $f$ 在 $S$ 上取得最大与最小值。
Thm 6.3：对 $X\subseteq\mathbb{R}^n$、$Y\subseteq\mathbb{R}^m$，若 $f:X\times Y\to\mathbb{R}$ 关于 $x,y$ 连续，$\Gamma:X\to Y$ 紧值且连续，则 $h(x)=\max_{y\in\Gamma(x)}f(x,y)$ 连续，且 $G(x)=\{y\in\Gamma(x):f(x,y)=h(x)\}$ 非空、紧值、上半连续。Thm 6.2: For continuous $f:S\to\mathbb{R}$, if $S$ is a compact subset of $\mathbb{R}^m$, then $f$ attains a maximum and minimum value on $S$.
Thm 6.3: For $X\subseteq\mathbb{R}^n$, $Y\subseteq\mathbb{R}^m$, if $f:X\times Y\to\mathbb{R}$ is continuous in $x,y$ and $\Gamma:X\to Y$ is compact-valued and continuous, then $h(x)=\max_{y\in\Gamma(x)}f(x,y)$ is continuous, and $G(x)=\{y\in\Gamma(x):f(x,y)=h(x)\}$ is non-empty, compact-valued, and upper hemi-continuous.

为方便记号，定义超额需求。

For the purpose of notation, let's define excess demand.

Important

Definition 6.4（Excess demand for exchange economy）价格 $\mathbf{p}$ 下商品 $k$ 的超额需求为 $z_k(\mathbf{p})=\sum_{i\in\mathcal{I}}x_k^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)-\sum_{i\in\mathcal{I}}e_k^i$，并定义 $\mathbf{z}(\mathbf{p})\equiv(z_1(\mathbf{p}),z_2(\mathbf{p}),\dots,z_n(\mathbf{p}))$。Excess demand for good $k$ at price $\mathbf{p}$ is $z_k(\mathbf{p})=\sum_{i\in\mathcal{I}}x_k^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)-\sum_{i\in\mathcal{I}}e_k^i$, and we define $\mathbf{z}(\mathbf{p})\equiv(z_1(\mathbf{p}),z_2(\mathbf{p}),\dots,z_n(\mathbf{p}))$.

Important

Theorem 6.4（Properties of excess demand）若每个 $u^i$ 满足 Assumption 1，则 $\mathbf{z}(\mathbf{p})$ 对 $\forall\mathbf{p}\in\mathbb{R}_{++}^n$ 良定义，且：(1)（连续性）$\mathbf{z}(\cdot)$ 在 $\mathbb{R}_{++}$ 上连续；(2)（零次齐次）$\mathbf{z}(\lambda\mathbf{p})=\mathbf{z}(\mathbf{p})$，$\forall\lambda>0$、$\forall\mathbf{p}\in\mathbb{R}_{++}^n$；(3)（Walras 律）$\mathbf{p}\cdot\mathbf{z}(\mathbf{p})=0$，$\forall\mathbf{p}\in\mathbb{R}_{++}^n$。If each $u^i$ in $\{u^i\}_{i\in\mathcal{I}}$ satisfies Assumption 1, then $\mathbf{z}(\mathbf{p})$ is well-defined for $\forall\mathbf{p}\in\mathbb{R}_{++}^n$, and: (1) (continuity) $\mathbf{z}(\cdot)$ is continuous on $\mathbb{R}_{++}$; (2) (homogeneity of degree zero) $\mathbf{z}(\lambda\mathbf{p})=\mathbf{z}(\mathbf{p})$ for $\forall\lambda>0$, $\forall\mathbf{p}\in\mathbb{R}_{++}^n$; (3) (Walras' law) $\mathbf{p}\cdot\mathbf{z}(\mathbf{p})=0$ for $\forall\mathbf{p}\in\mathbb{R}_{++}^n$.

Note

证明 / Proof (Theorem 6.4)

$\mathbf{z}(\cdot)$ 在 $\mathbf{p}$ 处连续，因 $x_k^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)$ 连续且减去常数 $e_k^i$ 保持连续，故各分量连续，$\mathbf{z}(\cdot)$ 连续。$\mathbf{z}(\cdot)$ 零次齐次，因价格乘 $\lambda>0$ 不改变预算集。因 $u^i$ 强单调，每个个体的预算约束取等号 $\mathbf{p}\cdot\mathbf{x}^i=\mathbf{p}\cdot\mathbf{e}^i$，故

$\mathbf{z}(\cdot)$ is continuous at $\mathbf{p}$ because $x_k^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)$ is continuous and subtracting the constant $e_k^i$ preserves continuity; so each element is continuous and thus $\mathbf{z}(\cdot)$ is continuous. $\mathbf{z}(\cdot)$ is homogeneous of degree zero because multiplying prices by $\lambda>0$ does not change the budget set. Since $u^i$ is strongly increasing, each agent's budget constraint holds with equality $\mathbf{p}\cdot\mathbf{x}^i=\mathbf{p}\cdot\mathbf{e}^i$, hence

$$ > \begin{aligned} > \sum_{i\in\mathcal{I}}\mathbf{p}\cdot\mathbf{x}^i=\sum_{i\in\mathcal{I}}\mathbf{p}\cdot\mathbf{e}^i&\Leftrightarrow\mathbf{p}\cdot\sum_{i\in\mathcal{I}}\mathbf{x}^i=\mathbf{p}\cdot\sum_{i\in\mathcal{I}}\mathbf{e}^i\\ > &\Leftrightarrow\mathbf{p}\cdot\left(\sum_{i\in\mathcal{I}}\mathbf{x}^i-\sum_{i\in\mathcal{I}}\mathbf{e}^i\right)=0\\ > &\Leftrightarrow\mathbf{p}\cdot\mathbf{z}(\mathbf{p})=0.\quad\blacksquare > \end{aligned} > $$

Important

Definition 6.5（Walrasian equilibrium, equivalent）向量 $\mathbf{p}^\star\in\mathbb{R}_{++}^n$ 为瓦尔拉斯均衡，若 $\mathbf{z}(\mathbf{p}^\star)=\mathbf{0}$。A vector $\mathbf{p}^\star\in\mathbb{R}_{++}^n$ is a Walrasian equilibrium if $\mathbf{z}(\mathbf{p}^\star)=\mathbf{0}$.

Tip

注记 6.1（局部 vs 一般均衡）/ Remark 6.1 (Partial vs general equilibrium) 任一市场的超额需求 $z_k(\mathbf{p})$ 可能依赖于所有其他市场的现行价格，故市场系统可能完全或部分相互依赖。若 $z_k(\mathbf{p})=0$ 称单一市场 $k$ 有局部均衡，但完全可能某商品 $k'$ 仍有 $z_{k'}(\mathbf{p})\ne0$。只有当 $\mathbf{z}(\mathbf{p})=\mathbf{0}$ 时，才称市场系统处于一般均衡。Excess demand in any particular market $z_k(\mathbf{p})$ may depend on the prevailing prices in every other market. So the system could be completely or partially interdependent. We say there is a partial equilibrium in single market $k$ if $z_k(\mathbf{p})=0$, but it is entirely possible that $z_{k'}(\mathbf{p})\ne0$ for some other good $k'$. Only when $\mathbf{z}(\mathbf{p})=\mathbf{0}$ can we say the system of markets is in a general equilibrium.

6.4 Existence of a Walrasian equilibrium

Important

Theorem 6.5 设 $\mathbf{z}:\mathbb{R}_{++}^n\to\mathbb{R}$ 满足：(1)（连续性）$\mathbf{z}(\cdot)$ 在 $\mathbb{R}_{++}^n$ 上连续；(2)（Walras 律）$\mathbf{p}\cdot\mathbf{z}(\mathbf{p})=0$，$\forall\mathbf{p}\in\mathbb{R}_{++}^n$；(3) 若 $\{\mathbf{p}^n\}$ 为 $\mathbb{R}_{++}^n$ 中价格序列且 $\mathbf{p}^n\to\bar{\mathbf{p}}\ne\mathbf{0}$，$\bar p_k=0$（某商品 $k$，即收敛到边界但非原点），则 $\exists k'$ 使 $\bar p_{k'}=0$ 且序列 $\{z_{k'}(\mathbf{p}^n)\}$ 上无界（取 $k,k'$ 分开，是因为允许多个商品价格趋零，但只能保证其中一个有无穷超额需求）。则 $\exists\mathbf{p}^\star\in\mathbb{R}_{++}^n$ 使 $\mathbf{z}(\mathbf{p}^\star)=\mathbf{0}$。Suppose $\mathbf{z}:\mathbb{R}_{++}^n\to\mathbb{R}$ satisfies: (1) (continuity) $\mathbf{z}(\cdot)$ continuous on $\mathbb{R}_{++}^n$; (2) (Walras' law) $\mathbf{p}\cdot\mathbf{z}(\mathbf{p})=0$ for $\mathbf{p}\in\mathbb{R}_{++}^n$; (3) if $\{\mathbf{p}^n\}$ is a sequence of prices in $\mathbb{R}_{++}^n$ and $\mathbf{p}^n\to\bar{\mathbf{p}}\ne\mathbf{0}$, $\bar p_k=0$ for some good $k$ (converges to the boundary but not the origin), then $\exists k'$ s.t. $\bar p_{k'}=0$ and the sequence $\{z_{k'}(\mathbf{p}^n)\}$ is unbounded above (we take $k,k'$ separately because we allow several goods' prices to converge to zero but can only guarantee one of them has infinite excess demand). Then $\exists\mathbf{p}^\star\in\mathbb{R}_{++}^n$ such that $\mathbf{z}(\mathbf{p}^\star)=\mathbf{0}$.

Important

Theorem 6.6（Brouwer's Fixed Point Theorem）设 $f:C\to C$ 为连续的自映射，$C$ 为 $\mathbb{R}^n$ 的非空紧凸子集。则 $\exists x^\star\in C$ 使 $f(x^\star)=x^\star$。Let $f:C\to C$ be a continuous self-mapping, where $C$ is a nonempty compact convex subset of $\mathbb{R}^n$. Then $\exists x^\star\in C$ such that $f(x^\star)=x^\star$.

Note

证明 / Proof (Theorem 6.5)

固定 $\varepsilon\in(0,1)$，考虑一组远离零边界的价格。因 $\mathbf{z}(\mathbf{p})$ 零次齐次，可把价格限制在 $[0,1]$：$S_\varepsilon=\left\{\mathbf{p}\in[0,1]^n:\sum_{k=1}^n p_k=1\text{ and }p_k\ge\tfrac{\varepsilon}{1+2n}\ \forall k\right\}$。$S_\varepsilon$ 紧（闭，不等式皆弱；有界，$\mathbf{p}\in[0,1]^n$），非空（取 $p_k=\tfrac{2+1/n}{1+2n}$，则 $\sum p_k=\tfrac{2n+1}{1+2n}=1$ 且 $\ge\tfrac{\varepsilon}{1+2n}$）且凸（任取 $\mathbf{p},\mathbf{p}'\in S_\varepsilon$，凸组合 $\tilde{\mathbf{p}}=t\mathbf{p}+(1-t)\mathbf{p}'$ 满足 $\sum\tilde p_k=1$ 且各分量 $\ge\tfrac{\varepsilon}{1+2n}$）。定义

Fix $\varepsilon\in(0,1)$ and consider prices kept away from the zero boundary. Since $\mathbf{z}(\mathbf{p})$ is h.o.d. 0, restrict prices to $[0,1]$: $S_\varepsilon=\left\{\mathbf{p}\in[0,1]^n:\sum_{k=1}^n p_k=1\text{ and }p_k\ge\tfrac{\varepsilon}{1+2n}\ \forall k\right\}$. $S_\varepsilon$ is compact (closed since the inequalities are all weak; bounded since $\mathbf{p}\in[0,1]^n$), non-empty (take $p_k=\tfrac{2+1/n}{1+2n}$, then $\sum p_k=\tfrac{2n+1}{1+2n}=1$ and $\ge\tfrac{\varepsilon}{1+2n}$), and convex (for any $\mathbf{p},\mathbf{p}'\in S_\varepsilon$, the convex combination $\tilde{\mathbf{p}}=t\mathbf{p}+(1-t)\mathbf{p}'$ satisfies $\sum\tilde p_k=1$ and each component $\ge\tfrac{\varepsilon}{1+2n}$). Define

$$f_k(\mathbf{p})=\frac{\varepsilon+p_k+\max\{\tilde z_k(\mathbf{p}),0\}}{n\varepsilon+1+\sum_{m=1}^n\max\{\tilde z_m(\mathbf{p}),0\}}$$

其中 $\tilde z_k(\mathbf{p})=\min\{z_k(\mathbf{p}),1\}$。令 $f(\mathbf{p})=(f_1(\mathbf{p}),\dots,f_n(\mathbf{p}))$ 为新价格向量。

where $\tilde z_k(\mathbf{p})=\min\{z_k(\mathbf{p}),1\}$. Let $f(\mathbf{p})=(f_1(\mathbf{p}),\dots,f_n(\mathbf{p}))$ be the new price vector.

Important

Claim 6.1 $f(\mathbf{p})$ 把 $S_\varepsilon$ 连续地映入自身。$f(\mathbf{p})$ maps $S_\varepsilon$ continuously into itself.

证明：$x^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)$ 连续、加总保连续，故 $z_k(\mathbf{p})$ 连续；$\min\{z_k,1\}$、$\max\{z_k,0\}$ 保连续，故每个 $f_k(\mathbf{p})$ 连续，$f(\mathbf{p})$ 连续。再证 $\mathbf{p}\in S_\varepsilon\Rightarrow f(\mathbf{p})\in S_\varepsilon$：$f_k(\mathbf{p})\ge0$（$\varepsilon>0$、$p_k\ge0$、$\max\ge0$）；且 $\sum_{k=1}^n f_k(\mathbf{p})=\tfrac{n\varepsilon+\sum_k p_k+\sum_k\max\{\tilde z_k,0\}}{n\varepsilon+1+\sum_m\max\{\tilde z_m,0\}}=1$（由构造）；又因 $\max\{\tilde z_k,0\}\le1$，

Proof: $x^i(\mathbf{p},\mathbf{p}\cdot\mathbf{e}^i)$ is continuous and adding up preserves continuity, so $z_k(\mathbf{p})$ is continuous; $\min\{z_k,1\}$ and $\max\{z_k,0\}$ preserve continuity, so each $f_k(\mathbf{p})$ is continuous and $f(\mathbf{p})$ is continuous. Now show $\mathbf{p}\in S_\varepsilon\Rightarrow f(\mathbf{p})\in S_\varepsilon$: $f_k(\mathbf{p})\ge0$ (since $\varepsilon>0$, $p_k\ge0$, $\max\ge0$); and $\sum_{k=1}^n f_k(\mathbf{p})=\tfrac{n\varepsilon+\sum_k p_k+\sum_k\max\{\tilde z_k,0\}}{n\varepsilon+1+\sum_m\max\{\tilde z_m,0\}}=1$ (by construction); and since $\max\{\tilde z_k,0\}\le1$,

$$f_k(\mathbf{p})=\frac{\varepsilon+p_k+\max\{\tilde z_k(\mathbf{p}),0\}}{n\varepsilon+1+\sum_{m=1}^n\max\{\tilde z_m(\mathbf{p}),0\}}\ge\frac{\varepsilon+0+0}{n\varepsilon+1+n}=\frac{\varepsilon}{1+(\varepsilon+1)n}\ge\frac{\varepsilon}{1+2n}$$

对 $\forall k$ 成立。$\blacksquare$

for $\forall k$. $\blacksquare$

由 Brouwer 不动点定理（Thm 6.6），$\exists\mathbf{p}^\varepsilon\in S_\varepsilon$ 使 $f(\mathbf{p}^\varepsilon)=\mathbf{p}^\varepsilon$。故 $f_k(\mathbf{p}^\varepsilon)=p_k^\varepsilon$，整理（从两边减 $p_k^\varepsilon$）：

By Brouwer's Fixed Point Theorem (Thm 6.6), $\exists\mathbf{p}^\varepsilon\in S_\varepsilon$ such that $f(\mathbf{p}^\varepsilon)=\mathbf{p}^\varepsilon$. So $f_k(\mathbf{p}^\varepsilon)=p_k^\varepsilon$, and rearranging (subtract $p_k^\varepsilon$ from both sides):

$$p_k^\varepsilon\left[n\varepsilon+\sum_{m=1}^n\max\{\tilde z_m(\mathbf{p}^\varepsilon),0\}\right]=\varepsilon+\max\{\tilde z_k(\mathbf{p}^\varepsilon),0\}\tag{6.1}$$

现想令 $\varepsilon\to0$ 以放松 $\varepsilon>0$。但 $z_k(\mathbf{p})$ 在 $p_k=0$ 处无定义，故须先证 $\mathbf{p}^\varepsilon\to\mathbf{p}^\star\in\mathbb{R}_{++}^n$。考虑 $\mathbf{p}^\varepsilon\in S_\varepsilon\subseteq S\equiv\{\mathbf{p}\in[0,1]^n:\sum_{k=1}^n p_k=1\}$，$S$ 闭且有界故紧。$\{\mathbf{p}^\varepsilon\}$ 或不收敛，但必有收敛子列；WLOG 设 $\{\mathbf{p}^\varepsilon\}$ 自身收敛，记 $\mathbf{p}^\star=\lim_{\varepsilon\to0}\mathbf{p}^\varepsilon$，紧性给出 $\mathbf{p}^\star\in S$。

Now we want to relax $\varepsilon>0$ by $\varepsilon\to0$. But $z_k(\mathbf{p})$ is not defined at $p_k=0$, so first prove $\mathbf{p}^\varepsilon\to\mathbf{p}^\star\in\mathbb{R}_{++}^n$. Consider $\mathbf{p}^\varepsilon\in S_\varepsilon\subseteq S\equiv\{\mathbf{p}\in[0,1]^n:\sum_{k=1}^n p_k=1\}$; $S$ is closed and bounded, hence compact. $\{\mathbf{p}^\varepsilon\}$ may not converge, but there is a convergent subsequence; WLOG assume $\{\mathbf{p}^\varepsilon\}$ itself converges, denote $\mathbf{p}^\star=\lim_{\varepsilon\to0}\mathbf{p}^\varepsilon$; compactness implies $\mathbf{p}^\star\in S$.

Important

Claim 6.2 $\mathbf{p}^\star\in\mathbb{R}_{++}^n$。$\mathbf{p}^\star\in\mathbb{R}_{++}^n$.

证明：反证。设 $\exists k$ 使 $p_k^\star=0$。因 $\mathbf{p}^\varepsilon\to\mathbf{p}^\star\ne\mathbf{0}$ 且 $\mathbf{p}^\varepsilon\in\mathbb{R}_{++}^n$，由 Thm 6.5 条件 3，$\exists k'$ 使 $\{z_{k'}(\mathbf{p}^\varepsilon)\}$ 上无界（$\varepsilon\to0$）。但 (6.1) 对 $\forall k$ 成立，故

Proof: by contradiction. Suppose $\exists k$ s.t. $p_k^\star=0$. Since $\mathbf{p}^\varepsilon\to\mathbf{p}^\star\ne\mathbf{0}$ and $\mathbf{p}^\varepsilon\in\mathbb{R}_{++}^n$, by condition 3 in Thm 6.5, $\exists k'$ s.t. $\{z_{k'}(\mathbf{p}^\varepsilon)\}$ is unbounded above (as $\varepsilon\to0$). However, (6.1) is true for $\forall k$, so

$$\underbrace{p_{k'}^\varepsilon}_{\to0}\bigg[\underbrace{n\varepsilon}_{\to0}+\underbrace{\sum_{m=1}^n\max\{\tilde z_m(\mathbf{p}^\varepsilon),0\}}_{\le n}\bigg]=\underbrace{\varepsilon}_{\to0}+\max\{\tilde z_{k'}(\mathbf{p}^\varepsilon),0\}\tag{6.2}$$

因 $\{z_{k'}(\mathbf{p}^\varepsilon)\}$ 上无界，$\tilde z_{k'}(\mathbf{p}^\varepsilon)$ 无穷多次取值 1，故 $\max\{\tilde z_{k'}(\mathbf{p}^\varepsilon),0\}$ 不收敛到 0，(6.2) 右边不收敛到 0；但左边收敛到 0，矛盾。$\blacksquare$

Since $\{z_{k'}(\mathbf{p}^\varepsilon)\}$ is unbounded above, $\tilde z_{k'}(\mathbf{p}^\varepsilon)$ infinitely often takes value 1, so $\max\{\tilde z_{k'}(\mathbf{p}^\varepsilon),0\}$ does not converge to 0, and the RHS of (6.2) does not converge to 0; but the LHS converges to 0 — contradiction. $\blacksquare$

现取极限 $\varepsilon\to0$，把 (6.1) 改写为 $p_k^\star\sum_{m=1}^n\max\{\tilde z_m(\mathbf{p}^\star),0\}=\max\{\tilde z_k(\mathbf{p}^\star),0\}$（$\forall k$）。两边乘 $z_k(\mathbf{p}^\star)$ 并对 $k$ 求和：

Now take the limit $\varepsilon\to0$ and rewrite (6.1) as $p_k^\star\sum_{m=1}^n\max\{\tilde z_m(\mathbf{p}^\star),0\}=\max\{\tilde z_k(\mathbf{p}^\star),0\}$ for $\forall k$. Multiply both sides by $z_k(\mathbf{p}^\star)$ and sum across $k$:

$$\underbrace{\left(\sum_{k=1}^n z_k(\mathbf{p}^\star)p_k^\star\right)}_{=0\text{ by Walras' law}}\left(\sum_{m=1}^n\max\{\tilde z_m(\mathbf{p}^\star),0\}\right)=\sum_{k=1}^n z_k(\mathbf{p}^\star)\max\{\tilde z_k(\mathbf{p}^\star),0\}$$

$$\Rightarrow 0=\sum_{k=1}^n z_k(\mathbf{p}^\star)\max\{\tilde z_k(\mathbf{p}^\star),0\}\tag{6.3}$$

须证 (6.3) 蕴含 $z_k(\mathbf{p}^\star)=0$（$\forall k$）。若 $\exists k$ 使 $z_k(\mathbf{p}^\star)>0$，则 $\max\{\tilde z_k,0\}>0$ 且该项 $>0$；而对 $z_k(\mathbf{p}^\star)<0$ 的 $k$，$\max\{\tilde z_k,0\}=0$ 该项为 0。故为使 (6.3) 成立，必有 $z_k(\mathbf{p}^\star)\le0$（$\forall k$）。再由 Walras 律，任一严格为负的 $z_k(\mathbf{p}^\star)$ 必伴随某 $k'$ 使 $z_{k'}(\mathbf{p}^\star)$ 严格为正，而后者已被排除。故 $z_k(\mathbf{p}^\star)=0$（$\forall k$）。$\blacksquare$

We must show (6.3) implies $z_k(\mathbf{p}^\star)=0$ for $\forall k$. If $\exists k$ s.t. $z_k(\mathbf{p}^\star)>0$, then $\max\{\tilde z_k,0\}>0$ and that term $>0$; while for any $k$ with $z_k(\mathbf{p}^\star)<0$, $\max\{\tilde z_k,0\}=0$ and the term is 0. So for (6.3) to hold, $z_k(\mathbf{p}^\star)\le0$ for $\forall k$. By Walras' law, any strictly negative $z_k(\mathbf{p}^\star)$ must be accompanied by some $k'$ with $z_{k'}(\mathbf{p}^\star)$ strictly positive, which has already been ruled out. Therefore $z_k(\mathbf{p}^\star)=0$ for $\forall k$. $\blacksquare$

现讨论保证 Thm 6.5 条件成立的交换经济假设。

Now let's discuss the assumptions on the exchange economy that insure the conditions in Theorem 6.5 are satisfied.

Important

Theorem 6.7（Existence of Walrasian equilibrium）设 $\mathcal{E}=(u^i,\mathbf{e}^i)_{i\in\mathcal{I}}$ 中每个 $u^i$ 满足 Assumption 1 且 $\sum_{i\in\mathcal{I}}\mathbf{e}^i\gg\mathbf{0}$。则 $\mathbf{z}(\cdot)$ 满足 Thm 6.5 的条件，从而存在至少一个瓦尔拉斯均衡价格 $\mathbf{p}^\star\gg\mathbf{0}$ 使 $\mathbf{z}(\mathbf{p}^\star)=\mathbf{0}$。Suppose each $u^i$ in $\mathcal{E}=(u^i,\mathbf{e}^i)_{i\in\mathcal{I}}$ satisfies Assumption 1 and $\sum_{i\in\mathcal{I}}\mathbf{e}^i\gg\mathbf{0}$. Then the conditions on $\mathbf{z}(\cdot)$ in Thm 6.5 are satisfied, and thus there exists at least one Walrasian equilibrium price $\mathbf{p}^\star\gg\mathbf{0}$, i.e. at least one price vector $\mathbf{p}^\star\gg\mathbf{0}$ such that $\mathbf{z}(\mathbf{p}^\star)=\mathbf{0}$.

Note

证明 / Proof (Theorem 6.7)

由 Thm 6.4，条件 (1)(2) 已满足，只需证条件 (3)。设 $\{\mathbf{p}^m\}$ 为 $\mathbb{R}_{++}^n$ 中序列，$\mathbf{p}^m\to\bar{\mathbf{p}}\ne\mathbf{0}$ 且 $\bar p_k=0$（某 $k$）。须证 $\exists k'$ 使 $\bar p_{k'}=0$ 且 $\{z_{k'}(\mathbf{p}^m)\}$ 上无界，用反证法。因 $\sum_{i\in\mathcal{I}}\mathbf{e}^i\gg\mathbf{0}$ 且 $\bar{\mathbf{p}}\ne\mathbf{0}$，

By Thm 6.4, conditions (1) and (2) are already satisfied, so it remains to show condition (3). Suppose $\{\mathbf{p}^m\}$ is a sequence in $\mathbb{R}_{++}^n$, $\mathbf{p}^m\to\bar{\mathbf{p}}\ne\mathbf{0}$ and $\bar p_k=0$ for some $k$. We must show $\exists k'$ s.t. $\bar p_{k'}=0$ and $\{z_{k'}(\mathbf{p}^m)\}$ is unbounded above, by contradiction. Since $\sum_{i\in\mathcal{I}}\mathbf{e}^i\gg\mathbf{0}$ and $\bar{\mathbf{p}}\ne\mathbf{0}$,

$$\bar{\mathbf{p}}\cdot\sum_{i\in\mathcal{I}}\mathbf{e}^i=\sum_{i\in\mathcal{I}}\bar{\mathbf{p}}\cdot\mathbf{e}^i>0$$

故 $\exists i\in\mathcal{I}$ 使 $\bar{\mathbf{p}}\cdot\mathbf{e}^i>0$，即至少一个个体收入严格为正。考虑该个体 $i$，令 $\mathbf{x}^m=\mathbf{x}^i(\mathbf{p}^m,\mathbf{p}^m\cdot\mathbf{e}^i)$，先证 $i$ 对某商品 $k'$ 的需求上无界。反证设 $\mathbf{x}^m$ 有界，则有收敛子列 $\mathbf{x}^m_{\text{sub}}\to\mathbf{x}^\star$，对应价格 $\mathbf{p}^m_{\text{sub}}$。因 $u^i$ 强单调，最优处预算约束取等号 $\mathbf{p}^m_{\text{sub}}\cdot\mathbf{x}^m_{\text{sub}}=\mathbf{p}^m_{\text{sub}}\cdot\mathbf{e}^i$，取 $m\to\infty$：$\bar{\mathbf{p}}\cdot\mathbf{x}^\star=\bar{\mathbf{p}}\cdot\mathbf{e}^i>0$。考虑束 $\tilde{\mathbf{x}}=\mathbf{x}^\star+(0,\dots,0,1,0,\dots,0)$（第 $k$ 分量加 1）。因 $\bar p_k=0$，$\bar{\mathbf{p}}\cdot\tilde{\mathbf{x}}=\bar{\mathbf{p}}\cdot\mathbf{x}^\star>0$。又 $u^i$ 强单调，$u^i(\tilde{\mathbf{x}})>u^i(\mathbf{x}^\star)$。由 $u^i$ 连续，对足够大 $t\in(0,1)$：$\bar{\mathbf{p}}\cdot t\tilde{\mathbf{x}}<\bar{\mathbf{p}}\cdot\mathbf{x}^\star$、$u^i(t\tilde{\mathbf{x}})>u^i(\mathbf{x}^\star)$。又因 $\mathbf{p}^m_{\text{sub}}\to\bar{\mathbf{p}}$、$\mathbf{x}^m_{\text{sub}}\to\mathbf{x}^\star$ 且 $u^i$ 连续，对足够大 $m$：$\mathbf{p}^m_{\text{sub}}\cdot t\tilde{\mathbf{x}}<\mathbf{p}^m_{\text{sub}}\cdot\mathbf{x}^m_{\text{sub}}$、$u^i(t\tilde{\mathbf{x}})>u^i(\mathbf{x}^m_{\text{sub}})$，与 $\mathbf{x}^m_{\text{sub}}$ 为最优需求矛盾。故 $\mathbf{x}^m$ 上无界。因 $\mathbf{x}^m$ 非负，上无界蕴含 $\exists k'$ 使 $\{x_{k'}^m\}$ 上无界。然而 $i$ 的收入收敛到有限的 $\bar{\mathbf{p}}\cdot\mathbf{e}^i$，故 $\{\mathbf{p}^m\cdot\mathbf{e}^i\}$ 有界，结合 $\{x_{k'}^m\}$ 上无界得 $p_{k'}^m\to0$。最后，因总禀赋有限且各个体对 $k'$ 的需求非负，$\{x_{k'}^m\}$ 上无界蕴含 $\{z_{k'}(\mathbf{p}^m)\}$ 上无界。$\blacksquare$

So $\exists i\in\mathcal{I}$ s.t. $\bar{\mathbf{p}}\cdot\mathbf{e}^i>0$, i.e. at least one agent has strictly positive income. Consider this agent $i$, let $\mathbf{x}^m=\mathbf{x}^i(\mathbf{p}^m,\mathbf{p}^m\cdot\mathbf{e}^i)$; we first show $i$'s demand for some good $k'$ is unbounded above. By contradiction, suppose $\mathbf{x}^m$ is bounded; then there is a convergent subsequence $\mathbf{x}^m_{\text{sub}}\to\mathbf{x}^\star$ with corresponding prices $\mathbf{p}^m_{\text{sub}}$. Since $u^i$ is strongly increasing, the budget constraint holds with equality at the optimum, $\mathbf{p}^m_{\text{sub}}\cdot\mathbf{x}^m_{\text{sub}}=\mathbf{p}^m_{\text{sub}}\cdot\mathbf{e}^i$; taking $m\to\infty$: $\bar{\mathbf{p}}\cdot\mathbf{x}^\star=\bar{\mathbf{p}}\cdot\mathbf{e}^i>0$. Consider the bundle $\tilde{\mathbf{x}}=\mathbf{x}^\star+(0,\dots,0,1,0,\dots,0)$ (adding 1 to the $k$th component). Since $\bar p_k=0$, $\bar{\mathbf{p}}\cdot\tilde{\mathbf{x}}=\bar{\mathbf{p}}\cdot\mathbf{x}^\star>0$. Moreover $u^i$ strongly increasing gives $u^i(\tilde{\mathbf{x}})>u^i(\mathbf{x}^\star)$. By continuity of $u^i$, for sufficiently large $t\in(0,1)$: $\bar{\mathbf{p}}\cdot t\tilde{\mathbf{x}}<\bar{\mathbf{p}}\cdot\mathbf{x}^\star$ and $u^i(t\tilde{\mathbf{x}})>u^i(\mathbf{x}^\star)$. Since $\mathbf{p}^m_{\text{sub}}\to\bar{\mathbf{p}}$, $\mathbf{x}^m_{\text{sub}}\to\mathbf{x}^\star$ and $u^i$ continuous, for sufficiently large $m$: $\mathbf{p}^m_{\text{sub}}\cdot t\tilde{\mathbf{x}}<\mathbf{p}^m_{\text{sub}}\cdot\mathbf{x}^m_{\text{sub}}$ and $u^i(t\tilde{\mathbf{x}})>u^i(\mathbf{x}^m_{\text{sub}})$, contradicting that $\mathbf{x}^m_{\text{sub}}$ is optimal. So $\mathbf{x}^m$ is unbounded. Since $\mathbf{x}^m$ is non-negative, unboundedness implies $\exists k'$ s.t. $\{x_{k'}^m\}$ is unbounded above. However $i$'s income converges to the finite $\bar{\mathbf{p}}\cdot\mathbf{e}^i$, so $\{\mathbf{p}^m\cdot\mathbf{e}^i\}$ is bounded, and with $\{x_{k'}^m\}$ unbounded above this implies $p_{k'}^m\to0$. Finally, since the aggregate endowment is finite and all agents demand a non-negative amount of good $k'$, $\{x_{k'}^m\}$ unbounded above implies $\{z_{k'}(\mathbf{p}^m)\}$ unbounded above. $\blacksquare$

6.5 Efficiency of Walrasian equilibrium allocation

下面给出若干有用定义。

6.5 Efficiency of Walrasian equilibrium allocation

Below are a couple of useful definitions.

Important

Definitions 6.6–6.10（WEA / feasible / Pareto / blocking / core） Def 6.6（WEA）：若 $\mathbf{p}^\star$ 为瓦尔拉斯均衡价格，消费者 $i$ 在 $\mathbf{p}^\star\mathbf{x}^i\le\mathbf{p}^\star\mathbf{e}^i$ 下最大化 $u^i$ 于 $\hat{\mathbf{x}}^i$ 且 $\sum_{i\in\mathcal{I}}\hat{\mathbf{x}}^i=\sum_{i\in\mathcal{I}}\mathbf{e}^i$，则 $\hat{\mathbf{x}}=(\hat{\mathbf{x}}^1,\dots,\hat{\mathbf{x}}^I)$ 为瓦尔拉斯均衡配置。
Def 6.7（可行配置集）：$F(\mathbf{e})=\{\mathbf{x}=(\mathbf{x}^1,\dots,\mathbf{x}^I):\sum_{i\in\mathcal{I}}\mathbf{x}^i=\sum_{i\in\mathcal{I}}\mathbf{e}^i\}$。
Def 6.8（Pareto 效率）：可行配置 $\mathbf{x}\in F(\mathbf{e})$ Pareto 有效，当且仅当 $\nexists\mathbf{y}\in F(\mathbf{e})$ 使 $u^i(\mathbf{y}^i)\ge u^i(\mathbf{x}^i)$（$\forall i$）且至少一处严格。
Def 6.9（阻塞联盟）：$S\subseteq\mathcal{I}$ 阻塞 $\mathbf{x}\in F(\mathbf{e})$，若存在 $\mathbf{y}$ 在联盟内可行（$\sum_{i\in S}\mathbf{y}^i=\sum_{i\in S}\mathbf{e}^i$）且 $u^i(\mathbf{y}^i)\ge u^i(\mathbf{x}^i)$（$\forall i\in S$）至少一处严格。
Def 6.10（核 core）：交换经济的核是所有未被阻塞的可行配置之集 $C(\mathbf{e})\subseteq F(\mathbf{e})$。Def 6.6 (WEA): if $\mathbf{p}^\star$ is a Walrasian equilibrium price, consumer $i$ maximizes $u^i$ s.t. $\mathbf{p}^\star\mathbf{x}^i\le\mathbf{p}^\star\mathbf{e}^i$ at $\hat{\mathbf{x}}^i$ and $\sum_{i\in\mathcal{I}}\hat{\mathbf{x}}^i=\sum_{i\in\mathcal{I}}\mathbf{e}^i$, then $\hat{\mathbf{x}}=(\hat{\mathbf{x}}^1,\dots,\hat{\mathbf{x}}^I)$ is a Walrasian equilibrium allocation.
Def 6.7 (feasible allocation set): $F(\mathbf{e})=\{\mathbf{x}=(\mathbf{x}^1,\dots,\mathbf{x}^I):\sum_{i\in\mathcal{I}}\mathbf{x}^i=\sum_{i\in\mathcal{I}}\mathbf{e}^i\}$.
Def 6.8 (Pareto efficiency): a feasible $\mathbf{x}\in F(\mathbf{e})$ is Pareto efficient iff $\nexists\mathbf{y}\in F(\mathbf{e})$ s.t. $u^i(\mathbf{y}^i)\ge u^i(\mathbf{x}^i)$ for all $i$ with at least one strict inequality.
Def 6.9 (blocking coalition): $S\subseteq\mathcal{I}$ blocks $\mathbf{x}\in F(\mathbf{e})$ if there is $\mathbf{y}$ feasible within the coalition ($\sum_{i\in S}\mathbf{y}^i=\sum_{i\in S}\mathbf{e}^i$) and $u^i(\mathbf{y}^i)\ge u^i(\mathbf{x}^i)$ for all $i\in S$ with at least one strict inequality.
Def 6.10 (core): the core of an exchange economy is the set of all unblocked feasible allocations $C(\mathbf{e})\subseteq F(\mathbf{e})$.

Tip

注记 6.2 / Remark 6.2 未被阻塞指配置不被任何联盟阻塞。由于 Pareto 最优等价于不被大联盟（全体 $\mathcal{I}$）阻塞，故核中配置必皆 Pareto 最优，但 Pareto 最优不必在核中。若记全部 Pareto 最优配置为 $P(\mathbf{e})$，则 $C(\mathbf{e})\subseteq P(\mathbf{e})$。Unblocked means the allocation is not blocked by any coalition. Since Pareto optimality is equivalent to not being blocked by the grand coalition (the set $\mathcal{I}$ of all consumers), allocations in the core must all be Pareto optimal, but Pareto optimal allocations are not necessarily in the core. If we denote the set of all Pareto optimal allocations as $P(\mathbf{e})$, then $C(\mathbf{e})\subseteq P(\mathbf{e})$.

Important

Lemma 6.1 & Definition 6.11（strictly increasing） Lemma 6.1：对任意 $\mathbf{p}\in\mathbb{R}_+^n$，设 $u^i$ 严格单调。若 $\tilde{\mathbf{x}}^i$ 在 $\mathbf{p}\mathbf{x}^i\le\mathbf{p}\mathbf{e}^i$ 下最大化，则对任意 $\mathbf{x}^i\in\mathbb{R}_+^n$：(1) $u^i(\mathbf{x}^i)>u^i(\tilde{\mathbf{x}}^i)\Rightarrow\mathbf{p}\mathbf{x}^i>\mathbf{p}\tilde{\mathbf{x}}^i$；(2) $u^i(\mathbf{x}^i)\ge u^i(\tilde{\mathbf{x}}^i)\Rightarrow\mathbf{p}\mathbf{x}^i\ge\mathbf{p}\tilde{\mathbf{x}}^i$。
Def 6.11：$u$ 严格单调，若 $\mathbf{x}\ge\mathbf{y}\Rightarrow u(\mathbf{x})\ge u(\mathbf{y})$ 且 $\mathbf{x}>\mathbf{y}\Rightarrow u(\mathbf{x})>u(\mathbf{y})$。强单调必严格单调，反之不必。Lemma 6.1: for any $\mathbf{p}\in\mathbb{R}_+^n$, suppose $u^i$ is strictly increasing. If $\tilde{\mathbf{x}}^i$ maximizes s.t. $\mathbf{p}\mathbf{x}^i\le\mathbf{p}\mathbf{e}^i$, then for any $\mathbf{x}^i\in\mathbb{R}_+^n$: (1) $u^i(\mathbf{x}^i)>u^i(\tilde{\mathbf{x}}^i)\Rightarrow\mathbf{p}\mathbf{x}^i>\mathbf{p}\tilde{\mathbf{x}}^i$; (2) $u^i(\mathbf{x}^i)\ge u^i(\tilde{\mathbf{x}}^i)\Rightarrow\mathbf{p}\mathbf{x}^i\ge\mathbf{p}\tilde{\mathbf{x}}^i$.
Def 6.11: $u$ is strictly increasing if $\mathbf{x}\ge\mathbf{y}\Rightarrow u(\mathbf{x})\ge u(\mathbf{y})$ and $\mathbf{x}>\mathbf{y}\Rightarrow u(\mathbf{x})>u(\mathbf{y})$. Strongly increasing is always strictly increasing, but not vice versa.

Note

证明 / Proof (Lemma 6.1)

(1) 直接来自带约束的最大化问题的定义。证 (2)：对任意 $\varepsilon>0$，$u^i(\mathbf{x}^i+(\varepsilon,\dots,\varepsilon))>u^i(\mathbf{x}^i)\ge u^i(\tilde{\mathbf{x}}^i)$，由 (1) $\mathbf{p}\cdot(\mathbf{x}^i+(\varepsilon,\dots,\varepsilon))>\mathbf{p}\cdot\tilde{\mathbf{x}}^i$，取 $\varepsilon\to0$：$\mathbf{p}\cdot\mathbf{x}^i\ge\mathbf{p}\cdot\tilde{\mathbf{x}}^i$。$\blacksquare$

(1) is straight from the definition of the constrained maximization problem. Prove (2): for any $\varepsilon>0$, $u^i(\mathbf{x}^i+(\varepsilon,\dots,\varepsilon))>u^i(\mathbf{x}^i)\ge u^i(\tilde{\mathbf{x}}^i)$, so by (1) $\mathbf{p}\cdot(\mathbf{x}^i+(\varepsilon,\dots,\varepsilon))>\mathbf{p}\cdot\tilde{\mathbf{x}}^i$; take $\varepsilon\to0$: $\mathbf{p}\cdot\mathbf{x}^i\ge\mathbf{p}\cdot\tilde{\mathbf{x}}^i$. $\blacksquare$

6.5.1 First Welfare Theorem

证明 WEA 为 Pareto 有效有两种方式：一是先证 WEA 在核中，再由定义得核中配置皆 Pareto 有效；二是直接证 WEA Pareto 有效。先用第一种。

6.5.1 First Welfare Theorem

There are two ways of proving WEA is Pareto efficient. One: prove WEA is in the core, then by definition every allocation in the core is Pareto efficient. The other: directly prove WEA is Pareto efficient. Start with the first way.

Important

Theorem 6.8 若 $u^i$ 严格单调，则每个 WEA 都在核中。If $u^i$ is strictly increasing, then every WEA is in the core.

Note

证明 / Proof (Theorem 6.8)

反证。设有 WEA $\hat{\mathbf{x}}$ 不在核中。因 $\hat{\mathbf{x}}$ 可行，必 $\exists S\subseteq\mathcal{I}$ 及配置 $\tilde{\mathbf{x}}$ 在联盟内可行（$\sum_{i\in S}\tilde{\mathbf{x}}^i=\sum_{i\in S}\mathbf{e}^i$）且至少一名成员严格更好，即 $u^i(\tilde{\mathbf{x}}^i)\ge u^i(\hat{\mathbf{x}}^i)$（$\forall i\in S$）至少一处严格。设 WE 价格 $\mathbf{p}$，由 Lemma 6.1，$\mathbf{p}\tilde{\mathbf{x}}^i\ge\mathbf{p}\hat{\mathbf{x}}^i$ 至少一处严格。对联盟内求和：$\mathbf{p}\sum_{i\in S}\mathbf{e}^i=\mathbf{p}\sum_{i\in S}\tilde{\mathbf{x}}^i>\mathbf{p}\sum_{i\in S}\hat{\mathbf{x}}^i$。因每个 $u^i$ 严格单调，每人预算约束取等号 $\mathbf{p}\hat{\mathbf{x}}^i=\mathbf{p}\mathbf{e}^i$，故 $\mathbf{p}\sum_{i\in S}\hat{\mathbf{x}}^i=\mathbf{p}\sum_{i\in S}\mathbf{e}^i$，矛盾。故此联盟不存在，所有 WEA 在核中。$\blacksquare$

By contradiction. Suppose there is a WEA $\hat{\mathbf{x}}$ not in the core. Since $\hat{\mathbf{x}}$ is feasible, $\exists S\subseteq\mathcal{I}$ and an allocation $\tilde{\mathbf{x}}$ feasible within the coalition ($\sum_{i\in S}\tilde{\mathbf{x}}^i=\sum_{i\in S}\mathbf{e}^i$) with at least one member strictly better off, i.e. $u^i(\tilde{\mathbf{x}}^i)\ge u^i(\hat{\mathbf{x}}^i)$ for $\forall i\in S$ with at least one strict. Let the WE price be $\mathbf{p}$; by Lemma 6.1, $\mathbf{p}\tilde{\mathbf{x}}^i\ge\mathbf{p}\hat{\mathbf{x}}^i$ with at least one strict. Summing within the coalition: $\mathbf{p}\sum_{i\in S}\mathbf{e}^i=\mathbf{p}\sum_{i\in S}\tilde{\mathbf{x}}^i>\mathbf{p}\sum_{i\in S}\hat{\mathbf{x}}^i$. Since each $u^i$ is strictly increasing, each agent is up against the budget constraint $\mathbf{p}\hat{\mathbf{x}}^i=\mathbf{p}\mathbf{e}^i$, so $\mathbf{p}\sum_{i\in S}\hat{\mathbf{x}}^i=\mathbf{p}\sum_{i\in S}\mathbf{e}^i$ — contradiction. So such a coalition doesn't exist, and all WEA are in the core. $\blacksquare$

于是由核的定义，WEA Pareto 有效。下面不借核直接证明。

Thus by the definition of the core, WEA is Pareto efficient. Now directly prove it without the core.

Important

Theorem 6.9（First Welfare Theorem）对交换经济 $\mathcal{E}=(u^i,\mathbf{e}^i)_{i\in\mathcal{I}}$，若每个 $u^i$ 在 $\mathbb{R}_+^n$ 上严格单调，则每个 WEA 都是 Pareto 有效配置。For the exchange economy $\mathcal{E}=(u^i,\mathbf{e}^i)_{i\in\mathcal{I}}$, if each $u^i$ is strictly increasing on $\mathbb{R}_+^n$, then every WEA is a Pareto efficient allocation.

Note

证明 / Proof (Theorem 6.9)

反证。设 $\hat{\mathbf{x}}$ 为 WE 价格 $\mathbf{p}^\star$ 下的 WEA 但非 Pareto 有效。因 $\hat{\mathbf{x}}$ 可行，$\exists\mathbf{y}\in F(\mathbf{e})$ 使 $u^i(\mathbf{y}^i)\ge u^i(\hat{\mathbf{x}}^i)$（$\forall i$）至少一处严格。由 Lemma 6.1，$\mathbf{p}^\star\mathbf{y}^i\ge\mathbf{p}^\star\hat{\mathbf{x}}^i$ 至少一处严格，故 $\sum_{i\in\mathcal{I}}\mathbf{p}^\star\mathbf{y}^i>\sum_{i\in\mathcal{I}}\mathbf{p}^\star\hat{\mathbf{x}}^i\Rightarrow\mathbf{p}^\star\sum\mathbf{y}^i>\mathbf{p}^\star\sum\hat{\mathbf{x}}^i$，与 $\mathbf{y}$、$\hat{\mathbf{x}}$ 皆可行（$\sum\mathbf{y}^i=\sum\mathbf{e}^i=\sum\hat{\mathbf{x}}^i$）矛盾。故 $\hat{\mathbf{x}}$ Pareto 有效。$\blacksquare$

By contradiction. Suppose $\hat{\mathbf{x}}$ is a WEA at WE price $\mathbf{p}^\star$ but not Pareto efficient. Since $\hat{\mathbf{x}}$ is feasible, $\exists\mathbf{y}\in F(\mathbf{e})$ s.t. $u^i(\mathbf{y}^i)\ge u^i(\hat{\mathbf{x}}^i)$ for $\forall i$ with at least one strict. By Lemma 6.1, $\mathbf{p}^\star\mathbf{y}^i\ge\mathbf{p}^\star\hat{\mathbf{x}}^i$ with at least one strict, so $\sum_{i\in\mathcal{I}}\mathbf{p}^\star\mathbf{y}^i>\sum_{i\in\mathcal{I}}\mathbf{p}^\star\hat{\mathbf{x}}^i\Rightarrow\mathbf{p}^\star\sum\mathbf{y}^i>\mathbf{p}^\star\sum\hat{\mathbf{x}}^i$, contradicting that $\mathbf{y}$ and $\hat{\mathbf{x}}$ are both feasible ($\sum\mathbf{y}^i=\sum\mathbf{e}^i=\sum\hat{\mathbf{x}}^i$). Therefore $\hat{\mathbf{x}}$ is Pareto efficient. $\blacksquare$

6.5.2 Second Welfare Theorem

Important

Theorem 6.10（No Trade theorem）设 Assumption 1 成立且 $\sum_{i\in\mathcal{I}}\mathbf{e}^i\gg\mathbf{0}$。若禀赋 $\mathbf{e}=(\mathbf{e}^1,\dots,\mathbf{e}^I)$ 本身 Pareto 有效，则 $\mathbf{e}$ 即唯一的 WEA。Suppose Assumption 1 holds and $\sum_{i\in\mathcal{I}}\mathbf{e}^i\gg\mathbf{0}$. If $\mathbf{e}=(\mathbf{e}^1,\dots,\mathbf{e}^I)$ is Pareto efficient, then $\mathbf{e}$ is the unique WEA.

Note

证明 / Proof (Theorem 6.10)

由 Thm 6.7，Assumption 1 与 $\sum\mathbf{e}^i\gg\mathbf{0}$ 共同保证 WE 与 WEA 存在，即存在 WE 价格 $\mathbf{p}^\star$ 下的 WEA $\hat{\mathbf{x}}\in F(\mathbf{e})$，故 $u^i(\hat{\mathbf{x}}^i)\ge u^i(\mathbf{e}^i)$（$\forall i$）。因 $\mathbf{e}$ 本身 Pareto 有效，必对 $\forall i$ 取等号 $u^i(\hat{\mathbf{x}}^i)=u^i(\mathbf{e}^i)$。又由 Assumption 1，$\mathbf{p}^\star$ 下消费者最大化问题的解唯一，故 $\hat{\mathbf{x}}^i=\mathbf{e}^i$（$\forall i$）。$\blacksquare$

By Thm 6.7, Assumption 1 and $\sum\mathbf{e}^i\gg\mathbf{0}$ together imply the existence of WE and WEA, i.e. there is a WEA $\hat{\mathbf{x}}\in F(\mathbf{e})$ at WE price $\mathbf{p}^\star$, so $u^i(\hat{\mathbf{x}}^i)\ge u^i(\mathbf{e}^i)$ for $\forall i$. Since $\mathbf{e}$ is itself Pareto efficient, it must be equality for $\forall i$: $u^i(\hat{\mathbf{x}}^i)=u^i(\mathbf{e}^i)$. By Assumption 1, the solution to the consumer's maximization at $\mathbf{p}^\star$ is unique, so $\hat{\mathbf{x}}^i=\mathbf{e}^i$ for $\forall i$. $\blacksquare$

Important

Theorem 6.11（Second Welfare Theorem）设 Assumption 1 成立且 $\sum_{i\in\mathcal{I}}\mathbf{e}^i\gg\mathbf{0}$。若 $\bar{\mathbf{x}}\in F(\mathbf{e})$ Pareto 有效，且把禀赋重新分配为 $\bar{\mathbf{x}}$，则新经济的唯一 WEA 是 $\bar{\mathbf{x}}$。
变体：若在 $\bar{\mathcal{E}}=(u^i,\bar{\mathbf{x}}^i)_{i\in\mathcal{I}}$ 中支撑 $\bar{\mathbf{x}}$ 的 WE 价格为 $\mathbf{p}^\star$，则把禀赋 $\mathbf{e}$ 重新分配为任意满足 $\mathbf{p}^\star\tilde{\mathbf{e}}=\mathbf{p}^\star\bar{\mathbf{x}}$ 的 $\tilde{\mathbf{e}}$ 也可，新经济的唯一 WEA 仍是 $\bar{\mathbf{x}}$。Suppose Assumption 1 holds and $\sum_{i\in\mathcal{I}}\mathbf{e}^i\gg\mathbf{0}$. If $\bar{\mathbf{x}}\in F(\mathbf{e})$ is Pareto efficient and endowments are redistributed to $\bar{\mathbf{x}}$, then the unique WEA of the new economy is $\bar{\mathbf{x}}$.
Variation: if the WE price that supports $\bar{\mathbf{x}}$ in $\bar{\mathcal{E}}=(u^i,\bar{\mathbf{x}}^i)_{i\in\mathcal{I}}$ is $\mathbf{p}^\star$, then redistributing endowment $\mathbf{e}$ to any $\tilde{\mathbf{e}}$ such that $\mathbf{p}^\star\tilde{\mathbf{e}}=\mathbf{p}^\star\bar{\mathbf{x}}$ will also do, and the unique WEA of the new economy is still $\bar{\mathbf{x}}$.

Note

证明 / Proof (Theorem 6.11)

第二福利定理的第一种陈述直接来自 Thm 6.10。看变体：因禀赋 $\mathbf{e}$ 被重分为 $\tilde{\mathbf{e}}$ 使 $\mathbf{p}^\star\tilde{\mathbf{e}}=\mathbf{p}^\star\bar{\mathbf{x}}$，配置 $\bar{\mathbf{x}}$ 可由个体间交易达成。因 $\bar{\mathbf{x}}$ Pareto 有效，故 $\bar{\mathbf{x}}$ 是 WEA。又 Assumption 1 保证每人效用最大化问题解唯一，故 $\bar{\mathbf{x}}$ 是唯一 WEA。$\blacksquare$

The first version of the statement is straight from Thm 6.10. For the variation: since endowment $\mathbf{e}$ is redistributed to $\tilde{\mathbf{e}}$ such that $\mathbf{p}^\star\tilde{\mathbf{e}}=\mathbf{p}^\star\bar{\mathbf{x}}$, the allocation $\bar{\mathbf{x}}$ is achievable by trading among agents. Since $\bar{\mathbf{x}}$ is Pareto efficient, $\bar{\mathbf{x}}$ is a WEA. Also Assumption 1 implies uniqueness of the solution to each agent's utility maximization, which makes $\bar{\mathbf{x}}$ the only WEA. $\blacksquare$