about this course; first look at rationality, equilibrium and efficiency; in-class illustration of a market (CH 1)

• Allocation (分配): an arrangement of who gets what
• Pareto-efficiency (帕累托有效): an allocation is PE if there is no other feasible allocation that makes someone better off and no one worse off

Resource Allocation

Basic

An economy has one machine and $400 cash. The machine can produce $50 worth of goods if given to Producer A, $100 worth of goods if given to Producer B. Both producers want to maximize wealth. Feasible allocations are $(m_A,m_B,c_A,c_B)$ where $m_A,m_B\in\{0,1\}$, $m_A+m_B=1,c_A,c_B\in\mathbb{R},c_A+c_B=400$.

M-N setup

Now we have $N$ producers, $M$ machines and $\text{\$}X$ cash (transferable), and producer $i$ can turn (at most) one machine into $\text{\$}x_i$.

• Theorem 1.1: An allocation is PE if and only if $M$ producers with the highest $x_i$’s receive the machines.
  • Proof (Only if): Suppose producer $k$ receives a machine, producer $i$ does not, and $x_k < x_i$, then the allocation obtained by giving $k$’s machine to $i$ and transferring $\frac{x_k + x_i}{2}$ from $i$ to $k$ makes someone better off and nobody worse off, so the original allocation is not PE.
  • Proof (if): Suppose for contradiction another allocation makes someone better off and no one worse off. If the two allocations give the machines to the same producers, the new allocation must involve taking money from one producer without rewarding her, a contradiction. Let $K$ be the set of producers who lost a machine and $I$ the set who earned a machine. The new allocation must reward those who lost a machine a minimum of $\sum_{k\in K}x_{k}$, and can only take from those who earned a machine a maximum of $\sum_{i\in I}x_{i}$. Since $\$\ x_{k}\geq x_{i}$ for all $k\in K$ and $\$\ i\in I$, either $\sum_{k\in K}x_{k}>\sum_{i\in I}x_{i}$ and this is impossible or $\sum_{k\in K}x_{k}=\sum_{i\in I}x_{i}$ and nobody in $I\cup K$ can be better off without someone becoming worse off.