Cobb-Douglas函数

Cobb-Douglas function

Y = x_{1}^{α} x_{2}^{β}

核心特征：

\frac{Expenditure on x_{1}}{Expenditure on x_{2}} = \frac{α}{β}

效用最大化

\begin{aligned} max & u = x_{1}^{α} x_{2}^{β} \\ s.t. & p_{1} x_{1} + p_{2} x_{2} = m \end{aligned}

The Lagrangian for this problem is

L = x_{1}^{α} x_{2}^{β} + λ (m - p_{1} x_{1} - p_{2} x_{2})

F.O.C.

\begin{aligned} \frac{\partial L}{\partial x_{1}} & = α x_{1}^{α - 1} x_{2}^{β} - λ p_{1} = 0 \\ \frac{\partial L}{\partial x_{2}} & = β x_{1}^{α} x_{2}^{β - 1} - λ p_{2} = 0 \end{aligned}

which implies

\frac{p_{1} x_{1}}{p_{2} x_{2}} = \frac{α}{β}

代入预算约束解得 Marshallian demand function

\begin{aligned} x_{1}^{*} & = \frac{\frac{α}{α + β} m}{p_{1}} \\ x_{2}^{*} & = \frac{\frac{β}{α + β} m}{p_{2}} \end{aligned}

也就是说 $\frac{α}{α + β}$ 比例的预算用于购买 $x_{1}$ ， $\frac{β}{α + β}$ 比例的预算用于购买 $x_{2}$

Cobb-Douglas 函数是一种位似函数，其特征是：

所有商品需求的收入弹性均为 1
所有商品的的平均消费倾向（average propensity to consume, APC）和边际消费倾向（marginal propensity to consume, MPC）为常数：

\begin{aligned} A P C_{1} & \equiv \frac{p_{1} x_{1}}{m} = \frac{α}{α + β} = \frac{\partial p_{1} x_{1}}{\partial m} \equiv M P C_{1} \\ A P C_{2} & \equiv \frac{p_{2} x_{2}}{m} = \frac{β}{α + β} = \frac{\partial p_{2} x_{2}}{\partial m} \equiv M P C_{2} \end{aligned}

这符合代表性消费者的特征。

Code

import sympy as sp
x1,x2,a,b,p1,p2,m,la=sp.symbols('x_{1},x_{2},\\alpha,\\beta,p_{1},p_{2},m,\lambda',real=True)
u=x1**a*x2**b
g=m-(p1*x1+p2*x2)
L=u+la*g
FOC=[sp.diff(L,i) for i in [x1,x2,la]]
df=sp.solve(FOC,[x1,x2,la],dict=True)
dx1=sp.Eq(x1,df[0][x1])
dx2=sp.Eq(x2,df[0][x2])
print(sp.latex(dx1),sp.latex(dx2),sep='\n')

支出最小化

\begin{aligned} min & p_{1} x_{1} + p_{2} x_{2} \\ s.t. & x_{1}^{α} x_{2}^{β} = \bar{u} \end{aligned}

The Lagrangian of this problem is

L = p_{1} x_{1} + p_{2} x_{2} + μ (\bar{u} - x_{1}^{α} x_{2}^{β})

F.O.C

\begin{aligned} \frac{\partial L}{\partial x_{1}} & = p_{1} - μ α x_{1}^{α - 1} x_{2}^{β} = 0 \\ \frac{\partial L}{\partial x_{2}} & = p_{2} - μ β x_{1}^{α} x_{2}^{β - 1} = 0 \end{aligned}

which implies

\frac{p_{1} x_{1}}{p_{2} x_{2}} = \frac{α}{β}

代入预算约束解得希克斯需求函数（解法同下文成本最小化一节）

\begin{aligned} x_{1}^{H} & = {\bar{u}}^{\frac{1}{α + β}} {(\frac{α / β}{p_{1} / p_{2}})}^{\frac{β}{α + β}} \\ x_{2}^{H} & = {\bar{u}}^{\frac{1}{α + β}} {(\frac{p_{1} / p_{2}}{α / β})}^{\frac{α}{α + β}} \end{aligned}

利润最大化

max_{x_{i}} p x_{1}^{α} x_{2}^{β} - (w_{1} x_{1} + w_{2} x_{2})

F.O.C.

\begin{aligned} p α x_{1}^{α - 1} x_{2}^{β} - w_{1} & = 0 \\ p β x_{1}^{α} x_{2}^{β - 1} - w_{2} & = 0 \end{aligned}

which implies

\frac{w_{1} x_{1}}{w_{2} x_{2}} = \frac{α}{β}

这个方程直接求解比较困难，可以先将 F.O.C.转化为对数形式

\begin{aligned} (α - 1) \ln x_{1} + β \ln x_{2} & = \ln w_{1} - \ln α - \ln p \\ α \ln x_{1} + (β - 1) \ln x_{2} & = \ln w_{2} - \ln β - \ln p \end{aligned}

这就得到了关于变量 $\ln x_{1}$ 和 $\ln x_{2}$ 的线性方程组，其扩展矩阵为

[\begin{matrix} (α - 1) & β & \ln w_{1} - \ln α - \ln p \\ α & β - 1 & \ln w_{2} - \ln β - \ln p \end{matrix}]

根据 Cramer's Law 可得

\begin{aligned} (1 - α - β) \ln x_{1} & = (β - 1) (\ln w_{1} - \ln α - \ln p) - β (\ln w_{2} - \ln β - \ln p) \\ (1 - α - β) \ln x_{2} & = (α - 1) (\ln w_{2} - \ln β - \ln p) - α (\ln w_{1} - \ln α - \ln p) \end{aligned}

初步化简

\begin{aligned} (1 - α - β) \ln x_{1} & = (β - 1) (\ln \frac{w_{1}}{α}) - β (\ln \frac{w_{2}}{β}) + \ln p \\ (1 - α - β) \ln x_{2} & = (α - 1) (\ln \frac{w_{2}}{β}) - α (\ln \frac{w_{1}}{α}) + \ln p \end{aligned}

继续化简

\begin{aligned} (1 - α - β) \ln x_{1} & = \ln [p \frac{{(\frac{w_{1}}{α})}^{β - 1}}{{(\frac{w_{2}}{β})}^{β}}] \\ (1 - α - β) \ln x_{2} & = \ln [p \frac{{(\frac{w_{2}}{β})}^{α - 1}}{{(\frac{w_{1}}{α})}^{α}}] \end{aligned}

最终可得要素需求函数

\begin{aligned} x_{1}^{*} & = {[p \frac{{(\frac{w_{1}}{α})}^{β - 1}}{{(\frac{w_{2}}{β})}^{β}}]}^{\frac{1}{1 - α - β}} \\ x_{2}^{*} & = {[p \frac{{(\frac{w_{2}}{β})}^{α - 1}}{{(\frac{w_{1}}{α})}^{α}}]}^{\frac{1}{1 - α - β}} \end{aligned}

注意：当 $α + β = 1$ 即规模报酬不变时，要素需求无限大，这意味着厂商的规模无法确定。

扩展

宏观经济学常用 Cobb-Douglas 函数 $Y = K^{α} L^{1 - α}$ 的紧凑形式：

y = \frac{Y}{L} = {(\frac{K}{L})}^{α} = k^{α}

利润最大化问题为

max_{k} p k^{α} - (r k + w)

F.O.C.仅有单个条件

p α k^{α - 1} = r

将其代回目标函数后需要结合竞争性条件（零利润）得到第二条件

(1 - α) p k^{α} = w

成本最小化

\begin{aligned} min_{x_{i}} & w_{1} x_{1} + w_{2} x_{2} \\ s.t. & x_{1}^{α} x_{2}^{β} = q \end{aligned}

The Lagrangian for this problem is

L = w_{1} x_{1} + w_{2} x_{2} + λ (q - x_{1}^{α} x_{2}^{β})

F.O.C

\begin{array}{r} \frac{\partial L}{\partial x_{1}} = w_{1} - λ α x_{1}^{α - 1} x_{2}^{β} = 0 \\ \frac{\partial L}{\partial x_{2}} = w_{2} - λ β x_{1}^{α} x_{2}^{β - 1} = 0 \end{array}

which implies

\frac{w_{1} x_{1}}{w_{2} x_{2}} = \frac{α}{β}

这个方程直接求解比较困难，可以先将约束条件和上述条件转化为对数形式

\begin{aligned} α \ln x_{1} + β \ln x_{2} & = \ln q \\ \ln x_{1} - \ln x_{2} & = \ln \frac{α}{β} - \ln \frac{w_{1}}{w_{2}} \end{aligned}

这就得到了关于变量 $\ln x_{1}$ 和 $\ln x_{2}$ 的线性方程组，其扩展矩阵为

[\begin{matrix} α & β & \ln q \\ 1 & - 1 & \ln \frac{α}{β} - \ln \frac{w_{1}}{w_{2}} \end{matrix}]

根据 Cramer's Law 可得

\begin{aligned} (- α - β) \ln x_{1} & = - \ln q - β \ln (\frac{α / β}{w_{1} / w_{2}}) \\ (- α - β) \ln x_{2} & = - \ln q + α \ln (\frac{α / β}{w_{1} / w_{2}}) \end{aligned}

最终可得条件要素需求函数

\begin{aligned} x_{1}^{c} & = q^{\frac{1}{α + β}} {(\frac{α / β}{w_{1} / w_{2}})}^{\frac{β}{α + β}} \\ x_{2}^{c} & = q^{\frac{1}{α + β}} {(\frac{w_{1} / w_{2}}{α / β})}^{\frac{α}{α + β}} \end{aligned}

边际成本

边际成本是最重要的厂商特征，标准的求解路径为：成本最小化→条件要素需求函数→成本函数→边际成本，然而这一路径过于繁琐。因此，了解如何从生产函数“直达”边际成本很有帮助。

成本最小化问题为

\begin{aligned} min_{x_{i}} & w_{1} x_{1} + w_{2} x_{2} \\ s.t. & x_{1}^{α} x_{2}^{β} = q \end{aligned}

根据生产平衡等式可得方程组

\begin{aligned} M C & = \frac{w_{1}}{M P_{1}} \\ M C & = \frac{w_{2}}{M P_{2}} \end{aligned}

已知成本函数变量包含 $q$ ，故通过边际产出引入可得

\begin{aligned} M P_{1} & = α x_{1}^{α - 1} x_{2}^{β} = \frac{α q}{x_{1}} \\ M P_{2} & = β x_{1}^{α} x_{2}^{β - 1} = \frac{β q}{x_{2}} \end{aligned}

代回生产平衡等式得到

\begin{aligned} x_{1} & = \frac{α q}{w_{1}} M C \\ x_{2} & = \frac{β q}{w_{2}} M C \end{aligned}

代入约束条件得解

\begin{aligned} {(\frac{α q}{w_{1}} M C)}^{α} {(\frac{β q}{w_{2}} M C)}^{β} & = q \\ {(\frac{α}{w_{1}})}^{α} {(\frac{β}{w_{2}})}^{β} M C^{α + β} & = q^{1 - α - β} \\ M C & = q^{\frac{1 - α - β}{α + β}} {(\frac{w_{1}}{α})}^{\frac{α}{α + β}} {(\frac{w_{2}}{β})}^{\frac{β}{α + β}} \end{aligned}