考虑连续时间的social planner最优化问题, 就是在考虑:
$$\int_{0}^{\infty}e^{-pt}u(c(t))dt,p>0\\ s.t.\begin{cases}k(t)=f(k(t))-\delta k(t)-c(t)\\k(0)=k_{0}\\c(t)\geq0\end{cases}$$
得到Hamiltonian: $H(c,k,\lambda)\equiv\underbrace{u(c)}_{utility}+\underbrace{\lambda}_{ multiplier}\underbrace{(f(k)-\delta k-c)}_{Budget\ Constraint}$.
有optimal condition:$\begin{cases}H_{c}=0\\H_{k}= \rho\lambda(t)- \dot{\lambda}(t)\\H_{\lambda}= \dot{k}(t)\end{cases},\begin{cases}k(0)=k_{0}\\c(t)\geq0\\\lim_{T \rightarrow\infty}e^{-pT}u'(c(T))k(T)=0\end{cases}$, 分别对$\left(\begin{matrix}c\\k\\\lambda\end{matrix} \right)$偏导, 可得:$\begin{cases}H_{c}=u^{\prime}(c)-\lambda\\H_{k}=\lambda(f^{\prime}(k)-\delta)\\H_{\lambda}=f(k)-\delta k-c\end{cases}$.
$\to \begin{cases}u^{\prime}(c(t))=\lambda(t)\\ \dot{\lambda}(t)=( \rho-(f^{\prime}(k(t))-\delta))\lambda(t)\\ \dot{k}(t)=f(k(t))-\delta k(t)-c(t)\end{cases}$. 消去$ \rho$: $\begin{cases}\lambda(t)=u^{\prime}(c(t))\to \dot\lambda(t)= \frac{\partial\lambda(t)}{\partial t}=u^{\prime\prime}(c(t)) \cdot \dot{c}(t)\\ \dot\lambda(t)=[ \rho-(f^{\prime}(k(t))-\delta]\lambda(t)\end{cases}$.
$\to \frac{ \dot{c}(t)}{c(t)}= \frac{f^{\prime}(k(t))- \rho-\delta}{- \frac{u^{\prime\prime}(c(t))c(t)}{u^{\prime}(c(t))}}$. 令$\sigma(c)=- \frac{u^{\prime\prime}(c(t))c(t)}{u^{\prime}(c(t))}\to \frac{ \dot{c}(t)}{c(t)}= \frac{f^{\prime}(k(t))-p-\delta}{\sigma(c)}$.
Interpretation of $ \frac{ \dot{c}(t)}{c(t)}$: 消费的增长率等于marginal product of capital 减去depreciation rate和discount rate of capital (impatience), adjusted for intertemporal elasticity of substitution.
注意一下$\sigma(c)$是跨期代替弹性, 也就是大名鼎鼎的Arrow/Pratt coefficient of relative risk aversion.
假设isoelastic utility: $u(c)= \frac{c^{1-\sigma}-1}{1-\sigma},\sigma>0$, 其跨期替代弹性为$\sigma(c)=\sigma \in R$. 它是常数的话就能简化运算了.
$\to \frac{ \dot{c}(t)}{c(t)}= \frac{f^{\prime}(k(t))-\delta- \rho}{\sigma}$.
注意一下这个增长率和离散时间下的一样: $\begin{align} \frac{c_{t+1}}{c_{t}}=(\beta[f^{\prime}(k_{t+1})+1-\delta])^{ \frac{1}{\sigma}}\xrightarrow{\log} \Delta\log(c_{t+1})= \frac{\log[f^{\prime}(k_{t+1})-\delta+1]-\log \frac{1}{\beta}}{\sigma}\\ \approx \frac{f^{\prime}(k_{t+1}-\delta)-\log( \frac{1}{\beta}-1+1)}{\sigma}\\ \approx \frac{1}{\sigma}[f^{\prime}(k_{t+1})-\delta-( \frac{1}{\beta}-1)]\\ \approx \frac{f^{\prime}(k_{t+1})-\delta- \rho}{\sigma}, \frac{1}{\beta}-1=p \end{align}$.
因为是一样的, 所以连续时间和离散时间模型的economic interpretation都是一样的.
用以上信息能够得出differential equation system:
$\begin{cases} \frac{ \dot{c}(t)}{c(t)}= \frac{f^{\prime}(k(t))-\delta-p}{\sigma(c(t))}\\\\ \dot{k}(t)=f(k(t))-\delta k(t)-c(t)\end{cases}$, s.t.$\begin{cases}\lim_{T\to\infty}e^{-pT}u^{\prime}(c(T))k(T)=0\\\\k(0)=k_{0}\end{cases}$.
假设存在稳定点, 则在稳定点处满足$\begin{cases}c(t)=\overline{c}\\ \dot{c}(t)=0\end{cases} \rightarrow \frac{f(\overline{k})-\delta- \rho}{\sigma(\overline{c})}=0\\ \rightarrow f^{\prime}(\overline{k})-\delta- \rho=0 \rightarrow f^{\prime}(\overline{k})=\delta+ \rho $.
能够通过system判断在稳定点任何位置的点的路径. eg: 当$k\lt\overline{k}$时, $f^{\prime}(k)>f^{\prime}(\bar{k})=\delta+p$, 同时消费增长率大于0. 这意味着当人均资本比较少的时候, 多牺牲一些现期消费的时候在下一期增加的消费能够突破impatience的影响, 所以social planner决定当期少消费一些.
所以能画出相图:
因为我们现在得出的system是非线性的, 所以需要把它转化为线性system:
$\begin{cases} \dot{c}=g_{1}(c,k) \approx \frac{\partial g_{1}}{\partial c}(c-\overline{c})+ \frac{\partial g_{2}}{\partial k}(k-\bar{k})\\ \dot{k}=g_{2}(c,k) \approx \frac{\partial g_{2}}{c}(c-\bar{c})+ \frac{\partial g_{2}}{k}(k-\bar{k})\end{cases}$. $\to \begin{pmatrix} \dot{c}(t)\\\\ \dot{k}(t)\end{pmatrix}=\begin{pmatrix} \frac{\partial}{\partial c}g_{1}& \frac{\partial}{\partial k}g_{1}\\\\ \frac{\partial}{\partial c}g_{2}& \frac{\partial}{\partial k}g_{2}\end{pmatrix}\begin{pmatrix}c(t)-\overline{c}\\\\k(t)-\overline{k}\end{pmatrix}$.
for $\begin{cases}g_{1}= \frac{f'(k(t))-\delta- \rho}{\sigma(c(t))} \cdot c(t)\\\\g_{2}=f(k(t))-\delta k(t)-c(t)\end{cases}$.
for $\begin{cases} \frac{\partial g_{1}}{\partial c}= \frac{f^{\prime}(k)- \rho-\delta}{\sigma}=0,at\ k=\bar{k}\\ \frac{\partial g_{1}}{\partial k}= \frac{f^{\prime\prime}(k)}{\sigma}c<0\\ \frac{\partial g_{2}}{\partial c}=-1\\ \frac{\partial g_{2}}{\partial k}=f^{\prime}(k)-\delta= \rho,at\ k=\bar{k}\end{cases}$. 因为线性化考虑的是稳定点邻域内的, 所以矩阵的值可近似于取稳定点的值. 则有:
$A=\left(\begin{matrix}{0}&{{ \frac{f^{\prime\prime}(k)\overline{C}}{\sigma}}}\\{-1}&{ \rho}\\\end{matrix} \right)$. 记住这是为了方便计算, 所有这一切都发生在稳定点的邻域内.
有: $\begin{cases}\det(A)=\lambda_{1}\lambda_{2}= \frac{f^{\prime\prime}(\overline{k})\overline{c}}{\sigma}<0\\tr(A)=\lambda_{1}+\lambda_{2}= \rho>0\end{cases}\to \lambda_{1}<0<\lambda_{2}$. 说明相图符合saddle path.
用特征多项式解:
$p(\lambda)=\lambda^{2}-tr(A)\lambda+det(\lambda)\to \begin{cases}\lambda_{1}= \frac{ \rho-\sqrt{ \rho^{2}-4 \frac{f^{\prime\prime}\bar{c}}{\sigma}}}{2}\\\\\lambda_{z}= \frac{ \rho+\sqrt{ \rho^{2}-4 \frac{f^{\prime\prime}\bar{c}}{\sigma}}}{2}\end{cases}$.
用guess 解: 假设$ \dot{k}=\lambda(k-\bar{k})$
$ \dot{k}=-(c-\overline{c})+ \rho(k-\overline{k})\to \ddot{k}=- \dot{c}+ \rho \dot{k}\to \ddot{k}=-[ \frac{f^{\prime\prime}}{\sigma}\overline{c}(k-\overline{k})]+p \dot{k}=\lambda \dot{k}=\lambda^{2}(k-\bar{k})$.
$\to \lambda^{2}- \rho\lambda+ \frac{f^{\prime\prime}\overline{c}}{\sigma}=0$.
两种方法的解都一样. 选$\lambda<0$的那个解作为stable root. 带入$ \dot{k}$中, 可得:$\lambda(k-\overline{k})=-(c-\overline{c})+ \rho(k-\overline{k})\to ( \rho-\lambda)(k-\bar{k})=c-\bar{c}$.
可得出在$\left(\begin{matrix}\overline{c}\\\\\overline{k}\end{matrix} \right)$处stable arm的导数大于$ \dot{k}=0$曲线的导数.
Decentralized Problem
现在将模型中加入household, household决定生产, 消费和工作的多少, 而不是social planner决定. 当去中心化之后, 经济系统中就发生了交换.
Assumption
- 每个家庭有初始资本$k_{0}>0$, 折旧率$\delta$.
- 每个家庭有一单位劳动力禀赋: $l=1$
- 每个家庭向竞争市场中的公司提供$(k_{t},1)$的要素组合, 并产生资本租金和工资, 为$(R(t),w(t))$.
- 每个家庭的净资产有利息: 令资产价值为$a(t)$, 有: $ \dot{a}(t)=\underbrace{\underbrace{r(t)a(t)}_{return\ of\ investment}+\underbrace{w(t)}_{wage}}_{total\ income}-\underbrace{c(t)}_{consumption}$. (full budget constraint, 意味着net asset possession change over time)
- physical capital and other assets are perfect substitutes, so no arbitrage: $\underbrace{R(t)}_{capital\ price}=\underbrace{r(t)}_{net\ return}+\delta\to \underbrace{r(t)}_{capital\ return}=R(t)-\delta$.
Question for household and firm
则每个家庭都要解一个最优化问题: $\begin{aligned}U&=\int_{0}^{\infty}e^{- \rho t}u(c(t))dt\\s.t.\ &\begin{cases} \dot{a}(t)=r(t)a(t)+w(t)-c(t)\\\lim_{T\to\infty}q(T)a(T)\geq0,q(t)\equiv e^{-\int_{0}^{t}r(s)ds},\ (Ponzi\ condition)\end{cases}\end{aligned}$.
和social planner的区别: 1. 这里的constraint是budget constraint而不是resource constraint. 2. no Ponzi game constraint. 这个等式相当于是未来远期asset (a)的价格折现到第0期的时候非负, 其中$q(T)$是intertemporal price of consumption. 如果违反, 就相当于在未来的某期开始一直在借款活着,而且是贷滚贷.
得哈密尔顿函数: $\begin{cases}H_{c}=u'(c)-\lambda=0\\H_{a}=\lambda r= \rho\lambda(t)- \dot{\lambda}(t)\\H_{\lambda}=ra+w-c= \dot{a}(t)\end{cases}\to \begin{cases}u^{\prime}(c(t))=\lambda(t)\\ \dot{\lambda}(t)=( \rho-r(t))\lambda(t)\\ \dot{a}(t)=r(t)a(t)+w(t)-c(t)\end{cases}$
Household Side:
$\to \begin{cases} \dot{\lambda}(t)=(p-r(t))u^{\prime}(c(t))\\u^{\prime\prime}(c(t)) \cdot c^{\prime}(t)= \dot{\lambda}(t)\\\end{cases} \rightarrow \dot{c}(t)=[ \rho-r(t)] \frac{u^{\prime}(c(t))}{u^{\prime\prime}(c(t))}\to \frac{ \dot{c}(t)}{c(t)}=[ \rho-r(t)] \frac{u^{\prime}(c(t))}{u^{\prime\prime}(c(t))} \cdot \frac{1}{c(t)}= \frac{r- \rho}{\sigma(c(t))}$.
$= \frac{r- \rho}{\sigma}$
for $\sigma (c(t))$ 为intertemporal elasticity of substitution.
Firm Side:
profit equation: $\Pi=F(K,L)-RK-wL\to \begin{cases}F_{k}=R\\ F_{L}=w\end{cases}$.
竞争市场下的最大化利润决定了两件事:$\begin{cases}\pi= \frac{\Pi}{L}=0\\\\\pi^{\prime}=FOC=0\end{cases}\to \begin{cases}f^{\prime}(k)=R=r+\delta\\\ f(k)+f^{\prime}(k)k=w\end{cases}$.
Equilibrium
均衡指3件事:
- 家庭最大化utility when 价格给定
- 厂商最大化利润for given price
- 市场出清.$\to$ 劳动力供给=需求, 资本供给=需求. $\to\begin{cases}L=1, (???就一个家庭???)\\\\k=a,(a为net\ asset,就是capital\ saving)\end{cases}$.
则有:$ \frac{ \dot{c}(t)}{c(t)}= \frac{r(t)- \rho}{\sigma}= \frac{f^{\prime}(k)-\delta- \rho}{\sigma}$, for $r(t)=f^{\prime}(k)-\delta $. 这和social planner下的方程一样, 意味着: 去中心化的情况下个人会选择和social planner的命令一样的行为.
还有: $ \dot k(t)= \dot a(t)=r(t)a(t)+w(t)-c(t)=[f^{\prime}(k(t))-\delta]k(t)+[f(k(t))-f^{\prime}(k(t))k(t)]-c(t)$
$=f(k(t))-\delta k(t)-c(t)$. 居然和social planner下的resource constraint一样.
结论: 如果没有market failure或别的影响家庭决策行为的因素, 有没有social planner都是一样的.
Alternative Method to solve Household Problem
利用每一个household的在生命周期中的总消费等于总budget来得到最优效用的constraint:
$\int_{0}^{\infty}q(t)c(t)dt=a(0)+\int_{0}^{\infty}q(t)w(t)dt$
其中$q(t)$为价格, 在之前的no Ponzi Game中提及. 有: $q(t)\equiv exp(-\int_{0}^{t}r(s)ds)\to \dot{q}(t)=-r(t)q(t)$.
则可以不用哈密尔顿, 用拉格朗日来解决问题:
$L=\int_{0}^{\infty}e^{- \rho t}u(c(t))dt+\lambda(a(0)+\int_{0}^{\infty}q(t)[w(t)-c(t)]dt)$
我们要做的就是$max_{c(t)}L$.
FOC: $e^{-pt}u^{\prime}(c(t))=\lambda q(t)\to \lambda = ...$.
对t全微分: $-pe^{-pt}u^{\prime}(c(t))+e^{-pt}u^{\prime\prime}(c(t)) \dot{c}(t)=\lambda \dot{q}(t)$.
$\to \frac{ \dot{c}(t)}{c(t)}= \frac{r(t)- \rho}{\sigma}$.