RBC

标注 ※ 的是必考题库

※ Example 1 标准 RBC 模型：分散化经济

经济中有一个代表性居民，一个代表性厂商。其中居民持有资本，且经济中仅有实物资本。

居民的效用函数为

$${\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})$$

其中 $N_t$ 为居民的劳动供给，$C_t$ 为居民的消费，$\partial u/\partial C>0,{\partial }^{2}u/\partial C^2<0,\partial u/\partial N<0,{\partial }^{2}u/\partial N^2<0$。

居民的融资约束为

$$C_t+I_t\le W_t{N}_t+R_t{K}_t+{\mathrm{\Pi }}_t+T_t$$

其中 $I_t$ 为居民的实际实物投资支出，$K_t$ 为居民持有的资本，$R_t$ 为资本边际回报，$W_t$ 为劳动的工资，${\mathrm{\Pi }}_t$ 为企业利润，$T_t$ 为政府的转移支付。

当资本投资支出为 $I_t$，资本存量为 $K_t$ 时，新生成的资本数目为 $Q^I(K_t,I_t)$，资本积累方程为

$$K_{t+1}=Q^I(K_t,I_t)+(1-\delta )K_t$$

企业的利润函数为

$${\mathrm{\Pi }}_t=Y_t-R_t{K}_t-W_t{N}_t$$

企业的生产技术为

$$F(K_t,L_t;A_t)$$

市场出清

$$\begin{gathered} K_t=K_t \\ {N}_t=N_t \\ {Y}_t=C_t+I_t \\ {T}_t^{\text{ax}}=G_t+T_t=0 \end{gathered}$$

技术变化 $\{A_t{\}}_{t=0}^{+\mathrm{\infty }}$ 是一个外生的随机过程。

A. 标准模型

令 $Q(K_t,I_t)=I_t;F(K_t,N_t;A_t)=A_t{K}_t^{\alpha }{N}_t^{1-\alpha };\ln A_{t+1}=\rho \ln A_t+{\epsilon }_t^A,{\epsilon }_t^A\sim N(0,{\sigma }_A^2)$。

（1） 求解此时居民效用最大化问题。

解：

$$\begin{array}{rl}V(K_t;R_t,W_t)\stackrel{\mathrm{△}}{=}\underset{\{C_{t+k},N_{t+k},K_{t+k+1}{\}}_{k=0}^{+\mathrm{\infty }}}{max}& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{K}_{t+1}=W_t{N}_t+R_t{K}_t+{\mathrm{\Pi }}_t-C_t+(1-\delta )K_t\end{array}$$$$\begin{array}{rl}V(K_t;R_t,W_t)\stackrel{\mathrm{△}}{=}\underset{C_t,N_t,K_{t+1}}{max}& u(C_t,N_t)+{\mathbb{E}}_t\beta V(K_{t+1};R_{t+1},W_{t+1})\\ & \text{s.t.}{K}_{t+1}=W_t{N}_t+R_t{K}_t+{\mathrm{\Pi }}_t-C_t+(1-\delta )K_t\end{array}$$$$V(K_t;R_t,W_t)=\underset{C_t,N_t,K_{t+1}}{max}u(C_t,N_t)+{\mathbb{E}}_t\beta V(K_{t+1};R_{t+1},W_{t+1})+{\lambda }_t\{K_{t+1}+C_t-W_t{N}_t-R_t{K}_t-{\mathrm{\Pi }}_t-(1-\delta )K_t\}$$

FOC

$$\begin{array}{rl}& \frac{\partial u}{\partial C}(C_t,N_t)\stackrel{\mathrm{△}}{=}{u}_C(C_t,N_t)={\lambda }_t\\ & \frac{\partial u}{\partial N}(C_t,N_t)\stackrel{\mathrm{△}}{=}{u}_N(C_t,N_t)=-{\lambda }_t{W}_t\end{array}$$

对 $K_{t+1}$

$$-{\lambda }_t=\beta {\mathbb{E}}_t\frac{\partial V}{\partial K}(K_{t+1};R_{t+1},W_{t+1})$$

包络定理

$$\frac{\partial V}{\partial K}(K_t;R_t,W_t)=-{\lambda }_t(R_t+1-\delta )$$

整理得

$$\begin{array}{}\text{(E-L)}& {\mathbb{E}}_t\beta \frac{u_C(C_{t+1},N_{t+1})}{u_C(C_t,N_t)}[R_{t+1}+1-\delta ]=1\end{array}$$$$\begin{array}{}\text{(A-S)}& -\frac{u_N(C_t,N_t)}{u_C(C_t,N_t)}=W_t\end{array}$$

（2） 企业利润最大化问题求解。

解：

$$\begin{array}{rl}{\mathrm{\Pi }}_t=\underset{K_t,N_t}{max}& Y_t-R_t{K}_t-W_t{N}_t\\ & \text{s.t.}{Y}_t=F(K_t,N_t;A_t)\end{array}$$

解得

$$\begin{array}{}\text{(K-D)}& \frac{\partial F}{\partial K}(K_t,N_t;A_t)=R_t\end{array}$$$$\begin{array}{}\text{(A-D)}& \frac{\partial F}{\partial N}(K_t,N_t;A_t)=W_t\end{array}$$

（3） 若 $u(C_t,N_t)=\frac{C_t^{1-\sigma }}{1-\sigma }-{\eta }_L\frac{N_t^{1+\varphi }}{1+\varphi }$，求解稳态的各项变量值。

解：

联立 (A-S), (A-D), (K-D),(E-L) 以及生产技术方程 (Y): $Y_t=A_t{K}_t^{\alpha }{N}_t^{1-\alpha }$ 和资本积累方程 (K-A): $K_{t+1}=I_t+(1-\delta )K_t$ 和市场出清方程 $Y_t=C_t+I_t$ 以及技术外生冲击，可以得到市场均衡的表述。

$$\begin{array}{}\text{(E-L)}& & {\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }[R_{t+1}+1-\delta ]=1\end{array}$$$$\begin{array}{}\text{(A-S)}& & W_t={\eta }_L{N}_t^{\varphi }{C}_t^{\sigma }\end{array}$$$$\begin{array}{}\text{(A-D)}& & W_t{N}_t=(1-\alpha )Y_t\end{array}$$$$\begin{array}{}\text{(K-D)}& & R_t{K}_t=\alpha Y_t\end{array}$$$$\begin{array}{}\text{(Y)}& & Y+t=A_t{K}_t^{\alpha }{N}_t^{1-\alpha }\end{array}$$$$\begin{array}{}\text{(K-A)}& & K_{t+1}=I_t+(1-\delta )K_t\end{array}$$$$\begin{array}{}\text{(Market Clear)}& & Y_t=C_t+I_t\end{array}$$$$\begin{array}{}\text{(TFP shock)}& & \ln A_{t+1}=\rho \ln A_t+{\epsilon }_t^A\end{array}$$

因为 $\mathbb{E}{\epsilon }_t^A=0$，考虑稳态

$$\ln \overline{A}=\rho \ln \overline{A}\Rightarrow \overline{A}=1$$

则在稳态下有

$$\begin{array}{rl}& K_{t+1}=K_t=K^{\ast }\\ & C_{t+1}=C_t=C^{\ast }\\ & I_{t+1}=I_t=I^{\ast }\\ & N_{t+1}=N_t=N^{\ast }\\ & W_{t+1}=W_t=W^{\ast }\\ & R_{t+1}=R_t=R^{\ast }\end{array}$$

由 Euler 方程可得

$$\begin{array}{}\text{(1)}& \beta (R^{\ast }+1-\delta )=1\Rightarrow R^{\ast }=\frac{1}{\beta }+\delta -1\end{array}$$

由 K-D 方程可得

$$R^{\ast }{K}^{\ast }=\alpha Y^{\ast }\Rightarrow Y^{\ast }=\frac{R^{\ast }}{\alpha }{K}^{\ast }$$

代入 (Y) 可得

$$N^{\ast }={\left(\frac{\alpha }{R}\right)}^{-\alpha /(1-\alpha )}{Y}^{\ast }$$

由 A-D 方程可得

$$W^{\ast }=(1-\alpha ){\left(\frac{\alpha }{R^{\ast }}\right)}^{\alpha /(1-\alpha )}$$

由 K-A 方程可得

$$C^{\ast }=(1-\frac{\alpha \delta }{R^{\ast }})Y^{\ast }$$

代入 (A-S) 可得

$$Y^{\ast }={\left[\frac{(1-\alpha ){\left(\frac{\alpha }{R^{\ast }}\right)}^{\alpha (1+\varphi )/(1-\alpha )}}{{\eta }_L{(1-\frac{\alpha \delta }{R^{\ast }})}^{\sigma }}\right]}^{1/(\varphi +\sigma )}$$

最终可得

$$\begin{array}{rl}{Y}^{\ast }& =\frac{(1-\alpha ){\left(\frac{\alpha }{R^{\ast }}\right)}^{\alpha /(1-\alpha )}}{{\left[{\eta }_L{\left(\frac{\alpha }{R^{\ast }}\right)}^{\alpha \varphi /(1-\alpha )}{(1-\frac{\alpha \delta }{R^{\ast }})}^{\sigma }\right]}^{1/(\varphi +\sigma )}}\\ R^{\ast }& =\frac{1}{\beta }+\delta -1\\ w^{\ast }& =(1-\alpha ){\left(\frac{\alpha }{R^{\ast }}\right)}^{\alpha /(1-\alpha )}\\ N^{\ast }& ={\left(\frac{\alpha }{R}\right)}^{\alpha /(1-\alpha )}{Y}^{\ast }\\ K^{\ast }& =\frac{\alpha Y^{\ast }}{R^{\ast }}\\ I^{\ast }& =\frac{\alpha \delta }{R^{\ast }}{Y}^{\ast }\\ C^{\ast }& =(1-\frac{\alpha \delta }{R^{\ast }})Y^{\ast }\end{array}$$

（4） 获取稳态下的对数线性化。

解：

对 $X_{t+1}$ 而言，${\hat{x}}_{t+1}=(X_{t+1}-X^{\ast })/X^{\ast }$，给出下列方程体系：

1. Euler 方程$$-\sigma [{\mathbb{E}}_t{\hat{x}}_{t+1}-{\hat{x}}_t]+\beta R^{\ast }{\mathbb{E}}_t{\hat{r}}_{t+1}=0$$$${\hat{x}}_t={\mathbb{E}}_t{\hat{x}}_{t+1}-\frac{1}{\sigma }\beta R^{\ast }{\mathbb{E}}_t{\hat{r}}_{t+1}$$
2. 利率水平$${\hat{r}}_t+{\hat{k}}_t={\hat{y}}_t$$
3. 生产函数$${\hat{y}}_t={\hat{a}}_t+\alpha {\hat{k}}_t+(1-\alpha ){\hat{n}}_t$$
4. 资本积累方程$${\hat{k}}_{t+1}=(1-\delta ){\hat{k}}_t+\delta {\hat{i}}_t$$
5. 市场出清条件$$\alpha \frac{\delta }{R^{\ast }}{\hat{i}}_t+(1-\alpha \frac{\delta }{R^{\ast }}){\hat{c}}_t={\hat{y}}_t$$
6. 劳动市场供给$${\hat{w}}_t=\varphi {\hat{n}}_t+\sigma {\hat{c}}_t$$
7. 劳动市场需求$${\hat{w}}_t+{\hat{n}}_t={\hat{y}}_t$$
8. 技术冲击$${\hat{a}}_{t+1}=\rho {\hat{a}}_t+{\epsilon }_t^A$$

（5） 定义

$$\begin{array}{rl}{\epsilon }_L& \triangleq \frac{1}{1/\varphi +1/\alpha }\\ {\epsilon }_L^D& \triangleq \frac{1}{\alpha }{\epsilon }_L\\ {\epsilon }_L^S& \triangleq \frac{1}{\varphi }{\epsilon }_L\\ {\rho }_1& \triangleq \frac{R^{\ast }}{\varphi }{\epsilon }_L^S\sigma +\frac{R^{\ast }}{\alpha }-\delta \\ {\rho }_2& \triangleq R^{\ast }+1-\delta +R^{\ast }\frac{1-\alpha }{\alpha }{\epsilon }_L^D\\ {\rho }_3& \triangleq 1+\frac{\alpha \beta R^{\ast }{\epsilon }_L^S}{\varphi }\\ \rho & \triangleq \frac{1-\alpha }{\sigma }\beta R^{\ast }{\epsilon }_L^D\end{array}$$

写出 $\left[\begin{array}{c}{\mathbb{E}}_t{\hat{c}}_{t+1}\\ {\hat{k}}_{t+1}\end{array}\right]$ 的方程。

解：

资本积累

• 由上问 5 式可得

  $$\delta {\hat{i}}_t=\frac{R^{\ast }}{\alpha }{\hat{y}}_t-(\frac{R^{\ast }}{\alpha }-\delta ){\hat{c}}_t$$

• 由上问 4 式可得

  $$\begin{array}{}\text{(5.1 方程)}& {\hat{k}}_{t+1}=(1-\delta ){\hat{k}}_t+\frac{R^{\ast }}{\alpha }{\hat{y}}_t-(\frac{R^{\ast }}{\alpha }-\delta ){\hat{c}}_t\end{array}$$

劳动供需

• 劳动需求$${\hat{w}}_t+{\hat{n}}_t={\hat{y}}_t={\hat{a}}_t+\alpha {\hat{k}}_t+(1-\alpha ){\hat{n}}_t$$$$\alpha {\hat{n}}_t={\hat{a}}_t+\alpha {\hat{k}}_t-{\hat{w}}_t$$$${\hat{n}}_t=\frac{1}{\alpha }{\hat{a}}_t+{\hat{k}}_t-\frac{1}{\alpha }{\hat{w}}_t$$
• 劳动供给$${\hat{n}}_t=\frac{1}{\varphi }{\hat{w}}_t-\frac{\sigma }{\varphi }{\hat{c}}_t$$

结合之后可得

$$\frac{1}{\varphi }{\hat{w}}_t-\frac{1}{\varphi }\sigma {\hat{c}}_t=\frac{1}{\alpha }{\hat{a}}_t+{\hat{k}}_t-\frac{1}{\alpha }{\hat{w}}_t$$$$\begin{array}{rl}{\hat{w}}_t& =\frac{1/\alpha }{1/\alpha +1/\alpha }{\hat{a}}_t+\frac{1}{\frac{1}{\varphi }+\frac{1}{\alpha }}{\hat{k}}_t+\frac{\frac{1}{\varphi }}{\frac{1}{\varphi }+\frac{1}{\alpha }}\sigma {\hat{c}}_t\\ & ={\epsilon }_L^S{\hat{a}}_t+{\epsilon }_L^D\sigma {\hat{c}}_t+{\epsilon }_L{\hat{k}}_t\end{array}$$$${\hat{n}}_t=\frac{1}{\varphi }{\epsilon }_L^S{\hat{a}}_t+{\epsilon }_L^S{\hat{k}}_t-\frac{1}{\varphi }{\epsilon }_L^S\sigma {\hat{c}}_t$$$${\hat{y}}_t={\epsilon }_L^S{\hat{a}}_t+[(1-\alpha ){\epsilon }_L^S+\alpha ]{\hat{k}}_t-\frac{\alpha }{\varphi }{\epsilon }_L^S\sigma {\hat{c}}_t$$$${\hat{r}}_t={\epsilon }_L^S{\hat{a}}_t-(1-\alpha ){\epsilon }_L^D{\hat{k}}_t-\frac{\alpha }{\varphi }{\epsilon }_L^S\sigma {\hat{c}}_t$$

从而有

$${\hat{k}}_{t+1}=\frac{R^{\ast }}{\alpha }{\epsilon }_L^S{\hat{a}}_t-\underset{{\rho }_1}{\underbrace{(\frac{R^{\ast }}{\varphi }{\epsilon }_L^S\sigma +\frac{R^{\ast }}{\alpha }-\delta )}}{\hat{c}}_t+\underset{{\rho }_2}{\underbrace{[R^{\ast }+R^{\ast }\frac{1-\alpha }{\alpha }{\epsilon }_L^S+1-\delta ]}}{\hat{k}}_t$$$${\hat{c}}_t={\mathbb{E}}_t{\hat{c}}_{t+1}-\frac{1}{\sigma }\beta R^{\ast }[{\mathbb{E}}_t{\epsilon }_L^S{\hat{a}}_{t+1}-(1-\alpha ){\epsilon }_L^D{\mathbb{E}}_t{\hat{k}}_{t+1}-\frac{\alpha }{\varphi }{\epsilon }_L^S\sigma {\mathbb{E}}_t{\hat{c}}_{t+1}]$$$${\hat{c}}_t=\underset{{\rho }_3}{\underbrace{(1+\frac{\alpha \beta R^{\ast }{\epsilon }_L^S}{\varphi })}}{\mathbb{E}}_t{\hat{c}}_{t+1}+\underset{{\rho }_4}{\underbrace{\left(\frac{1-\alpha }{\sigma }\beta R^{\ast }{\epsilon }_L^D\right)}}{\mathbb{E}}_t{\hat{k}}_{t+1}-\frac{1}{\sigma }\beta R^{\ast }{\epsilon }_L^S{\mathbb{E}}_t{\hat{a}}_{t+1}$$

由 ${\hat{a}}_{t+1}=\rho {\hat{a}}_t+{\epsilon }_t^A$ 可得

$${\mathbb{E}}_t{\hat{c}}_{t+1}=\frac{1}{{\rho }_3}{\hat{c}}_t-\frac{{\rho }_4}{{\rho }_3}{\mathbb{E}}_t{\hat{k}}_{t+1}+\frac{\rho }{\sigma {\rho }_3}\beta R^{\ast }{\epsilon }_L^S{\hat{a}}_t$$$${\hat{k}}_{t+1}=-\rho {\hat{c}}_t+\rho {\hat{k}}_t+\frac{R^{\ast }}{\alpha }{\epsilon }_L^S{\hat{a}}_t$$

令 $a_{1,1}=\frac{1}{{\rho }_3}$，$a_{1,2}=-\frac{{\rho }_4}{{\rho }_3}$，$a_{2,1}=-{\rho }_1$，$a_{2,2}={\rho }_2$，$b_1=\frac{\rho }{\sigma {\rho }_3}\beta R^{\ast }$，$b_2=\frac{R^{\ast }}{\alpha }{\epsilon }_L^S$，可得

$$\left[\begin{array}{c}{\mathbb{E}}_t\\ {\hat{k}}_{t+1}\end{array}\right]=\left[\begin{array}{cc}{a}_{1,1}& a_{1,2}\\ a_{2,1}& a_{2,2}\end{array}\right]\left[\begin{array}{c}{\hat{c}}_t\\ {\hat{k}}_t\end{array}\right]+\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]{\hat{a}}_t$$

B. 去增速

令 $Q(K_t,I_t)=I_t$，$F(K_t,N_t;A_t)=A_t{K}_t^{\alpha }{N}_t^{1-\alpha }$，$\ln A_{t+1}=\ln {\overline{A}}_{t+1}+\ln {\hat{A}}_{t+1}$，其中 $\ln {\overline{A}}_{t+1}=g_A+\ln {\overline{A}}_t$，$\ln {\hat{A}}_{t+1}=\rho \ln {\hat{A}}_t+{\epsilon }_t^A$，$u(C_t,N_t)=\frac{C_t^{1-\sigma }}{1-\sigma }-\frac{N_t^{1+\varphi }}{1+\varphi }$。

（1） 获取居民的一阶条件

（2） 求企业的一阶条件

（3） 求解市场的均衡和稳态方程

解：

经济中的市场结构如下

$$\begin{gathered} \begin{array}{}\text{(E-L)}& {\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_{{}_t}}\right)}^{-\sigma }[R_{t+1}+1-\delta ]=1 \\ \end{array} \end{gathered}$$$$\begin{gathered} \begin{array}{}\text{(A-S)}& W_t={\eta }_L{N}_t^{\varphi }{C}_t^{\sigma } \\ \end{array} \end{gathered}$$$$\begin{gathered} \begin{array}{}\text{(A-D)}& W_t{N}_t=(1-\alpha )Y_t \\ \end{array} \end{gathered}$$$$\begin{gathered} \begin{array}{}\text{(K-D)}& R_t{K}_t=\alpha Y_t \\ \end{array} \end{gathered}$$$$\begin{gathered} \begin{array}{}\text{(Y)}& Y_t=A_t{K}_t^{\alpha }{N}_t^{1-\alpha } \\ \end{array} \end{gathered}$$$$\begin{gathered} \begin{array}{}\text{(K-A)}& K_{t+1}=I_t+(1-\delta )K_t \\ \end{array} \end{gathered}$$$$\begin{gathered} \begin{array}{}\text{(Market Clear)}& Y_t=C_t+I_t \\ \end{array} \end{gathered}$$

在平衡增长路径下，$A_t={\overline{A}}_t$，增速为$g_A$，因此平衡路径上 $g_C,g_K,g_Y,g_I,g_N,g_W,g_R$ 均为常增速，不难得到 $g_C=g_I=g_K=g_Y$。

从而可得

$$g_W=\varphi g_N+\sigma g_C=\varphi g_N+\sigma g_Y$$$$g_W+g_N=g_Y\Rightarrow (1-\sigma )g_Y=(\varphi +1)g_N\Rightarrow g_N=\frac{1-\sigma }{1+\varphi }{g}_Y$$$$g_Y=g_A+\alpha g_K+(1-\alpha )g_N\Rightarrow g_Y=\frac{1}{1-\alpha }{g}_A+g_N=\frac{1}{1-\alpha }{g}_A+\frac{1-\sigma }{1+\varphi }{g}_Y$$$$\frac{\sigma +\varphi }{1+\varphi }{g}_Y=\frac{1}{1-\alpha }{g}_A$$$$g_Y=\frac{1}{1-\alpha }\frac{1+\varphi }{\sigma +\varphi }{g}_A=g_C=g_I=g_K$$$$g_N=\frac{1}{1-\alpha }\frac{1-\sigma }{\sigma +\varphi }{g}_A$$$$g_W=\frac{1}{1-\alpha }{g}_A$$$$y_t\triangleq \frac{Y_t}{A_t^{\frac{1}{1-\alpha }\frac{1+\varphi }{\sigma +\varphi }}}$$$$c_t\triangleq \frac{C_t}{A_t^{\frac{1}{1-\alpha }\frac{1+\varphi }{\sigma +\varphi }}}$$$$i_t\triangleq \frac{I_t}{A_t^{\frac{1}{1-\alpha }\frac{1+\varphi }{\sigma +\varphi }}}$$$$k_t\triangleq \frac{K_t}{A_t^{\frac{1}{1-\alpha }\frac{1+\varphi }{\sigma +\varphi }}}$$$$w_t\triangleq \frac{W_t}{A_t^{\frac{1}{1-\alpha }}}$$$$n_t\triangleq \frac{N_t}{A_t^{\frac{1}{1-\alpha }\frac{1-\sigma }{\sigma +\varphi }}}$$$$g_{Y,t}={\left(\frac{A_{t+1}}{A_t}\right)}^{\frac{1}{1-\alpha }\frac{1+\varphi }{\sigma +\varphi }}$$

可以将上述方程写为

$$\begin{array}{}\text{(1)}& \beta {\mathbb{E}}_t(g_{Y,t}{)}^{-\sigma }{\left(\frac{c_{t+1}}{c_t}\right)}^{-\sigma }(R_{t+1}+1-\delta )=1\end{array}$$$$\begin{array}{}\text{(2)}& \frac{1}{{\eta }_L}{w}_t=n_t^{\varphi }{c}_t^{\sigma }\end{array}$$$$\begin{array}{}\text{(3)}& y_t=k_t^{\alpha }{n}_t^{1-\alpha }\end{array}$$$$\begin{array}{}\text{(4)}& R_t{k}_t=\alpha y_t\end{array}$$$$\begin{array}{}\text{(5)}& w_t{n}_t=(1-\alpha )y_t\end{array}$$$$\begin{array}{}\text{(6)}& c_t+i_t=y_t\end{array}$$$$\begin{array}{}\text{(7)}& g_{Y,t}{k}_{t+1}=(1-\delta )k_t+i_t\end{array}$$

同时有

$$\ln A_{t+1}-\ln A_t=\rho [\ln A_t-\ln A_{t-1}]+(1-\rho )g_A+({\epsilon }_t^A-{\epsilon }_{t-1}^A)$$

可得

$$\begin{array}{}\text{(8)}& \ln g_{Y,t}=\rho \ln g_{Y,t-1}+(1-\rho )\underset{g_y}{\underbrace{g_A\frac{1+\varphi }{\sigma +\varphi }\frac{1}{1-\alpha }}}+\underset{{\epsilon }_t^y}{\underbrace{\frac{1+\varphi }{\sigma +\varphi }\frac{1}{1-\alpha }({\epsilon }_{t+1}^A-{\epsilon }_t^A)}}\end{array}$$$$\ln g_{Y,t}=\rho \ln g_{Y,t-1}+(1-\rho )g_y+{\epsilon }_t^y$$

稳态下

$$(R^{\ast }+1-\delta )=\frac{1}{\beta }{g}_Y^{\sigma }\Rightarrow R^{\ast }=\frac{1}{\beta }{g}_Y^{\sigma }-1+\delta$$$$R^{\ast }k=n^{1-\alpha }{k}^{\alpha }\alpha \Rightarrow \{\begin{array}{l}{k}^{\ast }=\frac{\alpha }{R^{\ast }}{y}^{\ast }\\ n^{\ast }=(\frac{R^{\ast }}{\alpha }{)}^{\frac{1}{1-\alpha }}{k}^{\ast }=(\frac{R^{\ast }}{\alpha }{)}^{\frac{\alpha }{1-\alpha }}{y}^{\ast }\end{array}$$$$w^{\ast }{n}^{\ast }=(1-\alpha )y^{\ast }\Rightarrow w^{\ast }=(1-\alpha )(\frac{\alpha }{R^{\ast }}{)}^{\frac{\alpha }{1-\alpha }}$$$$i^{\ast }+(1-\delta )k^{\ast }=g_Y{k}^{\ast }\Rightarrow i^{\ast }=(g_Y-1+\delta )k^{\ast }=\frac{\alpha (g_Y-1+\delta )}{R^{\ast }}{y}^{\ast }$$

有

$$c^{\ast }=[1-\frac{\alpha (g_Y-1+\delta )}{R^{\ast }}]y^{\ast }$$

代入 (2) 可得

$$\frac{1}{{\eta }_L}{w}^{\ast }=[1-\frac{\alpha (g_Y-1+\delta )}{R^{\ast }}{]}^{\sigma }{\left(\frac{R^{\ast }}{\alpha }\right)}^{\frac{\alpha \varphi }{1-\alpha }}{\left(y^{\ast }\right)}^{\sigma +\varphi }$$

最终可得

$$\begin{array}{rl}{y}^{\ast }& ={\left[\frac{w^{\ast }}{{\eta }_L(1-\frac{\alpha (g_Y-1+\delta )}{R^{\ast }}{)}^{\sigma }{\left(\frac{R^{\ast }}{\alpha }\right)}^{\frac{\alpha \varphi }{1-\alpha }}}\right]}^{1/(\varphi +\sigma )}\\ n^{\ast }& ={\left(\frac{k^{\ast }}{\alpha }\right)}^{\alpha /(1-\alpha )}{y}^{\ast }\\ k^{\ast }& =\frac{\alpha y^{\ast }}{R^{\ast }}\\ i^{\ast }& =\frac{\alpha (g_Y-1+\delta )}{R^{\ast }}{y}^{\ast }\\ c^{\ast }& =(1-\frac{\alpha (g_Y-1+\delta )}{R^{\ast }})y^{\ast }\end{array}$$

（4） 对数线性化

解：

对 $X_{t+1}$ 而言，${\hat{x}}_{t+1}=(x_{t+1}-x^{\ast })/x^{\ast }$，给出下列方程体系：

1. Euler 方程$$-\sigma [{\mathbb{E}}_t{\hat{c}}_{t+1}-{\hat{c}}_t]-\sigma {\mathbb{E}}_t{\hat{g}}_{Y,t+1}+\beta R^{\ast }{\mathbb{E}}_t{\hat{r}}_{t+1}=0$$$${\mathbb{E}}_t{\hat{c}}_{t+1}={\hat{c}}_t-{\mathbb{E}}_t{\hat{g}}_{Y,t+1}+\frac{1}{\sigma }\beta R^{\ast }{\mathbb{E}}_t{\hat{r}}_{t+1}$$
2. 利率水平$${\hat{r}}_t+{\hat{k}}_t={\hat{y}}_t$$
3. 生产函数$${\hat{y}}_t={\hat{a}}_t+\alpha {\hat{k}}_t+(1-\alpha ){\hat{n}}_t$$
4. 资本积累方程$${\hat{k}}_{t+1}+{\hat{g}}_{Y,t+1}=\frac{(1-\delta )}{g_Y}{\hat{k}}_t+(1-\frac{1-\delta }{g_Y}){\hat{i}}_t$$
5. 市场出清条件$$\alpha \frac{\delta }{R^{\ast }}{\hat{i}}_t+(1-\alpha \frac{\delta }{R^{\ast }}){\hat{c}}_t={\hat{y}}_t$$
6. 劳动市场供给$${\hat{w}}_t=\varphi {\hat{n}}_t+\sigma {\hat{c}}_t$$
7. 劳动市场需求$${\hat{w}}_t+{\hat{n}}_t={\hat{y}}_t$$
8. 技术冲击$${\hat{g}}_{Y,t+1}=\rho {\hat{g}}_{Y,t}+{\epsilon }_t^y$$

C. 投资品质量变化

$Q(I_t,K_t)=Q_t{I}_t$，$F(A_t,K_t,N_t)$，其中 $Q_t={\overline{Q}}_t{\hat{Q}}_t$，$\ln {\hat{Q}}_{t+1}={\rho }_Q\ln {\hat{Q}}_t+{\epsilon }_t^Q$，$\ln {\overline{Q}}_t=g_Q\cdot t$,$A_t={\overline{A}}_t{\hat{A}}_t$，$\ln {\overline{A}}_t=g_A\cdot t$，$\ln {\hat{A}}_{t+1}={\rho }_A\ln {\hat{A}}_t+{\epsilon }_t^A$，$u(C_t,N_t)=\frac{C_t^{1-\sigma }}{1-\sigma }-{\eta }_L\frac{N_t^{1+\varphi }}{1+\varphi }$。

（1） 求解居民问题的一阶条件。

解：

$$\begin{array}{rl}V(K_t;R_t,W_t,Q_t)\triangleq \underset{\{N_{t+k},C_{t+k}{\}}_{k=0}^{+\mathrm{\infty }}}{max}& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_t,N_t)\\ & \text{s.t.}{K}_{t+k+1}=Q_{t+k}[R_{t+k}{K}_{t+k}+W_{t+k}{N}_{t+k}-C_{t+k}]+(1-\delta )K_{t+k}\end{array}$$

Bellman 方程

$$\begin{array}{rl}V(K_t;R_t,W_t,Q_t)=\underset{N_t,C_t}{max}& u(C_t,N_t)+\beta {\mathbb{E}}_{t}V(K_{t+1};R_{t+1},W_{t+1},Q_{t+1})\\ & \text{s.t.}{K}_{t+1}=Q_t[R_t{K}_t+W_t{N}_t-C_t]+(1-\delta )K_t\end{array}$$

无约束最优化

$$\begin{array}{rl}V(K_t;R_t,W_t,Q_t)=\underset{N_t,C_t}{max}& u(C_t,N_t)+\beta {\mathbb{E}}_{t}V(K_{t+1};R_{t+1},W_{t+1},Q_{t+1})\\ & -{\lambda }_t[K_{t+1}-Q_t(R_t{K}_t+W_t{N}_t-C_t)-(1-\delta )K_t]\end{array}$$

FOCs

$$\frac{\partial u}{\partial C}(C_t,N_t)={\lambda }_t{Q}_t$$$$-\frac{\partial u}{\partial N}(C_t,N_t)={\lambda }_t{Q}_t{W}_t$$$${\lambda }_t=\beta {\mathbb{E}}_t\frac{\partial V}{\partial K}(K_{t+1};Q_{t+1};R_{t+1};W_{t+1})$$

包络定理

$$\frac{\partial V}{\partial K}(K_t;W_t,Q_t,R_t)={\lambda }_t{Q}_t[R_t+\frac{1-\delta }{Q_t}]$$

从而可得

$$\frac{\partial u}{\partial C}(C_t,N_t)\frac{1}{Q_t}=\beta {\mathbb{E}}_t\frac{\partial u}{\partial C}(C_{t+1},N_{t+1})[R_{t+1}+\frac{1-\delta }{Q_{t+1}}]$$$$\begin{array}{}\text{(E-L)}& {\mathbb{E}}_t\underset{M_{t,t+1}}{\underbrace{\beta \frac{\partial u/\partial C(C_{t+1},N_{t+1})}{\partial u/\partial C(C_t,N_t)}}}\underset{\triangleq R_{t+1}^k}{\underbrace{[R_{t+1}+\frac{1-\delta }{Q_{t+1}}]Q_t}}=1\end{array}$$

劳动供给方程

$$-\frac{\partial u/\partial N(C_t,N_t)}{\partial u/\partial C(C_t,N_t)}=W_t$$

（2） 企业行为的利润最大化一阶条件。

（3） 市场均衡和稳态解。

解：

市场均衡条件如下

$$\begin{array}{}\text{(E-L)}& & {\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_{{}_t}}\right)}^{-\sigma }[R_{t+1}{Q}_t+(1-\delta )\frac{Q_t}{Q_{t+1}}]=1\end{array}$$$$\begin{array}{}\text{(A-S)}& & W_t={\eta }_L{N}_t^{\varphi }{C}_t^{\sigma }\end{array}$$$$\begin{array}{}\text{(A-D)}& & W_t{N}_t=(1-\alpha )Y_t\end{array}$$$$\begin{array}{}\text{(K-D)}& & R_t{K}_t=\alpha Y_t\end{array}$$$$\begin{array}{}\text{(Y)}& & Y_t=A_t{K}_t^{\alpha }{N}_t^{1-\alpha }\end{array}$$$$\begin{array}{}\text{(K-A)}& & K_{t+1}=I_t{Q}_t+(1-\delta )K_t\end{array}$$$$\begin{array}{}\text{(Market Clear)}& & Y_t=C_t+I_t\end{array}$$

在平衡增长下，$g_{\overline{Q}},g_{\overline{A}}$ 已知。

由 (M-C) 可得

$$g_C=g_Y=g_I$$

由 (K-A) 可得

$$g_K=g_I+g_Q$$

由 (Y) 可得

$$g_Y=g_A+\alpha g_K+(1-\alpha )g_N$$

由 (A-D)，(K-D) 可得

$$g_W+g_N=g_Y$$$$g_R+g_K=g_Y$$

由 (A-S) 可得

$$g_W=\varphi g_N+\sigma g_C$$

由

$$\begin{gathered} g_K=g_Y+g_{\overline{Q}} \\ {g}_I=g_C=g_Y \\ {g}_K+g_R=g_Y \end{gathered}$$

可得

$$\begin{gathered} g_R+g_{\overline{Q}}=0 \\ {g}_W=g_Y-g_N \end{gathered}$$

代入 $g_W=\varphi g_N+\sigma g_C$ 可得

$$g_Y(1-\sigma )=(\varphi +1)g_N$$$$\{\begin{array}{l}{g}_N=\frac{1-\sigma }{1+\varphi }{g}_Y\\ g_W=\frac{\varphi +\sigma }{1+\varphi }{g}_Y\end{array}$$

代入 $g_Y=g_A+\alpha g_K+(1-\alpha )g_N$ 可得

$$g_Y=\frac{1}{(1-\alpha )\frac{\varphi +\sigma }{1+\varphi }}(g_{\overline{A}}+\alpha g_{\overline{Q}})$$$$g_Y=\frac{1+\varphi }{\varphi +\sigma }\frac{1}{1-\alpha }(g_{\overline{A}}+\alpha g_{\overline{Q}})$$

可得

$$g_Y=g_C=g_I=\frac{1+\varphi }{\varphi +\sigma }\frac{1}{1-\alpha }(g_{\overline{A}}+\alpha g_{\overline{Q}})$$$$g_N=\frac{1-\sigma }{\varphi +\sigma }\frac{1}{1-\alpha }(g_{\overline{A}}+\alpha g_{\overline{Q}})$$$$g_W=\frac{1}{1-\alpha }(g_{\overline{A}}+\alpha g_{\overline{Q}})$$$$g_K=g_Y+g_Q$$$$g_R=-g_Q$$

令

$$g_{Y,t+1}\triangleq {\left(\frac{A_{t+1}}{A_t}\right)}^{\frac{1+\varphi }{\varphi +\sigma }\frac{1}{1-\alpha }}{\left(\frac{Q_{t+1}}{Q_t}\right)}^{\frac{1+\varphi }{\varphi +\sigma }\frac{\alpha }{1-\alpha }}$$$$g_{Q,t+1}=\frac{Q_{t+1}}{Q_t}$$

记

$$y_t=\frac{Y_t}{A_t^{\frac{1+\varphi }{\sigma +\varphi }\frac{1}{1-\alpha }}{Q}_t^{\frac{1+\varphi }{\sigma +\varphi }\frac{\alpha }{1-\alpha }}}$$$$i_t=\frac{I_t}{A_t^{\frac{1+\varphi }{\sigma +\varphi }\frac{1}{1-\alpha }}{Q}_t^{\frac{1+\varphi }{\sigma +\varphi }\frac{\alpha }{1-\alpha }}}$$$$c_t=\frac{C_t}{A_t^{\frac{1+\varphi }{\sigma +\varphi }\frac{1}{1-\alpha }}{Q}_t^{\frac{1+\varphi }{\sigma +\varphi }\frac{\alpha }{1-\alpha }}}$$$$k_t=\frac{K_t}{A_t^{\frac{1+\varphi }{\sigma +\varphi }\frac{1}{1-\alpha }}{Q}_t^{\frac{1+\varphi }{\sigma +\varphi }\frac{\alpha }{1-\alpha }}{Q}_t}$$$$w_t=\frac{W_t}{A_t^{\frac{1}{1-\alpha }}{Q}_t^{\frac{\alpha }{1-\alpha }}}$$$$n_t=\frac{N_t}{A_t^{\frac{1-\sigma }{\sigma +\varphi }\frac{1}{1-\alpha }}{Q}_t^{\frac{1-\sigma }{\sigma +\varphi }\frac{\alpha }{1-\alpha }}}$$$$r_t=\frac{R_t}{Q_t^{-1}}$$

改写上述方程

$$\begin{array}{}\text{(K-A)}& & g_{Y,t+1}{k}_{t+1}=i_t+(1-\delta )k_t\end{array}$$$$\begin{array}{}\text{(A-S)}& & w_t={\eta }_L{n}_t^{\varphi }{c}_t^{\sigma }\end{array}$$$$\begin{array}{}\text{(A-D)}& & w_t{n}_t=(1-\alpha )y_t\end{array}$$$$\begin{array}{}\text{(K-D)}& & r_t{k}_t=\alpha y_t\end{array}$$$$\begin{array}{}\text{(M-C)}& & y_t=c_t+i_t\end{array}$$$$\begin{array}{}\text{(E-L)}& & {\mathbb{E}}_t\beta {\left(\frac{c_{t+1}}{c_{{}_t}}\right)}^{-\sigma }{g}_{Y,t+1}^{-\sigma }[r_{t+1}+(1-\delta )]g_{Q,t+1}^{-1}=1\end{array}$$$$\begin{array}{}\text{(Y)}& & y_t=k_t^{\alpha }{n}_t^{1-\alpha }\end{array}$$

稳态下冲击

$$\ln g_{Y,t+1}=\frac{1+\varphi }{\varphi +\sigma }\frac{1}{1-\alpha }[\ln g_{t+1}^A+\alpha \ln g_{t+1}^Q]$$$$\ln g_{t+1}^A=(1-{\rho }_A)g_{\overline{A}}+{\rho }_A\ln g_t^A+\underset{{\tilde{\epsilon }}_t^A}{\underbrace{{\epsilon }_{t+1}^A-{\epsilon }_t^A}}$$$$\ln g_{t+1}^Q=(1-{\rho }_Q)g_{\overline{Q}}+{\rho }_Q\ln g_t^Q+\underset{{\tilde{\epsilon }}_t^Q}{\underbrace{{\epsilon }_{t+1}^Q-{\epsilon }_t^Q}}$$

在稳态下，(E-L) 变成

$$R^{\ast }=\frac{1}{\beta }{g}_{\overline{Q}}{g}_{\overline{Y}}-1+\delta$$

其他与 B 中相同。

（4） 对数线性化方程

$$\begin{array}{}\text{(1)}& {\hat{g}}_{Y,t+1}+{\hat{k}}_{t+1}=(1-\frac{1-\delta }{g_{\overline{Y}}}){\hat{i}}_t+\frac{1-\delta }{g_Y}{\hat{k}}_t\end{array}$$$$\begin{array}{}\text{(2)}& {\hat{w}}_t=\sigma {\hat{c}}_t+\varphi {\hat{n}}_t\end{array}$$$$\begin{array}{}\text{(3)}& {\hat{w}}_t+{\hat{n}}_t={\hat{y}}_t\end{array}$$$$\begin{array}{}\text{(4)}& {\hat{r}}_t+{\hat{k}}_t={\hat{y}}_t\end{array}$$$$\begin{array}{}\text{(5)}& \frac{c^{\ast }}{y^{\ast }}{\hat{c}}_t+\frac{i^{\ast }}{y^{\ast }}{\hat{i}}_t={\hat{y}}_t\end{array}$$$$\begin{array}{}\text{(6)}& -\sigma {\mathbb{E}}_t{\hat{c}}_{t+1}-\sigma {\mathbb{E}}_t{\hat{g}}_{Y,t+1}+\beta R^{\ast }{\mathbb{E}}_t{\hat{r}}_{t+1}-{\mathbb{E}}_t{\hat{g}}_{Q,t+1}=0\end{array}$$$$\begin{array}{}\text{(7)}& \alpha {\hat{k}}_t+(1-\alpha ){\hat{n}}_t={\hat{y}}_t\end{array}$$

D. 调整成本计算

将 A 中 $Q(I_t,K_t)=I_t$ 改为 $Q(I_t,K_t)=I_t[1-\frac{\varphi }{2}{(\frac{I_t}{I_{t-1}}-1)}^2]$，其余保持不变。

（1） 求居民的一阶条件。

解：

$$\begin{array}{rl}V(K_t,I_{t-1};R_t,W_t)\triangleq max& {\mathbb{E}}_t\sum _{k=0}^{\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{K}_{t+1}=I_t[1-\frac{\varphi }{2}{(\frac{I_t}{I_{t-1}}-1)}^2]+(1-\delta )K_t\\ & C_t+I_t=R_t{K}_t+W_t{N}_t\end{array}$$

Bellman 方程

$$\begin{array}{rl}V(K_t,I_{t-1};R_t,W_t)\triangleq max& u(C_t,N_t)+\beta {\mathbb{E}}_{t}V(K_{t+1},I_t;R_{t+1},W_{t+1})\\ & \text{s.t.}{K}_{t+1}=I_t[1-\frac{\varphi }{2}{(\frac{I_t}{I_{t-1}}-1)}^2]+(1-\delta )K_t\\ & C_t+I_t=R_t{K}_t+W_t{N}_t\end{array}$$

无约束最优化

$$\begin{array}{rl}V(K_t,I_{t-1};R_t,W_t)\triangleq \underset{C_t,N_t,I_t,K_{t+1},{\lambda }_t^K,{\lambda }_t^B}{max}& u(C_t,N_t)+\beta {\mathbb{E}}_{t}V(K_{t+1},I_t;R_{t+1},W_{t+1})\\ & -{\lambda }_t^K\{K_{t+1}-I_t[1-\frac{\varphi }{2}{(\frac{I_t}{I_{t-1}}-1)}^2]-(1-\delta )K_t\}\\ & -{\lambda }_t^B[C_t+I_t-R_t{K}_t-W_t{N}_t]\end{array}$$

FOCs

$$\begin{array}{}\text{(1)}& \frac{\partial u}{\partial C}(C_t,N_t)={\lambda }_t^B\end{array}$$$$\begin{array}{}\text{(2)}& -\frac{\partial u}{\partial N}(C_t,N_t)={\lambda }_t^B{W}_t\end{array}$$$$\begin{array}{}\text{(3)}& {\lambda }_t^K=\beta {\mathbb{E}}_t\frac{\partial V}{\partial K}(K_{t+1},I_{t+1};R_{t+1},W_{t+1})\end{array}$$$$\begin{array}{}\text{(4)}& {\lambda }_t^B={\lambda }_t^K[1-\frac{\varphi }{2}{(\frac{I_t}{I_{t-1}}-1)}^2-\frac{I_t}{I_{t-1}}(\frac{I_t}{I_{t-1}}-1)\varphi +\beta {\mathbb{E}}_t\frac{\partial V}{\partial I}(K_{t+1},I_{t+1};R_{t+1},W_{t+1})]\end{array}$$

${\lambda }_t^D$ 为边际效用，而 ${\lambda }_t^K$ 为资本乘子，有差异为调整成本。

包络定理

$$\frac{\partial V}{\partial K}(K_t,I_{t-1};R_t,W_t)={\lambda }_t^k(1-\delta )+{\lambda }_t^B{R}_t$$$$\frac{\partial V}{\partial I}(K_t,I_{t-1};R_t,W_t)={\lambda }_t^K\left(\frac{I_t}{I_{t-1}}\right)(\frac{I_t}{I_{t-1}}-1)\varphi \frac{1}{I_{t-1}}{I}_t$$

代入 (3) 有

$$\beta {\mathbb{E}}_t\frac{{\lambda }_{t+1}^B}{{\lambda }_t^B}[R_{t+1}+(1-\delta )\frac{{\lambda }_{t+1}^K}{{\lambda }_{t+1}^B}]\frac{{\lambda }_t^B}{{\lambda }_t^K}=1$$

令 $q_t=\frac{{\lambda }_t^K}{{\lambda }_t^B}$ 表示因资本调成成本导致的损失，可得

$$\begin{array}{}\text{(E-L)}& {\mathbb{E}}_t\underset{M_{t,t+1}}{\underbrace{\beta \frac{\partial u/\partial C(C_{t+1},N_{t+1})}{\partial u/\partial C(C_t,N_t)}}}[R_{t+1}+(1-\delta )q_{t+1}]=q_t\end{array}$$$$\begin{array}{}\text{(A-S)}& -\frac{\partial u/\partial N(C_t,N_t)}{\partial u/\partial C(C_t,N_t)}=W_t\end{array}$$

对 $I_t$ 的一阶方程为

$$[1-\frac{\varphi }{2}{(\frac{I_t}{I_{t-1}}-1)}^2-\varphi \frac{I_t}{I_{t-1}}(\frac{I_t}{I_{t-1}}-1)]\times \frac{{\lambda }_t^K}{{\lambda }_t^B}+\beta {\mathbb{E}}_t\frac{\frac{\partial V}{\partial I}(k_{t+1},I_t;R_{t+1},W_{t+1})}{{\lambda }_t^B}=1$$

由对 $I_t$ 的包络定理，延后一期可获得

$$\frac{\partial V}{\partial I}(K_{t+1},I_t;R_{t+1},W_{t+1})=\varphi {\left(\frac{I_{t+1}}{I_t}\right)}^2(\frac{I_{t+1}}{I_t}-1){\lambda }_{t+1}^K$$

因为 ${\lambda }_{t+1}^K=\frac{{\lambda }_{t+1}^K}{{\lambda }_{t+1}^B}{\lambda }_{t+1}^B$，代入后可得 $q_t$ 的运动方程

$$1=q_t[1-\frac{\varphi }{2}{(\frac{I_t}{I_{t+1}}-1)}^2-\varphi \frac{I_t}{I_{t+1}}(\frac{I_t}{I_{t+1}}-1)]+{\mathbb{E}}_t{M}_{t,t+1}{q}_{t+1}\varphi (\frac{I_{t+1}}{I_t}-1){\left(\frac{I_{t+1}}{I_t}\right)}^2$$

（2） 企业行为

（3） 稳态值

因稳态下 $q_t\equiv q_{t+1}=1$，从而均与 A 中相同，仅变动 Euler 方程同时增加上述动态方程。

（4） 对数线性化

$$1=q_t-\frac{\varphi }{2}{q}_t{\left(\frac{I_t}{I_{t-1}}\right)}^2+\varphi q_t\frac{I_t}{I_{t-1}}-\varphi q_t{\left(\frac{I_t}{I_{t+1}}\right)}^2+\varphi q_t\frac{I_t}{I_{t-1}}+{\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }{q}_{t+1}\varphi {\left(\frac{I_{t+1}}{I_t}\right)}^3-{\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }{q}_{t+1}\varphi {\left(\frac{I_{t+1}}{I_t}\right)}^2$$

稳态下

$$\frac{I_{t+1}}{I_t}=\frac{I_t}{I_{t-1}}=1$$$$\frac{C_{t+1}}{C_t}=1$$

对数线性化之后有

$$0={\hat{q}}_t-\frac{3\varphi }{2}{\hat{q}}_t-3\varphi [{\hat{i}}_t-{\hat{i}}_{t-1}]+2\varphi {\hat{q}}_t+2\varphi ({\hat{i}}_t-{\hat{i}}_{t-1})+\beta \varphi [-\sigma {\mathbb{E}}_t{\hat{c}}_{t+1}+\sigma {\hat{c}}_t+{\mathbb{E}}_t{\hat{q}}_{t+1}+3{\mathbb{E}}_t{\hat{i}}_{t-1}-3{\hat{i}}_t]-\beta \varphi [-\sigma {\mathbb{E}}_t{\hat{c}}_{t+1}+\sigma {\hat{c}}_t+{\mathbb{E}}_t{\hat{q}}_{t+1}+2{\mathbb{E}}_t{\hat{i}}_{t+1}-2{\hat{i}}_t]$$

整理得

$$0=(1+\frac{\varphi }{2}){\hat{q}}_t-\varphi ({\hat{i}}_t-{\hat{i}}_{t-1})+\beta \varphi {\mathbb{E}}_t{\hat{i}}_{t+1}-\beta \varphi {\hat{i}}_t$$

从而有调整成本比价 $q_t$ 的变化方程

$$(1+\frac{\varphi }{2}){\hat{q}}_t=\varphi [({\hat{i}}_t-{\hat{i}}_{t+1})-\beta ({\mathbb{E}}_t{\hat{i}}_{t+1}-{\hat{i}}_t)]$$

之后 Euler 方程变为

$$-\sigma ({\mathbb{E}}_t{\hat{c}}_{t+1}-{\hat{c}}_t)+\beta R^{\ast }{\mathbb{E}}_t{\hat{r}}_{t+1}+\beta (1-\delta ){\mathbb{E}}_t{\hat{q}}_{q+1}={\hat{q}}_t$$

其余维持不变。

E. 居民消费习惯问题

将 $u(C_t,N_t)$ 换为 $\frac{(C_t-r{C}_{t-1}{)}^{1-\sigma }}{1-\sigma }-{\eta }_L\frac{N_t^{1+\varphi }}{1+\varphi }$，其余均如同 A 中所示。

（1） 居民问题。

解：

$$\begin{array}{rl}V(K_t,C_{t-1};R_t,W_t)\triangleq max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},C_{t+k-1},N_{t+k})\\ & \text{s.t.}{K}_{t+1}=R_t{K}_t+(1-\delta )K_t+W_t{N}_t-C_t\end{array}$$

Bellman 方程

$$\begin{array}{rl}V(K_t,C_{t-1};R_t,W_t)\triangleq max& u(C_t,C_{t-1},N_t)+\beta {\mathbb{E}}_{t}V(K_{t+1},C_t;R_{t+1},W_{t+1})\\ & \text{s.t.}{K}_{t+1}=R_t{K}_t+(1-\delta )K_t+W_t{N}_t-C_t\end{array}$$

无约束最大化问题

$$\begin{array}{rl}V(K_t,C_{t-1};R_t,W_t)\triangleq max& u(C_t,C_{t-1},N_t)+\beta {\mathbb{E}}_{t}V(K_{t+1},C_t;R_{t+1},W_{t+1})\\ & -{\lambda }_t[K_{t+1}-R_t{K}_t-(1-\delta )K_t-W_t{N}_t+C_t]\end{array}$$

FOCs

$$\frac{\partial u}{\partial C_t}(C_t,C_{t-1},N_t)+\beta {\mathbb{E}}_t\frac{\partial V}{\partial C_t}(K_{t+1},C_t;R_{t+1},W_{t+1})={\lambda }_t$$$$-\frac{\partial u}{\partial N}(C_t,C_{t-1};N_t)={\lambda }_t{W}_t$$$${\lambda }_t=\beta {\mathbb{E}}_t\frac{\partial V}{\partial K}(K_{t+1},C_t;R_{t+1},W_{t+1})$$

包络定理

$$\frac{\partial V}{\partial C_{t-1}}(K_t,C_{t-1};R_t,W_t)=\frac{\partial u}{\partial C_{t-1}}(C_t,C_{t-1},N_t)$$$$\frac{\partial V}{\partial K_t}(K_t,C_{t-1};R_t,W_t)={\lambda }_t[R_t+(1-\delta )]$$

上述方程可以简化为

$$\begin{array}{r}{\lambda }_t=\frac{\partial u}{\partial C_t}u(C_t,C_{t-1},N_t)+\beta {\mathbb{E}}_t\frac{\partial u}{\partial C_t}(C_{t+1},C_t,N_{t+1})\end{array}$$$$\begin{array}{}\text{(2)}& -\frac{\partial u}{\partial N_t}(C_t,C_{t-1},N_t)={\lambda }_t{W}_t\end{array}$$$$\begin{array}{}\text{(3)}& {\lambda }_t=\beta {\mathbb{E}}_t{\lambda }_{t+1}(R_{t+1}+1-\delta )\end{array}$$

（2）

同之前

（3）

略

（4）

略

F. 企业持有资本，居民持债向企业投资以及中央计划者问题

（1） 中央计划者问题

$$\begin{array}{rl}V(K_t,A_t)\triangleq \underset{\{C_{t+k},N_{t+k}{\}}_{k=0}^{+\mathrm{\infty }}}{max}& \mathbb{E}\sum _{k=0}^{\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{K}_{t+1}=A_t{K}_t^{\alpha }{N}_t^{1-\alpha }+(1-\delta )K_t-C_t\end{array}$$

说明解与 A 相同，并分析福利经济学第一二定理。

（2） 企业持有资本、居民用债券向企业注资

居民行为

$$\begin{array}{rl}max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{C}_{t+k}+B_{t+k+1}\le R_{t+k}^{\beta }{B}_{t+k}+W_{t+k}{N}_{t+k}\end{array}$$

令 $M_{t,t+1}=\beta \frac{u_C(C_{t+1},N_{t+1})}{u_C(C_t,N_t)}$，$M_{t,t+k}={\beta }^k\frac{\partial u/\partial C(C_{t+k},N_{t+k})}{\partial u/\partial C(C_t,N_t)}$ 为贴现因子。

企业行为

$$\begin{array}{rl}max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{M}_{t,t+k}[Y_{t+k}-W_{t+k}{N}_{t+k}+D_{t+k+1}-R_{t+k}^{\beta }{D}_{t+k}-I_t]\\ & \text{s.t.}{K}_{t+1}=(1-\delta )K_t+I_t\\ & Y_t=A_t{K}_t^{\alpha }{N}_t^{1-\alpha }\end{array}$$

虚拟资本市场出清

$$B_t=D_t$$

实物产品市场出清

$$Y_t=C_t+I_t$$

说明上述过程的解与 A 相同。

G. 需求冲击

考虑一个中央计划者问题

$$\begin{array}{rl}V(K_t,{\xi }_t)=max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^k{\xi }_{t+k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{K}_{t+1}=K_t^{\alpha }{N}_t^{1-\alpha }-C_t+(1-\delta )K_t\end{array}$$$$\ln {\xi }_{t+1}={\rho }_D\ln {\xi }_t+{\epsilon }_t^D,{\epsilon }_t^D\sim N(0,{\sigma }_D^2)$$

（1） 解 Euler 方程。

解：

$$\begin{array}{rl}V(K_t,{\xi }_t)=\underset{C_t,N_t}{max}& u(C_t,N_t){\xi }_t+\beta {\mathbb{E}}_{t}V(K_{t+1},{\xi }_{t+1})\\ & \text{s.t.}{K}_{t+1}=K_t^{\alpha }{N}_t^{1-\alpha }+(1-\delta )K_t-C_t\end{array}$$$$\begin{array}{rl}V(K_t,{\xi }_t)=\underset{C_t,N_t}{max}& u(C_t,N_t){\xi }_t+\beta {\mathbb{E}}_{t}V(K_{t+1},{\xi }_{t+1})\\ & -{\lambda }_t[K_{t+1}-K_t^{\alpha }{N}_t^{1-\alpha }-(1-\delta )K_t+C_t]\end{array}$$

FOCs

$${\xi }_t\frac{\partial u}{\partial C}(C_t,N_t)={\lambda }_t$$$$-{\xi }_t\frac{\partial u}{\partial N}(C_t,N_t)={\lambda }_t\frac{\partial (K_t^{\alpha }{N}_t^{1-\alpha })}{\partial N}$$$${\lambda }_t=\beta {\mathbb{E}}_t\frac{\partial V}{\partial K}(K_t,{\xi }_{t+1})$$

包络定理

$$\frac{\partial V}{\partial K}(K_t,{\xi }_t)={\lambda }_t[\frac{\partial }{\partial K}(K_{t+1}^{\alpha }{N}_t^{1-\alpha })+1-\delta ]$$

可得 Euler 方程

$$\begin{array}{}\text{(1)}& {\mathbb{E}}_t\underset{\triangleq M_{t,t+1}}{\underbrace{\beta \frac{{\xi }_{t+1}}{{\xi }_t}\frac{\partial u/\partial C(C_{t+1},N_{t+1})}{\partial u/\partial C(C_t,N_t)}}}[\frac{\partial }{\partial K}(K_{t+1}^{\alpha }{N}_t^{1-\alpha })+1-\delta ]=1\end{array}$$

劳动供给方程

$$\begin{array}{}\text{(2)}& -\frac{\partial u/\partial N(C_t,N_t)}{\partial u/\partial C(C_t,N_t)}=\frac{\partial }{\partial N}(K_t,N_t^{1-\alpha })\end{array}$$

（2） 解均衡。

解：

配合

$$\begin{array}{}\text{(3)}& K_{t+1}=K_t^{\alpha }{N}_t^{1-\alpha }+(1-\delta )K_t-C_t\end{array}$$

获得方程。

由于

$${\xi }_{t+1}={\rho }_D{\xi }_t+{\epsilon }_t^D,{\epsilon }_t^D\sim N(D,{\sigma }_D^2)$$

所以 $\xi$ 在稳态下为 1，则其解与 A 相同。

（3） 对数线性化。

解：

令 $R^{\ast }=\frac{1}{\beta }-1+\delta$

$$\begin{array}{}\text{(1)}& -\sigma ({\mathbb{E}}_t{\hat{c}}_{t+1}-c_t)+({\mathbb{E}}_t{\hat{\xi }}_{t+1}-{\hat{\xi }}_t)+\beta R^{\ast }(1-\alpha ){\mathbb{E}}_t({\hat{n}}_{t+1}-{\hat{k}}_{t+1})=0\end{array}$$$$\begin{array}{}\text{(2)}& \alpha ({\hat{k}}_t-{\hat{n}}_t)=\sigma {\hat{c}}_t+\varphi {\hat{n}}_t\end{array}$$$$\begin{array}{}\text{(3)}& {\hat{k}}_{t+1}=\frac{1}{\beta }{\hat{k}}_t+(1-\alpha )\frac{Y^{\ast }}{K^{\ast }}{\hat{n}}_t-\frac{C^{\ast }}{K^{\ast }}{\hat{c}}_t\end{array}$$$$\begin{array}{}\text{(4)}& {\hat{\xi }}_{t+1}={\rho }_D{\hat{\xi }}_t+{\epsilon }_t^D\end{array}$$

Example 2 虚拟资本情况分析

本模型下仅考虑分散化经济

A. 股票

居民决策

$$\begin{array}{rl}max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{X}_{t+1}{P}_t+C_t=X_t(P_t+{\mathrm{\Pi }}_t)+W_t{N}_t\end{array}$$

企业决策

$$\begin{gathered} {\mathrm{\Pi }}_t\triangleq Y_t-W_t{N}_t \\ {Y}_t=A_t{N}_t^{1-\alpha } \end{gathered}$$

市场出清

$$Y_t=C_t$$$$X_t=1$$

(1) 求居民的利润最大化条件

(2) 企业的利润最大化条件

(3) 写出均衡方程和稳态值

(4) 写出价格 $p_t$ 的对数线性化

(5) 考虑 $\ln A_{t+1}=(1-\rho ){\overline{g}}_A+\rho \ln A_t+{\epsilon }_t^A$ 时的 ${\hat{p}}_t$ 的分析。

B. 企业的垄断势力的分析

居民决策

$$\begin{array}{rl}max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k})\\ & \text{s.t.}{X}_{t+1}{P}_t+C_t=X_t(P_t+{\mathrm{\Pi }}_t)+W_t{N}_t\end{array}$$

企业决策

$$\begin{gathered} {\mathrm{\Pi }}_t\triangleq P_t{Y}_t-W_t{N}_t \\ {Y}_t=A_t{N}_t \end{gathered}$$$$P_t=\mu M{C}_t=\mu \frac{W_t}{A_t}$$$$P_t=1$$

市场出清

$$Y_t=C_t$$

C. 短期债券

居民决策

$$\begin{array}{rl}max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{B}_{t+1}+C_t=R_t{B}_t+W_t{N}_t\end{array}$$

企业决策

$$\begin{gathered} {\mathrm{\Pi }}_t\triangleq Y_t-W_t{N}_t \\ {Y}_t=A_t{N}_t \end{gathered}$$

市场出清

$$Y_t=C_t$$$$B_t=0$$

D. 长期债券

居民决策

$$\begin{array}{rl}max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{B}_{t+1}-\kappa B_t+C_t=R_t{B}_t+W_t{N}_t\end{array}$$

企业决策

$$\begin{gathered} {\mathrm{\Pi }}_t\triangleq Y_t-W_t{N}_t \\ {Y}_t=A_t{N}_t \end{gathered}$$

市场出清

$$Y_t=C_t$$$$B_t=0$$

※ Example 3 垄断势力：微观基础

A. 分散化经济

居民端

$$\begin{array}{rl}max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k}),u=\frac{C^{1-\sigma }}{1-\sigma }-{\eta }_L\frac{N^{1+\varphi }}{1+\varphi }\\ & \text{s.t.}{C}_t+K_{t+1}-(1-\delta )K_t=R_t{K}_t+W_t{N}_t+{\mathrm{\Pi }}_t\end{array}$$

最终产品部门

$$\begin{array}{r}{Y}_t={[{\int }_0^{M_t}{y}_t(i{)}^{1-\frac{1}{\epsilon }}di]}^{\frac{\epsilon }{\epsilon -1}}\end{array}$$

其中 $M_t$ 为企业数，$\epsilon /(\epsilon -1)=\mu$。

$$\mathrm{\Pi }=Y_t-{\int }_0^{M_t}{p}_t(i)y_t(i)di$$

企业端

在已知 $y_t(i)=D(p_t(i))$ 时，求解

$${\mathrm{\Pi }}_t(i)=p_t(i)y_t(i)-R_t{k}_t(i)-W_t{n}_t(i)$$$$y_t(i)=A_t{k}_t^{\alpha }(i)n_t^{1-\alpha }(i)$$

市场出清

$${\mathrm{\Pi }}_t={\int }_0^{M_t}{\mathrm{\Pi }}_t(i)di$$$$N_t={\int }_0^{M_t}{n}_t(i)di$$$$K_t={\int }_0^{M_t}{k}_t(i)di$$

（1） 求居民的 Euler 方程和劳动供给方程。

解：

$${\mathbb{E}}_t\beta \frac{\partial u/\partial C(C_{t+1},N_{t+1})}{\partial u/\partial C(C_t,N_t)}[R_{t+1}+1-\delta ]=1$$$$-\frac{\partial u/\partial N}{\partial u/\partial C}(C_t,N_t)=W_t$$

（2） 最终厂商的情况

$${\mathrm{\Pi }}_t=Y_t-{\int }_0^{M_t}{p}_t(i)y_t(i)di\Rightarrow \frac{\partial Y_t}{\partial y_t(i)}=p_t(i),\mathrm{\forall }i\in [0,M_t]$$$$Y_t={[{\int }_0^{M_t}{y}_t(i{)}^{1-\frac{1}{\epsilon }}di]}^{\frac{\epsilon }{\epsilon -1}}$$$$\frac{\partial Y_t}{\partial y_t(i)}=Y_t^{1/\epsilon }{y}_t(i{)}^{-1/\epsilon }$$

可得需求函数

$$y_t(i)=Y_t{p}_t(i{)}^{-\epsilon }$$

（3） 已知 $D(p_t(i))=Y_t{p}_t(i{)}^{-\epsilon }$，求企业的利润最大化。

$${\mathrm{\Pi }}_t(i)=p_t(i)y_t(i)-R_t{k}_t(i)-W_t{n}_t(i)$$$$P_t(i)=Y_t^{1/\epsilon }{Y}_t(i{)}^{-1/\epsilon }$$$$y_t(i)=A_t{k}_t(i{)}^{\alpha }{n}_t(i{)}^{1-\alpha }$$

可得

$${\mathrm{\Pi }}_t(i)=\underset{p_t(i)y_t(i)}{\underbrace{Y_t{A}_t^{1-1/\epsilon }{k}_t(i{)}^{\alpha (1-1/\epsilon )}{n}_t(i{)}^{(1-\alpha )(1-1/\epsilon )}}}-R_t{k}_t(i)-W_t{n}_t(i)$$

可得

$$\{\begin{array}{l}\alpha (1-\frac{1}{\epsilon })p_t(i)y_t(i)=R_t{k}_t(i)\\ (1-\alpha )(1-\frac{1}{\epsilon })p_t(i)y_t(i)=W_t{n}_t(i)\\ y_t(i)=A_t{k}_t(i{)}^{\alpha }{n}_t(i{)}^{1-\alpha }\end{array}\Rightarrow p_t(i)=\frac{\epsilon }{\epsilon -1}\frac{1}{A_t}{\left(\frac{R_t}{\alpha }\right)}^{\alpha }{\left(\frac{W_t}{1-\alpha }\right)}^{1-\alpha },\mathrm{\forall }i$$

因此，$\mathrm{\forall }i\in [0,M_t],p_t(i)$ 相同，从而 $y_t(i)$ 与 $n_t(i),k_t(i)$ 均相同，结合

$$N_t={\int }_0^{M_t}{n}_t(i)di$$$$K_t={\int }_0^{M_t}{k}_t(i)di$$$$Y_t={\int }_0^{M_t}{p}_t(i)y_t(i)di$$

可得

$$R_t{K}_t=(1-\frac{1}{\epsilon })\alpha Y_t$$$$W_t{N}_t=(1-\frac{1}{\epsilon })(1-\alpha )Y_t$$

且

$$k_t(i)=\frac{K_t}{M_t}$$$$n_t(i)=\frac{N_t}{M_t}$$$$y_t(i)=\frac{A_t{K}_t^{\alpha }{N}_t^{1-\alpha }}{M_t}$$

可得

$$Y_t={[{\int }_0^{M_t}{y}_t(i{)}^{1-1/\epsilon }di]}^{\frac{\epsilon }{\epsilon -1}}=M_t^{\frac{1}{\epsilon -1}}{A}_t{K}_t^{\alpha }{N}_t^{1-\alpha }$$$$p_t(i)y_t(i)=\frac{Y_t}{M_t}=\frac{A_t{K}_t^{\alpha }{N}_t^{1-\alpha }}{M_t^{\frac{\epsilon }{\epsilon -1}}}$$$${\mathrm{\Pi }}_t(i)=\frac{1}{\epsilon }{p}_t(i)y_t(i)=\frac{1}{\epsilon }\frac{A_t{K}_t^{\alpha }{N}_t^{1-\alpha }}{M_t^{\frac{\epsilon }{\epsilon -1}}}$$

（4） 市场均衡，若 $\ln M_{t+1}={\rho }_M\ln M_t+(1-{\rho }_A)\ln \overline{M}+{\epsilon }_t^M$；$\ln A_{t+1}={\rho }_A\ln A_t+{\epsilon }_t^A$。${\epsilon }_t^A\sim N(0,{\sigma }_A^2)$；${\epsilon }_t^M\sim N_{(}0,{\sigma }_M^2)$；$\overline{M}=1$。求稳态。

解：

I-S 曲线

$$\begin{array}{}\text{(1)}& {\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }[R_{t+1}+1-\delta ]=1\end{array}$$

A-S 曲线

$$\begin{array}{}\text{(2)}& W_t={\eta }_L{C}_t^{\sigma }{N}_t^{\varphi }\end{array}$$

劳动报酬占比（劳动需求）

$$\begin{array}{}\text{(3)}& W_t{N}_t=(1-\frac{1}{\epsilon })(1-\alpha )Y_t\end{array}$$

资本报酬占比（资本需求）

$$\begin{array}{}\text{(4)}& R_t{K}_t=(1-\frac{1}{\epsilon })\alpha Y_t\end{array}$$

生产技术

$$\begin{array}{}\text{(5)}& Y_t=M_t^{1/(\epsilon -1)}{A}_t{K}_t^{\alpha }{N}_t^{1-\alpha }\end{array}$$

市场出清

$$\begin{array}{}\text{(6)}& Y_t=C_t+I_t\end{array}$$

资本积累

$$\begin{array}{}\text{(7)}& K_{t+1}=(1-\delta )K_t+I_t\end{array}$$

由 (1) 可得

$$R^{\ast }=\frac{1}{\beta }-1+\delta$$

代入 (4)，令 $\tilde{\alpha }=(1-\frac{1}{\epsilon })\alpha$ 可得

$$K^{\ast }=\frac{\tilde{\alpha }{Y}^{\ast }}{R^{\ast }}$$

代入 (5) 可得

$$\begin{array}{}\text{(8)}& N^{\ast }={\left(\frac{R^{\ast }}{\tilde{\alpha }}\right)}^{\frac{\alpha }{1-\alpha }}{Y}^{\ast }\end{array}$$

代入 (3) 可得

$$\begin{array}{}\text{(9)}& W^{\ast }=(1-\frac{1}{\epsilon })(1-\alpha ){\left(\frac{R^{\ast }}{\tilde{\alpha }}\right)}^{\alpha /(1-\alpha )}\end{array}$$

代入 (7) 可得

$$I^{\ast }=\delta K^{\ast }=\frac{\tilde{\alpha }\delta }{R^{\ast }}{Y}^{\ast }$$

代入 (6) 可得

$$\begin{array}{}\text{(10)}& C^{\ast }=(1-\frac{\tilde{\alpha }\delta }{R^{\ast }})Y^{\ast }\end{array}$$

将 (7)(8)(9) 代入 $W^{\ast }={\eta }_L(C^{\ast }{)}^{\sigma }(N^{\ast }{)}^{\varphi }$ 可得 $Y^{\ast }$。

之后可获得 $I^{\ast },C^{\ast },N^{\ast },K^{\ast }$。

（5） 对数线性化

$$\begin{array}{}\text{(1)}& -\sigma {\mathbb{E}}_t{\hat{c}}_{t+1}+\sigma {\hat{c}}_t+\beta R^{\ast }{\mathbb{E}}_t{\hat{r}}_{t+1}=0\end{array}$$$$\begin{array}{}\text{(2)}& {\hat{w}}_t=\sigma {\hat{c}}_t+\varphi {\hat{n}}_t\end{array}$$$$\begin{array}{}\text{(3)}& {\hat{w}}_t+{\hat{n}}_t={\hat{y}}_t\end{array}$$$$\begin{array}{}\text{(4)}& {\hat{r}}_t+{\hat{k}}_t={\hat{y}}_t\end{array}$$$$\begin{array}{}\text{(5)}& {\hat{y}}_t=\frac{1}{\epsilon -1}{\hat{m}}_t+{\hat{a}}_t+\alpha {\hat{k}}_t+(1-\alpha ){\hat{n}}_t\end{array}$$$$\begin{array}{}\text{(6)}& {\hat{c}}_t\frac{C^{\ast }}{Y^{\ast }}+{\hat{i}}_t\frac{I^{\ast }}{Y^{\ast }}={\hat{y}}_t\end{array}$$$$\begin{array}{}\text{(7)}& {\hat{k}}_{t+1}=(1-\delta ){\hat{k}}_t+{\hat{i}}_t\end{array}$$$$\begin{array}{}\text{(8)}& {\hat{a}}_{t+1}={\rho }_A{\hat{a}}_t+{\epsilon }_t^A\end{array}$$$$\begin{array}{}\text{(9)}& {\hat{m}}_{t+1}={\rho }_M{\hat{m}}_t+{\epsilon }_t^M\end{array}$$

（6） 写出中央计划者问题，并分析与分散决策的差别。

解：

$$\begin{array}{rl}V(K_t,M_t,A_t)\triangleq max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{K}_{t+1}=Y_t-C_t+(1-\delta )K_t\\ & {\int }_0^{M_t}{k}_t(i)di=K_t\\ & {\int }_0^{M_t}{n}_t(i)di=N_t\\ & Y_t={[{\int }_0^{M_t}(y_t(i){)}^{1-1/\epsilon }di]}^{\epsilon /(\epsilon -1)}\\ & y_t(i)=A_t{k}_t^{\alpha }(i)n_t^{1-\alpha }(i)\end{array}$$

无约束最优化

$$\begin{array}{rl}V(K_t,M_t,A_t)\triangleq max& u(C_t,N_t)+\beta {\mathbb{E}}_{t}V(K_{t+1},M_{t+1},A_{t+1})\\ & -{\lambda }_t^B[K_{t+1}-Y_t+C_t-(1-\delta )K_t]\\ & -{\lambda }_t^Y\{Y_t-{[{\int }_0^{M_t}(y_t(i){)}^{1-1/\epsilon }di]}^{\epsilon /(\epsilon -1)}\}\\ & -{\int }_0^{M_t}{\lambda }_t(i)[y_t(i)-A_t{k}_t^{\alpha }(i)n_t^{1-\alpha }(i)]di\\ & -{\lambda }_t^K[{\int }_0^{M_t}{k}_t(i)di-K_t]\\ & -{\lambda }_t^N[{\int }_0^{M_t}{n}_t(i)di-N_t]\end{array}$$

FOCs

$$\begin{array}{}\text{(1)}& \frac{\partial u}{\partial C}(C_t,N_t)={\lambda }_t^B\end{array}$$$$\begin{array}{}\text{(2)}& {\lambda }_t^B={\mathbb{E}}_t\beta \frac{\partial V}{\partial K}(K_{t+1};A_{t+1},M_{t+1})\end{array}$$

包络定理

$$\frac{\partial V}{\partial K_t}={\lambda }_t^B(1-\delta )+{\lambda }_t^K$$$$\begin{array}{}\text{(3)}& {\lambda }_t^B={\lambda }_t^Y\end{array}$$$$\begin{array}{}\text{(4)}& Y_t^{1/\epsilon }{y}_t(i{)}^{-1/\epsilon }{\lambda }_t^Y={\lambda }_t(i)\end{array}$$$$\begin{array}{}\text{(5)}& {\lambda }_t(i)A_t{k}_t(i{)}^{\alpha }{n}_t(i{)}^{1-\alpha }\alpha ={\lambda }_t^K{k}_t(i)\end{array}$$$$\begin{array}{}\text{(6)}& {\lambda }_t(i)A_t{k}_t(i{)}^{\alpha }{n}_t(i{)}^{1-\alpha }(1-\alpha )={\lambda }_t^N{n}_t(i)\end{array}$$$$\begin{array}{}\text{(7)}& -\frac{\partial u}{\partial N}(C_t,N_t)={\lambda }_t^N\end{array}$$

令 $W_t\triangleq {\lambda }_t^N/{\lambda }_t^Y$；$R_t\triangleq {\lambda }_t^K/{\lambda }_t^Y$；$p_t(i)\triangleq {\lambda }_t(i)/{\lambda }_t^Y$，改写上述方程

$$\begin{array}{}\text{(5)}& \alpha p_t(i)y_t(i)=R_t{k}_t(i)\end{array}$$$$\begin{array}{}\text{(6)}& (1-\alpha )p_t(i)y_t(i)=W_t{n}_t(i)\end{array}$$$$\begin{array}{}\text{(4)}& y_t(i)=p_t(i{)}^{-\epsilon }{Y}_t\end{array}$$

对 $K_t$ 进行包络

$$\frac{\partial V}{\partial K}(K_t;M_t,A_t)=R_t+1-\delta$$

代入 (1) 和 (2) 可得

$${\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }(R_{t+1}+1-\delta )=1$$$$\begin{array}{}\text{(7)}& -\frac{\partial u/\partial N(C_t,N_t)}{\partial /\partial C(C_t,N_t)}=W_t\end{array}$$

配合

$$\begin{array}{rl}& K_{t+1}=Y_t-C_t+(1-\delta )K_t\\ & {\int }_0^{M_t}{k}_t(i)di=K_t\\ & {\int }_0^{M_t}{n}_t(i)di=N_t\\ & Y_t={[{\int }_0^{M_t}(y_t(i){)}^{1-1/\epsilon }di]}^{\epsilon /(\epsilon -1)}\\ & y_t(i)=A_t{k}_t^{\alpha }(i)n_t^{1-\alpha }(i)\end{array}$$

可以化为加总方程

I-S 曲线

$$\begin{array}{}\text{(1)}& {\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }[R_{t+1}+1-\delta ]=1\end{array}$$

A-S 曲线

$$\begin{array}{}\text{(2)}& W_t={\eta }_L{C}_t^{\sigma }{N}_t^{\varphi }\end{array}$$

劳动报酬占比

$$\begin{array}{}\text{(3)}& W_t{N}_t=(1-\alpha )Y_t\end{array}$$

资本报酬占比

$$\begin{array}{}\text{(4)}& R_t{K}_t=\alpha Y_t\end{array}$$

生产技术

$$\begin{array}{}\text{(5)}& Y_t=M_t^{1/(\epsilon -1)}{A}_t{K}_t^{\alpha }{N}_t^{1-\alpha }\end{array}$$

市场出清

$$\begin{array}{}\text{(6)}& Y_t=C_t+I_t\end{array}$$

资本积累

$$\begin{array}{}\text{(7)}& K_{t+1}=(1-\delta )K_t+I_t\end{array}$$

B. 虚拟资本，债券形式

居民

$$\begin{array}{rl}max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k}),u(C_t,N_t)=\frac{C_t^{1-\sigma }}{1-\sigma }-{\eta }_L\frac{N_t^{1+\varphi }}{1+\varphi }\\ & \text{s.t.}{C}_t+B_{t+1}=R_t{B}_t+W_t{N}_t+{\mathrm{\Pi }}_t\end{array}$$

最终厂商

$$Y_t={[{\int }_0^{M_t}{y}_t(i{)}^{1-1/\epsilon }di]}^{\epsilon /(\epsilon -1)}$$$${\mathrm{\Pi }}_t=Y_t-{\int }_0^{M_t}{p}_t(i)y_t(i)di$$

企业行为

$${\mathrm{\Pi }}_t(i)=p_t(i)y_t(i)-w_t{n}_t(i)$$$$y_t(i)=A_t{n}_t(i)$$

市场出清

$${\mathrm{\Pi }}_t={\int }_0^{M_t}{\pi }_t(i)di$$$$N_t={\int }_0^{M_t}{n}_t(i)di$$$$B_t=0$$$$C_t=Y_t$$

解：

市场均衡方程

欧拉方程

$$\begin{array}{}\text{(1)}& {\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }{R}_{t+1}=1\end{array}$$

劳动供给

$$\begin{array}{}\text{(2)}& W_t={\eta }_L{C}_t^{\sigma }{N}_t^{\varphi }\end{array}$$

市场出清

$$C_t=Y_t$$

劳动工资

$$\begin{array}{}\text{(4)}& W_t{N}_t=(1-\frac{1}{\epsilon })Y_t\end{array}$$

生产函数

$$\begin{array}{}\text{(5)}& Y_t=A_t{M}_t^{1/(\epsilon -1)}{N}_t\end{array}$$

后略

C. 股票定价

居民

$$\begin{array}{rl}max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k}),u(C,N)=\frac{C^{1-\sigma }}{1-\sigma }-{\eta }_L\frac{N^{1+\varphi }}{1+\varphi }\\ & \text{s.t.}{\int }_0^{M_t}{x}_{t+1}(i)p_t(i)di+C_t={\int }_0^{M_t}{x}_t(i)({\pi }_t(i)+p_t(i))di+W_t{N}_t\end{array}$$

最终厂商

$$Y_t={({\int }_0^{M_t}{y}_t(i{)}^{1-1/\epsilon }di)}^{\epsilon /(\epsilon -1}$$$${\mathrm{\Pi }}_t^F=Y_t-{\int }_0^{M_t}{p}_t(i)y_t(i)di$$

企业

$${\pi }_t(i)=p_t(i)y_t(i)-W_t{n}_t(i)$$$$y_t(i)=A_t{n}_t(i)$$

市场出清

$$x_t(i)=1,\mathrm{\forall }i\in [0,M_t]$$$$C_t=Y_t$$$${\int }_0^{M_t}{n}_t(i)di=N_t$$

解：

$$\mathrm{\forall }i\in [0,M_t],{\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }\frac{p_{t+1}+{\pi }_{t+1}(i)}{p_t(i)}=1$$

$\mathrm{\forall }i\in [0,M_t]$，$p_t(i),{\pi }_t(i)$ 相同。令 $p_t(i)\equiv p_t,{\pi }_t(i)\equiv {\pi }_t$，可得市场均衡

$$\begin{array}{}\text{(1)}& {\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }\frac{p_{t+1}+{\pi }_{t+1}}{p_t}=1\end{array}$$$$\begin{array}{}\text{(2)}& {\mathrm{\Pi }}_{t+1}=\frac{1}{\epsilon }\frac{Y_t}{M_t}\end{array}$$$$\begin{array}{}\text{(3)}& Y_t=A_t{M}_t^{1/(\epsilon -1)}{N}_t\end{array}$$$$\begin{array}{}\text{(4)}& W_t=\theta C_t^{\sigma }{N}_t^{\varphi }\end{array}$$$$\begin{array}{}\text{(5)}& Y_t=C_t\end{array}$$$$\begin{array}{}\text{(6)}& W_t{N}_t=(1-\frac{1}{\epsilon })Y_t\end{array}$$

D. 内生企业数目

居民

$$\begin{array}{rl}max& {\mathbb{E}}_t\sum _{k=0}^{\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k}),u(C,N)=\frac{C^{1-\sigma }}{1-\sigma }-{\eta }_L\frac{N^{1+\varphi }}{1+\varphi }\\ & \text{s.t.}{X}_{t+1}(M_t+M_{t,E})p_t+C_t=({\pi }_t+p_t)M_t{X}_t+W_t{N}_t\\ & M_{t+1}=(M_t+M_{t,E})(1-\delta )\end{array}$$

企业

$$C_t=M_t^{1/(\epsilon -1)}{A}_t{N}_t^P$$$${\pi }_t=\frac{1}{\epsilon }\frac{C_t}{M_t}$$$$W_t=(1-\frac{1}{\epsilon })\frac{C_t}{N_t^P}$$

市场出清

$$\begin{array}{}\text{(进入成本)}& p_t=\frac{f_E{W}_t}{Z_t}\end{array}$$$$\begin{array}{}\text{(劳动力市场出清)}& N_t^P+M_{E,t}\frac{f_E}{Z_t}=N_t\end{array}$$

解：

Euler 方程

$$\begin{array}{}\text{(1)}& {\mathbb{E}}_t\beta (1-\delta ){\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }\frac{p_{t+1}+{\pi }_{t+1}}{p_t}=1\end{array}$$

劳动供给

$$\begin{array}{}\text{(2)}& W_t={\eta }_L{C}_t^{\sigma }{N}_t^{\varphi }\end{array}$$

消费

$$\begin{array}{}\text{(3)}& C_t=A_t{M}_t^{1/(\epsilon -1)}{N}_t^P\end{array}$$

利润

$$\begin{array}{}\text{(4)}& {\mathrm{\Pi }}_t=\frac{1}{\epsilon }\frac{C_t}{M_t}\end{array}$$

工资

$$\begin{array}{}\text{(5)}& W_t=(1-\frac{1}{\epsilon })A_t{M}_t^{1/(\epsilon -1)}\end{array}$$

企业数目

$$\begin{array}{}\text{(6)}& M_{t+1}=(1-\delta )(M_t+M_{t,E})\end{array}$$

企业进入条件

$$\begin{array}{}\text{(7)}& \frac{f_E{W}_t}{Z_t}=p_t\end{array}$$

劳动市场

$$\begin{array}{}\text{(8)}& N_t^P+\frac{M_{E,t}{f}_E}{Z_t}=N_t\end{array}$$