NK 模型

带货币的 RBC 模型

我们列出带货币的 RBC；MIU 模型和 CIA 模型，说明真实商业周期模型下依旧有以下几种可能。

1. 带有货币
2. 存在扭曲或者无效率
3. 存在内生的扭曲或无效率
4. 存在货币的非中性作用

但以上模型中即使有货币的供需问题，依旧是“价格中性”的。

Example 1

已知有如下的居民效用函数：${\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})$；$C_{t+k}$ 为 $t+k$ 时的居民消费，$N_{t+k}$ 为 $t+k$ 时的劳动供给，经济中并不存在实物资本。

已知此时经济中的生产技术如下 $Y_t=A_t{N}_t^{\alpha }$，市场出清为 $Y_t=C_t$，技术服从 $\ln A_{t+1}={\rho }_A\ln A_t+{\epsilon }_{t+1}$。

(1) 求解此时的中央计划者经济。

(2) 若满足效用函数 $u(C_t,N_t)=\frac{C_t^{1-\sigma }}{1-\sigma }-{\varphi }_L\frac{N_t^{1+\varphi }}{1+\varphi }$，求解上述方程的稳态与一阶差分（对数线性化）近似（当 ${\varphi }_L=\alpha$ 时）。

Example 3

考虑如下的分散化经济

居民效用

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

其中 $C_{t+k}$ 为 $t+k$ 时的居民消费，$N_{t+k}$ 为 $t+k$ 时的劳动供给，经济中并不存在实物资本。

居民融资约束

$$P_t{C}_t+Q_t{B}_{t+1}=B_t+W_t{N}_t+T_t$$

其中 $P_t$ 为消费品价格，$C_t$ 为消费品产量，$W_t$ 为名义工资，$N_t$ 为劳动供给，$T_t$ 为居民的转移支付，$B_t$ 为时刻 $t$ 通过国债获得的收益，$B_{t+1}$ 为 $t+1$ 时刻的国债收益，$Q_t$ 为时刻 $t+1$ 获得 $1$ 单位的名义收入的短期国债，在时刻 $t$ 出售的名义价格，因此 $R_{t+1}^N=\frac{1}{Q_t}$，$R_{t+1}^N$ 为债券的名义利率。

最终产品厂商利润

$${\mathrm{\Pi }}_t^F=P_t{Y}_t-W_t{N}_t$$

其中 $P_t$ 为最终产品价格，$Y_t$ 为最终产出，$W_t$ 为名义工资，$N_t$ 为劳动需求。

生产函数为

$$Y_t=A_t{N}_t^{\alpha }$$

市场出清方程

1. 实物市场产品出清$$Y_t=C_t$$
2. 国债市场出清$$B_t=0$$

冲击设定

$$\ln A_{t+1}=\rho \ln A_t+{\epsilon }_{t+1}^A$$

稳态下 $\ln A_{t+1}=\ln A_t=1$

（1） 写出居民的欧拉方程和劳动供给方程。

解：

居民值函数

$$\begin{array}{rl}V(B_t;Q_t,P_t,W_t,T_t)\triangleq max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{P}_t{C}_t+Q_t{B}_{t+1}=B_t+W_t{N}_t+T_t\end{array}$$$$\begin{array}{rl}V(B_t;Q_t,P_t,W_t,T_t)\triangleq max& u(C_t,N_t)+\beta {\mathbb{E}}_{t}V(B_{t+1};Q_{t+1},P_{t+1},W_{t+1},T_{t+1})\\ & \text{s.t.}{P}_t{C}_t+Q_t{B}_{t+1}=B_t+W_t{N}_t+T_t\end{array}$$

无约束最优化

$$\begin{array}{rl}V(B_t;Q_t,P_t,W_t,T_t)\triangleq \underset{C_t,N_t}{max}& u(C_t,N_t)+\beta {\mathbb{E}}_{t}V(B_{t+1};Q_{t+1},P_{t+1},W_{t+1},T_{t+1})\\ & -{\lambda }_t(P_t{C}_t+Q_t{B}_{t+1}-B_t-W_t{N}_t-T_t)\end{array}$$

FOCs

$$\frac{\partial u}{\partial C}(C_t,N_t)={\lambda }_t{P}_t$$$$\frac{\partial u}{\partial N}(C_t,N_t)=-{\lambda }_t{W}_t$$

可得劳动供给方程

$$-\frac{\partial u/\partial N(C_t,N_t)}{\partial u/\partial C(C_t,N_t)}=\frac{W_t}{P_t}$$$$\beta {\mathbb{E}}_t\frac{\partial V_{t+1}}{\partial B_{t+1}}(B_{t+1};S_{t+1})={\lambda }_t{Q}_t$$

包络定理

$$\frac{\partial V}{\partial B_t}(B_t;{\overrightarrow{S}}_t)={\lambda }_t$$$${\lambda }_t{Q}_t=\beta {\mathbb{E}}_t\frac{\partial V_{t+1}}{\partial B_{t+1}}(B_{t+1};{\overrightarrow{S}}_{t+1})=\beta {\mathbb{E}}_t{\lambda }_{t+1}\Rightarrow {\mathbb{E}}_t\beta \frac{{\lambda }_{t+1}}{{\lambda }_t}\frac{1}{Q_t}=1$$

代入 ${\lambda }_t=\frac{\partial u/\partial C(C_t,N_t)}{P_t}$ 可得欧拉方程（跨期投资方程）

$${\mathbb{E}}_t\frac{\partial u/\partial C(C_{t+1},N_{t+1})}{\partial u/\partial C(C_t,N_t)}\frac{P_t}{P_{t+1}}\frac{1}{Q_t}=1$$

结果

$$\begin{array}{}\text{(A-S)}& -\frac{\partial u/\partial N(C_t,N_t)}{\partial u/\partial C(C_t,N_t)}=\frac{W_t}{P_t}\end{array}$$$$\begin{array}{}\text{(Euler)}& {\mathbb{E}}_t\frac{\partial u/\partial C(C_{t+1},N_{t+1})}{\partial u/\partial C(C_t,N_t)}\frac{P_t}{P_{t+1}}\frac{1}{Q_t}=1\end{array}$$

（2） 结合企业的利润最大化方程和市场出清方程，分析经济中的实物变量，说明此方程是价格中性的。

解：

企业利润方程

$$max{\mathrm{\Pi }}_t=P_t{A}_t{N}_t^{\alpha }-W_t{N}_t$$

FOC

$$\begin{array}{}\text{(A-D)}& \frac{W_t}{P_t}=\frac{\partial Y_t}{\partial N_t}=\alpha A_t{N}_t^{\alpha -1}\end{array}$$

结合 (A-D) 和 (A-S) 有

$$-\frac{\partial u/\partial N(C_t,N_t)}{\partial u/\partial C(C_t,N_t)}=\frac{W_t}{P_t}=\frac{\partial Y_t}{\partial N_t}=\alpha A_t{N}_t^{\alpha -1}=\alpha \frac{Y_t}{N_t}$$

生产技术与市场出清有

$$Y_t=C_t=A_t{N}_t^{\alpha }$$

技术冲击方程

$$\ln A_{t+1}=\rho \ln A_t+{\epsilon }_{t+1}^A$$

说明此时的市场配置是独立于物价的。

（3） 若 $u(C_t,N_t)=\frac{C_t}{1-\sigma }-\alpha \frac{N_t}{1+\varphi }$，且 $W^{\ast }=P^{\ast }$，$Q^{\ast }=\beta$，求上述问题的对数线性化。

解：

对 ${\hat{y}}_t,{\hat{n}}_t,{\hat{c}}_t$ 类似问题2，可以获得

$$\begin{array}{}\text{(1)}& \varphi {\hat{n}}_t+\sigma {\hat{c}}_t={\hat{a}}_t+(\alpha -1){\hat{n}}_t\end{array}$$$$\begin{array}{}\text{(2)}& {\hat{c}}_t={\hat{y}}_t={\hat{a}}_t+\alpha {\hat{n}}_t\end{array}$$$$\begin{array}{}\text{(3)}& {\hat{a}}_{t+1}={\rho }_A{\hat{a}}_t+{\epsilon }_{t+1}^A\end{array}$$

同时有

$$\frac{W_t}{P_t}=\alpha \frac{Y_t}{A_t}\Rightarrow {\hat{w}}_t-{\hat{p}}_t={\hat{y}}_t-{\hat{a}}_t$$

对 ${\mathbb{E}}_t\frac{\partial u/\partial C(C_{t+1},N_{t+1})}{\partial u/\partial C(C_t,N_t)}\frac{P_t}{P_{t+1}}\frac{1}{Q_t}=1$，定义如下：

$${\hat{i}}_t=\ln R_{t+1}^N-\ln R^{N\ast }=\frac{R_{t+1}^N-R^{N\ast }}{R^{N\ast }}$$$${\hat{w}}_t=\ln W_t-\ln W^{\ast }=\frac{W_t-W^{\ast }}{W^{\ast }}$$$${\hat{p}}_t=\ln P_t-\ln P^{\ast }=\frac{P_t-P^{\ast }}{P^{\ast }}$$

可得

$$-\sigma {\mathbb{E}}_t{\hat{c}}_{t+1}+\sigma {\hat{c}}_t+{\hat{p}}_t-{\mathbb{E}}_t{\hat{p}}_{t+1}+{\hat{i}}_t=0$$

引入

$${\mathbb{E}}_t{\hat{p}}_{t+1}-{\hat{p}}_t\triangleq {\mathbb{E}}_t{\hat{\pi }}_{t+1}$$

因 $\ln \frac{P_{t+1}}{P_t}-\ln \frac{P^{\ast }}{P^{\ast }}=\ln {\pi }_{t+1}-\ln {\pi }^{\ast }={\hat{\pi }}_{t+1}$

可得

$${\hat{c}}_t={\mathbb{E}}_t{\hat{c}}_{t+1}-\frac{1}{\sigma }({\hat{i}}_t-{\mathbb{E}}_t{\hat{\pi }}_{t+1})$$

有

$${\hat{w}}_t-{\hat{p}}_t={\hat{y}}_t-{\hat{a}}_t$$$${\hat{c}}_t={\mathbb{E}}_t{\hat{c}}_{t+1}-\frac{1}{\sigma }({\hat{i}}_t-{\mathbb{E}}_t{\hat{\pi }}_{t+1})$$

Example 4 Simple MIU

居民效用

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

融资约束

$$P_t{C}_t+Q_t{B}_{t+1}+M_{t+1}=B_t+W_t{N}_t+T_t+M_t$$

企业利润

$${\mathrm{\Pi }}_t^F=P_t{Y}_t-W_t{N}_t$$

生产函数

$$Y_t=A_t{N}_t^{\alpha }$$

市场出清方程

1. 实物市场产品出清$$Y_t=C_t$$
2. 国债市场出清$$B_t=0$$

冲击设定

$$\ln A_{t+1}=\rho \ln A_t+{\epsilon }_{t+1}^A$$

稳态技术下

$$\ln A_{t+1}=\ln A_t=1$$

（1） 写出居民的欧拉方程和劳动供给方程、货币需求方程。

解：

$$\begin{array}{rl}V(B_t,M_{t-1};Q_t,P_t,W_t,T_t)\triangleq max& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k},\frac{M_{t+k}}{P_{t+k}})\\ & \text{s.t.}{M}_t+P_t{C}_t+Q_t{B}_{t+1}=B_t+W_t{N}_t+T_t+M_{t-1}\end{array}$$

FOCs

$$\frac{\partial u}{\partial C}={\lambda }_t{P}_t$$$$\frac{\partial u}{\partial N}=-{\lambda }_t{W}_t$$$$\beta {\mathbb{E}}_t\frac{\partial V_{t+1}}{\partial B_{t+1}}={\lambda }_t{Q}_t$$

可得劳动供给方程

$$-\frac{\partial u/\partial N}{\partial u/\partial C}=\frac{W_t}{P_t}$$

包络定理可得跨期投资方程

$${\mathbb{E}}_t\beta \frac{\partial u/\partial C(C_{t+1},N_{t+1},\frac{M_{t+1}}{P_{t+1}})}{\partial u/\partial C(C_t,N_t,\frac{M_t}{P_t})}\frac{1}{Q_t}=1$$

对于 FOC: 对 $M_t$ 有

$$\beta {\mathbb{E}}_t{V}_M(B_{t+1},M_t;{\overrightarrow{S}}_{t+1})+\frac{\partial u}{\partial (M/P)}(C_t,N_t,\frac{M_t}{P_t})\frac{1}{P_t}={\lambda }_t$$

包络： 对 $M_{t-1}$ 有

$$\frac{\partial V_t}{\partial M_{t-1}}(B_t,M_{t-1};\overrightarrow{S_t})={\lambda }_t$$

代入之后

$$\frac{\frac{\partial u}{\partial (M/P)}(C_t,N_t,\frac{M_t}{P_t})\frac{1}{P_t}}{\frac{\partial u}{\partial C}(C_t,N_t,\frac{M_t}{P_t})\frac{1}{P_t}}=1-{\mathbb{E}}_t\beta \frac{{\lambda }_{t+1}}{{\lambda }_t}=1-Q_t$$

可得货币的需求方程

$$\frac{\frac{\partial u}{\partial (M/P)}(C_t,N_t,\frac{M_t}{P_t})}{\frac{\partial u}{\partial C}(C_t,N_t,\frac{M_t}{P_t})}=1-Q_t$$

（2） 令 $u(C,N,\frac{M}{P})=\frac{C^{1-\sigma }}{1-\sigma }-\alpha \frac{N^{1+\varphi }}{1+\varphi }+\frac{(M/P{)}^{1-\sigma }}{1-\sigma }$，求实际产出和货币供给、需求方程。

解：

代入之后，除去货币需求方程外均与 3 中相同，稳态下 $C^{\ast }=Y^{\ast }=N^{\ast }=1$，同时又货币供给、需求方程为

$$\frac{M_t}{P_t}=C_t(1-Q_t{)}^{-1/\sigma }$$

从而有

$${\hat{m}}_t-{\hat{p}}_t={\hat{c}}_t-\frac{1}{\sigma }\frac{Q^{\ast }}{1-Q^{\ast }}{\hat{i}}_t$$

令 $\eta =\frac{1}{\sigma }\frac{Q^{\ast }}{1-Q^{\ast }}$ 即与 3 中货币外生供给相同，又 $Q^{\ast }=\beta$，有

$$\eta =\frac{{\beta }^{\ast }}{1-{\beta }^{\ast }}\frac{1}{\sigma }$$

(3) 若 $u(C_t,N_t,\frac{M_t}{P_t})={[(1-\theta )C_t^{1-r}+\theta {\left(\frac{M_t}{P_t}\right)}^{1-r}]}^{\frac{1-\sigma }{1-r}}-{\varphi }_L\frac{N_t^{1+\varphi }}{1+\varphi }$，若 ${\varphi }_L=\frac{\alpha }{1-\theta },\theta =\frac{1-\beta }{2-\beta }$，求稳态与对数线性化。

作为一个不可分效用的MIU模型，有两点启示：

1. RBC 模型未必是货币中性的，实际货币量 $M_t/P_t$ 对经济的配置亦有影响；从而用上述不可分情形分析货币政策的价格型和数量型议题是经典方案。
2. RBC 模型中，价格水平的作用是中性的，这里若无货币的工序方程 LM 曲线，要素是由实际货币供给变动和技术冲击、需求冲击等引起变化的。

NK 基准模型

居民

$$\begin{array}{rl}& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\beta }^{k}u(C_{t+k},N_{t+k})\\ & \text{s.t.}{P}_t{C}_t+B_{t+1}{Q}_t=B_t+W_t{N}_t+T_t\end{array}$$

最终产品厂商

$$Y_t={[{\int }_0^1{y}_t(i{)}^{1-\frac{1}{\epsilon }}di]}^{\epsilon /(\epsilon -1)}$$$${\mathrm{\Pi }}_t^F=\underset{\{y_t(i){\}}_{i\in [0,1]}}{max}{P}_t{Y}_t-{\int }_0^1{p}_t(i)y_t(i)di$$

每个时刻有 $(1-\theta )$ 的企业可调价，有 $\theta$ 的不可调价。

（1） 写出居民和最终厂商的最终条件。

解：

见题 3 有居民的最终行为条件如下

$$\{\begin{array}{l}{\mathbb{E}}_t\frac{\partial u/\partial C(C_{t+1},N_{t+1})}{\partial u/\partial C(C_t,N_t)}\frac{P_t}{P_{t+1}}\frac{1}{Q_t}=1\\ -\frac{\partial u/\partial N(C_t,N_t)}{\partial u/\partial C(C_t,N_t)}=\frac{W_t}{P_t}\end{array}$$

最终企业的最终条件：见题 5

$$\begin{array}{rl}{\mathrm{\Pi }}_t^F=\underset{\{y_t(i)\}}{max}& P_t{Y}_t-{\int }_0^1{p}_t(i)y_t(i)di\\ & Y_t={[{\int }_0^1{y}_t(i{)}^{1-\frac{1}{\epsilon }}di]}^{\frac{\epsilon }{\epsilon -1}}\end{array}$$

FOC 可得

$$\begin{gathered} \frac{\epsilon }{\epsilon -1}{[{\int }_0^1{y}_t(i)di]}^{\frac{\epsilon }{\epsilon -1}-1}\times P_t\times (1-\frac{1}{\epsilon })y_t(i{)}^{-\frac{1}{\epsilon }}=p_t(i) \\ \Rightarrow P_t{Y}_t^{\frac{1}{\epsilon }}{y}_t(i{)}^{-\frac{1}{\epsilon }}=p_t(i)\Rightarrow \frac{y_t(i)}{Y_t}={\left(\frac{p_t(i)}{P_t}\right)}^{-\epsilon } \\ \text{ 又有 }{P}_t{Y}_t={\int }_0^1{p}_t(i)y_t(i)di={\int }_0^1{p}_t(i){\left(\frac{p_t(i)}{P_t}\right)}^{-\epsilon }{Y}_{t}di \\ \Rightarrow P_t^{1-\epsilon }={\int }_0^1{p}_t(i{)}^{1-\epsilon }di \end{gathered}$$

从而可得

$$y_t(i)=(p_t(i)/P_t{)}^{-\epsilon }{Y}_t$$$$P_t={[{\int }_0^1{p}_t(i{)}^{1-\epsilon }di]}^{1/(1-\epsilon )}$$

（2） 当 $u(C,N)=\frac{C^{1-\sigma }}{1-\sigma }-\alpha \frac{N^{1+\varphi }}{1+\varphi }$，$y_t(i)=A_t{n}_t(i{)}^{\alpha }$时，写出在时刻 t 可以调价的企业的最大化目标和最终条件。

解：

$$T{C}_{t+k}(p_t^{\ast }(i))\triangleq {\left[\frac{(p_t^{\ast }(i)/P_{t+k}{)}^{-\epsilon }{Y}_{t+k}}{A_{t+k}}\right]}^{1/\alpha }{W}_{t+k}$$$$M_{t,t+k}^N={\beta }^k{\left(\frac{C_{t+k}}{C_t}\right)}^{-\sigma }\frac{P_t}{P_{t+k}}$$

企业的最大化目标

$$\underset{p_t^{\ast }(i)}{max}{\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\theta }^k{M}_{t,t+k}^N[{\left(\frac{p_t^{\ast }(i)}{P_{t+k}}\right)}^{1-\epsilon }{P}_{t+k}{Y}_{t+k}-T{C}_{t+k}(p_t^{\ast }(i))]$$

最终表示如下

$$\{{\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\theta }^k{M}_{t,t+k}^N{P}_{t+k}^{\epsilon }{Y}_{t+k}{p}_t^{\ast }(i{)}^{-\epsilon }\}{p}_t^{\ast }(i)$$$$=\frac{1}{\alpha }\frac{\epsilon }{\epsilon -1}\sum _{k=0}^{+\mathrm{\infty }}{\theta }^k{M}_{t,t+k}^N{W}_{t+k}{\left[\frac{(p_t^{\ast }(i)/P_{t+k}{)}^{-\epsilon }{Y}_{t+k}}{A_{t+k}}\right]}^{1/\alpha }$$

由求解可知，$p_t^{\ast }(i)$ 对 $\mathrm{\forall }i$ 可调价均相同，后面记为 $p_t^{\ast }$。

记 $T{C}_{t+k}(p_t^{\ast })=T{C}_{t+k}(y_{t+k|t}(p_t^{\ast })$，可得

$$\frac{dT{C}_{t+k}}{d{p}_t^{\ast }}=\frac{dT{C}_{t+k}}{d{y}_{t+k|t}}\frac{d{y}_{t+k|t}}{d{p}_t^{\ast }}$$$$M{C}_{t+k|t}^N\triangleq \frac{\partial T{C}_{t+k}}{\partial y_{t+k|t}}$$

因

$$y_{t+k|t}=P_{t+k}^{\epsilon }{Y}_{t+k}{p_t^{\ast }}^{-\epsilon }$$

有

$$M{C}_{t+k|t}^N=W_{t+k}\frac{1}{\alpha }(\frac{y_{t+k|t}}{A_{t+k}}{)}^{1/\alpha -1}\frac{1}{A_{t+k}}$$

上述表示可写为

$$\left({\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\theta }^k{M}_{t,t+k}^N{y}_{t+k|t}\right)p_t^{\ast }(i)=\frac{\epsilon }{\epsilon -1}{\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}{\theta }^k{M}_{t,t+k}^N{y}_{t+k|t}M{C}_{t+k|t}^N$$

（3） 因为 $N_t={\int }_0^1{n}_t(i)di$，写出此时的劳动需求方程。

解：

$$y_t(i)={\left(\frac{p_t(i)}{P_t}\right)}^{-\epsilon }{Y}_t$$$$n_t(i)={\left(\frac{y_t(i)}{A_t}\right)}^{1/\alpha }$$$$n_t(i)=\frac{p_t(i{)}^{-\epsilon /\alpha }}{P_t^{-\epsilon /\alpha }}{\left(\frac{Y_t}{A_t}\right)}^{1/\alpha }$$$$N_t={\int }_0^1{n}_t(i)di={\int }_0^1\frac{p_t(i{)}^{-\epsilon /\alpha }}{P_t^{-\epsilon /\alpha }}di\times {\left(\frac{Y_t}{A_t}\right)}^{1/\alpha }$$

（4） 已知 $\ln A_{t+1}=\rho \ln A_t+{\epsilon }_{t+1}^A$。写出上述问题下的市场出清条件，并给出当 $P_{t+1}=P_t=1$，$A_{t+1}=A_t=1$，且无垄断时的稳态解。

解：

居民的劳动供给

$$\begin{array}{}\text{(1)}& \alpha C_t^{\sigma }{N}_t^{\varphi }=\frac{W_t}{P_t}\end{array}$$

居民的跨期行为

$$\begin{array}{}\text{(2)}& {\mathbb{E}}_t\beta {\left(\frac{C_{t+1}}{C_t}\right)}^{-\sigma }\frac{P_t}{P_{t+1}}\frac{1}{Q_t}=1\end{array}$$

最终产品定价

$$\begin{array}{}\text{(3)}& P_t^{1-\epsilon }={\int }_0^1{p}_t(i{)}^{1-\epsilon }di=\theta P_{t-1}^{1-\epsilon }+(1-\theta ){P_t^{\ast }}^{1-\epsilon }\end{array}$$

市场出清

$$\begin{array}{}\text{(4)}& Y_t=C_t=\frac{A_t{N}_t^{\alpha }}{d_t}\end{array}$$

扭曲定义

$$\begin{array}{}\text{(5)}& d_t={\left[{\int }_0^1{\left(\frac{p_t(i)}{P_t}\right)}^{-\epsilon /\alpha }di\right]}^{\alpha }\end{array}$$

最优定价行为

$$\begin{array}{}\text{(6)}& P_t^{\ast }{\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k{\left(\frac{C_{t+k}}{C_t}\right)}^{-\sigma }\frac{P_t}{P_{t+k}}{y}_{t+k|t}=\frac{\epsilon }{\epsilon -1}{\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k{\left(\frac{C_{t+k}}{C_t}\right)}^{-\sigma }\frac{P_t}{P_{t+k}}{y}_{t+k|t}M{P}_{t+k|t}^N\end{array}$$

定义产出

$$\begin{array}{}\text{(7)}& y_{t+k|t}={\left(\frac{P_t^{\ast }}{P_{t+k}}\right)}^{-\epsilon }{Y}_{t+k}\end{array}$$

定义名义成本

$$\begin{array}{}\text{(8)}& M{C}_{t+k|t}^N=W_{t+k}\frac{1}{\alpha }(\frac{y_{t+k|t}}{A_{t+k}}{)}^{1/\alpha -1}\frac{1}{A_{t+k}}\end{array}$$

泰勒规则

略给定

冲击设定

$$\ln A_{t+1}={\rho }_A\ln A_t+{\epsilon }_{t+1}^A$$

$\ln A_t\equiv 0$，$P_{t+1}\equiv 1$ 时可得稳态

$$C^{\ast }=Y^{\ast }=d^{\ast }=P^{\ast }=p_t^{\ast }(i)=P_t^{\ast }=1$$$$W^{\ast }=\alpha$$$$Q^{\ast }=\beta$$

（5） 对 $d_t$ 对数线性化，并结合 (4) 和 (2) 获得 I-S 曲线。

解：

$${\hat{p}}_t={\int }_0^1{\left(\frac{P_t^{\ast }(i)}{P^{\ast }}\right)}^{1-\epsilon }{\hat{p}}_t(i)di={\int }_0^1{\hat{p}}_t(i)di$$$$d_t={\left[{\int }_0^1{\left(\frac{p_t(i)}{P_t}\right)}^{-\epsilon /\alpha }di\right]}^{\alpha }$$$${\hat{d}}_t=-\frac{\epsilon }{d^{\ast }}({\int }_0^1{\hat{p}}_t(i)di-{\hat{P}}_t)=0$$

将 ${\hat{y}}_t={\hat{c}}_t$ 代入 (2) 可以获得

$${\mathbb{E}}_t{\hat{y}}_{t+1}-({\hat{i}}_t-{\mathbb{E}}_t{\hat{\pi }}_{t+1})\frac{1}{\sigma }={\hat{y}}_t$$$${\tilde{y}}_t\triangleq {\hat{y}}_t-{\hat{y}}_t^n$$

又因 ${\hat{y}}_t={\rho }_y{\hat{a}}_t$，而 ${\hat{r}}_t^n=\sigma {\rho }_y[{\mathbb{E}}_t{\hat{a}}_{t+1}-{\hat{a}}_t]$ 代入之后可得 I-S 曲线

$${\tilde{y}}_t={\mathbb{E}}_t{\tilde{y}}_{t+1}-\frac{1}{\sigma }({\hat{i}}_t-{\mathbb{E}}_t{\hat{\pi }}_{t+1}-{\hat{r}}_t^n)$$

（6） 定义 $M{C}_{t+k|t}^N/P_{t+k}=M{C}_{t+k|t}$，用 ${\widehat{mc}}_{t+k|t}$ 与 ${\hat{p}}_{t+k}-{\hat{p}}_{t-1}$ 表示 ${\hat{p}}_t^{\ast }-{\hat{p}}_{t-1}$。

解：

对于 (6)，可以获得

$$\begin{array}{rl}& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )(-\sigma {\hat{c}}_{t+k}+\sigma {\hat{c}}_t+{\hat{p}}_t-{\hat{p}}_{t+k}+{\hat{y}}_{t+k|t}+{\hat{p}}_t^{\ast }\\ =& {\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )(-\sigma {\hat{c}}_{t+k}+\sigma {\hat{c}}_t+{\hat{p}}_t-{\hat{p}}_{t+k}+{\hat{y}}_{t+k|t}+{\widehat{mc}}_{t+k|t}^N)\end{array}$$

可得

$${\hat{p}}_t^{\ast }={\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )({\widehat{mc}}_{t+k|t}^N)$$$${\hat{p}}_t^{\ast }-{\hat{p}}_{t-1}={\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )({\widehat{mc}}_{t+k|t})+{\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )({\hat{p}}_{t+k}-{\hat{p}}_{t-1})$$

（7） 因为 $M{C}_{t+k|t}={\left(\frac{P_t^{\ast }}{P_{t+k}}\right)}^{-\frac{(1-\alpha )\epsilon }{\alpha }}M{C}_{t+k}$，将上式结合 $P_t^{1-\epsilon }=(1-\theta )(P_t^{\ast }{)}^{1-\epsilon }+\theta P_{t-1}^{1-\epsilon }$ 获得通胀与物价的关系。

解：

$$\begin{array}{rl}& P_t^{1-\epsilon }=(1-\theta )(P_t^{\ast }{)}^{1-\epsilon }+\theta P_{t-1}^{1-\epsilon }\\ \Rightarrow & (1-\epsilon ){\hat{p}}_t=[(1-\theta ){\hat{p}}_t^{\ast }+\theta {\hat{p}}_{t-1}](1-\epsilon )\\ \Rightarrow & {\hat{p}}_t-{\hat{p}}_{t-1}=(1-\theta )({\hat{p}}_t^{\ast }-{\hat{p}}_{t-1})\\ \Rightarrow & {\hat{p}}_t^{\ast }-{\hat{p}}_{t-1}=\frac{1}{1-\theta }({\hat{p}}_t-{\hat{p}}_{t-1})\end{array}$$

而

$${\widehat{mc}}_{t+k|t}={\widehat{mc}}_{t+k}-\frac{(1-\alpha )\epsilon }{\alpha }({\hat{p}}_t^{\ast }-{\hat{p}}_{t+k})$$$$\begin{gathered} M{C}_{t+k|t}=\frac{W_{t+k}}{P_{t+k}}{\left(\frac{P_t^{\ast }(i{)}^{-\epsilon }{P}_{t+k}^{\epsilon }{Y}_{t+k}}{A_{t+k}}\right)}^{1/\alpha -1}\frac{1}{\alpha }\frac{1}{A_{t+k}} \\ =\underset{M{C}_{t+k}}{\underbrace{\frac{1}{A_{t+k}}\frac{W_{t+k}}{P_{t+k}}{\left(\frac{Y_{t+k}}{A_{t+k}}\right)}^{1/\alpha -1}\frac{1}{\alpha }}}{\left[\frac{p_t^{\ast }(i)}{P_{t+k}}\right]}^{-\epsilon (1/\alpha -1)} \end{gathered}$$

可得

$$\begin{gathered} {\widehat{mc}}_{t+k|t}={\widehat{mc}}_{t+k}-\frac{(1-\alpha )\epsilon }{\alpha }({\hat{p}}_t^{\ast }-{\hat{p}}_{t+k}) \\ ={\widehat{mc}}_{t+k}-\frac{(1-\alpha )\epsilon }{\alpha }({\hat{p}}_t^{\ast }-{\hat{p}}_{t-1})+\frac{(1-\alpha )\epsilon }{\alpha }({\hat{p}}_{t+k}-{\hat{p}}_{t-1}) \end{gathered}$$

代入 ${\hat{p}}_t^{\ast }-{\hat{p}}_{t-1}={\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )({\widehat{mc}}_{t+k|t})+{\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )({\hat{p}}_{t+k}-{\hat{p}}_{t-1})$ 可得

$${\hat{p}}_t^{\ast }-{\hat{p}}_{t-1}={\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta ){\widehat{mc}}_{t+k}-\frac{(1-\alpha )\epsilon }{\alpha }({\hat{p}}_t^{\ast }-{\hat{p}}_{t-1})+{\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )[1+\frac{(1-\alpha )\epsilon }{\alpha }]({\hat{p}}_{t+k}-{\hat{p}}_{t-1})$$

令 $\mathrm{\Theta }=\frac{\alpha }{\alpha +(1-\alpha )\epsilon }\le 1$，可得

$${\hat{p}}_t^{\ast }-{\hat{p}}_{t-1}={\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta ){\widehat{mc}}_{t+k}\mathrm{\Theta }+{\mathbb{E}}_t\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )({\hat{p}}_{t+k}-{\hat{p}}_{t-1})$$

有

$${\hat{p}}_{t+1}^{\ast }-{\hat{p}}_t={\mathbb{E}}_{t+1}\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta ){\widehat{mc}}_{t+1+k}\mathrm{\Theta }+{\mathbb{E}}_{t+1}\sum _{k=0}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )({\hat{p}}_{t+k+1}-{\hat{p}}_t)$$$${\mathbb{E}}_t({\hat{p}}_{t+1}^{\ast }-{\hat{p}}_t)\theta \beta ={\mathbb{E}}_t\sum _{k=1}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )\mathrm{\Theta }{\widehat{mc}}_{t+k}+{\mathbb{E}}_t\sum _{k=1}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )({\hat{p}}_{t+k}-{\hat{p}}_t)$$

又因为

$$\begin{gathered} {\hat{p}}_t^{\ast }-{\hat{p}}_{t-1}=(1-\theta \beta )\mathrm{\Theta }{\widehat{mc}}_t+(1-\theta \beta )({\hat{p}}_t-{\hat{p}}_{t-1})+{\mathbb{E}}_t\sum _{k=1}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )\mathrm{\Theta }{\widehat{mc}}_{t+k}+{\mathbb{E}}_t\sum _{k=1}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )({\hat{p}}_{t+k}-{\hat{p}}_{t-1}) \\ =(1-\theta \beta )\mathrm{\Theta }\widehat{mc}+({\hat{p}}_t-{\hat{p}}_{t-1})+{\mathbb{E}}_t\sum _{k=1}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )\mathrm{\Theta }{\widehat{mc}}_{t+k}+{\mathbb{E}}_t\sum _{k=1}^{+\mathrm{\infty }}(\theta \beta {)}^k(1-\theta \beta )({\hat{p}}_{t+k}-{\hat{p}}_t) \end{gathered}$$

从而有

$${\hat{p}}_t^{\ast }-{\hat{p}}_{t-1}=(1-\theta \beta )\mathrm{\Theta }\widehat{mc}+({\hat{p}}_t-{\hat{p}}_{t-1})+{\mathbb{E}}_t\theta \beta ({\hat{p}}_{t+1}^{\ast }-{\hat{p}}_t)$$

有

$${\hat{p}}_t^{\ast }-{\hat{p}}_{t-1}=\theta \beta {\mathbb{E}}_t({\hat{p}}_{t+1}^{\ast }-{\hat{p}}_t)+{\hat{\pi }}_t+(1-\theta \beta )\mathrm{\Theta }{\widehat{mc}}_t$$

又因为

$${\hat{p}}_t^{\ast }-{\hat{p}}_{t-1}=\frac{1}{1-\theta }({\hat{p}}_t-{\hat{p}}_{t-1})=\frac{1}{1-\theta }{\hat{\pi }}_t$$$$\frac{1}{1-\theta }{\hat{\pi }}_t=\theta \beta {\mathbb{E}}_t{\hat{\pi }}_{t+1}\frac{1}{1-\theta }+{\hat{\pi }}_t+(1-\theta \beta )\mathrm{\Theta }{\widehat{mc}}_t$$$$\frac{\theta }{1-\theta }{\hat{\pi }}_t=\frac{\theta }{1-\theta }\beta {\mathbb{E}}_t{\hat{\pi }}_{t+1}+(1-\theta \beta )\mathrm{\Theta }{\widehat{mc}}_t$$$${\hat{\pi }}_t=\beta {\mathbb{E}}_t{\hat{\pi }}_{t+1}+\frac{(1-\theta \beta )\mathrm{\Theta }(1-\theta )}{\theta }{\widehat{mc}}_t$$

令 $\lambda \triangleq \frac{(1-\theta \beta )\mathrm{\Theta }(1-\theta )}{\theta }$ 可得

$${\pi }_t=\beta {\mathbb{E}}_t{\hat{\pi }}_{t+1}+\lambda {\widehat{mc}}_t$$

（8） 因 $Y_t=A_t{N}_t^{\alpha }/d_t$，且 $M{C}_t=\frac{W_t}{P_t}\frac{\partial N_t}{\partial Y_t}=\frac{1}{\alpha }\frac{W_t}{P_t}{\left(\frac{d_t{Y}_t}{A_t}\right)}^{1/\alpha -1}\frac{d_t}{A_t}$，结合 $\frac{\epsilon -1}{\epsilon }\alpha N_t^{\varphi }{C}_t^{\sigma }=W_t/P_t$，分析上述方程。

解：

因

$${\widehat{mc}}_t={\hat{w}}_t-{\hat{p}}_t+(\frac{1}{\alpha }-1)({\hat{y}}_t-{\hat{a}}_t+{\hat{d}}_t{)}_{+}{\hat{d}}_t-{\hat{a}}_t$$$${\hat{d}}_t=0$$

可得

$$\{\begin{array}{l}{\widehat{mc}}_t={\hat{w}}_t-{\hat{p}}_t-(1-\frac{1}{\alpha })({\hat{y}}_t-{\hat{a}}_t)-{\hat{a}}_t\\ {\hat{y}}_t={\hat{a}}_t+\alpha {\hat{n}}_t={\hat{c}}_t\\ {\hat{w}}_t-{\hat{p}}_t=\varphi {\hat{n}}_t+\sigma {\hat{c}}_t\end{array}$$

可得

$$\begin{array}{rl}{\widehat{mc}}_t& =({\hat{w}}_t-{\hat{p}}_t)-(1-\frac{1}{\alpha })({\hat{y}}_t-{\hat{a}}_t)-{\hat{a}}_t\\ & =\varphi {\hat{n}}_t+\sigma {\hat{c}}_t-(1-\frac{1}{\alpha })({\hat{y}}_t-{\hat{a}}_t)-{\hat{a}}_t\\ & =\frac{\varphi }{\alpha }({\hat{y}}_t-{\hat{a}}_t)+\sigma {\hat{y}}_t-(1-\frac{1}{\alpha })({\hat{y}}_t-{\hat{a}}_t)-{\hat{a}}_t\\ & =(\sigma +\frac{\varphi +1-\alpha }{\alpha }){\hat{y}}_t-\left(\frac{\varphi +1}{\alpha }\right)\widehat{a_t}\end{array}$$

又因为

$$\frac{\varphi +1}{\alpha }{\hat{a}}_t=(\sigma +\frac{\varphi +1-\alpha }{\alpha }){\hat{y}}_t^n$$$${\rho }_m\triangleq \sigma +\frac{\varphi +1-\alpha }{\alpha }$$$${\widehat{mc}}_t={\rho }_m({\hat{y}}_t-{\hat{y}}_t^n)$$

从而可得菲利普斯曲线

$${\hat{\pi }}_t=\beta {\mathbb{E}}_t{\hat{\pi }}_{t+1}+\lambda {\widehat{mc}}_t=\beta {\mathbb{E}}_t{\hat{\pi }}_{t+1}+\underset{\triangleq \kappa }{\underbrace{\lambda {\rho }_m}}\underset{\triangleq {\tilde{y}}_t}{\underbrace{({\hat{y}}_t-{\hat{y}}_t^n)}}$$

从而可得 I-S 曲线

$${\tilde{y}}_t={\mathbb{E}}_t{\tilde{y}}_{t+1}-\frac{1}{\sigma }({\hat{i}}_t-{\mathbb{E}}_t{\hat{\pi }}_{t+1}-{\hat{r}}_t^n)$$

P-C 曲线

$${\hat{\pi }}_t=\beta {\mathbb{E}}_t{\hat{\pi }}_{t+1}+\kappa {\tilde{y}}_t$$

LM 曲线用 Taylor 规则替代

基本型

$${\hat{i}}_t={\varphi }_{\pi }{\hat{\pi }}_t+{\varphi }_y{\tilde{y}}_t$$

前瞻型

$${\hat{i}}_t={\varphi }_{\pi }^F{\mathbb{E}}_t{\hat{\pi }}_{t+1}+{\varphi }_y^F{\mathbb{E}}_t{\tilde{y}}_{t+1}$$

(9) 居民的福利变动

$$u(C_t,N_t)-u(C^{\ast },N^{\ast })=u_C\overline{C}\frac{C_t-\overline{C}}{C}+\dots$$