DSGE-非李嘉图主体

非李嘉图主体是无法借贷、没有投资（收入全部用于消费）的消费者。假设消费者中比例为 $ω$ 的部分是李嘉图主体，比例为 $1 - ω$ 的部分是非李嘉图主体。

消费者

李嘉图消费者

和 DSGE Model 完全一致，均衡条件为

\begin{aligned} \frac{1 - γ}{γ} \frac{C_{i, t}}{1 - L_{i, t}} & = w_{t} \\ \frac{C_{i, t}}{C_{i, t - 1}} & = β (1 + r_{t} - δ) \end{aligned}

非李嘉图消费者

效用最大化问题为

\begin{aligned} max & E_{t} \sum_{t = 0}^{\infty} β^{t} U (C_{j, t}, O_{j, t}) \\ s.t. & U (C_{j, t}, O_{j, t}) = γ \ln C_{j, t} + \ln (1 - γ) O_{j, t} \\ O_{j, t} + L_{j, t} \equiv 1 \\ C_{j, t} = w_{t} L_{j, t} \end{aligned}

等价于

\begin{aligned} max & E_{t} \sum_{t = 0}^{\infty} β^{t} [γ \ln C_{j, t} + (1 - γ) \ln (1 - L_{j, t})] \\ s.t. & C_{j, t} = w_{t} L_{j, t} \end{aligned}

Lagrangian is

\begin{aligned} L = & E_{t} {\sum_{t = 0}^{\infty} β^{t} [γ \ln C_{j, t} + (1 - γ) \ln (1 - L_{j, t})] + λ_{t} [w_{t} L_{j, t} - C_{j, t}]} \end{aligned}

F.O.C

\begin{aligned} \frac{\partial L}{\partial C_{j, t}} & = β^{t} [\frac{γ}{C_{j, t}} - λ_{t}] = 0 \\ \frac{\partial L}{\partial L_{j, t}} & = β^{t} [- \frac{1 - γ}{1 - L_{j, t}} + λ_{t} w_{t}] = 0 \end{aligned}

which implies

\frac{1 - γ}{γ} \frac{C_{j, t}}{1 - L_{j, t}} = w_{t}

消费者加总

假设

\begin{aligned} C_{t} & \equiv ω C_{i, t} + (1 - ω) C_{j, t} \\ L_{t} & \equiv ω L_{i, t} + (1 - ω) L_{j, t} \\ K_{t} & \equiv ω K_{i, t} \\ I_{t} & \equiv ω I_{i, t} \end{aligned}

厂商

和 [DSGE Model]]完全一致

一般均衡

\begin{aligned} (M1) & \frac{1 - γ}{γ} \frac{C_{i, t}}{1 - L_{i, t}} & = w_{t} \\ (M2) & \frac{C_{i, t}}{C_{i, t - 1}} & = β (1 + r_{t} - δ) \\ (M3) & \frac{1 - γ}{γ} \frac{C_{j, t}}{1 - L_{j, t}} & = w_{t} \\ (M4) & C_{t} & \equiv ω C_{i, t} + (1 - ω) C_{j, t} \\ (M5) & L_{t} & \equiv ω L_{i, t} + (1 - ω) L_{j, t} \\ (M6) & K_{t} & \equiv ω K_{i, t} \\ (M7) & I_{t} & \equiv ω I_{i, t} \\ (M8) & r_{t} & = α A_{t} K_{t}^{α - 1} L_{t}^{1 - α} = \frac{α Y_{t}}{K_{t}} \\ (M9) & w_{t} & = (1 - α) A_{t} K_{t}^{α} L_{t}^{- α} = \frac{(1 - α) Y_{t}}{L_{t}} \\ (M10) & Y_{t} & = C_{t} + I_{t} \\ (M11) & Y_{t} & = A_{t} K_{t}^{α} L_{t}^{1 - α} \\ (M12) & I_{i, t} & = K_{i, t + 1} - (1 - δ) K_{i, t} \end{aligned}

此外假设

\ln A_{t} = (1 - ρ_{A}) \ln \bar{A} + ρ_{A} \ln A_{t - 1} + ε_{t}^{A}, ε_{t}^{A} \sim N (0, σ_{A}^{2})