DSGE-公共支出

消费者

假设消费者同时受益于私人消费和公共消费。

效用最大化问题为

\begin{aligned} max_{{C_{P, t}, L_{t}, K_{t}}} & E_{t} \sum_{t = 0}^{\infty} β^{t} U (B_{t}, O_{t}) \\ s.t. & U (B_{t}, O_{t}) = γ \ln B_{t} + \ln (1 - γ) O_{t} \\ B_{t} = [ω C_{P, t}^{η} + (1 - ω) C_{G, t}^{η}]^{\frac{1}{η}} \\ O_{t} + L_{t} \equiv 1 \\ (1 + τ_{t}^{c}) C_{P, t} + I_{t} = (1 - τ_{t}^{l}) w_{t} L_{t} + [(1 - τ_{t}^{k}) r_{t} + τ_{t}^{k} δ] K_{t} + G_{t} \\ I_{t} = K_{t + 1} - (1 - δ) K_{t} \end{aligned}

等价于

\begin{aligned} max & E_{t} \sum_{t = 0}^{\infty} β^{t} {γ \ln [ω C_{P, t}^{η} + (1 - ω) C_{G, t}^{η}]^{\frac{1}{η}} + (1 - γ) \ln (1 - L_{t})} \\ s.t. & (1 + τ_{t}^{c}) C_{P, t} + K_{t + 1} = (1 - τ_{t}^{l}) w_{t} L_{t} + [1 + (1 - τ_{t}^{k}) (r_{t} - δ)] K_{t} + G_{t} \end{aligned}

Lagrangian is

\begin{aligned} L = & E_{t} (\sum_{t = 0}^{\infty} β^{t} {γ \ln [ω C_{P, t}^{η} + (1 - ω) C_{G, t}^{η}]^{\frac{1}{η}} + (1 - γ) \ln (1 - L_{t})} \\ + λ_{t} [(1 - τ_{t}^{l}) w_{t} L_{t} + [1 + (1 - τ_{t}^{k}) (r_{t} - δ)] K_{t} + G_{t} - (1 + τ_{t}^{c}) C_{P, t} - K_{t + 1}]) \end{aligned}

F.O.C.

\begin{aligned} (1) & \frac{\partial L}{\partial C_{P, t}} & = β^{t} [\frac{γ ω C_{P, t}^{η - 1}}{[ω C_{P, t}^{η} + (1 - ω) C_{G, t}^{η}]^{\frac{1}{η}}} - λ_{t} (1 + τ_{t}^{c})] = 0 \\ (2) & \frac{\partial L}{\partial L_{t}} & = β^{t} [- \frac{1 - γ}{1 - L_{t}} + λ_{t} (1 - τ_{t}^{l}) w_{t}] = 0 \\ (3) & \frac{\partial L}{\partial K_{t}} & = β^{t} λ_{t} [1 + (1 - τ_{t}^{k}) (r_{t} - δ)] + β^{t - 1} λ_{t - 1} (- 1) = 0 \end{aligned}

由 $(1) (2)$ 可得

\frac{\frac{1 - γ}{1 - L_{t}}}{\frac{γ ω C_{P, t}^{η - 1}}{[ω C_{P, t}^{η} + (1 - ω) C_{G, t}^{η}]^{\frac{1}{η}}}} = \frac{(1 - τ_{t}^{l}) w_{t}}{1 + τ_{t}^{c}} ⟹ \frac{1 - γ}{γ} \frac{[ω C_{P, t}^{η} + (1 - ω) C_{G, t}^{η}]^{\frac{1}{η}}}{(1 - L_{t}) γ ω C_{P, t}^{η - 1}} = \frac{(1 - τ_{t}^{l}) w_{t}}{1 + τ_{t}^{c}}

由 $(1)$ 得到欧拉方程再利用 $(3)$ 消去 $λ$ 可得

\frac{\frac{C_{P, t - 1}^{η - 1}}{(1 + τ_{t - 1}^{c}) [ω C_{P, t - 1}^{η} + (1 - ω) C_{G, t - 1}^{η}]^{\frac{1}{η}}}}{\frac{C_{P, t}^{η}}{(1 + τ_{t}^{c}) [ω C_{P, t}^{η} + (1 - ω) C_{G, t}^{η}]^{\frac{1}{η}}}} = β [1 + (1 - τ^{k} t) (r_{t} - δ)]

和 DSGE Model 完全一致

τ_{t}^{c} C_{P, t} + τ_{t}^{l} w_{t} L_{t} + τ_{t}^{k} (r_{t} - δ) K_{t} = G_{t} + C_{G, t}

\begin{aligned} (M1) & \frac{1 - γ}{γ} \frac{[ω C_{P, t}^{η} + (1 - ω) C_{G, t}^{η}]^{\frac{1}{η}}}{(1 - L_{t}) γ ω C_{P, t}^{η - 1}} & = \frac{(1 - τ_{t}^{l}) w_{t}}{1 + τ_{t}^{c}} \\ (M2) & \frac{\frac{C_{P, t - 1}^{η - 1}}{(1 + τ_{t - 1}^{c}) [ω C_{P, t - 1}^{η} + (1 - ω) C_{G, t - 1}^{η}]^{\frac{1}{η}}}}{\frac{C_{P, t}^{η}}{(1 + τ_{t}^{c}) [ω C_{P, t}^{η} + (1 - ω) C_{G, t}^{η}]^{\frac{1}{η}}}} & = β [1 + (1 - τ^{k} t) (r_{t} - δ)] \\ (M3) & r_{t} & = α A_{t} K_{t}^{α - 1} L_{t}^{1 - α} = \frac{α Y_{t}}{K_{t}} \\ (M4) & w_{t} & = (1 - α) A_{t} K_{t}^{α} L_{t}^{- α} = \frac{(1 - α) Y_{t}}{L_{t}} \\ (M5) & Y_{t} & = C_{P, t} + C_{G, t} + I_{t} \\ (M6) & Y_{t} & = A_{t} K_{t}^{α} L_{t}^{1 - α} \\ (M7) & I_{t} & = K_{t + 1} - (1 - δ) K_{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})