\documentclass[a4paper,11pt]{article}
\title{Leisure RBC model}
\author{Eric Scheffel}
\begin{document}

\maketitle



The problem is set up as a social planner's problem in the following way:

\begin{eqnarray*}
L & = & E_0\sum_{t=0}^\infty\beta^t\left\lbrace\frac{\left(c_t\left(1-l_t\right)^A\right)^{\left(1-\eta\right)}}{1-\eta}\right\rbrace\\
  & + & \lambda_t\left[z_tk_t^\rho l_t^{1-\rho}-c_t+\left(1-\delta\right)k_t-k_{t+1}\right]
\end{eqnarray*}
The shock process is given by:
\[z_{t+1}=\bar{z}^{1-\psi}z_t^\psi\epsilon_{t+1}\]
Market equilibrium requires:
\[z_tk_t^\rho l_t^{1-\rho} = c_t+k_{t+1}-\left(1-\delta\right)k_t\]
which is just a repetition of the budget constraint.
The first order conditions to this problem are:
\[
c_t^{-\eta}\left(1-l_t\right)^{A\left(1-\eta\right)}=\lambda_t;
\]
\[
Ac_t^{1-\eta}\left(1-l_t\right)^{A\left(1-\eta\right)-1}=\lambda_tz_t\left(1-\rho\right)k_t^\rho l_t^{-\rho};
\]
\[
\beta E_t\lambda_{t+1}\left[z_{t+1}\rho k_{t+1}^{\rho-1}l_t^{1-\rho}+\left(1-\delta\right)\right]=\lambda_t;
\]
But using the following definitions for the gross real interest and wage rate:
\[
R_t=z_t\rho k_t^{\rho-1}l_t^{1-\rho}+\left(1-\delta\right)
\]
\[
w_t=z_t\left(1-\rho\right)k_t^\rho l_t^{-\rho}
\]
and recalling market equilibrium and the error process, the whole system is equal to:
\begin{equation}
z_tk_t^\rho l_t^{1-\rho}-c_t-k_{t+1}+\left(1-\delta\right)k_t=0;
\end{equation}
which is market equilibrium
\begin{equation}
z_t\rho k_t^{\rho-1}l_t^{1-\rho}+\left(1-\delta\right)-R_t=0;
\end{equation}
which is the definition of the gross real interest rate
\begin{equation}
z_t\left(1-\rho\right)k_t^\rho l_t^{-\rho}-w_t=0;
\end{equation}
which is the definition of the wage rate
\begin{equation}
c_t^\eta\left(1-l_t\right)^{-A\left(1-\eta\right)}\beta E_t c_{t+1}^{-\eta}\left(1-l_{t+1}\right)^{A\left(1-\eta\right)}R_{t+1}-1=0;
\end{equation}
which is the consumption Euler equation
\begin{equation}
z_t k_t^\rho l_t^{1-\rho}-y_t=0;
\end{equation}
which is the definition for output
\begin{equation}
c_t^{-\eta}\left(1-l_t\right)^{A\left(1-\eta\right)}\left[w_t-\left(1-l_t\right)^{-1}\right]=0;
\end{equation}
which is the FOC for labour (where we have substituted out for $\lambda$)
\begin{equation}
\bar{z}^{1-\psi}z_t^\psi\epsilon_{t+1}-z_{t+1}=0;
\end{equation}
which is the error process. Notice that the steady state is computed numerically according to:
\[\bar{z}\bar{k}^\rho\bar{l}^{1-\rho}-\bar{c}-\delta\bar{k}=0;\]
\[\bar{z}\rho\bar{k}^{\rho-1}\bar{l}^{1-\rho}+\left(1-\delta\right)-\bar{R}=0;\]
\[\bar{z}\left(1-\rho\right)\bar{k}^\rho\bar{l}^{-\rho}-\bar{w}=0;\]
\[\beta\bar{R}-1=0;\]
\[\bar{z}\bar{k}^\rho l_t^{1-\rho}-\bar{y}=0;\]
\[\bar{c}^{-\eta}\left(1-l_t\right)^{A\left(1-\eta\right)}\left[\bar{w}-\bar{c}\left(1-\bar{l}\right)^{-1}\right]=0;\]


\end{document}