\documentclass[10pt]{article} 
\usepackage{array,natbib,times}
\usepackage{amsmath, amsthm, amssymb}

%\usepackage[pdftex,colorlinks]{hyperref}

\begin{document}

\title{Implementation of Ramsey Optimal Policy in Dynare++, Timeless Perspective}

\author{Ondra Kamen\'\i k}

\date{June 2006}
\maketitle

\textbf{Abstract:} This document provides a derivation of Ramsey
optimal policy from timeless perspective and describes its
implementation in Dynare++.

\section{Derivation of the First Order Conditions}

Let us start with an economy populated by agents who take a number of
variables exogenously, or given. These may include taxes or interest
rates for example. These variables can be understood as decision (or control)
variables of the timeless Ramsey policy (or social planner). The agent's
information set at time $t$ includes mass-point distributions of these
variables for all times after $t$. If $i_t$ denotes an interest rate
for example, then the information set $I_t$ includes
$i_{t|t},i_{t+1|t},\ldots,i_{t+k|t},\ldots$ as numbers. In addition
the information set includes all realizations of past exogenous
innovations $u_\tau$ for $\tau=t,t-1,\ldots$ and distibutions
$u_\tau\sim N(0,\Sigma)$ for $\tau=t+1,\ldots$. These information sets will be denoted $I_t$.

An information set including only the information on past realizations
of $u_\tau$ and future distributions of $u_\tau\sim N(0\sigma)$ will
be denoted $J_t$. We will use the following notation for expectations
through these sets:
\begin{eqnarray*}
E^I_t[X] &=& E(X|I_t)\\
E^J_t[X] &=& E(X|J_t)
\end{eqnarray*}

The agents optimize taking the decision variables of the social
planner at $t$ and future as given. This means that all expectations
they form are conditioned on the set $I_t$. Let $y_t$ denote a vector
of all endogenous variables including the planer's decision
variables. Let the number of endogenous variables be $n$. The economy
can be described by $m$ equations including the first order conditions
and transition equations:
\begin{equation}\label{constr}
E_t^I\left[f(y_{t-1},y_t,y_{t+1},u_t)\right] = 0.
\end{equation}
This lefts $n-m$
the planner's control variables. The solution of this problem is a
decision rule of the form:
\begin{equation}\label{agent_dr}
y_t=g(y_{t-1},u_t,c_{t|t},c_{t+1|t},\ldots,c_{t+k|t},\ldots),
\end{equation}
where $c$ is a vector of planner's control variables.

Each period the social planner chooses the vector $c_t$ to maximize
his objective such that \eqref{agent_dr} holds for all times following
$t$. This would lead to $n-m$ first order conditions with respect to
$c_t$. These first order conditions would contain unknown derivatives
of endogenous variables with respect to $c$, which would have to be
retrieved from the implicit constraints \eqref{constr} since the
explicit form \eqref{agent_dr} is not known.

The other way to proceed is to assume that the planner is so dumb that
he is not sure what are his control variables. So he optimizes with
respect to all $y_t$ given the constraints \eqref{constr}. If the
planner's objective is $b(y_{t-1},y_t,y_{t+1},u_t)$ with a discount rate
$\beta$, then the optimization problem looks as follows:
\begin{align}
\max_{\left\{y_\tau\right\}^\infty_t}&E_t^J
\left[\sum_{\tau=t}^\infty\beta^{\tau-t}b(y_{\tau-1},y_\tau,y_{\tau+1},u_\tau)\right]\notag\\
&\rm{s.t.}\label{planner_optim}\\
&\hskip1cm E^I_\tau\left[f(y_{\tau-1},y_\tau,y_{\tau+1},u_\tau)\right]=0\quad\rm{for\ }
\tau=\ldots,t-1,t,t+1,\ldots\notag
\end{align}
Note two things: First, each constraint \eqref{constr} in
\eqref{planner_optim} is conditioned on $I_\tau$ not $I_t$. This is
very important, since the behaviour of agents at period $\tau=t+k$ is
governed by the constraint using expectations conditioned on $t+k$,
not $t$. The social planner knows that at $t+k$ the agents will use
all information available at $t+k$. Second, the constraints for the
planner's decision made at $t$ include also constraints for agent's
behaviour prior to $t$. This is because the agent's decision rules are
given in the implicit form \eqref{constr} and not in the explicit form
\eqref{agent_dr}.

Using Lagrange multipliers, this can be rewritten as
\begin{align}
\max_{y_t}E_t^J&\left[\sum_{\tau=t}^\infty\beta^{\tau-t}b(y_{\tau-1},y_\tau,y_{\tau+1},u_\tau)\right.\notag\\
&\left.+\sum_{\tau=-\infty}^{\infty}\beta^{\tau-t}\lambda^T_\tau E_\tau^I\left[f(y_{\tau-1},y_\tau,y_{\tau+1},u_\tau)\right]\right],
\label{planner_optim_l}
\end{align}
where $\lambda_t$ is a vector of Lagrange multipliers corresponding to
constraints \eqref{constr}. Note that the multipliers are multiplied
by powers of $\beta$ in order to make them stationary. Taking a
derivative wrt $y_t$ and putting it to zero yields the first order
conditions of the planner's problem:
\begin{align}
E^J_t\left[\vphantom{\frac{\int^(_)}{\int^(\_)}}\right.&\frac{\partial}{\partial y_t}b(y_{t-1},y_t,y_{t+1},u_t)+
\beta L^{+1}\frac{\partial}{\partial y_{t-1}}b(y_{t-1},y_t,y_{t+1},u_t)\notag\\
&+\beta^{-1}\lambda_{t-1}^TE^I_{t-1}\left[L^{-1}\frac{\partial}{\partial y_{t+1}}f(y_{t-1},y_t,y_{t+1},u_t)\right]\notag\\
&+\lambda_t^TE^I_t\left[\frac{\partial}{\partial y_{t}}f(y_{t-1},y_t,y_{t+1},u_t)\right]\notag\\
&+\beta\lambda_{t+1}^TE^I_{t+1}\left[L^{+1}\frac{\partial}{\partial y_{t-1}}f(y_{t-1},y_t,y_{t+1},u_t)\right]
\left.\vphantom{\frac{\int^(_)}{\int^(\_)}}\right]
 = 0,\label{planner_optim_foc}
\end{align}
where $L^{+1}$ and $L^{-1}$ are one period lead and lag operators respectively.

Now we have to make a few assertions concerning expectations
conditioned on the different information sets to simplify
\eqref{planner_optim_foc}. Recall the formula for integration through
information on which another expectation is conditioned, this is:
$$E\left[E\left[u|v\right]\right] = E[u],$$
where the outer expectation integrates through $v$. Since $J_t\subset
I_t$, by easy application of the above formula we obtain
\begin{eqnarray}
E^J_t\left[E^I_t\left[X\right]\right] &=& E^J_t\left[X\right]\quad\rm{and}\notag\\
E^J_t\left[E^I_{t-1}\left[X\right]\right] &=& E^J_t\left[X\right]\label{e_iden}\\
E^J_t\left[E^I_{t+1}\left[X\right]\right] &=& E^J_{t+1}\left[X\right]\notag
\end{eqnarray}
Now, the last term of \eqref{planner_optim_foc} needs a special
attention. It is equal to
$E^J_t\left[\beta\lambda^T_{t+1}E^I_{t+1}[X]\right]$. If we assume
that the problem \eqref{planner_optim} has a solution, then there is a
deterministic function from $J_{t+1}$ to $\lambda_{t+1}$ and so
$\lambda_{t+1}\in J_{t+1}\subset I_{t+1}$. And the last term is equal
to $E^J_{t}\left[E^I_{t+1}[\beta\lambda^T_{t+1}X]\right]$, which is
$E^J_{t+1}\left[\beta\lambda^T_{t+1}X\right]$. This term can be
equivalently written as
$E^J_{t}\left[\beta\lambda^T_{t+1}E^J_{t+1}[X]\right]$. The reason why
we write the term in this way will be clear later. All in all, we have
\begin{align}
E^J_t\left[\vphantom{\frac{\int^(_)}{\int^(\_)}}\right.&\frac{\partial}{\partial y_t}b(y_{t-1},y_t,y_{t+1},u_t)+
\beta L^{+1}\frac{\partial}{\partial y_{t-1}}b(y_{t-1},y_t,y_{t+1},u_t)\notag\\
&+\beta^{-1}\lambda_{t-1}^TL^{-1}\frac{\partial}{\partial y_{t+1}}f(y_{t-1},y_t,y_{t+1},u_t)\notag\\
&+\lambda_t^T\frac{\partial}{\partial y_{t}}f(y_{t-1},y_t,y_{t+1},u_t)\notag\\
&+\beta\lambda_{t+1}^TE^J_{t+1}\left[L^{+1}\frac{\partial}{\partial y_{t-1}}f(y_{t-1},y_t,y_{t+1},u_t)\right]
\left.\vphantom{\frac{\int^(_)}{\int^(\_)}}\right]
 = 0.\label{planner_optim_foc2}
\end{align}
Note that we have not proved that \eqref{planner_optim_foc} and
\eqref{planner_optim_foc2} are equivalent. We proved only that if
\eqref{planner_optim_foc} has a solution, then
\eqref{planner_optim_foc2} is equivalent (and has the same solution).

%%- \section{Implementation}
%%- 
%%- The user inputs $b(y_{t-1},y_t,y_{t+1},u_t)$, $\beta$, and agent's
%%- first order conditions \eqref{constr}. The algorithm has to produce
%%- \eqref{planner_optim_foc2}.
%%- 
\end{document}