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%	TITLE PAGE
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\title[]{Macro II: Difference Equations} % The short title appears at the bottom of every slide, the full
% title is only on the title page

\author{Professor Griffy} % Your name
\institute[University at Albany, SUNY] % Your institution as it will appear on the bottom of
% every slide, may be shorthand to save space
{
UAlbany  \ % Your institution for the title page
}
\date{Spring 2021} % Date, can be changed to a custom date

\begin{document}

\begin{frame}
  \titlepage % Print the title page as the first slide
\end{frame}




% ----------------------------------------------------------------------------------------
%	PRESENTATION SLIDES
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% ------------------------------------------------
\section{Introduction} % Sections can be created in order to organize your presentation into discrete blocks, all sections and subsections are automatically printed in the table of contents as an overview of the talk
% ------------------------------------------------

\begin{frame}
  \frametitle{Introduction}
  \begin{itemize}
  \item Today: review linear algebra/difference equations.
  \item Apply to time series/macroeconomics.
  \end{itemize}
\end{frame}

% ------------------------------------------------


\section{Difference equations and linear algebra}

% ------------------------------------------------

\begin{frame}
  \frametitle{A linear difference equation}
\begin{itemize}
\item Simple first-order linear difference equation:
  \begin{align*}
    x_{t + 1} &= A x_{t} + C w_{t + 1}
  \end{align*}
\item We might think of $x_{t}$ as a vector of states (capital, assets, etc.)
\item and $w_{t + 1}$ as a vector of shocks.
\item note that $w_{t + 1}$ is not known at time-t.
\item Thus, a stochastic difference equation.
\end{itemize}
\end{frame}

% ------------------------------------------------

\begin{frame}
  \frametitle{A linear difference equation}
\begin{itemize}
\item Simple first-order linear difference equation:
  \begin{align*}
    x_{t + 1} &= A x_{t} + C w_{t + 1}
  \end{align*}
\item $w_{t + 1}$ as a vector of shocks:
  \begin{itemize}
  \item A1: iid $w_{t + 1}\sim N(0, I)$
  \item A2 (A1'):
    \begin{align*}
      E[w_{t+1}|J_{t}] &= 0\\
      E[w_{t+1}w_{t+1}'|J_{t}] &= I\\
      J_{t} = [w_{t},...,w_{1},x_{0}]
    \end{align*}
  \item A3 (A1''):
    \begin{align*}
      E[w_{t + 1}] &= 0\\
      E[w_{t}w_{t-j}'] &= I\;if\;j=0\;and\;0\;otherwise
    \end{align*}
  \end{itemize}
\end{itemize}
\end{frame}

% ------------------------------------------------

\begin{frame}
  \frametitle{A linear difference equation}
\begin{itemize}
\item Simple first-order linear difference equation:
  \begin{align*}
    x_{t + 1} &= A x_{t} + C w_{t + 1}\\
    y_{t} &= G x_{t}
  \end{align*}
\item We can think of $y_{t}$ as some type of measurement equation.
\item This is called a state-space formulation.
\item We could also think of $y_{t}$ as a choice variable (more on this later).
\end{itemize}
\end{frame}

% ------------------------------------------------

\begin{frame}
  \frametitle{Eigenvalues and eigenvectors}
\begin{itemize}
\item eigenvector: the direction a system moves.
\item eigenvalue: the distance it moves in that direction.
\item Simple first-order linear difference equation:
  \begin{align*}
    x_{t + 1} &= A x_{t} + C w_{t + 1}\\
    \begin{bmatrix}
      x_{1,t+1}\\
      x_{2,t+1}
    \end{bmatrix} &=
    \begin{bmatrix}
      1 & 0\\
      0 & \tilde{A}
    \end{bmatrix}
    \begin{bmatrix}
      x_{1,t}\\
      x_{2,t}
    \end{bmatrix}
    +
    \begin{bmatrix}
      0 \\
      \tilde{C}
    \end{bmatrix}w_{t + 1}
  \end{align*}
\item This says that a subset $x_{1}$ of the state is always at its initial value, $x_{1,t} = x_{1,0}$.
\item i.e., it has a unit eigenvalue: solution of $(A_{11} - 1)x_{1,0}$ is any $x_{1,0}$.
\item For this to be \textit{covariance stationary}, the eigenvalues of $\tilde{A}$ must all be less than 1.
\item i.e., the solution to $(A - \lambda I)v = 0$ is $|\lambda| < 1$ or $v = 0$ and $\lambda = 1$.
\end{itemize}
\end{frame}

% ------------------------------------------------

\section{Lag Operators}

\begin{frame}\frametitle{Lag operators: preliminaries}
\begin{itemize}
\item Let $\mathbf{S}$ be a set of stochastic processes. Define the lag
operator $L^{n}:\mathbf{S}\rightarrow \mathbf{S}$, $n$ an integer, by
\begin{equation*}
L^{n}\left\{ X_{t}\right\} _{t=-\infty }^{\infty }=\left\{ X_{t-n}\right\}
_{t=-\infty }^{\infty }.
\end{equation*}

\item Lag operator is linear%
\begin{equation*}
L\left( aX_{t}+bL^{n}X_{t}\right) =(aL+bL^{n+1})X_{t},
\end{equation*}%
so that lag operations can be manipulated like polynomials.
\end{itemize}
\end{frame}


\begin{frame}\frametitle{Preliminaries II}
\begin{itemize}
\item Some geometry
\item Because the lag operator is linear (everything nets out),
\begin{equation*}
\hspace{-0.25in}\left( 1-\phi L^{n}\right) \left(
\sum\nolimits_{j=0}^{J}\left( \phi L^{n}\right) ^{j}\right) X_{t}=\left(
1-\left( \phi L^{n}\right) ^{J+1}\right) X_{t},
\end{equation*}%
and if $\left( \phi L^{n}\right) ^{J+1}X_{t}$ and $\left(
\sum\nolimits_{j=0}^{J}\left( \phi L^{n}\right) ^{j}\right) X_{t}$
\textquotedblleft converge\textquotedblright ---which might be true even if $%
\left\vert \phi \right\vert >1$---we get%
\begin{equation*}
\frac{1}{1-\phi L^{n}}X_{t}=\left( \sum\nolimits_{j=0}^{\infty }\left( \phi
L^{n}\right) ^{j}\right) X_{t},
\end{equation*}%
the inverse of the operation ${1-\phi L^{n}}$
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Preliminaries II}
\begin{itemize}
\item Suppose $X_{t}=c$, $\forall t$. Then
\begin{equation*}
L^{n}c=L^{n}X_{t}=c.
\end{equation*}

\item The lag operator does not shift information sets%
\begin{equation*}
L^{n}E_{t}\left( X_{t+j}\right) =E_{t}\left( X_{t+j-n}\right) \neq
E_{t-n}\left( X_{t+j-n}\right) .
\end{equation*}%
\end{itemize}
\end{frame}

% \begin{frame}\frametitle{Note}
% \begin{itemize}
% \item Warning: Although%
% \begin{align*}
% w_{t}=y_{t}& \Rightarrow E_{t}\left( w_{t}\right) =E_{t}\left( y_{t}\right)
% \\
% & \Rightarrow L^{-1}E_{t}\left( w_{t}\right) =L^{-1}E_{t}\left( y_{t}\right)
% \\
% & \Rightarrow E_{t}\left( w_{t+1}\right) =E_{t}\left( y_{t+1}\right) ,
% \end{align*}%
% the opposite does not hold, i.e,
% \begin{align*}
% E_{t}\left( w_{t+1}\right) =E_{t}\left( y_{t+1}\right) & \nRightarrow
% LE_{t}\left( w_{t+1}\right) =LE_{t}\left( y_{t+1}\right) \\
% & \Rightarrow w_{t}=y_{t}.
% \end{align*}%
% \item $w_{t}$ and $y_{t}$ are already in our information sets--i.e., cannot change them using lag operator.
% \end{itemize}
% \end{frame}

\section{Homogeneous first-order LEDE}


\begin{frame}\frametitle{Linear difference equations again}
\begin{itemize}
\item Another way to write it
\begin{align*}
& E_{t}\left( b_{t+1}\right) =\lambda b_{t} \\
\Leftrightarrow \;\;& E_{t}\left( \left( 1-\lambda L\right) b_{t+1}\right)
=0.
\end{align*}

\item Rewrite this as
\begin{eqnarray*}
b_{t+1} &=&\lambda b_{t}+\varepsilon _{t+1}, \\
\varepsilon _{t+1} &\equiv &b_{t+1}-E_{t}\left( b_{t+1}\right) .
\end{eqnarray*}

\item As a forecast error, $\varepsilon _{t}$ forms a martingale difference
sequence, i.e.%
\begin{equation*}
E_{t}\left( \varepsilon _{t+1}\right) =0
\end{equation*}
\end{itemize}
\end{frame}

\begin{frame}\frametitle{LEDE II}
\begin{itemize}
\item Generalize%
\begin{eqnarray*}
b_{t+1}-c\lambda ^{t+1} &=&\lambda b_{t}-\lambda c\lambda ^{t}+\varepsilon
_{t+1}, \\
\left( 1-\lambda L\right) \left( b_{t+1}-c\lambda ^{t+1}\right)
&=&\varepsilon _{t+1}, \\
b_{t+1} &=&c\lambda ^{t+1}+\frac{1}{1-\lambda L}\varepsilon _{t+1},
\end{eqnarray*}%
where $c$ is a constant
\item Solution tells us $b_{t}$ at any time, $t$.
\item Goal: find (solve for) the set of admissible $\left\{ \varepsilon _{t}\right\} $
and $c$
\item Two approaches:
  \begin{itemize}
  \item Backward solution: follow sequence from past to now to find current value.
  \item Foward solution: start in future and work backwards to pin down path.
  \end{itemize}
\end{itemize}
\end{frame}

\subsection{Backward Solution}


\begin{frame}\frametitle{Backward solution}
\begin{itemize}
\item If time starts at $-\infty $, the backward solution (if well-defined)
is%
\begin{equation*}
b_{t}=c\lambda ^{t}+\sum\nolimits_{j=0}^{\infty }\lambda ^{j}\varepsilon
_{t-j}.
\end{equation*}

\item If time starts at $0,$ the backward solution is%
\begin{equation*}
b_{t}=b_{0}\lambda ^{t}+\sum\nolimits_{j=0}^{t-1}\lambda ^{j}\varepsilon
_{t-j},
\end{equation*}%
where $b_{0}$ is a (possibly) random variable
\end{itemize}
\end{frame}

\subsubsection{Restrictions to reduce the solution set}

\begin{frame}\frametitle{Solution set restrictions}
\begin{itemize}
\item Initial conditions:
\begin{itemize}
\item $\left\{ \varepsilon _{t}\right\} $ and $b_{0}$
are given.
\item i.e., Perfect foresight $\varepsilon _{t}=0,\forall
t $
\end{itemize}
\item Non-explosiveness:
\begin{align*}
\underset{j\rightarrow \infty }{\lim }& \;E_{t}\left( b_{t+j}\right)
=0,\quad \forall t, \\
\underset{t}{\sup }& \;V\left( b_{t}\right) <\infty .
\end{align*}%
Note that
\begin{eqnarray*}
E_{t}\left( b_{t+2}\right) &=&E_{t}\left( E_{t+1}\left( b_{t+2}\right)
\right) \\
&=&E_{t}\left( \lambda b_{t+1}\right) =\lambda \left( \lambda b_{t}\right) ,
\\
\Rightarrow E_{t}\left( b_{t+j}\right) &=&\lambda ^{j}b_{t}.
\end{eqnarray*}
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Restrictions II}
\begin{itemize}
\item If $|\lambda |<1,$ there are many $c$ and $\left\{ \varepsilon
_{t}\right\} $ where the non-explosiveness conditions do not restrict

\item But if $|\lambda |\geq 1,$ the only admissible solution is $%
\varepsilon _{t}=c=0$, so that $b_{t}=0,\forall t$
\item Because if any deviation from steady-state, will explode over time.

\item Note that if $|\lambda |\geq 1,$ then $b_{t}$ cannot generally satisfy both an
initial condition and a non-explosiveness condition
\end{itemize}
\end{frame}


\subsubsection{General LEDE}

\begin{frame}\frametitle{Nonhomogeneous differential equations}
\begin{itemize}
\item Wish to solve%
\begin{equation*}
E_{t}\left( x_{t+1}\right) =\lambda x_{t}+z_{t},
\end{equation*}%
where $\left\{ z_{t}\right\} $ is a stochastic \underline{forcing process}.

\item Generalize by adding a bubble term%
\begin{align*}
& E_{t}\left( x_{t+1}-b_{t+1}\right) =\lambda x_{t}+z_{t}-\lambda b_{t} \\
\Leftrightarrow \;& E_{t}\left( \left( 1-\lambda L\right) \left(
x_{t+1}-b_{t+1}\right) \right) =z_{t},
\end{align*}%
where $b_{t+1}$ is a \textquotedblleft \underline{bubble term}%
\textquotedblright\ that solves
\begin{equation*}
E_{t}\left( b_{t+1}\right) =\lambda b_{t}.
\end{equation*}%
\item i.e., a process unrelated to the fundamental term, $x_{t}$.
\end{itemize}
\end{frame}


\begin{frame}
  \frametitle{General LEDE II}
\begin{itemize}
\item The general problem is
\begin{align}
& x_{t+1}-b_{t+1}=\lambda \left( x_{t}-b_{t}\right) +\widetilde{\eta }%
_{t+1}+z_{t},  \notag \\
& \widetilde{\eta }_{t+1}\equiv \left( x_{t+1}-b_{t+1}\right) -E_{t}\left(
x_{t+1}-b_{t+1}\right) ,  \notag \\
& \left( 1-\lambda L\right) \left( x_{t+1}-b_{t+1}\right) =\widetilde{\eta }%
_{t+1}+z_{t}.  \tag{GP}
\end{align}

\item Goal: find the set of admissible $\left\{ \widetilde{\eta }%
_{t}\right\} $ and $\left\{ b_{t}\right\} $
\item $\tilde{\eta}_{t + 1}$: expectational errors.
\item $b_{t}$: bubble term (non-fundamental value).
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Backward solution}
\begin{itemize}
\item $\left\{ \widetilde{\eta }_{t}\right\} $ and $\left\{ b_{t}\right\} $
cannot be identified separately

\item If time starts at $-\infty ,$ backward solution (if well-defined) is%
\begin{eqnarray*}
x_{t+1} &=&\sum\nolimits_{j=0}^{\infty }\lambda ^{j}\left( z_{t-j}+%
\widetilde{\eta }_{t+1-j}\right) +b_{t+1} \\
&=&\sum\nolimits_{j=0}^{\infty }\lambda ^{j}z_{t-j}+\widetilde{b}_{t+1}, \\
\widetilde{b}_{t+1} &\equiv &b_{t+1}+\sum\nolimits_{j=0}^{\infty }\lambda
^{j}\widetilde{\eta }_{t+1-j}.
\end{eqnarray*}

\item $\widetilde{b}_{t+1}$ is a bubble term

\item Fundamental (sometimes called particular) solution is
\begin{equation*}
x_{t+1}=\sum\nolimits_{j=0}^{\infty }\lambda ^{j}z_{t-j}
\end{equation*}%
\item i.e., must reflect sequence of shocks (stochastic forcing process).
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Backwards solution II}
\begin{itemize}
\item If time starts at $0,$ the backward solution can be written as%
\begin{align*}
x_{t+1}=\;& \sum\nolimits_{j=0}^{t}\lambda
^{j}z_{t-j}+\sum\nolimits_{j=0}^{t}\lambda ^{j}\widetilde{\eta }_{t+1-j} \\
& +\left( x_{0}-b_{0}\right) \lambda ^{t+1}+b_{t+1},
\end{align*}%
which becomes
\begin{eqnarray*}
\hspace{-0.25in}x_{t+1} &=&\sum\nolimits_{j=0}^{t}\lambda
^{j}z_{t-j}+\sum\nolimits_{j=0}^{t}\lambda ^{j}\eta _{t+1-j}+x_{0}\lambda
^{t+1}, \\
\eta _{t} &\equiv &\widetilde{\eta }_{t}+b_{t}-E_{t-1}\left( b_{t}\right) \\
&=&x_{t}-E_{t-1}\left( x_{t}\right) .
\end{eqnarray*}
\item $x_{t}$ is stochastic, will have errors.
\item $b_{t}$ is deterministic. Cannot be wrong or will be systematic.
\end{itemize}
\end{frame}

\subsection{Forward Solution}

\begin{frame}\frametitle{Forward solution}
\begin{itemize}
\item First, rewrite \begin{align*}
& \left( 1-\lambda L\right) \left( x_{t+1}-b_{t+1}\right) =\widetilde{\eta }%
_{t+1}+z_{t}
\end{align*}
as
\begin{align*}
\left( \frac{1-\lambda L}{-\lambda L}\right) \left( -\lambda L\right) &
\left( x_{t+1}-b_{t+1}\right) =\widetilde{\eta }_{t+1}+z_{t}, \\
\left( 1-\lambda ^{-1}L^{-1}\right) & \left( x_{t}-b_{t}\right) =-\frac{1}{%
\lambda }\left( z_{t}+\widetilde{\eta }_{t+1}\right) .
\end{align*}%
To ensure that $x_{t}$ is a function only of variables known at time $t$,
write this as
\begin{equation*}
E_{t}\left( \left( 1-\lambda ^{-1}L^{-1}\right) \left( x_{t}-b_{t}\right)
\right) =-\frac{1}{\lambda }E_{t}\left( z_{t}+\widetilde{\eta }_{t+1}\right)
.
\end{equation*}%
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Forward solution II}
\begin{itemize}
\item Invert the lag operator%
\begin{equation*}
E_{t}\left( x_{t}-b_{t}\right) =-\frac{1}{\lambda }E_{t}\left( \frac{1}{%
1-\lambda ^{-1}L^{-1}}\left( z_{t}+\widetilde{\eta }_{t+1}\right) \right) ,
\end{equation*}%
\begin{eqnarray*}
x_{t} &=&-\frac{1}{\lambda }E_{t}\left( \sum\nolimits_{j=0}^{\infty }\left(
\frac{1}{\lambda }\right) ^{j}\left( z_{t+j}+\widetilde{\eta }%
_{t+j+1}\right) \right) +b_{t}, \\
&=&-\frac{1}{\lambda }E_{t}\left( \sum\nolimits_{j=0}^{\infty }\left( \frac{1%
}{\lambda }\right) ^{j}z_{t+j}\right) +b_{t},
\end{eqnarray*}%
because $E_{t}\left( \widetilde{\eta }_{t+j}\right) =0,\forall j\geq 1$%
\item note: $\frac{1}{L}^{j} = L^{-j}$ subsumed into $z_{t + j}$ (bc negative exponent on lag operator equals lead operator)
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Forward solution III}
\begin{itemize}
\item The fundamental (particular) solution is
\begin{equation*}
x_{t}=-\frac{1}{\lambda }E_{t}\left( \sum\nolimits_{j=0}^{\infty }\left(
\frac{1}{\lambda }\right) ^{j}z_{t+j}\right)
\end{equation*}

\item Note that $\widetilde{\eta }_{t}$ depends only on the forcing process $%
z_{t}$%
\begin{align*}
\widetilde{\eta }_{t}=-\frac{1}{\lambda }\Bigg[& E_{t}\left(
\sum\nolimits_{j=0}^{\infty }\left( \frac{1}{\lambda }\right)
^{j}z_{t+j}\right) \\
& -E_{t-1}\left( \sum\nolimits_{j=0}^{\infty }\left( \frac{1}{\lambda }%
\right) ^{j}z_{t+j}\right) \Bigg],\forall t.
\end{align*}%
\end{itemize}
\end{frame}

\subsection{Summary}

\begin{frame}\frametitle{Summing up}
\begin{itemize}
\item Forward solution%
\begin{equation*}
x_{t}=-\frac{1}{\lambda }E_{t}\left( \sum\nolimits_{j=0}^{\infty }\left(
\frac{1}{\lambda }\right) ^{j}z_{t+j}\right) +b_{t}.
\end{equation*}

\item Backward solution%
\begin{equation*}
x_{t+1}=\sum\nolimits_{j=0}^{\infty }\lambda ^{j}z_{t-j}+\tilde{b}_{t+1},
\end{equation*}%
or
\begin{equation*}
x_{t+1}=\sum\nolimits_{j=0}^{t}\lambda
^{j}z_{t-j}+\sum\nolimits_{j=0}^{t}\lambda ^{j}\eta _{t+1-j}+x_{0}\lambda
^{t+1}.
\end{equation*}%
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Restrictions}
\begin{itemize}
\item Initial conditions:
\begin{itemize}
\item $x_{0}$ and $\left\{ \widetilde{\eta }_{t}\right\}
_{t=1}^{\infty }$ are directly given, for example with capital accumulation%
\begin{align*}
& k_{t+1}=\left( 1-\delta \right) k_{t}+i_{t}, \\
& k_{0}\ \text{given}, \\
& k_{t+1}-E_{t}\left( k_{t+1}\right) =0,\;\forall t.
\end{align*}
\end{itemize}
\item Non-Explosiveness (boundary condition):
\begin{align*}
\underset{j\rightarrow \infty }{\lim }& E_{t}\left( x_{t+j}\right) =0,\quad
\forall t, \\
\underset{t}{\sup }\;& V\left( x_{t}\right) <\infty .
\end{align*}%
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Solutions}
\begin{itemize}
\item If $\left\vert \lambda \right\vert <1$, for \textquotedblleft
well-behaved\textquotedblright\ $\left\{ z_{t}\right\} $ (e.g, ARMA
processes), one solves $\left( 1-\lambda L\right) ^{-1}$ backwards to get
\begin{equation*}
x_{t+1}=\sum\nolimits_{j=0}^{\infty }\lambda ^{j}z_{t-j}+\tilde{b}_{t+1},
\end{equation*}%
with a large number of permissable $\left\{ \tilde{b}_{t}\right\} $.

\item But if $\left\vert \lambda \right\vert >1$, for \textquotedblleft
typical\textquotedblright\ $\left\{ z_{t}\right\} $ (e.g, ARMA processes),
we must solve $\left( 1-\lambda L\right) ^{-1}$ forward and set $b_{t}=0$,
so that
\begin{equation*}
x_{t}=-\frac{1}{\lambda }E_{t}\left( \sum\nolimits_{j=0}^{\infty }\left(
\frac{1}{\lambda }\right) ^{j}z_{t+j}\right) .
\end{equation*}

\item If $\left\vert \lambda \right\vert >1,$ cannot satisfy both initial
conditions and non-explosiveness
\end{itemize}
\end{frame}



\begin{frame}
  \frametitle{Rule of Thumb}
\begin{itemize}
\item If $\left\vert \lambda \right\vert <1,$ set
\begin{equation*}
x_{t+1}=\sum\nolimits_{j=0}^{t}\lambda
^{j}z_{t-j}+\sum\nolimits_{j=0}^{t}\lambda ^{j}\eta _{t+1-j}+x_{0}\lambda
^{t+1}.
\end{equation*}%
and use initial conditions to pin down $x_{0}$ and $\left\{ \eta
_{t}\right\} $

\item If $\left\vert \lambda \right\vert >1,$ set
\begin{equation*}
x_{t}=-\frac{1}{\lambda }E_{t}\left( \sum\nolimits_{j=0}^{\infty }\left(
\frac{1}{\lambda }\right) ^{j}z_{t+j}\right) .
\end{equation*}

\item If $\left\vert \lambda \right\vert =1,$ consider case by case
\end{itemize}
\end{frame}

\section{Conclusion}
% ------------------------------------------------


\begin{frame}
  \frametitle{Next Time}
  \begin{itemize}
  \item Discuss rational expectations and Lucas Critique.
  \item See my webpage for new homework.
  \end{itemize}
\end{frame}


\end{document}