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%	TITLE PAGE
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\title[]{Macro II} % 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 2023} % 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
%----------------------------------------------------------------------------------------

%------------------------------------------------
\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{Announcements}
  \begin{itemize}
  \item Today: Start discussing solution techniques.
  \item Focus on linearization \& its problems.
  \item Midterm on Thursday! Will not cover today's lecture.
  \item HW4 due tonight.
  \end{itemize}
\end{frame}

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

\section{Standard RBC Model} % 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{Motivation}
\begin{itemize}
\item Models are hard to solve globally.
\item Requires a lot of grid points, entails curse of dimensionality, takes a long time.
\item A linearized system, by contrast, is easy to solve.
\item Need to pick a place to linearize around.
\item Pick the steady state.
\item Underlying assumption: economy will stay close to the steady-state.
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Empirical Motivation}
\begin{itemize}
\item Standard RBC: all fluctuations of hours worked on the {\it intensive} margin, i.e. average number of hours worked.
\item Data: little fluctuation in average hours worked; lots of fluctuation in whether or not people are working ({\it extensive} margin).
\item Standard RBC: missed badly on labor fluctuations (Frisch Elasticity, i.e. response of labor to change in wage too low).
\item Solution: Modify model to have extensive margin with high Frisch Elasticity.
\item Now: households pick the {\it probability} of working, but have to work a set number of hours.
\item This is a {\it nonconvexity} in that it forces individuals to work either 0 or h hours.
\end{itemize}
\end{frame}

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


\begin{frame}
\frametitle{Hansen (1985)}
\begin{itemize}
\item Neoclassical growth model with labor-leisure lottery.
\item A social planner maximize the following:
\begin{equation}
E(\sum_{t = 0}^{\infty}\beta^{t}[ln(C_{t}) - \gamma H_{t}])
\end{equation}
\item Subject to the following constraints:
\begin{equation}
Y_{t} = A_{t}K_{t}^{\theta}(\eta^{t}H_{t})^{1 - \theta}
\end{equation}
\begin{equation}
ln(A_{t}) = (1 - \rho)ln(A) + \rho ln(A_{t - 1}) + \epsilon_{t}, \hspace{3 mm} \epsilon_{t}\sim N(0,\sigma_{\epsilon}^{2})
\end{equation}
\item The goods market clears and capital evolves in a predetermined fashion.
\item Here, we assume that per capita labor productivity grows at rate $\eta$.
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Equilibrium}
\begin{itemize}
\item First step: detrend appropriate variables by per capita growth to get stationarity: i.e. $y_{t} = Y_{t}/\eta^{t}$.
\item The system of equations that characterize the equilibrium are:
\begin{equation}
y_{t} = a_{t}k_{t}^{\theta}h_{t}^{1 - \theta}
\end{equation}
\begin{equation}
ln(a_{t}) = (1 - \rho)ln(A) + \rho ln(a_{t - 1}) + \epsilon_{t}
\end{equation}
\begin{equation}
y_{t} = c_{t} + i_{t}
\end{equation}
\begin{equation}
\eta k_{t + 1} = (1 - \delta)k_{t} + i_{t}
\end{equation}
\item Combine FOC[c] and FOC[h]:
\begin{equation}
\gamma c_{t}h_{t} = (1 - \theta)y_{t}
\end{equation}
\item Euler Equation:
\begin{equation}
\frac{\eta}{c_{t}} = \beta E_{t}[\frac{1}{c_{t + 1}}(\theta(\frac{y_{t + 1}}{k_{t + 1}}) + 1 - \delta)]
\end{equation}
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Solving for the Steady-State}
\begin{itemize}
\begin{equation*}
ln(a^{*}) = (1 - \rho)ln(A) + \rho ln(a^{*})
\end{equation*}
\begin{equation}
\Rightarrow ln(a^{*}) = ln(A)
\end{equation}
\item Euler Equation:
\begin{equation*}
\frac{\eta}{c^{*}} = \beta E_{t}[\frac{1}{c^{*}}(\theta(\frac{y^{*}}{k^{*}}) + 1 - \delta)]
\end{equation*}
\begin{equation*}
\Rightarrow \frac{\eta}{\beta} = \theta\frac{y^{*}}{k^{*}} + 1 - \delta
\end{equation*}
\begin{equation}
\Rightarrow k^{*} = (\frac{\theta}{\frac{\eta}{\beta} - 1 + \delta})y^{*}
\end{equation}
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Solving for the Steady-State}
\begin{itemize}
\item Use the previous to solve for investment
\begin{equation*}
\eta k^{*} = (1 - \delta)k^{*} + i^{*}
\end{equation*}
\begin{equation*}
\Rightarrow (\eta - 1 + \delta)k^{*} = i^{*}
\end{equation*}
\begin{equation}
\Rightarrow i^{*} = (\frac{\theta(\eta - 1 + \delta)}{\frac{\eta}{\beta} - 1 + \delta})y^{*}
\end{equation}
\item FOC[c] and FOC[h]:
\begin{equation*}
\gamma c^{*}h^{*} = (1 - \theta)y^{*}
\end{equation*}
\begin{equation*}
\Rightarrow \gamma [1 - (\frac{\theta(\eta - 1 + \delta)}{\frac{\eta}{\beta} - 1 + \delta})]y^{*}h^{*} = (1 - \theta)y^{*}
\end{equation*}
\begin{equation}
\Rightarrow h^{*} = (\frac{1 - \theta}{\gamma})[1 - (\frac{\theta(\eta - 1 + \delta)}{\frac{\eta}{\beta} - 1 + \delta})]^{-1}
\end{equation}
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Solving for the Steady-State}
\begin{itemize}
\item Finally, solve for output.
\begin{equation*}
y^{*} = a^{*}k^{*\theta}h^{*1 - \theta}
\end{equation*}
\begin{equation*}
y^{*} = a^{*}((\frac{\theta}{\frac{\eta}{\beta} - 1 + \delta})y^{*})^{\theta}[(\frac{1 - \theta}{\gamma})[1 - (\frac{\theta(\eta - 1 + \delta)}{\frac{\eta}{\beta} - 1 + \delta})]^{-1}]^{1 - \theta}
\end{equation*}
\begin{equation*}
y^{*1 - \theta} = a^{*}(\frac{\theta}{\frac{\eta}{\beta} - 1 + \delta})^{\theta}[(\frac{1 - \theta}{\gamma})[1 - (\frac{\theta(\eta - 1 + \delta)}{\frac{\eta}{\beta} - 1 + \delta})]^{-1}]^{1 - \theta}
\end{equation*}
\begin{equation}
y^{*} = a^{*\frac{1}{1 - \theta}}(\frac{\theta}{\frac{\eta}{\beta} - 1 + \delta})^{\frac{\theta}{1 - \theta}}[(\frac{1 - \theta}{\gamma})[1 - (\frac{\theta(\eta - 1 + \delta)}{\frac{\eta}{\beta} - 1 + \delta})]^{-1}]^{1 - \theta}
\end{equation}
\item All variables now a function of parameters.
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Steady-States}
\begin{itemize}
\item In steady-state $y_{t} = y_{t + 1} = y^{*}$.
\begin{equation}
 ln(a^{*}) = ln(A)
\end{equation}
\begin{equation}
k^{*} = (\frac{\theta}{\frac{\eta}{\beta} - 1 + \delta})y^{*}
\end{equation}
\begin{equation}
 i^{*} = (\frac{\theta(\eta - 1 + \delta)}{\frac{\eta}{\beta} - 1 + \delta})y^{*}
\end{equation}
\begin{equation}
c^{*} = [1 - (\frac{\theta(\eta - 1 + \delta)}{\frac{\eta}{\beta} - 1 + \delta})]y^{*}
\end{equation}
\begin{equation}
 h^{*} = (\frac{1 - \theta}{\gamma})[1 - (\frac{\theta(\eta - 1 + \delta)}{\frac{\eta}{\beta} - 1 + \delta})]^{-1}
\end{equation}
\begin{equation}
y^{*} = a^{*\frac{1}{1 - \theta}}(\frac{\theta}{\frac{\eta}{\beta} - 1 + \delta})^{\frac{\theta}{1 - \theta}}[(\frac{1 - \theta}{\gamma})[1 - (\frac{\theta(\eta - 1 + \delta)}{\frac{\eta}{\beta} - 1 + \delta})]^{-1}]^{1 - \theta}
\end{equation}
\item These steady-states will be used for calibration/solving.
\end{itemize}
\end{frame}

%------------------------------------------------
\section{Solving the Model} % 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{Overview}
\begin{itemize}
\item Broadly, two methods of solving models:
\begin{enumerate}
\item Local linear methods.
\item Global non-linear methods.
\end{enumerate}
\item Tradeoff: accuracy (global non-linear) for speed and simplicity (local linear).
\item My preference: global methods (linear methods involve linearizing Euler Equation, distorting choices over risk).
\item Here: Discuss log linearization and Blanchard and Kahn's Method.
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Local Linear Methods}
\begin{itemize}
\item Log-linearize the system around the steady-state, then proceed.
\item First have to solve the system for stability:
\begin{enumerate}
\item Klein's Method (2000): Used for singular matrices.
\item Sim's Method (2001): Used when it is unclear which variables are states and controls.
\item Blanchard and Kahn's Method (1980): First solution method for rational expectations models.
\end{enumerate}
\item Here, we will use Blanchard and Kahn's Method.
\end{itemize}
\end{frame}

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

\begin{frame}\frametitle{Log-Linearizing the System}
\item We first wish to rewrite $\tilde{x}_{t} = ln(x_{t}) - ln(x)$ in two convenient ways:

\begin{equation*}
\tilde{x}_{t} = ln(\frac{x_{t}}{x})
\end{equation*}

\item Then, the first-order Taylor Approximation to this equation yields:

\begin{equation*}
\tilde{x}_{t} \approxeq \tilde{x}_{t}(x) + \der{\tilde{x}_{t}}{x_{t}}(x)(x_{t} - x)
\end{equation*}

\begin{equation*}
\Rightarrow \tilde{x}_{t} \approxeq ln(1) + \frac{1}{x}(x_{t} - x)
\end{equation*}
\item We can also rewrite the equation for $\tilde{x}_{t}$ as

\begin{equation}
x_{t} = xe^{\tilde{x}_{t}}
\end{equation}
\end{frame}

\begin{frame}\frametitle{Log-Linearizing the System}
\item From equilibrium conditions:

\begin{equation}
y_{t} = a_{t}k_{t}^{\theta}h_{t}^{1 - \theta}
\end{equation}

\begin{equation*}
\Rightarrow ln(y_{t}) = ln(a_{t}) + \theta ln(k_{t}) + (1 - \theta)ln(h_{t})
\end{equation*}

\begin{equation*}
ln(y) = ln(a) + \theta ln(k) + (1 - \theta)ln(h)
\end{equation*}
\begin{eqnarray*}
\Rightarrow \tilde{y}_{t} = ln(y_{t}) - ln(y) = ln(a_{t}) &+& \theta ln(k_{t}) + (1 - \theta)ln(h_{t})\\ &-& (ln(a) + \theta ln(k) + (1 - \theta)ln(h))
\end{eqnarray*}

\begin{equation}
\Rightarrow \tilde{y}_{t} = \tilde{a}_{t} + \theta\tilde{k}_{t} + (1 - \theta)\tilde{h}_{t}
\end{equation}
\end{frame}

\begin{frame}\frametitle{Log-Linearizing the System}
\begin{equation*}
ln(a_{t}) = (1 - \rho)ln(A) + \rho ln(a_{t - 1}) + \epsilon_{t}
\end{equation*}

\begin{equation*}
ln(a) = (1 - \rho)ln(A) + \rho ln(a)
\end{equation*}

\begin{equation}
\Rightarrow \tilde{a}_{t} = \rho \tilde{a}_{t - 1} + \epsilon_{t}
\end{equation}
\end{frame}

\begin{frame}\frametitle{Log-Linearizing the System}
\begin{equation*}
y_{t} = c_{t} + i_{t}
\end{equation*}

\begin{equation*}
\Rightarrow \tilde{x}_{t} \approxeq ln(1) + \frac{1}{x}(x_{t} - x) = (\frac{x_{t}}{x} + 1)
\end{equation*}

\begin{equation*}
\Rightarrow y(\tilde{y}_{t} + 1) = c(\tilde{c}_{t} + 1) + i(\tilde{i}_{t} + 1)
\end{equation*}

\begin{equation*}
\tilde{y}_{t} = \frac{c}{y}\tilde{c}_{t} + \frac{i}{y}\tilde{i}_{t}
\end{equation*}
\end{frame}

\begin{frame}
\frametitle{Log-Linearizing the System}
\begin{itemize}
\item Let $\tilde{y}_{t} = ln(y_{t}) - ln(y^{*})$. Then, using Taylor Series approximations, the system characterizing the equilibrium becomes:
\begin{equation}
\tilde{y}_{t} = \tilde{a}_{t} + \theta\tilde{k}_{t} + (1 - \theta)\tilde{h}_{t}
\end{equation}
\begin{equation}
\tilde{a}_{t} = \rho \tilde{a}_{t - 1} + \epsilon_{t}
\end{equation}
\begin{equation}
(\frac{\eta}{\beta} - 1 + \delta)\tilde{y}_{t} = [\frac{\eta}{\beta} - 1 + \delta - \theta(\eta - 1 + \delta)]\tilde{c}_{t} + \theta(\eta - 1 + \delta)\tilde{i}_{t}
\end{equation}
\begin{equation}
\eta\tilde{k}_{t + 1} = (1 - \delta)\tilde{k}_{t} + (\eta - 1 + \delta)\tilde{i}_{t}
\end{equation}
\begin{equation}
\tilde{y}_{t} = \tilde{c}_{t} + \tilde{h}_{t}
\end{equation}
\begin{equation}
0 = \frac{\eta}{\beta}\tilde{c}_{t} + E[(\frac{\eta}{\beta} - 1 + \delta)(\tilde{y}_{t + 1} - \tilde{k}_{t + 1}) - \frac{\eta}{\beta}\tilde{c}_{t + 1}]
\end{equation}
% \item {\hyperlink{Math}{Lots of math to get to this point: \beamerbutton{here}}}
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Log-Linearizing the System}
\begin{itemize}
\item We can now write the system as:
\begin{equation}
\Psi_{1}\zeta_{t} = \Psi_{2}\xi_{t} + \Psi_{3}\tilde{a}_{t}\tag{ME}
\end{equation}
\begin{equation}
\Psi_{4}E_{t}(\xi_{t + 1}) = \Psi_{5}\xi_{t} + \Psi_{6}\zeta_{t} + \Psi_{7}\tilde{a}_{t}\tag{TE}
\end{equation}
\item $\zeta_{t}$ are static predetermined and nonpredetermined variables, $[\tilde{y}_{t}, \tilde{i}_{t}, \tilde{h}_{t}]'$.
\item $\xi_{t}$ are dynamic predetermined and nonpredetermined variables, $[\tilde{k}_{t}, \tilde{c}_{t}]'$.
\item $\tilde{a}_{t}$ is the technology process.
\item Why is $\tilde{c}_{t}$ among the dynamic variables?
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Matrices}
\begin{equation*}
\kappa = \eta/\beta - 1 + \delta
\end{equation*}

\begin{equation*}
\lambda = \eta - 1 + \delta
\end{equation*}
\begin{center}
\includegraphics[width=5in]{matrices.PNG}
\end{center}

\end{frame}

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


\begin{frame}
\frametitle{Solving the Model - Blanchard and Kahn (1980)}

\centering\includegraphics[width = 3in]{blanchardkahn_1.png}
\begin{itemize}
\item Select $\tilde{c}_{0}$ st the system isn't explosive (optimal control!).
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Solving the Model - Cont.}
\begin{itemize}
\item Solve systems (TE and ME) so that $\xi_{t + 1}$ is only a function on $\xi_{t}$ and $\tilde{a}_{t}$:
\begin{equation}
\Psi_{1}\zeta_{t} = \Psi_{2}\xi_{t} + \Psi_{3}\tilde{a}_{t}
\end{equation}
\begin{equation}
\Psi_{4}E_{t}(\xi_{t + 1}) = \Psi_{5}\xi_{t} + \Psi_{6}\zeta_{t} + \Psi_{7}\tilde{a}_{t}
\end{equation}
\begin{equation*}
\Rightarrow \zeta_{t} = \Psi_{1}^{-1}[\Psi_{2}\xi_{t} + \Psi_{3}\tilde{a}_{t}]
\end{equation*}
\item Plug into transition equation:
\begin{equation*}
\Psi_{4}E_{t}(\xi_{t + 1}) = \Psi_{5}\xi_{t} + \Psi_{6}\Psi_{1}^{-1}[\Psi_{2}\xi_{t} + \Psi_{3}\tilde{a}_{t}] + \Psi_{7}\tilde{a}_{t}
\end{equation*}
\begin{equation}
\Rightarrow E_{t}(\xi_{t + 1}) = \Psi_{4}^{-1}[\Psi_{5} + \Psi_{6}\Psi_{1}^{-1}\Psi_{2}]\xi_{t} + \Psi_{4}^{-1}[\Psi_{7} + \Psi_{6}\Psi_{1}^{-1}\Psi_{3}]\tilde{a}_{t}
\end{equation}
\item Desired result!
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Solving the Model - Cont.}
\begin{itemize}
\item Having solved systems on previous slide so that $\xi_{t + 1}$ is only a function on $\xi_{t}$ and $\tilde{a}_{t}$:
\begin{equation}
\begin{bmatrix}
\tilde{k}_{t+1}\\
E_{t}(\tilde{c}_{t+1})
\end{bmatrix}= \Lambda^{-1}J\Lambda\begin{bmatrix}
\tilde{k}_{t}\\\tilde{c}_{t}\end{bmatrix}+ E\tilde{a}_{t}\end{equation}
\item $\Lambda^{-1}J\Lambda$ is the Jordan Decomposition.
\item Subsume $\Lambda$ into the model variables, denoted by hats:
\begin{equation}
\hat{c}_{t} = \Lambda_{12}\tilde{k}_{t} + \Lambda_{22}\tilde{c}_{t}
\end{equation}
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Solving the Model - Cont.}
\begin{itemize}
\item Subsume $\Lambda$ into the model variables, denoted by hats.
\begin{equation}
\begin{bmatrix}
\hat{k}_{t+1}\\
E_{t}(\hat{c}_{t+1})
\end{bmatrix}= \begin{bmatrix}J_{1} & 0\\ 0 & J_{2}\end{bmatrix}\begin{bmatrix}
\hat{k}_{t}\\\hat{c}_{t}\end{bmatrix}+ D\tilde{a}_{t}\end{equation}
\begin{equation}
E_{t}(\hat{c}_{t + 1}) = J_{2}\hat{c}_{t}  + D_{2}\tilde{a}_{t}
\end{equation}
\item $J_{2} > 1 \rightarrow$ bad choice of $c_{t}$ and this explodes.
\item Solution: pick $c_{t}$ so that it isn't a function of $c_{t - 1}$!
\item Rearranging:
\begin{equation}
\hat{c}_{t} = J_{2}^{-1}E_{t}(\hat{c}_{t + 1}) - J_{2}^{-1}D_{2}\tilde{a}_{t}
\end{equation}
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Solving the Model - Cont.}
\begin{itemize}
\item Iterating on previous equation:
\begin{equation}
\hat{c}_{t + 1} = J_{2}^{-1}E_{t}(\hat{c}_{t + 2}) - J_{2}^{-1}D_{2}\tilde{a}_{t + 1}
\end{equation}
\begin{equation*}
\Rightarrow \hat{c}_{t} = J_{2}^{-1}E_{t}(J_{2}^{-1}E_{t}(\hat{c}_{t + 2}) - J_{2}^{-1}D_{2}\tilde{a}_{t + 1}) - J_{2}^{-1}D_{2}\tilde{a}_{t}
\end{equation*}
\begin{equation}
\Rightarrow \hat{c}_{t} = J_{2}^{-2}E_{t}(\hat{c}_{t + 2})) - J_{2}^{-2}D_{2}\rho\tilde{a}_{t } - J_{2}^{-1}D_{2}\tilde{a}_{t}
\end{equation}
\item Impose transversality condition (i.e. $E_{t}(\hat{c}_{t + i})) = 0$ for large enough i):
\begin{equation}
\Rightarrow \hat{c}_{t} = - \sum_{i = 0}^{\infty}J_{2}^{-(i + 1)}D_{2}\rho\tilde{a}_{t}
\end{equation}
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Solving the Model - Cont.}
\begin{itemize}
\item Iterating on (33):
\begin{equation*}
\hat{c}_{t} = \Lambda_{12}\tilde{k}_{t} + \Lambda_{22}\tilde{c}_{t}
\end{equation*}

\begin{equation*}
\Rightarrow \Lambda_{22}\tilde{c}_{t} = -\Lambda_{12}\tilde{k}_{t} - \sum_{i = 0}^{\infty}J_{2}^{-(i + 1)}D_{2}\rho\tilde{a}_{t}
\end{equation*}
\item Solving this yields:
\begin{equation}
\Rightarrow c_{t} = -\Lambda_{22}^{-1}\Lambda_{12}\tilde{k}_{t} + (1/\Lambda_{22})(\frac{D_{2}}{\rho - J_{2}})\tilde{a}_{t}
\end{equation}
\item The system will now be saddle-path stable.
\end{itemize}
\end{frame}

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

\section{Conclusion}

\begin{frame}
  \frametitle{Next Time}
  \begin{itemize}
  % \item Calibration and RBC extensions.
  \item Midterm!
  \item Start value function iteration next week.
  \item See my website for homework.
  \end{itemize}
\end{frame}

\end{document}
