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\begin{document}
\begin{flushright}
Jaime Frade\\
ECO5282-Dr. Garriga\\
Homework\:$\#2$\\
\end{flushright}
\begin{enumerate}

\item Consider an economy with a single consumer.  There is one good in the economy, which arrives in the form of an exogenous endowement obeying, $y_{t+1} =\lambda_{t+1}\,y_{t},$ where $y_t$ is the endowment at time $t$ and $\{\lambda_{t+1}\}$ is governed by a two-state Markov chain with transition matrix, 
$P = \left[ \begin{array}{c c}
p_{11} & 1-p_{11}  \\
1-p_{22} & p_{22}
\end{array} \right],$ and intial distribution $\pi_{\lambda} = [\pi_0, 1-\pi_0]$.  The value of $\lambda_t$ is given by $\overline{\lambda_1} = 0.98$ in state 1 and $\overline{\lambda_2} = 1.03$ in state 2.  Assume that the history of $y_s$ and $\lambda_s$, up to $t$ is observed at time $t$.  The consumer has endowment process $\{y_t\}$ and has preferences over consumption streams that are ordered by $E_0\,\sum^\infty_{t=0}\,\beta^t\,\dfrac{c^{1-\gamma}}{1-\gamma}$, here $\beta \in (0,1)$ and $\gamma \geq 1$.


\begin{description}
\item[(a)]  Define a competitive equilibrium.  
\vspace{.25cm}
\\
\textbf{Definition}  A allocation is said to be a \underline{feasible allocation} if it satisfies
\begin{eqnarray}\label{eq:feas2}
\sum^2_{i=1} c^i_t(s^t) = \sum^2_{i=1} y^i_t(s_t)
\\
\textrm{Since one agent} \rightarrow c_t(s^t) = y_t(s_t) \label{eq:feas3}
\end{eqnarray}

\textbf{Definition} A \underline{competitive equilibrium} is a feasible allocation, $\{c^i\}^2_{i=1} = \{\{c^i_t(s^t)\}^\infty_{t=0}\}^2_{i=1}$, and a price system, $\{p^0_t(s^t)\}^\infty_{t=0}$, such that the allocation solves each household problem.
\vspace{.25cm}
\\
For a given household $i$ solves 
\begin{displaymath}
U(c^i) = \max_{\{c^T_t(s^t)\}} E_0 \sum^\infty_{t=0}\beta^t\,u(c^i_t)
\end{displaymath}
\begin{equation}\label{eq:max1}
U(c^i) = \max_{\{c^T_t(s^t)\}}\sum^\infty_{t=0}\sum_{s^t}\beta^t\,\pi(s^t|s_0)\,u(c^i_t(s^t)) 
\end{equation}

\begin{equation}\label{eq:bc1}
\textrm{subject to} \hspace{.25cm} \sum^\infty_{t=0}\sum_{s^t}\,p^0_t(s^t)c^i_t(s^t) = \sum^\infty_{t=0}\sum_{s^t}\,p^0_t(s^t)y^i_t(s_t), \hspace{.25cm} \textrm{and} \hspace{.25cm} c^1_t(s^t),c^2_t(s^t) \geq 0 
\end{equation}


\item[(b)]  From (\ref{eq:feas3}), \textbf{consumption} = \textbf{endowment}, since only one agent obtain the following:
\begin{equation}\label{eq:ceqe}
c_t = y_t
\end{equation}

In (\ref{eq:ceqe}), $y$ is defined as $y_{t+1}= \lambda_{t+1}\,y_{t}$, where ${\lambda_{t+1}}$ is defined as a two state Markov Chain.  Using the properites of Markov process, 
\begin{equation}\label{eq:income1}
y_{t+1}= \prod^t_{i=0}\lambda_{i}\,y_{0}
\end{equation}

Substituting (\ref{eq:ceqe}) into the first order conditions of the household problem, into (\ref{eq:max3}), will obtain an equation that will solve the price system for any asset that will lead to competitive equilibrium.
\begin{equation}
p^0_t(s^t) = \beta^t\,\pi(s^t|s_0)\,\dfrac{u\,'(c^i_t(s^t))}{u\,'(c^i_0(s^0))} \hspace{.75in} \textrm{from (\ref{eq:max3})}
\end{equation}
\begin{equation}\label{eq:ans2}
p^0_t(s^t) = \beta^t\,\pi(s^t|s_0)\,\dfrac{u\,'(y_t)}{u\,'(y_0)} \hspace{.75in} \textrm{using (\ref{eq:ceqe})}
\end{equation}
\vspace{.25in}
\\

\newpage
Suppose $p_{11} = .8, p_{22}=.85, \pi_0=.5, \beta =.96,$ and $\gamma = 2$.  Suppose economy starts off with $\lambda_0=.98$ and $y_0 = 1$.

Also, $P = \left[ \begin{array}{c c}
p_{11} & 1-p_{11}  \\
1-p_{22} & p_{22}
\end{array} \right] =
\left[ \begin{array}{c c}
.8 & .2  \\
.15 & .85
\end{array} \right]$ \hspace{.05cm} and intial distribution $\pi_{\lambda} = [\pi_0, 1-\pi_0] = [0.5, .5]$. 
  The value of $\lambda_t$ is given by $\overline{\lambda_1} = 0.98$ in state 1 and $\overline{\lambda_2} = 1.03$ in state 2.


\item[(c)]  




I will use the given conditions in the setup of the problem, stated above, and using the properties of a Markov chain, (that is, as time $\rightarrow \infty$, the probability of each state occuring will converge to a stationary distrubition) to solve for the average growth rate of consumption.
\\
\textbf{Goal:}  Find a stationary distribution such that,
\begin{equation}
\sum a_i = 1 \hspace{.15cm} \textrm{and} \hspace{.15cm} a = aP, \hspace{.15cm} \textrm{where} \hspace{.15cm} a=[\alpha, 1-\alpha]
\end{equation}

Using $P$, setup a system of equations to solve for $\alpha$.  
\vspace{.15cm}
\\
\begin{tabular}{l l}
& \hspace{.75cm} $a = aP \longrightarrow [\alpha, 1-\alpha] = [\alpha, 1-\alpha] \, \left[\begin{array}{c c} p_{11} & 1-p_{11}  \\ 1-p_{22} & p_{22} \end{array} \right]$
\end{tabular}
\vspace{.5cm}
\\
\begin{tabular}{l l l}
\textbf{From top equation, (Eq1)} & \hspace{.05cm}  & \hspace{.05cm} \textbf{From bottom equation, (Eq2)}
\vspace{.25cm}
\\
$\alpha = \alpha\,p_{11} + (1-\alpha)(1-p_{22})$  & \hspace{.05cm} and  & \hspace{.05cm} $(1-\alpha) = \alpha\,(1-p_{11}) + (1-\alpha)p_{22}$
\vspace{.25cm}
\\
$\alpha = \alpha\,p_{11} + 1-\alpha-p_{22}+ \alpha\,p_{22}$  & \hspace{.05cm}   & \hspace{.05cm} $(1-\alpha)+ (1-\alpha)p_{22} = \alpha\,(1-p_{11})$
\vspace{.25cm}
\\
$\alpha + \alpha - \alpha\,p_{11} -  \alpha\,p_{22}= 1- p_{22}$  & \hspace{.05cm}   & \hspace{.05cm} $(1-\alpha)(1-p_{22}) = \alpha\,(1-p_{11})$
\vspace{.25cm}
\\
$\alpha(2-p_{11} -p_{22})= 1- p_{22}$  & \hspace{.05cm}   & \hspace{.05cm} $\dfrac{(1-\alpha)}{\alpha} = \dfrac{(1-p_{11})}{(1-p_{22})}$
\vspace{.25cm}
\\
$\alpha = \dfrac{1- p_{22}}{(2-p_{11} -p_{22})}$  & \hspace{.05cm}   & \hspace{.05cm} $\dfrac{1}{\alpha} = \dfrac{(1-p_{11})}{(1-p_{22})}+1 = \dfrac{(1-p_{11}+1-p_{22})}{(1-p_{22})}$
\end{tabular}

Solving in both equations for $\alpha$ obtain equalvalent realtions, which is suppose to happen since it is a stationary distribution.


\begin{equation}\label{eq:station}
\textbf{From Eq1:}\; \alpha = \dfrac{1- p_{22}}{2-p_{11} -p_{22}}  \hspace{.5in} \textbf{From Eq2:}\; \alpha = \dfrac{1-p_{22}}{2-p_{11}-p_{22}} 
\end{equation}

Therefore subsituting the values for $p_{11}=.8$ and $p_{22}=.85$ in (\ref{eq:station}) will obtain:
\begin{displaymath}
\alpha = \dfrac{1- p_{22}}{2-p_{11} -p_{22}} = \dfrac{1- .85}{2-.80 -.85} = \dfrac{1- .85}{2-.80 -.85} = \dfrac{3}{7} \end{displaymath}
\vspace{.25cm}
\\
\begin{equation}\label{eq:station1}
a=[\alpha, 1-\alpha] =\left[\dfrac{3}{7}\,, \dfrac{4}{7} \right] 
\end{equation}

By properties of Markov process, as time $\rightarrow \infty$, the probability of the each state occuring will converge to the corresponding probabilities in the stationary distribution.  Hence:

\begin{equation}\label{eq:prob}
P(\lambda_t = \overline{\lambda_i}\,) = a \hspace{.25cm} \textrm{where i=1,2}
\end{equation}

Therefore will obtain
\begin{equation}\label{eq:prob}
P(\lambda_t = \overline{\lambda_1}\,) = \dfrac{3}{7} \quad \textrm{and} \quad P(\lambda_t = \overline{\lambda_2}\,) = \dfrac{4}{7}
\end{equation}

Using (\ref{eq:prob}), able to compute the average growth rate of consumption, before having observed $\lambda_0$.

{\setlength\arraycolsep{2pt}
\begin{eqnarray}
E[\lambda] & = & P(\lambda_t = \overline{\lambda_1}\,)\overline{\lambda_1} + P(\lambda_t = \overline{\lambda_2}\,)\overline{\lambda_2}
\nonumber\\
& = &  \left(\dfrac{3}{7}\right)(.98)+ \left(\dfrac{4}{7}\right)(1.03) \nonumber\\
& = &  1.0086 \label{eq:ansC}
\end{eqnarray}}



In order to compute the time-$0$ prices for the for each of the three bonds promising to pay one unit of time-$j=0,1,2$.
\\Assume that given intial state, $\lambda_0 = .98 = \overline{\lambda_1}$.  Thus, has corresponding values, $p_{11}=.8$, and $1-p_{11}=.2$ on first branch of tree.  

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\put(0,0){\line(2,-1){4}}
\put(0,-4.5){$j=0$}
\put(4,-4.5){$j=1$}
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\put(1.5,1.75){$p_{11}=.8$}
\put(.85,-1.95){$1-p_{11}=.2$}
\put(5.5,2.95){$.8$}
\put(5.5,.75){$.2$}
\put(4.85,-1.75){$p_{22}=.85$}
\put(4.25,-3.5){$1-p_{22}=.15$}
\put(-.75,.5){state $1$}
\put(-.75,-.75){$\lambda_0 = .98$}
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\put(3.5,1.25){$\overline{\lambda_1} = .98$}
\put(2.75,.85){price $=0.7997$}
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\put(3.5,-1.5){state $2$}
\put(3,-2.75){$\overline{\lambda_1} = 1.03$}
\put(3,-3.1){price $=.1810$}

\put(7.5,4){state $1$}
\put(7.5,3.5){$\overline{\lambda_2} = .98$}
\put(7.5,3){price $=0.6394$}

\put(7.5,1.75){state $2$}
\put(7.5,1.25){$\overline{\lambda_2} = 1.03$}
\put(7.5,.75){price $=0.1447$}

\put(7.5,0){state $1$}
\put(7.5,-.5){$\overline{\lambda_1} = .98$}
\put(7.5,-1){price $=0.027$}

\put(7.5,-2.5){state $2$}
\put(7.5,-3){$\overline{\lambda_2} = 1.03$}
\put(7.5,-3.5){price $=0.1392$}


\end{picture}
\vspace{3.5in}
\\
From (\ref{eq:ans2}), using the following given information, $u(c)=\dfrac{c^{1-\gamma}}{1-\gamma}$, where $\gamma=2$, $\beta=.96$, and $y_0=1$. \\
Also known from (\ref{eq:income1}), $y_t= \prod^{t}_{i=1}\lambda_i\,y_0$ \hspace{.05cm} $\rightarrow$ \hspace{.05cm} $y_t= \prod^{t}_{i=1}\lambda_i$ \hspace{.05cm} \\
\\
From the given conditions, will obtain
\begin{eqnarray}\label{eq:equa1}
p\,^0_t(s^t) & = & \beta^t\,\pi(s^t|s_0)\,\dfrac{u\,'(y_t)}{u\,'(y_0)} \nonumber
\vspace{.5cm}
\\
p\,^0_t(s^t) & = & (.96)^t\,\pi(s^t|s_0)\,\left(\dfrac{y_0}{y_t}\right)^2 \label{eq:equa2}
\end{eqnarray}

I will use (\ref{eq:equa2}), to compute the price at each time $t=0,1,2$

\newpage


\begin{itemize}
\item[t=1]
\item $p\,^0_1(\overline{s^1}|s_0) = (.96)\,(.8)(.2)\,\left(\dfrac{1}{.98}\right)^2 = .7997$
\item $p\,^0_1(\underline{s^1}|s_0) = (.96)\,(.2)\,\left(\dfrac{1}{1.03}\right)^2 = .1810$

\item[t=2]
\item $p\,^0_2(\overline{s^2}|s_0) = (.96)^2\,(.8)(.8)\,\left(\dfrac{1}{(.98)(.98)}\right)^2 = .6394$
\item $p\,^0_2(\overline{\underline{s^2}}|s_0) = (.96)^2\,(.2)(.15)\,\left(\dfrac{1}{(1.03)(.98)}\right)^2 = .027$
\item $p\,^0_2(\underline{s^2}|s_0) = (.96)^2\,(.8)(.2)\,\left(\dfrac{1}{(1.03)(.98)}\right)^2 = .1447$
\item $p\,^0_2(\underline{\underline{s^2}}|s_0) = (.96)^2\,(.85)(.15)\,\left(\dfrac{1}{(1.03)(1.03)}\right)^2 = .1392$

\end{itemize}


From Cochrane, (pg.52 1ed.), using to what he refers to as the \textbf{happy-meal theorem (3.1):}

\begin{equation}\label{eq:happy6}
p(x) =  \sum_s \,pc(s)x(s)
\end{equation}

where $x(s)$ denotes an asset's payoff in state of nature $s$, $pc$ is the price of a contingent claim and $(s)$ is used to denote which state $s$ the claim pays off.  In problems \textbf{(d)-(f)}, the bond promises to pay one unit, $\mathbf{x(s)=1}$.  From (\ref{eq:happy6}), able to compute the price of the bond promising to pay one unit of time-$j$ comsumption for $j=0,1,2$. 
\vspace{.33cm}
\\


\item[(d)]  Compute the time-$0$ prices of three risk-free discount bonds, in particular, those promising to pay one unit of time-$j$ consumption for $j=0,1,2$, respectively.


\begin{itemize}
\item[t= 0]
\item $p(x) = 1$
\item[t= 1]
\item $p(x) = .7997 + .1810 = .9807$
\item[t= 2]
\item $p(x) = .6394+.027+.1447+.1392 = .9508$
\end{itemize}


\item[(e)] Compute the time-$0$ prices of three bonds, in particular, those promising to pay one unit of time-$j$ contingent consumption on $\lambda_j = \overline{\lambda_1}$ for $j=0,1,2$, respectively.


\begin{itemize}
\item[t= 0]
\item $p(x) = 1$ \hspace{.25cm} from $\lambda_0 = \overline{\lambda_1}$
\item[t= 1]
\item $p(x) = .7997$
\item[t= 2]
\item $p(x) = .6394+.027 = .666$
\end{itemize}

\newpage

\item[(f)] Compute the time-$0$ prices of three bonds, in particular, those promising to pay one unit of time-$j$ contingent consumption on $\lambda_j = \overline{\lambda_2}$ for $j=0,1,2$, respectively.


\begin{itemize}
\item[t= 0]
\item $p(x) = 0$ \hspace{.25cm} from $\lambda_0 = \overline{\lambda_1}$
\item[t= 1]
\item $p(x) = .1810$
\item[t= 2]
\item $p(x) = .1447+.1392 = .2839$
\end{itemize}


\item[(g)]  Compare the prices that you computed in parts \textbf{(d)-(f)}.
\vspace{.25cm}
\\
From above, can see that at each time $t$, the sum of each price in answers \textbf{(e)} and \textbf{(f)}, is equal to the price of the risk free discount bond, \textbf{(d)}.  
\vspace{.25cm}
\\
At each time, $t=0,1,2$ \textbf{(e)+(f)=(d)}.

\newpage

\end{description}



\end{enumerate}

\end{document}