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\begin{document}
\begin{flushright}
Jaime Frade\\
ECO5282-Dr. Garriga\\
Homework\:$\#3$\\
\end{flushright}
\begin{enumerate}
\item Consider and economy with a representative consumer with preferences described by $E_0\sum^\infty_{t=0}\beta^t\,u(c_t)$ where $u(c_t) = \ln(c_t + \gamma)$ where $\gamma \geq 0$ and $c_t$ denotes consumption of the fruit in period $t$.  The sole source of the single good is an everlasting tree that produces $d_t$ units of the consumption good in period $t$.  The dividend process $d_t$ is Markov, with $prob\{d_{t+1} \leq d' |d_t=d\} = F(d',d).$  Assume the conditional density $f(d',d)$ of $F$ exists.  There are competitive markets in the title of trees and in state-contingent claims.  Let $p_t$ be the price at $t$ of a title to all future dividends from the tree.
\begin{description}
\item[(a)]  Prove that the equilibrium prices $p_t$ satisfies
\begin{equation}\label{eq:ques1a}
p_t = (d_t + \gamma) \sum^\infty_{j=1} \beta^t\,E_t\left(\dfrac{d_{i+j}}{d_{i+j}+\gamma}\right)
\end{equation}
For a given household, the comsumer solves
\begin{equation}\label{eq:max1a}
\max E_0\sum^{\infty}_{t=0}\beta^t\,u(c_t)
\end{equation}
Subject to the following budget constraint
\begin{equation}\label{eq:bc1a}
A_{t+1}=R_t(A_t+y_t-c_t)
\end{equation}
From (\ref{eq:bc1a}), 
\begin{eqnarray}\nonumber
c_t & \Longrightarrow & \textrm{consumption of an agent at time} \hspace{.15cm} t \nonumber
\vspace{.25cm}
\\
y_t & \Longrightarrow & \textrm{agent's labor income}\nonumber
\vspace{.25cm}
\\
A_t & \Longrightarrow & \textrm{amount of a single asset valued in the units of the } \nonumber
\\
 & &\textrm{consumption good held at start of time} \hspace{.15cm} t \nonumber
\\
& & \textrm{(Note:}\hspace{.15cm} A_0 \hspace{.15cm} \textrm{is given)} \nonumber
\vspace{.25cm}
\\
R_t & \Longrightarrow & \textrm{real gross rate of return on asset between time} \hspace{.15cm} t \hspace{.15cm} \textrm{and}\hspace{.15cm} t+1 \nonumber
\end{eqnarray}

By using the Legrange multipliers, combining the first order conditions in (\ref{eq:max1a}) we obtain the \textbf{Euler equation} that is necessary condition of optimality for  $t=0, 1, \ldots$.
\begin{equation} \label{eq:euler1a}
u\:'(c_t) = E_t \beta\,R_t  u\,'(c_{t+1})
\end{equation}



However (\ref{eq:euler1a}) does not applicable to models with complete general equilibrium.  In the Lucas's asset-pricing model, general equilibrium is used.  This model assumes that the labor income is zero, $y_t =0$, $\forall$ $t$.  It also assumes that the only durable good in the economy is a set of trees, (which equal the number of people in economy).  The tree also yields fruit (dividends) in the amount of $d_t$ to its owner ar the beginning of time $t$.  This fruit is also nonstorable.   Each consumer starts a $t=0$ with, one tree and initial  dividend of good, (fruit).  

\begin{eqnarray}\nonumber
\textrm{Let} \hspace{.15cm} p_t & \Longrightarrow & \textrm{price of a tree at time} \hspace{.15cm} t \nonumber
\vspace{.25cm}
\\
R_t & = & \dfrac{p_{t+1}+d_{t+1}}{p_t} \label{eq:sub1}
\end{eqnarray}

Using (\ref{eq:sub1}) subsitute into (\ref{eq:euler1a}), will obtain the following:
\begin{equation} \label{eq:euler1a-2}
1 = E_t\,\beta\,\dfrac{u\,'(c_{t+1})}{u\,'(c_{t})}\left(\dfrac{p_{t+1}+d_{t+1}}{p_t} \right)
\end{equation}

\newpage

In this economy, equilibrium must exist because all consumers are identical, the utility function is strictly increasing, and there is only one place for goods, period's dividends.  Therefore obtain:

\begin{equation} \label{eq:equal}
c_t = d_t
\end{equation}

Substituting (\ref{eq:equal})  into (\ref{eq:euler1a-2}) and solve for $p_t$, will obtain the following:
\begin{eqnarray}\label{eq:euler1a-3}
p_t & = & E_t\,\beta\,\dfrac{u\,'(d_{t+1})}{u\,'(d_{t})}\left(p_{t+1}+d_{t+1}\right)
\end{eqnarray}

Given $u(c_t) = \ln(c_t + \gamma)$ where $\gamma \geq 0$, and using (\ref{eq:equal}) and differentiating  to plug into (\ref{eq:euler1a-3}) will obtain:
\begin{equation}\label{eq:price1}
p_t  = E_t\,\beta\,\dfrac{(d_t+\gamma)}{(d_{t+1}+\gamma)}\left(\underbrace{p_{t+1}}_{\Uparrow}+d_{t+1}\right)
\end{equation}

Using recursions on (\ref{eq:price1}) for the value $p_{t+1}$, will obtain:
\begin{eqnarray}\nonumber
p_t  &=& E_t\,\left[\beta\,\dfrac{(d_t+\gamma)}{(d_{t+1}+\gamma)}\left(d_{t+1} +\underbrace{\beta\,\dfrac{(d_{t+1}+\gamma)}{(d_{t+2}+\gamma)}\left(p_{t+2}+d_{t+2}\right)}_{= p_{t+1}}\right)\right]
\end{eqnarray}

\begin{eqnarray}\nonumber
p_t  &=& E_t\,\left[\beta\,\dfrac{(d_t+\gamma)}{(d_{t+1}+\gamma)}\left(d_{t+1} 
+\beta\,\dfrac{(d_{t+1}+\gamma)}{(d_{t+2}+\gamma)}\left(d_{t+2} + \beta\,\dfrac{(d_{t+2}+\gamma)}{(d_{t+3}+\gamma)}\left(d_{t+3}+ p_{t+3}\right) \right)\right)\right]
\end{eqnarray}

\begin{eqnarray}\nonumber
p_t  &=& E_t\,\left[\beta\,\dfrac{(d_t+\gamma)}{(d_{t+1}+\gamma)}d_{t+1} 
+\beta^2\,\dfrac{(d_t+\gamma)}{(d_{t+2}+\gamma)}d_{t+2} +\beta^3\,\dfrac{(d_t+\gamma)}{(d_{t+3}+\gamma)}d_{t+3} + \cdots \right]
\end{eqnarray}

From above equation, can see a pattern forming, obtain
\begin{eqnarray}\nonumber
p_t  &=&\, (d_t+\gamma) E_t\sum^{\infty}_{j=0} \left[\beta^j\,\left(\dfrac{d_{t+j}}{d_{t+j}+\gamma}\right)\right] + \lim_{j \rightarrow \infty}E_t\left[\,\beta^j\,\left(\dfrac{d_t+\gamma}{d_{t+j}+\gamma}\right)p_{t+j}\right]
\end{eqnarray}

As $j \rightarrow \infty$ $\lim$ $\rightarrow$ 0, will prove result for pricing equation.
\begin{eqnarray}\label{eq:Ans1a}
p_t  &=&\, (d_t+\gamma) E_t\sum^{\infty}_{j=0} \beta^j\,\left(\dfrac{d_{t+j}}{d_{t+j}+\gamma}\right)
\end{eqnarray}



\newpage

\item[(b)]  
Modify the Lucas Model to a market that consist of one- and two period perfectly safe loans, which bear gross rates of return $R_{1t}$ and $R_{2t}$.  $R^{-1}_{1t}$ is the price of a perfectly sure claim to one unit of consumption at time $(t+1)$.  $R^{-1}_{2t}$ is the price of a perfectly sure claim to one unit of consumption at time $(t+2)$.  

The agent needs to solve the following maximum

\begin{equation}\label{eq:max1b}
\max_{c_t,L_{1t+1},L_{2t+1}} E_0 \sum^{\infty}_{t=0} \beta^t\,u(c_t)
\end{equation}

Subject to the following budget constraint, where $L_{jt}$ is the amount lent for $j$ period at time $t$.
\begin{equation}\label{eq:bc2}
c_t + L_{1t} + L_{2t} \leq d_t + R_{1t-1}L_{1t-1} + R_{2t-2}L_{2t-2}
\end{equation}

Solving (\ref{eq:max1b}), using Legrange Multiplier, where $\{\lambda_t\}$ is a sequence of random Legrange Multipliers.  Will obtain the following:

\begin{equation}\label{eq:L1}
L=E_0\sum^\infty_{t=0}\beta^t\,\left[u(c_t)+\lambda_t(d_t+L_{1t-1}R_{1t-1}+L_{2t-2}R_{2t-2}-c_t-L_{1t}-L_{2t})\right]
\end{equation}
\underline{First Order Conditions (\ref{eq:max1b})} with respect to $\{c_t,L_{1t},L_{2t}\}$.  Also $E_0=E_t$ since the Markov property.
\vspace{.25in}
\\
\begin{tabular}{c c c c}
& \hspace{.25cm} $\left[c_t\right]:$ &  \hspace{.25cm}  $E_0 \beta^t (u'(c_t)-\lambda_t)=0$ &  $\Longrightarrow$ \hspace{.5cm} $\lambda_t=u'(c_t)$
\vspace{.25cm}
\\
& \hspace{.25cm} $\left[L_{1t} \right]:$ &  \hspace{.25cm}  $E_0 \beta^t (\beta \lambda_{t+1} R_{1t} -\lambda_t)=0$ & \hspace{.95cm}    $\Longrightarrow$ \hspace{.5cm} $\lambda_t=E_t (\beta \lambda_{t+1} R_{1t} )$
\vspace{.25cm}
\\
& \hspace{.25cm} $\left[L_{2t} \right]:$ &  \hspace{.25cm}  $E_0 \beta^t (\beta^2 \lambda_{t+2} R_{2t} -\lambda_t)=0$ & \hspace{.95cm}   $\Longrightarrow$ \hspace{.5cm} $\lambda_t=E_t (\beta^2 \lambda_{t+2} R_{2t})$
\end{tabular}

Combining the first order conditions above will obtain the following \textbf{Euler Equations}:
\begin{eqnarray}
	E_t \left [ \beta \dfrac{u'(c_{t+1})}{u'(c_t)}R_{1t} \right] =1 \label{eq:euler1}
	\\
	E_t \left [ \beta^2 \dfrac{u'(c_{t+2})}{u'(c_t)}R_{2t} \right] =1 \label{eq:euler2}
\end{eqnarray}

Given that both assets are riskless, rewrite (\ref{eq:euler1}) and (\ref{eq:euler2}) as:
\begin{eqnarray}
	R^{-1}_{1t} & =&  E_t \left[ \beta \dfrac{u'(c_{t+1})}{u'(c_t)} \right] \label{eq:euler1-1}
	\\
	R^{-1}_{2t} & = &  E_t \left [ \beta^2 \dfrac{u'(c_{t+2})}{u'(c_t)} \right] \label{eq:euler2-2}
\end{eqnarray}




\begin{itemize}
\item[(i)]  Find a formula for the risk-free one-period interest rate $R_{1t}$.  
\vspace{.25cm}
\\
From (\ref{eq:euler1-1}), using given $u(c_t) = \ln(c_t+\gamma)$ and (\ref{eq:equal}) in equilibrium, will obtain a formula for th risk-free one-period interest rate $R_{1t}$.
\begin{eqnarray}
	R^{-1}_{1t} & =&  E_t \left[ \beta \dfrac{(d_t+\gamma)}{d_{t+1}+ \gamma} \right] \label{eq:ans1b}
\end{eqnarray}




\item[(ii)]  Prove that in the special case in which $\{d_t\}$ is independently and identically distributed, $R_{1t}$ is given by $R^{-1}_{1t} = \beta\,k\,(d_{t}+\gamma)$, where $k$ is a constant.  
\vspace{.25cm}
\\
If assume that $\{d_t\}$ is $i.i.d$, then (\ref{eq:ans1b}) can be rewritten:
\begin{eqnarray}
	R^{-1}_{1t} & =& (d_t+\gamma) \beta\, E_t \left[  \dfrac{1}{d_{t+1}+ \gamma} \right] \nonumber
	\\
		R^{-1}_{1t} & =& (d_t+\gamma) \beta\, E_t \left[  \dfrac{1}{d_{t+1}+ \gamma} \right] \hspace{.5cm} \Longrightarrow \textrm{Let} \hspace{.25cm} k =  E_t  \dfrac{1}{d_{t+1}+ \gamma} \nonumber
		%\\
		%	& &  \nonumber
	  \\  R^{-1}_{1t} & =& (d_t+\gamma) \beta\, k \label{eq:ans1b-2}
\end{eqnarray}


\item[(iii)]Give a formula for $k$.
\vspace{.25cm}
\\
From (\ref{eq:ans1b-2}), and assumption that $\{d_t\}$ is $i.i.d$, obtain a formula for $k$, which is not a function of time:
\begin{eqnarray}
 k =  E_t \left[ \dfrac{1}{d_{t+1}+ \gamma} \right] \label{eq:ans1b-3}
\end{eqnarray}

\end{itemize}

\item[(c)]  
\begin{itemize}
\item[(i)]  Find a formula for the risk-free two-period interest rate $R_{2t}$.  
\vspace{.25cm}
\\
From (\ref{eq:euler2-2}), using given $u(c_t) = \ln(c_t+\gamma)$ and (\ref{eq:equal}) in equilibrium, will obtain a formula for th risk-free one-period interest rate $R_{2t}$.
\begin{eqnarray}
	R^{-1}_{2t} & =&  E_t \left[ \beta^2 \dfrac{(d_t+\gamma)}{d_{t+2}+ \gamma} \right] \label{eq:ans1c-1}
\end{eqnarray}

\item[(ii)]  Prove that in the special case in which $\{d_t\}$ is independently and identically distributed, $R_{2t}$ is given by $R^{-1}_{2t} = \beta^2\,k\,(d_{t}+\gamma)$, where $k$ is the same constant you found in part \textbf{(b)}. 
\vspace{.25cm}
\\
If assume that $\{d_t\}$ is $i.i.d$, then (\ref{eq:ans1b}) can be rewritten:
\begin{eqnarray}
	R^{-1}_{2t} & =& (d_t+\gamma) \beta^2\, E_t \left[  \dfrac{1}{d_{t+2}+ \gamma} \right] \nonumber
	\\
		R^{-1}_{2t} & =& (d_t+\gamma) \beta^2\, E_t \left[  \dfrac{1}{d_{t+2}+ \gamma} \right] \hspace{.5cm} \Longrightarrow \textrm{Let} \hspace{.25cm} j =  E_t  \dfrac{1}{d_{t+2}+ \gamma} \nonumber
		%\\
		%	& &  \nonumber
	  \\  R^{-1}_{2t} & =& (d_t+\gamma) \beta^2\, j \label{eq:ans1c-2}
\end{eqnarray}


\item Prove that $j$ from (\ref{eq:ans1c-2})  is same $k$ from (\ref{eq:ans1b-3}).  From $\{d_t\}$ is $i.i.d$ and given that the dividend process $d_t$ is Markov, with $prob\{d_{t+1} \leq d' |d_t=d\} = F(d',d).$
\vspace{.25cm}
\\
\begin{eqnarray}
 j &=&  E_t \left[ \dfrac{1}{d_{t+2}+ \gamma} \right] \nonumber
 \\
 	&=&  E_{t+1} \left[ \dfrac{1}{d_{t+2}+ \gamma} \right] \nonumber
 	\\
 		&=& k \nonumber
\end{eqnarray}



 
\end{itemize}
\end{description}

%%

\newpage


\item Consider the following version of the Lucas's tree economy.  There are two kinds of trees.  The first kind is ugly and gives no direct utility to consumers, but yields a stream of fruit $\{d_{1t}\}$, where $d_{1t}$ denotes a positive random process obeying a first-order Markov process.  The second tree is beautiful and yields utilty on itself.  This tree also yields a stream of the same kind of fruit $d_{2t}$, where it happens that $d_{2t} = d_{1t}= \left(\frac{1}{2}\right)d_t$ $\forall$ $t$, so that the physical yields of the two kinds of trees are equal.  There is one of each tree for each $N$ individuals in the economy.  Trees last forever, but the fruit is not storable.  Trees are the only source of fruit.  
\vspace{.15cm}
\\
Each of the $N$ individuals in the economy has preferences described by 
\begin{equation}\label{eq:ques2}
E_0\sum^\infty_{t=0}\beta^t\,u(c_t, s_{2t})
\end{equation}
where $u(c_t, s_{2t})= \ln c_t + \gamma \ln(s_{2t})$ where $\gamma \geq 0$, $c_{t}$ denotes consumption of the fruit in period $t$ and $s{2t}$ is the stock of beautiful trees owned at the beginning of the period $t$.  The owner of a tree of either kind $i$ at the start of the period receives the fruit $d_{it}$ produced by the tree during that period. 
\vspace{.15cm}
\\
Let $p_{it}$ be the price of a tree of type $i$ (where $i=1,2)$ during period $t$.  Let $R_{it}$ be the gross rate of returns of tree $i$ during that period held from period $t$ to $t+1$.

\begin{description}
\item[(a)]  Write down the consumer optimization problem in sequential and recursive form.
\vspace{.25cm}
\\
\textbf{Consumer optimization problem in \underline{sequential form}}

\begin{equation} \label{eq:2aseq}
\max_{\{c_t,s_{1t+1},s_{2t+1}\}} E_0\,\sum^\infty_{t=0}\beta^t\, (\ln(c_t)+\gamma \ln (s_{2t}))
\end{equation}

\begin{equation} \label{eq:2aseqbc}
\textrm{\textbf{subject to}} \hspace{.25cm} c_t+p_{1t}s_{1t+1}+p_{2t}s_{2t+1} \leq (d_{1t}+p_{1t})s_{1t}+(d_{2t}+p_{2t})s_{2t}.
\end{equation}

\textbf{Consumer optimization problem in \underline{recursive form}} 
\vspace{.25cm}
\\
The Bellman's equation is given by
\begin{equation} \label{eq:2arec}
v(d_t,s_{1t},s_{2t})=\max_{\{c_t,s_{1t+1},s_{2t+1}\}} 
  \left( \ln(c_t)+\gamma \ln(s_{2t}) +E_t\, \beta\, v(d_t,s_{1t+1},s_{2t+1}) \right)
\end{equation}

\begin{equation} \label{eq:2arecbc}
\textrm{\textbf{subject to}} \hspace{.25cm} c_t+p_{1t}s_{1t+1}+p_{2t}s_{2t+1} \leq (d_{1t}+p_{1t})s_{1t}+(d_{2t}+p_{2t})s_{2t}.
\end{equation}
  

\item[(b)]  Define a rational expectations equilibrium.
\vspace{.25cm}
\\
\textbf{Defintion:} A \underline{rational expectations equilibrium} is an allocation $\{\{c^i_t\}^\infty_{t=0}\}_{i=1}^{I}$, $\{\{s^i_{1t},s^i_{2t}\}^\infty_{t=0}\}_{i=1}^{I}$, and a price system, $\{p_{1t},p_{2t}\}^\infty_{t=0}$, such it solves each household problem and satisfies the market clear condition.
\vspace{.33cm}
\\
\textbf{Defintion:} For the \underline{market clearing condition} to exist, 
\vspace{.15cm}
\\
$\sum^I_{i=1} c^i_t = \sum^I_{i=1} d^i_t$
, \hspace{.25cm} $\sum^I_{i=1} s^i_{1t} = \sum^I_{i=1} s^i_{10}=I,$ \hspace{.25cm} $ \sum^I_{i=1} s^i_{2t} = \sum^I_{i=1} s^i_{20}=I$, 
\vspace{.15cm}
\\
where $s^i_{10}$ and $s^i_{20}$ are each agent's number of trees at initial time. 
\vspace{.33cm}
\\
The \underline{household problem} is defined sequentially by each agent's utility optimization problem:  

\begin{equation}\label{eq:2b-equ}
\max_{\{c_t,s_{1t+1},s_{2t+1}\}} E_0\,\sum^\infty_{t=0}\beta^t\, (\ln(c_t)+\gamma \ln (s_{2t}))
\end{equation}

\begin{equation}\label{eq:2b-BC}
\textrm{\textbf{subject to}} \hspace{.25cm} c_t+p_{1t}s_{1t+1}+p_{2t}s_{2t+1} \leq (d_{1t}+p_{1t})s_{1t}+(d_{2t}+p_{2t})s_{2t}
\end{equation}

\newpage

\item[(c)]  Find the pricing functions mapping the state of the ecomony at $t$ unto $p_{1t}$ and $p_{2t}$ (give precise formulas).  [Hint:  You should be able to directly derive $p_{1t}$ from the example seen in class, then since pricing function have to be linear you can guess a pricing function $p_{2t} = kd_t$ and solve for $k$ parameter using Euler equation of the second stock.]
\vspace{.25cm}
\\
From (\ref{eq:2b-equ}), using Legrange multipilers with the given budget constraints, (\ref{eq:2b-BC}), we obtain the following:

\begin{eqnarray}\label{eq:L2}
L&=& E_0\,\sum^\infty_{t=0}\beta^t\, ((\ln(c_t)+\gamma \ln (s_{2t})) \nonumber \\
&+&\lambda_t ((d_{1t}+p_{1t})s_{1t}+(d_{2t}+p_{2t})s_{2t} - c_t-p_{1t}s_{1t+1}-p_{2t}s_{2t+1} ))
\end{eqnarray}

\underline{First Order Conditions (\ref{eq:L2})} with respect to $\{c_t,s_{1t+1},s_{2t+1}\}$.  Also using $E_0=E_t$ since the Markov property.
\vspace{.25in}
\\
\begin{tabular}{c c c c}
& \hspace{.05cm} $\left[c_t\right]:$ &  \hspace{.75cm}  $E_0\, \beta^t\, (\dfrac{1}{c_t}-\lambda_t)=0$ & \hspace{.5cm} $\Longrightarrow$ \hspace{.1cm} $\dfrac{1}{c_t}=\lambda_t$
\end{tabular}
\vspace{.25cm}
\\
\begin{tabular}{c c c c}
& \hspace{.05cm} $\left[s_{1t+1}\right]:$ &  \hspace{.25cm}  $E_0\, \beta^t\, (\beta \lambda_{t+1} (d_{1t+1}+p_{1t+1}) -\lambda_t p_{1t})=0$ & \hspace{.65cm}    $\Longrightarrow$ \hspace{.5cm} $ p_{1t}=E_t \beta \dfrac{\lambda_{t+1}}{\lambda_{t}}(d_{1t+1}+p_{1t+1})$
\end{tabular}
\vspace{.25cm}
\\
\begin{tabular}{c c c c}
& \hspace{.05cm} $\left[s_{2t+1}\right]:$ &  \hspace{.25cm}  $E_0\, \beta^t\, \left(\beta \lambda_{t+1} (d_{2t+1}+p_{2t+1}) -\lambda_t p_{2t}+\beta \dfrac{\gamma}{s_{2t+1}}\right)=0$ 
\end{tabular}
\vspace{.25cm}
\\
\begin{tabular}{c c c c}
&   &    \hspace{2in} $\Longrightarrow$  $p_{2t}=E_t \beta \dfrac{\lambda_{t+1}}{\lambda_{t}}(d_{2t+1}+p_{2t+1})+\beta \dfrac{\gamma}{s_{2t+1}\lambda_t}$
\end{tabular}
\vspace{.25cm}
\\
Combining the first order conditions above will obtain the following \textbf{pricing formula}:
\begin{eqnarray}
 	p_{1t}&=&E_t\, \beta\, \dfrac{c_{t}}{c_{t+1}}(d_{1t+1}+p_{1t+1})\label{eq:2price1} 
 	\\
	p_{2t}&=&E_t\, \beta\, \dfrac{c_{t}}{c_{t+1}}\left(d_{2t+1}+p_{2t+1}\right) +\beta \dfrac{\gamma c_t}{s_{2t+1}}\label{eq:2price2}
 \label{eq:}
\end{eqnarray}


In this economy, for general equilibrium to exist:

\begin{equation} \label{eq:2equal}
c_t = d_t
\end{equation}

From (\ref{eq:2equal}), substitute into the pricing formulas in (\ref{eq:2price1}) and (\ref{eq:2price2}), obtain:
\begin{eqnarray}
	p_{1t}&=&E_t\, \beta\, \dfrac{d_{t}}{d_{t+1}}(d_{1t+1}+p_{1t+1}) \label{eq:2price1-1}
	\\
	p_{2t}&=&E_t\, \beta\, \dfrac{d_{t}}{d_{t+1}}\left(d_{2t+1}+p_{2t+1}\right) +\beta \dfrac{\gamma d_t}{s_{2t+1}} \label{eq:2price2-2}
\end{eqnarray}	

For $p_{1t}$, if assuming a the pricing function have to be of linear form obtain:

\begin{equation} \label{eq:linearprice1}
p_{1t}=k_{1t}d_t
\end{equation}

From (\ref{eq:linearprice1}), substitute this into (\ref{eq:eq:2price1-1}) to obtain:

\begin{eqnarray}
k_{1t}d_t&=&E_t\,\beta\, \dfrac{d_{t}}{d_{t+1}}(d_{1t+1}+k_{1t+1}d_{t+1}) \nonumber
\\
k_{1t}&=&E_t\,\beta\, \underbrace{k_{1t+1}}_{\Uparrow} + E_t\,\beta\, \dfrac{d_{1t+1}}{d_{t+1}} \label{eq:k1form}
\\
& & \Longrightarrow \hspace{.05cm} \textrm{apply recursions on} \hspace{.05cm}  k_{1t+1} \nonumber
\\
k_{1t}& = &E_t\, \sum_{j=1}^{\infty} \beta^j \dfrac{d_{1t+j}}{d_{t+j}} \label{eq:k1formans}
\end{eqnarray}

From $d_{2t} = d_{1t}= \left(\frac{1}{2}\right)d_t$ $\forall$ $t$,  rewrite (\ref{eq:k1formans}) to obtain:

\begin{eqnarray}
	k_{1t}&=&\dfrac{1}{2}\,E_t\, \sum_{j=1}^{\infty}\, \beta^j \nonumber
	\\
	&=&\dfrac{\beta}{2(1-\beta)} \label{eq:k1formans-1}
\end{eqnarray} 

For $p_{2t}$, if assuming a the pricing function have to be of linear form obtain:

\begin{equation} \label{eq:linearprice2}
p_{2t}=k_{2t}d_t
\end{equation}

From (\ref{eq:linearprice2}), substitute this into (\ref{eq:eq:2price2-2}) to obtain:

\begin{eqnarray}
k_{2t}d_t&=&E_t\,\beta\, \dfrac{d_{t}}{d_{t+1}}(d_{2t+1}+k_{2t+1}d_{t+1})+\beta\, \dfrac{\gamma d_t}{s_{2t+1}} \nonumber
\\
k_{2t}&=&E_t\,\beta\, \underbrace{k_{2t+1}} + E_t\,\beta\, \left ( \dfrac{d_{2t+1}}{d_{t+1}}+\dfrac{\gamma}{s_{2t+1}} \right) \label{eq:k2form}
\\
& & \Longrightarrow \hspace{.05cm} \textrm{apply recursions on} \hspace{.05cm}  k_{2t+1} \nonumber
\\
k_{2t}=E_t \sum_{j=1}^{\infty} \beta^j \left ( \dfrac{d_{2t+j}}{d_{t+j}}+\dfrac{\gamma}{s_{2t+1}} \right ) \label{eq:k2formans}
\end{eqnarray}

From the problem, $d_{2t}=\frac{1}{2}d_t$ and $s_{2t+1}=1$ in equilibrium,  rewrite (\ref{eq:k2formans}) to obtain:

\begin{eqnarray}
	k_{2t}&=&E_t \sum_{j=1}^{\infty} \beta^j \left (\dfrac{1}{2}+ \gamma \right) \nonumber
	\\
	&=&\left(\dfrac{1}{2}+ \gamma \right) \dfrac{\beta}{1-\beta} \label{eq:k2formans-2}
\end{eqnarray}



Subsititing (\ref{eq:k1formans-1}) and (\ref{eq:k2formans-2}) into (\ref{eq:2price1-1}) and (\ref{eq:2price1-1}), will obtain the pricing functions:
\begin{eqnarray}
	p_{1t}&=&\left(\dfrac{\beta}{2(1-\beta)}\right)d_t \label{eq:ans2b-1}
	\\
	p_{2t}&=&\left(\left(\dfrac{1}{2}+ \gamma \right) \dfrac{\beta}{1-\beta}\right)d_t \label{eq:ans2b-2}
\end{eqnarray}

\newpage

\item[(d)]  Prove that if $\gamma > 0$, then $R_{1t} > R_{2t}$ $\forall$ $t$
\vspace{.25cm}
\\
From (\ref{eq:sub1}), derive the following definitions for $R_{1t}$ and  $R_{2t}$.
\begin{eqnarray}
R_{1t} &=&\dfrac{p_{1t+1}+d_{1t+1}}{p_{1t}} \label{eq:r1}
\\
R_{2t} & =& \dfrac{p_{2t+1}+d_{2t+1}}{p_{2t}} \label{eq:r2}
\end{eqnarray}

Substiting (\ref{eq:ans2b-1}) and  (\ref{eq:ans2b-2}) the pricing functions, into (\ref{eq:r1}) and (\ref{eq:r1}) can rewrite equation for $R_{1t}$ and  $R_{2t}$ as:

\begin{eqnarray}
R_{1t}&=&\dfrac{\dfrac{1}{2}\left(\dfrac{\beta}{1-\beta}\right)d_{t+1}+\dfrac{1}{2}\,d_{t+1}}{\dfrac{1}{2}\left(\dfrac{\beta}{1-\beta}\right)d_{t}} \label{eq:r1-1}
\\
R_{2t}&=&\dfrac{\left(\dfrac{1}{2}+\gamma \right) \left(\dfrac{\beta}{1-\beta}\right)d_{t+1}+\dfrac{1}{2}\,d_{t+1}}{\left(\dfrac{1}{2}+\gamma \right)\left(\dfrac{\beta}{1-\beta}\right)d_{t}} \label{eq:r2-2}
\end{eqnarray}

Analyzing difference in the returns, $R_{1t}$ and  $R_{2t}$, is:
\begin{eqnarray}
R_{1t}-R_{2t} &=& \left[\dfrac{1}{\frac{1}{2}\frac{\beta}{1-\beta}} - \dfrac{1}{\left(\gamma+\frac{1}{2}\right)\frac{\beta}{1-\beta}}\right]\,\dfrac{d_{t+1}}{d_{t}} \nonumber
\\
&=&\dfrac{1-\beta}{\beta}\,\left(1-\dfrac{1}{1+2\gamma}\right)\dfrac{d_{t+1}}{d_t} \label{eq:diffr}
\end{eqnarray}

From (\ref{eq:diffr}), if let $\gamma>0$, $\Longrightarrow$ \hspace{.5cm} $R_{1t}-R_{2t} > 0$.  Able to obtain:
\begin{equation}
R_{1t} > R_{2t} \label{eq:ans2c}
\end{equation}




\end{description}
\end{enumerate}

\end{document}