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\begin{enumerate}
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\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 utility 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}
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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. 
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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.
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\textbf{Consumer optimization in a recursive form}
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The Bellman's equation is given by
\[v(d_t,s_{1t},s_{2t})=\max_{\{c_t,s_{1t+1},s_{2t+1}\}} 
  \left( \ln(c_t)+\gamma \ln(s_{2t}) +E_t \beta v(d_t,s_{1t+1},s_{2t+1}) \right)\]
\\where $c_t+p_{1t}s_{1t+1}+p_{2t}s_{2t+1} \leq (d_{1t}+p_{1t})s_{1t}+(d_{2t}+p_{2t})s_{2t}$.
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\textbf{Consumer optimization in a sequential form}
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The sequential form is given by
\[\max_{\{c_t,s_{1t+1},s_{2t+1}\}} E_0\sum^\infty_{t=0}\beta^t\, (\ln(c_t)+\gamma \ln (s_{2t})) \]
\\where $c_t+p_{1t}s_{1t+1}+p_{2t}s_{2t+1} \leq (d_{1t}+p_{1t})s_{1t}+(d_{2t}+p_{2t})s_{2t}$.
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\item[(b)]  Define a rational expectations equilibrium.
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\textbf{Definition} The following is called the \textit{market clear condition}
\begin{equation}
\sum^I_{i=1} c^i_t = \sum^I_{i=1} d^i_t
\end{equation}
\begin{eqnarray*}
\sum^I_{i=1} s^i_{1t}&=&\sum^I_{i=1} s^i_{10}=I\\
\sum^I_{i=1} s^i_{2t}&=&\sum^I_{i=1} s^i_{20}=I 	
\end{eqnarray*}
where $s^i_{10}$ and $s^i_{20}$ are each agent's number of trees at initial time. 
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\textbf{Definition} \textit{A sequential household problem} is defined by each agent's utility optimization problem:  
\[\max_{\{c_t,s_{1t+1},s_{2t+1}\}} E_0\sum^\infty_{t=0}\beta^t\, (\ln(c_t)+\gamma \ln (s_{2t})) \]
\\where $c_t+p_{1t}s_{1t+1}+p_{2t}s_{2t+1} \leq (d_{1t}+p_{1t})s_{1t}+(d_{2t}+p_{2t})s_{2t}$.
\vspace{.25cm}
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\textbf{Definition} A \textit{rational competitive equilibrium} is an allocation,\\ 
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\item[(c)]  Find the pricing functions mapping the state of the economy 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.]
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I am going to use the sequential form to find a solution. Using the Lagrange Multiplier,
\begin{eqnarray*}
L&=& E_0\sum^\infty_{t=0}\beta^t\, ((\ln(c_t)+\gamma \ln (s_{2t})) \\
&+&\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*}
By taking differentiations with respect to $\{c_t,s_{1t+1},s_{2t+1}\}$.
\begin{eqnarray*}
	{c_t}&:& E_0 \beta^t (\dfrac{1}{c_t}-\lambda_t)=0 \\
	{s_{1t+1}}&:& E_0 \beta^t (\beta \lambda_{t+1} (d_{1t+1}+p_{1t+1}) -\lambda_t p_{1t})=0 \\
	{s_{2t+1}}&:& 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{eqnarray*}
Using the Markov property $E_0=E_t$.
\begin{eqnarray*}
	{c_t}&:& \dfrac{1}{c_t}=\lambda_t \\
	{s_{1t+1}}&:& p_{1t}=E_t \beta \dfrac{\lambda_{t+1}}{\lambda_{t}}(d_{1t+1}+p_{1t+1}) \\
	{s_{2t+1}}&:& 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{eqnarray*}	
Combining the first-order conditions, we can obtain the pricing formula.
\begin{eqnarray*}
	p_{1t}&=&E_t \beta \dfrac{c_{t}}{c_{t+1}}(d_{1t+1}+p_{1t+1}) \\
	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}}
\end{eqnarray*}	
Recall $c_t=d_t$ in a general equilibrium.
\begin{eqnarray*}
	p_{1t}&=&E_t \beta \dfrac{d_{t}}{d_{t+1}}(d_{1t+1}+p_{1t+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}}
\end{eqnarray*}	
Let us assume the linear form of the first tree's pricing function.
\[p_{1t}=k_{1t}d_t\]
Next apply the above linear form to the Euler equation.
\begin{eqnarray*}
	k_{1t}d_t&=&E_t\beta \dfrac{d_{t}}{d_{t+1}}(d_{1t+1}+k_{1t+1}d_{t+1})\\
	k_{1t}&=&E_t\beta k_{1t+1} + E_t\beta \dfrac{d_{1t+1}}{d_{t+1}}
\end{eqnarray*}
Recursively we obtain $k_{1t}$.
\[k_{1t}=E_t \sum_{j=1}^{\infty} \beta^j \dfrac{d_{1t+j}}{d_{t+j}}\]
Since we are given $d_{1t}=\frac{1}{2}d_t$.
\begin{eqnarray*}
	k_{1t}&=&\dfrac{1}{2}E_t \sum_{j=1}^{\infty} \beta^j \\
	&=&\dfrac{\beta}{2(1-\beta)}
\end{eqnarray*}
Let us assume the linear form of the second tree's pricing function.
\[p_{2t}=k_{2t}d_t\]
Next apply the above linear form to the Euler equation.
\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}}\\
	k_{2t}&=&E_t\beta k_{2t+1} + E_t\beta \left ( \dfrac{d_{2t+1}}{d_{t+1}}+\dfrac{\gamma}{s_{2t+1}} \right)
\end{eqnarray*}
Recursively we obtain $k_{2t}$.
\[k_{2t}=E_t \sum_{j=1}^{\infty} \beta^j \left ( \dfrac{d_{2t+j}}{d_{t+j}}+\dfrac{\gamma}{s_{2t+1}} \right )\]
Since we are given $d_{2t}=\frac{1}{2}d_t$ and $s_{2t+1}=1$ in equilibrium.
\begin{eqnarray*}
	k_{2t}&=&E_t \sum_{j=1}^{\infty} \beta^j \left (\dfrac{1}{2}+ \gamma \right)\\
	&=&\left(\dfrac{1}{2}+ \gamma \right) \dfrac{\beta}{1-\beta}
\end{eqnarray*}
Finally, we have got the pricing equations.
\begin{eqnarray*}
	p_{1t}&=&\left(\dfrac{\beta}{2(1-\beta)}\right)d_t\\
	p_{2t}&=&\left(\left(\dfrac{1}{2}+ \gamma \right) \dfrac{\beta}{1-\beta}\right)d_t
\end{eqnarray*}
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\item[(d)]  Prove that if $\gamma > 0$, then $R_{1t} > R_{2t}$ $\forall$ $t$
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The returns $R_{1t},R_{2t}$ are defined as:
\[R_{1t}=\dfrac{p_{1t+1}+d_{1t+1}}{p_{1t}}\]
\[R_{2t}=\dfrac{p_{2t+1}+d_{2t+1}}{p_{2t}}\]
From the derived pricing equations,
\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}}\\
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}}
\end{eqnarray*}
By rearranging the equations,
\[R_{1t}-R_{2t}=\left(1-\dfrac{1}{1+2\gamma}\right)\dfrac{1-\beta}{\beta}\dfrac{d_{t+1}}{d_t}\]
If $\gamma>0$, then
\[R_{1t}>R_{2t}\]
\end{description}
\end{enumerate}

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