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%TCIDATA{Created=Tuesday, September 11, 2007 11:02:45}
%TCIDATA{LastRevised=Wednesday, September 12, 2007 17:00:51}
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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}


\begin{eqnarray*}
\frac{1}{c_{1}} &=&\lambda \left( 1+\tau _{1}\right)  \\
\frac{1}{l_{1}} &=&\lambda w\left( 1-\tau _{l}\right)  \\
\beta \frac{1}{c_{2}} &=&\lambda \frac{1+\tau _{2}}{1+r} \\
\beta \frac{1}{l_{2}} &=&\lambda w\frac{1-\tau _{l}}{1+r} \\
c_{1}\left( 1+\tau _{1}\right) +\frac{c_{2}\left( 1+\tau _{1}\right) }{1+r}
&=&w\left( 1-l_{1}\right) \left( 1-\tau _{1}\right) +\frac{w\left(
1-l_{2}\right) \left( 1-\tau _{1}\right) }{1+r}
\end{eqnarray*}%
, Solution is: 

\begin{eqnarray*}
\frac{l_{1}}{c_{1}} &=&\frac{1+\tau _{1}}{w\left( 1-\tau _{l}\right) } \\
\frac{l_{2}}{c_{2}} &=&\frac{1+\tau _{2}}{w\left( 1-\tau _{l}\right) } \\
\frac{c_{2}}{\beta c_{1}} &=&\frac{1+\tau _{1}}{1+\tau _{2}}1+r \\
c_{1}\left( 1+\tau _{1}\right) +\frac{c_{2}\left( 1+\tau _{2}\right) }{1+r}
&=&w\left( 1-l_{1}\right) \left( 1-\tau _{l}\right) +\frac{w\left(
1-l_{2}\right) \left( 1-\tau _{l}\right) }{1+r} \\
g+\frac{g}{1+r} &=&c_{1}\tau _{1}+\frac{c_{2}\tau _{2}}{1+r}+w\left(
1-l_{1}\right) \tau _{l}+\frac{w\left( 1-l_{2}\right) \tau _{l}}{1+r}
\end{eqnarray*}%
, Solution is: $\left\{ 
\begin{array}{c}
l_{2}=\frac{\frac{1}{2}\left( 1+\tau _{1}+r+r\tau _{2}\right) \left(
r+2\right) \beta }{1+\tau _{1}r+\beta \tau _{1}+\tau _{1}+r+\beta r\tau
_{2}+\beta r+\beta }, \\ 
c_{1}=\frac{1}{2}\frac{\left( r+2\right) \left( w\tau _{2}-2g+w-2g\tau
_{2}\right) }{r\tau _{2}+r+\beta r\tau _{2}+\beta r+\beta \tau _{1}+\tau
_{2}+1+\beta }, \\ 
c_{2}=\beta \frac{\frac{1}{2}\left( 1+\tau _{1}+r+r\tau _{2}\right) \left(
w-2g\right) \left( r+2\right) }{r\tau _{2}+r+\beta r\tau _{2}+\beta r+\beta
\tau _{1}+\tau _{2}+1+\beta }, \\ 
l_{1}=\frac{1}{2}\frac{\tau _{1}r+2\tau _{1}+2+r}{1+\tau _{1}r+\beta \tau
_{1}+\tau _{1}+r+\beta r\tau _{2}+\beta r+\beta }%
\end{array}%
\right\} \allowbreak \allowbreak $ is true is true is true is false is false

$\allowbreak $

\bigskip 

\[
\beta =.98,r=0.05,g=.1,
\]
\begin{eqnarray}
\frac{l_{1}}{c_{1}} &=&\frac{1}{w\left( 1-\tau _{l}\right) } \\
\frac{l_{2}}{c_{2}} &=&\frac{1}{w\left( 1-\tau _{l}\right) }  \nonumber \\
\frac{c_{2}}{\left( 1+r\right) \beta c_{1}} &=&1-\tau _{s}  \nonumber \\
c_{1}+\frac{c_{2}}{\left( 1+r\right) \left( 1-\tau _{s}\right) } &=&w\left(
1-l_{1}\right) \left( 1-\tau _{l}\right) +\frac{w\left( 1-l_{2}\right)
\left( 1-\tau _{l}\right) }{\left( 1+r\right) \left( 1-\tau _{s}\right) } \\
g+\frac{g}{\left( 1+r\right) \left( 1-\tau _{s}\right) } &=&w\left(
1-l_{1}\right) \tau _{l}+\frac{w\left( 1-l_{2}\right) \tau _{l}}{\left(
1+r\right) \left( 1-\tau _{s}\right) }
\end{eqnarray}%
, Solution is:$\tau _{l}=2\frac{g}{w}$ $\left\{ 
\begin{array}{c}
c_{2}=\frac{1}{2}\left( -2+\tau _{s}-r+r\tau _{s}\right) \beta \frac{2g-w}{%
1+\beta } \\ 
c_{1}=-\frac{1}{2}\left( -2+\tau _{s}-r+r\tau _{s}\right) \frac{2g-w}{r\tau
_{s}-r-1+\tau _{s}-\beta -\beta r+\tau _{s}\beta +r\tau _{s}\beta } \\ 
l_{1}=\frac{1}{2}\frac{-2+\tau _{s}-r+r\tau _{s}}{r\tau _{s}-r-1+\tau
_{s}-\beta -\beta r+\tau _{s}\beta +r\tau _{s}\beta } \\ 
l_{2}=-\frac{1}{2}\left( -2+\tau _{s}-r+r\tau _{s}\right) \frac{\beta }{%
1+\beta }%
\end{array}%
\right\} \allowbreak $%
\[
\left[ -\frac{1}{2}\left( -2+\tau _{s}-r+r\tau _{s}\right) \frac{2g-w}{r\tau
_{s}-r-1+\tau _{s}-\beta -\beta r+\tau _{s}\beta +r\tau _{s}\beta }\right]
_{\tau _{s}=0}=\left[ \frac{1}{2}\frac{\left( r+2\right) \left( w\tau
_{2}-2g+w-2g\tau _{2}\right) }{r\tau _{2}+r+\beta r\tau _{2}+\beta r+\beta
\tau _{1}+\tau _{2}+1+\beta }\right] _{\tau _{1}=\tau _{2}}
\]
is true : $-\frac{1}{2}\left( 2g-w\right) \frac{r+2}{r+\beta r+\beta +1}$ : $%
-\frac{1}{2}\left( 2g-w\right) \frac{r+2}{r+\beta r+\beta +1}$

\bigskip 
\[
c_{1}=-\frac{1}{2}\frac{2g+gr-2w-wr}{r+\beta r+\beta +1}-\left( -\frac{1}{2}%
\left( 2g-w\right) \frac{r+2}{r+\beta r+\beta +1}\right) 
\]
: $\left[ \frac{1}{2}g\frac{r+2}{r+\beta r+\beta +1}\right]
_{r=0}=\allowbreak \frac{g}{1+\beta }$

\begin{eqnarray}
\frac{l_{1}}{c_{1}} &=&\frac{1}{w} \\
\frac{l_{2}}{c_{2}} &=&\frac{1}{w}  \nonumber \\
\frac{c_{2}}{\left( 1+r\right) \beta c_{1}} &=&1  \nonumber \\
g+\frac{g}{\left( 1+r\right) }+c_{1}+\frac{c_{2}}{\left( 1+r\right) }
&=&w\left( 1-l_{1}\right) +\frac{w\left( 1-l_{2}\right) }{\left( 1+r\right) }
\end{eqnarray}%
, Solution is: $\left\{ l_{1}=-\frac{1}{2}\frac{2g+gr-2w-wr}{w\left( r+\beta
r+\beta +1\right) },c_{2}=-\frac{1}{2}\left( 2g+gr-2w-wr\right) \frac{\beta 
}{1+\beta },c_{1}=-\frac{1}{2}\frac{2g+gr-2w-wr}{r+\beta r+\beta +1},l_{2}=-%
\frac{1}{2}\left( 2g+gr-2w-wr\right) \frac{\beta }{\left( 1+\beta \right) w}%
\right\} \allowbreak $

$l_{1}=\left[ \frac{1}{2}\frac{-2+\tau _{s}-r+r\tau _{s}}{r\tau
_{s}-r-1+\tau _{s}-\beta -\beta r+\tau _{s}\beta +r\tau _{s}\beta }\right]
_{\tau _{s}=0}-\left( -\frac{1}{2}\frac{2g+gr-2w-wr}{w\left( r+\beta r+\beta
+1\right) }\right) =\allowbreak \frac{1}{2}g\frac{r+2}{w\left( r+\beta
r+\beta +1\right) }$

\[
c_{1}\left( 1+\tau _{c,1}\right) +c_{2}\frac{\left( 1+\tau _{c,2}\right) }{%
\left( 1+r\right) \left( 1-\tau _{s}\right) }=w\left( 1-l_{1}\right) \left(
1-\tau _{l,1}\right) +\frac{w\left( 1-l_{2}\right) \left( 1-\tau
_{l,2}\right) }{\left( 1+r\right) \left( 1-\tau _{s}\right) }
\]

The government has a budget constraints%
\begin{eqnarray*}
g_{1} &=&\tau _{c,1}c_{1}+\tau _{l,1}w\left( 1-l_{1}\right) +b \\
g_{2} &=&\tau _{c,2}c_{2}-b\left( 1+r\right) \left( 1-\tau _{s}\right)
+i\left( 1+r\right) \tau _{s}
\end{eqnarray*}%
giving 
\[
g_{1}+\frac{g_{2}}{b\left( 1+r\right) \left( 1-\tau _{s}\right) }=\tau
_{c,1}c_{1}+\tau _{l,1}w\left( 1-l_{1}\right) +\frac{\tau _{c,2}c_{2}}{%
b\left( 1+r\right) \left( 1-\tau _{s}\right) }+\frac{\tau _{c,2}c_{2}}{%
b\left( 1+r\right) \left( 1-\tau _{s}\right) }+\frac{i\tau _{s}}{b\left(
1-\tau _{s}\right) }
\]

\end{document}