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Thread: A dynamic problem with Pontryagin's Maximum Principle

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    A dynamic problem with Pontryagin's Maximum Principle

    Let the typical individual maximizes lifetime utility function be:

    $\displaystyle U(t)= \int ^ \infty _t e^{-(p- \lambda )( \tau - t)} \log u( \tau )d \tau \ , \ \ \ \ p > \lambda \geq 0 $

    where $\displaystyle \log u = \log C + \gamma \log(1-l)+ \mu \log G , \ \ \ \gamma , \mu > 0 $

    Flow budget constraint is given by:

    $\displaystyle \frac {dA}{dt} = \dot {A}= [r(1-t_A)- \lambda ]A+1- t_L)Wl-(1+t_E)E+T $

    Question: Find the growth rate of expenditure $\displaystyle \frac {dE}{dt} \frac {1}{E} = \frac { \dot {E}}{E} $

    Solution so far:

    Now, I look ahead in the paper, and I see $\displaystyle \frac { \dot {E}}{E}= r(1-t_A)-p $

    First, I set up the Hamiltonian for this problem as follows:

    $\displaystyle J = e^{-(p- \lambda )( \tau - t)} \log (u) + v \{ [r(1-t_A)- \lambda ]A+1- t_L)Wl-(1+t_E)E+T \} $

    I assume that $\displaystyle C = E $, consumption is the same as expenditure here.

    So I have $\displaystyle J = e^{-(p- \lambda )( \tau - t) } \log (u) + v \{ [r(1-t_A)- \lambda ]A+1- t_L)Wl-(1+t_E)C+T \} $

    By the Pontryagin's Maximum Principle, we have the following conditions:

    i. $\displaystyle \frac { \partial J}{ \partial u } = 0 $ and ii. $\displaystyle \frac { \partial J }{ \partial A } = - \dot {v} $

    Now, by i., I know that $\displaystyle \frac {1}{C}e^{-(p- \lambda )( \tau - t)} -v(1+t_E) = 0 $.

    So I then have $\displaystyle v = \frac {e^{-(p- \lambda )( \tau - t)}}{C}( \frac {1}{1+t_E}) $
    $\displaystyle v = ( \frac {1}{1+t_E}) (e^{-(p- \lambda )( \tau - t)}C^{-1}) $

    Taking the derivative with respect to time, I have:

    $\displaystyle \dot {v} = ( \frac {1}{1+t_E})(e^{-(p- \lambda )( \tau - t)}(-C^{-2} \dot {C} + e^ {p- \lambda } C^{-1} ) $

    $\displaystyle \dot {v} = ( \frac {1}{1+t_E})(e^{p- \lambda })( \frac {1}{C} - e^{- ( \tau - t)} \frac { \dot {C} }{C^2}) $

    by ii., I have $\displaystyle \frac { \partial J }{ \partial A} = v[r(1-t_A)- \lambda ] = - \dot {v} $

    $\displaystyle v[r(1-t_A)- \lambda ] = -( \frac {1}{1+t_E})(e^{p- \lambda })( \frac {1}{C} - e^{- ( \tau - t)} \frac { \dot {C} }{C^2}) $

    $\displaystyle ( \frac {1}{1+t_E}) (e^{-(p- \lambda )( \tau - t)}C^{-1})[r(1-t_A)- \lambda ] = ( \frac {1}{1+t_E})(e^{p- \lambda })( e^{- ( \tau - t)} \frac { \dot {C} }{C^2} - \frac {1}{C}) $

    $\displaystyle (e^{-( \tau - t)}C^{-1})[r(1-t_A)- \lambda ] = ( e^{- ( \tau - t)} \frac { \dot {C} }{C^2} - \frac {1}{C}) $

    $\displaystyle (e^{-( \tau - t)})[r(1-t_A)- \lambda ] = ( e^{- ( \tau - t)} \frac { \dot {C} }{C} - 1)$

    $\displaystyle r(1-t_A)- \lambda + e^{ \tau - t } = \frac { \dot {C} }{C}$

    So I guess somehow $\displaystyle e^{ \tau - t } - \lambda = p $, the interest rate?

    In the attachment I have a copy of the paper that this equation is on, they are on page 4 of the pdf file.

    Thank you very much!!!
    Attached Files Attached Files
    Last edited by tttcomrader; Dec 10th 2008 at 06:11 AM.
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