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Dp_ln

# Dp_ln - Dynamic Programming Econ720 Fall 2009 Lutz...

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Dynamic Programming Econ720. Fall 2009. Lutz Hendricks. 1 Basic Idea We are given a dynamic optimization problem in discrete time of the form max X T t =1 u ( k t ; c t ; t ) + V T +1 ( k T +1 ) subject to k t +1 = g ( k t ; c t ; t ) with the initial condition k 0 given and the terminal value of k given, call it V T +1 ( k T +1 ) . We call the variable with the equation of motion the state variable and the other choice variable the control variable . One way of solving this problem is to set up a Lagrangean ° = max X T t =1 u ( k t ; c t ; t ) + X T t =1 ° t [ g ( k t ; c t ; t ) ° k t +1 ] + V T +1 ( k T +1 ) (1) The °rst sum is simply the objective function, while the second sum is a conve- nient way of collecting all constraints. The FOCs are u c ( t ) = ° ° t g c ( t ) u k ( t ) = ° ° t g k ( t ) + ° t ° 1 ° T = V 0 T +1 ( k T +1 ) This is a perfectly good approach, but DP is an alternative that is sometimes more convenient. The basic idea is to restart the problem at some date ± . First note that we can think of the maximized Lagrangean as an indirect utility function. After maximizing out all future values of c and k , ° becomes a function only of k t and t which we write as V ° ( k ° ) . Now move to period ± + 1 and restart the problem again. Given a suitable structure of the problem (whatever that means!), the solution to the period ± + 1 problem will be the same as that of the period ± problem, except for the period ± values of course, which are in the past from the ± + 1 perspective. In other words, if the solution to the date ± problem is (^ c t ; ^ k t +1 ) ; t = ±; : : : ; T , then the solution to the date ± + 1 problem is (^ c t ; ^ k t +1 ) ; t = ± + 1 ; : : : ; T . The decision maker does not revise his date ± plan at some later time. The problem is time consistent . For this to work, we must be able to rearrange the terms in the Lagrangean so that the problem is divided into two sets of equations. The °rst set only contains 1

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variables prior to date ± and k ° . The second set contains only variables from date ± onwards. Here: ° = max X ° ° 1 t =1 u ( k t ; c t ; t ) + X ° ° 1 t =1 ° t [ g ( k t ; c t ; t ) ° k t +1 ] + X T t = ° u ( k t ; c t ; t ) + X T t = ° ° t [ g ( k t ; c t ; t ) ° k t +1 ] + V T +1 ( k T +1 ) More generally, ° ° ± ( k ° ± ) = max X ° ° 1 t = ° ± u ( k t ; c t ; t ) + X ° ° 1 t = ° ± ° t [ g ( k t ; c t ; t ) ° k t +1 ] + ° ° ( k ° ) (2) If this is true, then we can write the date ± problem as V ° ( k ° ) = max c ° ; k ° +1 f u ( k ° ; c ° ; ± ) + ° ° [ g ( k ° ; c ° ; ± ) ° k ° +1 ] + V ° +1 ( k ° +1 ) g (3) You can convince yourself that this is true by expanding V ( : ) on the right hand side repeatedly. You will recover exactly (1), except that there is a ± max²for each date. But since the decision maker does not change his mind, these ± max²³s are irrelevant. The interpretation is simple. Given that behavior is optimal from date ± + 1 onwards, the value of k ° +1 is given by the indirect utility function V ° +1 ( k ° +1 ) . The date ± decision can then be made by trading o/ current utility against next period utility.
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Dp_ln - Dynamic Programming Econ720 Fall 2009 Lutz...

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