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 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 1

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variables prior to date and k . The second set contains only variables from date onwards. Here: 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 irrelevant. The interpretation is simple. Given that behavior is optimal from date
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## This note was uploaded on 10/29/2009 for the course ECON 720 at UNC.

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

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