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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/29/2009 for the course ECON 720 at UNC.

Page1 / 10

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online