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Unformatted text preview: ECO 475 HW4 Solution Yu LIU September 29, 2010 1 Neoclassical Growth Model max { c t ,k t +1 } t =0 X t =0 t u ( c t ) s.t. c t + k t +1 = F ( k t ) + (1 ) k t k t +1 ,c t t k > is given (1) Control variable: c State variable: k Bellman equation v ( k ) = max c { u ( c ) + v ( k ) } s.t. k = F ( k ) + (1 ) k c c F ( k ) + (1 ) k From the general recursive problem: v ( x ) = max a ( x ) { U ( x,a ) + v [ f ( x,a )] } x = k a = c U ( k,c ) = u ( c ) f ( k,c ) = F ( k ) + (1 ) k c ( k ) = [0 ,F ( k ) + (1 ) k ] (2) We need assumptions that guarantee 1) value function exists and is continuously di erentiable and 2) FOC is su cient and necessary. Assumptions: u is strictly increasing, strictly concave, and continuously di erentiable; F is increasing, concave, and continuously di erentiable; (0 , 1) ; 1 F (0) = 0 , lim k F ( k ) = + , lim k F ( k ) <  1 + 1 , lim c u ( c ) = + . With those assumptions, the maximization in v has a unique solution which is interior and thus charac terized by Euler equation First order condition: u ( c ) = v [ F ( k ) + (1 ) k c ] Envelope condition: v ( k ) = v [ F ( k ) + (1 ) k c ] [ F ( k ) + 1 ] Envelope condition one period forward: v [ F ( k ) + (1 ) k c ] = v [ F ( k ) + (1 ) k c ] [ F ( k ) + 1 ] (1) First order condition one period forward: u ( c ) = v [ F ( k ) + (1 ) k c ] (2) Then plugging equation (1) and equation (2) into the original rst order condition, we can get the Euler equation with c as the control variable. u ( c ) = u ( c ) [ F ( k ) + 1 ] (3) (3) Euler equation when k is used as a control variable: u [ F ( k ) + (1 ) k k ] = u [ F ( k ) + (1 ) k k 00 ] [ F ( k ) + 1 ] u [ F ( k ) + (1 ) k ( k )] = u [ F ( k ) + (1 ) k ( k )] [ F ( k ) + 1 ] u [ F ( k ) + (1 ) k ( k )] = u [ F ( ( k )) + (1 ) ( k ) ( ( k ))] [ F ( ( k )) + 1 ] (4) From equation (3) , u ( g c ( k )) = u ( g c ( k )) [ F ( k ) + 1 ] u ( F ( k ) + (1 ) k g k ( k )) = u [ F ( g k ( k )) + (1 ) g k ( k ) g k ( g k ( k ))] [ F ( g k ( k )) + 1 ] (5) Thus, these two equations (4) and (5) are the same functional equations, and thus give the same set of solutions. If we impose the transversality condition, the solution is unique and we can say g k ( k ) = ( k ) . To show ( k * ) < 1 , we rst need to show that ( k ) > , k . Show that ( k ) > : From the rst order condition: u [ F ( k ) + (1 ) k k ] = v ( k ) Take the derivative of k with respect to k , by assuming u and v are twice continuously di erentiable and u 00 < and v 00 < : ( k ) = dk dk = u 00 [ F ( k ) + (1 ) k k ][ F ( k ) + 1 ] u 00 [ F ( k ) + (1 ) k k ] v 00 ( k ) = u 00 [ F ( k ) + (1 ) k k ][ F ( k ) + 1 ] u 00 [ F ( k ) + (1 ) k k ] + v...
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This note was uploaded on 09/06/2011 for the course ECO 475 taught by Professor Hong during the Fall '07 term at Rochester.
 Fall '07
 Hong

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