homework2_sol

homework2_sol - ECO 475 HW2 Solution Yu LIU 1 Value...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECO 475 HW2 Solution Yu LIU September 16, 2010 1 Value Function Iteration & Guess and Verify max { c t ,k t +1 } ∞ t =0 ∞ X t =0 β t u ( c t ) s.t. k t +1 = F ( k t ) + (1- δ ) k t- c t , ∀ t k t +1 ≥ , ∀ t c t ≥ , ∀ t k is given (1) Bellman equation: V ( k ) = max c,k { u ( c ) + βV ( k ) } s.t. k = F ( k ) + (1- δ ) k- c k ≥ c ≥ Assume u ( c ) = log( c ) ,F ( k ) = Ak α where A > ,α ∈ (0 , 1) ,β ∈ (0 , 1) and δ = 1 . (2) Rewriting Bellman equation: V ( k ) = max c,k { log( c ) + βV ( k ) } s.t. k = Ak α- c Derive V 1 ( k ) : Since V ( k ) = 0 , V 1 ( k ) = max k log( Ak α- k ) Since we must have k ≥ , k 1 = 0 and c 1 = Ak α . V 1 ( k ) = log( Ak α ) = α log k + log A Derive V 2 ( k ) : V 2 ( k ) = max k log( Ak α- k ) + βV 1 ( k ) = max k log( Ak α- k ) + β [ α log k + log A ] 1 First order condition for k : 1 Ak α- k = βα k k 2 = βαAk α 1 + βα Solving for consumption: c 2 = Ak α- βαAk α 1 + βα = Ak α 1 + βα Then we have: V 2 ( k ) = log c + β ( α log k + log A ) = log Ak α 1 + βα + βα log βαAk α 1 + βα + β log A = ( α + βα 2 )log k + (1 + β + βα )log A- (1 + βα )log (1 + βα ) + βα log βα Derive V n ( k ) : For the following iterations (applying the expression for a sum of terms of a geometric progression): k n = βα + β 2 α 2 + · · · + β n- 1 α n- 1 1 + βα + β 2 α 2 + · · · + β n- 1 α n- 1 Ak α = βα- β n α n 1- β n α n Ak α c n = 1 1 + βα + β 2 α 2 + · · · + β n- 1 α n- 1 Ak α = 1- βα 1- β n α n Ak α V n ( k ) = α (1- β n α n ) 1- βα log k + n- 1 X i =0 β i α i 1- β n- i 1- β log A- n X i =2 β n- i 1- β i α i 1- βα log 1- β i α i 1- βα + n X i =2 β n- i βα- β i α i 1- βα log βα- β i α i 1- βα (3) Derive V ( k ) by n → ∞ : 2 lim n →∞ V n ( k ) = α 1- βα log k + ∞ X i =0 β i α i 1 1- β log A- lim n →∞ n X i =2 β n- i- β n α i 1- βα log 1- β i α i 1- βα + lim n →∞ n X i =2 β n- i +1 α- β n α i 1- βα log βα- β i α i 1- βα = α 1- βα log k + log A (1- βα )(1- β )- lim n →∞ n X i =2 β n- i 1- βα log 1- β i α i 1- βα + lim n →∞ n X i =2 β n- i +1 α 1- βα log βα- β i α i 1- βα = α 1- βα log k + log A (1- βα )(1- β ) + lim n →∞ ( n X i =2 β n- i 1- βα log (1- βα )- n X i =2 β n- i 1- βα log ( 1- β i α i ) ) + lim n →∞ n X i =2 β n- i +1 α 1- βα log βα 1- βα + n- 1 X j =1 β n- j α 1- βα log ( 1- β j α j ) = α 1- βα log k + log A (1- βα )(1- β ) + 1 (1- β )(1- βα ) log (1- βα ) + βα (1- βα )(1- β ) log βα 1- βα + lim n →∞ ( ( α- 1) n- 1 X i =2 β n- i 1- βα log ( 1- β i α i ) + β n- 1 α 1- βα log (1- βα )- 1 1- βα log (1- β n α n ) ) = α 1- βα log k + log A (1- βα )(1- β ) + 1 (1- β )(1- βα ) log (1- βα ) + βα (1- βα )(1- β ) log βα 1- βα = α 1- βα log k + log A + (1- βα )log (1- βα ) + βα log βα...
View Full Document

This note was uploaded on 09/06/2011 for the course ECO 475 taught by Professor Hong during the Fall '07 term at Rochester.

Page1 / 12

homework2_sol - ECO 475 HW2 Solution Yu LIU 1 Value...

This preview shows document pages 1 - 4. Sign up to view the full document.

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