homework2_sol

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

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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...
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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.

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

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