handoutrbc08final - Solving RBC Models 1 Planner's Problem...

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Solving RBC Models 1 Planner’s Problem Owing to the second welfare theorem, the allocation we look for solves the planner’s problem: max { C t ,K t +1 } t =0 E 0 " X t =0 β t ln( C t ) # subject to C t + K t +1 = Z t K θ t ¡ (1 + γ ) t ¢ 1 θ +(1 δ ) K t (1) Z t =(1 ρ )+ ρZ t 1 + ε t (2) where ε t is drawn from N (0 2 ε ) and K 0 is given. Necessary and su cient conditions for solution of this problem are: 1 C t = βE t " £ θZ t +1 (1 + γ ) ( t +1)(1 θ ) K θ 1 t +1 δ ) ¤ C t +1 # (3) Transversality condition: lim T →∞ E 0 β T K T +1 C T =0 . (4) 2 Verifying balanced growth Conjecture that along a balanced growth path ( Z t =1 , t ), Y t +1 Y t , K t +1 K t , C t +1 C t all grow at a common constant rate. To see this is consistent with our model, consider the conditions that characterize a solution to the model (1)-(4). First, assume that C t +1 C t grows at some constant rate κ C (we’ll have to come back and verify this in the end). Then the f.o.c. (3) implies C t +1 C t = β " θ (1 + γ ) ( t +1) K t +1 ¸ (1 θ ) +1 δ # = κ C . (5) But the only way for this to hold is for K t +1 (1+ γ ) ( t +1) = κ. The initial condition K 0 = κ (1 + γ ) 0 pins down κ. Hence K t +1 K t =(1+ γ ) . (6) 1
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Taking the ratios of the production function gives Y t +1 Y t = (1 + γ ) ( t +1)(1 θ ) K θ t +1 (1 + γ ) t (1 θ ) K θ t =(1+ γ ) 1 θ μ K t +1 K t θ γ ) (7) where the last equality follows from (6). Next dividing the resource feasibility constraint (1) through by K t we have C t K t = K t +1 K t + Y t K t +(1 δ )= κ r where the last equality follows since (6) and (7) imply Y t K t = Y t +1 K t +1 in which case both K t +1 K t and Y t K t are constants so C t K t has to be a constant. But then κ r = C t +1 K t +1 = C t +1 C t = K t +1 K t γ ) so that we veri f ed the original assumption. Notice also that under these conditions, the transversality condition is satis f ed since K T +1 C T is independent of t (i.e. K t +1 C t = 1+ γ κ r ). If all the variables grow at rate (1 + γ ) , then construct new “detrended” variables e X t = X t (1+ γ ) t . In that case the problem max E 0 X t =0 β t ln( C t ) s.t.C t + K t +1 = Z t (1 + γ ) t (1 θ ) K θ t δ ) K t can be written max E 0 X t =0 β t ln( e C t ) (8) s.t. e C t +(1+ γ ) e K t +1 = Z t e K θ t δ ) e K t since X t =0 β t ln( e C t X t =0 β t [ln( C t ) t ln(1 + γ )] = X t =0 β t ln( C t ) ln(1 + γ ) 1 β · β 1 β and C t (1 + γ ) t + K t +1 (1 + γ ) t = Z t (1 + γ ) t (1 θ ) K θ t (1 + γ ) t (1 θ ) (1 + γ ) θt δ ) K t (1 + γ ) t ⇐⇒ e C t γ ) e K t +1 = Z t e K θ t δ ) e K t . 2
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3C a l i b r a t i o n The parameters (technology= { θ,δ,ρ,γ,σ ε } , preferences= { β } )arematched o f of long run growth “ f rst” moments. The model is tested against short run F uctuation “second” moments.
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handoutrbc08final - Solving RBC Models 1 Planner's Problem...

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