Real Business Cycles

Jesús Fernández-Villaverde
University of Pennsylvania

1
 Household Problem

• Preferences:
∞
X
max E β {log ct + ψ log (1 − lt)}
t=0

• Budget constraint:

ct + kt+1 = wtlt + rtkt + (1 − δ) kt, ∀ t > 0

2
 Problem of the Firm

• Neoclassical production function:

yt = ktα (ezt lt)1−α

• By profit maximization:

αktα−1 (ezt lt)1−α = rt

(1 − α) ktα (ezt lt)1−α lt−1 = wt

3
 Evolution of the technology

• zt changes over time.

• It follows the AR(1) process:

zt = ρzt−1 + εt
εt ∼ N (0, σ)

• Interpretation of ρ.

4
 A Competitive Equilibrium

• We can define a competitive equilibrium in the standard way.

• The competitive equilibrium is unique.

• This economy satisfies the conditions that assure that both welfare
theorems hold.

• Why is this important? We could solve instead the Social Planner’s

Problem associated with it.

• Advantages and disadvantages of solving the social planner’s problem.

5
 The Social Planner’s Problem

• It has the form:

∞
X
max E β {log ct + ψ log (1 − lt)}
t=0

ct + kt+1 = ktα (ezt lt)1−α + (1 − δ) kt, ∀ t > 0

zt = ρzt−1 + εt, εt ∼ N (0, σ)

• This is a dynamic optimization problem.

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 Computing the RBC

• The previous problem does not have a known “paper and pencil” so-
lution.

• We will work with an approximation: Perturbation Theory.

• We will undertake a first order perturbation of the model.

• How well will the approximation work?

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 Equilibrium Conditions

From the household problem+firms’s problem+aggregate conditions:

( )
1 1 ³ ´
= βEt 1 + αktα−1 (ezt lt)1−α − δ
ct ct+1
ct
ψ = (1 − α) ktα (ezt lt)1−α lt−1
1 − lt
ct + kt+1 = ktα (ezt lt)1−α + (1 − δ) kt
zt = ρzt−1 + εt

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 Finding a Deterministic Solution

• We search for the firts component of the solution.

• If σ = 0, the equilibrium conditions are:

1 1 ³ α−1 1−α
´
=β 1 + αkt lt −δ
ct ct+1
ct
ψ = (1 − α) ktαlt−α
1 − lt
ct + kt+1 = ktαlt1−α + (1 − δ) kt

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 Deterministic Steady State

• The equilibrium conditions imply a steady state:

1 1³ α−1 1−α
´
= β 1 + αk l −δ
c c
c
ψ = (1 − α) kαl−α
1−l
c + δk = kαl1−α

• Or symplifying:
1
= 1 + αkα−1l1−α − δ
β
c
ψ = (1 − α) kαl−α
1−l
c + δk = kαl1−α
10
 Solving the Steady State

Solution:
µ
k =
Ω + ϕµ
l = ϕk
c = Ωk
y = kαl1−α

³ ³ ´´ 1
where ϕ = α β − 1 + δ 1−α , Ω = ϕ1−α − δ and µ = ψ1 (1 − α) ϕ−α.
1 1

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 Linearization I

• Loglinearization or linearization?

• Advantages and disadvantages

• We can linearize and perform later a change of variables.

12
 Linearization II

We linearize:

( )
1 1 ³ ´
= βEt 1 + αktα−1 (ezt lt)1−α − δ
ct ct+1
ct
ψ = (1 − α) ktα (ezt lt)1−α lt−1
1 − lt
ct + kt+1 = ktα (ezt lt)1−α + (1 − δ) kt
zt = ρzt−1 + εt

around l, k, and c with a First-order Taylor Expansion.

13
 Linearization III

We get:
( y )
1 1
− c (ct+1 − c) + α (1 − α) β k zt+1+
− (ct − c) = Et
c α (α − 1) β ky2 (kt+1 − k) + α (1 − α) β kl
y
(lt+1 − l)
1 1 α α
(ct − c) + (lt − l) = (1 − α) zt + (kt − k) − (lt − l)
c (1 − l) k l
 µ ¶ 
 y (1 − α) z + α (k − k) + (1−α) (l − l) 
t k t l t
(ct − c) + (kt+1 − k) =
 
+ (1 − δ) (kt − k)
zt = ρzt−1 + εt

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 Rewriting the System I

Or:

α1 (ct − c) = Et {α1 (ct+1 − c) + α2zt+1 + α3 (kt+1 − k) + α4 (lt+1 − l)}

α
(ct − c) = α5zt + c (kt − k) + α6 (lt − l)
k
(ct − c) + (kt+1 − k) = α7zt + α8 (kt − k) + α9 (lt − l)

zt = ρzt−1 + εt

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 Rewriting the System II

where

α1 = − 1c α2 = α (1 − α) β ky
α3 = α (α − 1) β ky2 y
α4 = α (1 − α) β kl
µ ¶
α5 = (1 − α) c 1
α6 = − αl + (1−l) c
α7 = (1 − α) y α8 = y α
k + (1 − δ)
(1−α)
α9 = y l y = kαl1−α

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 Rewriting the System III

After some algebra the system is reduced to:

A (kt+1 − k) + B (kt − k) + C (lt − l) + Dzt = 0

Et (G (kt+1 − k) + H (kt − k) + J (lt+1 − l) + K (lt − l) + Lzt+1 + Mzt) = 0

Etzt+1 = ρzt

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 Guess Policy Functions

We guess policy functions of the form (kt+1 − k) = P (kt − k) + Qzt and

(lt − l) = R (kt − k) + Szt, plug them in and get:

A (P (kt − k) + Qzt) + B (kt − k)

+C (R (kt − k) + Szt) + Dzt = 0

G (P (kt − k) + Qzt) + H (kt − k) + J (R (P (kt − k) + Qzt) + SN zt)

+K (R (kt − k) + Szt) + (LN + M ) zt = 0

18
 Solving the System I

Since these equations need to hold for any value (kt+1 − k) or zt we need
to equate each coefficient to zero, on (kt − k):

AP + B + CR = 0
GP + H + JRP + KR = 0
and on zt:

AQ + CS + D = 0
(G + JR) Q + JSN + KS + LN + M = 0

19
 Solving the System II

• We have a system of four equations on four unknowns.

1 (AP + B) = − 1 AP − 1 B
• To solve it note that R = − C C C

• Then:
µ ¶
2 B K GC KB − HC
P + + − P+ =0
A J JA JA

a quadratic equation on P .

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 Solving the System III

• We have two solutions:

 õ !0.5
¶
1  B K GC B K GC 2 KB − HC 
P =− − − + ± + − −4
2 A J JA A J JA JA
one stable and another unstable.

1 (AP + B) we have to a
• If we pick the stable root and find R = − C
system of two linear equations on two unknowns with solution:
−D (JN + K) + CLN + CM
Q =
AJN + AK − CG − CJR
−ALN − AM + DG + DJR
S =
AJN + AK − CG − CJR

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 Calibration

• What does it mean to calibrate a model?

• Our choices

Calibrated Parameters
Parameter β ψ α δ ρ σ
Value 0.99 1.75 0.33 0.023 0.95 0.01

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