Lecture 16. Introduction to Real Business Cycle Theory (RBC)
1. Business Cycle Facts 2. Detrending Procedures 3. Basic Model (Long and Plosser 1983 JPE) 4. Stability of a system of difference equations 5. Log Linearize a system of equations 6. King and Rebelo (RCER 2000) 7. Farmer and Guo (JET 1994) 1. Business Cycle Facts: Comovement; Persistence; Recurrent but not period; Most
variables are procyclical except unemployment (countercyclical) and real interest rate (acyclical)
2. Leading variables: stock prices; residential investment etc. 3. Lagging variables: inflation; nominal interest rate 4. Detrending I: linear detrending (log linear) 5. Detrending II: Piecewise linear detrending (log linear) 6. Detrending III: Hodrick-Prescott filter (HP filter)
( )
( ) ( )
( ) ( )
1
2
1
1 2
1 1
2
1 22
1 1
1 2
min ( )
subject to:
( )
FOCs are:
0
NT
t
t N
T
t t
Y t
t N
T T T T
t t t t
t
t N t N
T T T T T
t t t t t t
t t
T
t
Y Y
Y Y Y Y
L Y Y Y Y Y Y
L
Y
µ
! µ
=
=
= "
+ "=
= = "
+ "= =
"
# $" " " %& '
( )# $= " " " " " "* +& ', -
.=
.
/
/
/ /
7. ! is a function of µ . Selecting either ! or µ is equivalent. Is '( ) 0 or '( ) 0?! µ ! µ" #
8. Think about the following: If µ = +! If 0µ = If 0 µ< < +! For quarterly data, ! is recommended to be set to 1600.
9. The idea of RBC is to model business fluctuation as responses of rational individuals to productivity shocks. The simple model of one-sector Long and Plosser (1983 JPE) demonstrate this possibility.
10. In Long and Plosser, the representative agent is assumed to have log utility function:
[ ]0
ln (1 ) lnt t tE c l
!! !
!
" # #$
+ +
=
+ %&
11. The resource constraint is given by, 1
1t t t t tK z K n c
! !
" " " " "
#
+ + + + + += # In other words, 100 percent depreciation is assumed.
12. With these assumptions, the model has an explicit solution. To proceed, write down the Lagrangian and derive the first order conditions,
( )1
1
0
ln (1 ) lnt t t t t t t t t
L E c l z K n c K! " "
! ! ! ! ! ! ! !!
# $ $ %&
'+ + + + + + + + +
=
( )* += + ' + ' ', -. /0 12
13. FOCs are (for all ! ):
( )1 1
1 1 1 1
0
1(1 ) 0
1
0
t t
t
t t t t t
t
t t t t t t
Ec
E z K nn
E z K n
!
!
" "! ! ! !
!
" "! ! ! ! !
#$
#$ "
$ %$ "
+
+
&+ + + +
+
& &+ + + + + + + + +
' (& =) *
+ ,
' (&& & =) *
&+ ,
& =
14. Thus, the same has to be true for 0! = , which yields:
( )1 1
1 1 1 1
1(1 )
1
t
t
t t t t
t
t t t t t t
c
z K nn
E z K n
! !
! !
"#
"# !
# !$ #
%
% %
+ + + +
=
%= %
%
=
and 1
1t t t t tK z K n c
! !"
+ = " plus the transversality condition:
1TVC: lim 0
t t tE K
!
! !!
" # + + +$%
=
15. The solution method is the same as we used for the case without uncertainty, namely, guess and check:
1
1
1
1
1 1
(1 )
1
(1 )
1 and 1
t t t t
t t t t
t
t
t t t
t
t t
c az K n
K a z K n
YE
c c K
EaY a a Y
a a
! !
! !
" "!#
" "!#
!# !#
$
$+
+
+ +
=
= $
% &= ' (
) *
% &= ' (
$) *
$ = = $
16. Need to check the labor-leisure equation is also satisfied: 1
(1 )1
(1 )
(1 )
Thus, constant:
(1 )
(1 ) (1 )(1 )
t
t
t t
t
t t
t
t
Y
n n
Y
c n
an
n n
n
!" #
!#
! #
! #
! # ! #$
%= %
%
= %
%=
= =
%=
% + % %
17. Also need to verify the transversality condition. 18. Final result:
( )
1
1
1
2
1 1
1
ln constant + ln ln
Thus, if ln , , which is i.i.d, then,
ln (1) stationary
If ln is itself an (1), namely,
ln ln , is i.i.d.
Then ln is (2), so ar
t t t
t t t
t z
t
t
t t t t
t
K z K n
K K z
z
K AR
z AR
z z
K AR
! !!"
!
µ #
$ % %
&
+
+
+ +
+
=
= +
= +
e ln and lnt tc Y
19. This example shows that a simple general equilibrium model with productivity shocks may general business cycle phenomena.
1. Linear Difference Equations. 2. Simplest different equation:
1
0Solution:
t t
t
t
x x
x x
!
!
+ =
=
3. Slightly more difficult one: 1t t
x x b!+ = + 4. Find the steady state first (to economize on notation, denote the steady state by x)
, ( 1)1
bx x b x! !
!= + " = #
$
5. Let t
t
x xx
x
!=% denote the percentage deviation from the steady state, then
1 0
t
t t tx x x x! !+ = " =% % % %
6. Higher dimension case: 1 0
t
t t tx Ax x A x+ = ! =
7. But how to compute tA and its properties?
8. Look at easy case first. If 1
1
0
.
.
.
Then
.
.
.
n
t
t
t
n
A
x x
!
!
!
!
" #$ %$ %
= $ %$ %$ %$ %& '
" #$ %$ %$ %=$ %$ %$ %& '
9. A little bit more complicated case. If matrix A has n distinct eigenvalues. Let i!
be the thi eigenvalue. Namely,
[ ]det 0, 1,2,...iI A i n! " = =
10. Then there exists a matrix Q such that 1
1
.
.
.
n
Q AQ
!
!
"
# $% &% &
= % &% &% &% &' (
11. Redefine 1ˆt tx Q x
!= , then
1 1 1 1
1
1
1
1
1
0
1
.
ˆ ˆ.
.
.
ˆ = .
.
t t t
t t
n
t
t
n
Q x Q Ax Q AQQ x
x x
x
!
!
!
!
" " " "+
+
+
+
= =
# $% &% &
= % &% &% &% &' (
# $% &% &% &% &% &% &' (
12. Then we can obtain ˆt tx Qx=
13. Stability Issues. 14. Case 1.
1t tx x!+ =% %
0
0
| | 1 unstable diverges to unless 0
| | 1 stable: converges to zero for any
| | 1 periodic cycles
t
t
x x
x x
!
!
!
> ±" =
<
=
% %
% %
15. Case 2. 1t t
x Ax+ =% % 16. Then: 17. Case of 1 2 33 and | | 1,| | 1,| | 1n ! ! != > < < . Saddle point stable. Need one linear
constraint on 0x% .
18. What if 1 2 33 and | | 1,| | 1,| | 1n ! ! != > > < ? 19. Suppose that 1 22, and | | 1 and | | 1n ! != < < , then what happens? (Draw
diagrams)
How to log linearize an equation
1. Again define (thus, (1 ) )t
t t t
x xx x x x
x
!= = +% %
2. Why is it called log linearize? 3. If
t t ty x z= , then
t t ty x z= +% % %
4. If t ty x
!= , then
t ty x!=% %
5. If '( )( ), then
( )t t t t
f x xy f x y x
f x
! "= = # $
% &% %
6. If , then t t t t t t
x zy x z y x z
y y= + = +% % %
7. Here is a practice question: how to log linearize 1t t t
k Ak c!
+ = " ?