Technical Appendix: The Role of Transitory and
Persistent Shocks in the Consumption Correlation and
International Comovement Puzzles∗
Tatsuma Wada†
Wayne State University
This Version: April 26, 2012
Abstract
This technical appendix provides the details of modeling, computations, construction of the data set, and estimation.
JEL Classification Number: E32, F41, F43.
Keywords: Two-country model, International Real Business Cycle Model, Learning, Kalman filter.
∗
I am grateful to Mario Crucini, Ana María Herrera, Jean-François Rouillard, Katheryn Russ, two anonymous referees, an associate editor, and participants at the 2012 Canadian Economic Association Meeting for
their useful comments. I thank Simon Gilchrist for conversations in the very early stages. This paper was
previously circulated with the title “Structural Breaks and Learning in a Two-Country Model” and “Trend
and Cycles in a Two-Country Model.”
†
Department of Economics, Wayne State University, Faculty and Administration Building, 656 W.Kirby
St., Detroit, MI, 48202 (tatsuma.wada@wayne.edu, Tel. 313-577-3001).
1
Model
• Utility Functions
0
∞
X

=0
0
∞
X
=0

1 £  1− ¤1−
 
1−  
1 £ ∗ ∗1− ¤1−
 
1−  
• Normalized Labor
1 −  −  ≥ 0
1 − ∗ − ∗ ≥ 0
• Technology
Capital used in production in a specific economy is not necessarily owned by the residents
of that country.
 =  (  )1−
∗ = ∗ (∗ ∗ )1−

 (  )1− −1

=


−1

1−
 =  (   ) =  1−  −

• Adjustment costs and investment
+1
∗
+1
¶


= (1 − )  + 

µ ∗¶

∗
∗
= (1 − )  + 
∗

µ
1
•
¶
µ
+1
 −1
 −1
 
= (1 − )
+

−1 
  −1 
¶  
µ
1
1

+1 = (1 − )  + 
  



• Resource constraint
 ( −  −  ) + (1 − ) (∗ − ∗ − ∗ ) ≥ 0
(1)
To understand this condition, consider
1 ( −  −  ) + 2 (∗ − ∗ − ∗ ) ≥ 0
there are 1 people in home and 2 people in foreign. If one divides both sides by 1 + 2 ,
(1) is obtained.
• Riskless Bond
 +1 +  +  =  + 
∗
 +1
+ ∗ + ∗ = ∗ + ∗
where  = (1 +  )−1 . Note that  is a discounted bond: it pays 1 unit of consumption goods at  + 1, and each individual can buy 1 unit of bond at . In case
+1 +  +  =  + (1 +  )  
the price of the bond at  is 1; and this bond is repaid (1 + +1 ) at  + 1.
•

+1 


 −1
+
+
=
+

 
 −1 
1
 +1 +  +  =  + 

• (Incomplete markets) Bond in the equilibrium
 + (1 − ) ∗ = 0
2
• Driving forces

=   =  
−1
µ
¶
−1 

=



 = −1 
and
∗
=  ∗ = ∗ ∗
∗
−1
µ ∗ ¶
−1
∗
=
∗
∗
∗
∗
∗ = ∗
 
•
 =
 =
 =
 =
 =



−1





−1
 +1 +  +  =  +
with the transversality condition:
e  +1 = 0
lim 
→∞
e =  (1−) and  is a Lagrange multiplier.
where 
• Balanced Growth path
 =  =  = 
3


Closed economy
0
∞
X
=0
e { (   ) +  (1 −  −  ) +  [(1 − )  − (  +1 −  (    )  )]

+ [ (   ) −  −  −  ]}
1  (   ) −  = 0
2  (   ) −  = 0
(2)
 2  (   ) −  = 0
(3)
  (    )   −  = 0
e +1 + 
e  +1 1  (+1  +1 ) −    = 0
  (+1 +1 ) 
£ ¡
¢
¡
¢
e  +1   +1 +1 +1 −   +1 +1 +1  +1 +1 +1 + (1 − )
−   + 
(4)
Complete Markets
(7)
1 −  −  = 0
(8)
  +1 = (1 − )  +  (    ) 
(9)
 −  −  = 0
(10)
e
+1
¶
´ µ
1− e
1 −  ³e
 − 

=
e + 1 −


Without tax, government purchases and transfer:
L = max 0
(6)
+ +1 1  (+1  +1 )] = 0
thus (9) is
2
(5)
∞
X
=0
e {[ (   ) + (1 − )  (∗  ∗ )]



+ (1 −  −  ) + (1 − ) ∗ (1 − ∗ − ∗ )
+ [(1 − )  − (  +1 −  (    )  )]
¡ ∗
£
¢¤
+ (1 − ) ∗ (1 − ) ∗ −  ∗ +1
−  ( ∗ ∗ ∗ ) ∗
+ [ ( (   ) −  −  ) + (1 − ) ( (∗  ∗ ) − ∗ − ∗ )]}
FOCs (assuming  = ∗ = 05)
4
1  (   ) −  = 0
(11)
1  (∗  ∗ ) −  = 0
(12)
2  (   ) −  = 0
(13)
2  (∗  ∗ ) − ∗ = 0
(14)
 2  (   ) −  = 0
(15)
 2  (∗  ∗ ) − ∗ = 0
(16)
  (    )   −  = 0
(17)
∗  ( ∗ ∗ ∗ )  ∗ −  = 0
e +1 + 
e  +1 1  (+1  +1 ) −    = 0
  (+1 +1 ) 
¡
¡ ∗
¢ ∗
¢
∗
∗
∗
∗ ∗
e
e
  ∗+1 +1

+1 +  +1 1  +1  +1 −    = 0
1 −  −  = 0
(18)
(19)
(20)
(21)
1 − ∗ − ∗ = 0
(22)
  +1 = (1 − )  +  (    ) 
(23)
∗
 ∗ +1
= (1 − ) ∗ +  ( ∗ ∗ ∗ ) ∗
( (   ) −  −  ) + ( (∗  ∗ ) − ∗ − ∗ ) = 0
where
 (    ) =  (    ) −  (    )

 + (1 − ) 
 
The first order condition with respect to +1 :
¤
£
£ ¡
¢
¡
¢
e  +1   +1 +1 +1 −   +1 +1 +1  +1 +1 +1 + (1 − )
−   + 
++1 1  (+1  +1 )] = 0
(and Foreign correspondence) is rewritten as above.
We have 15 equations and 15 variables:
¡
¢
∗
        +1  ∗  ∗  ∗  ∗  +1
   ∗    ∗   
hence, there is no determinacy problem.
5
(24)
(25)
(26)
3
Incomplete Market
L = max 0
∞
X
=0
½
∙
¶ ¶¸
µ
µ


e

  (   ) +  (1 −  −  ) +  (1 − )  −   +1 −   

µ
¶¾


+  (   ) + −  +1 −  − 

The first order conditions
1  (   ) −  = 0
(27)
2  (   ) −  = 0
(28)
 2  (   ) −  = 0
(29)
  (    )   −  = 0
(30)
e +1 + 
e  +1 1  (+1  +1 ) −    = 0
  (+1 +1 ) 
(31)
¤
£
£ ¡
¢
¡
¢
e  +1   +1 +1 +1 −   +1 +1 +1  +1 +1 +1 + (1 − )
or −    + 
(32)
++1 1  (+1  +1 )] = 0
∙
¸

+1

e 
=0
−  + 
 +1
1 −  −  = 0
  +1 = (1 − )  +  (    ) 

 (   ) + −  +1 −  − 

(33)
(34)
(35)
(36)
(37)
where

 + (1 − ) 
 
In this case, we have 9 equations and we need to solve for 10 variables:
¢
¡
        +1  +1         
 (    ) =  (    ) −  (    )
(38)
The corresponding set of equations are obtained for Foreign, and we need to solve for
¡ ∗ ∗ ∗ ∗ ∗
¢
        +1  ∗+1  ∗  ∗  ∗   
With the bonds market clearing condition:
 + (1 − ) ∗ = 0
We have 19 equations and 19 variables.
6
(39)
3.1
The Steady State and Its Indeterminacy
In the steady state (where growth rates are   =  ∗ = ),
∙
¸
e
e



+1

e 
= 0 ⇒  = = ∗
−  + 
 +1


This implies that we have 18 equations to solve for 19 variables.
This problem can be seen from another view point:
If we eliminate  from the system, equations (37) and its foreign correspondence, together
with (39) become
 ( ) + (1 − )  ( ∗   ∗ ) =  + ∗ +  + ∗ 
we have 16 equations and 17 variables (19- and ∗ ), yet the problem remains. Setting
 = ∗ = 0 is arbitrarily choosing a steady state.
To ensure determinacy, several approaches have been proposed. Among others,
1. Endogenous discount factor
1-.Endogenous discount factor without internalization (only two first order conditions,
namely 40 and 41 are different from the above)
2. Debt elastic interest rate
3. Portfolio adjustment costs
3.2
A Model with an Endogenous Discount Factor
Endogenous discount factor:
+1 =  (   )  
£
¤−
 (   ) = 1 +  1−

with 0 = 1 and  ≥ 0.
The maximization problem is then:
½
∙
¶ ¶¸
µ
µ
∞ ∙
X

  (   ) +  (1 −  −  ) +  (1 − )  −   +1 −   

L = max 0

=0
µ
¶¾
¸


+  (+1 −  (   )  ) 
+  (   ) + −  +1 −  − 

with a new control variable +1 .
7
3.2.1
 :
The FOCs
1  (   ) −  ( ) −  = 0
 :
  )
2  (   ) −  (
−  = 0

 :
 2  (   ) −  = 0
 :
  (    )   −  = 0
+1 :
  (+1 +1 )  (   ) +1 +  (   )  +1 1  (+1  +1 ) −    = 0
ª
£
© ¡
¢
¡
¢
−   +  (   )  +1   +1 +1 +1 −   +1 +1 +1  +1 +1 +1 + (1 − )

+1 :
++1 1  (+1  +1 )] = 0
h
i
−  +  (   )  +1 = 0
+1
 :
 +  [ (+1  +1 ) − +1  (+1  +1 )] = 0
 :
1 −  −  = 0
 :
  +1 = (1 − )  +  (    ) 
 :
 :
 (   ) +


−  +1 −  −  = 0
+1 −  (   )  = 0
8
 (   )
−  = 0

 (   )
 : 2  (   ) − 
−  = 0

 :  2  (   ) −  = 0
(41)
 :   (    )   −  = 0
(43)
 : 1  (   ) − 
(40)
(42)
+1 :   (+1 +1 )  (   ) +1 +  (   )  +1 1  (+1  +1 ) −    = 0
ª
£
© ¡
¢
¡
¢
or −    +  (   )  +1   +1 +1 +1 −   +1 +1 +1  +1 +1 +1 + (1 − )
(44)
++1 1  (+1  +1 )] = 0
∙
¸
+1

=0
+1 : −  +  (   ) 
 +1
 :  +  [ (+1  +1 ) − +1  (+1  +1 )] = 0
(45)
 : 1 −  −  = 0
(47)
 :   +1 = (1 − )  +  (    ) 

 :  (   ) + −  +1 −  −  = 0

 : +1 −  (   )  = 0
(48)
(46)
(49)
(50)
With the bonds market clearing condition:
 + (1 − ) ∗ = 0
(51)
we have 23 equations and 23 variables:
¡
¢
∗
        +1  +1  +1  ∗  ∗  ∗  ∗  +1
 ∗+1  ∗+1          ∗  ∗  ∗  ∗   
Contrary to the previous case, in the steady state, we have
 =
 (∗  ∗ )
 ( )
=


∗
Since  =  ∗ in the steady state,  ( ) =  (∗  ∗ ) determines the steady state values.
(We shall see this in the following subsection.)
3.2.2
Finding the steady state
1. The steady state level of the labor supply:
9
Equations (40), (41) and (42) imply
  )
+  2  (   )
 (
2  (   )


=
1  (   )
 (  ) + 

The left hand side is
(1 − ) 
2  (   )

=
1  (   )

For the right hand side, we have
¡
¢
+  2  (   )
− (1 − )  1+ 1−  −

 
=
¡
¢
−  1+ 1− −1
1−
+ 


  )
 (
+  2  (   )

 ( ) + 


 +  2  (   )
=
 
 1−
+ 

where
¡  − ¢

  
 1−
1 +  
Then, by equating the left hand side with the right hand side,
 = − (1 − ) 
 +  2  (   )
(1 − ) 
=

 

 1−
+ 

By dividing both sides by (1 − )   , we have
1=
 
 
+  2  (   ) 1−
 1−


 
 1−

+ 
or
⇒

(1 − ) 
= 2  (   ) = 


(1 − ) 
1−
=

 

By defining  =  , we have
 =
=
1
1+
1− 
 
1−
1 −  +  1−

(52)
since  = 1 − . Note:  is determined later. In practice, we set the steady state
level of the labor supply, and then we recover  so that it is consistent with (52).
Then,  is


=
 +  1−

10
2. Finding capital stock
Since 0 = 1 in ss,  =  in the steady state (by 43). Then, equation (44) implies
∙
¸
 ( )

 −  + (1 − ) +  ( ) 1  ( ) − 1 = 0


∙
¸
1−
 ( )
+  ( ) 1  ( ) − 1 = 0

 ( ) (1 −  + 1  ( )) = 
³
´

 −  (1 − )
= ⇒ =
 ( ) 1 −  + 



since by setting 0 = 1 in the steady state, the values in the steady state with the
adjustment costs are the same as those in the steady state without adjustment costs.
(See Baxter and Crucini p.828). Therefore, the capital accumulation equation without
adjustment costs implies:
+1 = (1 − )
⇒

+ 


 =  − (1 − ) =  () ;

also, note
 =

1  ( ) 
⇒ 1  ( ) =  
 ( )

Then,



=
=



 −  (1 − )
µ ¶1−


=

Hence, we obtain the steady state value of capital
µ ¶1

 −
 = 

¸−1
∙
 −  (1 − )
= 
 −



=



=  − (1 − )
 =

 
 


=
= 
= = [ − (1 − )]


 

 −  (1 − )
11
3. The bond price
From (45)
 =

1

⇒
= 

1+

4. Output
From (45) and (44)
− 
+  (   ) 
∙
¸
∙
¸
 (   )
+1
+1
= 0 ⇒  =

 +1
 +1

ª
£
© ¡
¢
¡
¢
−   +  (   )  +1   +1 +1 +1 −   +1 +1 +1  +1 +1 +1 + (1 − )
++1 1  (+1  +1 )] = 0;
we have (with    =  from 43)
∙
ª
¢
¡
¢
+1 © ¡
  +1 +1 +1 −   +1 +1 +1  +1 +1 +1 + (1 − )
0 = − +  (   ) 
 +1
++1 1  (+1  +1 )]
ª
£
© ¡
¢
¡
¢
= − +      +1 +1 +1 −   +1 +1 +1  +1 +1 +1 + (1 − )
¤
+   +1 1  (+1  +1 )
ª
£
© ¡
¢
¡
¢
= −1 +     +1 +1 +1 −   +1 +1 +1  +1 +1 +1 + (1 − )
¤
+  +1 1  (+1  +1 )
1
= (1 − ) +  1−

Note that
µ


¶−1
∙
¸1(−1)


(1 − )
=
−





∗
= ∗


and hence

=
∗
µ

∗
¶ µ
12

∗
¶1−
=


∗
(53)
5. Consumption and Labor input
In the steady state, from (45),
 ( ) =  (∗  ∗ ) ⇒  1− = ∗ ∗1− 
(54)
From Equations (40), (41) and (42),
  )
2  (   ) −  (

1  (   ) −  ( )
= 2  (   ) 
The left hand side is
n£
£
¤−
¤−−1 o
(1 − )  −  1−
+   1 +  1−
(1 − ) 
n
o =
;

−1 1− [ 1− ]− +   [1 +  1− ]−−1
thus, the optimality condition (MRS=MPL) is
(1 − ) 

= (1 − ) 


From (54),

=

Then,
1−

µ
∗

µ
∗

¶1
¶1
(55)
∗

∗
∗

= (1 − )
∗


and the foreign corresponding (55) is
∗
(1 − ) ∗
=
(1
−
)

∗
∗
(56)
Therefore,
µ
∗

¶1
=
 ∗

 ∗
from (53),  = ∗ ,  = ∗ ,  =  ∗ ,  = ∗ ,  = ∗ = 0. Note that because of
the condition (54), our steady state is unique, unlike a model without an endogenous
discount factor.
13
6. The discount factor parameter. We set  so that the steady state condition is satisfied
by:
1
 ( )
 (∗  ∗ )

 =

=
=
1+

∗
To this end, we set
¡  ¢
ln 1+
=−

ln (1 +  1− )
3.2.3
Summary — Finding Steady State Values
The set of parameters used in this paper is:
 = 2
 = 1004
 = 058
 = 00654
 = 0025
 = 02
” 
 =
 ≡ −025
0 
Then, the steady state levels of the variables are given as follows.
 = [ − (1 − )]

 −  (1 − )
 = 1 − 

 =
 +  1−

¸1(−1)
∙
(1 − )

−
 = 


 1− −
 =   
 − (1 − )
 = 

 =  
¡  ¢
ln 1+

⇒ =−
 ( ) =
1+
ln (1 +  1− )
The Lagrange multipliers
14
 =

1−
 ( )

 ( )
 = 2  ( ) − 


 =

 = 1  ( ) − 
The derivatives:
£  1− ¤1−
 
1  (   ) =  

£  1− ¤1−
 
2  (   ) = (1 − ) 

£  1− ¤1−
 
11  (   ) =  [ (1 − ) − 1] 

£  1− ¤1−
 
12  (   ) =  (1 − ) (1 − ) 

£  1− ¤1−
 
21  (   ) =  (1 − ) (1 − ) 

£  1− ¤1−
 
22  (   ) = (1 − ) [(1 − ) (1 − ) − 1] 

£
¤−−1 ¡ −1 1− ¢
1  (   ) = − 1 +  1−
 

¡ −1 1− ¢

= −
 
 1−
1 +  
 (   )
= −
 + 1−
−1


¡  − ¢

2  (   ) = − (1 − ) 
 
1−
1 +  
£
¤
 (   )
11  (   ) =  £
( + 1)  1−
−2
1−
+ (1 − )
¤



2
−1
 + 1−


12  (   ) = 21  (   )
¤
£  1−
 (   )
−1 −
= (1 − )  £



−
1

¤




2
−1
 + 1−


¤
£
 (   )
 −−1
 1−
22  (   ) = (1 − )  £



+

{
(1
−
)
+
1}

¤




2
−1
 + 1−


15
4
Linearized Equations
e
 + 12  ( ) 
11  ( ) e
e −e
−1  (   ) e − 11  (   ) e
 − 12  (   ) 
 = 0
|
{z
}
e
21  ( ) e
 + 22  ( ) 
(57)
e −
−2  (   ) e − 21  (   ) e
 − 22  (   ) 
(58)
e = 0
|
{z
}
21  ( ) e
22  ( ) e
 +
 +
e = 0
(59)
e − e
 − 
2  ( )
2  ( )
µ
¶
”  e ”  e
” 
e
 + 0   − 0   − e + 1 + 0  
(60)
e = 0
 
 
 
µ 2
¶

2
 e
  2 ” + 11  ( )  e
+1 −  2 ”e+1 +
+1 + 1  ( ) e+1



∙
¸
2
e − 
e+1 − 
+ −1  ( ) −  2 ” 
e+1 + 12  ( )  
e

½
¾
½
¾


e = 0
(61)
+ 1  (   )
 + 2  (   )
+ 1  ( ) e
+ 1  ( ) 


|
{z
}
1  (   )
2  (   ) e
 e+1 − 
e
 +
e+1 − e − e +
(62)
 = 0


|
{z
}
e

h
i
e+1 −  ( ) e
e = 0
+ 1  ( ) e
+1 + 2  ( ) 
+1 − 1  ( ) e
 − 2  ( ) 
(63)
e = 0
e +  
(1 − ) 
¶
´ µ
1 −  ³e
1− e
e
e + 1 −
 − 

+1 =


1  (   )  e
2  (   ) 

1
 +

e − e
 −
 +
0 = e − e




(64)
(65)
e
 −  e+1

(66)
Since we have 10 equations (Home) and 10 equations (foreign), plus the market clearing
condition:
e∗ =  e 

−1
there are 21 equations to solve for 21 variables:
¢
¡
∗
 ∗+1          ∗  ∗  ∗  ∗   
        +1  +1  ∗  ∗  ∗  ∗  +1
16
5
5.1
Learning
Shock Process
⎡
⎣

 ∗
⎤
⎡
| {z }

 =
⎣

⎤
⎦ = ⎣
 ∗
| {z }

⎡
⎤
|
⎡
1



⎢
⎢  ⎥
⎢
⎢
⎥
⎢  ⎥ = ⎢ 0
⎣
⎣
⎦
0
∗
|
| {z }

5.2
⎡
⎡
⎤
1 0 ⎢
⎦⎢
⎢
0 1 ⎣
{z
}

|
⎤⎡
0 0
⎥⎢
⎥⎢
0 0 ⎥⎢
⎦⎣
0 0
{z
}|

⎡
⎦ + ⎣

| {z }


∗
⎤
⎦
  + 
|{z} −1

Notations by KW
⎡
1
⎦ = ⎣
⎤

⎤
⎥
⎥
 ⎥
⎦
∗

{z }

−1
−1
⎤
⎥ ⎢
⎥ ⎢
⎥+⎢
⎦ ⎣
∗−1
{z }
 −1
⎡
12

|
© ª © ª
Learning: How to Get | =1 , | =1
0
0
0
12
11
12
12
0
21
{z
12
12
22

⎤⎡
⎤

⎥⎢  ⎥
⎥⎢
⎥
⎥⎢  ⎥
⎦⎣
⎦
∗
}| {z }

Note that: the Kalman filter yields
¡
¢
| =  −1|−1 +   −  −1|−1
= ( −  )  −1|−1 +   
and
| ≡  − | 
Hence, we need to compute  at steady state. As the main text explains, this model
has the unique steady state Kalman gain. To compute it,
17
1. Get Λ
⎡
Λ=⎣
0
0
−1
 +   
−1
− −1 0
0

0
−1
−  
−1
 −1
⎤
⎦
2. Compute the Schur decomposition of Λ. For example, one can use a Matlab function
"schur."
3. Reorder the diagonal elements of  from smallest to largest. Get the corresponding
orthogonal matrix , where
⎤
⎡
11 12
⎦
 =⎣
21 22
such that  0 Λ =  is quasi-upper-traiangular:
⎡
⎤
11 12
⎦
=⎣
0 22
The diagonal elements of  are 1  2 (eigenvalues of Λ) in any order, but 11 has only
stable eigenvalues. For example, Algorithm 7.6.1 in Golub and Van Loan (1996) with
the Givens rotation can be used.
−1
. This is the unique solution to the ARE.
4. Compute  = 21 11
5. The steady state Kalman Gain is then  =   0 (  0 + )−1 .
A Matlab file “steady_kalmangain.m” computes the steady state Kalman gain following
the above steps (1 through5).
5.3
Impulse Responses
The solution to the rational expectations model is given by:
18
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣










∗
∗
∗
∗
∗
∗
∗
∗
∗


⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎡
⎥
⎥
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥ = Π⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎥
⎥
⎥
|
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦

⎤
⎥
⎥
⎥
⎥
⎥
 ⎥
⎥
⎥
 ⎥
⎥
⎥
 ⎥
⎦
∗

{z }
∗

 ∗
| {z }

⎡
⎡
⎤


⎢ −1
⎢  ⎥
⎢ ∗
⎢ ∗ ⎥
⎢ −1
⎢  ⎥
⎢
⎢
⎥
⎢
⎢
⎥
⎢ 
⎢  ⎥
⎢
⎥ =  ⎢ −1
⎢
⎢
⎥
⎢ −1
⎢  ⎥
⎢
⎢
⎥
⎢
⎢
⎥
⎢ −1
⎢  ⎥
⎣
⎣
⎦
∗

∗−1
| {z }
| {z

−1
⎤
⎡
0
0
0
⎥ ⎢
⎥ ⎢
⎥ ⎢ 0
0
0
⎥ ⎢
⎥ ⎢
⎥ ⎢ 0
0
0
⎥+⎢
⎥ ⎢ 12
⎥ ⎢ 
0
0
⎥ ⎢
⎥ ⎢
12
12
⎥ ⎢ 0
11 12
⎦ ⎣
12
12
21 22
19 0
} |
{z

⎤
⎥
⎥⎡
⎤
⎥
⎥ 
⎥⎢
⎥
⎥⎢
⎥⎢  ⎥
⎥⎣  ⎥
⎦
⎥
⎥ ∗
⎥

⎥
⎦
}
Example:
Suppose that a persistent growth rate shock occurs:
• Perfect Information
⎡
• Imperfect Information
⎢
⎢
⎢
⎢
⎢
⎢
1 = ⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎢
⎢
⎢
1 = ⎢
⎢
⎢
⎢
⎢
⎢
⎣
0
⎤
⎥
⎥
0 ⎥
⎥
⎥
0 ⎥
⎥
⎥
1 ⎥
⎥
⎥
0 ⎥
⎦
0
0
⎤
⎥
⎥
0 ⎥
⎥
⎥
0 ⎥
⎥
⎥
1|1 ⎥
⎥
⎥
1|1 ⎥
⎦
∗
1|1
Then
 = Π
 = −1
Computation note
Step 1.
© ª © ª n o
Compute the steady state Kalman gain and then compute | =1 , | =1 , ∗|
=1
For every  = 1 to  − 1, iterate the following steps.
Step 2.
20
Plug | and | into the corresponding elements in  .
Step 3.
Compute  = Π .
Step 4.
Update +1 =  .
For the perfect information case, one can omit steps 1 and 2.
5.4
Simulation
Given our model/solution, one can exercise the simulation of the model. Here are the steps
for it:
(Persistent change)
Set the Kalman gain and
Step 1
Set 0 = 0 and generate 1 , then construct 1
Step 2
Compute  1 = 1 + 1
Step 3
Compute 1|1 and plug into 1 .
Step 4.
Compute  = Π .
Step 5.
Update +1 =  .
Iterate.
5.5
Growth Rate Change — Growth rate of macro variables
Consider change in the growth rate of variable  .
log  −1 =   + 
e + 
e − 
e−1
where 
e is the deviation of the new trend from its old one, and 
e is the deviation of 
from its old trend, which is obtained by the impulse response function.
21
For example, the growth rate of capital stock is
 + 
e + e
 − e
−1
and the growth rate of consumption is
 + e
  + e
 − e
−1
6
Data Appendix
=  + 
e + e
 − e
−1 
Following Boileau and Normandin (2008), we construct the data as follows.
1. Output: Real output for country  is computed as:
 =
∗


where ∗ is nominal output (in terms of national currency) and  is the all-item
consumer price index. From the Penn World Table (Heston et al. 2011), using real
GDP per capita (“rgdpch”, in terms of U.S. dollars, in 1995) and population (“pop”
in 1995),we compute a weight for country :
where e is real output:
e
 = P 
e

e =  ×  
Then, the European aggregate of real output is
X 
∗ =
  

and U.S. real output is
 =    
2. Employment
 =  ×  
where  is the civilian employment index (“Employment, all persons, seasonally
adjusted, base year=2005” from Labour Force Statistics in OECD Main Economic
22
Indicators) and  is the population in 1995. As for the European aggregate, employment is computed as:
X
   
∗ =

where

  = P
=
and for the U.S.,

;
 =  
3. Consumption
Using nominal consumption ∗ and the price index,  , real consumption for country
 can be computed as:
∗
 =  

Then, the European aggregate of real consumption is
∗ =
X
 

and U.S. real consumption is
 =   
4. Investment
Using nominal investment ∗ and the price index,  , real investment for country  can
be computed as:
∗
 =  

Then, the European aggregate of real investment is
∗ =
X
  

and U.S. real investment is
 =   
23
5. Productivity
For both the U.S. and the European aggregate, we compute the (log of) productivity
as follows:
ln  = ln  − (1 − ) ln 
where  is real output;  is labor input; and  = 036
7
Dealing with a Special Case:  = 1
In our model,  = 1 would result in a degeneration of , hence the steady-state Kalman
gain, derived in Appendix is not applied. In such a case, we rewrite the state space model:
⎡
⎣

 ∗
| {z

⎡

⎣ 

| {z

⎤
⎡
⎦ = ⎣
1 1
⎤⎡
⎦⎣
 
| {z }|

⎡
⎤⎡
 0
⎦ = ⎣
⎦⎣
0 0
}
| {z }|
}
⎤


⎤
⎦

{z }

−1
⎤
⎦+⎣
−1
{z }
−1
⎡
|
12
0

0

{z

12
⎤⎡
⎦⎣

⎤
⎦

}| {z }

In this case, we can confirm that there is a unique steady-state Kalman gain. the steady
state Algebraic matrix Riccati Equation (ARE) becomes
i
h
−1
 =   −   0 (  0 )   0 + 0 
Provided  6=  ( =  is trivial because in such a case,  ∗ =   , and hence, the model
degenerate to a univariate model),
h
i
−1
−1
 =   −1   −   0 (  0 )   0 ( 0 )  0 + 0
= 0 
Hence,
 =  −1 
© ª © ª n o
, we can use the formula:
To compute | =1 , | =1 , ∗|
=1
24
⎡
⎣
|
|
⎤
⎡
⎦ = ⎣
⎤

 ∗
⎡
=  −1 ⎣
⎡
⎦ + ( − )  ⎣

 ∗
⎤
−1|−1
−1|−1
⎤
⎦
⎦
The Solution to this model
Using the notations of King-Watson, first we set the model as
⎡
⎣

 ∗
| {z

⎡

⎣ 

| {z

⎤
⎡
⎦ = ⎣
1 1
⎤⎡
⎦⎣
 
| {z }|

⎡
⎤⎡
 0
⎦⎣
⎦ = ⎣
0 0
| {z }|
}
}
⎤


⎤
⎦

{z }

−1
⎤
⎦+⎣
−1
{z }
 −1
⎡
|
12
Suppose that a persistent growth rate shock occurs:
• Perfect Information
⎡
• Imperfect Information
⎤
0
⎢ ⎥
⎢ ⎥
⎢ 0 ⎥
⎢ ⎥
⎢ ⎥
1 = ⎢ 0 ⎥
⎢ ⎥
⎢ ⎥
⎢ 1 ⎥
⎣ ⎦
0
25
0
0

{z

12
⎤⎡
⎦⎣

⎤
⎦

}| {z }

⎡
⎤
0
⎥
⎢
⎥
⎢
⎢ 0 ⎥
⎥
⎢
⎥
⎢
1 = ⎢ 0 ⎥
⎥
⎢
⎥
⎢
⎢ 1|1 ⎥
⎦
⎣
1|1
Then
 = Π
 = −1
In this case, agents in the both countries immediately identify the persistence of the
shock, because all shocks affect the two countries simultaneously, and any shocks that hit
the two countries with the size of (1 ) is persisten, if not, transitory. In other words, the
size of the shock is informative enough for the agents to realize the persistence of the shock.
7.1
Anthor special case  = 
In case of  = , all shocks are symmetric in the two countries. The only question the agents
have is whether the shock is perstsent or transitory. Knowing that  ∗ =   , it is followed
that the ARE becomes
µ
¶
2
2
+ 0 
 = −
 +
and hence,
q
h
i
2
0 − 1 (1 − 2 ) + 1 + 12 (1 − 2 )2 + 2 + 2 + 1
 =
2
and
q
2
 − (1 − 2 ) +  12 (1 − 2 )2 + 2 + 2 + 1
q
=

2
2 +  − (1 − 2 ) +  12 (1 − 2 )2 + 2 + 2 + 1
Note that  = 0 ,  = 1,  = .
© ª © ª n o
To compute | =1 , | =1 , ∗|
, we can use the formula:
=1
| =   + (1 − ) −1|−1
| =   − | 
26
Then, the impulse-responses are computed by the same way as the previous case.
8
Hicksian Decomposition
8.1
The decomposition
We compute the Hicksian decomposition proposed and utilized by King (1990) and Baxter
and Crucini (1995). There are: (1) the interest rate effect; (2) the wage rate effect; (3)
wealth effect. Although we follow Baxter and Crucini (1995) closely, due to our own setup of
the maximization problem that are slightly different from that of Baxter and Crucini (1995),
the Lagrange multipliers reflect the direct effect of the growth rate change. Therefore, we
need (4) the adjustment term. To clarify this point, our Lagrangian is
½
∙
¶ ¶¸
µ
µ
∞ ∙
X

  (   ) +  (1 −  −  ) +  (1 − )  −   +1 −   

L = max 0


=0
µ
¶¾
¸


+  (+1 −  (   )  ) 
+  (   ) + −  +1 −  − 

whereas models assuming a constant growth rate of technlogy have their Lagrangian:
½
∙
¶ ¶¸
µ
µ
∞ ∙
X


L = max 0
   (   ) +

(1 −  −  ) +  (1 − )  −   +1 −   




=0
¤
¡
¢ª
+  (   ) +  −  +1 −  −  +  (+1 −  (   )  ) 
where  =   0 .
Hence, it should be noticed that
 =   
To maintain consistency, we compute the interest rate and the wage rate effects using
variations in  , not in  . In other words, we exclude the changes in  from the variations
in the Lagramge multiplier  in order to compute both the interest rate and the wage rate
effects.
8.2
The Wealth Effect
The main idea of the computation of the wealth effect is as follows.
1. Compute the discounted present value of the change in utility caused by the altered
time path of consumption and leisure.
27
The change in utility (measured in consumption good unit) due to the changes in
consumption and leisure paths is:
∞
∞
X

X 
1−
= _ =
(1 − )
 + +
  + 

 =0

=0
This is due to the fact that the utility change is
∙
¸
∞
X

 
+ +
+

 =


=0
∙
¸
∞
X
 +
  +


+

=




=0
∙
½
¾
¸
∞
X

 +
  +

+
=




  


=0
∙
¸
∞
X
 + (1 − )   +
 (1 − ) 

+
since
=

=







=0
∙
¸
∞
X
 +
  +
(1 − ) 


+ (1 − )
since
= (1 − )

=
 
 


=0
∙
¸
∞
X
 +
1 −  +
+ (1 − )

=
 


=0
Recall that our system is:
 = Π
 = −1 + 
Thus, the total change in the observable variables + due to any shocks in  can be
28
computed by
∞
X
=0

 + =
∞
X
  Π  
=0
∞
X
    
= Π
=0
= Π
"∞
X
⎡
#
()  −1 
=0
⎢
⎢
= Π ⎢
⎣
1 (1 − 1 )
0
0
...
0
0
0
0
1 (1 −  )
The decomposition of matrix M is:
⎤
⎥
⎥ −1
⎥   
⎦
(67)
 =   −1
where the diagonals of  : 1  · · ·   are eigenvalues of .
2. Compute the constant consumption and leisure profiles that yield the same change in
utility, without changing the interest rate and wages.
Consider constant changes in  and  so that utility will change by  _. That is, we
need to find  and  :
∙
¸

1− 
1
+
 
 _ =
1− 
 
Since any variable other than  and  remaind unchanged, the time paths of  and 
must satisfy the following two conditions from the log-linearizations.
12  ( ) − 12  ( )
11  ( ) − 11  ( )
+
=0
1  ( ) − 1  ( )
1  ( ) − 1  ( )
21  ( ) − 21  (   )
22  ( ) − 22  (   )
+
=0
2  ( ) − 2  ( )
2  ( ) − 2  ( )
Then, changes in  and  satisfy:
=
 21 −  11

 12 −  22
29
(68)
(69)
where
11  ( ) − 11  ( )
21  ( ) − 21  (   )
;  21 =
1  ( ) − 1  ( )
2  ( ) − 2  ( )
12  ( ) − 12  ( )
22  ( ) − 22  (   )
;  22 =
=
1  ( ) − 1  ( )
2  ( ) − 2  ( )
 11 =
 12
Thus, we need to know:
¸
∙
 1−
 21 −  11
1

+
(1 − )
_ =
1− 

 12 −  22
or  that satisfies
∙
 1−
 −  11
 = (1 − ) _
+
(1 − ) 21


 12 −  22
¸
and  that satisfies
¸
∙
 21 −  11
 21 −  11
 1−

=
+
(1 − )
× (1 − ) _
 12 −  22


 12 −  22
In the paper, we do not compute the wealth effect directly. Rather, the wealth effect is
calculated as the residual of the total effects less the interest rate effect, the wage rate effect
and the adjustment term.
8.3
The Interest Rate Effect
The interest rate effect is more complicated. It is the response due to the change in the
interest rate: mainly because of the intertemporal substitution. A simple two period model
implies that a rise in the real interest rate requires an increase in the marginal rate of
substitution between today’s consumption and tomorrow’s; the growth rate of consumption
has to change. To compute these substitution effects, one needs to distinguish two effects:
(i) the wealth or income effect arising from an increase in the net worth (which is caused
by the interest rate change), without changing the price ratio; and (ii) the pure substitution
effect arising from the change in the price ratio, but not altering the real wealth.
As is well known, we can always decompose the effect of a price ratio change into two
effects.
Total change of consumption (wealth) due to a change in interest rate (wage)
= Substitution effect + Wealth Effect.
30
8.3.1
Wealth Effect
First, we compute the wealth effect, in terms of the percent deviation from a variabe’ steady
state value.
Suppose that there is no change in the price ratio, but the wealth increases: that is,
( ) increases by (1 1) (both shadow prices go up by 1%). The increases in consumption
and leisure can be computed as:
⎡
⎣
1
1
⎤
⎡
⎦=⎣
 11  12
 21  22
⎤−1 ⎡
⎦
⎣
1
1
⎤
⎦
where 1 and 1 are percent deviations from their steady state values when both shadow
prices go up by 1%.
Then, the wealth (measured in utility) at time  increases by:
1−

1 +
(1 − ) 1 


Since consumption and leisure increase by
1−

1 and
(1 − ) 1 


respectively, one unit (measured in utility) of an additional increase in wealth at t raises
consumption and leisure by


 1
+


 1
1−
(1

− ) 1
and
1−
(1 − ) 1



1−

+
(1
−
)

1
1


respectively. Suppose that the increase in wealth is divided by the wealth that is attained
through constant increases in  and , say 1 and 1 . Then, changes in  and  are
1
1−
1
1−
£
 +
 1
1−



 1
1−

(1 − ) 1
¤ × change in wealth
(1 − ) 1
¤ × change in wealth.
 + 1−
(1 − ) 1
 1

£
Next, consider how much wealth will increase due to actual changes in shadow prices.
(1 − ) 1 .
Suppose that ( ) = ( ). The wealth increases by  1  + 1−

Therefore, the total change in consumption due to the wealth effect is
¸
∙

1
1−


£
¤×
1  +
(1 − ) 1 
1


 + 1−
(1 − ) 1
1−  1

31
and the change in leisure is
¸
∙
(1 − ) 1
1−

£
¤×
1  +
(1 − ) 1 
1
1−



+
(1
−
)

1
1
1− 

1−

in consumption good unit. If one wants to compute the percent deviations, then
(
¸)
∙


1
−




 1
£
¤

(1
−
)

=

+


×
1
1
1

 



 + 1−
(1 − ) 1
1−  1

¸
∙
1−
1

£
¤×
1  +
(1 − ) 1 
=
1
1−


 +  (1 − ) 1
1−  1
and

=
 
1
1−
£

1
1 +
1−

¸
1−

¤×
1  +
(1 − ) 1 


(1 − ) 1
∙
Finally, we can derive the following matrix representation:
⎡
where
⎣

 

 
⎤
⎦=
1
1−
1
£
 +
 1
1−

(1 − ) 1
⎡
¤⎣
1 1 1
2 1 1
1 1 1 2 1 1
⎤⎡
⎦⎣


⎤
⎦

1−
, 2 =
(1 − ) 


Then, the real interest rate and the wage effects can be computed as follows.
For the interest rate effect, one can set ( ) = ( ), which is computed by (67).
For the wage effect, one can set ( ) = (0  − ), which is computed by (67).
1 =
8.3.2
Substitution Effect
The total effect (in terms of percent deviation) is given by
⎡
⎣


⎤
⎡
⎦=⎣
 11  12
 21  22
the substitution effect can be computed as:
⎡
⎣

 

 
⎤
⎡
⎦=⎣


⎤
⎤−1 ⎡
⎦
⎡
⎦−⎣
32
⎣



 

 
⎤
⎦
⎤
⎦
(70)
8.4
Computation: The Hicksian Decomposition
The decomposition focuses on: (i) the interest rate effect (inter-temporal price change);
(ii) the wage effect (intra-temporal price change); (iii) the growth rate effect and (iv) the
adjustment term.
1. “The interest rate effect” (the effect of inter-temporal price changes excluding the
change in the level of technology):
⎧⎡
⎤⎫∞
⎨  − ⎬

⎦
⎣ 
⎩  − ⎭


=0
⎡
⎤
less the effect that should be included in the wealth effect: ⎣

 

 
⎦
2. “The wage effect” (the effect of intra-temporal price changes excluding the change in
the level of technology):
⎧⎡
⎤⎫∞
⎬
⎨
0
⎦
⎣
⎩  − ( −  ) ⎭



=0
⎤
⎡
less the effect that should be included in the wealth effect: ⎣

 

 
⎦
3. “The growth rate effect” (the effect of growth rate change that causes price  to
change):
⎧⎡
⎤⎫∞
⎨  ⎬
⎣  ⎦
⎩ 0 ⎭
=0
⎡
4. The adjustment of the wealth effect: ⎣

 

 
⎤
⎦ arising from
⎧⎡
⎤⎫∞
⎨  ⎬
⎣  ⎦

⎩ 0 ⎭
=0
(4) is meant to avoid to count the wealth effect arising from the variations in the Lagrange
maultiplier  due to changes in the level of technology.
33
The adjustment term is the sum of (3) and (4). Therefore, we arrive at:
Total Effects = Interest Rate Effect + Wage Effect + Wealth Effect + Growth Rate
adjustments (3+4)
9
Estimation
Our model
 =  + 
 =  −1 +  
is identified. To see this, note that the model satisfies:
1. Minimum representation (i.e., observable and controllable; See p.279 in Gourieroux
and Monfort, 1997). This is proven in Appendix.
2. Any linear transformation by a nonsingular matrix  :
 ∗ =  −1
 ∗ =    −1
∗ =  
∗ =   0
produces the identical likelihood function value only if  =  = 1. This is obvious
since
⎡ ⎤
1
 = ⎣ ⎦

34
Table
Variable
Explanation
Source
∗
Nominal GDP; “Gross Domestic Product-expenditure approach, national currency, current prices, seasonally adjusted”
QNA

Price level; “Consumer price index, all items, base year=2005”
MEI
Real GDP per capita in 1995
PWT

Total population in 1995
PWT

Civilian employment index; “Employment, all persons, seasonally adjusted, base year=2005”
MEI
∗
Nominal Consumption; “Private final consumption expenditure, national currency, current prices, seasonally adjusted”
QNA
∗
Nominal Investment; “Gross fixed capital formation, national currency, current prices, seasonally adjusted”
QNA

Notes: QNA: OECD Quarterly National Accounts;
MEI: OECD Main Economic Indicators;
PWT: Penn World Table 7.0, Heston et al (2011)