ARE/ECN 240C Time Series Analysis
Winter 2004
P
ROBLEM
S
ET
4
–
S
OLUTIONS
Professor Òscar Jordà
Economics, U.C. Davis
Part I – Analytical Questions
Problem 1: Consider a stationary autoregressive process A(L)X t
=
 t
and its corresponding moving average representation, X t
= C(L)
 t
, where C ( L )
   i

0
C i
L i
.
(a) Find the moving average coefficients for an VAR(1) process.
Solution
Because this is a VAR(1), calculation of the MA representation is quite easy. Thus, if
X t
= A
1
X t-1
+
 t
, then C i
= A
1 i
.
(b) Show that the moving average coefficients for a VAR(2) can be found recursively by C
0

I ; C
1

A
1
; and C i

A
1
C i

1

A
2
C i

2 for i

2 ,...
Solution
A stationary VAR has a moving average representation given by
Plugging this formula into that of a VAR(2) such as, X t

A
1
X t

1
X t

A
2

X
  i

0

C i
 t
 i
.
, we t

2
 t find,
  i

0
C i
 t
 i

A
1
  i

0
C i
 t
 i

1

A
2
  i

0
C i
 t
 i

2
  t
, which can be rewritten as,
0

( C
0

I )
 t

( C
1

A
1
C
0
)
 t

1

( C
2

A
1
C
1

A
2
C
0
)
 t

2

...
, which delivers the coefficients for each of the epsilon. Since the epsilon can take on any value in
 p
, each of these coefficients must equal zero. Hence, C
0
= I , C
1
= A
1
, and
C t

A
1
C t

1

A
2
C t

2 t

2 , 3 ,...
1

ARE/ECN 240C Time Series Analysis
Winter 2004
Problem 2: Consider the following bivariate VAR, y t m t


 yy
 my y t

1 y t

1
  ym m t

1
  mm m t

1

 u yt u mt with E ( u t u t
' )





2 y


2 m

 .
Professor Òscar Jordà
Economics, U.C. Davis
(a) Find a matrix H , which is lower triangular and ensures that if Hu t
E (
 t
 t
' )

D where D is a diagonal matrix.
  t
, then
Solution
For example, H


1


2 y
0
1
(b) Given this matrix H calculate the structural representation of this VAR.
Solution y m t t




2 y y t

(
 my



 yy
2 y y t
 yy

1
)
  y t

1 ym
 m t

1
(
 mm
 
 yt


2 y
 ym
) m t

1
  mt
(c) Calculate the VMA representation for the reduced form of this VAR (notice that it is very simple in this case – don’t apply the usual formulas mechanically!)
Solution

 y m t t




 u yt u mt





 yy
 my
 ym
 mm



 u yt

1 u mt

1



...



 yy
 my
 ym
 mm

 k 
 u u yt
 k mt
 k



...
2

ARE/ECN 240C Time Series Analysis
Winter 2004
Professor Òscar Jordà
Economics, U.C. Davis
(d) Calculate the VMA representation of the structural form of the VAR.
Solution

 y m t t





1

 y
2

1


2 y
0
1
0
1


 yy
 my

 u yt u mt



 ym
 mm

1


2 y
0
1

 k 
 u yt
 k u mt
 k



...



 yy my
 ym
 mm



 u yt

1 u mt

1



...
(e) Under what conditions will the reduced form and the structural form produce identical impulse response functions?
Solution: The obvious one is

= 0 . Less obvious,
 yy
  ym

0 .
(f) Suppose you obtained the structural form as in part (a) but for a system that had the variable m ordered first. Under what conditions would these two structural identification schemes deliver the same impulse responses?
Solution: Notice that the matrix H is in this case,
H





1


2 m
Naturally,

= 0, would work, but also, either
 yy
0
1



  ym

0 or
 my
  mm

0 .
Problem 3: Consider the following bivariate VAR y
1 t y
2 t


0 .
3 y
1 t

1
0 .
9 y
1 t

1


0 .
8 y
2 t

1
0 .
4 y
2 t

1




1 t
2 t with and E
E (

(

1 t
1 t


1

2

)
)


0
1 for t =

and 0 other wise, E (

2 t

2

)

2 for t =
for all t , and

. Answer the following questions:

and 0 other wise,
3

ARE/ECN 240C Time Series Analysis
Winter 2004
(a) Is this system covariance-stationary?
Solution
Professor Òscar Jordà
Economics, U.C. Davis
To answer this question, verify the roots of the polynomial


1
0
0
1





0 .
3
0 .
9
0 .
8
0 .
4

 z

( 1

0 .
3 z )( 1

0 .
4 z )

( 0 .
8 z )( 0 .
9 z )

1

0 .
7 z

0 .
6 z
2
The roots are 0.833 and 2, hence the system is not-stationary.
(b) Calculate
 s


 y
 t
 t s
'
for s = 0, 1, and 2 . What is the limit as s
 
?
Solution

0



1
0
0
1

 ;

1



0 .
3
0 .
9
0 .
8
0 .
4

 ;

2



0 .
81
0 .
63
0 .
56
0 .
88


Clearly, since
 s



0 .
3
0 .
9
0 .
8
0 .
4

 s
and the process is not stationary, then
 s
 
.
(c) Calculate the fraction of the MSE of the two period-ahead forecast error for variable 1, E
 y
1 , t

2
 ˆ
( y
1 , t

2
| y t
, y t

1
,...)

2
, that is due to

1 , t

1 and

1 , t

2
Solution
E
 y
1 , t

2
 ˆ
( y
1 , t

2
| y t
, y t

1
,...)
2


E


1 , t

2

0 .
3

1 , t

1

0 .
8

2 , t

1

2 
1

0 .
3
2 
0 .
8
2 
2

2 .
37
The fraction due to

1
is (1 + 0.3
2
)/2.37 = 0.46 or 46%.
Problem 4: Consider the process y t z t


 z t
 z t

1


 y t

1

 
1 t
|
2 t
|


|

|

1
1
 t
~ N




0
0

 ;




11
12


12




22
4

ARE/ECN 240C Time Series Analysis
Winter 2004
Professor Òscar Jordà
Economics, U.C. Davis
(a) Derive E ( x t function and
| x t

1 x t
'

); V (
( y t z t x t
).
| x t

1
) and D ( x t
| x t

1
) where D denotes the density
Hint : the system can be rewritten in matrix form as


1
0

1
 


 y z t t






0
0




 y t

1 z t

1







1 t
2 t


Solution
Pre-multiply the system by A =


1
0

1

 , the inverse of the contemporaneous correlation matrix, to obtain,

 y t z t






0





 y t

1 z t

1




 u
1 t u
2 t

 ;

 u
1 t u
2 t





1
0

1






1 t
2 t


Thus,
E ( x t
| x t

1
)
E ( u t u t
' )


E (



0
A
 t
 t






' A ' ) y t

1 z t

1





1
0

1






11
12

12

22




1

0
1


  x t

1
;




11


2

12

12



22
2 
22
Given this expressions for the conditional mean and the variance, and noting that the u
’s are linear combinations of the 
and therefore are normally distributed, the conditional distribution D(x t
| x t-1
) is multivariate normal with conditional mean and variance given by the expressions derived above.
(b) Assume that x t
is stationary. Derive E ( x t
); V ( x t
) and show that
V ( x t
)

V ( x t
| x t

1
) is positive definite. What are the implications of this result?
Solution
By stationarity,
V
V
( x t
)
( x t
)
E ( x t
)
 
E ( x t

1
)

0 and therefore, E ( x t
)

0 . Similarly,
 
V

V
( x t

1
( x t
|
)
 x t

1
)
'

W

. Since in (a) we calculated that V ( x t

V ( x t
)

'
| x t

1
)

W , it follows that
which is a quadratic form and therefore positive definite for V ( x t
)

0
. The rational is that the conditional variance uses “more information”
(hence the conditioning) than the unconditional variance.

12
 
22

22



W .
5

ARE/ECN 240C Time Series Analysis
Winter 2004
Problem 5: Consider the Gaussian linear regression model, y t
 x t
'
  u t
Professor Òscar Jordà
Economics, U.C. Davis with u t
~ i.i.d. N (0,

2
) and u t
independent of x  for all t and

. Define
  log of the likelihood of ( y
1
, …, y
T
) conditional on ( x
1
,…, x
T
) is given by
(

' ,

2
)'.
The
L (

)
 
( T / 2 ) log( 2

)

( T / 2 ) log(

2
)
 t
T 

1
( y t
 x t
'

)
2
/( 2

2
)
(a) Show that the estimate
ˆ
T
'

 g (

 
; Y
T
'
)
   ˆ
T

( 1 / T is given by u
ˆ t

( y t

ˆ
T
'
 x t
'
 ˆ
T
) t
T 

1
 h (

 
, Y
T
'
)

( 1 / T ) t
T 

1
 2 log f (
  y
 t

'
| Y t

1
;

)






1
T
) and
   ˆ
T t
T 

1
 ˆ
T x t x t
'
0
/
 ˆ
T
2
1
T t
T 

1


0
1
2
 ˆ
T
4
 u
ˆ t
2
 ˆ 6
T


 





’ where
and
 ˆ 2
T
denote the maximum likelihood estimates.
   ˆ
T
Solution
The proof is straightforward by direct differentiation of the likelihood and noticing that t
T 

1 u
ˆ t

0 from the first order conditions.
(b) Show that the estimate S
ˆ
T
S
ˆ
T







1
T
1
T
T  t

1 u
ˆ t
2 x t x t t
T 

1

 u
ˆ t
3
2
 ˆ x
6
T t
' /
 ˆ
T
4


1
T

( 1 / T ) t
T 

1 h

(
 ˆ
, Y t
)
 h (
 ˆ
, Y t
) '

is given by
1
T t
T 

1 t
T 

1

 u
ˆ t
2
2
 ˆ
T
4
ˆ t
3
2
 ˆ x
6
T t

1
2
 ˆ
T
2


2






.
Solution: this proof is also straightforward once you realize h (
 ˆ
, Y t
) is the score of the log-likelihood evaluated at the MLE estimates.
6

ARE/ECN 240C Time Series Analysis
Winter 2004
Professor Òscar Jordà
Economics, U.C. Davis
(c) Show that the p lim( S
ˆ
T
)
  p lim(
T
)
 
where
 


Q /

2
0
0
1 /( 2
 4
)

 for
Q

Solution: p lim( 1 / T )

T t

1 x t x t
'.
The proof requires the following intermediate results: p lim
T
1  T t

1 u
ˆ t
2   2
; p lim
1  t
T

1 u
ˆ t
T desired result.

0 . Direct application of conventional asymptotic results delivers the
(d) Consider a set of m linear restrictions on

of the form R

= r for R a known
( m
 k ) matrix and r a known ( m

1 ) vector. Show that for
ˆ
T
  ˆ
T
, the Wald test statistic given by T g (
 ˆ
T


 g (

 
'
)
   ˆ
T





T

1


 g (

)
 
'
   ˆ
T


' 


1 g (
 ˆ
T
is identical to the Wald test form of the OLS
( R
  ˆ
T r )'
 s
T
2
R ( x
T
' x
T
)

1
R '


1
( R
  ˆ
T r )

2
test given by
with the OLS estimate of the variance s
T
2 replaced with the MLE
 ˆ 2
T
.
(e) Show that when the lower left and upper right blocks of S
ˆ
T
are set to their plim of zero, then the quasi-maximum likelihood Wald test of R

= r is identical to the heteroskedasticity-consistent form of the OLS
( R
  ˆ
T r )'

R (

1
T T

1
T
/ T ) R '


1
( R
  ˆ
T r )

2
test given by
Problem 6: Consider the following DGP for the cointegrated random variables z and y
( 1

L )

 y t z t



( 1

0 .
4 L )

1


1

0 .
0 .
8
1 L
L 0 .
8 L
1

0 .
6 L






1 t
2 t

 where

~ N(0, I) with z
0
= y
0
= 0.
(a) Obtain the autoregressive representation of this DGP.
(b) Obtain the error-correction representation of this DGP.
(c) Deduce the long-run relation between z and y.
(a) Directly inverting the lagged matrices on the right-hand side, we get
7

ARE/ECN 240C Time Series Analysis
Winter 2004


1


0 .
6
0 .
1 L
L
1


0 .
8 L
0 .
8 L



 y z t t







1 t
2 t


Professor Òscar Jordà
Economics, U.C. Davis
(b) From the autoregressive representation


1


0
0 .
.
6
1 L
L
1


0
0
.
.
8 L
8 L





1


L

0
0 .
1 L
.
4 L


1

0
L
Hence

1
0

L



( 0 .
4




0 .
4 L
0 .
1 L

0 .
1 )' ;


0 .
0 .
8
2 L
L



( 1

 
I
2 )'
1


L
0 .

8 L
0 .
2 L






0 .
4
0 .
1



1


2

L
(c) From the ECM, the long run solution is y = 2z .
Problem 7: Consider the following DGP x t x t
  y t
  y t

 u
1 t
; where u
1 t u
2 t
; where u
2 t


 u
1 t
 u
2

1 t

1



1 t

2 t with |

| < 1, and




1 t
2 t

 ~ D




0
0

 ,




1
2


2
2



 where D denotes a generic distribution.
(a) Derive the degree of integratedness of the two series, x t depend on any restrictions on the values of

,

, and

and y t
. Do your results
? Discuss how.
If θ = 1, then x t and y t are I(1) since x t
  y
 u
1 t
~ I ( 1 ) . In addition, given |

| < 1 one needs to impose the condition
 t
 
. The same linear combination of x t and y t
 cannot be simulatenously I(0) and I(1).
(b) Under what coefficient restrictions are x t
and y t cointegrated? What are the cointegrating vectors in such cases?

1 ;
 
1 . Cointegrating vector (1

).
(c) Choose a particular set of coefficients that ensures x t
and y t
are cointegrated and derive the following representations:
I.
The moving-average.
8

ARE/ECN 240C Time Series Analysis
Winter 2004
II.
The autoregressive.
III.
The error-correction.
I. MA


1
1





 x y t t



1



1
1


1
L

L

1 t

2 t



;

 x t y t





1
1
Note :


1
1





1


1
 



1
  

1 
Hence :

 x t
 y t
 




 i

0 i

0


1 t
1 t
 i
 i

 
 
 i

0 i



0 i


2 t i

 i
2 t
 i





1


 1
1

1

1
L

L

1 t

2 t



Professor Òscar Jordà
Economics, U.C. Davis
Aside :
( 1

) is the cointegrat ing vector, while (1

) is a serial correlatio n feature
II. AR

 x t x t x t y t




 y t
 y t




1


1 t
1


2
1
 t
L
L





1



1
1




1




1
1



 x t y t




(



1
1


1 )





(



1 )



 x t

1 y t

1

 x t

1 y t

1




1
 




1





1 t
2 t




1






1 t
2 t


III. ECM
Define the error correction term z t

1
 x t

1
  y t

1
, then
 x t
 y t



(
 
1 )

( 1

 

)
   z t

1 z t

1

 u
1 t u
2 t with u
1 t


1
 
(
 
1 t
 
2 t
) u
2 t


1
 
(

1 t
 
2 t
)
9

ARE/ECN 240C Time Series Analysis
Winter 2004
Professor Òscar Jordà
Economics, U.C. Davis
(d) Can all the cointegrated systems be represented as an error-correction model?
What are the problem/s of analyzing a VAR in the differences when the system is cointegrated?
From the Granger representation theorem we know the answer is yes. Analyzing the
VAR in the differences omits the error correction term in the specification. Therefore we have the classical problems of omitted variable bias.
(e) Suppose that economic theory suggests that x t cointegrating vector [1

+ 0.5
t
]. Describe:
and y t
should be cointegrated with
I.
How would you test whether this is indeed a cointegrating vector?
Run the OLS regression x t
 
(
 
0 .
5 t
) y t
 v t and test the residuals with an ADF test.
II.
What is the likely outcome of the test in short samples? Why?
In short-samples, the cointegrating vector (1

+0.5
t
) will differ from the cointegrating vector (1

). However, as the sample size gets larger, note that the bias 0.5
t disappears very quickly.
III.
What is the likely outcome of the test asymptotically? Why?
Asymptotically the bias disappears sufficiently quickly.
Problem 8: Consider the bivariate VECM
 y t
 c
 
' y t

1
  t
 it iid
~ ( 0 ,
 2
) where
 
(

1
, 0 )' and
 
( 1 ,
 
2
)'.
Equation by equation, the system is given by
 y
1 t
 y
2 t
 c
1
 c
2



1
(

2 t y
1 t

1
 
2 y
2 t

1
)
 
1 t
Answer the following questions:
(a) From the VECM representation above, derive the VECM representation
 y t
 c
  y t

1
  t and the VAR(1) representation
10

ARE/ECN 240C Time Series Analysis
Winter 2004 y t
 



1
0
 c

Ay t

1
  t
 
1

2
0

 ; A




1
0

1
Professor Òscar Jordà
Economics, U.C. Davis
 
1

2
1


(b) Based on the given values of the elements in

and

, determine


,


, such that

'



0 and

'



0 .





0 k


;






2 k k


; k

0
(c) Using the Granger representation theorem determine that

( 1 )
 

(


' I
2


' )

1


, where

( L ) is the moving average polynomial corresponding to the VECM system above and I
2
is the identity matrix of order 2.
Hint: you may show this result by showing that

( 1 ) is orthogonal to the cointegrating space.
Using the hint:
 (1)’
= 0. It is easy to show that


(


' I
2


' )

1





0
0

2
1

 and therefore



1
0
 
1

2
0




0
0

2
1



0
(d) Using the Beveridge-Nelson decomposition and the result in (c), determine the common trend in the VECM system.
All you need to remember is that from the B-N decomposition, the trends are the linear combinations captured in

(1)y t
, which in this case turns out to be

2 y
1t
+ y
2t
.
Notice that this combination is orthogonal to the cointegrating vector.
(e) Show that

' y t
follows an AR(1) process and show that this AR(1) is stable provided that

2
 
1

0 . What can you say about the system when

1
= 0?
Let z t have
 y
1 t
 
2 y
2 t
be the cointegrating vector. From the equations for y
1 and y
2 we z t
 c
1

(

1

1 ) y
1 t

1
 
1

2 y
2 t

1
 
1 t
 
2 c
2
 
2 y
2 t

1
 
2

2 t
11

ARE/ECN 240C Time Series Analysis
Winter 2004
Combining terms z t

( c
1
 
2 c
2
)

(

1

1 ) z t

1
 v t
; v t
 
1 t
Professor Òscar Jordà
Economics, U.C. Davis
 
2

2 t which is an AR(1) whose stationarity requires that |

1
+ 1| < 1 or the equivalent condition -2 <

1
< 0. When

1 cointegration for any value of

= 0, z t is no longer stationary, so there is no
2
. y
1 and y
2
are in this case two independent random walks.
12

ARE/ECN 240C Time Series Analysis
Winter 2004
Problem 9: Consider the following VAR y t x t


( 1

 y t

1


) y t

1
( 1

 
 x t

1
) x t

1

 
1 t

2 t
(a) Show that this VAR is not-stationary.
Stationarity requires that the values of z satisfying


1
0
0
1





1



( 1




)

 z

0 lie outside the unit circle. For z = 1, notice
 
 


     
0
Professor Òscar Jordà
Economics, U.C. Davis
(b) Find the cointegrating vector and derive the VECM representation.
Notice that

( 1 )












(
   
)' ( 1
 
) so that
 y t
 x t

 

(
( y t

1 y t

1


 x t

1
 x t

1
)
)




2 t
1 t
(c) Transform the model so that it involves the error correction term (call it z ) and a difference stationary variable (call it
 w t
). w will be a linear combination of x and y but should not contain z . Hint: the weights in this linear combination will be related the coefficients of the error correction terms.
Given the ECM in part (b), notice
   y t
 w t


 
  x t
 y t

   z t

1
  x t

 
 
1 t
1 t

  z t

1

2 t
 
2 t
Next
13

ARE/ECN 240C Time Series Analysis
Winter 2004 y t x t

 y t

1 x t

1

 

(
( y t

1 y t

1



 x t

1 x t

1
)
)




1 t
2 t
Professor Òscar Jordà
Economics, U.C. Davis
( y t z t

  x t
( 1

)



( y t

1


) z t

1
 x t

1
 
1 t
)


 z
 t

1
  z t

1
 
1 t
 
2 t

 z t
 w t





1
 
0
 
0
0



 z t

1
 w t

1






1

 







1
2 t t


(d) Verify that y and x can be expressed as a linear combination of w and z . Give an interpretation as a decomposition of the vector ( y x
)’ into permanent and transitory components.
From part (c)

 z w t t






1

 




 x t y t

 taking the inverse
 
1






1



 z w t t




 x y t t

 and therefore y t x t





 

  z t
 z t




 

1
 w t w t w t
is I(1) and z t
is I(0), which is a version of the Beveridge-Nelson decomposition proposed by Gonzalo and Granger (1995).
14