ARE/ECN 240C Time Series Analysis
Winter 2006
Professor Òscar Jordà
Economics, U.C. Davis
PROBLEM SET 5 – DUE: FEBRUARY 21
Instructions
This problem set is divided into two parts: (1) Analytical Questions, and (2) Applied
Questions.
Please try to answer the questions rigorously by stating any implied assumptions and
ensuring all the steps to your conclusion have been properly verified.
Part I – Analytical Questions
Problem 1: Consider the following DGP
with || < 1 and
where D denotes a generic distribution.
(a) Derive the degree of integratedness of the two series, xt and yt, explicitly stating the
parameter restrictions required in each case.
(b) Under what coefficient restrictions are x and y cointegrated? What are the
cointegrating vectors in such cases?
(c) Choose a particular set of coefficients that ensures x and y are cointegrated and derive
the following representations:
(i) the moving-average: that is (xt yt)’ on the left hand side, (1t 2t)’ and its lags on
the right hand side.
(ii) The autoregressive in the levels: that is, (xt yt)’ on the right hand side, and
lags of (xt yt)’ and (1t 2t)’ on the right hand side.
(iii) The Error-Correction Representation: that is, (xt yt)’ as a function of zt-1
and residuals (no need to be specific about the residuals).
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ARE/ECN 240C Time Series Analysis
Winter 2006
Professor Òscar Jordà
Economics, U.C. Davis
(d) [5 points] In general, discuss the pros and cons of obtaining impulse responses from a
VAR estimated in the levels, as in part (c)-(i) and one in vector error correction form, as
in part (c)-(iii).
Solution:
(a) If   1 then xt  yt ~ I (1) but xt  yt ~ I (0) . Hence    and:


If   0 , xt ~ I (1) and yt ~ I (1) but (1 ) is a cointegrating vector.
If  = 0, then xt ~ I (0) and yt ~ I (1)
If |  | 1 then both x and y are stationary.
(b)   1,   0,   0 ; (1 )
(c) (i)
1
  1t 
0
 xt  1    (1  L) 1
  =
   
 
1 
0
(1  L)   2 t 
 y t  1   
1
xt 
    1t     t  2 t


yt 
1




   1t    t  2 t 




(ii)
     1 0  xt 1 
1       1t 
 




 
 1  1 0   yt 1      1  1  2t 
1
 xt 1  yt 1  1t  2t 
xt 

1
x  yt 1   1t   2t 
yt 
   t 1
 xt 
1
  
 yt    
(iii)
 (1   )
z 
   t 1 1t
(1   )
y t  
z 
   t 1 2t
xt 
(d) the VAR in levels is guaranteed to be consistent but not as efficient as when we
impose the cointegrating restrictions. However, imposing incorrect cointegrating
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ARE/ECN 240C Time Series Analysis
Winter 2006
Professor Òscar Jordà
Economics, U.C. Davis
restrictions can lead to misspecification. Additionally, it is even more complicated to
obtain standard errors for impulse responses from cointegrated VARs.
Problem 2: Consider the following D.G.P.
xt  y t  vt ,
vt (1  1L)  1t
2 xt  yt  ut , ut (1   2 L)   2t
and
 0    2 0  
  1t 

  ~ N  ,  1
2 
 0   0  2 
  2t 
(a) Determine whether this system is stationary, non-stationary, or non-stationary but
cointegrated according to the following scenarios:
(i) | 1 | 1, | 2 | 1
(ii) 1  1, | 2 | 1
(iii) 1  1,  2  1
Solution:
(i)
(ii)
(iii)
Stationary
Cointegrated
Nonstationary
(b) Obtain the reduced-form, autoregressive representation of the system in the levels
when it is cointegrated.
Solution:
(1  L) xt  (1  L) yt  1t
2(1   2 L) xt  (1   2 L) yt   2t
rearranging
 1 1 xt   1

   
 2 1 yt   2 2
1  xt 1   1t 
 

 2  yt 1    2t 
Finally, the reduced form is obtained by inverting the matrix of contemporaneous
correlations
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ARE/ECN 240C Time Series Analysis
Winter 2006
Professor Òscar Jordà
Economics, U.C. Davis
 xt   2 2  1  2  1  xt 1    1t   2t 
   
  


 yt   2(1   2 ) 2   2  yt 1   21t   2t 
(c) Given the representation that you just found, calculate the reduced-form, impulse
response function coefficient matrices, s for periods s = 0, 1, and 2 for 2 = 0.5. What is
s as s   ? Given the reduced form impulse response matrices, calculate the
structural impulse responses for periods 0, 1, 2. What is happening as s   ? Explain
this result.
Solution:
Reduced Form
 1 0
 0  

 0 1
 0  0.5 
 1  

 1 1.5 
  0.5  0.75 
 2  

1.75 
 1.5

Structural
 1 0  1 1  1 1
 0  

  

 0 1  2 1  2 1
 1 1 0  0.5  1 1 
 1  

  

 2 1 1 1.5  1 0.5 
1 
 1 1  0.5  0.75   1
 2  

  

1.75   0.5 0.25 
 2 1 1.5

 s  , s  
 s  , s   but only for xt
(d) Find the moving-average representation of the system and the cointegrating vector
when it is cointegrated.
Solution
 2t
(1  L) (1   2 L)
21t
 2t
yt 

(1  L) (1   2 L)
xt  
1t

hence the obvious cointegrating vector is (2 1)’
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ARE/ECN 240C Time Series Analysis
Winter 2006
Professor Òscar Jordà
Economics, U.C. Davis
(e) Describe how you could estimate the cointegrating vector in a regression of yt on xt
which is well behaved in small samples.
Solution:
In large samples the regression of of yt on xt will deliver an asymptotically consistent
estimate of the coefficient  in the regression
yt  xt  ut
 2t
. Typically this is of unknown
(1   2 L)
form so a recommended strategy to correct for small sample bias is to use the Saikkonen,
Phillips and Loretan, or Stock and Watson approach of including lags and leads of xt.
Here, because the source of the correlation in the residuals is known, we have that an
AR(1) correction would solve the problem since,
however, notice that the error term ut is ut  
(1   2 L) yt  2 xt (1   2 L)   2t
Problem 3: Consider the following bivariate VAR
y1t  0.3 y1t 1  0.8 y 2t 1   1t
y 2t  0.9 y1t 1  0.4 y 2t 1   2t
E ( 1t 1 )  1
E ( 2t 2 )  2
with
for t =  and 0 other wise,
for t =  and 0 other wise, and
E ( 1t 2 )  0
for all t, and . Answer the following questions:
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ARE/ECN 240C Time Series Analysis
Winter 2006
Professor Òscar Jordà
Economics, U.C. Davis
(a) Is this system covariance-stationary?
Solution
To answer this question, verify the roots of the polynomial
 1 0   0.3 0.8 

  
 z  (1  0.3z )(1  0.4 z )  (0.8 z )( 0.9 z ) 
 0 1   0.9 0.4 
1  0.7 z  0.6 z 2
The roots are 0.833 and 2, hence the system is not-stationary.
s 
(b) Calculate
Solution
y t  s
 t ' for s = 0, 1, and 2. What is the limit as s   ?
 1 0
 0.3 0.8 
 0.81 0.56 
 0  
; 1  
; 2  

 0 1
 0.9 0.4 
 0.63 0.88 
s
 0.3 0.8 
 s  

0
.
9
0
.
4

 and the process is not stationary, then  s   .
Clearly, since
(c) Calculate the fraction of the MSE of the two period-ahead forecast error for
2
 and  1,t 2
E y1,t 2  Eˆ ( y1,t 2 | yt , yt 1 ,...)
variable 1,
, that is due to 1,t 1


Solution


2
2
E y1,t 2  Eˆ ( y1,t 2 | yt , yt 1 ,...)  E1,t 2  0.31,t 1  0.8 2,t 1   1  0.32  0.82  2  2.37
The fraction due to 1 is (1 + 0.32)/2.37 = 0.46 or 46%.
Problem 4: Consider the following VAR
y t  (1   ) y t 1  xt 1   1t
xt  y t 1  (1   ) xt 1   2 t
(a) Show that this VAR is not-stationary.
Stationarity requires that the values of z satisfying
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ARE/ECN 240C Time Series Analysis
Winter 2006
Professor Òscar Jordà
Economics, U.C. Davis
 1 0  1  

  
 0 1  
  
z  0
(1   ) 
lie outside the unit circle. For z = 1, notice



      0

(b) Find the cointegrating vector and derive the VECM representation.
Notice that
 
 (1)  
 
 
  ( 
 
  )' (1   )
so that
yt   ( yt 1  xt 1 )   1t
xt   ( yt 1  xt 1 )   2 t
(c) Transform the model so that it involves the error correction term (call it z) and a
difference stationary variable (call it wt). 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
 yt  xt  zt 1  1t  zt 1  2t
wt  yt  xt  1t  2t
Next
7
ARE/ECN 240C Time Series Analysis
Winter 2006
Professor Òscar Jordà
Economics, U.C. Davis
y t  y t 1   ( y t 1  xt 1 )   1t
xt  xt 1   ( y t 1  xt 1 )   2 t
( y t  xt )  ( y t 1  xt 1 )  zt 1  zt 1   1t   2 t
zt  (1     ) zt 1   1t  
 zt  1    

  
0
 wt  
0  zt 1   1


0  wt 1    
    1t 
 
   2 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)
 zt   1
   
 wt    
   xt 
 
  yt 
taking the inverse
1
  
    zt   xt 

    
  1  wt   yt 
and therefore

z 
   t 

xt 
z 
   t 
yt 

wt
 
1
wt
 
wt is I(1) and zt is I(0), which is a version of the Beveridge-Nelson decomposition
proposed by Gonzalo and Granger (1995).
Problem 5: Consider the bivariate VECM
y t  c   ' y t 1   t
iid
 it ~ (0,  2 )
where   (1 ,0)' and   (1,  2 )'. Equation by equation, the system is given by
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ARE/ECN 240C Time Series Analysis
Winter 2006
Professor Òscar Jordà
Economics, U.C. Davis
y1t  c1  1 ( y1t 1   2 y 2t 1 )   1t
y 2t  c2   2t
Answer the following questions:
(a) From the VECM representation above, derive the VECM representation
yt  c  yt 1   t
and the VAR(1) representation
yt  c  Ayt 1   t

   1
0
 1  2 
   1  1  2 
; A   1

0 
1 
 0
(b) Based on the given values of the elements in  and , determine   ,   , such that
 '    0 and  '    0.
 0
 k
    ;     2 ; k  0
k 
 k 
(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 I2 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
 0 2 

1 
  (  ' I 2   ' ) 1   
0
and therefore
1  1  2  0  2 


0
0  0 1 
0
(d) Using the Beveridge-Nelson decomposition and the result in (c), determine the
common trend in the VECM system.
9
ARE/ECN 240C Time Series Analysis
Winter 2006
Professor Òscar Jordà
Economics, U.C. Davis
All you need to remember is that from the B-N decomposition, the trends are the
linear combinations captured in (1)yt, which in this case turns out to be 2y1t + y2t.
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 zt  y1t   2 y 2t be the cointegrating vector. From the equations for y1 and y2 we
have
zt  c1  (1  1) y1t 1  1  2 y2t 1   1t   2 c2   2 y2t 1   2 2t
Combining terms
zt  (c1   2 c2 )  (1  1) zt 1  vt ; vt   1t   2 2t
which is an AR(1) whose stationarity requires that |1 + 1| < 1 or the equivalent condition
-2 < 1 < 0. When 1 = 0, zt is no longer stationary, so there is no cointegration for any
value of 2. y1 and y2 are in this case two independent random walks.
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