1
HG
Jan 2014 ECON 5101
Exercises II
- 10 Feb 2014
(with answers)
Exercise 1.
Read section 8 in lecture notes 3 (LN3) on the common factor problem in ARMA-processes.
Consider the following process
(1)
Yt  0.4Yt 1  0.45Yt 2   t   t 1  0.25 t 2
where
 t ~ WN (0, 2 ) .
This is an ARMA process of the form
A.
 ( L)Yt   ( L) t
without constant term.
Investigate if the two lag-polynomials have a common factor. If so, reformulate the
difference equation specification for
B.
Yt  to
a proper ARMA specification.
Is there a causal stationary solution for
Yt  ?
If yes, write up the solution as a
MA(  ) process. Is the MA specification invertible? If so, write up the
t
AR() solution for
.
ANSWER:
A:
The companion polynomials:
Roots
 ( z )  1  0.4 z  0.45 z 2
-2
10 9  1.1111...
 ( z )  1  z  0.25z 2
-2
-2
Factorized:
 ( z )  (1  0.5 z )(1  0.9 z)
 (z)  (1  0.5z)2
2
To have a proper ARMA we should cancel the common factor, giving new lag polynomials
 ( L)  1  0.9 L
 ( L)  1  0.5L
The proper ARMA(1,1) specification
(1.1)
Yt  0.9Yt 1   t  0.5 t 1
B:
The solution of (1.1) is causal stationary since
  0.9  1 .
1  0.5L
t
1  0.9 L
We need to find the  ' s in
Yt 
1  0.5 z
 1  1 z  2 z 2 
1  0.9 z
(1.2)
or
1  0.5 z  (1  0.9 z ) 1  1 z  2 z 2 
Multiplying out the right side
(putting

  0.9,   0.5 )
1   z  1  1 z  2 z 2  3 z 3 
  z   1 z 2   2 z 3 

 1  ( 1   ) z  ( 2   1 ) z 2  ( 3   2 ) z 3 
Hence (using the uniqueness of power series coefficients (see appendix to the exercises II))
( 1   )  
 1    
 j   j 1   j   j 1 1 
  j 1 (   )  (0.9) j 1 (1.4) for j  2,3,
3
The process is invertible since
t 
Putting
  0.5  1 . The
1  0.9 L
Yt  (1  1 L   2 L2 
1  0.5L
AR(  ) solution is
)Yt
  0.9,   0.5 , and using the companion forms, we get
1   z  (1   z )(1  1 z   2 z 2 
)
 1  1 z   2 z 2   3 z 3 
  z  1 z 2   2 z 3 

 1  (1   ) z  ( 2  1 ) z 2  ( 3   2 ) z 3 
Hence
1      1  (   )  1.4
 j  ( ) j 11  ( ) j 1 (   )  (0.5) j 1 (1.4) for j  2,3,
Some values:
j
0
1
2
3
4
5
10
20
40
j
1
1.4
1.26
1.134
1.021
0.919
0.542
0.189
0.023
j
1
-1.4
0.70
-0.350
0.175
-0.0875
0.003 2.67E-06 2.55E-12
Exercise 2.
A.
Suppose that Yt ~ ARMA( p, q) and causal, satisfying )  ( L)Yt  0   ( L) t , where
 ( L)  1  1L 
  p Lp ,  ( L)  (1  1L 
 q Lq ) . Introduce the centered series,
yt  Yt   , leading to  ( L) yt   ( L) t without the constant  0 , and where E ( yt )  0 (see
exercise 3 of seminar1 ).
Explain why statement (7) on page 5 of LN3, saying
 (L) (h)  0 and  ( L)  (h)  0 for h  max( p, q 1) , is true, where  (h),  (h)
are the autocovariance function and acf respectively.
[Hint. Assume
h  max( p, q  1) and write
4
 (h)  E ( yt h yt )  E 1 yt h1 yt 
  p yt h p yt   t h yt 
 q t hq yt 
etc. ]
ANSWER:
Continuing the hint we get
(*)  (h)  E ( yt h yt )  1 (h  1) 
  p (h  p)  E ( t h yt ) 
  q E ( t hq yt )
h  p , the “answer” I gave on the seminar was not good. The point is rather:
Suppose, e.g., h  p  1 . Then, remembering that  (1)   (1) , (*) becomes
(i)
If
 ( h)  E ( yt  h yt )  1 ( h  1) 
  p  2 (1)   p 1 (0)   p  ( 1)  E ( t  h yt ) 
 1 ( h  1) 
  p 2 (1)   p 1 (0)   p  ( 1)  E ( t  h yt ) 
 1 ( h  1) 
 ( p 2   p ) (1)   p 1 (0)  E ( t  h yt ) 
and we see that the difference equation part for
longer the same). . Therefore, we must have
 ( h)


has changed (the coefficients are no
h  p , for the difference equation for  (h) to
hold.
(ii)
Since the (causal) solution for
yt depends on  t ,  t 1 ,  t 2 ,
, we must have
h  q  1 (or h  q  1) in order that all the last  -terms become zero.
Hence
h  max( p, q  1) .
B.
Introduction. If
Yt ~ AR(1)
with
yt   yt 1   t
and
  1, we get from page
h
3 in LN2 that the autocovariance function,  1 (h)   1 (h)  
12
h
that the acf is  (h)   (h)   for h  0,1, 2,
2
which implies
We are interested in this exercise to find out what the effect is on the AR(1) acf,
 ( h)
by
yt   yt 1   t .
yt   yt 1   t   t 1 , i.e., an ARMA(1,1), where
adding a MA-term,  t 1 to
Assume therefore that
  1,
  1, and  t ~ WN (0, 2 ) . We need the
 ( h)
yt
for
in this new situation. First
 ( h) :
autocovariance,
 ( h)
and acf
Using (8) on page 5 in LN3 and A., we have
5
(i)
 (h)   (h  1)  0 for h  2,3,
with initial conditions
(ii)
 (0)   (1)   2 ( 0  1 )
, i.e.,
 (0)   (1)   2 ( 0  1 )
 (1)   (0)   2 ( 0 )
1 z
where  0 , 1 are found from
  0  1 z   2 z 2 
1 z
(iii)
Question.
Show that
 0  1 and 1    
[Hint. Write the lag-equation (in terms of companion series)
1   z  (1   z )( 0  1 z   2 z 2 
sufficiently to determine  0 , 1 . ]
) . Then multiply out the right side
ANSWER: This was done in exercise 1B
C.
Show first from ii. and iii. that
1  2   2
 (0)  
12
(   )(1   )
 (1)   2
12
2 (   )(1   ) h 1
and then that  (h)  
 for h  1 .
12
2
 0  1, 1     ,  (h)   (h  1) for h  2,3,
 (h)   h1 (1) for h  2,3,
ANSWER:
From ii., iii. we get
gives
6
(*)  (0)   2 (1   (   ))   (1)
 (1)   2   (0)   2   (1   (   )) 2   2 (1)

(1   2 ) (1)   2      (   )    2 (   )(1   )

 (1)   2
(   )(1   )
12
Substituting in (*) gives

 (0)   2 1   (   )  
(   )(1   ) 

12


2
2
2 1     (   )(1   )   (   )(1   )


12

2
1   2  (   )  (1   2 )   (1   ) 
12

2
2
2
1   2  (   ) 2
2 1      2  


12
12
2
Hence
1  2   2
(   )(1   )
, and
 (0)  
,  (1)   2
2
1
12
2
(**)
 ( h)   2
D.
(   )(1   ) h1

for h  2,3,
12
Show that the acf of
where
1 (h)
yt
can be expressed as
 ( h) 
is the acf of the AR(1) process.
(   )(1   )
 ( h) ,
 (1  2   2 ) 1
7
ANSWER:
From (**) we get
 ( h) 
Considering,
E.
(   )(1   ) h1
 for h  1, 2,3,
(1  2   2 )
1 (h)   h , we get  (h) 
The constant factor in front of
(   )(1   )
 ( h) .
 (1  2   2 ) 1
1 (h) , characterize the effect 
has on the AR(1) acf,
1 (h) . Look at this effect (factor) in the two special cases,
  0.9 and   0.2 , i.e., for which values of  will the AR(1) acf increase
and for which values will it decrease?
1 (h) is not so easy to discuss analytically. The
best way, in my view, is simply to plot the factor as a function of  for
1    1 in the two cases. The easiest is probably to plot it with a computer.
[Hint. The factor in front of
This is, for example, easy in Stata. The following command, for example,
twoway function y=2*x^2-1, range(-1 2.5)
plots the function
ANSWER:
y  2 x 2  1 for 1  x  2.5 ]
The factor in front of
1 (h)
represents the effect the MA-term has on the acf
of AR(1). Writing x for  , the factor is the function
g ( x) 
Case
  0.9 :
( x   )(1   x)
 (1  2 x  x 2 )
Stata-plot :
twoway function y=(x+.9)*(1+.9*x)/(.9*(1+2*.9*x+x^2)), range(-1 1)
.6
.8
1
8
0
.2
.4
y
Phi = 0.9
-1
-.5
0
x
.5
1
gmax  g (1)  (1   ) 2  1.056, gmin  g (1)   (1   ) 2  0.056
(We see that if  increases from 0 (e.g.),  (h) will increase also from 1 (h) , but not
 0.2 :
3
much.)
Case 
-2
-1
0
y
1
2
Phi = -0.2
-1
-.5
0
x
.5
1
9
F.
For any 
with   1 , is there any value of 
that turns
 yt  into white noise? If
so, for which value?
ANSWER:
From the figures we see that there is a  that makes
From the formula we see that this happens for 
yt 
 (h)  0 for h  0 .
  . We also see from the solution series
1 L
t
1L
that there is a common factor when 
  , so that  yt  is white noise for this value of  .
(For the appendix on power series see the original exercise text)