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Some exercises on ARMA (not compulsory)2015 0311173925
Time Series Econometrics (Università Commerciale Luigi Bocconi)
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Some Exercises on ARMA Processes
February 2015
Abstract
These exercises do not represent outright study material but simply an attempt to encourage a
deeper understanding of the subject.
1. Exercise 1
Consider the following  (2) process:
 =  + 24−1 + 08−2 
 ∼   (0  2 )
(1)
1a. Re-write this process using the lag operator.
1b. Is the MA process invertible? Explain why it is (or it is not) invertible.
1c. Derive the first two unconditional moments of the process.
Consider now the following (2) process:
 = 11−1 − 018−2 +  
 ∼  (0 2 )
(2)
1d. Re-write this process using the lag operator.
1e. Is it possible to rewrite the process as an  (∞) process? Why? If possible, re-write it as an
 (∞) process.
1f. Is it possible to apply Wold’s decomposition theorem? Why is this (or is this not) possible?
1g. Compute ( | −1 ), ( | −2 ),  ( | −1 ) and  ( | −2 )
2. Exercise 2
In order to illustrate a practical application of an   process, suppose you win 1 if by throwing
a fair coin, it shows head and you lose 1 if it shows tail. Denote the outcome on toss  by   The
sequence { } is a white noise process. However, for each coin toss , your average payoff on the last
four tosses is given by:
 = 025 + 025−1 + 025−2 + 025−3   ∼   (0  2 )
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2a. Can you recognize any of the models shown in class in the process for  ?
2b. Find the expected value of your winnings (i.e., ( )). Find the expected value of  given
that −2 = −3 = 1 (i.e., ( | −2 = −3 = 1)).
2c. Find  ( ) and  ( | −2 = −3 = 1)
2d. Find (  −1 ) (  −2 ), and (  −3 ).
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Solutions
1a. Using the lag operator, (1) can be re-written as:
 = (1 + 24 + 082 ) 
 ∼   (0  2 )
1b. An   process is said to be invertible if all the roots of the characteristic equation lie outside
the unit circle. Starting from the expression derived under 1a,
 = (1 + 24 + 082 ) ,
we can verify whether the roots of the characteristic equation lie outside the unit circle. The characteristic equation is given by:
(1 + 24 + 08 2 ) = 0
whose roots are:
1 = −25 and 2 = −05
Because |2 |  1 we conclude that the process is not invertible.
1c.
( ) = ( + 24−1 + 08−2 )
= ( ) + 24(−1 ) + 08(−2 )
Because
( ) = 0 ∀
it follows that
( ) = ( ) + 24(−1 ) + 08(−2 ) = 0
 ( ) is equal to:
 ( ) =  ( + 24−1 + 08−2 )
then, because  ∼  (which implies that all cross product expectations are zero), we can write:
 ( ) =  ( + 24−1 + 08−2 )
=  ( ) + 242  (−1 ) + 082  (−2 )
Moreover, we know that
 ( ) =  2  ∀
so that
 ( ) =  ( ) + (24)2  (−1 ) + (08)2  (−2 )
=  2 + 576 2 + 064 2 = 740 2 
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The autocovariances of  are given by:
(  −1 ) = ( + 24−1 + 08−2  −1 + 24−2 + 08−3 ) = 432 2 ;
(  −2 ) = ( + 24−1 + 08−2  −2 + 24−3 + 08−4 ) = 082 ;
(  − ) = 0 for   2
1d. Using the lag operator, (2) can be re-written as:
(1 − 11 + 0182 ) =  
 ∼   (0 2 )
1e . Wold’s decomposition theorem states that any stationary process can be expressed as the
sum of two processes, a deterministic part and a stochastic part, which will be a  (∞) process.
A simpler way of stating this in the context of  modeling is that any stationary () process
with no constant and no other terms can be expressed as a  (∞) process. In order to answer the
question it is necessary to determine whether (2) is stationary (i.e., to verify whether all the roots of
the characteristic equation lie outside the unit circle). Starting from the equation derived in 1d,
(1 − 11 + 0182 ) =  
the characteristic equation is given by:
1 − 11 + 018 2 = 0
whose roots are:
1 = 5 and 2 = 111
Because both |1 |  1 and |2 |  1, we can conclude that (2) is stationary and it can be written as a
 (∞) processes.
In order to express (2) as an  (∞) process, first, we re-write:
(1 − 11 + 0182 ) = 
as
Φ() = 
then, the Wold’s decomposition is:
 = Ψ()
where Ψ() = Φ()−1 .
1f. See answer 1e.
1g. ( |−1 ) is given by:
( |−1 ) = (11−1 − 018−2 +  | −1 )
= 11(−1 |−1 ) − 018(−2 |−1 ) + ( |−1 )
= 11−1 − 018−2 
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In order to find ( |−2 ) we need to express  as function of −2  First, we express −1 as function
of −2 :
−1 = 11−2 − 018−3 + −1 
Now we can express  as function of −2 :
 = 11(11−2 − 018−3 + −1 ) − 018−2 + 
(3)
= 103−2 − 020−3 +  + 11−1 
Taking expectations of both sides of (3) gives:
( |−2 ) = (103−2 − 020−3 +  + 11−1 | −2 ) = 103−2 − 020−3 
 ( |−1 ) and  ( |−2 ) are given by:
 ( |−1 ) =  (11−1 − 018−2 +  | −1 ) =  2
 ( |−2 ) =  (103−2 − 020−3 +  + 11−1 | −1 ) = (12 + 112 ) 2 
In order to find  ( |−2 ), we haved used the expression of  as function of −2 found in (3).
2a. It is an  (3) process.
2b.
( ) = (025 + 025−1 + 025−2 + 025−3 ) = 0;
(
|
−2 = −3 = 1) = (025 + 025−1 + 025−2 + 025−3 | −2 = −3 = 1)
= 025(0 + 0 + 1 + 1) = 050
For a proof of the first result, see answer 1c. The second result depends on the fact that while
(−2 | −2 = −3 = 1) = 1 and (−2 | −2 = −3 = 1) = 1 ( | −2 = −3 = 1) = ( ) = 0
and (−1 | −2 = −3 = 1) = (−1 ) = 0.
2c.  ( ) and  ( | −2 = −3 = 1) are equal to:
 ( ) =  (025 + 025−1 + 025−2 + 025−3 )
= 0252  ( ) + 0252  (−1 ) + 0252  (−2 ) + 0252  (−3 ) = 025 2
 (
|
−2 = −3 = 1) =  (025 + 025−1 + 025−2 + 025−3 | −2 = −3 = 1) = 013 2 
For a demonstration of the first result, see answer 1c. The second result depends on the fact that
 (−2 | −2 = −3 = 1) = 0 and  (−3 | −2 = −3 = 1) = 0, while  ( | −2 = −3 =
1) =  ( ) =  2 and  (−1 | −2 = −3 = 1) =  (−1 ) =  2 
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2d. The autocovariances of  are given by:
(  −1 ) = (025 + 025−1 + 025−2 + 025−3  025−1 + 025−2 + 025−3 + 025−4 )
= 3(0252 ) 2
(  −2 ) = (025 + 025−1 + 025−2 + 025−3  025−2 + 025−3 + 025−4 + 025−5 )
= 2(0252 ) 2
(  −3 ) = (025 + 025−1 + 025−2 + 025−3  025−3 + 025−4 + 025−5 + 025−6 )
= (0252 ) 2 
All these results depend on the fact that (  ) =
(
 2 when  = 
.
0 when  6= 
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