1
Economics 240C
Due 5-28-2002
Answers to Problem Set Three
1. This MA(1) process can be expressed as:
x(t) = [ 1 - (-0.5Z)] e(t)
or equivalently as:
[ 1 - (-0.5Z)]-1x(t) = e(t),
[1 + (-0.5Z) + (-0.5Z)2 + (-0.5Z)3 + ...] x(t) = e(t),
x(t) -0.5 x(t-1) + 0.25 x(t-2) - 0.125 x(t-3) + ... = e(t)
2. This ARMA(1, 1) process can be expressed as:
x(t) - 0.9 Z x(t) = e(t) - 0.5 Z e(t),
[ 1 - 0.9 Z] x(t) = [ 1 - 0.5 Z] e(t)
or equivalently as:
x(t) = [ 1 - 0.9 Z]-1 [ 1 - 0.5 Z] e(t)
or
x(t)=[1 +0.9Z + 0.81Z2 + 0.729Z3 + ...] [ 1 - 0.5 Z] e(t)
and multiplying the two polynomials in Z,
[1 +0.9Z + 0.81Z2 + 0.729Z3 + ...]
[ 1 - 0.5 Z]
--------------------------------------------
x(t) = [ 1 + 0.4 Z + 0.36 Z2 + 0.324 Z3 - ... ] e(t)
3. The ARMA(1,1) process can be expressed as
[ 1- a Z] x(t) = [ 1- b Z] e(t)
As a approaches b the polynomials cancel and x(t) approaches e(t).
4. For the ARIMA(1,1) process:
x(t) = 0.5 x(t-1) + e(t) - 0.9 e(t-1)
or equivalently,
(1)
x(t) = [e(t) - 0.9 e(t-1)]/[1-0.5Z]
and hence x(t-2) will depend upon e(t-2) and earlier realizations of the
orthogonal process e(t). Thus, multiplying Eq. (1) by x(t-2),
x(t) x(t-2) = 0.5 x(t-1) x(t-2) + x(t-2) e(t) - 0.9 x(t-2) e(t-1)
and taking covariances,
xx(2) = 0.5 xx(1)
or
2
Economics 240C
Due 5-28-2002
Answers to Problem Set Three
xx(2) = 0.5 xx(1) .
Similarly,
xx(j) = 0.5 xx(j-1) for j ≥ 2.
Of course xx(0) = 1. That just leaves xx(1).
Multiplying Eq. (1) by x(t-1),
x(t) x(t-1) = 0.5 x(t-1) x(t-1) + x(t-1) e(t) - 0.9 x(t-1) e(t-1)
and taking covariances,
xx(1) = 0.5 xx(0) - 0.9 ex(0).
Multiplying Eq. (1) by e(t):
x(t) e(t) = 0.5 x(t-1) e(t) + e(t) e(t) - 0.9 e(t) e(t-1)
and taking covariances, ex(0) = e2, thus
xx(1) = 0.5 xx(0) - 0.9 e2.
(2)
Multiplying Eq. (1) by x(t),
x(t) x(t) = 0.5 x(t-1) x(t) + x(t) e(t) - 0.9 x(t) e(t-1)
and taking covariances,
xx(0) = 0.5 xx(1) + ex(0) - 0.9 xe(-1)
or
xx(0) = 0.5 xx(1) + e2 - 0.9 xe(-1)
Multiplying Eq. (1) by e(t-1):
x(t) e(t-1) = 0.5 x(t-1) e(t-1) + e(t) e(t-1) - 0.9 e(t-1) e(t-1)
and taking covariances,
xe(-1) = 0.5 ex(0) - 0.9e2
or
xe(-1) = 0.5 e2 - 0.9e2 = -0.4 e2
and hence
xx(0) = 0.5 xx(1) + e2 - 0.9 (-0.4 e2)
or
xx(0) = 0.5 xx(1) + 1.36 e2
Solving equations (2) and (3),
xx(1) = - 0.293 e2
and
xx(0) = 1.21 e2
and thus
xx(1) = -0.293/1.21 = -0.242
(3)
3
Economics 240C
Due 5-28-2002
Answers to Problem Set Three
Note the autocorrelation function has a negative spike at lag 1,
capturing the MA(1) part of the ARMA(1,1) and then decays out like an
AR(1).
5. For the moving average process, x(t) = -0.8 e(t-1) + e(t)
where e(t) is white noise, square both sides of the expression for x(t) and
take expectations;
E[x(t)x(t)]= E[0.64e(t)e(t)-0.8e(t-1)e(t)-0.8e(t-1)e(t)+e(t)e(t)]
x,x(0) = 0.64  + 0 + 0 +  = 1.64 
Lagging the expression for x(t) by one: x(t-1) = -0.8 e(t-2) + e(t-1) and
multiplying x(t) by x(t-1) and the corresponding expressions on the right
sides of the equality and taking expectations,
E[x(t)x(t-1)]= E[0.64e(t-1)e(t-2)-0.8e(t-2)e(t)-0.8e(t-1)e(t-1)+e(t)e(t-1)]
x,x(1) = 0 + 0 -0.8  = -0.8 

Thus x,x(1) =-0.8/1.64 = -0.488
Lagging the expression for x(t) by two: x(t-2) = -0.8 e(t-3) + e(t-2) and
multiplying x(t) by x(t-2) and the corresponding expressions on the right
sides of the equality and taking expectations,
E[x(t)x(t-2)]= E[0.64e(t-1)e(t-3)-0.8e(t-3)e(t)-0.8e(t-1)e(t-2)+e(t)e(t-2)]
x,x(2) = 0 + 0 + 0  = 0

Thus x,x(2) =0/1.64 = 0 and autocorrelations will be zero, as will
autocorrelations at higher lags.
4
Economics 240C
Due 5-28-2002
Answers to Problem Set Three
Autoc orre lation
2
1
0
-1
0
1
2
Lag
3
The MA(1) time series, x(t) = -0.8 e(t-1) + e(t)
can be simulated on TSP with:
»
CREATE U 1000
»
Genr wn = nrnd
»
Smpl 1 1
»
Genr maone = wn
»
Smpl 2 1000
»
Genr maone = -0.8*WN(-1) + WN
»
Smpl 1 50
»
PLOT (A) MAONE
»
Smpl 1 1000
»
IDENT (24) MAONE
»
4
5
5
Economics 240C
Due 5-28-2002
Answers to Problem Set Three
Trace of MAONE: = -0.8 WN(-1) + WN
3
2
1
0
-1
-2
-3
10
20
30
40
50
TIME
MAONE