Question (Total Marks: 4)
Consider the ARMA(1,1) process
Xt = ϕ Xt−1 + Zt + θ Zt−1 ,
where {Zt } is white noise with mean 0 and variance σ 2 , |ϕ| < 1, and the MA
part is invertible.
(a) (2 marks) Write the process in lag operator form and derive its infinite
MA representation
∞
X
Xt =
ψj Zt−j .
j=0
Find ψ0 , ψ1 , and ψ2 in terms of ϕ and θ.
(b) (2 marks) Using your result from part (a), derive a closed-form expression
for Var(Xt ).
Hint: Use the lag operator and note that (1 − ϕL)−1 is invertible.
1
Solution
(a) Infinite MA Representation (2 marks)
We begin by writing the ARMA(1,1) process in lag operator form. Recall that
the lag operator L is defined by LXt = Xt−1 . Thus, we can rewrite:
Xt = ϕXt−1 + Zt + θZt−1
Xt − ϕXt−1 = Zt + θZt−1 ,
=⇒
or equivalently,
(1 − ϕL)Xt = (1 + θL)Zt .
Since |ϕ| < 1, the operator (1 − ϕL) is invertible, and we have
Xt = (1 − ϕL)−1 (1 + θL)Zt .
Using the power series expansion
(1 − ϕL)−1 =
∞
X
ϕj Lj ,
j=0
we obtain

Xt = 
∞
X

ϕj Lj  (1 + θL)Zt .
j=0
Expanding this product yields:
Xt =
∞
X
ϕj Zt−j + θ
j=0
∞
X
ϕj Zt−j−1 .
j=0
By a change of index in the second sum (letting k = j + 1):
θ
∞
X
ϕj Zt−j−1 = θ
j=0
∞
X
ϕk−1 Zt−k .
k=1
Thus, combining the sums we have:
Xt = Zt +
∞
X
ϕj + θ ϕj−1 Zt−j .
j=1
That is, the infinite MA representation is
Xt =
∞
X
ψj Zt−j ,
j=0
with
ψ0 = 1,
ψj = ϕj + θ ϕj−1
2
for j ≥ 1.
In particular,
ψ0 = 1,
ψ1 = ϕ + θ,
ψ2 = ϕ2 + θ ϕ.
(b) Variance of Xt (2 marks)
Since {Zt } is white noise with Var(Zt ) = σ 2 and the series has the representation
Xt =
∞
X
ψj Zt−j ,
j=0
the variance of Xt is given by
Var(Xt ) = σ 2
∞
X
ψj2 .
j=0
Substituting the coefficients from part (a), we have:


∞
X
2
ϕj + θ ϕj−1  .
Var(Xt ) = σ 2 1 +
j=1
Notice that the sum for j ≥ 1 can be simplified by a change of index:
∞
X
j=1
ϕj + θ ϕj−1
2
= (ϕ + θ)2
∞
X
ϕ2(j−1) = (ϕ + θ)2
j=1
∞
X
ϕ2k = (ϕ + θ)2 ·
k=0
since |ϕ| < 1.
Thus, the closed-form expression for the variance is:
(ϕ + θ)2
Var(Xt ) = σ 2 1 +
.
1 − ϕ2
3
1
,
1 − ϕ2