MP, MS, DT.
F70TS2 – Time Series
Exercise Sheet 3 – MA(∞), AR(∞), ARMA and ARIMA
Solution 1 Yt = 0.6Yt−1 + Zt − 0.3Zt−1 − 0.1Zt−2. Take σZ2 = 1 for convenience. Note that E[YtZt] = σZ2 = 1. We multiply the defining equation of the model by each of Yt, Yt−1 and Yt−2 in turn and take expectations to obtain:
• Multiplying by Yt: γ0 = 0.6γ1 + 1 − 0.3E[YtZt−1] − 0.1E[YtZt−2]. • Multiplying by Yt−1: γ1 = 0.6γ0 − 0.3 − 0.1E[Yt−1Zt−2].
• Multiplying by Yt−2: γ2 = 0.6γ1 − 0.1.
Now, multiplying the equation of the model by Zt−1 and Zt−2 and taking expectations, we have
E[YtZt−1] E[YtZt−2]
Solving the above equations gives
γ0 =1.1,
= 0.6 − 0.3 = 0.3
= 0.6(0.3) − 0.1 = 0.08
γ1 =0.33, γ2 =0.098
Hence, ρ1 = 0.3 and ρ2 = 0.089.
Now, multiplying the equation of our model by Yt−k (for k ≥ 3) and taking expectations,
we get that γk = 0.6γk−1. So, ρk = 0.6ρk−1 for k ≥ 3. Solving this gives ρk = a(0.6)k for k ≥ 2. The value of a may be found using ρ2 = 0.089 (so that 0.36a = 0.089). This gives a = 89 .
360
Hence we have ρ1 = 0.3 and ρk = 89 (0.6)k for k ≥ 2. A plot of the autocorrelation function 360
is given in Figure 1.
Figure 1: ACF for Q1
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Solution 2 We consider the process
Yt =Zt +a(Zt−1 +Zt−2 +···).
Clearly E[Yt] = 0 for all t. We have
Var(Yt)=σZ2(1+a2 +a2 +a2 +···)
which diverges as t increases. Hence, Var(Yt) increases with time. So, the model is non– stationary.
DYt =Yt−Yt−1
= Zt +a(Zt−1 +Zt−2 +···)−Zt−1 −a(Zt−2 +Zt−3 +···) = Zt +(a−1)Zt−1
This is an MA(1) process, and is hence stationary. For {DY }, ρ = a−1 and ρ = 0 for
k ≥ 2.
Solution 3 We have to check that the three conditions on the MA coefficients ψ1 and ψ2, so that Xt is invertible, hold. Again, these conditions are certainly fulfilled, if |ψ1| + |ψ2| < 1.
a)Wehaveψ1 =−0.9,ψ2 =0.2. Andψ1+ψ2 >−1,ψ1−ψ2 <1and−1<ψ2 <1. Hence this process is invertible.
b) We have ψ1 = 0.3, ψ2 = −0.6. Hence this process is clearly invertible, because |ψ1|+|ψ2| < 1.
c) Now we have an MA(3) process. The explicit conditions for the invertibility in this case are not given. Hence we have to try to find a factorisation of ψ(z). For this special example we have
ψ(z) = 1 − 1.5z + 0.75z2 − 0.125z3 = (1 − 0.5z)3 with the roots z1 = z2 = z3 = 2. This process is invertible.
Solution 4 We have to check whether the AR part is causal stationary (φ1 +φ2 < 1, φ2 −φ1 < 1 and−1<φ2 <1)andwhethertheMApartisinvertible(ψ1+ψ2 >−1,ψ1−ψ2 <1and −1<ψ2 <1).
a) This process is causal stationary and invertible, because all of the above mentioned conditions are fulfilled.
b) This process is causal stationary, because φ1 and φ2 satisfy the required conditions. But it is non-invertible, because the MA(1) part is with coefficient ψ = 1.2.
c) This process is invertible, because |ψ1|+|ψ2| < 1 for the MA(2) part. But it is not causal stationary, because φ1 + φ2 > 1 for the AR(2) part.
d) This process is neither causal stationary, because φ1 + φ2 > 1 for the AR(2) part, nor invertible, because ψ1 − ψ2 > 1 for the MA(2) part.
t 1 1+(a−1)2 k
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Solution 5 For Xt = φ1Xt−1 + εt with |φ1| > 1 we have Xt+1 = φ1Xt + εt+1. It follows that Xt = φ∗1(Xt+1 + ε∗t+1),
where φ∗1 = 1/φ1 with |φ∗1| < 1 and ε∗t+1 = −εt+1. (Note that ε∗t+i is also an i.i.d. series with E(ε∗t ) = 0 and Var(ε∗t ) = σε2/φ21). Further expansion leads to
Xt = φ∗1(φ∗1(Xt+2 + ε∗t+2) + ε∗t+1) ∞∞
= ... = (φ∗1)iε∗t+i = αiεt+i, i=1 i=1
where αi = −(φ∗1)i = −(1/φ)i for i = 1,2,..., which are absolutely summable. Hence Xt is stationary, but non-causal, since it depends on future information (εt+i).
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