MP, MS, DT.
F70TS2 – Time Series
Solution Exercise Sheet 2 – Moving Average and Autoregressive Processes
Solution 1
1. The acfs for the two considered MA processes are given by (i) ρ = β1+β1β2 ,ρ = β2 andρ =0fork≥3.
1 1+β12 +β2 2 1+β12 +β2 k
(ii) ρ =β1+β1β2+β2β3,ρ = β2+β1β3 ,ρ = β3 andρ =0fork≥4.
1 1+β12 +β2 +β32 2 1+β12 +β2 +β32 3 1+β12 +β2 +β32 k
2. Figure 1 below shows the acfs for the coefficient values specified in the exercise sheet
(i) ρ1 =120 =40,ρ2 = 50 189 63 189
(ii)ρ1=60 =20,ρ2=−64,ρ3=−30 =−10 189 63 189 189 63
Figure 1: Autocorrelation functions for Q1(b).
Solution2FortheMA(1)processY=Z+βZ ,wehaveρ= β .Hence ttt−1 11+β2
dρ1 1−β2
dβ =(1+β2)2 =0 forβ=±1 (⇒ρ1 =±1/2)
setting ∂ρ1 = ∂ρ1 = 0 gives ∂β1 ∂β2
0=(1+β2)(1−β12 +β2)=β1(1+β12 −β2 −2β2) The only valid solutions are
So |ρ1| ≤ 1/2.
For the MA(2) process Y = Z + β Z + β Z , ρ = β1(1+β2) and ρ = β2
. Hence
t t 1 t−1 2 t−2 1 1+β12 +β2 2 1+β12 +β2
(i) β1 =0,β2 =−1whichgivesρ1 =0
(ii)1−β2+β2=0,1+β2−β2−2β =0,i.e.β =±√2andβ =1.Thisgivesρ =±1 .
12 122 1 2 1√2 1
Sosupρ = 1 ,attainedatβ =√2,β =1(whenwehaveρ = 1). Also,infρ =−1 1√2√12 241√2
attainedatβ1 =− 2,β2 =1(whenwealsohaveρ2 =1).
For ρ2, setting ∂ρ2 = ∂ρ2 = 0 gives ∂β1 ∂β2
4
β 1 β 2 = 1 + β 12 − β 2 2 = 0 whichgivesβ1 =0,β2 =±1,andρ2 =±1.
2
Sosupρ2 = 1,attainedatβ1 =0,β2 =1(whenwehaveρ1 =0). Also,infρ2 =−1 attained 22
atβ1 =0,β2 =−1(whenwealsohaveρ1 =0).
Solution 3 We have to check that the three conditions on α1 and α2, so that Xt is causal stationary, hold and then calculate ρ(k) following the general formulas ρ(0) = 1, ρ(1) = α1/(1− α2), ρ(k) = α1ρ(k − 1) + α2ρ(k − 2) for k ≥ 2. Note that, the causal stationary conditions are certainly fulfilled, if |α1| + |α2| < 1.
a)Wehaveα1 =−1.4,α2 =−0.65,forwhichwehaveα1+α2 <1,α2−α1 <1and −1 < α2 < 1. Hence this process is causal stationary. Using the above formulas we obtain ρ(k) for k = 0,1,...,9:
Lagk 0 1 2 3 4 5 6 7 8 9 ρ(k) 1.000 -0.848 0.537 -0.201 -0.067 0.225 -0.271 0.233 -0.150 0.059
b) We have α1 = 0.45, α2 = 0.25. This process is causal stationary, because |α1| + |α2| < 1. Using the above formulas we obtain ρ(k) for k = 0, 1, ..., 9:
Lagk 0 1 2 3 4 5 6 7 8 9 ρ(k) 1.000 0.600 0.520 0.384 0.302 0.232 0.180 0.139 0.107 0.083
c)Wehaveα1 =1.2,α2 =−0.75,forwhichwehaveα1+α2 <1,α2−α1 <1and −1 < α2 < 1. Hence this process is causal stationary. Using the above formulas we obtain ρ(k) for k = 0,1,...,9:
Lagk 0 1 2 3 4 5 6 7 8 9 ρ(k) 1.000 0.685 0.072 -0.426 -0.566 -0.360 -0.006 0.261 0.319 0.186
Solution 4 The first two Yule–Walker equations are ρ1 = α1 +α2ρ1 and ρ2 = α1ρ1 +α2. Hence
Similarly,
Solution 5
α1 α12
ρ1 = 1 − α , ρ2 = α2 + 1 − α .
22
α1 = ρ1(1−ρ2), α2 = ρ2 −ρ21 . 1 − ρ 21 1 − ρ 21
1. Wehave0.14z2 +0.5z−1=0,sowehavetherootsz1 =−5andz2 =10/7.
|zi| > 1 for i = 1, 2 and so the process is stationary.
Hence, from the hint, the solution of the Yule–Walker equations is of the form ρk = az−k + bz−k, subject to ρ = 1 and ρ = −0.5 + 0.14ρ (from the first Yule–Walker
12011
equation). Hence ρ1 = −25. 43
Solvinga+b=1and0.2a−0.7b=−25 givesa= 17 andb=112.Therefore 43 129 129
ρk = 17 (0.2)k + 112(−0.7)k 129 129
for k ≥ 0. Plotting this we obtain the correlogram in Figure 2. 2
2. We have −0.6z2 − 1 = 0 which has roots z = ±i5/3 outside the unit circle. Hence we
have
subject to a+b = 1 (since ρ0 = 1) and ρ1 = −0.6ρ1 (from the first Yule–Walker equation),
ρk = a(i√0.6)k + b(−i√0.6)k = (0.6k/2)ik{a + b(−1)k} so ρ1 = 0. Hence a − b = 0 and so a = b = 1 . Therefore
2
fork≥0. AplotoftheacfisgiveninFigure3.
2
ρk = 1ik(0.6)k/2{1 + (−1)k}
Figure 2: ACF for Q4(a)
Figure 3: ACF for Q4(b)
3