Stochastic Modelling Solutions to Exercises on Time Series∗

Q6. (i) EW t = EX t +EY t is constant over time. The au<strong>to</strong>covariance of the process {W t } is
Cov [W t , W t−k ] = Cov [X t + Y t , X t−k + Y t−k ] = Cov [X t , X t−k ] + Cov [Y t , Y t−k ] and
depends on lag k and not on time t.
(ii) X t = ln CPI t − ln CPI t−1 and hence ln CPI t = ∑ t
j=−∞ X j. Since X t is an AR(1)
process, it is clear that ln CPI t is ‘integrated’ once and is ARIMA(1, 1, 0). Alternatively,
note that (1 − B)(1 − αB) ln CPI t = Z t + µ(1 − α) and a unit root
occurs.
<strong>Solutions</strong> <strong>to</strong> Past Exam Questions
Q1. (i) (a) γ 0 = Var(e t +β 1 e t−1 ) = (1+β 2 1 )σ2 e and γ 1 = Cov [e t + β 1 e t−1 , e t−1 + β 1 e t−2 ] =
β 1 σ 2 e, with γ k = 0 for k > 1.
This gives ρ 0 = 1, ρ 1 = β 1 /(1 + β 2 1 ), ρ k = 0 otherwise.
(b) Invertibility requires that |β 1 | < 1, so that the sum X t −β 1 X t−1 +β 2 1 X t−2 +· · ·
converges. µ and σ e are irrelevant.
(ii) We need <strong>to</strong> solve (1 + β 2 1 )σ2 e = 14.5, β 1 σ 2 e = 5.0. Eliminating σ 2 e, we have 1 + β 2 1 =
2.9β 1 , or β 1 = 1 2 (2.9 ± √ (2.9 2 − 4)) = 2.5 or 0.4.
β 1 = 2.5 corresponds <strong>to</strong> σ 2 e = 2, whereas β 1 = 0.4 corresponds <strong>to</strong> σ 2 e = 12.5.
For invertibility, solve 1 + β 1 z = 0. In the first case, z = −0.4 (no good); in the
second, z = −2.5 (OK).
(Exam Board of the Institute and Faculty of Actuaries)
Q2. (i) (a) γ 1 = Cov [X t , X t−1 ] = Cov [α 1 X t−1 + α 2 X t−2 + e t , X t ] = α 1 γ 0 +α 2 γ 1 +0, since
e t is independent of X t−1 .
(b) Similarly γ 2 = α 1 γ 1 + α 2 γ 0 and γ 0 = α 1 γ 1 + α 2 γ 2 + Cov [X t , e t ]. A further
application of the same technique gives Cov [X t , e t ] = σe. 2 Thus,
γ 1 = α (
)
1
γ 0 , γ 2 = α 2 + α2 1
γ 0 .
1 − α 2 1 − α 2
(c) ρ k is found by the relation ρ k = γ k /γ 0 .
(ii) We have ˆα 1 = r 1 (1 − ˆα 2 ) and ˆα 2 + ˆα 2 1 /(1 − ˆα 2) = r 2 , which are solved by
ˆα 1 = r 1(1 − r 2 )
1 − r 2 1
, ˆα 2 = r 2 − r1
2
1 − r1
2 .
(Exam Board of the Institute and Faculty of Actuaries)
Q3. (i) γ 1 = 0.8γ 0 − 0.4γ 1 + 0 and γ 2 = 0.8γ 1 − 0.4γ 0 + 0.
Hence, γ 1 = 4 7 γ 0 and γ 2 = 2 35 γ 0, implying that ρ 1 = 4 7 and ρ 2 = 2
φ 1 = ρ 1 = 4 7 and φ 2 = ρ 2−ρ 2 1
= −0.4.
1−ρ 2 1
(ii) The acf will reduce <strong>to</strong> zero as k increases; the pacf, however, will be equal <strong>to</strong> zero
for all k > 2.
35 .
(Exam Board of the Institute and Faculty of Actuaries)
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