Stochastic Modelling Solutions to Exercises on Time Series∗

Taking expectation on both sides yields
E [X t X t−k ] + 0.5E [X t−1 X t−k ] − 0.1E [X t−2 X t−k ] = 0
and dividing by VarX t on both sides gives ρ k + 0.5ρ k−1 − 0.1ρ k−2 = 0.
When k = 1, ρ 1 + 0.5ρ 0 − 0.1ρ 1 = 0 which solves <strong>to</strong> ρ 1 = −0.5/0.9 = −5/9
(noting that ρ k = ρ −k and ρ 0 = 1 by definition).
When k = 2, ρ 2 + 0.5ρ 1 − 0.1ρ 0 = 0 yielding ρ 2 = 0.1 − 0.5(−5/9) = 17/45.
When k = 3, ρ 3 +0.5ρ 2 −0.1ρ 1 = 0 yielding ρ 3 = 0.1(−5/9)−0.5(17/45) = −11/45.
(ii) {X t } in X t + 0.6X t−2 = Z t will clearly have zero au<strong>to</strong>correlation at odd lags.
ρ k + 0.6ρ k−2 = 0 for k ≥ 1.
When k = 1, ρ 1 + 0.6ρ 1 = 0 and ρ 1 = 0.
When k = 2, ρ 2 + 0.6ρ 0 = 0 and ρ 2 = −0.6.
When k = 3, ρ 3 + 0.6ρ 1 = 0 and ρ 3 = 0.
(iii) ρ k − 1.1ρ k−1 + 0.18ρ k−2 = 0 for k ≥ 1.
When k = 1, ρ 1 − 1.1ρ 0 + 0.18ρ 1 = 0 and ρ 1 = 1.1/1.18 = 55/59.
When k = 2, ρ 2 − 1.1ρ 1 + 0.18ρ 0 = 0 and ρ 2 = 1.1(55/59) − 0.18 = 1247/1475.
When k = 3, ρ 3 − 1.1ρ 2 + 0.18ρ 1 = 0 and ρ 3 = 1.1(1247/1475) − 0.18(55/59) =
5621/7375.
(iv) ρ k + αρ k−1 + α 2 ρ k−2 + α 3 ρ k−3 = 0 for k ≥ 1.
When k = 1, ρ 1 (1 + α 2 ) + α + α 3 ρ 2 = 0.
When k = 2, ρ 1 (α + α 3 ) + α 2 + ρ 2 = 0.
(Note in the above that ρ 0 = 1 and ρ k = ρ −k ).
Hence ρ 1 = −α/(1 + α 2 ) and ρ 2 = 0.
When k = 3, ρ 3 + αρ 2 + α 2 ρ 1 + α 3 ρ 0 = 0.
Hence, ρ 3 = −α 5 /(1 + α 2 ).
Q3. (i) Y t = Z t − βZ t−1 .
VarY t = VarZ t + β 2 VarZ t−1 = σ 2 Z (1 + β2 ).
Cov [Y t , Y t−1 ] = Cov [Z t − βZ t−1 , Z t−1 − βZ t−2 ] = −βσ 2 Z .
Cov [Y t , Y t−k ] = 0 for k ≥ 2.
Since ρ k = γ k /γ 0 , ρ 1 = −β/(1 + β 2 ) and ρ k = 0 for k ≥ 2.
(ii) Y t = Z t + 2.4Z t−1 + 0.8Z t−2 .
VarY t = σ 2 Z (1 + 2.42 + 0.8 2 ) = 7.4σ 2 Z .
Cov [Y t , Y t−1 ] = Cov [Z t + 2.4Z t−1 + 0.8Z t−2 , Z t + 2.4Z t−1 + 0.8Z t−2 ] = (2.4+0.8×
2.4)σ 2 Z = 7.4σ2 Z .
Cov [Y t , Y t−2 ] = 0.8σ 2 Z .
Cov [Y t , Y t−k ] = 0 for k ≥ 3.
Since ρ k = γ k /γ 0 , ρ 1 = 4.32/7.4 = 0.5838, ρ 2 = 0.8/7.4 = 0.1081 and ρ k = 0 for
k ≥ 3.
Q4. (i) The correlogram is a plot of au<strong>to</strong>correlation function against lag, i.e. ρ k vs. |k|. An
MA(q) process has a correlogram with a cu<strong>to</strong>ff at lag q, whereas a stationary AR
process has a correlogram that tapers off gradually (possibly with some damped
oscillations) <strong>to</strong> zero as the lag increases.
Comments. Plotting the correlogram is therefore a way of distinguishing MA from
AR time series, although in practice estimation errors may mean that it is not so
easy <strong>to</strong> discern a difference.
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