Stochastic Modelling Solutions to Exercises on Time Series∗

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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. 2
(ii) The process is assumed <strong>to</strong> be stationary and we can proceed <strong>to</strong> find the au<strong>to</strong>correlation function by using the usual method for stationary time series. The time series is shifted by µ and EX t = µ ∀ t. Let Y t = X t − µ. Then, EY t = 0 ∀ t. The au<strong>to</strong>covariance function is γ k = Cov [X t , X t−k ] = Cov [Y t , Y t−k ] = E[Y t Y t−k ]. Recall that γ k = γ −k . Also, EZ t = 0 and VarZ t = σ 2 Z ∀ t. Multiply Y t − αY t−1 = Z t − βZ t−1 by Y t−k on both sides and take expectations <strong>to</strong> obtain γ k − αγ k−1 = E[Y t−k Z t ] − βE[Y t−k Z t−1 ]. Observe that Z t is statistically independent of Y t−k , k ≥ 1. Therefore, when k = 0, γ 0 − αγ 1 = E[Y t Z t ] − βE[Y t Z t−1 ], (1) when k = 1, γ 1 − αγ 0 = −βE[Y t Z t−1 ], (2) when k > 1, γ k − αγ k−1 = 0. (3) The non-zero cross-correlations involving Y t and Z t are easily found. Multiply Y t − αY t−1 = Z t − βZ t−1 by Z t on both sides and take expectations <strong>to</strong> obtain E[Y t Z t ] = σZ 2 ∀ t, noting that {Z t} is a sequence of independent and identically distributed random variables. Also, multiply Y t − αY t−1 = Z t − βZ t−1 by Z t−1 on both sides and take expectations <strong>to</strong> obtain E[Y t Z t−1 ] − αE[Y t−1 Z t−1 ] = −βσZ 2 . Hence, E[Y t Z t−1 ] = σZ 2 (α − β). Equation (1) may be simplified <strong>to</strong> γ 0 − αγ 1 = σZ 2 [1 − αβ + β2 ] and equation (2) simplifies <strong>to</strong> γ 1 − αγ 0 = −βσZ 2 . Solving these two equations simultaneously yields: γ 0 = σ2 Z (1 − 2αβ + β2 ) 1 − α 2 , γ 1 = σ2 Z (α − β)(1 − αβ) 1 − α 2 . Finally, equation (3) solves <strong>to</strong> γ k = α k−1 γ 1 , for k > 1. Since ρ k = γ k /γ 0 by definition, ⎧ ⎪⎨ 1 k = 0 ρ k = (α − β)(1 − αβ)/(1 − 2αβ + β ⎪⎩ 2 ) k = 1 α k−1 ρ 1 k > 1. (iii) The correlogram of the ARMA(1, 1) time series has a ‘kink’ at lag 1. At lag 1, both the MA and AR components affect the correlogram. From lag 2 onwards, it looks rather like the correlogram of an AR series and it decays exponentially. Q5. Comments. If the process were known (or assumed) <strong>to</strong> be stationary over time, then EX t = 0 and ρ k = 0.4 |k| , using the properties of a stationary AR(1) process. But the process has a finite his<strong>to</strong>ry (with a given initial condition) and so it will be nonstationary for t < ∞. Solution. (i) In terms of the backward shift opera<strong>to</strong>r, X t = Z t /(1 − 0.4B) = Z t + 0.4Z t−1 + 0.4 2 Z t−2 + · · · 3
Page 1: Stochastic
Page 5 and 6: Q6. (i) EW t = EX t +EY t is consta
Page 7: We see that X n+1 − ˆx n (1) = e

Stochastic Modelling Solutions to Exercises on Time Series∗

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