ECON 3350/7350: Applied Econometrics for Macroeconomics and Finance
Tutorial 2: Univariate Time Series – I
This tutorial aims to get you familiar with the fundamental features of univariate time series models.
1. Derive the expected value, variance, covariance, autocorrelation function (ACF), and partial autocorrelation function (PACF) for the time series yt having the fol- lowing data generating processes (DGP):
(a) AR(1):yt =a0 +a1yt−1 +εt,0≤|a1|<1. (b) MA(1): yt = β0 + β1εt−1 + εt.
(c) ARMA(1,1):yt =a0 +a1yt−1 +β1εt−1 +εt,0≤|a1|<1.
Solution:
(a) • The Expected Value
yt =a0 +a1yt−1 +εt; 0≤|a1|<1
E{yt} = μ = a0 + a1E{yt−1} + E{εt} μ= a0 ; sinceE{yt−1}=μ
1−a1
• The Variance
V {yt} = γ0 = a21V {yt−1} + V {εt} + +2cov{a1yt−1, εt}
σ2
γ0 = 1−a2; sinceV{yt−1}=γ0, cov(yt−1,εt)=0
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• Covariance:
– Set a0 = 0 without loss of generality
cov{yt, yt−k} = γk = E{ytyt−k}
= E{(a1yt−1 + εt)yt−k}
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– γ1 (k = 1)
– γ2 (k = 2)
– γk (k > 2)
• Autocorrelation: – ρ1
– ρ2
– ρk,k>2
• Partial Autocorrelation
– φ11 – φ22
– φ33
γ1 = E{(a1yt−1 + εt)yt−1}
σ2 =a11−a2 =a1γ0
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γ2 = E{(a1yt−1 + εt)yt−2}
2σ2 2 =a11−a2 =a1γ0
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γk = E{(a1yt−1 + εt)yt−k}
kσ2 k =a11−a2 =a1γ0
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ρ1 = γ1 = a1 γ0
ρ 2 = γ 2 = a 21 γ0
ρ k = γ k = a k1 γ0
φ11 =ρ1 =a1
φ22 = (ρ2 − ρ21)/(1 − ρ21) = (a21 − a21)/(1 − a21) =0
ρ3 − 3−1 φ2,j ρ3−j j=1
φ33= 1−3−1φ3−1,jρj j=1
a31 − φ21ρ2 − φ22ρ1 = 1−3−1φ3−1,jρj
j=1
a 31 − a 1 a 21 + 0
= 1 − 3−1 φ3−1,j ρj j=1
since
=0
φ21 = φ1,1 − φ22φ1,1 = φ1,1
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(b) •
The Expected Value
E{yt} = β0 + β1E{εt−1} + E{εt} =μ
• The Variance
V{yt}=γ0 =V{β0}+β12V{εt−1}+V{εt}+2cov{εt,εt−1}
γ0 =σ2(1+β12)
• Covariance:
– Set μ = 0 without loss of generality cov{yt, yt−k} = γk = E{ytyt−k}
= E{(β1εt−1 + εt)yt−k} cov(yt,yt−k) > 0 for k = 1, cov(yt,yt−k) = 0 for k > 1
– γ1 (k = 1)
γ1 = E{(β1εt−1 + εt)yt−1}
= E{β1εt−1(β1εt−2 + εt−1) + εtyt−1} = β1σ2
= β1 ×γ0; sinceσ2 =γ0/(1+β12) 1 + β 12
– γ2 (k = 2)
γ2 = E{(β1εt−1 + εt)yt−2}
– γk (k > 2)
• Autocorrelation: – ρ1
– ρk,k>1
• Partial Autocorrelation
= 0; since yt−2 is not a function of εt or εt−1
γk = E{(β1εt−1 + εt)yt−k} =0
ρ1=γ1= β1
γ 0 1 + β 12
ρk = γk = 0 γ0
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– φ11 – φ22
– φ33
φ11 = ρ1
φ22 = (ρ2 − ρ21)/(1 − ρ21) = (0 − ρ21)/(1 − ρ21) = −ρ21/(1 − ρ21)
ρ3 − 3−1 φ2,j ρ3−j j=1
1 − 3−1 φ2,jρj j=1
φ33 = =
ρ3 − φ21ρ2 − φ22ρ1 1 − 3−1 φ2,jρj
j=1 ρ31/(1 − ρ21)
= 1−3−1φ ρ ; sinceρ2 =ρ3 =0 j=1 2,j j
(c) • The Expected Value
E{yt} = a0 + a1E{yt−1} + β1E{εt−1} + E{εt}
μ= a0 ; sinceE{yt}=E{yt−1}=μ 1−a1
• The Variance
V{yt}=γ0 =V{a0}+a21V{yt−1}+β12V{εt−1}+V{εt}
+ 2cov{a1yt−1, β1εt−1} + 2cov{a1yt−1, εt} + 2cov{β1εt−1, εt} γ0 = 1 + β12 + 2a1β1 σ2, since cov(a1yt−1, β1εt−1) = a1β1E(ε2t−1)
1 − a 21
– To show cov(a1yt−1, β1εt−1) = a1β1E(ε2t−1) you can proceed as fol-
lows
cov(a1yt−1, β1εt−1) = E[(a1yt−1)(β1εt−1)]
= E{[a1(a1yt−2 + β1εt−2 + εt−1)](β1εt−1)}
= E{a1εt−1β1εt−1}
is the only non-zero expected value. • Covariance:
– Set μ = 0 without loss of generality
cov{yt, yt−k} = γk = E{ytyt−k}
= E{(a1yt−1 + β1εt−1 + εt)yt−k}
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– γ1 (k = 1)
– γk (k > 2)
• Autocorrelation:
– ρ1
– ρk,k≥2
γ1 = (1 + a1β1)(a1 + β1)σ2 1 − a 21
γ2 = a1γ1
ρ1 = (1 + a1β1)(a1 + β1) 1 + β 12 + 2 a 1 β 1
ρk = a1ρk−1
– Autoregressive pattern dominates from k > 1
• Partial Autocorrelation – φ11
– φ22
– φ33
where
φ11 = ρ1
φ22 = (ρ2 − ρ21)/(1 − ρ21)
φ33 = =
j=1
1 − 3−1 φ2,jρj
= (a1ρ1 − ρ21)/(1 − ρ21) ρ3 − 3−1 φ2,j ρ3−j
j=1
a21ρ1 − φ21a1ρ1 − φ22ρ1
1 − 3−1 φ2,jρj j=1
φ21 = φ11 − φ22φ11
= ρ1[1 − (a1ρ1-ρ21)/(1-ρ21)]
– Moving Average pattern dominates after k > 1
2. Compute the true ACF values for the following DGPs:
(1) DGP1: yt = 0.75yt−1 + εt
(2) DGP2: yt = −0.75yt−1 + εt
(3) DGP3: yt = 0.95yt−1 + εt
(4) DGP4: yt = 0.5yt−1 + 0.25yt−2 + εt (5) DGP5: yt = 0.25yt−1 − 0.5yt−2 + εt
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(6) DGP6: yt = 0.75εt−1 + εt
(7) DGP7: yt = 0.75εt−1 − 0.5εt−2 + εt (8) DGP8: yt = 0.75yt−1 + 0.5εt−1 + εt
Solution:
(1) ρ0 = 1, ρ1 = 0.75, …, ρk = 0.75k. The ACF will decay geometrically.
(2) ρ0 = 1, ρ1 = −0.75, …, ρk = (−1)k0.75k. The ACF will decay in a damp-
ened oscillatory path.
(3) ρ0 = 1, ρ1 = 0.95, …, ρk = 0.95k. The ACF will decay geometrically but at a much slower rate than DGP1.
(4) ForAR(2)modelyt =a0 +a1yt−1 +a2yt−2 +εt,ρ0 =1,ρ1 =a1/(1−a2), …, ρk = a1ρk−1 +a2ρk−2. Thus, ρ0 = 1, ρ1 = 2/3, ρ2 = 7/12, …, ρk = a1ρk−1 + a2ρk−2 for k ≥ 2.
(5) ρ0 =1,ρ1 =1/6,ρ2 =−11/24,…,ρk =a1ρk−1 +a2ρk−2 fork≥2.
(6) ForMA(q)modelyt =β0 +β1εt−1 +···+βqεt−q +εt,,theACFcutsoffat k=q,i.e.,ρk =0forallk>q.Thus,ρ0 =1,ρ1 =β1/(1+β12)=12/25, ρk =0,fork≥2.
(7) ρ0 = 1andρk = 0fork ≥ 3. ρ1 = β1(1+β2)/(1+β12 +β2) = 6/29and ρ2 = β2/(1 + β12 + β2) = −8/29.
(8) ForARMA(1,1)modelyt = a0 +a1yt−1 +β1εt−1 +εt,ρ0 = 1,ρ1 = (1+ a1β1)(a1 +β1)/(1+β12 +2a1β1), ρk = a1ρk−1 for all k ≥ 2. Thus, ρ0 = 1, ρ1 = 0.859, ρ2 = 0.645, …
3. Thedatafilearma.csvcontains(simulated)dataforeachoftheDGPsinQuestion 2. Import the data to Stata and use the variable t to declare time series. Compute, plot, and describe the behavior of the ACF and PACF of each DGP. Discuss the effects of parameter signs. Hint: Use the ac and pac commands, respectively.
Solution: See the do-file tutorial2.do.
(1) (a)
(b) PACF: One non-zero peak.
ACF: Decay geometrically as parameter is positive.
ACF: Decay in a dampened oscillatory path as parameter is nega- tive.
(2) (a)
(b) PACF: One non-zero peak.
(3) (a)
ACF: Decay geometrically but slower than DGP1.
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(b) PACF: One non-zero peak.
(4) (a)
(b) PACF: Two non-zero peaks.
ACF: Decay geometrically as parameter is positive.
(5) (a)
(b) PACF: Two non-zero peaks.
ACF: Decay in a oscillatory path as one parameter is negative (and large in absolute value).
(6) (a)
(b) PACF: Decay in a oscillatory path.
ACF: One non-zero peak.
(7) (a)
(b) PACF: Decay in a oscillatory path.
ACF: Two non-zero peak.
(8) (a)
(b) PACF: Decay in a oscillatary path from k = 2 as MA(1) dominates.
ACF: Decay geometrically from k = 2 onwards as AR(1) process dominates.
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