Module 5. February 14, h 0min
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- Βάαλ Καζαντζής
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1 Module 5 Stationary Time Series Models Part 2 AR and ARMA Models and Their Properties Class notes for Statistics 451: Applied Time Series Iowa State University Copyright 2015 W. Q. Meeker. February 14, h 0min 5-1
2 Module 5 Segment 1 AR(p) Mean and AR(1) Properties 5-2
3 Mean the AR(p) Model Model: Z t = θ 0 +φ 1 Z t 1 + +φ p Z t p +a t, a t nid(0,σ 2 a) Mean: µ Z E(Z t ) E(Z t ) = E(θ 0 + φ 1 Z t φ p Z t p + a t ) = E(θ 0 ) + φ 1 E(Z t 1 ) + + φ p E(Z t p ) + E(a t ) = θ 0 + (φ 1 + +φ p )E(Z t ) = θ 0 1 φ 1 φ p The last step requires several algebraic steps, but is straightforward. 5-3
4 Variance of the AR(1) Model Model: Z t = θ 0 +φ 1 Z t 1 +a t, a t nid(0,σ 2 a) Variance: γ 0 Var(Z t ) E[(Z t µ Z ) 2 ] = E(Ż 2 ) γ 0 = E(Żt 2) = E[(φ 1 Ż t 1 +a t ) 2 ] = E[(φ 2 1Ż2 t 1 + 2φ 1Żt 1a t + a 2 t )] = φ 2 1 E(Ż2 t 1 ) + 2φ 1E(Ż t 1 a t ) + E(a 2 t ) = φ 2 1 γ σa 2 = σ2 a 1 φ 2 1 or Var(Z t ) = Var(a t) 1 φ
5 Simulated AR(1) Data with φ 1 = 0.9,σ a = 1 Showing ±2 σ Z limits plot(arima.sim(n=250, model=list(ar=0.90)), ylab= ) Time 5-5
6 Simulated AR(1) Data with φ 1 = 0.9,σ a = 1 Showing ±2 σ Z limits plot(arima.sim(n=250, model=list(ar=-0.90)), ylab= ) Time 5-6
7 Autocovariance and Autocorrelation Functions for the AR(1) Model Autocovariance: γ k Cov(Z t,z t+k ) E(Ż t Ż t+k ) γ 1 E(Ż t Ż t+1 ) = E[Ż t (φ 1 Ż t +a t+1 )] = φ 1 E(Żt 2) + E(Ż t a t+1 ) = φ 1 γ = φ 1 γ 0 Thus ρ 1 = γ 1 γ 0 = φ 1. γ 2 E(Ż t Ż t+2 ) = E[Ż t (φ 1 Ż t+1 + a t+2 )] = φ 1 E(Ż t Ż t+1 ) + E(Ż t a t+2 ) = φ 1 γ = φ 1 (φ 1 γ 0 ) = φ 2 1 γ 0 Thus ρ 2 = γ 2 γ 0 = φ 2 1. In general, for the AR(1) model, ρ k = γ k γ 0 = φ k 1 for 1 < φ 1 <
8 True and P for AR(1) Model with φ 1 = 0.95 True ar: 0.95 True True P ar: 0.95 True P
9 True and P for AR(1) Model with φ 1 = 0.95 True ar: True True P ar: True P
10 Simulated Realization (AR(1),φ 1 = 0.95,n = 75) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-10
11 Simulated Realization (AR(1),φ 1 = 0.95,n = 300) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-11
12 Simulated Realization (AR(1),φ 1 = 0.95,n = 75) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-12
13 Simulated Realization (AR(1),φ 1 = 0.95,n = 300) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-13
14 Module 5 Segment 2 AR(1) Stationarity Conditions 5-14
15 Using the Geometric Power Series to Re-express the AR(1) Model as an Infinite MA (1 φ 1 B)Ż t = a t Ż t = (1 φ 1 B) 1 a t = (1+φ 1 B+φ 2 1 B2 + )a t = φ 1 a t 1 +φ 2 1 a t 2 + +a t Thus the AR(1) can be expressed as an infinite MA model. If 1 < φ 1 < 1, then the weight on the old residuals is decreasing with age. This is the condition of stationarity for an AR(1) model. 5-15
16 Using the Back-substitution to Re-express the AR(1) Model as an Infinite MA Ż t = φ 1 Ż t 1 +a t Ż t 1 = φ 1 Ż t 2 +a t 1 Ż t 2 = φ 1 Ż t 3 +a t 2 Ż t 3 = φ 1 Ż t 4 +a t 3 Substituting, successively, Ż t 1,Ż t 2,Ż t 3..., shows that Ż t = φ 1 a t 1 +φ 2 1 a t 2 +φ 3 1 a t 3 + +a t This methods works, more generally, for higher-order AR(p) models, but the algebra is tedious. 5-16
17 Notes on the AR(1) model Z t = φ 1 Z t 1 +a t, a t nid(0,σ 2 a) Because ρ 1 = φ 1, we can estimate φ 1 by φ 1 = ρ 1. The root of (1 φ 1 B) = 0 is B = 1/φ 1 and thus if 1 < φ 1 < 1, then the root is outside [ 1,1] so that AR(1) will be stationary. φ 1 = 1 implies Z t = Z t 1 +a t, the random walk model. If φ 1 > 1, then Z t is explosive. If φ 1 < 1, then Z t is oscillating explosive. 5-17
18 True and P for AR(1) Model with φ 1 = True ar: True True P ar: True P
19 Simulated Realization (AR(1),φ 1 = ,n = 75) Graphical Output from Function iden Simulated data w= Data w time Range-Mean Plot Range Partial P Mean 5-19
20 Simulated Realization (AR(1),φ 1 = ,n = 300) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-20
21 Re-expressing the AR(p) Model as an Infinite MA More generally, any AR(p) model can be expressed as φ p (B)Ż t = a t Ż t = 1 φ p (B) a t = ψ(b)a t = ψ 1 a t 1 +ψ 2 a t 2 + +a t = k=1 ψ k a t k +a t Values of ψ 1,ψ 2,... depend on φ 1,...,φ p. For the AR model to be stationary, the ψ j values should not remain large as j gets large. Formally, the stationarity condition is met if j=1 ψ j < Stationarity of an AR(p) model can be checked by finding the roots of the polynomial φ p (B) (1 φ 1 B φ 2 B 2 φ p B p ). All p roots must lie outside of the unit-circle. 5-21
22 Module 5 Segment 3 AR(2) Stationarity and Model Properties 5-22
23 Checking the Stationarity of an AR(2) Model Stationarity of an AR(2) model can be checked by finding the roots of the polynomial (1 φ 1 B φ 2 B 2 ) = 0. Both roots must lie outside of the unit-circle. Roots have the form B = φ 1 ± φ φ 2 2φ 2 z = x+iy where i = 1. A root is outside of the unit circle if z = x 2 +y 2 > 1 This method works for any p, but for p > 2 it is best to use numerical methods to find the p roots. 5-23
24 Checking the Stationarity of an AR(2) Model Simple Rule Both roots lying outside of the unit-circle implies φ 2 +φ 1 < 1 φ 2 < 1 φ 1 φ 2 φ 1 < 1 φ 2 < 1+φ 1 1 < φ 2 < 1 1 < φ 2 < 1 and defines the AR(2) triangle. 5-24
25 Autocovariance and Autocorrelation Functions for the AR(2) Model γ 1 E(Ż t Ż t+1 ) = E[Ż t (φ 1 Ż t + φ 2 Ż t 1 + a t+1 )] = φ 1 E(Ż 2 t ) + φ 2E(Ż t Ż t 1 ) + E(Ż t a t+1 ) = φ 1 γ 0 + φ 2 γ γ 2 E(Ż t Ż t+2 ) = E[Ż t (φ 1 Ż t+1 + φ 2 Ż t + a t+2 )] = φ 1 E(Ż t Ż t+1 ) + φ 2 E(Ż 2 t ) + E(Ż t a t+2 ) = φ 1 γ 1 + φ 2 γ Then using ρ k = γ k /γ 0 and ρ 0 = 1, gives the AR(2) ρ 1 = φ 1 + φ 2 ρ 1 ρ 2 = φ 1 ρ 1 + φ 2 ρ 3 = φ 1 ρ 2 + φ 2 ρ 1... ρ k = φ 1 ρ k 1 + φ 2 ρ k
26 True and P for AR(2) Model with φ 1 = 0.78,φ 2 = 0.2 True ar: 0.78, 0.2 True True P ar: 0.78, 0.2 True P
27 True and P for AR(2) Model with φ 1 = 1,φ 2 = 0.95 True ar: 1, True True P ar: 1, True P
28 Simulated Realization (AR(2),φ 1 = 0.78,φ 2 = 0.2,n = 75) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-28
29 Simulated Realization (AR(2),φ 1 = 0.78,φ 2 = 0.2,n = 300) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-29
30 Simulated Realization (AR(2),φ 1 = 1,φ 2 = 0.95,n = 75) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-30
31 Simulated Realization (AR(2),φ 1 = 1,φ 2 = 0.95,n = 300) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-31
32 AR(2) Triangle Relating ρ 1 and ρ 2 to φ 1 and φ
33 AR(2) Triangle Showing and P Functions 5-33
34 Module 5 Segment 4 Yule-Walker Equations and Their Applications and AR(p) Variance 5-34
35 Yule-Walker Equations (Correlation Form) AR(1) Yule-Walker Equation ρ 1 = φ 1 AR(2) Yule-Walker Equations ρ 1 = φ 1 + ρ 1 φ 2 ρ 2 = ρ 1 φ 1 + φ 2 AR(3) Yule-Walker Equations ρ 1 = φ 1 + ρ 1 φ 2 + ρ 2 φ 3 ρ 2 = ρ 1 φ 1 + φ 2 + ρ 1 φ 3 ρ 3 = ρ 2 φ 1 + ρ 1 φ 2 + φ
36 Yule-Walker Equations (Correlation Form) AR(4) Yule-Walker Equations ρ 1 = φ 1 + ρ 1 φ 2 + ρ 2 φ 3 + ρ 3 φ 4 ρ 2 = ρ 1 φ 1 + φ 2 + ρ 1 φ 3 + ρ 2 φ 4 ρ 3 = ρ 2 φ 1 + ρ 1 φ 2 + φ 3 + ρ 1 φ 4 ρ 4 = ρ 3 φ 1 + ρ 2 φ 2 + ρ 1 φ 3 + φ 4 For given ρ 1,...,ρ 4, this is four linear equations with four unknowns. 5-36
37 AR(p) Yule-Walker Equations (Correlation Form) ρ 1 = φ 1 +ρ 1 φ 2 +ρ 2 φ 3 + +ρ p 1 φ p ρ 2 = ρ 1 φ 1 +φ 2 +ρ 1 φ 3 + +ρ p 2 φ p. ρ p = ρ p 1 φ 1 +ρ p 2 φ 2 +ρ p 3 φ 3 + +φ p For given ρ 1,...,ρ p, this is p linear equations with p unknowns. Solving the Yule Walker equations for an AR(p) for φ p gives the P value φ pp. 5-37
38 Applications of the Yule-Walker Equations Provides the true of an AR(p) model (given φ 1,φ 2,...,φ p, compute ρ 1,ρ 2,...ρ p recursively) Provides estimate of φ 1,φ 2,...,φ p of an AR(p) model (given sample values ρ 1, ρ 2,..., ρ p, substitute in Y-W equations and solve the set of simultaneous equations for φ 1, φ 2,..., φ p ) Provides the sample P φ 1,1, φ 2,2,..., φ kk for any realization. (substitute ρ 1, ρ 2,..., ρ k into the AR(k) Y-W equations and solve for φ k ). Same as Durbin s formula. Provides the true P φ 1,1,φ 2,2,...,φ kk for any ARMA model. (substitute ρ 1,ρ 2,...,ρ k into the AR(k) Y-W equations and solve for φ k ). Same as Durbin s formula. 5-38
39 Variance of the AR(p) model Ż t = φ 1 Ż t 1 + φ 2 Ż t φ p Ż t p + a t Multiplying through by Ż t gives Ż 2 t = φ 1 Ż t Ż t 1 + φ 2 Ż t Ż t φ p Ż t Ż t p + Ż t a t Taking expectations and noting that E(Ż t a t ) = σ 2 a gives γ 0 = Var(Z t ) = E(Ż 2 t ) = φ 1γ 1 + φ 2 γ φ p γ p + σ 2 a Then dividing through by γ 0 and solving for γ 0 gives γ 0 = σ 2 a 1 φ 1 ρ 1 φ 2 ρ 2 φ p ρ p Note: E(Ż t a t ) = E(φ 1 Ż t 1 a t +φ 2 Ż t 2 a t + +φ p Ż t p a t +a 2 t) = σ 2 a 5-39
40 Module 5 Segment 5 ARMA(1,1) Model Properties 5-40
41 Properties of the ARMA(1, 1) Model φ 1 (B)Z t = θ 0 +θ 1 (B)a t Z t = θ 0 +φ 1 Z t 1 θ 1 a t 1 +a t Noting that µ Z = E(Z t ) = θ 0 /(1 φ 1 ) φ 1 (B)Ż t = θ 1 (B)a t Ż t = φ 1 Ż t 1 θ 1 a t 1 +a t Omitting derivations, γ 0 = Var(Z t ) = [(1 2φ 1 θ 1 +θ1 2 )/(1 φ2 1 )]σ2 a ρ 1 = (1 φ 1 θ 1 )(φ 1 θ 1 )/(1+θ1 2 2φ 1θ 1 ) ρ 2 = φ 1 ρ 1 ρ k = φ 1 ρ k
42 True and P for ARMA(1,1) Model with φ 1 = 0.95,θ 1 = 0.95 True ma: ar: 0.95 True True P ma: ar: 0.95 True P
43 Simulated Realization (ARMA(1,1),φ 1 = 0.95,θ 1 = 0.95,n = 75) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-43
44 True and P for ARMA(1,1) Model with φ 1 = 0.95,θ 1 = 0.95 True ma: 0.95 ar: True True P ma: 0.95 ar: True P
45 Simulated Realization (ARMA(1,1),φ 1 = 0.95,θ 1 = 0.95,n = 75) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-45
46 True and P for ARMA(1,1) Model with φ 1 = 0.9,θ 1 = 0.5 True ma: 0.5 ar: 0.9 True True P ma: 0.5 ar: 0.9 True P
47 Simulated Realization (ARMA(1,1),φ 1 = 0.9,θ 1 = 0.5,n = 75) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-47
48 True and P for ARMA(1,1) Model with φ 1 = 0.9,θ 1 = 0.5 True ma: -0.5 ar: -0.9 True True P ma: -0.5 ar: -0.9 True P
49 Simulated Realization (ARMA(1,1),φ 1 = 0.9,θ 1 = 0.5,n = 75) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-49
50 ARMA(1,1) Square Relating ρ1 and ρ2 to φ1 and θ1 5-50
51 ARMA(1,1) Square Showing and P Functions 5-51
52 Module 5 Segment 6 Stationarity and Invertibility of the ARMA(1,1) Model 5-52
53 Stationarity of an ARMA(1, 1) Model Ż t = (1 θ 1B) (1 φ 1 B) a t (1 φ 1 B)Ż t = (1 θ 1 B)a t = (1 θ 1 B)(1 φ 1 B) 1 a t = (1 θ 1 B)(1+φ 1 B+φ 2 1 B2 + )a t = (φ 1 θ 1 )a t 1 +φ 1 (φ 1 θ 1 )a t 2 +φ 2 1 (φ 1 θ 1 )a t 3 + +a t = j=1 φ j 1 1 (φ 1 θ 1 )a t j +a t = j=0 ψ j a t j where ψ j = φ j 1 1 (φ 1 θ 1 ) for j > 0, ψ 0 = 1, and φ ARMA(1,1) model is stationary if 1 < φ 1 < 1. ARMA(1,1) model is white noise (trivial model) if φ 1 = θ
54 Invertibility of an ARMA(1, 1) Model a t = (1 φ 1B) (1 θ 1 B)Żt (1 φ 1 B)Ż t = (1 θ 1 B)a t = (1 φ 1 B)(1 θ 1 B) 1 Ż t = (1 φ 1 B)(1+θ 1 B+θ 2 1 B2 + )Ż t = Ż t +(θ 1 φ 1 )Ż t 1 +θ 1 (θ 1 φ 1 )Ż t 2 +θ 2 1 (θ 1 φ 1 )Ż t 3 + Ż t = (φ 1 θ 1 )Ż t 1 +θ 1 (φ 1 θ 1 )Ż t 2 +θ 2 1 (φ 1 θ 1 )Ż t 3 + +a t = j=1 θ j 1 1 (φ 1 θ 1 )Ż t j +a t = j=1 where π j = θ j 1 1 (φ 1 θ 1 ),j > 0 and π 0 = 1. π j Ż t j +a t ARMA(1,1) model is invertible if 1 < θ 1 <
55 True and P for ARMA(1,1) Model with φ 1 = 0.9,θ 1 = 0.9 True ma: 0.9 ar: 0.9 True True P ma: 0.9 ar: 0.9 True P
56 Simulated Realization (ARMA(1,1),φ 1 = 0.9,θ 1 = 0.9,n = 75) Graphical Output from Function iden Simulated data w= Data Range-Mean Plot w time Range Partial P Mean 5-56
57 Properties of ARMA(p, q) model φ p (B)Z t = θ 0 +θ q (B)a t φ p (B)Ż t = θ q (B)a t (1 φ 1 B φ 2 B 2 φ p B p )Ż t = (1 θ 1 B θ 2 B 2 θ q B q )a t µ Z E(Z t ) = θ 0 1 φ 1 φ p An ARMA(p,q) model is stationary if all of the roots of φ p (B) lie outside of the unit circle. An ARMA(p,q) model is invertible if all of the roots of θ q (B) lie outside of the unit circle. 5-57
58 Expressing an ARMA(p,q) Model as an Infinite MA φ p (B)Ż t = θ q (B)a t Ż t = [φ p (B)] 1 θ q (B)a t = ψ(b)a t φ p (B)ψ(B) = θ q (B) Given φ p (B) and θ q (B), solve for ψ(b), giving ψ 1,ψ 2,... Similar method for expressing an ARMA(p, q) model as an infinite AR φ p (B) = π(b)θ q (B) Given φ p (B) and θ q (B), solve for π(b), giving π 1,π 2,
59 Expressing and ARMA(p,q) Model as an Infinite MA Using ARMA(2,2) as a particular example φ 2 (B)ψ(B) = θ 2 (B) (1 φ 1 B φ 2 B 2 )(1+ψ 1 B+ψ 2 B 2 +ψ 3 B 3 + ) = (1 θ 1 B θ 2 B 2 ) Multiplying out gives 1+ψ 1 B +ψ 2 B 2 +ψ 3 B 3 + φ 1 B ψ 1 φ 1 B 2 ψ 2 φ 1 B 3 φ 2 B 2 ψ 1 φ 2 B 3 = 1 θ 1 B θ 2 B 2 Equating terms with the same power of B gives B : ψ 1 φ 1 = θ 1 ψ 1 = θ 1 +φ 1 B 2 : ψ 2 ψ 1 φ 1 φ 2 = θ 2 ψ 2 = θ 2 +ψ 1 φ 1 +φ 2 B 3 : ψ 3 ψ 2 φ 1 ψ 1 φ 2 = 0 ψ 3 = ψ 2 φ 1 +ψ 1 φ 2... B k : ψ k ψ k 1 φ 1 ψ k 2 φ 2 = 0 ψ k = ψ k 1 φ 1 +ψ k 2 φ
60 Module 5 Segment 7 ARMA Identification Exercises 5-60
61 Simulated Time Series #5 Graphical Output from Function iden(simsta5.d) Simulated Time Series #5 w= Sales Range-Mean Plot w time Range Partial P Mean
62 Simulated Time Series #6 Graphical Output from Function iden Simulated Time Series #6 w= Sales Range-Mean Plot w time Range Partial P Mean
63 Simulated Time Series #7 Graphical Output from Function iden Simulated Time Series #7 w= Deviation Range-Mean Plot w time Range Partial P Mean
64 Simulated Time Series #8 Graphical Output from Function iden Simulated Time Series #8 w= decibles Range-Mean Plot w time Range Partial P Mean
65 Simulated Time Series #11 Graphical Output from Function iden Simulated Time Series #11 w= Deviation Range-Mean Plot w time Range Partial P Mean
66 Simulated Time Series #13 Graphical Output from Function iden Simulated Time Series #13 w= Pressure Range-Mean Plot w time Range Partial P Mean
67 Simulated Time Series #15 Graphical Output from Function iden Simulated Time Series #15 w= Pressure Range-Mean Plot w time Range Partial P Mean
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