# AR(p) + MA(q) = ARMA(p, q)

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1 AR(p) + MA(q) = ARMA(p, q)

2 Outline 1 3.4: ARMA(p, q) Model 2 Homework 3a Arthur Berg AR(p) + MA(q) = ARMA(p, q) 2/ 12

3 ARMA(p, q) Model Definition (ARMA(p, q) Model) A time series is ARMA(p, q) if it is stationary and satisfies Z t = α + φ 1 Z t φ p Z t p + θ 1 a t θ q a t q +a t ( ) } {{ } } {{ } AR part MA part Using the AR and MA operators, we can rewrite ( ) as φ(b)ż t = θ(b)a t The textbook defines the ARMA process with a slightly different parametrization given by Z t = α + φ 1 Z t φ p Z t p θ 1 a t 1 θ q a t q + a t ( ) Arthur Berg AR(p) + MA(q) = ARMA(p, q) 3/ 12

4 ARMA(p, q) Model Definition (ARMA(p, q) Model) A time series is ARMA(p, q) if it is stationary and satisfies Z t = α + φ 1 Z t φ p Z t p + θ 1 a t θ q a t q +a t ( ) } {{ } } {{ } AR part MA part Using the AR and MA operators, we can rewrite ( ) as φ(b)ż t = θ(b)a t The textbook defines the ARMA process with a slightly different parametrization given by Z t = α + φ 1 Z t φ p Z t p θ 1 a t 1 θ q a t q + a t ( ) Arthur Berg AR(p) + MA(q) = ARMA(p, q) 3/ 12

5 ARMA(p, q) Model Definition (ARMA(p, q) Model) A time series is ARMA(p, q) if it is stationary and satisfies Z t = α + φ 1 Z t φ p Z t p + θ 1 a t θ q a t q +a t ( ) } {{ } } {{ } AR part MA part Using the AR and MA operators, we can rewrite ( ) as φ(b)ż t = θ(b)a t The textbook defines the ARMA process with a slightly different parametrization given by Z t = α + φ 1 Z t φ p Z t p θ 1 a t 1 θ q a t q + a t ( ) Arthur Berg AR(p) + MA(q) = ARMA(p, q) 3/ 12

6 Causality and Invertibility of ARMA(p, q) Facts: The ARMA(p, q) model given by φ(b)z t = θ(b)a t is causal if and only if φ(z) 0 when z 1 i.e. all roots including complex roots of φ(z) lie outside the unit disk. The ARMA(p, q) model given by φ(b)z t = θ(b)a t is invertible if and only if θ(z) 0 when θ 1 i.e. all roots including complex roots of θ(z) lie outside the unit disk. Arthur Berg AR(p) + MA(q) = ARMA(p, q) 4/ 12

7 Causality and Invertibility of ARMA(p, q) Facts: The ARMA(p, q) model given by φ(b)z t = θ(b)a t is causal if and only if φ(z) 0 when z 1 i.e. all roots including complex roots of φ(z) lie outside the unit disk. The ARMA(p, q) model given by φ(b)z t = θ(b)a t is invertible if and only if θ(z) 0 when θ 1 i.e. all roots including complex roots of θ(z) lie outside the unit disk. Arthur Berg AR(p) + MA(q) = ARMA(p, q) 4/ 12

8 Causality and Invertibility of ARMA(p, q) Facts: The ARMA(p, q) model given by φ(b)z t = θ(b)a t is causal if and only if φ(z) 0 when z 1 i.e. all roots including complex roots of φ(z) lie outside the unit disk. The ARMA(p, q) model given by φ(b)z t = θ(b)a t is invertible if and only if θ(z) 0 when θ 1 i.e. all roots including complex roots of θ(z) lie outside the unit disk. Arthur Berg AR(p) + MA(q) = ARMA(p, q) 4/ 12

9 Parameter Redundancy The process Z t = a t ( ) is equivalent to.5z t 1 =.5a t 1 ( ) which is also equivalent to ( ) ( ), i.e. Z t.5z t 1 = a t.5a t 1 Z t in the last representation is still white noise, but has an ARMA(1,1) representation. We remove redundancies in an ARMA model φ(b)z t = θ(b)a t by canceling the common factors in θ(b) and φ(b). Arthur Berg AR(p) + MA(q) = ARMA(p, q) 5/ 12

10 Parameter Redundancy The process Z t = a t ( ) is equivalent to.5z t 1 =.5a t 1 ( ) which is also equivalent to ( ) ( ), i.e. Z t.5z t 1 = a t.5a t 1 Z t in the last representation is still white noise, but has an ARMA(1,1) representation. We remove redundancies in an ARMA model φ(b)z t = θ(b)a t by canceling the common factors in θ(b) and φ(b). Arthur Berg AR(p) + MA(q) = ARMA(p, q) 5/ 12

11 Parameter Redundancy The process Z t = a t ( ) is equivalent to.5z t 1 =.5a t 1 ( ) which is also equivalent to ( ) ( ), i.e. Z t.5z t 1 = a t.5a t 1 Z t in the last representation is still white noise, but has an ARMA(1,1) representation. We remove redundancies in an ARMA model φ(b)z t = θ(b)a t by canceling the common factors in θ(b) and φ(b). Arthur Berg AR(p) + MA(q) = ARMA(p, q) 5/ 12

12 Parameter Redundancy The process Z t = a t ( ) is equivalent to.5z t 1 =.5a t 1 ( ) which is also equivalent to ( ) ( ), i.e. Z t.5z t 1 = a t.5a t 1 Z t in the last representation is still white noise, but has an ARMA(1,1) representation. We remove redundancies in an ARMA model φ(b)z t = θ(b)a t by canceling the common factors in θ(b) and φ(b). Arthur Berg AR(p) + MA(q) = ARMA(p, q) 5/ 12

13 Parameter Redundancy The process Z t = a t ( ) is equivalent to.5z t 1 =.5a t 1 ( ) which is also equivalent to ( ) ( ), i.e. Z t.5z t 1 = a t.5a t 1 Z t in the last representation is still white noise, but has an ARMA(1,1) representation. We remove redundancies in an ARMA model φ(b)z t = θ(b)a t by canceling the common factors in θ(b) and φ(b). Arthur Berg AR(p) + MA(q) = ARMA(p, q) 5/ 12

14 Example Removing Redundancy Example Consider the process which is equivalent to Z t =.4Z t Z t 2 + a t + a t a t 2 (1.4B.45B 2 ) Z } {{ } t = (1 + B +.25B 2 ) } {{ } φ(b) θ(b) a t Is this process really ARMA(2,2)? Factoring φ(z) and θ(z) gives φ(z) = 1.4z.45z 2 = (1 +.5z)(1.9z) θ(z) = (1 + z +.25z 2 ) = (1 +.5z) 2 Removing the common term (1 +.5z) gives the reduced model Z t =.9Z t 1 +.5a t 1 + a t Arthur Berg AR(p) + MA(q) = ARMA(p, q) 6/ 12

15 Example Removing Redundancy Example Consider the process which is equivalent to Z t =.4Z t Z t 2 + a t + a t a t 2 (1.4B.45B 2 ) Z } {{ } t = (1 + B +.25B 2 ) } {{ } φ(b) θ(b) a t Is this process really ARMA(2,2)? Factoring φ(z) and θ(z) gives φ(z) = 1.4z.45z 2 = (1 +.5z)(1.9z) θ(z) = (1 + z +.25z 2 ) = (1 +.5z) 2 Removing the common term (1 +.5z) gives the reduced model Z t =.9Z t 1 +.5a t 1 + a t Arthur Berg AR(p) + MA(q) = ARMA(p, q) 6/ 12

16 Example Removing Redundancy Example Consider the process which is equivalent to Z t =.4Z t Z t 2 + a t + a t a t 2 (1.4B.45B 2 ) Z } {{ } t = (1 + B +.25B 2 ) } {{ } φ(b) θ(b) a t Is this process really ARMA(2,2)? Factoring φ(z) and θ(z) gives φ(z) = 1.4z.45z 2 = (1 +.5z)(1.9z) θ(z) = (1 + z +.25z 2 ) = (1 +.5z) 2 Removing the common term (1 +.5z) gives the reduced model Z t =.9Z t 1 +.5a t 1 + a t Arthur Berg AR(p) + MA(q) = ARMA(p, q) 6/ 12

17 Example Removing Redundancy Example Consider the process which is equivalent to Z t =.4Z t Z t 2 + a t + a t a t 2 (1.4B.45B 2 ) Z } {{ } t = (1 + B +.25B 2 ) } {{ } φ(b) θ(b) Is this process really ARMA(2,2)? No! Factoring φ(z) and θ(z) gives φ(z) = 1.4z.45z 2 = (1 +.5z)(1.9z) θ(z) = (1 + z +.25z 2 ) = (1 +.5z) 2 Removing the common term (1 +.5z) gives the reduced model Z t =.9Z t 1 +.5a t 1 + a t a t Arthur Berg AR(p) + MA(q) = ARMA(p, q) 6/ 12

18 Example Removing Redundancy Example Consider the process which is equivalent to Z t =.4Z t Z t 2 + a t + a t a t 2 (1.4B.45B 2 ) Z } {{ } t = (1 + B +.25B 2 ) } {{ } φ(b) θ(b) Is this process really ARMA(2,2)? No! Factoring φ(z) and θ(z) gives φ(z) = 1.4z.45z 2 = (1 +.5z)(1.9z) θ(z) = (1 + z +.25z 2 ) = (1 +.5z) 2 Removing the common term (1 +.5z) gives the reduced model Z t =.9Z t 1 +.5a t 1 + a t a t Arthur Berg AR(p) + MA(q) = ARMA(p, q) 6/ 12

19 Example Removing Redundancy Example Consider the process which is equivalent to Z t =.4Z t Z t 2 + a t + a t a t 2 (1.4B.45B 2 ) Z } {{ } t = (1 + B +.25B 2 ) } {{ } φ(b) θ(b) Is this process really ARMA(2,2)? No! Factoring φ(z) and θ(z) gives φ(z) = 1.4z.45z 2 = (1 +.5z)(1.9z) θ(z) = (1 + z +.25z 2 ) = (1 +.5z) 2 Removing the common term (1 +.5z) gives the reduced model Z t =.9Z t 1 +.5a t 1 + a t a t Arthur Berg AR(p) + MA(q) = ARMA(p, q) 6/ 12

20 Example Causality and Invertibility Example (cont.) The reduced model is φ(b)z t = θ(b)a t where φ(z) = 1.9z θ(z) = 1 +.5z There is only one root of φ(z) which is z = 10/9, and 10/9 lies outside the unit circle so this ARMA model is causal. There is only one root of θ(z) which is z = 2, and 2 lies outside the unit circle so this ARMA model is invertible. Arthur Berg AR(p) + MA(q) = ARMA(p, q) 7/ 12

21 Example Causality and Invertibility Example (cont.) The reduced model is φ(b)z t = θ(b)a t where φ(z) = 1.9z θ(z) = 1 +.5z There is only one root of φ(z) which is z = 10/9, and 10/9 lies outside the unit circle so this ARMA model is causal. There is only one root of θ(z) which is z = 2, and 2 lies outside the unit circle so this ARMA model is invertible. Arthur Berg AR(p) + MA(q) = ARMA(p, q) 7/ 12

22 Example Causality and Invertibility Example (cont.) The reduced model is φ(b)z t = θ(b)a t where φ(z) = 1.9z θ(z) = 1 +.5z There is only one root of φ(z) which is z = 10/9, and 10/9 lies outside the unit circle so this ARMA model is causal. There is only one root of θ(z) which is z = 2, and 2 lies outside the unit circle so this ARMA model is invertible. Arthur Berg AR(p) + MA(q) = ARMA(p, q) 7/ 12

23 ARMA Mathematics Start with the ARMA equation φ(b)z t = θ(b)a t To solve for Z t, we simply divide! That is Z t = θ(b) φ(b) } {{ } ψ(b) Similarly, solving for a t gives a t = φ(b) θ(b) } {{ } π(b) The key is to figuring out the ψ j and π j in ψ(b) = 1 + ψ j B j ψ(b) = 1 + ψ j B j a t Z t MA( ) representation AR( ) representation Arthur Berg AR(p) + MA(q) = ARMA(p, q) 8/ 12

24 ARMA Mathematics Start with the ARMA equation φ(b)z t = θ(b)a t To solve for Z t, we simply divide! That is Z t = θ(b) φ(b) } {{ } ψ(b) Similarly, solving for a t gives a t = φ(b) θ(b) } {{ } π(b) The key is to figuring out the ψ j and π j in ψ(b) = 1 + ψ j B j ψ(b) = 1 + ψ j B j a t Z t MA( ) representation AR( ) representation Arthur Berg AR(p) + MA(q) = ARMA(p, q) 8/ 12

25 ARMA Mathematics Start with the ARMA equation φ(b)z t = θ(b)a t To solve for Z t, we simply divide! That is Z t = θ(b) φ(b) } {{ } ψ(b) Similarly, solving for a t gives a t = φ(b) θ(b) } {{ } π(b) The key is to figuring out the ψ j and π j in ψ(b) = 1 + ψ j B j ψ(b) = 1 + ψ j B j a t Z t MA( ) representation AR( ) representation Arthur Berg AR(p) + MA(q) = ARMA(p, q) 8/ 12

26 ARMA Mathematics Start with the ARMA equation φ(b)z t = θ(b)a t To solve for Z t, we simply divide! That is Z t = θ(b) φ(b) } {{ } ψ(b) Similarly, solving for a t gives a t = φ(b) θ(b) } {{ } π(b) The key is to figuring out the ψ j and π j in ψ(b) = 1 + ψ j B j ψ(b) = 1 + ψ j B j a t Z t MA( ) representation AR( ) representation Arthur Berg AR(p) + MA(q) = ARMA(p, q) 8/ 12

27 Example MA( ) Representation Example (cont.) Note that ψ(z) = θ(z) φ(z) = 1 +.5z 1.9z = (1 +.5z)(1 +.9z z z 3 + ) for z 1 = 1 + (.5 +.9)z + (.5(.9) )z 2 + (.5(.9 2 ) )z 3 + for z 1 = 1 + (.5 +.9)z + (.5 +.9)(.9)z 2 + (.5 +.9)(.9 2 )z 3 + for z 1 = 1 + (.5 +.9).9 j 1 z j for z 1 } {{ } 1.4 This gives the following MA( ) representation: Z t = θ(b) φ(b) a t = j 1 B j a t = a t j 1 a t j Arthur Berg AR(p) + MA(q) = ARMA(p, q) 9/ 12

28 Example MA( ) Representation Example (cont.) Note that ψ(z) = θ(z) φ(z) = 1 +.5z 1.9z = (1 +.5z)(1 +.9z z z 3 + ) for z 1 = 1 + (.5 +.9)z + (.5(.9) )z 2 + (.5(.9 2 ) )z 3 + for z 1 = 1 + (.5 +.9)z + (.5 +.9)(.9)z 2 + (.5 +.9)(.9 2 )z 3 + for z 1 = 1 + (.5 +.9).9 j 1 z j for z 1 } {{ } 1.4 This gives the following MA( ) representation: Z t = θ(b) φ(b) a t = j 1 B j a t = a t j 1 a t j Arthur Berg AR(p) + MA(q) = ARMA(p, q) 9/ 12

29 Example AR( ) Representation Example (cont.) Similarly, one can show π(z) = φ(z) θ(z). = 1.9z 1 +.5z = (.5) j 1 z j for z 1 This gives the following AR( ) representation: Z t = 1.4 (.5) j 1 a t j + a t Arthur Berg AR(p) + MA(q) = ARMA(p, q) 10/ 12

30 Example AR( ) Representation Example (cont.) Similarly, one can show π(z) = φ(z) θ(z). = 1.9z 1 +.5z = (.5) j 1 z j for z 1 This gives the following AR( ) representation: Z t = 1.4 (.5) j 1 a t j + a t Arthur Berg AR(p) + MA(q) = ARMA(p, q) 10/ 12

31 Outline 1 3.4: ARMA(p, q) Model 2 Homework 3a Arthur Berg AR(p) + MA(q) = ARMA(p, q) 11/ 12

32 Homework 3a Read 4.1 and 4.2. Derive the equation for π(z) on slide 10. Do exercise #3.14(a) only for models (i), (ii), and (iv) Do exercise #3.14(b,c) only for model (ii) Arthur Berg AR(p) + MA(q) = ARMA(p, q) 12/ 12

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