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

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Outline

1 §3.4: ARMA(p,q) Model

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§3.4: ARMA(p,q) Model Homework 3a

ARMA(p

,

q) Model

Definition (ARMA(p,q) Model)

A time series is ARMA(p,q) if it is stationary and satisfies

Zt =α+φ1Zt−1+· · ·+φpZt−p | {z } AR part +θ1at−1+· · ·+θqat−q | {z } MA part +at (?)

Using the AR and MA operators, we can rewrite(?)as

φ(B) ˙Zt =θ(B)at

The textbook defines the ARMA process with a slightly different parametrization given by

Zt=α+φ1Zt−1+· · ·+φpZt−p−θ1at−1− · · · −θqat−q+at (?)

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ARMA(p

,

q) Model

Using the AR and MA operators, we can rewrite(?)as φ(B) ˙Zt =θ(B)at

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ARMA(p

,

q) Model

Using the AR and MA operators, we can rewrite(?)as φ(B) ˙Zt =θ(B)at

Zt=α+φ1Zt−1+· · ·+φpZt−p−θ1at−1− · · · −θqat−q+at (?)

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Causality and Invertibility of ARMA(p

,

q)

Facts:

The ARMA(p,q) model given byφ(B)Zt =θ(B)at iscausalif and only if

|φ(z)| 6=0 when|z| ≤1

i.e. all rootsincluding complex rootsofφ(z)lie outside the unit disk. The ARMA(p,q) model given byφ(B)Zt =θ(B)at isinvertibleif and only

if

|θ(z)| 6=0 when|θ| ≤1

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Causality and Invertibility of ARMA(p

,

q)

Facts:

|φ(z)| 6=0 when|z| ≤1

if

|θ(z)| 6=0 when|θ| ≤1

i.e. all rootsincluding complex rootsofθ(z)lie outside the unit disk.

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Causality and Invertibility of ARMA(p

,

q)

Facts:

|φ(z)| 6=0 when|z| ≤1

if

|θ(z)| 6=0 when|θ| ≤1

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Parameter Redundancy

The process Zt =at (?) is equivalent to .5Zt−1=.5at−1 (??)

which is also equivalent to(?)−(??), i.e.

Zt−.5Zt−1 =at−.5at−1

Ztin the last representation is still white noise, but has an ARMA(1,1)

representation.

We remove redundancies in an ARMA modelφ(B)Zt =θ(B)atby

canceling the common factors inθ(B)andφ(B).

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Parameter Redundancy

The process Zt =at (?) is equivalent to .5Zt−1=.5at−1 (??) which is also equivalent to(?)−(??), i.e.

representation.

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Parameter Redundancy

representation.

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Parameter Redundancy

representation.

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Parameter Redundancy

representation.

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Example – Removing Redundancy

Example

Consider the process

Zt=.4Zt−1+.45Zt−2+at+at−1+.25at−2 which is equivalent to (1−.4B−.45B2) | {z } φ(B) Zt= (1+B+.25B2) | {z } θ(B) at

Is this process really ARMA(2,2)?

No!

Factoringφ(z)andθ(z)gives φ(z) =1−.4z−.45z2= (1+.5z)(1−.9z)

θ(z) = (1+z+.25z2) = (1+.5z)2

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Example – Removing Redundancy

Example

No!

θ(z) = (1+z+.25z2) = (1+.5z)2

Removing the common term(1+.5z)gives the reduced model

Zt=.9Zt−1+.5at−1+at

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Example – Removing Redundancy

Example

No!

θ(z) = (1+z+.25z2) = (1+.5z)2

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Example – Removing Redundancy

Example

Is this process really ARMA(2,2)? No!Factoringφ(z)andθ(z)gives φ(z) =1−.4z−.45z2= (1+.5z)(1−.9z)

θ(z) = (1+z+.25z2) = (1+.5z)2

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Example – Removing Redundancy

Example

Is this process really ARMA(2,2)? No!Factoringφ(z)andθ(z)gives

φ(z) =1−.4z−.45z2= (1+.5z)(1−.9z)

θ(z) = (1+z+.25z2) = (1+.5z)2

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Example – Removing Redundancy

Example

Is this process really ARMA(2,2)? No!Factoringφ(z)andθ(z)gives

φ(z) =1−.4z−.45z2= (1+.5z)(1−.9z)

θ(z) = (1+z+.25z2) = (1+.5z)2

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Example – Causality and Invertibility

Example (cont.)

The reduced model isφ(B)Zt =θ(B)atwhere

φ(z) =1−.9z

θ(z) =1+.5z

There is only one root ofφ(z)which isz=10/9, and|10/9|lies outside the unit circle so this ARMA model iscausal.

There is only one root ofθ(z)which isz=−2, and| −2|lies outside the unit circle so this ARMA model isinvertible.

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Example – Causality and Invertibility

Example (cont.)

φ(z) =1−.9z

θ(z) =1+.5z

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Example – Causality and Invertibility

Example (cont.)

φ(z) =1−.9z

θ(z) =1+.5z

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ARMA Mathematics

Start with the ARMA equation

φ(B)Zt =θ(B)at

To solve forZt, we simply divide! That is

Zt = θ(B) φ(B) | {z } ψ(B) at “MA(∞)representation”

Similarly, solving forat gives

at= φ(B) θ(B) | {z } π(B) Zt “AR(∞)representation”

The key is to figuring out theψjandπjin

ψ(B) =1+ ∞ X j=1 ψjBj ψ(B) =1+ ∞ X j=1 ψjBj

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ARMA Mathematics

φ(B)Zt =θ(B)at

ψ(B) =1+ ∞ X ψBj ψ(B) =1+ ∞ X ψBj

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ARMA Mathematics

φ(B)Zt =θ(B)at

ψ(B) =1+ ∞ X j=1 ψjBj ψ(B) =1+ ∞ X j=1 ψjBj

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ARMA Mathematics

φ(B)Zt =θ(B)at

ψ(B) =1+ ∞ X ψBj ψ(B) =1+ ∞ X ψBj

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Example – MA(

∞

) Representation

Example (cont.) Note that ψ(z) = θ(z) φ(z) = 1+.5z 1−.9z = (1+.5z)(1+.9z+.92z2+.93z3+· · ·) for|z| ≤1 =1+ (.5+.9)z+ (.5(.9) +.92)z2+ (.5(.92) +.93)z3+· · · for|z| ≤1 =1+ (.5+.9)z+ (.5+.9)(.9)z2+ (.5+.9)(.92)z3+· · · for|z| ≤1 =1+ (.5+.9) | {z } 1.4 ∞ X j=1 .9j−1zj for|z| ≤1

This gives the following MA(∞) representation:

Zt = θ(B) φ(B)at =  1+1.4 ∞ X j=1 .9j−1Bj  at =at+1.4 ∞ X j=1 .9j−1at−j Arthur Berg AR(p) + MA(q) = ARMA(p,q) 9/ 12

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Example – MA(

∞

) Representation

Example (cont.) Note that ψ(z) = θ(z) φ(z) = 1+.5z 1−.9z = (1+.5z)(1+.9z+.92z2+.93z3+· · ·) for|z| ≤1 =1+ (.5+.9)z+ (.5(.9) +.92)z2+ (.5(.92) +.93)z3+· · · for|z| ≤1 =1+ (.5+.9)z+ (.5+.9)(.9)z2+ (.5+.9)(.92)z3+· · · for|z| ≤1 =1+ (.5+.9) | {z } 1.4 ∞ X j=1 .9j−1zj for|z| ≤1

This gives the following MA(∞) representation: θ(B)  ∞ X _j−₁ _j  ∞ X _j−₁

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Example – AR(

∞

) Representation

Example (cont.)

Similarly, one can show π(z) = φ(z) θ(z) = 1−.9z 1+.5z .. . =1−1.4 ∞ X j=1 (−.5)j−1zj for|z| ≤1

This gives the following AR(∞) representation:

Zt =1.4 ∞

X

j=1

(−.5)j−1at−j+at

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Example – AR(

∞

) Representation

Example (cont.)

Similarly, one can show π(z) = φ(z) θ(z) = 1−.9z 1+.5z .. . =1−1.4 ∞ X j=1 (−.5)j−1zj for|z| ≤1

This gives the following AR(∞) representation:

∞

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Outline

1 §3.4: ARMA(p,q) Model

2 Homework 3a

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Homework 3a

Read §4.1 and §4.2.

Derive the equation forπ(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)