AR part MA part - Pennsylvania State University · §3.4: ARMA(p;q) Model Homework 3a Example –...

AR(p) + MA(q) = ARMA(p, q)§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︸︷︷︸AR part

+ θ1at−1 + · · ·+ θqat−q︸︷︷︸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 differentparametrization given by

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

Arthur Berg AR(p) + MA(q) = ARMA(p, q) 3/ 12

§3.4: ARMA(p, q) Model Homework 3a

Causality and Invertibility of ARMA(p, q)

Facts:The ARMA(p,q) model given by φ(B)Zt = θ(B)at is causal if andonly if

|φ(z)| 6= 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)Zt = θ(B)at is invertible if andonly if

|θ(z)| 6= 0 when |θ| ≤ 1

i.e. all roots including complex roots of θ(z) lie outside the unit disk.



Parameter Redundancy

The processZt = at (?)

is equivalent to.5Zt−1 = .5at−1 (??)

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

Zt − .5Zt−1 = at − .5at−1

Zt in the last representation is still white noise, but has an ARMA(1,1)representation.

We remove redundancies in an ARMA model φ(B)Zt = θ(B)at bycanceling the common factors in θ(B) and φ(B).



Example – Removing Redundancy

ExampleConsider the process

Zt = .4Zt−1 + .45Zt−2 + at + at−1 + .25at−2

which is equivalent to

(1− .4B − .45B2)︸︷︷︸φ(B)

Zt = (1 + B + .25B2)︸︷︷︸θ(B)

at

Is this process really ARMA(2,2)? No! Factoring φ(z) and θ(z) gives

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

θ(z) = (1 + z + .25z2) = (1 + .5z)2

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

Zt = .9Zt−1 + .5at−1 + at



Example – Causality and Invertibility

Example (cont.)

The reduced model is φ(B)Zt = θ(B)at where

φ(z) = 1− .9zθ(z) = 1 + .5z

There is only one root of φ(z) which is z = 10/9, and |10/9| liesoutside the unit circle so this ARMA model is causal.There is only one root of θ(z) which is z = −2, and | − 2| lies outsidethe unit circle so this ARMA model is invertible.



ARMA Mathematics

Start with the ARMA equation

φ(B)Zt = θ(B)at

To solve for Zt , we simply divide! That is

Zt =θ(B)

φ(B)︸︷︷︸ψ(B)

at “MA(∞) representation”

Similarly, solving for at gives

at =φ(B)

θ(B)︸︷︷︸π(B)

Zt “AR(∞) representation”

The key is to figuring out the ψj and πj in

ψ(B) = 1 +∞∑

j=1

ψjBj ψ(B) = 1 +∞∑

j=1

ψjBj



Example – MA(∞) RepresentationExample (cont.)Note that

ψ(z) =θ(z)

φ(z)=

1 + .5z1− .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)︸︷︷︸1.4

∞∑j=1

.9j−1z j for |z| ≤ 1

This gives the following MA(∞) representation:

Zt =θ(B)

φ(B)at =

1 + 1.4∞∑

j=1

.9j−1Bj

at = at + 1.4∞∑

j=1

.9j−1at−j



Example – AR(∞) Representation

Example (cont.)Similarly, one can show

π(z) =φ(z)

θ(z)

=1− .9z1 + .5z

...

= 1− 1.4∞∑

j=1

(−.5)j−1z j for |z| ≤ 1

This gives the following AR(∞) representation:

Zt = 1.4∞∑

j=1

(−.5)j−1at−j + at



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)


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