8 Mixed Arma Models

The MA and AR models can be combined to obtain an ARMA model. A mixed ARMA with p AR terms and q MA terms is an ARMA process of order \((p,q)\) \[X_t=\alpha _1 X_{t-1}+\alpha _2 X_{t-2}+\cdots \cdots +\alpha _{p} X_{t-p}+Z_t+\beta _1 Z_{t-1}+\cdots \cdots +\beta _q Z_{t-q}\] or
\(\phi (B) X_t=\theta (B) Z_t \) where \(\phi (B)\) and \(\theta (B)\) are polynomials of order p and q respectively and \[\phi (B)= 1-\alpha _1 B-\alpha _2 B^2- ------\alpha _p B^p\] \[\theta (B)=1+\beta _1 B+\beta _2B^2 +.........+\beta _q B^q\] For AR the values of \(\{\alpha _i \}\) which makes the process stationary are such that the roots of \(\phi (B) =0\) lie outside the unit circle.
Similarly for the MA process, the values of \(\{\beta _i \}\) which make the process stationary are such that the roots of \(\theta (B) =0\) lie outside the unit circle.
ARMA process are important in that a stationary time series may be described by an ARMA model involving few parameters than the pure MA or AR by itself.
It is useful to express ARMA as pure MA process in the form \(X_t=\psi (B) Z_t\), where \(\psi (B) =\sum \psi _i \beta ^i\). \(\psi (B)\) is an MA operator and maybe of finite or infinite order.

From \(\phi (B) X_t=\theta (B) Z_t\) we have \(\psi (B) =\dfrac {\theta (B)}{\phi (B)}\) and \(\pi (B) X_t=Z_t\) where \(\pi (B) =\dfrac {\phi (B)}{\theta (B)}\)

That \(\pi (B) =1- \sum \pi _i B^i \), since \(X_t=\sum \limits ^{\infty }_{i=0} \pi _i X_{t-i}+Z_t\)

Thus from \(X_t=\psi (B) Z_t\) and \(\pi (B) X_t=Z_t\), we have \(\psi (B)\,\cdot \,\pi (B) =1\) \[\implies \hspace {1cm}\frac {\theta (B)}{\phi (B)}\,\cdot \,\frac {\phi (B)}{\theta (B)}=1\] the \(\psi \) weight and \(\pi \) weight maybe computed by equating the process of B in the equation \(\psi (B) \phi (B) =\theta (B)\).

\(\phi (B)X_t=(1-0.5B)X_t=X_t-0.5X_t\)

\(\theta (B) =1-0.3............(2)\)

\(\phi (B) =0\) and \(\theta (B) =0\) have roots outside the unit circle, so the process is stationary and invertible.

(1) \(\phi (B) =1-0.5B=0\)

\(\implies B=\dfrac {1}{0.5}=2\)

\(\psi (B)=\dfrac {\theta (B)}{\phi (B)}=\dfrac {1-0.3B}{1-0.5B}\)

\(\psi (B) =(1-0.3B)(1-0.5B)^{-1}\)

\(\psi (B)=(1-0.3B)(1+0.5B+ 0.5^2B^2 +0.5^3B^3+\cdots \cdots )\)

\(\implies \psi (B)=1+0.2B+0.1B^2+0.005B^3+\cdots \cdots \)

Hence \(\psi _i = 0.2\times 0.5^{i-1}\) for \(i=1,2,3\cdots \cdots \cdots \)

Similarly show that(2) \(\pi _i =0.2\times 0.3^{i-1}\) for \(i= 1,2,\cdots \cdots \)

Questions on this section

Stuck on something here? Ask below and it stays attached to this topic.