# Showing a Time Series is Regular

Note: this is not an assessed question, it is a practice question from a past exam.

If we are given the following time series: $X_t=\alpha^2X_{t-1}+Z_t-\alpha Z_{t-1}+2\alpha^2$

I am asked to find the values for $\alpha$ for which this is regular (i.e. the MA part is invertible and the AR part is causal, moreover the lack of a common root means that the process may not be simplified). Is what I have done correct?

I write the time series in the form $\phi(B)X_t=\theta(B)Z_t$ with $\phi(B)=(1-\alpha^2)$ and $\theta(B)=(1-\alpha B- \alpha B^2)$, where $B$ is the backshift operator.

We can then see that this is an $ARMA(1,2)$ model, and for an ARMA model we know that it is regular if the roots of both $\phi(B)=0$ and $\theta(B)=0$ lie outside the unit circle and they share no common roots.

Now the roots of $\phi(B)=0$ are $B=\frac{1}{\alpha^2}$ so this lies outside the unit circle if $|\alpha| < 1$

The roots of $\theta(B)$ are given by $\frac{-1}{\alpha},\frac{-1}{2\alpha}$ and these lie outside the unit circle is $|\alpha|< 1$ and $|\alpha |<\frac{1}{2}|$

So in order for this process to be regular we require that $|\alpha| < \frac{1}{2}$, where we note that the roots will never be shared in this range.

Is the above correct/ the correct approach?

• By "regular" do you mean "stationary"? – Scortchi - Reinstate Monica May 4 '15 at 12:23
• @Scortchi My definition of regular is that $\phi(B)=0$ and $\theta(B)=0$ have roots outside the unit circle and do not share a common root. That is the MA part is invertible and the AR part is casual, moreover the lack of a common root mean that the process may not be simplified. Does this help? – hmmmm May 4 '15 at 12:28
• Yes - I think "regular" is more often used of the sampling interval, to contrast with irregular. – Scortchi - Reinstate Monica May 4 '15 at 12:37
• how do you get $\alpha B^2$ in corresponding equation? otherwise your solution is correct. – Hemant Rupani May 4 '15 at 16:09