1
$\begingroup$

Say we have the time series $z_t=z_{t-2}/2+a_t$ where $a_t$ is white noise.

then we have $(1+B^2/4)z_t=a_t$, where B is the backward shift operator.

We can solve for the roots of $(1+B^2/4)$ and obtain $B=2i$ or $B=-2i$

I'm interested in the stationarity and causality of the time series, so I would like to know where these roots lie on the unit circle.

Is it as simple as plotting B on the imaginary axis? In which case, both roots would lie outside the unit circle.

$\endgroup$
  • $\begingroup$ I believe you made a mistake in your question in writing out the lag operator. Shouldn't it be (1-B^2/2)*zt=at instead? $\endgroup$ – ColorStatistics Nov 13 '18 at 19:24
6
$\begingroup$

Both $2 i$ and $-2 i$ are outside the unit circle. A pure imaginary number $b i$ is outside the unit circle if $|b| > 1$.

More generally, a complex number, $a + b i$ is outside the unit circle if its magnitude is greater than $1$, i.e., $\sqrt{a^2 + b^2} > 1$. A point is inside, on, or outside the unit circle, if its magnitude is $< 1$, $= 1$, or $> 1$ respectively.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.