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.

  • $\begingroup$ 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? $\endgroup$ Nov 13, 2018 at 19:24

1 Answer 1


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.


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