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I'm sorry if this is a duplicate, but I can't seem to find the answer to this.

If $Z_t$ is a white noise process and $X_t$ satisfies

$$ \phi(B) X_t = \theta(B) Z_t $$

(where $B$ is the lag operator), then when is $X_t$ stationary?

If $\theta(z)$ and $\phi(z)$ have no common roots and $\phi(z)$ has a root with $|z| = 1 $ then $X_t$ is necessarily non-stationary [Brockwell and Davis, Remark 3 in Chapter 3].

I also know that a moving average of a stationary process with absolutely summable coefficients is also stationary [Brockwell an Davis, Proposition 3.1.2]. So if $\phi(z)$ has no roots in with $|z| \leq 1$ then the Maclaurin series for $1/\phi(z)$ has radius of convergence greater than 1. Therefore $X_t$ is a moving average of $\theta(B) Z_t$ with absolutely summable coefficients, and hence stationary.

By stationarity, I mean weak (=second order) stationarity.

The lecturer has a habit of putting this on the final, so I'd like to know if there's a general theorem governing it.

From a past paper:

enter image description here

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You are almost correct. The condition for stationarity is that $\phi(z)$ should not to have any roots on $|z|=1$. (There is also a requirement that $\phi(z)$ and $\theta(z)$ should not have any common roots, but it deals not with the stationarity but with the uniqueness of the definition of ARIMA process.)

The reasoning is the same, i.e. convergence of Laurent (you've confused Maclaurin with Laurent) series. If $\phi(z)$ does not have any roots on $|z|=1$, this means that there is an annulus $r<|z|<R$ with $r<1<R$ on which $\theta(z)/\phi(z)$ has no roots and is analytic, hence it has a Laurent series representation:

$$\theta(z)/\phi(z)=\sum_{j=-\infty}^{\infty}\psi_jz^j$$

and since it is analytic on annulus which includes $|z|=1$ the series $\sum_{j}\psi_j$ is absolutely convergent, which makes $\sum_j\psi_jZ_{t-j}$ convergent.

Note that the summation can have negative values of $j$, hence the resulting ARMA solution $X_t$ depends on future values of $Z_t$, which is not always desirable. The stationary solution $X_t$ which does not depend on future values of $Z_t$ is called causal.

The necessary condition for stationary solution $X_t$ to be causal is that $\phi(z)$ is nonzero for $|z|\le 1$.

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    $\begingroup$ Thank you @mpiktas, that was very clear. I know it's not part of the original question, but could you add the uniqueness condition for completeness? $\endgroup$ – user80379 Dec 17 '15 at 14:16
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    $\begingroup$ What do you mean add the condition? If $\phi(z)$ and $\theta(z)$ do not have common roots, the ARIMA process is defined uniquely, that is it. $\endgroup$ – mpiktas Dec 17 '15 at 14:28
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    $\begingroup$ I have seen in text books that for stationary $\phi(z)$ does not have any roots on $|z|=1$. Why it can't be $\phi(z)$ does not have any roots on $|z| \leq1$ $\endgroup$ – shani Aug 5 '18 at 11:42

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