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Let's assume that we have process $X_t$ that follows $\phi(B)X_t=\theta(B)Z_t$, where $Z_t$ is a white noise process, and $\deg(\phi)=p$, $\deg(\theta)=q$.

Assuming that $\phi$ has no unit roots, we can state that $X_t=\frac{\theta(B)}{\phi(B)}Z_t$ is the unique stationary solution to the above equation.

However, if we don't restrict ourselves to stationarity, then we can find infinite solutions, depending on how we initialize the first $p$ values of $X_t$, in a deterministic way. This is precisely what I don't get, in an intuitive way ...

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  • $\begingroup$ Could you give some background? You say $X_t=\frac{\theta(B)}{\phi(B)}Z_t$ is a solution; but with respect to what is it a solution? What is treated as given and what are the variables? $\endgroup$
    – Richard Hardy
    Commented Dec 4, 2016 at 9:55
  • $\begingroup$ @RichardHardy I'm saying that $X_t$ is a solution, when $\phi, \theta$ polynomials, and white noise process $Z_t$ are given. $\endgroup$
    – An old man in the sea.
    Commented Dec 4, 2016 at 14:28
  • $\begingroup$ @RichardHardy In van der vaart time series notes (freely available on the net), page 130, in the section 8.7 on ARMA stability. He writes on this, but I don't understand exactly what he means... $\endgroup$
    – An old man in the sea.
    Commented Dec 4, 2016 at 15:42
  • $\begingroup$ Interesting. I would think that for given $\theta(B)$, $\phi(B)$ and $Z_t$, $\frac{\theta(B)}{\phi(B)}Z_t$ generates a unique $X_t$. Probably $Z_t$ is not fixed and can be any white noise process? $\endgroup$
    – Richard Hardy
    Commented Dec 4, 2016 at 15:57
  • $\begingroup$ @RichardHardy that's the thing. For a given $X_t$, we can find different polynomials and a different white noise, from the original equation, such that $X_t$ is still a solution. Read the notes from staff.fnwi.uva.nl/p.j.c.spreij/onderwijs/master/… If you manage to understand that section, then please help me. ;) thanks $\endgroup$
    – An old man in the sea.
    Commented Dec 4, 2016 at 16:29

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There is an answer to a similar question here discussing the relationship between stationarity and the recursive equation for a model. Some additional remarks are pertinent to your case.

Firstly, it is important to understand that the recursive equation defining the ARMA($p,q$) process is not sufficient to fully specify the process, even with specification of the distribution of the underlying noise series $\boldsymbol{Z} = \{ Z_t| t \in \mathbb{Z} \}$. The recursive equations lock in the auto-correlation of the process, and specification of the distribution of the underlying noise series $\boldsymbol{Z}$ is then sufficient to give an asymptotic marginal distribution for the observable values. However, the full joint distribution for the observable series $\boldsymbol{X} = \{ X_t| t \in \mathbb{Z} \}$ also depends on the distribution of the "starting values". In some cases this is specifically defined, and in cases where it is not, it is usual to take this to mean that you are using the unique stationary distribution.

ARMA($p,q$) model with white noise: You have specified an underlying white noise series so you have distribution $Z_t \sim \text{IID N}(0, \sigma^2)$ for the underlying series. Assuming that $\max| \phi_i | <1$ you have asymptotic stationarity with limiting distribution:

$$X_\infty \sim \text{N} \Bigg( 0, \sigma^2 \sum_{i=0}^\infty \psi_i^2 \Bigg) \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \psi(B) \equiv \frac{\phi(B)}{\theta(B)}.$$

To get an ARMA process with strict stationarity, you would set $X_1, \cdots , X_p$ to have a joint normal distribution with zero mean, the above variance, and auto-correlation that is consistent with your specified ARMA model. This would give you strict stationarity, in the sense that the joint distribution of observable values does not change when shifted in time.

If you were to use a different joint distribution for your "starting values", this would lead to a non-stationary model. It would still be asymptotically stationary with the above limiting distribution, but the condition of strict stationarity would not hold.

Intuitive explanation: The recursive formula for an ARMA model specifies the relationship that each observable value has with previous values. There is an infinite class of possible models that are consistent with this recursive specification (each represented by a full joint distribution for which that recursive equation holds). For a model with auto-regressive coefficients inside the unit circle, there is a unique strictly stationary model in this class, but there are also an infinite number of other models that are non-stationary, but still have an asymptotic limiting distribution. The basic thing to remember for the intuitive explanation is that the ARMA equation, which is merely a recursive equation does not fully specify the model.

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  • $\begingroup$ Ben, thanks for the answer. Just a doubt. If I use the above variance of the limiting distribution for the initial values, how can I use the covariance function for the ARMA process for those same values? $\endgroup$
    – An old man in the sea.
    Commented Mar 13, 2018 at 8:44
  • $\begingroup$ You should have $\mathbb{C}(X_t, X_{t+k}) = \sigma^2 \sum_{i=0}^\infty \psi_i \psi_{i+k}$ as the general form for the covariance values. So any vector of observable values will have a joint normal distribution with zero mean and variance matrix formed using that covariance expression. $\endgroup$
    – Ben
    Commented Mar 13, 2018 at 10:57

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