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For this question I am interested in rigorous proofs or in references explaining details.

Let us suppose we have an ARIMA process of order $(p,d,q)=(p,0,q)$:

$$A(L)X_t=B(L)\epsilon_t [1],$$

where $A$ and $B$ are polynomials in the lag operator $L$ of order $p$ and $q$ respectively. $\epsilon_t$ are i.i.d. normal variables of variance $\sigma$ and zero mean.

  1. Would it be possible for the same choice of the parameters, i.e. A, B and $\sigma$, to have two different stationary distributions ?

To clarify the question, we can define stationary distribution for this question this way. Equation [1] can be considered a way to propagate the time series given some initial conditions. If propagating according to [1] we converge to a constant probability distribution than we call this a stationary distribution. In Markov chains we have a similar setting for example, but the stationary distribution can be not unique (unless some assumptions are verified), and dependent on the initial conditions. There are ARMA process withouth stationary distribution (for example in the presence of a unit root). My question is if there are ARMA process having more than one stationary distributions.

  1. Would it be possible to have a limit cyclostationary distribution ?See https://en.wikipedia.org/wiki/Cyclostationary_process for definition.

For "limit cyclostationary distribution" I mean an ARMA process that, given some initial configuration, will behave in a long run as a (wide-sense) cyclostationary process, with a mean and variance changing cyclically over time.

  1. Now let's take $d \ge 1$, so that the differentiated process $Y_t=(1-L)^d X_t$ is an $ARIMA(p,0,q)$ process. Suppose now that the $ARMA(p,q)$ process $Y_t$ has a stationary distribution. Is it true that $X_t$ will be of the form $X_t=f_t +Y_t$ where $Y_t$ is stationary and $f_t$ is deterministic and respects $(1-L)^d f_t=0$ (so that the trend must be linear, quadratic, cubic as $d$ increases)?
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    $\begingroup$ Thanks a lot for helping improve the question. I hope I edited it with a consistent notation. Do not hesitate to suggest other improvements. $\endgroup$
    – Thomas
    May 7, 2020 at 19:29
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    $\begingroup$ Please motivate -1 . Bad habit of putting -1 without explanation... This habit damages stackexchange sites. $\endgroup$
    – Thomas
    May 9, 2020 at 11:10
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    $\begingroup$ What a nice person would do is : (1) explain why the question is not well posed (considering that any question is a good question, if well posed) ; (2) wait for changes by the OP ; (3) If these changes are not performed, put a -1 . Everything else in my opinion is just unpolite . $\endgroup$
    – Thomas
    May 9, 2020 at 11:15
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    $\begingroup$ Agreed. But maybe put a -1 for a bad question and retract the vote once the question is improved? After all, it is only logical to vote on the question as is, not as it was or will/might be. (I did not downvote.) $\endgroup$ May 9, 2020 at 11:35
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    $\begingroup$ I tried to improve the question. I would expect than to see either further comments or a removal of the -1 ;) $\endgroup$
    – Thomas
    May 9, 2020 at 13:56

1 Answer 1

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  1. Question does not quite make sense. Stationary distributions are not identified by ARMA parameters of the series. For the same ARMA parameter $A$ and $B$ (with stationary solution), any i.i.d. $\epsilon_t$ with finite second moment would give a strictly stationary series. Different marginal distribution of $\epsilon_t$ would give different stationary distribution. For a trivial example, take i.i.d. white noise with different distributions.

  2. What is "limit cyclostationary distribution"...? An example of cyclostationary series is seasonal ARIMA time series---e.g. the seasonal AR(1) specification
    $$ (1- \phi L^{12}) X_t = \epsilon_t. $$ In SARIMA notation, this is an SARIMA(0,0,0)$\times$(1,0,0)$_{12}$ model. Its ACF is $\rho(12h) = \phi^h$, and zero at other lags.

  3. No, clearly this is not true---e.g. take the unit root $(1-L) X_t = \epsilon_t$. In fact, given the way you've specified ARMA series (with not intercept), there can be no deterministic trend.

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  • $\begingroup$ Thanks for your answer. Here some points. (1) I am considering $\epsilon_i$ normal i.i.d. variables with a variance $\sigma$. My question is if such a ARMA model can have different stationary distributions at fixed A, B and $\sigma$ . I think no but I just wanted to be sure if there is a proof of this. $\endgroup$
    – Thomas
    May 8, 2020 at 20:55
  • $\begingroup$ (2) The model you write is $X_t=\phi X_{t-12}+\epsilon_t$ with no intercept. So essentially there are 12 indipendent AR(1) processes. Each process has than zero mean, which is a bit of a trivial case of a cyclostationary signal, where the expectation value $E[X(t)]$ can be non-constant and oscillate (see the wikipedia page). Can such cyclostationary signals be modelled via an ARMA process ? $\endgroup$
    – Thomas
    May 8, 2020 at 20:56
  • $\begingroup$ (3) I am not getting your answer. If we have $A(L)X_t=B(L) \epsilon_t$ and $Y_t=X_t-k_1$ with $k_1$ constant, than $A(L)Y_t=A(L)X_t-k_2$ where the constant $k_2$ is $k_1$ times the number of non-zero terms of $A$. This shows that the intercept term causes a constant mean shift but not a trend. Further an ARIMA model with $d \ge 1$ has been defined exactly so that deterministic trends can be removed by the differentiation operation, or am I wrong? Therefore such models can contain deterministic trends. $\endgroup$
    – Thomas
    May 8, 2020 at 21:12
  • $\begingroup$ 1) With i.i.d. normal error term, then question is trivial---Gaussian time series are characterized by the autocovariance function. $\endgroup$
    – Michael
    May 8, 2020 at 21:40
  • $\begingroup$ 2) Covariance-stationarity implies mean stationary. In particular, ARMA processes are mean-stationary---one can always demeaning, a general wide-sense cyclostationary process. ("Cyclostationarity" is not really time series vernacular.) $\endgroup$
    – Michael
    May 8, 2020 at 21:48

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