3
$\begingroup$

Let $(a_j)_{j \in \mathbb{N}_0}$ be a real-valued sequence such that $\sum_{j = 0}^\infty a_j^2 < \infty$. Further, define an infinite moving average time series $X = \{ X(t), t \in \mathbb{Z}\}$ via $$ X(t) = \sum_{j = 0}^\infty a_j \varepsilon_{t - j}, \quad t \in \mathbb{Z}, $$ where $\varepsilon_j$ are zero-mean, finite variance, uncorrelated random variables. These latter random variables are often referred to as innovations. In this setting, the innovations are not necessarily independent or identically distributed.

Is $X$ strictly stationary (i.e. all finite-dimensional distributions are translation-invariant) if and only if the innovations are strictly stationary?

The direction "stationarity of the innovations implies stationarity of $X$" should follow from the continuous mapping theorem. But I can't figure out how to prove the other direction.

EDIT: I suspect that this may be related to whether the moving average is invertible. I believe that the MA is invertible if the corresponding polynomial $$ \theta(z) = \sum_{j = 0}^\infty a_j z^j $$ fulfills $\theta(z) \neq 0$ for all $\vert z \vert \leq 1$ (I think this follows from a theorem in Brockwell & Davis). But I am not quite sure how the non-zero condition of the polynomial relates to the square-summability of the coefficients $a_j$.

$\endgroup$
2
  • $\begingroup$ have you looked at the standard material in, say, Brockwell & Davis on expressing MA series as infinite ARs and vice versa. The issue is where the zeros of a fourier series made out of the coefficients lie. $\endgroup$
    – mike
    Commented Aug 11, 2022 at 13:32
  • $\begingroup$ Yeah, I just noticed today that if the zeroes are outside of the unit circle, then the MA is invertible. Then, the innovations could be expressed as a MA of the X's. But in my particular setting the only information I have is that the coefficients are square-summable. Thus, I don't whether the MA is invertible. $\endgroup$
    – AlbertRapp
    Commented Aug 11, 2022 at 19:08

0

You must log in to answer this question.

Browse other questions tagged .