Stationarity of MA infinity process

Any MA process is weakly stationary be definition. This applies also to the following $$\text{MA}(\infty)$$ infinity process:

$$X_{t} = \sum_{s=0}^{\infty} \theta^s Z_{t-s}$$, where $$Z_{t}$$ is a mean zero process with variance $$\sigma_{Z}^2$$. This process is weakly stationary and invertible for $$0.

The variance is finite given by $$\text{Var}(X_{t}) = \frac{\sigma_{Z}^2}{1-\theta^2}$$, which comes from the infinite geometric sum.

However, in real life we do not have infinite datasets, so the sum becomes a finite one, and therefore if we apply a pseudo $$\text{MA}(\infty)$$ model (pseudo in the sense that it encorporates all the available data up to $$t=0$$, meaning that it would look like $$X_{t} = \sum_{s=0}^{t} \theta^s Z_{t-s}$$), then the mean and variance would change with time, i.e. rendering the process weakly non-stationary.

Does this mean that a series containing all recorded realisations up to $$t=0$$ can never be stationary?

• The sense in which you use "process" appears to oscillate between a kind of model and a kind of dataset. This makes your question a little difficult to interpret. The vagueness of the verb "use" is particularly problematic: what is that intended to mean?
– whuber
Commented Jun 15, 2022 at 19:43
• I have made some edits to the post and hope this clarifies your questions. The verb "use" was supposed to mean that a realisation at a given time/lag also appears in the model, i.e. that the specific lag was still in the model order. Therefore, if a model "uses" all available realisations, the model is of order $MA(q=t)$, where t increases with time.
– DLTS
Commented Jun 16, 2022 at 6:17
• I still find it confusing, because (among other things) you seem to refer to the $Z_{t-s}$ as "datasets," but they are not: they are model constructs only. Even with a finite dataset (1) the model still posits an infinite sequence of Z's and (2) the data are only a partial realization of the time series process. Your question appears to rest on failing to make these distinctions.
– whuber
Commented Jun 16, 2022 at 11:06

We have: \begin{align} var(X_{t+1}) &= var(\theta X_t + Z_{t+1})\\ &= var(\theta X_t) + var(Z_{t+1})\\ &= \theta^2 var(X_t) + \sigma^2_z, \end{align} because the random variables $$Z_t$$ are all independent.
For this to be constant, i.e. for $$var(X_{t+1}) = var(X_t)$$, we would have to have: $$var(X_t) = \frac{\sigma^2_z}{1-\theta^2},$$ for all $$t$$, in particular also for $$t=0$$. But since $$X_0 = Z_0$$ has variance $$\sigma^2_z$$, this is only possible for $$\theta^2=0$$.
In summary, the only way to have $$X_t$$ be stationary is by setting $$\theta=0$$ which means $$X_t = Z_0$$ for all $$t$$.