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 $0<s<1$.

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?

Thank you for your help!

  • $\begingroup$ 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? $\endgroup$
    – whuber
    Commented Jun 15, 2022 at 19:43
  • $\begingroup$ 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. $\endgroup$
    – DLTS
    Commented Jun 16, 2022 at 6:17
  • $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Jun 16, 2022 at 11:06

1 Answer 1


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 $t$.


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