Stationarity of MA infinity process

Question

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!

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 $t$.