Is the MA($\infty$) process with i.i.d. noise strictly stationary? I have a MA($\infty$) process defined by
$$ 
X_t = \sum_{k=0}^\infty \alpha_{k} \epsilon_{t-k}, \qquad t\in\mathbb{Z}
$$
where the sums converge a.s. and the $\epsilon_t$ are i.i.d. centered noise with Var($\epsilon_t$) = $\sigma^2< \infty$.
There are plenty of proofs that this process is weakly stationary in the literature.
Is this process strictly stationary?
 A: This process is always strictly stationary by definition.
Recall that the process is (strictly) stationary when all $n$-variate distributions formed by selecting any pattern $(s_1,s_2,\ldots,s_{n-1})$ of (integral) indexes are identical: that is, for all $n\ge 1$ and integral $s$ and $t,$
$$(X_s, X_{s+s_1}, \ldots, X_{s+s_{n-1}}) \sim (X_t, X_{t+s_1}, \ldots, X_{t+s_{n-1}}).$$
But that is trivially the case due to the iid assumption on the $\epsilon_t.$  One merely substitutes "$\epsilon_{s-k}$" for "$\epsilon_{t-k}$" in the definition of the process $(X_t).$
A: Strict stationarity do not imply weak essentially because is possible that the first two moments are not finite. If we add that
$V[\epsilon_t]=\sigma^2< \infty$
strictly stationarity imply weak also.
Seems me useful to note that stationarity is strongly related to ergodicity and, then, memory (this discussion can help: Stationarity and Ergodicity - links). You assume independence among $\epsilon_t$, so any memory problem depend only from the $MA$ parameters. Note that in $MA(q)$, for finite q case, parameters restrictions do not need, for infinite case absolute summability of parameters need. Moreover stationarity deal with unchangeable moments and distributional form. You assume identicity in distribution.
Keep in mind that if $\epsilon_t$ in also Gaussian, strict stationarity surely hold. However It seem me that under iid condition, that you invoke, strict stationarity is implied regardless distributional assumption. Considering that the finiteness of variance are added, also weak stationarity hold.
Your assumptions are very strong for time series. Is not a surprise that strict stationarity hold.
I add some detail in order to rend more clear what I said above. In the $MA(q)$ process we have that
$V[X_t] = \sigma^2 \displaystyle\sum_{k=0}^{q} a_k^2$
$COV(X_t,X_{t-s})= \sigma^2 \displaystyle\sum_{k=0}^{q-s} a_{k+s}a_k$;  for ($1 \leq s \leq q$)
under absolute summability of coefficients, necessary condition for stationarity (weak and/or strict), the above formulas can be used also in the $MA(\infty)$ case; both term converge to a finite quantities. Moreover if the errors $(\epsilon_t)$ are not only iid but also Normal, the distribution of $X_t$ is Normal too (variance given above and mean $0$). All the possible joint distributions $(X_t,X_{t-1},…,X_{t-s})$ are jointly Normal, with any single terms of the covariance matrix that depend of the above formulas. If we shift the joint of $j$ step, we have $(X_{t+j},X_{t-1+j},…,X_{t-s+j})$ but this distribution remain the previous. There are no reason for its modifications. The memory decays with $s$, only this term matters.
If we rule out the Gaussian assumption among $\epsilon_t$, we don’t known more the form of the distributions, marginals and joints, of $X$ too. However there is no reason because the joint distributions of $X$ have to change under shift, $iid$ assumption matters here, therefore the process remain strict stationary (considering the finiteness of $\sigma^2$ also weak stationarity hold).
As counter example we can consider the case where identically distribution among $\epsilon_t$ not hold; more in particular it change at any realization. So, even if the moments above holds yet, we cannot find two joint distribution that share exactly the same form. Therefore strict stationarity clearly not hold, however we have to note that weak stationarity hold yet. This fact can happen under standard white noise condition for $\epsilon_t$.
