Strict stationarity of a MA(1) process From the definition we know that a stochastic process $(X_t)_{t\in\mathbb{Z}}$ is called strictly stationary if for all $h\in \mathbb{N}$ the distribution of $(X_t,X_{t+1},\ldots,X_{t+h})$ is independent of $t$. 
Now consider that $X_t$ is a MA(1) process so we can actually write $X_t = Z_t + \theta Z_{t-1}$ for some other doubly infinite sequence $Z_t$ in other words $t\in\mathbb{Z}$.
It should be fairly clear that if $Z_t$ is strictly stationary then also is $X_t$ but I want to make it explicit. So if we write out:
$$
(X_t,\ldots,X_{t+h}) = (Z_t + \theta Z_{t-1},\ldots, Z_{t+h} + \theta Z_{t+h-1}).
$$
My question is: How is this in distribution independent of $t$? I know that independence of $t-1$ should imply independence of $t$ as we have $Z_t + \theta Z_{t-1} \sim Z_{t+1} + \theta Z_{t}$ for all $t\in\mathbb{Z}$. But how can we see that? Should the vector the splitted into $(Z_t,\ldots,Z_{t+h}) + \theta(Z_{t-1},\ldots,Z_{t+h-1})$? Any help is greatly appreciated.
 A: 
A stochastic process $\left\{X_t\right\}_{t\in\mathbb{Z}}$ is strictly stationary if, for all $~k\in\mathbb{N}~$ and all $~t_1,\ldots,t_k, \tau\in\mathbb{Z}$,
$$
P\left( X_{t_1+\tau}\leqslant x_1, \ldots,  X_{t_k+\tau}\leqslant x_k\right){}={}P\left( X_{t_1}\leqslant x_1, \ldots,  X_{t_k}\leqslant x_k\right)\,.
$$


That is, all of the finite-dimensional distributions are translation invariant. This is the sense in which it is said the process is "independent" of $\tau$.
Weak-sense stationarity, on the other hand, requires merely that the $1$st moments and autocovariances of the processes constituent random variables are translation invariant:

For all $~t_1, t_2, \tau \in \mathbb{Z}~$, 
$$
\mathbb{E}\left[ X_{t_1} \right] = \mathbb{E}\left[ X_{t_1+\tau} \right],\quad\mathbb{C}ov\left( X_{t_1+\tau}, X_{t_2+\tau} \right) = \mathbb{C}ov\left( X_{t_1}, X_{t_2} \right)\,.
$$


This is also a sense in which the properties of the process are said to be "independent" of $\tau$. One can see how both these invariances would be the case for the process you defined if, for instance, one assumes the $Z_t$ to be continuous i.i.d. random variables, with mean $0$ and finite variance $\sigma^2$. For then, we have the following:

(i)$\quad \left(Z_{t_1+\tau}, \ldots, Z_{t_k+\tau}\right)\sim\left(Z_{t_1}, \ldots, Z_{t_k}\right);$


(ii)$\quad$Interpret the following integrals as being taken with respect to the joint pdf of the unique set of $Z_t$ that define $X_{t_1}, \ldots, X_{t_k}$, so that 
$\ \\ P\left( X_{t_1+\tau}\leqslant x_1, \ldots,  X_{t_k+\tau}\leqslant x_k\right){}={}\displaystyle \int {\bf 1}_{\left\{ X_{t_1+\tau}\leqslant x_1, \ldots,  X_{t_k+\tau}\leqslant x_k \right\}} f_{Z_{t_1+\tau}}\ldots f_{Z_{t_k-1}+\tau} ~\mathrm dz_{{t_1+\tau}}\ldots \mathrm dz_{{t_k-1+\tau}}{}={}
\displaystyle \int {\bf 1}_{\left\{ Z_{t_1+\tau}+\theta Z_{t_1-1+\tau}\leqslant x_1, \ldots,  Z_{t_k+\tau}+\theta Z_{t_k-1+\tau}\leqslant x_k \right\}} f_{Z_{t_1+\tau}}\ldots f_{Z_{t_k-1}+\tau}~\mathrm dz_{{t_1+\tau}}\ldots \mathrm dz_{{t_k-1+\tau}}{}={}
\displaystyle \int {\bf 1}_{\left\{ Z_{t_1}+\theta Z_{t_1-1}\leqslant x_1, \ldots,  Z_{t_k}+\theta Z_{t_k-1}\leqslant x_k \right\}} f_{Z_{t_1}}\ldots f_{Z_{t_k-1}}~\mathrm dz_{{t_1}}\ldots \mathrm dz_{{t_k-1}}\\={}
\displaystyle \int {\bf 1}_{\left\{ X_{t_1}\leqslant x_1, \ldots,  X_{t_k}\leqslant x_k \right\}} f_{Z_{t_1}}\ldots f_{Z_{t_k-1}}~\mathrm dz_{{t_1}}\ldots \mathrm dz_{{t_k-1}}\\ \ \\={}
P\left( X_{t_1}\leqslant x_1, \ldots,  X_{t_k}\leqslant x_k\right).
$



That is, the process is strictly stationary. Weak sense stationarity follows, as we can explicitly see in the following:

$$
\mathbb{E}\left[ X_{t+\tau}\right] = 0, \quad \mathbb{C}ov\left( X_{t_1+\tau}, X_{t_2+\tau} \right) = \left\{\begin{array}{ll} \left(1+\theta^2\right) \sigma^2 & : \mbox{if }~t_1 = t_2\ \\ \theta \sigma^2 & : \mbox{if }~\left|t_1-t_2\right| = 1 \\ 0& : \mbox{if }~\left|t_1-t_2\right| > 1\,.\end{array} \right.\,
$$

Notice that the r.h.s. of each set of equations above are "independent" of $\tau$.
