Linear transform of a strictly stationary time series

Question

First, let me clarify what I mean by a strictly stationary time series. Let $(X_t)_{t\in \mathbb{Z}}$ be a sequence of random variables on some probability space. If it holds that $$(X_t, X_{t+1},\ldots,X_{t+h}) \stackrel{d}{=} (X_s, X_{s+1},\ldots,X_{s+h})$$ for every $s,t \in \mathbb{Z}$ and every $h \in \mathbb{Z}_{\geq 0}$.

Next, I will explain what I mean by linear transform. Given a sequence of reals $(\psi_j)_{j\in \mathbb{Z}}$ such that $\sum_j\lvert\psi_j\rvert < \infty$, linear transform of $(X_t)$, which we denote by $(Y_t)$, is defined as

$$Y_t = \sum_j\psi_jX_{t-j}$$

Now suppose that the sum above is well-defined, i.e. $\sum_j\psi_jX_{t-j}$ is finite almost surely. ($E\lvert X_t\rvert < \infty$ would be sufficient for this but we don't assume that.)

How do I then show that $(Y_t)$ is also strictly stationary? Intuitively speaking, $$\sum_j\psi_jX_{t-j} \stackrel{d}{=} \sum_j\psi_jX_{s-j}$$ since $(X_t)$ is strictly stationary. But shifting a finite sequence is not quite the same as ``shifting an infinite sequence". How do I make this rigorous?

Answer 1

Fix some $s,t \in \Bbb Z$. Note that since $(X_t, X_{t+1},\ldots,X_{t+h}) \stackrel{d}{=} (X_s, X_{s+1},\ldots,X_{s+h})$ for all $h$, we have that $$(X_{t+k})_{k \in \Bbb Z} \stackrel{d}{=} (X_{s+k})_{k \in \Bbb Z}$$ since both processes have the same finite-dimensional distributions. If you are not familiar with this fact from elementary measure theory, see for example Theorem 23 here (there $\Xi$ denotes the space of functions endowed with product $\sigma$-algebra) or alternatively Proposition 3.1 here.

Now define the map $\psi: \Bbb R^{\Bbb Z} \to \Bbb R^{\Bbb Z}$ by sending $(x_j)_{j \in \Bbb Z} \mapsto (\sum_j \psi_j x_{k-j})_{k \in \Bbb Z}$ and note that it is measurable on the product $\sigma$-algebra (since the individual components are). Fix some $s,t \in \Bbb Z$, and since $(X_{t+k})_{k \in \Bbb Z} \stackrel{d}{=} (X_{s+k})_{k \in \Bbb Z}$ we easily get that $$(Y_{t+k})_{k\in\Bbb Z}=\psi\big((X_{t+k})_{k \in \Bbb Z}\big) \stackrel{d}{=} \psi\big((X_{s+k})_{k \in \Bbb Z}\big)=(Y_{s+k})_{k\in\Bbb Z}$$ which precisely means stationarity of $Y$.

Answer 2

Similar to what @user1952009 was saying, we can fix a positive integer M and then define the finite linear transformation:

$$(Y_t^M,\dots,Y_{t+h}^M) = (\sum_{|j|\le M}\psi_jX_{t-j},\dots, \sum_{|j|\le M}\psi_jX_{t+h-j})$$

Here we can without problem make the argument:

$$(\sum_{|j|\le M}\psi_jX_{t-j},\dots, \sum_{|j|\le M}\psi_jX_{t+h-j}) \stackrel{d}{=} (\sum_{|j|\le M}\psi_jX_{s-j},\dots, \sum_{|j|\le M}\psi_jX_{s+h-j})$$

Assuming that $(Y_t^M,\dots,Y_{t+h}^M)$ is integrable, we can apply the dominated convergence theorem to conclude that in the limit as $M \to \infty$, $(Y_t^M,\dots,Y_{t+h}^M) \to (Y_t,\dots,Y_{t+h})$ a.s. (or definitely at least in distribution, which is all that is necessary for our purposes).

Likewise $(\sum_{|j|\le M}\psi_jX_{t-j},\dots, \sum_{|j|\le M}\psi_jX_{t+h-j}) \to (\sum_{j}\psi_jX_{t-j},\dots, \sum_{j}\psi_jX_{t+h-j})$ and $(\sum_{|j|\le M}\psi_jX_{s-j},\dots, \sum_{|j|\le M}\psi_jX_{s+h-j}) \to (\sum_{j }\psi_jX_{s-j},\dots, \sum_{j}\psi_jX_{s+h-j})$ by the dominated convergence theorem, therefore the two sequences, since they are equal in distribution for every $M$, must approach the same limit in distribution, i.e. we have the desired conclusion:

$$(\sum_{j}\psi_jX_{t-j},\dots, \sum_{j}\psi_jX_{t+h-j}) \stackrel{d}{=} (\sum_{j }\psi_jX_{s-j},\dots, \sum_{j}\psi_jX_{s+h-j})$$

I hope this helps even though I glossed over some details.

EDIT: For any measurable $g$, if $Z_1, Z_2$ are two random vectors with the same distribution $\mathbb{P}( \cdot)$, then both $g(Z_1)$ and $g(Z_2)$ have the distribution $\mathbb{P}(g^{-1}(\cdot))$, i.e. it follows immediately that $g(Z_1)$ and $g(Z_2)$ have the same distribution. See for example Kallenberg's book on probability theory -- if I recall correctly, he treats strictly stationary sequences extensively in the context of ergodic theory.