# Linear transform of a strictly stationary time series

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?

• you can clamp all the $X_t$ with the function $f_M(x) = x$ if $|x| < M$, $f_M(x) = \pm M$ otherwise, clearly $Y_t(M) = \sum_j \psi_j f_M(X_{t-j})$ is stationary, and when $M \to \infty$, $Y_t(M)$ converges in distribution to $Y_t$ Apr 24, 2016 at 21:52
• @user1952009 What do you mean by "clamp all the.."? Apr 24, 2016 at 21:55
• it's not clear ? and I guess if there are other ways, they won't be fundamentally different. and what you call a linear transform is called a linear time-invariant transformation : en.wikipedia.org/wiki/LTI_system_theory (aka. a filtering) Apr 24, 2016 at 22:03

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 $Y$.

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.

• Please read the definition of strict stationarity I gave in the question carefully. You need to consider a vector of finite length of $Y$. What I wrote at the end of my post concerns only the marginal distribution of $Y_t$. May 4, 2016 at 13:40
• I know what the definition of stationarity (strict and weak) is. However my argument generalizes immediately to that case. May 4, 2016 at 13:42
• I edited the argument to make my point clearer. May 4, 2016 at 13:48
• I think I get the proof and it seems OK to me. Denoting the vector at the top by $Y_t^M$, we have $Y_t^M \stackrel{d}{=} Y_s^M$. Furthermore, $Y_t^M \to Y_t$ and $Y_s^M \to Y_s$ a.s.. Hence also in distribution. By the uniqueness of limits, $Y_t \stackrel{d}{=} Y_s$. Thanks for the clarification. May 4, 2016 at 14:34
• How much extra effort would it require to show that $Y_t = g(X_t,X_{t-1},\ldots)$ is strictly stationary for any measurable function $g$? May 5, 2016 at 19:25