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Imagine we have two time-series processes, which are stationary, producing: $x_t,y_t$.

Is $z_t=\alpha x_t +\beta y_t$, $\forall \alpha, \beta \in \mathbb{R}$ also stationary?

Any help would be appreciated.

I would say yes, since it has an MA representation.

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    $\begingroup$ Why is it guaranteed to be MA? there are stable AR processes. Either way though, if you are talking about BIBO stability, then yes the sum is trivially stable because you can calculate the new bounds. Asymptotic stability also holds because $\lim_{t \to \infty} z_t = \alpha\lim_{t \to \infty} x_t + \beta\lim_{t \to \infty} y_t$ $\endgroup$ – Steve Cox Mar 30 '18 at 14:51
  • $\begingroup$ Related to some extend: Note in numerical analysis, you use what is called a preconditioner (a particular linear transformation) to gain in stability so I doubt the answer is yes. $\endgroup$ – Surb Mar 30 '18 at 16:50
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Perhaps surprisingly, this is not true. (Independence of the two time series will make it true, however.)

I understand "stable" to mean stationary, because those words appear to be used interchangeably in millions of search hits, including at least one on our site.

For a counterexample, let $X$ be a non-constant stationary time series for which every $X_t$ is independent of $X_s$, $s\ne t,$ and whose marginal distributions are symmetric around $0$. Define

$$Y_t = (-1)^t X_t.$$

![Figure 1: plots of X, Y, and (X+Y)/2 over time

These plots show portions of the three time series discussed in this post. $X$ was simulated as a series of independent draws from a standard Normal distribution.

To show that $Y$ is stationary, we need to demonstrate that the joint distribution of $(Y_{s+t_1}, Y_{s+t_2}, \ldots, Y_{s+t_n})$ for any $t_1\lt t_2 \lt \cdots \lt t_n$ does not depend on $s$. But this follows directly from the symmetry and independence of the $X_t$.

Figure showing some cross-scatterplots of Y

These lagged scatterplots (for a sequence of 512 values of $Y$) illustrate the assertion that the joint bivariate distributions of $Y$ are as expected: independent and symmetric. (A "lagged scatterplot" displays the values of $Y_{t+s}$ against $Y_{t}$; values of $s=0,1,2$ are shown.)

Nevertheless, choosing $\alpha=\beta=1/2$, we have

$$\alpha X_t + \beta Y_t = X_t$$

for even $t$ and otherwise

$$\alpha X_t + \beta Y_t = 0.$$

Since $X$ is non-constant, obviously these two expressions have different distributions for any $t$ and $t+1$, whence the series $(X+Y)/2$ is not stationary. The colors in the first figure highlight this non-stationarity in $(X+Y)/2$ by distinguishing the zero values from the rest.

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    $\begingroup$ Independence of the two time series is obviously a sufficient condition. But would't the weaker requirement of joint stationarity also suffice? $\endgroup$ – Dilip Sarwate Apr 1 '18 at 15:58
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    $\begingroup$ Yes, that's right @Dilip. Thank you for that observation. $\endgroup$ – whuber Apr 1 '18 at 22:50
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Consider the two-dimensional process

$$w_t = (x_t, y_t)$$

If it is strictly stationary, or alternatively, if the processes $(x_t)$ and $(y_t)$ are jointly strictly stationary, then a process formed by any measurable function $f:= f(x_t,y_t), f:\mathbb R^2 \to \mathbb R$ will also be strictly stationary.

In @whuber's example we have

$$w_t = (x_t, (-1)^t x_t)$$

To examine whether this $w_t$ is strictly stationary, we have to first obtain its probability distribution. Assume the variables are absolutely continuous. For some $c \in \mathbb R$, we have

$$\text{Prob}(X_t \leq c,(-1)^t X_t \leq c)= \cases {\text{Prob}(X_t \leq c, X_t \leq c)\;\;\;\; \text{t is even}\\ \\ \text{Prob}(X_t \leq c, -X_t \leq c)\;\;\;\; \text{t is odd}}$$

$$= \cases {\text{Prob}(X_t \leq c)\;\;\;\; \text{t is even}\\ \\ \text{Prob}(-c\leq X_t \leq c)\;\;\;\; \text{t is odd}}$$

$$\implies \text{Prob}(X_t \leq c,(-1)^t X_t \leq c)= \cases {\text{Prob}(X_t \leq c)\;\;\;\; \text{t is even}\\ \\ \text{Prob}( |X_t| \leq c)\;\;\;\; \text{t is odd}}$$

Sticking with whuber's example, the two branches are different probability distributions because $x_t$ has a distribution symmetric around zero.

Now to examine strict stationarity, shift the index by a whole number $k>0$. We have

$$\text{Prob}(X_{t+k} \leq c,(-1)^t X_{t+k} \leq c)= \cases {\text{Prob}(X_{t+k} \leq c)\;\;\;\; \text{t+k is even}\\ \\ \text{Prob}( |X_{t+k}| \leq c)\;\;\;\; \text{t+k is odd}}$$

For strict stationarity, we must have

$$\text{Prob}(X_t \leq c,(-1)^t X_t \leq c)=\text{Prob}(X_{t+k} \leq c,(-1)^t X_{t+k} \leq c),\;\;\; \forall t,k$$

And we don't have this equality $\forall t,k$, because, say, if $t$ is even and $k$ is odd, then $t+k$ is odd, in which case

$$\text{Prob}(X_t \leq c,(-1)^t X_t \leq c) = \text{Prob}(X_t \leq c) $$

while

$$ \text{Prob}(X_{t+k} \leq c,(-1)^t X_{t+k} \leq c) = \text{Prob}( |X_{t+k}| \leq c)= \text{Prob}( |X_{t}| \leq c)$$

So we do not have joint strict stationarity, and then we have no guarantees about what will happen to a function of $f(x_t,y_t)$.

I have to point out that the dependence between $x_t$ and $y_t$, is a necessary but not a sufficient condition for the loss of joint strict stationarity. It is the additional assumption of dependence of $y_t$ on the index that does the job.

Consider

$$q_t = (x_t, \theta x_t),\;\;\; \theta \in \mathbb R$$

If one does the previous work for $(q_t)$ one will find that joint strict stationarity holds here.

This is good news because for a process to depend on the index and be strictly stationary is not among the modelling assumptions we need to make very often. In practice therefore, if we have marginal strict stationarity, we expect also joint strict stationarity even in the presence of dependence (although we should of course check.)

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I would say yes, since it has an MA representation.

One observation. I think that having a MA representation implies weak stationarity, not sure if it implies strong stationarity.

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    $\begingroup$ Re "I can't imagine": please see my answer for a counterexample. $\endgroup$ – whuber Mar 30 '18 at 14:10
  • $\begingroup$ oneloop, remove the part related to strict stationarity, and just leave that related to weak stationarity. I'll give you a +1, since it also helped me. ;) $\endgroup$ – An old man in the sea. Mar 30 '18 at 17:33
  • $\begingroup$ @Anoldmaninthesea. Like this? $\endgroup$ – oneloop Mar 30 '18 at 17:36
  • $\begingroup$ yes, like that. MA representation implies weak stationarity, indeed. $\endgroup$ – An old man in the sea. Mar 30 '18 at 17:41
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    $\begingroup$ This is being automatically flagged as low quality, probably because it is so short. At present it is more of a comment than an answer by our standards. Can you expand on it? You can also turn it into a comment. $\endgroup$ – gung - Reinstate Monica Apr 24 '18 at 14:53

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