By "stationary" I mean "weakly stationary".

Consider a "stationary" AR(1) equation:

$$X_t=\varphi X_{t-1}+\varepsilon_t,$$ where $t\in\mathbb{Z}$ are discrete time moments, $\varepsilon_t$ a zero-mean white noise (just some iid sequence), $\varphi\in(-1,1)$. It is well known that there is a stationary solution (that is, a discrete time series satisfying the equation). Denote it by $X_t.$ However, we can introduce another time series $Y_t=X_t+\varphi^t$, which appears to be a nonstationary solution for the "stationary" equation (clearly, $\mathbb{E}[Y_t]$ is not free of $t$, since $X_t$ is evidently zero-mean).

Given more general stationary AR($p$) process, is it possible to somehow damage the weak stationarity property? Or, in general, is it true that any stationary discrete time AR (or even ARMA) equation has a nonstationary solution?

  • $\begingroup$ Could expand a little bit? Could you explain how $Y_t=X_t+\phi^t$ appears to be a nonstationary solution for $X_t=\varphi X_{t-1}+\varepsilon_t$? (Perhaps a better practice would be not to use $\phi$ and $\varphi$ in the same exercise because both are "phi", which may make it confusing.) $\endgroup$ – Richard Hardy Mar 20 '16 at 11:32
  • $\begingroup$ What do you mean by a "solution", what kind of object is that? (Like a constant, a stochastic process, ...) Could you elaborate on that, perhaps expand that section of the post? $\endgroup$ – Richard Hardy Mar 20 '16 at 12:35
  • $\begingroup$ Richard, the solution is assumed to be a time series, of course. I've added it to the post. $\endgroup$ – Nikita Mar 20 '16 at 12:39
  • $\begingroup$ Could you show fully or partly that $Y_t$ is a (nonstationary) solution? Also, maybe I am being picky, but I am not that used to the terminology and so having $X_t$ in the general form AR(1) equation AND as a solution to it is a bit confusing. Could we somehow distinguish between the two notationally? (But perhaps it is standard to use such notation, then just ignore my comment.) $\endgroup$ – Richard Hardy Mar 20 '16 at 12:46
  • $\begingroup$ I meant that the fact $Y_t$ is a solution is easy to check. $\endgroup$ – Nikita Mar 20 '16 at 12:55

If you let your process go on, then you'll notice how the term $\varphi^t$ disappears: $$\lim_{t\to\infty}\varphi^t=0$$

So, $$E[Y_t]=E[X_t]+E[\varphi^t]=_{t\to\infty}0$$ this is despite the right hand side having being dependent on finite $t$.

So the answer to your question is that your process $Y_t$ is not non-stationary. Hence, it doesn't serve as a counter-example.

Additional thoughts. You formulated your question in terms of solutions of the stochastic processes. Look at what is the solution of the AR(1) process.

For instance, if you forecast $\tau$ steps ahead you get: $$X_{t+\tau}=\varphi^\tau (X_t+\sum_{s=1}^\tau \varepsilon_{t+s}\varphi^{-s})$$

You can see how it simply collapses to the noise around zero as $\tau$ grows, no matter what was the initial $X_t$. When you add your $\varphi^\tau$ term it also disappears, so the stable solution is the same: noise around zero:

$$X_{t+\tau}+\varphi^\tau=\varphi^\tau (X_t+\sum_{s=1}^\tau \varepsilon_{t+s}\varphi^{-s}+\varphi^s)$$

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  • $\begingroup$ Does that answer the questions? I have trouble seeing the connection clearly (although I do see some connection). $\endgroup$ – Richard Hardy Mar 15 '17 at 15:38
  • $\begingroup$ I updated the answer. $\endgroup$ – Aksakal Mar 15 '17 at 15:39
  • $\begingroup$ Thank you. I find both the question and the answer interesting, but it takes some effort to wrap my head around them. Element by element these are easy, but the relations between them can be deceptive :) $\endgroup$ – Richard Hardy Mar 15 '17 at 15:43
  • $\begingroup$ @RichardHardy, I'm being lazy here. Maybe I should have describe this all in the framework of SDE, then it would be clearer. $\endgroup$ – Aksakal Mar 15 '17 at 15:44
  • $\begingroup$ Also, are we allowed to consider the case $t\rightarrow\infty$ and sort of hand-wave finite $t$ (especially in the expectation in the second formula)? I can see where you are going if you look at the fixed points of SDEs, but is that really what we need here? I guess that depends on the definition of what a solution is, and the OP seems to be interested not in fixed points but in processes that satisfy some property/equation (see the comments under the OP). $\endgroup$ – Richard Hardy Mar 15 '17 at 15:46

Terminology being used in the question is not quite correct. You're mixing up the model (or equations) and solution to the model.

It does not make sense to speak of an equation (in this case, a system of stochastic difference equations) being stationary or non-stationary. Stationarity, of lack thereof, is a property of a solution. An equation can have stationary or non-stationary solutions.

What you have found are two solutions, one stationary and one non-stationary, to the AR(1) equation when AR parameter $|\phi| \neq 1 $. (If $|\phi| > 1$, substitute $-t$ for $t$ in your example.) In contrast, when $|\phi| = 1$, there are only non-stationary solutions.

The answer to your question is, yes, this generalizes to the AR(p) case. The AR(p) equation(s) $$ \Phi(L)X_t = \epsilon_t, \; t = \cdots -1, 0, 1, \cdots $$ has both stationary and non-stationary solutions if the polynomial $\Phi(z^{-p})$ has no roots on the unit circle and all roots are real.

For example, suppose the AR(2) model $$ X_t = \phi_1 X_{t-1} + \phi_2 X_{t-2} + \epsilon_t $$ has a stationary solution $(X_t)$ and $z^2 - \phi_1 z - \phi_2$ has two real roots $a$ and $b$, then $$ X_t + a^t + b^{t-1} $$ is a non-stationary solution.

Notice setting $\phi_2 = 0$ and considering $z - \phi_1 = 0$ recovers your AR(1) example.

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