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I just started working on time series with the book from Brockwell and Davis. I'm still not that familiar with stationary time series. The book says that a time series is stationary if:

1 $\mathbb{E}[X_t] = \mu$ is independent of $t$.

2 $\operatorname{Cov}(X_{t+h},X_t)$ is independent of $t$ for each $h$.

So if I have the following time series: $X_t=a+bt+Z_t$ with {${Z_t}$} ~ $WN(0,\sigma^2)$

Can you simply say that ${X_t}$ is not a stationary time series because: $\mu_{x_t} = \mathbb{E}[a+bt+Z_t]=a+\mathbb{E}[bt + 0]=a+b \mathbb{E}[t]=a+ bt$, and this obviously is dependent on $t$?

Edit: Forget to say that $t=1,...,n$ in my time series.

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  • $\begingroup$ Welcome to CV! I assume you made a typo of "independent" when you meant "dependent" at the end, but if you don't like my edit feel free to revert it. Thanks for taking the time to mark up your maths nicely. (Also note that there are different flavours of stationarity, "weak" and "strong". But as you have observed, there is a trend in mean level, which rules out even the weaker forms.) $\endgroup$
    – Silverfish
    Sep 29, 2015 at 22:06
  • $\begingroup$ By the way, why is it $\frac{bt}{2}$ at the end? Is that a typo? I wonder if you have been treating $t$ as if it is stochastic. What do you think $\mathbb{E}(t)$ is? $\endgroup$
    – Silverfish
    Sep 29, 2015 at 22:27
  • $\begingroup$ I think I did. Check my edit, I thought dat $\mathbb{E}[n]=\frac{t}{2}$ made sense. But you have a point that it isn't stochastic. So I guess you should just can't calculate $\mathbb{E}[t]$? $\endgroup$
    – Alfons Ingomar
    Sep 29, 2015 at 22:33
  • $\begingroup$ In your comment I think you mean $\mathbb{E}(t)=\frac{n}{2}$ which is the mistake I thought you were making (I assumed you might be trying to average the possible values of $t$, from 1 to $n$). But that would still be wrong. You can certainly calculate it \mathbb{E}(t)! If $t$ isn't stochastic, then what would $\mathbb{E}(t)$ be? $\endgroup$
    – Silverfish
    Sep 29, 2015 at 22:35
  • $\begingroup$ Sorry, I guess I shouldn't stay up a lot longer;) I guess $\mathbb{E}(t)=t$? $\endgroup$
    – Alfons Ingomar
    Sep 29, 2015 at 22:41

2 Answers 2

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There are several different flavours of stationarity. The type described in your definition is weak-sense stationarity, also known as wide-sense stationarity, covariance stationarity, or second-order stationarity.

Your definition is not quite complete: a preliminary condition for weak stationarity is that the mean and covariance must exist and be finite, but this is satisfied here. As you noted, weak stationarity further requires the mean to be constant over time and $\operatorname{Cov}(X_{t+h},X_t)$ to be independent of $t$ for each $h$ i.e. the autocovariance at each lag $h$ is constant over time. The fact that $\mathbb{E}(t)=a+bt$ shows the first of these conditions is not met. Even the mean is not stationary.

Another form of stationarity is strong stationarity, also called strict stationarity or just stationarity. This requires the joint distribution function of the joint distribution of $X_t$ taken at any $k$ times $t_1, t_2, \dots , t_k$ is the same when lagged by any $\tau$. Technically, for any $k$ and any $\tau$, and for any $t_1, t_2, \dots, t_k$ we require

$$F_X(x_{t_1+\tau}, \dots, x_{t_k+\tau}) = F_X(x_{t_1}, \dots, x_{t_k})$$

This does not just imply that the mean and covariance at any given lag (if either exists) must stay constant over time, but that every conceivable property one can derive from the distribution is invariant under a time shift, in which sense its conditions are "stricter" (though note that if mean or covariance are not finite, we can have a strongly stationary series which does not fulfil the preliminary condition for weak stationarity). Since your mean is not constant over time, then your process can't be strongly stationary either. In general, so long as the mean and covariance exist, we can conclude that a process that is not weakly stationary will not be strongly stationary either. The converse is not true: just because a process is not strongly stationary, doesn't mean it can't be weakly stationary, since it is possible to be weakly stationary yet not strongly stationary.

There is a sense in which your $X_t$ is "stationary": it is trend stationary.

This means that the trend in your time series can be expressed as a function of $t$; if we strip this trend out then what we are left with is a stationary process. In particular, $X_t$ is trend stationary if we can express it as

$$X_t = f(t) + Y_t$$

where $f$ is a deterministic function of time $t$ and $\{Y_t\}$ is a stationary process. In your case we can take $f(t)=a + bt$ and $Y_t = Z_t$; the idea is that by "stripping out" the trend $f(t)$ we would have $X_t - (a + bt) = Z_t$ which is stationary (because it's white noise).

Alternatively we could have taken $f(t)=bt$ and $Y_t= a + Z_t$.

Note that although in this case we could take a process that didn't have a constant mean, then can subtract a deterministic trend to obtain a stationary process, doesn't mean we can do this on any such process. Consider a random walk with drift, for instance. Here the mean is not constant over time, but we can't simply subtract a deterministic trend to obtain a stationary process. So a random walk with drift would not be a trend stationary process. See the question "Difference between series with drift and series with trend" for an illustration.

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    $\begingroup$ The following is not true: the fact your process isn't weakly stationary means it won't be strongly stationary either. There exist strictly stationary processes that are not weakly stationary. That happens with processes that do not have a second moment, for example (the moment is infinity). $\endgroup$
    – Richard Hardy
    Sep 30, 2015 at 5:17
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    $\begingroup$ @RichardHardy is correct. An iiid Cauchy series is strictly stationary but not covariance stationary. See p. 40-41 for some scenarios where these two terms are the same: quant-econ.net/_downloads/time_series_book.pdf $\endgroup$
    – Plissken
    Sep 30, 2015 at 8:41
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    $\begingroup$ @RichardHardy you are of course quite right, I was leaving it implicit that it was not "weakly stationary in this way" (i.e. because the moments were not constant, rather than because they didn't exist). But that makes it look like a sweeping and incorrect generalisation so I have endeavoured to tidy things up. This also gives me the opportunity to point out a slight omission in the OP's definition. $\endgroup$
    – Silverfish
    Sep 30, 2015 at 12:38
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    $\begingroup$ @Plissken, indeed $X_t \overset{iid} \sim t_1$ is clearly not even stationary in first order moments (or whatever the appropriate term is)! Let me know if you think there's anything I haven't tightened up enough. As an aside it would be nice if there different terms for "not stationary because the moments existed but were not constant" versus "not stationary because the moments did not necessarily exist", which would have let me express what I wanted in a simpler fashion. Or perhaps there are such terms, and I just don't know them! $\endgroup$
    – Silverfish
    Sep 30, 2015 at 13:11
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    $\begingroup$ (+1) A random walk with drift would be a difference-stationary process (because the 1st differences are stationary). $\endgroup$
    – Scortchi - Reinstate Monica ♦
    Sep 30, 2015 at 13:33
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The definition that you claim is from Brockwell and Davis is actually the definition of weakly stationary or wide-sense-stationary time series and not that of stationary series (also called strictly stationary) time series. Strictly stationary series (with finite second moment) are a subclass of weakly stationary series and thus also enjoy properties 1. and 2. (See also this answer of mine on dsp.SE; for the sake of keeping matters as simple as possible, I had not mentioned finite second moments in the earlier version of my answer. Also, those getting ready to nitpick that it is also necessary to assume that the first moment is finite should bear in mind that $|x| < 1+x^2$ for all $x\in\mathbb R$ and so $E[|X|] < 1+E[X^2]$ from which it follows that $E[X]$ is finite whenever $E[X^2]$ is assumed to be finite.

The two conditions that you state make the series strictly stationary if we assume (in addition to 1. and 2.) that all the $X_i$ are jointly normal random variables.

So, yes, the series that you propose is not weakly stationary because the mean depends on $t$, and so it cannot be strictly stationary either.

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    $\begingroup$ The following is not true: Strictly stationary series are a subclass of weakly stationary series. There exist strictly stationary processes that are not weakly stationary. That happens with processes that do not have a second moment, for example (the moment is infinity). Also, are you sure that stationary and strictly stationary are the same? I am afraid it might not hold universally. $\endgroup$
    – Richard Hardy
    Sep 30, 2015 at 5:15
  • $\begingroup$ @RichardHardy As for whether "stationary" and "strictly stationary" are the same, I guess that's a matter of definition. When I've seen a definition of "stationary" per se, it's always been the "strict stationary" definition. But it's common in economics books, and quite possibly in other fields, to define both weak and strong forms, then say something like "for the purposes of these notes, we shall use stationary to refer to weakly stationary processes, unless otherwise specified". $\endgroup$
    – Silverfish
    Sep 30, 2015 at 14:01
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    $\begingroup$ @RichardHardy I have incorporated your comment into my answer. As to whether stationary and strictly stationary are the same, I stand with Humpty Dumpty in Through the Looking Glass by Lewis Carroll: "When I use a word," Humpty Dumpty said, in rather a scornful tone, "it means just what I choose it to mean—neither more nor less." "The question is," said Alice, "whether you can make words mean so many different things." "The question is," said Humpty Dumpty, "which is to be master—that's all." $\endgroup$
    – Dilip Sarwate
    Sep 30, 2015 at 21:21

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