# Talk:Stationary process

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## Style

No Wikipedia article should begin by saying "A stationary process is one in which...". Some context setting is needed first, at least by saying which discipline is being discussed; e.g., is this about law, about software, etc.? The first sentence seems confused to me. But I'm not sure what the person who wrote it had in mind. Michael Hardy 01:50, 29 Sep 2004 (UTC)

... maybe a stochastic process with stationary increments?? Michael Hardy 01:51, 29 Sep 2004 (UTC)

I believe the {{technical}}, which has been up for two years, is outdated. The comments by Michael Hardy have already been corrected. --Zvika 15:09, 13 October 2006 (UTC)

## stationary signal

"in signal processing, a stationary signal is a signal whose frequency content does not change over time." Is this true? Is it the same thing? - Omegatron 23:29, Sep 29, 2004 (UTC)

## Stationary process in statistics

(EDIT - DELETED NONSENSE) Alternative examples of non-stationary processes are stock prices, economic aggregates (e.g. GDP) and the position of a gas particle in space. In all of these examples the long run distribution of the level/position depends on the current level/position violating the constancy of the (EDIT) unconditional (EDIT) distribution required for stationarity.

there are many ways to extend this article with brief discusison of: history, applications, differencing to induce stationarity (orders of non-stationarity), linear systems (ARMA processes, VARs, GARCH processes etc), difference between trending and non-stationary series, "unit roots", spurious regressioon, link to cointegration, link to brownian motion etc. --Cripes 21:28, 27 March 2006 (UTC)

In books on stochastic processes and on time series, I have seen stationarity defined as meaning the probability distribution of X(t) is the same for all values of t. And you seem to contradict yourself concerning the cymbal: if it decays, then its probability distribution approaches a point mass as t approaches ∞; therefore it is stationary according to the definition you gave us. Michael Hardy 21:38, 27 March 2006 (UTC)
you're right about the definition, apologies - i've edited out the nonsensical parts of my previous post. I was mis-interpreting your statement of "distribution" as the "conditional distribution" (which is not necessarily constant). since i am finding it necessary at work to brush up on this stuff i might offer some additional content for the page (as per the list) if you are interested in editing it --Cripes 22:40, 27 March 2006 (UTC)

## Stationarity implies WSS?

Please try to work things out in the talk page rather than reverting each other's edits. IMO, User:128.214.205.6 is technically correct: A process cannot be said to be WSS if its mean or variance do not exist, but it could still be stationary (e.g. an iid process where each time sample is Cauchy distributed). However, I think this is something of a pathological case; in most cases of interest, the moments exist and so stationary implies WSS. So perhaps User:128.214.205.6's sentence is more confusing than instructive. I tried to change the wording, I hope you will like it. --Zvika 19:10, 16 October 2006 (UTC)

This was my first revision of the material in Wikipedia. Hence please excuse my ignorance and reverting "each other's edits". I hope editing this talk page is the proper method to discuss revisions. (Is it?) Now to the topic itself: The above mentioned iid Cauchy series is a well-known theoretical case in point. Strictly stationary while not weakly stationary processes can arise while trying to design empirically relevant processes as well. Such a well-known (in the econometrics literature) example is an integrated GARCH process (D.B. Nelson, 1990, Stationarity and Persistence in the GARCH(1,1) Model, Econometric Theory 6: 318-34). Thus the caveat (strict stationarity does not imply weak stationarity) is certainly in order. The new sentence "Any strictly stationary process which has a mean and a covariance is also WSS." is fine. However, I think that the present descriptions of strict and weak stationarity (in Wikipedia) still inadequately suggest dominance of strict stationarity over weak stationarity (e.g. "A weaker form of stationarity commonly employed - - "). It is possible that a process is weakly stationary while not strictly stationary. This could happen if the first two moments were time invariant while the third or fourth, say, moment were not. This should be pointed out in the text as well IMO. (PJP) —Preceding unsigned comment added by 128.214.205.4 (talkcontribs) 17 October 2006
Thanks -- apologies for the revert, I wanted to get an example to be sure that what you were saying was correct. You are absolutely right that strict stationarity does not imply weak stationarity -- I had not realised this until now. Thanks for providing the example here. --Richard Clegg 15:12, 17 October 2006 (UTC)
Is it proper to equate wide-sense stationarity (WSS) with second-order stationarity. I think there is a difference. A process can be WSS without being second-order stationary. The definition in the article is that of WSS; second order stationary is different. (of course it depends on your definition, but that must be clarified.) See Section 6.2 of Peebles Jr., Peyton: Probability, Random Variables, and Random Signal Principles (2/e) Dakshayani 05:39, 4 November 2006 (UTC)
Could you quote that definition for us? I don't have the book. --Zvika 07:31, 4 November 2006 (UTC)
There is a difference between second order stationarity and WSS. A process is second order ( according to this def) if the second order density function satisfies $~f_X(x_1 ,x_2 : t_1, t_2 ) = f_X(x_1 ,x_2 : t_1 + \Delta, t_2 +\Delta )~$ for all $t_1 , t_2 ~and~ \Delta$ . Such a process will be WSS if the mean and corelation functions are finite. A process can be WSS without being Second Order Stationary. The definition of Second Order Stationarity can be generalised to Nth order and strict stationary means stationary of all orders. Dakshayani 09:56, 4 November 2006 (UTC)
You're probably right then. Sounds like this is related to the previous discussion (above). I removed second-order stationarity from the definition. --Zvika 13:03, 4 November 2006 (UTC)

## not all stationary discrete-time processes on {0,1} are bernoulli processes?!

The introduction strongly suggests that all stationary discrete-time random processes are Bernoulli. I don't think that is correct. For example, isn't a random process where the distribution of Xn is determined by Xn-k, Xn-k+1, ..., Xn-1 (in other words, an order k Markov chain) stationary? —Preceding unsigned comment added by 77.162.102.4 (talk) 14:02, 21 December 2007 (UTC)

This certainly looks like a mistake to me. I've gone ahead and fixed it, and also the related mistake in the page Bernoulli scheme. If anyone thinks differently, please correct me. --Zvika (talk) 19:32, 22 December 2007 (UTC)
No, the original statement was correct. Just take the n'th product of the of the Markov chain, where n is the length of the longest period. I agree, though, additional detail would be useful. In short, though, the Bernoulli shift is isomorphic to pretty much everything, including many/most dynamical systems. See, for example D.S. Ornstein (2001), "Ornstein isomorphism theorem", in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4 for one statement of this. Err, well, when its infinite, you just get the Koopman operator and its adjoint, the transfer operator; in either case, you're still dealing with a shift operator, of which Markov chains and subshifts of finite type are more-or-less, in a hand-waving way, are special cases. linas (talk) 17:21, 21 November 2010 (UTC)

## more precision in intro?

A recent revision (no real objection to this) has diverted an attempt to make the intro material more accurate ... the present version seems to me to imply that only the marginal distribution need be fixed. Also, the rearrangement into sections means that weak stationarity is not mentioned till much later ... it would be good to have both weak and strict refered to in the intro. Can someone find some appropriate wording to deal with these points. Melcombe (talk) 11:11, 9 April 2008 (UTC)

I realize now what you were trying to say. Sorry for reverting you, but I think the sentence you added was more confusing than helpful. The lead should be accurate, but not verbose. I tweaked it about just now -- tried to emphasize the fact that we're talking about the joint distribution. What do you think? --Zvika (talk) 11:52, 9 April 2008 (UTC)

## The sound of a cymbal crashing

"However, the sound of a cymbal crashing [ like one hand clapping?! ] is not stationary because the acoustic power of the crash (and hence its variance) diminishes with time." I don't think this is correct: The sound is not the process; the process is one that always reverts to zero decibels. Seems like a stationary AR(1) process to me, with the occasional crashing being the additive stochastic term. Duoduoduo (talk) 18:51, 2 December 2010 (UTC)

## Seasonal data, and Xt = Y

Hi, Melcombe!

• I stand corrected on seasonal data — you're right. Sorry about that.
• I still think there's a notational problem with the Xt = Y example. Y seems to stand for many different things: the name of a random variable; the Y value that determines $X_{\tau}$; the Y value that determines $X_{\tau -1}$; etc. Can you find a notational system that distinguishes these? I tried to, and I'm not sure why you didn't like it.
• In the Xt = Y example, the article says "...realisations consist of a series of constant values, with a different constant value for each realisation." But to me "constant values" means "values that do not change over time", which presumably is not what's meant here. Would it be better to say "non-stochastic values" instead of "constant values"?

Thanks for all your good work on the stat articles! Duoduoduo (talk) 17:04, 3 December 2010 (UTC)

"Values that do not change over time" is exactly what is meant here" and any single realisation of the process {$\scriptstyle{X_{t}}$} correspnds to a single realisation of the random variable Y. Thus the single value of Y determines every single value of X ... $X_{\tau}=Y$ and $X_{\tau -1}=Y$, etc. for the same value of Y. It follows from this that the marginal distribution $X_{\tau}$ is the same as the marginal distribition of Y, and hence is the same for any τ, which is part of the requirement for strict stationarity.
The change you made converted the example to an entirtely different case, with independent values for each time-point ... in which case the end of the example (about the "law of large numbers" would be incorrect, as it would usuualy apply).
As for notation, it might be slightly clearer to those who already know the notation if we made use of the probability-space notation that might appear as {$\scriptstyle{ X_{t}(\omega)}$}, where ω denotes the point in the underlying probability space. Then
$X_{t}(\omega)=Y(\omega) \qquad \text{for all } t.$
However this may just be confusing to those not familiar with the notation, and the notation seems not to be used much in other Wikipedia articles.
Melcombe (talk) 11:18, 7 December 2010 (UTC)
I think the notation $X_{t}(\omega)=Y(\omega) \qquad \text{for all } t.$ is much better. Could you put it in? Duoduoduo (talk) 15:59, 7 December 2010 (UTC)

## Stationary vs Cyclostationary

Early in the article we have: "An important type of non-stationary process that does not include a trend-like behavior is the cyclostationary process"

But then later on referencing a random variable Y: "...Let Y have a uniform distribution on (0,2π] and define the time series { Xt } by

$X_t=\cos (t+Y) \quad \text{ for } t \in \mathbb{R}.$

Then { Xt } is strictly stationary."

how is it not cyclostationary and therefore non-stationary?

watson (talk) 22:27, 12 July 2011 (UTC)

If Y is treated as a random variable when deriving the joint distributions of the Xt, then these distributions satisfy the conditions for the process to be strictly stationary. For example the distribution function of a single Xt does not depend on t. However, if Y is treated as having a single realisation y, then the process is not stationary, but is cyclostationary (although I am not sure we have a definition for that). This is an important point, and is the reason that these sorts of examples are included here and in text books. Possibly it illustrates a weakness in the usefulness of the concept of "stationarity". But it illustrates that "stationarity" (according to this definition) may not mean what one thinks it means. However there are instances where a time series or stochastic process is most naturally defined with respect to a fixed time point, such that the process is formally non-stationary, where adding a random time shift to the whole series (as in the example here) makes the process stationary and hence allows a much simpler description of its properties (moments, correlations) than would otherwise be possible: an example here is the "broken line process". Melcombe (talk) 09:02, 14 July 2011 (UTC)

## Inconsistent "definition"

"Stationarity, is defined as a quality of a process in which the statistical parameters (mean and standard deviation) of the process do not change with time. The most important property of a stationary process is that the auto-correlation function (ACF) depends on lag alone and does not change with the time at which the function was calculated."

A process with mean 0 and standard deviation 1 may have very different ACFs. Thus, such a process may switch at some (nonrandom) time from one ACF to another ACF. Then it does satisfy the first phrase above, but violate the second one. Not a valid definition. I revert. Boris Tsirelson (talk) 17:09, 27 December 2012 (UTC)

## "Stationary process" and "process with a stationary distribution"

The lede currently says

Note that a "stationary process" is not the same thing as a "process with a stationary distribution".

But the link stationary distribution says

Stationary distribution may refer to...[t]he set of joint probability distributions of a stationary process....

So it seems to me that the passage in this article contradicts its own link. I'll remove it unless there is an objection. Duoduoduo (talk) 14:08, 11 January 2013 (UTC)

Or maybe the article "Stationary distribution" should be changed? I never saw that notion applied to non-stationary processes, but someone could say that a nonlinear deterministic time change of a stationary process leads to a non-stationary process with a stationary (marginal, single-time) distribution... I am not sure. Boris Tsirelson (talk) 14:42, 11 January 2013 (UTC)

## Wide sense stationarity and Hilbert space techniques

1. This new paragraph treats the discrete-time case, while the rest of the article is about continuous time.

2. If the discrete time runs from -∞ to +∞ then "the Hilbert subspace of L2(μ) generated by {e-2πiλ⋅t}" is the whole L2(μ), isn't it?

Boris Tsirelson (talk) 11:18, 13 January 2013 (UTC)

Fixed to be consistent with rest of article. My bad. Mct mht (talk) 23:28, 13 January 2013 (UTC)

A. Bochner's theorem requires continuity of the positive definite function. It is not obvious that this criterion is satisfied. A reference would be helpful. — Preceding unsigned comment added by Dunham08 (talkcontribs) 16:34, 17 September 2014 (UTC)

Continuity of the (positive definite) autocovariation function can be violated, but only if the Hilbert space is nonseparable. Boris Tsirelson (talk) 16:55, 17 September 2014 (UTC)

## Examples

Is it just me, or are the examples involving Y misleading? I understand the context in which they are stationary, but I think they might do more to confuse than inform. For evidence, just look at the debate on this page. I'd suggest either removing them entirely, or adding some sectioning above them to give a clearer indication that they are (in my opinion) "pathological examples".

briardew (talk) 16:35, 20 November 2013 (UTC)

## embedded maths notation formatting (possible Opera bug)

as an example:

for me running Opera, the maths in:

    Formally, let $\left\{X_t\right\}$ be a stochastic process...


looks all fucked up, wheras the maths in:

    and let $F_{X}(x_{t_1 + \tau}, \ldots, x_{t_k + \tau})$ represent the cumulative distribution function of the joint distribution of...


looks fine.

by all fucked up I mean lots of visual noise making the image almost unreadable.

woo-hoo my first constructive wikipedia edit, up until now i've just vandalised things ocasionally.

• What version of opera (and what platform, I guess) are you running? I just downloaded it and it looks the same to me (latest, on a mac). Protonk (talk) 23:58, 10 October 2014 (UTC)