Intuitively, a stationary stochastic process needs to be mean-reverting. This should follow immediately from the definition of stationarity: the mean of the process needs to be constant over time, so when the process deviates from the mean, it should go back to it sooner or later.

Is this reasoning correct? How can one prove it formally?

  • 1
    $\begingroup$ The reasoning is correct. You can prove it using the law of large numbers. $\endgroup$
    – Kian
    Apr 15, 2015 at 4:30
  • $\begingroup$ Not a formal answer but i would say that the stationarity process has a special dynamic with a certain mean and that all other innovations are iid and zero expected value so it is necessarily mean reverting since innovations cancel out. $\endgroup$
    – Malick
    Apr 15, 2015 at 9:22
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    $\begingroup$ What is your definition of "mean reverting"? $\endgroup$
    – q.t.f.
    Apr 15, 2015 at 18:54
  • $\begingroup$ This is a good question! Actually, I don't have a precise definition of "mean-reverting process". Can you help and suggest a good reference for it? $\endgroup$
    – Abramo
    Apr 15, 2015 at 22:35

4 Answers 4


The concept of 'mean reversion' is tricky in continuous time. Most people would call 'mean reverting' a process where the drift pulls back towards a long run mean, and I assume that this is what you also mean. Something like the drift of an OU process.

However, in continuous time the 'pull' can be generated by the volatility. For example the process $$ dX_t = dt+X_t^2 dW_t $$ is stationary although the drift seems to be pushing the paths towards infinity. The first place I saw that behaviour was the Conley et al 1997 paper (bottom of pg 12).

In these processes the 'pull' is caused by the volatility, and in this example it is sufficient to overcome the drift. For general processes $X_t = \mu(X_t)+\sigma(X_t)dW_t$ this 'pull' is quantified by the scale density $$s(x)=\exp\left(-\int^x \frac{\mu(u)}{\sigma^2(u)}du\right)$$

I do not think that these things happen for discrete time processes.

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    $\begingroup$ As it is not clear why this process is stationary, can you please provide more details? $\endgroup$
    – Gordon
    Apr 15, 2015 at 16:02
  • $\begingroup$ Seems weakly stationary to me. $\endgroup$
    – John
    Apr 15, 2015 at 17:34
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    $\begingroup$ Is it obvious, that this process is well-defined for all t > 0? Otherwise, I wouldn't call it a legit counter-example. $\endgroup$
    – LazyCat
    Apr 28, 2015 at 21:38
  • $\begingroup$ As some questions were asked on the original post, I edited to make the answer clearer, put a better example, and added reference. $\endgroup$
    – Kiwiakos
    Apr 28, 2015 at 23:20
  • $\begingroup$ Thanks for the reference to the paper. It's very helpful - I wasn't aware of this phenomenon. $\endgroup$
    – LazyCat
    Apr 29, 2015 at 0:49

Let's consider the following example: the process is initialized randomly with $\pm1$ and then stays there forever. Seems stationary to me, but it would never cross its mean.

  • $\begingroup$ Nice one! But in this example the long-run mean is not convergent... right? $\endgroup$
    – Abramo
    May 6, 2015 at 17:00
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    $\begingroup$ Yes. Such a stationary process though is not ergodic (which is why an assumption of ergodicity is often added). $\endgroup$ Aug 31, 2018 at 16:20

Suppose we estimate the regression model

$$\triangle y_{t}=\alpha + \beta y_{t-1}+\varepsilon_{t}$$

This is actually quite similar to the Dickey-Fuller test. If $\beta=0$, then the process has a unit root. Let's proceed assuming that $\beta<0$, i.e. that the process is stationary.

The first equation is also similar to the continuous time Ornstein-Uhlenbeck process

$$dy_{t}=k\left(m-y_{t}\right)+\sigma dW_{t}$$

where the mean-reverting level is $m$ and the speed by which it mean-reverts is $k$.

For the first equation, it can be shown that


So long as $\beta\neq 0$, which we already assumed, we know that it has a mean-reverting level that corresponds to the Ornstein-Uhlenbeck process.

  • 2
    $\begingroup$ Hi, I am not sure you are answering the question. You chose a very specific stationary process and claimed that it is mean reverting. That says very little about stationary processes in general. $\endgroup$
    – AFK
    Apr 15, 2015 at 19:46

The answer is no, because although a mean-reverting process has necessarily to be stationary, it is not true the opposite, that is a stationary process has to be mean-reverting, as you stated in the question.

Look at this article for a formal definition of a mean-reverting process.

Now, think about one of the most famous mean-reverting process: the Ornstein–Uhlenbeck; the assumption underlying such process are the following:

  • Stationarity;
  • Normality;
  • Markovianity;

The stationarity is a necessary assumption in order that such process is mean-reverting, but it is not true the opposite. The mean reverting process assumption is not a necessary condition such that a process is stationary.

For reference, look at the books I cited to have an idea of the application of such concepts in finance; you can find them online for free in pdf format. Since this is a quantitative finance site, I suppose the question indirectly refers to a pair-trading strategy, so, I think it is important to suggest you to read:

Chan, Ernest P. "Quantitative Trading." New Jersey (2008).

Particularly, read the paragraph about stationarity condition and cointegration (chp. 7). There, the author suggests:

[]. You can find a pair of stocks such that if you long one and short the other, the market value of the pair is stationary. If this is the case, then the two individual time series are said to be cointegrated. They are so described because a linear combination of them is integrated of order zero. []

Again, it is not true that a stationary process implies necessarily that the time series composing the process are cointegrated, but it is true the opposite ones; that is, if 2 time series are cointegrated, necessarily their linear combination (in the right proportions) is a stationary process.

Said that, in quantitative finance terms, it is not necessary that the 2 processes you are considering have to be stationary, but that it necessary their market value is stationary; in such case, the process resulting from the linear combination of those two time series is a good candidate for a pair-trading strategy.

Moreover, for a formal proof, look at:

Alexander, Carol. Market models: a guide to financial data analysis. John Wiley & Sons, 2001.

He provides a formal proof of what I wrote down above and answers exactly to your question.

  • 1
    $\begingroup$ Hi Quantopic. Please, what is your definition of "mean-reverting"? Moreover, why does the fact that 2 cointegrated processes give rise to a stationary one implies that the answer to my question is "no"? Please expand on this point. $\endgroup$
    – Abramo
    Apr 28, 2015 at 20:38
  • $\begingroup$ Hi @Abramo! I updated the answer; I hope everything will be clearer; I suggest to read the link about the formal definition of a mean-reverting process that's contain also some application to financial markets. I think it could be of interest for you, if you're still interested to the topic. $\endgroup$
    – Quantopik
    Apr 28, 2015 at 21:31
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    $\begingroup$ Thanks for the references, they are surely interesting. However, the "paper" you claim to give a formal definition does not. It only introduces 3 classes of models which are mean-reverting, but there's no formal definition of the concept itself $\endgroup$
    – Abramo
    Apr 29, 2015 at 7:14

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