5
$\begingroup$

What is the relationship between autocorrelation and non-stationary?

Is it true that non-zero autocorrelation $\implies$ non-stationary, but not vice versa?

$\endgroup$
  • $\begingroup$ What do you mean with "Is it true that autocorrelation"? as you stated it for me the crucial information/property about autocorrelation is missing: do you mean autocorrelation is = 0 or != 0, does (not) exist and is finite, decays slowly/fast, depends on the point in time rather than lag, ...? what property of the autocorrelation do you want to relate to non-stationarity? $\endgroup$ – Georg M. Goerg Aug 19 '15 at 3:03
  • $\begingroup$ What I meant was does autocorrelation imply non-stationary? $\endgroup$ – JPN Aug 19 '15 at 10:50
  • $\begingroup$ @GeorgM.Goerg there is an implication arrow in the statement. Maybe there is an issue with MathJax rendering in your browser? $\endgroup$ – ekvall Aug 19 '15 at 13:33
  • $\begingroup$ I do see the implication arrow of A => B, where A = "autocorrelation" and B = "non-stationary". however, for me just "autocorrelation" is not a statement alone; "autocorrelation = 0" or "autocorrelation != 0" is a statement. but now I guess OP meant that autocorrelation is non-zero, as is in the statement "time series with autocorrelation" where it;s implied you mean it's non-zero, as otherwise you wouldnt say it. I'll add that to original post to clarify $\endgroup$ – Georg M. Goerg Aug 19 '15 at 21:08
9
$\begingroup$

Neither is true. Consider the two following examples:

(1) Let $\xi \sim N(0,1)$ and define the stochastic process $X_i=\xi, i=1,2,\dots$. it's easy to check that this process is (strongly) stationary, while at the same time $\mathrm{Cov}(X_i, X_j)=1, \forall i,j$.

(2) Consider a stochastic process consisting of independent Gaussian variables $X_i \sim N(0, i), i=1,2,\dots$. This process is clearly not stationary, but the autocorrelation is zero for all lags since the variables are independent.

$\endgroup$
  • 2
    $\begingroup$ +1 You might prefer to write "zero autocorrelation," since many people would naturally interpret "no autocorrelation" as "the autocorrelation is not defined." $\endgroup$ – whuber Aug 18 '15 at 20:39
5
$\begingroup$

Autocorrelation doesn't cause non-stationarity. Non-stationarity doesn't require autocorrelation. I won't say they're not related, but they're not related the way you stated.

For instance, AR(1) process is autocorrelated, but it's stationary: $$x_t=c+\phi_1 x_{t-1}+\varepsilon_t$$ $$\varepsilon_t\sim\mathcal{N}(0,\sigma)$$

You can see that the unconditional mean is $E[x_t]=\frac{c}{1-\phi_1}$, i.e. stationary.

I(1) is non-stationary: $$x_t=x_{t-1}+\varepsilon_t$$ Here, errors $\varepsilon_t$ are not autocorrelated, but $x_t$ are autocorrelated: $\operatorname{cov}[x_t,x_{t-1}]\ne 0$.

Obviously, we want $|\phi_1|<1$, otherwise, AR(1) will blow up, as Richard Hardy noted.

$\endgroup$
  • $\begingroup$ @RichardHardy, for a math markup, they should really have \cov as an operator. $\endgroup$ – Aksakal Aug 18 '15 at 18:52
  • $\begingroup$ They don't seem to. But I am a very basic TeX user, so I might not be aware of the right way to write things. $\endgroup$ – Richard Hardy Aug 18 '15 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.