# Autocorrelation vs Non-stationary

What is the relationship between autocorrelation and non-stationary?

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

• 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? – Georg M. Goerg Aug 19 '15 at 3:03
• What I meant was does autocorrelation imply non-stationary? – JPN Aug 19 '15 at 10:50
• @GeorgM.Goerg there is an implication arrow in the statement. Maybe there is an issue with MathJax rendering in your browser? – ekvall Aug 19 '15 at 13:33
• 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 – Georg M. Goerg Aug 19 '15 at 21:08

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.

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

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.

• @RichardHardy, for a math markup, they should really have \cov as an operator. – Aksakal Aug 18 '15 at 18:52
• 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. – Richard Hardy Aug 18 '15 at 18:56