8
$\begingroup$

I am a student studying time series econometric online and looking for help talking around some of the principles. There is one point I am trying to understand better.

I've come across some literature online which talks about how non-stationary data series have slowly decaying ACF. The graph of the ACF of VWAP in this post is a good example: https://coolstatsblog.com/2013/08/07/how-to-use-the-autocorreation-function-acf/

Could someone explain why a slowly decaying ACF is an indication that a series is non stationary? I think hearing an explanation will help me get a better understanding of the material.

Thanks very much, Adrian

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ You've asked four questions and gotten a load of answers. Could you see if your questions have actually been answered and flag a correct one if they have? $\endgroup$
    – Thorst
    Commented Nov 16, 2016 at 6:24
  • 1
    $\begingroup$ Be careful though. To point out the obvious; because an ACF is decaying slowly doesn't necessarily mean it is non-stationary. You may get a slowly decaying ACF because the series has high persistence even though it lacks a unit root overall, e.g. if it is near unit root but I(0). $\endgroup$
    – Robert Brown
    Commented Dec 23, 2016 at 0:42

2 Answers 2

Reset to default
2
$\begingroup$

The (simplified) definition of a stationary process is that the mean and variance of the process are constant over time.

If the ACF is slowly decaying, that means future values of the series are correlated / heavily affected by past values. If past values of the series are high, the future values should also be high. If the value goes up, future values will also go up (assuming positive auto-correlation here). This means the mean will change over time, which means that the process is non-stationary.

Improve this answer
$\endgroup$
1
$\begingroup$

To answer directly why the slow decay in the ACF is an indication that a series is non-stationary, it is showing that the ratio of γs and γ0 (see below equations) is not approaching 0. The s in γs is what is represented on the x axis on the correlogram.

γs and γ0 are represented by:

γ0 = σ^2/[1-(α1)^2] γs = σ^2(α1)^s/[1-(α1)^2]

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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