1
$\begingroup$

I am analysing a logarithmic returns series only to find the ADF result to signify the stationarity of the series. I understand that this is a way of differencing the original price series, however I am confused regarding how an I(0) process can be used to explain the presence of volatility clustering as this characteristic accounts for unequal variance as time progresses. Would the explanation have something to do with weak stationarity? And if so, how and why? Thanks in advance

$\endgroup$
  • 2
    $\begingroup$ An ARMA process can be stationary. A GARCH-type conditional variance equation is sort of the same as ARMA-type of conditional mean equation. $\endgroup$ – Richard Hardy Mar 27 '18 at 11:41
1
$\begingroup$

A stationary process can exhibit volatility clustering. This has nothing to do with weak stationarity, it is true for strict statioarity as well.

As you say, a stationary process must have the same variance for every point in time. But this doesn't apply to the variance of the distribution at some point in time conditioned on realized values on other points in time.

By volatility clustering one refers to either a conditional relationship or a feature of the sample paths. It doesn't mean that unconditional variances will differ from one point to another.

To see the difference, consider the following example:

Say you have a process: $...,X_{-2},X_{-1},X_0,X_1,X_2,...$ which only has the sample paths: $$...,1,1,0,0,0,1,1,0,0,0,1,1,0,0,0,...$$ $$...,0,1,1,0,0,0,1,1,0,0,0,1,1,0,0,...$$ $$...,0,0,1,1,0,0,0,1,1,0,0,0,1,1,0,...$$ $$...,0,0,0,1,1,0,0,0,1,1,0,0,0,1,1,...$$ $$...,1,0,0,0,1,1,0,0,0,1,1,0,0,0,1,...$$

Where each of the sample paths has equal probability.

This process is strictly stationary, while there is "clustering" of ones in its sample paths.

$\endgroup$

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.