2
$\begingroup$

The time series is used in a regression (OLS) and then the diagnostics are been run.

Or does stationarity imply homoskedasticity in all cases?

I get heteroskedasticity through a breusch pagan test but I get stationarity from a Unit Root (Dickey Fuller Test). Is this possible? Are these two mutually exclusive?

$\endgroup$
6
  • $\begingroup$ This question would make some sense if the word "regression" didn't appear in the title. But with it there, we need to know something about how you conceive of regression as applying to concepts of stationarity and heteroscedasticity in time series. $\endgroup$
    – whuber
    Dec 1, 2020 at 17:32
  • $\begingroup$ You run a regression and then do the diagnostics. Which part of my question is not clear? $\endgroup$
    – adrCoder
    Dec 1, 2020 at 17:34
  • $\begingroup$ Running a regression and doing diagnostics ordinarily have nothing to do with time series; and the concept of "stationarity" simply doesn't apply to regression. I'm guessing you might be trying to ask something about the Durbin-Watson test, but that's purely a guess. $\endgroup$
    – whuber
    Dec 1, 2020 at 17:35
  • 1
    $\begingroup$ The Durbin-Watson test has nothing to do with heteroskedasticity. It is for first-order autocorrelation. $\endgroup$ Dec 1, 2020 at 18:16
  • 1
    $\begingroup$ I meant the breusch pagan test. apologies. $\endgroup$
    – adrCoder
    Dec 1, 2020 at 19:42

1 Answer 1

2
$\begingroup$

Answering to the general question in the title of this post:

Assuming that by "stationarity" the OP means "second-order", "covariance-", "weak-" stationarity, i.e. the stationarity concept that relates to a constant mean and variance throughout a stochastic process, then the answer depends on whether the heteroskedasticity is conditional or unconditional.

If it is unconditional, it is incompatible with covariance-stationarity, because it negates it directly.

If it is conditional on certain co-variates, then, if these covariates are themselves stationary of the appropriate order (depending on how they enter the heteroskedastic function), the unconditional variance will be constant, and so conditional heteroskedasticity given such covariates is compatible with second-order stationarity.

But the above are relations at population-level.

Statistical tests are based on finite samples, and so the phenomenon where two statistical tests may provide conflicting indications is nothing new.

$\endgroup$
14
  • 1
    $\begingroup$ @adrCoder Yes it is. $\endgroup$ Dec 1, 2020 at 19:44
  • 1
    $\begingroup$ @adrCoder The Breusch-Pagan test test for heteroskedasticity that depends on a selection of variables determined by the researcher. So it is always "conditional". Dickey Fuller tests for the existence of a unit root. If it exists, it makes the unconditional variance non-stationary. $\endgroup$ Dec 1, 2020 at 20:00
  • 1
    $\begingroup$ A naive question: how can heteroskedasticity be unconditional? It seems that the very definition requires conditioning. E.g. our tag definition says Heteroscedasticity refers to the property of a random process that has non-constant variance along some continuum where it seems the conditioning is done w.r.t. this continuum. $\endgroup$ Dec 1, 2020 at 20:19
  • 1
    $\begingroup$ @RichardHardy The "continuum" part I don't really understand. "Heteroskedasticity" is a descriptive term, describing a situation where in a collection of a random variables, collected together for any reason, these variables have different variance. $\endgroup$ Dec 2, 2020 at 2:55
  • 1
    $\begingroup$ @adrCoder Nothing comes to mind, but this is a general and known issue: with finite samples, anything goes, because we never really know how actually well a sample represents the population (even if it is an i.i.d. sample). $\endgroup$ Dec 2, 2020 at 10:15

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.