0
$\begingroup$

I apologize in advance if this proves to be a nonsensical question but it is something I have been struggling with.

Is there a way to prove/discern that covariance stationarity is not an attribute of the sample at hand and is an attribute of the underlying process?

edit:

The motivation of the question comes from unit root testing; when we perform unit root testing on a series, we may accept the null ie that the series in question, has a unit root on one sample, but reject the null on a different (sub-)sample.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Covariance stationarity is described as constancy over time of the first two moments of $X_t$. This is clearly a property of the underlying process: the moments are a process property, not a sample property.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Since samples have moments too, this reasoning would seem not to answer the question. $\endgroup$
    – whuber ♦
    Commented Jan 12, 2018 at 15:22
  • $\begingroup$ Let me add, then: "Clearly, the constancy implied by stationarity rules out that the moments are sample moments, for randomness would prevent these from being constant." $\endgroup$
    – F. Tusell
    Commented Jan 12, 2018 at 16:04
  • $\begingroup$ That helps. I'm not sure it addresses the spirit of the question, though. The question seems to concern making inferences about a process from a sample. It appears to ask, "how confidently may we conclude that the process is stationary based on properties of a sample?" To me, that looks like a version of the basic question, "how can we test any statistical hypothesis based on a sample?" From this point of view, observing that the sample can vary because it's random is correct but not terribly useful. $\endgroup$
    – whuber ♦
    Commented Jan 12, 2018 at 16:16
  • 1
    $\begingroup$ May be you are right. I thought I understood the question on my first reading. I am no longer sure, so I will refrain from making additional comments that rather than being of help may obfuscate the matter. $\endgroup$
    – F. Tusell
    Commented Jan 12, 2018 at 16:42

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.