1
$\begingroup$

enter image description here

source: Embrechts pg 165, 354 (3.52) G is Generalized Pareto Distribution

Base on that theorem, could I conclude that 1) if I have a data and the excess fit with Generalized Pareto Distribution, then the data fit Generalized Extreme Value Distribution? 2) the shape parameter of Generalized Pareto Distribution is the same as Generalized Extreme Value Distribution?

$\endgroup$

2 Answers 2

2
$\begingroup$

enter image description here

enter image description here

Hi,

See Coles(2001) "An Introduction to Statistical Modeling of Extreme Values" Pgs. 75-76.

I think it can help. enter image description here

$\endgroup$
2
  • 1
    $\begingroup$ This looks potentially useful, Maria, but without the context of the referenced equations $(4.2)$ and $(4.3)$ it's not sufficiently clear to answer the question. Perhaps you could add an explanation of the quoted text? $\endgroup$
    – whuber
    Oct 30, 2019 at 14:00
  • $\begingroup$ @whuber♦: condition X>u is threshold_exceeding-condition & 4.2 is CDF under such case - here page2. Further Maria A. describes constraints for y (as is for lim for integration), as I assume... $\endgroup$
    – JeeyCi
    Jun 18, 2023 at 12:11
0
$\begingroup$

look to An Introduction to Extreme Value Statistics

  1. seems, not - because "GP function can be approximated as a tail of a GEV" - in any case it often is used to approximate tails of ANY another distribution (Even Normal, but GEV is mostly used for imbalanced with rare-events cases). Besides, you should have enough data, because "with extreme Outliers GEV model seems to do worse than GP models"
  2. seems, yes - however, "Temporal dependence has a much stronger effect in the GP approximation." hence this can skew the distribution of excesses. But entropy is calculated a little bit differently for GEV & GP -- so the difference [usually not for truly extremes, but for outliers i.e.] in fitting could be caused by fitting algo itself , not by parameter shape - I tend to consider it to have the same meaning, though still didn't work with these models
$\endgroup$

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.