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I don't have a stats background let alone one in extreme value theory, and I have what I imagine is a simple question but one I that haven't been able to find the answer to. The cumulative distribution function for a GEV distribution is:

$$F(x;\mu,\sigma,\xi)=\exp\left\{-\left[1+\xi\left(\frac{x-\mu}{\sigma}\right)\right]^{-1/\xi}\right\}$$

where $\xi, \mu$ and $\sigma$ represents a shape, location, and scale, respectively.

For this equation to work, $1 + \xi(x-\mu)/\sigma$ must be greater than zero.

My question is, what does one do when $1 + \xi(x-\mu)/\sigma$ is less than or equal to zero?

Given the GEV is a unification of the Gumbel, Fréchet and Weibull distributions, can I simply use the CDF for one of these distributions when the correct criteria apply. For example, if my shape parameter is negative (which seems to be the source of my issue) could I use the CDF for the Weibull distribution (e.g. see wikipedia)?

I'm repeating this process many times. I'm using R and the fevd function to fit my parameters, so I'm also not sure if my parameters are compatible with the three sub-families. I know there are functions to calculate the CDF but I'd like to do this manually.

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For example, if my shape parameter is negative (which seems to be the source of my issue) could I use the CDF for the Weibull distribution?

Sure you can, keeping in mind that the parametrizations used by packages may not be the ones you expect, so I suggest looking at the documentation.

Note also that $1+\xi(\frac{x-\mu}{\sigma})$ cannot be $0$ or negative, if the support is respected. The domain of the function is defined to be that way. When the said quantity is $0$ or less, then we are out of the support of GEV for that specific shape parameter. So the full definition of the GEV CDF would be, in the case of $\xi < 0$

$$F(x;\mu, \sigma, \xi)= \exp \left \{ -\left ( 1+\xi\left ( \frac{x-\mu}{\sigma} \right ) \right )^{\left\{ -\frac{1}{\xi} \right \}} \right \} \quad \text{for} \quad x\in(-\infty, \mu -\frac{\sigma}{\xi}) $$

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    $\begingroup$ Or in some conventions using two lines to specify $F(x)$ for the entire range $x \in [-\infty, \infty]$ $$F(x;\mu, \sigma, \xi)= \begin{cases} \exp ( - ( 1+\xi \frac{x-\mu}{\sigma} )^{ -\frac{1}{\xi} }) & \quad \quad \text{for} \quad x\in[-\infty, \mu -\frac{\sigma}{\xi}) \\ 1 & \quad \quad \text{for} \quad x\in[\mu -\frac{\sigma}{\xi}, \infty] \end{cases}$$ $\endgroup$
    – Sextus Empiricus
    Commented Aug 8, 2019 at 11:36
  • $\begingroup$ To be honest, this is the convention that I wanted to use in the first place. I was scared of being a bit redundant though, and went with the chosen one instead. I then took the chance to investigate more on CDF conventions, and I found this nice related thread on the matter, in which you @MartijnWeterings also participate. I suggest the read for anybody interested. $\endgroup$
    – Easymode44
    Commented Aug 8, 2019 at 12:52
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    $\begingroup$ Yes, my comment was exactly with that old thread in mind. As you can see from my point of view in that thread, I, Personally, would accept your shorthanded single line expression. (Although this question is an example why the convention to define cumulative distribution functions on the entire real line is useful. It is an example of the case that Alex R noted "A common mistake is to assume F(x)=x for all x which will give nonsense results.") $\endgroup$
    – Sextus Empiricus
    Commented Aug 8, 2019 at 13:05
  • $\begingroup$ Thanks @Easymode44 and Martijn for your input. To clarify, the GEV is only a unification of the three families under certain conditions, and when those conditions aren't met then one must refer back to whichever of those three families suits to get - in my case - the CDF? Also, should I expect parameters (like shape) that are fit for a GEV to be valid when using equations (like CDF) from one of the sub-families? If this should be posted as a new question please let me know. $\endgroup$
    – Nick
    Commented Aug 9, 2019 at 3:15
  • $\begingroup$ I guess I'm struggling to know what to do when 1 + ξ(x-μ)/σ <= 0, or even what it means (does my data not fit a GEV distribution?). $\endgroup$
    – Nick
    Commented Aug 9, 2019 at 4:33

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