I'm trying to understand the functional form of the Generalized Pareto distribution (GPD) presented in Wikipedia. In the "Definition" section location parameter $\mu$ does not appear in the function, whilst in the "Characterization" section it does. My question is:

  1. how the GPD form presented in Wikipedia can be reconciled with GPD form presented in other sources. (See pictures below). In the third picture, a sketch derivation of the GPD from the Generalized Extreme Value (GEV) distribution would suggest that $\frac{x-\mu}{\sigma}$ should not appear in that form in the expression of the GPD as presented in Wikipedia.

Picture 3 features an extract from Extreme Value Modeling and Risk Analysis: Methods and Applications, Dipak K. Dey, Jun Yan

Picture 1. Picture 2. Picture 3.


Basically, you have to read the article more carefully. They are all the same.

Wikipedia begins by stating:

It is specified by three parameters: location $\mu$, scale $\sigma$, and shape $\xi$. Sometimes it is specified by only scale and shape and sometimes only by its shape parameter.

(Emphasis mine.) Then in the Definition,

The standard cumulative distribution function (cdf) of the GPD is defined by $$F_\xi(z) = \begin{cases} 1 - (1+\xi z)^{-1/\xi}, & \xi \ne 0 \\ 1 - e^{-z}, & \xi = 0. \end{cases}$$

The use of the word "standard" in this context is analogous to the way we call $Z = \operatorname{Normal}(\mu = 0, \sigma = 1)$ a standard or standardized normal distribution; i.e., the location parameter is zero and the scale parameter is unity.

The article proceeds to generalize the above (again, as we do with the normal distribution) by replacing $z$ with $(x-\mu)/\sigma$ with location $\mu$ and scale $\sigma$. This three-parameter family is the same as what we see in your attached images, after accommodating for the emphasized sentence above.

  • $\begingroup$ That's great! many thanks @heropup. The last thing I still find confusing in Picture 3 is that $\mu$ appears in the GEV distribution and during derivation of GPD it disappears from numerator, and appears in the denominator. Can we just re-introduce it again to generalize GPD? Are the $\mu$'s the same in GEV and GPD? $\endgroup$ – Alex Feb 19 '18 at 10:31
  • $\begingroup$ @Alex No, they are not the same $\mu$. As you can see, they postulate from Equation 1.1 the existence of a parameter vector $\theta$ that fixes a $\mu$, $\sigma$, $\xi$ such that the following approximation for $F^n (z)$ holds. Then the conditional survival distribution is generalized Pareto, with shape $\xi$ and scale $\sigma_u$, and location $0$, with respect to the variable $y$. $\endgroup$ – heropup Feb 19 '18 at 20:25

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