I'm trying to understand the functional form of the Generalized Pareto distribution (GPD). In the "D

dumnealorjavgj

dumnealorjavgj

Answered question

2022-04-06

I'm trying to understand the functional form of the Generalized Pareto distribution (GPD). In the "Definition" section location parameter μ does not appear in the function, whilst in the "Characterization" section it does.
1. how the GPD form can be reconciled with GPD form presented in other sources. Derivation of the GPD from the Generalized Extreme Value (GEV) distribution would suggest that x μ σ should not appear in that form in the expression of the GPD.

Answer & Explanation

radcas87gex5r

radcas87gex5r

Beginner2022-04-07Added 13 answers

Basically, you have to read the article more carefully. They are all the same.
It is specified by three parameters: location μ, scale σ, and shape ξ. Sometimes it is specified by only scale and shape and sometimes only by its shape parameter.
Then in the Definition,
The standard cumulative distribution function (cdf) of the GPD is defined by
F ξ ( z ) = { 1 ( 1 + ξ z ) 1 / ξ , ξ 0 1 e z , ξ = 0.
The use of the word "standard" in this context is analogous to the way we call Z = Normal ( μ = 0 , σ = 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 μ ) / σ with location μ and scale σ.

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