Processing math: 100%

Generalized Pareto Distribution

Shape parameter c0 and defined for x0 for all c and x<1|c| if c is negative.

f(x;c)=(1+cx)11cF(x;c)=11(1+cx)1/cG(q;c)=1c[(11q)c1]
M(t)={(tc)1cetc[Γ(11c)+Γ(1c,tc)πcsc(πc)/Γ(1c)]c>0(|c|t)1/|c|Γ[1|c|,t|c|]c<0
μn=(1)ncnnk=0(nk)(1)k1ckcn<1
μ1=11cc<1μ2=2(12c)(1c)c<12μ3=6(1c)(12c)(13c)c<13μ4=24(1c)(12c)(13c)(14c)c<14

Thus,

μ=μ1μ2=μ2μ2γ1=μ33μμ2μ3μ3/22γ2=μ44μμ36μ2μ2μ4μ223
h[X]=1+cc>0

Implementation: scipy.stats.genpareto