El Weibull MLE solo tiene solución numérica:
Deje
conβ,
fλ,β(x)={βλ(xλ)β−1e−(xλ)β0,x≥0,x<0
.
β,λ>0
1) Likelihoodfunction :
Lx^(λ,β)=∏i=1Nfλ,β(xi)=∏i=1Nβλ(xiλ)β−1e−(xiλ)β=βNλNβe−∑Ni=1(xiλ)β∏i=1Nxβ−1i
log-Likelihoodfunction :
ℓx^(λ,β):=lnLx^(λ,β)=Nlnβ−Nβlnλ−∑i=1N(xiλ)β+(β−1)∑i=1Nlnxi
2) Problema MLE :
3) Maximizaciónpor0-gradientes:
∂ l
max(λ,β)∈R2s.t.λ>0β>0ℓx^(λ,β)
0
Sigue:
-Nβ1∂l∂λ∂l∂β=−Nβ1λ+β∑i=1Nxβi1λβ+1=Nβ−Nlnλ−∑i=1Nln(xiλ)eβln(xiλ)+∑i=1Nlnxi=!0=!0
−Nβ1λ+β∑i=1Nxβi1λβ+1−β1λN+β1λ∑i=1Nxβi1λβ−1+1N∑i=1Nxβi1λβ1N∑i=1Nxβi=0=0=0=λβ
⇒λ∗=(1N∑i=1Nxβ∗i)1β∗
λ∗ into the second 0-gradient condition:
⇒β∗=[∑Ni=1xβ∗ilnxi∑Ni=1xβ∗i−lnx¯¯¯¯¯¯¯¯]−1
This equation is only numerically solvable, e.g. Newton-Raphson algorithm. β^∗ can then be placed into λ∗ to complete the ML estimator for the Weibull distribution.