Solution:
Given,
f(x;θ)={θxθ−1,0<x<1,θ>0 0, elsewhere
L(θ)=Πi=1n f(xi;θ)=Πi=1n θxiθ−1=θx1θ−1.θx2θ−1.θx3θ−1....θxnθ−1=θnxiθ−1
⇒logL(θ)=log(θnxiθ−1)=log(θn)+log(xiθ−1)=nlogθ+(θ−1)logxi
⇒dθd logL(θ)=θn+logxi
Put dθd logL(θ)=0
θn+logxi=0⇒θn=−logxi⇒θ=−logxin=log(nxi.n)−n=(logxˉ)+(logn)−n
Thus, MLE =θ^=(logxˉ)+(logn)−n
Comments