Question #202773

Let X1,X2,.....,Xbe random sample of size n from a distribution with probability 

density function 

f(X; ,0)={ θX^(θ-1), 0<X<1, θ>0; 0, elsewhere

Obtain a maximum likeyhood estimator of θ.


1
Expert's answer
2021-06-04T11:21:15-0400

Solution:

Given,

f(x;θ)={θxθ1,0<x<1,θ>0                        0, elsewheref(x; θ)=\{ θx^{θ-1}, 0<x<1, θ>0 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0, \ elsewhere

L(θ)=Πi=1n f(xi;θ)=Πi=1n θxiθ1=θx1θ1.θx2θ1.θx3θ1....θxnθ1=θnxiθ1L(θ)= \Pi_{i=1}^n \ f(x_i; θ)= \Pi_{i=1}^n \ θx_i^{θ-1} \\=θx_1^{θ-1}.θx_2^{θ-1}.θx_3^{θ-1}....θx_n^{θ-1} \\=θ^n{x_i}^{θ-1}

logL(θ)=log(θnxiθ1)=log(θn)+log(xiθ1)=nlogθ+(θ1)logxi\Rightarrow \log L(θ)=\log (θ^n{x_i}^{θ-1})=\log (θ^n)+\log ({x_i}^{θ-1}) \\=n \logθ+(θ-1)\log x_i

d logL(θ)dθ=nθ+logxi\Rightarrow \dfrac{d \ \log L(θ)}{dθ}=\dfrac nθ+\log x_i

Put d logL(θ)dθ=0\dfrac{d \ \log L(θ)}{dθ}=0

nθ+logxi=0nθ=logxiθ=nlogxi=nlog(xin.n)=n(logxˉ)+(logn)\dfrac nθ+\log x_i=0 \\ \Rightarrow \dfrac nθ=-\log x_i \\ \Rightarrow θ=\dfrac{n}{-\log x_i}=\dfrac{-n}{\log (\dfrac{x_i}{n}.n)}=\dfrac{-n}{(\log \bar x)+(\log n)}

Thus, MLE =θ^=n(logxˉ)+(logn)=\hat θ=\dfrac{-n}{(\log \bar x)+(\log n)}


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