91Ó°ÊÓ

\({X_1},...,{X_n}\) form a random sample from the beta distribution with parameters α and β.

We need to prove that the statistic T is a sufficient statistic for β

\(T = \frac{1}{n}\left( {\sum\limits_{i = 1}^n {\log \frac{1}{{1 - {X_i}}}} } \right)\)

\({\bf{T = }}\frac{{\bf{1}}}{{\bf{n}}}\left( {\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{\bf{log}}\frac{{\bf{1}}}{{{\bf{1 - }}{{\bf{X}}_{\bf{i}}}}}} } \right)\)

Fisher-Neyman Factorization Theorem

\({X_1},...,{X_n}\) be i.i.dr.v. with pdf \(f\left( {x;\theta } \right)\) and let \(T = r\left( {{X_1},...,{X_n}} \right)\) be a statistic.T is sufficient statistic for \(\theta \)iff

\(f\left( {x;\theta } \right) = u\left( x \right)\nu \left( {r\left( x \right);\theta } \right)\) where \(u\,\,\,{\rm{and}}\,\,\,\nu \) are non-negative functions.

The joint pdf is given by

\(f\left( {x;\theta } \right) = {\alpha ^n}\left( \theta \right){e^{c\left( \theta \right)t}}\prod\limits_{i = 1}^n {b\left( {{x_i}} \right)} \,\,\,\,\,\,,t = \sum\limits_{i = 1}^n {d\left( {{x_i}} \right)} \)

By Fisher-Neyman Factorization Theorem we get

\(\begin{align}\nu \left( {r\left( x \right);\theta } \right) &= {\alpha ^n}\left( \theta \right){e^{c\left( \theta \right)t}};\\u\left( x \right) &= \prod\limits_{i = 1}^n {b\left( {{x_i}} \right)} \end{align}\)

It follows that \(T = \sum\limits_{i = 1}^n {d\left( {{X_i}} \right)} \) is a sufficient statistic for θ.

For the beta distribution with known parameter \(\alpha \) and unknown parameter \(\beta \) we have the following

\(\begin{align}f\left( {x;\beta } \right) &= \frac{{\left ( {{\left( {\alpha + \beta } \right) \,}} \right. }}{{\left ( {{\left( \alpha \right) \,}} \right. }}{x^{\alpha - 1}}{\left( {1 - x} \right)^{\beta - 1}}\\ &= \frac{{\left ( {{\left( {\alpha + \beta } \right) \,}} \right. }}{{\left ( {{\left( \alpha \right) \,}} \right. }}{x^{\alpha - 1}}\exp \left\{ {\left( {\beta - 1} \right)\ln \left( {1 - x} \right)} \right\}\end{align}\)

Hence the statistic

\(T' = \sum\limits_{i = 1}^n {\log \left( {1 - x} \right)} \) is a sufficient statistic.

Now \(T = \frac{1}{n}\left( {\sum\limits_{i = 1}^n {\log \frac{1}{{1 - {X_i}}}} } \right)\)is a one-one transformation of \(T' = \sum\limits_{i = 1}^n {\log \left( {1 - x} \right)} \)then we can conclude that \(T = \frac{1}{n}\left( {\sum\limits_{i = 1}^n {\log \frac{1}{{1 - {X_i}}}} } \right)\)is a sufficient statistic for \(\beta \)