91Ó°ÊÓ

The given pdf is \(Beta\left( {\theta ,1} \right)\)

\(f\left( x \right) = \left\{ \begin{array}{l}\theta {x^{\theta - 1}},\,\,0 < x < 1\\0,\,\,{\rm{otherwise}}\end{array} \right.\)

Defining the likelihood function:

\(\begin{array}{c}L\left( \theta \right) = \prod\limits_{i = 1}^n {f\left( {{x_i}} \right)} \\ = \prod\limits_{i = 1}^n {\frac{{{x^{\alpha - 1}}{{\left( {1 - x} \right)}^{\beta - 1}}}}{{\beta \left( {\alpha ,\beta } \right)}}} \\ = \frac{{\prod\limits_{i = 1}^n {{x^{\alpha - 1}}} \prod\limits_{i = 1}^n {{{\left( {1 - x} \right)}^{\beta - 1}}} }}{{\beta {{\left( {\alpha ,\beta } \right)}^n}}}\end{array}\)

The log-likelihood is:

\(\ln L\left( \theta \right) = \left( {\alpha - 1} \right)\sum {\log {x_i} \times } \left( {\beta - 1} \right)\sum {\log \left( {1 - {x_i}} \right)} - n\log \beta \left( {\alpha ,\beta } \right)\)

Setting its derivative with respect to parameter:

\(\begin{array}{c}\frac{\partial }{{\partial \alpha }}\ln L\left( \theta \right) = \frac{\partial }{{\partial \alpha }}\left( {\alpha - 1} \right)\sum {\log {x_i} \times } \left( {\beta - 1} \right)\sum {\log \left( {1 - {x_i}} \right)} - n\log \beta \left( {\alpha ,\beta } \right)\\ = \sum {\log {x_i}} + 0 + 0\\ = \sum {\log {x_i}} \end{array}\)

\(\begin{array}{c}\frac{\partial }{{\partial \beta }}\ln L\left( \theta \right) = \frac{\partial }{{\partial \beta }}\left( {\alpha - 1} \right)\sum {\log {x_i} \times } \left( {\beta - 1} \right)\sum {\log \left( {1 - {x_i}} \right)} - n\log \beta \left( {\alpha ,\beta } \right)\\ = 0 + \sum {\log \left( {1 - {x_i}} \right)} + 0\\ = \sum {\log \left( {1 - {x_i}} \right)} \end{array}\)

Equating both of the equations to 0

\(\begin{array}{l}\sum {\log {x_i}} = 0\\\sum {\log \left( {1 - {x_i}} \right)} = 0\end{array}\)

Applying antilog on both sides we get

\(\prod {{X_i}} \,\,{\rm{and}}\,\,\prod {\left( {1 - {X_i}} \right)} \)

The log likelihood function is a decreasing function and it is maximised at \(\begin{array}{l}\alpha = \prod {{X_i}} \\\beta = \prod {\left( {1 - {X_i}} \right)} \end{array}\)

Since,

\(\alpha = \prod {{X_i}} \), in our case,\(\alpha = \theta \), therefore the MLE is

\(\widehat \theta = \prod {{X_i}} \)

Since all MLEs are consistent estimators, therefore, is proved.