91Ó°ÊÓ

\({X_1},...,{X_n}\) form a random sample from a beta distribution for which both parameters α and β are unknown. We need to show that \(\frac{{\Gamma '\left( \alpha \right)}}{{\Gamma \left( \alpha \right)}} - \frac{{\Gamma '\left( \beta \right)}}{{\Gamma \left( \beta \right)}} = \frac{1}{n}\sum\limits_{i = 1}^n {\log \frac{{{X_i}}}{{1 - {X_i}}}} \)

Let us assume that \({X_1},...,{X_n}\)has joint pdf \(f\left( {{x_1},...,{x_n};{\theta _1},...,{\theta _n}} \right)\)

The likelihood function of the beta distribution is given by the formula

\(f\left( {x;\alpha ,\beta } \right) = {\left[ {\frac{{\Gamma \left( {\alpha + \beta } \right)}}{{\Gamma \left( \alpha \right)\Gamma \left( \beta \right)}}} \right]^n}{\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)^{\alpha - 1}}{\left[ {\prod\limits_{i = 1}^n {\left( {1 - {x_i}} \right)} } \right]^{\beta - 1}}\)

Taking log of the likelihood function we get the following

\(\begin{array}{c}L\left( {\alpha ,\beta } \right) = \log f\left( {x;\alpha ,\beta } \right)\\ = n\log \Gamma \left( {\alpha + \beta } \right) - n\log \Gamma \left( \alpha \right) - n\log \Gamma \left( \beta \right)\\ + \left( {\alpha - 1} \right)\sum\limits_{i = 1}^n {\log {x_i}} + \left( {\beta - 1} \right)\sum\limits_{i = 1}^n {\log \left( {1 - {x_i}} \right)} \end{array}\)

Now, taking partial derivative and equating it with zero we get

\(\begin{array}{c}\frac{{\partial L\left( {\alpha ,\beta } \right)}}{{\partial \alpha }} = n\frac{1}{{\Gamma \left( {\alpha + \beta } \right)}}\Gamma '\left( {\alpha + \beta } \right) - n\frac{1}{{\Gamma \left( \alpha \right)}}\Gamma '\left( \alpha \right) + \sum\limits_{i = 1}^n {\log {x_i}} \\ = 0\end{array}\) and also

\(\begin{array}{c}\frac{{\partial L\left( {\alpha ,\beta } \right)}}{{\partial \beta }} = n\frac{1}{{\Gamma \left( {\alpha + \beta } \right)}}\Gamma '\left( {\alpha + \beta } \right) - n\frac{1}{{\Gamma \left( \beta \right)}}\Gamma '\left( \beta \right) + \sum\limits_{i = 1}^n {\log \left( {1 - {x_i}} \right)} \\ = 0\end{array}\)

Now solving the system of equation we get

\(\begin{array}{c}n\frac{1}{{\Gamma \left( {\alpha + \beta } \right)}}\Gamma '\left( {\alpha + \beta } \right) - n\frac{1}{{\Gamma \left( \alpha \right)}}\Gamma '\left( \alpha \right) + \sum\limits_{i = 1}^n {\log {x_i}} \\ = n\frac{1}{{\Gamma \left( {\alpha + \beta } \right)}}\Gamma '\left( {\alpha + \beta } \right) - n\frac{1}{{\Gamma \left( \beta \right)}}\Gamma '\left( \beta \right) + \sum\limits_{i = 1}^n {\log \left( {1 - {x_i}} \right)} \\\end{array}\)

On simplification we get

\(\frac{{\Gamma '\left( \alpha \right)}}{{\Gamma \left( \alpha \right)}} - \frac{{\Gamma '\left( \beta \right)}}{{\Gamma \left( \beta \right)}} = \frac{1}{n}\sum\limits_{i = 1}^n {\log \frac{{{X_i}}}{{1 - {X_i}}}} \)

Hence the proof.