91Ó°ÊÓ

A gamma distribution for which both parameters α and β are unknown (α > 0 and β > 0), one need to check if \({T_1} = \prod\limits_{i = 1}^n {{X_i}} \) and \({T_2} = \sum\limits_{i = 1}^n {{x_i}} \)are jointly sufficient statistics.

By Fisher-Neyman factorization theorem,

Suppose \({X_1},...,{X_n}\) be i.i.d. random samples, 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, \(u{\rm{ }}and{\rm{ }}\nu \) are non-negative functions.

The pdf of the gamma distribution is

\(f\left( {x;\alpha ,\beta } \right) = \frac{1}{{\Gamma \left( \alpha \right)}}{\beta ^\alpha }{x^{\alpha - 1}}{e^{ - \beta x}};x \ge 0,\alpha > 0,\beta > 0\).

The joint pdf of the gamma distribution with unknown parameters \(\alpha \,\,\,\,{\rm{and}}\,\,\,\,\beta \) is

\(\begin{align}f\left( {x;\alpha ,\beta } \right) &= \prod\limits_{i = 1}^n {\left( {\frac{1}{{\Gamma \left( \alpha \right)}}{\beta ^\alpha }{x^{\alpha - 1}}{e^{ - \beta x}}} \right)} \\ &= \frac{{{\beta ^{n\alpha }}}}{{{{\left( {\Gamma \left( \alpha \right)} \right)}^n}}}{\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)^{\alpha - 1}}{e^{ - \beta \sum\limits_{i = 1}^n {{x_i}} }}\\ &= \frac{{{\beta ^{n\alpha }}}}{{{{\left( {\Gamma \left( \alpha \right)} \right)}^n}}}{\left( {{t_1}} \right)^{\alpha - 1}}{e^{ - \beta \sum\limits_{i = 1}^n {{x_i}} }}\end{align}\)

Now by Fisher-Neyman factorization theorem,

\(\nu \left( {{r_1}\left( x \right),{r_2}\left( x \right);\alpha ,\beta } \right] = \frac{1}{{{{\left ( {\left( \alpha \right) \}} \right. }^n}}}t_1^{\alpha - 1}{\beta ^{n\alpha }}{e^{ - \beta {t_2}}}\) and \(u\left( x \right) = 1\)

It follows that \({T_1} = \prod\limits_{i = 1}^n {{x_i}} \) and \({T_2} = \sum\limits_{i = 1}^n {{x_i}} \)are jointly sufficient statistics.