91Ó°ÊÓ

Let \({X_1},...{X_n}\) be the random sample of size n from the specified distribution with density function \(f\left( {x,\theta } \right)\) where parameter \(\theta \).is unknown.

An estimator \(T = T\left( {{X_1},...{X_n}} \right)\) is said to be the sufficient estimator of \(\theta \) If conditional joint distribution\({X_1},...{X_n}\)given any value t of estimator is independent of \(\theta \)

Let \({X_1},...{X_n}\) be the random sample of size n from the specified distribution with density function \(f\left( {x,\theta } \right)\), here \(\theta \) is unknown An estimator \(T = T\left( {{X_1},...{X_n}} \right)\) is said to be the sufficient estimator of \(\theta \) if \(L\left( {x,\theta } \right) = g\left( {t,\theta } \right)h\left( x \right)\)

Here\(L\left( {x,\theta } \right)\)is the likelihood function of\({X_1},...{X_n}\).

\(g\left( {t,\theta } \right)\)is the function of\({X_1},...{X_n}\)which depends on\(\theta \)

\(h\left( x \right)\) is the s function which is independent on \(\theta \).

Let a continuous random variable x has the gamma distribution with parameter \(\alpha > 0\) and \(\beta > 0\)

Pdf of gamma distribution is given as:

\(f\left( x \right) = \frac{{{\beta ^\alpha }}}{{\left| {{\alpha \,}} \right. }}{x^{\alpha - 1}}{e^{ - \beta x}}\)

Using factorization theorem

\(L\left( p \right) = {\rm P}\left( {{X_1} = {x_1},{X_2} = {x_2}...{X_n} = {x_n}} \right)\)

\(\begin{align}L\left( p \right) &= \prod\limits_{i = 1}^n {\frac{{{\beta ^\alpha }}}{{\left ( {{\alpha \,}} \right. }}{x^{\alpha - 1}}{e^{ - \beta x}}} \\ &= \left\{ {\frac{{{\beta ^{n\alpha }}}}{{{{\left( {\left ( {{\alpha \,}} \right. } \right)}^n}}}{{\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)}^{\alpha - 1}}} \right\}\left\{ {{e^{ - \beta \sum\limits_{i = 1}^n {{x_i}} }}} \right\}\end{align}\)

One can observe that the first bracket is depending on\(\alpha \).

Hence by using factorization theorem, one can say thatThe statistic\(T = \prod\limits_{i = 1}^n {{X_i}} \)is sufficient statistic for the parameter\(\alpha \).