91Ó°ÊÓ

The Probability Density Function (PDF) of Gamma Distribution is given by \[f(x|\theta,6)= \frac{1}{6^{\theta}\Gamma(\theta)}x^{\theta-1}e^{-\frac{x}{6}} \] for \( x > 0\) and \( \theta > 0\). Given a random sample \(X_1, X_2,...,X_n\) from the gamma distribution, the joint pdf would be product of the individual pdfs. \[f(x_1,x_2,...,x_n|\theta) = \prod \limits _{i=1} ^{n} f(x_i|\theta,6)\]

Apply natural logarithm to the joint pdf and simplify. \[ ln\: f(x_1,x_2,...,x_n|\theta) = \sum \limits _{i=1} ^{n} ln\: f(x_i|\theta,6) = n[\theta\: ln\:6 - ln\Gamma(\theta)]+ (\theta - 1) \sum \limits _{i=1} ^{n} ln\:x_i-\sum \limits _{i=1} ^{n} \frac{x_i}{6}\]

The factorization theorem claims that a statistic T(X) is a sufficient statistic for parameter \(\theta\) if the joint pdf \(f(x_1,x_2,...,x_n|\theta)\) can be written as a product of two functions g(T(x), \(\theta\)) and h(x) such that: g function does not contain x except through the function T(x) and h function does not contain \(\theta\). After having done the simplifications in the previous step, you can notice that we can factorize the result into a function of sum of \(ln\:x_i\) and a function of the sample observations. Therefore, the sum of \(ln\:x_i\) is a sufficient statistic for \(\theta\). As a product of sample observations is equivalent to an exponentiated sum of log of observations, \[ \prod \limits _{i=1} ^{n} x_i \] is a sufficient statistic for \(\theta\).