91Ó°ÊÓ

It is given that \(f\left( {x|\theta } \right) = a\left( \theta \right)b\left( x \right)\exp \left( {\sum\limits_{i = 1}^k {{c_i}\left( \theta \right){d_i}\left( x \right)} } \right)\)

Here, a and \({c_1},...,{c_k}\) are arbitrary functions of θ, and b and \({d_1},...,{d_k}\) are arbitrary functions of x. Suppose now that \({X_1},...,{X_n}\) form a random sample from a distribution which belongs to a k-parameter exponential family of this type, and define the k statistics \({T_1},...,{T_k}\) as follows:

\({T_i} = \sum\limits_{j = 1}^n {{d_i}\left( {{X_j}} \right)} \)

We need to show that the statistics \({T_1},...,{T_k}\)are jointly sufficient statistics for θ.

Fisher-Neyman Factorization Theorem

\({X_1},...,{X_n}\) be i.i.dr.v. 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 \(u\,\,\,{\rm{and}}\,\,\,\nu \) are non-negative functions.

The joint pdf of random variables can be written as

\(a{\left( \theta \right)^n}\exp \left( {\sum\limits_{i = 1}^k {{c_i}\left( \theta \right)\sum\limits_{j = 1}^k {{d_i}\left( {{x_j}} \right)} } } \right)\prod\limits_{j = 1}^n {b\left( {{x_j}} \right)} \)

Let us consider the test statistics \({T_i} = {r_i}\left( X \right)\,\,\,\,i = 1,2,....,k\) are the jointly sufficient statistics for \(\theta \) if and only if the joint probability density function \(f\left( {x|\theta } \right)\) can be factored as follows for all values of \(\theta \in \Omega \)

\(f\left( {x|\theta } \right) = u\left( x \right)\nu \left( {\left( {{r_1}\left( x \right){r_2}\left( x \right)...{r_n}\left( x \right)} \right),\theta } \right)\)

Here the functions u and v are non-negative functions.

Now, by the Fisher-Neyman Factorization Theorem

\(\nu \left( {{r_1}\left( x \right),...,{r_n}\left( x \right);\theta } \right)\)=\(a{\left( \theta \right)^n}\exp \left( {\sum\limits_{i = 1}^k {{c_i}\left( \theta \right)\sum\limits_{j = 1}^k {{d_i}\left( {{x_j}} \right)} } } \right)\) and \(u\left( x \right) = \prod\limits_{j = 1}^n {b\left( {{x_j}} \right)} \)

It follows that the statistics \({T_1},...,{T_k}\)are jointly sufficient statistics for θ.