91Ó°ÊÓ

\({X_1},...,{X_n}\) form a random sample from a distribution that belongs to an exponential family of distributions as defined in Exercise 23 of Sec. 7.3. We need to prove that

\(T = \sum\limits_{i = 1}^n {d\left( {{X_i}} \right)} \) is a sufficient statistic for θ.

Fisher-Neyman Factorization Theorem

\({X_1},...,{X_n}\) be i.i.d r.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.

Now, in the question \({X_1},...,{X_n}\) form a random sample from a distribution that belongs to an exponential family of distributions. The joint pdf is given by

\(f\left( {x;\theta } \right) = {\alpha ^n}\left( \theta \right){e^{c\left( \theta \right)t}}\prod\limits_{i = 1}^n {b\left( {{x_i}} \right)} \,\,\,\,\,\,,t = \sum\limits_{i = 1}^n {d\left( {{x_i}} \right)} \)

By Fisher-Neyman Factorization Theorem we get

\(\begin{align}\nu \left( {r\left( x \right);\theta } \right) &= {\alpha ^n}\left( \theta \right){e^{c\left( \theta \right)t}};\\u\left( x \right) &= \prod\limits_{i = 1}^n {b\left( {{x_i}} \right)} \end{align}\)

It follows that \(T = \sum\limits_{i = 1}^n {d\left( {{X_i}} \right)} \) is a sufficient statistic for θ.