91Ó°ÊÓ

\({X_1},...,{X_n}\) is drawn from the Pareto distribution with parameters \({x_0}\,\,\,\,{\rm{and}}\,\,\,\alpha \) .

We need to calculate

a. If \({x_0}\) is known and \(\alpha > 0\) unknown, find a sufficient statistic

b. If \(\alpha \) is known and \({x_0}\) unknown, find a sufficient statistic.

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 the Pareto distribution with parameters \({x_0}\,\,\,\,{\rm{and}}\,\,\,\alpha \) is

\(f\left( {x;\alpha } \right) = {\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)^{ - \left( {\alpha + 1} \right)}}{\alpha ^n}x_0^{\alpha n}\)

When \(t = \prod\limits_{i = 1}^n {{x_i}} \)by Fisher-Neyman Factorization Theorem we get

\(\begin{align}\nu \left( {r\left( x \right);\alpha } \right) &= {t^{ - \left( {\alpha + 1} \right)}}{\alpha ^n}x_0^{\alpha n};\\u\left( x \right) &= 1\end{align}\)

It follows that \(T = \prod\limits_{i = 1}^n {{X_i}} \)is a sufficient statistic for parameter \(\alpha \)

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

The joint pdf of the Pareto distribution with parameters \({x_0}\,\,\,\,{\rm{and}}\,\,\,\alpha \) is

\(f\left( {x;\alpha } \right) = {\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)^{ - \left( {\alpha + 1} \right)}}{\alpha ^n}x_0^{\alpha n}\)

When \(t = \min \left\{ {{x_1},...,{x_n}} \right\}\) by Fisher-Neyman Factorization Theorem

\(\begin{align}\nu \left( {r\left( x \right);{x_0}} \right) &= x_0^{\alpha n}\,\,\,\,\,,t \ge {x_0}\\u\left( x \right) &= {\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)^{ - \left( {\alpha + 1} \right)}}{\alpha ^n}\end{align}\)

It follows that the statistic \(T = \min \left\{ {{X_1},...,{X_n}} \right\}\)is sufficient statistic for parameter \({x_0}\)