91Ó°ÊÓ

The vectors \(\left( {{X_1},{Y_1}} \right),\left( {{X_2},{Y_2}} \right),...,\left( {{X_n},{Y_n}} \right)\) form a random sample of two-dimensional vectors from a bivariate normal distribution. We need to prove that the following

\(\sum\limits_{i = 1}^n {{X_i}} ,\sum\limits_{i = 1}^n {{Y_i}} ,\sum\limits_{i = 1}^n {X_i^2} ,\sum\limits_{i = 1}^n {Y_i^2} \,\,\,\,{\rm{and}}\,\,\,\,\sum\limits_{i = 1}^n {{X_i}{Y_i}} \)are jointly sufficient statistics

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 bivariate normal distribution with unknown parameters \({\mu _1},{\mu _2},{\sigma _1},{\sigma _2},\rho \) is

\(f\left( {x,y} \right) = \frac{1}{{2\pi \sqrt {1 - {\rho ^2}} {\sigma _1}{\sigma _2}}}\exp \left\{ { - \frac{1}{{2\left( {1 - {\rho ^2}} \right)}}\left( {\frac{{{{\left( {x - \mu } \right)}^2}}}{{{\sigma _1}^2}} - 2\rho \frac{{\left( {x - \mu } \right)}}{{{\sigma _1}}}\frac{{\left( {y - \mu } \right)}}{{{\sigma _2}}} + \frac{{{{\left( {y - \mu } \right)}^2}}}{{{\sigma _2}^2}}} \right)} \right\}\)

The above joint distribution is

\(\begin{align}\prod {f\left( {x,y} \right) = } {\left( {\frac{1}{{2\pi {\sigma _1}{\sigma _2}\sqrt {1 - {\rho ^2}} }}} \right)^n}\exp \left( { - \frac{1}{{2\left( {1 - {\rho ^2}} \right)}}\left( {\frac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - {\mu _1}} \right)}^2}} }}{{\sigma _1^2}} - 2\rho \frac{{\sum\limits_{i = 1}^n {\left( {{x_i} - {\mu _1}} \right)} }}{{{\sigma _1}}}\frac{{\sum\limits_{i = 1}^n {\left( {{y_i} - {\mu _2}} \right)} }}{{{\sigma _2}}} + \frac{{\sum\limits_{i = 1}^n {{{\left( {{y_i} - {\mu _2}} \right)}^2}} }}{{\sigma _2^2}}} \right)} \right)\\ = {\left( {\frac{1}{{2\pi {\sigma _1}{\sigma _2}\sqrt {1 - {\rho ^2}} }}} \right)^n}\exp \left( { - \frac{1}{{2\left( {1 - {\rho ^2}} \right)}}\left( {\frac{{\sum\limits_{i = 1}^n {x_i^2} - 2{\mu _1}\sum\limits_{i = 1}^n {{x_i}} + \mu _1^2}}{{\sigma _1^2}} - 2\rho \frac{{\sum\limits_{i = 1}^n {{x_i}{y_i}} - {\mu _1}\sum\limits_{i = 1}^n {{y_i}} - {\mu _2}\sum\limits_{i = 1}^n {{x_i}} + {\mu _1}{\mu _2}}}{{{\sigma _1}{\sigma _2}}} + \frac{{\sum\limits_{i = 1}^n {y_i^2} - 2{\mu _2}\sum\limits_{i = 1}^n {{y_i}} + \mu _2^2}}{{\sigma _2^2}}} \right)} \right)\\ = {\left( {\frac{1}{{2\pi {\sigma _1}{\sigma _2}\sqrt {1 - {\rho ^2}} }}} \right)^n}\exp \left( { - \frac{1}{{2\left( {1 - {\rho ^2}} \right)}}\left( {\frac{{\sum\limits_{i = 1}^n {x_i^2} }}{{\sigma _1^2}} - \frac{{2{\mu _1}\sum\limits_{i = 1}^n {{x_i}} }}{{\sigma _1^2}} + \frac{{\mu _1^2}}{{\sigma _1^2}} - 2\rho \left( {\frac{{\sum\limits_{i = 1}^n {{x_i}{y_i}} }}{{{\sigma _1}{\sigma _2}}} - \frac{{{\mu _1}\sum\limits_{i = 1}^n {{y_i}} }}{{{\sigma _1}{\sigma _2}}} - \frac{{{\mu _2}\sum\limits_{i = 1}^n {{x_i}} }}{{{\sigma _1}{\sigma _2}}} + \frac{{{\mu _1}\mu }}{{{\sigma _1}{\sigma _2}}}} \right) + \frac{{\sum\limits_{i = 1}^n {y_i^2} }}{{\sigma _2^2}} - \frac{{2{\mu _2}\sum\limits_{i = 1}^n {{y_i}} }}{{\sigma _2^2}} + \frac{{\mu _2^2}}{{\sigma _2^2}}} \right)} \right)\end{align}\)

From the statistic it is clear that it is sufficient to prove that the sums depend only on the following five statistics.

\(\sum\limits_{i = 1}^n {{{\left( {{x_i} - {\mu _1}} \right)}^2}} ,\sum\limits_{i = 1}^n {{{\left( {{y_i} - {\mu _2}} \right)}^2},} \sum\limits_{i = 1}^n {\left( {{x_i} - {\mu _1}} \right)\left( {{y_i} - {\mu _2}} \right)} \)

This is because

\(\begin{align}{\sum\limits_{i = 1}^n {\left( {{x_i} - {\mu _1}} \right)} ^2} &= \sum\limits_{i = 1}^n {{x_i}^2 - 2{\mu _1}} \sum\limits_{i = 1}^n {{x_i} + n} {\mu _1}^2\\{\sum\limits_{i = 1}^n {\left( {{y_i} - {\mu _2}} \right)} ^2} &= \sum\limits_{i = 1}^n {{y_i}^2 - 2{\mu _2}} \sum\limits_{i = 1}^n {{y_i} + n} {\mu _2}^2\\\sum\limits_{i = 1}^n {\left( {{x_i} - {\mu _1}} \right)} \left( {{y_i} - {\mu _2}} \right) &= \sum\limits_{i = 1}^n {{x_i}{y_i}} - {\mu _2}\sum\limits_{i = 1}^n {{x_i}} - {\mu _1}\sum\limits_{i = 1}^n {{y_i} + n{\mu _1}{\mu _2}} \end{align}\)

Now, it is obvious that

\(\sum\limits_{i = 1}^n {{X_i}} ,\sum\limits_{i = 1}^n {{Y_i}} ,\sum\limits_{i = 1}^n {X_i^2} ,\sum\limits_{i = 1}^n {Y_i^2} \,\,\,\,{\rm{and}}\,\,\,\,\sum\limits_{i = 1}^n {{X_i}{Y_i}} \)are jointly sufficient statistic.