Editing Sufficient statistic (section)

===Proof===
Due to Hogg and Craig.<ref name="HoggCraig">{{cite book | last = Hogg | first = Robert V. |author2=Craig, Allen T.  | title = Introduction to Mathematical Statistics | publisher=Prentice Hall | year = 1995 | isbn=978-0-02-355722-4}}</ref> Let <math>X_1, X_2, \ldots, X_n</math>,  denote a random sample from a distribution having the [[Probability density function|pdf]] ''f''(''x'',&nbsp;''θ'') for ''ι''&nbsp;<&nbsp;''θ''&nbsp;<&nbsp;''δ''. Let ''Y''<sub>1</sub>&nbsp;=&nbsp;''u''<sub>1</sub>(''X''<sub>1</sub>,&nbsp;''X''<sub>2</sub>,&nbsp;...,&nbsp;''X''<sub>''n''</sub>) be a statistic whose pdf is ''g''<sub>1</sub>(''y''<sub>1</sub>;&nbsp;''θ''). What we want to prove is that ''Y''<sub>1</sub>&nbsp;=&nbsp;''u''<sub>1</sub>(''X''<sub>1</sub>, ''X''<sub>2</sub>,&nbsp;...,&nbsp;''X''<sub>''n''</sub>) is a sufficient statistic for ''θ'' if and only if, for some function ''H'',

:<math> \prod_{i=1}^n f(x_i; \theta) = g_1 \left[u_1 (x_1, x_2, \dots, x_n); \theta \right] H(x_1, x_2, \dots, x_n). </math>

First, suppose that
:<math> \prod_{i=1}^n f(x_i; \theta) = g_1 \left[u_1 (x_1, x_2, \dots, x_n); \theta \right] H(x_1, x_2, \dots, x_n). </math>

We shall make the transformation ''y''<sub>''i''</sub>&nbsp;=&nbsp;''u''<sub>i</sub>(''x''<sub>1</sub>,&nbsp;''x''<sub>2</sub>,&nbsp;...,&nbsp;''x''<sub>''n''</sub>), for ''i''&nbsp;=&nbsp;1,&nbsp;...,&nbsp;''n'', having inverse functions ''x''<sub>''i''</sub>&nbsp;=&nbsp;''w''<sub>''i''</sub>(''y''<sub>1</sub>,&nbsp;''y''<sub>2</sub>,&nbsp;...,&nbsp;''y''<sub>''n''</sub>), for ''i''&nbsp;=&nbsp;1,&nbsp;...,&nbsp;''n'', and [[Jacobian matrix and determinant|Jacobian]] <math> J = \left[w_i/y_j \right] </math>. Thus,

:<math>
 \prod_{i=1}^n f \left[ w_i(y_1, y_2, \dots, y_n); \theta \right]  = 
 |J| g_1 (y_1; \theta) H \left[ w_1(y_1, y_2, \dots, y_n), \dots, w_n(y_1, y_2, \dots, y_n) \right].
</math>

The left-hand member is the joint pdf ''g''(''y''<sub>1</sub>, ''y''<sub>2</sub>, ..., ''y''<sub>''n''</sub>; θ) of ''Y''<sub>1</sub> = ''u''<sub>1</sub>(''X''<sub>1</sub>, ..., ''X''<sub>''n''</sub>), ..., ''Y''<sub>''n''</sub> = ''u''<sub>''n''</sub>(''X''<sub>1</sub>, ..., ''X''<sub>''n''</sub>).  In the right-hand member, <math>g_1(y_1;\theta)</math> is the pdf of <math>Y_1</math>, so that <math>H[ w_1, \dots , w_n] |J|</math> is the quotient of <math>g(y_1,\dots,y_n;\theta)</math> and <math>g_1(y_1;\theta)</math>; that is, it is the conditional pdf <math>h(y_2, \dots, y_n \mid  y_1; \theta)</math> of <math>Y_2,\dots,Y_n</math> given <math>Y_1=y_1</math>.

But <math>H(x_1,x_2,\dots,x_n)</math>, and thus <math>H\left[w_1(y_1,\dots,y_n), \dots, w_n(y_1, \dots, y_n))\right]</math>, was given not to depend upon <math>\theta</math>. Since <math>\theta</math> was not introduced in the transformation and accordingly not in the Jacobian <math>J</math>, it follows that <math>h(y_2, \dots, y_n \mid y_1; \theta)</math> does not depend upon <math>\theta</math> and that <math>Y_1</math> is a sufficient statistics for <math>\theta</math>.

The converse is proven by taking:

:<math>g(y_1,\dots,y_n;\theta)=g_1(y_1; \theta) h(y_2, \dots, y_n \mid y_1),</math>

where <math>h(y_2, \dots, y_n \mid y_1)</math> does not depend upon <math>\theta</math> because <math>Y_2 ... Y_n</math> depend only upon <math>X_1 ... X_n</math>, which are independent on <math>\Theta</math> when conditioned by <math>Y_1</math>, a sufficient statistics by hypothesis. Now divide both members by the absolute value of the non-vanishing Jacobian <math>J</math>, and replace <math>y_1, \dots, y_n</math> by the functions <math>u_1(x_1, \dots, x_n), \dots, u_n(x_1,\dots, x_n)</math> in <math>x_1,\dots, x_n</math>. This yields

:<math>\frac{g\left[ u_1(x_1, \dots, x_n), \dots, u_n(x_1, \dots, x_n); \theta \right]}{|J^*|}=g_1\left[u_1(x_1,\dots,x_n); \theta\right] \frac{h(u_2, \dots, u_n \mid u_1)}{|J^*|}</math>

where <math>J^*</math> is the Jacobian with <math>y_1,\dots,y_n</math> replaced by their value in terms <math>x_1, \dots, x_n</math>. The left-hand member is necessarily the joint pdf <math>f(x_1;\theta)\cdots f(x_n;\theta)</math> of <math>X_1,\dots,X_n</math>. Since <math>h(y_2,\dots,y_n\mid y_1)</math>, and thus <math>h(u_2,\dots,u_n\mid u_1)</math>, does not depend upon <math>\theta</math>, then

:<math>H(x_1,\dots,x_n)=\frac{h(u_2,\dots,u_n\mid u_1)}{|J^*|}</math>

is a function that does not depend upon <math>\theta</math>.