Editing Sufficient statistic (section)

===Normal distribution===

If <math>X_1,\ldots,X_n</math> are independent and [[Normal distribution|normally distributed]] with expected value <math>\theta</math> (a parameter) and known finite variance <math>\sigma^2,</math> then

:<math>T(X_1^n)=\overline{x}=\frac1n\sum_{i=1}^nX_i</math>

is a sufficient statistic for <math>\theta.</math>

To see this, consider the joint [[probability density function]] of <math>X_1^n=(X_1,\dots,X_n)</math>. Because the observations are independent, the pdf can be written as a product of individual densities, i.e.

:<math>\begin{align}
f_{X_1^n}(x_1^n)
  & = \prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}} \exp \left (-\frac{(x_i-\theta)^2}{2\sigma^2} \right ) \\ [6pt]
   &= (2\pi\sigma^2)^{-\frac{n}{2}} \exp \left ( -\sum_{i=1}^n \frac{(x_i-\theta)^2}{2\sigma^2} \right ) \\ [6pt]
  & = (2\pi\sigma^2)^{-\frac{n}{2}} \exp \left (-\sum_{i=1}^n \frac{ \left ( \left (x_i-\overline{x} \right ) - \left (\theta-\overline{x} \right ) \right )^2}{2\sigma^2} \right ) \\ [6pt]
  & = (2\pi\sigma^2)^{-\frac{n}{2}} \exp \left( -{1\over2\sigma^2} \left(\sum_{i=1}^n(x_i-\overline{x})^2 + \sum_{i=1}^n(\theta-\overline{x})^2 -2\sum_{i=1}^n(x_i-\overline{x})(\theta-\overline{x})\right) \right) \\ [6pt] 
  &= (2\pi\sigma^2)^{-\frac{n}{2}} \exp \left( -{1\over2\sigma^2} \left (\sum_{i=1}^n(x_i-\overline{x})^2 + n(\theta-\overline{x})^2 \right ) \right ) && \sum_{i=1}^n(x_i-\overline{x})(\theta-\overline{x})=0 \\ [6pt]
  &= (2\pi\sigma^2)^{-\frac{n}{2}} \exp \left( -{1\over2\sigma^2} \sum_{i=1}^n (x_i-\overline{x})^2 \right ) \exp \left (-\frac{n}{2\sigma^2} (\theta-\overline{x})^2 \right )

\end{align}</math>

The joint density of the sample takes the form required by the Fisher–Neyman factorization theorem, by letting

:<math>\begin{align}
h(x_1^n) &= (2\pi\sigma^2)^{-\frac{n}{2}} \exp \left( -{1\over2\sigma^2} \sum_{i=1}^n (x_i-\overline{x})^2 \right ) \\[6pt]
g_\theta(x_1^n) &=  \exp \left (-\frac{n}{2\sigma^2} (\theta-\overline{x})^2 \right )
\end{align}</math>

Since <math>h(x_1^n)</math> does not depend on the parameter <math>\theta</math> and <math>g_{\theta}(x_1^n)</math> depends only on <math>x_1^n</math> through the function

:<math>T(X_1^n)=\overline{x}=\frac1n\sum_{i=1}^nX_i,</math>

the Fisher–Neyman factorization theorem implies <math>T(X_1^n)</math> is a sufficient statistic for <math>\theta</math>.

If <math> \sigma^2 </math> is unknown and since <math>s^2  = \frac{1}{n-1} \sum_{i=1}^n \left(x_i - \overline{x} \right)^2 </math>,  the above likelihood can be rewritten as

:<math>\begin{align}
f_{X_1^n}(x_1^n)= (2\pi\sigma^2)^{-n/2} \exp \left( -\frac{n-1}{2\sigma^2}s^2 \right) \exp \left (-\frac{n}{2\sigma^2} (\theta-\overline{x})^2 \right ) .
\end{align}</math>

The Fisher–Neyman factorization theorem still holds and implies that <math>(\overline{x},s^2)</math> is a joint sufficient statistic for <math> ( \theta , \sigma^2) </math>.