Editing Sufficient statistic (section)

===Another proof===
A simpler more illustrative  proof is as follows, although it applies only in the discrete case.

We use the shorthand notation to denote the joint probability density of <math>(X, T(X))</math> by <math>f_\theta(x,t)</math>. Since <math>T</math> is a deterministic function of <math>X</math>, we have <math>f_\theta(x,t) = f_\theta(x)</math>, as long as <math>t = T(x)</math> and zero otherwise. Therefore:

:<math>
\begin{align}
f_\theta(x) & = f_\theta(x,t) \\[5pt]
& = f_\theta (x\mid t) f_\theta(t) \\[5pt]
& = f(x\mid t) f_\theta(t)
\end{align}
</math>

with the last equality being true by the definition of sufficient statistics. Thus <math>f_\theta(x)=a(x) b_\theta(t)</math> with <math>a(x) = f_{X \mid t}(x)</math> and <math>b_\theta(t) = f_\theta(t)</math>.

Conversely, if <math>f_\theta(x)=a(x) b_\theta(t)</math>, we have

:<math>
\begin{align}
f_\theta(t) & = \sum _{x : T(x) = t} f_\theta(x, t) \\[5pt]
& = \sum _{x : T(x) = t} f_\theta(x) \\[5pt]
& = \sum _{x : T(x) = t} a(x) b_\theta(t) \\[5pt]
& = \left( \sum _{x : T(x) = t} a(x) \right) b_\theta(t).
\end{align}</math>

With the first equality by the [[Probability density function#Densities associated with multiple variables|definition of pdf for multiple variables]], the second by the remark above, the third by hypothesis, and the fourth because the summation is not over <math>t</math>.

Let <math>f_{X\mid t}(x)</math> denote the conditional probability density of <math>X</math> given <math>T(X)</math>. Then we can derive an explicit expression for this:
:<math>
\begin{align}
f_{X\mid t}(x)
& = \frac{f_\theta(x, t)}{f_\theta(t)} \\[5pt]
& = \frac{f_\theta(x)}{f_\theta(t)} \\[5pt]
& = \frac{a(x) b_\theta(t)}{\left( \sum _{x : T(x) = t} a(x) \right) b_\theta(t)} \\[5pt]
& = \frac{a(x)}{\sum _{x : T(x) = t} a(x)}.
\end{align}</math>

With the first equality by definition of conditional probability density, the second by the remark above, the third by the equality proven above, and the fourth by simplification. This expression does not depend on <math>\theta</math> and thus <math>T</math> is a sufficient statistic.<ref>{{cite web | url=http://cnx.org/content/m11480/1.6/ | title=The Fisher–Neyman Factorization Theorem}}. Webpage at Connexions (cnx.org)</ref>