Factorization theorem (Sufficient statistics)

$t = T(X=x)$ is a sufficient statistic $\iff$ $p(X=x | \theta) \equiv p_\theta(X = x) = h(x) \, g_\theta\left( T(x)\right)$

where $x = \set{x_1, \dots x_N}$ is a set of realizations, $X = \set{X_1, \dots, X_N}$ is a set of random variables and $p(X=x) \equiv p(X_1=x_1, \dots X_N=x_N)$ .

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Consider when $\exist x$, $T(X=x) = t$. In this case, $p_\theta(X=x) = p_\theta(X=x, T=t)$.

($\Rightarrow$)

$$\begin{align*} p_\theta(X=x) &= p_\theta(X=x, T=t) \\ &= p_\theta(X=x | T=t) p_\theta(T=t) \\ &= \underbrace{p(X=x | T=t)}_{h(x)} \, \underbrace{p_\theta(T=t)}_{g_\theta(T(x) = t)} \\ \end{align*}$$

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($\Leftarrow$)

$$\begin{align*} p_\theta(X=x|T=t) &= \frac{p_\theta(X=x, T=t)}{p_\theta(T=t)} \\ &= \frac{p_\theta(X=x)}{ \sum_{\set{x|T(X=x)=t}} p_\theta(X=x, T=t)} \\ &= \frac{p_\theta(X=x)}{ \sum_{\set{x|T(X=x)=t}} p_\theta(X=x)} \\ &= \frac{h(x) \cancel{g_\theta(t)}}{\sum_{\set{x|T(X=x)=t}} h(x) \cancel{g_\theta(t)}} \\ \end{align*}$$

therefore,

$p_\theta(X=x|T=t)$ does not depend on $\theta$.