Sufficient Statistic

Study guide: Exam MAS-I

A Sufficient Statistic $T(X_1, \ldots, X_n)$ for a parameter $\theta$ is a statistic that captures all the information in the sample about $\theta$ . Once $T$ is known, the conditional distribution of the sample given $T$ does not depend on $\theta$ .

Fisher–Neyman Factorization Theorem: $T$ is sufficient for $\theta$ if and only if the joint density (or PMF) factors as: $f(x_1, \ldots, x_n \mid \theta) = g(T(x_1,\ldots,x_n),\, \theta) \cdot h(x_1, \ldots, x_n)$

$g$ depends on the data only through $T$ ; $h$ does not depend on $\theta$
The sufficient statistic summarizes the sample without loss of information about $\theta$
For exponential family distributions, a natural sufficient statistic always exists (e.g., $\sum X_i$ for the Poisson and Exponential families)
A complete sufficient statistic combined with the Lehmann–Scheffé theorem yields the UMVUE (uniformly minimum variance unbiased estimator)

Distribution	Parameter	Sufficient Statistic
$\text{Poi}(\lambda)$	$\lambda$	$\sum_{i=1}^n X_i$
$N(\mu, \sigma^2)$	$\mu$ (known $\sigma^2$ )	$\bar{X}$
$\text{Exp}(\theta)$	$\theta$	$\sum_{i=1}^n X_i$

Verifying a Sufficient Statistic via Factorization Let $X_1, \ldots, X_n \stackrel{\text{iid}}{\sim} \text{Poi}(\lambda)$ . Show that $T = \sum X_i$ is sufficient for $\lambda$ .

Answer The joint PMF is: $f(\mathbf{x} \mid \lambda) = \prod_{i=1}^n \frac{e^{-\lambda}\lambda^{x_i}}{x_i!} = \frac{e^{-n\lambda}\lambda^{\sum x_i}}{\prod x_i!}$ Set $g(T, \lambda) = e^{-n\lambda}\lambda^T$ and $h(\mathbf{x}) = \dfrac{1}{\prod x_i!}$ . Since $g$ depends on the data only through $T = \sum x_i$ , the factorization theorem confirms $T$ is sufficient.