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A statistic $T$ is a sufficient statistic for $\theta$ if the conditional distribution of the data given $T$ does not depend on $\theta$. For a dominated family with density or mass function $f_{\theta }(x)$, the Fisher-Neyman factorization theorem states that this is equivalent towhere $h$ is nonnegative and independent of $\theta$, and $g_{\theta }$ is nonnegative and depends on the data only through $T$.

For a discrete sample space, suppose first that the factorization holds. If ${\mathbb{P}}_{\theta }(T=t)>0$, summing over the fibre gives ${\mathbb{P}}_{\theta }(T=t)=g_{\theta }(t){\sum }_{T(x)=t}h(x)$, soThis conditional distribution is independent of $\theta$, proving that $T$ is a sufficient statistic. Conversely, let $q_t(x)$ be the common conditional mass function guaranteed by the sufficient statistic property, for every fibre attainable under at least one parameter. Set $h(x)=q_{T(x)}(x)$ and $g_{\theta }(t)={\mathbb{P}}_{\theta }(T=t)$. Then ${\mathbb{P}}_{\theta }(X=x)=g_{\theta }(T(x))h(x)$. Fibres never attained under any parameter can be assigned any conditional distribution, since their multiplying probability is always zero. This proves both directions of the discrete criterion, including possible parameter-dependent supports.

Write $L={\min }_i{X}_i$, $R={\max }_i{X}_i$ and $M={\max }_i|X_i|=\max \{-L,R\}$. The joint probability density factors asBy the Fisher-Neyman factorization theorem, $T=(L,R)$ is a sufficient statistic. The same factorization shows that the smaller statistic $M$ is also a sufficient statistic. This gives the sufficient statistic for a symmetric uniform sample.

In fact, the likelihood-ratio criterion for minimal sufficiency identifies $M$ as a minimal sufficient statistic: two sample points with the same $M$ have identical likelihoods for all $\theta$; with different $M$, their likelihoods have different parameter supports, and so are not proportional as functions of $\theta$. Equivalently, the condition that $f_{\theta }(x)=c{f}_{\theta }(y)$ for every $\theta$ and some positive constant $c$ holds exactly when their absolute maxima agree.

To see rigorously that $T$ cannot be recovered from $M$ almost surely, fix any $\theta >0$. The sample has a distribution invariant under simultaneous sign reversal. If $T=h(M)$ almost surely, the same identity would hold for the sign-reversed sample on another probability-one set. Since $M$ is unchanged by sign reversal, it would follow thatand hence $L=-R$ almost surely. But this event has probability zero for a finite sample from a continuous uniform distribution: it entails $X_i=-X_j$ for some pair of indices, including the possible case $X_i=0$. Therefore $T$ is not a function of the sufficient statistic $M$. The pair of extrema is not minimally sufficient, for every $n\ge 1$, even though it is sufficient.

A sufficient statistic has a conditional distribution of the full data given its value that is independent of the unknown parameter. A minimal sufficient statistic is sufficient and is a function of every other sufficient statistic, up to null sets. For a dominated family with common positive support, the likelihood-ratio criterion for minimal sufficiency states that $T$ is minimal sufficient if $T(x)=T(y)$ holds exactly when $p_{\theta }(x)/p_{\theta }(y)$ is independent of $\theta$.

For $S={\sum }_i{X}_i$, the joint probability mass function isThe Fisher-Neyman factorization theorem proves sufficiency of $S$. For two samples, the likelihood ratio is $(1-1/\theta {)}^{S(x)-S(y)}$, independent of $\theta >1$ exactly when their sums agree. Thus $S$ is minimal sufficient. Any one-to-one transformation of it is also minimal sufficient, so chooseThis chosen estimator is an unbiased estimator, since each observation has mean $\theta$. If instead the statistic is reported as $S$, its expectation is $n\theta$, and as an estimator of $\theta$ it has bias $(n-1)\theta$ (zero only when $n=1$). Being a minimal sufficient statistic is unchanged by this rescaling; being an unbiased estimator is not.

Since $\mathbb{P}(X_1=0)=e^{-\lambda }$, the indicator $Y$ is an unbiased estimator of $\theta$. The joint probability mass function isThe Fisher-Neyman factorization theorem therefore makes $T={\sum }_i{X}_i$ a sufficient statistic. More explicitly, conditioning on $T=t$ gives a multinomial distribution with $t$ trials and cell probabilities $1/n$, so $X_1\mid T=t$ has the binomial distribution $\mathrm{Bin}(t,1/n)$. Consequently conditioning by the Rao-Blackwell theorem givesFor $n=1$, interpret the value at $T=0$ as $1$ and at $T>0$ as $0$; thus the same expression uses $0^0=1$ in this probability convention. For $\lambda =0$ all observations and $T$ are zero almost surely. For $n>1$ and $\lambda >0$, the reduction in variance is strict: given $T=1$, the event $X_1=0$ still has a nondegenerate conditional probability. Directly, using the probability generating function of $T\sim \mathrm{Poisson}(n\lambda )$, the new variance is $e^{-2\lambda }(e^{\lambda /n}-1)$.

For $k=1,\dots ,C$, define the class degree sumand define the number of within-college friendships byEach between-college edge contributes once to the degree sum of each endpoint's college, while each within-college edge contributes twice to its college's degree sum. The model-3 likelihood can therefore be writtenThe Fisher-Neyman factorization theorem shows that $T=(T_0,T_1,\dots ,T_C)$ is sufficient.

For two networks $y,y^{\prime }$, the likelihood ratio isBecause the parameters range independently over ${\mathbb{R}}^{C+1}$, this ratio is constant in $\beta$ exactly when every displayed difference vanishes. The likelihood-ratio criterion for minimal sufficiency therefore gives the answerThis is the degree-sum sufficient statistic for an additive logistic network model.

A statistic $T$ is sufficient for $\theta$ if the conditional distribution of the full sample given $T$ does not depend on $\theta$. It is minimal sufficient if it is a function of every sufficient statistic.

The likelihood factors asso the Fisher-Neyman factorization theorem shows that $M$, and hence its one-to-one transform $\hat{\theta }=M/2$, is sufficient. Moreover, for two samples $\mathbf{x}$ and $\mathbf{y}$, the ratio $L(\theta ;\mathbf{x})/L(\theta ;\mathbf{y})$ is independent of $\theta$ exactly when their maxima agree. The likelihood-ratio criterion for minimal sufficiency therefore shows that $M$ and $\hat{\theta }$ are minimal sufficient.

For $n\ge 2$, the sample mean $\tilde{\theta }$ is not sufficient: two samples can have the same mean but different maxima, and their likelihood ratio then depends on $\theta$ through the support indicators. It is consequently not minimal sufficient. For the degenerate special case $n=1$, the sample mean and maximum coincide and both conclusions reverse.

Writing $B_i$ for block $i$ and using ${\sum }_{j\in B_i}{\tilde{y}}_j={\bar{y}}_i/{\sigma }^2$, the replicated likelihood factors asThe second factor is independent of $\beta$, while the first depends on the data only through $\bar{y}$. The Fisher-Neyman factorization theorem therefore makes $\bar{Y}$ a sufficient statistic for $\beta$ in the replicated experiment.

A sufficient statistic $T$ is one for which the conditional distribution of the full sample $(X_1,\dots ,X_n)$ given $T$ does not depend on $p$. TakeFor a binary sample $x$ with ${\sum }_i{x}_i=t$, its likelihood function iswhich depends on the data only through $t$. By the Fisher-Neyman factorization theorem, $T$ is sufficient. Equivalently, conditionally on $T=t$, the sample is uniform over the $\left(\genfrac{}{}{0ex}{}{n}{t}\right)$ binary vectors containing $t$ ones, independently of $p$.

Expanding the exponent givesThe Fisher-Neyman factorization theorem therefore shows that $T=X_1+X_2$ is sufficient.

It is also minimal sufficient. For two sample points $x,y\in {\mathbb{R}}^2$,where $C$ is independent of $\mu$. This ratio is independent of $\mu$ exactly when $T(x)=T(y)$. The likelihood-ratio criterion for minimal sufficiency applies, proving the claim and the general normal sample sum with known variance result in this case.

For $X=x$, the last flip must be the $r$th head. Among the preceding $x+r-1$ flips, exactly $x$ are tails and $r-1$ are heads. There are $\left(\genfrac{}{}{0ex}{}{x+r-1}{x}\right)$ choices for their positions, and every resulting sequence has probability $(1-p{)}^x{p}^r$. The negative binomial stopping argument therefore givesso $X$ has the failures-before-the-$r$th-success negative binomial distribution.

To put the mass function into exponential family form, setSince $p=1-e^{\theta }$,whereThus the natural parameter of an exponential family is $\theta$, and the Fisher-Neyman factorization theorem shows that $T(X)=X$ is a sufficient statistic. This is the negative binomial exponential family.

The exponential-family derivative identities now giveand

A statistic $T(X)$ is sufficient for a parameter $\theta$ when the conditional distribution of the full sample $X$ given $T$ does not depend on $\theta$. For a dominated model, the Fisher-Neyman factorization theorem says this is equivalent to a factorizationwhere $h$ is independent of $\theta$.