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For independent affected sib pairs with IBD counts $n_0,n_1,n_2$, the likelihood function of the recombination fraction is proportional to $s^{2n_0+n_1}(1-s{)}^{2n_2+n_1}$, with $s=2\theta (1-\theta )$. At fixed sample size, the total shared-copy count $2n_2+n_1$ is a sufficient statistic by the Fisher-Neyman factorization theorem. Thus a one-sided likelihood-ratio test has the same ordering as a test for excess mean IBD sharing. This equivalence depends on the stated fully penetrant recessive model and does not extend to arbitrary disease models.

// Target: biology.bigb

For independent multinomial distributions with fixed stratum totals $N_s$ and baseline logits $\log (p_{sl}/p_{sL})=x_s^T{\beta }_l$, the sufficient statistics are the displayed response-specific design-weighted counts, $l<L$. Expanding the log-likelihood gives ${\sum }_{l<L}{\beta }_l^T{T}_l-{\sum }_s{N}_s\log (1+{\sum }_{l<L}{e}^{x_s^T{\beta }_l})$, plus a data-only term. The Fisher-Neyman factorization theorem therefore proves sufficiency; with an unrestricted identifiable canonical parameter, a likelihood-ratio argument proves minimal sufficiency. Factor indicators give the corresponding response/covariate margins.

Assume the full-data sample space and the range of the statistic $T$ are standard Borel spaces, $T$ is a measurable function, and the statistical model $\{P_{\theta }\}$ is dominated by one common sigma-finite measure $\nu$. Write $f_{\theta }=d{P}_{\theta }/d\nu$ for its densities. Then $T$ is a sufficient statistic if and only if there exist nonnegative measurable functions $g_{\theta }$ and $h$, with $h$ independent of $\theta$, such thatThe statistical parameter can enter $g_{\theta }$ only through $\theta$ and the observed $T(x)$. ^[ref]

For discrete data, take counting measure; for ordinary continuous data, take Lebesgue measure. Independence, identical sampling laws, smooth dependence on $\theta$, and a fixed positive support of a probability distribution are not requirements of this theorem. Parameter-dependent support must nevertheless be included in $f_{\theta }$, usually through an indicator function, and must fit the displayed factorization. The sufficient statistic for a symmetric uniform sample illustrates exactly this: its indicator function depends on the unknown scale through the observed absolute maximum. The null sets in the theorem concern versions of densities; the factors are not generally themselves normalized probability distributions.

For a discrete sample space, the cancellation behind the theorem is particularly transparent. On a group with positive probability, factorization givesConversely, multiply a common conditional probability mass function by $P_{\theta }(T=T(x))$ to obtain the required factorization. For continuous data, a sum over a group is not a density integral over the original Lebesgue measure: the group may have measure zero. Use the regular conditional distribution formulation and the Fisher-Neyman factorization theorem instead of that discrete calculation.

For independent normal distribution observations with mean $\mu >0$ and variance ${\mu }^2$, put $S_1={\sum }_i{X}_i$, $S_2={\sum }_i{X}_i^2$. The Fisher-Neyman factorization theorem makes $(S_1,S_2)$ a sufficient statistic, since the likelihood is proportional to ${\mu }^{-n}\exp [-S_2/(2{\mu }^2)+S_1/\mu ]$. Its logarithmic derivative has numerator $S_2-S_1\mu -n{\mu }^2$. When $S_2>0$, this changes sign once on the positive half-line and gives the displayed global maximum-likelihood estimator. The zero sample is a probability-zero exception with unbounded likelihood as $\mu \downarrow 0$.

Put $S_1={\sum }_{i=1}^n{X}_i$, $S_2={\sum }_{i=1}^n{X}_i^2$. The joint normal distribution density can be written asThe Fisher-Neyman factorization theorem says that, for a family dominated by a common measure, a statistic $T$ is a sufficient statistic when its densities have the form $g_{\mu }(T(\mathbf{x}))h(\mathbf{x})$ with $h$ independent of the parameter. The displayed factorization therefore shows that $(S_1,S_2)$ is a two-dimensional sufficient statistic.

For maximum likelihood estimation, discard parameter-independent terms from the log-likelihood:If $S_2>0$, the quadratic equation has exactly one positive root. The numerator is positive before that root and negative after it, while the likelihood tends to zero at both ends of $\mu >0$. Hence the normal likelihood with variance equal to squared mean gives the global maximizerThis holds almost surely. The exceptional all-zero sample has $S_2=0$ and likelihood proportional to ${\mu }^{-n}$, so its supremum is approached as $\mu \downarrow 0$ and there is no positive maximizer.

For the standard additive multinomial logistic regression, use the final satisfaction category as baseline and writeSet a reference effect to zero for each covariate factor. Define ${\eta }_{ijkl}={\alpha }_l+a_{il}+b_{jl}+c_{kl}$ for $l<L$, and ${\eta }_{ijkL}=0$. The probabilities areSubstitution into the conditional log-likelihood givesThe second term depends on the parameters and the fixed stratum totals, not on any additional within-stratum cell allocation. The Fisher-Neyman factorization theorem therefore proves thatThese margins contain linear redundancies; after reference constraints an equivalent independent statistic has dimensionHence its residual degrees of freedom are $24(3-1)-14=34$. In the equivalent Poisson surrogate for a conditional multinomial model, include a separate stratum intercept ${\gamma }_{ijk}$ and response interactions with each individual covariate. The 24 nuisance intercepts plus 14 response parameters give 38 parameters in 72 Poisson cells, again 34 residual degrees of freedom. Since the stratum totals are fixed in the conditional experiment, they need not be included as random sufficient statistics there; they do belong to the unconditional Poisson model's sufficient margins.

For a dominated family of sample distributions, a statistic $T$ is a sufficient statistic precisely when the probability mass function or density factors aswhere $h$ does not depend on the parameter. This is the Fisher-Neyman factorization theorem.

Here is the discrete proof in both directions. Suppose the factorization holds, and write $H(t)={\sum }_{x:T(x)=t}h(x)$. Then ${\mathbb{P}}_{\theta }(T=t)=g_{\theta }(t)H(t)$. Whenever this probability is positive, $H(t)$ is finite and positive andis independent of $\theta$. Thus the conditional distribution of the whole sample given $T$ is parameter-free, which is the definition of a sufficient statistic.

Conversely, if $T$ is sufficient, let $q_t(x)$ denote this common conditional mass function on each possible fiber. ThenTake $g_{\theta }(t)={\mathbb{P}}_{\theta }(T=t)$ and $h(x)=q_{T(x)}(x)$ to obtain the factorization. Fibers impossible for every parameter can be assigned $h=0$; conditional probabilities are required only for fibers with positive probability.

For the independent Poisson sample the joint mass function isAll parameter dependence is throughso this is a one-dimensional sufficient statistic. The sample mean contains the same information, since $n$ is known.

Put $S_1={\sum }_i{Y}_i$, $S_2={\sum }_i{Y}_i^2$ and $R=\sqrt{S_2}$. Apart from a data-only constant, the log-likelihood isThe natural parameters in the ambient two-parameter exponential family of normal distributions are ${\eta }_1=1/\mu$ and ${\eta }_2=-1/(2{\mu }^2)$, restricted to the nonaffine curve ${\eta }_2=-{\eta }_1^2/2$, ${\eta }_1>0$. Its parametrization has rank one. Thus this is a one-dimensional curved exponential family inside a two-dimensional full family.

The Fisher-Neyman factorization theorem makes $(S_1,S_2)$ sufficient. For two samples $y,z$, the parameter-dependent part of their log density ratio is $\Delta S_1/\mu -\Delta S_2/(2{\mu }^2)$. A polynomial in $1/\mu$ is constant on $(0,\infty )$ only if both nonconstant coefficients vanish. The likelihood-ratio criterion for minimal sufficiency therefore proves $(S_1,S_2)$ is minimal sufficient.

Write $Y_i=\mu Z_i$ with independent $Z_i\sim N(1,1)$. Since $\mu >0$, $a=\sqrt{n}\sqrt{{\sum }_i{Z}_i^2}/{\sum }_i{Z}_i$ has a probability distribution independent of $\mu$: it is an ancillary statistic. The denominator is nonzero almost surely. The Cauchy-Schwarz inequality gives $|a|\ge 1$; on the requested positive branch, $S_1>0$ and $a\ge 1$.

The likelihood equation is $n{\mu }^2+S_1\mu -S_2=0$. Its positive root isThere is exactly one positive root. The sign of the score function changes from positive to negative there, and $\ell \to -\infty$ at both ends of $(0,\infty )$ when $S_2>0$, so it is the unique maximum-likelihood estimator. The useful identities areSubstitution yields, dropping the constant $-n/2$,Differentiating twice and using $q/a=q^2-1$ at the fitted value gives the observed informationThe scalar P-star approximation uses $\sqrt{j(v)}\exp \{\ell (\mu ;v,a)-\ell (v;v,a)\}$. Consequently, with its normalizing factor explicitly retained,Here $C_n(a)$ is chosen to make the density integrate to one, and is independent of $\mu$. To see this without assuming an approximate normalizer, defineThe integral is finite and positive, and normalized $p^{\ast }$ is exactlyIndeed $C_n(a)=\{\sqrt{n(1+q^2)/(2\pi )}\exp [n{q}^2/2-nq/a]J_n(a){\}}^{-1}$.

There is a direct check of this conditional scale density in a curved Gaussian family. In polar coordinates $y=Ru$, the sample density times volume element is proportional to ${\mu }^{-n}{R}^{n-1}\exp [-R^2/(2{\mu }^2)+R{\sum }_i{u}_i/\mu ]dRd\omega (u)$. Conditioning on $a$ fixes ${\sum }_i{u}_i=\sqrt{n}/a$. The remaining angular factor is independent of $R,\mu$, while $R=q\sqrt{n}v$. Thus the exact conditional radial density has the same normalized form. For $n=1$ the positive ancillary branch fixes the positive sign and the same one-dimensional radial calculation applies. In this scale model the normalized $p^{\ast }$ expression equals the exact conditional density, not merely its leading approximation.

For $t>0$, compute its conditional distribution by one-dimensional quadrature:It is zero for $t\le 0$; with the conditional-density calculation just given, this normalized formula is exact for almost every admissible $a$. Numerically, subtract the maximum log integrand before exponentiating to avoid overflow. The sampling probability still depends on the stipulated true $\mu$; replacing it by a fitted value would be a separate plug-in inference step.

Put $N=n_0+n_1+n_2$ and $s=2x=2\theta (1-\theta )$. The usual recombination fraction range $0\le \theta \le 1/2$ gives $0\le s\le 1/2$, and the preceding likelihood function becomesTo see its relation to nonparametric linkage analysis, use Bayes' theorem and the constant disease probability $1/16$ to obtain the affected-pair marker-sharing probabilitiesTheir multinomial likelihood has exactly the same $s$-dependent kernel, since its extra factor $2^{n_1}$ is constant. Set $A=2n_0+n_1$ and $B=2n_2+n_1$. Then $A+B=2N$, and $B$ is the total number of shared ancestral copies. At fixed $N$, the Fisher-Neyman factorization theorem makes $B$ a sufficient statistic. The observed mean shared proportion is $\hat{p}=B/(2N)$, and the binomial likelihood kernel givesThe unlinked null is $s=1/2$, whereas linkage corresponds to $s<1/2$, or expected shared proportion $1-s>1/2$. If $\hat{p}\le 1/2$, the constrained likelihood is maximized at the unlinked null. For $\hat{p}>1/2$, its log likelihood ratio iswhose derivative in $\hat{p}$ is $2N\log [\hat{p}/(1-\hat{p})]>0$. Hence it orders samples exactly by excess IBD sharing. Under this fully penetrant recessive model, parametric linkage testing and a one-sided test of the overall IBD-sharing proportion use the same information. This is allele-sharing sufficiency for recessive linkage, not a general equivalence for all disease models. Use limiting values for terms such as $0\log 0$, and assume $N>0$ for the estimators.

A sufficient statistic $T(X)$ for $\theta$ is a function of the observations whose conditional distribution of $X$ given $T(X)$ does not depend on $\theta$, whenever that conditioning event is possible. It retains all parameter dependence of the likelihood.

In the discrete case the Fisher-Neyman factorization theorem provides a direct criterion: $f(x\mid \theta )=g(T(x),\theta )h(x)$, where $h$ is independent of the parameter. To justify sufficiency, for a value $t$ with positive probability, sum this expression over its fiber and divide:The parameter-dependent factor cancels. Conversely, parameter-independent conditional probabilities $h_t(x)$ give the factorization $f(x\mid \theta )={\mathbb{P}}_{\theta }(T=t)h_t(x)$ with $t=T(x)$. Thus the criterion is both necessary and sufficient, with conditional laws only required on fibers of positive probability.

For the shifted exponential sample the likelihood isIt factors through $T=X_{(1)}={\min }_i{X}_i$, so the sample minimum is a sufficient statistic. Writing $Y_i=X_i-\theta$, independence gives for $t\ge 0$Therefore $T-\theta$ is exponential with rate $n\lambda$. Its mean is $(n\lambda {)}^{-1}$ and its variance is $(n\lambda {)}^{-2}$. The unbiased endpoint estimator for a shifted exponential sample is consequentlyAlthough the parameter is nonnegative, this unbiased estimator can be negative. Truncating it at zero would change its expectation and lose the requested property of an unbiased estimator.

An exponential family has density $h(x)\exp \{\varphi \cdot T(x)-k(\varphi )\}$ on a parameter-independent support. For an independent and identically distributed sample, its likelihood factorizes through ${\sum }_{i}T(X_i)$, so the Fisher-Neyman factorization theorem gives a sufficient statistic whose dimension is the fixed dimension of $T$, independent of sample size.

Expanding the normal log-density gives ${\varphi }_1=\mu /v$, ${\varphi }_2=-1/(2v)$, andThe mean-value parameter is $\nabla k=(\mathbb{E}X,\mathbb{E}{X}^2)$, henceFor $n\ge 2$ and positive sample variance, $\hat{\mu }=\bar{X}$ and $\hat{v}=n^{-1}{\sum }_i(X_i-\bar{X}{)}^2$, so ${\hat{\Phi }}_1=\bar{X}/\hat{v}$. For this normal natural-mean parameter ratio estimator, the central limit theorem gives asymptotic independent errors for $\hat{\mu }$ and $\hat{v}$, with variances $v/n$ and $2v^2/n$. Applying the delta method to $\mu /v$ yieldsIf $v=v_0$ is known, ${\Phi }_2=-1/(2v_0)$ is fixed and ${\hat{\Phi }}_1=\bar{X}/v_0$ has exact distribution $\boxed{N({\Phi }_1,1/(n{v}_0))}$. The variance-estimation contribution $2{\mu }^2/v^2$ disappears.

Let $F_n(x)=n^{-1}{\sum }_i{1}_{\{X_i\le x\}}$. The Kolmogorov-Smirnov statistic is $D_n={\sup }_x|F_n(x)-F(x)|$, often reported as $\sqrt{n}{D}_n$. Under the simple null, the probability integral transform makes $U_i=F(X_i)$ independent uniform variables on $[0,1]$. Strict monotonicity and continuity identify the empirical distribution under this change of coordinates, soIts null law is therefore independent of $F$.

For the composite null, the joint density is ${\theta }^{-n}{\prod }_i{1}_{\{0\le x_i\le \theta \}}$. Its parameter dependence is only through $M={\max }_i{X}_i$, proving sufficiency by the Fisher-Neyman factorization theorem; the larger statistic $(J,M)$ is sufficient too. Ties have probability zero. For fixed $j$, the density of $(J=j,M=\xi )$ is ${\xi }^{n-1}/{\theta }^n$, $0<\xi <\theta$, because the other coordinates range independently over $[0,\xi ]$. Dividing the joint density by this marginal gives conditional densityThus the remaining observations are conditionally independent $U[0,\xi ]$.

For $n\ge 2$, discard the maximum and apply the Kolmogorov-Smirnov test to the $n-1$ ratios $X_i/M$, $i\ne J$, against $U[0,1]$. Use the ordinary null critical values for sample size $n-1$. Its conditional, hence unconditional, level is independent of $\theta$. Keeping the fitted maximum and using the ordinary $n$-sample critical values would not have that justification.

A complete statistic $T$ for a family $(P_{\theta })$ if every measurable $g$ with ${\mathbb{E}}_{\theta }|g(T)|<\infty$ and ${\mathbb{E}}_{\theta }g(T)=0$ for all $\theta$ satisfies $g(T)=0$ almost surely under every $P_{\theta }$. It is boundedly complete if the same implication is required only for bounded measurable $g$. Completeness therefore implies bounded completeness.

The vector $X$ is multivariate normal with mean zero and covarianceIts determinant is $(1-\theta {)}^2(1+2\theta )$ and its inverse is $(1-\theta {)}^{-1}I-\theta [(1-\theta )(1+2\theta ){]}^{-1}J$. Hence the joint density isThe Fisher-Neyman factorization theorem gives sufficiency. For minimality use the likelihood-ratio criterion for minimal sufficiency. The density ratio for $x,x^{\prime }$ is independent of $\theta$ precisely whenis constant. Multiplying by $2(1-\theta )(1+2\theta )$ shows that its constant would produce a quadratic term on the right whereas the left is linear, so that constant is zero. The remaining coefficients give $\Delta T_1=\Delta T_2=0$. Thus $T$ is minimal sufficient statistic.

Each $X_i$ is standard normal, so ${\mathbb{E}}_{\theta }{T}_1=3$. The integrable, nonzero function $T_1-3$ has zero expectation for every parameter, proving that $T$ is not complete.

By sufficiency, choose a version $S(T)={\mathbb{P}}_{\theta }(X_1^2\le 1\mid T)$ common to all parameters. The tower property givesOn the event $T_1<1$ we necessarily have $X_1^2<1$, so $S=1$ there almost surely. This event has positive probability under every parameter because the joint normal density is strictly positive on a ball about the origin. Thus $S$ is not identically $c$. The bounded function $S-c$ has zero expectation for every parameter but is not almost surely zero. Consequently

Given the complete stopped sample $(x_1,\dots ,x_n)$, the likelihood function isThe stopping factor is positive and independent of $\theta$. The Fisher-Neyman factorization theorem makes $(N,S)$ sufficient. For two possible stopped samples of lengths $n,m$ and sums $S,S^{\prime }$, their likelihood ratio is a parameter-free factor times ${\theta }^{n-m}{e}^{-\theta (S-S^{\prime })}$. This is independent of $\theta$ exactly when $n=m$ and $S=S^{\prime }$: differentiating its logarithm gives $(n-m)/\theta -(S-S^{\prime })=0$ for all $\theta >0$. By the likelihood-ratio criterion,The likelihood principle says proportional likelihoods give the same evidence about the parameter. Thus with the full sample observed, this parameter-free stopping rule adds no likelihood evidence beyond the realized values and length; it gives the same likelihood-based inference as a fixed-length sample with those values. This statement does not assert that frequentist sampling distributions are unchanged.

Let $q(\theta )={\int }_0^{\infty }\alpha (x)\theta e^{-\theta x}dx\in (0,1)$ be the chance of stopping at a given stage. Independence of successive stages givesTherefore $a(y)=\log \alpha (y)$ and $k(\theta )=\log {\int }_0^{\infty }{e}^{a(y)-\theta y}dy$ give the requested exponential family form.

Writing $p=p_Y$, differentiation in $y$ gives $p^{\prime }=(a^{\prime }-\theta )p$. Integrating and using $p(0)=p(\infty )=0$ shows ${\mathbb{E}}_{\theta }{a}^{\prime }(Y)=\theta$. LikewiseThe stipulated vanishing of $p^{\prime }$ at both boundaries givesThese integrations require the displayed moments and integrals to exist, the usual regularity interpretation of the question; when establishing a finite variance this assumption should be explicit.

Differentiating the normalization in $\theta$ gives $k^{\prime }(\theta )=-{\mathbb{E}}_{\theta }Y$ and $k^{\prime \prime }(\theta )={\mathrm{Var}}_{\theta }Y$. The score function is $-k^{\prime }(\theta )-Y$, so the Fisher information is $I(\theta )=k^{\prime \prime }(\theta )$. The Cramér-Rao bound states that a regular unbiased estimator of $\theta$ has variance at least $1/I(\theta )$. Apply it to $a^{\prime }(Y)$:Equivalently differentiate ${\mathbb{E}}_{\theta }{a}^{\prime }(Y)=\theta$ and use the Cauchy-Schwarz inequality to get $1\le \sqrt{\mathrm{Var}(a^{\prime }(Y))\mathrm{Var}(Y)}$, which gives the same conclusion.

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.