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Bounded completeness requires the defining implication for a complete statistic only for bounded measurable functions. Completeness implies it, but the converse need not hold. If a sufficient statistic $T$ gives a parameter-independent event probability $c$ and the conditional probability $S(T)$ of that event is nonconstant, then $S-c$ is a bounded zero-mean obstruction to bounded completeness.

A finite-dimensional exponential family has common dominating measure and densityHere $\eta$ is the natural parameter of an exponential family, and normalization determines the cumulant function of an exponential family $\kappa$. A full regular family lets $\eta$ vary over an open natural parameter domain. A curved exponential family restricts $\eta$ to a lower-dimensional submanifold, while a minimal representation removes redundant affine combinations of the statistic.

Differentiating the normalization under regularity givesConsequently $\kappa$ is convex and its Hessian is the Fisher information matrix in natural coordinates. For an independent size-$n$ sample, $S={\sum }_{i}T(Y_i)$ is a sufficient statistic: the joint density factors into ${\prod }_{i}h(y_i)$ times $\exp \{{\eta }^{\mathsf{T}}S-n\kappa (\eta )\}$, so the conditional law given $S$ loses its parameter dependence. An interior maximum-likelihood estimator solvesmatching empirical and theoretical sufficient-statistic means. Under minimality the Hessian is positive definite, so the log-likelihood is strictly concave; existence of an interior maximizer still needs the observed statistic to be in the appropriate interior convex support.

For a full open natural family, the canonical sufficient statistic is a complete statistic: if ${\mathbb{E}}_{\eta }g(S)=0$ throughout the open domain, exponential tilting about an interior point makes the Laplace transform of the integrable signed measure $g(s)P_{{\eta }_0}(ds)$ zero on a neighborhood. Transform uniqueness makes that measure zero, so $g(S)=0$ almost surely. This requires integrability under the family. It is not an automatic property of curved restrictions. The sufficient-statistic reduction, explicit information and cumulant identities make these families especially useful for exact and higher-order inference. Normal distributions, Poisson distributions and gamma distributions supply familiar examples.

A transformation model is built differently: a group acts on the sample and parameter spaces, and its probability laws satisfyIn a transitive model all laws are transforms of a fixed reference law. The location family is generated by translations, with density $f_{\mu }(y)=f_0(y-\mu )$. The positive-scale family has density $f_{\sigma }(y)={\sigma }^{-1}{f}_0(y/\sigma )$, and the location-scale family combines them asThe Jacobian factor is essential. For independent observations its power is ${\sigma }^{-n}$. The base density need not have moments, and a transformation model need not have fixed support.

An equivariant estimator respects the parameter action. With a well-defined equivariant fit $T(y)$ valued in the transformation group, the normalized sampleis a maximal invariant. Equivariance gives $a(gy)=a(y)$; if normalized samples agree, $y^{\prime }$ is obtained from $y$ by $T(y^{\prime })T(y{)}^{-1}$. In a transitive transformation model its law is parameter-free, so it is an ancillary statistic. For location and positive scale, centered standardized residuals provide this construction on nondegenerate samples. If the parameter action has a nontrivial stabilizer, normalization must retain its quotient action rather than pretending that a unique group element is determined.

This geometry motivates conditioning on an ancillary and using the P-star approximation for the fitted parameter. Ancillarity means parameter-free distribution; it does not by itself assert independence from the estimator. In a normal location model with known scale, the sample mean and centered residuals are independent by their joint Gaussian law and zero covariance. In an exponential scale model, the sample sum and its normalized proportions are independent, again an additional model property.

The two classes overlap but are distinct. A normal location-scale family is both an exponential family and a transformation model; an exponential mean family is both a one-dimensional natural exponential family and a scale model. A general Cauchy distribution location family is a transformation model without a fixed finite-dimensional regular exponential-family sufficient-statistic reduction. Exponential-family structure emphasizes sufficient statistics and natural-parameter convexity; transformation structure emphasizes equivariance, group orbits and ancillary reduction.

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

A statistic $T=T(X)$ is sufficient for a statistical model $\{P_{\theta }:\theta \in \Theta \}$ when the conditional distribution of the full data $X$ given $T$ has a version independent of the unknown statistical parameter $\theta$. Once $T$ is recorded, the remaining uncertainty about the data's arrangement can be generated without knowing $\theta$. ^[ref]

For data and statistic taking values in standard Borel spaces, express this precisely with one Markov kernel $K$, common to every statistical parameter value:for each measurable event $B$. Thus $K(t,\cdot )$ is a regular conditional distribution of $X$ given $T=t$. Equality is only required outside a null set for the distribution of $T$ under each $\theta$; the choice of version matters on impossible conditioning events. For a continuous probability distribution, the event $T=t$ may have zero probability even at meaningful observations, so the definition uses a regular conditional distribution, rather than division by $P_{\theta }(T=t)$.

A sufficient statistic is a property of a specified statistical model, rather than of a formula for summarizing data in isolation. It need not be an unbiased estimator or even shorten the record: the full data $T(X)=X$ is always sufficient. An estimator instead targets a quantity such as a mean. An ancillary statistic has a parameter-independent marginal probability distribution, whereas a sufficient statistic has a parameter-independent conditional full-data law. A minimal sufficient statistic permits the greatest reduction among sufficient statistics. A complete statistic satisfies a separate condition about zero expectations; a complete sufficient statistic must satisfy both conditions.