Exponential families and completeness

A one-parameter natural exponential family has density $h(x)\exp\{\eta T(x)-A(\eta)\}$ with respect to a fixed measure. Its natural parameter domain contains the values where the normalizing integral is finite. The data-only factor includes a fixed support restriction; a parameter-dependent support needs special care and may prevent this representation. For independent observations, the exponents add, producing the sufficient statistic equal to the sum of the individual statistics. This is a factorization argument, not a claim about every possible model.

For a Bernoulli observation with interior probability, choose the log odds $\eta=\log[p/(1-p)]$. Solving gives $p=e^\eta/(1+e^\eta)$, and the log normalizer is $A(\eta)=\log(1+e^\eta)$. Every real natural parameter is allowed. The endpoint Bernoulli laws are limits of this representation rather than finite values of $\eta$. Working with the open family is enough for the completeness argument below, and any almost-sure conclusion for its finite support also extends to the endpoint laws.

Differentiate the identity integrating the density to one. When differentiation can pass through the integral, the first derivative of the log normalizer equals the mean of the natural statistic. A second differentiation gives its variance. For a finite sample space these interchanges are automatic finite-sum manipulations. In continuous families, an open neighborhood of finite exponential moments supplies the needed domination. Do not take such interchanges for granted at the edge of a natural parameter domain.

In the Bernoulli family these derivatives yield $p$ and $p(1-p)$. For $n$ independent observations, the log normalizer becomes $nA$ and the count variance is $np(1-p)$. Convexity of the log normalizer follows from the nonnegative variance. A zero variance can signal a redundant statistic direction; strict convexity needs a nondegeneracy condition. These identities link probability structure to the curvature used in estimation, but curvature is expressed in the chosen natural coordinate, not invariant as an unqualified second derivative.

A statistic is complete for a family if every measurable function $g$ that is integrable under every model law and has expectation zero for every parameter is zero almost surely under every model law. Sufficiency concerns the conditional data distribution; completeness concerns functions of the statistic and expectations across the parameter space. They are different statements. A standard theorem gives completeness of the natural statistic in a full finite-dimensional exponential family when its natural parameter space contains a nonempty open subset of the full Euclidean parameter space, with integrability as stated. Curved subfamilies need not inherit this conclusion.

For a fixed binomial sample size, completeness has an elementary proof. Set the expectation of $g(K)$ to zero, divide by $(1-p)^n$, and write the odds as a positive variable $z$. The result is a polynomial in $z$ that vanishes for every positive value. All coefficients must be zero, and the positive binomial coefficients force every value $g(k)$ to be zero. Restricting the parameter to a few isolated probabilities would not justify this polynomial argument; the extent of the parameter space matters.