What is the broadest context where $\bar{X}$ is complete sufficient for $\mathbb E(X)$?

Question

$\bar{X}$ is complete sufficient for $\mathbb E(X)$ if $X$ is Normal with known standard deviation $\sigma$. Are there broader contexts? Like for exponential families in general or more general than that.

Answer 1

First, for $\bar X_n$ to be sufficient for all $n$'s, under an iid assumption on the observations, the distribution of the data must belong to an exponential family by the Darmois-Pitman-Koopman lemma. (The case when the support depends on $\theta$ can be eliminated as $\bar X_n$ would then cease to be sufficient.) This means there exists a measurable function $h(\cdot)$ and a normalising constant $C(\theta)$ such that$$f(x|\theta) = C(\theta)h(x)\exp\{\theta x\}$$(where $\theta$ is the natural parameter of the exponential family). The statistic $\bar X_n$ is then (minimal) sufficient. When the parameter space$$\Theta=\left\{\theta; \int h(x)\exp\{\theta x\}\text{d}x <\infty\right\}$$contains an open set, this makes $\bar X_n$ complete as well.

Note also that the answer can be generalised to any transform $T(X)$ of $X$, in which case the exponential family would be defined in terms of $t(x)$ rather than $x$, which makes the number of cases infinitely larger!

If the question is meant to apply to a single value of $n$, the most general case when $\bar X_n$ is sufficient is for a joint density of the form $$h(x_1,\ldots,x_n) \times m(\bar x_n,\theta)$$ but this is not particularly interesting since the observations are then dependent (except in the exponential case).