Completeness of a statistic - Open ball I was studying the slides of the course in statistics, but there is a theorem that is not clear for me. This chapter was about finding a complete statistic, and it explains that it can be found with an exponential family. The problem is that it is not enough to find it, it is needed also to check for an "open ball". I did not understand what is that concept, and how to apply it to find the estimator. For example, I didnt know how to check for the open ball when finding the estimators for a two-parameter exponential. It would be nice to show an example as I am new in maths and statistics... I uploaded the example when finding complete estimators for a gamma distribution. Thanks so much



 A: When formally defining a probability density as
$$f_\theta(x)=c(\theta)h(x)\exp\left\{\sum_{j=1}^k q_j(\theta)t_j(x)\right\}\tag{1}$$
it is always possible to add useless terms in the expression, as for instance in
\begin{align}f_\theta(x)&=c(\theta)h(x)\exp\left\{[q_1(\theta)-q_{k+1}(\theta)]t_1(x)+[q_2(\theta)-q_{k+1}(\theta)]t_2(x))\right.\\
&\left.\quad+\sum_{j=3}^k q_j(\theta)t_j(x) + q_{k+1}(\theta)[t_1(x)+t_2(x)]\right\}\\
\end{align}
for an arbitrary function $q_{k+1}(\theta)$. It thus also writes as
$$f_\theta(x)=c(\theta)h(x)\exp\left\{\sum_{j=1}^{k+1} q'_j(\theta)t'_j(x)\right\}\tag{2}$$
Another example is to write (1) as
$$f_\theta(x)=c(\theta)h(x)\exp\left\{\sum_{j=1}^k q_j(\theta)t_j(x)+0.
t_{k+1}(x)\right\}\tag{3}$$
and to set $q_{k+1}(\theta)=0$, with $t_{k+1}(x)$ an arbitrary statistic.
This means that the integer $k$ in (1) is not unique unless some constraints are set on the terms in (1). This leads to the notion of a full-rank exponential family whose density is of the type (1) with constraints that

*

*the functions $q_j(\cdot)$ are linearly independent

*the functions $t_j(\cdot)$ are linearly independent, which rules out the above counterexample

These two conditions are not enough for completeness: if the vector $z(\theta)=(q_1(\theta),\ldots,q_k(\theta))$ is not varying freely enough when $\theta$ varies in $\Theta$, completeness of $\tau(x)=(t_1(x),\ldots,t_k(x))$ does not always hold. See for instance this example of a translated exponential (which is not an exponential family).
The sufficient constraint for ensuring completeness is that the set $\mathcal Z=z(\Theta)$ of $z(\theta)$'s is large enough, which translates as, for every $\zeta\in\mathcal Z$, there exists a ball $\mathcal B$ centred at $\zeta$ and with a small enough positive radius such that $\mathcal B\subset\mathcal Z$.
A: Your book is presenting a simple concept in very formal mathematics. An "open ball" is a concept in mathematics referring to sets which  do not contain their boundary points. This is a very general concept in mathematics, but we will usually work with real number ($\mathbb{R}^n$) and use Euclidean distance in statistics. It is helpful to think in terms of this specific case.
In 1-dimension these are open sets, i.e $(0, 1)$ or ($-\infty, \infty)$ or $(a,b)$. In 2-dimensions these are disks that don't contain their boundary. This also can be the entire Cartesian plane.
Open sets are often useful when maximizing a likelihood function to get a maximum likelihood estimate (MLE). A maximum likelihood estimate will not be asymptotically normal if the derivative of the likelihood function at the MLE is nonzero. This can make it difficult to understand the variance of your test statistic. For example, if $X_i \sim N(\mu, 1)$ and we know $\mu > 0$, you can show the MLE is
$$ \hat{\mu} = \left\{\begin{array}{lr}
\bar{X} & \text{when } \bar{X} \geq 0 \\
0 & \text{when } \bar{X} < 0
\end{array}\right\}. $$
This maximum likelihood estimator isn't asymptotically Normal. It can't be; it can't take negative values.
