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 $\mathcal B\subset\mathcal Z$.
