# Definition of $k$-parameter exponential family

I am currently studying the concept of sufficient statistics in mathematical statistics. The following definition is presented:

Definition: $$k$$-parameter exponential family

Let $$\mathbf{Y} \sim f_\theta (\mathbf{y})$$, where $$\theta = (\theta_1, \dots, \theta_k)$$ belongs to an open set. We say that $$f_\theta$$ belongs to the $$k$$-parameter exponential family if

(1) the support $$\text{supp}(f_\theta)$$ does not depend on $$\theta$$

(2) $$f_\theta (\mathbf{y}) = \exp \{ \sum_{j = 1}^k c_j(\theta) T_j(\mathbf{y}) + d(\theta) + S(\mathbf{y}) \}, \mathbf{y} \in \text{supp}(f_\theta)$$, for some known functions $$c_j(\cdot), T_j (\cdot), j = 1, \dots, k$$; $$d(\cdot)$$ and $$S(\cdot)$$

The functions $$c(\theta) = (c_1(\theta), \dots, c_k(\theta))$$ are the natural parameters of the distribution.

There's some critical information missing here: Do we require (1) and (2), or is it (1) or (2)?

• In mathematical definitions with a list of requirements, you can safely assume and, unless explicitely stated otherwise! Apr 16, 2021 at 16:13

They're both required.

If the support of $$f(x \mid \theta)$$ depended on $$\theta$$, part of the pdf would be an indicator function with both $$\theta$$ and $$x$$ (for example, the pdf for $$X \sim \text{Uniform}(0, \theta)$$ is $$f(x \mid \theta) = \dfrac{1}{\theta} I_{(0, \theta)}(x)$$.

There's no way to work that indicator function into the form in (2); there's no way to separate that $$x$$ and $$\theta$$. So in that sense, they're the same requirement.