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)?

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

1 Answer 1


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.


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