1
$\begingroup$

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)?

I would greatly appreciate it if someone would please take the time to clarify this.

$\endgroup$
0
$\begingroup$

They're both required.

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

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.