7
$\begingroup$

The exponential family is defined (in many sources) as:

$$p(x | \theta) = h(x) \exp\{\theta^TT(x) - A(\theta)\}$$

where:

  • $T(x)$ is a sufficient statistic,
  • $\theta$ is a canonical parameter, and
  • $A(\theta)$ is a cumulant function

What is $h(x)$? Does it have a name or specific interpretation?

$\endgroup$
0

1 Answer 1

9
$\begingroup$

The $h(x)$ function in the exponential family is known as the "underlying measure." It serves to ensure $x$ is in the right space. For many functions, this correction is unnecessary (i.e. it is set to $1$ or $1/\sqrt{2}$). It does play a strong role in defining many functions, however. Since the role is function-specific beyond the definition above, I will link to a part of the Wikipedia page for "exponential family" with a few helpful examples (in table form) of the role of $h(x)$ in common distributions.

Link: https://www.wikiwand.com/en/Exponential_family#/Table_of_distributions

$\endgroup$
3
  • 2
    $\begingroup$ The link you provide uses the term "base measure." Are both "base measure" and "underlying measure" in common usage? I appreciate that, in a common parlance, the difference in plain meaning between "underlying measure" and "base measure" is small-to-non-existent, but since this question is particularly about terminology it seems important to give a full accounting of common synonyms and avoid confusing the matter. $\endgroup$
    – Sycorax
    Jun 5, 2019 at 17:25
  • 1
    $\begingroup$ That's a fair point. I will say that I have seen "underlying measure" used more frequently in academic settings and figure for someone who wants to know the term it would be better to supply the more accepted one rather than perpetuating the confusion of two. $\endgroup$ Jun 5, 2019 at 17:27
  • 1
    $\begingroup$ (+1) I would add that $h(\cdot)$ is the determinantal element of the family. Once the reference measure is chosen, plus the sufficient statistic $T(\cdot)$, the parameter space is naturally defined as made of these $\theta$'s for which the density is well-defined and $A(\theta)$ is the logarithm of the normalising constant. $\endgroup$
    – Xi'an
    Jun 5, 2019 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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