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Let $$ f(y | \eta) = h(y) \exp\left( \eta^\top T(y) + A(\eta) \right)$$ be the exponential family with base density/pmf $h$, sufficient statistic $T$, log partition function $A$ and natural parameter $\eta$. If it helps, we may restrict it to the case where $\eta$ is scalar. Obviously $h$ must be nonnegative. One can show that $A$ must also be convex.

The Wikipedia page states:

The function $A(\eta)$ ... is automatically determined once the other functions have been chosen, since it must assume a form that causes the distribution to be normalized (sum or integrate to one over the entire domain).

Bold s on functions by me for emphasis.

My Question:

Given some fixed $T$, am I free to choose any convex $A$ such that there exists an $h$ such that the resulting function is a valid pdf/pmf? More precisely, given some $T$ consider the set $$\mathcal{A}_T = \{ A:\mathbb{R} \to \mathbb{R} \mid \exists \, h:\mathbb{R} \to \mathbb{R} \mid A(\eta) = \log\int h(y)\exp \left(\eta T(y) \right)\, dy\}.$$ Is there a useful alternative characterisation of this set?

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I do not understand the "free" in the question. Or the question at all..

As rightly pointed by the Wikipedia page, $A(\cdot)$ is completely determined by the choice of the triplet

dominating measure $\text dy$ x base density $h(\cdot)$ x statistic $T(\cdot)$

Up to a constant, $A(\cdot)$ is a Laplace transform and hence satisfies all properties of Laplace transforms, including identifying the associated distribution (of $T$).

Choosing first $A(\cdot)$ out of the blue and then looking for an associated triplet does not seem like a fruitful endeavour.

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  • $\begingroup$ I appreciate your response, even though my question is evidently not clear. I understand that $A(\cdot)$ is must satisfy $A(\eta) = \log \int h(y) \exp\left( \eta^\top T(y) \right) \, d y$ and can see how it can be used to identify the cumulants of $T$. The reason I would like to choose $A$ first is that I find myself in the situation where I am forced to use a specific $A$ and am wondering whether the associated subset of exponential families is useful. Is there anything I can add to my question to make it better? $\endgroup$
    – Student
    Jul 15 at 8:11
  • $\begingroup$ A useful item of information would be to explain why you choose $A(\cdot)$ first. Otherwise, I do not for one see how one can answer the question. $\endgroup$
    – Xi'an
    Jul 15 at 15:56

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