Under what conditions are the cumulants of the sufficient statistic finite for an exponential family?

If we have $$ p(x \mid \theta) = \exp(\theta \cdot T(x) - A(\theta)) $$ then the derivatives of $A(\theta)$ wrt $\theta$ give the cumulants of $T(x)$. If $A(\theta)$ is finite, then are all of the cumulants of $T(x)$ finite? Ideally, I'm looking for a reference or proof.

I am trying to establish the conditions under which $A(\theta)$ can be approximated by its Taylor polynomial.

  • 2
    $\begingroup$ I suspect the answer is, under virtually no conditions, "always." Note that $\exp(A(\theta))$ is the moment generating of function of $X$ when $\theta = 0$, and moment generating functions are always infinitely differentiable at $0$. This only works if $\int 1 \cdot \nu(dx) < \infty$ where $\nu(dx)$ is the relevant dominating measure, so evidently this argument doesn't go through in general, but I have to imagine there is some way around this. $\endgroup$
    – guy
    Jan 3 '19 at 18:01
  • $\begingroup$ @MichaelHardy When the moment generating function exists in a neighborhood of zero then all the moments exist and are given by the derivatives of the moment generating function. If the MGF is not finite in a neighborhood of zero, I would usually say it does not exist. $\endgroup$
    – guy
    Jan 4 '19 at 4:14

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