# Are the cumulants of sufficient statistics finite for the exponential family?

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.

• 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.
– guy
Jan 3 '19 at 18:01
• @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.
– guy
Jan 4 '19 at 4:14