I am curious about this question, because in definitions I have never seen this property. Is it true? If yes, why?
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$\begingroup$ Don’t they have finite moments? $\endgroup$– BjörnCommented Dec 5, 2017 at 17:51
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1 Answer
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As fairly well-explained on Wikipedia, for any exponential family, there exists a parameterisation such that the density of the family is$$f(x|\theta)=\exp\{\theta\cdot T(x)-\Psi(\theta)\}$$wrt a constant measure $\text{d}\mu(x)$, where the components of $T(\cdot)$ are linearly independent. In this representation, the moment generating function of the random variable $T(X)$ is given by$$M(\upsilon)=\exp\{\Psi(\theta+\upsilon)-\Psi(\theta)\}$$and therefore$$\Psi(\theta+\upsilon)-\Psi(\theta)$$ is the cumulant generating function for T. This implies that all order moments of $T(X)$ can be derived from the successive derivative of $\Psi(\theta)$, provided $\theta$ is within the interior of the natural parameter space$$\Theta=\{\theta;\ |A(\theta)|<\infty\}$$
However, if one is interested in the moments of $X$ itself, with density$$f(x|\theta)=\exp\{\theta\cdot T(x)-\Psi(\theta)\}$$there is no reason those moments are always well-defined. The generic reason is that the density of an arbitrary one-to-one transform $Y=\Xi(X)$ is then $$g(y|\theta)=\exp\{\theta\cdot (T\circ\Xi^{-1})(y)-\Psi(\theta)\}$$again the measure $$\text{d}\xi(y) = \left| \frac{\text{d}\Xi{-1}}{\text{d}y}\right|\text{d}\mu(\Xi^{-1}(y)$$ Both $Y$ and $X$ thus share the same sufficient statistic in that $$(T\circ\Xi^{-1})(Y)=T(X)$$ as a random variable. The properties of this exponential family are thus characteristic of and characterised by the sufficient statistic, not $X$ or $Y$.
As exemplified by jbowman in a comment below, for the $X\sim\text{Ga}(\alpha,\beta)$ distribution, the intrinsic representation & sufficient statistic is $T(X)=\log X$. While $X$ has all (positive) moments finites, $Y=\exp\{X\}$ does not. And as pointed out by Glen_b, for the $X\sim\text{Pa}(\alpha,\beta)$ distribution [which is indeed an exponential family when the lower bound $\beta$ is fixed], the moments are only defined up to the $\alpha-1$ order.
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3$\begingroup$ It's not totally clear to me that just because the sufficient statistics all have finite moments of all orders that $X$ itself does too. It certainly seems plausible, but... think of the Gamma with known scale parameter. The SS is $\Sigma \log x$. Although in the case of the Gamma we know that $x$ has finite moments of all order, so this isn't a counterexample, the fact that $\log x$ does doesn't imply that $x$ itself does. I could imagine some exponential family distribution with a similar result, but for which $X$ doesn't have finite moments of all orders. $\endgroup$– jbowmanCommented Dec 5, 2017 at 22:07
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3$\begingroup$ Which is why my reply is only for $T(X)$. The question is meaningless without specifying which transform of $X$ has or has not a finite first moment, because there are transforms with infinite moments, at least for continuous families. Just consider that any bijective transform of $X$ has the same sufficient statistic $T(X)$. $\endgroup$– Xi'anCommented Dec 5, 2017 at 22:11
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4$\begingroup$ I like the answer; it's clean and to the point, but perhaps this distinction being discussed in comment could be more explicitly mentioned in the post. [One potential example -- for the Pareto, where $T(X)=\log(X)$; we'd have that $\log(X)$ has all finite moments (which is true), but $X$ may not.] $\endgroup$– Glen_bCommented Dec 5, 2017 at 22:22
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$\begingroup$ So, for example, if I will show, that distribution(in the original form) doesn't have finite expectation, then this distribution cannot belong to the exponential family? $\endgroup$– vitsukCommented Dec 6, 2017 at 12:36
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$\begingroup$ What do you mean by "original form"? Did you read the answer till the very end? $\endgroup$– Xi'anCommented Dec 6, 2017 at 12:39