Does exponential family of distributions have finite expected value? I am curious about this question, because in definitions I have never seen this property. Is it true? If yes, why?
 A: 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.

