Do all order of moments exist for the log-exponential family?

Question

Do all order of moments exist for the log-exponential family (both multivariate and single-variate)? I already know they exists for log-normal distribution: single variate $EX^n=e^{n\mu+\frac{n^2\sigma^2}{2}}$ and multivariate $E(X_1^{k_1}\cdots X_n^{k_n})=e^{k'\mu+\frac{1}{2}k'\Sigma k}$.

If not, how about the exponential family without log?

How about log-normal-mixture or or log-RBF-kernel-density or RBF-kernel density estimate without log?

Answer 1

If by "log-exponential" you mean a random variable $Y$ for which $\log Y \sim \operatorname{Exponential}(\lambda)$, then it is trivial to show $$f_Y(y) = \begin{cases}\lambda y^{-\lambda - 1}, & y \ge 1 \\ 0, & \text{otherwise}, \end{cases}$$ for the usual $\lambda > 0$. That is to say, $Y \sim \operatorname{Pareto}(1,\lambda)$ which is Pareto Type I with scale parameter $1$ and shape $\lambda$. We can easily observe that only finitely many positive integer moments exist, since $$\operatorname{E}[Y^k] = \int_{y = 1}^\infty y^k \lambda y^{-\lambda - 1} \, dy$$ exists if and only if $k - \lambda - 1 < -1$, i.e., $k < \lambda$.

Answer 2

If by exponential family you mean the class of exponential distributions, that is $\mathcal{E}(\lambda)$ ($\lambda>0$), with density $$f_X(x)=\lambda e^{-\lambda x} I_{(0,+\infty)}(x)$$ ($I_A(x)$ is $1$ if $x\in A$, $0$ otherwise), then yes. Is easy to show (by definition or using moment generating functions or other transforms) that for every $k\in \mathbb N$ $$E(X^k)=\frac{k!}{\lambda^k}.$$

If by exponential family you mean an exponential family of distributions (what @heropup showed you in the link he posted), then the answer is no.

For instance, the class of absolutely continuous distributions with densities $$f_X(x)=\frac\alpha{x^{\alpha+1}}I_{(1,+\infty)}(x)\qquad (\alpha>0)$$ is an exponential family since you can express this density as $$f_X(x)=e^{-(\alpha +1)\log x +\log\alpha}I_{(1,+\infty)}(x)=e^{\eta(\alpha)T(x)-A(\alpha)}h(x)$$ for $\eta(\alpha)=-(\alpha+1)$, $T(x)=\log x$, $A(\alpha)=-\log(\alpha)$ and $h(x)=I_{(1,+\infty)}(x)$.

Nevertheless, is easy to see that $E(X^k)$ is finite if and only if $\alpha +1-k>1$ $\iff$ $k<\alpha$, so this is a possible counterexample for your intended claim.