On moment generating function of generalized gamma distribution

I'm reading An Intermediate Course in Probability by Gut. I am confused about a statement made concerning the generalized gamma distribution and its existence of a moment generating function.

I quote:

Another class of distributions that possesses moments of all orders but not a moment generating function is the class of generalized gamma distributions whose densities are $$f(x)=Cx^{\beta-1}e^{-x^{\alpha}},\quad x>0,\tag1$$where $$\beta>-1$$, $$0<\alpha<1$$, and $$C$$ is a normalization constant (that is chosen such that the total mass equals $$1$$).

It is clear that all moments exist, but, since $$\alpha<1$$, we have $$\int_{0}^\infty e^{tx}x^{\beta-1}e^{-x^\alpha}\, dx=\infty\tag2$$ for all $$t>0$$, so that the moment generating function does not exist.

First, I don't think it is clear that all moments exist. Integrating $$(1)$$ and making the substitution $$y=x^\alpha$$, and rewriting the integral in terms of a gamma integral, I get that $$C=\Gamma(\beta/\alpha)$$. The only condition I get is $$\beta/\alpha>0$$ (this is so the gamma integral converges), so I don't see that the restriction $$\beta>-1$$ makes sense. I also don't see why $$\alpha<1$$ would make sense. According to Wikipedia, $$\alpha$$ and $$\beta$$ are simply positive, but in view of my finding that $$\beta/\alpha>0$$, could they both be negative too?

Second, why does $$(2)$$ diverge to infinity for all $$t>0$$? If we now modify so that $$\alpha$$ is not bounded above by $$1$$, does the integral still not converge?

• Thanks. Yes, looking at my solution again, I got $C=\alpha/\Gamma(\beta/\alpha)$ as well. I still don't quite see how $\beta>-1$ is needed; if $(\beta+r)/\alpha$ shows up, we require that to be positive, so $\beta>-r$ and $\alpha>0$, or?
– psie
Commented Jul 20 at 11:48
• The statement is that "all" moments exist. Thus $r$ could be 1, i.e., the first moment. Commented Jul 20 at 11:55

I got $$C=\alpha/\Gamma(\beta/\alpha)$$.
I think $$\beta>-1$$ is needed for the $$r$$th moment existence, where $$(\beta+r)/\alpha$$ will show up in place of $$\beta/\alpha$$. For example, if $$\beta=-1$$ and $$r=1$$, we'll have problems with the $$\Gamma$$.
For $$\beta$$ and $$\alpha$$ both being negative, it could be a valid distribution, with a very common example being the inverse gamma distribution, frequently used in Bayesian statistics.
For the $$\alpha<1$$, I think the quote meant that it is only a condition for (2) to be $$\infty$$, not a condition for existence of moments. When $$\alpha<1$$, for sufficiently large $$x$$, the $$tx^1$$ will dominate $$x^\alpha$$ in the exponent. Thus, the integrand of (2) will keep increasing in its tale, making the integral $$\infty$$.
• For the pdf to converge, however, we do need $\beta>0$, right? If $\alpha$ and $\beta$ would both be negative, we'd have problems with the moments existing, I believe.