I've seen that the moment generating function of the Beta Distribution is the following:

$${\displaystyle 1+\sum _{k=1}^{\infty }\left(\prod _{r=0}^{k-1}{\frac {\alpha +r}{\alpha +\beta +r}}\right){\frac {t^{k}}{k!}}}$$

However, nowhere have I seen it specified for what values of $t$ this moment generating function is valid. Is it only valid for positive $t$? All real numbers $t$? I'm not sure, nor do I know how to show what values it would be valid for.

  • $\begingroup$ Since Beta distribution has bounded support, its mgf is an entire function. $\endgroup$ – zhoraster Feb 16 '18 at 15:42

Using the Cauchy Root Test we can establish that the radius of convergence of this moment generating function is infinite, and so therefore it is defined for all $t \in \mathbb C$.

Recall, that the Cauchy Root Test implies that the radius of convergence, $R$, of a power series $A(z) = \sum_{n=0}^\infty a_n z^n$ satisfies $$ \frac{1}{R} = \limsup_{n \rightarrow \infty} |a_n|^{\frac1n}.$$

So applying this to the Moment generating function of a Beta distribution with $\alpha, \beta > 0$ \begin{align*} C & = \limsup_{n \rightarrow \infty}|a_n|^{\frac1n} \\ & = \limsup_{n \rightarrow \infty} \left| \frac{1}{n!} \prod_{r=0}^{n-1} \frac{\alpha + r}{\alpha + \beta + r} \right|^{\frac1n} \\ & \leq \limsup_{n \rightarrow \infty} \left| \frac{1}{n!} \prod_{r=0}^{n-1} 1\right|^{\frac1n} \\ & = \limsup_{n \rightarrow \infty} \left| \frac{1}{n!}\right|^{\frac1n} \\ & \leq \limsup_{n\rightarrow \infty} \left( \frac{1}{n^n} \right)^{\frac1n} \\ & = \limsup_{n \rightarrow \infty} (n^{-n})^{\frac1n}\\ & = \limsup_{n\rightarrow \infty} n^{-1}\\ & = 0. \end{align*} It follows (from the Cauchy Root Test) that the radius of convergence is $R = \infty$.


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