# Moment Generating Function

Suppose X is a random variable with a Beta distribution and x in (0,1)

How can I prove moment generating function exist

• Beta mgf : proofwiki.org/wiki/… – GAGA May 6 '19 at 2:12
• The sum manifestly has infinitely many non-zero terms. It also obviously converges everywhere because, term by term, the absolute values are only $1/(2k+1)$ as great as the terms in the series for $e^t,$ which converges everywhere. Could you therefore explain what you mean by the sum being "finite"? – whuber May 6 '19 at 14:36
• statlect.com/probability-distributions/beta-distribution in this example if you click on "proof" in the moment generating function of beta (general case) proof it says that moment generating function exists because , the integral is guaranteed to exist and be finite. In a similar way am trying to check if the integral exists for a= 1/2 and b =1 => the mgf exists – GAGA May 6 '19 at 16:10

First, since the Beta distributions have support on $$[0,1]$$, the mgf exists; that is, it is finite (for all parameters $$(a,b)$$). $$\DeclareMathOperator{\E}{\mathbb{E}} M_X(t)=\E e^{t X} =\int_0^1 e^{tx} f_{\text{Beta}}(x)\; dx$$. But for $$x \in [0,1]$$ we have $$e^{-|t|}\le e^{tx}\le e^{|t|},$$ so always $$e^{-|t|}\le M_X(t)\le e^{|t|}$$. So if your sum is a correct representation of the mgf it would have to converge ...
The looking at your sum $$\sum_{k=0}^\infty \frac{t^k}{k! (2k+1)}$$ the factor $$(2k+1) \ge 1;$$ and dropping it from the sum, we get the sum for $$e^t$$, so that will be an upper bound. So convergence is clear.
• $M_X(t)$ = $\mathbb{E}[e^{tX}]$ =$\frac{\Gamma(\frac{1}{2} +1)}{\Gamma(\frac{1}{2} ) +\Gamma(1)} \int_0^1 e^{tX} x^{\frac{1}{2}-1} (1-x)^{1-1}\ dx$= ${\frac{1}{2}}$$\sum_{k=0}^\infty$ $\int_0^1 \frac{{t^kx}^k}{k!}$ $x^{-\frac{1}{2}} \ dx$ = $\frac{1}{2}$ $\sum_{k=0}^\infty \frac{t^k}{k!}$ $\int_0^1 {}$ $x^{k-\frac{1}{2}} \ dx$ = $\sum_{k=0}^\infty \frac{t^k}{k! (2k+1)}$ Do you think there is something wrong with my calculation? – GAGA May 7 '19 at 16:46