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

How can I prove moment generating function exist

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    $\begingroup$ Beta mgf : proofwiki.org/wiki/… $\endgroup$ – GAGA May 6 at 2:12
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    $\begingroup$ 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"? $\endgroup$ – whuber May 6 at 14:36
  • $\begingroup$ 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 $\endgroup$ – GAGA May 6 at 16:10
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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.

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  • $\begingroup$ $ 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? $\endgroup$ – GAGA May 7 at 16:46
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    $\begingroup$ No, because your calculation is identical to the general formula shown on Wikipedia for the mgf of the Beta Distribution. $\endgroup$ – whuber May 7 at 16:57

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