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Given $X$ is a random variable ~ $Beta ( a , b)$ distribution and $X$ belongs in (0,1)

Does the (MGF ) $E[e^{tx}]$ exist for every value of $a , b$ ?

(Mgf must not be equal to infinity in order to exist)

thus , is $E[e^{tx}]$ finite ?

Update

what if $Beta ( a = \frac{1}{2} , b =1 ) $

the moment generating function is calculated as below

$ 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 $=

After - Using Taylor series expansion & interchanging the summation and integration of the taylor series

$ M_X(t) $ = $ \sum_{k=0}^\infty \frac{t^k}{k! (2k+1)}$

where k is non negative integer and $t \in \mathbb{R}$

How can I prove that this last sum is finite or infinite , is there a theorem ? (my math background is limited)

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For any random variable $X$ with $|X| \leq C$ with probability $1$ we have $Ee^{tX}\leq eE^{|t||X|} \leq e^{C|t|} <\infty$ for any real number $t$. In particular, Beta random variables are bounded so their MGF's exist.

About the series $\sum \frac {t^{k}} {(k!)(2k+1)}$ just observe that $|\frac {t^{k}} {(k!)(2k+1)}| <\frac {|t|^{k}} {(k!)}$ so the series is convergent for all $t$. ($\sum \frac {|t|^{k}} {(k!)}=e^{|t|}$).

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  • $\begingroup$ thank you , I update the question with a small answer I would really appreciate if you could tell me if we can get any assumption for the last Sum $ \sum_{k=0}^\infty \frac{t^k}{k! (2k+1)}$ $\endgroup$ – Pedros May 7 at 7:03
  • $\begingroup$ @Pedros I have updated my answer. $\endgroup$ – Kavi Rama Murthy May 7 at 7:19
  • $\begingroup$ Do you think there is an error with my calculations? $ 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)}$ >Used Taylor series expansion & interchanging the summation and integration of the taylor series $\endgroup$ – Pedros May 7 at 16:53
  • $\begingroup$ @Pedros Looks fine to me. $\endgroup$ – Kavi Rama Murthy May 7 at 23:05
  • $\begingroup$ Where C a constant ? correct? also $ eE^{|t||X|} is ={[eE(1)]}^{|t||X|} $ ? where E is expectation. $\endgroup$ – Pedros May 9 at 3:59

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