# Moment Generating Function of beta ( Hard )

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)

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 $$t$$. ($$\sum \frac {|t|^{k}} {(k!)}=e^{|t|}$$).
• 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 $\sum_{k=0}^\infty \frac{t^k}{k! (2k+1)}$ – Pedros May 7 at 7:03
• 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 – Pedros May 7 at 16:53
• Where C a constant ? correct? also $eE^{|t||X|} is ={[eE(1)]}^{|t||X|}$ ? where E is expectation. – Pedros May 9 at 3:59