Distribution of sum of independent random variables using MGF Assume you have  $x_i \sim \operatorname{Bernoulli}(p_i)$ with $p_i \sim \operatorname{Beta}(\alpha,\beta)$. 
and let $Z=X_1+ \dots +X_n$
and I wanted to show that $Z$, $Z \sim \operatorname{Binomial}(n,\frac{\alpha}{\alpha+\beta})$ using momemnt genrating function approch. 
Here what I have so far:
\begin{align}
M_X(t) =& E(e^{tZ})\\
&= E(e^{t(X_{1}+X_{2}+...+X_{n})})\\
&= E(e^{t X_{1}}). E(e^{t X_{2}}) .....E(e^{t X_{n}})  
\end{align}
I am not sure where to go from here. Any help would be great. 
 A: $\newcommand{\E}{\text{E}}$I am assuming the $p_i$ are iid so this is a graphical model for this process:

You're on the right track but there's an issue with the factorization as written. Marginally the $X_i$ are not independent. This is how I would approach it: by the tower property,
$$
\E(e^{tZ}) = \E_{p}\left(\E_{Z|p}\left[e^{tZ}\right]\right).
$$
Conditioned on $p$ we have that $Z|p$ is a Poisson Binomial as @Glen_b says in their comment, and its MGF can be shown to be
$$
M_{Z|p}(t) = \prod_{i=1}^n \left(1-p_i + p_i e^t\right)
$$
and then the outer expectation integrates this over $p\in[0,1]^n$. The $p_i$ are assumed iid so this means the expectation factorizes and therefore
$$
\E_Z(e^{tZ}) = \E_p\left(M_{Z|p}(t)\right) = \prod_{i=1}^n \E_{p_i}\left(1-p_i + p_i e^t\right) \\
= \left(1 - \frac{\alpha}{\alpha+\beta} + \frac{\alpha}{\alpha+\beta}e^t\right)^n
$$ 
which is the MGF of a $\text{Bin}\left(n, \frac{\alpha}{\alpha+\beta}\right)$ RV as desired.

If the $p_i$ are not iid then I don't think this would simplify much in general. 
